Post by: jakesilove on January 28, 2016, 06:51:01 pm
HSC MATHEMATICS Q&A THREAD
To go straight to posts from 2018, click here.
What is this thread for?
If you have general questions about the HSC Mathematics course or how to improve in certain areas, this is the place to ask! 👌
Who can/will answer questions?
Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.
Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding!
There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you. So you may even get multiple answers from different people offering their insights - very cool.
To ask a question or make a post, you will first need an ATAR Notes account. You probably already have one, but if you don't, it takes about four seconds to sign up - and completely free!
OTHER MATHEMATICS RESOURCES
CLICK ME!
* Free Mathematics notes
* How to Get a Band 6 in HSC Mathematics
* Mathematics Question Thread
* Compilation of solutions to HARD past HSC papers (2U)
* Mathematics MEGA THREAD
* Mathematics Challenge Marathon
* 5 Methods for Studying Mathematics
* Basic Arithmetic and Algebra in 2U: How It Can Come Back to Bite You
* Sequences and Series: Tackling the Financial Math Questions
* A Geometry Guide: Tackling Proofs in the HSC
* Plane Geometry in 2 Unit and Extension One
* Trigonometry: Part 1 (2 Unit)
* HSC Dysfunction: The Functions You'll See in 2/3 Unit
* Differentiation in HSC Mathematics: What You Need to Know!
* Integration in the HSC: What You Should Know!
* A Guide on Probability in the HSC You Probably (Definitely) Should Read
* Common Mistakes in Mathematics
* HSC Revision Videos: Mathematics
* 12 Avoidable Mistakes Students Make in Math Exams
* LaTeX Guide
* Verbs and Maths
* How to Get a Band 6 in HSC Mathematics
* Mathematics Question Thread
* Compilation of solutions to HARD past HSC papers (2U)
* Mathematics MEGA THREAD
* Mathematics Challenge Marathon
* 5 Methods for Studying Mathematics
* Basic Arithmetic and Algebra in 2U: How It Can Come Back to Bite You
* Sequences and Series: Tackling the Financial Math Questions
* A Geometry Guide: Tackling Proofs in the HSC
* Plane Geometry in 2 Unit and Extension One
* Trigonometry: Part 1 (2 Unit)
* HSC Dysfunction: The Functions You'll See in 2/3 Unit
* Differentiation in HSC Mathematics: What You Need to Know!
* Integration in the HSC: What You Should Know!
* A Guide on Probability in the HSC You Probably (Definitely) Should Read
* Common Mistakes in Mathematics
* HSC Revision Videos: Mathematics
* 12 Avoidable Mistakes Students Make in Math Exams
* LaTeX Guide
* Verbs and Maths
Original post.
Before you can ask a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)
Hey everyone!
A lot of you will have met me at the HSC Head Start lectures, where I lectured in 2U and 3U Maths, Physics and Chemistry.
My role on these forums is to help you. The HSC syllabus is tricky, nuanced and pretty damn huge. To help you out, I thought it would be a great idea to have a forum where you can just post questions, and myself or other forum members can post answers!
This is a community, so we want you to feel like you can post any type of 2U Mathematics question, no matter how "basic" you might think it is. Remember, IF YOU'RE HAVING TROUBLE WITH A TOPIC, THERE ARE THOUSANDS OF OTHERS HAVING THE SAME ISSUE. The best way to learn Maths is by looking through practice questions, and their associated answers. I honestly think a forum like this, and a place where I could always go to have difficult questions answered would have helped me in my HSC year.
I got an ATAR of 99.80, and a mark of 96 in the Mathematics course. There are similar forums for a bunch of other subjects, so make sure to take a look at them as well!
Hey everyone!
A lot of you will have met me at the HSC Head Start lectures, where I lectured in 2U and 3U Maths, Physics and Chemistry.
My role on these forums is to help you. The HSC syllabus is tricky, nuanced and pretty damn huge. To help you out, I thought it would be a great idea to have a forum where you can just post questions, and myself or other forum members can post answers!
This is a community, so we want you to feel like you can post any type of 2U Mathematics question, no matter how "basic" you might think it is. Remember, IF YOU'RE HAVING TROUBLE WITH A TOPIC, THERE ARE THOUSANDS OF OTHERS HAVING THE SAME ISSUE. The best way to learn Maths is by looking through practice questions, and their associated answers. I honestly think a forum like this, and a place where I could always go to have difficult questions answered would have helped me in my HSC year.
I got an ATAR of 99.80, and a mark of 96 in the Mathematics course. There are similar forums for a bunch of other subjects, so make sure to take a look at them as well!
Post by: liiz on February 03, 2016, 05:03:10 pm
Post by: Happy Physics Land on February 03, 2016, 06:30:16 pm
Hey Jake - I'm struggling with this question from my maths textbook: "A rectangular box is to have a square base and no top. If it's volume is 500cm3, find the least area of sheet metal of which it can be made." I'm not understanding how all this maximum and minimum stuff is applied to these kind of questions, I thought it just had to do with graphs? Thanks!! :)
Hey Liiz:
HEY Im jake (obviously not haha). Im a year 12 student who completed my 2u HSC mathematics course in year 11 and lm just happy to help out here. Now, this type of question is amongst one of the most difficult ones in geometrical applications of calculus, and unfortunately in HSC exams there WILL be harder ones. But don't worry, once you have practised enough, you will begin to seize some tricks to approach these questions.
Before I begin answering your question, just a few generally tips to help you answer questions like these where only one number value is provided:
BEFORE YOU DO ANYTHING, DRAW A DIAGRAM WITH LABELS
1. Highlight all USEFUL INFORMATIONS (in this case, highlight rectangular box, square base, no top, 500cm^3 and least area)
2. Appoint two variables to the unknown sides (in this case, I named the side length of the square base as x, and the height of the rectangular box as L)
3. There will be at least one number quantity in every one of these questions in 2u mathematics, so the first equation you should construct, using your name variables to construct an equation that uses the numbers provided by the question
4. Draw out the relationship between the two variables through this equation that you have constructed
5. Construct another equation using your variables and the subject that is asked for in the question (In this case, for example, we constructed an Area equation which directly relates to what we are asked to find)
6. Substitute in the equivalent expression of a variable (In this case, for example, L = 500/x^2, so we substitute any L we see with 500/x^2) to reduce the total number of values down to one, so that we can construct an equation entire out of only one variable, which then allows us to perform differentiation
7. Clean up the equation after the substitution to make life easier
8. Differentiate the equation
9. Let this derivative = 0 to find any stationary points (In an Exam, YOU MUST STATE "LET dy/dx = 0 TO FIND ANY STATIONARY POINTS, OTHERWISE MARKS MAYBE DEDUCTED!!!)
10. Solve the derivative equation and find a value for your variable which will be your stationary point
11. Test both sides to show that a local minimum/maximum occurs at your stationary points
12. Substitute your minimum value back into the area equation (or maximum value if the question asks for maximum area) to find the minimum area (or the maximum area if you substitute in the maximum value)
So here is my solution:
(http://i.imgur.com/t6wtIk1.jpg)
Hope you find my solutions clear and useful! If you are confused with anything, dont hesitate to ask!!! :D
Best Regards
Jacky He
Post by: jakesilove on February 03, 2016, 06:40:09 pm
Hey Jake - I'm struggling with this question from my maths textbook: "A rectangular box is to have a square base and no top. If it's volume is 500cm3, find the least area of sheet metal of which it can be made." I'm not understanding how all this maximum and minimum stuff is applied to these kind of questions, I thought it just had to do with graphs? Thanks!! :)
Hey Liiz! I see that HPL has already posted an answer, but I'll throw mine in as well in case any of the methodology is different. Hope it helps, and thanks for the questions!
(http://i.imgur.com/NubpZ8v.png?1)
(http://i.imgur.com/0beu0z5.png?1)
Thanks for your post!
If anyone else wants to ask or answer questions, I would really appreciate the contribution! Before you can ask a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)
Post by: Happy Physics Land on February 03, 2016, 07:25:29 pm
Hey Liiz! I see that HPL has already posted an answer, but I'll throw mine in as well in case any of the methodology is different. Hope it helps, and thanks for the questions!
(http://i.imgur.com/NubpZ8v.png?1)
(http://i.imgur.com/0beu0z5.png?1)
Thanks for your post!
If anyone else wants to ask or answer questions, I would really appreciate the contribution! Before you can ask a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)
Great work Jake, I think it was a really good idea for you to first state the equation of how to calculate the volume of the rectangular box (cant believe l forgot to include that) and to state that x does not equal to 0 which is the condition which must be met despite obvious in this case. Great combined effort and a really good question Liiz!!!
Post by: brenden on February 03, 2016, 08:21:06 pm
Great work Jake, I think it was a really good idea for you to first state the equation of how to calculate the volume of the rectangular box (cant believe l forgot to include that) and to state that x does not equal to 0 which is the condition which must be met despite obvious in this case. Great combined effort and a really good question Liiz!!!You have the best handwriting I've ever seen.
Post by: Happy Physics Land on February 03, 2016, 09:03:38 pm
You have the best handwriting I've ever seen.
This is the best forum I have ever seen! :)
Post by: liiz on February 08, 2016, 08:50:06 pm
Post by: Jacqui105 on February 09, 2016, 12:56:43 am
Okay, so I've got a (possibly silly) question to ask, but I actually - GENUINELY - do not have any idea and need help:
How does one study for Maths?
I seem to have a really consistent system for each of my subjects as to approaching study in the most efficient way possible, but I can not seem to find an effective way of studying for Maths. I've tried everything I can think of. Please help.
Thank you
Post by: jamonwindeyer on February 09, 2016, 08:40:05 am
Hi,
Okay, so I've got a (possibly silly) question to ask, but I actually - GENUINELY - do not have any idea and need help:
How does one study for Maths?
I seem to have a really consistent system for each of my subjects as to approaching study in the most efficient way possible, but I can not seem to find an effective way of studying for Maths. I've tried everything I can think of. Please help.
Thank you
Hey Jacqui105!
Firstly, absolutely not a silly question. I guarantee a large portion of the students reading this would be wanting to ask something similar.
This is a huge question that demands a huge response. We are planning a resource specifically answering this exact question for you, so stay tuned! ;D
In the meantime, I'll direct you to this guide I wrote a little while ago that has helped quite a few people. It contains a whole bunch of tips for everything Math related, including study tips!
Good luck :D
Post by: caninesandy on February 14, 2016, 02:39:43 pm
I have a math differentiation and integration assessment soon and was just wondering if you could help me on how to set answers out, especially with lengthy questions. Here is one question I have recently, recived the correct answers but was not sure on how to structure my answer exactly:
A magazine advertisement is to contain 50cm^2 of lettering with clear margins of 4cm each at the top and bottom and 2cm at each side. Find the overall dimensions if the total area of the advertisement is to be a minimum.
Thank you so much for your help!!!!!!
;D ;D ;) :D :)
Post by: jakesilove on February 14, 2016, 04:24:24 pm
Hello HSC Math geniuses :D
I have a math differentiation and integration assessment soon and was just wondering if you could help me on how to set answers out, especially with lengthy questions. Here is one question I have recently, recived the correct answers but was not sure on how to structure my answer exactly:
A magazine advertisement is to contain 50cm^2 of lettering with clear margins of 4cm each at the top and bottom and 2cm at each side. Find the overall dimensions if the total area of the advertisement is to be a minimum.
Thank you so much for your help!!!!!!
;D ;D ;) :D :)
Hey Caninesandy!
This was actually quite a difficult question, which required me to draw the picture in order to figure out what was going on! My general tips for these questions are
1. Figure out the equations that you have
2. Sub in the equations, so that you have a differentiatable one
3. Differentiate, and find the required minimum
4. Sub the found value into the original, to find the solution.
Here is the solution I came up with. Hope it helps! This is exactly how I would set out any question similar to this :)
(http://i.imgur.com/eyyUD6Q.jpg?1)
(http://i.imgur.com/FbVA77a.jpg?1)
Jake
Post by: Happy Physics Land on February 15, 2016, 08:59:38 am
Hey Caninesandy!
This was actually quite a difficult question, which required me to draw the picture in order to figure out what was going on! My general tips for these questions are
1. Figure out the equations that you have
2. Sub in the equations, so that you have a differentiatable one
3. Differentiate, and find the required minimum
4. Sub the found value into the original, to find the solution.
Here is the solution I came up with. Hope it helps! This is exactly how I would set out any question similar to this :)
(http://i.imgur.com/hY8GLxv.png?1)
(http://i.imgur.com/hQTKRDX.png?1)
Jake
Sorry to challenger your authority jake, but did you mean h = 50/(x-4) + 8??? (Is it a transcription error?)
Post by: jamonwindeyer on February 15, 2016, 10:06:53 am
Sorry to challenger your authority jake, but did you mean h = 50/(x-4) + 8??? (Is it a transcription error?)
Definitely a transcription error HPL :D
Post by: jakesilove on February 15, 2016, 11:38:18 am
Sorry to challenger your authority jake, but did you mean h = 50/(x-4) + 8??? (Is it a transcription error?)
Yep, definitely an error there, thanks for pointing that out! I'll fix up the proof and update the answer in a second.
Jake
Post by: aqsarana_ on February 20, 2016, 03:51:20 pm
At the moment I'm doing HSC Trigonometry and I'm really confused about when to change the calculator's mode to Degrees or Radians. How can i be able to tell when to change the mode?
Thank You.
Post by: jamonwindeyer on February 20, 2016, 05:35:10 pm
Hi!
At the moment I'm doing HSC Trigonometry and I'm really confused about when to change the calculator's mode to Degrees or Radians. How can i be able to tell when to change the mode?
Thank You.
Hey aqsarana! Awesome question. Your calculator should be in radian measure by default, since you must use radians for trigonometric calculus. The only time you switch to Degree measure is if degrees are used in the question, that is, values are given in degrees. This normally happens with geometry questions. For these questions, switch to degree measure, then switch back immediately when you are finished.
But yes, by default, leave your calculator in Radian measure. Only switch to Degrees if you see degree measurements in the question.
Hope this helps ;D
Post by: RuiAce on February 20, 2016, 05:44:51 pm
You will find that in questions that ask you to be in degrees, there will be a degree symbol (°) present. As soon as that goes missing, naturally assume radians.
Post by: Happy Physics Land on February 20, 2016, 05:48:51 pm
Post by: jamonwindeyer on February 20, 2016, 06:42:41 pm
And by personal experience there will definitely be cases where you would forget to change from radian mode to degrees modes and this is gonna be very disastrous. So some of my friends decided to bring two calculators, one labelled degrees and another labelled radians. If you are struggling to remember to change to radian mode then you might like to consider that method.
I also heard of students doing this! One person in my year did it, it's not a bad idea if you are so inclined!
Post by: jakesilove on February 21, 2016, 12:44:24 pm
Hi!
At the moment I'm doing HSC Trigonometry and I'm really confused about when to change the calculator's mode to Degrees or Radians. How can i be able to tell when to change the mode?
Thank You.
Definitely agree with all of the above: If the question mentions degrees, use degrees. If the question doesn't mention degrees, use Radians. Simple as that! If you sometimes get confused, maybe it is worth getting into the habit of highlighting every time you see a degrees symbol in a question, so you know you need to put your calculator into degrees mode.
Jake
Post by: Phillorsm on February 21, 2016, 09:57:47 pm
Integrate (4-(x-8)^2)^(1/2)
I know its a semicircle and the limits are 10 to 6, but I keep getting zero.
Is there a special way to integrate a semicircle and find the area? (Other than (pi*r^2)/2 )
Post by: Happy Physics Land on February 21, 2016, 10:01:45 pm
Hey jake, having trouble integrating this.
Integrate (4-(x-8)^2)^(1/2)
I know its a semicircle and the limits are 10 to 6, but I keep getting zero.
Is there a special way to integrate a semicircle and find the area? (Other than (pi*r^2)/2 )
Hey Phillorsm:
Is your question a definite or indefinite integral?
Assuming that your question is an indefinite integral, the answer should not be zero because after all, area cannot be zero right? I like your approach of drawing a semi-circle and you have correctly identified the domain of the semi-circle. So all you really had to do was apply pi(r^2)/2 to find the area of the semi circle, and that will give you the integral (l believe you have identified that r=2 because 4 = 2^2). Considering that this is an indefinite integral question, you will need to add +C at the end.
(sorry just ignore the sintheta stuff in the background, they are irrelevant)
(http://i.imgur.com/azgnjfQ.jpg)
If you have any further questions dont hesitate to ask!
Best Regards
Happy Physics Land
Post by: jakesilove on February 21, 2016, 10:07:51 pm
Hey jake, having trouble integrating this.
Integrate (4-(x-8)^2)^(1/2)
I know its a semicircle and the limits are 10 to 6, but I keep getting zero.
Is there a special way to integrate a semicircle and find the area? (Other than (pi*r^2)/2 )
Hey Phillorsm!
If you are looking to find the area underneath that curve, you've identified one great way to do it: by understanding that the graph is a semicircle, use the formula (pi*r^2)/2 !
However, you should also be able to integrate as normal and find the required result. I'm not at home at the moment, so can't write out a full solution, but hopefully someone else on the forums can!
Hope this helps! Let me know if there's anything you're unsure of.
Jake
Post by: Loki98 on February 23, 2016, 08:56:39 pm
Post by: jakesilove on February 24, 2016, 12:09:02 am
Could someone please help with question 10a of the 2003 2u mathematics hsc paper :)
Hey Loki!
This is one of the more difficult Finance question I've seen. Once you learn the general method, however, you will have no problem! Below is my solution :)
(http://i.imgur.com/MpSJxr4.png?1)
(http://i.imgur.com/IctwnG7.png?1)
(http://i.imgur.com/0GBauHq.png?1)
Jake
Post by: Phillorsm on February 26, 2016, 09:23:55 pm
Firstly I wasn't super strong on logs and exponentials from the beginning, and secondly I spent way too long trying to work out a question earlier in the exam that i was sure I could get. It was a question where you had to find the equation of a parabola given its vertex and x intercepts.
Based on all this, how do you think I should move on. What can I learn from this experience, and just because I got a 70% on this exam, does that mean i'm doomed to around a 70 for the rest of my 2u math career? Like, this is the real thing now, is it still possible to recover?
Post by: Happy Physics Land on February 26, 2016, 10:15:25 pm
Hey Jake, I just got my results back for my first 2u exam for year 12. It was on sequence and series, applications of calculus, integration and logs and exponentials. I got a 55/78 because I didn't finish the last section (logs and expos) of the exam. My teacher did say I smashed the first three parts though.
Firstly I wasn't super strong on logs and exponentials from the beginning, and secondly I spent way too long trying to work out a question earlier in the exam that i was sure I could get. It was a question where you had to find the equation of a parabola given its vertex and x intercepts.
Based on all this, how do you think I should move on. What can I learn from this experience, and just because I got a 70% on this exam, does that mean i'm doomed to around a 70 for the rest of my 2u math career? Like, this is the real thing now, is it still possible to recover?
Hey Phillorsm:
Im clearly not Jake (obviously) but I just wanna to be here to tell you that getting around a 70 for your first 2u assessment is really nothing to worry about. Im not just saying this to boost you with blind confidence, Im saying this because it's too early a stage to give up. If you've ever been to one of Jake's lectures, he told us his experience not being able to finish TWO PAGES of his trial exam for maths ext 1 upon his return from the international science forum. Yet he still got a remarkable 98. So there is absolutely nothing for you to worry about.
Another real life example from me: one of my classmates for accelerated maths only got 60 percent in his first exam, and throughout the entire course his mark fluctuated around 70 percent. Yet in the end he scored a mark of 85 for his 2u mathematics. So if you look at yourself now, you are in a much better situation. You know majority of your stuff in the first three sections and you are in a better starting point than my friend! As a person who's gone through 2u hsc mathematics before, I can tell you that in the end, after the tonnes of practise papers you have done, you will begin to realise how much you have improved and how surprisingly better you will be end up doing in your hsc exam. So dont get defeated by this one exam, because to be honest, I have failed 2 of my first assessments this year as well and I found out that in the end, they only contribute to 5% (sometimes not even) of your final mark.
Ok hmmm regarding exam techniques, I think you might have been stuck on a question for too long and that may have caused your inefficiency in time management. Of course you will hear people saying to you "skip the question if its taking too long", but I do empathise with you because sometimes we are fairly certain we have seen the particular question before and it shouldnt take us much time to complete it. Unfortunately sometimes we can make a mistake that we wouldnt realise because its too trivial or the teacher could have added something extra in the question to make the question one step harder. As a general rule of thumb for me, if there is a time-consuming question that you know you definitely can solve in a multiple choice section, SKIP IT AND COME BACK LATER. Surely, if we solve the question it boosts our ego, but its not worth spending 5 minutes on a question that merely awards you with 1 mark. If a question is in the free-response section and you have looked at it for 3 minutes not knowing where you should be starting, then I would also suggest to skip it and come back later.
Logarithmic function and exponentials definitely can be hard, perhaps one of the hardest concepts in 2u. I would recommend, ofc, to do past trial papers (you can find links in atarnotes to access trial papers) and practise those questions on the topic. Also practise some past HSC questions.
Tips I would give on logarithmic functions:
- (This is especially common with difficult integration questions involving logs). KNOW YOUR YEAR 8 GEOMETRY WELL! If you look up question 16b) on 2015 paper you will realise that you have to draw a rectangle and use integration about y-axis to figure out the area encompassed by the y-axis, the scope of the rectangle and the curve first, then use your area of rectangle to subtract that area to find the area you were asked to find. This is a VERY USEFUL TECHNIQUE.
- I often struggled with the idea of changing the bases, simply because its not frequently tested. But if the teacher wants to make a hard question, you would need to change the base as a technique to approach your final answer, so be familiar with the process
- I remember in one of my earlier exams, I was asked to find the area under a log curve and I accidentally wrote 1/x as an integral of ln x. Sometimes when the exam pressure kicks in, you really can be making those sorts of mistakes. So just remember that at your level, you definitely wouldnt be required to integrate ln x, and if you did end up having to integrate ln x, you know there must be an easier way.
- There is also that kind of typical question that's involved in both integration and the integration of logarithmic functions that involve several parts. The first part would most likely ask you to differentiate and then in the second step ask you to integrate. Keep in mind if you see this sort of question in integration/integration of logs and exponentials, you know that you can use part I in part II. You can somehow manipulate what you are required to integrate in part II into the outcome of differentiation in part I and then just integrate it saying "using part I, we know that _____ is a primitive function to _____, hence the integral is ______"
I will say this much for now, but yes definitely do not give up. The game has just started and sometimes its good to be on a lower end of a ladder because it avoids arrogance and it allows you to put in more effort to do better in the next exams.
Good luck!
Best Regards
Happy Physics Land
Post by: jakesilove on February 26, 2016, 11:15:57 pm
Hey Jake, I just got my results back for my first 2u exam for year 12. It was on sequence and series, applications of calculus, integration and logs and exponentials. I got a 55/78 because I didn't finish the last section (logs and expos) of the exam. My teacher did say I smashed the first three parts though.
Firstly I wasn't super strong on logs and exponentials from the beginning, and secondly I spent way too long trying to work out a question earlier in the exam that i was sure I could get. It was a question where you had to find the equation of a parabola given its vertex and x intercepts.
Based on all this, how do you think I should move on. What can I learn from this experience, and just because I got a 70% on this exam, does that mean i'm doomed to around a 70 for the rest of my 2u math career? Like, this is the real thing now, is it still possible to recover?
Hi Phillorsm!
Firstly, I just wanted to congratulate you on your fantastic 2u result! 71% is nothing to be worried about: in fact, I would say that the fact that you were able to do so well in the sections you completed puts you in good stead for the HSC (more on that later).
Next, I want to echo the words of HPL. I absolutely agree with his advice, and won't repeat anything he has said. Specifically, the idea of skipping questions and coming back and knowing specific techniques to answer difficult questions.
The only point that I would add, when discussing ways to speed up exam completion, is DO PAST PAPERS! Do them again, and again, and again. This is the best way to speed up answering questions. The reason is that, if you can spend one minute less per question figuring out what you actually need to do (ie. by recognizing that an answer should be identical to a practice response you did the day before), you will gain so much extra time. This is extremely valuable in finishing assessment tasks!
So, my main tip for you is to just do hundreds of past papers, in timed conditions, again and again. There are never too many papers you can complete!
The fact that you know your stuff is a really good indicator of how you are going in the year. Timing can always be improved, and the more past papers you do the faster you will get in completing individual questions. Knowing the content in-depth, however, is a far more difficult task. So, congratulations for where you are at! Just keep doing past papers, and keep in mind the comments made by HPL.
Jake :)
Post by: amandali on February 27, 2016, 08:10:47 am
how do i find the angles in a simple way without getting the angles confused?
Post by: RuiAce on February 27, 2016, 08:43:13 am
cos(2x+pie/6)=1/root2 -pie<=x<=pie
how do i find the angles in a simple way without getting the angles confused?
Edit: Holy crap the mistakes.
Post by: Phillorsm on February 27, 2016, 05:03:16 pm
Post by: amandali on February 27, 2016, 07:55:06 pm
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land? ans: 8km
Post by: RuiAce on February 27, 2016, 09:09:39 pm
how do you do this ques thanks ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land? ans: 8km
Weakly lacking information. There are three cases:
1. We want to minimise the time of the entire journey (this sounds extremely preferable to the others)
2. We want to minimise the distance of the entire journey (sounds very unfeasible as there is actually a very obvious answer)
3. We want to minimise only a part of the journey (e.g. reduce some walking
If the earlier answer didn't make sense, all the hints were given.
a) Rewrite boundaries because otherwise you don't know what you're trying to get at
b) Always use ASTC
c) Negative angles work in reverse (clockwise around ASTC)
Exact values are different. Use the formula sheet for exact values.
Post by: jakesilove on February 27, 2016, 09:12:34 pm
how do you do this ques thanks ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land? ans: 8km
Hey!
This was actually quite a difficult question! First, you need to construct the relevant equation. Then, the differentiation part is quite tricky for a 2U question. Finally, even solving the equation is a bit of work! Below is my answer, I hope you find it helpful! In the case of questions like this, just think through what is actually happening in the example. Then, always create some sort of equation that you can differentiate!
(http://i.imgur.com/sujfvuQ.png?1)
(http://i.imgur.com/FxQw2xR.jpg?1)
Jake
Post by: RuiAce on February 27, 2016, 09:20:05 pm
(http://uploads.tapatalk-cdn.com/20160227/37fd2b8c68f2b54865edec0515867baf.jpg)
P.S. Woah Jake what are you doing with the trig!
Post by: Happy Physics Land on February 27, 2016, 09:22:05 pm
how do you do this ques thanks ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land? ans: 8km
Hey Amanda:
Another great question from you amanda! Ok, so when I first looked at the question, my instinct tells me I will have to draw a diagram, because it involves distance and speed and even worse theres also time. And I will have to admit, despite my distaste towards drawing diagrams, it is essential for you to draw one because graphics help you to visualise stuff. So next time when you see these types of questions, definitely draw a diagram, its worth the time.
Ok and then I manipulated pythagorus theorem, made an algebraic expression for the time to travel distance MS, made another algebraic expression for the time to travel distance BS. The time taken to travel the total distance now becomes T(MS) + T(BS), where T =time. So after establishing such a relationship, the rest is straight forward: simply repeat the conventional process of simplifying, differentiating, make derivative = 0, and then find stationary point. Afterwards you test a value on both sides of stationary point to determine its nature (I.e. maximum or minimum) and then make a concluding statement for your answer. Actually it is quite crucial to include a concluding statement, because that makes it clearer for the marker what your final result is and you would less likely to be deducted a mark on not stating the result clearly.
Anyways, my solution as below:
(http://i.imgur.com/M88wbvs.jpg)
(http://i.imgur.com/zIqIQMu.jpg)
Sorry for the messy working btw, if you have any further questions dont hesitate to ask! :)
Best Regards
Happy Physics Land
Post by: jakesilove on February 27, 2016, 09:26:50 pm
Assumed we wanted to minimise time:
(http://uploads.tapatalk-cdn.com/20160227/37fd2b8c68f2b54865edec0515867baf.jpg)
P.S. Woah Jake what are you doing with the trig!
So, this is a wayyyy easier solution. There are so many ways to answer questions like these, so whichever makes more sense to you! Still, thanks everyone for participating and answering questions on the forums!
My method is definitely more tricky, because I have a tendency to over-use trig. If you can use algebra, that's usually a better and simpler solution.
Jake
Post by: 16ebond on March 07, 2016, 06:09:18 pm
So we have been doing e in class and I am a bit stuck on how to answer this question.
I have attached the question, so hopefully the file will open.
Thanks heaps
Em :)
Post by: RuiAce on March 16, 2016, 04:50:39 pm
Pick an arbitrary point (a,0) to be the first vertex of the rectangle. By default (a, e^(-a^2)) is also on the rectangle. But because it HAS TO BE A RECTANGLE and we have an even function
(-a,0) and (-a, e^(-a^2)) must lie on the rectangle. This is most easily shown with a diagram.
(http://i.imgur.com/fzE9TxN.png)
So the breadth and length of the triangle are:
Hence we can combine these to give an expression for area:
Set dA/dx=0 to maximise:
Reject x=0 though, because if x=0 we have no rectangle (we can't have a width of 0).
So just show that the remaining value gives a maxima using a table of values or the second derivative. Note that because the curve is an even function, the negative term can be ignored.
I would've used latex but I'm thoroughly lazy to on a phone (Hello, Jamon here, I went through and added LaTex ;D)
Post by: jamonwindeyer on March 17, 2016, 01:05:32 pm
I'm not sure if this was meant to be an unanswered question but here's a solution
Pick an arbitrary point (a,0) to be the first vertex of the rectangle. By default (a, e^(-a^2)) is also on the rectangle. But because it HAS TO BE A RECTANGLE and we have an even function
(-a,0) and (-a, e^(-a^2)) must lie on the rectangle. This is most easily shown with a diagram.
(http://i.imgur.com/fzE9TxN.png)
So the breadth and length of the triangle are: = 2x) )
Hence we can combine these to give an expression for area:
Set dA/dx=0 to maximise:\neq 0)
Reject x=0 though, because if x=0 we have no rectangle (we can't have a width of 0).
So just show that the remaining value gives a maxima using a table of values or the second derivative. Note that because the curve is an even function, the negative term can be ignored.
I would've used latex but I'm thoroughly lazy to on a phone (Hello, Jamon here, I went through and added LaTex ;D)
LEGEND! Thanks RuiAce! I went through and added LaTex and a diagram for you, cheers for the solution! Jake has been on holiday and I've been trying to keep an eye on all his forums, I must have missed this, you are a champion ;D
Post by: kal.123 on March 19, 2016, 11:43:17 am
Post by: jamonwindeyer on March 19, 2016, 12:00:28 pm
I was wondering how to draw the x-intercepts and vertex??? For Y= 3-2xsquared
Sure! So we have:
The x-intercepts occur when y=0, so we just substitute and solve. This is a quadratic, so we can expect anywhere from o to 2 answers:
Now, for the vertex, we are actually taught that the x-coordinate of the vertex is given by:
Now for your quadratic, a=2, b=0 and c=-3 (when rearranged), so the x-coordinate of the vertex will actually be zero. The y-coordinate will then be y=3 by substitution of x=0.
So, we have intercepts and a vertex, we can now draw a diagram. Hope this helps!
(http://i.imgur.com/jem7eQy.png?1)
Post by: RuiAce on March 19, 2016, 05:53:28 pm
LEGEND! Thanks RuiAce! I went through and added LaTex and a diagram for you, cheers for the solution! Jake has been on holiday and I've been trying to keep an eye on all his forums, I must have missed this, you are a champion ;D
Ah sweet. Thanks for the aid too Jamon!
Post by: imtrying on March 21, 2016, 04:26:50 pm
This is a Geometrical Applications of Calculus question I'm stuck on (I've attached a photo of the question and answer). I know the quotient and product rules, but for some reason I keep ending up with some ridiculous answer. A step by step explanation would be hugely helpful :)
Post by: jamonwindeyer on March 21, 2016, 09:39:11 pm
Hey :)
This is a Geometrical Applications of Calculus question I'm stuck on (I've attached a photo of the question and answer). I know the quotient and product rules, but for some reason I keep ending up with some ridiculous answer. A step by step explanation would be hugely helpful :)
Sure thing!!
The quotient rule says:
Note that our differentiation of that last term is done with the chain rule!
So the first derivative by substitution is:
Through cancellation and then expansion, we obtain:
Now for the second derivative, it works the exact same way!
Which through expansion and factorisation will turn into your solution for Part B!
I hope this helps!
Post by: Marlo365 on March 28, 2016, 06:57:44 pm
Post by: jakesilove on March 28, 2016, 07:37:43 pm
Hey Jake, so i got my 2u half yearly tomrrow and not sure what i'm supposed to be doing this very moment wether its past papers of relaxing haha, any advice would be great
Hey Marlo!
I think that it's totally up to you! Generally, I would recommend that you just keep doing past papers until a set time. Say you want to get 8 hours of sleep: figure out when you need to stop, and GO TO BED!
If you're feeling nervous, do a few past papers. If you find that this is making you more nervous, call it quits for the night! I really don't have any particular suggestion as to what you should do: just whatever makes you feel the most comfortable. If you're totally indifferent, maybe have a crack at one more paper (who know if it will help you tomorrow!) but definitely don't stay up all night.
Absolute best of luck, I'm sure you'll do great!
Jake
Post by: imtrying on March 29, 2016, 06:11:02 pm
The smith family buys a car for $38 000, paying a 10% deposit and taking out a loan for the balance. if the loan is over 5 years with interest of 1.5% monthly, find the amount of each monthly (a) loan repayment the total amount that the (b) smith family paid for the car.
Answer:
(a) $592.00 (b) $39 319.89
Post by: liiz on March 29, 2016, 06:16:09 pm
Thankyou!
Post by: jamonwindeyer on March 29, 2016, 06:37:49 pm
Hello, needing some help to integrate this: x / 2x2 - 3
Thankyou!
Hey Liiz! Sure thing!
Okay, so this question takes a bit of experience to spot the trick for. The integral of this will be a logarithm.
Remember, the derivative of the log of some function is:
For your question, we almost have the derivative of the bottom on top of the fraction, which is what is required. To make it work, we multiply by 4 on the numerator, then multiply the integral by a quarter, thus balancing everything out. You could also think of it as taking out a factor of 1/4:
This isn't super easy to spot, it might take some practice to see it quickly! Hope this helps ;D
Post by: jamonwindeyer on March 29, 2016, 06:44:01 pm
Loan Repayments Question
The smith family buys a car for $38 000, paying a 10% deposit and taking out a loan for the balance. if the loan is over 5 years with interest of 1.5% monthly, find the amount of each monthly (a) loan repayment the total amount that the (b) smith family paid for the car.
Answer:
(a) $592.00 (b) $39 319.89
I didn't forget you imtrying! I'm just not at my desk at the moment, I need to write out the series for your question (I think), so I'll tackle it tonight once I have some desk space! Unless someone else would like to have a crack ;D
Post by: imtrying on March 29, 2016, 06:58:30 pm
I didn't forget you imtrying! I'm just not at my desk at the moment, I need to write out the series for your question (I think), so I'll tackle it tonight once I have some desk space! Unless someone else would like to have a crack ;D
All good :)
Post by: RuiAce on March 30, 2016, 11:18:21 am
Loan Repayments Question
The smith family buys a car for $38 000, paying a 10% deposit and taking out a loan for the balance. if the loan is over 5 years with interest of 1.5% monthly, find the amount of each monthly (a) loan repayment the total amount that the (b) smith family paid for the car.
Answer:
(a) $592.00 (b) $39 319.89
Post by: liiz on March 30, 2016, 04:59:15 pm
Hey Liiz! Sure thing!
Okay, so this question takes a bit of experience to spot the trick for. The integral of this will be a logarithm.
Remember, the derivative of the log of some function is:
For your question, we almost have the derivative of the bottom on top of the fraction, which is what is required. To make it work, we multiply by 4 on the numerator, then multiply the integral by a quarter, thus balancing everything out. You could also think of it as taking out a factor of 1/4:
This isn't super easy to spot, it might take some practice to see it quickly! Hope this helps ;D
Ah okay, yep I understand! Thanks so much
Post by: SimonaB on March 30, 2016, 09:54:08 pm
The question is: Find the area bounded by y=sin x and y=cos 2x for π/6 ≤ x ≤ 5π/6
Thanks!
Post by: jamonwindeyer on March 30, 2016, 10:50:49 pm
I've got a question that was on my exam and I got it wrong after literally 10 attempts.
The question is: Find the area bounded by y=sin x and y=cos 2x for π/6 ≤ x ≤ 5π/6
Thanks!
Hey Simona! Welcome to the forums!! ;D Let's have a look.
The first step for an area question of this nature is to draw a quick sketch, just to get a better picture of what we are dealing with. Wolfram Alpha helped me out here, but you don't need any great level of accuracy, we just want the position of the curves with relation to each other. Note also that this sketch from Wolfram Alpha has the axes centred at the first point of intersection, totally irrelevant to this question, just ignore the coordinates! ;D
(http://i.imgur.com/nTy2OY1.png)
Okay! So it is clear from a quick sketch that the sinx curve is above the cos2x curve in this domain.
There is a really neat trick here. Even when the curves go above or below the x axis, any weird combination, we can always find the area by finding the area under the upper curve and subtracting the area under the lower curve, in the given domain. It's nice, and eliminates the need for weird sums of absolute values of integrals. Literally, the integral of the upper curve subtract the integral of the lower curve, works every time ;D
So, the working would be:
I hope this helps!!
Post by: SimonaB on March 30, 2016, 11:12:37 pm
Post by: Maz on April 03, 2016, 04:47:10 pm
can u please help me with how to differentiate this...i really don't know how to...
i'd really appreciate it :)
Post by: RuiAce on April 03, 2016, 04:56:02 pm
hey
can u please help me with how to differentiate this...i really don't know how to...
i'd really appreciate it :)
Can't see your image. If your integral was something along the lines of this then in NSW it belongs under MX2.
Also, you said differentiate? Did you mean you wanted to put d/dx in front of that integral? (or d/dt inside it?)
Post by: Maz on April 03, 2016, 04:59:49 pm
Can't see your image. If your integral was something along the lines of this then in NSW it belongs under MX2.yes that is the integral...and d/dx belongs in front of it...sorry...
Also, you said differentiate? Did you mean you wanted to put d/dx in front of that integral? (or d/dt inside it?)
and ill go put it in MX2 now then :)
Post by: Johny1234567 on April 03, 2016, 08:46:21 pm
Find the point T(a,lna) on y=lnx where the tangent passes through the origin, i.e. find the value of a.
Post by: RuiAce on April 03, 2016, 08:56:47 pm
Hello y'all. How do I solve this question:
Find the point T(a,lna) on y=lnx where the tangent passes through the origin, i.e. find the value of a.
Post by: aqsarana_ on April 10, 2016, 05:11:45 pm
So currently for Maths 2U and for Ext 1 maths, i am using the maths in focus textbook (Margaret Grove) but i want to purchase another textbook as this one is not that great. I was wondering if you could give suggestions as to what textbook is really good. I'm debating between the Fitzpatrick book and Cambridge book. Please help!
Thank you :)
Post by: jakesilove on April 10, 2016, 05:30:32 pm
Hi,
So currently for Maths 2U and for Ext 1 maths, i am using the maths in focus textbook (Margaret Grove) but i want to purchase another textbook as this one is not that great. I was wondering if you could give suggestions as to what textbook is really good. I'm debating between the Fitzpatrick book and Cambridge book. Please help!
Thank you :)
Hey!
I really think it depends on what you want a textbook for.
I had the Cambridge textbook, which (if I recall correctly) does a fairly good job of explaining concepts etc. It has a good range of past questions, which will challenge and extend any 2U student.
However, most students don't use their textbook to explain concepts. Usually, students who actually USE their textbook are the super keen ones. In that case, they just want a billion past questions to smash through.
I would always recommend doing HSC past questions rather than textbook ones. They give you a better idea of where you're at, what you need to prepare for, and exactly how you'll be assessed. However, if you've gone through all the past papers so many times that you feel like looking at them, or just want some variety, a textbook is a great option.
If you're just looking for questions, I love the Coroneos textbooks. They are simple, to the point. There are a trillion billion questions to answer. The style takes a bit of getting used to (its like they used a type writer), but the range of questions is really great. It'll extend you as well as test your basic knowledge.
But again, it depends on why you want the textbook. I don't know much about Fitzpatrick, but I would generally lean towards Cambridge of the two.
Hope this helps! I'm sure others on here will have different opinions.
Jake
Post by: RuiAce on April 10, 2016, 08:14:37 pm
Hi,
So currently for Maths 2U and for Ext 1 maths, i am using the maths in focus textbook (Margaret Grove) but i want to purchase another textbook as this one is not that great. I was wondering if you could give suggestions as to what textbook is really good. I'm debating between the Fitzpatrick book and Cambridge book. Please help!
Thank you :)
Comparison between Fitzpatrick and Cambridge is somewhat hard. This is because the two textbooks are kinda on par with question quality.
However, the Cambridge textbook's prime advantage is that it splits into foundation, development and extension, whereas Fitzpatrick doesn't mix up the topic order. Keep that in mind.
Of course, when it comes exam time, as mentioned always use past papers.
Post by: IkeaandOfficeworks on April 14, 2016, 07:34:45 pm
Post by: RuiAce on April 14, 2016, 08:28:33 pm
Hi guys!, I seem to struggle with this integration question.Can you explain why the answer is 1/6(2x-1)3/2 + 1/2(2x-1)1/2 + C ?. Thank you!
EDIT BY JAMON: Hey IkeaandOfficeworks! Glad you got an answer to your question, for any 2U students looking at this, don't stress, this is Extension 1 content!
Post by: jamesq on April 24, 2016, 02:14:23 am
For the AP: 100 + 97 + 94 + ...
Show that 68 is the minimum number of terms in which Sn is negative.
Post by: RuiAce on April 24, 2016, 08:45:44 am
I need help with a rates and finance question.
For the AP: 100 + 97 + 94 + ...
Show that 68 is the minimum number of terms in which Sn is negative.
Post by: Phillorsm on May 08, 2016, 05:34:27 pm
Some help with this question please?
https://goo.gl/3ZYQOC
Post by: jakesilove on May 08, 2016, 06:07:16 pm
Hey Guys,
Some help with this question please?
https://goo.gl/3ZYQOC
Hey! Below are my answers :) The whole tan(2x) thing was a little bit nuanced, but really it ended up being just about playing around with what you're given, and trying to figure out what they're actually asking you to do. Worst case scenario with questions like this; you can always test it on a calculator, figure out the answer without the whole 'hence' part, and fudge the reasoning.
(http://i.imgur.com/MiyII5t.png?1)
(http://i.imgur.com/L1oN7Xc.png?1)
Jake
Post by: Phillorsm on May 08, 2016, 08:23:12 pm
Thanks for the help! But i'm a little confused, isn't there another intersection at -3pi/8? Which is before the endpoint at -pi/2...
Post by: RuiAce on May 08, 2016, 08:43:29 pm
Post by: jakesilove on May 08, 2016, 09:09:24 pm
Hey Jake,
Thanks for the help! But i'm a little confused, isn't there another intersection at -3pi/8? Which is before the endpoint at -pi/2...
Hmm this is true, not sure how I missed that (was cracking out an answer on the bus, so I'll blame it on that...). I'm unfortunately not home at the moment, but will just admit my mistake and let someone else correct me. Sorry about that! You may even be able to get to an answer based roughly on my method (although hopefully more correctly).
Jake
Post by: Phillorsm on May 08, 2016, 09:35:06 pm
Post by: jakesilove on May 09, 2016, 10:25:06 am
No worries Jake, yeah I used your method and was able to get a solution I believe to be right. You had the right idea, just missed one little thing. No problem, HAPPY BIRTHDAY BY THE WAY!
Thanks Phillorsm! Appreciate it :)
Post by: Phillorsm on May 09, 2016, 02:55:56 pm
https://goo.gl/photos/wTkpMHqM6dVkFfSr7
Post by: jakesilove on May 09, 2016, 03:13:06 pm
No problem Jake, another question for you. I can do up to part iii, but part IV I am stuck.
https://goo.gl/photos/wTkpMHqM6dVkFfSr7
There ya go! Just think about what you need to find, use some random triangle to try and get the answer out!
(http://i.imgur.com/HPOel05.png?1)
Jake
Post by: Phillorsm on May 10, 2016, 05:17:15 pm
https://goo.gl/Hg97px
Post by: RuiAce on May 10, 2016, 06:30:43 pm
Hey Jake, wondering if we could revisit this question. For the final part, (the tan inequality) I seem to be getting different results from different places. The one I am currently thinking is correct is -pi/2<=x<=-3pi/8, -pi/4<x<=pi/8, pi/4<x<=pi/2
https://goo.gl/Hg97px
Refer to my post on the previous page. I have already addressed this stating that yes that IS the solution, except for one thing.
pi/4<x<pi/2
x can't equal to pi/2
Post by: Maz on May 10, 2016, 10:15:32 pm
can someone please explain how to do this question for me?
Q: Police set up a road block to check cars for seatbelt use. From experience, they estimate that 80% of drivers wear seat belts. Use this estimate to determine what the probability is that the second unbelted driver is in the 8th car stopped?
there is a second part of this question that i know how to do...but i don't know how to do this part..
any help will be much appreciated :)
Post by: RuiAce on May 10, 2016, 10:27:45 pm
hey
can someone please explain how to do this question for me?
Q: Police set up a road block to check cars for seatbelt use. From experience, they estimate that 80% of drivers wear seat belts. Use this estimate to determine what the probability is that the second unbelted driver is in the 8th car stopped?
there is a second part of this question that i know how to do...but i don't know how to do this part..
any help will be much appreciated :)
Post by: jakesilove on May 10, 2016, 10:28:02 pm
hey
can someone please explain how to do this question for me?
Q: Police set up a road block to check cars for seatbelt use. From experience, they estimate that 80% of drivers wear seat belts. Use this estimate to determine what the probability is that the second unbelted driver is in the 8th car stopped?
there is a second part of this question that i know how to do...but i don't know how to do this part..
any help will be much appreciated :)
Hey again!
My solution is below. Essentially, you need to break up the question into its constituent parts (first, 1 in 7 drivers, then, 8th driver), find the probability of each part, and multiply them together :)
(http://i.imgur.com/PlhUQpW.png?1)
Jake
Post by: jakesilove on May 10, 2016, 10:29:10 pm
Looks like Rui beat me too it :)
Post by: RuiAce on May 10, 2016, 10:30:12 pm
Looks like Rui beat me too it :)
Ninjas. Ah sorry :)
But yeah I basically also considered the binomial probability distribution. I just didn't exactly apply it since this question wasn't posted under the Extension 1 section of the forum.
Post by: Maz on May 10, 2016, 10:47:27 pm
Ninjas. Ah sorry :)
But yeah I basically also considered the binomial probability distribution. I just didn't exactly apply it since this question wasn't posted under the Extension 1 section of the forum.
Looks like Rui beat me too it :)
thank you both of you :)
i am doing binomial distribution but since I'm on the opposite side of the country as you guys i didn't know where to put which question :)
sorry ???
but thank you again :)
Post by: jamonwindeyer on May 11, 2016, 12:01:06 am
thank you both of you :)
i am doing binomial distribution but since I'm on the opposite side of the country as you guys i didn't know where to put which question :)
sorry ???
but thank you again :)
All good! 2 Unit students in NSW can do this question with their knowledge anyway, just a little more difficult than it would be for yourself or a NSW Extension 1 student ;D
Post by: jamesq on May 13, 2016, 11:19:39 pm
a) Find a formula for the balance on the loan after n months
b) How many years does it take to repay the loan
Thank you!
Post by: RuiAce on May 14, 2016, 10:05:38 am
A man takes out a $250 000 mortgage. Interest is 7.2% per annum, compounded monthly. Repayments are $3 000 monthly.
a) Find a formula for the balance on the loan after n months
b) How many years does it take to repay the loan
Thank you!
Post by: jamonwindeyer on May 14, 2016, 05:37:45 pm
Optionally, you can add a comment about the fact that a $250 000 dollar mortgage is totally unrealistic for first home buyers in Sydney ;)
Totally kidding, though it might give the marker a laugh.
Post by: ccarolineb on May 23, 2016, 08:32:57 pm
- a population in a certain city is growing at a rate proportional to the population itself. after 3 years the population increases by 20%. How long will it take for the population to double?
Post by: RuiAce on May 23, 2016, 08:58:10 pm
Struggling to understand how to do this!Inserted an edit to make Jamon's point a bit clearer. Let A=P0
- a population in a certain city is growing at a rate proportional to the population itself. after 3 years the population increases by 20%. How long will it take for the population to double?
Post by: jakesilove on May 23, 2016, 09:20:24 pm
Optionally, you can add a comment about the fact that a $250 000 dollar mortgage is totally unrealistic for first home buyers in Sydney ;)
Totally kidding, though it might give the marker a laugh.
This caused me much pain
Post by: jamonwindeyer on May 23, 2016, 09:31:33 pm
As an elaboration for you Caroline, in case you are still getting used to this style of problem, 'A' represents the initial value of the quantity in question (in this case, the population). You can prove this by substituting t=0 if you so choose, but Rui's solution is what you'd do in the HSC exam, a very standard process ;D
Post by: angela99 on May 24, 2016, 06:43:22 pm
Sub-topic: Exponential growth and decay
Formulas: Q = Ae^kt, dQ/dt = kQ
Can you help with this question please? not having obvious numbers to substitute into the formula is really throwing me off ???
The half-life of radium is
1600 years.
(a) Find the percentage of radium that will be decayed after 500 years.
(b) Find the number of years that it will take for 75% of the radium to decay.
Thank you! :) :)
Post by: jakesilove on May 24, 2016, 07:51:43 pm
Topic: Applications of Calculus to the Physical World
Sub-topic: Exponential growth and decay
Formulas: Q = Ae^kt, dQ/dt = kQ
Can you help with this question please? not having obvious numbers to substitute into the formula is really throwing me off ???
The half-life of radium is
1600 years.
(a) Find the percentage of radium that will be decayed after 500 years.
(b) Find the number of years that it will take for 75% of the radium to decay.
Thank you! :) :)
Hey! This is a standard method for exponential growth/decay questions that don't give you actual numbers, but rather percentages. You need to just LET Q equal something at time t=0 (eg. 100, like I used, or 1, which is what I usually used). Then, as time progresses, you just use an appropriate percentage of your original value. The explanation in the answer below should explain what I mean; once you've seen it once, you'll be able to do it again in a heartbeat!
(http://i.imgur.com/7MoE2dZ.jpg)
Jake
Post by: angela99 on May 24, 2016, 08:10:28 pm
Hey! This is a standard method for exponential growth/decay questions that don't give you actual numbers, but rather percentages. You need to just LET Q equal something at time t=0 (eg. 100, like I used, or 1, which is what I usually used). Then, as time progresses, you just use an appropriate percentage of your original value. The explanation in the answer below should explain what I mean; once you've seen it once, you'll be able to do it again in a heartbeat!
(http://i.imgur.com/7MoE2dZ.jpg)
Jake
Thank you!!
Post by: angela99 on May 27, 2016, 12:08:51 pm
Find, correct to 2 decimal places, the area enclosed by the curve
y = log2 x, the x-axis and the lines x = 1 and x = 3 by using simpson’s rule with 3 function values.
Post by: angela99 on May 27, 2016, 12:11:43 pm
Post by: jakesilove on May 27, 2016, 01:08:38 pm
PLEASE HELP!! THANK YOU!
Find, correct to 2 decimal places, the area enclosed by the curve
y = log2 x, the x-axis and the lines x = 1 and x = 3 by using simpson’s rule with 3 function values.
Hey! I've attached my answer below :) It's really important that you are comfortable using Simpson's rule, and I think it is also on your formula sheet in case you forget it. Just break up your area into the number of subdivisions you require, figure out the distance between each division, and then sub the values straight into the formula! For a comparison, the 'real' answer should be 1.87, so this is pretty damn close!
(http://i.imgur.com/v5DAlKt.jpg)
Jake
Post by: Anika1098 on May 28, 2016, 10:28:16 pm
1 b ii
8 a iii
10 b
14 c ii
16 b iii
17 iii Iv
18 b i ii
20
22 a iii d ii e i ii iii
I understand there are heaps so please just pick one or two that suit you best?
Here are thre questions; sorry about my dodgy af collage.
Post by: jakesilove on May 28, 2016, 11:12:11 pm
Hey Jake, sorry to bother you but we got this massive revision booklet for my test recently, but we are not allowed to ask our teachers for awnsers or anything and there are a few questions (a lot) that I'm really struggling to Understand?
1 b ii
8 a iii
10 b
14 c ii
16 b iii
17 iii Iv
18 b i ii
20
22 a iii d ii e i ii iii
I understand there are heaps so please just pick one or two that suit you best?
Here are thre questions; sorry about my dodgy af collage.
Hey!
Unfortunately I can't decipher the questions you've sent over, they are way too blurry. If you could either send higher quality pictures, or just type up the important ones, that would be great! I'm happy to help with things that you're struggling with, but maybe do try to limit the questions to like one of each type. That way, hopefully I can help explain the general way to solve those kinds of questions, and then you can figure out the others yourself.
Jake
Post by: RuiAce on May 28, 2016, 11:31:05 pm
Hey Jake, sorry to bother you but we got this massive revision booklet for my test recently, but we are not allowed to ask our teachers for awnsers or anything and there are a few questions (a lot) that I'm really struggling to Understand?
1 b ii
8 a iii
10 b
14 c ii
16 b iii
17 iii Iv
18 b i ii
20
22 a iii d ii e i ii iii
I understand there are heaps so please just pick one or two that suit you best?
Here are thre questions; sorry about my dodgy af collage.
Hey!Or maybe just a larger collage.
Unfortunately I can't decipher the questions you've sent over, they are way too blurry. If you could either send higher quality pictures, or just type up the important ones, that would be great! I'm happy to help with things that you're struggling with, but maybe do try to limit the questions to like one of each type. That way, hopefully I can help explain the general way to solve those kinds of questions, and then you can figure out the others yourself.
Jake
This is the solution to the only clear question.
Post by: itswags98 on June 26, 2016, 09:12:59 pm
Hey Jake, sorry to bother you but we got this massive revision booklet for my test recently, but we are not allowed to ask our teachers for awnsers or anything and there are a few questions (a lot) that I'm really struggling to Understand?
Must be a pretty dodgy teacher if ur not allowed to ask them questions
Post by: jakesilove on June 26, 2016, 09:26:00 pm
Must be a pretty dodgy teacher if ur not allowed to ask them questions
You got that right, surely that's just their entire job?
Post by: RuiAce on June 26, 2016, 09:34:07 pm
Must be a pretty dodgy teacher if ur not allowed to ask them questions
You got that right, surely that's just their entire job?Yep. People like them should be fired.
Post by: itswags98 on June 27, 2016, 08:45:09 pm
Yep. People like them should be fired.I thank the moon, stars and whatever the hell else is up there that i have the best maths teacher ever who puts in so much effort for her students. ;D
The difference between teachers can really make or break a subject IMO
Post by: jamonwindeyer on June 27, 2016, 09:01:29 pm
I thank the moon, stars and whatever the hell else is up there that i have the best maths teacher ever who puts in so much effort for her students. ;D
The difference between teachers can really make or break a subject IMO
Absolutely this. And the thing is, it's not just intelligence. I've had lecturers who were absolute wizards, knew their shit like absolutely nothing I've ever seen, but I still didn't like them as a lecturer. Then I have had tutors who know their course content really well and not too much beyond that, but holy crap they just knew how to explain it!! Being able to explain something well is just so important, something which I fear is overlooked in favour of pure academic skill.
The best teachers in my experience are the ones who love their job, because it shows ;D
Post by: RuiAce on June 27, 2016, 09:12:55 pm
I thank the moon, stars and whatever the hell else is up there that i have the best maths teacher ever who puts in so much effort for her students. ;D
The difference between teachers can really make or break a subject IMO
My MX2 teacher did a Ph. D in some field of pure mathematics.
His explanations were pretty much otherworldly. Maths was magic.
Post by: conic curve on June 27, 2016, 09:22:20 pm
I want to do it without inspection
Also how do I answer people's maths questions if I were to answer them?
Post by: conic curve on June 27, 2016, 09:33:17 pm
How do I find the domain of y=(3-x)squareroot of x
Also how do I type up maths questions on this forum?
Post by: jakesilove on June 27, 2016, 09:57:45 pm
Let me retype my question above
How do I find the domain of y=(3-x)squareroot of x
Also how do I type up maths questions on this forum?
Hey! So first, I'll answer your Maths question. To find the domain of
We just need to consider which values of x are essentially 'not allowed'. Generally, to do this we just set whatever is in the square root to be greater than zero (noting that the square root of a negative number does not exist!). So, this one turns out the be easy; the (3-x) bit doesn't do anything, but the square root of x means that x must be greater than 0! In maths;
Not too difficult!
Now, to answer your second question, you started on the right track (using the LaTeX button in the toolbar). However, various functions have various codes attributable to them. Check out Rui's incredible guide HERE to find out how to use it! I'm pretty shit at it, but Rui and Jamon are pros.
Jake
Post by: jakesilove on June 27, 2016, 09:59:36 pm
How do I find the domain of y=(3-x)?
I want to do it without inspection
Also how do I answer people's maths questions if I were to answer them?
It's great to here that you want to answer maths questions! When you see something you want to take a crack at, just hit "reply" on their post, and type out a response. You can hand write an answer and attach it using the mona lisa button in the toolbar (you'll need to upload the image to Imgr, as you need a hyperlink, but that doesn't take too long) or use LaTeX to type using maths, as I've explained above!
Looking forward to seeing more of you on the forum :)
Jake
Post by: conic curve on June 27, 2016, 10:07:35 pm
:-\
It's great to here that you want to answer maths questions! When you see something you want to take a crack at, just hit "reply" on their post, and type out a response. You can hand write an answer and attach it using the mona lisa button in the toolbar (you'll need to upload the image to Imgr, as you need a hyperlink, but that doesn't take too long) or use LaTeX to type using maths, as I've explained above!
Looking forward to seeing more of you on the forum :)
Jake
Thanks Jake
Post by: conic curve on June 27, 2016, 10:14:40 pm
Hey! So first, I'll answer your Maths question. To find the domain of
We just need to consider which values of x are essentially 'not allowed'. Generally, to do this we just set whatever is in the square root to be greater than zero (noting that the square root of a negative number does not exist!). So, this one turns out the be easy; the (3-x) bit doesn't do anything, but the square root of x means that x must be greater than 0! In maths;
Not too difficult!
Now, to answer your second question, you started on the right track (using the LaTeX button in the toolbar). However, various functions have various codes attributable to them. Check out Rui's incredible guide HERE to find out how to use it! I'm pretty shit at it, but Rui and Jamon are pros.
Jake
Once again thanks
So basically (3-x) has no effect on the domain? If so why?
Post by: jamonwindeyer on June 27, 2016, 10:15:34 pm
Once again thanks
So basically (3-x) has no effect on the domain? If so why?
Precisely! Because 3-x has no restrictions, we can put anything into it and it makes sense. It's only the square root term that causes issues (think Math error on your calculator) ;D
Post by: conic curve on June 27, 2016, 10:17:31 pm
Precisely! Because 3-x has no restrictions, we can put anything into it and it makes sense. It's only the square root term that causes issues (think Math error on your calculator) ;D
Thanks Jamon
Post by: itswags98 on June 27, 2016, 10:55:04 pm
Absolutely this. And the thing is, it's not just intelligence. I've had lecturers who were absolute wizards, knew their shit like absolutely nothing I've ever seen, but I still didn't like them as a lecturer. Then I have had tutors who know their course content really well and not too much beyond that, but holy crap they just knew how to explain it!! Being able to explain something well is just so important, something which I fear is overlooked in favour of pure academic skill.Theres that famous saying that goes along the lines of "If you cant explain it to a 6 year old in simple terms, you don't know it well enough" ahah
The best teachers in my experience are the ones who love their job, because it shows ;D
Post by: RuiAce on June 28, 2016, 12:20:02 am
Theres that famous saying that goes along the lines of "If you cant explain it to a 6 year old in simple terms, you don't know it well enough" ahah
Try explaining linear algebra to a 6 year old
Post by: jamonwindeyer on June 28, 2016, 09:10:51 am
Try explaining linear algebra to a 6 year old
Coordinates are steps to the left or right, or up and down. If you wanted the parametric version of a line you could explain it in terms of walking to a starting point, then walking so many steps in a given direction. Mappings would be taking longer steps, shorter steps, swapping the direction, etc. Vector spaces would be like the map of the park where you are walking. Etc.
Of course none of this would give them any actual useful skills, but kind of fun to think about!! ;) I'm a big believer than you can teach anyone, anything (albeit at a very simple level) provided they have the right dedication. Of course, this relies on 6 year olds having the dedication to try and understand linear algebra... So maybe not ;) anyway, back to the math!! ;D
Post by: RuiAce on June 28, 2016, 09:44:52 am
Coordinates are steps to the left or right, or up and down. If you wanted the parametric version of a line you could explain it in terms of walking to a starting point, then walking so many steps in a given direction. Mappings would be taking longer steps, shorter steps, swapping the direction, etc. Vector spaces would be like the map of the park where you are walking. Etc.
Of course none of this would give them any actual useful skills, but kind of fun to think about!! ;) I'm a big believer than you can teach anyone, anything (albeit at a very simple level) provided they have the right dedication. Of course, this relies on 6 year olds having the dedication to try and understand linear algebra... So maybe not ;) anyway, back to the math!! ;D
You can probably explain a coordinate system but try to explain cross product hahaha
Or more fun try explaining Gaussian elimination!
Post by: itswags98 on June 29, 2016, 07:03:48 pm
Everywhere i look at Maths resources, i find study notes of maths... which is not something i want. Ive looked all over the place and i cant seem to find simple formula sheets of the stuff in the course. Unfortunately, this is something i really suck at doing. Ive written study notes for all my subjects but for maths I prefer a simple, clear list of formulas. Any idea where i can find something like this?
Post by: RuiAce on June 29, 2016, 07:06:50 pm
Im not sure if this is the right place to ask this but eh...
Everywhere i look at Maths resources, i find study notes of maths... which is not something i want. Ive looked all over the place and i cant seem to find simple formula sheets of the stuff in the course. Unfortunately, this is something i really suck at doing. Ive written study notes for all my subjects but for maths I prefer a simple, clear list of formulas. Any idea where i can find something like this?
You're literally given a formula sheet by BOSTES now for the exam.
Post by: itswags98 on June 29, 2016, 07:16:37 pm
You're literally given a formula sheet by BOSTES now for the exam.sure, but theres way too many things missing from it.
Post by: jamonwindeyer on June 29, 2016, 07:20:59 pm
sure, but theres way too many things missing from it.
Totally agree! If there is nothing out there, I'll make it my business to get one of these formula sheets written in the next week or two and uploaded to the site! Great idea ;D
Post by: RuiAce on June 29, 2016, 07:47:07 pm
sure, but theres way too many things missing from it.
Totally agree! If there is nothing out there, I'll make it my business to get one of these formula sheets written in the next week or two and uploaded to the site! Great idea ;DWhat's missing that you guys want to see so badly?
Post by: jamonwindeyer on June 29, 2016, 08:29:00 pm
What's missing that you guys want to see so badly?
It's not something I'd want to see in an exam scenario (well this is given since I did without) but even with my tutoring students off the top of my head some missing ones:
- Area of a Secant
- Multiple Applications of Trapezoidal/Simpson's Rule
- General Forms of the Trig Expansions
I'm sure there are more ;D it's like Physics and Chem where you definitely have enough to get by, but it's nice to have something comprehensive around that covers absolutely everything!! And in a form that is easier to study from (I'm thinking this is a study resource more than anything) ;D
Post by: RuiAce on June 29, 2016, 08:37:37 pm
It's not something I'd want to see in an exam scenario (well this is given since I did without) but even with my tutoring students off the top of my head some missing ones:Well I guess the HSC basically just gave you all the simplified formulae that you need cause
- Area of a Secant
- Multiple Applications of Trapezoidal/Simpson's Rule
- General Forms of the Trig Expansions
I'm sure there are more ;D it's like Physics and Chem where you definitely have enough to get by, but it's nice to have something comprehensive around that covers absolutely everything!! And in a form that is easier to study from (I'm thinking this is a study resource more than anything) ;D
Area of a segment = Area of a sector minus area of a triangle
Trap/Simpson's rule can just be done recursively (tbh, the general Simpson's rule depends on if you start counting at x0 or x1)
Trig is idk I just know it
But yeah fair enough if you want sophistication I can see where you're coming at. But at least they give you the building blocks in the HSC now instead of just an integrals sheet!
Post by: jamonwindeyer on June 29, 2016, 10:14:10 pm
Well I guess the HSC basically just gave you all the simplified formulae that you need cause
Area of a segment = Area of a sector minus area of a triangle
Trap/Simpson's rule can just be done recursively (tbh, the general Simpson's rule depends on if you start counting at x0 or x1)
Trig is idk I just know it
But yeah fair enough if you want sophistication I can see where you're coming at. But at least they give you the building blocks in the HSC now instead of just an integrals sheet!
Oh absolutely, yes the current formula sheet is more than enough!! I'm interested to see whether the difficulty of the exam is any different in compensation for the sheet, or whether the sorts of questions will change at all :)
PS - Different counting spots for Simpson's Rule was so annoying in my HSC year I literally refused to look at any formula sheets with it on it but my own posters :P
Post by: conic curve on June 30, 2016, 07:20:00 pm
Post by: jamonwindeyer on June 30, 2016, 07:39:03 pm
For trigonometry limits, why do we have to multiply by the fraction?
Hey conic, not quite sure what you mean! Could you give an example?? ;D
Post by: anotherworld2b on July 03, 2016, 10:47:06 am
I was wondering if i could get help in understanding what i can and can not do to prove trig identities
and how i should interpret the proofs
I also seem to do illegal steps in getting one side equal the other and never seem to understand how I should 'visualise' and understand what to do when given a proof
Post by: RuiAce on July 03, 2016, 10:55:39 am
Hi at school we've started trigonometry identities proofsBe clearer about the word "illegal". Only certain things are illegal and it always depends on what you're trying to prove and restrictions.
I was wondering if i could get help in understanding what i can and can not do to prove trig identities
and how i should interpret the proofs
I also seem to do illegal steps in getting one side equal the other and never seem to understand how I should 'visualise' and understand what to do when given a proof
Post by: wyzard on July 03, 2016, 11:35:07 am
Hi at school we've started trigonometry identities proofs
I was wondering if i could get help in understanding what i can and can not do to prove trig identities
and how i should interpret the proofs
I also seem to do illegal steps in getting one side equal the other and never seem to understand how I should 'visualise' and understand what to do when given a proof
What exactly do you mean by 'illegal'?
If it means doing something that is mathematically wrong like dividing by zero on both sides on a equation, or assuming what you are trying to prove, then that's a big no no.
However if you're using theorems, results and formulas that are outside of the syllabus and not taught covered in school, that's a big yes as it shows you have the initiative to learn more than you're expected to.
Post by: RuiAce on July 03, 2016, 11:52:48 am
What exactly do you mean by 'illegal'?Not necessarily in the HSC. Concepts such as L'Hopital's rule that used to get abused by students are now strictly forbidden and subject to being penalised.
If it means doing something that is mathematically wrong like dividing by zero on both sides on a equation, or assuming what you are trying to prove, then that's a big no no.
However if you're using theorems, results and formulas that are outside of the syllabus and not taught covered in school, that's a big yes as it shows you have the initiative to learn more than you're expected to.
Post by: anotherworld2b on July 03, 2016, 12:55:14 pm
I was referring to my habit of constantly doing something that is mathematically wrong/not possible.
How do you know what you can and can not do in regards to the trig identities?
Are there particular rules that must be followed when proving trig identities?
Post by: RuiAce on July 03, 2016, 12:58:22 pm
I apologize for not being clear about what I meant by 'illegal'The rule that everything is not a mathematical mistake or misuse in assumption? Show me an example of where you messed up and did something "illegal" for a better idea.
I was referring to my habit of constantly doing something that is mathematically wrong/not possible.
How do you know what you can and can not do in regards to the trig identities?
Are there particular rules that must be followed when proving trig identities?
Post by: wyzard on July 03, 2016, 03:54:08 pm
Not necessarily in the HSC. Concepts such as L'Hopital's rule that used to get abused by students are now strictly forbidden and subject to being penalised.
Damn didn't realize the use L'Hopital rule got caught on, it's such a useful theorem to use though ;D Yeah I agree with you though, be mindful of what concepts you can't use too.
I apologize for not being clear about what I meant by 'illegal'
I was referring to my habit of constantly doing something that is mathematically wrong/not possible.
How do you know what you can and can not do in regards to the trig identities?
Are there particular rules that must be followed when proving trig identities?
Unfortunately there's no easy way out of this, the process of learning mathematics is by making mistakes, and learning from them. So the more mistake you make, the better. It's best learned by experience, just jump in there, make many silly "illegal" moves, and study why you can't do that. Over time, you'll become more sensitive to the mistake and you'll avoid them.
Post by: RuiAce on July 03, 2016, 06:03:30 pm
Damn didn't realize the use L'Hopital rule got caught on, it's such a useful theorem to use though ;D Yeah I agree with you though, be mindful of what concepts you can't use too.Yep
One of the more minor reasons as to why I wanted to do uni maths already was just to abuse the crap out of L'H ;D
But the HSC is just the way it is. Always try to use what's in the syllabuses. I think one of the most lenient things they do is allow implicit differentiation in Ext 1...when it's really just properly taught in Ext 2.
Post by: jamonwindeyer on July 03, 2016, 07:59:05 pm
Yep
One of the more minor reasons as to why I wanted to do uni maths already was just to abuse the crap out of L'H ;D
But the HSC is just the way it is. Always try to use what's in the syllabuses. I think one of the most lenient things they do is allow implicit differentiation in Ext 1...when it's really just properly taught in Ext 2.
You'll get marks for the correct answer (if they are available based on the criteria) but not the working ;D you're totally right, stick to the method taught in that course, no point playing with fire :P
Would like to say that L'Hopital was my favourite thing in first year math, now it's probably Cauchy-Goursat's Theorem for use in complex contour integration (beautiful math) ;)
Post by: conic curve on July 04, 2016, 09:08:06 am
I don't seem to understand things like x^2+x+1=a(x-1)^2+b(x-1)+c
How is a=1 c=3 b=3 in this case
Same with this case 3x^2-5x-2=a(x-1)^2+b(x+1)+c
How is a=3 c=-6 and b=1
If anyone here is good at explaining clearly then could you please help me
Thanks ;D
Post by: RuiAce on July 04, 2016, 09:26:27 am
Okay so I am having trouble with "quadratic identities"
I don't seem to understand things like x^2+x+1=a(x-1)^2+b(x-1)+c
How is a=1 c=3 b=3 in this case
Same with this case 3x^2-5x-2=a(x-1)^2+b(x+1)+c
How is a=3 c=-6 and b=1
If anyone here is good at explaining clearly then could you please help me
Thanks ;D
Post by: wyzard on July 04, 2016, 11:08:45 am
Would like to say that L'Hopital was my favourite thing in first year math, now it's probably Cauchy-Goursat's Theorem for use in complex contour integration (beautiful math) ;)
Oh that's such a sweet result :-* I'd say my the Residue Theorem's (again from Complex Analysis) my favorite, and how it can be used to find really nasty integrals and find inverse Laplace Transforms so efficiently.
Post by: jamonwindeyer on July 04, 2016, 08:41:07 pm
Oh that's such a sweet result :-* I'd say my the Residue Theorem's (again from Complex Analysis) my favorite, and how it can be used to find really nasty integrals and find inverse Laplace Transforms so efficiently.
Definitely up there, love the Residue Theorem, after I learned it I was like "Why did you even bother teaching us Cauchy's Integral Theorem then?"
Post by: conic curve on July 11, 2016, 01:38:12 pm
6. A motorbike and car leave a service station at the same time. The motorbike travels on a bearing of 080 degrees and the car travels for 15.7km on a bearing of 108 degrees until the bearing of the motorbike from the car is 310 degrees. How far, correct to 1 decimal place, has the motorbike travelled?
7. A plane flies from Dubbo on a bearing of 139 degrees for 852 km then turns and flies on a bearing of 285 degrees until it is due west of Dubbo. How far from Dubbo is the place, to the nearest km?
8. Stig leaves home and trave;s on a bearing of 248 degrees for 109.8 km. He then tuens and travels for 271.8 km on a bearing of 143 degrees. Stif then turns and travels home on a bearing of "a"
a. How far does he travel on the final part of his journey?
b. Evaluate a
9. Two cars leave an intersection at the same time, one travelling at 70km/h along one road and he other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection?
Post by: RuiAce on July 11, 2016, 05:12:41 pm
Need help with the following questions:The others are lecturing right now and I'm afraid I don't have paper on me at this moment. Have you tried drawing diagrams?
6. A motorbike and car leave a service station at the same time. The motorbike travels on a bearing of 080 degrees and the car travels for 15.7km on a bearing of 108 degrees until the bearing of the motorbike from the car is 310 degrees. How far, correct to 1 decimal place, has the motorbike travelled?
7. A plane flies from Dubbo on a bearing of 139 degrees for 852 km then turns and flies on a bearing of 285 degrees until it is due west of Dubbo. How far from Dubbo is the place, to the nearest km?
8. Stig leaves home and trave;s on a bearing of 248 degrees for 109.8 km. He then tuens and travels for 271.8 km on a bearing of 143 degrees. Stif then turns and travels home on a bearing of "a"
a. How far does he travel on the final part of his journey?
b. Evaluate a
9. Two cars leave an intersection at the same time, one travelling at 70km/h along one road and he other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection?
If you could upload images of diagram I can tell you if it looks correct, and then just sketch out the method to attempting these questions.
Post by: happy_turtle on July 12, 2016, 05:14:46 pm
Need help with the following questions:
6. A motorbike and car leave a service station at the same time. The motorbike travels on a bearing of 080 degrees and the car travels for 15.7km on a bearing of 108 degrees until the bearing of the motorbike from the car is 310 degrees. How far, correct to 1 decimal place, has the motorbike travelled?
7. A plane flies from Dubbo on a bearing of 139 degrees for 852 km then turns and flies on a bearing of 285 degrees until it is due west of Dubbo. How far from Dubbo is the place, to the nearest km?
8. Stig leaves home and trave;s on a bearing of 248 degrees for 109.8 km. He then tuens and travels for 271.8 km on a bearing of 143 degrees. Stif then turns and travels home on a bearing of "a"
a. How far does he travel on the final part of his journey?
b. Evaluate a
9. Two cars leave an intersection at the same time, one travelling at 70km/h along one road and he other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection?
I have question 9... I'm not sure about the others though
Post by: RuiAce on July 12, 2016, 07:03:58 pm
I have question 9... I'm not sure about the others though
This looks correct. I had glanced at Q9 and realised that it would throw some people off because it makes you go back to speed=distance/time, but judging by that you catered it correctly (and the cosine rule does not look incorrectly substituted) it appears correct
Post by: jamonwindeyer on July 12, 2016, 09:36:01 pm
I have question 9... I'm not sure about the others though
Thanks for that answer happy_turtle, and welcome to the forums!! Absolute legend ;D conic I'll definitely come and give you an explanation of this style of question tomorrow, just need to finish this lecture series first ;) I'll run through one besides Question 9 ;D
Post by: conic curve on July 12, 2016, 09:38:37 pm
Thanks for that answer happy_turtle, and welcome to the forums!! Absolute legend ;D conic I'll definitely come and give you an explanation of this style of question tomorrow, just need to finish this lecture series first ;) I'll run through one besides Question 9 ;D
Okay cheers
I need someone to comment below so then I can post a couple of new questions down here. If I post consecutively, it will be labelled spam and I may get a warning
Post by: HighTide on July 13, 2016, 10:38:28 am
Okay cheersJust going to highlight that you can post multiple qs in the same comment and avoid a warning. You can also edit your comment by clicking modify. You don't need to post consecutively to ask multiple qs :)
I need someone to comment below so then I can post a couple of new questions down here. If I post consecutively, it will be labelled spam and I may get a warning
Post by: RuiAce on July 13, 2016, 10:41:32 am
Just going to highlight that you can post multiple qs in the same comment and avoid a warning. You can also edit your comment by clicking modify. You don't need to post consecutively to ask multiple qs :)Can I just say, on a person note I really do not wish to be bombarded by a continuous streak of questions especially of the same topic. It begins to feel like a chore, instead of a generous act, in helping.
Post by: conic curve on July 13, 2016, 11:07:40 am
Can I just say, on a person note I really do not wish to be bombarded by a continuous streak of questions especially of the same topic. It begins to feel like a chore, instead of a generous act, in helping.
Well, I asked only like 4 questions here. Also you seem to spend a lot of time on this online forum. May I ask but why is that the case? Whenever I'm online (supposedly trying to increase my post count in order to qualify for more free essays/creatives to mark) I tend to see you online here as well?
Post by: RuiAce on July 13, 2016, 12:59:55 pm
Well, I asked only like 4 questions here. Also you seem to spend a lot of time on this online forum. May I ask but why is that the case? Whenever I'm online (supposedly trying to increase my post count in order to qualify for more free essays/creatives to mark) I tend to see you online here as well?It's becoming pushy when it's the same trigonometry based questions. If your diagram is wrong, then you don't make any progress altogether.
Your answer: Because I leave the website open in its own separate tab. Sometimes I look through it, sometimes it sits there like my Facebook.
As for the bit in the brackets, that is most likely assumed by some people because you seem to go a bit excessively on and on over it. Starts to feel like your only reason to be on here - boosting up the post count to have 50 or so essays marked. I'm sure that's not the case, but I reckon you wouldn't be too obsessed over your post count otherwise.
Post by: heids on July 13, 2016, 01:24:09 pm
I need someone to comment below so then I can post a couple of new questions down here. If I post consecutively, it will be labelled spam and I may get a warning
Well, I asked only like 4 questions here. Also you seem to spend a lot of time on this online forum. May I ask but why is that the case? Whenever I'm online (supposedly trying to increase my post count in order to qualify for more free essays/creatives to mark) I tend to see you online here as well?
I see you. This sounds quite like... 'elaborate trolling' to me...
We don't want to be too strict but if I see you post about your post count or random useless irrelevancies, I'll just delete rather than discussing it, because it clutters the place irrelevantly so others can't give/receive useful help so easily. Fair? :) This is one last request to be respectful of others. The help RuiAce and others are giving is free, fast and simply amazing and it's incredibly rude for someone to start taking it as a given and keep demanding them to do more and more for that user alone; they aren't your private, paid tutors.
So this is the last of us discussing your posting habits in public; please PM me if you think we're treating you unfairly :)
Post by: conic curve on July 13, 2016, 02:08:17 pm
For a graph like x^3 if you were to differentiate it, it would be 3x^2 however whenever graphed here: https://www.desmos.com/calculator it doesn't seem to be the case
Can someone here help me understand the concept of differnetiation and when it comes to cubic equations like this
Thanks
Post by: Floatzel98 on July 13, 2016, 02:34:12 pm
For a graph like x^3 if you were to differentiate it, it would be 3x^2 however whenever graphed here: https://www.desmos.com/calculator it doesn't seem to be the caseNothing changes about differentiating cubic functions compared to any other power functions. I don't know exactly what you graphed, but the two graphs should be the same.
Post by: EEEEEEP on July 13, 2016, 02:42:48 pm
Whenever you differentatiate, you are finding the gradient of the tangent/finding the rate of change, right?
For a graph like x^3 if you were to differentiate it, it would be 3x^2 however whenever graphed here: https://www.desmos.com/calculator it doesn't seem to be the case
Can someone here help me understand the concept of differnetiation and when it comes to cubic equations like this
Thanks
You differentiate and then sub in values to find the rate of change from A to B (or whatever)..
EDIT By Jamon: Ensure all discussion is civil and respects the right of others to ask questions :)
Post by: conic curve on July 13, 2016, 02:45:11 pm
Nothing changes about differentiating cubic functions compared to any other power functions. I don't know exactly what you graphed, but the two graphs should be the same.
x^3 is a cubic and 3x^2 is a parabola. They are two separate graphs that do not touch each other except for when they touch each other at the origin
Post by: WLalex on July 13, 2016, 02:51:12 pm
Whenever you differentatiate, you are finding the gradient of the tangent/finding the rate of change, right?
For a graph like x^3 if you were to differentiate it, it would be 3x^2 however whenever graphed here: https://www.desmos.com/calculator it doesn't seem to be the case
Can someone here help me understand the concept of differnetiation and when it comes to cubic equations like this
Thanks
I don't quite understand your question...they are obviously going to be different when graphed as the derivative and original function are now different, one a cubic and one a parabola?
I'd like to be of more help :P
Alex
Post by: Floatzel98 on July 13, 2016, 02:54:46 pm
x^3 is a cubic and 3x^2 is a parabola. They are two separate graphs that do not touch each other except for when they touch each other at the originYes x^3 and 3x^2 are indeed different graphs. When you differentiate x^3 you get 3x^2.
If you graph x^3 against 3x^2 you will get different graphs. If you graph d/dx(x^3) against 3x^2 you will get the exact same graph.
Post by: RuiAce on July 13, 2016, 02:57:14 pm
Do not, under any circumstances, assume that the graph of the derivative is supposed to be a consequence of the actual y-coordinates of the original function. The derivative is only a measure of the GRADIENT.
Post by: conic curve on July 13, 2016, 03:01:10 pm
Post by: RuiAce on July 13, 2016, 03:05:42 pm
Just in case I am confusing, I hope this helps everyone outEDIT by Jamon: Ensure all discussion is civil and respects others right to ask questions.
Draw a tangent on y=x3. The GRADIENT of the TANGENT, is the y-coordinate on the derivative.
Post by: EEEEEEP on July 13, 2016, 03:06:44 pm
Just in case I am confusing, I hope this helps everyone outTHe original equation is different from the first differentiation.
The first differentiation which is 3x^2, shows the gradients (which also happens to be the rate of change). Both equations should look different.
What else don't you get?
Post by: conic curve on July 13, 2016, 03:09:12 pm
I told you, do NOT relate the y-coordinates of the derivative, to the y-coordinates of the original function.
Draw a tangent on y=x3. The GRADIENT of the TANGENT, is the y-coordinate on the derivative.
Using this: http://www.teacherschoice.com.au/Maths_Library/Calculus/tangents_and_normals.htm
Post by: RuiAce on July 13, 2016, 03:10:52 pm
Using this: http://www.teacherschoice.com.au/Maths_Library/Calculus/tangents_and_normals.htmAccording to the website you gave, the derivative is a measure of the GRADIENT of the TANGENT at the point (to the curve). Anything else?
Obviously the gradient of the tangent at the point does not have to resemble that of the actual curve. The tangent to y=x2 at (1,1) is 2. What does that have to do with the original curve? Nothing.
The derivative is not a tilted graph of the original thing. It ONLY measures the gradient of the tangent. If you want a function whose derivative is itself y=ex is your answer and you do that in Yr 12.
Post by: conic curve on July 13, 2016, 03:16:00 pm
According to the website you gave, the derivative is a measure of the GRADIENT of the TANGENT at the point (to the curve). Anything else?
Obviously the gradient of the tangent at the point does not have to resemble that of the actual curve. The tangent to y=x2 at (1,1) is 2. What does that have to do with the original curve? Nothing.
Okay I think I am very confusing
ANyways, going back to basics, if you were to differentaite y=x, you would get y=1 right? This is finding the gradient of the tangent (i.e. the rate of change). As you can see, the gradient of the tangent is "visible" on this graph
For the example I posted originally, it was y=x^3 and if that was differentiated (i.e. the rate of change was found) it was y=3x^2 and that did not touch the graph at all except for the origin. Is this case shouldn't of 3x^2 of had been the gradient of the tanngent to x^3
If I am very confusing then just forget about this because I feel bad making all of you guys do this to me
Post by: RuiAce on July 13, 2016, 03:18:09 pm
Okay I think I am very confusingOk. Let's clear one thing up.
ANyways, going back to basics, if you were to differentaite y=x, you would get y=1 right? This is finding the gradient of the tangent (i.e. the rate of change). As you can see, the gradient of the tangent is "visible" on this graph
For the example I posted originally, it was y=x^3 and if that was differentiated (i.e. the rate of change was found) it was y=3x^2 and that did not touch the graph at all except for the origin. Is this case shouldn't of 3x^2 of had been the gradient of the tanngent to x^3
If I am very confusing then just forget about this because I feel bad making all of you guys do this to me
x is not necessarily a number. x is a variable. We have to let x = a certain number, to get the gradient of the tangent, at that number.
The plot of the derivative shows the gradient of the tangent, EVERYWHERE on the original curve.
It is not immediately obvious that 3x2 is the derivative of x3. The reason why this is the case is that
Post by: conic curve on July 13, 2016, 03:22:11 pm
Ok. Let's clear one thing up.
x is not necessarily a number. x is a variable. We have to let x = a certain number, to get the gradient of the tangent, at that number.
The plot of the derivative shows the gradient of the tangent, EVERYWHERE on the original curve.
It is not immediately obvious that 3x2 is the derivative of x3. The reason why this is the case is that
Oh....Right, I get it now thank you thank you thank you ;D
Sorry for making you go through the pain in answering this :'(
Post by: RuiAce on July 13, 2016, 03:24:22 pm
Oh....Right, I get it now thank you thank you thank you ;DGlad that we eventually got there
Sorry for making you go through the pain in answering this :'(
Post by: shivali on July 13, 2016, 07:21:06 pm
My working:
T0 = 100
T80 = 90
Tn=T0e^(kn)
T80 = T0e^(80k)
90 = 100e^(80k)
k = (ln(0.9))/80
T500 = 100e^500k
T500 = 51.72655
Therefore the substance will decay by 51.7% in 500 years?
Post by: RuiAce on July 13, 2016, 07:39:40 pm
Question: "A radioactive substance decays by 10% after 80 years. By how much will it decay after 500 years?"
My working:
T0 = 100
T80 = 90
Tn=T0e^(kn)
T80 = T0e^(80k)
90 = 100e^(80k)
k = (ln(0.9))/80
T500 = 100e^500k
T500 = 51.72655
Therefore the substance will decay by 51.7% in 500 years?
Permit k to be negative in your working out.
Post by: shivali on July 13, 2016, 07:48:02 pm
Permit k to be negative in your working out.
The correct answer was 48.2% ? In the textbook
Post by: RuiAce on July 13, 2016, 08:15:41 pm
51.76...% of the original mass is whatever is left behind.
100% - 51.76%... of the original mass is what has actually decayed.
Post by: jakesilove on July 14, 2016, 09:04:17 am
Edit: No never mind. This is exactly what happens when you try to rush.
51.76...% of the original mass is whatever is left behind.
100% - 51.76%... of the original mass is what has actually decayed.
Can we all just appreciate that Rui has answered about 100 questions in the past three days. An Atar Notes legend,who deserves all of our thanks. You're an absolute king, and hundreds of HSC students are better off with you around!
Jake
Post by: ehatton2016 on July 14, 2016, 12:12:21 pm
I have like 2 and a bit weeks to trials and have been pumping our past 2U papers but my marks in each has not improved at all!!
I've done the whole, do extra revision on stuff you get wrong but IT ISN'T WORKING!!! :(
What do I do? I'm usually a band 6 student in maths, so this is really annoying me!!
Thank you!!
Post by: RuiAce on July 14, 2016, 12:14:53 pm
Hey Jake!When you do past papers, do you struggle on questions? Because doing past papers "ineffectively" is the same thing as ineffective studying habits.
I have like 2 and a bit weeks to trials and have been pumping our past 2U papers but my marks in each has not improved at all!!
I've done the whole, do extra revision on stuff you get wrong but IT ISN'T WORKING!!! :(
What do I do? I'm usually a band 6 student in maths, so this is really annoying me!!
Thank you!!
Or are you saying that you struggle out of exam pressure.
Post by: jakesilove on July 14, 2016, 12:17:35 pm
Hey Jake!
I have like 2 and a bit weeks to trials and have been pumping our past 2U papers but my marks in each has not improved at all!!
I've done the whole, do extra revision on stuff you get wrong but IT ISN'T WORKING!!! :(
What do I do? I'm usually a band 6 student in maths, so this is really annoying me!!
Thank you!!
Hey!
Firstly, welcome to the forums! Nice job studying for trials; sounds like you're really putting the hours in.
Firstly, two and a bit weeks is ages. You could learn the entire Maths curriculum from scratch in that time.
It looks like you've identified what areas you are weak in, because you have tried doing revision for topics that you lose marks in. What kind of sections are you struggling with? At the end of the day, if there is a section you're not very confident with, and you just do a billion questions of that type, you might not improve at all. That's because, in those sections, it might come down to UNDERSTANDING, before repetition. If you struggle to understand a concept, doing it wrong 100 times won't improve your mark.
Past papers is really all you can do. Write out questions you get incorrect, and redo them once a day for a few days until you can do them with ease. This will solidify the methodology in your mind, and make sure that none of your time is wasted. Again, if you let us know where you are specifically struggling with, maybe we can help!
Keep at it. You have so much time left, and doing past papers DOES help even if it feels like it doesn't. There's no like secret trick or anything, nothing I can tell you right now that will get you 100% in every paper you do. But past papers helps you improve, and identifying areas of weakness will help you improve. If you have any specific questions, post them here, and we can talk them through with you in a way that will hopefully help you understand the fundamental underlying mathematics as well as the answer itself.
You'll be fine! Keep at it.
Jake
Post by: bing3 on July 14, 2016, 03:22:36 pm
Post by: RuiAce on July 14, 2016, 03:26:41 pm
How do I do part (b) of this question?
Your answers should be somewhere around the value here.
Post by: ehatton2016 on July 14, 2016, 07:05:20 pm
Hey!
Firstly, welcome to the forums! Nice job studying for trials; sounds like you're really putting the hours in.
Firstly, two and a bit weeks is ages. You could learn the entire Maths curriculum from scratch in that time.
It looks like you've identified what areas you are weak in, because you have tried doing revision for topics that you lose marks in. What kind of sections are you struggling with? At the end of the day, if there is a section you're not very confident with, and you just do a billion questions of that type, you might not improve at all. That's because, in those sections, it might come down to UNDERSTANDING, before repetition. If you struggle to understand a concept, doing it wrong 100 times won't improve your mark.
Past papers is really all you can do. Write out questions you get incorrect, and redo them once a day for a few days until you can do them with ease. This will solidify the methodology in your mind, and make sure that none of your time is wasted. Again, if you let us know where you are specifically struggling with, maybe we can help!
Keep at it. You have so much time left, and doing past papers DOES help even if it feels like it doesn't. There's no like secret trick or anything, nothing I can tell you right now that will get you 100% in every paper you do. But past papers helps you improve, and identifying areas of weakness will help you improve. If you have any specific questions, post them here, and we can talk them through with you in a way that will hopefully help you understand the fundamental underlying mathematics as well as the answer itself.
You'll be fine! Keep at it.
Jake
Thank you! I will definitely keep doing questions, it's my nerdy form of procrastination! By the way, your chem lecture was great!
Post by: jakesilove on July 14, 2016, 07:31:22 pm
Thank you! I will definitely keep doing questions, it's my nerdy form of procrastination! By the way, your chem lecture was great!
Really appreciate the positive feedback! Hope you learnt a thing or two, and definitely keep up the interaction on the forums!
Post by: conic curve on July 15, 2016, 12:57:52 pm
x+y, x-y, x-y+3, y-3, x-6, etc (using no graphing calculator)
Thanks
Post by: EEEEEEP on July 15, 2016, 01:05:22 pm
Just wondering but how do you graph lines (without plotting the points) for the following (I don't remember learning this at school):You can do it by letting x or y =0 and then find the intercepts.
x+y, x-y, x-y+3, y-3, x-6, etc (using no graphing calculator)
Thanks
E.g. If we had an equation say... y = x + 1...
Y intercept
> >let x = 0
y =1
X intercept
>> let y = 0
x = -1
Mark the points (-1, 0), (0, 1) and the join the two dots with a line.
*Note, the lines can be joined in that fashion as the equation is linear.*
Post by: jamonwindeyer on July 15, 2016, 01:15:56 pm
Just wondering but how do you graph lines (without plotting the points) for the following (I don't remember learning this at school):
x+y, x-y, x-y+3, y-3, x-6, etc (using no graphing calculator)
Thanks
You can also use the gradient-intercept form for a line if it happens to be in the correct form, or is easy to get there. If we have an equation that looks like this:
Then it is just a line with a y-intercept of b, and a gradient of m (remember gradient is just rise over run) ;D
Post by: conic curve on July 15, 2016, 02:05:02 pm
You can also use the gradient-intercept form for a line if it happens to be in the correct form, or is easy to get there. If we have an equation that looks like this:
Then it is just a line with a y-intercept of b, and a gradient of m (remember gradient is just rise over run) ;D
So for the equations, I plotted above you just assume they're all equal to 0
E.g. x-y+3, does this mean x-y+3=0?
Post by: jamonwindeyer on July 15, 2016, 02:30:00 pm
So for the equations, I plotted above you just assume they're all equal to 0
E.g. x-y+3, does this mean x-y+3=0?
We wouldn't assume it in all cases, but it is very common for equations to be expressed with everything on the left hand side equal to zero, called the general form. So that's how nerd1 interpreted it above ;D
Post by: conic curve on July 15, 2016, 02:33:06 pm
We wouldn't assume it in all cases, but it is very common for equations to be expressed with everything on the left hand side equal to zero, called the general form. So that's how nerd1 interpreted it above ;D
So if it's an equation like graph x+3 then would it be x+3=0 or y=x+3 because in the original question there's no y value?
Post by: jamonwindeyer on July 15, 2016, 02:34:28 pm
So if it's an equation like graph x+3 then would it be x+3=0 or y=x+3 because in the original question there's no y value?
The question itself is flawed, they'd say which, but I'd interpret it as y=x+3 in that case :)
Post by: conic curve on July 15, 2016, 03:12:37 pm
The question itself is flawed, they'd say which, but I'd interpret it as y=x+3 in that case :)
What if it was y+3?
Post by: EEEEEEP on July 15, 2016, 03:23:20 pm
Post by: RuiAce on July 15, 2016, 04:05:16 pm
Let's just clear one thing up. If in the exam they wanted you to graph, something is always EQUAL to something.
Otherwise, as Jamon said, the question is flawed.
Post by: conic curve on July 15, 2016, 04:38:55 pm
@conic
Let's just clear one thing up. If in the exam they wanted you to graph, something is always EQUAL to something.
Otherwise, as Jamon said, the question is flawed.
Oh okay. I just tend to freak out at these questions because I've seen them in the past and don't know how to graph it
Post by: RuiAce on July 15, 2016, 04:48:23 pm
An equation forces one variable to be strictly defined, or form a relation between two or more variables. This is why it must be present to form a graph.
(Or at least something similar to an equal sign. The < symbol will give us an inequality)
Post by: conic curve on July 15, 2016, 09:24:56 pm
Post by: jamonwindeyer on July 15, 2016, 09:27:53 pm
Has anyone here learnt a topic called the unit circle where you know the exact values of the trig ratios and where you learnt the fundamental formula (I.e. +k360)
Yep I think I can help there ;D if/when you have questions let me know!!
Post by: RuiAce on July 15, 2016, 10:07:18 pm
Yep I think I can help there ;D if/when you have questions let me know!!Oh ok :P
Post by: jamonwindeyer on July 15, 2016, 10:14:18 pm
Oh ok :P
I can now anyway! I went to a pretty decent 2U lecture on Trigonometry on Monday, lecturer was pretty good, think his name was James, or, Jameen, or something... ;)
Post by: jakesilove on July 15, 2016, 10:27:44 pm
I can now anyway! I went to a pretty decent 2U lecture on Trigonometry on Monday, lecturer was pretty good, think his name was James, or, Jameen, or something... ;)
It was aright I guess
Post by: conic curve on July 16, 2016, 04:05:16 pm
Post by: RuiAce on July 16, 2016, 04:15:37 pm
Anyone here know how to solve this trigonometry equation:
Post by: jakesilove on July 16, 2016, 04:16:55 pm
Anyone here know how to solve this trigonometry equation:
Hey! That was very algebra intensive and required the use of trig identities; I'm sure there are heaps of ways to get to an answer, some of which will be quicker than mine. Still, when all I really did was sub in a bunch of identities, brute force doesn't take too long!
(http://i.imgur.com/faFHmif.png?1)
I also didn't get to the final answer (ie. what x/theta actually is); I'll leave that to you
Jake
Post by: jakesilove on July 16, 2016, 04:17:33 pm
Dammit Rui
Post by: conic curve on July 16, 2016, 04:20:46 pm
COuldn't of you have divided by 3 first?
Post by: RuiAce on July 16, 2016, 04:22:13 pm
Damm, it required that much working out. Woah, you guys are awesome. Cheers ;DThat would've made the factorisation process suck
COuldn't of you have divided by 3 first?
Also if you think that's a lot of working out, check out the Ext 2 qn thread :P
Post by: conic curve on July 16, 2016, 04:25:02 pm
Post by: jakesilove on July 16, 2016, 04:27:29 pm
Jake what is that typing program you use to type up the answer to the question?
It's a nifty little program called Word; I suck at LaTeX (at least when it comes to slightly more complicated stuff, or typing quickly), so I prefer to use Word, screenshot, upload to Imgur and attach it here. Totally impractical, but I'm an idiot when it comes to coding-related languages.
Post by: conic curve on July 16, 2016, 04:29:54 pm
It's a nifty little program called Word; I suck at LaTeX (at least when it comes to slightly more complicated stuff, or typing quickly), so I prefer to use Word, screenshot, upload to Imgur and attach it here. Totally impractical, but I'm an idiot when it comes to coding-related languages.
You mean microsoft word?
Post by: RuiAce on July 16, 2016, 04:31:58 pm
You mean microsoft word?Uh huh.
Post by: conic curve on July 16, 2016, 04:43:25 pm
I tried to upload an image from my phone but it won't let me :(
Post by: jakesilove on July 16, 2016, 04:45:24 pm
How do I upload photos to this forum when the file is too big?
I tried to upload an image from my phone but it won't let me :(
You might have to do what I do, and upload the file to Imgur. I'm not super techy though, so I'll leave it to someone else to answer properly
Post by: conic curve on July 16, 2016, 05:09:01 pm
You might have to do what I do, and upload the file to Imgur. I'm not super techy though, so I'll leave it to someone else to answer properly
Do you need a imgur account?
Also if anyone can help me with these Q's it would be great:
4. g(x)=2x-1.find x if
a. g(x+1)=3
b. [g(x)]^2=3
Post by: RuiAce on July 16, 2016, 05:11:58 pm
Do you need a imgur account?
Also if anyone can help me with these Q's it would be great:
4. g(x)=2x-1.find x if
a. g(x+1)=3
b. [g(x)]^2=3
The second one could've been done by quadratic formula
Also, no.
Post by: conic curve on July 16, 2016, 05:13:34 pm
The second one could've been done by quadratic formula
Also, no.
Thanks
Also sorry for the overburdening of questions:
5. a. g(x)= x +(1/x)
Prove that g(x)g(x+(1+x))=g(x^2)+3
b. f(x)= (a^x+a^(-x))/2
Show that f(2x)=2[f(x)]^2-1
c.g(x)=log((1+x)/(1-x) .Show that g(2x/(1+x^2 ))=2g(x)
d. f(x)=log((x+1)/x)
Show that:
I. F(1)+f(2)+f(3)=log4
II. F(1)+f(2)+…+f(n)=log(n+1)
Post by: RuiAce on July 16, 2016, 05:14:26 pm
ThanksBefore tackling these questions, can I ask about if you've revised how functions actually work?
Also sorry for the overburdening of questions:
5. a. g(x)= x +(1/x)
Prove that g(x)g(x+(1+x))=g(x^2)+3
b. f(x)= (a^x+a^(-x))/2
Show that f(2x)=2[f(x)]^2-1
c.g(x)=log((1+x)/(1-x) .Show that g(2x/(1+x^2 ))=2g(x)
d. f(x)=log((x+1)/x)
Show that:
I. F(1)+f(2)+f(3)=log4
II. F(1)+f(2)+…+f(n)=log(n+1)
Post by: jakesilove on July 16, 2016, 05:16:17 pm
The second one could've been done by quadratic formula
Also, no.
Given that Rui has now shown you very clearly how these kinds of questions work, I would recommend giving the new ones a go yourself. Try to work through them slowly, and if you really need more help then upload the work you've done. Then, we can help you figure out the next step, so you can keep working through them yourself. Giving you a billion answers won't help you learn, if you don't try to do some of the work yourself.
Post by: conic curve on July 16, 2016, 05:21:16 pm
Before tackling these questions, can I ask about if you've revised how functions actually work?
I did it last term for school but these were unanswered questions from a while ago which I found out today. The reason why it was unanswered was because I wasn't sure of how to do it back then
Post by: RuiAce on July 16, 2016, 05:23:42 pm
I did it last term for school but these were unanswered questions from a while ago which I found out today. The reason why it was unanswered was because I wasn't sure of how to do it back thenSo have you tried looking over your resources and figuring how to do them now instead of blindly giving us your unfinished homework?
Post by: conic curve on July 16, 2016, 05:25:36 pm
So have you tried looking over your resources and figuring how to do them now instead of blindly giving us your unfinished homework?
I did back then. Just curious but do these questions seem easy to you all?
Post by: jakesilove on July 16, 2016, 05:28:03 pm
I did back then. Just curious but do these questions seem easy to you all?
Obviously to Rui and myself, who did Mathematics a while ago, have continued to do Maths at uni, and studied pretty damn hard for the HSC, year 11 questions will be pretty straight forward. That doesn't mean it should be, or would be, easy for you. Ideally, it should become easy over time.
Post by: jamonwindeyer on July 16, 2016, 09:04:40 pm
A friendly reminder to all users that posts which are clearly "spam" in nature, and do not contribute to discussion (EG - literally repeating what others have said, or simply replying "ha") will be deleted ;D
Post by: EEEEEEP on July 16, 2016, 10:52:51 pm
I did back then. Just curious but do these questions seem easy to you all?That's because we have all studied it before.
I recommend you grab your textbook and read over things before you try any questions.
Feeding answers does not help.
Post by: dreamdog10 on July 17, 2016, 11:48:20 am
For example, when the discriminant = 0, it says theres one solution but two real equal roots and i have a diagram and all but i dont really understand that.
Post by: RuiAce on July 17, 2016, 11:57:29 am
Hi! So i was wondering what the difference between a solution and a root is.
For example, when the discriminant = 0, it says theres one solution but two real equal roots and i have a diagram and all but i dont really understand that.
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Post by: EEEEEEP on July 17, 2016, 12:20:34 pm
Hi! So i was wondering what the difference between a solution and a root is.Alternative solution/explanation to rui's, but I think it is more friendly to beginners.
For example, when the discriminant = 0, it says theres one solution but two real equal roots and i have a diagram and all but i dont really understand that.
Look at it this way to find the roots (you find out the x intercepts) or what x values make it equal to 0
For an equation with an discriminant of 0, such as (x +2)^2 ... DELTA = 16 - 4 x 1 x 2 = 0
Lets look at the roots
(x+2)(x+2) = 0
That means the first bracket equals 0 or the second bracket equals 0, so either x=-2 or x=-2. Again, you really only have one root here: 1. But in this situation you can call it a "repeating root" or "equal roots".
"Real roots" simply mean that you HAVE roots.
Post by: jamonwindeyer on July 17, 2016, 02:13:06 pm
Hi! So i was wondering what the difference between a solution and a root is.
For example, when the discriminant = 0, it says theres one solution but two real equal roots and i have a diagram and all but i dont really understand that.
As stated above, a solution is just a value (or values) that satisfies a given condition (EG - an equation). A root signifies a value that puts an expression equal to zero, usually, an x-value where a given function is equal to zero. When you look at a function on the number plane, the roots represent the x-intercepts!!
I totally understand the confusion about the discriminant. Let me explain it this way. The discriminant is a particular tool we can use to determine the number of roots of a quadratic function. This stems directly from the quadratic formula (the thing inside the square root sign):
If the discriminant is positive, then the quadratic formula has two solutions, and thus the quadratic has two roots (two x-intercepts). If the discriminant is equal to zero, then the plus and minus in the formula means nothing (since we are adding and subtracting zero), it has a single solution. This means that the quadratic has a single root, the parabola would just be resting on the x-axis, touching with its vertex at a single point. Finally, if the discriminant is negative, we can't have the square root of a negative, and thus the formula is invalid (at least, for our level of study). Thus, at this level of Mathematics, we say that there are no solutions to the equation, and the parabola will never touch the x-axis (no roots). This corresponds to positive/negative definite expressions.
Just from your question, when it says two real equal roots, that really means just one root. Like, there are two roots, but they are the same root, so we say there is a single root. Saying that there are two is kind of confusing if they are the same! ;D
I hope these answers help!! Mine is probably a solid conceptual basis, nerd gave a great practical example, and Rui gave you the more complex terminology and ideas ;D
Post by: dreamdog10 on July 17, 2016, 02:44:19 pm
Post by: anotherworld2b on July 22, 2016, 01:01:52 am
I have tried to answer this question but i didnt understand how to do it
Post by: RuiAce on July 22, 2016, 01:10:12 am
Hi, at school we have started probability. I was wondering would anyone have tips on how to remember whether event A and B are independent or mutually exclusive. I am confused about the probability rules in general. There are so many so how would you know which rule to use? TTT ^ TTTGenerally, you can use your intuition to determine if multiple events are mutually exclusive or not. If two events are mutually exclusive, they cannot occur at the same time (simultaneously). If you want, you can provide some examples and I will attempt to explain which are and which are not.
I have tried to answer this question but i didnt understand how to do it
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Before I try the Venn diagram proof, can I have a look at what you have learnt thus far? Because the peculiar nature of the question makes me uncertain as to where I should start and how limited my mathematical communication should be. Note that in 2U Venn Diagram analysis are basically NEVER asked.
Post by: anotherworld2b on July 23, 2016, 01:51:27 am
It takes me a while to understand a topic so i will mosy likely require further explanation and guidance of that is okay.
I will try to upload some questions that we have done tomorrow :)
Post by: RuiAce on July 23, 2016, 08:52:03 am
Examples of mutually exclusive events are the probabilities it is raining and sunny at the same time.
Examples of statistically independent events include turning on the computer but also playing a PSP game simultaneously.
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Post by: jackcornwell on July 23, 2016, 12:54:03 pm
i let u = e^x then found u but cant remember where to go from here Thanks
Post by: RuiAce on July 23, 2016, 12:56:21 pm
solve 2e^2x -7e^x +3I can assume you meant e^(2x) but more importantly is the expression equal to 0?
i let u = e^x then found u but cant remember where to go from here Thanks
Without it being equal to something it's just an expression, not an equation.
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If so
Post by: jamonwindeyer on July 23, 2016, 01:36:35 pm
solve 2e^2x -7e^x +3
i let u = e^x then found u but cant remember where to go from here Thanks
Welcome to the forums Jack! Glad Rui could help you out! Let me know if you need any help finding things, be sure to stop by our New User's Lounge if you haven't already ;D
Post by: jamie anderson on July 23, 2016, 10:57:10 pm
Thanks
Post by: RuiAce on July 23, 2016, 11:10:49 pm
Hey, this may be a simple question but i cant seem to grasp the idea of domain and range, for example a common question would be the multiple choice ones where it says find the value for x that satisfy this equation ( sorry dont have one on me ) i can solve it but i wont know what the domain is e.g 4<= x <= 2, 4< x while x> 2etcFind the values for x that satisfy this equation sounds like just solving something such as 5x+3>5, i.e. an equation (or inequality)
Thanks
Domain and range is much more different. Domain is the set of x-values that are allowed for a function you're given.
E.g. y=x: you can allow for any x-value whatsoever, so you have all real x
E.g. y=x^2: you can allow for any x-value as well, so you also have all real x
E.g. y=sqrt(x): you can only square root a positive number, so you have x≥0
E.g. y=1/x: you can't divide by zero, so you have all real x except x≠0
If you just started doing domain and range, draw graphs. Going by intuition at the very start is a dangerous idea.
The range on the other hand is the set of y-values allowed FOR the x-values you're given.
E.g. y=x: you can have any y-value whatsoever, so you have all real y
E.g. y=x2: when you square something it can no longer be negative. So you have y≥0
E.g. y=sqrt(x): taking a square root cannot spit out a negative value. So you have y≥0
E.g. y=1/x: 0=1/x has no solution, so you hvae all real y except y≠0
Once again, draw graphs
The concept of restricting the domain is what might get some people. Restricting the domain is when you limit the domain of something even further than what it already is.
E.g. y=x for x≥1: You chopped off every bit of the graph where x is less than 1. That no longer exists.
If you sketch the graph of y=x, you'll also find a new range of y≥1
E.g. y=x2 for -1≤x≤1: You chopped off every bit of the graph where x is greater than 1, and less than -1. You only have a tiny u-shape left.
If you sketch this graph, you'll also find a new range of 0≤y≤1
If you need more specific help, please find some questions.
Post by: dtinaa on July 24, 2016, 02:50:53 pm
Post by: Jakeybaby on July 24, 2016, 03:20:30 pm
Is anyone else struggling with Applications of Calculus?? Does anyone have any tips on how to do those describe the motion of a particle questions??I'll just go through what I know from SACE, it may be a little different for you, but I'm sure that the fundamentals are still there.
You'll always be given the function for s, displacement, at the stage of the course. This tells you the displacement (how far away) the particle is from the origin. When:
s(t) = 0, the particle is at the origin
s(t) < 0, the particle is to the left of the origin
s(t) > 0, the particle is to the right of the origin
Hopefully you know that the derivative of the displacement function results in the velocity function. This is due to the velocity being how quickly the particle is moving / how quickly it's displacement is changing. Hence:
s'(t) = v(t)
When:v(t) = 0, the particle is at rest/not moving
v(t) < 0, the particle is moving to the left
v(t) > 0, the particle is moving to the right
And the rate of which the velocity changes, is the acceleration, as if I am accelerating in the same direction as I am travelling, my velocity will also increase at said rate. Therefore acceleration is how quickly the velocity of the particle is increasing/decreasing.
v'(t) = a(t)
a(t) = 0, velocity may be at a maximum or minimumNow, the most fundamental concept in one of these questions is determining :
(https://i.gyazo.com/34a6e612aeb9646f23ce8b2e215f037a.png)
Another key concept is the be able to determine using sign tests, when the speed is increasing or decreasing.
If the signs of v(t) and a(t) are the same, then the speed of particle P is increasing
If the signs of v(t) and a(t) are different, then the speed of P is decreasing.
This is due to the fact that if I am travelling to the left, but I am being pushed to the right, my speed will decrease as I won't be able to walk to the left for much longer.
Post by: RuiAce on July 24, 2016, 03:43:21 pm
That aside, here's the add-on.
Is anyone else struggling with Applications of Calculus?? Does anyone have any tips on how to do those describe the motion of a particle questions??
Post by: Jakeybaby on July 24, 2016, 04:13:56 pm
Oh alright, just different notation between the states :)
Post by: RuiAce on July 24, 2016, 06:27:07 pm
Oh alright, just different notation between the states :)Yep. All good here; I'm not entirely why sure why x, r and s get juggled to represent displacement. (Incidentally the latter two are found in the HSC physics courses.)
But yeah, at the 2U level the information you provided formed the bulk of what is necessary.
The one other thing to keep in mind (regarding notation) though is how v as a function of t really gets defined. In Extension 1, because they also do velocity as a function of displacement (so v(x)), we might have problems if we just say v(t)=2t so v(x)=2x which, depending on the situation can easily be mistaken.
It's probably why Leibniz's notation (or just Newton's notation - the dot) is more favourable. (Of course, in 2U this is hardly problematic in my opinion)
Post by: isaacdelatorre on July 24, 2016, 07:18:55 pm
This question might be super simple, but I've been having some trouble with this question, mainly part ii.
Could someone please explain how to do it?
Post by: RuiAce on July 24, 2016, 07:49:18 pm
Hey,
This question might be super simple, but I've been having some trouble with this question, mainly part ii.
Could someone please explain how to do it?
For part (i) just gotta be careful when factorising by completing the square. In point gradient form the tangent has equation
y-(t-b)^2 = 2(t-b)(x-t)
So y = (t-b)(t-b)+(t-b)(2x-2t)
So y=(t-b)(t-b+2x-2t)
=(t-b)(2x-t-b)
Post by: isaacdelatorre on July 25, 2016, 06:42:13 pm
For part (i) just gotta be careful when factorising by completing the square. In point gradient form the tangent has equation
y-(t-b)^2 = 2(t-b)(x-t)
So y = (t-b)(t-b)+(t-b)(2x-2t)
So y=(t-b)(t-b+2x-2t)
=(t-b)(2x-t-b)
Wow, thanks RuiAce.
I don't know why I was struggling with it.
Thanks again :D
Post by: Deng on July 28, 2016, 11:23:43 pm
Part ii for the area under the curve questions
Thanks!
Post by: jamonwindeyer on July 29, 2016, 01:32:15 am
Hey was looking for help with a couple questions ive been stuck on
Part ii for the area under the curve questions
Thanks!
Hey Deng!!
Interestingly, I'd actually approach that first question not as an integration area question, but with some geometry! Let's have a think. The area we want is actually just the area of the triangle enclosed by the axes and the tangent, then subtract the area of the quarter circle! Let me know if that's unclear what I mean there! ;D
So, we need the intercepts of the tangent so we can find the area of that triangle. Use whatever method you like, I'll use substitution:
So the triangle has a width of 5 and a height of 15/4, let's use that to find the area we need:
I'll let you take it from there ;D
For Question 2, we can massively simplify that expression by recognising that sine squared plus cosine squared of the same value is just 1. So, using that Pythagorean Identity, the expression becomes:
Now we just need to think about this intuitively (easier than rigorous mathematics). We want the smallest value of this expression over the given domain, which means we want the denominator as large as possible. Well, the maximum value of sin^2 is 1, so that means the minimum value of the expression is:
Part (ii) of that last question is a little tricky! Basically, we can get it by understanding that the integral of the first derivative will get us back to the function itself!
I'll leave you to have the full attempt. But let me get you started. So, from your sketch in Part (i), you know that the graph of f'(x) has intercepts at x=-2, x=-1, and x=3. First, let's find the area enclosed between x=3 and x=-1. Use our regular method, and remember, we are integrating a derivative! The result will be our original function:
We now just need to evaluate this at 3 and -1, which we can do, because we are given these values in the graph:
The same principle would apply to the other part of the area between x=-1 and x=-2, but with an absolute value sign as is normal for these sorts of questions ;D then just add the two areas together to get your answer!
That last one was a bit of a doozy, does that make sense Deng? :D
Post by: Deng on July 29, 2016, 01:14:39 pm
Thanks!
Post by: conic curve on July 29, 2016, 02:59:52 pm
I searched it up and it said that "magnitude is the size of a mathematical object" but then, I'm thinking, what? This is so confusing
Thanks :D
Post by: jakesilove on July 29, 2016, 03:04:37 pm
What does it mean by "magnitude" in maths?
I searched it up and it said that "magnitude is the size of a mathematical object" but then, I'm thinking, what? This is so confusing
Thanks :D
Magnitude is literally the 'size' of something. The magnitude of 2 is 2. It just strips a vector value of it's direction; so, say you had a value which was 2km East. The MAGNITUDE of that value is 2km. The DIRECTION is East.
Post by: anotherworld2b on July 30, 2016, 02:12:10 am
Post by: RuiAce on July 30, 2016, 08:17:21 am
These diagrams are pretty small and I can't see everything typed on Q11. Don't shrink diagrams to upload them; just upload them to imgur or something and post the links to the images next time please. Also, try not to ask too many questions on the exact same topic; it's burdening
Post by: dreamdog10 on July 30, 2016, 02:09:24 pm
dv/dt=1−(√2)sin(πt/60)
between 0<=t<=120
Post by: RuiAce on July 30, 2016, 02:21:24 pm
Hi! just wondering how you would go about graphing
dv/dt=1−(√2)sin(πt/60)
between 0<=t<=120
Post by: dreamdog10 on July 30, 2016, 02:55:34 pm
Post by: vamshimadas on July 30, 2016, 04:58:46 pm
Find the radius of the circle that has its centre at the origin and a tangent with equation given by 4x - 3y - 5 =0
Post by: jakesilove on July 30, 2016, 05:03:16 pm
Hi, I'm just having trouble working through methodology of this question. By substitution i can assume the answer, but I want to know how to actually work it out.
Find the radius of the circle that has its centre at the origin and a tangent with equation given by 4x - 3y - 5 =0
I think this question requires the use of the distance between a line and a point formula! Since the radius will be the line perpendicular to the tangent, to the origin, you can literally plug in the values and a radius should pop out.
(http://images.tutorcircle.com/cms/images/108/formula-of-distance-from-a-point-to-line.png)
Where A=4, B=-3, C=-5, x=0 and y=0
I think the radius turns out to be 1!
Jake
Post by: RuiAce on July 30, 2016, 05:15:52 pm
On the other hand, it would be more easier to identify for a 3U student due to the circle geometry theorem "the radius is perpendicular to the tangent drawn to point of contact". The formalised proof is found here.
An alternate informal proposal (that is, however, probably much easier to understand,) is that since the tangent only meets the circle once, the distance from the tangent to the centre is forcibly the radius. Basically, the tangent meets the circumference.
Post by: jamonwindeyer on July 30, 2016, 08:45:57 pm
I think this question requires the use of the distance between a line and a point formula! Since the radius will be the line perpendicular to the tangent, to the origin, you can literally plug in the values and a radius should pop out.
(http://images.tutorcircle.com/cms/images/108/formula-of-distance-from-a-point-to-line.png)
Where A=4, B=-3, C=-5, x=0 and y=0
I think the radius turns out to be 1!
Jake
This is definitely the correct method, since we know the centre of the circle we just need the radius, which can be obtained through this formula, and it does indeed end up as 1!
Rui's method above is a cool proof, but far beyond what is required here. At this level, intuition is absolutely fine. The perpendicular distance to a line is the shortest distance to said line. Since it is the shortest distance, a circle with a radius equal to that distance will clearly only touch the line once.
So basically, in a 2 Unit Exam, just the formula would suffice, but if you are doing 3 unit or above Rui's proof is good to understand ;D
Post by: conic curve on July 31, 2016, 08:19:42 am
if secθ-tanθ= 3/5 show sinθ=8/17
Post by: RuiAce on July 31, 2016, 09:04:45 am
Post by: conic curve on July 31, 2016, 09:07:35 am
Why is cos theta not equal to zero at the beginning of the question?
Post by: RuiAce on July 31, 2016, 09:11:17 am
Why is cos theta not equal to zero at the beginning of the question?
Post by: conic curve on August 01, 2016, 08:31:44 am
Thanks I get it now
In this equation: (x-3)(x+5)=<0 I got x=<5 and x>=-3 as the answer but the answer says -3=<x=<5. What am I doing wrong?
For this question 4x^2-12x+10>0 how is the answer all real x
Thanks guys ;D
Post by: RuiAce on August 01, 2016, 08:39:53 am
Thanks I get it nowFor the first one your solution is actually the same as the answer. But the answer makes it tidier, which is good practice that you should be getting use to.
In this equation: (x-3)(x+5)=<0 I got x=<5 and x>=-3 as the answer but the answer says -3=<x=<5. What am I doing wrong?
For this question 4x^2-12x+10>0 how is the answer all real x
Thanks guys ;D
Note how x≤5 can stay as x≤5. But x≥-3 can become -3≤x.
Just combine them together to get -3≤x≤5.
This answer is tidier because you have an enclosed domain, not one that's going on infinitely and forever. It shows that x is between two things where possible.
Post by: conic curve on August 01, 2016, 01:10:21 pm
For the first one your solution is actually the same as the answer. But the answer makes it tidier, which is good practice that you should be getting use to.
Note how x≤5 can stay as x≤5. But x≥-3 can become -3≤x.
Just combine them together to get -3≤x≤5.
This answer is tidier because you have an enclosed domain, not one that's going on infinitely and forever. It shows that x is between two things where possible.
Thanks
How do you do the question I attached below:
I know it involves substituting in the function but I find it confusing and very hard to understand
Post by: RuiAce on August 01, 2016, 01:21:15 pm
Thanks
How do you do the question I attached below:
I know it involves substituting in the function but I find it confusing and very hard to understand
Post by: jamonwindeyer on August 01, 2016, 02:06:43 pm
Thanks
How do you do the question I attached below:
I know it involves substituting in the function but I find it confusing and very hard to understand
And remember that you can always check yourself/give yourself some idea of where to go by just differentiating the function normally ;D
Post by: conic curve on August 01, 2016, 03:53:53 pm
Thanks ;D
Post by: RuiAce on August 01, 2016, 04:02:39 pm
Could you please elaborate further for what I have attached below because I don't seem to understand thatApplied the definition of a function?
Thanks ;D
Where there's x I replace it with x+h?
If f(x)=x^2
Then f(a)=a^2
f(2)=2^2 = 4
f(x+h)=(x+h)^2 = x^2+2xh+h^2
And then I just expanded the whole thing out.
Post by: isaacdelatorre on August 01, 2016, 07:21:20 pm
I'm just having a bit of trouble with this question. Could someone please give me an explanation of how to do this. Thank you!!!
Let g(t) be another function defined for all t ≥ 0. The gradient function is given by g′(t) = 33/10 te−kt .
It is given that g′(t) attains the greatest value at t = 7.5 and g(0) = 0. g′(t)
where k is a positive constant. The diagram below shows a sketch of g′(t) against t.
(i) Show that k = 2/15
(ii) Use the trapezoidal rule with four sub-intervals to estimate the shaded area in the diagram above.
(iii) Explain why your answer to (ii) is an estimate for g(12).
Post by: RuiAce on August 01, 2016, 09:37:00 pm
Hey,
I'm just having a bit of trouble with this question. Could someone please give me an explanation of how to do this. Thank you!!!
Let g(t) be another function defined for all t ≥ 0. The gradient function is given by g′(t) = 33/10 te−kt .
It is given that g′(t) attains the greatest value at t = 7.5 and g(0) = 0. g′(t)
where k is a positive constant. The diagram below shows a sketch of g′(t) against t.
(i) Show that k = 2/15
(ii) Use the trapezoidal rule with four sub-intervals to estimate the shaded area in the diagram above.
(iii) Explain why your answer to (ii) is an estimate for g(12).
Post by: liiz on August 02, 2016, 05:22:08 pm
Post by: RuiAce on August 02, 2016, 06:06:05 pm
Hey there, wondering whether someone could please help me out with part b) of this question that I've attached. Not quite sure how to approach it! Thanks so much :))
Post by: wesadora on August 03, 2016, 11:27:42 pm
Post by: RuiAce on August 04, 2016, 07:11:20 am
2U Question. Halp.This does not look like a 2U question given that I do not believe spirals are so easy to analyse. However this is the best I could come up with.
An assumption is made that the inclination of the string is constant. The basis of this is that the midpoint is collinear to the endpoints of the string, and the line between them is perpendicular to the base.
Post by: jamonwindeyer on August 04, 2016, 09:03:45 am
2U Question. Halp.
Just to extend on Rui's answer (I get the same outcome), you could consider it in a similar way. This is definitely 2U content, but a really weird one, they like to do strange stuff with Pythag for some reason :P it could show up in an Extension exam I suppose, but I see it more as a Band 6 2U question :)
The string is wrapped around a cylinder right, so what we can almost do (bear with me) is assign a cylindrical coordinate system. This sounds complicated, but it's actually easy if we think about it practically.
Picture grabbing the end of the string at the top and unravelling it from the cylinder by pulling directly outward, not changing it's height, and leaving the other end fixed on the ground. This forms a right angled triangle where the string is the hypotenuse (and hence comes back in Rui's method, but I'm considering the whole cylinder at once).
The height of this triangle is still the same as the height of the cylinder (remember I said don't change the height), so that is h. The base is whatever distance the string travelled around the circumference. What I've almost done is initially considered the x (horizontal) axis as wrapped around the cylinder, and then when I unravel it, I've moved back to our regular Cartesian system. But the distance is the same, that string has wrapped itself around that cylinder twice, so the 'horizontal' distance travelled is double the circumference. I've taken the horizontal distance as measured around the cylinder (double the circumference), and stretched it out into a straight line.
Then, as Rui said, we use Pythagoras:
To simplify all that, I copied Rui's method but considered the whole cylinder at once, just to make it more practical. You COULD consider it in terms of changing coordinate systems, but that's unnecessary if you can picture unravelling the string. I threw it in to make myself sound smart ;)
Post by: conic curve on August 04, 2016, 06:51:37 pm
Post by: RuiAce on August 04, 2016, 06:55:25 pm
How do I differentiate this?
Post by: jakesilove on August 04, 2016, 06:56:15 pm
Post by: leila_ameli on August 05, 2016, 08:38:56 pm
I need some help with how i should approach the following;
The parabola P has the equation y^2=8(x+2).
i) write down the coordinates of the vertex of P.
ii) find the coordinates of the focus of P.
iii) write down the equation of the directrix of P.
iv) sketch the parabola P showing the features found above, as well as any x or y intercepts.
cheers
Post by: jamonwindeyer on August 05, 2016, 08:50:03 pm
Heyyy :) :)
I need some help with how i should approach the following;
The parabola P has the equation y^2=8(x+2).
i) write down the coordinates of the vertex of P.
ii) find the coordinates of the focus of P.
iii) write down the equation of the directrix of P.
iv) sketch the parabola P showing the features found above, as well as any x or y intercepts.
cheers
Hey there!! No worries I'll give you a hand! ;D
Before we do, we notice that the equation is in the typical form of a 'sideways' parabola:
If we consider it this way, we can immediately read off the coordinates of the vertex (h,k) as (-2, 0):
We can also determine the focal length:
Now, remember that the focus is 'a' units away from the vertex. This is a right facing parabola (try some random points to check), so it is 2 units to the right of the vertex:
The directrix is 2 units the other way (remember, in this case, the directrix is a vertical line since the parabola is sideways):
All of this should help you sketch ;D does that help?
Post by: leila_ameli on August 05, 2016, 10:30:59 pm
Hey there!! No worries I'll give you a hand! ;D
Before we do, we notice that the equation is in the typical form of a 'sideways' parabola:
If we consider it this way, we can immediately read off the coordinates of the vertex (h,k) as (-2, 0):
We can also determine the focal length:
Now, remember that the focus is 'a' units away from the vertex. This is a right facing parabola (try some random points to check), so it is 2 units to the right of the vertex:
The directrix is 2 units the other way (remember, in this case, the directrix is a vertical line since the parabola is sideways):
All of this should help you sketch ;D does that help?
Post by: conic curve on August 06, 2016, 02:30:13 am
Post by: RuiAce on August 06, 2016, 08:48:19 am
If they a function is monotonic increasing does that mean f(x)>0 or f(x)>=0? If so why, I can't seem to distinguish between the twof'(x)≥0
Don't forget the dash
A stationary point doesn't necessarily impede it being "monotonic" increasing. f(x)>0 still satisfies monotonic increasing but more specifically it means strictly increasing.
Post by: conic curve on August 06, 2016, 05:03:59 pm
f'(x)≥0
Don't forget the dash
A stationary point doesn't necessarily impede it being "monotonic" increasing. f(x)>0 still satisfies monotonic increasing but more specifically it means strictly increasing.
I don't get it. Could you please elaborate on f'(x)>0 and f'(x)>=0 (and try to give me examples)
Thanks
Post by: jamonwindeyer on August 06, 2016, 05:11:51 pm
I don't get it. Could you please elaborate on f'(x)>0 and f'(x)>=0 (and try to give me examples)
Thanks
Post by: conic curve on August 06, 2016, 05:13:15 pm
Finally this cleared things up for me ;D
Thanks
When would you actually use it (with an example question)
Post by: RuiAce on August 06, 2016, 05:14:28 pm
I don't get it. Could you please elaborate on f'(x)>0 and f'(x)>=0 (and try to give me examples)
Thanks
Further reading: Wikipedia
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Edited on:
Post by: jamonwindeyer on August 06, 2016, 05:16:35 pm
Finally this cleared things up for me ;D
Thanks
When would you actually use it (with an example question)
Check that post again, I made a slight error, fixed now :) (and Rui goes into more detail if it helps). No problem, to be honest it is a tiny thing, it is unlikely to be pressed hard. If you are asked whether a function is monotonic increasing (they probably wouldn't use this language in 2U tbh), then just remember the dash, that's all ;D
Post by: olivercutbill on August 07, 2016, 12:50:41 pm
Thank you!
Post by: zsteve on August 07, 2016, 01:38:07 pm
Can someone explain 'b)' and 'c)'?
Thank you!
For (b), you'll have to differentiate for the velocity as a function of time. Then solve for zeros.
For (c), you'll have to use a definite integral. Assuming $t$ can only be positive, $x$ is defined over the interval \([0, 7]\), so 'altogether' means from t=0 to t=7.
These are standard procedures in kinematics :)
Post by: RuiAce on August 07, 2016, 01:41:27 pm
Can someone explain 'b)' and 'c)'?
Thank you!
For (b), you'll have to differentiate for the velocity as a function of time. Then solve for zeros.
For (c), you'll have to use a definite integral. Assuming $t$ can only be positive, $x$ is defined over the interval \([0, 7]\), so 'altogether' means from t=0 to t=7.
These are standard procedures in kinematics :)
But we don't need to. We can just rely on x(t).
Post by: jakesilove on August 07, 2016, 01:45:27 pm
Can someone explain 'b)' and 'c)'?
Thank you!
Dammit guys, I was just posting a solution. But yeah got to the same thing; I better pick up my game
Post by: olivercutbill on August 07, 2016, 07:32:26 pm
Problem was I forgot to use the product rule in part b). Classic error.
Post by: conic curve on August 08, 2016, 10:51:04 am
Post by: jakesilove on August 08, 2016, 11:02:02 am
Need help with this Q
Let's start by simplifying some stuff
By difference of two squares. But, p+q=1, so
Now, let's use sum of two cubes
As p+q=1
As required
Post by: conic curve on August 08, 2016, 01:04:27 pm
Let's start by simplifying some stuff
By difference of two squares. But, p+q=1, so
Now, let's use sum of two cubes
As p+q=1
As required
Thanks Jake
How would I do question iii and iv
Post by: RuiAce on August 08, 2016, 01:08:11 pm
Thanks Jake
How would I do question iii and iv
Post by: conic curve on August 09, 2016, 10:58:46 am
Post by: jakesilove on August 09, 2016, 11:24:45 am
How do I differentiate each function with respect to x?
Far out buddy, those are some killer questions. Still, both are simple applications of the Chain rule: I'll only do the second, because it is substantially harder, and understanding it should let you differentiate the first one.
We want to differentiate
Now, we know that if
So, for the question above, we have to let
At first, I was thinking we needed to do two iterations of the Chain rule. But, since it's only raised to the power of 2, let's just expand it. This will make the solution much easier to come to; if you had a power higher than 2, you would have to chain rule again.
So, for the entire thing,
Post by: Jakeybaby on August 09, 2016, 07:19:39 pm
Could someone assist me with the integral of (e2x-1 +3)2 dx?
Thankyou
Post by: RuiAce on August 09, 2016, 07:43:05 pm
Sorry, but I haven't become familiar with LaTeX yet, but I will do soon.
Could someone assist me with the integral of (e2x-1 +3)2 dx?
Thankyou
Post by: Jakeybaby on August 09, 2016, 07:58:33 pm
Alright, did it a different way, would you always recommend expanding?
Post by: RuiAce on August 09, 2016, 08:04:52 pm
Alright, did it a different way, would you always recommend expanding?
Post by: conic curve on August 10, 2016, 01:18:10 pm
Post by: jakesilove on August 10, 2016, 01:46:33 pm
How do I find the deriative of the following?
Hey Conic,
A lot of your questions appear to be fairly normal applications of the Chain Rule, Product rule or Quotient rule. The rules are described below.
(http://mathcs.slu.edu/history-of-math/images/history/thumb/7/7f/Pqr-rules.jpg/350px-Pqr-rules.jpg)
I don't think that it is very helpful if we just keep answering your questions, because that doesn't seem to be helping you very much. Rather, I think I would prefer if you post up your attempt at a solution, so that we can correct you and help you as you go. My main advice is to (for product/quotient) identify u, v, u' and v', write it all down then simply apply the formula. Give these questions a go; will be happy to help once I can see that you've made an attempt!
Post by: jamonwindeyer on August 10, 2016, 01:49:32 pm
How do I find the deriative of the following?
Hey conic! To show you what Jake means, if you do the derivatives separately like so:
Note that u' in the question above was obtained with the chain rule! Pull the power out the front, subtract one from the power, the multiply by the derivative of the inside. You then literally sub this into the chain rule formula:
If you try breaking your big problems down into smaller ones, they will become much easier ;D as Jake said feel free to come back to us with your working if it isn't quite coming together and we can help you figure out what went wrong!
Post by: RuiAce on August 10, 2016, 02:20:37 pm
Hey Conic,This.
A lot of your questions appear to be fairly normal applications of the Chain Rule, Product rule or Quotient rule. The rules are described below.
(http://mathcs.slu.edu/history-of-math/images/history/thumb/7/7f/Pqr-rules.jpg/350px-Pqr-rules.jpg)
I don't think that it is very helpful if we just keep answering your questions, because that doesn't seem to be helping you very much. Rather, I think I would prefer if you post up your attempt at a solution, so that we can correct you and help you as you go. My main advice is to (for product/quotient) identify u, v, u' and v', write it all down then simply apply the formula. Give these questions a go; will be happy to help once I can see that you've made an attempt!
And I also feel that in the previous accounts you've posted up a solution, most of the mistakes were in algebraic manipulation, not the actual computation. You should put more emphasis on the basics because your grasp over the advanced concepts is superior to those as evidenced.
Post by: conic curve on August 10, 2016, 04:15:26 pm
Post by: RuiAce on August 10, 2016, 04:37:21 pm
(http://i.imgur.com/CTH9ycL.jpg)
Post by: conic curve on August 10, 2016, 05:35:29 pm
Oh yeah I was meant to use the product rule. Apologies.
Anyways here is my working to the question
Spoiler
(http://i.imgur.com/EqDdbz2.jpg)
(http://i.imgur.com/zGV36Yr.jpg)
(http://i.imgur.com/zGV36Yr.jpg)
Someone please correct me if I am wrong
Moderator edit: Put images in spoiler
Post by: RuiAce on August 10, 2016, 06:46:00 pm
Oh yeah I was meant to use the product rule. Apologies.Here's my feedback:
Anyways here is my working to the questionSpoiler(http://i.imgur.com/EqDdbz2.jpg)
(http://i.imgur.com/zGV36Yr.jpg)
Someone please correct me if I am wrong
Moderator edit: Put images in spoiler
In lines 2 and 3 you quoted the product rule correctly
In line 4, suddenly all the x's turned into u's. Please fix this. (There are several instances where this occurs)
In line 6, your derivatives are correct.
In lines 7 and 8, your substitution is correct
In line 9, your factorisation is correct. However you can go (at least one step) further.
__________________________
Once again, too many of those x's look like u's.
Your answer is certainly correct according to WolframAlpha
Except they factorised the 4 out.
Post by: olivercutbill on August 13, 2016, 04:39:01 pm
Can anyone help me understand this question?
Post by: RuiAce on August 13, 2016, 05:30:46 pm
Hey guys-
Can anyone help me understand this question?
Post by: olivercutbill on August 15, 2016, 07:17:27 pm
Thanks for that guys!
Post by: conic curve on August 17, 2016, 10:52:50 am
Post by: RuiAce on August 17, 2016, 11:13:02 am
Help with iv and v
Post by: conic curve on August 17, 2016, 01:04:05 pm
y= -x^3/6 + x^2/3 - x/6
I did y' and got -3x^2/6 +4x/6 -1/6=0 and when I try to solve for the x intercepts, I can't seem to find them
Also I'm having trouble drawing a y dash table in order to graph it, like what are the steps to doing that and how do I fill it in?
Thanks guys :D
Post by: jamonwindeyer on August 17, 2016, 01:42:19 pm
Need help for sketching the following:
y= -x^3/6 + x^2/3 - x/6
I did y' and got -3x^2/6 +4x/6 -1/6=0 and when I try to solve for the x intercepts, I can't seem to find them
Also I'm having trouble drawing a y dash table in order to graph it, like what are the steps to doing that and how do I fill it in?
Thanks guys :D
Hey conic! Derivative looks good, remember that solving for y'=0 isn't solving for x-intercepts, it is solving for turning points:
We now test their nature. You can use one of the y' tables as you call it, or more conveniently, we can use the second derivative test:
Is this a test you are familiar with? What you suggested is comparing the sign of the derivative either side of the turning point, but there is an easier way. Basically, it is using a concavity test to determine the nature of a stationary point. If it is concave down, it is a maximum, if it is concave down, it is a minimum ;D
Finding the x-intercepts is done by equating the initial expression to 0. Remember, the x intercept is where y=0:
Putting all of that (x-intercepts and turning points) together, you should be able to sketch something like this ;D
Hope that makes sense ;D
Post by: anotherworld2b on August 19, 2016, 12:58:22 am
Could I please have help with solving these three questions? I apologise if i have put too many questions at once
Post by: conic curve on August 19, 2016, 06:34:13 am
Post by: RuiAce on August 19, 2016, 08:08:23 am
(http://i.imgur.com/oDcI5Ll.jpg)Nice, but I have a few comments.
Also for part 3, I'm not sure if you picked up on it but you don't have to FOIL out the whole thing. It's just difference of two squares.
Post by: anotherworld2b on August 19, 2016, 11:22:38 pm
I was wondering how exactly did the surds disappear to equal 2/8?
I also wanted to ask how do you know when to use a plus or minus sign? I've only learnt that its only for when the power is to 2?
(http://i.imgur.com/oDcI5Ll.jpg)Mod edit: Swapped over the quote.
Post by: RuiAce on August 19, 2016, 11:50:18 pm
thank you for the help :D
I was wondering how exactly did the surds disappear to equal 2/8?
I also wanted to ask how do you know when to use a plus or minus sign? I've only learnt that its only for when the power is to 2?
Mod edit: Took out the quote. I'm not the one who deserves the thanks here.
Post by: anotherworld2b on August 20, 2016, 12:31:07 am
Post by: jamonwindeyer on August 20, 2016, 12:59:06 am
I was also wondering how would you do these 2 questions? I am not sure how to approach them
You might make both a bit easier by making a substation for yourself!
In the first:
Have a go at a similar thing in the second one! Make a substitution to turn it into a quadratic equation, and see if you can follow the same procedure as me ;D let me know if you need more of a hand with it ;D
Post by: anotherworld2b on August 20, 2016, 01:48:05 am
I also wanted to ask for a question like this how should you approach it? Im not sure where i should and how to begin ???
You might make both a bit easier by making a substation for yourself!
In the first:
Have a go at a similar thing in the second one! Make a substitution to turn it into a quadratic equation, and see if you can follow the same procedure as me ;D let me know if you need more of a hand with it ;D
Post by: conic curve on August 20, 2016, 07:55:48 am
thank you for the help conic curve :D
I was wondering how exactly did the surds disappear to equal 2/8?
I also wanted to ask how do you know when to use a plus or minus sign? I've only learnt that its only for when the power is to 2?Mod edit: Swapped over the quote.
nws
Usually if there is a constant (number) to the power of an even index (i.e. power) it is usually even
e.g. -1 to the power of 4 is equal to 1 right because -1 x -1 x -1 x-1 =1 (remember two negatives make a positive)
if it was -1 to the power of 5 the answer would be -1 (obviously)
2Squareroot of 2x=8
Post by: RuiAce on August 20, 2016, 08:49:49 am
Thank you for your help :D i was able to get the answer for the second one by following your example.
I also wanted to ask for a question like this how should you approach it? Im not sure where i should and how to begin ???
Post by: anotherworld2b on August 20, 2016, 10:15:46 am
I have never used or hear about lim before
Post by: RuiAce on August 20, 2016, 11:10:05 am
How do you know that lim a^x = 2?This does get forgotten by heaps of 2U students to be fair but it's absolutely crucial.
I have never used or hear about lim before
The exponential a^x (doesn't have to specifically be e^x) has a horizontal asymptote at y=0, for very large NEGATIVE x. We say that the limit as x goes to negate infinity is therefore 0.
Because fact is if you raise a number to a really negative number it just goes down to 0
However, if you've never heard of what a limit is, then there's something that you weren't taught that belongs in 2U. I do suspect you're a VCE student but in here I use anything that belongs to the 2U course.
Post by: jamonwindeyer on August 20, 2016, 12:37:01 pm
This does get forgotten by heaps of 2U students to be fair but it's absolutely crucial.
The exponential a^x (doesn't have to specifically be e^x) has a horizontal asymptote at y=0, for very large NEGATIVE x. We say that the limit as x goes to negate infinity is therefore 0.
Because fact is if you raise a number to a really negative number it just goes down to 0
However, if you've never heard of what a limit is, then there's something that you weren't taught that belongs in 2U. I do suspect you're a VCE student but in here I use anything that belongs to the 2U course.
Another world is a WACE student ;D you can think about limits intuitively without the notation! Like, for the limit of something as x approaches infinity, you mean, What happens as you pop in MASSIVE values of x. It's just a formal notation for saying, when x is HUGE, or x is HUGELY NEGATIVE (-infinity), or (less important for you), when x is REALLY CLOSE to some other value :)
Post by: RuiAce on August 20, 2016, 12:49:38 pm
Another world is a WACE student ;D you can think about limits intuitively without the notation! Like, for the limit of something as x approaches infinity, you mean, What happens as you pop in MASSIVE values of x. It's just a formal notation for saying, when x is HUGE, or x is HUGELY NEGATIVE (-infinity), or (less important for you), when x is REALLY CLOSE to some other value :)Hmm. Yeah I thought that VCE also do have limits.
But I had a look in the WACE courses. There's an explicit reference to limits in relation to the definition of the derivative in the methods course.
Post by: jamonwindeyer on August 20, 2016, 03:02:46 pm
Hmm. Yeah I thought that VCE also do have limits.
But I had a look in the WACE courses. There's an explicit reference to limits in relation to the definition of the derivative in the methods course.
Eh, maybe not covered yet, s'all gee in the bee :o
Post by: olivercutbill on August 24, 2016, 05:09:24 pm
Indefinite integral of (3)^(2x-1)
Thanks guys
Post by: RuiAce on August 24, 2016, 05:26:41 pm
Got a foolish question.
Indefinite integral of (3)^(2x-1)
Thanks guys
Post by: olivercutbill on August 24, 2016, 07:32:14 pm
ah - more complicated than I thought but I got it. Awesome, you're a legend Rui!
Post by: jamonwindeyer on August 24, 2016, 11:23:22 pm
ah - more complicated than I thought but I got it. Awesome, you're a legend Rui!
Yeah that was definitely not a foolish question (not that there are any foolish questions, but that one is actually pretty tough to do unless you know the trick) ;D
Post by: olivercutbill on August 27, 2016, 12:00:15 pm
Yeah that was definitely not a foolish question (not that there are any foolish questions, but that one is actually pretty tough to do unless you know the trick) ;D
Yeah 100% - that is one of those things to look out for when you are stuck I guess. Goes into the list of:
-equations reducible to quadratics
-factoring
-simultaneous equations
-the limiting sum
-graphing the problem
etc etc
That being said, I have another question for probability:
a) Given a standard deck of cards, a player is dealt five cards. What is the probability that 4 aces are drawn?
b) Dealt five cards, what is the probability of all cards being of the same suit?
I've been trying a four of a kind +1 kind of approach but it hasn't worked.
Cheers guys
Post by: RuiAce on August 27, 2016, 12:36:17 pm
Yeah 100% - that is one of those things to look out for when you are stuck I guess. Goes into the list of:
-equations reducible to quadratics
-factoring
-simultaneous equations
-the limiting sum
-graphing the problem
etc etc
That being said, I have another question for probability:
a) Given a standard deck of cards, a player is dealt five cards. What is the probability that 4 aces are drawn?
b) Dealt five cards, what is the probability of all cards being of the same suit?
I've been trying a four of a kind +1 kind of approach but it hasn't worked.
Cheers guys
Post by: olivercutbill on August 27, 2016, 02:48:53 pm
Weird
Post by: RuiAce on August 27, 2016, 02:54:11 pm
The point: When ordering isn't specified, the number of favourable outcomes is changed. (I had a feeling my answer was wrong but I couldn't see where so blindly.)
Post by: olivercutbill on August 27, 2016, 05:19:26 pm
The point: When ordering isn't specified, the number of favourable outcomes is changed. (I had a feeling my answer was wrong but I couldn't see where so blindly.)
Ok! I thought so but didn't full grasp the logic.
Post by: Deng on August 30, 2016, 02:24:17 pm
Also, when a question asks for maximum velocity e.g attached how would i do part (ii)
Thanks
Post by: jamonwindeyer on August 30, 2016, 02:38:33 pm
Hey guys i was wondering how to effectively study maths in the sense that i do past hsc papers and review my mistakes but i feel like it isnt necessarily improving my maths skills. Like even doing maths everyday i manage to forget small details and concepts of questions, im not sure how to explain it but an example would be like one of my maths exams i did well in, when i sat the same exam again 2 months later i manage to forget how to do some of the questions despite doing at least 30mins to 2 hours of maths everyday.
Also, when a question asks for maximum velocity e.g attached how would i do part (ii)
Thanks
Hey Deng! To address your question first, the answer is to differentiate! We seek a maxima for the velocity function, so:
Any subsequent values will be irrelevant because they will give the same answer. Test those values in the original function:
Clearly, the second is larger, and you can use typical tests to deduce that there is a maxima at t equal to pi.
Note that you could, in this case, do this conceptually. A cosine function will always range between -1 and 1, and so, a -2cost function will always range between -2 and 2. Pick the case which maximises velocity, and you get the same answer as above:
As for your study dilemma, let me ask you something. When you do those papers and forget concepts, when was the last time you had studied that specific topic, rather than Mathematics in general? Often, if you haven't done something in a while, you'll mess things up. I do it all the time if/when I revisit old math concepts. Hell, not just math, it applies to pretty much everything (writing a bit of code in a coding language you've not used in ages, or driving a manual after a year of driving an auto, for example). Your brain will forget things, are you consistently revising the old concepts? :)
Post by: jakesilove on August 30, 2016, 02:39:25 pm
Hey guys i was wondering how to effectively study maths in the sense that i do past hsc papers and review my mistakes but i feel like it isnt necessarily improving my maths skills. Like even doing maths everyday i manage to forget small details and concepts of questions, im not sure how to explain it but an example would be like one of my maths exams i did well in, when i sat the same exam again 2 months later i manage to forget how to do some of the questions despite doing at least 30mins to 2 hours of maths everyday.
Also, when a question asks for maximum velocity e.g attached how would i do part (ii)
Thanks
Hey! I'm on the run so can only quickly answer your Maths question, will get to your study question a bit later (unless someone else jumps in!). I'll just quickly say that your experience with maths is not abnormal, and there are definitely techniques to help you out!
For your question, you just need to recall that
In this case, velocity will be at a MAXIMUM when cos(t) is at its most NEGATIVE (as it is a negative term). Therefore, when cos(t) is -1, velocity will be 3. This is the maximum value!
Post by: Deng on August 30, 2016, 02:50:50 pm
I havent touched the topics individually since ive been taught them so i guess it does make sense, should probably get down with the basics before doing the HSC papers to be more effective?
@Jake, Thanks, but i suck with inequalities so ill stick with Jamon's way
Post by: olivercutbill on August 30, 2016, 03:48:29 pm
This one is a series/probability one, not sure how to get to the answer.
Thanks!
Post by: jamonwindeyer on August 30, 2016, 03:57:08 pm
@Jamon, ah i assumed you had to differentiate since it said maximum however i had never seen a question like this so i was confused on what to do
I havent touched the topics individually since ive been taught them so i guess it does make sense, should probably get down with the basics before doing the HSC papers to be more effective?
@Jake, Thanks, but i suck with inequalities so ill stick with Jamon's way
That absolutely makes sense then, so yep, perhaps tackle some old chapter reviews before hitting HSC past papers? ;D I always did Paper first, review second if required, but it might work better the other way around for you!! ;D
Post by: jamonwindeyer on August 30, 2016, 04:08:24 pm
Specifically iii) please!
This one is a series/probability one, not sure how to get to the answer.
Thanks!
Hey! This is a fairly common ace up the sleeve for BOSTES, the same style of question was at the end of my MX1 exam:
First question is fairly easy, remember that there are 36 possible outcomes for throwing two dice, and only one yields a win!
For the second bit, remember that for Pat to have a second turn, Chandra must not roll a double six on her turn, otherwise she would win, AND Pat must not roll a double six on his first turn either. This creates the following:
Now we get technical. For Pat to eventually win the game, we need to consider an infinite number of turns. After all, the game could go on forever, couldn't it? Let's look at the series for just the 1st-3rd turns:
Notice what's inside the brackets? An infinite series! We can find it's sum using the formula:
Note that we are allowed to use this formula because the common ratio is less than 1 ;D
Using that formula should get the answer you need! ;D
Post by: RuiAce on August 30, 2016, 10:57:27 pm
@Jake, Thanks, but i suck with inequalities so ill stick with Jamon's way
Jamon's method is your standard method that you turn to when you're stuck. But something like this is something you should be able to cope with.
Hey! This is a fairly common ace up the sleeve for BOSTES, the same style of question was at the end of my MX1 exam:This looks good (y)
First question is fairly easy, remember that there are 36 possible outcomes for throwing two dice, and only one yields a win!
For the second bit, remember that for Pat to have a second turn, Chandra must not roll a double six on her turn, otherwise she would win, AND Pat must not roll a double six on his first turn either. This creates the following:
Now we get technical. For Pat to eventually win the game, we need to consider an infinite number of turns. After all, the game could go on forever, couldn't it? Let's look at the series for just the 1st-3rd turns:
Notice what's inside the brackets? An infinite series! We can find it's sum using the formula:
Note that we are allowed to use this formula because the common ratio is less than 1 ;D
Using that formula should get the answer you need! ;D
Post by: anotherworld2b on September 02, 2016, 07:10:19 pm
Post by: jakesilove on September 02, 2016, 07:18:11 pm
Hi i was wonering how do I solve these questions?
I'll just help you out with the first, one, because once you've done one you can definitely figure out the rest yourself! You need to start with the relevant formula for the type of sequence (ie. geometric or arithmetic). For arithmetic sequences;
Where n is the term number, a is the first term, and d is the difference between subsequent terms. So, we know two pieces of information.
We can quickly subtract one equation from the other (to eliminate the a term)
So, subbing back into an equation above
Giving us the final formula
We've already found the first term (a=2000). The 51st term will be
The same approach (simultaneous equations to fill in the relevant equation) should be taken for your other questions. If you can't get them, post up your attempt at a solution and I'll help you out!
Jake
Post by: anotherworld2b on September 02, 2016, 11:27:24 pm
I'll just help you out with the first, one, because once you've done one you can definitely figure out the rest yourself! You need to start with the relevant formula for the type of sequence (ie. geometric or arithmetic). For arithmetic sequences;
Where n is the term number, a is the first term, and d is the difference between subsequent terms. So, we know two pieces of information.
We can quickly subtract one equation from the other (to eliminate the a term)
So, subbing back into an equation above
Giving us the final formula
We've already found the first term (a=2000). The 51st term will be
The same approach (simultaneous equations to fill in the relevant equation) should be taken for your other questions. If you can't get them, post up your attempt at a solution and I'll help you out!
Jake
Post by: jamonwindeyer on September 02, 2016, 11:39:06 pm
thank you for your help with Q29 but for Q30 I'm not sure how to do it. I got a really large number that is completely off the correct answer
Show us your working mate! ;D wait, are you definitely using geometric sequence terms?
Try solving that simultaneously to find the parameters, then proceed! If you still have trouble then definitely pop up your working and we'll see if we can spot the problem ;D
Post by: anotherworld2b on September 02, 2016, 11:50:24 pm
Post by: jamonwindeyer on September 03, 2016, 12:19:11 am
I found why I was getting it wrong yay ;D
Well done! ;D glad to hear it :)
Post by: anotherworld2b on September 03, 2016, 12:58:46 am
Post by: onepunchboy on September 03, 2016, 08:15:36 pm
Hey guys im not very good at doing probability qs can someone help me out with this one?
Thanks !
Post by: jakesilove on September 03, 2016, 10:21:35 pm
(http://uploads.tapatalk-cdn.com/20160903/9c8e20f1a0050a56205d57886442b93f.jpg)
Hey guys im not very good at doing probability qs can someone help me out with this one?
Thanks !
Hey! So the smartest way to consider this question is to think about what happens if NONE of the runners finish the race in under 10 seconds. Then, we can take 1-P(none) to find the probability that AT LEAST one finishes within 10 seconds! Let me know if you need me to explain why in any more detail.
Now, the probability of NONE of the runners finishing within 10 seconds will be each of their respective probability (of not completing) multiplied together. If we want events A AND B AND C to happen simultaneously, then their probability will be P(A)*P(B)*P(C). The probability of not finishing will be 1-P(finishing)
Therefore, the probability of AT LEAST ONE of the runners finishing within 10 seconds will be
Post by: Ali_Abbas on September 03, 2016, 10:29:53 pm
[IMG]http://uploads.tapatalk-cdn.com/20160903/9c8e20f1a0050a56205d57886442b93f.jpg[/IMG
Hey guys im not very good at doing probability qs can someone help me out with this one?
Thanks !
I just noticed that this question has already been answered by Jake but since mine has already been typed up I will still post it as an additional reference to potentially consolidate your understanding.
The probabilities of each competitor finishing the race in under 10 seconds are given as:
respectively. So these are the probabilities of "success".
To solve this question, we make note of the sample space i.e. the list of all possible outcomes (keeping in mind the probabilities of all instances sum to 1). These are:
No successes and three failures, one success and two failures, two successes and one failure, three successes and no failures. Notice however, that the latter three cases can be captured collectively by the single statement that at least one success occurs. We thus obtain the relationship: + P(no \space successes) = 1)
 = 1 - P(no \space successes) = 1 - P(all \space failures))
To calculate the probability that each competitor fails, we simply take the complements of p, q, and r and multiply them together. This is using two applications of the rule that P(A and B) = P(A)P(B). Thus,  = 1 - (1 - \frac{1}{4})(1 - \frac{1}{6})(1 - \frac{2}{5}) = ... = \frac{5}{8})
and the answer is D.
The probabilities of each competitor finishing the race in under 10 seconds are given as:
To solve this question, we make note of the sample space i.e. the list of all possible outcomes (keeping in mind the probabilities of all instances sum to 1). These are:
No successes and three failures, one success and two failures, two successes and one failure, three successes and no failures. Notice however, that the latter three cases can be captured collectively by the single statement that at least one success occurs. We thus obtain the relationship:
Post by: jakesilove on September 04, 2016, 12:01:27 pm
Post by: onepunchboy on September 08, 2016, 06:50:13 pm
Hey guys im not really sure how to do question i) , i got out cos(pi on 6) = -1/2 but not really sure where to go from there..
Thanks !
Post by: RuiAce on September 08, 2016, 07:01:22 pm
(http://uploads.tapatalk-cdn.com/20160908/fbd35324c152443c57020f766686223e.jpg)
Hey guys im not really sure how to do question i) , i got out cos(pi on 6) = -1/2 but not really sure where to go from there..
Thanks !
Post by: onepunchboy on September 11, 2016, 11:09:17 am
Hello again :), i was wondering how to get the equation for example 10.. dont really know what the question is asking. Thanks !!
Im not sure if this is past hsc
Post by: RuiAce on September 11, 2016, 11:16:00 am
(http://uploads.tapatalk-cdn.com/20160911/db59b46aafce63ccd1e9b1a018625847.jpg)
Hello again :), i was wondering how to get the equation for example 10.. dont really know what the question is asking. Thanks !!
Im not sure if this is past hsc
Post by: conic curve on September 13, 2016, 11:42:43 am
(http://i.imgur.com/DiZub5k.jpg)
Post by: jakesilove on September 13, 2016, 12:03:11 pm
How do you do this question
(http://i.imgur.com/DiZub5k.jpg)
You just need to think about the physical interpretation of the derivative. If the derivative is positive (ie. above the x axis), the gradient will be positive. If the derivative is negative (ie. below the x-axis), the gradient will be negative. If the derivative is zero (ie. on the x axis), the gradient will be zero, and so there will be a turning point. Have a go at sketching the function, and I'll let you know if you're correct!
Post by: conic curve on September 13, 2016, 12:26:25 pm
I think it may be wrong
(http://i.imgur.com/gOFmDSR.jpg)
Post by: RuiAce on September 13, 2016, 12:36:39 pm
This is what I gotYou didn't understand what Jake said.
I think it may be wrong
(http://i.imgur.com/gOFmDSR.jpg)
Think about what the derivative is. If the derivative f'(x) is negative (below the x-axis), then the original function f(x) is DECREASING.
If the derivative is positive (above the x-axis), then the original function f(x) is INCREASING.
If the derivative is 0 (x-intercept), then the original function has a STATIONARY POINT.
(Also it doesn't matter where your graphs start. They can start however low and however high you like. All we care about is that your curve's stationary points are placed at a correct x-coordinate and it increases/decreases where it should be.)
As an exercise, you may wish to sketch y=4x2(x-3) and an antiderivative y=x3(x-4). Use GeoGebra or just Desmos to ensure your graph is correct.
Compare where the DERIVATIVE is ABOVE the x-axis, to where the ORIGINAL CURVE is INCREASING.
Post by: jakesilove on September 13, 2016, 02:19:41 pm
This is what I got
I think it may be wrong
(http://i.imgur.com/gOFmDSR.jpg)
As an example, I've plotted the first two derivatives of the basic cubic graph. Can you see what's going on?
(http://i.imgur.com/tK6XbBG.png)
Jake
Post by: conic curve on September 13, 2016, 05:08:17 pm
As an example, I've plotted the first two derivatives of the basic cubic graph. Can you see what's going on?
(http://i.imgur.com/tK6XbBG.png)
Jake
Nah not really. Which one would be f'(x)?
Post by: RuiAce on September 13, 2016, 05:26:27 pm
Nah not really. Which one would be f'(x)?
Post by: conic curve on September 13, 2016, 05:53:50 pm
They interesect at the origin
Post by: RuiAce on September 13, 2016, 06:12:11 pm
They interesect at the originNot the point.
Here I might have a favouritism towards the graphs I suggested (Jake picked the most coincidental) but read the above posts 5 times.
Where f(x) is INCREASING, f'(x) is ABOVE THE X-AXIS etc.
Know what you're comparing. To know what you're comparing, read all of the above posts 5 times. Read and compare simultaneously
Post by: conic curve on September 13, 2016, 06:24:02 pm
Not the point.
Here I might have a favouritism towards the graphs I suggested (Jake picked the most coincidental) but read the above posts 5 times.
Where f(x) is INCREASING, f'(x) is ABOVE THE X-AXIS etc.
Know what you're comparing. To know what you're comparing, read all of the above posts 5 times. Read and compare simultaneously
So imagine f'(x) was the cubic in the image I have attached above
Is it the fact that f(x), f'(x) and f''(x) are all monotonic increasing
Also for this question below, for 16 and 17 I know how to do it in my head but am not sure how to geometrically prove it
Also for 18, the answer is 6,6 but I got 3
(http://i.imgur.com/W3VpSJs.jpg)
Post by: RuiAce on September 13, 2016, 06:29:11 pm
So imagine f'(x) was the cubic in the image I have attached aboveFor the sake of this question, STOP considering f"(x).
Is it the fact that f(x), f'(x) and f''(x) are all monotonic increasing
Also for this question below, for 16 and 17 I know how to do it in my head but am not sure how to geometrically prove it
Also for 18, the answer is 6,6 but I got 3
(http://i.imgur.com/W3VpSJs.jpg)
Like I said, WHEN f(x) is INCREASING, the DERIVATIVE f'(x) is POSITIVE.
ALL I want you to note is when f(x) is increasing/decreasing/stationary that f'(x) is above/below/intersecting the x-axis respectively. I think you do not understand what our explanations meant.
For Q18, the correct answer should technically be 6,3. 6,6 is not right in my opinion as there's double counting.
Post by: conic curve on September 13, 2016, 06:39:02 pm
If it is above the x axis, it is monotonic increasing and if it is below the x axis, it is monotonic decreasing
(http://i.imgur.com/bAbUgni.jpg)
Post by: RuiAce on September 13, 2016, 06:57:49 pm
To prevent confusion I decided to redo the question given the information you and Jake gave meHow you labelled your +,s and -'s are correct but the graph you ended up drawing makes no sense.
If it is above the x axis, it is monotonic increasing and if it is below the x axis, it is monotonic decreasing
(http://i.imgur.com/bAbUgni.jpg)
(http://uploads.tapatalk-cdn.com/20160913/5102c53b9e2230b3dd561d295cba8f43.jpg)
Post by: conic curve on September 13, 2016, 09:32:53 pm
Post by: RuiAce on September 13, 2016, 09:38:43 pm
WHat is the answer to thisWell what was your attempt?
Post by: conic curve on September 13, 2016, 10:04:59 pm
Well what was your attempt?
I found the domain first and got x is not equal to 4 and subbed back in to the original equation in order to find the range but then I get 1/0 as the range (which is infinity which is all y)
Am I right?
Post by: RuiAce on September 13, 2016, 10:06:36 pm
I found the domain first and got x is not equal to 4 and subbed back in to the original equation in order to find the range but then I get 1/0 as the range (which is infinity which is all y)That's only a part of the answer.
Am I right?
Anything under a square root also has to be positive.
Post by: kaufoou on September 17, 2016, 08:52:00 am
Post by: RuiAce on September 17, 2016, 08:59:30 am
Hey! So I was wondering if it is worth buying a Mathomat or Math-Aid to take into the exam? (btw they're permitted by BOSTES)They are most certainly permitted and I bought one in for each of my maths exams.
But their value isn't that great. They just help you draw a really neat sine curve and circles and what not. I'm fairly positive BOSTES is lenient on certain diagrams being decent enough.
If you spend too much time fussing over graphing like me, however, then you may as well because they're not "expensive" either. The math-aid is more powerful than the mathomat, but the latter comes with a protractor
Post by: olivercutbill on September 17, 2016, 04:00:42 pm
Answer for v) is (40)^1/2
Thanks
Post by: RuiAce on September 17, 2016, 04:31:33 pm
How am I supposed to do vi)
Answer for v) is (40)^1/2
Thanks
Post by: lha on September 19, 2016, 06:50:33 pm
Post by: jamonwindeyer on September 19, 2016, 06:56:19 pm
How many past papers shpuld we complete before hsc?
I think that depends on the person! I personally did over 20, but that was probably excessive, do however many you think you need to!! Your aim should be to be able to know how to do almost any question on sight ;D
Post by: jakesilove on September 19, 2016, 07:05:15 pm
I think that depends on the person! I personally did over 20, but that was probably excessive, do however many you think you need to!! Your aim should be to be able to know how to do almost any question on sight ;D
Like Jamon says, it's as many as you think works. That being said, if you did two past papers a day (for different subjects), 5 days a week, for two weeks, that's already 20 past papers. You've got way more than two weeks for most of your subjects, so plenty of time to do heaps of papers!
Post by: conic curve on September 19, 2016, 07:38:10 pm
Post by: jamonwindeyer on September 19, 2016, 08:08:14 pm
Just curious but how do I change my username?
You message a national moderator (like me), but it does need to be a good reason for us to agree to change it for you :)
Post by: itswags98 on September 21, 2016, 09:41:41 pm
Post by: RuiAce on September 21, 2016, 09:58:18 pm
How do i attempt this question? Havent done anything like it before. Dumb brain immediately thought of pythag to work out the slope lol
__________________________
__________________________
Answers to the 2015 HSC can be found here.
Post by: jakesilove on September 21, 2016, 10:00:50 pm
__________________________
__________________________
Answers to the 2015 HSC can be found here.
Alternatively (or, rather, using the exact same method but without the actual integration), you can remember that the area under the curve is the change in displacement. Assuming that, prior to t=4, the particle continues to move away from the origin, you can easily find the area under the curve (16) and add that to the original displacement (2). I would have done Rui's way, and they're exactly the same.
Post by: RuiAce on September 21, 2016, 10:01:57 pm
Alternatively (or, rather, using the exact same method but without the actual integration), you can remember that the area under the curve is the change in displacement. Assuming that, prior to t=4, the particle continues to move away from the origin, you can easily find the area under the curve (16) and add that to the original displacement (2). I would have done Rui's way, and they're exactly the same.I was going to do it that way, but then I remembered the original displacement was 2 and that's just a bugger. Don't feel safe just adding 2 on for most questions.
It's safer here cause most of the graph is positive. In general though, the absolute extrema might change due to the new ordinates of the stationary points
Post by: itswags98 on September 21, 2016, 10:18:23 pm
__________________________
__________________________
Answers to the 2015 HSC can be found here.
I appreciate it. Makes sense and the other way suggested makes sense too. Thanks :3. Hope you wont get sick of me asking questions until HSC time ahah
Post by: lozil on September 25, 2016, 03:26:56 pm
Post by: RuiAce on September 25, 2016, 04:04:44 pm
Hi, I was just wondering how you find the limit approaching infinity (I get how to do it for the limit approaching an actual number). E.g. Idk just a simple equation like lim x->infinity (x+1)/x
Post by: lozil on September 25, 2016, 04:27:29 pm
Ok, thanks!
Post by: olivercutbill on September 27, 2016, 12:32:36 pm
Post by: RuiAce on September 27, 2016, 12:40:44 pm
How is this done?
Post by: lha on September 27, 2016, 02:21:49 pm
Like Jamon says, it's as many as you think works. That being said, if you did two past papers a day (for different subjects), 5 days a week, for two weeks, that's already 20 past papers. You've got way more than two weeks for most of your subjects, so plenty of time to do heaps of papers!
Okay thank you!
Can anyone help me with question 9c)iii) in the 2010 CSSA HSC Trial paper? Its not letting me attach a picture but the paper should be online.
Mod Edit: I've already told you how to edit posts. If you have further issues such as images, take the time to read this article.
How to Use the ATAR Notes Forums
And it shouldn't be, because CSSA papers are copyrighted. If you find a copy online (which may happen), it was illegally uploaded.
Post by: Vinhtran on September 28, 2016, 06:38:35 pm
Determine the equation of the tangent to the graph of y = x^(2)e^(x) + 1, x ≥ 0 at any point where x = a
Post by: MightyBeh on September 28, 2016, 06:59:20 pm
I'm confused on how they found c in the tangent equation, please help.Derivative of y at x=a will give you the gradient of the tangent. Subbing x=a back into y will give you a point, that the tangent will also pass through. Subbing this point into your tangent equation will give you c.
Determine the equation of the tangent to the graph of y = x^(2)e^(x) + 1, x ≥ 0 at any point where x = a
Post by: RuiAce on September 28, 2016, 06:59:52 pm
I'm confused on how they found c in the tangent equation, please help.
Determine the equation of the tangent to the graph of y = x^(2)e^(x) + 1, x ≥ 0 at any point where x = a
Post by: Vinhtran on September 28, 2016, 07:16:29 pm
Post by: samuels1999 on September 29, 2016, 09:21:25 pm
I just wanted to know what text books you used for physics (jacaranda?) and maths (cambridge?)
Thanks
Post by: RuiAce on September 29, 2016, 09:27:55 pm
Hi jamonwindeyer and jakeisloveYour textbook for mathematics doesn't matter at the end of the day, because you're going to be glued to past papers.
I just wanted to know what text books you used for physics (jacaranda?) and maths (cambridge?)
Thanks
Though with that out of the way, in general Cambridge is the recommended textbook. This is due to its wide range of coverage and its suitability in addressing all levels of student capability. It starts off nice and easy for people to start grasping the basics and still has a range of challenging problems for the very keen student.
In general, I do advocate for it. But I didn't use it. I used Fitzpatrick for 2U and Maths in Focus (this textbook is actually strongly advised against under normal situations) for 3U.
Note that every textbook has its pros and cons. An answer cannot be made definitively as it is entirely based off personal preference. Cambridge just happens to be the stronger option in general.
With physics, Jacaranda and Physics in Focus are the two recommended. Not the place to be discussing it though.
Jake's surname is "Silove"
Post by: jakesilove on September 29, 2016, 09:29:03 pm
Your textbook for mathematics doesn't matter at the end of the day, because you're going to be glued to past papers.
Though with that out of the way, in general Cambridge is the recommended textbook. This is due to its wide range of coverage and its suitability in addressing all levels of student capability. It starts off nice and easy for people to start grasping the basics and still has a range of challenging problems for the very keen student.
In general, I do advocate for it. But I didn't use it. I used Fitzpatrick for 2U and Maths in Focus (this textbook is actually strongly advised against under normal situations) for 3U.
Note that every textbook has its pros and cons. An answer cannot be made definitively as it is entirely based off personal preference. Cambridge just happens to be the stronger option in general.
With physics, Jacaranda and Physics in Focus are the two recommended. Not the place to be discussing it though.
Jake's surname is "Silove"
Agree completely with the above
Post by: jamonwindeyer on September 29, 2016, 09:32:08 pm
Hi jamonwindeyer and jakeislove
I just wanted to know what text books you used for physics (jacaranda?) and maths (cambridge?)
Thanks
To answer the question, Physics in Focus, and I used a very old textbook by Jones and Couchman for 3 Unit Mathematics (called, incidentally, 3 Unit Mathematics). In my opinion, actually better than Maths in Focus/Cambridge, based on the exposure I've had to the other two :)
But yeah, really doesn't matter which you get, all have pros and cons :) :)
Post by: pels on September 30, 2016, 11:44:46 am
Quick question here
Evaluate lim θ -> 0
Of (sin2θ)/3θ
Im not sure how to arrive at the final answer.
Any help would be appreciated.
cheers :^)
Post by: RuiAce on September 30, 2016, 11:58:54 am
Hey all.
Quick question here
Evaluate lim θ -> 0
Of (sin2θ)/3θ
Im not sure how to arrive at the final answer.
Any help would be appreciated.
cheers :^)
Post by: pels on September 30, 2016, 04:45:28 pm
Since sinx/x = 0, i still don't understand the above
Post by: call.me.blueberry on September 30, 2016, 06:00:26 pm
Can you explain how sin2θ/2θ = 1
Since sinx/x = 0, i still don't understand the above
for the limit of sin(x)/(x), is x approaching 0?
I check with wolframalpha the limit is =1 too: https://www.wolframalpha.com/input/?i=lim+x-%3E+0+sin(x)%2F(x)
Post by: RuiAce on September 30, 2016, 06:03:32 pm
Can you explain how sin2θ/2θ = 1My apologies, I made a mistake in typing. That 0 is meant to be a 1. I will fix that now
Since sinx/x = 0, i still don't understand the above
for the limit of sin(x)/(x), is x approaching 0?Yes. x goes to 0.
I check with wolframalpha the limit is =1 too: https://www.wolframalpha.com/input/?i=lim+x-%3E+0+sin(x)%2F(x)
Post by: lozil on September 30, 2016, 08:01:14 pm
Thanks!
Post by: jakesilove on September 30, 2016, 09:15:11 pm
I'd appreciate if someone could help me with this one:
Thanks!
Hey! So, essentially, we need to find a value for k such that the area under the curve from -6 to 4 equals the area under the curve from 4 to k (that way, the negative and positive areas cancel out!
You can find the area under the first section in many ways; there's a formula, which I don't know, so I'll divide it into a triangle and a rectangle. The rectangle has area 10*2, and the triangle has area 10*(1/2)*3, so the total area will be 35 units squared.
The area of the second section can be found in the same way; the rectangle will have area (k-4)(4) and the triangle will have area (1/2)(k-4)(2). So, the total area will be (k-4)(4)+(1/2)(k-4)(2). Now, we need these two areas to be the same, so
As required.
Jake
Post by: lozil on September 30, 2016, 09:54:51 pm
Hey! So, essentially, we need to find a value for k such that the area under the curve from -6 to 4 equals the area under the curve from 4 to k (that way, the negative and positive areas cancel out!ok, thanks :)
You can find the area under the first section in many ways; there's a formula, which I don't know, so I'll divide it into a triangle and a rectangle. The rectangle has area 10*2, and the triangle has area 10*(1/2)*3, so the total area will be 35 units squared.
The area of the second section can be found in the same way; the rectangle will have area (k-4)(4) and the triangle will have area (1/2)(k-4)(2). So, the total area will be (k-4)(4)+(1/2)(k-4)(2). Now, we need these two areas to be the same, so
As required.
Jake
Post by: lozil on October 01, 2016, 12:28:20 pm
So concerning the attachment (the question is express with a rational denominator), for my answer i got the second last line: 8(2-sqrt7)/3 (which is just factorised from the final answer). If i left it in the factorised form, would i still get full marks?
Post by: RuiAce on October 01, 2016, 12:29:46 pm
Hi again :PYes that last step wouldn't be required. Still a rational denominator.
So concerning the attachment (the question is express with a rational denominator), for my answer i got the second last line: 8(2-sqrt7)/3 (which is just factorised from the final answer). If i left it in the factorised form, would i still get full marks?
Post by: lozil on October 01, 2016, 03:12:49 pm
Yes that last step wouldn't be required. Still a rational denominator.
ok, thanks!
Post by: olivercutbill on October 03, 2016, 04:32:06 pm
How does (d-6)^2 - 4 < 0
become 4 < d < 8 ?
im confused with the individual steps
Post by: RuiAce on October 03, 2016, 04:41:55 pm
Hey guys,
How does (d-6)^2 - 4 < 0
become 4 < d < 8 ?
im confused with the individual steps
Post by: jakesilove on October 03, 2016, 04:43:22 pm
Hey guys,
How does (d-6)^2 - 4 < 0
become 4 < d < 8 ?
im confused with the individual steps
Hey! There are heaps of ways of going about this, but this is how I generally work through these kinds of questions. Let me know if you don't understand any steps!
Now, I usually work out inequalities by first solving an equation. As such,
So, we now have two solutions for d. The range will either be OUTSIDE (ie. d>8, d<4) or INSIDE (ie. 4<d<8 ). You can figure this out in a number of ways; the easiest is to sub in a point to the original equation. Let's say that d is equal to 5.
Clearly, the relationship is true for d=5, which is between 4 and 8. Therefore, the solution for d is
As required!
Jake
Post by: kevin217 on October 03, 2016, 04:45:09 pm
Hey guys,
How does (d-6)^2 - 4 < 0
become 4 < d < 8 ?
im confused with the individual steps
Rearrange the equation to get:
(d-6)2 < 4
(d-6) < 2 or (d-6) > - 2
d < 8 or d > 4
therefore 4 < d < 8
I hope this helped
Post by: Vinhtran on October 03, 2016, 09:29:54 pm
The probability of a seventeen year old boy having a part-time job while at school is 0.4
In a sample of 80 seventeen year old boys, what is the probability that between 20 and 25 of them have a part-time job.
The answer key says the answer is 0.07
Post by: RuiAce on October 03, 2016, 09:43:12 pm
Please help me on this question. So I managed to get the mean and standard deviation correct, but i didn't understand why they made certain choices.Statistics is not in the HSC mathematics course.
The probability of a seventeen year old boy having a part-time job while at school is 0.4
In a sample of 80 seventeen year old boys, what is the probability that between 20 and 25 of them have a part-time job.
The answer key says the answer is 0.07
Post by: olivercutbill on October 04, 2016, 09:59:45 pm
Hey! There are heaps of ways of going about this, but this is how I generally work through these kinds of questions. Let me know if you don't understand any steps!
Now, I usually work out inequalities by first solving an equation. As such,
So, we now have two solutions for d. The range will either be OUTSIDE (ie. d>8, d<4) or INSIDE (ie. 4<d<8 ). You can figure this out in a number of ways; the easiest is to sub in a point to the original equation. Let's say that d is equal to 5.
Clearly, the relationship is true for d=5, which is between 4 and 8. Therefore, the solution for d is
As required!
Jake
This makes more sense to me than the solution posted above this. How are quadratic inequalities different from linear ones? In the first method posted by RuiAce, I don't understand the movement from the first line to the second.
Thanks guys!
Post by: RuiAce on October 04, 2016, 10:10:22 pm
Ignore method 1 if it makes no sense and focus on method 2. Because it is the exact same as Jake's method
________________________
________________________
Quadratic inequalities (nothing's an identity here) are different from linear ones in that for starters, parabolas have a turning point. This messes things up. A linear function is just a straight line.
Secondly, every quadratic function has two roots. (You just don't get taught that in 2U because you don't know what complex numbers are)
Post by: IkeaandOfficeworks on October 04, 2016, 10:30:12 pm
A piece of wire of length 50 cm is cut into two sections. One section is used to construct a rectangle whose
dimensions are in the ratio 3 : 1; the other section is used to construct a square. Find the dimensions of the
rectangle and the square so that the total enclosed area is a minimum.
I'm confused with the 3:1 ratio and how am I going to incorporate this in another similar question that talks about ratio? Thank you :)
Post by: RuiAce on October 04, 2016, 10:48:45 pm
Hi guys! I'm a bit confused with this question:
A piece of wire of length 50 cm is cut into two sections. One section is used to construct a rectangle whose
dimensions are in the ratio 3 : 1; the other section is used to construct a square. Find the dimensions of the
rectangle and the square so that the total enclosed area is a minimum.
I'm confused with the 3:1 ratio and how am I going to incorporate this in another similar question that talks about ratio? Thank you :)
Post by: fizzy.123 on October 05, 2016, 01:21:24 am
Can someone please help me out
Post by: RuiAce on October 05, 2016, 08:40:22 am
A particle is moving in a straight life, starting from the origin. At time t seconds the particle has a displacement of x metres from the origin and a velocity v m s^-1. the displacement is given by x=2t - 3ln(t+1). Find the distance travelled by the particle in the first 2 seconds. (HSC Paper 2000).
Can someone please help me out
I had a look: The question actually asks for the first THREE seconds. This will mean that my final answer won't match with the actual answer since you said two seconds (presumably a typo). Also, please indicate any previous parts if there happens to be some in the future.
Post by: lha on October 05, 2016, 08:52:37 am
Post by: fizzy.123 on October 05, 2016, 09:56:34 am
What has been the main topic that they have assessed from the prelim course in the hsc exam?
From what ive seen, locus & parabola. Also, just basic algebra and quadratics too.
Post by: fizzy.123 on October 05, 2016, 09:58:01 am
I had a look: The question actually asks for the first THREE seconds. This will mean that my final answer won't match with the actual answer since you said two seconds (presumably a typo). Also, please indicate any previous parts if there happens to be some in the future.
Oh yes sorry! It was a typo. Thank you!
Post by: lha on October 05, 2016, 10:04:44 am
From what ive seen, locus & parabola. Also, just basic algebra and quadratics too.
Alright thank you! Do you know anywhere that I can find a whole bunch of locus questions from past papers (other than going through every past paper)?
Post by: jakesilove on October 05, 2016, 11:10:50 am
Alright thank you! Do you know anywhere that I can find a whole bunch of locus questions from past papers (other than going through every past paper)?
Purchase a past paper book that goes by the dotpoint. Other than that, you'll have have to scroll through papers! (Which honestly won't take too long)
:)
Post by: fizzy.123 on October 05, 2016, 11:33:16 am
I had a look: The question actually asks for the first THREE seconds. This will mean that my final answer won't match with the actual answer since you said two seconds (presumably a typo). Also, please indicate any previous parts if there happens to be some in the future.
in regards to this question, for case one, it is t(1/2) - t(0) & you got 3ln(3/2) - 1. how did you get that because im getting 1 - 3ln(3/2).
Post by: RuiAce on October 05, 2016, 12:06:24 pm
in regards to this question, for case one, it is t(1/2) - t(0) & you got 3ln(3/2) - 1. how did you get that because im getting 1 - 3ln(3/2).The absolute value brackets are there for a reason.
Distance is positive, not negative
Post by: Daliaradosevic on October 05, 2016, 07:59:56 pm
also i do so so so many practice questions and i still struggle in understanding them!
Post by: RuiAce on October 05, 2016, 08:17:25 pm
Hey guys so i was just curious as to how to study for the finance (money) area of maths because i am terrible at understanding the question and answering them and i find that theres alot of finance based questions in the exams so please any advice would be grateful!! :)Effectively, every finance question is just the same thing in different words.
also i do so so so many practice questions and i still struggle in understanding them!
1. We always establish our interest rate first.
2. We start writing our recursive relationship.
3. Make this go on for n years. There is ALWAYS a geometric progression that forms in these types of questions.
4. Use the GP formula
Then whatever last is always a matter of understanding the wording. You will need to provide examples here, clearly indicating where you are stuck in the wording.
And of course, they can hide the yearly/monthly/... repayment from you, the number of years from you or the interest rate from you. The process is still the exact same regardless of which is chosen.
Post by: IkeaandOfficeworks on October 06, 2016, 12:55:24 pm
Two dice with their faces numbered 1 to 6 are tossed.
i. By listing the possible outcomes find the probability of
a) one die showing a 6 with other showing at least 4
b) one die showing a 6 with the other showing 3 or less
ii) Deduce from the results in part (i) the probability that neither die shows a 6.
Post by: RuiAce on October 06, 2016, 01:05:39 pm
Hey guys! I'm having trouble with this question. Thank you! :DThe question asks to list the possibilities, hence you should. However I'll structure my answers as though they weren't listed
Two dice with their faces numbered 1 to 6 are tossed.
i. By listing the possible outcomes find the probability of
a) one die showing a 6 with other showing at least 4
b) one die showing a 6 with the other showing 3 or less
ii) Deduce from the results in part (i) the probability that neither die shows a 6.
Let the rows correspond to the first dice, and the second correspond to the second dice.
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpso2y3tyak.png)
The total possible outcomes is always 36
a) Suppose the first dice shows a 6. Then the second die shows either 4, 5 or 6. There are 3 possible choices (going down column 6)
Suppose the second dice shows a 6. Then the first die shows either 4, 5 or 6. There are 3 possible choices (going across row 6)
(6,6) is counted twice, so we deduct it once.
Therefore there are 5 favourable outcomes, so the answer is 5/36
b) Suppose again the first dice shows a 6. Then the second shows either 1, 2 or 3 instead. There are 3 possible choices
And similarly for the second dice
However here, nothing gets counted twice.
Therefore there are 6 favourable outcomes, so the answer is 6/36 = 1/6
For part ii), note that the sum of the above probabilities is 11/36.
Notice that in the above parts, we've catered for all the cases where 6 must appear at least once.
Hence the probability 6 doesn't appear at all is just the complement. This has a probability of 25/36
This can also be deduced from reading off the table, as we are basically asking for the probability that something not in row 6 and/or column 6 is chosen.
Post by: BPunjabi on October 06, 2016, 01:20:36 pm
(http://i63.tinypic.com/106mzac.png)
I would say for:
i) (triangle)ACB = (triangle)DCF
AC=DC (Given)
(angle)ACB=(angle)DCF (Angles bisect/intesect at 90' - Given)
- SAS somehow??
I dont know what else...
i
Post by: RuiAce on October 06, 2016, 01:28:15 pm
Guys I have no Idea how to answer this question at all. Its question 15 b) from the 2015 paper.
(http://i63.tinypic.com/106mzac.png)
I would say for:
i) (triangle)ACB = (triangle)DCF
AC=DC (Given)
(angle)ACB=(angle)DCF (Angles bisect/intesect at 90' - Given)
- SAS somehow??
I dont know what else...
i
AC is not equal to DC either. AD is equal to DC.
And a hint for part (ii) - you were given a pair of similar triangles in part (i). You'd be crazy to not use it.
Post by: BPunjabi on October 06, 2016, 11:22:25 pm
AC is not equal to DC either. AD is equal to DC.
And a hint for part (ii) - you were given a pair of similar triangles in part (i). You'd be crazy to not use it.
Thanks So MUCH!!!
Post by: BPunjabi on October 07, 2016, 12:05:26 am
AC is not equal to DC either. AD is equal to DC.
And a hint for part (ii) - you were given a pair of similar triangles in part (i). You'd be crazy to not use it.
RUI im still lost... can you help out please!
Post by: RuiAce on October 07, 2016, 12:19:07 am
Guys I have no Idea how to answer this question at all. Its question 15 b) from the 2015 paper.
(http://i63.tinypic.com/106mzac.png)
I would say for:
i) (triangle)ACB = (triangle)DCF
AC=DC (Given)
(angle)ACB=(angle)DCF (Angles bisect/intesect at 90' - Given)
- SAS somehow??
I dont know what else...
i
Post by: onepunchboy on October 07, 2016, 11:11:26 pm
Can someone help me with 15) ii. Im not sure about it
Post by: jakesilove on October 07, 2016, 11:44:16 pm
(http://uploads.tapatalk-cdn.com/20161007/58acb99cb53b5093233aba2f8a357806.jpg)
Can someone help me with 15) ii. Im not sure about it
Which ii) did you need help with? There are two in the image you've posted
Post by: RuiAce on October 07, 2016, 11:45:31 pm
(http://uploads.tapatalk-cdn.com/20161007/58acb99cb53b5093233aba2f8a357806.jpg)
Can someone help me with 15) ii. Im not sure about it
Essentially, 2 is the special number here, because \(y=|2x+3|\) has gradient either 2 or -2. (Which depends on if \( x < -\frac32\) or if \( x >\frac32\), but that's not too important.)
(Note: The critical point here is basically the corner of the absolute value function.)
BOSTES solutions has a diagram.
GeoGebra simulation
Edit: A remark for case 3 - the value of \( m = -\frac23\) was computed by essentially plugging the points \( (0,1) \) and \( \left( \frac32, 0\right) \) into the usual gradient formula \( m = \frac{y_2-y_1}{x_2-x_1} \)
Which ii) did you need help with? There are two in the image you've postedI just noticed that lol
Post by: onepunchboy on October 08, 2016, 12:11:04 am
Post by: nibblez16 on October 08, 2016, 02:29:42 pm
Post by: RuiAce on October 08, 2016, 02:32:39 pm
Hello, I am sort of having trouble with geometry, for similar triangles etc. How many proofs do we need for each?? ???What do you mean by "each"? You only prove two triangles are similar at a time, not three or four at a time.
Instead, there are three tests that you need to know:
- Equiangular: Two angles in both triangles are equal
- All three sides in proprotion
- Two sides in proportion, included angle equal
(Note: There is a similar one for RHS but it is not needed)
The similarity tests can be matched up with a congruence test
Equiangular VS AAS
3 sides in prop VS SSS
2 sides in prop + inc. angle VS SAS
Post by: nibblez16 on October 08, 2016, 02:36:31 pm
What do you mean by "each"? You only prove two triangles are similar at a time, not three or four at a time.
Instead, there are three tests that you need to know:
- Equiangular: Two angles in both triangles are equal
- All three sides in proprotion
- Two sides in proportion, included angle equal
(Note: There is a similar one for RHS but it is not needed)
The similarity tests can be matched up with a congruence test
Equiangular VS AAS
3 sides in prop VS SSS
2 sides in prop + inc. angle VS SAS
Okay Thanks! What I meant by 'how many proofs for each' was similar triangles, congruent triangles, alternate etc :)
Post by: RuiAce on October 08, 2016, 02:40:26 pm
Okay Thanks! What I meant by 'how many proofs for each' was similar triangles, congruent triangles, alternate etc :)These can be found in any textbook. They stem down from properties of the geometric figure we're interested in.
Parallel lines have corresponding angles equal, alternate angles equal and cointerior angles corresponding,
Hence, if one of the following are true:
1. If alternate angles equal
2. If corresponding angles equal
3. If cointerior angles add to 180deg
Then they are parallel
Congruent triangles look the exact same, so they must have the exact same sides and angles.
Therefore
1. SSS works because you just showed all 3 sides match up
2. SAS works because imagine you have two lines equal in size. Join them together. The only thing you can change is the angle. If you fix that angle, you're stuck.
3. AAS works because imagine you have two angles. If two angles are equal, the third is equal because a triangle only has three angles. So what can you do? Make the triangle larger or smaller. If you fix a side, you end up fixing all three sides.
4. RHS - a special case, and only works because right angled triangles are nice
Post by: :3 on October 08, 2016, 07:43:42 pm
Do any of you have any tips (besides practice as it doesn't seem to work) on doing geometry-related questions such as Question 10, B from the 2006 HSC exam (http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/maths_06.pdf)?
I often struggle to see the connections between what has been given and what is being asked.
Much appreciated. :P
Post by: RuiAce on October 08, 2016, 08:37:44 pm
Hey Mathematicians,
Do any of you have any tips (besides practice as it doesn't seem to work) on doing geometry-related questions such as Question 10, B from the 2006 HSC exam (http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/maths_06.pdf)?
I often struggle to see the connections between what has been given and what is being asked.
Much appreciated. :P
____________________________________________
____________________________________________
I will leave you to do the rest of the question. Answers may be found here.
Post by: jamonwindeyer on October 08, 2016, 11:40:46 pm
Hey Mathematicians,
Do any of you have any tips (besides practice as it doesn't seem to work) on doing geometry-related questions such as Question 10, B from the 2006 HSC exam (http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/maths_06.pdf)?
I often struggle to see the connections between what has been given and what is being asked.
Much appreciated. :P
To add to Rui's answer, here are a few of my tips for Geometry questions:
- Leave them until last. So this is more of a general thing, but if you struggle with a question type you need to leave it until last. Get the easy marks first!
- Use some scrap paper or a doodling box or something to write everything you know, and everything you can find out about the situation. Don't try and find the solution, try and find everything. See if any of this helps you spot a solution.
- Work backwards! If you need to prove a result, see if you can reverse engineer it to figure out what path you would need to take. If you need similarity, what angles do you need? What do you need to get those angles? Can you get that info?
- When in doubt, assign pro numerals! It can help make the problem more logical/algebraic :) :)
And practice, because I have to say it ;)
Post by: Fadwa9 on October 09, 2016, 10:21:01 pm
Can you please explain how the the solution to question 14 (d) is obtained and how exactly the graph is interpreted [2013 HSC] http://boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
Post by: RuiAce on October 09, 2016, 10:38:11 pm
Hi,
Can you please explain how the the solution to question 14 (d) is obtained and how exactly the graph is interpreted [2013 HSC] http://boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
Material added later on
having obtained the \( -(a-3) \) by again using the formula for the area of a rectangle.
Post by: isaacdelatorre on October 10, 2016, 01:18:42 pm
Could someone please help me with this question?
I'm not sure where to start and have been blankly staring at it for a while now.
Thanks in advance :)
Post by: RuiAce on October 10, 2016, 01:31:56 pm
Hey guys,
Could someone please help me with this question?
I'm not sure where to start and have been blankly staring at it for a while now.
Thanks in advance :)
____________________________
I want to let you have a go at (iii) first. Use a graph similar to the graph in (ii). Come back with whatever progress you have.
Post by: isaacdelatorre on October 10, 2016, 02:42:55 pm
____________________________(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-10-10%20at%201.29.11%20PM_zpsnz6vxwit.png)
I want to let you have a go at (iii) first. Use a graph similar to the graph in (ii). Come back with whatever progress you have.
Awesome!! Thanks for the help Rui.
It took a few tries... and apparently a whole hour; but I got it in the end.
Thanks :)
Post by: asd987 on October 10, 2016, 10:10:56 pm
find the stationary points on the curve y=(x-4)(x+2)^2 and hence sketch the curve
Post by: jakesilove on October 10, 2016, 10:18:21 pm
Hi, I'm can some please complete this question for me.
find the stationary points on the curve y=(x-4)(x+2)^2 and hence sketch the curve
Hey! So, we can expand the graph out to
First, we find the derivative.
We set it equal to zero, to find stationary point (where the gradient is zero)
We find the second derivative to determine the nature of the stationary points.
Therefore, x=2 is a min
Therefore, x=-2 is a max.
We know the x-intercepts (from the original graph, at x=4, and x=2). We can sub in our turning points to find y values. Finally, we know that it is a 'positive' cubic, is we get +x^3. Therefore, the graph will look like
(http://i.imgur.com/o3wyjDH.png)
Let me know if I can clarify anything!
Post by: asd987 on October 10, 2016, 11:03:06 pm
Post by: RuiAce on October 10, 2016, 11:07:27 pm
TYVM i understand. i just didn't know you had to expand the equation of the graph :-[
Post by: anotherworld2b on October 10, 2016, 11:50:32 pm
I was wondering if i could get help understanding how to graph cubics and tan graphs please
Post by: BPunjabi on October 11, 2016, 12:07:18 am
I was wondering how do we simply imaginary numbers? We just started doing them but im already confused.
I was wondering if i could get help understanding how to graph cubics and tan graphs please
Wait is this any maths course or just 2 Unit cause I swear I have seen nothing like that before.
Post by: RuiAce on October 11, 2016, 12:08:58 am
Wait is this any maths course or just 2 Unit cause I swear I have seen nothing like that before.It's not in 2U.
Post by: BPunjabi on October 11, 2016, 12:09:44 am
It's not in 2U.
Thank the heavens...
Post by: BPunjabi on October 11, 2016, 12:18:29 am
Hey Liiz:
HEY Im jake (obviously not haha). Im a year 12 student who completed my 2u HSC mathematics course in year 11 and lm just happy to help out here. Now, this type of question is amongst one of the most difficult ones in geometrical applications of calculus, and unfortunately in HSC exams there WILL be harder ones. But don't worry, once you have practised enough, you will begin to seize some tricks to approach these questions.
Before I begin answering your question, just a few generally tips to help you answer questions like these where only one number value is provided:
BEFORE YOU DO ANYTHING, DRAW A DIAGRAM WITH LABELS
1. Highlight all USEFUL INFORMATIONS (in this case, highlight rectangular box, square base, no top, 500cm^3 and least area)
2. Appoint two variables to the unknown sides (in this case, I named the side length of the square base as x, and the height of the rectangular box as L)
3. There will be at least one number quantity in every one of these questions in 2u mathematics, so the first equation you should construct, using your name variables to construct an equation that uses the numbers provided by the question
4. Draw out the relationship between the two variables through this equation that you have constructed
5. Construct another equation using your variables and the subject that is asked for in the question (In this case, for example, we constructed an Area equation which directly relates to what we are asked to find)
6. Substitute in the equivalent expression of a variable (In this case, for example, L = 500/x^2, so we substitute any L we see with 500/x^2) to reduce the total number of values down to one, so that we can construct an equation entire out of only one variable, which then allows us to perform differentiation
7. Clean up the equation after the substitution to make life easier
8. Differentiate the equation
9. Let this derivative = 0 to find any stationary points (In an Exam, YOU MUST STATE "LET dy/dx = 0 TO FIND ANY STATIONARY POINTS, OTHERWISE MARKS MAYBE DEDUCTED!!!)
10. Solve the derivative equation and find a value for your variable which will be your stationary point
11. Test both sides to show that a local minimum/maximum occurs at your stationary points
12. Substitute your minimum value back into the area equation (or maximum value if the question asks for maximum area) to find the minimum area (or the maximum area if you substitute in the maximum value)
So here is my solution:
(http://i.imgur.com/t6wtIk1.jpg)
Hope you find my solutions clear and useful! If you are confused with anything, dont hesitate to ask!!! :D
Best Regards
Jacky He
LOL WTF how were you guys so good at math? Did you constantly study!
Post by: Alize on October 11, 2016, 11:47:21 am
Could someone please explain to me how solve the following?
7sin3x=2x-1
Thank you in advance! ;D
Post by: RuiAce on October 11, 2016, 11:55:41 am
Hey,
Could someone please explain to me how solve the following?
7sin3x=2x-1
Thank you in advance! ;D
Your answers should be close to WolframAlpha's exact answers
Post by: Aussie1Italia2 on October 11, 2016, 12:05:25 pm
Post by: BPunjabi on October 11, 2016, 12:18:09 pm
So you make the equation first x^1/2 then differentiate. Then it becomes 1/2x^-1/2. Sub in x=4 and then find the gradient. Grad become 1/4 . With normal same principals but switch it around so repeat first steps. Differentiate x^2+5 which is 2x. Sub in -2, making it =-4. Then put the reciprocal which is 1/4. Thats the grad of a normal
Post by: RuiAce on October 11, 2016, 12:22:26 pm
Hey! I feel like this would be really obvious but I'm struggling with a few questions like find the gradient of the tangent to the curve: y= square root of x at (4,2) and find the gradient of the normal to the curve y=x2 + 5 at (-2,9)
___________
Post by: BPunjabi on October 11, 2016, 12:33:23 pm
___________
Rui I swear that the normal is the exact same method as tangent but you swap it to the reciprocal? Like in my method above^^
Post by: RuiAce on October 11, 2016, 12:38:51 pm
Rui I swear that the normal is the exact same method as tangent but you swap it to the reciprocal? Like in my method above^^*Negative reciprocal
But yeah nothing I said really contradicts the main idea you posed
Post by: BPunjabi on October 11, 2016, 12:41:06 pm
*Negative reciprocal
But yeah nothing I said really contradicts the main idea you posed
No I just meant the second part because I thought you were alluding that the answer was -1. Had no idea what you were trying to say
Post by: RuiAce on October 11, 2016, 12:43:40 pm
No I just meant the second part because I thought you were alluding that the answer was -1. Had no idea what you were trying to sayI never said that. I said that two lines that are perpendicular have their gradients multiply to -1.
Post by: BPunjabi on October 11, 2016, 12:45:26 pm
I never said that. I said that two lines that are perpendicular have their gradients multiply to -1.
Oh I understand aha, Sorry for the confusion!!
Post by: Alize on October 11, 2016, 01:26:46 pm
Your answers should be close to WolframAlpha's exact answers
Oh ok, that makes a lot more sense, thank you! ;D
Post by: asd987 on October 11, 2016, 11:49:47 pm
sketch y=√x+5 for −4≤x≤4 and find its max and min values. btw, the square root is over x+5, not just x
Post by: RuiAce on October 12, 2016, 12:03:29 am
i need help with this question ty
sketch y=√x+5 for −4≤x≤4 and find its max and min values. btw, the square root is over x+5, not just x
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps6wk2hleu.png)
Use brackets to show that you meant y=√(x+5) in the future.
Post by: wesadora on October 14, 2016, 04:56:04 pm
Post by: RuiAce on October 14, 2016, 04:57:19 pm
This question. This was a 1998 2 unit paper and I know I didn't get taught what the hell a table of possible outcomes is. Was this part of the older syllabus or something? Cause nowadays they just ask us to do probability trees.It means to draw a table listing out every single outcome.
An outcome is something that can happen. So for example you can get 4 on the blue die and 2 on the pink.
It is to be interpreted literally.
_________
And once you have a table you can literally just count the number of favourable outcomes for your probability.
Post by: wesadora on October 14, 2016, 05:01:39 pm
Post by: RuiAce on October 14, 2016, 05:03:09 pm
That being said I don't believe that this sort of thing is applicable for questions in 2U that are not about dice rolls
Post by: julzzz on October 15, 2016, 11:00:41 am
Differentiate p=2x+50/x with respect to x.
Find any stationary points on the curve and determine their nature.
Thanks
Post by: RuiAce on October 15, 2016, 11:06:53 am
Post by: asd987 on October 15, 2016, 11:14:10 am
Post by: RuiAce on October 15, 2016, 11:18:04 am
I'm stuck on part b) of the question, but this includes the whole thing: A farmer wants to make a rectangular paddock with an area of 4000m^2. However, fencing costs are high and she wants the paddock to have a minimum perimeter. a) Show that the perimeter is given by the equation P = 2x + 8000/x b) Find the dimensions of the rectangle that will give the minimum perimeter, correct to 1 decimal place.
Post by: julzzz on October 15, 2016, 11:29:42 am
In reference to my differentiation and nation of SP question, I tried testing points on either side but it doesn't match up to the answers or what the actual graph looks like.
Could you solve it and let me see if my answers match. The 2 SP I got were (5, 20) and (-5,-20). I thought the first was the minimum and the second a maximum.
Thanks in advance.
Post by: RuiAce on October 15, 2016, 11:34:32 am
(Sorry new to this - don't know how to reply to one comment on its own...)Can you please show me your working out? There's guides to how you can upload images onto this forum if you don't want to type it.
In reference to my differentiation and nation of SP question, I tried testing points on either side but it doesn't match up to the answers or what the actual graph looks like.
Could you solve it and let me see if my answers match. The 2 SP I got were (5, 20) and (-5,-20). I thought the first was the minimum and the second a maximum.
Thanks in advance.
Or you can upload through the ATARnotes app.
Post by: julzzz on October 15, 2016, 11:59:01 am
Can you please show me your working out? There's guides to how you can upload images onto this forum if you don't want to type it.
Or you can upload through the ATARnotes app.
Ok, so my differentiation is in my working.
I used the table in the picture to figure out if it was the maximum or minimum. For (5,20), because it is negative before and positive after, I thought that would make it a minimum but the answers have maximum. Turns out my diagram isn't right according to geogebra so ignore that btw.
I haven't learnt using second derivatives, but am keen to learn it if helps me understand the question.
Thanks again
Post by: RuiAce on October 15, 2016, 12:04:41 pm
Ok, so my differentiation is in my working.Its 2x+50/x right?
I used the table in the picture to figure out if it was the maximum or minimum. For (5,20), because it is negative before and positive after, I thought that would make it a minimum but the answers have maximum. Turns out my diagram isn't right according to geogebra so ignore that btw.
I haven't learnt using second derivatives, but am keen to learn it if helps me understand the question.
Thanks again
I'm pretty sure that (5,20) is a minimum and (-5,-20) is a maximum. GeoGebra should confirm that.
One thing to note: the graph has an asymptote at x=0. You have to analyse this separately.
Post by: BPunjabi on October 15, 2016, 01:17:42 pm
Need help for the whole question please!
(http://i63.tinypic.com/260d1xl.png)
Post by: RuiAce on October 15, 2016, 01:32:37 pm
Hey Rui!
Need help for the whole question please!
(http://i63.tinypic.com/260d1xl.png)
_________________________
_________________________
_________________________
Post by: justdoit on October 15, 2016, 01:39:40 pm
Thank youuu :)
Post by: RuiAce on October 15, 2016, 01:47:43 pm
Hi, can someone please help me with the 2015 q.16 c. Looked at the working out and cant understand it :/
Thank youuu :)
In the future please provide a link to the paper or the question
____________________________________
Post by: kevin217 on October 15, 2016, 02:28:02 pm
Post by: RuiAce on October 15, 2016, 02:36:43 pm
What are the steps to graphing this sine curve?
As an exercise, you should use Desmos to graph all of the mentioned curves for you. You should observe what happens - both how the graphs stretch, and how the x-intercepts change. GeoGebra is a good substitute for Desmos.
Post by: justdoit on October 15, 2016, 02:42:04 pm
Post by: RuiAce on October 15, 2016, 02:50:25 pm
Hi, thanks for the help but how did u get 2R/3
Post by: julzzz on October 15, 2016, 03:33:52 pm
I end up getting (9√2)/8 but the answer is (3√2)8
I'm so close yet so far... help!
Post by: RuiAce on October 15, 2016, 03:40:08 pm
I have to differentiate: 3x/√3x-4+√3x-4
I end up getting (9√2)/8 but the answer is (3√2)8
I'm so close yet so far... help!
Post by: julzzz on October 15, 2016, 03:43:27 pm
Oh I'm so sorry, I meant to add that f'(x) = 2
Post by: jakesilove on October 15, 2016, 03:50:24 pm
Oh I'm so sorry, I meant to add that f'(x) = 2
Using Rui's answer, and setting it equal to two, you find that x is a complex number. Potentially you haven't fully described the question properly; can you just post a screenshot?
Post by: RuiAce on October 15, 2016, 03:52:04 pm
Oh I'm so sorry, I meant to add that f'(x) = 2Doesn't mean anything because that has no (real) solutions.
Are you sure you haven't differentiated already and were trying to say that what you gave us WAS the derivative?
Post by: jakesilove on October 15, 2016, 03:53:42 pm
Doesn't mean anything because that has no (real) solutions.
Are you sure you haven't differentiated already and were trying to say that what you gave us WAS the derivative?
Nup, that still produces a complex solution
Post by: RuiAce on October 15, 2016, 03:55:14 pm
Nup, that still produces a complex solutionI have no clue either then. We need the actual question.
Post by: julzzz on October 15, 2016, 04:07:51 pm
I have no clue either then. We need the actual question.
Here you go :)
Sorry if I stuffed up the question
Post by: RuiAce on October 15, 2016, 04:10:52 pm
Here you go :)
Sorry if I stuffed up the question
Post by: julzzz on October 15, 2016, 04:17:07 pm
I understand how to get f'(x) but don't know how to get f''(x) which will let me finish the question as I can sub in 2 then.
Post by: RuiAce on October 15, 2016, 04:21:14 pm
I understand how to get f'(x) but don't know how to get f''(x) which will let me finish the question as I can sub in 2 then.
Post by: RuiAce on October 15, 2016, 04:30:18 pm
Sorry, your post came up with random symbols thrown into it making it very difficult to read...If you're in an app, go into web view. LaTeX does not get produced on the app; you need a proper web browser.
Otherwise I'm not sure at all what the problem is and you will need to provide sufficient evidence including screenshots for technical help.
Post by: julzzz on October 15, 2016, 04:38:44 pm
Sorry for being so naggy, but where did the 2 go at the end of the first differentiation?
Post by: RuiAce on October 15, 2016, 04:40:51 pm
Sorry for being so naggy, but where did the 2 go at the end of the first differentiation?Fixed, my bad on the accident there
Post by: julzzz on October 15, 2016, 04:56:06 pm
Fixed, my bad on the accident there
Cheers
Really appreciate all your help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Post by: BPunjabi on October 15, 2016, 05:52:33 pm
How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ?
Post by: jakesilove on October 15, 2016, 05:57:06 pm
How do you do these questions?
How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ?
Hey! For that equation to hold true, either
or
Looking to the first equation,
Looking to the second equation
So those should be our three solutions!
Post by: RuiAce on October 15, 2016, 06:14:28 pm
How do you do these questions?
How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ?
Post by: BPunjabi on October 15, 2016, 06:15:15 pm
Hey! For that equation to hold true, either
or
Looking to the first equation,
Looking to the second equation
So those should be our three solutions!
Sorry man the answers says 2. Its multiple choice question 7 for the 2014 paper. Also could you explain to me how you quickly worked out the fractions for the solutions like:
Post by: nibblez16 on October 15, 2016, 06:16:54 pm
Post by: RuiAce on October 15, 2016, 06:17:23 pm
Sorry man the answers says 2. Its multiple choice question 7 for the 2014 paper. Also could you explain to me how you quickly worked out the fractions for the solutions like:
Anyway, I explained his mistake.
Post by: BPunjabi on October 15, 2016, 06:19:22 pm
Anyway, I explained his mistake.
Sorry I quoted and said the wrong thing... I meant the answers for the second part. When I put those values into my calc it does not equal exactly zero, and is there a way to figuring it out fast?
Post by: RuiAce on October 15, 2016, 06:20:22 pm
Hello, can you please help me with HSC 2007 question 7a(i)... Its a little confusing....In the future, please provide a link to the paper.
Post by: RuiAce on October 15, 2016, 06:21:10 pm
Sorry I quoted and said the wrong thing... I meant the answers for the second part. When I put those values into my calc it does not equal exactly zero, and is there a way to figuring it out fast?2.03 and 5.18 are obviously rounded.
Post by: BPunjabi on October 15, 2016, 06:23:53 pm
2.03 and 5.18 are obviously rounded.
Yes but can you please tell me how you get that answer quickly... you dont just randomly test those points in your calculator.. there must be some method to getting those values.
Thanks,
Bobby
Post by: RuiAce on October 15, 2016, 06:25:13 pm
Yes but can you please tell me how you get that answer quickly... you dont just randomly test those points in your calculator.. there must be some method to getting those values.
Thanks,
Bobby
Post by: asd987 on October 15, 2016, 06:30:15 pm
Post by: nibblez16 on October 15, 2016, 06:34:40 pm
In the future, please provide a link to the paper.
Thank You so much!
Sorry, I'll provide a link next time... Thanks :)
Post by: RuiAce on October 15, 2016, 06:45:59 pm
help plz A rectangle is cut from a circular disc of radius 6 cm. Find the area of the largest rectangle that can be produced(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsffwxytg1.png)
Post by: RuiAce on October 15, 2016, 07:58:15 pm
hi, i always get stuck when it comes to drawing the derivative function of a graph, has anyone got any tips to draw these?Be more specific please? If you're given the function, say, f(x)=x3+3x2 then all you have to do is differentiate it and then continue.
Do you mean drawing the graph of the derivative, given the graph of the original thing?
Post by: nimasha.w on October 15, 2016, 11:23:36 pm
Be more specific please? If you're given the function, say, f(x)=x3+3x2 then all you have to do is differentiate it and then continue.
Do you mean drawing the graph of the derivative, given the graph of the original thing?
sorry for the confusion!
like with (iii) i could get it all except from t1 to t3
Post by: asd987 on October 15, 2016, 11:29:43 pm
A surfboard is in the shape of a rectangle and semicircle. The perimeter is to be 4m. Find the maximum area of the surfboard, correct to 2 decimal places.
Post by: RuiAce on October 15, 2016, 11:43:34 pm
I need help with this question pleaseDraw a diagram and show us your working out thus far, like how I used a diagram for your earlier question. You cannot simply do those questions without a diagram.
A surfboard is in the shape of a rectangle and semicircle. The perimeter is to be 4m. Find the maximum area of the surfboard, correct to 2 decimal places.
sorry for the confusion!
like with (iii) i could get it all except from t1 to t3
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Post by: kevin217 on October 16, 2016, 10:36:28 am
Post by: RuiAce on October 16, 2016, 10:43:03 am
Depending on whether a function is concave up or down, how does Simpsons rule over/underestimate the exact area?Recall that Simpson's rule uses a parabolic arc to estimate the area.
Given any three points, sketch a parabola through them. Look at whether the area under the parabola is greater than, or less than the original area.
Note that the concavity of the function itself isn't enough for Simpson's rule. For the trapezoidal rule it is all that's needed because you're estimating with just trapeziums (and hence you draw straight lines). For Simpson's rule, you need to physically consider what the parabola through the three points looks like to make a conclusion. An example of where this was needed was in the 2013 HSC.
Post by: kevin217 on October 16, 2016, 11:13:16 am
Recall that Simpson's rule uses a parabolic arc to estimate the area.I see, thanks
Given any three points, sketch a parabola through them. Look at whether the area under the parabola is greater than, or less than the original area.
Note that the concavity of the function itself isn't enough for Simpson's rule. For the trapezoidal rule it is all that's needed because you're estimating with just trapeziums (and hence you draw straight lines). For Simpson's rule, you need to physically consider what the parabola through the three points looks like to make a conclusion. An example of where this was needed was in the 2013 HSC.
Post by: MysteryMarker on October 16, 2016, 06:55:22 pm
Thanks.
Post by: RuiAce on October 16, 2016, 07:39:50 pm
For question 15c, could someone explain to me how to do it both graphically and algebraically. Isn't that if I equate the two expressions, square them and solve for their discriminant to = 0, i should get a value for m? (I tried and its not working, just curious as to why).The graphical method was done in post #454
Thanks.
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Post by: epherbertson on October 16, 2016, 07:56:53 pm
This is from question 10 of the 1995 paper and am having trouble with the working of the question. I got part I out but i am lost from there. any help would be largely appreciated...
Thanks
Emily :-\
Post by: imtrying on October 16, 2016, 08:13:59 pm
Post by: RuiAce on October 16, 2016, 08:23:40 pm
Hey:) Just need a hand with this question from the 2012 paper. I've tried looking at the solutions but they just arent making sense to me. Thanks!
Post by: imtrying on October 16, 2016, 08:28:38 pm
Thanks so much, your explanation really helped:)
Post by: RuiAce on October 16, 2016, 08:41:43 pm
HelpThe question says to show us part a) so please link part a) next time.
This is from question 10 of the 1995 paper and am having trouble with the working of the question. I got part I out but i am lost from there. any help would be largely appreciated...
Thanks
Emily :-\
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Part (iv) cannot be done because you don't know how to solve -6+2t-t2/2 = 4sin(t)
Hence, try a graphical approach.
You may find the solutions here
Post by: cajama on October 17, 2016, 09:24:10 am
Post by: RuiAce on October 17, 2016, 09:37:13 am
Hello, I'm not sure if this question has been answered before and I know it's been talked about in one of the ATARNotes Maths lectures but are we still allowed to use abbreviations in the reasonings (e.g. isosceles Δ abbreviated to isos.Δ / supplementary to supp.)? A friend of mine said that this year, we wouldn't be allowed to use them and I'm just a little confused.I haven't heard anything about this (maybe the two other mods will input further).
Personally I always chose when to abbreviate. At times, abbreviating is a convenient shorthand, but at other times it reduces the formality of your work and begins to read like laziness.
Post by: isaacdelatorre on October 17, 2016, 11:30:37 am
Just a question, regarding rounding.
I know that you shouldn't round any numbers until the last part of the question. But when it says round to 4dp for example, do you continue to use the whole decimal or just 4 decimal places?
E.g. rates of change when finding k - continue to use the whole thing or whatever decimal places they initially say
Post by: imtrying on October 17, 2016, 12:02:16 pm
I've done part i, the answer is (n+1)1/2 - n1/2.
Post by: jamonwindeyer on October 17, 2016, 12:13:15 pm
Hey guys,
Just a question, regarding rounding.
I know that you shouldn't round any numbers until the last part of the question. But when it says round to 4dp for example, do you continue to use the whole decimal or just 4 decimal places?
E.g. rates of change when finding k - continue to use the whole thing or whatever decimal places they initially say
Hey Isaac! If they say find to 4dp, and then the question continues, you can use that value!! :) but they would probably pay either or ;D
Post by: jakesilove on October 17, 2016, 12:53:19 pm
Just needing some help with part ii of this question from the 2011 HSC.
I've done part i, the answer is (n+1)1/2 - n1/2.
Hey! Cool question, check out my solution below and let me know if I can clarify anything!
(http://i.imgur.com/94nRCvj.png)
Jake
Annotation by Rui: In case his first line wasn't clear, Jake rationalised the denominator
Annotation by Jake of Annotation by Rui: Cheers buddy
Post by: imtrying on October 17, 2016, 01:49:22 pm
Hey! Cool question, check out my solution below and let me know if I can clarify anything!Thanks so much! :)
(http://i.imgur.com/94nRCvj.png)
Jake
Annotation by Rui: In case his first line wasn't clear, Jake rationalised the denominator
Post by: jamonwindeyer on October 17, 2016, 03:37:18 pm
Hello, I'm not sure if this question has been answered before and I know it's been talked about in one of the ATARNotes Maths lectures but are we still allowed to use abbreviations in the reasonings (e.g. isosceles Δ abbreviated to isos.Δ / supplementary to supp.)? A friend of mine said that this year, we wouldn't be allowed to use them and I'm just a little confused.
Sorry cajama I missed this! Why does your friend say specifically this year? What do they think has changed? That's the curious bit to me :)
In general, you want to make your work as readable for the marker as is humanly possible. I am reasonably sure that abbreviations are fine, and that they would pay you for those that you've provided, but I was always hesitant to use them. Not because they wouldn't recognise them and not pay you for it intentionally, but more along Rui's line of thinking. It's about clarity and formality of proof, and basically, I wanted to take no chances that what I'd written had been misinterpreted.
My thinking is this. The BOSTES sample solutions don't use those abbreviations, so if you want to be 100% sure that you are okay, then take the 1 second to write the full word. Of course, symbols like Δ and // are universal and definitely acceptable :)
Post by: jakesilove on October 17, 2016, 03:47:06 pm
Sorry cajama I missed this! Why does your friend say specifically this year? What do they think has changed? That's the curious bit to me :)
In general, you want to make your work as readable for the marker as is humanly possible. I am reasonably sure that abbreviations are fine, and that they would pay you for those that you've provided, but I was always hesitant to use them. Not because they wouldn't recognise them and not pay you for it intentionally, but more along Rui's line of thinking. It's about clarity and formality of proof, and basically, I wanted to take no chances that what I'd written had been misinterpreted.
My thinking is this. The BOSTES sample solutions don't use those abbreviations, so if you want to be 100% sure that you are okay, then take the 1 second to write the full word. Of course, symbols like Δ and // are universal and definitely acceptable :)
I definitely used abbreviations for anything that was absolutely obvious (alt. angles on || are =, for instance), however if I thought there was a chance of confusion I would avoid abbreviations all together. I wouldn't say this makes much of a difference though!
Post by: olivercutbill on October 17, 2016, 04:43:53 pm
Post by: jamonwindeyer on October 17, 2016, 05:01:18 pm
I've seen questions like this before but they've given a time constraint. How do I do 12?
Are you an Extension student Oliver? Because I think this has strayed into that territory of difficulty :P
So we have this expression for cost in terms of speed, per hour:
$$c=150+\frac{v^2}{80}$$
What we want is the total cost for a 500km trip. To find how long THIS takes, we use our speed formula:
$$\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\v=\frac{500}{t}\\\therefore t=\frac{500}{v}$$
So our trip takes that many hours. To find our total cost, we multiply the cost per hour by the amount of hours taken that we just found.
$$C_\text{Total}=ct=\frac{500}{v}\left(50+\frac{v^2}{80}\right)$$
And now we're back to familiar territory, differentiate and find your minima! Does that help? :)
Post by: olivercutbill on October 17, 2016, 05:45:32 pm
Are you an Extension student Oliver? Because I think this has strayed into that territory of difficulty :P
So we have this expression for cost in terms of speed, per hour:
$$c=150+\frac{v^2}{80}$$
What we want is the total cost for a 500km trip. To find how long THIS takes, we use our speed formula:
$$\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\v=\frac{500}{t}\\\therefore t=\frac{500}{v}$$
So our trip takes that many hours. To find our total cost, we multiply the cost per hour by the amount of hours taken that we just found.
$$C_\text{Total}=ct=\frac{500}{v}\left(50+\frac{v^2}{80}\right)$$
And now we're back to familiar territory, differentiate and find your minima! Does that help? :)
No 2u, this was in the textbook :)
I get 63km/h as a minimum - textbook answer says 110km/h.
What am I doing wrong?
Post by: jamonwindeyer on October 17, 2016, 05:51:04 pm
No 2u, this was in the textbook :)
I get 63km/h as a minimum - textbook answer says 110km/h.
What am I doing wrong?
Oh I mean like, it's strayed into that level of difficulty, definitely 2U content. I mean like, don't stress that it's hard for you, because it is a hard question, aha ;D
Snap a picture of your working? I'll see if I can spot ;D
Post by: olivercutbill on October 17, 2016, 06:25:28 pm
Oh I mean like, it's strayed into that level of difficulty, definitely 2U content. I mean like, don't stress that it's hard for you, because it is a hard question, aha ;D
Snap a picture of your working? I'll see if I can spot ;D
Figured it out, and did another similar question and got that too.
Thanks :)
Post by: Rikahs on October 17, 2016, 07:37:01 pm
Post by: RuiAce on October 17, 2016, 07:54:44 pm
Does anyone know how to answer the attached question (all parts please). Im super confused :(Part (iii) was already addressed in post #533
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Post by: nibblez16 on October 17, 2016, 09:59:50 pm
Post by: RuiAce on October 17, 2016, 10:01:39 pm
Hello, can you please explain the last question... I always get confused on those types... Thanks :)Can we please have a link to the paper or the answers to all the previous parts? For a question like this that might be relevant.
Post by: nibblez16 on October 17, 2016, 10:11:38 pm
Can we please have a link to the paper or the answers to all the previous parts? For a question like this that might be relevant.
Post by: RuiAce on October 18, 2016, 12:43:31 am
Post by: FallonXay on October 18, 2016, 09:04:14 am
with this question, why does the minimum of D occur at the same time as the minimum of D^2?
Also, why do you use the negative root in the volume question?
And, how do you solve c part ii (Using the graph... find the values of m... one solution) :D
Thanks!
Post by: kimmie on October 18, 2016, 09:49:54 am
Post by: RuiAce on October 18, 2016, 09:49:57 am
Hello,
with this question, why does the minimum of D occur at the same time as the minimum of D^2?
Also, why do you use the negative root in the volume question?
And, how do you solve c part ii (Using the graph... find the values of m... one solution) :D
Thanks!
Post by: RuiAce on October 18, 2016, 09:52:08 am
Hello,
with this question, why does the minimum of D occur at the same time as the minimum of D^2?
Also, why do you use the negative root in the volume question?
And, how do you solve c part ii (Using the graph... find the values of m... one solution) :D
Thanks!
Post by: RuiAce on October 18, 2016, 09:54:31 am
Hey can you please help me with this question. It's from the 2013 hsc paper. I did part (i) but I don't understand how to do part (ii)Already done in post #454
Post by: FallonXay on October 18, 2016, 10:00:47 am
Why do we specifically need to use the left branch though? Sorry - I still don't understand :P
Post by: RuiAce on October 18, 2016, 10:08:26 am
Why do we specifically need to use the left branch though? Sorry - I still don't understand :P
Post by: FallonXay on October 18, 2016, 10:20:43 am
Ahh k, I see. thx :)
Post by: kimmie on October 18, 2016, 10:28:28 am
Already done in post #454
I checked it and I'm still confused why would we use 2 for the gradient thing?
Post by: RuiAce on October 18, 2016, 10:33:07 am
I checked it and I'm still confused why would we use 2 for the gradient thing?
Post by: kimmie on October 18, 2016, 10:46:13 am
Post by: kimmie on October 18, 2016, 10:48:01 am
Post by: jakesilove on October 18, 2016, 11:36:30 am
Can you help me with part (iv) and (v) please
Hey! Cool question, I think this was from my year's paper :)
We know that the equation for Carp population will be
From the information given. This means that the rate of increase in Carp will be
The rate of decrease in Trout is
Now, for part iv), we want these two rates to be equal! We absolute value them, because one is decreasing and one is increasing. So,
For the next part, we just set the numbers equal to each other
This is a tricky solution. BOSTES give fairly straight forward answers to this part.
(http://i.imgur.com/9pkfwAr.png)
Let me know if I can clarify anything!
Post by: kevin217 on October 18, 2016, 11:52:46 am
Thanks
Post by: jakesilove on October 18, 2016, 12:06:20 pm
Hi, could someone provide me a solution for the second derivative in i).
Thanks
Hey! So we've got
To find the first derivative, we can just apply the chain rule to the second part
To find the second derivative, we can just apply the chain rule again.
As t is always positive, (1+t) will always be positive, and so (1+t)^(-3) will always be positive. Therefore, -2*positive will always be negative, as required.
Post by: kevin217 on October 18, 2016, 12:10:56 pm
Hey! So we've gotSweet, thanks for the solution.
To find the first derivative, we can just apply the chain rule to the second part
To find the second derivative, we can just apply the chain rule again.
As t is always positive, (1+t) will always be positive, and so (1+t)^(-3) will always be positive. Therefore, -2*positive will always be negative, as required.
Post by: nimasha.w on October 18, 2016, 03:04:34 pm
Post by: mfjw on October 18, 2016, 03:33:50 pm
Post by: RuiAce on October 18, 2016, 03:37:52 pm
Can someone help me with this question, please?
There. The indices are all positive now.
Post by: RuiAce on October 18, 2016, 03:39:08 pm
hey! could someone please explain how you would do this question :-)Im on a phone right now so I can't link but part i is already done.
Edit: Page 37 post 549 or something
Post by: mfjw on October 18, 2016, 03:41:43 pm
Post by: RuiAce on October 18, 2016, 03:43:04 pm
I'm a bit stuck on this question right now. Could someone please explain how to do Q4g and Q4h?The answer is the question itself. It's called a trick question; it is already answered for you
Post by: jakesilove on October 18, 2016, 03:43:24 pm
I'm a bit stuck on this question right now. Could someone please explain how to do Q4g and Q4h?
Remember that
So,
I'll let you try h)
Post by: RuiAce on October 18, 2016, 03:46:34 pm
Remember thatI nearly walked into that.
So,
I'll let you try h)
Jake, that index was already positive
Post by: jakesilove on October 18, 2016, 03:49:18 pm
I nearly walked into that.
Jake, that index was already positive
I know it looks positive, but I assumed the picture was just bad quality. Surely they wouldn't give an already-positive indice in numerous questions, as the image appears.
Post by: RuiAce on October 18, 2016, 03:50:38 pm
I know it looks positive, but I assumed the picture was just bad quality. Surely they wouldn't give an already-positive indice in numerous questions, as the image appears.Yeah it makes sense because there was a "space" there. Oh well.
Post by: jakesilove on October 18, 2016, 03:51:52 pm
Yeah it makes sense because there was a "space" there. Oh well.
Either way, questioner get's a comprehensive response!
Post by: zachary99 on October 18, 2016, 06:28:03 pm
Can you score 100 for a subject? or is the maximum hsc mark 99? (just like the max atar is 99.95 not 100)
Post by: RuiAce on October 18, 2016, 06:30:46 pm
just did last years paper and scored 100% ;DYes. Whilst it's extremely hard, it is possible to score 100.
Can you score 100 for a subject? or is the maximum hsc mark 99? (just like the max atar is 99.95 not 100)
1st place in 2U mathematics always gets 100.
Post by: zachary99 on October 18, 2016, 06:43:09 pm
yeah it will be very hard
Post by: jamonwindeyer on October 18, 2016, 07:33:25 pm
just did last years paper and scored 100% ;D
Can you score 100 for a subject? or is the maximum hsc mark 99? (just like the max atar is 99.95 not 100)
In case anyone is curious, the reason for this is because an ATAR of 100 means you beat 100% of students in the state. Including yourself ;)
Post by: RuiAce on October 18, 2016, 07:36:57 pm
Post by: Blissfulmelodii on October 18, 2016, 07:43:20 pm
In case anyone is curious, the reason for this is because an ATAR of 100 means you beat 100% of students in the state. Including yourself ;)
That's a fair point but in that case why stop at 99.95? why not go to 99.99?
Post by: jakesilove on October 18, 2016, 08:52:06 pm
That's a fair point but in that case why stop at 99.95? why not go to 99.99?
Policy reasons; an ATAR is one a tool to limit university intake on a supply/demand basis. There's really no point scaling people about 99.95, since only like 40 people in the state get that mark anyway!
Post by: Blissfulmelodii on October 18, 2016, 08:53:00 pm
Policy reasons; an ATAR is one a tool to limit university intake on a supply/demand basis. There's really no point scaling people about 99.95, since only like 40 people in the state get that mark anyway!
ahh okay. Fair enough
Post by: FallonXay on October 18, 2016, 09:01:36 pm
I'm having some trouble understanding how to go about doing part v of this question. ???
Thanks for the help btw!
Post by: RuiAce on October 18, 2016, 09:04:15 pm
Hiya all!This was answered somewhere in the last two pages (on my phone right now so can't link)
I'm having some trouble understanding how to go about doing part v of this question. ???
Thanks for the help btw!
Post by: FallonXay on October 18, 2016, 09:05:34 pm
This was answered somewhere in the last two pages (on my phone right now so can't link)
ahh k, found it. cheers ;D
Post by: bethjomay on October 18, 2016, 09:07:16 pm
I have a couple of questions! For the first one, I've done part I) but I'm having trouble with the transformation to reach the answer in part ii) and a bit lost after that. In the second question I got all but part three. The answers mentioned 'favourable outcomes' which seems quite simple but I'm not familiar, could someone enlighten me? Thank you! (http://uploads.tapatalk-cdn.com/20161018/5809966f3ef60c285d37eff0bfca1a82.jpg)(http://uploads.tapatalk-cdn.com/20161018/cb463a433b62ebdeb078691f6b64554e.jpg)
Post by: RuiAce on October 18, 2016, 09:42:38 pm
(http://uploads.tapatalk-cdn.com/20161018/ae3ca657354ac4d4ef18ca34ea27cb96.jpg)
I have a couple of questions! For the first one, I've done part I) but I'm having trouble with the transformation to reach the answer in part ii) and a bit lost after that. In the second question I got all but part three. The answers mentioned 'favourable outcomes' which seems quite simple but I'm not familiar, could someone enlighten me? Thank you! (http://uploads.tapatalk-cdn.com/20161018/5809966f3ef60c285d37eff0bfca1a82.jpg)(http://uploads.tapatalk-cdn.com/20161018/cb463a433b62ebdeb078691f6b64554e.jpg)
Some hints now:
For part (iii), recall your usual coordinate methods in geometry questions. You're given a length, and you want to find the area of a triangle with it. Usually, to use A=1/2 bh, we consider an associated perpendicular distance.
(iv) is just your brute force max/min question
Post by: RuiAce on October 18, 2016, 09:44:24 pm
Note that in this context, however, it's used to list the amount of favourable combinations of tickets.
Post by: :3 on October 18, 2016, 10:33:20 pm
(https://i.gyazo.com/801fbb3e27932847d4000021a9557c6d.png)
Thanks :P
Post by: RuiAce on October 18, 2016, 11:19:25 pm
2007, Q10: having trouble understanding the answer.
(https://i.gyazo.com/801fbb3e27932847d4000021a9557c6d.png)
Thanks :P
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Post by: fizzy.123 on October 19, 2016, 12:18:09 am
Post by: jamonwindeyer on October 19, 2016, 12:22:35 am
How do we solve these sort of questions??
Hey Fizzy! This one is a bit of intuition, a bit of Trial and Error, you are looking for a value of x on either side of the origin that puts as much area above the x-axis as below! This way, the areas cancel each other out and you get zero ;D
1 to -1: 2 square units above, 0 square units below
2 to -2: 3.5 square units above, 0 square units below
3 to -3: 3.75 square units above, 0.75 square units below
4 to 0 -4: 3.75 square units above, 2.75 square units below
And I'll let you think about the value we'd need to have them match ;) note that I am finding these areas using basic area formula for rectangles and triangles! ;D
Post by: RuiAce on October 19, 2016, 12:26:07 am
How do we solve these sort of questions??Covered in post #464 or alternatively just use what Jamon said
Post by: fizzy.123 on October 19, 2016, 01:00:39 am
how many solutions of the equation (sinx -1)(tanx +2) = 0 lie between 0 & 2pi
How do we solve these sort of questions?
Post by: ml125 on October 19, 2016, 04:06:25 am
thank you both! Also, another question.
how many solutions of the equation (sinx -1)(tanx +2) = 0 lie between 0 & 2pi
How do we solve these sort of questions?
Rui edit: Just fixed up some LaTeX for ya
Post by: smile123 on October 19, 2016, 09:07:26 am
I was wondering if some one can help with this question part (iii)
Post by: RuiAce on October 19, 2016, 09:18:45 am
Hey :)
I was wondering if some one can help with this question part (iii)
Post by: smile123 on October 19, 2016, 09:33:18 am
Post by: kimmie on October 19, 2016, 12:24:26 pm
Post by: jamonwindeyer on October 19, 2016, 12:31:28 pm
soz for barging in but can you explain why tanx= -2 has two solutions? im confused about that
No worries if you look at a graph of tanx like the one here, we can see that from 0 to \(2\pi\), there are two locations where \(\tan{x}=-2\) ;D it's a graphical approach that we take here (or you can use your ASTC rules to find the values exactly, if you want to) ;D
Post by: rinagee12 on October 19, 2016, 03:54:49 pm
Post by: jakesilove on October 19, 2016, 04:01:41 pm
Would someone please be able to help me with this question? :) Just with the last part, (iv).
Hey! To find the total distance traveled, you would usually just think to integrate, right? However, you need to think about different sections separately. We already found that velocity equals zero at Pi/3. Before that, the velocity is negative, and after that, the velocity is positive. Therefore, it is travelling in opposite directions in these two sectors.
To answer the question, just integrate between 0 and Pi/3, and the Pi/3 to Pi. Absolute value both solutions, and add them together! Let me know if you want a fully worked solution.
Jake
Post by: FallonXay on October 19, 2016, 04:14:33 pm
The answers say 2, but shouldn't it be 3???
Post by: RuiAce on October 19, 2016, 04:35:03 pm
2014 Q7: How many solutions of the equation (sin x – 1)(tan x + 2) = 0 lie between 0 and 2π ?
The answers say 2, but shouldn't it be 3???
That was the trap in that question.
Post by: FallonXay on October 19, 2016, 04:46:39 pm
Thanks. :)
Post by: RuiAce on October 19, 2016, 06:22:44 pm
Hiya. How do you do b) part (ii) of this question?
Thanks. :)
Post by: FallonXay on October 19, 2016, 07:33:28 pm
Thankyou!
Post by: RuiAce on October 19, 2016, 07:45:24 pm
With this question, how come the little 'triangle' section under the semi circle is shaded? Doesn't the region of the semicircle stop at the x axis?
Thankyou!
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsjszycfb4.png)
And then you're combining regions.
Post by: SimplyNikhil on October 19, 2016, 07:49:59 pm
Thanks in advance!
Post by: RuiAce on October 19, 2016, 07:52:20 pm
Hey, can someone please show how to do this question?
Thanks in advance!
Post by: RuiAce on October 19, 2016, 07:54:34 pm
Makes sense, but then why is the range for semicircles always stated between 0 and some number (i.e in this case stated as between 0<=y<=3 instead of just y<=3?)
Post by: FallonXay on October 19, 2016, 07:56:51 pm
Yeah, just realised. Thanks :P
Post by: Rikahs on October 19, 2016, 08:11:51 pm
I have attached a pic below (sorry for sloppy drawing). Let point D be the midpoint of hypotenuse BC. If in HSC you say CD=AD=AB (Midpoint of hypotenuse is equidistant from vertices of triangle ABC). Is the part in the brackets a valid proof or would you have to prove AD= CD or AD= DB another way.
Thanks
Post by: jakesilove on October 19, 2016, 08:17:55 pm
Hi everyone,
I have attached a pic below (sorry for sloppy drawing). Let point D be the midpoint of hypotenuse BC. If in HSC you say CD=AD=AB (Midpoint of hypotenuse is equidistant from vertices of triangle ABC). Is the part in the brackets a valid proof or would you have to prove AD= CD or AD= DB another way.
Thanks
Personally, I would suggest that that you prove it another way, because it isn't a 'classic' 2U proof. As such, markers may not be happy with you 'assuming' things beyond the curriculum.
Post by: RuiAce on October 19, 2016, 08:22:34 pm
Hi everyone,Definitely not. The midpoint on a hypotenuse theorem is not assumed and you need to prove it from what you're given
I have attached a pic below (sorry for sloppy drawing). Let point D be the midpoint of hypotenuse BC. If in HSC you say CD=AD=AB (Midpoint of hypotenuse is equidistant from vertices of triangle ABC). Is the part in the brackets a valid proof or would you have to prove AD= CD or AD= DB another way.
Thanks
Post by: rahaf on October 19, 2016, 09:02:11 pm
Post by: RuiAce on October 19, 2016, 09:16:44 pm
Hi! Can somebody help me with this please.Addressed in post #620
Post by: RuiAce on October 19, 2016, 10:09:28 pm
Saw that but was wondering how is there 2 solutions??? I understand sinx-1=0 but how do you solve the tan one. How do you know it has two solutions?
Post by: Deng on October 19, 2016, 11:32:15 pm
locus and parabola for y^2=4ax, is the directrix y=-a and the focus (x+ or - a, y). For some reason i cant grasp the y^2 side despite knowing how the x^2 side works
This may be hard to explain but how would i graph velocity graphs ? I can work out the intercepts and roots, but i get stumped when wondering how to draw them and i dont necessarily have a question on hand
The sum and product of roots and the variations outside of alpha + beta, alpha beta, there is one of them which i cant exactly remember ( which i know isnt helpful in trying to get help for) but its a more 'complex' one that requires a bit more algebraic manipulation
Thanks
Post by: RuiAce on October 19, 2016, 11:41:33 pm
Hey guys, i was wondering if anyone could help give me iron out some stuff: e.gRecall that the parabola x^2=4ay has directrix y=-a and focus (0,a)
locus and parabola for y^2=4ax, is the directrix y=-a and the focus (x+ or - a, y). For some reason i cant grasp the y^2 side despite knowing how the x^2 side works
This may be hard to explain but how would i graph velocity graphs ? I can work out the intercepts and roots, but i get stumped when wondering how to draw them and i dont necessarily have a question on hand
The sum and product of roots and the variations outside of alpha + beta, alpha beta, there is one of them which i cant exactly remember ( which i know isnt helpful in trying to get help for) but its a more 'complex' one that requires a bit more algebraic manipulation
Thanks
Hence, the parabola y^2=4ax has directrix x=-a and focus (a,0)
_________
Sorry but yes you will need to find a question. What you're saying isn't coming through. If you just meant derivative curve sketching that's somewhere in this thread already.
________
Post by: Deng on October 20, 2016, 12:02:55 am
Thanks again !
Post by: RuiAce on October 20, 2016, 12:12:29 am
This example isnt the best example because i know how to graph this one, essentially the motion questions where i have to graph the velocity as part of it, also for part V for questions like that when would i know i would have to integrate or i can use t = 0;x=2 , t=1;x=4 etcStill not sure what the problem is. If you're given the equation then like you said, you basically know how to draw it.
Thanks again !
You need to further clarify exactly what you mean by "you don't know how to sketch it". If you're given the actual equation of the velocity, you will virtually always know when and how to sketch it.
That one is basically sketching y=3x2-9x+6 except with y switched for v and x switched for t
_________________________
Part (v) can be made slightly easier with the answer to part (i). The idea is that the displacement takes into account direction, but the distance does not. So you need to watch out for whenever the particle turns around.
Of course, you CAN use integration for part (v). You just don't need to.
_________________________
Also, if that is a past HSC question, consider checking worked solutions.
Post by: Deng on October 20, 2016, 12:42:18 am
Would the focus be (k + a, h) and the directix y= k -a
So for the y^2 side would it essentially be the x value shifting up and down for the focus and the directix would be the x value minus the focal lenght?
Thanks
Post by: RuiAce on October 20, 2016, 08:10:47 am
For the equation (y-k)^2= 4a(x-h)^2Recall that for the parabola (x-h)^2 = 4a(y-k)
Would the focus be (k + a, h) and the directix y= k -a
So for the y^2 side would it essentially be the x value shifting up and down for the focus and the directix would be the x value minus the focal lenght?
Thanks
The vertex is at (h,k).
Since the parabola concaves UP, the focus is a units ABOVE the vertex, and the directrix is a units BELOW the vertex.
i.e. (h, k+a) and y=k-a
I never memorised these formulae because there was no point. I used common sense
So for the parabola (y-k)^2 = 4a(x-h)
The vertex is still at (h,k).
The parabola concaves to the RIGHT now. So the focus is a units TO THE RIGHT of the vertex, and the directrix is a units to the LEFT of the vertex
i.e. (h+a, k) and x=h-a
Post by: epherbertson on October 20, 2016, 09:19:21 am
Thanks,
Emily :-\
Post by: RuiAce on October 20, 2016, 10:47:53 am
This is a question from the 2005 paper that no one in my class has figured out yet so just looking for a bit of help to conquer this mess.Already done in post #466
Thanks,
Emily :-\
Post by: SimplyNikhil on October 20, 2016, 11:32:34 am
Post by: BPunjabi on October 20, 2016, 11:48:15 am
http://pasteboard.co/h2n5xrD5Q.png
(https://cdn.pbrd.co/images/h2n5xrD5Q.png)
Post by: Deng on October 20, 2016, 11:49:18 am
But to find the area it would be top curve - bottom curve
ln2x-lnx
By index laws that would become ln2
Since ln2 is a constant you can integrate it to xln2 and you just simply sub in b and a to solve
Hope i answered your question
Post by: BPunjabi on October 20, 2016, 11:50:34 am
@SimplyNikhil ( not sure how to tag your post )
But to find the area it would be top curve - bottom curve
ln2x-lnx
By index laws that would become ln2
Since ln2 is a constant you can integrate it to xln2 and you just simply sub in b and a to solve
Hope i answered your question
Lol I was thinking that too but then I choked and was like what the fuck? do we integrate ln2 or divide now??
Post by: Deng on October 20, 2016, 11:52:11 am
How i solve these ( which may not be correct, Rui may correct me after if im wrong)
Equation is Sin(2x+pi/3)
To find the where it crosses the x axis
let 2x+pi/3=0
2x = -pi/3
x = -pi/6
Therefore when we inspect A,B,C,D
A and B are eliminated
And between C and D , D is -pi/6
Again, not sure if this is the best way to do it but that is how ive always done it and it has worked for me
Post by: Deng on October 20, 2016, 11:56:03 am
The first derivative would be f'(a) < 0 since the gradient is negative
I just cant picture in my head what the second derivate would look like ( some reason i picture a horizontal line since f(x) looks like a parabola to me
Post by: RuiAce on October 20, 2016, 12:00:15 pm
???This is partly wrong. Please wait for a correction. This question is actually much harder than what it looks like.
Everything that's wrong starts from about here. But I'm not going to delete this post. I'm going to KEEP this mistake here, but also post up a new solution. (http://i1164.photobucket.com/albums/q566/Rui_Tong/tediousness_zpscmj0pqbu.png)
Post by: BPunjabi on October 20, 2016, 12:05:41 pm
@BPunjabiYou are a fucking Beast, probs to you man!!
How i solve these ( which may not be correct, Rui may correct me after if im wrong)
Equation is Sin(2x+pi/3)
To find the where it crosses the x axis
let 2x+pi/3=0
2x = -pi/3
x = -pi/6
Therefore when we inspect A,B,C,D
A and B are eliminated
And between C and D , D is -pi/6
Again, not sure if this is the best way to do it but that is how ive always done it and it has worked for me
Thanks so much, btw to reply click quote on the top right of every post if you want to reply to that message!
Post by: jamonwindeyer on October 20, 2016, 12:08:05 pm
@BPunjabi
How i solve these ( which may not be correct, Rui may correct me after if im wrong)
Equation is Sin(2x+pi/3)
To find the where it crosses the x axis
let 2x+pi/3=0
2x = -pi/3
x = -pi/6
Therefore when we inspect A,B,C,D
A and B are eliminated
And between C and D , D is -pi/6
Again, not sure if this is the best way to do it but that is how ive always done it and it has worked for me
Good method Deng! I'd personally suggest a more intuitive approach based on rules that you can find here, once you get the hang of doing it this way, purely for speed! ;D
@SimplyNikhil ( not sure how to tag your post )
But to find the area it would be top curve - bottom curve
ln2x-lnx
By index laws that would become ln2
Since ln2 is a constant you can integrate it to xln2 and you just simply sub in b and a to solve
Hope i answered your question
This is also correct, and extremely clever, well done Deng ;D
Post by: Deng on October 20, 2016, 12:12:28 pm
Would this not give the correct working
Area = integrate between e and 1 for ln2x-ln2
= ln 2 + ln x - ln x
= ln2 ( constant there is no x, therefore integratable )
Therefore, it becomes xln2 between e and 1
eln2 - ln2
= ln2 ( e - 1 )
I'm not sure if the method you provided gave same answer because i dont understand it
Post by: BPunjabi on October 20, 2016, 12:13:39 pm
Good method Deng! I'd personally suggest a more intuitive approach based on rules that you can find here, once you get the hang of doing it this way, purely for speed! ;D
So Jamon, we are basically looking at (Bx+C)
Post by: RuiAce on October 20, 2016, 12:14:15 pm
(http://i1164.photobucket.com/albums/q566/Rui_Tong/tediousness_zpsxuxhax8e.png)
I oversimplified it. You actually have to analyse THREE regions, not just TWO. I have labelled them in the diagram for reference.
Jamon please double check
Post by: jamonwindeyer on October 20, 2016, 12:15:03 pm
At Rui for the area under the curve integral question
Would this not give the correct working
...
I'm not sure if the method you provided gave same answer because i dont understand it
Jamon please double check
Yeah to reiterate above, Rui's method is overkill in this case (though I think correct), Deng your answer is definitely correct ;D
Post by: RuiAce on October 20, 2016, 12:15:42 pm
At Rui for the area under the curve integral questionAh. I missed the logarithm law.
Would this not give the correct working
Area = integrate between e and 1 for ln2x-ln2
= ln 2 + ln x - ln x
= ln2 ( constant there is no x, therefore integratable )
Therefore, it becomes xln2 between e and 1
eln2 - ln2
= ln2 ( e - 1 )
I'm not sure if the method you provided gave same answer because i dont understand it
Post by: BPunjabi on October 20, 2016, 12:16:21 pm
(https://cdn.pbrd.co/images/h2KkrmZyj.png)
Post by: jamonwindeyer on October 20, 2016, 12:16:34 pm
So Jamon, we are basically looking at (Bx+C)
Yeah precisely! ;D
Post by: RuiAce on October 20, 2016, 12:17:45 pm
Hey guys, i was wondering on how i would identify the second derivative for this qDo not picture the second derivative in your head. Use definitions.
The first derivative would be f'(a) < 0 since the gradient is negative
I just cant picture in my head what the second derivate would look like ( some reason i picture a horizontal line since f(x) looks like a parabola to me
The first derivative tells us whether it is increasing or decreasing. Since it is decreasing, f'(a) < 0
The second derivative tells us its concavity. Since it is concave up, f"(a) > 0
Post by: jamonwindeyer on October 20, 2016, 12:18:55 pm
anyone know how to do this?
(https://cdn.pbrd.co/images/h2KkrmZyj.png)
Edit: Woops, dictation error
The denominator of a function cannot be zero, so:
But also we can't square root a negative number:
Put these together to obtain \(x>-3\) ;D
Post by: Deng on October 20, 2016, 12:37:07 pm
I've been told to draw a parabola but i dont particularly understand this method and just want more clarification
Thanks
Post by: RuiAce on October 20, 2016, 12:39:05 pm
For this question, if i did the maths right, the x value where it is increasing is 2/3 and 2, but how would i know if its 2/3<x<2 or if its split
I've been told to draw a parabola but i dont particularly understand this method and just want more clarification
Thanks
Post by: BPunjabi on October 20, 2016, 12:42:23 pm
Edit: Woops, dictation error
The denominator of a function cannot be zero, so:
But also we can't square root a negative number:
Put these together to obtain \(x>-3\) ;D
Thanks Jamon but the answer says a?
Post by: RuiAce on October 20, 2016, 12:43:46 pm
Thanks Jamon but the answer says a?That's what he got. He said in the last line that hence the answer is x > -3
Post by: rinagee12 on October 20, 2016, 12:44:59 pm
For what values of k will the equation 2x^2 + 8kx - (1-k) = 0
have roots that differ by 2?
My first thought was to use the formula
α+β=-b/a
and αβ=c/a
I tried to replace 'β' with (α-2) but I'm not sure if I'm doing the right thing, I can't seem to get the answer
Post by: RuiAce on October 20, 2016, 12:48:12 pm
Hey I'm stuck on this question:
For what values of k will the equation 2x^2 + 8kx - (1-k) = 0
have roots that differ by 2?
My first thought was to use the formula
α+β=-b/a
and αβ=c/a
I tried to replace 'β' with (α-2) but I'm not sure if I'm doing the right thing, I can't seem to get the answer
That should give the answer. Please show us your working if something happened from here onwards
Post by: jakesilove on October 20, 2016, 12:49:54 pm
That should give the answer. Please show us your working if something happened from here onwards
Would you also need to do a case where beta = alpha + 2? As the question only asks for 'differs'? Like I dunno if you'd get different answers, but I suppose you would. Conceptually, it shouldn't make a difference, but since I think it would change the answer...
Post by: RuiAce on October 20, 2016, 12:52:06 pm
Would you also need to do a case where beta = alpha + 2? As the question only asks for 'differs'? Like I dunno if you'd get different answers, but I suppose you would. Conceptually, it shouldn't make a difference, but since I think it would change the answer...Those cases should all be addressed at once though... If we do beta = alpha + 2 we just get beta = -2k+1 and alpha = -2k-1
Post by: jakesilove on October 20, 2016, 12:54:40 pm
Those cases should all be addressed at once though... If we do beta = alpha + 2 we just get beta = -2k+1 and alpha = -2k-1
Yeah cool cool, just couldn't be bothered actually checking for consistency aha cheers
Post by: onepunchboy on October 20, 2016, 03:16:54 pm
Hey guys can someone help me with part iii) and iv)..
Thanks in adv !!
Post by: imtrying on October 20, 2016, 03:18:01 pm
A rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced
Thanks in advance:)
Post by: MysteryMarker on October 20, 2016, 03:23:30 pm
Post by: jakesilove on October 20, 2016, 03:33:34 pm
(http://uploads.tapatalk-cdn.com/20161020/f16fe5c9672d37242bfd4693afcab75d.jpg)
Hey guys can someone help me with part iii) and iv)..
Thanks in adv !!
Hey! For this question, you need to think about the area under the curve. The area under the curve, when positive, is movement (displacement) in the positive direction. The area under the curve, when negative, is movement (displacement) in the negative direction. You need to find the point at which the area in the NEGATIVE portion is equal to the area in the POSITIVE portion.
For the last part, just consider how integration works. When the gradient equals zero, (ie. dy/dx =0), what does that mean? Have a go, and if you still need help let me know!
Post by: jakesilove on October 20, 2016, 03:41:00 pm
Hey:) I'm sure this is fairly simple but i cant remember how to do it:
A rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced
Thanks in advance:)
Hey! We basically need to set up some sort of equation, and then differentiate to find a maximum value! Not a simple question at all.
There are actually heaps of ways to do this. I would probably draw a circle, and throw in the angles to your rectangle. You can break the problem into four triangles, and four minor segments. You can easily find the area of the four segments, and then find the MINIMUM area of the four minor segments using calculus. Then, when you find this MINIMUM external area, you can subtract that from the area of the circle to find the MAXIMUM area of the rectangle! Give it a go, and let me know if you can't get a solution out.
Post by: RuiAce on October 20, 2016, 03:42:06 pm
Hey:) I'm sure this is fairly simple but i cant remember how to do it:Done in post #538
A rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced
Thanks in advance:)
Hey guys, just a question from the 2005 HSC mathematics exam. I don't really understand this question, so i'd appreciate it if someone could explain how to do each part. Cheers :PDone in post #466
Post by: jamonwindeyer on October 20, 2016, 04:34:34 pm
Done in post #538Done in post #466
How the hell do you remember these? ;D
Post by: RuiAce on October 20, 2016, 04:39:39 pm
How the hell do you remember these? ;D(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsfzvianbw.png)
I just have to remember a few words I typed and the fact I answered it before :P
Post by: nibblez16 on October 20, 2016, 05:53:00 pm
Post by: jakesilove on October 20, 2016, 05:55:20 pm
Hello, can you please help me with the last question. I cannot understand properly... Thanks
Check out my answer HERE.
Post by: nibblez16 on October 20, 2016, 06:04:08 pm
Hey! Cool question, check out my solution below and let me know if I can clarify anything!
(http://i.imgur.com/94nRCvj.png)
Jake
Annotation by Rui: In case his first line wasn't clear, Jake rationalised the denominator
Annotation by Jake of Annotation by Rui: Cheers buddy
Thank You so muchh! Much more easier to understand :D
Quick question... Why does root 2 go infront and root 1 becomes negative?
Post by: RuiAce on October 20, 2016, 06:09:10 pm
Thank You so muchh! Much more easier to understand :D
Quick question... Why does root 2 go infront and root 1 becomes negative?
Post by: nibblez16 on October 20, 2016, 06:13:41 pm
Ohhh! Ok cool! :D
Post by: nibblez16 on October 20, 2016, 06:43:46 pm
Post by: jakesilove on October 20, 2016, 06:49:26 pm
Can you please help me with the last question
Hey! So we can say that Alex's salary grows like this
If they save a third of their funds, then they will save
etc. Now, we want this value to equal $87,500. You can use the sum of an arithmetic sequence (a formula you should definitely know!) to simplify the terms in brackets. Then, set that equal to 87,500 and solve for n! Let me know if you need a worked solution for this.
Jake
Post by: yasmineturner on October 20, 2016, 06:54:17 pm
I was just wondering if anyone could help me with this question: 10b from the 1995 10b (i, ii, iii, iv).
Thanks for your help :)
Post by: RuiAce on October 20, 2016, 06:56:46 pm
Hi everyone,Refer to post #522
I was just wondering if anyone could help me with this question: 10b from the 1995 10b (i, ii, iii, iv).
Thanks for your help :)
Post by: jakesilove on October 20, 2016, 06:57:41 pm
Hi everyone,
I was just wondering if anyone could help me with this question: 10b from the 1995 10b (i, ii, iii, iv).
Thanks for your help :)
Hey! Welcome to the forums! You can find velocity by differentiating displacement, so
To show that there are only two times where these velocities are the same, I would graph the two functions. There will only be two intercepts, thus showing that there are only two times in which the particles have the same velocity!
Annnnd I'm going to stop answering, because Rui is finding a solution he already wrote up. Bear with us! Good luck tomorrow :)
Post by: nibblez16 on October 20, 2016, 07:01:17 pm
Hey! So we can say that Alex's salary grows like this
If they save a third of their funds, then they will save
etc. Now, we want this value to equal $87,500. You can use the sum of an arithmetic sequence (a formula you should definitely know!) to simplify the terms in brackets. Then, set that equal to 87,500 and solve for n! Let me know if you need a worked solution for this.
Jake
Okay understandable! But what do we do with the 1/3? Thats where I am confused, how do we place it into formula?
Post by: jakesilove on October 20, 2016, 07:11:09 pm
Okay understandable! But what do we do with the 1/3? Thats where I am confused, how do we place it into formula?
We'll end up getting something that looks like
Where we know d, and we know a. Therefore, it should be fairly straight-forward to solve for n!
Post by: yasmineturner on October 20, 2016, 07:43:54 pm
I was just wondering if any could help as I am unsure of how to complete question 10a, iv of the 2010 paper. I have attached a copy of the question and my working for the other parts :) Thankyou
Post by: RuiAce on October 20, 2016, 07:45:54 pm
Hi again,Both ∆ABC and ∆ACD are isosceles.
I was just wondering if any could help as I am unsure of how to complete question 10a, iv of the 2010 paper. I have attached a copy of the question and my working for the other parts :) Thankyou
This implies that we can use base angles on isosceles triangles twice.
< ACD = < ABC (= common < A)
Post by: yasmineturner on October 20, 2016, 07:48:32 pm
Both ∆ABC and ∆ACD are isosceles.How do you do part iv? Thankyou :)
This implies that we can use base angles on isosceles triangles twice.
< ACD = < ABC (= common < A)
Post by: jakesilove on October 20, 2016, 07:52:08 pm
Hi again,
I was just wondering if any could help as I am unsure of how to complete question 10a, iv of the 2010 paper. I have attached a copy of the question and my working for the other parts :) Thankyou
Hey! Firstly, note that y must be positive. We know that cos(theta) can be between -1 and 1, so actually, the maximum value of 2cos(theta) is 2. However, (1-2cos(theta))=1-2=-1 is negative, which is not allowed for the size of a line. If you let cos(theta)=-1, you get 2cos(theta)=-2. So, the maximum value is a(1-2cos(theta))=a(1-2(-1))=3a. y will always have to be less than that!
This was really badly explained. Hope it made sense!
Post by: RuiAce on October 20, 2016, 07:53:11 pm
...Ok don't worry Jake beat me to it
Post by: MysteryMarker on October 20, 2016, 07:57:57 pm
Cheers :P
Post by: nibblez16 on October 20, 2016, 08:05:57 pm
Post by: jakesilove on October 20, 2016, 08:09:16 pm
Help please
Hey! So, we need to set up a geometric series. We can easily see that
So, which of the terms fit into this pattern? It's clearly looking at the 10th and 11th term (as that will get us a power of 10 and 11 for x).
Okay, not helpful
So, the answer is C
Post by: olivercutbill on October 20, 2016, 08:11:02 pm
Hey, I came across this question and realised that I have absolutely no idea what simple interest is. Could someone explain to me the last part of this question, the parts before are fine just the last part. The answers used some PRT/100 = I formula, which i have never used lol.
Cheers :P
what year is this from? :) If i check I can help
Post by: RuiAce on October 20, 2016, 08:14:55 pm
Hey, I came across this question and realised that I have absolutely no idea what simple interest is. Could someone explain to me the last part of this question, the parts before are fine just the last part. The answers used some PRT/100 = I formula, which i have never used lol.
Cheers :P
To illustrate the r and R, for a rate of 6% p.a.
R = 6
r = 0.06
_____________
That being said: To do that question, just deduct off the original principal and plug it into the formula. I = PrT = PRT/100 is assumed knowledge from Year 10.
Post by: caninesandy on October 20, 2016, 08:15:42 pm
I just have a quick question about an easy question but I always get confused...
Solve 3 tan(x) + 1= 0 for 0° ≤ x ≤ 360°
Are the answers: 150 and 210???
Thank you! :D
Post by: nibblez16 on October 20, 2016, 08:15:50 pm
Hey! So, we need to set up a geometric series. We can easily see that
So, which of the terms fit into this pattern? It's clearly looking at the 10th and 11th term (as that will get us a power of 10 and 11 for x).
Okay, not helpful
So, the answer is C
Alright! Thank You!
Btw for limiting some the absolute ratio should be less than 1 right? Because I saw a question where they used ratio more than one ???
Post by: jakesilove on October 20, 2016, 08:17:56 pm
Alright! Thank You!
Btw for limiting some the absolute ratio should be less than 1 right? Because I saw a question where they used ratio more than one ???
Yeah, ratio should be between -1 and 1. Sounds like a weird question!
Post by: RuiAce on October 20, 2016, 08:18:29 pm
Alright! Thank You!
Btw for limiting some the absolute ratio should be less than 1 right? Because I saw a question where they used ratio more than one ???
Post by: jakesilove on October 20, 2016, 08:19:30 pm
This would imply that you can have a complex series with a limiting sum? Interesting....
Post by: RuiAce on October 20, 2016, 08:20:08 pm
Hey guys! :D
I just have a quick question about an easy question but I always get confused...
Solve 3 tan(x) + 1= 0 for 0° ≤ x ≤ 360°
Are the answers: 150 and 210???
Thank you! :D
Post by: RuiAce on October 20, 2016, 08:21:00 pm
This would imply that you can have a complex series with a limiting sum? Interesting....What are you trying to pull here Jake lol I don't get it
Post by: caninesandy on October 20, 2016, 08:31:41 pm
Oh, yeah sorry I meant the root of 3 :)
Post by: RuiAce on October 20, 2016, 08:34:32 pm
Oh, yeah sorry I meant the root of 3 :)
Post by: caninesandy on October 20, 2016, 08:36:39 pm
Post by: nibblez16 on October 20, 2016, 08:41:09 pm
What are you trying to pull here Jake lol I don't get it
I suck at this part of series :P
Help on last q
Post by: olivercutbill on October 20, 2016, 08:43:17 pm
To illustrate the r and R, for a rate of 6% p.a.
R = 6
r = 0.06
_____________
That being said: To do that question, just deduct off the original principal and plug it into the formula. I = PrT = PRT/100 is assumed knowledge from Year 10.
Can you do the solution? I think I was taught a different way...
Post by: RuiAce on October 20, 2016, 08:46:52 pm
I suck at this part of series :PRefer to post #631
Help on last q
Post by: RuiAce on October 20, 2016, 08:47:19 pm
Can you do the solution? I think I was taught a different way...I don't have all the answers to the previous parts
Post by: imtrying on October 20, 2016, 09:03:10 pm
Thank you!
Post by: RuiAce on October 20, 2016, 09:06:02 pm
Hi, just wondering if someone would be able to explain to me how to do part ii) of this question?
Thank you!
Post by: jakesilove on October 20, 2016, 09:06:31 pm
Hi, just wondering if someone would be able to explain to me how to do part ii) of this question?
Thank you!
This is just something that you need to be comfortable using
The maximum value of velocity will clearly occur when cos(theta)=-1, thus
Post by: imtrying on October 20, 2016, 09:17:07 pm
This is just something that you need to be comfortable using
The maximum value of velocity will clearly occur when cos(theta)=-1, thus
Thanks, that helps:) Definitely wasn't thinking that one through properly!
Post by: Rikahs on October 20, 2016, 09:33:55 pm
Thanks, that helps:) Definitely wasn't thinking that one through properly!
Btw, there is an easy method typically known by 3u or 4u students, maximum velocity occurs when the acceleration is zero; if you just differentiat velocity then make it equal to zero to find t then just sub back into velocity you get max velocity. May seem a bit more confusing or may make you go like "oh yeahhhh I see" but it can sometimes make life easier especially if you get ugly equations.
Post by: RuiAce on October 20, 2016, 09:36:10 pm
Btw, there is an easy method typically known by 3u or 4u students, maximum velocity occurs when the acceleration is zero; if you just differentiat velocity then make it equal to zero to find t then just sub back into velocity you get max velocity. May seem a bit more confusing or may make you go like "oh yeahhhh I see" but it can sometimes make life easier especially if you get ugly equations.That works in general. 2U are taught this.
It just so happens that this is a 1 mark question and thus that method is not necessary because of the trick.
Post by: Rikahs on October 20, 2016, 09:40:42 pm
Post by: nibblez16 on October 20, 2016, 09:46:48 pm
Post by: Rikahs on October 20, 2016, 09:48:05 pm
Post by: nibblez16 on October 20, 2016, 09:55:09 pm
Pretty damn sure that is a rumour, they can't just take out the integral section without telling the students.
Yea! I guess people are making up stuff, I mean HSC is already stressful, y give more surprises lol
Post by: RuiAce on October 20, 2016, 09:56:47 pm
Wait I just realized something, did you do ur 2u hsc in year 10?Yes
The whole reference sheet is in the exam right? Coz ive been hearing rumours that theyre taking out integrals part... Just confirming that the rumour is a rumour lol
Pretty damn sure that is a rumour, they can't just take out the integral section without telling the students.
Yea! I guess people are making up stuff, I mean HSC is already stressful, y give more surprises lolOnly 2 or 3 of the standard integrals are not on the reference sheet. Every other standard integral has been transferred
The 2 (or 3) that have been taken out are also useless for 2U AND 3U.
Post by: nibblez16 on October 20, 2016, 10:12:00 pm
Post by: RuiAce on October 20, 2016, 10:21:28 pm
Heyy! How do we know the procedure to solve this?
Post by: nibblez16 on October 20, 2016, 10:24:12 pm
The answer is different, its 91/216
Post by: emilyf on October 20, 2016, 10:27:26 pm
The answers can be found here - http://www.pasthsc.com.au//HSC_Mathematics_files/Maths2U07WorkedSolutions.pdf - on page eight however I don't understand part (i).
For the first part, I got tan(x) =√3, yet I then used this to say that x = π/3, and that the intercepts occur at π/3 and 2π/3. I don't understand why the 2π/3 isn't right and why it's 4π/3 instead??
Post by: RuiAce on October 20, 2016, 10:31:07 pm
The answer is different, its 91/216
Post by: :3 on October 20, 2016, 10:33:24 pm
Sorry if this has been answered previously but could someone help me out with this question please?
The answers can be found here - http://www.pasthsc.com.au//HSC_Mathematics_files/Maths2U07WorkedSolutions.pdf - on page eight however I don't understand part (i).
For the first part, I got tan(x) =√3, yet I then used this to say that x = π/3, and that the intercepts occur at π/3 and 2π/3. I don't understand why the 2π/3 isn't right and why it's 4π/3 instead??
tan(x) is positive in the first and 3rd quadrant.
Therefore, the respective values for x are 60 (π/3) and 180 +60 (240, 4π/3; not 2π/3).
Post by: RuiAce on October 20, 2016, 10:33:31 pm
Sorry if this has been answered previously but could someone help me out with this question please?You're looking at the wrong quadrants.
The answers can be found here - http://www.pasthsc.com.au//HSC_Mathematics_files/Maths2U07WorkedSolutions.pdf - on page eight however I don't understand part (i).
For the first part, I got tan(x) =√3, yet I then used this to say that x = π/3, and that the intercepts occur at π/3 and 2π/3. I don't understand why the 2π/3 isn't right and why it's 4π/3 instead??
Tan is positive in the first and THIRD quadrants, not first and second. (Rule of ASTC.)
So the valuse are π/3, and π + π/3, which is 4π/3
Not π/3 with π - π/3 , which is 2π/3
Edit: Done above
Post by: nibblez16 on October 20, 2016, 10:36:08 pm
Okay! But y do we cube it?
Post by: RuiAce on October 20, 2016, 10:37:21 pm
Okay! But y do we cube it?They give you the probabilities for 1 week.
You want the probability for a 3 week period. Not just 1.
So you want the same event to happen 3 times. It's like saying I lost, and then I lost again, and then yet I lost one more time.
So if the probability I lose is, say, 1/2
The probability I lose 3 times is 1/2 times 1/2 times 1/2, which is 1/8
Successive events, if you will
Post by: nibblez16 on October 20, 2016, 11:18:24 pm
Post by: RuiAce on October 20, 2016, 11:25:18 pm
Does anyone have prelim formula sheet?What's a prelim formula sheet? You're going to be given that same reference sheet as you have been tomorrow
Post by: :3 on October 20, 2016, 11:47:31 pm
(https://i.gyazo.com/22dd1130d3f0673db48ce6351b24fa1d.png)
Also, any tips on drawing a displacement graph from a velocity graph or acceleration and vice versa? (without knowing the function itself) :P
Post by: RuiAce on October 20, 2016, 11:54:39 pm
For iii), I'm unsure as to why AP = AC (according to the answer, its because of part II).AP is not equal to AC. AP/AC = 1/2 according to (ii)
(https://i.gyazo.com/22dd1130d3f0673db48ce6351b24fa1d.png)
Also, any tips on drawing a displacement graph from a velocity graph or acceleration and vice versa? (without knowing the function itself) :P
Post by: :3 on October 21, 2016, 12:01:55 am
AP is not equal to AC. AP/AC = 1/2 according to (ii)
Here are the answers from BOSTE 2009 HSC exam.
(https://i.gyazo.com/f78e5a0f152aba594d4c4502e6e9cf27.png)
Post by: RuiAce on October 21, 2016, 12:02:36 am
Here are the answers from BOSTE 2009 HSC exam.That's CP. Not AC.
(https://i.gyazo.com/f78e5a0f152aba594d4c4502e6e9cf27.png)
Post by: Rikahs on October 21, 2016, 12:04:16 am
For iii), I'm unsure as to why AP = AC (according to the answer, its because of part II).
(https://i.gyazo.com/22dd1130d3f0673db48ce6351b24fa1d.png)
Also, any tips on drawing a displacement graph from a velocity graph or acceleration and vice versa? (without knowing the function itself) :P
First of all AC= 2AP
Then for iii) I would use congruent triangles, my proof:
In triangle AMP and traingle MPC
MP is common
AC = 2AP ( from ii))
Since AC = AP + PC
therefore AP = PC
Angle MPA = Angle MPC = 90 (MP bisects AC)
Therefore Triangle AMP is congruent to Triangle MPC (SAS)
Therefore Angle MAP = Angle MCP ( corresponding angles in congruent triangles)
Therefore Triangle AMC is Isos ( base angles are equal)
Post by: :3 on October 21, 2016, 12:11:40 am
My mind totally blanked out (probably because its geometry + late at night).
Post by: RuiAce on October 21, 2016, 12:17:35 am
Post by: Hua Fei on October 21, 2016, 12:27:10 am
I don't understand how to do these questions:
1) the answer is 4.5. I don't even know how to approach this...
2) the answer is 2. Wouldn't it be 3? Sin gives us one solution and tan gives us the 2nd?
Thank you so much in advance!
Post by: RuiAce on October 21, 2016, 12:39:30 am
Hi there!Addressed in post #464
I don't understand how to do these questions:
1) the answer is 4.5. I don't even know how to approach this...
2) the answer is 2. Wouldn't it be 3? Sin gives us one solution and tan gives us the 2nd?
Thank you so much in advance!
Addressed in post #620
Post by: Hua Fei on October 21, 2016, 12:44:40 am
Addressed in post #464
Addressed in post #620
Thank you! You've really helped me out :)
Post by: nibblez16 on October 21, 2016, 01:04:51 am
Post by: RuiAce on October 21, 2016, 01:08:53 am
Help please last part
Post by: nibblez16 on October 21, 2016, 06:37:03 am
For a question like (iii) we have to differentiate, so can be times the power 'k' with the whole number? In this case k= 0.0347 (4dp) but the whole number is 0.03465735903
Can we times it with the whole number? because the answer multplies it with only 4dp....
Post by: jamonwindeyer on October 21, 2016, 08:03:19 am
Good Morning :)
For a question like (iii) we have to differentiate, so can be times the power 'k' with the whole number? In this case k= 0.0347 (4dp) but the whole number is 0.03465735903
Can we times it with the whole number? because the answer multplies it with only 4dp....
Morning! You would be paid for either, but as a rule of thumb, if you find a rounded value in an earlier part of the question, carry it through the rest of the questions! It is cleaner and still yields the required level of accuracy ;D
Post by: nibblez16 on October 21, 2016, 08:34:58 am
Morning! You would be paid for either, but as a rule of thumb, if you find a rounded value in an earlier part of the question, carry it through the rest of the questions! It is cleaner and still yields the required level of accuracy ;D
Okay Thanks!
Post by: nibblez16 on October 21, 2016, 08:36:35 am
Post by: zemilyx on October 21, 2016, 09:07:26 am
with this time payment question, I don't understand why the expression for A1 ends up as Px1.04^7+5000 instead of Px1.04^8 +5000. Is there a general rule for working out the time period in time repayment questions where they don't just state over a period of n years but rather say between this year and that year etc because to me it seems the answer varies between different questions and I don't know why.
Thanks :))
Post by: BPunjabi on October 21, 2016, 09:09:43 am
hi
with this time payment question, I don't understand why the expression for A1 ends up as Px1.04^7+5000 instead of Px1.04^8 +5000. Is there a general rule for working out the time period in time repayment questions where they don't just state over a period of n years but rather say between this year and that year etc because to me it seems the answer varies between different questions and I don't know why.
Thanks :))
Its too late now, dont worry just wing it!
Post by: BPunjabi on October 21, 2016, 09:13:21 am
How do we know when to use a discriminant and which one to use (>, <, =)
Its too late now just wing it ;D.
But here:
1. if triangle is <0, roots are real, rational and equal
2. If triangle is =0 roots are unreal and unrational
3. If triangle is >0 roots are unreal, irrational and unequal
Post by: WLalex on October 21, 2016, 09:13:32 am
hi
with this time payment question, I don't understand why the expression for A1 ends up as Px1.04^7+5000 instead of Px1.04^8 +5000. Is there a general rule for working out the time period in time repayment questions where they don't just state over a period of n years but rather say between this year and that year etc because to me it seems the answer varies between different questions and I don't know why.
Thanks :))
Do not count 2007 for interest as the 5000 is deposited at the beginning therefore the P is only gaining interest until 2006 e.g. 7 times not 8 (8 would be at the end of 2007)
Post by: WLalex on October 21, 2016, 09:15:06 am
Its too late now just wing it ;D.
But here:
1. if triangle is <0, roots are real, rational and equal
2. If triangle is =0 roots are unreal and unrational
3. If triangle is >0 roots are unreal, irrational and unequal
If the discriminate is less than 0 there is no real roots...
= 0 is 1 real root
> 0 is 2 real, unequal roots
> and equal to means real roots
Post by: jamonwindeyer on October 21, 2016, 09:25:51 am
Its too late now, dont worry just wing it!
5 hours is heaps of time to get some things clarified mate :)
If the discriminate is less than 0 there is no real roots...
= 0 is 1 real root
> 0 is 2 real, unequal roots
> and equal to means real roots
This is the correct breakdown. Thanks WLalex! ;D and for your response above, awesome stuff ;D
Post by: RuiAce on October 21, 2016, 09:26:36 am
How do we know when to use a discriminant and which one to use (>, <, =)Note that f'(x) ≥ 0 means monotonic increasing
f'(x) ≤ 0 means monotonic decreasing
f'(x) = 0 for stationary points
f'(x) > 0 means strictly increasing
f'(x) < 0 means strictly decreasing
When they don't say monotonic, assume strictly
Post by: WLalex on October 21, 2016, 09:28:04 am
Note that f'(x) ≥ 0 means monotonic increasing
f'(x) ≤ 0 means monotonic decreasing
f'(x) = 0 for stationary points
f'(x) > 0 means strictly increasing
f'(x) < 0 means strictly decreasing
When they don't say monotonic, assume strictly
so on top of this for clarity you will have to differentiate f(x) and make the discriminate less than 0 (as it will have no roots)
Post by: RuiAce on October 21, 2016, 09:29:18 am
Hi, my answers say that d.) is correct but how did they get to this point?
Post by: WLalex on October 21, 2016, 09:30:56 am
Hi, my answers say that d.) is correct but how did they get to this point?
I would do the calculation the normal way (just add them all together) and then see which answer matches your answers via calculation
Post by: RuiAce on October 21, 2016, 09:31:24 am
so on top of this for clarity you will have to differentiate f(x) and make the discriminate less than 0 (as it will have no roots)In theory, if we were to fully complete the question
Not only do we want ∆ < 0, we want the leading coefficient to be positive. We basically want f'(x) to be positive definite.
Because if the leading coefficient were negative, and ∆ < 0, we would be strictly DECREASING.
However, it's easy to check that 1 > 0 and you are certainly right in using the discriminant here. Because f'(x) is indeed quadratic.
Post by: RuiAce on October 21, 2016, 09:32:03 am
I would do the calculation the normal way (just add them all together) and then see which answer matches your answers via calculationTakes far too long. That's what you do after you've finished the paper and it's time to check your answers, not during.
There are 13 terms to add up in the sum.
Post by: jamonwindeyer on October 21, 2016, 09:36:04 am
Takes far too long. That's what you do after you've finished the paper and it's time to check your answers, not during.
There are 13 terms to add up in the sum.
Eh, when in doubt go back to basics; the techniques are obviously more effective. But if you have a choice between spending 1 minute getting an answer you are 50% confident with, and 3 minutes getting an answer you KNOW is correct, then why not (assuming you know you've got the time). Plus, doing the calculator work for this with 13 terms isn't actually that bad :P
Post by: RuiAce on October 21, 2016, 09:39:26 am
Eh, when in doubt go back to basics; the techniques are obviously more effective. But if you have a choice between spending 1 minute getting an answer you are 50% confident with, and 3 minutes getting an answer you KNOW is correct, then why not (assuming you know you've got the time). Plus, doing the calculator work for this with 13 terms isn't actually that bad :PEven at the fastest typing speed on the calculator that still takes me a full 45 seconds to get through all the options whereas I could've evaluated it in 15-25 :P
They say to check your answers for reasons - get an answer on the paper first, and then calm the paranoia down when you find time to.
Of course, if you have a stress attack though then yeah that's when you should just go back to basics or you'll be overly paranoid
Post by: WLalex on October 21, 2016, 09:49:58 am
Even at the fastest typing speed on the calculator that still takes me a full 45 seconds to get through all the options whereas I could've evaluated it in 15-25 :P
They say to check your answers for reasons - get an answer on the paper first, and then calm the paranoia down when you find time to.
Of course, if you have a stress attack though then yeah that's when you should just go back to basics or you'll be overly paranoid
Yep valid point Rui but since it was a simple calculation thats just where my mind went :)
Post by: RuiAce on October 21, 2016, 09:51:24 am
Yep valid point Rui but since it was a simple calculation thats just where my mind went :)At least it's 100% not penalisable
Post by: lha on October 21, 2016, 09:58:31 am
Post by: kavinila on October 21, 2016, 09:59:58 am
Post by: RuiAce on October 21, 2016, 10:05:05 am
hey guys! i was just wondering if someone could please quickly run through locus? i don't think i understand it and i just noticed it in the syllabus 😳 please and thankyouu!!Find the section on locus
Post by: RuiAce on October 21, 2016, 10:09:04 am
You know those velocity graphs or acceleration graphs and it says describe the motion of the particle at t=whatever? Can someone give me a run down on how to do that please!Well if you need to describe the motion based off what you have.
Suppose you have the graph of the displacement.
Velocity is the first derivative of the displacement
Therefore if the graph is increasing, velocity is positive
If graph is decreasing, velocity is negative
If graph is at a stationary point, particle is at rest
Acceleration is the second derivative of the displacement
Therefore if the graph is concave up, acceleration is positive
If the graph is concave down, acceleration is positive
If we are at a point of inflexion, acceleration is constant (possibly 0 but we don't know).
And say we were given the graph of the velocity.
Velocity is the first derivative
Hence v > 0 means particle is travelling away from the origin (+'ve direction, to the right, however you memorised it)
v < 0 means particle is travelling in the other direction from the origin (-'ve direction, to the left etc.)
And those are just some examples. If you know what the derivatives actually MEAN then this is more intuitive than trying to rote learn a few dot points.
Post by: onepunchboy on October 21, 2016, 10:10:38 am
Post by: RuiAce on October 21, 2016, 10:12:16 am
Hey anyone know how to find the asymptotes for tan graphs? For normal tanx graph asymptote is at pi/2 and 3pi/2 but what if it was tan(1/2 x) or tan (2x) how to find new asymptote.. thanks
Post by: nibblez16 on October 21, 2016, 10:13:23 am
Post by: jakesilove on October 21, 2016, 10:16:51 am
How do we know how to solve the last 2 questions
Hey! For the second part, you need to think about the cases. Pat can win on the first turn (probability same as i) ), or Pat can win on the second turn. If he wins on the second turn, it will be (1-P(win on first turn))*P(win). Then, we add up these two cases!
You need to create some sort of 'series' which you can add up in a sum-to-infinity type equation for the last part. Can you see how?
Post by: BPunjabi on October 21, 2016, 10:20:31 am
Post by: Blissfulmelodii on October 21, 2016, 10:21:41 am
Guys have confidence, we will smash it!
Really good attitude to have :) :)
Post by: Rikahs on October 21, 2016, 10:26:01 am
I saw a post above that "monotonic increasing" is when f'(x) greater than or equal to zero.
Can someone please 100% confirm this, i always thought when it said "monotonic increasing" it meant f'(x) was just larger than 0 and not equal to it.
Thanks!!!
Post by: jamonwindeyer on October 21, 2016, 10:27:12 am
Hi,
I saw a post above that "monotonic increasing" is when f'(x) greater than or equal to zero.
Can someone please 100% confirm this, i always thought when it said "monotonic increasing" it meant f'(x) was just larger than 0 and not equal to it.
Thanks!!!
Confirming, monotonic means \(f'(x)\ge0\) ;D
Post by: Rikahs on October 21, 2016, 10:27:27 am
What kind of question would want you to make the discriminant greater and equal to zero?
Post by: RuiAce on October 21, 2016, 10:27:35 am
Hi,Ditto to Jamon.
I saw a post above that "monotonic increasing" is when f'(x) greater than or equal to zero.
Can someone please 100% confirm this, i always thought when it said "monotonic increasing" it meant f'(x) was just larger than 0 and not equal to it.
Thanks!!!
Monotonic means it can equal.
STRICTLY means it can't equal.
Post by: RuiAce on October 21, 2016, 10:28:19 am
Hi again,The discriminant is positive OR zero when ALL you care about is real roots.
What kind of question would want you to make the discriminant greater and equal to zero?
When it's strictly positive (greater than zero), i.e. >, it means your two real roots must be DISTINCT.
Post by: nibblez16 on October 21, 2016, 10:28:37 am
Hey! For the second part, you need to think about the cases. Pat can win on the first turn (probability same as i) ), or Pat can win on the second turn. If he wins on the second turn, it will be (1-P(win on first turn))*P(win). Then, we add up these two cases!
You need to create some sort of 'series' which you can add up in a sum-to-infinity type equation for the last part. Can you see how?
Its still confusing...
Post by: g98 on October 21, 2016, 10:33:53 am
1. If the question doesn't specify do we leave answer in exact form or decimal and would we get marked down for either?
2. If we answer a question and you could take out a common factor - should you do that or not - some past paper answers do it and some don't and would we get marked down either way?
3. What are acceptable abbreviations for proving geometrical questions?
4. do we leave our answer in improper fraction or mixed numeral - if not specified in question?
5. If not specified in question how many decimals should you round to?
Thank You!
Post by: Belindka on October 21, 2016, 10:35:22 am
Just wondering, would changing our calculators to radians by default change our calculations for other answers that don't need radians? I always have mine in degrees and change it to radians whenever i need it ???
Thanks!
Post by: jamonwindeyer on October 21, 2016, 10:35:47 am
Its still confusing...
All good! So:
For Pat to win on the second throw, he has to not win on his first throw, and then win on his second:
So the answer to B is:
If we went to three throws, it would be similar; he needs to lose TWO throws and then win on his third:
See the pattern? We have a geometric series forming, \(a=\frac{1}{36}\) and \(r=\frac{35}{36}\) ;D
Now this value of r lets us take an infinite series and get a value; so use that formula for infinite series from your reference sheet with those values, and that is how you answer Part C ;D
Post by: Blissfulmelodii on October 21, 2016, 10:38:43 am
Hi, I have a couple of questions:
1. If the question doesn't specify do we leave answer in exact form or decimal and would we get marked down for either?
2. If we answer a question and you could take out a common factor - should you do that or not - some past paper answers do it and some don't and would we get marked down either way?
3. What are acceptable abbreviations for proving geometrical questions?
4. do we leave our answer in improper fraction or mixed numeral - if not specified in question?
5. If not specified in question how many decimals should you round to?
Thank You!
From what I have been told by my teachers if the question doesn't specify you should always leave it in exact form but they probably won't mark you down for putting it into decimal form
Again with the second question, if its not specifies then they won't take marks of for how you have left it but they usually always want to simplified answer which mean taking out common factors and factorizing.
For the third question I've heard mixed fractions are better but again if not specified either works
And for your last question I would personally round to about 3 or 4 to keep accuracy.
Hope this helps
Post by: jamonwindeyer on October 21, 2016, 10:38:57 am
Hi, I have a couple of questions:
1. If the question doesn't specify do we leave answer in exact form or decimal and would we get marked down for either?
2. If we answer a question and you could take out a common factor - should you do that or not - some past paper answers do it and some don't and would we get marked down either way?
3. What are acceptable abbreviations for proving geometrical questions?
4. do we leave our answer in improper fraction or mixed numeral - if not specified in question?
5. If not specified in question how many decimals should you round to?
Thank You!
1. If it doesn't specify and you have the capability, exact form.
2. Factorising your final answer is something you can do, but it is unnecessary; you won't get marked down as long as you factorise don't divide
3. \(\Delta\) for a triangle, \\ for parallel lines, and the congruency and similarity symbols to name a few. Try not to abbreviate words like "isosceles" unless you are short on time (you will probably be fine even if you do)
4. Improper Fraction is nicer imo, but BOSTES does use mixed numeral at times. Either or :)
5. If not specified, pick a sensible number. General rule is the most amount of decimal places used in the question (but then I usually just add one to that just to be safe) :)
Post by: RuiAce on October 21, 2016, 10:40:32 am
Post by: jamonwindeyer on October 21, 2016, 10:40:47 am
Hey!
Just wondering, would changing our calculators to radians by default change our calculations for other answers that don't need radians? I always have mine in degrees and change it to radians whenever i need it ???
Thanks!
Yes, you need to be careful about this because it does affect your answers for, say, geometry questions that use degrees.
Leave your calculator in radians, then swap to degrees as soon as you see the degree symbol, then swap back as soon as that question is done. That's what I recommend since radians is the default unit of angular measure ;D
Post by: jamonwindeyer on October 21, 2016, 10:41:43 am
Some chaos occurs with mixed numerals. Don't smack me Jamon but I need to emphasise your point on this one :P preferably improper fractions please.
Happy to be emphasised; it does create room for more error! ;D (although BOSTES uses them in sample solutions, I never did, and was fine. Don't create more work for yourself) :)
Post by: rinagee12 on October 21, 2016, 10:58:02 am
Post by: RuiAce on October 21, 2016, 11:00:46 am
Good luck to everyone who is sitting the exam today ;D (although I'm probably the one that's going to need it ;) )Have faith, don't imply that about yourself
Post by: isaacdelatorre on October 21, 2016, 11:20:51 am
Just a question with rates of change.
If t equals something like 152.65 months, what is your final answer if it says when does P = x, occur. Do we leave it as a decimal? Or round to a month?
Post by: MysteryMarker on October 21, 2016, 11:21:17 am
Cheers, ALSO GOOD LUCK PEEPS and a MASSIVE shoutout to all the moderators and members that make revising for 2U that much easier. :P
Post by: RuiAce on October 21, 2016, 11:22:51 am
Hey guys,I recommend rounding so that your units are consistent, but I think unless they explicitly say how many years and months it isn't that much of a concern to use a decimal answer.
Just a question with rates of change.
If t equals something like 152.65 months, what is your final answer if it says when does P = x, occur. Do we leave it as a decimal? Or round to a month?
If they say explicitly how many years and months then you have to answer the question.
I'll let others comment on this one as well
Post by: g98 on October 21, 2016, 11:24:25 am
Post by: RuiAce on October 21, 2016, 11:24:55 am
Hey guys, this is more just a rule that i'm asking, but I remember that when there are parallel lines and two transversal intersect them, the ration between the two transversals are equal or something? Could someone just rephrase that properly or enlighten me on whether that is the correct rule?You can call it "Ratio of intercepts on transversals of parallel lines" when you write out the ratio relevant. Or just "intercepts on transversals of parallel lines"
Cheers, ALSO GOOD LUCK PEEPS and a MASSIVE shoutout to all the moderators and members that make revising for 2U that much easier. :P
TBH - I've never needed that theorem yet even though it's in the course
Glad we've helped ya
Post by: jakesilove on October 21, 2016, 11:25:35 am
is using the @ symbol ok?
In what sense? But probably not a good idea
Post by: RuiAce on October 21, 2016, 11:26:09 am
is using the @ symbol ok?That just stands for "at". It's not too much of a hassle to write two letters lol
When do you need it in 2/4U though? I don't really recall any scenario I came across it
Post by: g98 on October 21, 2016, 11:31:15 am
In what sense? But probably not a good idea
I tend to use when i sub points into a line...
find the tangent at (1,0) and the gradient equation was m=x-2
then I would write:
@ x=1 m=-1
...I also use it in stationary point questions..I guess i'll try not to in the exam though ;)
Post by: RuiAce on October 21, 2016, 11:33:08 am
Of course if you really forget to then don't stress over it when you come out of the exam either.
Post by: dtinaa on October 21, 2016, 11:40:02 am
Post by: lha on October 21, 2016, 11:43:02 am
Well if you need to describe the motion based off what you have.
Suppose you have the graph of the displacement.
Velocity is the first derivative of the displacement
Therefore if the graph is increasing, velocity is positive
If graph is decreasing, velocity is negative
If graph is at a stationary point, particle is at rest
Acceleration is the second derivative of the displacement
Therefore if the graph is concave up, acceleration is positive
If the graph is concave down, acceleration is positive
If we are at a point of inflexion, acceleration is constant (possibly 0 but we don't know).
And say we were given the graph of the velocity.
Velocity is the first derivative
Hence v > 0 means particle is travelling away from the origin (+'ve direction, to the right, however you memorised it)
v < 0 means particle is travelling in the other direction from the origin (-'ve direction, to the left etc.)
And those are just some examples. If you know what the derivatives actually MEAN then this is more intuitive than trying to rote learn a few dot points.
If velocity is positive, does that mean its going in thd positive direction?
Also if acceleration is positive, does that mean it is speeding up and if accelerstion is negative, its slowing down?
also, how do i know it has changed direction if it gives me a displacement, velocity or acceleration graph
Post by: imtrying on October 21, 2016, 11:43:15 am
Post by: jakesilove on October 21, 2016, 11:51:11 am
are both these formulas right for surface area of a cone?
Yep they are both correct!
Post by: BPunjabi on October 21, 2016, 11:51:38 am
If velocity is positive, does that mean its going in thd positive direction?
Also if acceleration is positive, does that mean it is speeding up and if accelerstion is negative, its slowing down?
also, how do i know it has changed direction if it gives me a displacement, velocity or acceleration graph
Imagine a ball thrown in the air, the exponential either moves to the right or the left. The velocity goes up but then comes to a stop. Then acceleration is applied. Apples.
Gyet Yit?
Post by: jakesilove on October 21, 2016, 11:52:24 am
If velocity is positive, does that mean its going in thd positive direction?
Also if acceleration is positive, does that mean it is speeding up and if accelerstion is negative, its slowing down?
also, how do i know it has changed direction if it gives me a displacement, velocity or acceleration graph
Yep, that's pretty much right. However, if the velocity is negative, and the acceleration is negative, the particle is actually speeding up in the negative direction!
It changes direction if the displacement starts to decrease (ie. go back towards the origin) or the velocity changes sign (ie. goes from positive to negative).
Post by: BPunjabi on October 21, 2016, 11:52:59 am
Yep they are both correct!
are we expected to know that?
Post by: BPunjabi on October 21, 2016, 11:54:26 am
Post by: jakesilove on October 21, 2016, 11:54:56 am
Please help!!
Hey! I think it's easier here to take cases where he DOESN'T win in the first three throws. This would only occur if the coins read HTH, or THT. The probability of HTH is (0.5)(0.5)(0.5), as is the probability of THT. Therefore, 2(0.5)(0.5)(0.5)=0.25. The probability of WINNING is 1-P(not winning), therefore the answer to i) is 1-0.25=0.75.
Fuck Joe. Why make such a complicated game? Now, we don't want him to win on the second go (even), or on any even goes. The probability of NOT winning on an even go must be something like
P(Not winning)=HTHTHTHT...+THTHTHTHTHTHT=2(0.5)^n for n throws, where n is even. Him WINNING on an odd turn is the same as NOT WINNING on an even term (sort of? In terms of infinity, where winning is assured? I'm not actually 100% about that, but let's go with it).
Therefore, we need to sum up 2(0.5)+2(0.5)^2+...+2(0.5)^n for n=infinity. Use your sum to infinity formula to get a solution, then your answer should be 1-ans!
Let me know if this works out,
Jake
Post by: jakesilove on October 21, 2016, 11:55:43 am
Good Luck Guys!! We will need it. See you in an hour.
Know whatever formulas you think you'll need. You won't need to know multiple variations of volume/surface area for the same shape
Post by: lha on October 21, 2016, 12:01:19 pm
Yep, that's pretty much right. However, if the velocity is negative, and the acceleration is negative, the particle is actually speeding up in the negative direction!
It changes direction if the displacement starts to decrease (ie. go back towards the origin) or the velocity changes sign (ie. goes from positive to negative).
So if velocity is negative but acceleration is positive, the particle is going in the negative direction but slowing down?
Post by: Hua Fei on October 21, 2016, 12:10:06 pm
So if velocity is negative but acceleration is positive, the particle is going in the negative direction but slowing down?
Yes. If velocity and acceleration are opposite in direction, the particle would be decelerating in the direction of the velocity (although, I'd wait for someone else to second that)
Also, I don't get how to do this Q
Post by: ml125 on October 21, 2016, 12:35:52 pm
Yes. If velocity and acceleration are opposite in direction, the particle would be decelerating in the direction of the velocity (although, I'd wait for someone else to second that)The displacement is equivalent to the area under the velocity curve: (8×4)/2=16. In the question, it states that at t=0, x=2. Therefore, the maximum displacement at t=4 would be x=2+16=18 :)
Also, I don't get how to do this Q
Post by: yogiburay on October 21, 2016, 02:43:56 pm
* Find the stationary points on the curve y=(3x-1)(x-2)^4
*Differentiate y=x√x+1. Hence find the stationary point on the curve, giving the exact value.
*The curve f(x)=ax^4-2x^3+7x^2-x+5 has a stationary point at x=1. Find the value of a.
* Show that f(x) =√x has no stationary points
*Show that f(x) = 1/x^3 has no stationary points.
I hope you can explain these questions to me. I really want to improve my maths. Thankyou. :)
Post by: jamonwindeyer on October 21, 2016, 03:15:04 pm
Hi. Sorry to ask silly questions, but I'm having trouble doing these questions. We just started these topic and I'm so confused. I just can't keep up with my teacher, he's just too fast. I hope you can help me. :'(
...
I hope you can explain these questions to me. I really want to improve my maths. Thankyou. :)
Hey Yogi! Welcome to the forums! Definitely don't be sorry for wanting to improve, good on you for working hard! ;D
Before we give you a hand with these, I'll get you to read this guide which covers some of the theory that you are confused with. Have a read, and see if it helps you understand the questions you've posted (it's brief, but it might help a tad). Otherwise, pick the question there that is confusing you the most and post it back, and we'll give you a hand! Best to start with the most confusing and then see if it helps you answer the others ;D
Post by: yogiburay on October 21, 2016, 03:20:58 pm
Hey Yogi! Welcome to the forums! Definitely don't be sorry for wanting to improve, good on you for working hard! ;D
Before we give you a hand with these, I'll get you to read this guide which covers some of the theory that you are confused with. Have a read, and see if it helps you understand the questions you've posted (it's brief, but it might help a tad). Otherwise, pick the question there that is confusing you the most and post it back, and we'll give you a hand! Best to start with the most confusing and then see if it helps you answer the others ;D
Thankyou :)
Post by: rinagee12 on October 21, 2016, 05:42:04 pm
Post by: samuels1999 on October 21, 2016, 08:33:16 pm
I am a 2017 Year 12 Student. This year I did accelerated 2U Mathematics and just 3 and a bit hours ago I did my HSC mathematics. It was ok- but I am certain I was way off 100%, which would probably mean I'll redo it next year. With that in mind, along with my experience of doing past HSC papers, I realised that It was pretty easy for me to get 70% to 80% in a Mathematics Paper since so much of that style of question I had seen before. I found it much trickier getting that last 15-20% since it involves far more application of skills and technique than the earlier questions.
I know this is a problem partially due to the fact I kinda rushed the course in one year, but with respect to the next year, I just wanted to know if you have any advice on how I could fine tune maths skills to the level I could actually do those harder questions...or even just have the confident mindset. Whether it's more past papers, or maybe going back to basic text book work, I just want to know how I should approach the next year keeping in mind the fact that I have bascially completed "course work" already.
Thanks,
Samuel
Post by: anotherworld2b on October 21, 2016, 08:51:54 pm
Post by: Blissfulmelodii on October 21, 2016, 08:56:21 pm
Hi could i get help with these questions?
You have been given the displacement equation and are being asked to find the velocity which means you have to differentiate your equations before you substitute
Post by: anotherworld2b on October 21, 2016, 11:06:55 pm
I was also wondering if i could get help with q15 and 16c-e
You have been given the displacement equation and are being asked to find the velocity which means you have to differentiate your equations before you substitute
Post by: Blissfulmelodii on October 22, 2016, 09:15:35 am
Thank you for the help
I was also wondering if i could get help with q15 and 16c-e
No worries. For Q15 you would differentiate the displacement equation to get the velocity equation and then let it equal to 0 to find t
Once you know what time the velocity is 0 you can go back and substitute the t into the displacement equation to find out the displacement at that time.
I hope that makes sense. Let me know if you need me to write up the solution
I'll look at 16 now
Post by: jamonwindeyer on October 22, 2016, 10:39:31 am
Hi Jake or anyone who wants to answer,
I am a 2017 Year 12 Student. This year I did accelerated 2U Mathematics and just 3 and a bit hours ago I did my HSC mathematics. It was ok- but I am certain I was way off 100%, which would probably mean I'll redo it next year. With that in mind, along with my experience of doing past HSC papers, I realised that It was pretty easy for me to get 70% to 80% in a Mathematics Paper since so much of that style of question I had seen before. I found it much trickier getting that last 15-20% since it involves far more application of skills and technique than the earlier questions.
I know this is a problem partially due to the fact I kinda rushed the course in one year, but with respect to the next year, I just wanted to know if you have any advice on how I could fine tune maths skills to the level I could actually do those harder questions...or even just have the confident mindset. Whether it's more past papers, or maybe going back to basic text book work, I just want to know how I should approach the next year keeping in mind the fact that I have bascially completed "course work" already.
Thanks,
Samuel
Hey Samuels!! It definitely sounds like the troubles you are having are completely down to just doing the course in one year. You don't quite have the experience to be able to tackle the weird stuff (and that is absolutely not an insult to your abilities or anything of the sort, it's purely just, you haven't been doing the course for two years). So for starters, you have nothing to worry about at all, and you might be happy with how you went this year! You never know! ;D
Practice papers of all types are the best way to improve here; and doing higher levels of Math (3U and 4U presumably?) will help there too. They will expose you to problem types that can be emulated in 2U exams to challenge the high achievers; the questions you are struggling with.
Practice makes perfect here Sam, but it sounds like you know that already. Keep doing past papers, and watch yourself improve! ;D
Post by: asd987 on October 22, 2016, 11:03:07 pm
Can someone please help me with Q16c ii from the 2015 paper. Thankyou
Post by: RuiAce on October 22, 2016, 11:05:38 pm
Hello,Addressed in post #505
Can someone please help me with Q16c ii from the 2015 paper. Thankyou
Post by: katnisschung on October 23, 2016, 02:08:20 pm
q6. Novy leaves her base camp, O and walks in the direction 150°T until she reaches point A on the bank of the river.
Unable to cross the river, she follows it on a bearing of 030°T and stops at point C. The bearing and distance of C from O are 120°T
and 4km respectively
ii)show that angle OAC=60°
I proved it by drawing the diagram which is the first part of the diagram
figured out a few angles
proof
(co-interior angle sum is 180 on parallel lines)
Q5. A person on top of a building 49m in height wishes to find the height of an adjacent building.
If the angle of depression of the adjacent building is 47° 21' from where the person is standing and the
buildings are a distance of 12 m apart.
ii) Calculate the height of the adjacent building (2 dp)
my answer=35.97m
simplify
i) sin^2x - sin^2x cos^x
Thank you!! :)
Post by: RuiAce on October 23, 2016, 02:26:23 pm
there are no answers >:( >:( for the practice test i didNot sure what cos^x is meant to mean there. If it was a typo for cos(x) then there are many ways of simplifying that and we cannot say for certain which is "the correct answer". Presumably sin^2x-sin^2x cos^2x which might make some sense.
q6. Novy leaves her base camp, O and walks in the direction 150°T until she reaches point A on the bank of the river.
Unable to cross the river, she follows it on a bearing of 030°T and stops at point C. The bearing and distance of C from O are 120°T
and 4km respectively
ii)show that angle OAC=60°
I proved it by drawing the diagram which is the first part of the diagram
figured out a few angles
proof
(co-interior angle sum is 180 on parallel lines)
Q5. A person on top of a building 49m in height wishes to find the height of an adjacent building.
If the angle of depression of the adjacent building is 47° 21' from where the person is standing and the
buildings are a distance of 12 m apart.
ii) Calculate the height of the adjacent building (2 dp)
my answer=35.97m
simplify
i) sin^2x - sin^2x cos^x
Thank you!! :)
Please put up diagrams that you used in your attempt to complete the other questions (and preferably also your working out for checking). ATARnotes has an app to upload photos or you can attach screenshots if the test was a pdf
_____________________
Post by: katnisschung on October 23, 2016, 03:40:02 pm
I've attached it here
Post by: RuiAce on October 23, 2016, 04:05:14 pm
thanks ruiAce(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps0bvz0oxt.png)
I've attached it here
And then part iii) can be done by using the angle sum of a triangle, and the sine rule.
Post by: asd987 on October 24, 2016, 06:40:12 pm
Post by: jakesilove on October 24, 2016, 07:11:10 pm
How would I integrate x^-3 / 3
Hey! To integrate something like that, you just use the normal anti-derivative formula
Apply this directly to the equation you've given us! You can bring the 1/3 outside the integral :)
Post by: asd987 on November 01, 2016, 09:57:36 pm
the area bounded by the curve y=x^2 and y=x+2 is rotated about the x axis. find the exact volume of the solid formed
Post by: RuiAce on November 01, 2016, 10:05:12 pm
hi i've got a question.
the area bounded by the curve y=x^2 and y=x+2 is rotated about the x axis. find the exact volume of the solid formed
Post by: Aussie1Italia2 on November 03, 2016, 12:08:05 pm
1. Find the discriminant of the quadratic equation:
A) -x2 - 3x = 0
B) x2 + 4 = 0
2. Find any values of b for which 2x2 + x + b + 1 = 0
Again, thanks!!! :D
Post by: RuiAce on November 03, 2016, 12:12:42 pm
Hey, I have a series of question I need help with; please and thank you.
1. Find the discriminant of the quadratic equation:
A) -x2 - 3x = 0
B) x2 + 4 = 0
2. Find any values of b for which 2x2 + x + b + 1 = 0
Again, thanks!!! :D
Your second question is inconclusive. Because the way you put it, it depends on your value for x. Suspecting something in the question wasn't typed because I can sort of tell what the question should be.
Post by: Aussie1Italia2 on November 03, 2016, 12:42:06 pm
I forgot the last bit of the question.
Find any values of b for which 2x2 + x + b + 1 = 0 has real roots.
Post by: jamonwindeyer on November 03, 2016, 12:44:04 pm
Oh yep thanks;
I forgot the last bit of the question.
Find any values of b for which 2x2 + x + b + 1 = 0 has real roots.
Cool! For real roots we require our discriminant to be greater than or equal to zero!
Post by: RuiAce on November 03, 2016, 12:45:45 pm
Cool! For real roots we require our discriminant to be greater than or equal to zero!Well damn I didn't type fast enough
Post by: Hplovers on November 03, 2016, 03:36:23 pm
Thankyou!
Post by: RuiAce on November 03, 2016, 03:44:48 pm
Would be great if you could help me out with 4 (ii) please :)
Thankyou!
Post by: Hplovers on November 03, 2016, 03:55:45 pm
Thankyou so so much!
Post by: teapancakes08 on November 04, 2016, 09:43:08 am
4. Simplify:
(h) 1 - sin^2 (180-x)
If it was just 1-sin^2x it would be cos^2x, but...
If someone could help clarify that would be great, thanks ^^
Post by: RuiAce on November 04, 2016, 09:55:49 am
Probably should be able to answer this, but the second term is throwing me off:
4. Simplify:
(h) 1 - sin^2 (180-x)
If it was just 1-sin^2x it would be cos^2x, but...
If someone could help clarify that would be great, thanks ^^
Post by: Aussie1Italia2 on November 04, 2016, 08:52:26 pm
1. Express x2 - 4x + 5 in the form Ax (x+2) + B (x +1) + C + 4
2. Show that x2 + 2x + 9 can be written in the form a (x-2)(x+3) + b (x-2) + c where a=1, b=1, and c=17
Post by: jakesilove on November 04, 2016, 08:55:23 pm
Hey!!! I have two new questions that I need help with. :-\ I think I know how to do them but I just need some reassurance; please and thank you!!!
1. Express x2 - 4x + 5 in the form Ax (x+2) + B (x +1) + C + 4
2. Show that x2 + 2x + 9 can be written in the form a (x-2)(x+3) + b (x-2) + c where a=1, b=1, and c=17
Hey! In both methods, you just need to expand the 'form' equation (ie. Ax (x+2) + B (x +1) + C + 4). Then, equate the x^2 terms, the x terms and the constant terms with the original question. For instance, if you expand, the only x^2 term is Ax^2. Therefore, for the first one, A=1! Let me know if there's something specific you're struggling with, or if you need the whole answer; sounds like you're on the right track!
Post by: Aussie1Italia2 on November 04, 2016, 09:04:40 pm
Post by: jakesilove on November 04, 2016, 09:11:25 pm
Can you please try explaining that another way? It didn't make sense at all to me. Thank you!!! :D
We equate x squared terms in each equation
Therefore, the expression can be written as
Post by: RuiAce on November 04, 2016, 09:24:48 pm
Can you please try explaining that another way? It didn't make sense at all to me. Thank you!!! :D
Post by: Aussie1Italia2 on November 05, 2016, 02:52:16 pm
Can we tell I suck at maths yet?
Thanks for the help!
Find values of A, B, and C if x2 + x - 2 = A (x-2)2 + Bx + C
Post by: RuiAce on November 05, 2016, 02:55:57 pm
Hey again!
Can we tell I suck at maths yet?
Thanks for the help!
Find values of A, B, and C if x2 + x - 2 = A (x-2)2 + Bx + C
Post by: katnisschung on November 05, 2016, 09:37:20 pm
Post by: jakesilove on November 05, 2016, 09:52:46 pm
need help solving this trig identity
Hey! The trick to a question like this is to let sin(x) be equal to a variable; I like using u.
Now, we can just solve this like any old quadratic! You can either use the quadratic equation, or any other method you can implement. It ends up simplifying to
Great! We can now let each brackets be equal to zero
Now, let's change our u's back into sin(x)s
Can you see any issues? You definitely will as you study this area more :) sin(x) oscilates between -1 and 1, so will never equal 3! If you don't believe me, try typing shift sin of 3 into your calculator :)
That leaves us with the equation sin(x)=1/2. We can solve this in the normal way, which gives us
So those will be our answers!
Post by: RuiAce on November 06, 2016, 08:58:23 am
need help solving this trig identityRegarding the website you used, that was a trigonometric equation; not proving an identity. So if you chose the next option on that list on the left you would've gotten the computerised solution(http://i.imgur.com/OrdUL13.png)
Post by: katnisschung on November 06, 2016, 12:31:40 pm
Post by: RuiAce on November 06, 2016, 12:59:53 pm
i cant find it in exact value form :(
Post by: katnisschung on November 07, 2016, 05:57:54 pm
question on series
The sum of the first 5 terms of an arithmetic series is 35 and the sum of the next 5 terms is 160
find the first term and the common difference
Post by: RuiAce on November 07, 2016, 06:08:03 pm
thanks in advance :)
question on series
The sum of the first 5 terms of an arithmetic series is 35 and the sum of the next 5 terms is 160
find the first term and the common difference
Post by: RuiAce on November 08, 2016, 02:13:30 pm
(http://i.imgur.com/jS1KfIa.png)
Any thoughts Jamon? :P
Post by: jamonwindeyer on November 08, 2016, 02:28:28 pm
I know that nobody really cares, but I just felt the need to share this exponential growth question which could've gotten 4-6 marks in the HSC... as opposed to in ACTL ;D
(http://i.imgur.com/jS1KfIa.png)
Any thoughts Jamon? :P
I think they'd have to step you through that one for it to be included in the HSC ;)
Post by: katnisschung on November 08, 2016, 08:30:46 pm
Ruiace thanks for your reply
but im having trouble grasping the second part of your answer.
would u mind elaborating on s=5/2 [2(a+5d) +4d]
why is it +4d when the formula is sn=n/2 [2a+(n-1)d]
i dont understand :P
Post by: RuiAce on November 08, 2016, 08:33:51 pm
Ruiace thanks for your replyT6, T7, T8, T9, T10
but im having trouble grasping the second part of your answer.
would u mind elaborating on s=5/2 [2(a+5d) +4d]
why is it +4d when the formula is sn=n/2 [2a+(n-1)d]
i dont understand :P
There are 5 terms
So n=5
So n-1=4
So (n-1)d=4d
Post by: katnisschung on November 08, 2016, 08:41:10 pm
Post by: RuiAce on November 08, 2016, 08:46:20 pm
thanks Ruiace got it !Excellent :)
Post by: teapancakes08 on November 08, 2016, 11:50:11 pm
(sin^2x tanx) + (cos^2x cotx) + 2sinxcox = tanx + cotx
Assuming that factorised it correctly (tbh, I sort of bludged it through brute force...), is it allowed to skip to:
(sin^2x + cos^2x)(tanx +cotx)
Or do you have to show the whole working out from A to B?
Post by: Dolphax on November 08, 2016, 11:59:28 pm
When proving that:
(sin^2x tanx) + (cos^2x cotx) + 2sinxcox = tanx + cotx
Assuming that factorised it correctly (tbh, I sort of bludged it through brute force...), is it allowed to skip to:
(sin^2x + cos^2x)(tanx +cotx)
Or do you have to show the whole working out from A to B?
I was taught that (and pretty sure that for the HSC):
"Prove that" questions: you can start off with LHS or RHS. You could even work on both sides and show that they are equal, although it's not recommended if not necessary.
"Show that" questions: you can only start off with LHS.
... however this seems to go against RuiAce's guide Verbs and Maths.
Post by: jamonwindeyer on November 09, 2016, 12:35:20 am
I was taught that (and pretty sure that for the HSC):
"Prove that" questions: you can start off with LHS or RHS. You could even work on both sides and show that they are equal, although it's not recommended if not necessary.
"Show that" questions: you can only start off with LHS.
... however this seems to go against RuiAce's guide Verbs and Maths.
Yeah, they don't make that distinction at the HSC level, prove and show are for all purposes identical ;D
When proving that:
(sin^2x tanx) + (cos^2x cotx) + 2sinxcox = tanx + cotx
Assuming that factorised it correctly (tbh, I sort of bludged it through brute force...), is it allowed to skip to:
(sin^2x + cos^2x)(tanx +cotx)
Or do you have to show the whole working out from A to B?
Hey pancakes! In this case, I would say that's a tad few steps to skip. If you are doing it anyway, you might as well write it down! Especially in proof questions, where the marker needs to verify that you didn't just fudge steps based on a guess/intuition, it's best to show all but the most trivial steps :) in this case, try and show how you got to that next result; you've skipped virtually all of the hard work :)
Post by: RuiAce on November 09, 2016, 04:30:36 am
I was taught that (and pretty sure that for the HSC):
"Prove that" questions: you can start off with LHS or RHS. You could even work on both sides and show that they are equal, although it's not recommended if not necessary.
"Show that" questions: you can only start off with LHS.
... however this seems to go against RuiAce's guide Verbs and Maths.
Post by: katnisschung on November 09, 2016, 07:10:33 am
Post by: RuiAce on November 09, 2016, 07:57:39 am
The sum of 12 terms of an arithmetic series is 186 and the 20th term is 83 find the sum of 40 term....I triedIs that supposed to mean the sum of the first 12 and the first 40 terms?
Post by: katnisschung on November 09, 2016, 09:48:43 am
Is that supposed to mean the sum of the first 12 and the first 40 terms?thats the issue i had RuiAce so i just assumed it was the first 12
Post by: RuiAce on November 09, 2016, 09:59:24 am
thats the issue i had RuiAce so i just assumed it was the first 12Well doing a check with WolframAlpha this gives you
a=-12
d=5
Which gives S40 = 3420
Is that meant to be the answer? (Where confusion arises supplying the answer helps.)
Post by: katnisschung on November 09, 2016, 12:13:35 pm
Well doing a check with WolframAlpha this gives you
a=-12
d=5
Which gives S40 = 3420
Is that meant to be the answer? (Where confusion arises supplying the answer helps.)
yes that is the correct answer
Post by: RuiAce on November 09, 2016, 12:35:24 pm
Post by: katnisschung on November 10, 2016, 06:16:56 pm
her wage increases by 5% each year.
Find
a) her wage after 6 years
b) her total earnings (before tax)
in 6 years
Post by: RuiAce on November 10, 2016, 06:21:14 pm
Lucia currently earns $25,000
her wage increases by 5% each year.
Find
a) her wage after 6 years
b) her total earnings (before tax)
in 6 years
Part b) is just weird. Although it looks to me like they meant the sum, their wording isn't clear. I'm not providing a solution to that one without knowing what's going on (so provide the answer if you want a solution)
Post by: jamonwindeyer on November 10, 2016, 06:26:25 pm
Lucia currently earns $25,000
her wage increases by 5% each year.
Find
a) her wage after 6 years
b) her total earnings (before tax)
in 6 years
I'll take a stab at what it means! This is a geometric series question, where \(a=25000\) and ](r=1.05\) (the 5% interest rate).
The wage after 6 years is just the 6th term (assuming the question means the sixth, like, the wage during the sixth year, which is why my answer differs from Rui's). The question makes a bit more sense this way.
Then the total earned is just the sum of the first six terms:
This topic is just about knowing your formula and knowing the numbers to sub :) hope this helps!
Post by: katnisschung on November 10, 2016, 06:28:25 pm
also i think the answers are wrong
the common ratio of a geometric series is 4 and the sum of
the first 5 terms is 3069 find the first term
Post by: RuiAce on November 10, 2016, 06:31:08 pm
I'll take a stab at what it means! This is a geometric series question, where \(a=25000\) and ](r=1.05\) (the 5% interest rate).Yeah the question was just irritating with its wording. Upsetting when you're studying business at uni haha
The wage after 6 years is just the 6th term (assuming the question means the sixth, like, the wage during the sixth year, which is why my answer differs from Rui's). The question makes a bit more sense this way.
Then the total earned is just the sum of the first six terms:
This topic is just about knowing your formula and knowing the numbers to sub :) hope this helps!
thanks Ruiace! it popped up in a series exercise?
also i think the answers are wrong
the common ratio of a geometric series is 4 and the sum of
the first 5 terms is 3069 find the first term
Post by: katnisschung on November 10, 2016, 06:41:03 pm
I'll take a stab at what it means! This is a geometric series question, where \(a=25000\) and ](r=1.05\) (the 5% interest rate).
The wage after 6 years is just the 6th term (assuming the question means the sixth, like, the wage during the sixth year, which is why my answer differs from Rui's). The question makes a bit more sense this way.
Then the total earned is just the sum of the first six terms:
This topic is just about knowing your formula and knowing the numbers to sub :) hope this helps!
question why did you use the arithmetic series formula?
a
Post by: RuiAce on November 10, 2016, 06:41:43 pm
question why did you use the arithmetic series formula?Haha lol I thought something was weird but I didn't pick up on it. I think he made a mistake there.
a
Post by: katnisschung on November 10, 2016, 06:47:36 pm
Haha lol I thought something was weird but I didn't pick up on it. I think he made a mistake there.
phew thats why there was a discrepancy between both your answers ;D
Post by: jamonwindeyer on November 10, 2016, 07:40:46 pm
phew thats why there was a discrepancy between both your answers ;D
Sorry just fixed it up! Note also that Rui and I interpreted Part (a) differently, which could lead to different answers for (b) (it's a bad question) :P
Post by: katnisschung on November 10, 2016, 08:32:15 pm
i know y'all did 3 or 4 unit and are maths geniuses
and jamon and jake advice against it
and i know the emphasis of solving problems over
writing notes but i personally find that writing notes
helps me to remember all aspects of the topic.
tips?
note i am not a maths genius i take 2u and get around 80s
(my teacher keeps on saying my results have fluctuated and
i have 'potential' ;D let's hope so)
oh and when i mean writing notes i mean like transcribing
everytime after i have a maths lesson
Post by: RuiAce on November 10, 2016, 08:35:23 pm
should i write notes for my maths testBack when there was no "reference sheet" I found it acceptable for some people to write a formula sheet or formula book. Things in there were meant to be really basic and serve no more purpose than be a refresher, and their only advantage was that they're going to be easier to use than textbooks.
i know y'all did 3 or 4 unit and are maths geniuses
and jamon and jake advice against it
and i know the emphasis of solving problems over
writing notes but i personally find that writing notes
helps me to remember all aspects of the topic.
tips?
The fact that the reference sheet has all the formulas you actually need really downs on this. I don't see any more that you can do than just printing it off and dropping in a few annotations here and there (A3 paper is fine for this though). I am with them in that in general doing "notes" for maths is not advisable, because it really doesn't help you remember as much as you think it would.
The furthest you should ever go with notes is about 5 dot points per topic, just highlighting some key areas of concern. If you ask me, that is.
Post by: jamonwindeyer on November 10, 2016, 09:01:00 pm
should i write notes for my maths test
i know y'all did 3 or 4 unit and are maths geniuses
and jamon and jake advice against it
and i know the emphasis of solving problems over
writing notes but i personally find that writing notes
helps me to remember all aspects of the topic.
tips?
note i am not a maths genius i take 2u and get around 80s
(my teacher keeps on saying my results have fluctuated and
i have 'potential' ;D let's hope so)
oh and when i mean writing notes i mean like transcribing
everytime after i have a maths lesson
You should write as much as you need to to remember your concepts! Like, if you need to have a set of notes specifically proving differentiation by first principles, then have that. It's all about whatever works for you! :)
That said, whatever your choice, you need to be doing practice as well. No one ever got a strong result in a HSC Math subject without doing any practice at all. Do what you need to do, but make sure practice is mixed in :)
Post by: jamesq on November 11, 2016, 11:17:11 pm
(a) What is the amount of each instalment?
Answer is $1835.68.
I tried the question and got $1712.87, any help is appreciated. Thanks in advance!
Post by: jamonwindeyer on November 11, 2016, 11:20:27 pm
A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.
(a) What is the amount of each instalment?
Answer is $1835.68.
I tried the question and got $1712.87, any help is appreciated. Thanks in advance!
Hey James! So two steps here; find the total amount owed, and then divide that into the 5 repayments :) I'd be happy to lend a hand, but we need to know whether this is simple interest or compound interest? Indeed, perhaps you used the wrong one in your working? Or did the question not say? :)
If you let me know whether its simple or compound I'd be happy to show you the working for it! ;D
Edit: Yeah actually, sorry, these are always compound. That said, reckon you could snap a pic of your working and pop it as an attachment?Since you are close I reckon you might have just made a small mistake, I'll try and spot it for you! :)
Post by: RuiAce on November 11, 2016, 11:33:02 pm
A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.
(a) What is the amount of each instalment?
Answer is $1835.68.
I tried the question and got $1712.87, any help is appreciated. Thanks in advance!
Hey James! So two steps here; find the total amount owed, and then divide that into the 5 repayments :) I'd be happy to lend a hand, but we need to know whether this is simple interest or compound interest? Indeed, perhaps you used the wrong one in your working? Or did the question not say? :)When you're an actuary and just chuck it into the present value formula before doing the question 8)
If you let me know whether its simple or compound I'd be happy to show you the working for it! ;D
Edit: Yeah actually, sorry, these are always compound. That said, reckon you could snap a pic of your working and pop it as an attachment?Since you are close I reckon you might have just made a small mistake, I'll try and spot it for you! :)
Post by: teapancakes08 on November 12, 2016, 09:06:36 am
25. The sum of the first six terms of an arithmetic sequence is -12 and the sum of it's first fourteen term is 196. Find (a) in the sum of n terms, (b) the smallest value of n if the sum is to exceed 250.
I got the part a fine by using the formula Sn = (n/2)[2a + (n-1)d] and simultaneous equations, but the answer I got was 2n^2 - 12n....and the answer at the back is 2n^2 - 14n.
In part b I'm having trouble following. I tried using quadradics but with either one I end up getting surds (which is probably the point since 250 isn't a factor that fits nicely with either of them). And I don't think it possible to do logs because there are three terms...
If anyone could help that would be much appreciated, thanks ^^
Post by: RuiAce on November 12, 2016, 09:21:48 am
Need to check for this question:
25. The sum of the first six terms of an arithmetic sequence is -12 and the sum of it's first fourteen term is 196. Find (a) in the sum of n terms, (b) the smallest value of n if the sum is to exceed 250.
I got the part a fine by using the formula Sn = (n/2)[2a + (n-1)d] and simultaneous equations, but the answer I got was 2n^2 - 12n....and the answer at the back is 2n^2 - 14n.
In part b I'm having trouble following. I tried using quadradics but with either one I end up getting surds (which is probably the point since 250 isn't a factor that fits nicely with either of them). And I don't think it possible to do logs because there are three terms...
If anyone could help that would be much appreciated, thanks ^^
Post by: jakesilove on November 12, 2016, 09:23:12 am
Need to check for this question:
25. The sum of the first six terms of an arithmetic sequence is -12 and the sum of it's first fourteen term is 196. Find (a) in the sum of n terms, (b) the smallest value of n if the sum is to exceed 250.
I got the part a fine by using the formula Sn = (n/2)[2a + (n-1)d] and simultaneous equations, but the answer I got was 2n^2 - 12n....and the answer at the back is 2n^2 - 14n.
In part b I'm having trouble following. I tried using quadradics but with either one I end up getting surds (which is probably the point since 250 isn't a factor that fits nicely with either of them). And I don't think it possible to do logs because there are three terms...
If anyone could help that would be much appreciated, thanks ^^
Hey! Just setting n=6, you'll see that the only equation that gets out the required value of -12 is the one at the back of the book. Potentially you messed up some algebra somewhere? Check it over, and if you still need help, post it here and I'll take a look!
As for the second part, you're 100% on the right track. Set your equation equal to 250, and find the surd value. Using the equation at the back of the book, n will equal -8.215 and 15.215. Now, obviously n can't be zero, so we take the positive value 15.215. Let's think about what that means for a second; if n=15, the answer will be LESS than 250. If n=16, the answer will be MORE than 250. So, 16 is out answer, because that's what the question wants!
Let me know if that all makes sense.
Jake
Post by: teapancakes08 on November 12, 2016, 09:45:28 am
Hey! Just setting n=6, you'll see that the only equation that gets out the required value of -12 is the one at the back of the book. Potentially you messed up some algebra somewhere? Check it over, and if you still need help, post it here and I'll take a look!
As for the second part, you're 100% on the right track. Set your equation equal to 250, and find the surd value. Using the equation at the back of the book, n will equal -8.215 and 15.215. Now, obviously n can't be zero, so we take the positive value 15.215. Let's think about what that means for a second; if n=15, the answer will be LESS than 250. If n=16, the answer will be MORE than 250. So, 16 is out answer, because that's what the question wants!
Let me know if that all makes sense.
Jake
I think I got it. I checked back and saw that I made an algebra mistake when subbing in the values for a and d. Ruiace also helped me out on that one. As for the part b, I should be good. The explanations helped clear things up a lot. Thanks so much for the help ^^
Post by: teapancakes08 on November 12, 2016, 09:59:38 am
Guess I did make an algebraic error ^^; but I suppose better now than during the test.
Thanks for the explanation, it cleared my understanding of it by a large margin. Thank you so much for the help ^^
Post by: julzzz on November 12, 2016, 10:45:12 am
This is a prelim question which i couldnt manage to do
Using indices, not log, thanks
Post by: jakesilove on November 12, 2016, 11:01:55 am
(1/27)^(3n+1)=sqrt{3}/81
This is a prelim question which i couldnt manage to do
Using indices, not log, thanks
So, you're question is
We need to convert everything into powers of three;
Therefore, we can let
So our answer is n=1/18! Hope that all makes sense :)
Post by: teapancakes08 on November 12, 2016, 09:51:00 pm
20. Show that the sum of the first n odd natural numbers is a perfect square.
The method I've tried is similar to that of induction, letting the sum equal to n^2 and solving it by plugging in terms, where n = the number of terms in current series and the difference being two...if that makes any sense (as in 1 = 1^2, 1 + 3 = 2^2....etc.). Is there a way to neatly show this without confusing yourself (and, by extension, probably the marker)? Or is there a better way to express this proof in general?
Post by: RuiAce on November 12, 2016, 09:54:22 pm
Wondering about this question:
20. Show that the sum of the first n odd natural numbers is a perfect square.
The method I've tried is similar to that of induction, letting the sum equal to n^2 and solving it by plugging in terms, where n = the number of terms in current series and the difference being two...if that makes any sense (as in 1 = 1^2, 1 + 3 = 2^2....etc.). Is there a way to neatly show this without confusing yourself (and, by extension, probably the marker)? Or is there a better way to express this proof in general?
Post by: RuiAce on November 12, 2016, 09:57:53 pm
Post by: samuels1999 on November 12, 2016, 11:15:54 pm
I got this question. And it is nothing like the usual ones I have done before. It is probably really easy, but I found it a little tricky.
Thanks,
Samuel
Post by: RuiAce on November 13, 2016, 06:03:59 am
Hi everyone
I got this question. And it is nothing like the usual ones I have done before. It is probably really easy, but I found it a little tricky.
Thanks,
Samuel
________________________
________________________
Post by: Aussie1Italia2 on November 13, 2016, 09:42:21 am
Alpha & beta are roots of the quadratic equation x2 - 6x + 3 = 0.
Find alpha x beta2 + alpha2 x beta
Post by: jamonwindeyer on November 13, 2016, 09:59:30 am
Hello, please and thank you for your help!
Alpha & beta are roots of the quadratic equation x2 - 6x + 3 = 0.
Find alpha x beta2 + alpha2 x beta
Hey Aussie!
And then you just sub and go to get 18 as your answer ;D
Post by: katnisschung on November 13, 2016, 11:55:58 am
Q) ap: 52,46,40
Find the smallest value of n so Sn<0
so i got n=19
i couldn't get it any bigger becos the limit for file size
Moderator Edit: Merged question w/ answer
Post by: RuiAce on November 13, 2016, 12:02:41 pm
so i got n=19Snipping tool (automatically in Windows 7 and above) works miracles for captures; saves you having to use PrtScn to get a full screenshot
i couldn't get it any bigger becos the limit for file size
A quick check on WolframAlpha confirms that answer.
Post by: katnisschung on November 13, 2016, 12:04:12 pm
Yeah I should start using wolfram alpha more
Post by: katnisschung on November 13, 2016, 12:25:50 pm
the terms of the sequence are all different and the sum of the 1st 2 two terms is equal to twice the third term
Find the 1st 3 terms of the sequence
So I interpreted this as
a=12
S2=2T3
So how do I know which formula to use for Sn becos I don't know what r is ? Or do I test them into both formulas
Post by: RuiAce on November 13, 2016, 01:08:30 pm
A gp has 1st term 12
the terms of the sequence are all different and the sum of the 1st 2 two terms is equal to twice the third term
Find the 1st 3 terms of the sequence
So I interpreted this as
a=12
S2=2T3
So how do I know which formula to use for Sn becos I don't know what r is ? Or do I test them into both formulas
Post by: teapancakes08 on November 13, 2016, 09:22:44 pm
31. The track of a gramophone record is in the shape of a spiral curve and may be considered as a number of concentric circles of inner and outer radius 5.25cm and 10.5cm respectively. The record rotates at 33 1/3 rev/min and takes 18 minutes to play. Find the length of the track.
I'm presuming you have to use Sn = (n/2)(a+l) as it's possible to find both values by using C = 2πr (if I'm doing this correctly...), but the other values throw me off. Do I use d = st to help solve the question, or is there a entirely different way of solving it? The question also suggests to take π = 22/7, although I'm not sure what to do with it...any suggestions? Thanks to anyone who can help ^^
Post by: jakesilove on November 13, 2016, 10:12:52 pm
Having trouble visualising this:
31. The track of a gramophone record is in the shape of a spiral curve and may be considered as a number of concentric circles of inner and outer radius 5.25cm and 10.5cm respectively. The record rotates at 33 1/3 rev/min and takes 18 minutes to play. Find the length of the track.
I'm presuming you have to use Sn = (n/2)(a+l) as it's possible to find both values by using C = 2πr (if I'm doing this correctly...), but the other values throw me off. Do I use d = st to help solve the question, or is there a entirely different way of solving it? The question also suggests to take π = 22/7, although I'm not sure what to do with it...any suggestions? Thanks to anyone who can help ^^
Weird question! Here's how I interpret it; if it takes 18 minutes to play, and the speed is 33 1/3 rev/min, then the number of revolutions will be (33.3333)*(18)=600 revolutions. So, we have 600 circles with radius' between 5.25cm and 10.5cm. That means that the radius will increase by (10.5-5.25)/600= 0.0525cm with each subsequent circle. This is a seriously weird question.
Our first circumference is 2π(5.25), then 2π(5.325) etc. etc. Clearly, this is increasing by 2*0.0525π with each circle, so that is going to be our value for d! We have n, a, l and d, so we can sum up the equation :)
Let me know if that all made sense!
Post by: teapancakes08 on November 13, 2016, 10:55:17 pm
Weird question! Here's how I interpret it; if it takes 18 minutes to play, and the speed is 33 1/3 rev/min, then the number of revolutions will be (33.3333)*(18)=600 revolutions. So, we have 600 circles with radius' between 5.25cm and 10.5cm. That means that the radius will increase by (10.5-5.25)/600= 0.0525cm with each subsequent circle. This is a seriously weird question.
Our first circumference is 2π(5.25), then 2π(5.325) etc. etc. Clearly, this is increasing by 2*0.0525π with each circle, so that is going to be our value for d! We have n, a, l and d, so we can sum up the equation :)
Let me know if that all made sense!
It seriously does...
My working out kind of looked like this:
full rev = speed*time
C = 2πr
a = 2π*5.25
l = 2π10.5
d (total revolutions) = speed*time
hence n = vt
n = 100/3 x 18
n = 600
...and then I pretty much plug in everything into Sn = (n/2) (a+l) and then BOOM. Answer. (Which is 296.88m for 2.d.p or 297m if you round it up to the nearest whole.)
To be honest the first thing that came to mind was physics, haha...
But yeah, it makes a lot more sense now - thanks so much for the clarifications ;D
Post by: jakesilove on November 14, 2016, 11:19:26 am
It seriously does...
My working out kind of looked like this:
full rev = speed*time
C = 2πr
a = 2π*5.25
l = 2π10.5
d (total revolutions) = speed*time
hence n = vt
n = 100/3 x 18
n = 600
...and then I pretty much plug in everything into Sn = (n/2) (a+l) and then BOOM. Answer. (Which is 296.88m for 2.d.p or 297m if you round it up to the nearest whole.)
To be honest the first thing that came to mind was physics, haha...
But yeah, it makes a lot more sense now - thanks so much for the clarifications ;D
No problem! Glad I could help :)
Post by: katnisschung on November 14, 2016, 05:03:00 pm
show that three numbers a-d, a, a+d are in arithmetic progression. Hence find three
numbers in arithmetic sequence whose sum is -12 and whose product is 36
Post by: RuiAce on November 14, 2016, 05:27:59 pm
sighs.... series
show that three numbers a-d, a, a+d are in arithmetic progression. Hence find three
numbers in arithmetic sequence whose sum is -12 and whose product is 36
Post by: teapancakes08 on November 14, 2016, 07:21:46 pm
40. The lengths of the sides of a right-angled triangle form the terms of an arithmetic sequence. If the hypotenuse is 15cm in length, what is the length of the other two sides?
I tried working it out using these two equations — 15^2 = a^2 + b^2 and 15 = 2b - a — but I have a sinking feeling that I'm looking at it the wrong way as when equating them I got irrational numbers, when the answers are definitely rational. Any suggestions for a less confusing (and potentially more efficient) way of solving this question?
Post by: RuiAce on November 14, 2016, 08:04:40 pm
Am enternally frustrated by this question:
40. The lengths of the sides of a right-angled triangle form the terms of an arithmetic sequence. If the hypotenuse is 15cm in length, what is the length of the other two sides?
I tried working it out using these two equations — 15^2 = a^2 + b^2 and 15 = 2b - a — but I have a sinking feeling that I'm looking at it the wrong way as when equating them I got irrational numbers, when the answers are definitely rational. Any suggestions for a less confusing (and potentially more efficient) way of solving this question?
Remark on the note:
If we used a, a+b and a+2b, then we have a+2b=15 so a=15-2b
Then, 225 = a^2 + (a+b)^2 so 225 = (15-2b)^2 + (15-b)^2 which gives b=3 anyway.
Post by: teapancakes08 on November 14, 2016, 08:49:10 pm
"44. For a potato race, a straight line is marked on the ground from point A, and point B, C, D,... are marked on the line so that AB = BC = CD =...2 meters, and a potato is placed at each of the points B, C, D,.... A runner has to start from A and bring each potato by seperate journey back to a basket at A. Find the number of potatoes so that the total distance run during the race will be 480 meters."
(Sorry if the working out looks a bit iffy in some parts...^^;; )
Post by: RuiAce on November 14, 2016, 09:21:52 pm
I think I might have made a carry-over error in this question:
"44. For a potato race, a straight line is marked on the ground from point A, and point B, C, D,... are marked on the line so that AB = BC = CD =...2 meters, and a potato is placed at each of the points B, C, D,.... A runner has to start from A and bring each potato by seperate journey back to a basket at A. Find the number of potatoes so that the total distance run during the race will be 480 meters."
(Sorry if the working out looks a bit iffy in some parts...^^;; )
A check on WolframAlpha showed that solving for n should give n as an integer, so looking at your final line I don't suspect there's been a mistake in algebra halfway through
Post by: teapancakes08 on November 14, 2016, 09:49:03 pm
A check on WolframAlpha showed that solving for n should give n as an integer, so looking at your final line I don't suspect there's been a mistake in algebra halfway through
Got it ;D
Geh, I got stuck thinking it was like physics and started at zero. Actually, I got confused by the question stating A twice. Now that I think about it does make a lot more sense since it does indicate the the first run is the distance travelled from A to the next point (B) and back. Guess I wasn't paying much attention and jumped ahead to quickly...^^;;
Thanks for pointing that out ^^
Post by: teapancakes08 on November 15, 2016, 07:45:46 pm
Which GS formula would I use to solve this one? I tried using both and got the really weird answers...to which I take it that I'm looking at the question wrong...
Any suggestions on how to approach it?
Post by: jakesilove on November 15, 2016, 09:09:58 pm
16. The population of a certain town is 10000; if its population will decrease each year by 10% of it population in the preceding year, find its population in 5 years time.
Which GS formula would I use to solve this one? I tried using both and got the really weird answers...to which I take it that I'm looking at the question wrong...
Any suggestions on how to approach it?
To be honest, I'd probably do this without bothering with the series, because doing the iteration 5 times isn't too difficult.
The first year, there are 10,000 people. Then next year (after one year), there will be 0,9*10,000=9,000 people. The next year (after two years), there will be 0,9*9,000=8,100 people. The next year (after three years) there will be 0.9*8,100=7,290 people. The next year (after four years), there will be 0.9*7290=6561 people. The next year (after five years), there will be 0.9*6561=5904.7 people (which is either 5905 or 5904 people, depending on how you choose to round).
You could have set up a GP, which the common ratio being 0.9 and a=10,000. Works just as well. Difficulty would be whether you set n=5 or n=6; 6 gets you the answer I posited above
Post by: teapancakes08 on November 15, 2016, 09:46:33 pm
To be honest, I'd probably do this without bothering with the series, because doing the iteration 5 times isn't too difficult.
The first year, there are 10,000 people. Then next year (after one year), there will be 0,9*10,000=9,000 people. The next year (after two years), there will be 0,9*9,000=8,100 people. The next year (after three years) there will be 0.9*8,100=7,290 people. The next year (after four years), there will be 0.9*7290=6561 people. The next year (after five years), there will be 0.9*6561=5904.7 people (which is either 5905 or 5904 people, depending on how you choose to round).
You could have set up a GP, which the common ratio being 0.9 and a=10,000. Works just as well. Difficulty would be whether you set n=5 or n=6; 6 gets you the answer I posited above
Thanks for clearing things up ^^
I went back and found out that using Tn = ar^n-1 worked for this equation. I think n would be 6 regardless as it's five years plus the first year (i found it after plugging in n = 5 and got the wrong answer). When plugging it into the formula it give 5904.5, but since you can't half a person I rounded it up to 5905. It gets the answer pretty fast too, although powering through five terms isn't too hard. Once again, thank you for the help ;D
Post by: teapancakes08 on November 15, 2016, 10:26:21 pm
23. Find the first three terms of a geometric sequence given that the sum of the first four terms is 21 2/3, and the sum to infinity is 27.
Post by: FallonXay on November 15, 2016, 11:21:55 pm
More GS questions...
23. Find the first three terms of a geometric sequence given that the sum of the first four terms is 21 2/3, and the sum to infinity is 27.
So from the given information, you know that }\ = 27)
(sum to infinity) and that
}{1-r}\ = \frac{65}{3}\<br />\end{equation}<br />)
(sum of the first 4 terms of the GS)
Now, you want to eliminate one of the variables so that you can solve for the other variable. To do this you want to find a common subject, the easiest way in this case is to rearrange for the common subject of
So rearranging the equation for the sum of the first 4 terms to this will give you
Once you have achieved this common subject, you can equate the sum to infinity formula and the newly rearranged formula giving you:
Then solving for r should give you a ratio of 2/3. Sub this back into one of your original equations (for example, the sum to infinity) and this should solve for 'a' giving you the value of 9. Then proceed to multiply this by your now known ratio of 2/3 to find the first 3 terms of the geometric series which will be 9, 6 and 4.
Hope this helps :)
(p.s trying out latex and I kind of eventually gave up so if something doesn't make sense feel free to ask ;) )
Post by: RuiAce on November 16, 2016, 08:17:41 am
(p.s trying out latex and I kind of eventually gave up so if something doesn't make sense feel free to ask ;) )If you wanna try LaTeX don't use my guide at the bottom of the math section for now. Try experimenting the basics with Daum Equation Editor and then when you're typing it if something seems weird ask Jamon or myself.
Anything's hard to learn but LaTeX is just better than word and has a ton of other advantages when you use it to write up pdfs that needs heaps of math equations. Can also be quite fun
Post by: FallonXay on November 16, 2016, 09:00:01 am
If you wanna try LaTeX don't use my guide at the bottom of the math section for now. Try experimenting the basics with Daum Equation Editor and then when you're typing it if something seems weird ask Jamon or myself.
Anything's hard to learn but LaTeX is just better than word and has a ton of other advantages when you use it to write up pdfs that needs heaps of math equations. Can also be quite fun
Ooooooh, I didn't even realise you had a guide!!! I'm gonna check it out now despite your word of warning hehehehe. Also, will check out Daum Equation Editor. Thanks for the advice :)
Post by: RuiAce on November 16, 2016, 09:07:40 am
Ooooooh, I didn't even realise you had a guide!!! I'm gonna check it out now despite your word of warning hehehehe. Also, will check out Daum Equation Editor. Thanks for the advice :)Thing with it is that after learning some more LaTeX I realised I overcomplicated it. I'm gonna be fixing it but I probably won't get started on that until Dec
Post by: teapancakes08 on November 16, 2016, 10:09:06 pm
I found the sum, which (a+1)/a, but I don't quite understand what to do for the second part. Any suggestions?
Post by: RuiAce on November 16, 2016, 10:13:51 pm
33. Find the sum of series 1 + 1/(a+1) + 1/(a+1)^2 +.... For what range of values of a has this infinite series a sum?
I found the sum, which (a+1)/a, but I don't quite understand what to do for the second part. Any suggestions?
Blech. Cheers Jamon (y)
Post by: jamonwindeyer on November 16, 2016, 10:15:45 pm
33. Find the sum of series 1 + 1/(a+1) + 1/(a+1)^2 +.... For what range of values of a has this infinite series a sum?
I found the sum, which (a+1)/a, but I don't quite understand what to do for the second part. Any suggestions?
Hey! So the idea here is that infinite series only converge when the absolute value of the common ratio is less than 1.
So, we form an absolute value inequality ;D
Edit: Woops, double, but I'l leave it, Rui I think you made a sign error from Line 4 to Line 5 :)
Post by: teapancakes08 on November 16, 2016, 10:24:50 pm
34. For the geometric sequence a, ar, ar^2, ...show that the sequence loga, log (ar), log (ar^2), .... is an arithmetic sequence.
36. Find, without using tables or calculator, the sum of 7 terms of log27 + log9 + log3 + ....
Post by: RuiAce on November 16, 2016, 10:33:09 pm
List of GS logs because I have little clue on how to do them...
34. For the geometric sequence a, ar, ar^2, ...show that the sequence loga, log (ar), log (ar^2), .... is an arithmetic sequence.
36. Find, without using tables or calculator, the sum of 7 terms of log27 + log9 + log3 + ....
_____________________________________________________
Post by: teapancakes08 on November 16, 2016, 11:07:04 pm
35. Find the sum of 10 terms of the series log3 + log6 + log12 + .... given that log 3 = 0.4771 and log2 = 0.3010
I'm sure I can figure it out (using logmn = logm + logn), but the answer I got was way off...and r was confusing to find...)
Post by: RuiAce on November 16, 2016, 11:10:08 pm
One more log...You can't treat these log ones like a geometric series. Because if you try putting everything inside one log you get this:
35. Find the sum of 10 terms of the series log3 + log6 + log12 + .... given that log 3 = 0.4771 and log2 = 0.3010
I'm sure I can figure it out (using logmn = logm + logn), but the answer I got was way off...and r was confusing to find...)
log(3*6*12*...)
Notice how it's all multiply. Not add. This is a product, not a sum.
Have to decompose it into
(log 3) + (log 3 + log 2) + (log 3 + 2 log 2) + ... which is an arithmetic series.
Post by: teapancakes08 on November 16, 2016, 11:18:51 pm
You can't treat these log ones like a geometric series. Because if you try putting everything inside one log you get this:
log(3*6*12*...)
Notice how it's all multiply. Not add. This is a product, not a sum.
Have to decompose it into
(log 3) + (log 3 + log 2) + (log 3 + 2 log 2) + ... which is an arithmetic series.
Oh! No wonder it wouldn't divide nicely...
Thanks ^^
Post by: teapancakes08 on November 16, 2016, 11:22:04 pm
(I'm so sorry for flooding the thread with so many of these...)
30. The sum of the first 8 terms of a geometric series 17 times the sum go its first 4 terms. Find the common ratio.
Post by: RuiAce on November 16, 2016, 11:34:45 pm
LAST SERIES. PROMISE.
(I'm so sorry for flooding the thread with so many of these...)
30. The sum of the first 8 terms of a geometric series 17 times the sum go its first 4 terms. Find the common ratio.
Post by: katnisschung on November 17, 2016, 10:03:58 pm
in a geometric sequence
Post by: RuiAce on November 17, 2016, 10:16:03 pm
insert two numbers between 64 and 27 so that the sequence of four numbers is
in a geometric sequence
Post by: f_tan on November 19, 2016, 10:35:00 pm
The rate of change of V with respect to t is given by dV/dt = (2t-1)2. If V=5 when t=1/2, find V when t=3.
Thanks!
Post by: RuiAce on November 19, 2016, 11:17:22 pm
Can anyone help me with this question?
The rate of change of V with respect to t is given by dV/dt = (2t-1)2. If V=5 when t=1/2, find V when t=3.
Thanks!
Post by: Thebarman on November 21, 2016, 01:20:27 pm
Q1) A series is given by: 4 + 8 + 16 + …. . The nth term of this series is given by:
A) 4 + (n - 1) x 4
B) 4 x 2^n
C) 4 x 2^n+1
D) 2^n+1
Why is the answer D?
2) How do you describe the locus of the point, P (x,y) if it has the equation (x-3)^2 = 4(y-3). (worth 1 mark)
3) For the parabola given by the equation below, find:
x^2 + 6x - 8y = -1
i) the co-ordinates of the vertex.
ii) the co-ordinates of the focus. (mainly confused about this)
Thanks!
Post by: RuiAce on November 21, 2016, 01:25:19 pm
Hey guys, couple of questions from a previous exam paper.
Q1) A series is given by: 4 + 8 + 16 + …. . The nth term of this series is given by:
A) 4 + (n - 1) x 4
B) 4 x 2^n
C) 4 x 2^n+1
D) 2^n+1
Why is the answer D?
2) How do you describe the locus of the point, P (x,y) if it has the equation (x-3)^2 = 4(y-3). (worth 1 mark)
3) For the parabola given by the equation below, find:
x^2 + 6x - 8y = -1
i) the co-ordinates of the vertex.
ii) the co-ordinates of the focus. (mainly confused about this)
Thanks!
__________________
This is just a parabola. Clearly, its vertex is at (3,3) and its focal length is a=1. So (bearing in mind its orientation) this is a parabola with focus at (3,4) and directrix y=2
__________________
Post by: Thebarman on November 21, 2016, 07:10:58 pm
How would I solve this? I'm getting everything but the constant right.
Find the equation of the parabola with focus (-1,3), directrix y=-1.
Focus (-4,1) and directrix y=-1.
Thanks again :)
Post by: RuiAce on November 21, 2016, 08:50:02 pm
Thanks!Here is the method.
How would I solve this? I'm getting everything but the constant right.
Find the equation of the parabola with focus (-1,3), directrix y=-1.
Focus (-4,1) and directrix y=-1.
Thanks again :)
Do a free-hand sketch clearly showing the focus and the directrix. Then, figure out the orientation of the parabola (does it concave upwards, downwards or sideways). Then work out the focal length, and finally work out the vertex (which you can write its equation afterwards).
Note: The focal length is the distance from the vertex to the focus.
The (perpendicular) distance from the focus to the directrix is twice the focal length
If it doesn't come out, post up your working
Post by: SSSS on November 22, 2016, 10:10:07 pm
Let R(1/5,-2) be a point on the parabola y^2 = 20x.
(a) Find the equation of the focal chord passing through R.
(b) Find the coordinates of the point Q where this chord cuts the directrix.
(c) Find the area of DOFQ where O is the origin and F is the focus. (d) Find the perpendicular distance from the chord to the point P^-1, -7h.
(e) Hence find the area of DPQR.
Thanks! ::)
Post by: RuiAce on November 22, 2016, 10:25:55 pm
Hey. I'm having issues with this question:Break it down.
Let R(1/5,-2) be a point on the parabola y^2 = 20x.
(a) Find the equation of the focal chord passing through R.
(b) Find the coordinates of the point Q where this chord cuts the directrix.
(c) Find the area of DOFQ where O is the origin and F is the focus. (d) Find the perpendicular distance from the chord to the point P^-1, -7h.
(e) Hence find the area of DPQR.
Thanks! ::)
a) Where is the focus? It is at (0,4).
And then you should realise that this is your classic find the gradient, then sub into y-y1=m(x-x1) problem
b) What is the equation of the directrix? y=-4. Question: WHY is it y=-4
And then what are you doing? You're just finding a point of intersection. So whatever your answer was for part a) just sub y=-4 in.
Check if you made a typo from here on. Because you haven't told us what D is.
Also, for a question like this a freehand diagram (or computer simulation) is highly recommended, if you haven't been provided one.
Post by: SSSS on November 22, 2016, 11:24:35 pm
Here is the method.
Do a free-hand sketch clearly showing the focus and the directrix. Then, figure out the orientation of the parabola (does it concave upwards, downwards or sideways). Then work out the focal length, and finally work out the vertex (which you can write its equation afterwards).
Yeah, I made quite a few typos.
C is actually find area of traiangle OFQ where o origin and f is focus
E is find area of triangle PQR
But after your explanation, it makes sense. I fear locus because I wasn't taught it properly [missed class so whole topic was kind of taught by students]. But I'll try it again. Thanks again!
Note: The focal length is the distance from the vertex to the focus.
The (perpendicular) distance from the focus to the directrix is twice the focal length
If it doesn't come out, post up your working
Post by: RuiAce on November 22, 2016, 11:44:15 pm
If you think about it, all that falls under the locus topic is just the focus and the directrix. Everything else is coordinate geometry.
With that being said, yeah have a go. You're welcome to come back if you remain stuck (or of course a new question), but post your working if you do
Post by: SSSS on November 23, 2016, 07:03:51 am
Post by: Thebarman on November 23, 2016, 04:35:01 pm
Here is the method.
Do a free-hand sketch clearly showing the focus and the directrix. Then, figure out the orientation of the parabola (does it concave upwards, downwards or sideways). Then work out the focal length, and finally work out the vertex (which you can write its equation afterwards).
Note: The focal length is the distance from the vertex to the focus.
The (perpendicular) distance from the focus to the directrix is twice the focal length
If it doesn't come out, post up your working
Thank you so much, everything is good now!
Mind if I ask another question?
The price of shares in a particular company is falling by an average of 2% each day. How many days would it take fo rhte shares to half in value, and after how many days will the shares be worth 25% of their value?
Thanks again :)
Post by: RuiAce on November 23, 2016, 04:44:09 pm
Thank you so much, everything is good now!
Mind if I ask another question?
The price of shares in a particular company is falling by an average of 2% each day. How many days would it take fo rhte shares to half in value, and after how many days will the shares be worth 25% of their value?
Thanks again :)
Post by: katnisschung on November 26, 2016, 04:19:59 pm
the second part is all under a square root....
Post by: RuiAce on November 26, 2016, 04:28:23 pm
find any stationary points on the curve y=(x-3) (4-x)^1/2
the second part is all under a square root....
Post by: katnisschung on November 26, 2016, 04:32:07 pm
Post by: RuiAce on November 26, 2016, 04:43:05 pm
ahaha i differentiated it wrong thanks ruiaceThese things happen haha. When in doubt, use WolframAlpha
Post by: Wales on November 26, 2016, 07:59:52 pm
When you differentiate a Root it ALWAYS :
dy/dx = u'/2root(u)
eg:
SQRT(3x^2+2x+3)
dy/dx = 6x+2/2SQRT(3x^2+2x+3)
=3x+1/SQRT(3x^2+2x+3)
Saves the time writing it all out. Just remember u'/2rootu
:)
Post by: RuiAce on November 26, 2016, 08:14:49 pm
Just a handy tip my tutor told me. Not sure if it's been here yet but
When you differentiate a Root it ALWAYS :
dy/dx = u'/2root(u)
eg:
SQRT(3x^2+2x+3)
dy/dx = 6x+2/2SQRT(3x^2+2x+3)
=3x+1/SQRT(3x^2+2x+3)
Saves the time writing it all out. Just remember u'/2rootu
:)
It's better if you can remember it as a mnemonic or some catchy rhyme
_________________
Personally I never showed working out for any chain rule. I just smashed it in one step for every single thing.
Post by: Happy Physics Land on November 26, 2016, 08:16:23 pm
Just a handy tip my tutor told me. Not sure if it's been here yet but
When you differentiate a Root it ALWAYS :
dy/dx = u'/2root(u)
eg:
SQRT(3x^2+2x+3)
dy/dx = 6x+2/2SQRT(3x^2+2x+3)
=3x+1/SQRT(3x^2+2x+3)
Saves the time writing it all out. Just remember u'/2rootu
:)
I still work everything out through first principles. It's safer that way.
Post by: Wales on November 26, 2016, 08:40:05 pm
It's better if you can remember it as a mnemonic or some catchy rhyme
_________________
Personally I never showed working out for any chain rule. I just smashed it in one step for every single thing.
Yeah the explanation is on point. I personally found it much easier to just remember it, 1 over 2rootu is pretty catchy in my head aha. Each to their own though.
I still work everything out through first principles. It's safer that way.
HUH... Wouldn't that literally take a entire page? I hated First Principals. It was pointless to me and never made any sense why we needed to know :(
Post by: jakesilove on November 26, 2016, 09:21:20 pm
HUH... Wouldn't that literally take a entire page? I hated First Principals. It was pointless to me and never made any sense why we needed to know :(
Yeah. He's weird. Ignore him.
Post by: Happy Physics Land on November 26, 2016, 09:46:03 pm
Yeah. He's weird. Ignore him.
Stfu Jake.
By first principle I meant like I like to differentiate it using the normal way, not using any tricks.
Like Im not talking about the f'(x) = f(x-h) - f(x) / h thing.
Post by: RuiAce on November 26, 2016, 09:50:38 pm
Post by: Wales on November 26, 2016, 10:16:08 pm
Stfu Jake.
By first principle I meant like I like to differentiate it using the normal way, not using any tricks.
Like Im not talking about the f'(x) = f(x-h) - f(x) / h thing.
Bit misleading aye :^)
That aside... Don't ever want to see First Principals again :( As Rui Said, building blocks
Post by: Happy Physics Land on November 26, 2016, 10:52:10 pm
Bit misleading aye :^)
That aside... Don't ever want to see First Principals again :( As Rui Said, building blocks
First Principal is Duncan.
True, dont want to see First Principal again :) .
Post by: jamonwindeyer on November 26, 2016, 11:10:01 pm
dont want to see First Principal again :) .
Differentiation by first principles is revisited in first year math courses 8)
Post by: RuiAce on November 26, 2016, 11:19:12 pm
Differentiation by first principles is revisited in first year math courses 8)It's one of the easier things there though :P
Screw epsilon-delta stuff
Post by: Wales on November 26, 2016, 11:29:54 pm
epsilon-delta stuff
That sounds scary...
Post by: feeah on November 30, 2016, 03:33:15 pm
I'm not sure if this the right thread to take this question to, but I'm looking for some advice. I currently do 14 units, including 2 unit mathematics. Math has always come really easily to me, so I'm doing relatively fine in the subject, but I have no motivation to study at all. Math was my second highest-scoring subject last but, but this year I've already messed up my first exam (which weighed 20% of the internal assessments) because I didn't study at all-- I estimate perhaps a 60-75% mark from it. I've been considering dropping math, primarily because I have no motivation to study, but also because of my low mark for the first exam.However, I'm hesitant to do so because I'm certain that if I do put the effort in, I could get really high marks. Additionally, I'm worried about only having 2 extension subjects as my back-ups, as I'm not doing really well in them, and I'm not sure if I'll even want to keep them throughout the year.
Do you think I should drop math right now, which gives me more time to study for my assignments that are due in the next few weeks, or should I wait a while until I'm sure dropping is the best decision?
Post by: jamonwindeyer on November 30, 2016, 04:20:17 pm
Hi there,
I'm not sure if this the right thread to take this question to, but I'm looking for some advice. I currently do 14 units, including 2 unit mathematics. Math has always come really easily to me, so I'm doing relatively fine in the subject, but I have no motivation to study at all. Math was my second highest-scoring subject last but, but this year I've already messed up my first exam (which weighed 20% of the internal assessments) because I didn't study at all-- I estimate perhaps a 60-75% mark from it. I've been considering dropping math, primarily because I have no motivation to study, but also because of my low mark for the first exam.However, I'm hesitant to do so because I'm certain that if I do put the effort in, I could get really high marks. Additionally, I'm worried about only having 2 extension subjects as my back-ups, as I'm not doing really well in them, and I'm not sure if I'll even want to keep them throughout the year.
Do you think I should drop math right now, which gives me more time to study for my assignments that are due in the next few weeks, or should I wait a while until I'm sure dropping is the best decision?
Hey feeah! Definitely the right spot ;D
Normally I would say that if you are lacking motivation for a subject early in the game, and you have the units to spare, then drop. Motivation only becomes harder to find the further through Year 12 you get, and so starting with none might be an indicator that you aren't in a subject that you are passionate about. That said, it doesn't sound like you dislike the subject, more just that you are in a bit of a rut?
My question to you is this; is Math something you enjoy? Is it something that realistically you will work hard to improve in? Yes, it's nice to say you can, but will you with that many units and that much on your plate?
Don't stress too much about results at the end, stress more about whether you have the time to put work into the subject or not I think ;D
If you think you can put the work in, then hang around in my opinion. Especially if you are iffy on your Extension subjects too ;D just try really hard to put some work into it over the holidays!
Post by: Thebarman on December 02, 2016, 12:47:19 am
The normal of the parabola x^2 = 18y at (-6,2) cuts the parabola again at Q. Find the coordinates of Q.
Thanks again!
Post by: jamonwindeyer on December 02, 2016, 01:45:50 am
Thank you so much for answering my math questions. You guys have been lifesavers! Would you mind helping me out with another question?
The normal of the parabola x^2 = 18y at (-6,2) cuts the parabola again at Q. Find the coordinates of Q.
Thanks again!
Sure! Let's find the equation of that normal. First we'll differentiate to find the gradient:
So that's the gradient of the tangent though, we want the normal!
So we now have a gradient of the normal, and a point it passes through, we use point-gradient form to find the equation:
So now we have the equation of the normal; that normal meets the parabola again at the point Q. How do we find Q? Well, we find it the same as any other point of intersection, by solving \(2x-3y+18=0\) and \(y=\frac{x^2}{18}\) simultaneously. I'll let you take that, but note it will have two solutions. One of them will be \(x=-6,y=2\), but this is the point we already have for the normal. We want the other solution ;D
I hope that helps! :)
Post by: julzzz on December 02, 2016, 06:49:50 pm
1. find the volume of the solid formed when the curve y=(x+5)^2 is rotated about the y-axis from y=1 to y=4
2a. differentiate (2x^2+1)/(3x^2-4)
b. hence evaluate (integral from 0 to 1) of x/(3x^2-4)^2 dx
P.s. for question 2 i can do part a but dont know how to do part b.
3. find the exact area bounded by the parabola y=x^2 and the line y=4-x
4. find the area enclosed between the curves y=(root)x and y=x^3
5. for the shaded region find the area and the volume around the x-axis. The region is of x^2=4ay coloured to (0,a)
These are from maths in focus. cheers for ANY help
Post by: RuiAce on December 02, 2016, 07:06:19 pm
hey doing a few questions... any help would be much apreciated!Hints.
1. find the volume of the solid formed when the curve y=(x+5)^2 is rotated about the y-axis from y=1 to y=4
2a. differentiate (2x^2+1)/(3x^2-4)
b. hence evaluate (integral from 0 to 1) of x/(3x^2-4)^2 dx
P.s. for question 2 i can do part a but dont know how to do part b.
3. find the exact area bounded by the parabola y=x^2 and the line y=4-x
4. find the area enclosed between the curves y=(root)x and y=x^3
5. for the shaded region find the area and the volume around the x-axis. The region is of x^2=4ay coloured to (0,a)
These are from maths in focus. cheers for ANY help
1. You're trying to find a volume with respect to the y axis. Hence you need to make x2 the subject, which takes quite a fair bit of rearranging.
2. What is the answer to 2a? Cause it's bound to be necessary for 2b
3. This is a classic area between the curves question. Sketch the curves, figure out which is the upper and lower curve, and integrate (upper - lower)
4. Same as above, but you need to work out the point of intersections first, which are 0 and 1
5. The region is bounded between x^2=4ay and y=a.
So figure out whether you want to integrate with respect to dx, or dy. Then show your progress.
(Note that if you integrate with respect to x you have to consider the area of a rectangle. If you do it w.r.t. y then you don't)
Post by: J.B on December 03, 2016, 10:33:25 am
The start of the question:
The speed of a train was recorded at intervals of one minute. The times, in minutes, and the corresponding speeds v, in kilometres per hour, are listed in the following table.
And i have attached the second part as a photo.
Thank you.
Post by: RuiAce on December 03, 2016, 10:52:33 am
Hi, I was wondering if anyone could help me with this question?
The start of the question:
The speed of a train was recorded at intervals of one minute. The times, in minutes, and the corresponding speeds v, in kilometres per hour, are listed in the following table.
And i have attached the second part as a photo.
Thank you.
It would make sense as to if part (ii) (or later) is a Simpson's rule or Trapezoidal rule computation. Because that table is only barely useful for part (i) but would be very very useful for a part (ii)
_____________________________
Consider this analogy: Suppose the particle starts at x=1. It travels at a constant speed of 15km/hr. Then 1 hour later, it will be at x=16
Hence, the distance travelled is 15.
Now suppose the particle starts at x=-1. It travels at a constant speed of 15km/hr. Then 1 hour later, it will be at x=14
Hence, the distance travelled is STILL 15.
Which is the point. The definite integral allows us to IGNORE any initial conditions, provided we have enough information on the velocity of the particle.
Essentially, what we are doing is integrating the velocity, AND working our way around the initial conditions.
The explanation is hard to juggle without a concrete example (this is just a table of values). I've tried my best but do voice any confusion
Post by: J.B on December 03, 2016, 11:32:56 am
It would make sense as to if part (ii) (or later) is a Simpson's rule or Trapezoidal rule computation. Because that table is only barely useful for part (i) but would be very very useful for a part (ii)
_____________________________
Consider this analogy: Suppose the particle starts at x=1. It travels at a constant speed of 15km/hr. Then 1 hour later, it will be at x=16
Hence, the distance travelled is 15.
Now suppose the particle starts at x=-1. It travels at a constant speed of 15km/hr. Then 1 hour later, it will be at x=14
Hence, the distance travelled is STILL 15.
Which is the point. The definite integral allows us to IGNORE any initial conditions, provided we have enough information on the velocity of the particle.
Essentially, what we are doing is integrating the velocity, AND working our way around the initial conditions.
The explanation is hard to juggle without a concrete example (this is just a table of values). I've tried my best but do voice any confusion
Thank you.
I'm still a bit confused. It's probably really stupid, but I don't understand why v=dx/dt. Like I understand that velocity = distance/time. Is this just the same thing? And then I don't understand how you bring the x to the other side.
Sorry this is probably really stupid!
Post by: RuiAce on December 03, 2016, 11:35:01 am
Thank you.Speed = distance/time, also the absolute value of the velocity
I'm still a bit confused. It's probably really stupid, but I don't understand why v=dx/dt. Like I understand that velocity = distance/time. Is this just the same thing? And then I don't understand how you bring the x to the other side.
Sorry this is probably really stupid!
Not sure what you mean by bring x to the other side
Post by: J.B on December 03, 2016, 11:37:34 am
Speed = distance/time, also the absolute value of the velocity
Not sure what you mean by bring x to the other side
What topic is it taught in? As I have only done Integration, Exp and Logs, Geometry 2, and Trig Functions?
Post by: RuiAce on December 03, 2016, 11:40:36 am
The last topic. Applications of calculus.
What topic is it taught in? As I have only done Integration, Exp and Logs, Geometry 2, and Trig Functions?
Post by: Happy Physics Land on December 03, 2016, 11:40:47 am
What topic is it taught in? As I have only done Integration, Exp and Logs, Geometry 2, and Trig Functions?
It will be taught later in the 2 unit course in physical application of calculus to the real world. But it should be general knowledge that:
Post by: jamonwindeyer on December 03, 2016, 11:41:56 am
It will be taught later in the 2 unit course in physical application of calculus to the real world. But it should be general knowledge that:
Only for Physicists is this general knowledge ;)
Post by: J.B on December 03, 2016, 11:42:46 am
Post by: anotherworld2b on December 06, 2016, 01:29:40 am
In regards to q12 i was particularly confused on how to do parts a and b
Post by: jamonwindeyer on December 06, 2016, 02:18:11 am
Hi i was wondering if i could please get help with these two questions. I have attempted to do q8 but im not sure whether i am right.
In regards to q12 i was particularly confused on how to do parts a and b
Hey!
For Question 8, I think parts of your working are correct. Let me give you a hand interpreting!
Part A: The two lines are each parallel to \(x+2y=0\). Now if you rearrange this line, you get \(y=\frac{-x}{2}\), you did this yourself. So this line has a gradient of \(m=\frac{-1}{2}\). The lines L1 and L2 just need to be any lines with a gradient of this, aka, any line of the form \(y=\frac{-x}{2}+c\). Just pick two random values for \(c\) ;D
Part B: Here, it is now easier to consider two general equations in gradient intercept form. Before we do though, the fact that the lines are perpendicular means that we can set the two gradients to be \(m\), and \(\frac{-1}{m}\). This is because, when two lines are perpendicular, \(m_1=\frac{-1}{m_2}\). So we can write the lines like this:
L1: \(y=mx+b_1\)
L2: \(y=\frac{-x}{m}+b_2\)
Note that the two intercepts are different, at least for now. In fact, both lines meet at the point (0, 4), which sits on the y-axis. So in actuality, both lines have a y-intercept of 4, so the equations are now:
L1: \(y=mx+4\)
L2: \(y=\frac{-x}{m}+4\)
Both of our conditions are now satisfied, so you can pick any value of \(m\) you like and substitute it here to get your answer :)
Part C: Just do a cheeky answer; any two parallel lines will do. Do y=0 and y=1, those will definitely not intersect! ;D
For Question 12 your equations look fine for Part B, that answer is correct! For Part A, your equation should be \(y=80x+100\), the $80 is the half-hourly rate and the $100 is the constant call out fee. Be careful, you have defined \(x\) as half an hour here but one hour in Part B. This is okay as long as you are aware of it!
Try subbing with this new equation, you should get $740.00 as your answer for Part A ;D
Post by: anotherworld2b on December 07, 2016, 12:28:14 am
I was wondering what would be the best way to do q12 d)?
Hey!
For Question 8, I think parts of your working are correct. Let me give you a hand interpreting!
Part A: The two lines are each parallel to \(x+2y=0\). Now if you rearrange this line, you get \(y=\frac{-x}{2}\), you did this yourself. So this line has a gradient of \(m=\frac{-1}{2}\). The lines L1 and L2 just need to be any lines with a gradient of this, aka, any line of the form \(y=\frac{-x}{2}+c\). Just pick two random values for \(c\) ;D
Part B: Here, it is now easier to consider two general equations in gradient intercept form. Before we do though, the fact that the lines are perpendicular means that we can set the two gradients to be \(m\), and \(\frac{-1}{m}\). This is because, when two lines are perpendicular, \(m_1=\frac{-1}{m_2}\). So we can write the lines like this:
L1: \(y=mx+b_1\)
L2: \(y=\frac{-x}{m}+b_2\)
Note that the two intercepts are different, at least for now. In fact, both lines meet at the point (0, 4), which sits on the y-axis. So in actuality, both lines have a y-intercept of 4, so the equations are now:
L1: \(y=mx+4\)
L2: \(y=\frac{-x}{m}+4\)
Both of our conditions are now satisfied, so you can pick any value of \(m\) you like and substitute it here to get your answer :)
Part C: Just do a cheeky answer; any two parallel lines will do. Do y=0 and y=1, those will definitely not intersect! ;D
For Question 12 your equations look fine for Part B, that answer is correct! For Part A, your equation should be \(y=80x+100\), the $80 is the half-hourly rate and the $100 is the constant call out fee. Be careful, you have defined \(x\) as half an hour here but one hour in Part B. This is okay as long as you are aware of it!
Try subbing with this new equation, you should get $740.00 as your answer for Part A ;D
Post by: Jakeybaby on December 07, 2016, 12:45:56 am
thank you very much for your help :DPlease correct me if I am wrong, but your equation for Bill seems incorrect to me personally.
I was wondering what would be the best way to do q12 d)?
If Bill charges a call-out fee of $100, initially, the cost will be $100 after 0 half-hours of work. Therefore, the y-intercept of your equation should be $100.
As he charges $80 per half-hour, if we allow x to be the number of hours worked, therefore:
y = 80(2x) + 100
We shall use 2x, as there are obviously two half hours in one full hour, meaning that per hour, Bill costs $160.
Therefore: y = 160x + 100
So for a):
For a job that takes exactly 4 hours,
y = 160(4) + 80
B is correct.
c)
A job that takes 3 hours and 20 minutes, this means that the number of hours worked is 3.33333.... hours, (i.e. 10/3 hours), as 20 minutes is one-third of an hour.
Therefore:
For Ian: y = 180(10/3)
Bill: y = 160(10/3) + 100
You can justify which one is cheaper for the job.
d)
I would find the point on the lines where they intersect, this shows you the length of job where they both cost the same amount of money.
Sketching the two equations with a graphics calculator will be best.
You can see both the point of intersection, one employee is cheaper than the other, and then after the point of intersection, the one that was cheaper, is now more expensive for the job.
(Hint: You need to state the domain (hours) of the job needed for Bill to be cheaper than Ian.)
Post by: jamonwindeyer on December 07, 2016, 12:51:34 am
thank you very much for your help :D
I was wondering what would be the best way to do q12 d)?
Just a warning that D is more complicated than what is shown above, it's a tricky one. Thanks for your post Jake! ;D
You could use an inequality, but the different way that the things change is tricky. To do it, you would have to consider both men in terms of their price per hour. That turns Bill's equation into \(y=100+160x\). Now, \(x\) represents hours for both men. So we want when Bill is cheaper:
Indeed, at any time above 5 hours, Bill costs less than Ian (you could check this with substitution). However, since Ian charges hourly and Bill charges half hourly, you have some extra options.
Consider a simple example.
For a single hour, Bill costs $260.00 ($100.00 plus $80.00 each for two half hour periods) and Ian costs $180.00. For anything above that, both plumbers have a price jump, since they charge for any part of an hour/half an hour. So into the next half hour, Ian charges $360.00 and Bill charges $340.00; Bill is cheaper between 1 and 1.5 hours.
However, say we jump once more to a period between 1.5-2 hours. Ian's price would stay the same, but Bill's would jump to $420.00; he is now more expensive. Once we exceed 2 hours, Bill jumps to $500.00 and Ian jumps to $540.00, Bill is again cheaper. Exceed two and a half, and Bill jumps to $580 and is again more expensive.
So, the full answer is that Bill is cheaper for any work exceeding 5 hours, or, any work exceeding 1 hour that is an odd number of half hour periods :) honestly, the intuitive approach is best here, because this is quite a tricky little problem :)
Post by: anotherworld2b on December 07, 2016, 12:34:03 pm
I am not sure how to find the exact x intercepts without using a calculator.
I am also not sure how to find the equation of a parabola given the diagram. There are 3 different forms? :-*
Post by: jamonwindeyer on December 07, 2016, 01:04:45 pm
Hi i was wondering if i could get help with these questions please.
I am not sure how to find the exact x intercepts without using a calculator.
I am also not sure how to find the equation of a parabola given the diagram. There are 3 different forms? :-*
The two questions you have about finding x-intercepts can be answered with the quadratic formula:
You may need to expand the quadratic first (the first question), but from there, just substituting into this formula is all you need! Do you learn it in WACE?
As for determining a formula given a diagram, you have a few options. In this case we have intercepts, so it is probably best to think of the parabola like this:
It needs to be in this form, because the roots need to be -1 and 3 (from the diagram). We can only multiply by some constant \(a\), so if we expand and then differentiate:
Now we need our turning point to be at (1,7). The first derivative definitely allows this, but we need the equation to yield \(y=7\) when \(x=1\);
So the equation of the parabola is \(y=-10(x+1)(x-3)\) ;D
There are heaps of ways to do this question, this is just my approach. I'm not sure what forms you learn, this is a way to do it from scratch :)
Post by: RuiAce on December 07, 2016, 01:15:43 pm
Hi i was wondering if i could get help with these questions please.
I am not sure how to find the exact x intercepts without using a calculator.
I am also not sure how to find the equation of a parabola given the diagram. There are 3 different forms? :-*
Post by: anotherworld2b on December 09, 2016, 04:44:54 pm
I am not sure how to draw the diagrams :-*
Post by: katnisschung on December 10, 2016, 05:19:59 pm
find the two numbers whose sum is 28 and whose product is a maximum
Post by: RuiAce on December 10, 2016, 05:28:16 pm
how to do this without guess and check
find the two numbers whose sum is 28 and whose product is a maximum
Incidentally, the correct answer is the answer where a=b. It can be proven that in general, if the perimeter of any rectangle is fixed, the area is maximised when the rectangle is a square. This is just the geometry put into numbers only.
Post by: jamonwindeyer on December 10, 2016, 05:42:05 pm
Could i please get help with these two questions?
I am not sure how to draw the diagrams :-*
Hey! I've popped the diagrams below for you; let me know how you go! :)
Images
(http://i.imgur.com/I0LAMyd.jpg)
Post by: Blissfulmelodii on December 10, 2016, 05:43:49 pm
how to do this without guess and check
find the two numbers whose sum is 28 and whose product is a maximum
Edit: This was deleted, but I've legit never seen this question done this way before. I love it, and I think everyone will benefit from seeing it. Bliss, if you don't want it here, feel free to re-delete but I think its great! Doesn't matter that Rui already answered ;D
I'm not sure how to write equations on here so I hope you will understand my working out.
Let x be the first number. The sum of the two numbers is 28 so 28-x is an expression for the second number.
Let y be the product of the two numbers
Then you will end up with an equation like this y=x(28-x)
Expand it: y=28x-x^2
Since it is a quadratic equation, it will graph a concave down parabola which means that the vertex will be the maximum value.
If the vertex of a parabola is the point (h,k) then h=-b/2a so you just have to substitute the 'a' and 'b' values from the previous equations
h= -28/2(-1)
h= 14
substitute this value into the quadratic to find the y-value which should come out as 196 and you end up with the vertex of the parabola having coordinates of (14,196)
Now if you go back to the start we created an expression for the second number with was 28-x. Sub in x=14 and you end up with the second number also being 14. Hence the two numbers are 14 and 14
Post by: RuiAce on December 10, 2016, 06:35:21 pm
Edit: This was deleted, but I've legit never seen this question done this way before. I love it, and I think everyone will benefit from seeing it.Yeah it's pretty standard. Just not common.
Most people vouch for the calculus approach because it's what they're used to. What they're used to is what works universally, i.e. for every question of the sort.
And because I don't know how HSC examiners mark, I've just been using it as well. I would've used this approach, if I had more reassurance with it, HEAPS more times in my life. It's clearly faster
The important thing, however, is to explicitly state the concavity (as Bliss did). If you don't state the concavity, then you don't justify why the turning point offers a maximum and I would deduct one mark for the working out.
Post by: Blissfulmelodii on December 10, 2016, 07:00:28 pm
Edit: This was deleted, but I've legit never seen this question done this way before. I love it, and I think everyone will benefit from seeing it. Bliss, if you don't want it here, feel free to re-delete but I think its great! Doesn't matter that Rui already answered ;D
No worries at all. I only deleted it because I had noticed afterwards that Rui had answered the question already so I figured mine was irrelevant hahaha
Post by: anotherworld2b on December 10, 2016, 08:01:36 pm
I am still a bit confused on how interpret info from a question.
I was also wondering if i could get help answering this question. I have tried to figure out the answer but i keep getting the wrong answer. :-\
Quote from: jamonwindeyer links=topic=164548.msg923580#msg923580 date=1481352125
Hey! I've popped the diagrams below for you; let me know how you go! :)Images(http://i.imgur.com/I0LAMyd.jpg)
Post by: jamonwindeyer on December 10, 2016, 11:47:10 pm
Could you explain the process of how you drew the diagrams please?
I am still a bit confused on how interpret info from a question.
I was also wondering if i could get help answering this question. I have tried to figure out the answer but i keep getting the wrong answer. :-\
The first one (Q6) is a standard angles of elevation question. Remember, an angle of elevation is the angle formed by your eye line when you look up at something. So at point A, that angle is 40 degrees. Then we move 4 metres closer to the tree that the ball in (as represented by that vertical line on the right), and the angle becomes 45 degrees. That's at the point B in the diagram ;D
The second one is harder (Q5); the line on the top represents the flight of the plane in two halves. First, the question identifies a point where the angle of depression is 40 degrees. That is the top left vertex of that bottom triangle. The plane then flies until it is directly above the farmhouse; that forms the right angled triangle at the bottom. After that it elevates at an angle of 15 degrees, that's where the line slants upwards into the right hand side of the diagram!
Try to work through the information one piece at a time. Draw the diagrams yourself, from scratch, piece by piece and see if you get something similar to mine! Ensure you have tried a few easier 'angles of elevation/depression' questions before tackling these, if you are having trouble :)
For that question you posted, do you learn the formula for finding the area of a minor segment of a circle? That will help me answer it in a way that makes sense for you ;D
Post by: katnisschung on December 11, 2016, 11:21:03 am
Post by: RuiAce on December 11, 2016, 11:23:48 am
stuck on the second part of this question
Post by: katnisschung on December 11, 2016, 11:32:28 am
Post by: RuiAce on December 11, 2016, 11:44:02 am
what de.. im lost
Post by: anotherworld2b on December 11, 2016, 12:55:57 pm
Would this diagram(below) be right?
In regards to the question i posted prior i have learnt the formula for finding tha area for a minor segement of a circle.
I also wanted to ask for this question attached below I'm not sure what I'm supposed to find
The first one (Q6) is a standard angles of elevation question. Remember, an angle of elevation is the angle formed by your eye line when you look up at something. So at point A, that angle is 40 degrees. Then we move 4 metres closer to the tree that the ball in (as represented by that vertical line on the right), and the angle becomes 45 degrees. That's at the point B in the diagram ;D
The second one is harder (Q5); the line on the top represents the flight of the plane in two halves. First, the question identifies a point where the angle of depression is 40 degrees. That is the top left vertex of that bottom triangle. The plane then flies until it is directly above the farmhouse; that forms the right angled triangle at the bottom. After that it elevates at an angle of 15 degrees, that's where the line slants upwards into the right hand side of the diagram!
Try to work through the information one piece at a time. Draw the diagrams yourself, from scratch, piece by piece and see if you get something similar to mine! Ensure you have tried a few easier 'angles of elevation/depression' questions before tackling these, if you are having trouble :)
For that question you posted, do you learn the formula for finding the area of a minor segment of a circle? That will help me answer it in a way that makes sense for you ;D
Post by: jamonwindeyer on December 11, 2016, 02:25:51 pm
For q5. How did you get 15 degrees?
Would this diagram(below) be right?
In regards to the question i posted prior i have learnt the formula for finding tha area for a minor segement of a circle.
I also wanted to ask for this question attached below I'm not sure what I'm supposed to find
15 degrees comes from the question itself, that looks okay to me! :)
For the circles one, you are finding two minor segments. The minor segment in circle 1 and the minor segment in circle 2, the formula should be something like this:
The radii are in the question; you've found the angles already! What this formula does is takes the area of the sector, and subtracts the area of the triangle, leaving only the area of the minor segment. So you don't need to find the areas of the triangles separately like you did in your earlier working; if you know this formula, you can just substitute it into here!
Unfortunately I can't see that diagram for the last question properly, too small :P
Post by: katnisschung on December 11, 2016, 08:29:10 pm
stuck on finding the third equation for this word problem
The sum of the dimensions of a box with a square base is 60cm.
Find the dimensions that will give the box a maximum volume
the two equations i have are
l+b+h=60
SA= 2lb+2lh+2bh
but there are 3 unknowns so presumably i need another one.....
sighs
Post by: RuiAce on December 11, 2016, 08:38:24 pm
hi again!The base is a square. Two sets of sides are equal (l=b is the third equation)
stuck on finding the third equation for this word problem
The sum of the dimensions of a box with a square base is 60cm.
Find the dimensions that will give the box a maximum volume
the two equations i have are
l+b+h=60
SA= 2lb+2lh+2bh
but there are 3 unknowns so presumably i need another one.....
sighs
And remember that you're interested in the volume. Not surface area.
Post by: katnisschung on December 11, 2016, 08:39:42 pm
im such an idiot :P
Post by: jamonwindeyer on December 11, 2016, 09:19:59 pm
ahaha thanks ruiace
im such an idiot :P
Definitely not! We're glad we can help you out too, don't ever feel less intelligent by asking questions. It is only unintelligent to keep questions to yourself :)
Post by: anotherworld2b on December 12, 2016, 01:27:25 pm
I was able to do q5 a and b but i wasnt able to do part c. Could i get some help with it please?
I reuploaded the question from before hopefully its clearer. I am not sure what i need to find
I also wanted to ask in exams is there a particular way to write the domain and range of a function?
15 degrees comes from the question itself, that looks okay to me! :)
For the circles one, you are finding two minor segments. The minor segment in circle 1 and the minor segment in circle 2, the formula should be something like this:
The radii are in the question; you've found the angles already! What this formula does is takes the area of the sector, and subtracts the area of the triangle, leaving only the area of the minor segment. So you don't need to find the areas of the triangles separately like you did in your earlier working; if you know this formula, you can just substitute it into here!
Unfortunately I can't see that diagram for the last question properly, too small :P
Post by: asd987 on December 15, 2016, 02:02:47 pm
can someone explain how to do part iii of Q15c of the 2016 hsc exam. thanks
Post by: RuiAce on December 15, 2016, 02:22:06 pm
Hi,
can someone explain how to do part iii of Q15c of the 2016 hsc exam. thanks
Post by: jamonwindeyer on December 15, 2016, 06:08:56 pm
thank you for your help
I was able to do q5 a and b but i wasnt able to do part c. Could i get some help with it please?
I reuploaded the question from before hopefully its clearer. I am not sure what i need to find
I also wanted to ask in exams is there a particular way to write the domain and range of a function?
Sorry I missed this! For that reuploaded question, draw the lines BF and AG. Those lines will intersect at some point, call it X. You want angle AXF! Hint, you'll only need the dimensions of the rectangle closest to us to get this question right ;D
For Part C of that question, good start on finding the distance travelled! What you need to consider is the bigger, non right angled triangle, formed by the points F, Farmhouse and End on your diagram! The angle on the left of the triangle, is 90+15=105 degrees. You are going to need to do some work on that triangle. Do the following:
1- Find the distance between the plane and the farmhouse using the cosine rule
2- Find the angle in the bottom left corner of the triangle, the one formed at the farmhouse.
Then, once you have that smaller angle, the angle of elevation you want is just the complementary angle to that, so just subtract 90 degrees ;D
That's a rough process, let me know how you go!
For domain and range, you can use either inequalities: \(x>5, y\le7\), or set notation, \(\left(5, \infty\right), \left(-\infty, 7\right]\), I'm not sure what is preferred in WACE but for NSW it is the former! ;D
Post by: asd987 on December 17, 2016, 05:49:45 pm
Post by: Jakeybaby on December 17, 2016, 06:00:50 pm
Hi, can i get some help with part iii of the 2015 hsc Q15b\begin{equation}\text{ } \\ 15~{b}~\text{iii} \\ {\frac{AB}{DF}}={\frac{AC}{DC}}~\text{(corresponding sides)} \\ \text{But }{D}~\text{is the midpoint of }AC \\ \therefore~{\frac{AB}{DF}}~=~{\frac{2}{1}} \\ AB~=~2DF \\ \therefore~AE~+~EB~=~2DF~① \\ \text{ } \\ \text{Now } \\ EB~=~EF~{\left({e}{q}{u}{a}{l}~\right.}\text{sides in Isos Triangle)} \\ EB~=~DF~+~{E}\Delta{B}~=~DF~+~AE~{\left({\operatorname{sin}{{c}{e}}}~AE~=~ED \right)} \\ \therefore~DF~=~EB~-~AE~② \\ \text{ } \\ \text{From }①~\text{and }② \\ AE~+~EB~=~2{\left(EB-AE \right)} \\ AE~+~EB~=~2EB~-~2AE \\ 3AE~=~EB \\ \text{ }\end{equation}
Post by: RuiAce on December 17, 2016, 06:08:28 pm
\begin{equation}\text{ } \\ 15~{b}~\text{iii} \\ {\frac{AB}{DF}}={\frac{AC}{DC}}~\text{(corresponding }\sigma\delta{s} \\ \text{But }{D}~\text{is the midpoint of }AC \\ \therefore~{\frac{AB}{DF}}~=~{\frac{2}{1}} \\ AB~=~2DF \\ \therefore~AE~+~EB~=~2DF~① \\ \text{ } \\ \text{Now } \\ EB~=~EF~{\left({e}{q}{u}{a}L~\right.}\text{sides in Isos Triangle)} \\ EB~=~DF~+~{E}\Delta{B}~=~DF~+~AE~{\left({\operatorname{sin}{{c}{e}}}~AE~=~ED \right)} \\ \therefore~DF~=~EB~-~AE~② \\ \text{ } \\ \text{From }①~\text{and }② \\ AE~+~EB~=~2{\left(EB-AE \right)} \\ AE~+~EB~=~2EB~-~2AE \\ 3AE~=~EB \\ \text{ }\end{equation}This is some...interesting...LaTeX... :o
Post by: Jakeybaby on December 17, 2016, 06:47:45 pm
This is some...interesting...LaTeX... :oI just used FX Equation 5 and then copied as LaTeX haha, as I don't know how to write in LaTeX myself :P
Post by: Yasminpotts1105 on December 23, 2016, 11:42:56 am
Post by: RuiAce on December 23, 2016, 11:51:39 am
"Find the radius and height, correct to 2 decimal places, of a cylinder that holds 200cm^3, if its surface area is to be a minimum."
Post by: Jakeybaby on December 23, 2016, 12:00:47 pm
I'm always too slow haha
Post by: Yasminpotts1105 on December 29, 2016, 10:39:56 am
Post by: RuiAce on December 29, 2016, 10:41:30 am
How do we differentiate something like y = 1/(x^2 + 2) or even something with just x in the numerator and an integer in the denominator?Do you mean an integer in the numerator and x in the denominator?
Post by: Yasminpotts1105 on December 29, 2016, 10:43:55 am
Post by: RuiAce on December 29, 2016, 10:57:09 am
Post by: Rathin on January 07, 2017, 11:40:17 pm
Post by: RuiAce on January 07, 2017, 11:50:34 pm
Find the area of the region bounded by the curve y=(x+1)(x-1)(x-3), the x axis and the ordinates at x=0 and x=2.
(http://i.imgur.com/AexlkjP.png)
(note: There IS symmetry going on, and this can be exploited. However, this should be done so carefully.)
Post by: Rathin on January 07, 2017, 11:58:04 pm
(http://i.imgur.com/AexlkjP.png)
(note: There IS symmetry going on, and this can be exploited. However, this should be done so carefully.)
Oh I see, I was trying to integrate from 0 to 2. Why do we have to separate the two integrals? and.. for the 1st half of the integral why is the boundary from 0 to 1 and not -1 to 0?
Post by: RuiAce on January 08, 2017, 12:02:52 am
Oh I see, I was trying to integrate from 0 to 2. Why do we have to separate the two integrals? and.. for the 1st half of the integral why is the boundary from 0 to 1 and not -1 to 0?To cater for the fact that a region is below the x-axis.
If we simply integrate from 0 to 2, we treat the area below the x-axis as what it ACTUALLY is - a negative area. This negative area will cancel out with the positive area from 0 to 1.
We don't want to treat it as a negative area; we want to keep it positive. Hence we split off the integral, and put an absolute value bracket around what's negative to make it positive again. This should be a procedure you're familiar with from classwork.
If it were from -1 to 0, you'd be going to the left of the blue line (the y-axis). That's not asked anywhere in the question because it is not a part of the region you're interested in.
Post by: jamonwindeyer on January 08, 2017, 12:04:03 am
Oh I see, I was trying to integrate from 0 to 2. Why do we have to separate the two integrals? and.. for the 1st half of the integral why is the boundary from 0 to 1 and not -1 to 0?
The integral is separated because one region sits above the x-axis, the other below. We need to take the absolute value of the integral representing the region below the x-axis, since that part of the integral will turn out negative. We do this for all regions sitting below the axis to make sure our area is correct ;D try finding the area from 1 to 2 without that absolute value, you'll get a negative! And that throws out our answer if we leave the integral undivided :)
As for why the limits of integration are 0 to 1, that is because the question specifies between the ordinates x=0 and x=2, we want the area on the right hand side of the y-axis!
Does that help? If not I can put together another diagram to show you what I mean? ;D
Edit: Rui won this one, I'll leave it here anyways, hearing it said two ways might help :)
Post by: RuiAce on January 08, 2017, 12:05:17 am
Edit: Rui won this one, I'll leave it here anyways, hearing it said two ways might help :)8)
Post by: Rathin on January 08, 2017, 12:09:09 am
Post by: RuiAce on January 08, 2017, 12:10:28 am
Thanks for the reply guys, just to clarify if the areas are on different sides of the x axis we have to split them up? and @rui I dont really have class notes haha..I am self learning this topic before school starts.Oh fair enough.
And basically yes to your question
Post by: Rathin on January 08, 2017, 02:11:14 pm
Post by: RuiAce on January 08, 2017, 02:14:06 pm
This is kinda a half rant/confusion..but whats the point of trapezoidal rule and simpson's rule when you can integrate the exact area directly? It just a very tedious process as the subintervals increase..Well, if you had to find the area of a pond, you'd probably take forever trying to approximate it with a function when you can just use one of those numeric methods to give an estimate.
Post by: jamonwindeyer on January 08, 2017, 02:34:32 pm
This is kinda a half rant/confusion..but whats the point of trapezoidal rule and simpson's rule when you can integrate the exact area directly? It just a very tedious process as the subintervals increase..
And further, even when you are given a function, some integrals are just plain disgusting. Sometimes these methods can be less tedious than doing the integral; obviously not the case for polynomials, but much stranger functions than that exist! So as long as you are happy to put up with a little inaccuracy (and have a solid understanding of why that inaccuracy exists and how large it should be) then the methods can prove really handy ;D
Post by: RuiAce on January 08, 2017, 02:39:16 pm
Post by: Rathin on January 08, 2017, 02:43:45 pm
Post by: jamonwindeyer on January 08, 2017, 02:44:58 pm
Is it mostly used for non-elementary integrals who don't have a primitive?
It can be, but that integral Rui provided has a primitive (right Rui?), it would just be flat disgusting to actually do that integral ;D but I mean yeah, pretty much! In the HSC it is mostly associated to physical scenarios where using a series of measurements to make an approximation is easier than fitting a function to the phenomena ;D
Post by: RuiAce on January 08, 2017, 02:46:35 pm
Is it mostly used for non-elementary integrals who don't have a primitive?Well yes, because if it has a primitive then provided we can assume the Fundamental Theorem of Calculus we have no reason to not use it.
It's important regardless though. Just not taught in high school in a useful manner.
There is a whole variety of things that can be known as "numeric methods". Another is the estimation of roots in MX1.
It can be, but that integral Rui provided has a primitive (right Rui?), it would just be flat disgusting to actually do that integral ;D but I mean yeah, pretty much! In the HSC it is mostly associated to physical scenarios where using a series of measurements to make an approximation is easier than fitting a function to the phenomena ;DIt does but it can't be expressed in terms of the elementary functions. According to Wolfram, it's what's called the Fresnel integral. Much like how this one is handled with the error function
Post by: jamonwindeyer on January 08, 2017, 02:49:45 pm
It does but it can't be expressed in terms of the elementary functions. According to Wolfram, it's what's called the Fresnel integral. Much like how this one is handled with the error function
I miss when elementary functions were just functions, and non-elementary functions were just magic :(
Post by: RuiAce on January 08, 2017, 02:51:05 pm
I miss when elementary functions were just functions, and non-elementary functions were just magic :(Ahahahahahaha well you can express them as a Taylor series if you want? :P
Post by: Rathin on January 08, 2017, 05:26:50 pm
Post by: RuiAce on January 08, 2017, 05:27:37 pm
I have found some good non elementary integrals. What do you think is a easier method? Trapezoidal rule or Simpson's rule?Um, easier is relative to each person. Which formula do you find easier?
Also, these are indefinite integrals. Numeric methods are for definite integrals.
Post by: Rathin on January 08, 2017, 05:35:27 pm
Um, easier is relative to each person. Which formula do you find easier?
Also, these are indefinite integrals. Numeric methods are for definite integrals.
Not too sure which one I find easy tbh
Oh and yeah..ill just add my own boundaries haha
Post by: RuiAce on January 08, 2017, 05:37:25 pm
Not too sure which one I find easy tbhWell in that case the answer to your question is probably that they're equally 'easy' :P
Oh and yeah..ill just add my own boundaries haha
Post by: Rathin on January 12, 2017, 12:10:01 pm
http://imgur.com/14AQTWj
Post by: RuiAce on January 12, 2017, 12:14:41 pm
Help with q4 and 5
http://imgur.com/14AQTWj
Post by: Rathin on January 12, 2017, 12:44:59 pm
For q4 I got the answer which you got for q5..why did you split the two areas in q4 and not q5?
Post by: RuiAce on January 12, 2017, 01:17:07 pm
For q4 I got the answer which you got for q5..why did you split the two areas in q4 and not q5?Because in Q5 the area was just sandwiched between two areas. BOTH of these areas ranged from 0 to 3
In Q4, one of the curves was ONLY above 0 to 1, whereas the other curve was ONLY above 1 to 3.
The key point is that the integral measures the area under the curve, but only of the SPECIFIED x-coordinates
I think maths in focus addresses this little section pretty well with examples
Post by: Rathin on January 12, 2017, 01:41:52 pm
Because in Q5 the area was just sandwiched between two areas. BOTH of these areas ranged from 0 to 3
In Q4, one of the curves was ONLY above 0 to 1, whereas the other curve was ONLY above 1 to 3.
The key point is that the integral measures the area under the curve, but only of the SPECIFIED x-coordinates
I think maths in focus addresses this little section pretty well with examples
Oh I see..the shaded area is for q5.
Post by: Rathin on January 12, 2017, 03:46:30 pm
Post by: jamonwindeyer on January 12, 2017, 04:02:18 pm
It's the little pocket underneath both curves, the weird triangular thing ;D (Rui provided that integral earlier, its correct)
Post by: RuiAce on January 12, 2017, 04:04:10 pm
Actually I am really confused..I dont even know what area I am finding for q4..I double checked. It's as I said - that weird triangle-like thing at the very bottom.
Look carefully. That tiny region is bounded by all four things at once: f(x), g(x), x-axis, x=3
Question 4 is a compound region:I doubted myself briefly :P whoops
It's the little pocket underneath both curves, the weird triangular thing ;D (Rui provided that integral earlier, its correct)
Post by: jamonwindeyer on January 12, 2017, 04:06:29 pm
(http://i.imgur.com/bNguLHe.png)
The answer is \(1\text{ units}^2\) ;D
Post by: Rathin on January 12, 2017, 04:44:46 pm
A diagram for clarity:
(http://i.imgur.com/bNguLHe.png)
The answer is \(1\text{ units}^2\) ;D
Oh I see..and for q5..why is the area between the two curves the shaded area?
Post by: jamonwindeyer on January 12, 2017, 04:47:28 pm
Oh I see..and for q5..why is the area between the two curves the shaded area?
That's because in that case, they don't specify the x-axis. That's the only difference! It's a trick of the wording ;D
Post by: Rathin on January 16, 2017, 11:57:55 am
Post by: jamonwindeyer on January 16, 2017, 12:14:02 pm
Is my reasoning sufficient?
This is a weird question - Given that the bounds are decided for you, I think what you actually need to prove is which curve is on top for \(1\le x\le4\). Why is it not the other way around in the integral? I'm pretty sure that is what you would need to show. Finding the point of intersection doesn't change anything, because the area is still only for \(1\le x\le4\), intersection or not :)
Not that this is hard to prove - Clearly for those x values one curve is always above the x-axis and one is always below, and so, one is clearly greater than the other :)
(You won't get this question in the HSC, it's a little awkward imo)
Post by: Rathin on January 16, 2017, 03:07:53 pm
Post by: jamonwindeyer on January 16, 2017, 03:37:45 pm
16c pls
That is simultaneous equations, use the coordinates at C and D to create your system:
Now the fact that we have \(-60^n\) in our equations means that \(n\) needs to be an integer (we'd be beyond the bounds of the HSC course if it wasn't). We can approach this several ways, but if we consider and check some of the easy possible answers (\(n=1,2,3\)), we find that \(n=2\) and \(a=0.005\) is the solution ;D
Post by: Rathin on January 16, 2017, 03:52:33 pm
That is simultaneous equations, use the coordinates at C and D to create your system:
Now the fact that we have \(-60^n\) in our equations means that \(n\) needs to be an integer (we'd be beyond the bounds of the HSC course if it wasn't). We can approach this several ways, but if we consider and check some of the easy possible answers (\(n=1,2,3\)), we find that \(n=2\) and \(a=0.005\) is the solution ;D
Legend, Thanks :)
Post by: katnisschung on January 19, 2017, 02:29:24 pm
i need help differentiating the attached
pls show all working out...im bad at maths
Post by: Syndicate on January 19, 2017, 02:58:40 pm
this one trips me up every freaking single time!
i need help differentiating the attached
pls show all working out...im bad at maths
You will need to apply the product and the chain rule to work out the derivative of this expression.
First of all, using the chain rule work out the derivative of the square root part.
Now use the product rule (which states that (f(x)g(x))' = f'(x) x g(x) + g'(x) x f(x) )
Just for this moment assume f(x) = x and g(x) = \( \sqrt{x-2} \)
so f'(x) = 1 and g'(x) = \( \frac{1}{2\sqrt{x-2}} \)
You can simplify this down to \( \frac{3x-4}{2\sqrt{x-2}} \)
Post by: katnisschung on January 19, 2017, 03:26:52 pm
idk why it was included in the function rule ex...
Post by: RuiAce on January 19, 2017, 04:05:42 pm
Thank you Syndicate!! this whole time i was using only the function rule...
idk why it was included in the function rule ex...
Post by: jamonwindeyer on January 19, 2017, 04:58:32 pm
Thank you Syndicate!! this whole time i was using only the function rule...
idk why it was included in the function rule ex...
Beyond Rui's reasoning, it is likely that it was in there because you are using the function rule within the product rule - You can be expected to use them in tandem after all :) a little mean if you've not learned the product rule yet though!
Post by: Rathin on January 19, 2017, 05:00:45 pm
Find the area bounded by the curves y=x^2(1-x) and y=x(1-x)^2.
Post by: jakesilove on January 19, 2017, 05:25:22 pm
I keep getting this wrong for some reason..
Find the area bounded by the curves y=x^2(1-x) and y=x(1-x)^2.
That's a bloody tough one. The problem is that there are TWO areas that are bounded by these curves. Between 0 and 0.5, the first graph is on top, and between 0.5 and 1, the second graph is on top. However, the areas will be equal, so I'm going to calculate one and then double it! Make sure to sketch the graph, so you know which is on top of which for the equation to work.
So, the intercept is at x=0.5 (can you see why?)
Now, we want to find the integral between zero and 0.5 of the first graph minus the second graph.
I can't latex integrals left. The integral we need to find is
Between zero and one. This expands to
From there, just integrate between 0 and 0.5, then double the result! Let me know if you need help with that part
Post by: jamonwindeyer on January 19, 2017, 05:28:58 pm
I keep getting this wrong for some reason..
Find the area bounded by the curves y=x^2(1-x) and y=x(1-x)^2.
A diagram of the scenario to complement Jake's answer above ;D
(http://i.imgur.com/J2javmg.png)
You can see the two area sections Jake is talking about, and by visual inspection (symmetry ;) ), they are equal :)
Post by: katnisschung on January 19, 2017, 09:25:23 pm
i got (8x+7)^2/16 +C
but presumably they want it expanded
becos the final answer is 4x^2+7x
but somehow my 49 is not cancelling out
Post by: Syndicate on January 19, 2017, 09:27:58 pm
stuck with this indefinite integral..
i got (8x+7)^2/16 +C
but presumably they want it expanded
becos the final answer is 4x^2+7x
but somehow my 49 is not cancelling out
hmm... I think you have made a mistake.
anti-diff of 8x is \( 4x^2 \) and 7 is 7x.
So your final answer should be \( 4x^2+7x+C \)
Not sure if this rule is thought in HSC, but it is quite helpful: \( \frac{ ax^{n+1}}{n+1} \)
Post by: katnisschung on January 19, 2017, 09:34:44 pm
i obviously went wrong somewhere
Post by: jamonwindeyer on January 19, 2017, 09:35:34 pm
Not sure if this rule is thought in HSC, but it is quite helpful: \( \frac{ ax^{n+1}}{n+1} \)
Just confirming that's definitely the rule for us too 8) cheers Syndicate!
Post by: RuiAce on January 19, 2017, 09:40:42 pm
thanks syndicate yeah i got the first part but i expanded to get 4x^2+7x+49/16??Remember that when integrating there is an extra +C
i obviously went wrong somewhere
The constant can be anything. If you let C = 49/16 + c, then you get 4x^2 + 7x + c
Which is the same thing that Syndicate said.
(That being said, your final answer should've been 4x^2 + 7x + 49/16 + C for it to work. If you drop out the +C without a good reason, something is indeed wrong.)
(The +C is important. For example, if you integrate 2x, then x^2+C is an answer, but so is x^2+4+C)
Post by: jamonwindeyer on January 19, 2017, 09:42:06 pm
thanks syndicate yeah i got the first part but i expanded to get 4x^2+7x+49/16??
i obviously went wrong somewhere
Edit: More detailed version of what Rui said :P
Ahh, I've figured out your mistake!
So you are using this rule:
So this rule is actually a little bit overkill for this question - You are way better off just tackling each term one at a time as Syndicate outlined above. If you do want to use this rule, it works, you get this:
But Syndicate (and the answers) get:
They are actually the same thing! It's just that the constant (\(C_1\)) in your answer is a different value to the constant (\(C_2\)) in Syndicate's. Both are arbitrary constants, so this is irrelevant.
You have definitely not taken the most efficient approach. Your working was likely about 4-5 lines - Doing it term at a time only takes one. But, you are correct nonetheless :)
Post by: RuiAce on January 19, 2017, 09:44:31 pm
Post by: jamonwindeyer on January 19, 2017, 09:48:48 pm
Well I mean, the rule \(\int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} \) can be used in one line as well. But yeah it's overkill; term by term integration should be favoured where no effort is needed (e.g. some binomial theorem expansion).
Not all of us skip expansion/cancellation steps ;)
Post by: Rathin on January 19, 2017, 10:09:18 pm
Post by: Rathin on January 19, 2017, 10:15:16 pm
stuck with this indefinite integral..
i got (8x+7)^2/16 +C
but presumably they want it expanded
becos the final answer is 4x^2+7x
but somehow my 49 is not cancelling out
They are actually the same thing.
You can Integrate them Individually or use the rule ∫(ax+b)^n = ((ax+b)^n+1)/a(n+1)
You used the latter rule.
EDIT: Jamon already mentioned this whoops
Post by: jamonwindeyer on January 19, 2017, 10:26:16 pm
I am probably gonna beat my self if I make a silly mistake but I tend to do all the Primitive Transformation and indices manipulation in my head cuz It looks ugly on paper and I ceebs..even do it for square roots and stuff..
I feel like everyone has a different balance that works for them, I'll skip certain things... But I'm probably more cautious than most, I've been stung too many times ;)
Post by: phebsh on January 26, 2017, 03:54:29 pm
What's the best way to study for math if you absolutely hate it and are terrified of it? What are the best techniques? I tend to put it away and avoid it for as long as possible and if I get a question wrong, I'll give up imediately... I'm more of an english person. :-\
Thanks! :)
Post by: RuiAce on January 26, 2017, 03:58:37 pm
Hi everyone,Before I reply, I must apologise for some unintended negativity that will inevitably flow out of this post.
What's the best way to study for math if you absolutely hate it and are terrified of it? What are the best techniques? I tend to put it away and avoid it for as long as possible and if I get a question wrong, I'll give up imediately... I'm more of an english person. :-\
Thanks! :)
Given that you hate it, is there a reason you chose to keep the subject?
Regardless of what subject it is, if you hate it you're not gonna try hard enough. You can push and attempt to focus, but it may not pay off.
Fact is, maths is all about practice. And getting questions wrong should not be surprising; if everyone in the state could do every maths question then we'd be questioning if our subjects were way too easy and need significant revamping. Most students push through it by eventually learning that they just have to accept that they must:
- Do some questions open book
- Look at answers if they're GENUINELY stuck
- Keep drilling in the past papers (note: emphasis on past papers over textbooks)
But if you hate it, you probably won't be able to push yourself to do all of that.
Post by: phebsh on January 26, 2017, 04:10:49 pm
Before I reply, I must apologise for some unintended negativity that will inevitably flow out of this post.
Given that you hate it, is there a reason you chose to keep the subject?
Regardless of what subject it is, if you hate it you're not gonna try hard enough. You can push and attempt to focus, but it may not pay off.
Fact is, maths is all about practice. And getting questions wrong should not be surprising; if everyone in the state could do every maths question then we'd be questioning if our subjects were way too easy and need significant revamping. Most students push through it by eventually learning that they just have to accept that they must:
- Do some questions open book
- Look at answers if they're GENUINELY stuck
- Keep drilling in the past papers (note: emphasis on past papers over textbooks)
But if you hate it, you probably won't be able to push yourself to do all of that.
The reason I kept it was because my school made math compulsory, which I'm obviously not happy about...
I'll definately put in the effort and do my absolute best. Thanks for your input!
Post by: RobertDyd on January 26, 2017, 05:13:06 pm
Hi everyone,
What's the best way to study for math if you absolutely hate it and are terrified of it? What are the best techniques? I tend to put it away and avoid it for as long as possible and if I get a question wrong, I'll give up imediately... I'm more of an english person. :-\
Thanks! :)
Hi there,
I used to hate math as well (I failed math in year 7 and I was 40th-ish in year 10, what I did was I used my absolute hate for maths to motivate myself to get good at it so that I'd never struggle again. It took a couple of months but I made it to 3rd place by term 1 of year 11 (Now in yr 12 I'm first in 4u and I love it). I find that we as people just hate things we aren't proficient at, like when my friend Raylan beats me in a 1v1 and I call the game absolute bull ****
The techniques I used in the first couple months to learn math were lots of paper and ink, but as for making math more enjoyable so that you could do well (Or pass it idk whats your case) I would recommend first learning how to absolutely smash algebra and wrap your head around it (I find that most questions in 2u you can mindlessly brute force it if you know your basics). Then with the topics (ie Trig), I would do the hardest questions everyday (just a few), take it to your teacher after school, and learn how they did them correctly, because if you can learn how to do math like your teachers (which is alot easier than reading a book) 90% of the rest of the topic is easy af.
If none of those work, just remember that the HSC is only one year and treat it like a bandaid, just do your best and rip it off as fast you can. Kind of like how you try to finish your hw in the last 10 minutes of class.
If you got this far into my reply then congratulations, I'm not a great writer and I'm currently fueled by anime rather than sleep (I have like another 2 series suggested by my friends to finish by these holidays)
Any questions just pm me and pray I don't fall asleep
Post by: kylesara on January 27, 2017, 10:46:53 pm
Post by: RuiAce on January 27, 2017, 10:48:37 pm
can anyone help with the primative function of (x-1)^2. Thanks
Post by: jamonwindeyer on January 28, 2017, 12:00:06 pm
can anyone help with the primative function of (x-1)^2. Thanks
Hey Kyle! I'll show you the full working two ways:
Remember, add one to the power and then divide by the new power :)
You could also use a rule on your reference sheet to find the primitive in one go:
Both are fine - That latter line uses the idea that the primitive of \((ax+b)^n\) is \(\frac{(ax+b)^{n+1}}{a(n+1)}\) - On your reference sheet ;D
Post by: Thebarman on January 29, 2017, 01:10:16 pm
1) If A = (1,5), B = (4,2), and C = (2,-3), find the coordinated of D if ABCD is a parallelogram.
2) The ratio of length to the breadth of a rectangle is 3:2. If the breadth of the rectangle is b units, find a formula for the area of the rectangle in terms of b.
3) Show that the points X (3,2) and Y (-1,0) are equal distances from the line 4x-3y-1=0.
Thanks!
Post by: Shadowxo on January 29, 2017, 01:59:22 pm
Hey guys, anyone able to help me out with these questions? I seem to have forgotten how to do these during the holidays.
1) If A = (1,5), B = (4,2), and C = (2,-3), find the coordinated of D if ABCD is a parallelogram.
2) The ratio of length to the breadth of a rectangle is 3:2. If the breadth of the rectangle is b units, find a formula for the area of the rectangle in terms of b.
3) Show that the points X (3,2) and Y (-1,0) are equal distances from the line 4x-3y-1=0.
Thanks!
1. I'd plot the points and go from there. I'm not sure if they'd expect you to do it mathematically, I'd just say AB = DC and AD = BC so D must be at (-1,0) as the x distances from A and B is equal to the x distance between D and C, so x(D) = 2-3 = -1. Same for y distance, y(D) = -3+3 = 0
2. Area = length*breadth. Length = 3b/2 (due to ratio) so area = 3b/2 * b = 3b2/2
3. Find the midpoint of those points. Midpoint is (2/2, 2/2) = (1,1)
x = 1, y=1, 4 - 3 - 1 =0 therefore the midpoint of X and Y, (1,1) lies on the line 4x-3y-1=0 therefore the distances from the line are equal.
Post by: RuiAce on January 29, 2017, 02:01:20 pm
Post by: kylesara on January 30, 2017, 04:55:40 pm
Post by: jamonwindeyer on January 30, 2017, 05:11:49 pm
Is the exact value needed when completing a table of values for stationary points, or just the sign ( - or +). Thanks :)
Hey! Only the sign is necessary, but you can write the exact value if you want to ;D
Post by: cindyung on January 31, 2017, 05:53:53 pm
I'm stuck in this dilemma if i should drop 2u of maths as i'm currently not doing well at all--
I have 13 units currently: English, Maths, Legal, Ancient, Chemistry, Biology and Ext History.
Its kind of stressful so i was wondering if i should drop 2U maths to have time to do other subjects?
Thank you!
Post by: RuiAce on January 31, 2017, 06:15:02 pm
Hi! Sorry this isn't really a question but a question relating to 2u maths --Following some conversations Jamon was involved in, the questions to ask are how much work is being put into 2U, as well as whether or not other courses suffer.
I'm stuck in this dilemma if i should drop 2u of maths as i'm currently not doing well at all--
I have 13 units currently: English, Maths, Legal, Ancient, Chemistry, Biology and Ext History.
Its kind of stressful so i was wondering if i should drop 2U maths to have time to do other subjects?
Thank you!
In the event that mathematics is something you happen to HATE, then you should get rid of it without any further question. Because if you hate it, you're not going to do anything for it.
Otherwise, think about how much effort you're putting into maths relative to other subjects, and think about why that's the case. If you're putting heaps of effort into 2U and it's beginning to impact on all the OTHER subjects then you should also drop. Because that's causing more harm than good. On the contrary, if your doing really well in everything else but not in mathematics, if you're not putting enough effort in that needs reflecting on. Or instead if you do put heaps of effort in it, then there's a possibility that the subject wasn't meant for you, and if too much hard work doesn't pay off then dropping seems reasonable.
(Seeing as though you have 13 units, dropping off 2 still gives you a safety net of 1 extra unit, so in that regard there's nothing wrong with it.)
Post by: kylesara on January 31, 2017, 08:13:39 pm
Question: If d2y/dx2 = 12x+6 and dy/dx = 1 at the point (-1,-2), find the equation of the curve
thanks!
Post by: RuiAce on January 31, 2017, 08:38:16 pm
Homework help please
Question: If d2y/dx2 = 12x+6 and dy/dx = 1 at the point (-1,-2), find the equation of the curve
thanks!
Post by: Adena on February 01, 2017, 10:26:44 pm
5x2 +10xy-25xy2
Thanks
Post by: sophiegmaher on February 01, 2017, 10:39:22 pm
Post by: jamonwindeyer on February 01, 2017, 10:42:35 pm
Hi, how do you completely factorise the following:
5x2 +10xy-25xy2
Thanks
Hey! You can start by taking out a common factor:
Unless there is a typo in the question, I think that's as far as you can go! It is almost a quadratic factorisation but the terms don't quite match up! Not what I expected to happen :P
Post by: RuiAce on February 01, 2017, 10:49:13 pm
Hi, how do you completely factorise the following:Are you sure that's -25xy2 and not just -25y2?
5x2 +10xy-25xy2
Thanks
Because if so...
Hey guys! Has anyone got any study resources in terms of practise questions? Like topic specific worksheets with solutions on half yearly content?You could find some ones you wish to target here.
However there's nothing wrong with simply extracting relevant questions from past trial papers. Half yearly exams follow a similar format.
Post by: laurenf58 on February 02, 2017, 05:55:53 pm
How do I find points of inflexion? (stationary points have already been found)
Really hoping someone can help! Thanks!
Post by: RuiAce on February 02, 2017, 06:01:57 pm
Hey!
How do I find points of inflexion? (stationary points have already been found)
Really hoping someone can help! Thanks!
Checklist:
1. Set second derivative to 0 to single out the candidate points
2. Verify a change in sign occurs
Post by: Hplovers on February 02, 2017, 08:39:25 pm
I'm getting confused with the wording of these two probability questions;
The ratio of girls to boys at a school is four to five. Two students are surveyed at random from the school. Find the probability that the students are
a) both boys
b) a girl and a boy
c) at least one girl
I have attached the probability tree i have used and i'm pretty sure its correct.
I have a ratio of 4/9 for girls and 5/9 for boys
However, what im confused about; am I supposed to use the same ratio for both the first AND second step?
For a), I have been doing ( 5/9 x 4/8 ) to find the probability of 'both boys', because after you take one boy out doesn't it then lower the numerator and denominator and numerator for the next boy to be picked?
My answers were;
a) 5/18
b) 5/9
c) 13/18
The textbooks answers were;
a) 25/81
b) 40/81
c) 56/81
( It appears they do not change the ratio for each additional student picked)
Am I reading it wrong? I thought this question was like the lottery ticket questions where it is without replacement, because in those questions they don't specify 'without replacement' they just assume it, and I assumed in this question the student was not replaced for the next random survey.
The second question I have the exact same issue with;
The number of cats to dogs at a pet hotel is in the ratio of 4 to 7. If 3 pets are chosen at random, find the probability that
a) they are all dogs
b) just one is a dog
c) at least one is a cat
Same issue, with the answers keeping the same ratio throughout, but I thought you were supposed to remove an animal from the numerator and denominator at each animal chosen.
Thankyou!!
Post by: RuiAce on February 02, 2017, 09:07:12 pm
Hey!The question's wording is very hard to decrypt.
I'm getting confused with the wording of these two probability questions;
The ratio of girls to boys at a school is four to five. Two students are surveyed at random from the school. Find the probability that the students are
a) both boys
b) a girl and a boy
c) at least one girl
I have attached the probability tree i have used and i'm pretty sure its correct.
I have a ratio of 4/9 for girls and 5/9 for boys
However, what im confused about; am I supposed to use the same ratio for both the first AND second step?
For a), I have been doing ( 5/9 x 4/8 ) to find the probability of 'both boys', because after you take one boy out doesn't it then lower the numerator and denominator and numerator for the next boy to be picked?
My answers were;
a) 5/18
b) 5/9
c) 13/18
The textbooks answers were;
a) 25/81
b) 40/81
c) 56/81
( It appears they do not change the ratio for each additional student picked)
Am I reading it wrong? I thought this question was like the lottery ticket questions where it is without replacement, because in those questions they don't specify 'without replacement' they just assume it, and I assumed in this question the student was not replaced for the next random survey.
The second question I have the exact same issue with;
The number of cats to dogs at a pet hotel is in the ratio of 4 to 7. If 3 pets are chosen at random, find the probability that
a) they are all dogs
b) just one is a dog
c) at least one is a cat
Same issue, with the answers keeping the same ratio throughout, but I thought you were supposed to remove an animal from the numerator and denominator at each animal chosen.
Thankyou!!
All we are saying is that the ratio of girls to boys at a school is 4:5.
But this doesn't mean there's only 9 students. Yes, if you treat it as exactly 9 students there, that's what you get.
But what if there's 900 students instead?
If the ratio of girls to boys at a school is 4:5, then there would be 400 girls and 500 boys.
So for a), technically the answer would be 500/900 * 499/899 now, right? Which is not equal to 5/9 * 4/8
The question is most certainly tricky, and the appearance of this sort of ambiguity is unlikely in the HSC. However, the ultimate purpose of this question is to not treat ratios as representative of the bigger picture - the total. Hence, the safest bet is to assume that in general, if we take one out, the ratio isn't varied much.
Note that something like 400:499 is approximately equal to 400:500 anyhow, which is 4:5. So since in general schools have quite a large population (and not just a mere 9 students), we assume that by choosing one, the ratio is not greatly disturbed (the disturbance is negligible).
Post by: Hplovers on February 03, 2017, 07:07:45 am
Post by: kiwiberry on February 03, 2017, 11:02:08 am
1) in the first year of joining a superannuation fund a women invests $400. The interest rate is 12% pa. Each year she invests an additional 10% of the amount invested in the previous year to allow for inflation. How much does she receive after 20 years?
2) A man put $500 savings into a bank for 2 years, where it earned interest at 6% pa, paid twice a year. He then changed to a credit union and his money earned 8% pa, paid quarterly. if he withdrew all his savings and had $633.75, how long was the money kept in the credit union?
Thank you :)
Post by: jakesilove on February 03, 2017, 11:27:10 am
Financial maths questions:
1) in the first year of joining a superannuation fund a women invests $400. The interest rate is 12% pa. Each year she invests an additional 10% of the amount invested in the previous year to allow for inflation. How much does she receive after 20 years?
2) A man put $500 savings into a bank for 2 years, where it earned interest at 6% pa, paid twice a year. He then changed to a credit union and his money earned 8% pa, paid quarterly. if he withdrew all his savings and had $633.75, how long was the money kept in the credit union?
Thank you :)
Hey!
It's easiest to tackle these questions 'by year'.
So, in the first year, we will have a deposit, and interest.
The second year will have another (greater) deposit, and more interest
And... the third year
Can you see a pattern? We can tell that
Pulling out the 400, we get a series
We're DIVIDING by 1.12, and MULTIPLYING by 1.1. This is a geometric series,
After 20 years, n=20, so
Therefore, the total will be 400*111.18=$44,473
Don't quite have time for Q2 (maybe someone else will step in), but try taking the same approach!
Post by: RuiAce on February 03, 2017, 11:38:09 am
2) A man put $500 savings into a bank for 2 years, where it earned interest at 6% pa, paid twice a year. He then changed to a credit union and his money earned 8% pa, paid quarterly. if he withdrew all his savings and had $633.75, how long was the money kept in the credit union?
Thank you :)
P.S. Jake assumed that the interest was compounded annually for Q1. This makes the most intuitive sense to me, however the question did not explicitly state this. The question therefore lacks information.
Post by: kiwiberry on February 03, 2017, 01:33:54 pm
Post by: Nialllovespie on February 03, 2017, 03:56:21 pm
Could someone please help me with this Integration question?
Find the volume of the paraboloid when y=x^2 is rotated about the y axis from y=0 to y=3
Thanks so much!!
Post by: jakesilove on February 03, 2017, 04:15:00 pm
Hiiii!!
Could someone please help me with this Integration question?
Find the volume of the paraboloid when y=x^2 is rotated about the y axis from y=0 to y=3
Thanks so much!!
Hey!
Our formula for volumes is
So, we start off by finding our limits. By subbing them into the equation, we know that they are 0 and the square root of three.
Between 0 and the square root of 3, which will be
I rotated around the X-AXIS, not the y-axis. Need to do what Rui did below instead!
Post by: RuiAce on February 03, 2017, 04:15:49 pm
Hiiii!!
Could someone please help me with this Integration question?
Find the volume of the paraboloid when y=x^2 is rotated about the y axis from y=0 to y=3
Thanks so much!!
Hey!Jake. What are you doing. Wrong axis.
Our formula for volumes is
So, we start off by finding our limits. By subbing them into the equation, we know that they are 0 and 9.
Between 0 and 9, which will be
Post by: joyg on February 03, 2017, 11:51:52 pm
3000 = 100e^n
Post by: Shadowxo on February 03, 2017, 11:54:27 pm
I was just wondering how to do this log question:
3000 = 100e^n
So 30 = en
loge(30) = n
Post by: jamonwindeyer on February 04, 2017, 12:04:14 am
I was just wondering how to do this log question:
3000 = 100e^n
Welcome to the forums! Shadow's got you covered - But if you've only just started Logs it might not be super clear how it works. Be sure to let us know if you need it clarified, or if you need help finding stuff around the site ;D
Post by: joyg on February 04, 2017, 12:21:51 am
Post by: asd987 on February 04, 2017, 03:31:11 pm
Post by: jamonwindeyer on February 04, 2017, 03:34:46 pm
Hi can i get some help with this question. thanks ;D
This is a bit of manipulation of the Log Laws! Namely, \(\log{a}+\log{b}=\log{ab}\):
Note that \(\log_7{7}=1\), by definition! Does this help? ;D
Post by: Shadowxo on February 04, 2017, 03:37:21 pm
Hi can i get some help with this question. thanks ;D
14 is 7*2
So log7(14) = log7(7*2) = log77 + log72 = 1 + log72
Does this help? Second question is done in same way as the first :)
Edit: someone beat me to it :P
Post by: joyg on February 04, 2017, 08:48:38 pm
Can you please help me with: 0.5M = Me0.116k
Thanks for your help :D
Post by: RuiAce on February 04, 2017, 08:50:04 pm
Another day and another question
Can you please help me with: 0.5M = Me0.116k
Thanks for your help :D
Note, if M=0 then any value of k would work. Generally not necessary for 2U though.
Post by: joyg on February 04, 2017, 08:55:24 pm
Post by: Hplovers on February 06, 2017, 07:39:30 pm
Express y in terms of x if:
dy/dx = 2/x2
I understand i need to find the primitive function but just not sure how to :)
Answer is y = -2/x + C
Thankyouu!
Post by: kiwiberry on February 06, 2017, 07:54:59 pm
How to solve this:
Express y in terms of x if:
dy/dx = 2/x2
I understand i need to find the primitive function but just not sure how to :)
Answer is y = -2/x + C
Thankyouu!
dy/dx = 2/x2 = 2x-2
\(\therefore \,y=2\times \frac{x^{-1}}{-1}+c= \frac{-2}{x}+c\) :)
Post by: RuiAce on February 06, 2017, 07:56:15 pm
How to solve this:
Express y in terms of x if:
dy/dx = 2/x2
I understand i need to find the primitive function but just not sure how to :)
Answer is y = -2/x + C
Thankyouu!
dy/dx = 2/x2 = 2x-2Go for the
\(\therefore \,y=2\times \frac{x^{-1}}{-1}+c= \frac{-2}{x}+c\) :)
Code: [Select]
[tex] [/tex]tags rather than the Code: [Select]
\( \) ones. Those bracket tags are more for when you have an equation in between ordinary text like \(this\)
Post by: kiwiberry on February 06, 2017, 08:14:29 pm
Go for theCode: [Select][tex] [/tex]tags rather than theCode: [Select]\( \)ones. Those bracket tags are more for when you have an equation in between ordinary text like \(this\)
Ah ok, sorry haha will do next time!
Post by: RuiAce on February 06, 2017, 08:19:16 pm
Ah ok, sorry haha will do next time!You get used to it. Starting LaTeX in high school ain't the easiest thing to do. Doesn't take too much effort but shouldn't come immediately either.
Post by: levendibigd on February 06, 2017, 08:45:13 pm
Post by: RuiAce on February 06, 2017, 08:48:44 pm
Hi there, can I please have help expressing |a+b| without any absolute values, by using the properties of absolute values
Post by: Shadowxo on February 06, 2017, 08:50:35 pm
Hi there, can I please have help expressing |a+b| without any absolute values, by using the properties of absolute values
Two ways of doing this:
|a+b| = \(\sqrt{(a+b)^2}\)
Or as Rui said, creating a hybrid function :)
Post by: RuiAce on February 06, 2017, 08:55:44 pm
Two ways of doing this:Yep. That being said though, AFAIK in the 2U course they're more relaxed about \( \sqrt{x^2}=|x|\). Some people are taught it, but others aren't and I've never seen it necessary in a 2U exam.
|a+b| = \(\sqrt{(a+b)^2}\)
Or as Rui said, creating a hybrid function :)
They need to know the piece-wise definition though.
Post by: QC on February 06, 2017, 11:34:23 pm
Post by: RuiAce on February 06, 2017, 11:48:15 pm
|a+b| = the distance of a+b to the origin which is sqrt((a+b)2)). Can also be writen as a piecemeal function as it is equal to a+b OR -a-ba+b is just a number.
"The distance of a+b to the origin" - where is a+b on the Cartesian plane? Unless by origin you were intending to refer to 0 on the number line.
Post by: QC on February 06, 2017, 11:58:57 pm
a+b is just a number.Yeah, I always refer to 0 as the origin in both the Cartesian plane and number line, apologies for confusion
"The distance of a+b to the origin" - where is a+b on the Cartesian plane? Unless by origin you were intending to refer to 0 on the number line.
Post by: joyg on February 07, 2017, 10:15:46 pm
Differentiate- e3x/x2 and e2x + 1/2x + 5
Post by: RuiAce on February 07, 2017, 10:23:01 pm
Hey, can someone please help me with the following to questions:Hard to tell what's going on with the second one, because of the absence of the brackets.
Differentiate- e3x/x2 and e2x + 1/2x + 5
Could be \( \frac{e^{2x+1}}{2x+5}\) or \( \frac{e^{2x+1}}{2x}+5\), to which the absence of brackets assumes the latter.
____________________________
Post by: laurenf58 on February 08, 2017, 07:24:37 pm
Thanks!
Post by: legorgo18 on February 08, 2017, 07:35:52 pm
Could someone please tell me how do I find the co-ordinates of the points A and B where the tangent (y=-4x+25) cuts the x and y axes respectively?
Thanks!
Hi, i havent done these for a long time but i think you just sub x and y=0
Sorry if im wrong
Post by: kiwiberry on February 08, 2017, 07:36:36 pm
Could someone please tell me how do I find the co-ordinates of the points A and B where the tangent (y=-4x+25) cuts the x and y axes respectively?
Thanks!
Post by: RuiAce on February 08, 2017, 07:42:09 pm
Hi, i havent done these for a long time but i think you just sub x and y=0Other way around so that it's respective to the question maybe, but the method is definitely correct (as kiwiberry demonstrated).
Sorry if im wrong
Keep going with the LaTeX. Advice here is to use \left( \right) if you have to put a bracket around a fraction.
(Which, Jake should probably do as well)
Post by: katnisschung on February 09, 2017, 10:14:08 am
(unfortunately i can't upload a pic..)
the diagram shows the cross-section of a creek with the depths
of the creek shown in metres at 4 metre intervals
the creek is 12 metres in width
i) use the trapezoidal rule to find an approximate value for the area of the cross section
=12 m^2
(im pretty sure its right... don't have answers but everyone got this)
ii) water flows through this section of the creek at a speed of 0.5 ms^-1
calculate the approx volume of water that flows past this section in one hour
need help... how do i find the volume??\
thanks!
Post by: RuiAce on February 09, 2017, 10:43:33 am
need help on this past hsc question on integration
(unfortunately i can't upload a pic..)
the diagram shows the cross-section of a creek with the depths
of the creek shown in metres at 4 metre intervals
the creek is 12 metres in width
i) use the trapezoidal rule to find an approximate value for the area of the cross section
=12 m^2
(im pretty sure its right... don't have answers but everyone got this)
ii) water flows through this section of the creek at a speed of 0.5 ms^-1
calculate the approx volume of water that flows past this section in one hour
need help... how do i find the volume??\
thanks!
Post by: anotherworld2b on February 09, 2017, 11:24:30 pm
Post by: kiwiberry on February 09, 2017, 11:42:51 pm
Could i have some help with this question please? I am a bit confused about how i differentiated it wrong
Use the chain rule to differentiate from here :)
Post by: laurenf58 on February 10, 2017, 04:22:56 pm
Thanks!
Post by: jamonwindeyer on February 10, 2017, 04:30:50 pm
How do I do this question?
Thanks!
Hey! This is a quadratic identity; for something to be equal for ALL values of \(x\) then the coefficients need to be the same:
So, \(a=3\) and \(b=6\). This makes the two quadratics the same and is the only way the condition can be satisfied ;D does that make sense?
Post by: Hplovers on February 10, 2017, 05:10:22 pm
Post by: asd987 on February 10, 2017, 05:26:14 pm
Thanks
Post by: kiwiberry on February 10, 2017, 05:46:11 pm
Ah this is confusing me so much, please help :)
Hey! If you split up the fraction like this:
You can integrate easily :)
Post by: Hplovers on February 10, 2017, 06:01:31 pm
Hey! If you split up the fraction like this:
You can integrate easily :)
I have gotten to that point but still end up with the wrong answer at the end :( , could you please finish the integration for me :)
After that step I got rid of the fractions to make them 1/3 . x^-3 + ... etc. but i'm not sure if that's correct. I thought that to use the integration( ax^n) rule, the function could not be a fraction. What is the rule for when you can integrate? :)
Post by: kiwiberry on February 10, 2017, 06:16:58 pm
Isn't it this?!!
Post by: RuiAce on February 10, 2017, 06:18:29 pm
Isn't it this?!!Aiya. I'll just blame the heat for my dead brain - mistake deleted
Post by: RuiAce on February 10, 2017, 06:23:58 pm
Hi, I have a couple of questions if some can please help me with them.
Thanks
_________________________________
I'm only going to do the first one. You can post up your working for the other if you get stuck because the process is the SAME; the only difference being the product rule used.
Post by: sophiegmaher on February 11, 2017, 10:31:05 am
"Find the maximum possible area if a straight 8m length of fencing is placed across a corner to enclose a triangular space"
Post by: RuiAce on February 11, 2017, 10:43:58 am
So I'm stuck on this question, I feel it's a simple answer but I'm not sure how to get it!This question is a bit unspecific in how it was worded. I'd assume that when they say "corner" they mean that the corner is right-angled, but they tell us nothing at all about that.
"Find the maximum possible area if a straight 8m length of fencing is placed across a corner to enclose a triangular space"
Where did this question come from, and does it come with a diagram?
Post by: sophiegmaher on February 11, 2017, 10:46:17 am
Post by: RuiAce on February 11, 2017, 11:01:28 am
Post by: sophiegmaher on February 11, 2017, 12:09:30 pm
Post by: Nialllovespie on February 13, 2017, 10:42:10 am
A circle with circumference 124mm has a chord cut off it that subtends an angle of 40 degrees at the centre. Find the length of the arc cut of by the chord.
Thanks so much!
Post by: RuiAce on February 13, 2017, 10:44:49 am
Can anyone help me with this question please?
A circle with circumference 124mm has a chord cut off it that subtends an angle of 40 degrees at the centre. Find the length of the arc cut of by the chord.
Thanks so much!
Post by: Shadowxo on February 13, 2017, 11:07:39 am
Can anyone help me with this question please?
A circle with circumference 124mm has a chord cut off it that subtends an angle of 40 degrees at the centre. Find the length of the arc cut of by the chord.
Thanks so much!
Great answer by RuiAce but here's an alternate way of doing it:
the length of the arc is a fraction of the circumference
The fraction is 40/360 = 1/9
So the length of the arc is 1/9th of the circumference,
= 1/9 * 124 = 124/9mm
Hope this helps :)
Post by: asd987 on February 13, 2017, 07:09:25 pm
Find the exact area between the curve y=1/x , the x axis and the lines y=x and x=2 in the first quadrant
Post by: RuiAce on February 13, 2017, 07:22:08 pm
Hi, Can i get some help with this question please:(http://i.imgur.com/1byLJ1o.png)
Find the exact area between the curve y=1/x , the x axis and the lines y=x and x=2 in the first quadrant
Post by: katnisschung on February 14, 2017, 03:47:17 pm
just want to check that i've got this stuff covered.
1. ordinates=function vales (i interpret this as how many individual values there actually are)
2. (sub) intervals= how many trapeziums there are (specific to the trapezium rule)
3. applications (how many times you apply the rule)
so for this questiton
b) integrate log 10 x dx (between 10 and 1) using 4 function values
using the trapezium rule...
so this would have 3 applications, function values being (1, 4, 7, 10)
correct?
thanks!
Post by: RuiAce on February 14, 2017, 03:54:46 pm
clarification of integration terminology..Looks good
just want to check that i've got this stuff covered.
1. ordinates=function vales (i interpret this as how many individual values there actually are)
2. (sub) intervals= how many trapeziums there are (specific to the trapezium rule)
3. applications (how many times you apply the rule)
so for this questiton
b) integrate log 10 x dx (between 10 and 1) using 4 function values
using the trapezium rule...
so this would have 3 applications, function values being (1, 4, 7, 10)
correct?
thanks!
Post by: kylesara on February 15, 2017, 04:36:14 pm
Find the areas enclosed by the curves x=y^2 and y=x^2
Thanks!
Post by: jakesilove on February 15, 2017, 04:55:36 pm
Hi, i need help with a question from hw.
Find the areas enclosed by the curves x=y^2 and y=x^2
Thanks!
Hey! First, we need to find both equations in terms of x.
We want the area enclosed. The only way to do that is to draw the function, so we know exactly what's going on. I'll leave that to you.
There is an area enclosed between y=x^2 and the positive part of the other function. By sketching, we know that y=x^2 is on the bottom of the enclosed section, therefore to find the area we simply subtract the bottom function from the top, and integrate. Clearly, the points of intersection are 0 and 1.
You should be able to do the rest!
Post by: brokenboxes on February 15, 2017, 06:33:19 pm
This may seem like a really foolish and unusual question but I have an infamous reputation for making silly mistakes in exams. Whether it be forgetting to change a sign or accidentally writing a 2 instead of a 7, I always get caught up by these despite the amount of practice I do. Would you have any tips for dealing with this?
Thank you :)
Post by: RuiAce on February 15, 2017, 06:40:30 pm
Hey there :)Well the answer to this question depends on if you make them AS you practice, or ONLY in the exam.
This may seem like a really foolish and unusual question but I have an infamous reputation for making silly mistakes in exams. Whether it be forgetting to change a sign or accidentally writing a 2 instead of a 7, I always get caught up by these despite the amount of practice I do. Would you have any tips for dealing with this?
Thank you :)
If you make the mistakes AS you practice, then the answer is somewhat simple. It means that your practice hasn't been sufficiently effective. This may or may not imply poor quantity of practice as well, but it means you have to look out more when you're practicing. Check your working every 5 or so lines, instead of at the end. And don't try to do too many things at once if it's too overloading (e.g. use trig identities on one side but then a lot of algebra on the other).
If you only bump into this issue in the exam, then there may be the exam stress factor fueling it. The first thing to do is to make sure you're not stressed and focused on the question. Similar to the above, check your working every 5 or so lines. But you should consider memorising this list that Jamon just has very handy up his sleeve. When going through your working out, recite all the steps you did (quickly though, because you don't want to go into time trouble) and make sure NONE of that nuisance happens. And for things such as writing a wrong number, I always think the number in my head. I think "seven-hundred-and-twenty-five" as I write 725. Like, not just process the number, think it as if I had to force it into my brain. (Also, make sure there's no handwriting problems.)
On top of Jamon's list, write down every single silly mistake you've encountered. You can make a mistake, then forget you ever made it and discard the entire thought. List them out, and have them laid out right in front of you when you do practice papers, so that they are always ringing a bell.
Post by: jakesilove on February 15, 2017, 06:42:18 pm
Hey there :)
This may seem like a really foolish and unusual question but I have an infamous reputation for making silly mistakes in exams. Whether it be forgetting to change a sign or accidentally writing a 2 instead of a 7, I always get caught up by these despite the amount of practice I do. Would you have any tips for dealing with this?
Thank you :)
Hey! Welcome to the forums!
I was absolutely notorious for the same thing. I'd smash out a question, and invariably forget a minus somewhere and lose a mark. However, this only really plagued me for the start of my 2U course. I managed to ween myself out of silly mistakes (at least, as much as you can; you'll always make one of two in every exam!).
There's really only two things you can do to get over silly mistakes. One is sheer attrition. Do a billion questions. Then do a billion more. The more you practice, the less likely you are to make stupid mistakes. It's just probability; the better you understand the content, the less likely you are to make a stupid mistake.
The second way I made sure I made as few silly mistakes as possible was by writing them out. I would keep a list of mistakes that I consistently made, and I'd try to make the list funny, sweary, and concise. Then, I'd have those sorts of mistakes in my head as I sat a paper. Keep updated the list, and you'll find yourself needing it less and less (which is the idea).
Really great question! Let me know if you need anything else :)
Post by: jakesilove on February 15, 2017, 06:43:17 pm
Well the answer to this question depends on if you make them AS you practice, or ONLY in the exam.
If you make the mistakes AS you practice, then the answer is somewhat simple. It means that your practice hasn't been sufficiently effective. This may or may not imply poor quantity of practice as well, but it means you have to look out more when you're practicing. Check your working every 5 or so lines, instead of at the end. And don't try to do too many things at once if it's too overloading (e.g. use trig identities on one side but then a lot of algebra on the other).
If you only bump into this issue in the exam, then there may be the exam stress factor fueling it. The first thing to do is to make sure you're not stressed and focused on the question. Similar to the above, check your working every 5 or so lines. But you should consider memorising this list that Jamon just has very handy up his sleeve. When going through your working out, recite all the steps you did (quickly though, because you don't want to go into time trouble) and make sure NONE of that nuisance happens. And for things such as writing a wrong number, I always think the number in my head. I think "seven-hundred-and-twenty-five" as I write 725. Like, not just process the number, think it as if I had to force it into my brain. (Also, make sure there's no handwriting problems.)
On top of Jamon's list, write down every single silly mistake you've encountered. You can make a mistake, then forget you ever made it and discard the entire thought. List them out, and have them laid out right in front of you when you do practice papers, so that they are always ringing a bell.
Typical Rui, beating me to the punch :(
Post by: biffi023 on February 15, 2017, 08:57:06 pm
Post by: RuiAce on February 15, 2017, 09:06:28 pm
What is the best way others have found to study Mathematics?! I have so much trouble remembering the formulas etc of older topics once we have been doing the more recent ones for a while!! (I do Mathematics but struggle with it! :( ) There is so much content that just physically remembering processes (especially when they didnt TRULY stick in your head when you learnt them) can get so confusing! Also, how do you overcome the way that different teachers and exam setters can word questions just slightly different enough to throw you?? :-\Simply memorising formulas is not enough. Everyone will say this - you must be consistently doing past papers and more past papers to study for mathematics. Maths is a skills-based course; hardly anything simply content based will appear on the day. You cannot prepare for an exam by simply going off content. You need to prepare for the exam by doing old past papers, because only they reflect what your exam will be like, and thus what you should be prepared for. By doing past papers, you train your brain to think the way you're expected to be thinking.
Do not stop the past papers. And do a whole variety of them so that you maximise your exposure to all the possible things they can ask.
Post by: Mathew587 on February 15, 2017, 09:36:16 pm
Can someone pls help me with q 15 b ii) iii) iv) from the 2unit 2014 paper. I dont understand how the marking guidelines got the answers like what specifically they used.
Ty peeps :)
Post by: RuiAce on February 15, 2017, 10:11:07 pm
Hiya everyone,Since you've presented the BOSTES answers (thanks - makes life easier), I'll just explain what they did a bit more slowly.
Can someone pls help me with q 15 b ii) iii) iv) from the 2unit 2014 paper. I dont understand how the marking guidelines got the answers like what specifically they used.
Ty peeps :)
I'll assume you got part (i) out. In part (ii), they commence by using part (i). Since you have a pair of similar triangles, you know that the ratios of their sides are in proportion to each other. Hence the original ratio DR/DF = DS/DE
Then, if you look at the diagram, the side DE is really the side DS AND the side SE joined together.
Hence DE = DS+SE. This is simply substituted into the above ratio.
And lastly, sub in your x's and y's
______________________________
Solving part (iii) needs a bit of thought and/or a tad looking ahead.
The question's interested in areas. Well, since we have triangles involved, we can use A=1/2 bh or A=1/2 ab sinC. But we have no height, so the latter gets picked.
A1 is the area of triangle DSR, which they tell you in the question. What do they do? Apply the formula 1/2 ab sin C straight into it.
And then A is the area of your large triangle. They apply the same formula here. This is what gets you the first mark.
The question wants you to find sqrt(A1/A). So we start by finding A1/A without the square root. That's just substitution.
But the reason 1/2 ab sin C gets used is that they realised "oh wow, sin D gets cancelled out!" And in fact, so does the 1/2. Which yields the result beautifully.
______________________________
You need to notice how in this part, triangle DSR is replaced by triangle SEQ. You need to somehow "see" that this is another pair of similar triangles. Hence, in a SIMILAR way to the above, they deduce that expression for sqrt(A2/A), which you obviously need because sqrt(A2) somehow appears in the final result.
Being able to see that if you repeat this whole process, except on a different triangle, is what earns you the first mark.
Lastly, direct substitution.
Post by: bananna on February 17, 2017, 06:32:09 am
thank you!
Post by: RuiAce on February 17, 2017, 08:26:11 am
Hi I'm having a bit of trouble with these questions (cambridge textbook)
thank you!
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Post by: RuiAce on February 17, 2017, 08:57:54 am
Hi I'm having a bit of trouble with these questions (cambridge textbook)This whole concept of retelling, especially for just a 2U level, is exactly why the extension Cambridge in the questions do not reflect the scope of the HSC, and purely exist for self interest. In general I will not do these questions. However, the previous one was not so difficult - it was only tricky because of the wording.
thank you!
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Post by: Hplovers on February 17, 2017, 05:58:56 pm
I am having trouble with these two questions:
1. Find the area bounded by the curve x = y^2 - 2y - 3 and the y-axis.
2. Find the area bounded by the curve x = -y^2 - 5y - 6
Thanks!
Post by: Kle123 on February 17, 2017, 07:52:51 pm
Hi!
I am having trouble with these two questions:
1. Find the area bounded by the curve x = y^2 - 2y - 3 and the y-axis.
2. Find the area bounded by the curve x = -y^2 - 5y - 6
Thanks!
Since i don't exactly know what is troubling you (may it be that it is the area with y-axis instead of x?), i just assumed you had the background knowledge to understand the following working.
Post by: jamonwindeyer on February 18, 2017, 12:15:19 am
Since i don't exactly know what is troubling you (may it be that it is the area with y-axis instead of x?), i just assumed you had the background knowledge to understand the following working.
Legend! Thanks for the awesome answer Kle123 ;D
Post by: bananna on February 18, 2017, 04:22:26 pm
had some trouble with this q:
A jar contains red buttons and white buttons in the ratio of 3:2. If three buttons are chosen at random from the jar, find the probability that:
a) exactly two are red
b) not more than one is white
thank you!!
Post by: RuiAce on February 18, 2017, 04:35:15 pm
hello!
had some trouble with this q:
A jar contains red buttons and white buttons in the ratio of 3:2. If three buttons are chosen at random from the jar, find the probability that:
a) exactly two are red
b) not more than one is white
thank you!!
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Post by: Kle123 on February 18, 2017, 05:08:18 pm
Hey Rui, great explanation. thanks for teaching me that in ratio questions we should treat it as replacement. But aren't three buttons being chosen? I haven't revised probability since prelims, but shouldn't the answer to part (a) be (3/5)*(3/5)*(2/5)*3?(times 3 as there are 3 combination)
______________________________________________
and part (b) 1-(P(3white) +P(2white,1red)=== 1-((2/5)^3+((2/5)^2*(3/5)*3)
Or
P(2red,1white)+P(3red)=((3/5)^2*(2/5)*3)+(3/5)^3
Post by: RuiAce on February 18, 2017, 05:12:09 pm
Hey Rui, great explanation. thanks for teaching me that in ratio questions we should treat it as replacement. But aren't three buttons being chosen? I haven't revised probability since prelims, but shouldn't the answer to part (a) be (3/5)*(3/5)*(2/5)*3?(times 3 as there are 3 combination)I might've misread the question. If 3 buttons are being picked then a) is definitely correct and at least the second answer to b) is correct (didn't check the complement)
and part (b) 1-(P(3white) +P(2white,1red)=== 1-((2/5)^3+((2/5)^2*(3/5)*3)
Or
P(2red,1white)+P(3red)=((3/5)^2*(2/5)*3)+(3/5)^3
Edit: Yep, misread. Looks great to me.
Post by: Mathew587 on February 19, 2017, 05:09:17 pm
Post by: jamonwindeyer on February 19, 2017, 05:24:38 pm
:( can someone tell me why i'm wrong...
Hey mate, I'm not 100% sure since I didn't see the debate on this question above, but it looks correct to me and Rui agrees? Why do you say it is incorrect? ;D
Post by: Mathew587 on February 19, 2017, 07:00:16 pm
Post by: jamonwindeyer on February 19, 2017, 07:52:05 pm
Oh I just realised that I misread part a) solution that Rui answered. When checking I kept typing (3/5)^3*3 instead of what it should have been. However didn't I do part b) wrong as (3/5)^2 * (2/5)^3 + (3/5)^3 = 747/3125 instead of it being my answer of 81/125?
No Part B looks okay in the version you posted! You've got:
That's correct! The 54/125 comes from Part A :)
Post by: Fahim486 on February 19, 2017, 08:57:40 pm
I was just wondering anytime we do working outs for integration, do we have to use the integral sign in a step, like for example if a question asks to integrate d^2y/dx^2 = x^2+3, do we have to write dy/dx = (integral sign) x^2+3 and then actually integrate or can we skip that step and integrate straight away. I wanted to ask this question because my tutor says you have to so that the HSC markers don't get confused but my school maths teacher has never said anything like this or taught to do this.
Thanks!
Post by: jamonwindeyer on February 19, 2017, 09:00:59 pm
Hi everyone,
I was just wondering anytime we do working outs for integration, do we have to use the integral sign in a step, like for example if a question asks to integrate d^2y/dx^2 = x^2+3, do we have to write dy/dx = (integral sign) x^2+3 and then actually integrate or can we skip that step and integrate straight away. I wanted to ask this question because my tutor says you have to so that the HSC markers don't get confused but my school maths teacher has never said anything like this or taught to do this.
Thanks!
Hey! You can definitely integrate straight away if you need to - The integral sign is not a bad thing but it doesn't overly help what you are doing, in terms of clarity at least - It is pretty clear you've integrated based on the LHS of your expression ;D
I always just integrated, if that puts you at ease :)
Post by: bananna on February 20, 2017, 07:03:38 am
I'm revising yr 11 work because I forgot a lot of content
Can someone pls help me with these q's
1) what is the solution to the equation 2cosβ = -√3 for 0° ≤ β ≤ 360° ?
2) what is the solution to the eqn cos(θ/2 + 20°) for 0° ≤ θ ≤ 90° ?
Thank you :)
Post by: kiwiberry on February 20, 2017, 07:56:59 am
Hi!
I'm revising yr 11 work because I forgot a lot of content
Can someone pls help me with these q's
1) what is the solution to the equation 2cosβ = -√3 for 0° ≤ β ≤ 360° ?
2) what is the solution to the eqn cos(θ/2 + 20°) for 0° ≤ θ ≤ 90° ?
Thank you :)
For 1), cosβ=-√3/2
Remember that cos is negative in the second and third quadrants
So β=150°, 210°!
For 2), did you forget to finish typing the equation? But use the same method as the first, solve for θ/2 + 20° and then rearrange to find θ :)
Post by: jamonwindeyer on February 20, 2017, 10:00:58 am
Hi!
I'm revising yr 11 work because I forgot a lot of content
Can someone pls help me with these q's
1) what is the solution to the equation 2cosβ = -√3 for 0° ≤ β ≤ 360° ?
2) what is the solution to the eqn cos(θ/2 + 20°) for 0° ≤ θ ≤ 90° ?
Thank you :)
Just to add to kiwi berry's answer, remember that the exact ratio results are on your reference sheet if you've forgotten them (the \(\sin{\theta}=\frac{1}{2}\implies\theta=30^\circ\) stuff!) ;D
I'd be happy to show the full working for that second example if you provide the full question - Just to show you how to tackle that sort of question that needs rearranging ;D
Post by: smile123 on February 20, 2017, 11:54:46 am
PLEASE HELP
Post by: RuiAce on February 20, 2017, 11:55:59 am
Equation of a line passing through (−5,7) and having infinite slope will be:A line with infinite slope is just a vertical line.
PLEASE HELP
All vertical lines are of the form x = something
So since the point (-5,7) passes through it, i.e. a point where x=-5, the line is also x=-5
Post by: katnisschung on February 20, 2017, 01:35:20 pm
i drew the graphs but i don't know how to find the point of intersection
using simultaneous equations...
i got x^3+3x-4=0... do 2 unit need to know how to simplify an equation with a cube power specific to this case?
Post by: katnisschung on February 20, 2017, 01:38:25 pm
find the area between y=x^2+2x-8
and y=2x+1
again i sketched the graphs,
found their point of intersection to be 3 and -3 using simultaneous equations
but i'm having trouble finding what i need to integrate??
the areas are not bounded by the x-axis, in fact they fall below... so
do i need to split it up becos some of the areas below would be negative?
confused...essentially pls show me the working out thanks :D
Post by: kiwiberry on February 20, 2017, 01:45:18 pm
find the area between the two curves: y=-3x+4 & y=x^3 and the x-axis
i drew the graphs but i don't know how to find the point of intersection
using simultaneous equations...
i got x^3+3x-4=0... do 2 unit need to know how to simplify an equation with a cube power specific to this case?
We can use the factor theorem and test some numbers to try and find a factor!
If P(x)=x3+3x-4, P(a)=0 when x-a is a factor
Subbing x=1, P(1)=1+3-4=0 so (x-1) is a factor
From here you can long divide to find the other factor and you'll get (x-1)(x2+x+4)=0
So the POI is at x=1 :)
Post by: RuiAce on February 20, 2017, 01:55:42 pm
We can use the factor theorem and test some numbers to try and find a factor!Problem. This is the 2U section, where polynomials are not taught. Therefore
If P(x)=x3+3x-4, P(a)=0 when x-a is a factor
Subbing x=1, P(1)=1+3-4=0 so (x-1) is a factor
From here you can long divide to find the other factor and you'll get (x-1)(x2+x+4)=0
So the POI is at x=1 :)
find the area between the two curves: y=-3x+4 & y=x^3 and the x-axisNo, you are not. At least not for weird ones.
i drew the graphs but i don't know how to find the point of intersection
using simultaneous equations...
i got x^3+3x-4=0... do 2 unit need to know how to simplify an equation with a cube power specific to this case?
Post by: kiwiberry on February 20, 2017, 01:58:14 pm
Problem. This is the 2U section, where polynomials are not taught.
Oh oops sorry, my bad 😅
Post by: itswags98 on February 20, 2017, 03:28:28 pm
Post by: RuiAce on February 20, 2017, 03:37:52 pm
Hey im trying to revise. Do you think you could help me out there with an explanation and working out of each of them? Thank you.
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:P
Ok, I think E is just a dud.
Post by: jakesilove on February 20, 2017, 03:39:33 pm
Hey im trying to revise. Do you think you could help me out there with an explanation and working out of each of them? Thank you.
Hey!
By cancelling terms.
To find the slope of a tangent, we have to find the first derivative.
Dammit Rui
Post by: RuiAce on February 20, 2017, 03:53:32 pm
Dammit RuiThis never gets old 8)
Post by: itswags98 on February 20, 2017, 05:25:27 pm
Dammit Rui
Thanks for your help guys :3 Especially Rui ahah
Post by: Hplovers on February 20, 2017, 05:47:01 pm
1. Find the area enclosed between the curve y=x^3 , the x-axis and the line y= -3x+4
Im not sure what my first equation should be... (not sure what is considered top or bottom curve, and I cant find the intersection point :'( )
Thanks!
Post by: smile123 on February 20, 2017, 05:53:10 pm
Post by: RuiAce on February 20, 2017, 06:52:22 pm
Please help! :o
1. Find the area enclosed between the curve y=x^3 , the x-axis and the line y= -3x+4
Im not sure what my first equation should be... (not sure what is considered top or bottom curve, and I cant find the intersection point :'( )
Thanks!
please help
Post by: jamonwindeyer on February 20, 2017, 09:07:54 pm
This never gets old 8)
We seriously need a dibs system :P
Post by: teapancakes08 on February 20, 2017, 09:49:24 pm
I've (sorta) drawn the graph but am stumped on finding the definite integrals. Should be fine with using the formula after that though. I think. Correct me if I'm wrong but would you use both the volume formula and the sums and differences of two areas (or volumes in this case)? Something like pi integral (x^2)^2 - pi integral ((x-2)^2)^2 (I have no idea what I'm doing, I'm sorry >.<)
Post by: jamonwindeyer on February 20, 2017, 09:51:21 pm
21. Find the volume of the solid of revolution formed if the area enclosed between the curves y = x^2 and y = (x - 2)^2 is rotated about the x-axis.
I've (sorta) drawn the graph but am stumped on finding the definite integrals. Should be fine with using the formula after that though. I think. Correct me if I'm wrong but would you use both the volume formula and the sums and differences of two areas (or volumes in this case)? Something like pi integral (x^2)^2 - pi integral ((x-2)^2)^2 (I have no idea what I'm doing, I'm sorry >.<)
Hey! Assuming they mean that little triangular looking region enclosed by the parabolas and the x-axis, just take the volume in two halves! One from 0 to 1 for \(y=x^2\), then the other for 1 to 2 for \(y=(x-2)^2\) ;D
Oh and never apologise for asking a question! Seriously all good, we are happy to help ;D
Post by: Hplovers on February 20, 2017, 10:09:33 pm
Sorry i'm still confused
I'm not sure how to find the point of intersection; please explain (2 unit level) and the crossing over of the curves makes it hard for me to know which curve i'm subtracting from which (which is one the bottom).
When you tell me the intersection is x = 1 I can solve it fine, but I just don't know which one is top and which is bottom.
(attached graph)
Also there's another question that i've tried so many times but keep getting confused (probably just with my algebra).
Find the exact area enclosed between the curve y= sqrt(4-x^2) and the line x-y+2=0
I would really appreciate the working out. :)
Thankyou so so much for your help but please don't assume that I know as much as you do, it's not obvious to me, otherwise I wouldn't have asked for the help; i've tried the questions many many time but I keep doing something wrong.
Thankyou :)
Post by: RuiAce on February 20, 2017, 10:16:40 pm
Sorry i'm still confused
I'm not sure how to find the point of intersection; please explain (2 unit level) and the crossing over of the curves makes it hard for me to know which curve i'm subtracting from which (which is one the bottom).
When you tell me the intersection is x = 1 I can solve it fine, but I just don't know which one is top and which is bottom.
(attached graph)
Also there's another question that i've tried so many times but keep getting confused (probably just with my algebra).
Find the exact area enclosed between the curve y= sqrt(4-x^2) and the line x-y+2=0
I would really appreciate the working out. :)
Thankyou so so much for your help but please don't assume that I know as much as you do, it's not obvious to me, otherwise I wouldn't have asked for the help; i've tried the questions many many time but I keep doing something wrong.
Thankyou :)
Post by: RuiAce on February 20, 2017, 10:21:14 pm
Post by: bananna on February 21, 2017, 06:23:30 am
No Part B looks okay in the version you posted! You've got:
That's correct! The 54/125 comes from Part A :)
awesome discussion on this q :)
everyone's working out gave me a great outline, but I think I got it when my cousin helped me-he drew a tree diagram
thanks to all :)
Post by: jamonwindeyer on February 21, 2017, 10:23:07 am
awesome discussion on this q :)
everyone's working out gave me a great outline, but I think I got it when my cousin helped me-he drew a tree diagram
thanks to all :)
Awesome! Tree diagrams are almost always the saviour ;)
Post by: smile123 on February 21, 2017, 12:24:16 pm
Post by: RuiAce on February 21, 2017, 12:31:55 pm
please helpThis is Extension 2 mathematics. Complex numbers are only a part of Extension 2 - I have no clue why it should appear here in 2U.
Post by: jakesilove on February 21, 2017, 12:33:40 pm
please help
So the answer is the last option :)
Post by: smile123 on February 21, 2017, 12:44:13 pm
Post by: kylesara on February 21, 2017, 02:13:07 pm
"Show that the volume of a sphere is given by the formula V= 4/3 pi.r^3 by rotating the semicircle y= (r^2-x^2)^1/2 about the x axis.
Thanks
Post by: laurenf58 on February 21, 2017, 02:16:32 pm
Thanks!
Post by: jamonwindeyer on February 21, 2017, 02:46:52 pm
How do you find the primitive function of (-x+1)/3 ???
Thanks!
Hey Lauren! You can consider it term by term:
Remember, to find the primitive, we add one to the power, then divide by the new power:
Don't forget the constant! ;D there are other ways of course but I think this is the simplest ;D
Post by: jamonwindeyer on February 21, 2017, 02:52:58 pm
Hi could i please have help with this hw question.
"Show that the volume of a sphere is given by the formula V= 4/3 pi.r^3 by rotating the semicircle y= (r^2-x^2)^1/2 about the x axis.
Thanks
Sure! So that is a typical semicircle, radius \(r\), sitting above the \(x\) axis. Therefore, it crosses the \(x\) axis at \(-r\) and \(r\): Those are the bounds of our integral!
If you know the volume formula and how it works, this should be reasonably clear on its own, but definitely let me know if it isn't ;D the only thing to note, besides the fact that we use the function squared as usual, is that we are integrating with respect to \(x\) ONLY - We treat \(r\) as a constant! :)
We also could have used symmetry to simplify the integral, by evaluating this instead:
You would get the same answer :) let me know if any of those steps needs clarifying!
Post by: teapancakes08 on February 21, 2017, 04:55:33 pm
Hey! Assuming they mean that little triangular looking region enclosed by the parabolas and the x-axis, just take the volume in two halves! One from 0 to 1 for \(y=x^2\), then the other for 1 to 2 for \(y=(x-2)^2\) ;D
Oh and never apologise for asking a question! Seriously all good, we are happy to help ;D
I got the answer ;D
Thank you for helping ^^
Post by: smile123 on February 21, 2017, 09:07:18 pm
Post by: Mathew587 on February 21, 2017, 11:20:18 pm
let o = theta
sin o = -1
o = -90 sin's positive in 1,2 quadrants
cos o = 0
o = 90 cos's positive in 1,4 quadrants
therefore 1st quadrant is constant
therefore o = 90 degrees = 180/2 = pi/2
:)
Post by: RuiAce on February 21, 2017, 11:21:13 pm
Post by: jamonwindeyer on February 22, 2017, 12:06:43 am
I got the answer ;D
Thank you for helping ^^
Woo! Nice work pancakes! ;D
Post by: Shadowxo on February 22, 2017, 12:12:56 am
please help
Disclaimer: I don't know the HSC content or what you know from 2U
As Rui said, I'd sketch a graph and use that. With a circle, you can find sin and cos by the angle. Sketch a unit circle. Let theta = ø
y=sinø, x=cosø, y=-1 and x=0 so you know the angle is 270º. Convert this into radians and you get 3π/2.
Post by: kylesara on February 22, 2017, 04:20:53 pm
Calculate area of the region bounded by x-axis and curves y= x^1/2 and y=6-x.
Thanks
Post by: Hplovers on February 22, 2017, 08:22:25 pm
Thankyou so so much for all your help on this! :)
Post by: RuiAce on February 22, 2017, 08:31:41 pm
Hi, could you please help me with finding the intercepts for these 2 graphs. I know how to find the area by integrating, its just this bit that is confusing me.
Calculate area of the region bounded by x-axis and curves y= x^1/2 and y=6-x.
Thanks
Post by: Zainbow on February 22, 2017, 10:23:29 pm
I'm a bit confused, how can you find the length of the arc only from this information?
Post by: RuiAce on February 22, 2017, 10:36:33 pm
A circle has a chord of 25 mm with an angle of π/6 subtended at the centre. Find, to 1 decimal place, the length of the arc cut off by the chord.Hint: Draw a diagram. You can use the cosine rule to find the radius of the sector you're interested in.
I'm a bit confused, how can you find the length of the arc only from this information?
Post by: jamonwindeyer on February 22, 2017, 10:36:52 pm
A circle has a chord of 25 mm with an angle of π/6 subtended at the centre. Find, to 1 decimal place, the length of the arc cut off by the chord.
I'm a bit confused, how can you find the length of the arc only from this information?
Hey! Bit of a tricky question, but what we need to do is form a triangle with the chord, and the two radii to the centre, and use the cosine rule to find the radii:
You can then use that, with the formula \(l=r\theta\), to find the answer!! Does that help? Happy to show more working if you need it ;D
Edit: What Rui said ;D
Post by: Shadowxo on February 22, 2017, 10:44:52 pm
25 = 2r*sin(π/12)
r = 12.5/sin(π/12)
Then l=rθ
Or use trig sin(θ)=O/H
sin(π/12)=12.5/r
r=12.5/sin(π/12)
Few different ways :)
Post by: jamonwindeyer on February 22, 2017, 10:47:04 pm
Or, alternatively, use chord length = 2rsin(θ/2)
25 = 2r*sin(π/12)
r = 12.5/sin(π/12)
Then l=rθ
Or use trig sin(θ)=O/H
sin(π/12)=12.5/r
r=12.5/sin(π/12)
Few different ways :)
Oh wow that last way is wayyy cleaner, thanks Shadow ;D ps - Is \(CL=2r\sin{\frac{\theta}{2}}\) a quotable result for VCE? Lucky devils ;)
Post by: Zainbow on February 22, 2017, 10:49:42 pm
Hint: Draw a diagram. You can use the cosine rule to find the radius of the sector you're interested in.
Hey! Bit of a tricky question, but what we need to do is form a triangle with the chord, and the two radii to the centre, and use the cosine rule to find the radii:
You can then use that, with the formula \(l=r\theta\), to find the answer!! Does that help? Happy to show more working if you need it ;D
Edit: What Rui said ;D
Or, alternatively, use chord length = 2rsin(θ/2)
25 = 2r*sin(π/12)
r = 12.5/sin(π/12)
Then l=rθ
Or use trig sin(θ)=O/H
sin(π/12)=12.5/r
r=12.5/sin(π/12)
Few different ways :)
Thanks everyone ;D this really helped
Post by: Shadowxo on February 22, 2017, 10:53:00 pm
Oh wow that last way is wayyy cleaner, thanks Shadow ;D ps - Is \(CL=2r\sin{\frac{\theta}{2}}\) a quotable result for VCE? Lucky devils ;)
Actually, I only had to do chords/arcs/circle equations in year 11, wasn't on the VCE year 12 course :P
\(CL=2r\sin{\frac{\theta}{2}}\) is basically a simplified version of using sin(θ)=O/H to find the radius. Pretty easy to use sin(θ)=O/H instead if the chord length formula isn't allowed
Post by: RuiAce on February 22, 2017, 11:01:59 pm
Edit: It's just a generalisation.
Post by: Shadowxo on February 23, 2017, 12:13:21 am
Uh. Ok. I will add that to the list of formulas I need to prove.
Edit: It's just a generalisation.
I always worked it out a different way, drawing it then splitting it into two right-angled triangles
Post by: RuiAce on February 23, 2017, 12:29:21 am
I always worked it out a different way, drawing it then splitting it into two right-angled trianglesMakes sense. I have a bias against splitting things up if unnecessary though.
Post by: jamonwindeyer on February 23, 2017, 12:54:11 am
Makes sense. I have a bias against splitting things up if unnecessary though.
Rui's all about togetherness ;)
Post by: smile123 on February 23, 2017, 08:53:45 am
Post by: jamonwindeyer on February 23, 2017, 11:15:01 am
Please help
Hey! So the expression that gives the number of infected cells after \(n\) days is:
Basically, you are solving for \(n\) when \(A=0\) - The problem is that it has no solutions! What you are looking for is the lowest \(n\) value in that equation for which the result gives a number of cells less than 1 - Meaning, the first time all the cells could feasibly be dead. At least, that is how I interpret it :) I think I'd go A?
Post by: smile123 on February 23, 2017, 11:58:07 am
Post by: RuiAce on February 23, 2017, 04:18:54 pm
Yet, if we're to be a bit technical here, we are using a continuous system to model a discrete system. It's physically impossible for there to be, say, 0.9 of an infected cell. Hence the actual answer is probably up to interpretation of the reader.
Post by: katnisschung on February 24, 2017, 08:40:07 pm
definite integrals?
Post by: jakesilove on February 24, 2017, 08:43:03 pm
can you use the integration power rule for
definite integrals?
If you mean does
Then yes, it certainly does! You just then have to sub in your limits of integration
Post by: katnisschung on February 24, 2017, 09:03:41 pm
If you mean does
Then yes, it certainly does! You just then have to sub in your limits of integration
not quite this one...
tried using it for this question kept getting the wrong answer
find the volume of the solid of revolution that is formed when the curve
y=x^2+2 is rotated about the x-axis from x=0 to x=2
Post by: anotherworld2b on February 24, 2017, 09:30:19 pm
Could i please have some help to understand it?
Post by: Shadowxo on February 24, 2017, 09:36:23 pm
not quite this one...
tried using it for this question kept getting the wrong answer
find the volume of the solid of revolution that is formed when the curve
y=x^2+2 is rotated about the x-axis from x=0 to x=2
that rule only works when it's x to the power of one, eg (7x + 4)5, x itself is only to the power of one but the whole bracket is to the power of 5 (or n)
Post by: RuiAce on February 24, 2017, 09:44:17 pm
I am struggling to understand how small changes in differentiation works/how to use it.Note to EVERYONE else - This is NOT in the HSC course. I have asked him to just post here for now.
Could i please have some help to understand it?
Post by: Shadowxo on February 24, 2017, 09:45:34 pm
For 1:
Gradient = f'(x)=2x+4
f'(4) = 2*4+4 = 12
Gradient = f'(x) = change in y divided by change in x = ∆y/∆x = 12
∆x = 0.02
∆y = 12*∆x = 12*0.02 = 0.24 approximately
If you find f(4.02)-f(4) aka change in y, it's 0.2404, which is quite close
Edit: Rui beat me by 23 seconds but I'll post it anyway in the hopes it'll help :)
Hope this helps :)
Post by: anotherworld2b on February 24, 2017, 11:25:10 pm
Could i also have help with this question please?
Edit: i have attached what i have tried to do
Post by: anotherworld2b on February 24, 2017, 11:39:39 pm
Post by: Shadowxo on February 24, 2017, 11:54:23 pm
Think of it like finding and drawing the tangent, and using that to approximate the change in y :)
With your question, for Surface area I think you meant
SA = 2*L2+4*L*w
Doing this, I'd use L instead of w so you don't have to square root it
Post by: anotherworld2b on February 25, 2017, 12:05:36 am
For Q7 would you differentiate SA and make it equal 0? Then solve it to find x?
I am not sure where to go from there.
We input x=4 as we want to find the gradient when x=4
Think of it like finding and drawing the tangent, and using that to approximate the change in y :)
With your question, for Surface area I think you meant
SA = 2*L2+4*L*w
(I'll continue looking over for anything else)
Post by: Shadowxo on February 25, 2017, 12:07:05 am
Post by: anotherworld2b on February 25, 2017, 12:10:40 am
thank you very much for your help ;D
I was wondering would you have any tips on dealing with questions on the application of differentiation? We were just taught this so I am still relatively a newbie that is slow to understand what to do.
We input x=4 as we want to find the gradient when x=4
Think of it like finding and drawing the tangent, and using that to approximate the change in y :)
With your question, for Surface area I think you meant
SA = 2*L2+4*L*w
Doing this, I'd use L instead of w so you don't have to square root it
Post by: RuiAce on February 25, 2017, 12:21:53 am
We input x=4 as we want to find the gradient when x=4Pretty sure this solution is correct. I'm just pointing out a technicality problem.
Think of it like finding and drawing the tangent, and using that to approximate the change in y :)
With your question, for Surface area I think you meant
SA = 2*L2+4*L*w
Doing this, I'd use L instead of w so you don't have to square root it
Not too fussed about it, but when differentiating, preferably use the parentheses: \(\frac{d(SA)}{dL}\). Otherwise, it causes confusion with this nonsense: \( \frac{dS}{dL}\times A\)
Post by: anotherworld2b on February 25, 2017, 12:40:43 am
Edit: I was working on this question but when i differentiated it i ended up 60? :o
Post by: jamonwindeyer on February 25, 2017, 12:50:21 am
I was working on this question but i ended up with 0 as an answer :o
Edit: I was working on this question but when i differentiated it i ended up 60? :o
Just a little error,
You divided instead of subtracting! ;D
Post by: jamonwindeyer on February 25, 2017, 12:53:18 am
ah I see :D
thank you very much for your help ;D
I was wondering would you have any tips on dealing with questions on the application of differentiation? We were just taught this so I am still relatively a newbie that is slow to understand what to do.
Oh, and on this, just lots of practice unfortunately. The process is always the same:
1 - Rearrange to get a single variable
2 - Use the first derivative to find critical point
It's just about doing these over and over until it becomes second nature, and you get used to all the different things that can get thrown at you ;D
Post by: Shadowxo on February 25, 2017, 01:02:26 am
Put things in terms of one thing (eg surface area in terms of L), if you can, use the easiest option - in this case putting it in terms of L so you don't have to use a square root
Make sure not to make any little mistakes. I saw with your attempt you made a little mistake saying 10001/2*6 = 6000^1/2, instead of 6√1000 = 60√10. I like to check over my working frequently to minimise mistakes
I think you basically have the hang of it. Can't think of much else to say, but if you want any advice or questions answered feel free to ask :D
Post by: anotherworld2b on February 25, 2017, 01:15:28 am
I found r= 40/pi
I'm not sure how to find y
Just a little error,
You divided instead of subtracting! ;D
Post by: jamonwindeyer on February 25, 2017, 01:18:29 am
I now see what I did wrong. Thank you :)
I found r= 40/pi
I'm not sure how to find y
Great work! Go back to the equation, \(y=120-2\pi r\), with that new value of \(r\) ;D
Post by: anotherworld2b on February 25, 2017, 01:28:44 am
Great work! Go back to the equation, \(y=120-2\pi r\), with that new value of \(r\) ;D
Post by: anotherworld2b on February 25, 2017, 01:48:25 am
Post by: jakesilove on February 25, 2017, 10:55:06 am
I am not sure what to do for this question ???
We'll have to do this without a graphing calculator. First, we need to find a function for the perimeter of the isosceles triangle. We know that the triangle has a fixed perimeter, so
Where x is the two equal sides, and y is the base (unequal) side. c is a constant. Now, we care about the area of the triangle. This can be calculated by
The base is going to be equal to y, and we can find the height using pythag.
So, our area function is going to be
We can rewrite our perimeter function like this
and sub every y for the function above.
Great! Now, we're looking for a maximum area. So, let's differentiate the function. Since you can use a graphic calculator, I'm going to use wolfram alpha.
Clearly, the turning point will occur when
I'll leave you to show that this is a maximum. Well, if two of the sides (ie. x) are equal to c/3, then the third side must be equal to c/3 (as the perimeter must add up to c). Therefore, the triangle is equilateral
Post by: anotherworld2b on February 25, 2017, 12:25:45 pm
We'll have to do this without a graphing calculator. First, we need to find a function for the perimeter of the isosceles triangle. We know that the triangle has a fixed perimeter, so
Where x is the two equal sides, and y is the base (unequal) side. c is a constant. Now, we care about the area of the triangle. This can be calculated by
The base is going to be equal to y, and we can find the height using pythag.
So, our area function is going to be
We can rewrite our perimeter function like this
and sub every y for the function above.
Great! Now, we're looking for a maximum area. So, let's differentiate the function. Since you can use a graphic calculator, I'm going to use wolfram alpha.
Clearly, the turning point will occur when
I'll leave you to show that this is a maximum. Well, if two of the sides (ie. x) are equal to c/3, then the third side must be equal to c/3 (as the perimeter must add up to c). Therefore, the triangle is equilateral
Post by: jakesilove on February 25, 2017, 12:34:12 pm
Like this?
Almost! Instead of 1/3 (which assumes that the perimeter of the triangle is 1), you should be writing c/3 (which assumes that the perimeter of the triangle is some length, c).
Post by: anotherworld2b on February 25, 2017, 12:48:48 pm
For these two questions i dont understand how to use the small changes rule?
I was able to find the radius for q11.
R= 6.180 but I'm not sure where to go from here
Almost! Instead of 1/3 (which assumes that the perimeter of the triangle is 1), you should be writing c/3 (which assumes that the perimeter of the triangle is some length, c).
Post by: jakesilove on February 25, 2017, 01:35:51 pm
Ah okay :)
For these two questions i dont understand how to use the small changes rule?
I was able to find the radius for q11.
R= 6.180 but I'm not sure where to go from here
Unfortunately, this isn't in the HSC! Sorry :(
Post by: anotherworld2b on February 25, 2017, 01:49:09 pm
Unfortunately, this isn't in the HSC! Sorry :(
Post by: jamonwindeyer on February 25, 2017, 02:05:05 pm
oh okay :)
Since you've started the 1st one, I'll show you the full working of the 2nd and hopefully it will help you finish the first!
Remember, the formula we use is:
This comes from rearranging the visually simpler approximation, \(\frac{\delta C}{\delta n}\approx\frac{dC}{dn}\) - Remember that the small change (on the left) is roughly equal to the derivative (on the right), when that derivative is evaluated at the correct point. Geometrically, this is the same as using a tangent to approximate a curve (as shadow said):
We also know that \(\delta n=1\), so:
So the additional cost will be $200, roughly. If you compare that to the actual cost found by substitution, it is $216, so we're reasonably close :)
So the steps are to find the derivative, and evaluate it at our base point (what we're increasing from), then use the formula ;D
Post by: anotherworld2b on February 25, 2017, 02:43:27 pm
I was wondering for the base point what would you do if it changes from 25 to 26? Would the base point be -1? (25-26?)
Since you've started the 1st one, I'll show you the full working of the 2nd and hopefully it will help you finish the first!
Remember, the formula we use is:
This comes from rearranging the visually simpler approximation, \(\frac{\delta C}{\delta n}\approx\frac{dC}{dn}\) - Remember that the small change (on the left) is roughly equal to the derivative (on the right), when that derivative is evaluated at the correct point. Geometrically, this is the same as using a tangent to approximate a curve (as shadow said):
We also know that \(\delta n=1\), so:
So the additional cost will be $200, roughly. If you compare that to the actual cost found by substitution, it is $216, so we're reasonably close :)
So the steps are to find the derivative, and evaluate it at our base point (what we're increasing from), then use the formula ;D
Post by: anotherworld2b on February 25, 2017, 03:18:55 pm
Post by: Shadowxo on February 25, 2017, 05:45:02 pm
The derivative at a certain point= dy/dx = gradient at that point= change in y divided by change in x (approximation, if that gradient stayed the same) = change in y per x
So you'd find f'(x), aka the gradient, sub in x=25
f'(25) = change in y divided by change in x = ∆y/∆x
x increases by 1 (x2-x1=1)
So f'(25)=∆y/1
∆y=f'(25)
This change in y is the approximation if the gradient were to stay the same from that point - like drawing a tangent at a point and using that to approximate the increase in y
Post by: smile123 on February 25, 2017, 05:51:00 pm
Post by: Shadowxo on February 25, 2017, 05:51:55 pm
I still dont really understand the steps to solve it ???
Remember, pi is a constant, not the variable r
So A'(r) = 2πr - only one step needed (also you should put it in terms of r : A'(r) instead of A'(x))
So where A=120, r=6.180
dA/dr = 2*π*6.180
Change in A = 1
Change in r : unknown
dA/dr = 1/∆r = 2*π*6.180
∆r = 1/(2π*6.180)
Hope this helps :)
Post by: Shadowxo on February 25, 2017, 05:57:59 pm
PLEASE HELP :) :)
|cos(2x)|=1
cos(2x) =1 or -1
2x = 0,π,2π,3π,4π,5π... (as cos of those values equals 1 or -1)
x = 0, π/2, π, 3π/2, 2π...
5 solutions within the domain
Let me know if you want me to explain anything further :)
Post by: anotherworld2b on February 25, 2017, 07:03:26 pm
Post by: Shadowxo on February 25, 2017, 09:05:31 pm
I am bit confused what to do for percentages
This is a bit of a tough question, and I'm not sure whether that would usually be tested. Also, remember you're differentiating A with respect to w (dA/dw) and k is a constant so you don't need to use the product rule to differentiate - treat is as though it is a number, like 4 or 7.
Anyway, here's the solution:
Post by: Fahim486 on February 25, 2017, 09:26:51 pm
Can someone pls help me with question 9 b) and c)
Thanks!
Post by: Fahim486 on February 25, 2017, 09:29:23 pm
Thanks again!!
Post by: Shadowxo on February 25, 2017, 10:07:25 pm
V is a max when dV/dx = 0
dV/dx = 12x2-440x+2400
12x2-440x+2400=0
Solve for x
for c) just substitute in the x value you found into V
5. We know that the rest of the perimeter, (not including the walls) equals 8
L+w=8 -> w=8-L
A = L*w = L*(8-L) = 8L-L2
Max area when dA/dL=0
dA/dL = 8-2L
8-2L=0
L = 4
w=8-L = 4
Max area when length =4, width =4
Post by: laurenf58 on February 26, 2017, 10:50:37 am
Post by: RuiAce on February 26, 2017, 11:03:03 am
Can someone please remind me how to do this? Thanks!
Post by: smile123 on February 26, 2017, 12:43:58 pm
Post by: RuiAce on February 26, 2017, 12:49:06 pm
PLEASE HELP
Post by: RuiAce on February 26, 2017, 01:19:51 pm
(http://uploads.tapatalk-cdn.com/20170225/aa514eea63d1a11191726f977f2910e6.jpg)
Post by: Shadowxo on February 26, 2017, 01:34:29 pm
PLEASE HELP
I already answered this question yesterday, next time please check if your question has been answered before reposting :)
If an answer doesn't explain it, say what you don't understand and we'll help!
Post by: Mathew587 on February 26, 2017, 08:13:29 pm
Everything is in the image
Thanks peeps :D
Post by: anotherworld2b on February 26, 2017, 08:17:53 pm
Post by: RuiAce on February 26, 2017, 08:33:41 pm
Hey allI'm not too sure what you did there. There's only one G.P. going on, not two.
Everything is in the image
Thanks peeps :D
Post by: Mathew587 on February 26, 2017, 08:40:47 pm
I'm not too sure what you did there. There's only one G.P. going on, not two.oh ok ty rui :)
Post by: Shadowxo on February 26, 2017, 09:27:40 pm
I'm not quite sure how to do these questions
18.That question's a bit of a tough one but I think this is how they want you to solve it.
Post by: anotherworld2b on February 27, 2017, 09:28:51 am
18.That question's a bit of a tough one but I think this is how they want you to solve it.
Post by: Arisa_90 on February 27, 2017, 09:50:04 am
Could someone help me understand the rules for trig and exponential differentiation? I don't know how to use the rules properly so I keep getting questions wrong.
Could someone also explain to me how to draw gradient function graphs as well as drawing the graph given the gradient function?
I would like to make step by step notes on these things
Post by: jamonwindeyer on February 27, 2017, 10:04:26 am
Hi there
Could someone help me understand the rules for trig and exponential differentiation? I don't know how to use the rules properly so I keep getting questions wrong.
Could someone also explain to me how to draw gradient function graphs as well as drawing the graph given the gradient function?
I would like to make step by step notes on these things
Hey Arisa! Welcome to the forums! ;D
So the rules for trig and exponential differentiation are on your reference sheet, but they all just involve multiplying by the derivative of the inside. So for example, consider \(y=\sin{x^2}\).
The 'inside' bit of the function is \(x^2\), the derivative of which is \(2x\). So we'll be multiplying by that. Now normally, the derivative of \(\sin\) is \(\cos\), so put it all together, and you get \(y'=2x\cos{x^2}\). Notice that the inside bit doesn't change!
This reflects the rule on your reference sheet, \(\frac{d}{dx}\sin{f(x)}=f'(x)\cos{f(x)}\) - Swap to cos and multiply by the inside derivative!
To improve your working there, why don't you upload some questions you've been struggling with and your attempts at solving them - We can try to show you the way to tackle them ;D
As for sketching gradient functions from regular graphs, again, it is just practice. But here are a few rules to help (possibly pop these in your notes):
- Maximum turning points turn into x-intercepts, going from positive to negative
- Minimum turning points turn into x-intercepts, going from negative to positive
- Inflexions turn into maxima/minima
- Intercepts are useless
I always approach these questions by marking the sign of the gradient at all points. That is, if it slopes up, mark with a '+'. If down, mark with a '-'. This will show you at a glance where the graph of your gradient function should be above the x-axis and where it should be below.
For going backwards, the rules are reversed a little:
- X-Intercepts turn into maxima/minima
- Turning points turn into inflexions
- Inflexions are useless
Again, the best way to improve here won't be to makes notes (though that is definitely a good thing to do). You need to practice! :)
Once again, welcome! Be sure to upload any specific questions you have so we can help you out! ;D
Post by: Arisa_90 on February 27, 2017, 10:17:59 am
I found some examples for trig differentiation that I am having trouble with
1. y= tanx^2
2. y= tan^3(2x)
3. y= sin^2(1+squareroot t)
4. y= square root cos(e^2t)
5. y= e^sin3t
6. y= (1+2x)^3 tan(1-squareroot x)
7. y=sint/cos2t
Hey Arisa! Welcome to the forums! ;D
So the rules for trig and exponential differentiation are on your reference sheet, but they all just involve multiplying by the derivative of the inside. So for example, consider \(y=\sin{x^2}\).
The 'inside' bit of the function is \(x^2\), the derivative of which is \(2x\). So we'll be multiplying by that. Now normally, the derivative of \(\sin\) is \(\cos\), so put it all together, and you get \(y'=2x\cos{x^2}\). Notice that the inside bit doesn't change!
This reflects the rule on your reference sheet, \(\frac{d}{dx}\sin{f(x)}=f'(x)\cos{f(x)}\) - Swap to cos and multiply by the inside derivative!
To improve your working there, why don't you upload some questions you've been struggling with and your attempts at solving them - We can try to show you the way to tackle them ;D
As for sketching gradient functions from regular graphs, again, it is just practice. But here are a few rules to help (possibly pop these in your notes):
- Maximum turning points turn into x-intercepts, going from positive to negative
- Minimum turning points turn into x-intercepts, going from negative to positive
- Inflexions turn into maxima/minima
- Intercepts are useless
I always approach these questions by marking the sign of the gradient at all points. That is, if it slopes up, mark with a '+'. If down, mark with a '-'. This will show you at a glance where the graph of your gradient function should be above the x-axis and where it should be below.
For going backwards, the rules are reversed a little:
- X-Intercepts turn into maxima/minima
- Turning points turn into inflexions
- Inflexions are useless
Again, the best way to improve here won't be to makes notes (though that is definitely a good thing to do). You need to practice! :)
Once again, welcome! Be sure to upload any specific questions you have so we can help you out! ;D
Post by: Shadowxo on February 27, 2017, 11:54:04 am
How do you know when its going from positive to negative and vice versa for the gradient function to the original function?
I found some examples for trig differentiation that I am having trouble with
1. y= tanx^2
2. y= tan^3(2x)
3. y= sin^2(1+squareroot t)
4. y= square root cos(e^2t)
5. y= e^sin3t
6. y= (1+2x)^3 tan(1-squareroot x)
7. y=sint/cos2t
Hi Arisa :)
When going from the gradient function to the original function, you don't know if the values on the original function are positive or negative - as when integrating you add +c. The gradient function shows you whether the function is going up or down - when the gradient is negative (below 0 on the gradient function), the original graph is going downwards, and when the gradient is positive (above 0 on the gradient function) it's going up. So the gradient function can help you determine the shape of the original graph, but not how far up or down it's translated (unless you're given a point on the original function).
So for your examples, you find the derivative of the outside and multiply by the derivative of the inside. If we represent these functions as g(f(x)), the derivative is g'(f(x)) * f'(x)
1. The derivative of tan is sec2, and the derivative of x2 is 2x. So the derivative of tan(x2) is sec2(x2) * 2x. If you want to graph this gradient function, you can see it's always negative when x is less than zero (sec2 is always positive) and positive when x>0, but the values fluctuate depending on the x value as cos2(x) is between 0-1
2. Same thing - derivative of tan3 is 3tan2*sec2, and the derivative of 2x is 2. So the derivative is
3tan2(2x)*sec2(2x)*2
Try some of the other questions and let me know if there's anything more you want to know / want clarification on :)
Post by: Hplovers on February 27, 2017, 05:22:26 pm
19. Find the volume of the solid formed when the line x+3y-1=0 is rotated about the x-axis from x=0 to x=8.
I keep getting the answer but in negative form and not sure what to do about this as I could simply add absolute value brackets but I know thats not what your supposed to do.
Thanks!!
Post by: RuiAce on February 27, 2017, 06:13:12 pm
Hi! I'd love your help, I keep getting confused with what do do when theres are area in the negative field of the graph...
19. Find the volume of the solid formed when the line x+3y-1=0 is rotated about the x-axis from x=0 to x=8.
I keep getting the answer but in negative form and not sure what to do about this as I could simply add absolute value brackets but I know thats not what your supposed to do.
Thanks!!
A volume should never end up negative. If this happens, you should post up your working so we can look out for mistakes.
Post by: cxmplete on February 27, 2017, 08:41:31 pm
how would i do question 14, im so confused!
Post by: RuiAce on February 27, 2017, 09:02:25 pm
hi!
how would i do question 14, im so confused!
(Note. We have 1 A0 paper. AND 2 A1 papers. AND 4 A2 papers. And so on. And we're stacking ALL of these papers on top of each other.)
Post by: cxmplete on February 27, 2017, 09:15:30 pm
Post by: anotherworld2b on February 27, 2017, 11:22:48 pm
Help is appreciated ;D
Post by: anotherworld2b on February 28, 2017, 12:14:48 am
Post by: Shadowxo on February 28, 2017, 12:24:01 am
I was also wondering how do you differentiate this function where g is a constant ?
You just need to remember that 2,π and g are all constants so you can just leave them out the front and only worry about the variable.
Post by: Arisa_90 on February 28, 2017, 12:50:45 am
Does constant simply mean that you do not change them when you differentiate them?
I also wanted to ask what would you do if you were ask to draw f(x) if given f''(x) and vice versa? Would this be a possible question that could be asked?
You just need to remember that 2,π and g are all constants so you can just leave them out the front and only worry about the variable.
Post by: Jakeybaby on February 28, 2017, 01:00:10 am
how do you know that 2 and π are constants?Constants are just numbers that stay the same forever (they aren't variable). The number 2 will always be 2, no matter what - this is the same for pi (3.1415....), pi never changes, it's always they same value. Hence, they are constant.
Does constant simply mean that you do not change them when you differentiate them?
I also wanted to ask what would you do if you were ask to draw f(x) if given f''(x) and vice versa? Would this be a possible question that could be asked?
Regarding your graphing question, this is a very common question within the SACE curriculum, not too sure about HSC, but I'd imagine so. With these questions, it is extremely important to understand what f(x), f'(x) and f''(x) represent.
For example: f'(x) represents the rate of change of some quantity (the slope) & f''(x) represents the rate of change of the slope. I find that it is most useful to think about what f'(x) and f''(x) mean when they are equal to 0.
When f'(x) = 0, you have either;
- Maximum point
- Minimum point
- or a Point of Inflection
Post by: Shadowxo on February 28, 2017, 01:08:39 am
how do you know that 2 and π are constants?
Does constant simply mean that you do not change them when you differentiate them?
I also wanted to ask what would you do if you were ask to draw f(x) if given f''(x) and vice versa? Would this be a possible question that could be asked?
Constants have values that do not change - their value is constant. 2 is equal to 2, a is equal to a, pi is equal to pi. The value of x can change depending on the x value - x is a variable (changes).
If you differentiate a constant, it equals zero, and the derivative of a constant times a function is equal to the constant times the derivative of the function. eg:
derivative of 2 =0
derivative of 2x3 = derivative of 2(x3) = 2*(3x2) = 6x2
With regards to the graphing, I'm not very familiar with the HSC curriculum but they wouldn't give you f''(x) and just say to graph it. They may ask you to find the points of inflection for a graph, or find f(x) / graph if you're given f''(x) plus when x=1, f'(x)=2 etc (in order to graph it, you'd need coordinates)
Jake does raise some good points though, you should know what each of f(x), f'(x) and f''(x) represent and how to use them.
Post by: anotherworld2b on February 28, 2017, 01:55:22 am
Post by: Arisa_90 on February 28, 2017, 02:00:41 am
Constants are just numbers that stay the same forever (they aren't variable). The number 2 will always be 2, no matter what - this is the same for pi (3.1415....), pi never changes, it's always they same value. Hence, they are constant.
Regarding your graphing question, this is a very common question within the SACE curriculum, not too sure about HSC, but I'd imagine so. With these questions, it is extremely important to understand what f(x), f'(x) and f''(x) represent.
For example: f'(x) represents the rate of change of some quantity (the slope) & f''(x) represents the rate of change of the slope. I find that it is most useful to think about what f'(x) and f''(x) mean when they are equal to 0.
When f'(x) = 0, you have either;
- Maximum point
- Minimum point
- or a Point of Inflection
Post by: jamonwindeyer on February 28, 2017, 08:20:18 am
I get incredibly confused about how to draw a graph from f '(x) to f(x)
Why don't you give us an example you are struggling with! We'll work through it for you to show you the process!
Post by: jamonwindeyer on February 28, 2017, 08:22:56 am
How would you do this question?
Hey! I'm on my phone so can't do working properly (someone else might tag in), but the idea is to substitute 1.06L in place of just L (to represent the 6% increase); rearrange the expression to get the regular fraction with a constant out the front - Pull the sqrt(1.06) out the front of the traction. That represents the change in T - Then convert to percentage
Post by: Arisa_90 on February 28, 2017, 10:20:02 am
Why don't you give us an example you are struggling with! We'll work through it for you to show you the process!
Post by: Shadowxo on February 28, 2017, 10:44:01 am
Help with these questions please
So for the first, question you know there's a turning point / point of inflection where x=1 and x=2 (dy/dx=0). They also tell us that the graph cuts the x axis at the origin, and no where else. Therefore there's only one x intercept, and it's at (0,0) - this is a point on the graph.
We also know dy/dx<0 between 1 and 2 and no where else, so we know between 1 and 2 the gradient is negative and therefore the graph is going down, and everywhere else it's going up. So the graph starts by going up, passing through (0,0), then starts going down where x=1 - so we know this is a local maximum turning point - then starts going up again where x=2 (before it hits the x axis) - so we know this is a local minimum turning point, then continues going up. Does that help with graphing that? :)
With the second question, you seem to have the right idea, but remember the graph of f(x) can be as far up and down as you like, so the first one doesn't have to be so far up. For these, just look at the values and how positive/negative they are to see how steep the graph is and use the gradient to get the general shape :)
How would you do this question?
Jamon's right, just sub in L=1.06L and divide the T values, the constants cancel out and you're left with √(1.06L)/√(L) = √1.06 = 1.03 therefore 3% change
Post by: Arisa_90 on February 28, 2017, 11:14:25 am
does it mean that on the f '(x) graph if its below the x axis its negative so the f (x) should be decreasing? how do you know what the gradient is? especially for the third image?
So for the first, question you know there's a turning point / point of inflection where x=1 and x=2 (dy/dx=0). They also tell us that the graph cuts the x axis at the origin, and no where else. Therefore there's only one x intercept, and it's at (0,0) - this is a point on the graph.
We also know dy/dx<0 between 1 and 2 and no where else, so we know between 1 and 2 the gradient is negative and therefore the graph is going down, and everywhere else it's going up. So the graph starts by going up, passing through (0,0), then starts going down where x=1 - so we know this is a local maximum turning point - then starts going up again where x=2 (before it hits the x axis) - so we know this is a local minimum turning point, then continues going up. Does that help with graphing that? :)
With the second question, you seem to have the right idea, but remember the graph of f(x) can be as far up and down as you like, so the first one doesn't have to be so far up. For these, just look at the values and how positive/negative they are to see how steep the graph is and use the gradient to get the general shape :)
Jamon's right, just sub in L=1.06L and divide the T values, the constants cancel out and you're left with √(1.06L)/√(L) = √1.06 = 1.03 therefore 3% change
Post by: jamonwindeyer on February 28, 2017, 11:20:48 am
I'm still confused about how to use the gradient.
does it mean that on the f '(x) graph if its below the x axis its negative so the f (x) should be decreasing? how do you know what the gradient is? especially for the third image?
Yeah that's it! So if \(f'(x)\) is below the x-axis (is negative), then \(f(x)\) has a negative gradient in that region and is thus decreasing!
Third image? I've only got the two, perhaps missing an attachment? ;D
Post by: Arisa_90 on February 28, 2017, 11:46:12 am
I meant the second diagram in the second image
Yeah that's it! So if \(f'(x)\) is below the x-axis (is negative), then \(f(x)\) has a negative gradient in that region and is thus decreasing!
Third image? I've only got the two, perhaps missing an attachment? ;D
Post by: jamonwindeyer on February 28, 2017, 12:04:31 pm
Does it matter where you draw it decreasing? Like above or below the x axis?
I meant the second diagram in the second image
Nope! There is no way to know whether it will be above or below - When you draw \(f(x)\) from \(f'(x)\), the axes mean nothing. You don't have the information to know where the intercepts are, because integration introduces an unknown constant.
Right! So in that image, moving from left to right, the gradient starts almost at zero (it is close to flat, but not quiiite flat, so it is a very small negative gradient). It then decreases, becoming more and more negative, approaching a vertical line. So, on the LHS of the graph, the gradient moves from 0 to \(-\infty\). So you just draw any line that moves from 0 to \(-\infty\), never touching either (that's the line in pencil).
On the RHS, it is reversed. We go from a huge positive gradient and slowly approach zero, never touching. This explains the shade of the other side of the graph.
Note that any graph that looks like it does in pencil, shifted UP or DOWN without changing the shape, is also correct.
I think you've got it pretty much nailed by the way you are talking! :)
Post by: Arisa_90 on February 28, 2017, 01:15:32 pm
Nope! There is no way to know whether it will be above or below - When you draw \(f(x)\) from \(f'(x)\), the axes mean nothing. You don't have the information to know where the intercepts are, because integration introduces an unknown constant.
Right! So in that image, moving from left to right, the gradient starts almost at zero (it is close to flat, but not quiiite flat, so it is a very small negative gradient). It then decreases, becoming more and more negative, approaching a vertical line. So, on the LHS of the graph, the gradient moves from 0 to \(-\infty\). So you just draw any line that moves from 0 to \(-\infty\), never touching either (that's the line in pencil).
On the RHS, it is reversed. We go from a huge positive gradient and slowly approach zero, never touching. This explains the shade of the other side of the graph.
Note that any graph that looks like it does in pencil, shifted UP or DOWN without changing the shape, is also correct.
I think you've got it pretty much nailed by the way you are talking! :)
Post by: jamonwindeyer on February 28, 2017, 01:58:28 pm
I see. I just wanted to ask well if there is an inflection in a f (x) graph does is simply mean it is a either a max or a min point depending on the gradient on either side for the f ' (x) graph? eg. if on the f(x) graph it is decreasing before the inflection point then increasing would it mean that on the f '(x) graph the line will be below the x axis touches the x axis(x intercept where the inflection point was) then goes above x axis?
Careful, you started correctly by saying that an inflexion point becomes a max or a min. Then you swapped to saying it is an intercept - Inflexions become critical points (maxima/minima!)
So, if your curve is concave down before the inflexion, and concave up after the inflexion, it becomes a maximum. Otherwise, it becomes a minimum :)
Post by: Arisa_90 on February 28, 2017, 02:13:08 pm
Would the maximum point touch the x axis? Example below.
Must all max and min points touch the x axis?
Careful, you started correctly by saying that an inflexion point becomes a max or a min. Then you swapped to saying it is an intercept - Inflexions become critical points (maxima/minima!)
So, if your curve is concave down before the inflexion, and concave up after the inflexion, it becomes a maximum. Otherwise, it becomes a minimum :)
Post by: jamonwindeyer on February 28, 2017, 02:59:06 pm
is maxima and minima simply maximum and minium points?
Would the maximum point touch the x axis? Example below.
Must all max and min points touch the x axis?
Yes they are!
Yes, maxima/minima on \(f(x)\) become x-intercepts on \(f'(x)\) ;D
Post by: Arisa_90 on February 28, 2017, 03:07:37 pm
Yes they are!
Yes, maxima/minima on \(f(x)\) become x-intercepts on \(f'(x)\) ;D
Post by: Shadowxo on February 28, 2017, 04:35:28 pm
does that mean all inflections on f(x) will be a max or min that touches the x axis?
There are two types of inflections:
Where f'(x)=0 - stationary point of inflection. For this, f'(x) = zero and f''(x)=0, so it'll be a max/min that touches the x axis on the graph y=f'(x).
Where f'(x)≠0 - non-stationary point of inflection. For this, f'(x)≠0 and f''(x)=0, so it'll be max/min that does not touch the x axis on the graph y=f'(x)
Post by: Arisa_90 on February 28, 2017, 05:15:30 pm
Also does the y intercept play any important part in the process of graphing f (x)?
There are two types of inflections:
Where f'(x)=0 - stationary point of inflection. For this, f'(x) = zero and f''(x)=0, so it'll be a max/min that touches the x axis on the graph y=f'(x).
Where f'(x)≠0 - non-stationary point of inflection. For this, f'(x)≠0 and f''(x)=0, so it'll be max/min that does not touch the x axis on the graph y=f'(x)
Post by: laurenf58 on February 28, 2017, 05:17:59 pm
Post by: kiwiberry on February 28, 2017, 05:39:06 pm
Could I have some help with this question please? Thanks!
Post by: jamonwindeyer on February 28, 2017, 05:50:01 pm
how do you know from the f '(x) graph? How do you know from the attachement above for f'(x) the last major point should touch the x axis?
Also does the y intercept play any important part in the process of graphing f (x)?
Because it is a horizontal point of inflexion, which you know because it flattens out then keeps going again! Think the shape of \(x^3\)!
Nope, y-intercept is pretty useless in these sorts of problems!
Post by: Mathew587 on February 28, 2017, 09:35:27 pm
help with these please :)
Post by: jamonwindeyer on February 28, 2017, 09:53:49 pm
Hiya,
help with these please :)
Hey! For the first one to have real and rational roots, we just need to prove that the discriminant (\\Delta\) is greater than or equal to zero, and that it is a perfect square.
Clearly, in that factorised form, this will never be a negative number (squared numbers are always positive). So, since the discriminant is never negative, there are always real roots. Since it is a perfect square, they are always rational ;D
For that second one, consider the general quadratic \(y=ax^2+bx+c\), and substitute the three pairs of coordinates in to get three equations in terms of \(a\), \(b\) and \(c\) - Solve those simultaneously if you can (otherwise post where you get stuck and we can guide you the rest of the way!) :)
Post by: Mathew587 on February 28, 2017, 10:48:15 pm
Hey! For the first one to have real and rational roots, we just need to prove that the discriminant (\\Delta\) is greater than or equal to zero, and that it is a perfect square.
Clearly, in that factorised form, this will never be a negative number (squared numbers are always positive). So, since the discriminant is never negative, there are always real roots. Since it is a perfect square, they are always rational ;D
For that second one, consider the general quadratic \(y=ax^2+bx+c\), and substitute the three pairs of coordinates in to get three equations in terms of \(a\), \(b\) and \(c\) - Solve those simultaneously if you can (otherwise post where you get stuck and we can guide you the rest of the way!) :)
I followed what you told me and got:
18-4a+2b-c=0 for (-2,18) and -2-9a-3b-c=0 for (3,-2)
I then equated c's and got 20=-5a-5b - (1)
I then got 0=a+b+c for (1,0), found a and subbed that into the (1) equation and got c=4 which is wrong.
Can you please help me :)
Post by: RuiAce on February 28, 2017, 10:53:36 pm
I followed what you told me and got:
18-4a+2b-c=0 for (-2,18) and -2-9a-3b-c=0 for (3,-2)
I then equated c's and got 20=-5a-5b - (1)
I then got 0=a+b+c for (1,0), found a and subbed that into the (1) equation and got c=4 which is wrong.
Can you please help me :)
Post by: Mathew587 on February 28, 2017, 11:04:33 pm
Yup i did that and continued by subbing in (1,0) into the quadratic as previously stated but it didn't work out. Any working out for that?
Post by: RuiAce on February 28, 2017, 11:23:14 pm
Which gets what you got. And when I checked on WolframAlpha c=4 is correct.
(From here, just sub c=4 into any two of the three other equations, and solve them to find a and b)
Post by: Mathew587 on February 28, 2017, 11:51:15 pm
Yuck. Sorry, seems I had made a transcription error in my working out.
Which gets what you got. And when I checked on WolframAlpha c=4 is correct.
(From here, just sub c=4 into any two of the three other equations, and solve them to find a and b)
Oh ok. My confusion came from the fact that the answers had a different set of solutions with a=4, b=-3, c=7. Does that still make sense with the equation?
Post by: RuiAce on March 01, 2017, 07:17:44 am
Oh ok. My confusion came from the fact that the answers had a different set of solutions with a=4, b=-3, c=7. Does that still make sense with the equation?I checked the textbook. You're looking at the answers to Q3. Not Q2.
You get \(a=1, b=-5, c=4\) which when you sub it all back in to \(ax^2+bx+c\) you have \(x^2-5x+4\)
Post by: bananna on March 01, 2017, 07:27:45 am
With this q attached, I understand everything until the graph
I'm not completely sure how to draw it
any help would be appreciated ..thank you!
Post by: RuiAce on March 01, 2017, 07:31:57 am
hi!Can you please list the answers to the previous parts? Because all of them (except maybe a) and f)) will be necessary to draw the final sketch
With this q attached, I understand everything until the graph
I'm not completely sure how to draw it
any help would be appreciated ..thank you!
Post by: bananna on March 01, 2017, 07:42:12 am
Can you please list the answers to the previous parts? Because all of them (except maybe a) and f)) will be necessary to draw the final sketch
yes! sorry about that
a) y'=e^x(1+x)
y''= e^x(2+x)
b) minimum Stat.P at (-1,-e^-1)
c) p.o.i at (-2,-2e^-2)
d) curve passes thru origin
e) as x --> infinity
y--> infinity
f) y --> 0
y ---> 0
thanks
Post by: Shadowxo on March 01, 2017, 08:23:03 am
yes! sorry about that
a) y'=e^x(1+x)
y''= e^x(2+x)
b) minimum Stat.P at (-1,-e^-1)
c) p.o.i at (-2,-2e^-2)
d) curve passes thru origin
e) as x --> infinity
y--> infinity
f) y --> 0
y ---> 0
thanks
Can't do a full solution with a graph, but we know
There are points at (-1,-e-1), (-2,-2e-2), and (0,0). We know as x becomes more negative, y approaches 0 from below, so y=0 is an asymptote. We know there's a point of inflection at (-2,-2e-2), and a minimum TP at (-1,-e-1), then it goes up, passes through the origin and as x->infinity y-> infinity. Also, as e>2, we know -2e-2< -e-1 (closer to y=0). Now we have all the information to graph it. Starts close to y=0 as x is very negative, then as x gets larger the y value goes down slightly, until (-2,-2e-2) where there's a small point of inflection, then continues down, then passes through (-1,-e-1) and starts going up, passes through (0,0) and continues upwards.
Post by: RuiAce on March 01, 2017, 11:09:33 am
yes! sorry about that(http://i.imgur.com/f533YZ4.png)
a) y'=e^x(1+x)
y''= e^x(2+x)
b) minimum Stat.P at (-1,-e^-1)
c) p.o.i at (-2,-2e^-2)
d) curve passes thru origin
e) as x --> infinity
y--> infinity
f) y --> 0
y ---> 0
thanks
Here is a simple GeoGebra simulation of the graph, including many important details you have discovered. Note clearly the intercept, turning point and inflexion point provided, as well as the fact that the remainder of the question builds up to the asymptote y=0 for really large negative values of x.
Post by: Mathew587 on March 01, 2017, 06:00:06 pm
I checked the textbook. You're looking at the answers to Q3. Not Q2.
You get \(a=1, b=-5, c=4\) which when you sub it all back in to \(ax^2+bx+c\) you have \(x^2-5x+4\)
Oh my god rui sorry about that :-[ :-[ Ive only had 5hrs of sleep and my heads on fire with the upcoming half yearlies. Thank you anyways :)
Mathew
Post by: katnisschung on March 01, 2017, 08:25:32 pm
A score for the roll is determined as the product of the two numbers on the two uppermost faces.
b) if the cubes are rolled twice and the scores for each roll are added, what is the probability of a
combined score of at least 41
my answer= 5/1296
actual answer= 7/1296
I don't understand why counted the scores of the following (2 are repeated?)
scores of (16,25), (25,16), (25,25), (20, 25), (25,20)
(20, 25), (25,20)--> why are these two repeated?? is this an error?
Post by: Arisa_90 on March 01, 2017, 08:35:31 pm
I am not sure what to do when asked to differentiate tan
Post by: RuiAce on March 01, 2017, 08:39:33 pm
Are there particular rules for differentiating trig and exponential?
I am not sure what to do when asked to differentiate tan
18) two identical cubes (similar to dice) each having faces 0-5 are rolled.
A score for the roll is determined as the product of the two numbers on the two uppermost faces.
b) if the cubes are rolled twice and the scores for each roll are added, what is the probability of a
combined score of at least 41
my answer= 5/1296
actual answer= 7/1296
I don't understand why counted the scores of the following (2 are repeated?)
scores of (16,25), (25,16), (25,25), (20, 25), (25,20)
(20, 25), (25,20)--> why are these two repeated?? is this an error?
Getting 4 and 5 on the first try and getting 5 and 5 on the second try, is DIFFERENT to getting 5 and 5 on the first try and 4/5 on the second here. They're two distinct ways of achieving the same final outcome.
I believe the repetition occurs because in just the first try, there's TWO ways of getting a 4 and a 5. The first die can roll the 4, but so can the second.
Whereas there's only one way of getting two 4's
Post by: katnisschung on March 01, 2017, 08:50:23 pm
(16,25) and (25,16) would have to be repeated?
as its different to rolling a (25,16)(16,25)
so there would be 9 possibilities??
Post by: RuiAce on March 01, 2017, 08:53:35 pm
yeah..... cos wouldn't that meanCheck my correction. I realised I messed up.
(16,25) and (25,16) would have to be repeated?
as its different to rolling a (25,16)(16,25)
so there would be 9 possibilities??
Post by: Arisa_90 on March 01, 2017, 10:15:18 pm
I keep getting the questions wrong especially when tan is invovled
Post by: RuiAce on March 01, 2017, 10:16:50 pm
what are the rules for differentiating trig and exponential?You have already received an answer regarding a VERY similar question recently.
I keep getting the questions wrong especially when tan is invovled
If you still have problems, please post up example questions and we will point out any mistakes + offer guidance.
Post by: bananna on March 01, 2017, 10:31:03 pm
(http://i.imgur.com/f533YZ4.png)
Here is a simple GeoGebra simulation of the graph, including many important details you have discovered. Note clearly the intercept, turning point and inflexion point provided, as well as the fact that the remainder of the question builds up to the asymptote y=0 for really large negative values of x.
thank you! :)
Post by: Arisa_90 on March 01, 2017, 10:39:09 pm
You have already received an answer regarding a VERY similar question recently.
If you still have problems, please post up example questions and we will point out any mistakes + offer guidance.
Post by: RuiAce on March 01, 2017, 10:41:56 pm
Oh sorry. I thought I didn't post this question properly so I accidently reposted it.It happens. Better check back though.
Post by: Arisa_90 on March 02, 2017, 01:23:04 am
So for your examples, you find the derivative of the outside and multiply by the derivative of the inside. If we represent these functions as g(f(x)), the derivative is g'(f(x)) * f'(x)
1. The derivative of tan is sec2, and the derivative of x2 is 2x. So the derivative of tan(x2) is sec2(x2) * 2x. If you want to graph this gradient function, you can see it's always negative when x is less than zero (sec2 is always positive) and positive when x>0, but the values fluctuate depending on the x value as cos2(x) is between 0-1
2. Same thing - derivative of tan3 is 3tan2*sec2, and the derivative of 2x is 2. So the derivative is
3tan2(2x)*sec2(2x)*2
Try some of the other questions and let me know if there's anything more you want to know / want clarification on :)
Post by: jamonwindeyer on March 02, 2017, 01:30:38 am
to differentiate tan doesn't it have to be tanx? = sec^2x x 1?
That's what you do when you apply the chain rule to the basic tangent function, yes!
Is that what you mean? :)
Post by: michaelalt on March 02, 2017, 08:51:43 am
Post by: RuiAce on March 02, 2017, 11:48:13 am
Hi can someone remind me of how to do q. 13 + 14 of these locus and parabola questions? I forget what everthing is.Very quickly with Q13 (I'm in a class)
Post by: jakesilove on March 02, 2017, 04:04:07 pm
Hi can someone remind me of how to do q. 13 + 14 of these locus and parabola questions? I forget what everthing is.
Let's look at 14. We want a point that is always 3 units from the line
When it asks for distance, it always means perpendicular distance. So, we pick any point on the line, and move 3 units away at a perpendicular angle. We can go in either directions, so there will be two points, three units away, for each x value of the initial function. As we move along the function, the dots will move as well. Can you tell what the locus will be?
Clearly, it will be two straight lines, parallel to the initial line, but displaced three units in the perpendicular direction. We use the perpendicular distance formula to find two points, and then use these two point to find the equation of the straight line.
We need this distance to be three, so
Here, we can be smart
This is also obvious from the definition of absolute values. So,
are our two solutions! You'd expect the gradient to be the same (which they are), but the intercepts the be different (again, they are).
Note: There is almost certainly an easier way. If you draw a triangle, you can just figure out how far to the left/right the new lines should shift. Oh well
Post by: cxmplete on March 02, 2017, 07:16:22 pm
Post by: Arisa_90 on March 02, 2017, 07:19:01 pm
Post by: Shadowxo on March 02, 2017, 07:47:50 pm
What would be the best way to differentiate these trig functions?
They want you to use the quotient rule
Hint: derivative of sin(t) is cos(t) and the derivative of cos2(t) is -2cos(t)sin(t) (make sure you know how to do these)
Post by: RuiAce on March 02, 2017, 07:49:04 pm
What would be the best way to differentiate these trig functions?
Post by: Arisa_90 on March 02, 2017, 08:12:21 pm
Post by: RuiAce on March 02, 2017, 08:13:55 pm
How do know when to use chain rule?Function of a function.
Here, the function cos then gets smacked with a square. i.e. if \(f(x) = \cos x\) and \(g(x) = x^2\), then here we have \((\cos x)^2\), which is \( f(g(x))\)
Recognising when to use the rules is a process you MUST be familiar with.
Post by: kiwiberry on March 02, 2017, 08:29:23 pm
Hi! I'd love your help with question 4)b), I'm confused with what do with the given formula
An is the amount they owe after the nth repayment, so if the loan is to be repaid after 288 months, then A288=0. We can use the given formula with n=288 to find M:
The part on the left is the sum of a GP, with a=1, r=1.0075 and n=288:
Post by: Arisa_90 on March 02, 2017, 08:47:07 pm
Can i do this?
Y= (cost)^2 /t
Dy/dx = (2 x cos t x -sin t x 1) x t - cos^2t x 1 / t^2
Post by: RuiAce on March 02, 2017, 08:49:49 pm
for y= cos^2t /t
Can i do this?
Y= (cost)^2 /t
Dy/dx = (2 x cos t x -sin t x 1) x t - cos^2t x 1 / t^2
Post by: Arisa_90 on March 02, 2017, 09:41:42 pm
(2tanx^2 )( sec^2 x^2 )(2x)
What is the difference to tan^2x?
Do you do dy/dx = (sec^2x 2x) (2)?
Also would you do when its
x^3tan^2 x 2x?
Post by: RuiAce on March 02, 2017, 09:54:48 pm
I see. What do you do for a equation like this? Tanx^2. Would you use the chain rule because its to a power? So dy/dx =That last question is unclear. Please insert more bracketing.
(2tanx^2 )( sec^2 x^2 )(2x)
What is the difference to tan^2x?
Do you do dy/dx = (sec^2x 2x) (2)?
Also would you do when its
x^3tan^2 x 2x?
Post by: Arisa_90 on March 02, 2017, 11:11:22 pm
Thank you RuiAce
I was wondering what would be the best process to tackle this question?
Post by: RuiAce on March 03, 2017, 06:26:17 pm
I get it now ;DNote that when x=-1, y=0. Hence the x-intercept at x=-1 is important.
Thank you RuiAce
I was wondering what would be the best process to tackle this question?
Else, all we care is that:
- The curve is decreasing for all x < 0, and decreasing for all 0 < x < 1
- There is a stationary point at x = 0. (Note, if you analyse it carefully it should become clear that at x=0 we have a horizontal point of inflexion)
- The curve is increasing for all x > 1. (Note, comparing this with the fact it's decreasing for x < 1, we should be able to infer that at x=1 we have a local minimum)
- The x-intercept at x = -1 should also be a point of inflexion (Note: ORDINARY point of inflexion, not the horizontal point of inflexion like at x=0)
Provided the conditions are met, the actual shape of the graph is totally irrelevant.
Post by: cxmplete on March 04, 2017, 09:22:31 am
Post by: RuiAce on March 04, 2017, 09:57:27 am
how would i do question 4
Post by: katnisschung on March 05, 2017, 10:50:00 am
(things like max and min stationary point obviously but are all type of inflection points counted?)
Post by: Rathin on March 05, 2017, 11:02:12 am
what's considered a turning point?
(things like max and min stationary point obviously but are all type of inflection points counted?)
A turning point is a stationary point where the gradient of the function changes (increasing or deceasing).
NOTE:
All turning points are stationary points,
BUT Not all stationary points are turning points (e.g. horizontal point of inflexion)
Post by: Shadowxo on March 05, 2017, 11:05:34 am
what's considered a turning point?
(things like max and min stationary point obviously but are all type of inflection points counted?)
A turning point is where the gradient changes sign - ie a local maximum or minimum, and f'(x) will =0 at a turning point.
A stationary point is a point where the gradient equals zero. This includes all turning points (maximum + minimum) and some points of inflection (the ones where the gradient equals zero). f'(x)=0 at a stationary point.
A point of inflection is where the gradient goes from increasing to decreasing, or decreasing to increasing. f''(x)=0 and for stationary points of inflection (where the gradient =0), f'(x) =0 too.
Hope this helps :)
Post by: jamesq on March 05, 2017, 09:52:02 pm
Post by: RuiAce on March 05, 2017, 10:03:08 pm
Hey guys, quick question, when doing the Simpson's and Trapezoidal rules, the question may 'using x function values' or 'x strips'. I know there are many synonyms for both and I sometimes get confused with them. So I was hoping you guys can help me list the synonyms of function values and strips that you have seen in past HSC and trial exams. Thanks!Function values refer to the amount of points you're interested in. This means it's related to how many coordinates you're actually taking. Tends to be associated with Simpson's rule because subintervals (strips) kind of makes no sense; Simpson's rule approximates using a parabola, and you use points to create a parabola, not intervals
Subintervals refer more to regions cut off by distinct lines. An interval is basically a section between two points. A sub-interval (which is what the strip is) is a smaller section OF that section. It's called that, because we're talking about a whole section now, and not just some points.
Comparison:
I always think of n strips, but n+1 function values. This is because, in a way, there is one strip between every two function values.
Post by: forevertired on March 06, 2017, 11:30:41 am
a) Is something like ex or xex always greater than 0?
b) Is there a rule that e-x (or to the power of negative anything) cannot equal to 0?
c) I did this question and got the answer but I'm not sure why it is?
Since x is approaching infinity I put in e-9999 etc. in the calculator which gave a 0, which is why I presumed the answer was 3 although I'm still a bit confused by it all
d) Similarly, there's also this question which asks you to solve for x
I'm also wondering if anyone knows whether Fitzpatrick (both 2/3u books) or Cambridge (just the 3U one, I have the 2U book as well but I pressume 2u is integrated/incorporated into the 3u book?) is the better option?
Thank you in advance! ;D
Post by: bananna on March 06, 2017, 11:52:54 am
I like to use the "quick" method--not the one on the formula sheet
the one where you multiply the odds by 4, and the evens by 2
but unfortunately, I didnt get the correct answer to this
thank you!!!! :)
Post by: jamonwindeyer on March 06, 2017, 12:04:59 pm
Hi! I'm a bit confused with logorithims and exponentials since I was absent for most lessons when it was taught and hoping that you guys could help me clear up some stuff.
Sure thing!
Quote
a) Is something like ex or xex always greater than 0?
\(e^x>0\) for all values of \(x\), but \(xe^x\) can be less than or equal to zero depending on the value of \(x\) ;D
Quote
b) Is there a rule that e-x (or to the power of negative anything) cannot equal to 0?
Yep, \(e^{-x}\neq0\) just like \(e^x\neq0\) ;D
Quote
c) I did this question and got the answer but I'm not sure why it is?
Since x is approaching infinity I put in e-9999 etc. in the calculator which gave a 0, which is why I presumed the answer was 3 although I'm still a bit confused by it all
So remember a limit just looks at what happens to a function as it approaches a given value. In this case, what happens as \(x\) gets really large. If you sketch a graph of \(e^{-x}\), you'll see it approaches zero for large values of \(x\). Hence, the value of the limit is 3, since that exponential term will vanish, if that makes sense? :)
Quote
d) Similarly, there's also this question which asks you to solve for x > 0 )
That exponential term can never equal zero, and will always be positive, so the inequality becomes completely reliant on the bracketed term:
Quote
I'm also wondering if anyone knows whether Fitzpatrick (both 2/3u books) or Cambridge (just the 3U one, I have the 2U book as well but I pressume 2u is integrated/incorporated into the 3u book?) is the better option?
Thank you in advance! ;D
Cambridge is the better option for questions imo, more challenging exercises that will better prepare you for exams :) both explain things quite well, though Cambridge takes a more theoretical approach that is a little tougher to comprehend for a lot of people :)
Post by: kiwiberry on March 06, 2017, 01:44:41 pm
hi can someone pls help me with this q?
I like to use the "quick" method--not the one on the formula sheet
the one where you multiply the odds by 4, and the evens by 2
but unfortunately, I didnt get the correct answer to this
thank you!!!! :)
Hey, I believe it's 4 times the evens and 2 times the odds (except the first and last), is this where you went wrong? :)
Post by: jamonwindeyer on March 06, 2017, 02:23:02 pm
Hey, I believe it's 4 times the evens and 2 times the odds (except the first and last), is this where you went wrong? :)
I think some people start counting from zero, and some from one, which throws things :P
Bananna, reckon you could show us your working for the question? We might be able to spot your error that way ;D
Post by: Aaron12038488 on March 08, 2017, 07:51:31 pm
Post by: RuiAce on March 08, 2017, 07:57:58 pm
When do I test absolute values. Is it when there's one absolute value on one side and nothing on the RHS.If you ever forget, ALWAYS test.
However, if you have something like |x+3|=5, i.e. only absolute values on one side AND x's on one side, you do not need to test.
Something like |x+5|=3x-8? Yeah, test.
Post by: bananna on March 10, 2017, 07:09:25 am
I'm having trouble shifting log graphs
please help with these q
maybe just do the first ones?
thank you so much! :)
Post by: ellipse on March 10, 2017, 07:21:33 am
hi
I'm having trouble shifting log graphs
please help with these q
maybe just do the first ones?
thank you so much! :)
note that the graph of f(x-k) is the f(x) shifted right by k units. For example, y=(x-2)^2 is the normal parabola x^2 shifted right by 2 units. If k is negative, then the graph is shifted to the left. Eg y=(x+3)^2 is the normal parabola x^2 shifted to the left by 3 units. In the case of logs, it is basically the same thing. log(x-1) is the logx curve shifted to the right by 1 units. So normal you would have a vertical asymptote at x=0. When you shift it to the right by 1 units, the asymptote is now at x=1. This also means that the new domain is all real x>1
Post by: bananna on March 10, 2017, 07:48:33 am
I think some people start counting from zero, and some from one, which throws things :P
Bananna, reckon you could show us your working for the question? We might be able to spot your error that way ;D
I got it!
thank you to all who helped :)
Post by: joyg on March 11, 2017, 01:03:16 am
Thanks in advance for the help
Post by: RuiAce on March 11, 2017, 07:52:15 am
Hi can someone please help me with this question, side question what is n0 ?
Thanks in advance for the help
I would like to see working out if you require further assistance as the approach to this question is very standard.
Post by: Shadowxo on March 11, 2017, 09:04:46 am
Hi can someone please help me with this question, side question what is n0 ?
Thanks in advance for the help
Just in addition to Rui's answer
0 usually means "initial" so N0 means initial number. N = (initial number)*e0.32t. You may also see A = A0ekt, A0 also means initial number, and A means current number. (side note - also works for other things, eg v0 is initial velocity but you don't need to know that)
Post by: Arisa_90 on March 11, 2017, 11:37:29 am
Post by: RuiAce on March 11, 2017, 11:46:38 am
Could i have some help with this question please?What are we doing with this expression?
Post by: J.B on March 11, 2017, 04:23:29 pm
(iii)
(http://uploads.tapatalk-cdn.com/20170310/07fa563213e4b26b3d72e76d60684d99.jpg)
Post by: jakesilove on March 11, 2017, 04:30:05 pm
I was just wondering how do you integrate this expression in respect to x?
(iii)
(http://uploads.tapatalk-cdn.com/20170310/07fa563213e4b26b3d72e76d60684d99.jpg)
We want to integrate
Now, we need to remember that when we differentiate an exponential function, the power stays the same, and we multiply be the derivative of the power. In formal terms;
This is the only way we know how to integrate exponentials; if the derivative of the power is in front of the exponential! Turning to the question at hand, let's differentiate the power.
Now, we don't quite have 6x in front of the exponential function. However, we can rewrite the question like this
Finally, we can integrate!
For questions like this, practice makes perfect :)
Post by: J.B on March 11, 2017, 04:35:18 pm
We want to integrate
Now, we need to remember that when we differentiate an exponential function, the power stays the same, and we multiply be the derivative of the power. In formal terms;
This is the only way we know how to integrate exponentials; if the derivative of the power is in front of the exponential! Turning to the question at hand, let's differentiate the power.
Now, we don't quite have 6x in front of the exponential function. However, we can rewrite the question like this
Finally, we can integrate!
For questions like this, practice makes perfect :)
Awesome thank you
Post by: RuiAce on March 11, 2017, 04:39:08 pm
Post by: jakesilove on March 11, 2017, 05:23:32 pm
Note that at the 2U level, you are NOT expected to do that sort of question without guidance.
I honestly didn't know that. Thanks Rui! In that case, they'd always get you to do something like differentiate the function, and then integrate a (slightly) different function.
Post by: Arisa_90 on March 11, 2017, 05:49:11 pm
What are we doing with this expression?
Post by: VydekiE on March 11, 2017, 07:28:45 pm
I'm not sure how to evaluate elog3
Thank you!!
Post by: jakesilove on March 11, 2017, 07:33:02 pm
Hi, it would be great if I could get some help on this question
I'm not sure how to evaluate elog3
Thank you!!
Let's let the value equal some value, x. Also, can I assume that you mean the natural log of 3?
Now, we can ln both sides
Using log laws,
Now, ln(e) is just 1, so
As a general rule,
Post by: jakesilove on March 11, 2017, 07:40:05 pm
Hi, it would be great if I could get some help on this question
I'm not sure how to evaluate elog3
Thank you!!
If you meant log base 10, then I don't think there's a straight forward answer.
From there, I don't think there are any basic operations you can use to move forward. Hmm...
Well... I guess that's something
Edit: That's sort of interesting, I've just shown that
Interesting, here, is obvious a relative term.
Post by: VydekiE on March 11, 2017, 08:02:54 pm
Let's let the value equal some value, x. Also, can I assume that you mean the natural log of 3?
Now, we can ln both sides
Using log laws,
Now, ln(e) is just 1, so
As a general rule,
Thank you so much!!
Post by: kinky_khan on March 11, 2017, 08:26:42 pm
Post by: jakesilove on March 11, 2017, 08:34:09 pm
Hi, I'm having some trouble with this question, I have answered it but I am unsure as to whether I am doing it correctly. Thank you for your time and help.
I like to think about this year-by-year.
In the first year, $1000 is deposited, and then 6% interest is added.
The second year, another $1000, another 6%
And again
See a pattern? For the nth year, we'll get
For n=18, we get
Using sums of geometric series, we get
For the second part, we invest a different amount each time.
This answers the second part. Can you see a pattern? Clearly,
So, for n=18,
That's one lucky son.
Post by: kinky_khan on March 11, 2017, 08:48:37 pm
I like to think about this year-by-year.
In the first year, $1000 is deposited, and then 6% interest is added.
The second year, another $1000, another 6%
And again
See a pattern? For the nth year, we'll get
For n=18, we get
Using sums of geometric series, we get
For the second part, we invest a different amount each time.
This answers the second part. Can you see a pattern? Clearly,
So, for n=18,
That's one lucky son.
Thanks heaps and I agree that is one lucky son
Post by: Shadowxo on March 11, 2017, 09:06:35 pm
Could i have some help with this question please?
Hi, since no one's answered this question I'll give you a hint
the derivative of x2+3x-1 is 2x+3
10x+15 = 5(2x+3)
Good luck, let me know if you're still stuck :)
Post by: RuiAce on March 11, 2017, 09:08:43 pm
Oops. the question asks to anti differentiate itNote that at the 2U level, you are NOT expected to do that sort of question without guidance.
So here's the guidance.
Jake recently answered a question that's of a similar kind. You can use that as a reference point.
Post by: Aaron12038488 on March 11, 2017, 09:27:29 pm
Post by: jakesilove on March 11, 2017, 09:35:36 pm
So i had a trig question I'm my maths test. It said a tree had an elevation of 55 degrees and had a 77m shadow. Does that mean u use sin to find the height?
I would expect that the expression you were meant to find was
As
The shadow would be 'along the floor', so between the Hypotenuse and the angle of elevation.
Post by: sophiegmaher on March 12, 2017, 12:40:00 pm
Post by: Mary_a on March 12, 2017, 04:45:19 pm
I was just wondering (half yearlies are coming up pretty soon!) what's the best way to approach a maths exam in terms of study and then actually receiving the paper? I'm always really nervous around exam time.
Thank you so much,
Mary x
Post by: RuiAce on March 12, 2017, 04:48:17 pm
Hey,By doing old exam papers. You cannot expect to succeed in maths without doing papers from previous years, because only they will reflect what you should expect to get on the day.
I was just wondering (half yearlies are coming up pretty soon!) what's the best way to approach a maths exam in terms of study and then actually receiving the paper? I'm always really nervous around exam time.
Thank you so much,
Mary x
This is technically true for every subject, but it is especially the case with maths. It is a skill-based subject and you have to constantly train your brain to adjust to all sorts of maths problems.
When you get given the paper, use the reading time appropriately. This does not mean do Q1-5 of the multiple choice; it means scan the papers. Look at which questions you definitely know how to do, and do them first. Also look out for questions which aren't obvious, but if you put your pen to the paper it may feel obvious soon. And then the rest is the usual; manage your time, don't waste too much time on a question, and maximise your marks.
Post by: Mary_a on March 12, 2017, 04:51:45 pm
By doing old exam papers. You cannot expect to succeed in maths without doing papers from previous years, because only they will reflect what you should expect to get on the day.
This is technically true for every subject, but it is especially the case with maths. It is a skill-based subject and you have to constantly train your brain to adjust to all sorts of maths problems.
When you get given the paper, use the reading time appropriately. This does not mean do Q1-5 of the multiple choice; it means scan the papers. Look at which questions you definitely know how to do, and do them first. Also look out for questions which aren't obvious, but if you put your pen to the paper it may feel obvious soon. And then the rest is the usual; manage your time, don't waste too much time on a question, and maximise your marks.
Thank you so much, I really appreciate your advice and will definitely use this for my upcoming exams!
Mary x
Post by: jamonwindeyer on March 12, 2017, 10:22:23 pm
Any ideas on what to do for the night before maths? I've tried studying the night before and ended up freaking out, and I've tried not studying the night before (as I had studied for weeks prior) yet I too freaked out. Is there a particular schedule that worked for any of you guys?
Hey Sophie! I have actually written a few guides on what to do in the lead up to a big Math exam. I wrote one on the last couple of days before Math, and one for the morning before. They might be of help to you (they are geared towards the HSC exam though)! ;D
Ultimately though, just do what feels right to you - What's in those guides might not work for you and that's okay! Hopefully it can at least give you some ideas :)
Post by: phebsh on March 13, 2017, 06:11:58 pm
It's frustrating because my school has made math compulsory for all seniors but there are some people who just can't do math! (ME!)
Thanks for reading :)
Post by: jamonwindeyer on March 13, 2017, 06:46:37 pm
Hi there, I have just gotten back my half yearly exam mark and am extremely disappointed... I tried my absolute best and studied as much as I could but I still didn't do well. My teacher has brought up the idea of me dropping down to general math but I'm afraid that I won't have time to catch up on all of the work that I've missed while simultaneously working on the new content... What should I do? And if I do remain in advanced, is there a way to recover my confidence and grades?
It's frustrating because my school has made math compulsory for all seniors but there are some people who just can't do math! (ME!)
Thanks for reading :)
Hey hey!
Definitely don't drop to General now. Way too much work to catch up on, it is not the best use of your time. The time you'd spend catching up would be better spent working super hard at 2U!
If your school has made math compulsory you should definitely stick with 2U. First of all, don't stress about a bad half yearly result. My half yearly for Extension 1 went awfully for me, just barely passed and ended up 2nd last in the cohort. I worked really hard and got back into 1st place by the end of Trials. It is never too late to pull things around and get a result you are proud of.
Just use the result as your drive - Work hard enough that you never get a result like that again. Let it motivate you, not break you down. You should also learn from it - Where did the marks go and why? Could you have studied more? Did the exam pressure get to you, so you need to do more practice exams under time? Every supposed 'failure' is really just a learning experience! :)
Keep working, use extra resources like this website to help you in understanding the content, and practice until your fingers fall off or you understand it confidently, whichever comes first ;) you CAN do it! But you need to have faith in yourself that it can be done, you won't get there if you have counted yourself out from the start (sorry that it is cheesy, but it is true) ;D
Post by: phebsh on March 13, 2017, 09:21:29 pm
Hey hey!
Definitely don't drop to General now. Way too much work to catch up on, it is not the best use of your time. The time you'd spend catching up would be better spent working super hard at 2U!
If your school has made math compulsory you should definitely stick with 2U. First of all, don't stress about a bad half yearly result. My half yearly for Extension 1 went awfully for me, just barely passed and ended up 2nd last in the cohort. I worked really hard and got back into 1st place by the end of Trials. It is never too late to pull things around and get a result you are proud of.
Just use the result as your drive - Work hard enough that you never get a result like that again. Let it motivate you, not break you down. You should also learn from it - Where did the marks go and why? Could you have studied more? Did the exam pressure get to you, so you need to do more practice exams under time? Every supposed 'failure' is really just a learning experience! :)
Keep working, use extra resources like this website to help you in understanding the content, and practice until your fingers fall off or you understand it confidently, whichever comes first ;) you CAN do it! But you need to have faith in yourself that it can be done, you won't get there if you have counted yourself out from the start (sorry that it is cheesy, but it is true) ;D
Thank you so much for that, I really did need it. I definitely need to have more faith in myself, work harder and not freak out under exam conditions! As far as this website helping me, it has! I've been doing well with biology and legal using the notes that I bought off you guys ;D Super helpful! So, thank you so much! :D
Post by: jamonwindeyer on March 13, 2017, 09:24:58 pm
Thank you so much for that, I really did need it. I definitely need to have more faith in myself, work harder and not freak out under exam conditions! As far as this website helping me, it has! I've been doing well with biology and legal using the notes that I bought off you guys ;D I'm considering buying the business and math books also... Super helpful! So, thank you so much! :D
Really glad they've been helpful for you!! ;D
We're with you every step of the way. Pop any Math questions here that are troubling you and we'll do our best to explain them for you :)
Post by: RuiAce on March 13, 2017, 09:27:35 pm
Thank you so much for that, I really did need it. I definitely need to have more faith in myself, work harder and not freak out under exam conditions! As far as this website helping me, it has! I've been doing well with biology and legal using the notes that I bought off you guys ;D Super helpful! So, thank you so much! :DThe forum isn't just here for any strictly content related questions. It's also about helping you get into shape.
What I would encourage you do is to discuss how you prepare for an exam, as well as what you feel as you're experiencing it. And also how you react when you see the paper and why. And maybe more, e.g. your thought processes in the exam.
Try identifying what you are like with these key characteristics and we can give pointers.
Post by: phebsh on March 14, 2017, 09:10:29 pm
Really glad they've been helpful for you!! ;D
We're with you every step of the way. Pop any Math questions here that are troubling you and we'll do our best to explain them for you :)
Thank you! :D
The forum isn't just here for any strictly content related questions. It's also about helping you get into shape.
What I would encourage you do is to discuss how you prepare for an exam, as well as what you feel as you're experiencing it. And also how you react when you see the paper and why. And maybe more, e.g. your thought processes in the exam.
Try identifying what you are like with these key characteristics and we can give pointers.
For exam prep, I mainly just go through each section and do questions from each topic and identify what I'm weak at, seek help and practice it.
My issue is that I feel somewhat confident when practising at home but when I get into the exam, I completely freak out and feel like I don't know anything... In the exam, I do the questions I know and guess the ones I think I don't know... Most of the time when I get back my paper, I am facepalming because I knew some the question but didn't do it correctly or made silly mistakes. It's so frustrating! :-\
Post by: RuiAce on March 14, 2017, 10:04:22 pm
Thank you! :DWhen you practice each topic, how exactly are you practicing? Because you need to let past papers gradually take over this necessity.
For exam prep, I mainly just go through each section and do questions from each topic and identify what I'm weak at, seek help and practice it.
My issue is that I feel somewhat confident when practising at home but when I get into the exam, I completely freak out and feel like I don't know anything... In the exam, I do the questions I know and guess the ones I think I don't know... Most of the time when I get back my paper, I am facepalming because I knew some the question but didn't do it correctly or made silly mistakes. It's so frustrating! :-\
It is perfectly fine to commence studying by taking out a textbook and making sure you know each corner of the topic. It lets you ensure you have all the foundations you need to go into the exam. But it is definitely insufficient. Questions in a textbook will never be the same as questions in a textbook; they can get close to, but they will never truly reflect what you will see on the day. You need to add lots and LOTS of past papers to this strategy.
If you are making silly mistakes, consider reading Jamon's article on common silly mistakes to look out for. Then, add to it any silly mistakes that relate to just you. Do past papers whilst looking at the silly mistakes, so that you know to not let that happen. Do enough past papers with this list in front of you so that by the time you walk into the exam, you already know what mistakes to watch out for. Then, when you're done writing and need to check your answers, you know what your checklist is.
Also, simulate exam conditions. Do some past papers as though you were in an exam - Find some place that's quiet and good for focusing. Set up your timer, and do the paper like you would do an exam. And mark it once you're done, clearly indicating to yourself where all the faults were.
Post by: phebsh on March 14, 2017, 10:58:31 pm
When you practice each topic, how exactly are you practicing? Because you need to let past papers gradually take over this necessity.
It is perfectly fine to commence studying by taking out a textbook and making sure you know each corner of the topic. It lets you ensure you have all the foundations you need to go into the exam. But it is definitely insufficient. Questions in a textbook will never be the same as questions in a textbook; they can get close to, but they will never truly reflect what you will see on the day. You need to add lots and LOTS of past papers to this strategy.
If you are making silly mistakes, consider reading Jamon's article on common silly mistakes to look out for. Then, add to it any silly mistakes that relate to just you. Do past papers whilst looking at the silly mistakes, so that you know to not let that happen. Do enough past papers with this list in front of you so that by the time you walk into the exam, you already know what mistakes to watch out for. Then, when you're done writing and need to check your answers, you know what your checklist is.
Also, simulate exam conditions. Do some past papers as though you were in an exam - Find some place that's quiet and good for focusing. Set up your timer, and do the paper like you would do an exam. And mark it once you're done, clearly indicating to yourself where all the faults were.
This is great advice, thank you very much. I will definately be using these techniques! :)
Post by: bananna on March 15, 2017, 03:33:05 pm
can someone pls help me with these qs:
q 9, q 13 (a and i) and q 10 (e)
thank you!
Post by: Kekemato_BAP on March 15, 2017, 08:24:20 pm
Post by: jamonwindeyer on March 16, 2017, 09:58:20 am
hi!
can someone pls help me with these qs:
q 9, q 13 (a and i) and q 10 (e)
thank you!
Hey! I certainly can! For Q9, the hard bit is the differentiation, which requires the Product Rule:
So we want the equation of the tangent at \(x=2\), which means we need the gradient at \(x=2\) - Substitute \(x=2\) into our derivative to get:
The y-coordinate is obtained by substituting into the original function, and we get \(y=0\) - Having a set of coordinates and a gradient, we can now use the point gradient formula:
Isn't that crazy! ;D
The first one from 13 is just product rule derivative like above! Hopefully you can follow the working (let me know if it doesn't make sense):
The next one is tougher, it's a chain rule derivative - Remember that the derivative of \(\ln{f(x)}\) is \(\frac{f'(x)}{f(x)}\) - That works even if \(f(x)\) happens to be another logarithm! ;D
Then as the question states, 10E is the quotient rule:
Let me know if you need me to go through any of these in a little more detail! ;D
Post by: jamonwindeyer on March 16, 2017, 10:12:50 am
Hi, I need help with question b. I don't get these max-min questions for calculus
Hey! No worries, I'll step through this for you!
These questions always have a setup (part (i) in this case) and then the actual calculus itself.
So we've been given a formula linking speed to fuel consumption per hour. So given a speed, we know how many litres we guzzle per hour. So, we can link that to cost per hour, but remember we need to pay the drivers $20 each ($40 in total) as well! So let the hourly cost be \(H\), and it will be:
Multiplying the litres by 0.50 is because each litre is 50 cents! From there, we need to know how long the trip is! The time taken to complete the trip can be related to the velocity and the speed by the speed/distance/time triangle:
So, the TOTAL cost will be the hourly cost, multiplied by the number of hours:
This matches the question, so we proceed! Remember that even if you don't get the proof in the first part of a question like this, you can still do the second bit!
The next bit is just a basic maxima question - We differentiate and put the derivative equal to zero. This allows us to find critical points (max/min) - We then use the second derivative test to classify them! In this question, we also have extra conditions to verify!
Setting \(y'=0\):
We need to check that this actually yields a minimum for the function - We do this using the second derivative.
When \(v=65.6\), \(y''>0\), so therefore, the critical point we have found is a minimum. But, does it satisfy the criteria? Well travelling at that speed, we can find that in 12 hours, we don't travel anywhere NEAR as fast as we need to to do the full 1000km. So this answer won't work.
From here, we have to simply deduce that we want to get as close to this minimum as possible. So, we should go as slow as possible to complete the trip in the allotted time of 12 hours. The velocity required, therefore, is:
I know that Calculus seems useless, but we have to do it to be allowed to make this final assumption!
I hope this makes sense - Let me know if you'd like any of it clarified ;D
Post by: laurenf58 on March 16, 2017, 11:19:04 am
Thank!
Post by: jakesilove on March 16, 2017, 11:34:33 am
How do I simplify this? I'm terrible with surds!
Thank!
Hey! So,
Looks like we'll be working in threes, so I'm gonna start by breaking up our 27.
Now, if we have a square number inside a square root, we can just bring the number outside the square root! So,
Great! Now, let's attack the first part of the problem
Let me know if I can clarify any steps!
Post by: laurenf58 on March 16, 2017, 11:45:07 am
Post by: 12070 on March 16, 2017, 05:29:53 pm
Post by: jakesilove on March 16, 2017, 05:47:14 pm
For points of inflexion. When do you need to prove that it actually is one using the table thing. My math teacher always seems to prove that it is whereas my tutor only sometimes does.
In the Mathematics course, I would recommend
Post by: 12070 on March 16, 2017, 06:02:19 pm
In the Mathematics course, I would recommendproving it. You need to show that the point is a POI, not just another stationary point. The only way to prove that is quantitatively, using a table or the double derivative. It doesn't take much longer, and you will be sure to get full marks.
Okay thanks. My tutor said you only need to prove it's a point of inflexion if the first and second derivative equals 0 for the same x value or something. All it has done is confused me though.
Post by: jakesilove on March 16, 2017, 06:04:08 pm
Okay thanks. My tutor said you only need to prove it's a point of inflexion if the first and second derivative equals 0 for the same x value or something. All it has done is confused me though.
Yeah exactly; if you prove it in certain cases only, you'll just end up confusing yourself (should I prove it now? What if I don't? Argh what was the rule again!??!?!?!). Instead, prove it every time; you'll get really good at it, and you'll always get the marks
Post by: RuiAce on March 16, 2017, 06:36:33 pm
For points of inflexion. When do you need to prove that it actually is one using the table thing. My math teacher always seems to prove that it is whereas my tutor only sometimes does.Honestly your teacher is right. You're supposed to always prove it (although a better word may be verify instead of prove).
Unless your tutor ignores it purely for saving some time, he/she is doing the maths incorrectly.
Post by: 12070 on March 16, 2017, 06:38:24 pm
1. Rationalising denominators of surds
2. Simple probability
3. Derivative of function
4. Minimum value of an expression
5. Equations reducible to Quadratics
6. Simpson’s Rule
7. Area enclosed between 2 functions
8. Geometric series: nth term and sum
9. Increasing & decreasing functions
10. Sign of first and second derivative
11. Limiting sum
12. Definite integrals
13. Second derivative and concavity
14. Finding primitives
15. Arithmetic series: nth term and sum
16. Tangent derivative
17. Trapezoidal rule using function values
18. Graphing functions
19. Integrals as exact values
20. Parabola vertex and focus
21. Volume of revolution
22. Find stationary points, determine nature,
sketch graph, state domain
23. Problem question on sequences
24. Maxima minima problem question
Hey again. Do you know any good sources of where I can get past papers for these sorts of questions.
Post by: 12070 on March 16, 2017, 06:41:50 pm
Honestly your teacher is right. You're supposed to always prove it (although a better word may be verify instead of prove).
Unless your tutor ignores it purely for saving some time, he/she is doing the maths incorrectly.
Yeah I think I'll just 'verify' it from now on but my tutor is pretty smart and probably has a logical reason why you can do it but I probably won't ask him because I'll get sucked back into his way and the cycle will continue.
Post by: 12070 on March 16, 2017, 06:46:10 pm
Post by: RuiAce on March 16, 2017, 07:00:52 pm
Follow up on my past paper question (Sorry for the plethora of questions). My teacher hasn't explicitly told us that this is the order of questions, but it seems as though they get progressively harder. Would you focus most of your study time on the latter questions assuming they will be extremely challenging?Not necessarily. Because if you don't put enough effort into the easy ones you make yourself VERY prone to either one of the following:
- Rushing it, so you make heaps of silly mistakes
- Overthinking it, because you lack practice with the standard difficulty questions.
If you want to put more effort into the later ones, that is fine. But you should be quite restrictive of how much time you put into it, because easy questions definitely matter.
Topics covered:Look just do past trial papers and single out the questions relating to those topics. Finding tests that have EXACTLY the same things you will be assessed on is extremely hard.
1. Rationalising denominators of surds
2. Simple probability
3. Derivative of function
4. Minimum value of an expression
5. Equations reducible to Quadratics
6. Simpson’s Rule
7. Area enclosed between 2 functions
8. Geometric series: nth term and sum
9. Increasing & decreasing functions
10. Sign of first and second derivative
11. Limiting sum
12. Definite integrals
13. Second derivative and concavity
14. Finding primitives
15. Arithmetic series: nth term and sum
16. Tangent derivative
17. Trapezoidal rule using function values
18. Graphing functions
19. Integrals as exact values
20. Parabola vertex and focus
21. Volume of revolution
22. Find stationary points, determine nature,
sketch graph, state domain
23. Problem question on sequences
24. Maxima minima problem question
Hey again. Do you know any good sources of where I can get past papers for these sorts of questions.
Post by: 12070 on March 16, 2017, 07:14:46 pm
Not necessarily. Because if you don't put enough effort into the easy ones you make yourself VERY prone to either one of the following:
- Rushing it, so you make heaps of silly mistakes
- Overthinking it, because you lack practice with the standard difficulty questions.
If you want to put more effort into the later ones, that is fine. But you should be quite restrictive of how much time you put into it, because easy questions definitely matter.
Look just do past trial papers and single out the questions relating to those topics. Finding tests that have EXACTLY the same things you will be assessed on is extremely hard.
True. Any tips on not making silly mistakes/overthinking. In prelim, I lost 12% in one test due to silly mistakes.
I have got the succes one HSC book with all the past HSC questions. Is that a good source? Any other recommendations?
Post by: jamonwindeyer on March 16, 2017, 08:10:15 pm
True. Any tips on not making silly mistakes/overthinking. In prelim, I lost 12% in one test due to silly mistakes.
I have got the succes one HSC book with all the past HSC questions. Is that a good source? Any other recommendations?
Here is a list of mistakes to watch out for! Being aware is half the battle - Besides that, lots of practice is really the best way to start ironing out those silly errors - Math is just muscle memory after all, and it takes ages to get to the point where you can kick 100 goals in soccer without missing one ;)
Topics covered:
...
Hey again. Do you know any good sources of where I can get past papers for these sorts of questions.
Those topics are Basic Arithmetic and Algebra, Probability, Quadratics, Geometrical Applications of Differentiation, Sequences and Series and Integration Perhaps go back and do the Chapter Reviews or cross reference with your Success One book? ;D
Post by: Kekemato_BAP on March 16, 2017, 09:19:08 pm
Is dx/x the same as 1/x? Is the integral of dx/x the same as integral 1/x? Thanks
Post by: jakesilove on March 16, 2017, 09:24:58 pm
Hi, I just want something cleared up.
Is dx/x the same as 1/x? Is the integral of dx/x the same as integral 1/x? Thanks
Hey! If you're asking whether
Then yes, it absolutely does!
Post by: RuiAce on March 16, 2017, 09:37:26 pm
Hi, I just want something cleared up.Mathematicians call \(\int \frac{dx}{x} \)an 'abuse of notation'. But they still mean the same thing like Jake said.
Is dx/x the same as 1/x? Is the integral of dx/x the same as integral 1/x? Thanks
Post by: K9810 on March 17, 2017, 06:48:58 pm
Q16, 4, 11 b
Thank you!
Post by: 1937jk on March 18, 2017, 08:36:14 am
Anyone feel free to explain this one, I have done up to what I understand from it. I think it goes back to basic trig rules but I've totally forgotten and feel very lost at the moment and don't know what to do from here
Thanks!
Post by: RuiAce on March 18, 2017, 08:39:47 am
HELP :'(
Anyone feel free to explain this one, I have done up to what I understand from it. I think it goes back to basic trig rules but I've totally forgotten and feel very lost at the moment and don't know what to do from here
Thanks!
Post by: 1937jk on March 18, 2017, 08:54:03 am
Wow! what a fast reply!
Ohhhhhhh yess i undersatnd now! Thank you so much!
Post by: jamonwindeyer on March 18, 2017, 02:21:29 pm
Hey, can you help me with these questions?
Q16, 4, 11 b
Thank you!
Hey!
For Question 4, we just need to find the value of \(m\) by putting the line in gradient intercept form:
So, \(m=\frac{4}{3}\), and you put that into the equation to find the answer ;D
For 16, we have 200 revolutions per minute, which is \(\frac{10}{3}\) revolutions per second. Each revolution moves any given point on the disc through an angle of \(2\pi\) radians. So, the angular velocity will be the number of radians it moves through per second, so:
Given a radius of 0.3 metres, and an angular change per unit time, we can use the arc length formula to find the answer to Part B:
As for 11B, it is very similar to 11A! You need to find the line inclined at an angle to the given line such that:
So, the new line will have a gradient that gives an angle of 18 degrees. So it is essentially 11A, with a different point and a different angle! ;D
Post by: marshmallow.on.fire on March 18, 2017, 09:40:26 pm
Thanks so much, atarnotes is incredible!
Post by: RuiAce on March 18, 2017, 10:00:44 pm
A 3 mark questions about series (attached screenshot)
Thanks so much, atarnotes is incredible!
__________________________
Post by: cxmplete on March 19, 2017, 08:42:33 am
A few days till the maths exam and I'm still really stuck on this point:
What is the difference between a point of inflection and a point of horizontal inflection? And how would you set the working out for both?
Post by: RuiAce on March 19, 2017, 09:04:59 am
Hi!A horizontal point of inflexion satisfies the added criteria of the first derivative equals to 0. It is a special case of the standard point of inflexion.
A few days till the maths exam and I'm still really stuck on this point:
What is the difference between a point of inflection and a point of horizontal inflection? And how would you set the working out for both?
Because it satisfies both \( \frac{dy}{dx}=0\) AND \(\frac{d^2y}{dx^2}=0 \) it is both a stationary point AND a point where concavity changes. A typical horizontal point of inflexion is found on the curve \(y=x^3\)
Post by: jamonwindeyer on March 19, 2017, 11:53:16 am
Thanks so much, atarnotes is incredible!
Just a PS -You are incredible! Thanks for your kind words ;D
Post by: Rathin on March 19, 2017, 01:04:01 pm
Post by: jakesilove on March 19, 2017, 01:14:42 pm
Express y in terms of x.
So, we have
Assume that all 'log's are base 10
Using our log laws;
Now, recall that
So,
So,
Post by: katnisschung on March 19, 2017, 02:02:57 pm
the derivative calculator is confusing the heck out of me so here i am!
2)find the second derivative of (first attached)
(second attached is my answer for first derivative
(third attached is my answer for the second derivative)
help can i simplify this more?!did i get it right?
thanks!
Post by: RuiAce on March 19, 2017, 02:08:14 pm
bonjour :)Your answer is correct
the derivative calculator is confusing the heck out of me so here i am!
2)find the second derivative of (first attached)
(second attached is my answer for first derivative
(third attached is my answer for the second derivative)
help can i simplify this more?!did i get it right?
thanks!
Post by: katnisschung on March 19, 2017, 02:36:24 pm
for the graph y=x^2(e^2x)
i used the second derivative and i'm getting it to be 0.... (and its not a horizontal pt
of inflexion... ???)
Post by: jakesilove on March 19, 2017, 02:59:13 pm
could anyone show the proof that there is a minimum at (0,0)
for the graph y=x^2(e^2x)
i used the second derivative and i'm getting it to be 0.... (and its not a horizontal pt
of inflexion... ???)
Hey!
Clearly, a turning point will occur at x=0 and x=-1.
Let's test some points around x=0.
Clearly, the gradient changes signs; it goes from negative to positive! So, the point must be a minimum :)
Post by: katnisschung on March 19, 2017, 03:09:49 pm
i just find that its faster to find the nature than to use the first...
( i know u can use either but i'm just trying to work out where i got it wrong)
Post by: jakesilove on March 19, 2017, 03:37:01 pm
hey! yeah would u be able to check that for 2nd derivative
i just find that its faster to find the nature than to use the first...
( i know u can use either but i'm just trying to work out where i got it wrong)
Sure!
So, at x=0,
Since the double derivative is positive, the point is a minimum
Post by: RuiAce on March 19, 2017, 05:06:20 pm
hey! yeah would u be able to check that for 2nd derivativeGonna make a note on this
i just find that its faster to find the nature than to use the first...
( i know u can use either but i'm just trying to work out where i got it wrong)
The second derivative is faster provided it's either easier to use, or you need to find points of inflexion later on.
If both of these conditions are not satisfied, you should just test both sides.
Post by: cxmplete on March 19, 2017, 06:16:07 pm
Post by: RuiAce on March 19, 2017, 06:53:43 pm
how would we do part iii?Addressed in post #351
Post by: f_tan on March 20, 2017, 09:58:09 pm
(http://i.imgur.com/6d6VG6q.png)
Answer is $78700
Post by: 1937jk on March 21, 2017, 04:14:33 pm
Log8x + log4 x =10
Post by: jakesilove on March 21, 2017, 04:28:49 pm
How do you solve,
Log8x + log4 x =10
Hey! So, let's first recall that
Now, looks like we want to be working in 2s, so lets give that a crack
So, we are at the stage
We can simply rearrange our log, such that
and we're done!
Post by: 1937jk on March 21, 2017, 04:54:37 pm
Hey! So, let's first recall that
Now, looks like we want to be working in 2s, so lets give that a crack
So, we are at the stage
We can simply rearrange our log, such that
and we're done!
Ohhhhhhh I see now! thank you so much!!! :)
Post by: jakesilove on March 21, 2017, 05:02:08 pm
Ohhhhhhh I see now! thank you so much!!! :)
No problem, always happy to help!
Post by: 1937jk on March 21, 2017, 05:32:23 pm
Find any stationary points of the curve y= x + e-x
I have the derivative
y'= 1 - e-x
And so to find stationary points it equals zero
So e-x = 1
I don't understand how to find x from this, if someone could please explain what I should know haha thanks!
Post by: RuiAce on March 21, 2017, 05:33:34 pm
Okay so another question,
Find any stationary points of the curve y= x + e-x
I have the derivative
y'= 1 - e-x
And so to find stationary points it equals zero
So e-x = 1
I don't understand how to find x from this, if someone could please explain what I should know haha thanks!
Post by: 1937jk on March 21, 2017, 05:38:48 pm
Oh of course! Thank you so much
Post by: cxmplete on March 21, 2017, 07:06:58 pm
Post by: jamonwindeyer on March 21, 2017, 07:15:33 pm
how would you solve 6b?
Hey! You would do an indefinite integral:
But we know at \(t=0\), we have no items yet, so:
Then just substitute \(t=10\) to get the answer of 110! Note you could also do a definite integral with limits 0 and 10, but this approach makes a little more sense on the surface, at least imo ;D
Post by: f_tan on March 21, 2017, 08:01:07 pm
Mrs Caine contributes $1000 on the day of her son's birth, and increases her annual contribution by 6% each year. Her investment also earns 6% compound interest per annum. Fine the total value of Mrs Caine's investment on her son's third birthday (just before she makes her fourth contribution.)
Thank you!
Post by: jamonwindeyer on March 21, 2017, 08:36:46 pm
How do I work this out?
Mrs Caine contributes $1000 on the day of her son's birth, and increases her annual contribution by 6% each year. Her investment also earns 6% compound interest per annum. Fine the total value of Mrs Caine's investment on her son's third birthday (just before she makes her fourth contribution.)
Thank you!
Hey f_tan! We can just step through this one:
On the day of her sons birth, we have \($1000\).
After 1 year, we have gained 6% interest - Going up to \(1.06($1000)=$1060\).
We then make another annual contribution, which is 6% higher than the previous contribution of \($1000\). So we add \(1.06(1000)=$1060\) to that total, giving a total of \($2120\)! That's after the child's first birthday!
Continue this process - Apply 6% interest, then make an annual contribution of 6% greater than previous - You need to do it twice more to get to the third birthday, but remember not to add the final contribution!! Does this help? :)
Post by: katnisschung on March 21, 2017, 09:19:31 pm
could somebody pls check my answer
simple integration of e but mine seems to be wrong?
Post by: jamonwindeyer on March 21, 2017, 09:45:24 pm
Super exciting to see one of the first ever threads for the forum community of NSW reaching such a cool milestone ;D
Post by: jamonwindeyer on March 21, 2017, 09:48:24 pm
hey hey!
could somebody pls check my answer
simple integration of e but mine seems to be wrong?
I think it is just the constants that are off! :)
Post by: dadwwdnnnn on March 22, 2017, 04:20:46 pm
For my half yearlies, this question was in the practice HSC and I find get the answers:
Question 1: (Pretty sure its just algebra)
(http://i67.tinypic.com/rji881.jpg)
Question 2 e): (I have completed a-d)
(http://i63.tinypic.com/168e55c.jpg)
Question 3: (Factorise)
(http://i64.tinypic.com/14tyk3b.jpg)
Thanks for the help!,
Thomas
Post by: RuiAce on March 22, 2017, 04:24:04 pm
Couple of Questions:I saw this whilst I'm in a lecture so I'm a bit busy to do it right now, but for Jamon/Jake/etc 's convenience please post up your answers to anything you've done in advance with Q2.
For my half yearlies, this question was in the practice HSC and I find get the answers:
Question 1: (Pretty sure its just algebra)
(http://i67.tinypic.com/rji881.jpg)
Question 2 e): (I have completed a-d)
(http://i63.tinypic.com/168e55c.jpg)
Question 3: (Factorise)
(http://i64.tinypic.com/14tyk3b.jpg)
Thanks for the help!,
Thomas
Post by: dadwwdnnnn on March 22, 2017, 04:27:21 pm
I saw this whilst I'm in a lecture so I'm a bit busy to do it right now, but for Jamon/Jake/etc 's convenience please post up your answers to anything you've done in advance with Q2.
final answers for q2 (dont know if they are correct, practice test does not come with answers)
a) (1,6)
b)3x+2y-38=0
c)7x-3y+11=0
d)(136/23,233/-23)
Post by: smile123 on March 22, 2017, 04:47:51 pm
Post by: 1937jk on March 22, 2017, 06:12:33 pm
Post by: jakesilove on March 22, 2017, 06:46:44 pm
How do I differentiate this question :)
Hey! So,
a is a constant, so goes to zero when we differentiate. The cos term is differentiated using the chain rule, as shown above.
Post by: jakesilove on March 22, 2017, 06:48:55 pm
How do you know when sketching a graph like y = sin(3x) which values (in radians) to put down on the x axis?
Hey! Personally, I could never remember stuff like that. So, I would often use my calculator to find zeros (for instance, 0, pi, 2pi and 3pi are all zeros that can be quickly found!) and the use those points as my x-axis. However, it really doesn't matter; choose something that makes sense to you (any regular increment is fine), then use your calculator to figure out what the y coordinate at that point will be.
Post by: f_tan on March 22, 2017, 07:13:20 pm
The cost of running a car at an average speed of v km/h is given by c = 150 + v^2/80 cents per hour. Find the everage speed, to the enarest km/h, at which the cost of a 500 km trip is a minimum.
Thanks!
Post by: dadwwdnnnn on March 22, 2017, 07:19:26 pm
Couple of Questions:
For my half yearlies, this question was in the practice HSC and I find get the answers:
Question 1: (Pretty sure its just algebra)
(http://i67.tinypic.com/rji881.jpg)
Question 2 e): (I have completed a-d)
(http://i63.tinypic.com/168e55c.jpg)
Question 3: (Factorise)
(http://i64.tinypic.com/14tyk3b.jpg)
Thanks for the help!,
Thomas
Could I please get some help?
Post by: RuiAce on March 22, 2017, 07:54:57 pm
Couple of Questions:I'll quickly do Q3.
For my half yearlies, this question was in the practice HSC and I find get the answers:
Question 1: (Pretty sure its just algebra)
(http://i67.tinypic.com/rji881.jpg)
Question 2 e): (I have completed a-d)
(http://i63.tinypic.com/168e55c.jpg)
Question 3: (Factorise)
(http://i64.tinypic.com/14tyk3b.jpg)
Thanks for the help!,
Thomas
Post by: dadwwdnnnn on March 22, 2017, 08:35:18 pm
I'll quickly do Q3.
soo this is wrong?
https://www.symbolab.com/solver/step-by-step/factorise%20%5Cleft(2x%5Cright)%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5Ccdot%5Cleft(x-2%5Cright)%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D-%5Cleft(5x%5Cright)%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%5Ccdot%5Cleft(x-2%5Cright)%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7D
Post by: jamonwindeyer on March 22, 2017, 08:40:21 pm
soo this is wrong?
It isn't mathematically incorrect, but it isn't doing the most efficient job, Rui's is neater ;D
Edit: Just fixed a little typo in the last line of Rui's response, check the updated!
Post by: dadwwdnnnn on March 22, 2017, 08:47:14 pm
It isn't mathematically incorrect, but it isn't doing the most efficient job, Rui's is neater ;D
Edit: Just fixed a little typo in the last line of Rui's response, check the updated!
cheers jamon, you mind helping with the other 2 questions, I have my 2 unit exam tomorrow. Thanks
Post by: jamonwindeyer on March 22, 2017, 08:55:28 pm
Couple of Questions:
For my half yearlies, this question was in the practice HSC and I find get the answers:
Question 1: (Pretty sure its just algebra)
1st Line to 2nd: Multiplying each term by \(fS_1S_2\)
2nd Line to 3rd: Move \(S_2\)'s to the LHS and factorise
3rd Line to Final: Divide to isolate \(S_2\)
For your second question, I think you can just find the slope of the two segments using the gradient formula on your reference sheet, and use that as your reasoning? :)
Post by: dadwwdnnnn on March 22, 2017, 09:03:17 pm
1st Line to 2nd: Multiplying each term by \(fS_1S_2\)
2nd Line to 3rd: Move \(S_2\)'s to the LHS and factorise
3rd Line to Final: Divide to isolate \(S_2\)
For your second question, I think you can just find the slope of the two segments using the gradient formula on your reference sheet, and use that as your reasoning? :)
Last question, sorry jamon, with question 1d) did I get the points of intersection wrong?
Post by: jamonwindeyer on March 22, 2017, 09:12:22 pm
Last question, sorry jamon, with question 1d) did I get the points of intersection wrong?
You should have gotten \((4,13)\) :)
Post by: 1937jk on March 23, 2017, 08:10:22 am
Hey! Personally, I could never remember stuff like that. So, I would often use my calculator to find zeros (for instance, 0, pi, 2pi and 3pi are all zeros that can be quickly found!) and the use those points as my x-axis. However, it really doesn't matter; choose something that makes sense to you (any regular increment is fine), then use your calculator to figure out what the y coordinate at that point will be.
Okay cool thank you! just to clarify, so you can pretty much use anything relevant to the question as long as their is a consistent increment? E.g To sketch Y =sin (3x) for the domain 0 < x < 2
Post by: Guest8373727727 on March 23, 2017, 02:41:49 pm
(http://i64.tinypic.com/dfdiip.jpg)
Post by: Guest8373727727 on March 23, 2017, 03:04:30 pm
(http://i63.tinypic.com/2vd66mf.jpg)
Post by: jakesilove on March 23, 2017, 06:48:03 pm
Need help with 4 b)
(http://i64.tinypic.com/dfdiip.jpg)
Hey! First, let's work out the initial domain and range. We know that x has to be greater than -1 (so that we don't have a square root of a negative!) and y will be between -4 and -infinity.
Now, let's consider f(x+2)+4
Great! For this new function, x will have to be greater than -3 (so, we have to shift across 2 units to the left) and y will always be greater than zero (so we shift four units up)
Let me know if I can clarify anything!
Post by: jakesilove on March 23, 2017, 06:49:18 pm
Question 2
(http://i63.tinypic.com/2vd66mf.jpg)
Hey! All you do here is break the graph up into two sections; one at x<=-3, and one where x>3. For the first section, sketch the straight line y=x+2. Then, ignore the first section completely, and in the second section sketch the other straight line specified. That's it!
Post by: RuiAce on March 23, 2017, 06:56:40 pm
Question 2It will look something like this
(http://i63.tinypic.com/2vd66mf.jpg)
(http://i.imgur.com/nLBmdXQ.png)
Post by: Annie657 on March 23, 2017, 09:41:19 pm
Post by: jamonwindeyer on March 24, 2017, 01:12:14 am
Hi Jake, I am stuck on Question 14 B :-[ Could you please help me? Thankyou :) :)
Hey Annie! Welcome to the forums! I reckon I can help here ;D
The idea is to calculate the odds of losing consecutive draws. The odds of losing \(n\) draws in a row can be given by:
Now we want to be 99% certain that a jackpot prize will have been won. We can consider this as a 1% chance that the Jackpot hasn't been won. So, how many times do we need to draw (aka, the value of \(n\)) for that probability above to reduce to 1% or less?
We can do some trial and error here, at you'll find that after \(228\) draws, the chance of 228 consecutive losses is less than 1%. So, we become 99% certain we have won at least a single jackpot ;D
I hope this makes sense! :)
Post by: laurenf58 on March 24, 2017, 08:52:22 am
Post by: jakesilove on March 24, 2017, 09:18:09 am
Could I get some help with the attached questions please?
Hey! So, the first question is really only asking you to find where the function is increasing. So, let's differentiate and find where its positive
There are any number of ways that you can solve this equation. I like to do it as follows
So, our two points of interest will be x=9, and x=-1. Let's test points to either side;
For x=0,
which is clearly less than zero. So, the important section isn't between the two points, but rather outside of it! Our answer for where the function is increasing will be
Now, to find where the curve is concave up, we find the double derivative and see where it is positive.
Clearly, this will occur for
Post by: Ellie__ on March 25, 2017, 08:41:08 am
So I've got my 1/2 yearlies on Tuesday for this, and well I mostly know my content but it takes me FOREVER to answer a question. Do you have any suggestions on how to be quick and efficient and maybe get 75% of the exam paper complete???
Thaankyou!!!
Post by: RuiAce on March 25, 2017, 09:12:31 am
Hey guys,When you are using past papers to study, how are you managing your time?
So I've got my 1/2 yearlies on Tuesday for this, and well I mostly know my content but it takes me FOREVER to answer a question. Do you have any suggestions on how to be quick and efficient and maybe get 75% of the exam paper complete???
Thaankyou!!!
Simply knowing content will not be enough for mathematics. You need to keep doing past papers so that you can constantly adapt to different types of questions that you may see on the day. And speeding up is another thing that comes with practice, which you need to do a lot of.
When doing the past paper, you should not be overanalysing questions until potentially the last few. With practice comes the ability to generate ideas more quickly. Reading time is the first instance where ideas can be generated. During reading time, you don't have to expect your ideas to actually work, but you should already have a brief idea about what to do. Come writing time, you start putting it to the test. The thing about maths is that if you have an idea that you're confident about, it's usually the right way to go about it. So write it down quickly. And whilst recklessness is common with calculators, there is no need to be too slow with punching in numbers either.
Whilst English was more stressful with continuously writing, maths was (potentially surprisingly) second to it. It's easy to run out of time in maths, so you need to develop quick handwriting abilities for it also.
And as always, follow typical exam guidelines. Get all the marks you can guarantee that you will get first, before moving on to the harder questions that require more thought. Never spend too long on a question and always check the clock if you feel you've just been staring the whole time.
Post by: katnisschung on March 25, 2017, 01:39:24 pm
needing help with integrating this
thanks!
Post by: RuiAce on March 25, 2017, 01:43:12 pm
hi :)As far as 2U goes you don't know how to integrate this.
needing help with integrating this
thanks!
Was the question given like this or were there other parts to the question?
Post by: katnisschung on March 25, 2017, 01:45:36 pm
it was the second part of
differentiate y=x^2e^x
what a legend ruiace! thank gosh i took 2u ;D
Post by: Shadowxo on March 25, 2017, 01:58:54 pm
other parts
it was the second part of
differentiate y=x^2e^x
what a legend ruiace! thank gosh i took 2u ;D
Using the product rule for this question,
dy/dx = 2xex + x2ex
= ex(2x+x2) which can be rewritten as
exx(2+x)
so the integral for exx(2+x) is just x2ex (the function we differentiated) +c!
Post by: RuiAce on March 25, 2017, 01:59:03 pm
other parts
it was the second part of
differentiate y=x^2e^x
what a legend ruiace! thank gosh i took 2u ;D
Post by: katnisschung on March 25, 2017, 02:05:01 pm
another q do i need to know how to integrate 5^x for 2u
Post by: RuiAce on March 25, 2017, 02:14:58 pm
thanks! yeah i realised that after i expanded it...
another q do i need to know how to integrate 5^x for 2u
Post by: katnisschung on March 25, 2017, 02:32:41 pm
Post by: RuiAce on March 25, 2017, 02:34:47 pm
stuck on another q
Post by: 12070 on March 25, 2017, 03:08:13 pm
Post by: RuiAce on March 25, 2017, 03:12:23 pm
Okay, so I got (64)/5 but apparently you need two volumes or something. I'm so confused. Please help :D
Post by: katnisschung on March 25, 2017, 05:38:07 pm
2u don't have to know how to integrate by parts correct?
example attached
Post by: jamonwindeyer on March 25, 2017, 05:40:45 pm
too lazy to check the syllabus..
2u don't have to know how to integrate by parts correct?
example attached
Correct :) but you don't need to integrate that one by parts:
Post by: Arisa_90 on March 25, 2017, 05:47:49 pm
Post by: anotherworld2b on March 25, 2017, 05:51:01 pm
Post by: katnisschung on March 25, 2017, 05:58:08 pm
i got
pi/2+3/4
Post by: Shadowxo on March 25, 2017, 06:10:11 pm
I am not sure how to antidifferentiate these questions. Can I please have help?
For those questions, you have to use substitution. For g), you know that the derivative of e-x is -e-x.
So use u=1+e-x, du/dx = -e-x
So you end up with
You can see that there's a function, and the derivative of that function in each question. So let u = the number to a power / more complicated part then find the derivative of u, and use that to substitute in to antidiff with respect to u.
stuck on a)
i got
pi/2+3/4
For a)
You can divide it up into 5 sections, I'll call them 1,2,3,4,5.
Section 3 and 4 cancel out as they're equal in area and opposite signs
Area 1 = positive 1/4 * π * 12 = π/4
Area 2 = negative 1/2 *(1*1) = -1/2
Area 5 = positive 1/4 * π * 12 = π/4
Area 1 + Area 2 + Area 5 = π/2 - 1/2
Edit: this is for b)
Area from 0-1 is positive, areas from 1-2 and 2-3 cancel out as one's positive and one's negative and they're equal, and area from 3-4 is negative
Area 1 = positive (1-1/4*π)
Area 4 = negative (1/4*π)
Area 1 + Area 4 = 1-1/4*π-1/4*π = 1-π/2
Post by: Ellie__ on March 25, 2017, 07:06:36 pm
When you are using past papers to study, how are you managing your time?
Simply knowing content will not be enough for mathematics. You need to keep doing past papers so that you can constantly adapt to different types of questions that you may see on the day. And speeding up is another thing that comes with practice, which you need to do a lot of.
When doing the past paper, you should not be overanalysing questions until potentially the last few. With practice comes the ability to generate ideas more quickly. Reading time is the first instance where ideas can be generated. During reading time, you don't have to expect your ideas to actually work, but you should already have a brief idea about what to do. Come writing time, you start putting it to the test. The thing about maths is that if you have an idea that you're confident about, it's usually the right way to go about it. So write it down quickly. And whilst recklessness is common with calculators, there is no need to be too slow with punching in numbers either.
Whilst English was more stressful with continuously writing, maths was (potentially surprisingly) second to it. It's easy to run out of time in maths, so you need to develop quick handwriting abilities for it also.
And as always, follow typical exam guidelines. Get all the marks you can guarantee that you will get first, before moving on to the harder questions that require more thought. Never spend too long on a question and always check the clock if you feel you've just been staring the whole time.
Ah thankyou Rui Ace!!
That makes sense, yeah I'm spending at least 10 mins on some 2-3 markers, so I'll give that a go :)
Post by: Ellie__ on March 25, 2017, 07:07:52 pm
Can someone please explain how to answer locus questions?? I'm so confused on what they are
Ah also, how do I complete the square???
Thankyou!!
Mod Edit [Aaron]: Posts merged. You are able to edit posts you have already made.
Post by: anotherworld2b on March 25, 2017, 10:29:46 pm
For those questions, you have to use substitution. For g), you know that the derivative of e-x is -e-x.
So use u=1+e-x, du/dx = -e-x
So you end up with
You can see that there's a function, and the derivative of that function in each question. So let u = the number to a power / more complicated part then find the derivative of u, and use that to substitute in to antidiff with respect to u.
For a)
You can divide it up into 5 sections, I'll call them 1,2,3,4,5.
Section 3 and 4 cancel out as they're equal in area and opposite signs
Area 1 = positive 1/4 * π * 12 = π/4
Area 2 = negative 1/2 *(1*1) = -1/2
Area 5 = positive 1/4 * π * 12 = π/4
Area 1 + Area 2 + Area 5 = π/2 - 1/2
Edit: this is for b)
Area from 0-1 is positive, areas from 1-2 and 2-3 cancel out as one's positive and one's negative and they're equal, and area from 3-4 is negative
Area 1 = positive (1-1/4*π)
Area 4 = negative (1/4*π)
Area 1 + Area 4 = 1-1/4*π-1/4*π = 1-π/2
Post by: jamonwindeyer on March 26, 2017, 02:22:15 am
Can i have help with these questions please?
Sure! For your first one, expand:
Then just integrate each piece ;D second one, split it into two terms:
Then integrate those pieces ;) when you can expand (ie, the power is a square) or otherwise separate into smaller pieces, do this! Almost always it will yield an easier integration! ;D
It could be because it is 2:30am, but I don't see a way to do that last one easily without just expanding that last bracket? It's a cubic, which isn't horrendous, you'll end up with 6 terms - You'll then be able to multiply all of them by the \(\frac{1}{x^2}\), group like terms then integrate each piece :)
Post by: jamonwindeyer on March 26, 2017, 02:40:12 am
Hey guys,
Can someone please explain how to answer locus questions?? I'm so confused on what they are
Ah also, how do I complete the square???
Thankyou!!
Check the bottom half of this guide! There is a quick summary/explanation of Locus - And feel free to pop any specific examples that are troubling you here and we can work through them for you! ;D
Completing the Square, let's do an example.
Solve \(2x^2-5x+3=0\)
Note: We could just factorise this normally, but I'll show you what it looks like with completing the square. The process is identical ;D
Step 1 - Remove the coefficient of \(x^2\). This is just the way we normally do it - Keeping it there isn't much harder at all but I'll just do what your textbook probably does.
So dividing by two:\(\implies x^2-\frac{5}{2}x+\frac{3}{2}=0\)
Step 2 - Move the constant to the other side.
Step 3 - Split the middle term in half.
Step 4 - Take the coefficient of one of those halves, and square it. In this case, \(\left(\frac{-5}{4}\right)^2=\frac{25}{16}\). Add this number to BOTH sides of the equation (we can't do just one, gotta keep the sides the same!)
Now look at the LHS - We have something in the form \(a^2+2ab+b^2\) - Exactly what we need to do a perfect square factorisation. So this becomes:
So that's the method - Take the middle number, halve it, and square it - Then add that to your quadratic. It will turn it into a complete square. As to WHY this works, the best explanation is visual - This is probably the best/most famous explanation for 2U students ;D
Post by: jamonwindeyer on March 26, 2017, 02:41:08 am
I was wondering how do you know when to use substitution when you are asked to antidifferentiate exponentials and trigonometry? Would it be when there is product rule involved?
If you can't do it with your basic rules, substitution is almost always the solution ;D (there is also integration by parts but if you don't learn it, obviously it won't be that ;))
Post by: 1937jk on March 26, 2017, 12:20:31 pm
Find d/dx (lnx/x)
Post by: asd987 on March 26, 2017, 12:34:18 pm
Post by: asd987 on March 26, 2017, 12:40:50 pm
So I'm having trouble figuring out what this question is asking and how to solve it:
Find d/dx (lnx/x)
Hi, you would have to use quotient rule
u=ln , u'=1/x
v=x , v' = 1
so (1 - lnx)/x^2
Post by: jakesilove on March 26, 2017, 12:43:18 pm
So I'm having trouble figuring out what this question is asking and how to solve it:
Find d/dx (lnx/x)
Yep, the answer above is absolutely correct! Just differentiate the function by using the quotient rule
Post by: jakesilove on March 26, 2017, 12:43:59 pm
Hi, is there anyway to solve this question without drawing the actual cos graph and y=+- 1?
Hey! You could solve the equation for cos(x)=+-1, however honestly it is much easier to just sketch the curve.
Post by: jamonwindeyer on March 26, 2017, 01:38:30 pm
Hey! You could solve the equation for cos(x)=+-1, however honestly it is much easier to just sketch the curve.
Ditto - Like, solving that equation is just a thing you learn, just the same as the exact ratios! But the graphs are the best way to jog your memory of them (the \(\sin\) and \(\cos\) curves equal to 1 or 0 are called the boundary angles in a lot of textbooks) ;D
Post by: katnisschung on March 26, 2017, 01:39:56 pm
need to check if my answer is right
bill wants to put a small rectangular vege garden
in his backyard using two existing walls (x,y) as part of the
border. He has 8m of garden edging for the other two sides
Find the dimensions of the garden bed that will give the greatest area
Thanks!
i got 4m by 4m (presuming that p=16m because x+y=8 then the other side also has
to be 8 as its a rectangle)
Post by: jamonwindeyer on March 26, 2017, 01:44:32 pm
hi hi!
need to check if my answer is right
bill wants to put a small rectangular vege garden
in his backyard using two existing walls (x,y) as part of the
border. He has 8m of garden edging for the other two sides
Find the dimensions of the garden bed that will give the greatest area
Thanks!
i got 4m by 4m (presuming that p=16m because x+y=8 then the other side also has
to be 8 as its a rectangle)
That's correct :)
Post by: katnisschung on March 26, 2017, 02:08:00 pm
no answers so i want to check..
A piece of wire 10m long is broken into two parts
which are bent into the shape of a rectangle as shown.
Find the dimensions x and y that make the total area
a minimum
basically the rectangle is x,y for its dimensions
and the square is x for its length
somehow i got 1.25 for both x and y which can't be correct
becos one is a rectangle?? ???
Post by: jamonwindeyer on March 26, 2017, 02:09:45 pm
hello again!
no answers so i want to check..
A piece of wire 10m long is broken into two parts
which are bent into the shape of a rectangle as shown.
Find the dimensions x and y that make the total area
a minimum
basically the rectangle is x,y for its dimensions
and the square is x for its length
somehow i got 1.25 for both x and y which can't be correct
becos one is a rectangle?? ???
Can't really tell without the diagram, I'd expect 2.5 maybe? Perhaps upload pics of your working so we can glance it :)
Post by: katnisschung on March 26, 2017, 02:45:14 pm
also attached the original q
2x+2y+4x=10
y=5-3x
let a=area
a= (5-3x) (x) + (x)^2
a=5x-2x^2
derivative of a=0 at stationary points
5-4x=0 therefore, x=1.25
used 2nd derivative to test nature
max as -4
sub x=1.25 into original equation
thus y=1.25
soz for the dodgy quality
Post by: jamonwindeyer on March 26, 2017, 02:56:20 pm
hey it won't let me attach my working so i'll try explain it best below
also attached the original q
2x+2y+4x=10
y=5-3x
let a=area
a= (5-3x) (x) + (x)^2
a=5x-2x^2
derivative of a=0 at stationary points
5-4x=0 therefore, x=1.25
used 2nd derivative to test nature
max as -4
sub x=1.25 into original equation
thus y=1.25
soz for the dodgy quality
Cheers! You are correct - There is nothing prohibiting \(y=x\) - A square is still a rectangle ;D
Post by: katnisschung on March 26, 2017, 03:01:24 pm
Post by: katnisschung on March 26, 2017, 04:49:22 pm
a rectangle is cut from a circular disc of 6cm
find the area of the largest rectangle that can be
produced
so i did
x^2+y^2=36
a=xy
rearranged the first to find y in terms of x for area
a=x(36-x^2)^1/2
derivate of a=0 at stationary points
found the derivative but having trouble finding x
could somebody pls show the full working out...terrible at algebra
this was as far as i got
(36-x^2)1/2 - (x^2)/(36x-x^2)^1/2 = 0
Thanks!
Post by: jakesilove on March 26, 2017, 05:11:11 pm
hi hi im stuck on this q
a rectangle is cut from a circular disc of 6cm
find the area of the largest rectangle that can be
produced
so i did
x^2+y^2=36
a=xy
rearranged the first to find y in terms of x for area
a=x(36-x^2)^1/2
derivate of a=0 at stationary points
found the derivative but having trouble finding x
could somebody pls show the full working out...terrible at algebra
this was as far as i got
(36-x^2)1/2 - (x^2)/(36x-x^2)^1/2 = 0
Thanks!
So, we have
So,
If we assume that x cannot equal six (which is clearly can't, else the denominator would equal zero) then we can just multiply this out
Then, you can use the quadratic formula etc. to solve for x
Post by: katnisschung on March 26, 2017, 05:23:36 pm
stuck on another ::)
a sufboard is in the shape of a rectangle and semi circle
(half a circle on top of a rectangle). The perimeter is to be 4m
find the max area of the surboard correct to 2 dp
im really having trouble differentiating this
quick run through of my working out
p=2l+b+1/2 2
found b as
b=4-2l/1+
substituted to find A
attached image
Post by: katnisschung on March 26, 2017, 05:27:50 pm
So, we have
So,
If we assume that x cannot equal six (which is clearly can't, else the denominator would equal zero) then we can just multiply this out
Then, you can use the quadratic formula etc. to solve for x
sorry what did u multiply out?? :-\
Post by: jakesilove on March 26, 2017, 05:32:16 pm
thanks jake!
stuck on another ::)
a sufboard is in the shape of a rectangle and semi circle
(half a circle on top of a rectangle). The perimeter is to be 4m
find the max area of the surboard correct to 2 dp
im really having trouble differentiating this
quick run through of my working out
p=2l+b+1/2 2
found b as
b=4-2l/1+
substituted to find A
attached image
It sounds like you need to go back and revise differentiation. We have the function
So,
by the chain rule for the second term. Then, set this equal to zero, and solve for l.
For the last question, I multiplied everything by the denominator.
Post by: Sukakadonkadonk on March 26, 2017, 05:47:12 pm
Could someone please explain to me why:
(2/3)(x)^3/2
equals to:
2(x)^3
??
Post by: RuiAce on March 26, 2017, 05:48:52 pm
Hey guysIt doesn't
Could someone please explain to me why:
(2/3)(x)^3/2
equals to:
2(x)^3
??
Where did you get this from
Post by: Mahan on March 26, 2017, 07:08:03 pm
Hi, is there anyway to solve this question without drawing the actual cos graph and y=+- 1?Well, it addition to using
Post by: Sukakadonkadonk on March 26, 2017, 08:28:52 pm
It doesn't
Where did you get this from
Ooops sorry, typo. I meant:
(2/3)(x)^3/2
Equals to:
2(x)^1/2
Post by: RuiAce on March 26, 2017, 08:41:19 pm
Ooops sorry, typo. I meant:They still don't equal.
(2/3)(x)^3/2
Equals to:
2(x)^1/2
Post by: Sukakadonkadonk on March 26, 2017, 08:54:44 pm
They still don't equal.
Ok so is this incorrect working then?
I've seen this twice already in two papers so I'm really confused.
The answer is boxed.
Thanks.
Post by: RuiAce on March 26, 2017, 09:04:43 pm
Ok so is this incorrect working then?Line 3 is wrong. It should not be there.
I've seen this twice already in two papers so I'm really confused.
The answer is boxed.
Thanks.
Post by: Sukakadonkadonk on March 26, 2017, 09:08:59 pm
Line 3 is wrong. It should not be there.
Oh damn, right. That's my mistake. ok thank you.
Post by: katnisschung on March 27, 2017, 09:45:15 am
Post by: RuiAce on March 27, 2017, 12:50:45 pm
hi stuck on this q thanks!
Post by: Fahim486 on March 27, 2017, 01:17:15 pm
Hi again, I've also got another question and I am having trouble understanding trapezoidal rules and can't figure this question out myself
Thanks!
Mod Edit: Post merge ;D
Post by: jamonwindeyer on March 27, 2017, 01:25:22 pm
Hi, how would you differentiate this?
Thanks!
Hey! This tricky way to do this is with product rule:
Buuut, you can rewrite it as \(x^\frac{5}{2}\) and get that last line straight away with the normal rule ;D
Post by: Fahim486 on March 27, 2017, 01:34:45 pm
Hey! This tricky way to do this is with product rule:
Buuut, you can rewrite it as \(x^\frac{5}{2}\) and get that last line straight away with the normal rule ;D
Thank you so much Jamon!!!
Post by: jamonwindeyer on March 27, 2017, 01:39:52 pm
Hi again, I've also got another question and I am having trouble understanding trapezoidal rules and can't figure this question out myself
Thanks!
For this one, just before I lend a hand, does this formula look familiar?
(There's two formula out there for this, just want to make sure I help you properly)
Post by: Fahim486 on March 27, 2017, 01:47:19 pm
For this one, just before I lend a hand, does this formula look familiar?
(There's two formula out there for this, just want to make sure I help you properly)
Yeah I've seen that formula before but I don't know how to apply it to the question
Post by: jamonwindeyer on March 27, 2017, 01:58:18 pm
Yeah I've seen that formula before but I don't know how to apply it to the question
So \(h\) in that formula represents the width of our intervals. For that question, this is 5 metres!
The \(y_k\) terms just represent the heights of our area at each value, counting from zero. It is first, plus last, plus double the rest. So, in this case it would be:
The answer is B! ;D
Post by: katnisschung on March 27, 2017, 01:58:34 pm
Hey ruiace the last bit of the answer didn't come through...
so why chain rule for integration? note i only take 2 unit maths
Post by: jamonwindeyer on March 27, 2017, 02:15:16 pm
Hey ruiace the last bit of the answer didn't come through...
so why chain rule for integration? note i only take 2 unit maths
Just fixed his last line ;D we're differentiating not integrating! Chain rule for differentiation ;D
Post by: kiiaaa on March 27, 2017, 06:37:33 pm
i was wondering if anyone had any way to get their head around or easier way to solve the loan repayment questions. I just don't get how to do it and often end up more confused than what i was lol :P
thank you
Post by: RuiAce on March 27, 2017, 06:56:17 pm
HelloMight wanna read Jamon's guides, in particular the series one.
i was wondering if anyone had any way to get their head around or easier way to solve the loan repayment questions. I just don't get how to do it and often end up more confused than what i was lol :P
thank you
Whilst the process is the same, there's no way you can get how to do it without knowing WHY your equations are true.
Post by: cxmplete on March 27, 2017, 08:01:57 pm
A factory assembles torches. Each torch requires one battery and one bulb. It is known that 6% of all batteries and 4% of all bulbs are defective. Find the probability that, in a torch selected at random, both the battery and the bulb are NOT defective. Give your answer in exact form.
Post by: kiiaaa on March 27, 2017, 08:02:45 pm
Post by: jamonwindeyer on March 27, 2017, 08:42:32 pm
Hi, so this was a question in our HY exam which i found pretty confusing:
A factory assembles torches. Each torch requires one battery and one bulb. It is known that 6% of all batteries and 4% of all bulbs are defective. Find the probability that, in a torch selected at random, both the battery and the bulb are NOT defective. Give your answer in exact form.
Hey! So if 6% of batteries are defective, then there is a 94% chance of a battery being NOT defective. Similarly, there is a 96% chance of a bulb being NOT defective. If we want bulb not defective AND battery not defective, we multiply these together ;D
Post by: jamonwindeyer on March 27, 2017, 08:48:23 pm
hi could anyone please help me answer this question? im super lost
Sure! Since \(BC=FG=x\), we know the other two sides must be what we have to add to get to 100. That is, if the sides BF and CG are \(y\):
To find the areas of the triangles, we need the side lengths besides the side from the rectangle. We do this with Pythagoras. For example, take the triangle FGH on the bottom. The hypotenuse is \(x\), so if the other sides are both \(d\) (they are isosceles remember!):
So the area of that triangle will be \(A=\frac{1}{2}bh=\frac{1}{2}d^2=\frac{1}{2}\times\frac{x^2}{2}=\frac{x^2}{4}\)
The triangle at the top, ABC, is the same area! ;D you can find a similar expression for the triangles on the side, then add em up! ;D
Post by: kiiaaa on March 27, 2017, 10:14:37 pm
Sure! Since \(BC=FG=x\), we know the other two sides must be what we have to add to get to 100. That is, if the sides BF and CG are \(y\):
To find the areas of the triangles, we need the side lengths besides the side from the rectangle. We do this with Pythagoras. For example, take the triangle FGH on the bottom. The hypotenuse is \(x\), so if the other sides are both \(d\) (they are isosceles remember!):
So the area of that triangle will be \(A=\frac{1}{2}bh=\frac{1}{2}d^2=\frac{1}{2}\times\frac{x^2}{2}=\frac{x^2}{4}\)
The triangle at the top, ABC, is the same area! ;D you can find a similar expression for the triangles on the side, then add em up! ;D
thank you very much!
would yo be able to help me in this question?
so i got my whole working correct legit like everything when i compare it to the answers but idk how when i feed the final answer for aii) i get -128/3 not 64/3 and i struggle to tell where i go wrong. i eliminated errors such as mixing up signs or not differentiating correctly but cant find the root cause of the issue =/
thank you so much
Post by: jamonwindeyer on March 27, 2017, 10:55:30 pm
thank you very much!
would yo be able to help me in this question?
so i got my whole working correct legit like everything when i compare it to the answers but idk how when i feed the final answer for aii) i get -128/3 not 64/3 and i struggle to tell where i go wrong. i eliminated errors such as mixing up signs or not differentiating correctly but cant find the root cause of the issue =/
thank you so much
Snap a picture of your working? I'll see if I can spot the error ;D
Post by: Sukakadonkadonk on March 27, 2017, 11:17:19 pm
A gold ball is dropped from a height of one metre. Each time it hits the ground it bounces to two-thirds of its previous height.
Calculate the distance that the golf ball travels before it comes to rest.
Thanks, I don't really get the solution they gave.
Post by: jakesilove on March 28, 2017, 09:16:07 am
Hey, could someone please help me with this series and sequences question?
A gold ball is dropped from a height of one metre. Each time it hits the ground it bounces to two-thirds of its previous height.
Calculate the distance that the golf ball travels before it comes to rest.
Thanks, I don't really get the solution they gave.
Hey! Let's think about what is physically happening. First, the ball will travel one meter to the ground.
Then, it will bounce UP 2/3 of a meter
Then, it will fall back down the same distance
Then, it will bounce UP 2/3 the distance of 2/3
Then, it falls back down again. Do you see a pattern? Eventually, the total distance will be
So, we can use the sum of a geometric series into infinity! The formula is
So, the total distance will be
Post by: katnisschung on March 28, 2017, 11:37:49 am
Post by: jakesilove on March 28, 2017, 11:41:29 am
is there any way to do this without substitution?
There absolutely is! However, you're essentially just doing substitution, but without the extra steps. Recall that
In this case, notice that
So, we can make a slight adjustment to the original integral,
and just write the answer, which is
Obviously, this is actually a definite integral, and you can evaluate the limits. However, you could easily have done this using substitution (which is literally the same method, which simplified algebra) by letting u=x^2 (or x^2-1 if you're clever!)
Post by: katnisschung on March 28, 2017, 12:01:35 pm
There absolutely is! However, you're essentially just doing substitution, but without the extra steps. Recall that
In this case, notice that
So, we can make a slight adjustment to the original integral,
and just write the answer, which is
Obviously, this is actually a definite integral, and you can evaluate the limits. However, you could easily have done this using substitution (which is literally the same method, which simplified algebra) by letting u=x^2 (or x^2-1 if you're clever!)
wait... substitution is not in the 2u course correct?
also did u mean to take out the 2 instead of half...
Post by: RuiAce on March 28, 2017, 01:36:57 pm
wait... substitution is not in the 2u course correct?He's only referring to substitution. He's not exactly saying it's in the 2u course.
also did u mean to take out the 2 instead of half...
Misread the question. Yes he intended to take the 2 out. I think he forgot that the original integrand had a 4 in it.
Post by: jakesilove on March 28, 2017, 02:01:12 pm
He's only referring to substitution. He's not exactly saying it's in the 2u course.
Misread the question. Yes he intended to take the 2 out. I think he forgot that the original integrand had a 4 in it.
Yep, definitely took out an incorrect factor!
Post by: Fahim486 on March 28, 2017, 02:53:25 pm
Thanks!
Post by: jamonwindeyer on March 28, 2017, 03:10:44 pm
Hi again, I'm sort of stuck on this question.
Thanks!
Hey! For the curve to be decreasing, the first derivative is negative, so:
You can draw a diagram of the parabola \(y=x^2-1\) to find that this occurs for \(-1<x<1\), so that is our answer!! Does that make sense? :)
Post by: Fahim486 on March 28, 2017, 03:18:21 pm
Hey! For the curve to be decreasing, the first derivative is negative, so:
You can draw a diagram of the parabola \(y=x^2-1\) to find that this occurs for \(-1<x<1\), so that is our answer!! Does that make sense? :)
Yes now I finally understand why the answer is -1<x<1
Thanks so much for your help!!! :D
Post by: katnisschung on March 28, 2017, 03:41:33 pm
in terms of the attached?
thanks :)
Post by: katnisschung on March 28, 2017, 03:58:26 pm
i did i) correctly and got ii)
as either
+ or - (R/3^1/2) (by differentiating once)
but i'm a little confused how to test their nature
i did the second differentiation and subbed it in
but confused....
Post by: Thebarman on March 28, 2017, 04:29:17 pm
Thanks
Post by: jakesilove on March 28, 2017, 05:47:44 pm
what does real solutions mean?
in terms of the attached?
thanks :)
Hey! It just means that there are three x-intercepts :) There are heaps of ways to answer this question; personally, I would sketch the function on the left hand side, and then see for what values of y=k (ie. horizontal line at y=k) the line hits three points of the cubic :)
Post by: jakesilove on March 28, 2017, 05:48:04 pm
another question...
i did i) correctly and got ii)
as either
+ or - (R/3^1/2) (by differentiating once)
but i'm a little confused how to test their nature
i did the second differentiation and subbed it in
but confused....
The attachment isn't coming up for me!
Post by: jakesilove on March 28, 2017, 05:53:04 pm
Hey guys, can someone explain to me how to prove that the function y=3 + (x-2)^2 is symmetric about the line x=2?
Thanks
So, we have the function
And we want to show that this is symmetrical around the vertical line x=2.
Personally, I would just sketch the function and show that it obviously is symmetrical around the line x=2. However, you can also just shift the function to the left two units, and prove that this function is even! Let's try that. The SHIFTED function will be
Now, to prove that something is even, we show that
Clearly, that is true for the above function.
Post by: RuiAce on March 28, 2017, 05:58:03 pm
Hey guys, can someone explain to me how to prove that the function y=3 + (x-2)^2 is symmetric about the line x=2?
Thanks
So, we have the function
And we want to show that this is symmetrical around the vertical line x=2.
Personally, I would just sketch the function and show that it obviously is symmetrical around the line x=2. However, you can also just shift the function to the left two units, and prove that this function is even! Let's try that. The SHIFTED function will be
Now, to prove that something is even, we show that
Clearly, that is true for the above function.
I don't feel this is really appropriate for 2U though
Post by: katnisschung on March 28, 2017, 06:23:41 pm
The attachment isn't coming up for me!
Post by: Jerylieiah on March 29, 2017, 03:34:54 pm
Thanks if someone can help
Post by: kiiaaa on March 29, 2017, 04:22:13 pm
so i was doing this question and when i looked at the solution i was completely lost on where did the '3' go? (look where there is the second tick) im just lost on what happened to it and how it happened and was wondering if someone could explain it to me please?
thank you
Post by: jakesilove on March 29, 2017, 06:09:42 pm
Hey! So, you got the actual roots, but you just weren't sure what to do from there? Remember that the root HAS to be positive, as we're talking about a length. Negative lengths don't make any sense! Once you assume that there is only one root, it is fair to assume that this root is a max.
Post by: jakesilove on March 29, 2017, 06:10:56 pm
Hey guys, how would you find the primitive function of y=25/(x-2)2
Thanks if someone can help
So we want to find
Here, we simply use the reverse chain rule as the function inside the brackets is linear
Post by: jakesilove on March 29, 2017, 06:12:04 pm
Hello
so i was doing this question and when i looked at the solution i was completely lost on where did the '3' go? (look where there is the second tick) im just lost on what happened to it and how it happened and was wondering if someone could explain it to me please?
thank you
Just expand the bracket
Post by: cxmplete on March 29, 2017, 07:24:00 pm
Post by: Jerylieiah on March 29, 2017, 07:44:00 pm
So we want to find
Here, we simply use the reverse chain rule as the function inside the brackets is linear
Thanks, but the question was actually:
If you could help with this one, it would be much appreciated
Post by: jakesilove on March 29, 2017, 08:13:40 pm
Thanks, but the question was actually:
If you could help with this one, it would be much appreciated
Answering this question is exactly the same! Just use a negative power.
Again, just an application of the reverse chain rule :)
Post by: jakesilove on March 29, 2017, 08:15:41 pm
How would I do question 14?
Hey! Recall that
Where x is displacement. So, in this case,
Imposing our initial conditions,
Post by: Youssk on March 30, 2017, 12:31:55 pm
A cylinderical can is to have a volume of 20πm^3. The material for the circular top and bottom costs $10 per m^2 and the material for the curve surface costs $8 per m^2.
i) given that there are no overlapping edges and that the surface area of a cylinder is given by SA=2πr^2 + 2πrh, show the cost of the cylinder is given by C= 20πr^2 + 320π/(divided by) r.
ii) find the minimum cost of the material for the can.
Thankyou! :)
Post by: jakesilove on March 30, 2017, 12:52:17 pm
Hey, could someone please help me with this question.
A cylinderical can is to have a volume of 20πm^3. The material for the circular top and bottom costs $10 per m^2 and the material for the curve surface costs $8 per m^2.
i) given that there are no overlapping edges and that the surface area of a cylinder is given by SA=2πr^2 + 2πrh, show the cost of the cylinder is given by C= 20πr^2 + 320π/(divided by) r.
ii) find the minimum cost of the material for the can.
Thankyou! :)
Hey! So, we know that the circular parts of the can cost $10 per square meter, so
given that there is a circular top, and a circular bottom. We also know that the cost of the curved surface is $8 per square meter, so
So, the total cost will be
We need to get rid of the h somehow, so we impose our initial conditions. The volume of a cylinder is
We have been given volume, so
Subbing this into our cost relationship,
Great! Now, we want to minimise this cost. So, we differentiate to find stationary points
Multiplying through by r squared and dividing by pi
Since there is only one stationary point, it's fair to assume it is a minimum. However, I would recommend testing points to either side using the first derivative, and finding that 1.9 yields a negative value, whilst 2.1 yields a positive value (showing that concavity is positive, and thus the point is a minimum. This is way easier than finding the second derivative in this case!)
Now, we plug r=2 back into our cost equation to get the total cost;
Post by: Thebarman on March 30, 2017, 02:47:20 pm
How would I find the equation of a parabola with a focus (3,2) and directrix y=-4?
Post by: jakesilove on March 30, 2017, 02:53:22 pm
Thanks for the help with the last question!
How would I find the equation of a parabola with a focus (3,2) and directrix y=-4?
Hey! Firstly, we can easily find the vertex of the parabola. It will be halfway between the directrix and the focus. So, vertex is (3, -1). Now, the equation of a parabola is
Where the vertex has coordinates (b,c). So,
The distance between the vertex and the directrix is a. So, a=3
Post by: Fahim486 on March 30, 2017, 03:01:54 pm
Thanks!
Post by: jakesilove on March 30, 2017, 03:12:37 pm
Hey, I'm having trouble with part i. and ii. so could someone pls show me how to do it
Thanks!
Hey,
It's a bit difficult for me to draw you a probability tree diagram. Basically, you will first have two branches stemming from a central point. These two branches represent Monday; one branch (labelled M for 'miss') has a 0.3 chance of occurring, whilst the other branch (labelled C for 'catch') has a 0.7 chance of occurring. From each of these branches, another two branches will be constructed. These four branches represented Tuesday; against, one of each branches is labelled M (probability 0.2) and the other is labelled C (probability 0.8).
The probability of missing the train on ONE of the days is going to be 1 minus the probability of missing the train on NO days, minus the probability of missing the train on BOTH days (note that it is not AT LEAST one, it is ONLY one). So,
We can calculate the above probabilities by just multiplying the two days together
Post by: Fahim486 on March 30, 2017, 03:16:08 pm
Hey,
It's a bit difficult for me to draw you a probability tree diagram. Basically, you will first have two branches stemming from a central point. These two branches represent Monday; one branch (labelled M for 'miss') has a 0.3 chance of occurring, whilst the other branch (labelled C for 'catch') has a 0.7 chance of occurring. From each of these branches, another two branches will be constructed. These four branches represented Tuesday; against, one of each branches is labelled M (probability 0.2) and the other is labelled C (probability 0.8).
The probability of missing the train on ONE of the days is going to be 1 minus the probability of missing the train on NO days, minus the probability of missing the train on BOTH days (note that it is not AT LEAST one, it is ONLY one). So,
We can calculate the above probabilities by just multiplying the two days together
ah okay now i see where i went wrong. Thanks so much for your help!!!
Post by: Fahim486 on March 30, 2017, 06:41:34 pm
Thanks again!!
Post by: Thebarman on March 30, 2017, 08:50:46 pm
Thank you!
Hi once again, for question ii. I've worked out the coordinates of the stationary point but how would you determine the nature? I'm confused as to whether you differentiate once to determine the nature or you have to double differentiate.Hey Fahim, I hope you don't mind if I take a look at your question.
Thanks again!!
To put it simply, we use the first derivative (y') to determine the gradient, as well as to find any stationary points.
The second derivative (y'') is used to determine concavity (is it curved upwards or downwards), and whether the curve has a minimum or maximum value. Essentially, this determines the shape of the curve.
So if you need to determine the nature, you could either create a table of values and look a little bit to the left and a little bit to the right of the stationary points to determine their value, OR you can use the second derivative. The first option is much more time consuming, so we'll opt for the second.
In your case, the stationary points would be (2,0) and (8/3, 4/27). [I think. Forgive me if my math is a little off...]
To determine the nature, you would then find the second derivative (i.e. y'' = -6x + 14), and then sub the two x-coordinates of the stationary points into this equation.
If the answer is greater than 0 (y'' > 0), then the curve is a minimum, as it is concave up (there is a minimum x-value).
If the answer is less than 0 (y'' < 0), then the curve is a maximum, as it is concave down (there is a maximum x-value).
So, when x=2,
y'' = -6(2) + 14 = 2
y'' > 0
Therefore, there is a maximum at (2,0)
Similarly, when x=8/3,
y'' = -6(8/3) + 14 = -2
y'' < 0
Therefore, there is a minimum at (8/3, 4/27)
And that's it! Your answer would simply be the last line in each explanation (i.e. max/min at (x,y)).
Hope this explanation was helpful!
Post by: Fahim486 on March 30, 2017, 09:52:45 pm
Hey Fahim, I hope you don't mind if I take a look at your question.
To put it simply, we use the first derivative (y') to determine the gradient, as well as to find any stationary points.
The second derivative (y'') is used to determine concavity (is it curved upwards or downwards), and whether the curve has a minimum or maximum value. Essentially, this determines the shape of the curve.
So if you need to determine the nature, you could either create a table of values and look a little bit to the left and a little bit to the right of the stationary points to determine their value, OR you can use the second derivative. The first option is much more time consuming, so we'll opt for the second.
In your case, the stationary points would be (2,0) and (8/3, 4/27). [I think. Forgive me if my math is a little off...]
To determine the nature, you would then find the second derivative (i.e. y'' = -6x + 14), and then sub the two x-coordinates of the stationary points into this equation.
If the answer is greater than 0 (y'' > 0), then the curve is a minimum, as it is concave up (there is a minimum x-value).
If the answer is less than 0 (y'' < 0), then the curve is a maximum, as it is concave down (there is a maximum x-value).
So, when x=2,
y'' = -6(2) + 14 = 2
y'' > 0
Therefore, there is a maximum at (2,0)
Similarly, when x=8/3,
y'' = -6(8/3) + 14 = -2
y'' < 0
Therefore, there is a minimum at (8/3, 4/27)
And that's it! Your answer would simply be the last line in each explanation (i.e. max/min at (x,y)).
Hope this explanation was helpful!
That was explained very clearly and I finally understood the question. Thanks so much for your help!!!
Post by: jakesilove on March 30, 2017, 09:58:26 pm
Thank you!
Hey Fahim, I hope you don't mind if I take a look at your question.
To put it simply, we use the first derivative (y') to determine the gradient, as well as to find any stationary points.
The second derivative (y'') is used to determine concavity (is it curved upwards or downwards), and whether the curve has a minimum or maximum value. Essentially, this determines the shape of the curve.
So if you need to determine the nature, you could either create a table of values and look a little bit to the left and a little bit to the right of the stationary points to determine their value, OR you can use the second derivative. The first option is much more time consuming, so we'll opt for the second.
In your case, the stationary points would be (2,0) and (8/3, 4/27). [I think. Forgive me if my math is a little off...]
To determine the nature, you would then find the second derivative (i.e. y'' = -6x + 14), and then sub the two x-coordinates of the stationary points into this equation.
If the answer is greater than 0 (y'' > 0), then the curve is a minimum, as it is concave up (there is a minimum x-value).
If the answer is less than 0 (y'' < 0), then the curve is a maximum, as it is concave down (there is a maximum x-value).
So, when x=2,
y'' = -6(2) + 14 = 2
y'' > 0
Therefore, there is a maximum at (2,0)
Similarly, when x=8/3,
y'' = -6(8/3) + 14 = -2
y'' < 0
Therefore, there is a minimum at (8/3, 4/27)
And that's it! Your answer would simply be the last line in each explanation (i.e. max/min at (x,y)).
Hope this explanation was helpful!
Absolute legend, cheers for answering the question! Just goes to show how well you know the topic area :)
Post by: Kekemato_BAP on March 30, 2017, 11:12:58 pm
Post by: kiwiberry on March 30, 2017, 11:35:23 pm
Hi, I am stuck on this series-sequences question (circled in pic). Could someone please explain how the answer is 180? Thanks!!
Hey!
We are given that S7 = 5T7 and T6 + T7 = 40
Looking at the first equation,
S7 = 5T7
7/2(2a + 6d) = 5(a + 6d)
2a - 9d = 0 ---(1)
Looking at the second equation,
T6 + T7 = 40
(a + 5d) + (a + 6d) = 40
2a + 11d = 40 ---(2)
Solving (1) and (2) simultaneously, (2)-(1): 20d = 40 so d=2. Subbing this back in we get a=9
So S10 = 10/2(2a + 9d) = 5(18 + 18) = 180 :)
Post by: Kekemato_BAP on March 31, 2017, 12:02:37 am
Hey!
We are given that S7 = 5T7 and T6 + T7 = 40
Looking at the first equation,
S7 = 5T7
7/2(2a + 6d) = 5(a + 6d)
2a - 9d = 0 ---(1)
Looking at the second equation,
T6 + T7 = 40
(a + 5d) + (a + 6d) = 40
2a + 11d = 40 ---(2)
Solving (1) and (2) simultaneously, (2)-(1): 20d = 40 so d=2. Subbing this back in we get a=9
So S10 = 10/2(2a + 9d) = 5(18 + 18) = 180 :)
Thank you!! That really helps :)
Post by: Sukakadonkadonk on March 31, 2017, 12:42:25 pm
Does anyone know if its possible to integrate ln(x) function?
Or if only way is to convert to exponential and have it restricted with y-axis and do it from there?
Thanks.
Post by: jamonwindeyer on March 31, 2017, 12:49:21 pm
Hey,
Does anyone know if its possible to integrate ln(x) function?
Or if only way is to convert to exponential and have it restricted with y-axis and do it from there?
Thanks.
Interesting question! You can do it, but for real numbers the integral is only defined for \(x>0\) as \(x\ln{x}-x\). You can prove that because differentiating this yields:
Once you try and do it for negative \(x\) values, you need to start using imaginary numbers. It is still doable though, and indeed, the form above works! But you need to do it in the complex plane, not the real plane ;D
Warning: Non assessable content for this course ;D
Post by: Sukakadonkadonk on March 31, 2017, 01:35:37 pm
Interesting question! You can do it, but for real numbers the integral is only defined for \(x>0\) as \(x\ln{x}-x\). You can prove that because differentiating this yields:
Once you try and do it for negative \(x\) values, you need to start using imaginary numbers. It is still doable though, and indeed, the form above works! But you need to do it in the complex plane, not the real plane ;D
Warning: Non assessable content for this course ;D
Ok thank you!
This was probably the wrong thread to post in ;D
Post by: Sukakadonkadonk on March 31, 2017, 01:50:25 pm
Another question I have,
Write the sum 5 + 9 + 13 + ... 4n-3
in sigma notation.
Could someone please explain? There was none in the solutions.
Thanks.
Post by: cxmplete on March 31, 2017, 02:47:53 pm
Post by: jakesilove on March 31, 2017, 03:21:33 pm
Hey guys,
Another question I have,
Write the sum 5 + 9 + 13 + ... 4n-3
in sigma notation.
Could someone please explain? There was none in the solutions.
Thanks.
Hey! All this is asking for is to write each term of the series in terms of an integer, n. So, we can write the series
as
In sigma notation, this looks like
Does that make sense?
Edit by Rui: can't use n as the summation index. Switched with k
Post by: RuiAce on March 31, 2017, 03:27:25 pm
Hey,There are actually many functions out there that are integrable. The only thing is that you aren't taught integration techniques in the HSC unless you do 4U, hence what Jamon means by not in this course.
Does anyone know if its possible to integrate ln(x) function?
Or if only way is to convert to exponential and have it restricted with y-axis and do it from there?
Thanks.
Hey guys,
Another question I have,
Write the sum 5 + 9 + 13 + ... 4n-3
in sigma notation.
Could someone please explain? There was none in the solutions.
Thanks.
Post by: jakesilove on March 31, 2017, 03:32:20 pm
how would i do question 4 parts c and d?
For c), we just sketch the exponential graph. You can do this by plotting points, or by shifting the regular exponential graph. We know the graph has been shifted UP 60, will be a negative exponential and goes through 100 when t=0. So,
(http://imgur.com/a/aGatd)
To calculate the distance, we need to integrate the function. We know that
So,
We can say that, at time t=0. x=0 (as the car is just passing A). So,
Now, subbing in 3 (three hours)
Post by: cxmplete on March 31, 2017, 04:00:41 pm
Post by: jamonwindeyer on March 31, 2017, 04:57:01 pm
so im doing question 6, and im really confused on how to do it
Sure! Just a note, this question type is super common, so make sure you understand it! Probably worth 5 marks in your HSC this stuff ;D
So we have this equation, \(N=Ae^{kt}\). Now we know that when \(t=0\), there were 124 million users. So we can substitute:
So now our equation is \(N=124000000e^{kt}\). Remember, the letters besides \(t\) are constants, so once we've found them, that is their value for the rest of the question. We also know that at the start of 2009, when \(t=8\), there were 220 million. So:
That's to four decimal places. So now we know all our constants!
To find the internet users at the start of 2012, that's \(t=11\), so:
That is to the nearest whole ;D then finally, just solve for \(N=300\text{ million}\):
So we'd expect it to exceed 300 million somewhere in the 13th year, so somewhere in 2013! ;D
Again, super important question type, so be sure to let me know if this doesn't make sense!
Post by: Kekemato_BAP on March 31, 2017, 07:56:33 pm
Post by: jamonwindeyer on March 31, 2017, 08:45:26 pm
Hi! Can someone help me on Question(ii) in the attached pic? How do I find $M based on what is given? Thanks!!
Hey Kekemato! What was your expression for \(A_{60}\)? You should be able to generalise it to the life of the loan and put it equal to zero. That is, instead of creating an expression for \(A_{60}\), craft a similar one of \(A_{180}\), and solve \(A_{180}=0\) ;D
You should have something like:
You can use the formula for a sum of a geometric series to tidy that series up ;D is this looking familiar based on your work in (i)? ;D
Post by: Sukakadonkadonk on March 31, 2017, 10:24:15 pm
Hey! All this is asking for is to write each term of the series in terms of an integer, n. So, we can write the series
as
In sigma notation, this looks like
Does that make sense?
Edit by Rui: can't use n as the summation index. Switched with k
Hey,
So are you guys saying that we need to our own intuition to figure what 'n' is in that series?
Post by: RuiAce on March 31, 2017, 10:29:49 pm
Deleted my post because I realised you were asking about n
Hey,
So are you guys saying that we need to our own intuition to figure what 'n' is in that series?
Write our the formula first. In Jake's example it was 4k-3
Then, it's not hard to just look at the first and last terms to deduce the lower and upper boundaries.
Lower: 5 = 4k-3 so k = 2
Upper: 4n-3 = 4k+3 so k = n
Post by: Kekemato_BAP on April 02, 2017, 12:34:43 am
Hey,
Does anyone know if its possible to integrate ln(x) function?
Or if only way is to convert to exponential and have it restricted with y-axis and do it from there?
Thanks.
It is possible although not in the 2-U syllabus. Intergration by Parts.
(I don't know how to use the fancy symbols, someone teach me please)
lnx = 1*lnx
Use the acronym LIATE(logs, inverse trig, algrebraic, trig, exponential) to find u and v'. The first order will become u.
u would equal lnx since lnx is a log.
v' would be 1 since it is algebraic
u' is 1/x
v is x
(integration)(uv')=uv-(integration)(vu')
= lnx*x - (integration)(x*1/x)
= xlnx- (integration)(1)
= xlnx - x
(Don't forget +C for every step too)
Anyways, that's a 3-4U topic and 2U don't need to know it. I never did 3-4U but was taught by this 3U kid who dropped into my 2U class.
Post by: jamonwindeyer on April 02, 2017, 12:53:56 am
Anyways, that's a 3-4U topic and 2U don't need to know it. I never did 3-4U but was taught by this 3U kid who dropped into my 2U class.
Thanks for the proof Kekemato! Just in case any 3 unit students read this and panic, it is only 4 unit students that could be expected to do this, only 4U learn integration by parts - 2U and 3U students could do it if they were asked to differentiate \(x\ln{x}-x\) first ;D
Here is a guide for the fancy symbols ;)
Post by: sophiegmaher on April 02, 2017, 09:47:31 am
*the limits are when x=2 and x=5*
I feel like the answer is simple but I'm stuck.
Post by: jamonwindeyer on April 02, 2017, 10:52:34 am
I really need some help on this question!!: "The area ln5/2 is rotated about the x-axis. Find the volume of the solid formed."
*the limits are when x=2 and x=5*
I feel like the answer is simple but I'm stuck.
Hey! So this one is interesting, because you are told the area but not the curve(s) being rotated - That's the tricky bit. You need to figure that out. Consider this:
So we know that whatever we integrate to get that area, our result is \(\ln{x}\) - Well that must be \(\frac{1}{x}\)! Our original integral to find that area must have been:
So therefore, to find the volume, we are doing:
And that is just a standard definite integral :) does that make sense? ;D
Post by: sophiegmaher on April 02, 2017, 11:01:03 am
Quote
And that is just a standard definite integral :) does that make sense? ;D
Yes it does! However, the answers say the volume is 3/10
Post by: jamonwindeyer on April 02, 2017, 11:08:55 am
Yes it does! However, the answers say the volume is 3/10
Perhaps some incorrect working? :)
Post by: cxmplete on April 02, 2017, 07:03:14 pm
Post by: K9810 on April 02, 2017, 08:00:07 pm
Can you please help me with this question?
Post by: sophiegmaher on April 02, 2017, 10:04:54 pm
Perhaps some incorrect working? :)haha I think so, I didn't bring the x^2 to the top! Thank you!!
Post by: jakesilove on April 03, 2017, 09:24:33 am
Hi, how would you do questions 30 and 31 (sorry that they're such large questions but i'm realy stuck)?
Hey! For the first question, we literally just sub in the relevant values;
From here, it gets a bit trickier. We need to find a nice pattern. Let's look at what happens in the third year, but without using the VALUE of the instalment, but rather using the relationship
Can you see a pattern? In the nth year, we find that
Now, letting n=20, we get $4306.38 (as required)
For the last part, we need to create a sum. This will look like
We use the sum of a geometric series to find that
Sub in n=20, and you get the correct answer out! Try using this approximate working out to get 31, and if you're still struggling we can walk you through it
Post by: jakesilove on April 03, 2017, 09:31:12 am
Hi,
Can you please help me with this question?
Hey! First, we find the angle by using the length of an arc formula
Now, there is just a formula for the area of the minor segment. But, since it isn't on your formula sheet, let's derive it!
First, we find the area of the whole segment. Thankfully, that formula IS on your formula sheet, and is
Now, let's find the area of the TRIANGLE.
So, the area of the minor segment will be
Post by: K9810 on April 03, 2017, 10:09:54 am
Hey! First, we find the angle by using the length of an arc formula
Now, there is just a formula for the area of the minor segment. But, since it isn't on your formula sheet, let's derive it!
First, we find the area of the whole segment. Thankfully, that formula IS on your formula sheet, and is
Now, let's find the area of the TRIANGLE.
So, the area of the minor segment will be
Hey! Thanks for helping.
Can I directly use the formula for the minor segment 1/2*r^2(theta-sintheta)? Or do I have to do it separately like the way you showed me?
Post by: jakesilove on April 03, 2017, 10:10:41 am
Hey! Thanks for helping.
Can I directly use the formula for the minor segment 1/2*r^2(theta-sintheta)? Or do I have to do it separately like the way you showed me?
If you know the formula, you can definitely just churn it out and get straight to the answer :)
Post by: K9810 on April 03, 2017, 10:41:59 am
If you know the formula, you can definitely just churn it out and get straight to the answer :)
Hey, sorry to bother you.
But when I used the minor segment formula I got
0.5(12)^2(pi/6-sinpi/6)=37.04
Post by: jakesilove on April 03, 2017, 10:45:11 am
Hey, sorry to bother you.
But when I used the minor segment formula I got
0.5(12)^2(pi/6-sinpi/6)=37.04
Is your calculator in radians? Google tells me the answer is like 1.7 :)
Post by: K9810 on April 03, 2017, 10:57:07 am
Is your calculator in radians? Google tells me the answer is like 1.7 :)
Thank you so much!
I forgot to put it in radians :)
Post by: K9810 on April 03, 2017, 01:27:06 pm
Can you please explain how you know whether the points are included/excluded?
Thanks!
Post by: jamonwindeyer on April 03, 2017, 01:36:49 pm
Hey!
Can you please explain how you know whether the points are included/excluded?
Thanks!
Hey! It's because the inequality for that bottom line is strict: \(y>4\), not \(y\ge4\). So, points on that line aren't included in the region ;D
Post by: chloeemadavis on April 04, 2017, 05:37:28 pm
Would someone be able to help me with a question?
1/ Evaluate the integral between 1 and 0 of 5/square root (e^x) dx
Thankyou!!
Post by: Bubbly_bluey on April 04, 2017, 06:25:23 pm
Hi!Hey! so that question looks pretty nasty with its root, so lets change that 5th root into index.
Would someone be able to help me with a question?
1/ Evaluate the integral between 1 and 0 of 5/square root (e^x) dx
Thankyou!!
Just for reference: sqroot (x)= x1/2
So: 5throot(x) = x1/5
Now you have an intergation of ex/5.
Intergation ex/5 dx = (ex/5) / (1/5)
now you can sub in your borders. Hope it makes sense :)
Post by: jamonwindeyer on April 04, 2017, 06:45:02 pm
Hi!
Would someone be able to help me with a question?
1/ Evaluate the integral between 1 and 0 of 5/square root (e^x) dx
Thankyou!!
Welcome to the forums Chloe! Hopefully I've interpreted your question correctly:
Hopefully this makes sense!! Feel free to shoot any other questions our way ;D
Post by: Leah_Mer on April 05, 2017, 10:40:35 am
Post by: jamonwindeyer on April 05, 2017, 11:10:35 am
Hey there, I'm wondering about school assessment marks and how badly they impact your final mark in the hsc! For adv maths, I'm ranked second in my class with one other above me who is really great at maths and we're usually a few marks from each other and always above 90%, however, in the half yearly I absolutely bombed the exam and am expecting a mark of about 30%-40% at the MOST (it was weighted 30%). What I'm wondering is, is it possible to get a band 6 with a school assessment mark of like 75? Whats the lowest raw school mark that still allows for a band 6 in maths, if you were to smash the HSC exam?
Hey Leah! Welcome to the forums!
You should give this guide a read - It goes through the entire process of how your marks are calculated. Essentially, you could get a Band 6 with a school assessment mark of 30. The mark is irrelevant - It's the rank that matters, and even that matters far less than a strong performance in the HSC exam. The guide contains the details, but the answer to your question is - Yes, you absolutely can!
A poor half yearly result isn't the end of the world - I got one, and it does suck, but it can be the motivator that pushes you to work harder and do really well in the more important exams at the end of the year! ;D keep your chin up, learn from your mistakes - This is just a bump in the road :)
Post by: Leah_Mer on April 05, 2017, 08:58:29 pm
Hey Leah! Welcome to the forums!
You should give this guide a read - It goes through the entire process of how your marks are calculated. Essentially, you could get a Band 6 with a school assessment mark of 30. The mark is irrelevant - It's the rank that matters, and even that matters far less than a strong performance in the HSC exam. The guide contains the details, but the answer to your question is - Yes, you absolutely can!
A poor half yearly result isn't the end of the world - I got one, and it does suck, but it can be the motivator that pushes you to work harder and do really well in the more important exams at the end of the year! ;D keep your chin up, learn from your mistakes - This is just a bump in the road :)
Thank you so much! Very good reassurance to hear that and will definitely be more motivated for maths study after this! Also thanks for getting back to me so quickly :)
Post by: laurenf58 on April 07, 2017, 01:30:04 pm
Thanks!
Post by: jamonwindeyer on April 07, 2017, 02:09:16 pm
Could I have some help with these questions please?
Thanks!
Hey! So even functions are symmetrical about the y-axis, so draw a mirrored copy of the curve on the other side of the axis! Just like, say, the parabola \(y=x^2\) is the same on both sides of the y-axis because it is even ;D
Odd functions have rotational symmetry, meaning you spin the curve 180 degrees around the origin to get the other side of the curve! To picture this, think the graph of \(y=x^3\) (see here) see how you can get the bottom left by spinning the top right around to the opposite corner? That's what you'll do for your curve :)
Hope that helps! ;D
Post by: cxmplete on April 09, 2017, 01:56:05 pm
Can someone please explain how to do exponential growth and decay questions (from ACPW)?
Post by: jakesilove on April 09, 2017, 03:08:53 pm
Hi,
Can someone please explain how to do exponential growth and decay questions (from ACPW)?
Could you give us an example of a question we can help you with?
Post by: cxmplete on April 09, 2017, 03:29:11 pm
Post by: jakesilove on April 09, 2017, 05:30:23 pm
question 15 from this paper https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-maths.pdf
Cool! So, the first part is pretty straight forward; we just sub the value of C that they gave us into the differential equation.
as required. Now, we are given that the initial quantity of caffeine is 130 mg. So, we sub in t=0 to find the value of A.
Then, to find the amount of caffeine in the body after 7 hours, we just sub in t=7
Finally, we want to find the time taken for the amount of caffeine in Lee's body to half. Thus, we want C to equal 130/2=65mg. So,
Post by: jakesilove on April 14, 2017, 09:55:28 am
Hello, I am new to this forum and am not sure if this has been posted or not.
This is from one of my schools past HSC trial paper. I need help with the whole question except for c) (I can do it myself after i find all the other answers)
(http://i64.tinypic.com/2vt7712.png)
Thank you cuties!
Hey! So, the equation we have been given is
and we want to rewrite it in standard form. First, we need to complete the square, as the final answer is in the form (x-h)^2. So,
And that's our answer! We got
Great! Now, we want to calculate the roots of the function. The easiest way of doing this is the quadratic equation.
Easy as :) Now, with all of the above information (and subbing in x=0, y=0 etc.) you can easily sketch the graph!
For the last part, we essentially want to know where f(x) is ABOVE the x-axis. We know the function is a positive parabola, so this will occur to either side of the roots (ie. to the left of the first x intercept, and to the right of the second x intercept). This will become super obvious once you draw the sketch :)
Post by: Mohammad130no on April 14, 2017, 11:38:54 am
Hey! So, the equation we have been given is
and we want to rewrite it in standard form. First, we need to complete the square, as the final answer is in the form (x-h)^2. So,
And that's our answer! We got
Great! Now, we want to calculate the roots of the function. The easiest way of doing this is the quadratic equation.
Easy as :) Now, with all of the above information (and subbing in x=0, y=0 etc.) you can easily sketch the graph!
For the last part, we essentially want to know where f(x) is ABOVE the x-axis. We know the function is a positive parabola, so this will occur to either side of the roots (ie. to the left of the first x intercept, and to the right of the second x intercept). This will become super obvious once you draw the sketch :)
Cheers for all the help.
So thats it for b) ?? the roots are just 4+-(sqrt)30?
Post by: RuiAce on April 14, 2017, 11:52:20 am
Cheers for all the help.Yes
So thats it for b) ?? the roots are just 4+-(sqrt)30?
Post by: Mohammad130no on April 14, 2017, 12:01:06 pm
Yes
Thank you guys so much
Post by: jakesilove on April 14, 2017, 12:46:23 pm
Sorry jake I am having a bit of difficulty when graphing this function. If I find the x intercepts mathematically it is (4+-[sqrt30])/2 , and y int is -7. Though if I put it into a graphing website it shows the y int as (2,-15)
The x intercepts isn't what you've written; you've divided by two for some reason. The y intercept can't be (2,-15) as it has to occur when x=0. (2, -15) is the vertex, not the y intercept.
Post by: Mohammad130no on April 14, 2017, 01:06:52 pm
The x intercepts isn't what you've written; you've divided by two for some reason. The y intercept can't be (2,-15) as it has to occur when x=0. (2, -15) is the vertex, not the y intercept.
all graphing wesbites show that as the x value. And you were right about the y and vertex, i messed up
Post by: kiwiberry on April 14, 2017, 01:24:59 pm
all graphing wesbites show that as the x value. And you were right about the y and vertex, i messed up
silly mistake with the quadratic formula haha :)
Post by: jakesilove on April 14, 2017, 01:33:48 pm
The x intercepts isn't what you've written; you've divided by two for some reason. The y intercept can't be (2,-15) as it has to occur when x=0. (2, -15) is the vertex, not the y intercept.
Aha true, true, you can see which line I messed up above (forget the 2*a in the quadratic equation!). So you're all sorted now?
Post by: jakesilove on April 14, 2017, 01:35:45 pm
Help Please, this is from the Christian 2011 boys High school Trial Paper
(http://i68.tinypic.com/297ajo.png)
Could you post this up on the Physics thread?
Post by: jakesilove on April 14, 2017, 02:05:43 pm
Sure, even though its from the 2 unit paper?
About the statement you said before where you made a mistake, are you refering to b) where you found the roots?
Aha that's crazy, I'm clearly having a bad day sorry about that! Just assumed it was Physics (Keplar's law is in the Space curriculum, and never seen a Maths question on it!). I'll have a crack now, and yet I just made a mistake in finding the roots.
Post by: jakesilove on April 14, 2017, 02:19:28 pm
Help Please, this is from the Christian 2011 boys High school Trial Paper
(http://i68.tinypic.com/297ajo.png)
So, for the first part (proportional to), we would write
As a full equation, we can replace the 'proportional to' with some constant, k.
Now, we know that the Earth will rotate around the sun in one year. We also know that the average distance is 150 million kilometers. So, we can write
Okay, so we have
Part c) doesn't actually have a question; I assume it wants us to find the period of Neptune? Well, the distance is 30*150 million km.
Google says that the orbital period is about 164 years, so this approximation works well!
Now, for the final part, we again know that
The k may be different, who knows. However, we assume that the k value is the same for planet X and planet Y. Let the distance between planet X and the Sun be
Then,
Now, Y is twice the distance away, so
So,
Thus, planet Y has an orbital period that is about 35% greater than planet X
Post by: JuliaPascale123 on April 15, 2017, 07:44:56 pm
Heeyyy, this question was ridiculous. Can I please have some help
Post by: jakesilove on April 15, 2017, 08:17:28 pm
(http://i64.tinypic.com/keaxiq.png)
Heeyyy, this question was ridiculous. Can I please have some help
Hey Julia! Welcome to the forums!
The first part is pretty straight forward, and I'm not sure how to construct a table on the forum. Basically, the quantity will decrease over time by 25% every 40 minutes. If t is minutes, and Q is in mg,
etc.
Now, we want to find an equation for Q. We know that what we're looking for here is some sort of exponential function. Once you've done these questions a number of times, you'll know that the form of the answer must be
Now, we know that the initial quantity is 300mg. So
We also know that the quantity has dropped to 225mg after 40 minutes. So,
So, the equation for the quantity of the medicine is
Next, we want to find the time it takes for the quantity of medicine to half. I won't bother guessing and checking, let's just find the actual answer.
Remember that this is in minutes. So, dividing by 60, we get t=9.04 hours.
Part e starts to get tricky. We could use an arbitrary time t=a. However, it's exactly the same thing to start at an arbitrary quantity, A. That is because, at time t=a, there will be some quantity A of medicine in the system. Then, we would look for when A halves!
is our new formula. t is still arbitrary, but so is A. Now, we need to show that, regardless of A, the halving time of this equation is a constant (ie. the time we proved in part d).
Our halving time will occur when the quantity is 0.5A (half of the initial, arbitrary amount). So,
But this is exactly the same equation as above, with the same solution! So, the halving time is a constant for any arbitrary starting point.
The next part is just a standard sketch of an exponential function. If you're not sure what it will look like, plot some points, or use an online graphing tool.
For the last part, we solve for
Which equals 50.99 hours. We should definitely round up, for the safety of the patient, giving us 51 hours.
Post by: itssona on April 16, 2017, 05:38:11 pm
with absolute value, I was doing problems and wanted to confirm;
if I have an inequality with x on both sides, do i square both sides??
like in
|x-2|<x
and how would you show this on a numberline:
|4x-2| >=3
thanks!! :)
Post by: RuiAce on April 16, 2017, 05:44:40 pm
Heey,
with absolute value, I was doing problems and wanted to confirm;
if I have an inequality with x on both sides, do i square both sides??
like in
|x-2|<x
and how would you show this on a numberline:
|4x-2| >=3
thanks!! :)
______________
Post by: RuiAce on April 16, 2017, 05:51:48 pm
Post by: itssona on April 16, 2017, 06:22:24 pm
thank you so much!!!!
I love this algebraic way you solved it :O I never knew this, thank you! :D :D
Post by: JuliaPascale123 on April 17, 2017, 07:45:01 pm
(http://i67.tinypic.com/ycq49.png)
Post by: RuiAce on April 17, 2017, 07:50:53 pm
Whoa!!This problem is not suitable for 2U purposes. It requires the use of a 4U technique known as implicit differentiation
(http://i67.tinypic.com/ycq49.png)
Post by: itssona on April 18, 2017, 10:51:07 am
5log(a) - loga(a)^4
so I would say 5loga-4loga(a)
but now im not sure,
like do i do 5/4 logaa ???
thank you :)
Post by: RuiAce on April 18, 2017, 10:57:37 am
hey would you be able to see if im doing this problem right- im not sure entirely
5log(a) - loga(a)^4
so I would say 5loga-4loga(a)
but now im not sure,
like do i do 5/4 logaa ???
thank you :)
Post by: itssona on April 18, 2017, 11:09:41 am
oh!! that passed me! so the answer is simply 5loga-4?
cant simplify it further yeah?
Post by: itssona on April 18, 2017, 11:11:48 am
oh wait, so 5-4
because that means both terms are 1
so the answer is 1?
Post by: RuiAce on April 18, 2017, 11:38:37 am
oh wait, so 5-4
because that means both terms are 1
so the answer is 1?
Post by: itssona on April 18, 2017, 11:50:27 am
ah that makes sense, thank you a tonne :D :D
Post by: itssona on April 18, 2017, 04:13:29 pm
thanksss!
why am i finding that 3u is easier than 2u for prelim HYs lol
Post by: RuiAce on April 18, 2017, 04:32:13 pm
hey how would i find the value of |x-4| when x<4
thanksss!
why am i finding that 3u is easier than 2u for prelim HYs lol
Post by: jamonwindeyer on April 18, 2017, 05:03:03 pm
hey how would i find the value of |x-4| when x<4
thanksss!
why am i finding that 3u is easier than 2u for prelim HYs lol
I found the same thing for most of Prelim, it's because 3U doesn't throw that much at you in Year 11 - Doesn't really get gnarly until Year 12 (Extension Trig was the only prelim topic that gave me strife) :)
Post by: Sukakadonkadonk on April 18, 2017, 06:45:11 pm
Thanks
Post by: itssona on April 18, 2017, 07:09:04 pm
oh! That makes sense, thank you so much :D :D
Post by: itssona on April 18, 2017, 07:12:59 pm
I found the same thing for most of Prelim, it's because 3U doesn't throw that much at you in Year 11 - Doesn't really get gnarly until Year 12 (Extension Trig was the only prelim topic that gave me strife) :)argh I wish they spaced it out better so they teach harder 3U concepts in yr11 too! Well I was looking at HSC trig questions and gave up right away XD
now I wonder- do they test you prelim topics in HSC?? Like in 2U HSC, is there absolute value and that kinda thing?
Post by: ellipse on April 18, 2017, 07:21:55 pm
argh I wish they spaced it out better so they teach harder 3U concepts in yr11 too! Well I was looking at HSC trig questions and gave up right away XD
now I wonder- do they test you prelim topics in HSC?? Like in 2U HSC, is there absolute value and that kinda thing?
Yes, they do test, up to 20% (or 30% not sure) prelim content in the HSC :)
Post by: itssona on April 18, 2017, 07:26:30 pm
Yes, they do test, up to 20% (or 30% not sure) prelim content in the HSC :)crap XD I guess it could've been worse but ugh
thank you for telling me! :)
Post by: RuiAce on April 18, 2017, 07:40:37 pm
Hi, could someone please help me with proving this?This is not appropriate for the 2U course. All approaches need minimum Extension 1 concepts.
Thanks
Post by: RuiAce on April 18, 2017, 07:42:01 pm
argh I wish they spaced it out better so they teach harder 3U concepts in yr11 too! Well I was looking at HSC trig questions and gave up right away XDIt doesn't have to be absolute value questions; it can be anything in the preliminary course. I've definitely seen it before, but absolute value stuff isn't considered 'common'.
now I wonder- do they test you prelim topics in HSC?? Like in 2U HSC, is there absolute value and that kinda thing?
Note that mathematics is the only subject in the HSC that you can be assessed on preliminary content.
Post by: itssona on April 18, 2017, 08:05:28 pm
It doesn't have to be absolute value questions; it can be anything in the preliminary course. I've definitely seen it before, but absolute value stuff isn't considered 'common'.Ah what a respite that maths is the only one where we're assessed on prelim content! Gahh that makes me happy ahah
Note that mathematics is the only subject in the HSC that you can be assessed on preliminary content.
Post by: jamonwindeyer on April 18, 2017, 08:51:19 pm
Ah what a respite that maths is the only one where we're assessed on prelim content! Gahh that makes me happy ahah
Quick Off Topic Aside
They can implicitly assess Prelim content in other subjects though! A lot of the basic ideas/principles in Moving About from Y11 can show up in Physics, for example. Criteria you use to evaluate legal mechanisms in Prelim Legal Studies is still useful in Y12. It all builds, so make sure you don't switch off entirely ;)
Post by: JuliaPascale123 on April 19, 2017, 11:44:56 am
(http://i63.tinypic.com/r70lma.png)
Post by: hanaacdr on April 19, 2017, 02:11:35 pm
Could i get some help on question 26 please(http://uploads.tapatalk-cdn.com/20170419/50f6616fed6c58952fc1129223ad0245.jpg)
Post by: jamonwindeyer on April 19, 2017, 02:36:30 pm
Question 5, Grandville 2014 Trial Paper:
(http://i63.tinypic.com/r70lma.png)
Hey!! So for the first one, stationary points occur when \(x=-2\), \(f(x)=3\) (from the y-value), and \(f'(x)=0\) because it is a stationary point! Make those substitutions: into the function and the first derivative, to get a couple of equations:
So the function we now know is \(f(x)=x^3-12x-13\). Now it is just standard calculus. To find the turning points, put the 1st derivative equal to zero:
To find the inflexion, put the second derivative equal to zero:
To find the y-coordinates of any of these points, just pop the x-value back into the original function! ;D
D is the hard bit - If you draw the curve you'll get something like this. What the question is asking is, where can you draw a horizontal line and cross that curve three times? That would mean you have three solutions to the equation \(f(x)=k\). For example, \(k=0\) works, because a horizontal line through \(y=0\) would cut the curve three times. The question is, what range of \(k\) values allow this to occur? At what point does it go too high or too low to cut the curve three times?
Hint: It is related directly to the coordinates of your turning points!
If that hint wasn't enough (pardon the edit Jamon!)
Specifically the y-coordinates
Hopefully this makes sense - Let me know if you need anything clarified ;D
Post by: jamonwindeyer on April 19, 2017, 02:41:07 pm
Hi
Could i get some help on question 26 please(http://uploads.tapatalk-cdn.com/20170419/50f6616fed6c58952fc1129223ad0245.jpg)
Sure! The sum of the first ten terms, we can express that as:
The sum of the next ten terms, the easiest way to express that is as \(S_{20}-S_{10}\). That will be:
We can therefore form two equations:
Solve those simultaneously - You'll get \(a=2,d=2\) :) hope that made sense!
Post by: RuiAce on April 19, 2017, 02:53:55 pm
The rest is identical to Jamon's description
Post by: asd987 on April 20, 2017, 09:24:26 pm
the rate of flow of water into a dam is given by R=500 + 20tLh^-1. If there is 15000L of water initially in the dam, how much water will there be in the dam ater 10 hours
Post by: RuiAce on April 20, 2017, 09:51:46 pm
Hi can I get some help with this question please
the rate of flow of water into a dam is given by R=500 + 20tLh^-1. If there is 15000L of water initially in the dam, how much water will there be in the dam ater 10 hours
Post by: mcheema on April 22, 2017, 08:55:28 pm
Post by: RuiAce on April 22, 2017, 10:30:58 pm
Hi, I'm struggling to do this question (the question is attached)
_________________________
Post by: katnisschung on April 23, 2017, 09:09:04 pm
q and a attached
Post by: RuiAce on April 23, 2017, 09:21:41 pm
i thought using the product rule was correct but apparently not. ???
q and a attached
Post by: jamonwindeyer on April 23, 2017, 09:48:54 pm
i thought using the product rule was correct but apparently not. ???
q and a attached
It would probably be slightly easier to expand that inside bit first, just to show another way ;D
Post by: katnisschung on April 24, 2017, 10:57:04 am
i have another question regarding differentiating logs (we had a sub for this and i didn't understand anything)
i tried differentiating the attached by applying change of base rule
and then did quotient rule (yeah i think i'm overthinking it)
can someone provide an explanation as to how to differentiate logs with a base other than e (thats all i remember being taught)
Post by: RuiAce on April 24, 2017, 11:19:09 am
thanks jamon yeah i did expand it haha i couldn't wrap my head around simplifying it ;D
i have another question regarding differentiating logs (we had a sub for this and i didn't understand anything)
i tried differentiating the attached by applying change of base rule
and then did quotient rule (yeah i think i'm overthinking it)
can someone provide an explanation as to how to differentiate logs with a base other than e (thats all i remember being taught)
Post by: laurenf58 on April 24, 2017, 02:08:09 pm
Post by: jakesilove on April 24, 2017, 02:32:31 pm
How do I find the focal length of the parabola x^2=8y-24? Thanks!
Hey! Recall that the focal length, 'a', is defined as when the equation of the parabola is
x and y can be shifted. In this case,
Clearly,
So a=2
Post by: mcheema on April 24, 2017, 08:41:51 pm
Can someone pls help me with q26
Post by: anotherworld2b on April 24, 2017, 10:45:39 pm
Post by: kiiaaa on April 25, 2017, 03:09:39 pm
i have a question and im soo confused how to do it especially when i look at how my teacher did im like wot?!?. could you please explain me the process in how to do and approach the question
thanks
QUESTION: find the equation to the tangent to the curve y'= 1+2e2x
MY TEACHERS SOLUTION: in the attactment below (i hope you could understand that dodgy working)
thanks once again
Post by: RuiAce on April 25, 2017, 03:23:48 pm
hi allThere is no attachment.
i have a question and im soo confused how to do it especially when i look at how my teacher did im like wot?!?. could you please explain me the process in how to do and approach the question
thanks
QUESTION: find the equation to the tangent to the curve y'= 1+2e2x
MY TEACHERS SOLUTION: in the attactment below (i hope you could understand that dodgy working)
thanks once again
Also was it actually given y'=1+2e2x and not y=1+2e2x?
Post by: jamonwindeyer on April 25, 2017, 03:30:41 pm
Can I please have some help with these 2 questions please? I got stuck on 20 d) and the questions onwards
Sure!
For 20E, you know the total is six. With 16 possible outcomes (some identical), the ones that match this are 24, 42, 33 and 33. Only one of these has 4 as the second spin. Meaning the answer is just the probability of spinning a 2 and then a 4 (use whatever value of \(k\) you obtained).
For 20F, product rule of probability - Multiply the probability of a 4, with the probability of 3, with the probability of a 2.
For 20G, consider the possible ways we can arrange the outcomes from 20F - There are \(3!=6\) arrangements. So we multiply the answer from 20F by 6!
20H, approach it in the same way as 20E.
20I, calculate the probabilites of three 1's, three 2's, three 3's and three 4's and add it together :)
For 21, do it piece by piece. First, what are the odds of choosing no faulty components? Well it means we pick four healthy components in a row:
For 1 faulty, we do a similar thing, multiplying by the number of ways we could possibly arrange the choices:
Do similar maths to find \(P_2,P_3\) and \(P_4\) - They will form your probability distribution (remember 5dp as specified!) :)
Post by: jamonwindeyer on April 25, 2017, 03:57:41 pm
(http://uploads.tapatalk-cdn.com/20170424/58b308d121764a5b811b2656e846aac8.jpg)
Can someone pls help me with q26
Hey! I've attached my working below - Pretty sure there is a more efficient way to find \(BF\), but this method only uses pythagoras and basic ideas of similarity ;D
Working
(http://i.imgur.com/SW4RMYz.png)
(http://i.imgur.com/SYjxtzK.png)
(http://i.imgur.com/SYjxtzK.png)
Post by: anotherworld2b on April 25, 2017, 06:55:03 pm
but the answer says it is 0.48
for 20H would it be
442, 424, 244, 334, 343, 433, = 6 possible outcomes
i'm not quite sure what to do from here.
I tried q21 but when i multiplied the values for P1 i got 1.848 but the answers says 0.30808
Sure!
For 20E, you know the total is six. With 16 possible outcomes (some identical), the ones that match this are 24, 42, 33 and 33. Only one of these has 4 as the second spin. Meaning the answer is just the probability of spinning a 2 and then a 4 (use whatever value of \(k\) you obtained).
For 20F, product rule of probability - Multiply the probability of a 4, with the probability of 3, with the probability of a 2.
For 20G, consider the possible ways we can arrange the outcomes from 20F - There are \(3!=6\) arrangements. So we multiply the answer from 20F by 6!
20H, approach it in the same way as 20E.
20I, calculate the probabilites of three 1's, three 2's, three 3's and three 4's and add it together :)
For 21, do it piece by piece. First, what are the odds of choosing no faulty components? Well it means we pick four healthy components in a row:
For 1 faulty, we do a similar thing, multiplying by the number of ways we could possibly arrange the choices:
Do similar maths to find \(P_2,P_3\) and \(P_4\) - They will form your probability distribution (remember 5dp as specified!) :)
Post by: jamonwindeyer on April 25, 2017, 07:51:53 pm
for 20E would it be 0.4 x 0.3 = 0.12?
but the answer says it is 0.48
for 20H would it be
442, 424, 244, 334, 343, 433, = 6 possible outcomes
i'm not quite sure what to do from here.
I tried q21 but when i multiplied the values for P1 i got 1.848 but the answers says 0.30808
Not quite, you need to adjust for the sample space because you KNOW the sum is 4, so you are actually only taking one quarter of your sample space (4 outcomes out of 16), so multiply your answer by 4!
For 20H, calculate the probability of each and add all of them together.
Woops, for P1 it is just 4, not 4 factorial (You can pick the faulty unit first, second, third or fourth - Four possible arrangements) that should hopefully fix it? Don't have my calculator on me so can't check
Post by: anotherworld2b on April 25, 2017, 08:24:37 pm
I was hoping to get some help with these questions please
Post by: mcheema on April 25, 2017, 09:06:42 pm
Hey! I've attached my working below - Pretty sure there is a more efficient way to find \(BF\), but this method only uses pythagoras and basic ideas of similarity ;DThanks really appreciate your help. I realised what my mistake wasWorking(http://i.imgur.com/SW4RMYz.png)
(http://i.imgur.com/SYjxtzK.png)
Post by: RuiAce on April 25, 2017, 09:32:24 pm
Thank you for your help :)Warning to everyone else: This is NOT HSC content.
I was hoping to get some help with these questions please
__________________
Post by: laurenf58 on April 26, 2017, 05:31:26 pm
Post by: RuiAce on April 26, 2017, 05:33:51 pm
Can I have some help with this question please?
Post by: anotherworld2b on April 26, 2017, 07:59:14 pm
Warning to everyone else: This is NOT HSC content.
__________________
Post by: JuliaPascale123 on April 26, 2017, 08:39:40 pm
Hey Julia! Welcome to the forums!
...
could you please label this as in which questions you are answering (a,b,c,d,e..)! I am getting so confused :D
Post by: jamonwindeyer on April 26, 2017, 08:55:35 pm
could you please label this as in which questions you are answering (a,b,c,d,e..)! I am getting so confused :D
I'll copy paste the response with headings :)
Hey Julia! Welcome to the forums!
Question A
The first part is pretty straight forward, and I'm not sure how to construct a table on the forum. Basically, the quantity will decrease over time by 25% every 40 minutes. If t is minutes, and Q is in mg,
etc.
Question B
Now, we want to find an equation for Q. We know that what we're looking for here is some sort of exponential function. Once you've done these questions a number of times, you'll know that the form of the answer must be
Now, we know that the initial quantity is 300mg. So
We also know that the quantity has dropped to 225mg after 40 minutes. So,
So, the equation for the quantity of the medicine is
Question D (C Skipped)
Next, we want to find the time it takes for the quantity of medicine to half. I won't bother guessing and checking, let's just find the actual answer.
Remember that this is in minutes. So, dividing by 60, we get t=9.04 hours.
Question E
Part e starts to get tricky. We could use an arbitrary time t=a. However, it's exactly the same thing to start at an arbitrary quantity, A. That is because, at time t=a, there will be some quantity A of medicine in the system. Then, we would look for when A halves!
is our new formula. t is still arbitrary, but so is A. Now, we need to show that, regardless of A, the halving time of this equation is a constant (ie. the time we proved in part d).
Our halving time will occur when the quantity is 0.5A (half of the initial, arbitrary amount). So,
But this is exactly the same equation as above, with the same solution! So, the halving time is a constant for any arbitrary starting point.
Question F
The next part is just a standard sketch of an exponential function. If you're not sure what it will look like, plot some points, or use an online graphing tool.
Question G
For the last part, we solve for
Which equals 50.99 hours. We should definitely round up, for the safety of the patient, giving us 51 hours.
Post by: JuliaPascale123 on April 27, 2017, 11:06:30 am
I'll copy paste the response with headings :)
Hey Julia! Welcome to the forums!
Question A
The first part is pretty straight forward, and I'm not sure how to construct a table on the forum. Basically, the quantity will decrease over time by 25% every 40 minutes. If t is minutes, and Q is in mg,
etc.
Question B
Now, we want to find an equation for Q. We know that what we're looking for here is some sort of exponential function. Once you've done these questions a number of times, you'll know that the form of the answer must be
Now, we know that the initial quantity is 300mg. So
We also know that the quantity has dropped to 225mg after 40 minutes. So,
So, the equation for the quantity of the medicine is
Question D (C Skipped)
Next, we want to find the time it takes for the quantity of medicine to half. I won't bother guessing and checking, let's just find the actual answer.
Remember that this is in minutes. So, dividing by 60, we get t=9.04 hours.
Question E
Part e starts to get tricky. We could use an arbitrary time t=a. However, it's exactly the same thing to start at an arbitrary quantity, A. That is because, at time t=a, there will be some quantity A of medicine in the system. Then, we would look for when A halves!
is our new formula. t is still arbitrary, but so is A. Now, we need to show that, regardless of A, the halving time of this equation is a constant (ie. the time we proved in part d).
Our halving time will occur when the quantity is 0.5A (half of the initial, arbitrary amount). So,
But this is exactly the same equation as above, with the same solution! So, the halving time is a constant for any arbitrary starting point.
Question F
The next part is just a standard sketch of an exponential function. If you're not sure what it will look like, plot some points, or use an online graphing tool.
Question G
For the last part, we solve for
Which equals 50.99 hours. We should definitely round up, for the safety of the patient, giving us 51 hours.
Thanks Jamon you f*&king Legend! You guys are awesome and amazing.
BTW for c) the answer would be -0.00128 right?
Post by: JuliaPascale123 on April 27, 2017, 11:26:39 am
Post by: RuiAce on April 27, 2017, 12:22:21 pm
(http://i63.tinypic.com/rmrbdg.png)
Post by: michaelalt on April 27, 2017, 01:40:52 pm
Post by: jamonwindeyer on April 27, 2017, 01:47:47 pm
Can someone please explain to me how to do q. 13? I don't quite understand. Thank you!
Hey Michael! These trig equations are applications of the angles of any magnitude system, you should have done this back when you first did trig equations in degrees (unless you've started in radians!) ;D
Let me show you the process with the first one:
So we first think, okay, what puts \(\cos{x}=\frac{1}{2}\)? Well if you check out your reference sheet and do some right angled trig, the answer to that is \(x=60\text{ degrees}=\frac{\pi}{3}\)
Now we need answers for \(0\le x\le2\pi\) - We need the All Stations to Central (ASTC) stuff. When is cosine positive? In the first quadrant, and the fourth quadrant. So the answers are:
Does that help at all? This is something that requires a lot of pre-knowledge, not sure what you know and what you don't ;D
Post by: michaelalt on April 27, 2017, 01:54:51 pm
Hey Michael! These trig equations are applications of the angles of any magnitude system, you should have done this back when you first did trig equations in degrees (unless you've started in radians!) ;D
Let me show you the process with the first one:
So we first think, okay, what puts \(\cos{x}=\frac{1}{2}\)? Well if you check out your reference sheet and do some right angled trig, the answer to that is \(x=60\text{ degrees}=\frac{\pi}{3}\)
Now we need answers for \(0\le x\le2\pi\) - We need the All Stations to Central (ASTC) stuff. When is cosine positive? In the first quadrant, and the fourth quadrant. So the answers are:
Does that help at all? This is something that requires a lot of pre-knowledge, not sure what you know and what you don't ;D
Hi Jamon! Yes that helps. We have done the trig stuff, and we're currently doing the radians unit. I got up to the step where you found pie/3, however, didn't know how to find the other 5pie/3 value. Also, i got confused on b. Are you able to do it for me? Thank you so much for your swift response!
Post by: jamonwindeyer on April 27, 2017, 01:59:14 pm
Hi Jamon! Yes that helps. We have done the trig stuff, and we're currently doing the radians unit. I got up to the step where you found pie/3, however, didn't know how to find the other 5pie/3 value. Also, i got confused on b. Are you able to do it for me? Thank you so much for your swift response!
Cool! No dramas ;D
So again, we consider the acute angle that would give us a value of \(\frac{1}{\sqrt{2}}\) (always ignore the negative). This is 45 degrees, or \(x=\frac{\pi}{4}\). This forms the 'base angle' that we use in our angles of any magnitude calculations.
So now, we think - Where is sine negative? By the ASTC diagram, we can see it is in the third and fourth quadrants! So the answers are:
Post by: michaelalt on April 27, 2017, 06:39:17 pm
Post by: jamonwindeyer on April 27, 2017, 07:58:05 pm
Thanks Jamon you f*&king Legend! You guys are awesome and amazing.
BTW for c) the answer would be -0.00128 right?
No worries! And not quite, so what the question wants is for you to be substituting values of \(t\) to guess how long it takes for the quantity to half. What this means is, estimate when:
So just try values of \(t\) and find the one that puts us closest to being true. It should be close to Jake's actual answer of 542 (which is a rounded answer, so in fact, your guess and check answer could also be 542 :) )
Post by: itssona on April 27, 2017, 09:20:17 pm
x^2 + kx + 25 is a perfect square
what are the values for k
so i tried using the completing the square formula and I did x^2 + kx + (k/2)^2 = -25 + (k/2)^2 but im kinda stuck now :/
Post by: jamonwindeyer on April 27, 2017, 09:24:10 pm
heeey please help
x^2 + kx + 25 is a perfect square
what are the values for k
so i tried using the completing the square formula and I did x^2 + kx + (k/2)^2 = -25 + (k/2)^2 but im kinda stuck now :/
Hey! So if the quadratic is a perfect square, that is another way of saying there is only one root ;D we use the discriminant, for one root we need \(\Delta=0\):
Does that help? ;D
Post by: Shadowxo on April 27, 2017, 09:29:57 pm
heeey please help
x^2 + kx + 25 is a perfect square
what are the values for k
so i tried using the completing the square formula and I did x^2 + kx + (k/2)^2 = -25 + (k/2)^2 but im kinda stuck now :/
Perfect square means it can be written as (a+b)2 or (a-b)2. in this case a=x and b=5
So 2ab=2*x*5=10x is equal to kx
k=10
or -2ab=-2*x*5=-10x is equal to kx
k=-10
Alternatively, if you had to show / prove it you might say
(k/2)2 = 25 and end up with the same result
Post by: itssona on April 27, 2017, 10:08:20 pm
Hey! So if the quadratic is a perfect square, that is another way of saying there is only one root ;D we use the discriminant, for one root we need \(\Delta=0\):
Does that help? ;D
oh!! I like this method, thank you Jamon!! It makes sense now :D
Perfect square means it can be written as (a+b)2 or (a-b)2. in this case a=x and b=5
So 2ab=2*x*5=10x is equal to kx
k=10
or -2ab=-2*x*5=-10x is equal to kx
k=-10
Alternatively, if you had to show / prove it you might say
(k/2)2 = 25 and end up with the same result
Thank you for showing this method too!! The concept is clearer now thanks to you and Jamon :) Thank you!
Mod Edit: Post merge :)
Post by: itssona on April 27, 2017, 10:31:35 pm
my sheet said the answer is 2??? in the solutions
isnt it 4
Post by: RuiAce on April 27, 2017, 10:35:44 pm
hey quick easy one but,
my sheet said the answer is 2??? in the solutions
isnt it 4
I disagree with both sources
Post by: jamonwindeyer on April 27, 2017, 10:37:07 pm
I disagree with both sources
It's the average of the two...
(https://res.cloudinary.com/teepublic/image/private/s--jPeOodc6--/t_Preview/b_rgb:ffffff,c_limit,f_jpg,h_630,q_90,w_630/v1463160661/production/designs/511217_1.jpg)
Post by: itssona on April 27, 2017, 11:04:12 pm
I disagree with both sources
oh!!! okay thanks :D i get it hmm
It's the average of the two...
(https://res.cloudinary.com/teepublic/image/private/s--jPeOodc6--/t_Preview/b_rgb:ffffff,c_limit,f_jpg,h_630,q_90,w_630/v1463160661/production/designs/511217_1.jpg)
AHAHAA
Post by: michaelalt on April 28, 2017, 09:56:33 am
Thanks!
Post by: jamonwindeyer on April 28, 2017, 11:09:36 am
Hi can someone help me with these? Q, 4 + 5, I've attached my diagrams + working. The correct answers are 4. 1343622km 5. 7367m
Thanks!
Hey Michael! Your only mistake is taking the given angles in degrees, they are in minutes! That little dash is the sign for minutes, so your angle for question four is \(0^\circ15'30''\)! The second one is \(0^\circ3'30''\) - You can also use decimals in the minutes place ;D
Post by: mcheema on April 28, 2017, 07:17:00 pm
Post by: mcheema on April 28, 2017, 07:20:52 pm
Post by: RuiAce on April 28, 2017, 07:56:57 pm
Also can I get help with this one, my answer is close to the one in the answers but I cant seem to figure out where I made the mistake.
Hopefully I didn't mess up whilst I had alcohol in my system
Post by: RuiAce on April 28, 2017, 08:00:53 pm
Can someone please help me with this question
Post by: mcheema on April 28, 2017, 09:03:13 pm
Hopefully I didn't mess up whilst I had alcohol in my system
Thanks for your help . I realised my mistake, i mixed the signs up
Post by: asd987 on April 29, 2017, 12:26:14 am
Post by: jamonwindeyer on April 29, 2017, 12:39:15 am
Hi, why do the rates have to be in absolute values?
Hey! If you think about it, how could a positive rate of change ever be directly equal to a negative rate of change? They are opposite in sign by definition, meaning they can't be equal! Positive numbers can't equal negatives - If you try doing it without absolute values, you'll have no answer :) we just want when the magnitude of the rates of change is the same! :)
Post by: bananna on April 29, 2017, 08:30:22 am
can someone please list the formulas you need to know in series and sequences?
I was away for the first lesson and don't understand the layout of my textbook
thank you!
Post by: RuiAce on April 29, 2017, 11:16:15 am
Hi!- General term of an AP
can someone please list the formulas you need to know in series and sequences?
I was away for the first lesson and don't understand the layout of my textbook
thank you!
- Partial sum of an AP
- General term of a GP
- Partial sum of a GP
- Limiting sum of a GP
Formulas-wise that's it. Processes-wise there's more
Post by: jamonwindeyer on April 29, 2017, 12:15:06 pm
Hi!
can someone please list the formulas you need to know in series and sequences?
I was away for the first lesson and don't understand the layout of my textbook
thank you!
And all of those are on your reference sheet! Bottom right of the first page ;D
Post by: anotherworld2b on April 29, 2017, 03:53:28 pm
Post by: jamonwindeyer on April 29, 2017, 05:22:18 pm
CanI have some help with this question please? I'm not sure where I went wrong
You've miscounted the double ups (the seven total is correct) ;D the total ways you can double up is two 1's, two 2's, two 3's, two 4's, two 5's or two 6's - That's six of the thirty six possible outcomes! So:
Post by: anotherworld2b on April 30, 2017, 12:07:14 am
I was also working on this question but I'm not sure what I did wrong again :-\
You've miscounted the double ups (the seven total is correct) ;D the total ways you can double up is two 1's, two 2's, two 3's, two 4's, two 5's or two 6's - That's six of the thirty six possible outcomes! So:
Post by: RuiAce on April 30, 2017, 12:09:51 am
Thank you for help :)Not too sure why you put the probability in the denominator as well
I was also working on this question but I'm not sure what I did wrong again :-\
Post by: anotherworld2b on April 30, 2017, 01:07:48 am
Thank you for your help
Not too sure why you put the probability in the denominator as well
Post by: anotherworld2b on April 30, 2017, 01:00:26 pm
Post by: jamonwindeyer on April 30, 2017, 03:20:12 pm
Im not sure how to do this question. Can i have help please?
I don't think you've considered the totals in that table properly!! It should look like this:
(http://i.imgur.com/LWj5wxf.png)
So Part (a), for example, would be:
Part (b), you'd look at the total winning amounts progressively, and go, "Okay, do 40% of my possible outcomes sit above this value?" 40% of 25 outcomes is 10, so find the value with 10 outcomes above it! I don't think you'll get exactly 40, but get as close as possible :)
Edit: Woops, there is a value where \(P(x>??)=0.4\), find it! ;D
Part (c), find the average win (it will be less than $5), that gives you the average profit per game (\(5-\mu\)) and you can go from there ;D
Post by: Thebarman on April 30, 2017, 09:21:17 pm
1) Find the area bounded by the curve x=y^2-2y-3 and the y-axis
2) Find the area bounded by the curve x=-y^2-5y-6 and the y-axis
Post by: RuiAce on April 30, 2017, 09:28:35 pm
Hey guys, can someone help me with the following questions please? I keep getting 0 as the answer, and i'm not sure why.
1) Find the area bounded by the curve x=y^2-2y-3 and the y-axis
2) Find the area bounded by the curve x=-y^2-5y-6 and the y-axis
Post by: Thebarman on April 30, 2017, 09:39:25 pm
After fixing that mistake, I answered both correctly. Thank you very much! ;D
Post by: BarnesK01 on May 01, 2017, 01:35:58 pm
Evaluate giving exact answers, for Sin(pi X) dx with the boundaries, x=0 and x=1/2
Post by: jamonwindeyer on May 01, 2017, 01:41:49 pm
Need help integrating a trig function!!!!
Evaluate giving exact answers, for Sin(pi X) dx with the boundaries, x=0 and x=1/2
Welcome to the forums Barnes!! Sure thing, here is the working, let me know if it doesn't make sense ;D
Post by: bananna on May 02, 2017, 04:36:57 pm
thank you!!!
Post by: jakesilove on May 02, 2017, 04:38:22 pm
hi can someone pls help with q 7.4, q 1, (f) and (j)?
thank you!!!
Not seeing an attachment!
Post by: bananna on May 02, 2017, 04:43:28 pm
i just realised i forgot to attach the screenshot
7.4 (f) and (j)
thank you! :D
Post by: RuiAce on May 02, 2017, 04:47:59 pm
haha so sorry!
i just realised i forgot to attach the screenshot
7.4 (f) and (j)
thank you! :D
Post by: bananna on May 02, 2017, 05:17:19 pm
thank you!!!
Post by: jakesilove on May 02, 2017, 05:17:55 pm
thank you!
sorry, i mostly didnt understand how to do (f)
Same methodology as Rui outlined above.
Post by: kylesara on May 02, 2017, 06:33:22 pm
A chord 8mm long is formed by an angle of 45* subtended at the centre of the circle. Find
a) The radius of the circle
b) the areas of the minor segment to 1 dp
Thanks :)
Post by: Sukakadonkadonk on May 02, 2017, 08:14:29 pm
What is the purpose of the differential of an exponential growth and decay equation? Can you use it to find something?
Thanks.
Post by: RuiAce on May 02, 2017, 10:19:05 pm
Hi,In history, the differential equation is where everything all began. The exponential form P=Ae^(kt) is the solution to the equation.
What is the purpose of the differential of an exponential growth and decay equation? Can you use it to find something?
Thanks.
Post by: RuiAce on May 02, 2017, 10:21:55 pm
Hi, could i please have help with this question.The area of the minor segment is just a formula you need to substitute in to.
A chord 8mm long is formed by an angle of 45* subtended at the centre of the circle. Find
a) The radius of the circle
b) the areas of the minor segment to 1 dp
Thanks :)
Post by: jamonwindeyer on May 02, 2017, 10:47:02 pm
Hi,
What is the purpose of the differential of an exponential growth and decay equation? Can you use it to find something?
Thanks.
In history, the differential equation is where everything all began. The exponential form P=Ae^(kt) is the solution to the equation.
And speaking practically, you use it to find the rate of change of the quantity at any instant! So say the differential form of an equation modelling population growth is \(\frac{dP}{dt}=0.1P\) - This means that the population is increasing at a rate equal to 10% of the total at any given time. Sometimes we don't just care how much there is, we care about how quickly it is going up/down ;D
Post by: gilliesb18 on May 03, 2017, 04:20:52 pm
And especially this one??
Find the range of each function over the given domain:
a) y= x^2 (x squared) for 0 < x < 3 (also i dont know how to make greater than or equal to signs, but in this question they are all greater than or equal to, if that makes sense)
Sorry about the muddly question too.....
Thanks!!!
B
Post by: RuiAce on May 03, 2017, 04:24:03 pm
Ahhhh can someone please help me understand domains and ranges??
And especially this one??
Find the range of each function over the given domain:
a) y= x^2 (x squared) for 0 < x < 3 (also i dont know how to make greater than or equal to signs, but in this question they are all greater than or equal to, if that makes sense)
Sorry about the muddly question too.....
Thanks!!!
B
Until you are experienced, you should never do such questions without a graph.
Post by: gilliesb18 on May 03, 2017, 04:33:08 pm
But I know what you mean about drawing a graph, especially as they get harder.
Thanks vv much.
Post by: gilliesb18 on May 03, 2017, 04:46:20 pm
I have even drawn it up, i believe my parabola is correct, but where do i go from there??
So same question as before, Find the range of each function over the given domain, but this time, the question is;
y= -x^2 + 4 for -1 < x < 2
Post by: RuiAce on May 03, 2017, 04:50:25 pm
Hey I'm getting stuck again (not a suprise :-[ ) but can someone help me on this one now??
I have even drawn it up, i believe my parabola is correct, but where do i go from there??
So same question as before, Find the range of each function over the given domain, but this time, the question is;
y= -x^2 + 4 for -1 < x < 2
Post by: JuliaPascale123 on May 03, 2017, 06:04:34 pm
Hey!! So for the first one, stationary points occur when \(x=-2\), \(f(x)=3\) (from the y-value), and \(f'(x)=0\) because it is a stationary point! Make those substitutions: into the function and the first derivative, to get a couple of equations:
So the function we now know is \(f(x)=x^3-12x-13\). Now it is just standard calculus. To find the turning points, put the 1st derivative equal to zero:
To find the inflexion, put the second derivative equal to zero:
To find the y-coordinates of any of these points, just pop the x-value back into the original function! ;D
D is the hard bit - If you draw the curve you'll get something like this. What the question is asking is, where can you draw a horizontal line and cross that curve three times? That would mean you have three solutions to the equation \(f(x)=k\). For example, \(k=0\) works, because a horizontal line through \(y=0\) would cut the curve three times. The question is, what range of \(k\) values allow this to occur? At what point does it go too high or too low to cut the curve three times?
Hint: It is related directly to the coordinates of your turning points!If that hint wasn't enough (pardon the edit Jamon!)Specifically the y-coordinates
Hopefully this makes sense - Let me know if you need anything clarified ;D
Hey,
I tried answering it but I have no idea what im doing, Would you mind telling me the answer please
Post by: jamonwindeyer on May 03, 2017, 06:46:09 pm
Hey,
I tried answering it but I have no idea what im doing, Would you mind telling me the answer please
No worries at all! So the line \(y=k\) will cut the curve three times as long as it sits above your minimum turning point, and below your maximum turning point. Outside this range is only cuts either once or twice :) the answer is
I'd recommend you try drawing some horizontal lines for \(y=-29\), \(y=3\), then values in between those and then not in between those! So you can spot the difference :)
Post by: Fahim486 on May 03, 2017, 07:22:50 pm
Thanks!
Post by: RuiAce on May 03, 2017, 07:28:20 pm
Hi I'm having trouble with this question and sort of forgot how to do geometrical applications of calculus.
Thanks!
Post by: Fahim486 on May 03, 2017, 07:45:27 pm
Ahh I see that makes so much sense now. Thank you so much !!!!!!!
Post by: bananna on May 04, 2017, 06:12:56 am
Use the formula Tn = ar^n−1 to find an expression for the 70th term of the GP with a=1andr=3
Also, I get really confused with the wording in these types of questions, like, do they want us to find Tn or n.
thanks!
Post by: RuiAce on May 04, 2017, 07:42:47 am
Hi, I'm having trouble with this q:The 70th term is just \(T_{70}\)
Use the formula Tn = ar^n−1 to find an expression for the 70th term of the GP with a=1andr=3
Also, I get really confused with the wording in these types of questions, like, do they want us to find Tn or n.
thanks!
Literally just substitute n=70, a=1 and r=30 straight into the formula to get \(3^{69}\) as your answer.
Look closely at the question. They clearly give you n.
Post by: JuliaPascale123 on May 04, 2017, 03:14:51 pm
I'll copy paste the response with headings :)
Hey Julia! Welcome to the forums!
Question A
The first part is pretty straight forward, and I'm not sure how to construct a table on the forum. Basically, the quantity will decrease over time by 25% every 40 minutes. If t is minutes, and Q is in mg,
etc.
Question B
Now, we want to find an equation for Q. We know that what we're looking for here is some sort of exponential function. Once you've done these questions a number of times, you'll know that the form of the answer must be
Now, we know that the initial quantity is 300mg. So
We also know that the quantity has dropped to 225mg after 40 minutes. So,
So, the equation for the quantity of the medicine is
Question D (C Skipped)
Next, we want to find the time it takes for the quantity of medicine to half. I won't bother guessing and checking, let's just find the actual answer.
Remember that this is in minutes. So, dividing by 60, we get t=9.04 hours.
Question E
Part e starts to get tricky. We could use an arbitrary time t=a. However, it's exactly the same thing to start at an arbitrary quantity, A. That is because, at time t=a, there will be some quantity A of medicine in the system. Then, we would look for when A halves!
is our new formula. t is still arbitrary, but so is A. Now, we need to show that, regardless of A, the halving time of this equation is a constant (ie. the time we proved in part d).
Our halving time will occur when the quantity is 0.5A (half of the initial, arbitrary amount). So,
But this is exactly the same equation as above, with the same solution! So, the halving time is a constant for any arbitrary starting point.
Question F
The next part is just a standard sketch of an exponential function. If you're not sure what it will look like, plot some points, or use an online graphing tool.
Question G
For the last part, we solve for
Which equals 50.99 hours. We should definitely round up, for the safety of the patient, giving us 51 hours.
Graphing weebsites dont show the answer for g
Post by: jakesilove on May 04, 2017, 03:53:00 pm
Graphing weebsites dont show the answer for g
We can do a quick sanity check to see if G was correct;
Let's set A=1
Now, we sub in 3059.65 (our answer)
Which is what we would expect! Remember, we were looking for 0.02*A which, in this case is 0.02*1=0.02.
Post by: itssona on May 05, 2017, 07:43:47 am
log base 2 of x is less than 3
solve
so i know log is defined for x>0 but idk how to do this, pls reply within an hour if u can before my exam aha, thank you :D
Post by: RuiAce on May 05, 2017, 07:55:45 am
heeey
log base 2 of x is less than 3
solve
so i know log is defined for x>0 but idk how to do this, pls reply within an hour if u can before my exam aha, thank you :D
Post by: itssona on May 05, 2017, 08:04:08 am
Omg THANK U RUI
Post by: gilliesb18 on May 05, 2017, 02:59:00 pm
so for ex. i am asked the domain in which the function is (i) increasing and (ii) decreasing, for these questions;
(a) y= |x-2|
(b) f(x)= |x| + 2
(c) f(x)= |2x- 3|
(d) y= 4 |x| -1
(e) f(x)= - |x|
for a) i know that (i) it is x > 2, and for (ii) x < 2
but im really not sure if this is correct and really how to get there with the next few ones.
Can someone help me?
Post by: jakesilove on May 05, 2017, 03:15:55 pm
what does it mean by a function increasing or decreasing over a domain??
so for ex. i am asked the domain in which the function is (i) increasing and (ii) decreasing, for these questions;
(a) y= |x-2|
(b) f(x)= |x| + 2
(c) f(x)= |2x- 3|
(d) y= 4 |x| -1
(e) f(x)= - |x|
for a) i know that (i) it is x > 2, and for (ii) x < 2
but im really not sure if this is correct and really how to get there with the next few ones.
Can someone help me?
An increasing function is a function with a positive gradient. A decreasing function is a function with a negative gradient.
So, let's look at
This looks like two straight lines, hitting the x-axis at x=2. The line moving off to the right has a positive gradient, and the line moving off to the left has a negative gradient. Thus, the function is increasing for x>2, and decreasing for x<2.
For all of these functions, I would personally just try to sketch them (on paper or in your head). Then, look at where the gradient of the line is positive. That is the domain (ie. x values) for which the function is increasing!
Post by: Dragomistress on May 06, 2017, 02:58:05 pm
How do I graph the attachment?
For an equation of solve graphically, do you have to label the x and y-intercepts?
How do I graph a hyperbola?
Post by: RuiAce on May 06, 2017, 03:03:57 pm
I would like some help with my functions.
How do I graph the attachment?
For an equation of solve graphically, do you have to label the x and y-intercepts?
How do I graph a hyperbola?
http://www.desmos.com
Please provide an example of a hyperbola you want to sketch.
Post by: Dragomistress on May 06, 2017, 03:24:49 pm
y=3/(x-1) -2
Post by: RuiAce on May 06, 2017, 03:29:15 pm
Let's say the sketch of a hyperbola is
y=3/(x-1) -2
Post by: Fahim486 on May 07, 2017, 06:29:51 pm
Thanks!!
Post by: RuiAce on May 07, 2017, 06:35:55 pm
Hey so I'm having trouble with question 11. b) so could someone please explain to me how to do this
Thanks!!
Thought: At x=e we may have a horizontal point of inflexion. But I have not verified this.
Post by: cxmplete on May 08, 2017, 06:01:12 pm
Post by: RuiAce on May 08, 2017, 06:25:26 pm
how to do both these questions (sorry it's slanted)Hint for second: \((1+2+3+\dots+100) - (6+12+\dots+96)\).
Post by: BarnesK01 on May 08, 2017, 07:16:53 pm
Post by: kiwiberry on May 08, 2017, 09:17:36 pm
I need help with the attached question.. Just have absolutely no clue where to start ???
Hey! The flow rate of water is given by \(\frac{dV}{dt}\), so to find the time when the flow rate is 0, set \(\frac{dV}{dt}=0\)
To find volume, we need to integrate \(\frac{dV}{dt}\) and sub in the initial conditions given to find c
Post by: Kle123 on May 08, 2017, 09:31:51 pm
Post by: gilliesb18 on May 11, 2017, 05:52:00 am
Post by: 1937jk on May 11, 2017, 06:36:11 am
Find the area bounded by the curve y = cos2x, the x axis and the lines x = 0 and x = pi
Post by: RuiAce on May 11, 2017, 06:42:55 am
Can someone help me understand the Hyperbola? How to sketch it and how to work out the domain and range?? I'm confused!! Thanks...Please provide an example.
(Alternatively there was an example done a few days ago but that may require going back a few pages, so you may provide your own example)
okay so i'm doing something very wrong here, I keep getting zero and can't figure out why,
Find the area bounded by the curve y = cos2x, the x axis and the lines x = 0 and x = pi
Post by: RuiAce on May 11, 2017, 07:21:47 am
Hi! Could i get help with this question. Not sure whether me or my peer is right :/
Post by: michaelalt on May 11, 2017, 08:56:42 pm
Post by: jamonwindeyer on May 11, 2017, 09:09:36 pm
Hi can someone else help with q.3?
Sure! So we first differentiate using the normal rule for tan:
Now we want to find the gradient at \(x=\frac{\pi}{9}\), we get that by substituting into our derivative:
Now we find the y-coordinate of our point by substitution into the original function - It ends up being \(y=\sqrt{3}\). So we put all that into the point gradient formula:
And that should be the answer! Does that make sense? :)
Post by: 1937jk on May 12, 2017, 08:20:39 am
[/quote]
ohhh Thank you so much!!!!
Post by: RuiAce on May 13, 2017, 06:51:21 pm
I need some help with this question.In the future please post your working so we can look for any potential errors.
It says to find the volume of the solid around the y-axis.
The answer I got was 64pi/15, but the textbook answer was 16pi/15.
I double checked my working out and can't find any errors.
Post by: asd987 on May 13, 2017, 11:50:21 pm
Post by: jamonwindeyer on May 13, 2017, 11:53:33 pm
Hi, can i get some help with ii)
Hey! So all you are doing here is finding when \(v=0\), using the expression from (i)!
Notice that I've taken only the smallest positive answer to the trig equation, because want the first time it has come to rest. Does this make sense? ;D
Post by: asd987 on May 14, 2017, 12:25:06 am
I don't understand how -1/2 becomes 7pi/12 when you move sin on the other side?
Post by: RuiAce on May 14, 2017, 12:29:14 am
so this is supposed to be the answer
I don't understand how -1/2 becomes 7pi/12 when you move sin on the other side?
Post by: jamonwindeyer on May 14, 2017, 12:30:31 am
so this is supposed to be the answer
I don't understand how -1/2 becomes 7pi/12 when you move sin on the other side?
so this is supposed to be the answer
I don't understand how -1/2 becomes 7pi/12 when you move sin on the other side?
Oh yep that's totally the answer, I made an error above (I'll go back and fix it up). So we are here:
What we do then is think, okay, what must \(2t\) be to give a result of \(\frac{-1}{2}\). We turn to angles of any magnitude, all stations to central. We know that sine is negative in the third and fourth quadrants (we only care about the third). We also know that \(\sin{\frac{\pi}{6}}=\frac{1}{2}\). Put that together:
This is something you first did with degrees in Year 11 :)
Post by: katnisschung on May 14, 2017, 12:39:45 pm
(e.g. cos2x + sin2x) i just do a table of values and put it into my calculator (i don't show working)
cos otherwise the table would be like 3 rows
thanks :)
also if a hsc q asks u to sketch a graph what details MUST u show (e.g. x-intercepts for trig?)
Post by: jamonwindeyer on May 14, 2017, 05:25:51 pm
quickest way to sketch superimposition of 2 graphs??
(e.g. cos2x + sin2x) i just do a table of values and put it into my calculator (i don't show working)
cos otherwise the table would be like 3 rows
thanks :)
Hey! Fastest way to sketch a superposition is to do a quick sketch of the two graphs in light pencil, then draw the superposition of the two in a darker pen (possibly erasing the other curves too) - For the example you provided for example, those are two curves that are fairly easy to sketch - It is faster to just quickly bump those out and then add them up visually rather than muck around with your calculator, at least in my opinion :)
Quote
also if a hsc q asks u to sketch a graph what details MUST u show (e.g. x-intercepts for trig?)
You should always show whatever details they ask for (at minimum), and besides that, whatever is necessary such that it is clear what graph you are sketching. For example, if asked to sketch a parabola, just the y-intercept isn't enough. You'd want either a turning point/vertex, or roots. You should also take into account the ease with which the points are found - For example, when sketching a cubic, you usually won't need to find x-intercepts unless they are made obvious to you, because finding roots of cubics is quite tedious for the most part ;D
Post by: Youssk on May 15, 2017, 04:21:38 pm
Show that 1/x-1 - 1/x+1 = 2/x^2-1. Hence or otherwise show that 3 2 ∫1/x^2 -1 dx = 1/2loge1.5
Thanks :D
Post by: RuiAce on May 15, 2017, 04:24:00 pm
Hey can someone please help me with this question, i'm confused as to where the 1/2 outside the log comes from.In the future, please use more bracketing (e.g. 1/(x-1) for \(\frac1{x+1}\)
Show that 1/x-1 - 1/x+1 = 2/x^2-1. Hence or otherwise show that 3 2 ∫1/x^2 -1 dx = 1/2loge1.5
Thanks :D
Post by: Hungry4Apples on May 15, 2017, 04:27:31 pm
Hey can someone please help me with this question, i'm confused as to where the 1/2 outside the log comes from.
Show that 1/x-1 - 1/x+1 = 2/x^2-1. Hence or otherwise show that 3 2 ∫1/x^2 -1 dx = 1/2loge1.5
Thanks :D
This is an integration by recognition question. You may not have the ability to integrate the function 1/(x^2+1) but you have been given enough information to do it by means you are already aware of. We found earlier that
2/(x^2+1) = 1/(x-1) - 1/(x+1)
so it would follow that if we divide both sides by two, we have 1/(x^2+1) in a form we can integrate. So the half outside comes from there, since we can only integrate 2/(x^2+1) we do so, while also dividing by a half to keep the same value of the function.
Post by: RuiAce on May 15, 2017, 04:29:21 pm
This is an integration by recognition question.A remark that this terminology is not used in the HSC.
Post by: kylesara on May 16, 2017, 06:24:23 pm
Find values of a and b if y''= ae^3x cos4x+ be^3x sin 4x, given y=e^3x cos4x
Thanks :)
Post by: jakesilove on May 16, 2017, 06:55:39 pm
Hi could i please have help with this question.
Find values of a and b if y''= ae^3x cos4x+ be^3x sin 4x, given y=e^3x cos4x
Thanks :)
Hey! So, the question tells us that
First, we differentiate our f(x). Note we have to use the chain rule, such that
So, here
Therefore
Then, we differentiate again! We apply the product rule, twice here.
Collecting like terms
From which you can extract your value of a and b :)
Post by: katnisschung on May 17, 2017, 09:22:37 pm
Post by: RuiAce on May 17, 2017, 09:35:25 pm
hi hi is there any other way to sketch an exponential other than table of valuesYou should know the shape of the exponential off by heart.
1. Write down the asymptote and put that in, clearly noting if it's approached as \(x\to \infty\) or \(x \to -\infty\)
2. Sketch any intercepts with the x and/or y axes
3. That's all the information you need. Sketch it.
Nobody says that your graph has to be to a perfect scale
Post by: sophiemacpherso on May 17, 2017, 09:49:25 pm
Post by: _____ on May 18, 2017, 09:50:00 am
Write the relation y=log2x in terms of the exponential function (base 2 to the pronumeral x).
The answer is:
y=lnx/ln2
yln2=lnx
lnx=yln2
Therefore x=e^(yln2)
What's going on in this last line? Think I understand the rest. Appreciate your help.
Post by: kiwiberry on May 18, 2017, 10:21:19 am
This is for a longer integration/volume question from a past paper but there's one part with logarithms that I don't understand.
Write the relation y=log2x in terms of the exponential function (base 2 to the pronumeral x).
The answer is:
y=lnx/ln2
yln2=lnx
lnx=yln2
Therefore x=e^(yln2)
What's going on in this last line? Think I understand the rest. Appreciate your help.
Remember that
In the same way, replacing y with yln2, we get
Post by: _____ on May 18, 2017, 10:54:32 am
Remember that
In the same way, replacing y with yln2, we get
Ahhhh now I understand. Thanks!
Post by: katnisschung on May 20, 2017, 03:35:16 pm
1. the tangent to the curve y=e^x at the point p meets the x-axis at an angle of 45 degrees. Find the coordinates of P
(i remember hearing somewhere the tangent to the point of an exponential curve will be 45 degrees at the y-intercept..is this true
for all exponentials like y=2e^4x)
2. find the maximum value of lnx/x (totally lost here...assuming its something to do with e)
Post by: RuiAce on May 20, 2017, 03:40:14 pm
hey hey i have 2 questions\text{Assume that }P\text{ is the point }(x,y)\\ y=e^x \implies \frac{dy}{dx} = e^x\\ \text{So the tangent at }P\text{ has gradient }e^x[/tex]
1. the tangent to the curve y=e^x at the point p meets the x-axis at an angle of 45 degrees. Find the coordinates of P
(i remember hearing somewhere the tangent to the point of an exponential curve will be 45 degrees at the y-intercept..is this true
for all exponentials like y=2e^4x)
2. find the maximum value of lnx/x (totally lost here...assuming its something to do with e)
__________________________
Post by: Kle123 on May 21, 2017, 03:01:28 pm
Post by: RuiAce on May 21, 2017, 03:14:46 pm
Could i get help with this question. I asked it a while back but didn't get a response :(
Although I should look at it more closely later
Post by: jakesilove on May 21, 2017, 04:26:08 pm
Although I should look at it more closely later
For the record, I started doing a tree diagram, and after about twenty minutes decided I had never wasted time as unbelievably as I was then.
So, I stopped.
(http://i.imgur.com/41TBj6B.png)
As for the real method... not really sure here. That tree diagram has me wanting to quit maths altogether.
Post by: RuiAce on May 21, 2017, 04:44:40 pm
For the record, I started doing a tree diagram, and after about twenty minutes decided I had never wasted time as unbelievably as I was then.Well seeing as though you used technology and not pen and paper to draw it.. welp..
So, I stopped.
(http://i.imgur.com/41TBj6B.png)
As for the real method... not really sure here. That tree diagram has me wanting to quit maths altogether.
Post by: jamonwindeyer on May 21, 2017, 04:48:41 pm
Could i get help with this question. I asked it a while back but didn't get a response :(
Hey! So as Rui said, I think this is a more natural question for combinatorics, but here's a long shot.
We can pick 9 numbers, which means we need to pick the 6 correct ones, and then just 3 others. We always pick 9, so really, it's just the probability of picking the 6 correct ones out of the 44. Which would mean:
This seems appropriate for a lotto - This is what I'd propose at least :)
Edit: Perhaps some sort of adjustment for the fact that, if we miss, we get three extra chances to nab them all. Not sure how I'd do without combinatorics :P
For the record, I started doing a tree diagram, and after about twenty minutes decided I had never wasted time as unbelievably as I was then.
So, I stopped.
As for the real method... not really sure here. That tree diagram has me wanting to quit maths altogether.
Save this to display with your vulgar study notes at next lectures ;)
Post by: katnisschung on May 21, 2017, 05:55:25 pm
so i have a couple of questions...which i will split into separate posts
1. find the area enclosed between the hyperbola y=3/x and the line y=-2x+5
ok so i know how to do this but i had a lot of trouble trying to determine which curve lied on top
becos i integrate to find the area by top curve-bottom curve. I mean I found it by putting the curves into geogebra
but im not gonna have that in the test rip. pls help
2. don't no how to do this...(so far i've only been exposed to sth like with no 'x's in the numerator or ones i could easily cancel out)
thanks! u guys are absolute legends
Post by: RuiAce on May 21, 2017, 06:02:34 pm
pls help last min study for my exam tmr
so i have a couple of questions...which i will split into separate posts
1. find the area enclosed between the hyperbola y=3/x and the line y=-2x+5
ok so i know how to do this but i had a lot of trouble trying to determine which curve lied on top
becos i integrate to find the area by top curve-bottom curve. I mean I found it by putting the curves into geogebra
but im not gonna have that in the test rip. pls help
2. don't no how to do this...(so far i've only been exposed to sth like with no 'x's in the numerator or ones i could easily cancel out)
thanks! u guys are absolute legends
Post by: jamonwindeyer on May 21, 2017, 06:05:24 pm
pls help last min study for my exam tmr
so i have a couple of questions...which i will split into separate posts
1. find the area enclosed between the hyperbola y=3/x and the line y=-2x+5
If you did it the wrong way around, you will get a negative answer - That's one easy way to tell! More generally, you can sketch the curves - This is the best approach. \(y=\frac{3}{x}\) is just a hyperbola and \(y=5-2x\) just a straight line, so by the HSC (even if not right now) those should be easily sketched for you. As a last minute fix, you can check the \(y\)-value for given \(x\) coordinates within the region of integration, see which one is bigger - Just watch out for intersections (which you can find with simulteneous)!
Quote
2. don't no how to do this...(so far i've only been exposed to sth like with no 'x's in the numerator or ones i could easily cancel out)
Just a little trick of addition and subtraction at the top there to make it a little nicer - It's a good one to remember! ;D
Edit: Beaten by Rui, except for the sign - Should be a positive logarithm ;D
Post by: katnisschung on May 21, 2017, 06:07:43 pm
(sorry it won't let me attach the image)
find all the values of theta for -pi is less than or equal to theta which is less than or equal to pi
for which sin theta= cos theta
Post by: jamonwindeyer on May 21, 2017, 06:13:57 pm
could someone pls explain this to me? thanks :)
(sorry it won't let me attach the image)
find all the values of theta for -pi is less than or equal to theta which is less than or equal to pi
for which sin theta= cos theta
Sure!
Now the tricky bit is adjusting to suit our domain - We know \(\frac{\pi}{4}\) is in there, so that is one answer. The other answer would normally be \(\frac{5\pi}{4}\), but that isn't in the given domain. To get it there, we just subtract \(2\pi\), remember that doesn't change the 'value' of our angle!
Does this last bit make sense? :)
Post by: katnisschung on May 21, 2017, 06:21:06 pm
Post by: jamonwindeyer on May 21, 2017, 06:52:41 pm
got it 100% thanks. u explain it way better than my textbook
You are too kind! I'll be sure to put that on my CV/Resume ;) aha awesome! Feel free to shoot any other questions our way :)
Post by: JuliaPascale123 on May 21, 2017, 09:21:16 pm
Post by: RuiAce on May 21, 2017, 09:25:44 pm
(http://i65.tinypic.com/11wdac2.png)
Post by: JuliaPascale123 on May 21, 2017, 09:37:41 pm
Post by: Fahim486 on May 21, 2017, 10:59:54 pm
Thanks!
Post by: RuiAce on May 21, 2017, 11:17:21 pm
Hey, I'm having trouble doing these questions because I suck at trigonometry.
Thanks!
Post by: RuiAce on May 21, 2017, 11:24:32 pm
(http://i68.tinypic.com/10ynmkn.png)
You can sketch the graph on software such as Desmos
_____________-
Post by: Rasika on May 22, 2017, 01:15:41 am
Post by: jamonwindeyer on May 22, 2017, 01:19:13 am
What's the best way to understand a hard concept?
Lots of practice! Maths is a skill best learned by putting your brain to work and just trying the question type you are struggling with over and over, having someone step you through initially and then taking the reigns yourself. Of course there are other things you can do, but it all ultimately comes back to practice makes perfect :)
Post by: Athos36 on May 22, 2017, 10:37:14 am
A parabola has form y=ax^2+bx+4.
a) its turning point is on the x-axis. Find "a" in terms of "b" (answer= b^2/16)
b) If turning point is at (-4,0), find values "a" and "b". )answer= a= 1/4; b= 2)
Thanks in advance!
Post by: RuiAce on May 22, 2017, 10:41:27 am
Hi there! (once again) Could you help me with a quadratic question please?
A parabola has form y=ax^2+bx+4.
a) its turning point is on the x-axis. Find "a" in terms of "b" (answer= b^2/16)
b) If turning point is at (-4,0), find values "a" and "b". )answer= a= 1/4; b= 2)
Thanks in advance!
b) is now taking \(x=-\frac{b}{2a}=4\)
Jake accidentally edited my post :P
As usual, Rui beats me to the punch. 'Different' methods, though, so hopefully it's helpful!
Post by: jakesilove on May 22, 2017, 10:42:29 am
Hi there! (once again) Could you help me with a quadratic question please?
A parabola has form y=ax^2+bx+4.
a) its turning point is on the x-axis. Find "a" in terms of "b" (answer= b^2/16)
b) If turning point is at (-4,0), find values "a" and "b". )answer= a= 1/4; b= 2)
Thanks in advance!
Hey! We know that the turning point is on the x-axis, which occurs when y=0. Therefore,
Now, we can use the quadratic formula to solve this equation!
However, we know that there can only be one solution to this! Think about it; if a parabola has a turning point on the x-axis, it can't have more than one x-intercept. Therefore, the square root must be equal to zero (do you understand why?)
As required
Now, we know that the turning point must occur at
To answer this part of the question, sub in our value of 'a' that we found in the first part of the question, and set x=-4. Then, solve for b, and for a! Let me know if you need more help with this part :)
Welcome to the forum!
Post by: 1937jk on May 22, 2017, 08:52:35 pm
The sum of 50 terms of an arithmetic series is 249 and the sum of 49 terms of the series is 233. Find the 50th term of the series.
Post by: RuiAce on May 22, 2017, 08:59:23 pm
Hey, just really stuck on this question,
The sum of 50 terms of an arithmetic series is 249 and the sum of 49 terms of the series is 233. Find the 50th term of the series.
Post by: RuiAce on May 24, 2017, 10:54:59 am
Hey! We know that the turning point is on the x-axis, which occurs when y=0. Therefore,
Now, we can use the quadratic formula to solve this equation!
However, we know that there can only be one solution to this! Think about it; if a parabola has a turning point on the x-axis, it can't have more than one x-intercept. Therefore, the square root must be equal to zero (do you understand why?)
As required
Now, we know that the turning point must occur at
To answer this part of the question, sub in our value of 'a' that we found in the first part of the question, and set x=-4. Then, solve for b, and for a! Let me know if you need more help with this part :)
Welcome to the forum!
Post by: Athos36 on May 24, 2017, 11:11:32 am
Thanks for your time.
Post by: RuiAce on May 24, 2017, 11:34:26 am
Post by: jamonwindeyer on May 24, 2017, 02:49:26 pm
I'm just finding it difficult to teach myself. You are way ahead of me since my math knowledge is rather small. I'll just have to muddle through the problem.
Thanks for your time.
Definitely let us know which bit doesn't quite click. I know it's frustrating not quite getting it right away, but we're happy to go into more detail where you need it :)
Post by: JuliaPascale123 on May 24, 2017, 04:16:01 pm
Thanks so much.
You can sketch the graph on software such as Desmos
_____________-
What would the answer for b) be?
http://i68.tinypic.com/10ynmkn.png
Post by: 1937jk on May 24, 2017, 04:16:46 pm
Okay cool! Thank you I see now!
Another question haha,
Which term of the series 8, -4 , + 2.... is 1/128?
I'm just particularly struggling with dealing with a negative r value and using log to solve for n
thanks!
Mod Edit [Aaron]: Merged double post. Please edit your previous post if you want to add something.
Post by: jakesilove on May 24, 2017, 05:21:13 pm
Okay cool! Thank you I see now!
Another question haha,
Which term of the series 8, -4 , + 2.... is 1/128?
I'm just particularly struggling with dealing with a negative r value and using log to solve for n
thanks!
Mod Edit [Aaron]: Merged double post. Please edit your previous post if you want to add something.
Hey! So we know that, for a geometric series,
Where a is the first term, r is the geometric difference between two terms, and n is the term number. Thus,
is the definition of each term for this series.
Now, we want to find
So,
The last line literally means -1/2 to the power of WHAT (ie. n-1) is equal to 1/1024 (at least, that's how I think of logs). Let's keep going
Actually, this is more difficult than it needs to be. Let's take a step back
The negative here doesn't matter; as long as n-1 is even, the term becomes positive.
Let's just check this in the original equation
As expected
Post by: RuiAce on May 24, 2017, 07:09:25 pm
Thanks so much.I told you what to sketch at the bottom of my post
What would the answer for b) be?
http://i68.tinypic.com/10ynmkn.png
Post by: JuliaPascale123 on May 24, 2017, 08:27:04 pm
I told you what to sketch at the bottom of my post
Thanks you!
Post by: HotDiggity on May 24, 2017, 08:32:06 pm
Sorry about the picture quality fam :-X .
(http://i68.tinypic.com/zom05.png)
Jennifur :p
Post by: RuiAce on May 24, 2017, 08:39:51 pm
Hi :) ,
Sorry about the picture quality fam :-X .
(http://i68.tinypic.com/zom05.png)
Jennifur :p
Technology.
Post by: 1937jk on May 25, 2017, 07:08:14 am
Hey! So we know that, for a geometric series,
Where a is the first term, r is the geometric difference between two terms, and n is the term number. Thus,
is the definition of each term for this series.
Now, we want to find
So,
The last line literally means -1/2 to the power of WHAT (ie. n-1) is equal to 1/1024 (at least, that's how I think of logs). Let's keep going
Actually, this is more difficult than it needs to be. Let's take a step back
The negative here doesn't matter; as long as n-1 is even, the term becomes positive.
Let's just check this in the original equation
As expected
awesome! thank you so much! I understand now
Post by: HotDiggity on May 25, 2017, 10:16:31 am
Technology.
Cheers Rui, but your link does not work for the graph and all the graphs im looking at dont intercept at 0.495
https://www.wolframalpha.com/input/?i=graph+(t%5E2%2Bcost)%2F(t%5E4%2B1)
Post by: jamonwindeyer on May 25, 2017, 10:51:48 am
Cheers Rui, but your link does not work for the graph and all the graphs im looking at dont intercept at 0.495
https://www.wolframalpha.com/input/?i=graph+(t%5E2%2Bcost)%2F(t%5E4%2B1)
Hey! Yeah I agree with your sketch, and that means there is a \(y\) intercept at \(\left(0,1\right)\), and there are no \(x\)-intercepts :)
Post by: RuiAce on May 25, 2017, 10:57:48 am
Cheers Rui, but your link does not work for the graph and all the graphs im looking at dont intercept at 0.495Not sure what was going on the link
https://www.wolframalpha.com/input/?i=graph+(t%5E2%2Bcost)%2F(t%5E4%2B1)
(@ Jamon as well) I believe the question asked to sketch the velocity, not the displacement.
You input the graph of s(t) into WolframAlpha in your link.
Post by: _____ on May 25, 2017, 11:12:57 am
"Newton's Law of Cooling states that the rate of decrease in temperature is proportional to the temperature difference between the body and its surroundings.
A batch of scones are removed from an oven whose temperature is 320 degrees Celsius. The rack onto which they are placed is surrounded by air at 20 degrees Celsius. They cool at a rate of 30% per minute.
What is the temperature of the scones after:
a) 2 minutes
b) 10 minutes"
I've done questions like this before where the temperatures would change from 320 > 0 for example. The equation if this were the case would be:
T = 320e^(0.3t). But how do I account for the temperature difference in this instance, where it is 20 degrees rather than 0 degrees in the new environment? I tried (320-20)e^(0.3t) but this returned ~16 degrees which must be wrong.
Post by: jamonwindeyer on May 25, 2017, 11:15:15 am
Got a dumb questions about ACPW. Currently on exponential growth/decay. The question is:
Hey! Definitely not a dumb question - That is a 3U question, do you do 3U? Happy to explain it if it is relevant to you ;D
Edit: If you want to have a crack at the Math, the formula should be:
Knowing where this comes from and actually using things like it in exams is left to the 3U students :)
Post by: _____ on May 25, 2017, 02:17:10 pm
Hey! Definitely not a dumb question - That is a 3U question, do you do 3U? Happy to explain it if it is relevant to you ;D
Edit: If you want to have a crack at the Math, the formula should be:
Knowing where this comes from and actually using things like it in exams is left to the 3U students :)
That's odd, I'm only doing 2 units. I've got to do this question anyway but I'll be sure to note that down (and look at the syllabus more closely).
Thanks for the equation, if you get the chance could you explain how you got 310e rather than using 300e or 320e? I understand why you'd add 20 but I thought you'd have to either keep the multiple the same (320) or reduce the multiple by 20 because the difference is 20 (300).
Post by: jamonwindeyer on May 25, 2017, 02:56:28 pm
That's odd, I'm only doing 2 units. I've got to do this question anyway but I'll be sure to note that down (and look at the syllabus more closely).
Thanks for the equation, if you get the chance could you explain how you got 310e rather than using 300e or 320e? I understand why you'd add 20 but I thought you'd have to either keep the multiple the same (320) or reduce the multiple by 20 because the difference is 20 (300).
You're absolutely right - I mistyped, it should be 300 as that second number, fixed it above! Sorry! :)
Post by: _____ on May 25, 2017, 03:35:41 pm
You're absolutely right - I mistyped, it should be 300 as that second number, fixed it above! Sorry! :)
NP thanks for the help!
Post by: sophiegmaher on May 25, 2017, 05:23:28 pm
- I'm confused about the "but not including" part? Would that mean that the final formula used for the final year would be 5000(1.06)^0 ?
Post by: RuiAce on May 25, 2017, 06:32:09 pm
Yooo so with superannuation questions: On 31st January 1990 Hayden joined a superannuation fund by investing $5000 at 6%p.a. compound interest. He added $5000 to the fund on 31st January each year and intends to continue to do so until, but not including, the day he intends to retire 31/1/2010. Find the total accumulated value of his investment, to the nearest dollar, on that day."Final formula" is ambiguous and it depends on whether you used the recursive approach, or the accumulation method.
- I'm confused about the "but not including" part? Would that mean that the final formula used for the final year would be 5000(1.06)^0 ?
What the question is saying is that in the last year, he does not deposit another $5000. (i.e. There is no cash flow that year.)
Post by: pacione on May 27, 2017, 10:33:26 am
Could anyone please help me with the attached question, it would be greatly appreciated.
Post by: RuiAce on May 27, 2017, 10:38:31 am
Hey,Which parts are you having trouble with?
Could anyone please help me with the attached question, it would be greatly appreciated.
(Hints in the meantime for all three)
a) That ray OR is drawn so that you have more than one triangle to consider. Use it.
b) A bit easier than part a), but follows the same idea of working around the diagram
c) Your typical maxima/minima problem. The derivative is annoying as it needs the product and chain rule, but then all you have to do is substitute theta=pi/12 in.
Post by: Youssk on May 27, 2017, 11:31:00 am
Thank you
Post by: RuiAce on May 27, 2017, 11:35:01 am
Hey could someone help me with this question please!
Thank you
Post by: cxmplete on May 27, 2017, 04:15:09 pm
How would I do question 8
Post by: jakesilove on May 27, 2017, 04:53:21 pm
Hi,
How would I do question 8
Hey! Make sure you've drawn this out; the minor segment will have perimeter AB plus the arc length of AB. We know that the arc length is
We can calculate the chord length of AB by the cosine rule.
Now, we add these two lengths together to get the total perimeter.
However, we know that this is equal to half the circumference, so
Now, from here it seems like you need to use a 3U trick; it this a 2U question? Recall that
As required.
Post by: RuiAce on May 27, 2017, 06:35:11 pm
Post by: bananna on May 27, 2017, 09:59:39 pm
pls help!!!
thank you so much
Ashley has just retired with $1 000 000 in her bank account. This account attracts interest at a rate of 5% per annum compounded annually. She intends to withdraw money in equal amounts at the end of each year, and wants the money in her account to last for exactly 20 years.
Let An be the amount of money in her account after n years.
(i) Write down an expression for the amount left after two withdrawals. 1
(ii) Calculate the amount of each withdrawal. 3
(iii) After her tenth withdrawal, the bank changes its interest rate to 3% p.a. on the balance remaining in her account. Interest is still compounded annually.
Ashley changes the amount of her withdrawals so that they last the remaining 10 years and are equal in value over these 10 years.
What will be the new amount of each withdrawal in these last 10 years? 3
Post by: RuiAce on May 27, 2017, 10:41:23 pm
Hi, I'm having a lot of trouble with this q
pls help!!!
thank you so much
Ashley has just retired with $1 000 000 in her bank account. This account attracts interest at a rate of 5% per annum compounded annually. She intends to withdraw money in equal amounts at the end of each year, and wants the money in her account to last for exactly 20 years.
Let An be the amount of money in her account after n years.
(i) Write down an expression for the amount left after two withdrawals. 1
(ii) Calculate the amount of each withdrawal. 3
(iii) After her tenth withdrawal, the bank changes its interest rate to 3% p.a. on the balance remaining in her account. Interest is still compounded annually.
Ashley changes the amount of her withdrawals so that they last the remaining 10 years and are equal in value over these 10 years.
What will be the new amount of each withdrawal in these last 10 years? 3
_________________
_________________
Post by: Fahim486 on May 28, 2017, 07:50:01 pm
Post by: jakesilove on May 28, 2017, 07:58:50 pm
Hey I'm having trouble understanding simpson's rule and trying to apply them. Thanks!
Hey! Simpson's rule states that
Where x0, x1 etc are evenly spaced intervals and h is the distance between such intervals. In this case, we only want three values, each spaced one unit apart. So,
Then, we simply plug in the values of the function (ie. f(2)=ln(2)) and use our calculator to find the answer!
Post by: Fahim486 on May 28, 2017, 08:44:02 pm
Post by: pikachu975 on May 28, 2017, 08:47:39 pm
Thanks Jake for that help and just as bad as I am at the Simpson's rule, I am equally bad at Trapezoidal rule so if you don't mind could you also show me how to do this question as well. Thanks!!!
4 strips means 5 function values, so x = 3, 3.5, 4, 4.5, 5 and h/2 = 1/4
A = 1/4 (f(3) + 2f(3.5) + 2f(4) + 2f(4.5) + f(5))
A = 1/4 (ln8 + 2ln11.25 + 2ln15 + 2ln19.25 + 24)
Sub into calculator for your answer
Post by: Fahim486 on May 28, 2017, 09:02:26 pm
4 strips means 5 function values, so x = 3, 3.5, 4, 4.5, 5 and h/2 = 1/4
A = 1/4 (f(3) + 2f(3.5) + 2f(4) + 2f(4.5) + f(5))
A = 1/4 (ln8 + 2ln11.25 + 2ln15 + 2ln19.25 + 24)
Sub into calculator for your answer
Thanks for your help but may I ask where you got h/2 = 1/4 from?
Post by: pikachu975 on May 28, 2017, 09:07:44 pm
Thanks for your help but may I ask where you got h/2 = 1/4 from?
Basically h is the difference between each function value.
The difference between each is 0.5, so 0.5/2 = 1/4!
Post by: Fahim486 on May 28, 2017, 09:43:50 pm
Basically h is the difference between each function value.
The difference between each is 0.5, so 0.5/2 = 1/4!
ooohhhhh okay thats makes so much sense. Thanks for your help!!!!!!!!!!!
Post by: laurenf58 on May 29, 2017, 06:33:19 pm
Thanks
Post by: Fahim486 on May 29, 2017, 07:46:17 pm
Post by: RuiAce on May 29, 2017, 08:32:49 pm
Hi really need help with this question. Thanks!!!
Post by: RuiAce on May 29, 2017, 08:37:13 pm
Can I have some help with these questions please?
Thanks
_________________
(http://i.imgur.com/g6mk1pW.png)
For convenience, y=5 was also plotted
_________________
This is just 1. That equation is formed when you try to find the point of intersection between the two graphs, and you already know from the previous part that there is only one.
Post by: Fahim486 on May 29, 2017, 09:33:27 pm
Post by: RuiAce on May 29, 2017, 09:37:58 pm
Thanks for the help RuiAce. I've got another question I also need help with and I can't figure out why I keep getting the wrong answer. Thanks!!!
Post by: Thebarman on May 30, 2017, 07:28:33 pm
y= 3cos(x) - cos(2x)
Post by: RuiAce on May 30, 2017, 08:43:31 pm
Does anyone have some tips on how to solve questions like these? I'm basically using guess and check in my calculator at this point, but there must be a better way to solve it. Any help would be appreciated :)What are you trying to solve? All you've specified is just an expression.
y= 3cos(x) - cos(2x)
Post by: Thebarman on May 31, 2017, 12:13:25 am
Post by: RuiAce on May 31, 2017, 12:15:30 am
My honest opinion - it's such an annoying process.
If you want to find the intercepts of 3cos(x) and cos(2x), upon equating you get 3cos(x) - cos(2x) = 0 so you just look at where all your x-intercepts are.
Post by: 1937jk on May 31, 2017, 03:58:37 pm
A loan of $6000 over 5 years at 15%, charged monthly, is paid back in 5 annual instalments.
a) what is the amount of each instalment?
b) how much is paid back altogether?
Post by: RuiAce on May 31, 2017, 04:18:11 pm
hi just stuck on this question,
A loan of $6000 over 5 years at 15%, charged monthly, is paid back in 5 annual instalments.
a) what is the amount of each instalment?
b) how much is paid back altogether?
______________________
Post by: K9810 on May 31, 2017, 09:48:47 pm
How can I differentiate this?
Post by: jamonwindeyer on May 31, 2017, 09:57:14 pm
Hey!
How can I differentiate this?
Oh this is cool! Do this:
You can then tackle that with the quotient rule! Let me know if you need a hand with it, but you should get:
Or some factorised form of it ;D
Post by: K9810 on May 31, 2017, 10:04:06 pm
Oh this is cool! Do this:
You can then tackle that with the quotient rule! Let me know if you need a hand with it, but you should get:
Or some factorised form of it ;D
Thanks for helping! :)
Post by: 1937jk on June 01, 2017, 08:24:59 am
______________________
Okay cool thank you! I understand now !
Post by: 1937jk on June 01, 2017, 08:27:22 am
so just trying to integrate tanx so sinx/cosx but I'm confused and don't understand how to do it
Post by: RuiAce on June 01, 2017, 09:10:55 am
hey,
so just trying to integrate tanx so sinx/cosx but I'm confused and don't understand how to do it
Post by: Fahim486 on June 01, 2017, 08:14:28 pm
Post by: RuiAce on June 01, 2017, 08:39:04 pm
Hey so I can't figure out how to do question 9. iii) and the answer says it's 7-3ln7 and I don't know how they figured it out. ThanksReading off your sketch, the maximum clearly occurs at the endpoint at x=7. So upon substituting in x=7 we have y=7-3ln7
Post by: Thebarman on June 01, 2017, 09:11:04 pm
I've attached the question in case my message is a bit confusing.
Thank you!
EDIT:
As far as the HSC goes, the only way to sketch that is to first sketch y=3cos(x) and y=-cos(2x) separately (AND to scale), and then add the y-coordinates together. This is essentially how superposition of waves work.
My honest opinion - it's such an annoying process.
If you want to find the intercepts of 3cos(x) and cos(2x), upon equating you get 3cos(x) - cos(2x) = 0 so you just look at where all your x-intercepts are.
Thanks for the reply! I definitely agree that it's an annoying process. Do you mind clarifying what you mean about looking at where all the x-intercepts are? Thanks again :)
Post by: RuiAce on June 02, 2017, 08:36:28 am
Hey guys, I'm working through a past paper right now and I'm slightly stuck on the second part of a question. I'm trying to simplify 15pi/2 - 9/2 sin(5pi/3). It needs to be in exact form, and I'm not sure how to simplify sin 5pi/3. According to the answers, that particular part simplifies to -root3/2. How do I get to this?
I've attached the question in case my message is a bit confusing.
Thank you!
EDIT:
Thanks for the reply! I definitely agree that it's an annoying process. Do you mind clarifying what you mean about looking at where all the x-intercepts are? Thanks again :)
______________________
That statement was mildly dependent on what you're intersecting. I took a guess as to what you were trying to intersect but I never knew what you meant, so perhaps ignore what I said and maybe make your question clearer for me please?
Post by: Thebarman on June 02, 2017, 09:51:01 am
Post by: 12070 on June 02, 2017, 04:33:14 pm
So I have a question on the mark allocation on one of my questions in a test.
The question states 'John has borrowed $200,000. The interest rate is 6% p.a compounded monthly.'
iii.) If John pays $10,000 to the loan each month, find how long it will take him to reduce the loan to $100,000.
I know how to do the question but just made a calculation error saying that 10,000/0.005 was 200,000 instead of 2 million.
My answer is attached; which I can re-write if you can't read it but my teacher wasn't giving me any marks for it which I'm a little surprised about so would love to hear your thoughts.
Post by: Wales on June 02, 2017, 06:02:02 pm
Hey guys,
So I have a question on the mark allocation on one of my questions in a test.
The question states 'John has borrowed $200,000. The interest rate is 6% p.a compounded monthly.'
iii.) If John pays $10,000 to the loan each month, find how long it will take him to reduce the loan to $100,000.
I know how to do the question but just made a calculation error saying that 10,000/0.005 was 200,000 instead of 2 million.
My answer is attached; which I can re-write if you can't read it but my teacher wasn't giving me any marks for it which I'm a little surprised about so would love to hear your thoughts.
How many marks was the question? Do you have the question paper? If it was 3 marks or so I would be curious as to why you did not receive at least 1 for carry on. At least that's what I've been told. Correct me if I am wrong.
Post by: 12070 on June 02, 2017, 06:56:49 pm
How many marks was the question? Do you have the question paper? If it was 3 marks or so I would be curious as to why you did not receive at least 1 for carry on. At least that's what I've been told. Correct me if I am wrong.
Yeah; it was worth 3 and I'm really gutted about it because I know how to do it and I got the same mark as someone who didn't even atempt the question.Normally I wouldn't mind too much but the test was worth 35% (because our mid-course got due to flooding in Lismore) and only out of 42 so that's almost a percent for every mark. Anyway, do you know how I should argue on Monday because if I had made a similar mistake towards the end i know for a fact that I would get 2 arks but because it was so early I got 0. Don't really understand it to be honest.
Post by: Wales on June 02, 2017, 07:20:04 pm
Yeah; it was worth 3 and I'm really gutted about it because I know how to do it and I got the same mark as someone who didn't even atempt the question.Normally I wouldn't mind too much but the test was worth 35% (because our mid-course got due to flooding in Lismore) and only out of 42 so that's almost a percent for every mark. Anyway, do you know how I should argue on Monday because if I had made a similar mistake towards the end i know for a fact that I would get 2 arks but because it was so early I got 0. Don't really understand it to be honest.
It does seem like you skipped multiple steps there, your teacher left 2 question marks. I've seen teachers mark people down for not showing working. My advice would be to ask the teacher and ask where the marks where allocated and check it over with what you've done.
I assume the mark allocation to be
1 for finding the equation 100000=200000x1.005^n-10000(1.005^n-1/0.005) (nope)
2 for reorganising for n (carry on? but you skipped like 3 lines of working)
3rd for the answer? (nope)
But I can't be sure. You've skipped far too many steps there. See what Rui says.
Post by: jamonwindeyer on June 02, 2017, 07:23:24 pm
Definitely worth following up - I'd wager you could get a single mark for what you did show :)
Post by: RuiAce on June 02, 2017, 08:06:12 pm
Hey guys,
So I have a question on the mark allocation on one of my questions in a test.
The question states 'John has borrowed $200,000. The interest rate is 6% p.a compounded monthly.'
iii.) If John pays $10,000 to the loan each month, find how long it will take him to reduce the loan to $100,000.
I know how to do the question but just made a calculation error saying that 10,000/0.005 was 200,000 instead of 2 million.
My answer is attached; which I can re-write if you can't read it but my teacher wasn't giving me any marks for it which I'm a little surprised about so would love to hear your thoughts.
Which, if you had noticed that, you probably could've saved a few marks by realising something isn't right.
I'm sorry to say but I would not give you any marks either, because I see multiple mistakes; not just the computational one.
Post by: Wales on June 02, 2017, 09:23:57 pm
Which, if you had noticed that, you probably could've saved a few marks by realising something isn't right.
I'm sorry to say but I would not give you any marks either, because I see multiple mistakes; not just the computational one.
Yeah it wouldn't hurt to ask the teacher for a pity mark but I don't think your chances are too high given you did skip out 4 lines of working. There some things markers can assume but in this case it seems not. Still worth a shot though, fight for that mark.
Post by: 12070 on June 02, 2017, 09:34:59 pm
Which, if you had noticed that, you probably could've saved a few marks by realising something isn't right.
I'm sorry to say but I would not give you any marks either, because I see multiple mistakes; not just the computational one.
I didn't even know I missed that last step and had so many chances to realise something isn't right as well. So that is probably fair enough. Thanks for the help in clearing things up though :)
Post by: Wales on June 02, 2017, 09:39:06 pm
I didn't even know I missed that last step and had so many chances to realise something isn't right as well. So that is probably fair enough. Thanks for the help in clearing things up though :)
You clearly know how to do the question. Just try not to skip out any working in the future. Still wouldn't hurt asking the teacher, I fought every mark I thought I had a chance to gain in every exam. It matters!
Post by: katnisschung on June 04, 2017, 06:32:38 pm
Mohammed's mother invests $200 for him each birthday up to and including his 18th birthday.
The money earns 6% pa How much money will Mohammed have on his 18th birthday
ans$6181.13
also q on small angles how do i find the limit for this...i've only ever seen sinx/x format not this..
3x/sin3x
Post by: f_tan on June 04, 2017, 09:02:39 pm
(http://i.imgur.com/oRTitim.png)
Mod Edit: Restored deleted post
Post by: Willba99 on June 04, 2017, 09:14:22 pm
Could someone explain why you subtract top curve and bottom curve? Why can't you just integrate sinx?
(http://i.imgur.com/oRTitim.png)
integrating sinx wouldn't find the shaded area under the x axis
Post by: jamonwindeyer on June 04, 2017, 09:22:53 pm
Could someone explain why you subtract top curve and bottom curve? Why can't you just integrate sinx?
(http://i.imgur.com/oRTitim.png)
Willba is right - Whenever you have an area between two curves, you always find it by taking the integral of top curve minus the bottom curve. This is a blanket rule that works whenever you have a single area bounded between two functions ;D
Post by: Willba99 on June 04, 2017, 09:39:38 pm
so long story short, i bit off more than i could chew with the take-home part of my specialist SAC and i have to solve
192600pi+60000=1254pi(x)+12500sin(pi(x)/100)-40000sin(pi(x)/200)
any help solving this would be greatly appreciated. Basically I'm looking for solutions of something in the form of a=bx+csin(dx)-esin(fx)
Post by: jakesilove on June 04, 2017, 11:00:55 pm
hey guys.
so long story short, i bit off more than i could chew with the take-home part of my specialist SAC and i have to solve
192600pi+60000=1254pi(x)+12500sin(pi(x)/100)-40000sin(pi(x)/200)
any help solving this would be greatly appreciated. Basically I'm looking for solutions of something in the form of a=bx+csin(dx)-esin(fx)
To be honest, this looks a bit absurd to me; doesn't look like there's a straight forward algebraic fix. I reckon it's more likely that you've made a mistake somewhere along the road. Also, note that this thread is for HSC students; there's a separate one for VCE questions!
Post by: RuiAce on June 04, 2017, 11:22:58 pm
also q on small angles how do i find the limit for this...i've only ever seen sinx/x format not this..
3x/sin3x
Post by: RuiAce on June 04, 2017, 11:27:36 pm
i feel like this is badly worded...
Mohammed's mother invests $200 for him each birthday up to and including his 18th birthday.
The money earns 6% pa How much money will Mohammed have on his 18th birthday
ans$6181.13
To be honest, this looks a bit absurd to me; doesn't look like there's a straight forward algebraic fix. I reckon it's more likely that you've made a mistake somewhere along the road. Also, note that this thread is for HSC students; there's a separate one for VCE questions!Yeah I agree. I'd want to see the progress to that stage before jumping in on it
Post by: Blazeee on June 05, 2017, 07:58:18 pm
Thanks! :D
Post by: RuiAce on June 05, 2017, 08:24:16 pm
Hey does anyone have an easy acronym or something to remember the justification for similar triangles? It's easy for the congruent triangles with the AAS and SAS and all that...but the similar triangles always tempt me to write these congruence acronyms..which we're not allowed to write ???You really aren't expected to use acronyms at all for similarity. You can develop your own ones to memorise them, however you cannot write them.
Thanks! :D
That being said, I ignored that. I learnt similarity when I found I could compare them to congruence
The similarity statements can be related to congruence:
- Equiangular: Related to AAS. Just chop out the side.
- All three sides in proportion: Related to SSS. Just replace equal sides with proportional sides
- Two sides in proportion, included angle equal: Related to SAS. Just replace equal sides with proportional sides again.
The tests are honestly almost the same. The only thing is:
- You keep the angles the same, but
- Whenever you have 2 or more equal sides, replace "equal" with proportion.
If you can write down the criteria for congruency, you can then easily write out the criteria for similarity. Force yourself to not use acronyms not permitted
Post by: Blazeee on June 05, 2017, 08:52:13 pm
You really aren't expected to use acronyms at all for similarity. You can develop your own ones to memorise them, however you cannot write them.
That being said, I ignored that. I learnt similarity when I found I could compare[i/] them to congruence
The similarity statements can be related to congruence:
- Equiangular: Related to AAS. Just chop out the side.
- All three sides in proportion: Related to SSS. Just replace equal sides with proportional sides
- Two sides in proportion, included angle equal: Related to SAS. Just replace equal sides with proportional sides again.
The tests are honestly almost the same. The only thing is:
- You keep the angles the same, but
- Whenever you have 2 or more equal sides, replace "equal" with proportion.
If you can write down the criteria for congruency, you can then easily write out the criteria for similarity. Force yourself to not use acronyms not permitted
Thanks very much! :)Hopefully i will remember in my exam tomorrow! ::)
Post by: jamonwindeyer on June 05, 2017, 09:06:33 pm
Thanks very much! :)Hopefully i will remember in my exam tomorrow! ::)
Similarity proofs in exams are way more commonly equiangular proofs rather than the ones involving proportional sides, so at the least, make sure you remember that one! Prove two angles equal for similarity ;D
Post by: RuiAce on June 05, 2017, 09:14:50 pm
Similarity proofs in exams are way more commonly equiangular proofs rather than the ones involving proportional sides, so at the least, make sure you remember that one! Prove two angles equal for similarity ;DWhilst I only came across one "all 3 sides in proportion" I swear 33% of the time I've seen two sides prop+angle and 66% of the time equiangular
Post by: Kekemato_BAP on June 05, 2017, 09:31:40 pm
"The line y=mx is the tangent to the curve y=e^3x. Find m"
Any help with step by step :)
Post by: jamonwindeyer on June 05, 2017, 09:57:55 pm
I'm stuck on this question on exponentials..
"The line y=mx is the tangent to the curve y=e^3x. Find m"
Any help with step by step :)
Hey! So since it is a tangent, the gradients of the line and the curve are the same, so we know that:
Further, they have to meet at the point where the gradients are the same, so:
This is a set of simultaneous equations. Via substitution:
Does that make sense? ;D
Post by: Kekemato_BAP on June 05, 2017, 10:05:46 pm
Post by: Kekemato_BAP on June 06, 2017, 12:15:45 am
How would I do this question?
I have a maths log/exp exam tomorrow :(
Post by: RuiAce on June 06, 2017, 08:21:42 am
(http://i.imgur.com/6ovpz0S.jpg)
How would I do this question?
I have a maths log/exp exam tomorrow :(
Post by: Fahim486 on June 06, 2017, 07:56:12 pm
Post by: RuiAce on June 06, 2017, 07:59:20 pm
Hey how would you solve y=2sin5(pi symbol)/3 - 5(pi symbol)/3 in exact form?Assuming you meant to simplify?
Side note: If you just type "pi" we will understand
Post by: Fahim486 on June 06, 2017, 08:08:23 pm
Post by: RuiAce on June 06, 2017, 08:16:10 pm
Post by: Fahim486 on June 06, 2017, 08:29:01 pm
oohhhh yes now that makes sense. Thank you!!!!!
Post by: kiiaaa on June 06, 2017, 08:36:45 pm
i was doing a past paper and this question appeared and i like lowkey know what to do but was slightly confused so i went to the solutions and their solutions they gave even confused my poor brain more. could you please help me out. it is question dii)
thank you :))
Post by: jakesilove on June 06, 2017, 08:43:50 pm
hello,
i was doing a past paper and this question appeared and i like lowkey know what to do but was slightly confused so i went to the solutions and their solutions they gave even confused my poor brain more. could you please help me out. it is question dii)
thank you :))
We know that
Now, we want to find
Note that the integral is ALMOST the same as the result of part i) (which is obviously not accident!). In fact, if we rearrange the above, we find that
If we then integrate both sides, we would get to the answer required. ie.
Note that the second term is literally the integral of a differential. ie. first, differentiate xtan(x), then integrate it. Obviously, you'll just get xtan(x) again, but with some constant C added to it. We can also perform the first integral, as it is just the negative natural logarithm relation.
As required! Does that make sense?
Post by: kiiaaa on June 06, 2017, 08:55:24 pm
We know that
Now, we want to find
Note that the integral is ALMOST the same as the result of part i) (which is obviously not accident!). In fact, if we rearrange the above, we find that
If we then integrate both sides, we would get to the answer required. ie.
Note that the second term is literally the integral of a differential. ie. first, differentiate xtan(x), then integrate it. Obviously, you'll just get xtan(x) again, but with some constant C added to it. We can also perform the first integral, as it is just the negative natural logarithm relation.
As required! Does that make sense?
heyy, im sort of confused on where did the 'cos' come from. your explanation made everything else make sense except where did the 'ln(cos)' come from
thank you so much! :))
Post by: RuiAce on June 06, 2017, 09:15:35 pm
heyy, im sort of confused on where did the 'cos' come from. your explanation made everything else make sense except where did the 'ln(cos)' come from
thank you so much! :))
Edit: It seems like Jake skipped the use of it, which is a bit surprising. In all honesty, I do believe that the integral of tan(x) is very unfair for 2U unless assisted, however that's the way of going about it. Rewrite \(\tan x = \frac{\sin x}{\cos x}\)
Post by: kiiaaa on June 07, 2017, 05:48:19 am
Edit: It seems like Jake skipped the use of it, which is a bit surprising. In all honesty, I do believe that the integral of tan(x) is very unfair for 2U unless assisted, however that's the way of going about it. Rewrite \(\tan x = \frac{\sin x}{\cos x}\)
ahh i get that now! thank you so much both to you and Jake :))
Post by: Wren on June 07, 2017, 06:22:39 pm
Post by: RuiAce on June 07, 2017, 06:26:42 pm
just wanted clarification when your using simpson's rule does would 3 function values mean a, (a+b)/2 and b?Yes
Post by: jakesilove on June 07, 2017, 06:36:04 pm
just wanted clarification when your using simpson's rule does would 3 function values mean a, (a+b)/2 and b?
Yep, 3 function values would mean
However, 3 intervals would require 4 function values, ie
Post by: RuiAce on June 07, 2017, 06:37:31 pm
Yep, 3 function values would meanAlbeit that can't happen with Simpson's rule though, that's only for trap rule
However, 3 intervals would require 4 function values, ie
Post by: jakesilove on June 07, 2017, 06:38:17 pm
Albeit that can't happen with Simpson's rule though, that's only for trap rule
Good point. Just trying to illustrate the differences of terminology used in questions!
Post by: K9810 on June 07, 2017, 09:21:11 pm
Need help with the attached question :)
Post by: RuiAce on June 07, 2017, 09:23:54 pm
Hey!
Need help with the attached question :)
Post by: jakesilove on June 07, 2017, 09:25:33 pm
Hey!
Need help with the attached question :)
Hey! So, we need to convert the 4 into some log form. We do this by recalling that
Literally means a to the power of c equals b. So, we know that
As 2 to the power for four equals 16! You can confirm this on your calculator. We are then left with
as when you add two logs with the same bases, you multiple what is being 'logified'
Post by: jakesilove on June 07, 2017, 09:26:04 pm
Post by: RuiAce on June 07, 2017, 09:39:57 pm
Post by: winstondarmawan on June 07, 2017, 11:13:22 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/19048562_1800159110300546_1670682986_o.jpg?oh=6f18ef0fd9d20bab6585b02e4b0039c1&oe=593A3C20
Post by: RuiAce on June 07, 2017, 11:34:32 pm
Would appreciate help for the following quesiton. TIAThere are 3 questions. Which one are you having difficulty with?
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/19048562_1800159110300546_1670682986_o.jpg?oh=6f18ef0fd9d20bab6585b02e4b0039c1&oe=593A3C20
Post by: Wales on June 07, 2017, 11:45:20 pm
Would appreciate help for the following quesiton. TIA
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/19048562_1800159110300546_1670682986_o.jpg?oh=6f18ef0fd9d20bab6585b02e4b0039c1&oe=593A3C20
Going to assume it's 34 but post back if it isn't
34) i) Simultaneous equations. You let x^2-6x+8 equal to 4x-x^2 and solve. You get x=1 and x=4. You know x=4 is already an intercept so the other is x=1. Sub x=1 into either Y equation and you get y=3. Therefore P is (1,3)
ii) The integral of the top function take away bottom function between 4 and 1 should give you the answer.
Let me know if anything is wrong or unclear or if it was the other questions.
Post by: RuiAce on June 07, 2017, 11:48:27 pm
Post by: Wales on June 07, 2017, 11:56:01 pm
34i) you could probably get away with by just subbing x=1 into both equations and checking that in both cases y=3
True. I found the coordinates rather then proved it. Both works but yours is more efficient
Post by: RuiAce on June 08, 2017, 12:00:50 am
Some people will sub in both together and check that it works. This is not acceptable.
Post by: _____ on June 08, 2017, 03:51:44 pm
1. Are the only possible conditions for a point of inflexion y'' = 0 and y'' changes sign at the point OR y' = 0 and y' does not change sign at the point? If I'm told to find points of inflexion should I typically be trying the y'' condition first?
2. I had one of those questions where a diagram is shown and you have to find the maximum value of the area by changing x. In this question we're told to show that A = 2x(200-x^2)^(1/2) for a rectangle. No problem with that. But to find the value of x, there's a hint that you should differentiate A^2 instead which kind of makes sense as differentiating 2x(200-x^2)^(1/2) is difficult. What does not make sense to me is that I was able to differentiate A^2 and find the value of x for which A is maximum without having to do anything to the final output which was x = 10. Why does squaring the entire equation not change the required value of x in this situation? I would have thought that I'd have to find x^(1/2) to ensure it is converted back since I changed A into A^2. Let me know if I need to explain this better.
3. Question I can't do:
A cylinder of radius r cm and height h cm is inscribed in a cone with base radius 3 cm and height 10 cm. By using similar triangles, show that the height of the cylinder is h = 10 - (10/3)r.
I can't work out how to express the relationship between h/r and 10/3 and change it into that expression. Let me know if you need the diagram scanned.
Thanks!
Post by: RuiAce on June 08, 2017, 04:00:12 pm
Going through GAD :( so I've got a few things to clear up.1. Yes. Because we know that possible points of inflexion ONLY occur when y" = 0. That's literally the definition.
1. Are the only possible conditions for a point of inflexion y'' = 0 and y'' changes sign at the point OR y' = 0 and y' does not change sign at the point? If I'm told to find points of inflexion should I typically be trying the y'' condition first?
2. I had one of those questions where a diagram is shown and you have to find the maximum value of the area by changing x. In this question we're told to show that A = 2x(200-x^2)^(1/2) for a rectangle. No problem with that. But to find the value of x, there's a hint that you should differentiate A^2 instead which kind of makes sense as differentiating 2x(200-x^2)^(1/2) is difficult. What does not make sense to me is that I was able to differentiate A^2 and find the value of x for which A is maximum without having to do anything to the final output which was x = 10. Why does squaring the entire equation not change the required value of x in this situation? I would have thought that I'd have to find x^(1/2) to ensure it is converted back since I changed A into A^2. Let me know if I need to explain this better.
3. Question I can't do:
A cylinder of radius r cm and height h cm is inscribed in a cone with base radius 3 cm and height 10 cm. By using similar triangles, show that the height of the cylinder is h = 10 - (10/3)r.
I can't work out how to express the relationship between h/r and 10/3 and change it into that expression. Let me know if you need the diagram scanned.
Thanks!
y' = 0 only gives you a stationary point. Therefore it does not necessarily cover all points of inflextion. Note that not every point need be a horizontal point of inflexion
______________________________________________
This can also be represented graphically.
As an exercise, you may wish to sketch the cubic y=f(x), where f(x) = (x-1)(x-2)(x-4)+10 on graphing software
and also sketch y=[f(x)]2, and then compare the x-coordinates of the stationary points.
Post by: RuiAce on June 08, 2017, 04:09:46 pm
Post by: Wales on June 08, 2017, 06:53:18 pm
(http://uploads.tapatalk-cdn.com/20170608/86ba352a23a04ace76a1b68c60469d7d.jpg)
You have a whiteboard on demand for answering questions.
Nice.
Also, how much of a pain is it to code your responses the way you do? It's so much cleaner.
Post by: jamonwindeyer on June 08, 2017, 07:10:25 pm
You have a whiteboard on demand for answering questions.
Nice.
Also, how much of a pain is it to code your responses the way you do? It's so much cleaner.
It's called LaTex! It's not super hard to learn, a little bit of a learning curve but you'll pick it up fast, Rui wrote a guide on it here! ;D
Post by: Wales on June 08, 2017, 07:12:21 pm
It's called LaTex! It's not super hard to learn, a little bit of a learning curve but you'll pick it up fast, Rui wrote a guide on it here! ;D
I'll take a look :) Awesome.
Post by: RuiAce on June 08, 2017, 07:14:20 pm
Post by: jamonwindeyer on June 08, 2017, 07:28:08 pm
My whiteboard was a birthday present lol
I have at least half a dozen whiteboards scattered around for tutoring ;D
Post by: K9810 on June 08, 2017, 08:13:47 pm
How can I solve for the points of intersection between y=x^2 and y=e^x
If I let x^2=e^x, how do I solve for x?
Post by: jamonwindeyer on June 08, 2017, 08:16:46 pm
Hey!
How can I solve for the points of intersection between y=x^2 and y=e^x
If I let x^2=e^x, how do I solve for x?
Hey! There's no easy way to find points of intersection in that case (even I personally don't know the Maths behind it) - Perhaps the question would suit an approximation obtained by guess and check? :)
Post by: K9810 on June 08, 2017, 08:17:46 pm
Hey! There's no easy way to find points of intersection in that case (even I personally don't know the Maths behind it) - Perhaps the question would suit an approximation obtained by guess and check? :)
The question was to find the area between the curves, I guess I should solve the points of intersection graphically?
Post by: jakesilove on June 08, 2017, 08:22:27 pm
The question was to find the area between the curves, I guess I should solve the points of intersection graphically?
I like don't think there are two points of intersection. Are you sure that you've interpreted the question correctly?
Post by: RuiAce on June 08, 2017, 08:54:52 pm
Post by: K9810 on June 08, 2017, 08:57:35 pm
I like don't think there are two points of intersection. Are you sure that you've interpreted the question correctly?
Yeah, I've realised I done it wrong. Sorry about that!
Post by: jamonwindeyer on June 08, 2017, 09:08:22 pm
Isn't there a special function for it?
Post by: jakesilove on June 08, 2017, 09:09:17 pm
Yeah, I've realised I done it wrong. Sorry about that!
No problemo!
Post by: RuiAce on June 08, 2017, 09:09:37 pm
Isn't there a special function for it?That's not really considered algebraic anymore though.
That being said, for something like this I'm not aware of a 'function' that fills in the gap here
Post by: jamonwindeyer on June 08, 2017, 09:11:24 pm
That's not really considered algebraic anymore though.
That being said, for something like this I'm not aware of a 'function' that fills in the gap here
Yeah I've got no clue of details, friend of mine once told me it existed and never really questioned (soz for the non-2U reference)
Post by: RuiAce on June 08, 2017, 09:14:28 pm
Not gonna talk about here though and intimidate all the 2U kids. Looks like it needs some elementary complex analysis to go with it. Definitely not considered "algebraic"
Post by: _____ on June 08, 2017, 09:20:04 pm
(http://uploads.tapatalk-cdn.com/20170608/86ba352a23a04ace76a1b68c60469d7d.jpg)
Thanks a bunch!
Post by: IronBark on June 09, 2017, 09:53:59 pm
Would be awesome if you guys could assist me in these questions. This was part of a 2015 paper and there are no answers to them (as staff cant give the answers out) as the questions get pretty repetitive.
(http://i68.tinypic.com/67sens.jpg)
I have no idea what to do for 3 a) or 3 b).
For 4 I concluded that the length would be x+2, whilst the width remains x.
Then you would have to set x+4:x-3, would you then plus 3 to each side and solve for x? I am not sure what to do there.
For 5 a), when you sub P(-2), a = -2. I have already worked that out. So this is our new eqn: x^3-4x^2-2x+20, but how do we factorise from there to show that (x+2) is a factor?. I used symbolab https://www.symbolab.com/solver/step-by-step/factorise%20x%5E%7B3%7D-4x%5E%7B2%7D-2x%2B20 to help explain it but I still could not gather what it was talking about.
Cheers for the help
Post by: RuiAce on June 09, 2017, 10:07:51 pm
\(\sin 2\theta\) requires a double angle identity taught in 3U and I will not address it here.
_______________
Q5a) is Extension 1 material and I will not address it here.
Post by: RuiAce on June 09, 2017, 10:14:56 pm
Post by: IronBark on June 11, 2017, 05:35:00 pm
Thanks so much for all of that. Sorry half our class is 3 Unit so they try and make us learn it and integrate it into exams.
What about questions that say like:
Evaluate this arithmic series:
1+4+7+10....+991 - Ok I get that the a is one, the difference is 3 and the last number is 991, but what are they asking me to do? They havnt asked for a n value so what do I sub in for n?
Post by: jakesilove on June 11, 2017, 05:39:25 pm
Thanks so much for all of that. Sorry half our class is 3 Unit so they try and make us learn it and integrate it into exams.
What about questions that say like:
Evaluate this arithmic series:
1+4+7+10....+991 - Ok I get that the a is one, the difference is 3 and the last number is 991, but what are they asking me to do? They havnt asked for a n value so what do I sub in for n?
It's asking you to sum up those values, using the formula
or
Now, you have the first and last term, but you don't have how many terms in the series. You could just logic it out, or you could do it formally. Create the equation for each term in the series,
and then sub in T=991. Solve for n, and you have your term number! Then, use that in the sum equation :)
Post by: IronBark on June 11, 2017, 06:31:59 pm
With this geo series (2+4+8+16+..2^n) how would you evaluate this.. please show working and answer:
So I said that the eqn is 2 x 2^(n-1)... now what do i do?
Post by: RuiAce on June 11, 2017, 06:36:12 pm
Post by: 12070 on June 11, 2017, 06:48:11 pm
Which expression is equivalent to 4 + logx2 ?
(A) log2(2x)
(B) log2(16 + x)
(C) 4log2(2x)
(D) log2(16x)
Sorry for the copy and paste issue but it's from the 2016 HSC paper; Question 10
Post by: RuiAce on June 11, 2017, 06:57:28 pm
How do you work this style of question out? I feel like it should be easy I don't know which laws to use.Addressed in posts #1990 and #1991
Which expression is equivalent to 4 + logx2 ?
(A) log2(2x)
(B) log2(16 + x)
(C) 4log2(2x)
(D) log2(16x)
Sorry for the copy and paste issue but it's from the 2016 HSC paper; Question 10
Post by: MisterNeo on June 11, 2017, 07:24:04 pm
How do you work this style of question out? I feel like it should be easy I don't know which laws to use.
Which expression is equivalent to 4 + logx2 ?
(A) log2(2x)
(B) log2(16 + x)
(C) 4log2(2x)
(D) log2(16x)
Sorry for the copy and paste issue but it's from the 2016 HSC paper; Question 10
Here's the proof for the question.
Note: You wrote logx2 when it was log2x :)
(http://i.imgur.com/GJQQT1j.jpg)
Looking at the options of multiple choice can help identify which log laws you need to apply.
Hope this helps :D
Post by: 12070 on June 11, 2017, 08:08:03 pm
Here's the proof for the question.
Note: You wrote logx2 when it was log2x :)
(http://i.imgur.com/GJQQT1j.jpg)
Looking at the options of multiple choice can help identify which log laws you need to apply.
Hope this helps :D
Thanks for that. I guess the hard part for me at least was realising that 4= log16 base 2. Tricky
Post by: IronBark on June 11, 2017, 11:25:15 pm
I need help with this question: As dry air moves upward, it expands and cools. The temp at a height of 1 km = 10(deg) and temp at 3 km (-5(deg))
1. Assuming there is a constant rate of change of temp with respect to height, express the temp in terms of the height h
2. What is the meaning of the slope?
3. What is the meaning of the horizontal intercept of the line
Can you please show working....
for 1 would you assume something like T= H(10).. I dont know
Post by: MisterNeo on June 12, 2017, 12:01:56 am
Thank you so much for your guys help.
I need help with this question: As dry air moves upward, it expands and cools. The temp at a height of 1 km = 10(deg) and temp at 3 km (-5(deg))
1. Assuming there is a constant rate of change of temp with respect to height, express the temp in terms of the height h
2. What is the meaning of the slope?
3. What is the meaning of the horizontal intercept of the line
Can you please show working....
for 1 would you assume something like T= H(10).. I dont know
I haven't done this topic, but I think this is it. Never heard of "meaning of.." :P
(http://i.imgur.com/hZmmfi9.jpg)
8)
Post by: jamonwindeyer on June 12, 2017, 12:11:19 am
Thank you so much for your guys help.
I need help with this question: As dry air moves upward, it expands and cools. The temp at a height of 1 km = 10(deg) and temp at 3 km (-5(deg))
1. Assuming there is a constant rate of change of temp with respect to height, express the temp in terms of the height h
2. What is the meaning of the slope?
3. What is the meaning of the horizontal intercept of the line
Can you please show working....
for 1 would you assume something like T= H(10).. I dont know
Just to help with 2 and 3, I think it wants more physical interpretations:
- The slope of the line represents the change in temperature with temperature, how fast the temperature drops with altitude.
- The horizontal intercept represents the altitude where the temperature reaches zero!
Post by: IronBark on June 12, 2017, 12:49:29 am
I haven't done this topic, but I think this is it. Never heard of "meaning of.." :PYour a legend Cheers. Sorry with this exam, it does not include answers so i dont know If I am getting these right or not. These are some of the questions, Which i am not sure if are correct.
(http://i.imgur.com/hZmmfi9.jpg)
8)
1. (integral sign)4xcos(x^2+1)dx
So I put that it = ((4x)^2)/2 x sin((x^2+1)^2)/2 x 2x
= 2x^3sin(x^2+1)/2 + c
2. Anti derivative of f(x)?
(f(x)^(n+1))/n
3. f(x) =4(root)x+(1)/((2x)^2), find f'(x)
f(x)= (4x)^1/2 + 2x^-2
f'(x)= ((2x)^-1/2)-4x^-3
= (2/(root)x) - 4/x^3
4. Explain what the function notation g: [0,1] (-1,1) means:
As 0,1 approach the x and y axis, they intercept the boundaries of -1,1 though not being limited to
Post by: MisterNeo on June 12, 2017, 01:28:06 am
Your a legend Cheers. Sorry with this exam, it does not include answers so i dont know If I am getting these right or not. These are some of the questions, Which i am not sure if are correct.
1. (integral sign)4xcos(x^2+1)dx
So I put that it = ((4x)^2)/2 x sin((x^2+1)^2)/2 x 2x
= 2x^3sin(x^2+1)/2 + c
2. Anti derivative of f(x)?
(f(x)^(n+1))/n
3. f(x) =4(root)x+(1)/((2x)^2), find f'(x)
f(x)= (4x)^1/2 + 2x^-2
f'(x)= ((2x)^-1/2)-4x^-3
= (2/(root)x) - 4/x^3
4. Explain what the function notation g: [0,1] (-1,1) means:
As 0,1 approach the x and y axis, they intercept the boundaries of -1,1 though not being limited to
Hmm...your post is a bit sketchy but I'll try to interpret it.
(http://i.imgur.com/51rMVHa.jpg)
Not sure about Q4 though...
Post by: Joseph41 on June 12, 2017, 08:40:45 am
Keep being awesome, ATAR Notes.
Post by: RuiAce on June 12, 2017, 08:50:27 am
Your a legend Cheers. Sorry with this exam, it does not include answers so i dont know If I am getting these right or not. These are some of the questions, Which i am not sure if are correct.Whilst Q4 is in the VCE course it is NOT in the HSC course.
1. (integral sign)4xcos(x^2+1)dx
So I put that it = ((4x)^2)/2 x sin((x^2+1)^2)/2 x 2x
= 2x^3sin(x^2+1)/2 + c
2. Anti derivative of f(x)?
(f(x)^(n+1))/n
3. f(x) =4(root)x+(1)/((2x)^2), find f'(x)
f(x)= (4x)^1/2 + 2x^-2
f'(x)= ((2x)^-1/2)-4x^-3
= (2/(root)x) - 4/x^3
4. Explain what the function notation g: [0,1] (-1,1) means:
As 0,1 approach the x and y axis, they intercept the boundaries of -1,1 though not being limited to
It means that the function takes values from the domain \(0\le x \le 1\) and maps them over to the codomain \(-1< x < 1\)
Asking a ton of questions especially when it's continuous can be quite offputting. Please relax the speed that you ask the questions. Additionally, if it's too hard to type please resort to simply posting images. (You could also try learning LaTeX, but that may be too time consuming and posting images would be more efficient)
Post by: RuiAce on June 12, 2017, 09:06:18 am
Hmm...your post is a bit sketchy but I'll try to interpret it.By the way, regarding your Q1 I personally would skip line 2 That assumes that you could divide by 2x even when there was no x term there to begin with which is false. Jump straight to the answer, or write out a derivative and alternatively compare it to the integral.
(http://i.imgur.com/51rMVHa.jpg)
Not sure about Q4 though...
Note: reversing the chain rule technically isn't required until Extension 2, but it's nice to know.
Post by: jamonwindeyer on June 12, 2017, 10:35:44 am
Asking a ton of questions especially when it's continuous can be quite offputting. Please relax the speed that you ask the questions. Additionally, if it's too hard to type please resort to simply posting images. (You could also try learning LaTeX, but that may be too time consuming and posting images would be more efficient)
What Rui means to say is, if you ask a lot of questions, it might take a bit to get to them all. So if you're really struggling with a specific thing, posting that one thing (or perhaps one question to check your understanding) would probably get you an answer faster ;D
But please ask as many questions as you like! ;D Ultimately, if Rui doesn't answer, I will. Or MisterNeo will. Or jakesilove will. Or kiwiberry will. Or ellipse will. Or Shadowxo will. Point being, heaps of people around to answer questions (such is the awesomeness of ATAR Notes), we'll get to all of them eventually ;D
Post by: cxmplete on June 12, 2017, 11:25:57 am
how would you graph a x-2lnx graph, and are there particular steps to graphing it?
Post by: jamonwindeyer on June 12, 2017, 11:34:25 am
hi,
how would you graph a x-2lnx graph, and are there particular steps to graphing it?
Hey! if you were looking for just a rough sketch (without calculus), then your best bet would be to use superposition. Draw a graph of \(y=x\), and then another of \(y=-2\ln{x}\) (that's just a regular looking log graph, flipped vertically). Adding the two function values at any given \(x\)-coordinate will give you the \(y-\)-coordinate, if that makes sense. Remember, if one of the functions doesn't exist (like the logarithm doesn't for negative values), your new one won't either!!
Otherwise, you'd apply the normal techniques based on differentiation (turning points, inflexions, etc.) ;D
Post by: samthemanfromacan on June 12, 2017, 01:17:26 pm
(http://i66.tinypic.com/2n6x73l.jpg)
Post by: RuiAce on June 12, 2017, 01:36:02 pm
2009 Chatswood Girls Paper
(http://i66.tinypic.com/2n6x73l.jpg)
Numbers subject to inaccuracy as they were typed up rapidly. Method is all there.
Post by: BarnesK01 on June 12, 2017, 05:01:38 pm
Post by: RuiAce on June 12, 2017, 05:21:24 pm
Just need a help with question 7 if I could. For part i) I am using SAS, so far I have used Angle EBC = Angle CDB (base angle of an isoscles triangle) and BC (common).To show that BE = DC, consider the following facts:
- BA = CA
- BE = 1/2 BA
- DC = 1/2 CA
Remember, the question tells you information about midpoints.
Post by: Wales on June 12, 2017, 08:14:30 pm
i) Show that e^1-ln2=e/2
ii) Find the equation of the tangent to the curve y=e^1-4x at x=ln2/4
I've had a look at the solutions but cant for the love of God figure out why x=e/2/. I get x=ln2/4.
http://puu.sh/whQDT/8cba8b73fd.jpg - Solution
Post by: RuiAce on June 12, 2017, 08:17:45 pm
A tad embarrassed to be asking something here.. Friend asked me a question and I cannot seem to get the answer.
i) Show that e^1-ln2=e/2
ii) Find the equation of the tangent to the curve y=e^1-4x at x=ln2/4
I've had a look at the solutions but cant for the love of God figure out why x=e/2/. I get x=ln2/4.
http://puu.sh/whQDT/8cba8b73fd.jpg - Solution
For the sake of clarity, please use brackets if you intend to describe e^(1-4x)
Post by: Wales on June 12, 2017, 09:14:29 pm
For the sake of clarity, please use brackets if you intend to describe e^(1-4x)
I understand that. What I don't understand is that in the final solution using the point gradient formula the X value is e/2. I got the same X value as you stated.
Apologies for the lack of brackets. I told my friend off for the same thing but forgot to correct it when copying the question over.
Post by: jamonwindeyer on June 12, 2017, 09:26:36 pm
I understand that. What I don't understand is that in the final solution using the point gradient formula the X value is e/2. I got the same X value as you stated.
Apologies for the lack of brackets. I told my friend off for the same thing but forgot to correct it when copying the question over.
It's an error - You are finding the equation of the tangent at \(x=\frac{\ln{2}}{4}\), so that has to be the x-value! ;D
Post by: RuiAce on June 12, 2017, 09:48:13 pm
Post by: Wales on June 12, 2017, 10:50:32 pm
It's an error - You are finding the equation of the tangent at \(x=\frac{\ln{2}}{4}\), so that has to be the x-value! ;D
At last I can rest :D Goddamn I spent far too long trying to figure that out,
Cheers
Post by: IronBark on June 13, 2017, 06:55:47 pm
I am really sorry for this but I need help with these two questions. My maths exam is tomorrow, so on the bright side, after today you will never hear from me again :).
(http://i67.tinypic.com/j61p2f.jpg)
a) I evaluated that 30(deg)=1/4 though I dont know what to do with the squaring...
b) I graphed but dont know how to solve or how to answer it.
Thanks Rui,
Im so sorry for all the trouble mate.
Post by: jamonwindeyer on June 13, 2017, 07:38:15 pm
Hi RuiAce,
I am really sorry for this but I need help with these two questions. My maths exam is tomorrow, so on the bright side, after today you will never hear from me again :).
(http://i67.tinypic.com/j61p2f.jpg)
a) I evaluated that 30(deg)=1/4 though I dont know what to do with the squaring...
b) I graphed but dont know how to solve or how to answer it.
Thanks Rui,
Im so sorry for all the trouble mate.
Don't be sorry friend! We'd love to keep hearing from you after your exam - We love having people around ;D we start that first one by taking the square root of both sides:
So this is just now two regular trig equations, \(\sin{x}=\frac{1}{2}\) and \(\sin{x}=-\frac{1}{2}\), each of which has two solutions you can find the normal way :)
Let me draw a graph for that second one, just a sec...
Post by: jamonwindeyer on June 13, 2017, 07:47:57 pm
Okay, so the graph above should match yours. Now, a few things we notice:
- The graphs intersect at two points in the domain we care about. The x-coordinates of these are actually the two solutions to \(\sin{x}=\frac{1}{2}\) we found in the previous problem, namely: \(\frac{\pi}{6},\frac{5\pi}{6}\).
- Between those two points, the sine curve is above the line. Everywhere else, it is below.
And that is the answer. We want when \(\sin{x}\le\frac{1}{2}\), when the line is below the curve. So we define it with two regions:
Read this through once or twice, hopefully it makes sense - Let me know! :)
Post by: Wales on June 13, 2017, 09:30:44 pm
Don't be sorry friend! We'd love to keep hearing from you after your exam - We love having people around ;D we start that first one by taking the square root of both sides:
So this is just now two regular trig equations, \(\sin{x}=\frac{1}{2}\) and \(\sin{x}=-\frac{1}{2}\), each of which has two solutions you can find the normal way :)
Let me draw a graph for that second one, just a sec...
*1 hour later*
Try out desmos to graph if you're struggling :)
http://puu.sh/wiORK/8df6de2f27.png
There's a sketch of y=1/2 and y=sinx between 0 and 2pi.
Post back if you need anymore help :)
Post by: jamonwindeyer on June 13, 2017, 09:50:49 pm
*1 hour later*
Try out desmos to graph if you're struggling :)
http://puu.sh/wiORK/8df6de2f27.png
There's a sketch of y=1/2 and y=sinx between 0 and 2pi.
Post back if you need anymore help :)
I think you might have missed the post I made just above yours, hidden on a new page ;) love your work though Wales, as always ;D
Post by: Wales on June 13, 2017, 09:51:51 pm
I think you might have missed the post I made just above yours, hidden on a new page ;) love your work though Wales, as always ;D
Gah D: I swear you hadn't responded yet ):
When you think you're one step ahead but you're actually 2 hours late.
Post by: jamonwindeyer on June 13, 2017, 10:02:54 pm
Gah D: I swear you hadn't responded yet ):
When you think you're one step ahead but you're actually 2 hours late.
#jamonwillnevergodown
Post by: Wales on June 13, 2017, 10:05:31 pm
#jamonwillnevergodown
Soon...Time to sharpen my 2U skills and help out more :P I'll catch up man...one day
Post by: Aaron12038488 on June 14, 2017, 10:38:08 pm
Find the equation of the line through the point of intersection of 2y - 3 = 5x and x+y+1 =0 which is perpendicular to 4x+ 2y-3 = 0.
thanks
Post by: RuiAce on June 14, 2017, 10:44:56 pm
help
Find the equation of the line through the point of intersection of 2y - 3 = 5x and x+y+1 =0 which is perpendicular to 4x+ 2y-3 = 0.
thanks
Alternatively use the k method. But in all honesty, until first year linear algebra that method is a bit of a mystery and this is more intuitive
Post by: katnisschung on June 15, 2017, 07:26:55 pm
i did a and then all i could do for be was figure out that Ao and A were the same at t=0 and
A=1/2Ao e^ -5750k (probably wrong becos there are too many unknowns) haha
if anyone could show the full step by step working out that would be awesome thanks!
Post by: RuiAce on June 15, 2017, 07:32:26 pm
hey I'm stuck on b..
i did a and then all i could do for be was figure out that Ao and A were the same at t=0 and
A=1/2Ao e^ -5750k (probably wrong becos there are too many unknowns) haha
if anyone could show the full step by step working out that would be awesome thanks!
Post by: RuiAce on June 15, 2017, 08:43:45 pm
Hey guys, I'm up to graphing trigonometric functions in class right now, and I'm confused as how to sketch them when they're added? For example, how do you sketch functions like y=sinx+sin2x and, on a separate axes, and y=cosx+sinx?Refer to post #2042 by Jamon for similar reasons
Post by: f_tan on June 15, 2017, 10:11:25 pm
(http://i.imgur.com/FNRtbqc.png)
Thanks!
Post by: jamonwindeyer on June 15, 2017, 10:20:43 pm
can anyone help me with this question?
(http://i.imgur.com/FNRtbqc.png)
Thanks!
Hey!! Would be happy to help, but this is a Extension 1 question. Still might be good to see though, not really that much more difficult than the 2U stuff and no new knowledge necessary, keen? :)
Post by: f_tan on June 15, 2017, 10:32:10 pm
Hey!! Would be happy to help, but this is a Extension 1 question. Still might be good to see though, not really that much more difficult than the 2U stuff and no new knowledge necessary, keen? :)
Sorry, I didn't realise! I've just figured out how to answer the question though, but thank you!
Post by: jamonwindeyer on June 15, 2017, 10:40:41 pm
Sorry, I didn't realise! I've just figured out how to answer the question though, but thank you!
Well done!! That's alright, the ones with the purple in that textbook are all Extension questions ;D that specific question type is a good one for you to tackle as a 2U student though! ;D
Post by: katnisschung on June 17, 2017, 12:06:36 pm
for b)
runthrough of my working out...files too big
i subbed in Q=1/5Qo
to find t cancelled out the Qo from both sides
used ln so that i could simpligy to find t and got
t=(ln1/5)/4/10^-4 (k found from previous answer)
i hope this makes sense or otherwise if someone could just post their working out...cos i got the magnitude right
just a neg??
Post by: jamonwindeyer on June 17, 2017, 12:16:38 pm
so i have no idea why i'm getting a negative... i know its not right cos its the time for decay which can't be neg
for b)
runthrough of my working out...files too big
i subbed in Q=1/5Qo
to find t cancelled out the Qo from both sides
used ln so that i could simpligy to find t and got
t=(ln1/5)/4/10^-4 (k found from previous answer)
i hope this makes sense or otherwise if someone could just post their working out...cos i got the magnitude right
just a neg??
You definitely divided by \(-k\) in that last step? Your denominator that you put into the calculator should be negative ;D
Post by: katnisschung on June 17, 2017, 12:22:46 pm
Post by: kylesara on June 17, 2017, 02:26:06 pm
The sum of the first 3 terms of an A.P is 33 and the sum of the seventh and eighth term is 49. Find the first term and common difference.
Thanks!
Post by: RuiAce on June 17, 2017, 02:29:44 pm
Hi could I please have help with this question
The sum of the first 3 terms of an A.P is 33 and the sum of the seventh and eighth term is 49. Find the first term and common difference.
Thanks!
Post by: katnisschung on June 19, 2017, 03:08:57 pm
Post by: jakesilove on June 19, 2017, 03:14:23 pm
stuck on b thanks
We turn everything into tans
Which gets to our answer
Post by: katnisschung on June 19, 2017, 03:14:52 pm
my answer... i need help in simplifying this further i think
q & ans attached
my answer... also i swear i remember my teacher telling me never to expand/simplify
the denominator for quotient rule is this true?
Post by: jakesilove on June 19, 2017, 03:25:41 pm
need help in differentiating this
my answer... i need help in simplifying this further i think
q & ans attached
my answer... also i swear i remember my teacher telling me never to expand/simplify
the denominator for quotient rule is this true?
Using the chain rule. No real need to simplify further.
using Quotient rule
By trig identities. Remember that
Post by: katnisschung on June 19, 2017, 03:30:10 pm
Post by: jakesilove on June 19, 2017, 03:32:54 pm
thanks jake :) seems that trig identities keep tripping me up. time to memorise them
Definitely need to memorise them! At least, memorise the above one, because you can derive all the others from there :)
Post by: katnisschung on June 20, 2017, 07:04:44 pm
i looked at the answers and it just integrates by the x-axis but isn't the area
bounded by the y-axis?
yeah i see that my perspective is flawed cos i can't see any way to rearrange the first equation in terms of x
but the area is bounded by the y-axis!! :(
confused. hope someone can make sense out of this.
Post by: RuiAce on June 20, 2017, 07:45:29 pm
so can somebody pls explain to me how this is not integrating by the y-axis...
i looked at the answers and it just integrates by the x-axis but isn't the area
bounded by the y-axis?
yeah i see that my perspective is flawed cos i can't see any way to rearrange the first equation in terms of x
but the area is bounded by the y-axis!! :(
confused. hope someone can make sense out of this.
Post by: itssona on June 21, 2017, 09:20:08 am
how would i graph y=5cos4x
I know amplitude is 5 and the period is pi/2 but i dont get how to label the x axis in terms of radians and then graph it
Thank you!
Post by: RuiAce on June 21, 2017, 09:27:45 am
HeeySketch the typical cosine curve, but then label your max and min values as 5 and -5 instead of 1 and -1. Then, you can just label the x-axis as normal
how would i graph y=5cos4x
I know amplitude is 5 and the period is pi/2 but i dont get how to label the x axis in terms of radians and then graph it
Thank you!
cos(x) has period 2pi, intercepts π/2,3π/2,...
So noting the period of Pi/2 our intercepts are π/8,3π/8,...
Post by: Aussie1Italia2 on June 21, 2017, 09:40:18 am
I need help sketching for 0<= x <= PI
y=7sinx/2
Thank you for the help.
Post by: MisterNeo on June 21, 2017, 09:41:53 am
Heey
how would i graph y=5cos4x
I know amplitude is 5 and the period is pi/2 but i dont get how to label the x axis in terms of radians and then graph it
Thank you!
Consine graphs like 5cos4x start at the amplitude value on the positive y-axis, which would be 5.
(http://i.imgur.com/ySS2NPc.jpg)
The first time it touches the x-axis will be (pi/8,0). After that, you repeat with the Pi/2 periods.
The easy way to graph is to remember:
Get the coefficient of the x (4) and multiply it by 2, then divide pi by it.
That is your first x-value for a cosine graph.
Hope this helps ;D
Post by: RuiAce on June 21, 2017, 09:51:50 am
Hello,This question is very similar to the one posted right before it, so all I'm gonna do is copy and paste the response and change numbers.
I need help sketching for 0<= x <= PI
y=7sinx/2
Thank you for the help.
_______
Sketch the typical sine curve, but then label your max and min values as 7 and -7 instead of 1 and -1. Then, you can just label the x-axis as normal
sin(x) has period 2π, intercepts 0,π,2π,...
So noting the period of 4π our intercepts are 0,2π,4π,...
Post by: Fahim486 on June 21, 2017, 04:08:50 pm
Post by: kiwiberry on June 21, 2017, 05:31:29 pm
Hey need help with this question. Thanks!!!
Given that population is growing at a rate proportional to the population:
then
Where \(P_o\) is the initial population. When the population increases by 20%, it will be 1.2 times the initial population. Therefore, when \(t=3, P=1.2P_o\)
Similarly, when the population has doubled, \(P=2P_o\). Sub this in and solve to find t :)
Post by: julia_warren13 on June 21, 2017, 05:35:36 pm
Hey need help with this question. Thanks!!!
Hopefully this makes sense! Let me know if you need me to explain it
P = Aekt
120% x A = Ae3k
1.2 = e3k
ln1.2 = lne3k
ln1.2 = 3k
k = 0.060773... (save in calculator memory!!)
200% x A = Aekt
2 = ekt
ln2 = lnekt
ln2 = kt
substituting in k = 0.060773...
t = 11.4 days
120% x A = Ae3k
1.2 = e3k
ln1.2 = lne3k
ln1.2 = 3k
k = 0.060773... (save in calculator memory!!)
200% x A = Aekt
2 = ekt
ln2 = lnekt
ln2 = kt
substituting in k = 0.060773...
t = 11.4 days
Post by: jamonwindeyer on June 21, 2017, 08:17:19 pm
Post by: katnisschung on June 25, 2017, 01:31:33 pm
14) in a particular game of chance, the probability of winning the only prize in any draw is 1 in 50.
ii) if it is to be 99% certain that the prize will have been won, how many consecutive draws must
be made?
so i already saw the fallacy of my reasoning when trying to prove it but i don't understand why u take the probability of not winning...
my reasoning
p(winning)=0.02
(0.02)^n=0.99
and used logs to solve but i could already see before doing it i would get a very small number (below 1) for the number
of draws which obviously doesn't make sense.
Answers:
p(not winning)=49/50 = 0.98
(0.98)^n ≤ 0.01
so could someone explain why we use the values for not winning to solve this question? I'm so lost
19) a game involving a single dice has the following rules
a player throws two ordinary dice repeatedly until the sum of the two numbers is either
a 6 or a 8. If the sum is a 8, the player wins. If the sum is 6, the player loses.
If the sum is any other number, the player continues to throw until it is 6 or 8.
i) Show the probability that the player wins on the first throw of the dice is 5/36
so i don't understand why it is 5/36.
technically isnt the probability= probability he throws an 8 and doesn't throw a 6 for every throw
so for the first throw isn't it 5/36 x 31/36
15) A local high school has a student population comprising of 52% female and 48% male.
a survey is carried out and two students are randomly selected to take part.
i) find the probability both students are male
so considering that one student is picked and another is picked without replacement
would this not affect the probability for the gender of the second student?
(becos one is removed so the percentage of females would increase and the males would decrease....
its probably incorrect but i was thinking in terms of fractions assuming the student body had 100 people or is it too large that taking one student out would have a minute effect on the percentages?)
thanksss :)
Post by: _____ on June 25, 2017, 03:06:06 pm
Post by: MisterNeo on June 25, 2017, 03:32:25 pm
Could you guys confirm whether differentiation and integration of a^x (a is any integer so not e) is in the 2 unit syllabus or not? The Cambridge math textbook covers it but it's not on the formula sheet and we didn't do it in class. Thanks.
I'm pretty sure it is since we did it in class a few weeks ago.
The easy way to do it is by taking log to both sides, then getting it as an exponential.
The same applies with integration:
Post by: RuiAce on June 25, 2017, 04:38:05 pm
Could you guys confirm whether differentiation and integration of a^x (a is any integer so not e) is in the 2 unit syllabus or not? The Cambridge math textbook covers it but it's not on the formula sheet and we didn't do it in class. Thanks.
Post by: _____ on June 25, 2017, 09:53:54 pm
Thanks guys.
So Rui you can quote x = e^lnx and then move to say 2^x = e^(xln2), correct? Just want to make sure I'm using that result correctly.
Post by: RuiAce on June 25, 2017, 09:57:06 pm
Thanks guys.
So Rui you can quote x = e^lnx and then move to say 2^x = e^(xln2), correct? Just want to make sure I'm using that result correctly.
Post by: RuiAce on June 25, 2017, 10:04:41 pm
hi i have 3 qs on probabilityThe prize may have been won on the first, second, third, fourth or fifth and so on (basically n-th) draw. We don't know what that is.
14) in a particular game of chance, the probability of winning the only prize in any draw is 1 in 50.
ii) if it is to be 99% certain that the prize will have been won, how many consecutive draws must
be made?
so i already saw the fallacy of my reasoning when trying to prove it but i don't understand why u take the probability of not winning...
my reasoning
p(winning)=0.02
(0.02)^n=0.99
and used logs to solve but i could already see before doing it i would get a very small number (below 1) for the number
of draws which obviously doesn't make sense.
Answers:
p(not winning)=49/50 = 0.98
(0.98)^n ≤ 0.01
so could someone explain why we use the values for not winning to solve this question? I'm so lost
19) a game involving a single dice has the following rules
a player throws two ordinary dice repeatedly until the sum of the two numbers is either
a 6 or a 8. If the sum is a 8, the player wins. If the sum is 6, the player loses.
If the sum is any other number, the player continues to throw until it is 6 or 8.
i) Show the probability that the player wins on the first throw of the dice is 5/36
so i don't understand why it is 5/36.
technically isnt the probability= probability he throws an 8 and doesn't throw a 6 for every throw
so for the first throw isn't it 5/36 x 31/36
15) A local high school has a student population comprising of 52% female and 48% male.
a survey is carried out and two students are randomly selected to take part.
i) find the probability both students are male
so considering that one student is picked and another is picked without replacement
would this not affect the probability for the gender of the second student?
(becos one is removed so the percentage of females would increase and the males would decrease....
its probably incorrect but i was thinking in terms of fractions assuming the student body had 100 people or is it too large that taking one student out would have a minute effect on the percentages?)
thanksss :)
So a direct approach would mean we'd have to add the probability of winning on the first draw, and on the second, AND on the third AND keep going until we finally get there. This is too much effort.
Hence we consider the complementary event. This is the probability that given n draws, the prize hasn't been won yet at all. Unlike the above, the one and ONLY way this can happen is if it doesn't get won, and doesn't get won again, and doesn't get won again and so on. Because we want a 99% confidence that the prize has been won, we want the (complement) probability of not winning the prize at all to be less than 1%.
Hence \(0.98^n \le 0.01\)
_____________________________________
He throws two dice at once, and the numbers shown must add up to 8. I'm not sure what you mean here because it appears as though you're implying that the two numbers can sum to both 8 and 6 at once, which obviously it can not.
Listing out the outcomes, the number sum to 8 only when you roll (6,2), (5,3), (4,4), (3,5) or (2,6) i.e. 5 possible ways. Hence 5/36.
_____________________________________
Yes. Always assume that the population size is too large so that taking one out affects the percentage only negligibly, so we ignore it. Only when the populations are given in proportions (or here, percentages) do we do this.
Post by: laurenf58 on June 27, 2017, 06:12:39 pm
Can I please have help with these two questions? I know how to do them but keep getting the wrong answers and want to see where i'm going wrong.
1) A sum of $1000 is invested at the end of each year for 22 years, at 9% pa. Find the amount of superannuation available at the end of 22 years.
2) What is the interest rate if a $5000 investment is worth $6511.30 after 6 years?
Thanks!
Post by: jamonwindeyer on June 27, 2017, 06:27:52 pm
Hey,
Can I please have help with these two questions? I know how to do them but keep getting the wrong answers and want to see where i'm going wrong.
1) A sum of $1000 is invested at the end of each year for 22 years, at 9% pa. Find the amount of superannuation available at the end of 22 years.
2) What is the interest rate if a $5000 investment is worth $6511.30 after 6 years?
Thanks!
Hey Lauren! Just on the way home ATM so can't do any LaTex, could you post a pic of your working for them? I reckon it is probably a really little thing, I'll try and spot the error for you!
Post by: MisterNeo on June 27, 2017, 07:06:53 pm
Hey,
Can I please have help with these two questions? I know how to do them but keep getting the wrong answers and want to see where i'm going wrong.
1) A sum of $1000 is invested at the end of each year for 22 years, at 9% pa. Find the amount of superannuation available at the end of 22 years.
2) What is the interest rate if a $5000 investment is worth $6511.30 after 6 years?
Thanks!
For Question 2...
Post by: laurenf58 on June 27, 2017, 09:56:41 pm
Hey Lauren! Just on the way home ATM so can't do any LaTex, could you post a pic of your working for them? I reckon it is probably a really little thing, I'll try and spot the error for you!
So far, i have a = 1000 x 1.09, r=1.09 and n=22
Not sure if i've got my numbers mixed up or if the textbook has the wrong answer!
Thanks!
Post by: laurenf58 on June 27, 2017, 09:57:44 pm
For Question 2...
Thanks so much!
Post by: RuiAce on June 27, 2017, 10:00:35 pm
So far, i have a = 1000 x 1.09, r=1.09 and n=22
Not sure if i've got my numbers mixed up or if the textbook has the wrong answer!
Thanks!
Post by: Thebarman on July 03, 2017, 12:00:09 pm
1. Given log7(2) = 0.36 and log7(5) = 0.86, find log7(35) and log7(98).
The answers were 1.83 and 2.36, respectively.
Post by: RuiAce on July 03, 2017, 12:06:26 pm
Hey guys, I'm doing some catchup work right now and I'm struggling with the following questions. Any help would be appreciated.The question (very horribly) made a typo in that log7(5) should be 0.83, not 0.86
1. Given log7(2) = 0.36 and log7(5) = 0.86, find log7(35) and log7(98).
The answers were 1.83 and 2.36, respectively.
Post by: Thebarman on July 03, 2017, 12:36:13 pm
Do you mind also helping me with this question? Use log laws to evaluate log5(50) - log5(2).
There shouldn't be any errors this time ;D
Post by: RuiAce on July 03, 2017, 12:43:06 pm
I didn't even notice the mistake. I'm so sorry about that!
Do you mind also helping me with this question? Use log laws to evaluate log5(50) - log5(2).
There shouldn't be any errors this time ;D
Post by: Thebarman on July 03, 2017, 01:23:10 pm
Any chance you could also solve this? Find the volume of the solid formed when the curve y=e^(-x) + 1 is rotated about the x-axis from x=1 to x-2
Post by: RuiAce on July 03, 2017, 01:49:41 pm
Thanks again, this is finally starting to make sense.Assuming a typo was made in "x-2"
Any chance you could also solve this? Find the volume of the solid formed when the curve y=e^(-x) + 1 is rotated about the x-axis from x=1 to x-2
Post by: Thebarman on July 03, 2017, 02:25:13 pm
Post by: RuiAce on July 03, 2017, 02:51:13 pm
Ah, when I first converted the equation to equal y^2, I had changed the entire equation, rather than just squaring the lot. I tried to integrate your answer from there as a whole, and I got ((e^-x +1)^3)/3. After subbing in the values, I'm still getting a negative answer (-0.37). Not sure where I'm going wrong
Post by: katnisschung on July 04, 2017, 12:36:38 pm
58. The limiting sum of the series
1-3t + 9t^2 - 27t^3 +.... is 2/3
Find the value of t
69. A pattern has been created that consists of a series of rectangles with a common centre. The first (innermost) rectangle is 4 units long and 2 units wide. The second rectangle is 2 units longer and 2 units wider than the first rectangle and every other rectangle is 2 units longer and 2 units wider than the previous one. The space between these rectangles is shaded.
The area inside the first rectangle,
A1 units is given by
A1= 4x2
=8
A2 is the area between the second and the third rectangle and so on.
A) show that An= 8n
B) find the first term and common difference of the arithmetic series formed by summing these areas
Post by: RuiAce on July 04, 2017, 01:22:38 pm
Hey hey need help on 2 series questions! Id really appreciate if u could show full working Thanks :)
58. The limiting sum of the series
1-3t + 9t^2 - 27t^3 +.... is 2/3
Find the value of t
Someone else will get to the second question if they're faster than me, but not too sure what the difficult part was here?
Post by: jamonwindeyer on July 04, 2017, 02:10:37 pm
69. A pattern has been created that consists of a series of rectangles with a common centre. The first (innermost) rectangle is 4 units long and 2 units wide. The second rectangle is 2 units longer and 2 units wider than the first rectangle and every other rectangle is 2 units longer and 2 units wider than the previous one. The space between these rectangles is shaded.
The area inside the first rectangle,
A1 units is given by
A1= 4x2
=8
A2 is the area between the second and the third rectangle and so on.
A) show that An= 8n
B) find the first term and common difference of the arithmetic series formed by summing these areas
Hey! So we know \(A1\), we get \(A2\) by subtracting the area of the smaller rectangle from the larger one (the area in between). So:
Similarly (notice the dimensions both jump by 2 each time):
So the series formed is \(8,16,24...\), which is given by \(A_n=8n\). By inspection, \(a=8\) and \(d=8\) ;D hope it helps!
Post by: phebsh on July 04, 2017, 05:47:53 pm
Post by: RuiAce on July 04, 2017, 05:51:33 pm
Hey there, my teacher said that I should practice trial papers considering they are more difficult than HSC and that I should be aiming for 40% on my trial paper. However, my tutor said to stick to past HSC papers... What do you think?I don't actually know why your tutor suggests this. I tutor as well and I agree with your teacher.
Unless your school writes their own past papers and they're literally questions stolen from HSC papers you should be doing trials. Trials best reflect the trials whereas the HSC past papers best reflects the HSC.
(If your school writes their own papers you should keep doing their's though. And if they use company papers you can always ask if they have past papers.)
Post by: jamonwindeyer on July 04, 2017, 09:06:26 pm
Hey there, my teacher said that I should practice trial papers considering they are more difficult than HSC and that I should be aiming for 40% on my trial paper. However, my tutor said to stick to past HSC papers... What do you think?
Just to add, you should definitely be (eventually, after some or maybe a lot of hard work) aiming for more than 40% in your Trial papers! Everyone is different obviously, but you should never be aiming for a mark as low as 40%! Aim high! ;D
Post by: hansolo9 on July 04, 2017, 11:40:16 pm
I really struggle with these maximum and minimum applications..
Thanks!! :)
Post by: legorgo18 on July 04, 2017, 11:48:08 pm
Hi can someone explain how to do part b?
I really struggle with these maximum and minimum applications..
Thanks!! :)
Hey! For part b you differentiate the function from Part A, then make it=0 for stationary points. After solving do a sign table to check for min/max and you are good to go!
Post by: hansolo9 on July 04, 2017, 11:59:42 pm
Hey! For part b you differentiate the function from Part A, then make it=0 for stationary points. After solving either do a sign table to check for min/max and you are good to go!
Oh wow! It was that simple lol
Thank you so much!!
Post by: RuiAce on July 05, 2017, 12:03:31 am
Oh wow! It was that simple lolThat process is (a bit boring but) very standard. Was there something else you thought you had to do?
Thank you so much!!
Post by: hansolo9 on July 05, 2017, 12:11:09 am
That process is (a bit boring but) very standard. Was there something else you thought you had to do?
I thought you needed to do something like sub into the area formula.
I struggle with the volume ones where you have variable h and variable r, then find max V of a cone.
Post by: RuiAce on July 05, 2017, 12:17:39 am
I thought you needed to do something like sub into the area formula.Max/Min problems are always related to the first derivative.
I struggle with the volume ones where you have variable h and variable r, then find max V of a cone.
The concepts were more formally taught in the geometry of the derivative (as the Cambridge textbook calls it).
Post by: hansolo9 on July 05, 2017, 05:51:05 pm
Am I wrong, or is the question wrong?
Post by: pikachu975 on July 05, 2017, 05:56:48 pm
This is a really dumb question, isn't the volume 360 ?
Am I wrong, or is the question wrong?
It is 420.
The area of the 2 top triangular prisms: 2 (1/2 * 6 * 6 * 5)
Area of bottom 2: 2 (1/2 * 8 * 6 * 5)
Post by: chelseam on July 05, 2017, 05:59:18 pm
This is a really dumb question, isn't the volume 360 ?Hey! This is what I got:
Am I wrong, or is the question wrong?
Each of the triangles at the top would be 6x6x0.5x5, and since there are 2 the total volume is 180.
The bottom part of the shape is missing the perpendicular height but you can find that by using pythag, and that will give you 8cm. So the volume of the bottom part will be 8x12x0.5x5, which gives you 240.
180+240=420 :)
Post by: hansolo9 on July 05, 2017, 06:02:58 pm
Hey! This is what I got:
Each of the triangles at the top would be 6x6x0.5x5, and since there are 2 the total volume is 180.
The bottom part of the shape is missing the perpendicular height but you can find that by using pythag, and that will give you 8cm. So the volume of the bottom part will be 8x12x0.5x5, which gives you 240.
180+240=420 :)
It is 420.
The area of the 2 top triangular prisms: 2 (1/2 * 6 * 6 * 5)
Area of bottom 2: 2 (1/2 * 8 * 6 * 5)
OHHH I SEE NOW. There's a pythagoras thing involved, didn't see that.
Thanks heaps :)
Post by: hansolo9 on July 05, 2017, 06:42:16 pm
(The correct answer is B)
Post by: RuiAce on July 05, 2017, 06:47:46 pm
How would I do this?This question lacks information. I believe that they forgot to state that we have an isosceles triangle up the top. That was the only way I was able to make it work.
(The correct answer is B)
Post by: Opengangs on July 05, 2017, 07:06:17 pm
Post by: Fahim486 on July 05, 2017, 09:17:13 pm
Do it matter how many sig figs you put your answer to because the question doesn't really indicate how many sig figs or decimal places to put your answer
Post by: jamonwindeyer on July 05, 2017, 09:18:20 pm
For C) ii. I only put my answer as -0.00425 but the BOSTES answer says -0.004252436
Do it matter how many sig figs you put your answer to because the question doesn't really indicate how many sig figs or decimal places to put your answer
Hey! Your answer would definitely be deemed correct - They'd just be providing that to indicate how many places would be likely to appear on your calculator, at a guess? :)
Post by: Kekemato_BAP on July 05, 2017, 09:29:07 pm
On the unity circle and the x-y graph?
Post by: pikachu975 on July 06, 2017, 01:28:59 am
Yup, the only way I could make it work is that we have an iso. triangle.
You can just angle chase to get the straight angle on the right to be 50 + x + 130 - x and just sub in each option to see which gives 180.
Post by: RuiAce on July 06, 2017, 08:36:20 am
You can just angle chase to get the straight angle on the right to be 50 + x + 130 - x and just sub in each option to see which gives 180.Clarify? Which angles are you chasing?
Also 50+x+130-x is always 180
Post by: michaelalt on July 06, 2017, 09:56:44 am
Post by: RuiAce on July 06, 2017, 10:14:44 am
hello ! does anyone know how to do question 18 & 19? its under exponential growth and decay, but I don't know how to do it when no initial value is given.. thank you!
Alternatively, we may W.L.O.G. suppose the initial value is 1.
Post by: jamonwindeyer on July 06, 2017, 10:15:11 am
hello ! does anyone know how to do question 18 & 19? its under exponential growth and decay, but I don't know how to do it when no initial value is given.. thank you!
Hey, to do this, define an initial amount as \(A_0\). Then you consider the amount at any given time as some multiple of \(A_0\), so for example, 10% more would be \(1.1A_0\). 50% more would be \(1.5A_0\) ;D
Doing it this way, you'll get the \(A_0\) cancelling out in your equations - Give it a go! ;D
Post by: 12070 on July 06, 2017, 12:01:55 pm
Post by: RuiAce on July 06, 2017, 12:12:46 pm
If you have 4m(m-15)< 0, why does the inequality change so m>0? It makes sense if you logically think about it but how do you know which inequality changes if you couldn't work it out?
Post by: 12070 on July 06, 2017, 12:27:21 pm
So when I sketched it I got 0<m<15 because it isn't it basically asking when is the function below the x axis?
Post by: RuiAce on July 06, 2017, 12:29:17 pm
So when I sketched it I got 0<m<15 because it isn't it basically asking when is the function below the x axis?Misread the symbol earlier for some reason. Fixing.
Post by: _____ on July 06, 2017, 10:32:46 pm
1. "Evaluate to 2 significant figures: sin (1/SQRT2)"
I look at this and think of using the sin^-1 button to work out the angle since there's no use of pi and the numbers are found on the sides of the standard 45 degree triangle. So I gave the answer as 45 degrees but it's actually 0.65 (considering it as radians). Does it need to say sin^-1 for it to refer to the sides rather than a radian measure?
2. "A circle has a circumference of 12pi cm. If an angle of pi/3 is subtended at the centre of the circle, find the exact area of the minor segment."
Their answer (having found radius to be 6cm):
(http://i.imgur.com/t8vE3Ah.png)
Why introduce sine rather than using the radian measure of the angle already given? I think I'm missing something here.
Thanks!
Post by: RuiAce on July 06, 2017, 10:35:32 pm
Going through past trials - I've given the answers so you don't need to work them out. I just don't get why they're right.You typed sin and not sin^-1. If it says sin you keep as sin. Of course, you may have typo'd it and actually meant sin^-1
1. "Evaluate to 2 significant figures: sin (1/SQRT2)"
I look at this and think of using the sin^-1 button to work out the angle since there's no use of pi and the numbers are found on the sides of the standard 45 degree triangle. So I gave the answer as 45 degrees but it's actually 0.65 (considering it as radians). Does it need to say sin^-1 for it to refer to the sides rather than a radian measure?
2. "A circle has a circumference of 12pi cm. If an angle of pi/3 is subtended at the centre of the circle, find the exact area of the minor segment."
Their answer (having found radius to be 6cm):
(http://i.imgur.com/t8vE3Ah.png)
Why introduce sine rather than using the radian measure of the angle already given? I think I'm missing something here.
Thanks!
Always assume radians unless specified otherwise. The radian is the standard unit of angular measure in mathematics and degrees just exist because of astronomy. You should never use degrees unless they specify it, or give the circle superscript that hints degrees.
For the second question, they want the area of the minor segment, not sector. Check that you did not attempt to use the wrong formula.
Post by: _____ on July 06, 2017, 10:55:57 pm
You typed sin and not sin^-1. If it says sin you keep as sin. Of course, you may have typo'd it and actually meant sin^-1
Always assume radians unless specified otherwise. The radian is the standard unit of angular measure in mathematics and degrees just exist because of astronomy. You should never use degrees unless they specify it, or give the circle superscript that hints degrees.
For the second question, they want the area of the minor segment, not sector. Check that you did not attempt to use the wrong formula.
Nah I didn't typo it said sin, I was just used to year 11 questions in that format I think. Thanks for clearing that up.
I thought segment == sector, so I was using A = 1/2 r^2 theta. Shouldn't be trusting my school to teach the entire syllabus like that >:(
Post by: RuiAce on July 06, 2017, 11:07:33 pm
Nah I didn't typo it said sin, I was just used to year 11 questions in that format I think. Thanks for clearing that up.A segment is not a sector.
I thought segment == sector, so I was using A = 1/2 r^2 theta. Shouldn't be trusting my school to teach the entire syllabus like that >:(
A sector looks like what happens when you slice a pizza. A segment is just one random slice. You might be able to find some images on google quite easily.
They can easily ask you for the sine of something that doesn't appear to be an angle. If it says sin and not sin^-1 it's likely it was intended to be that way to trick you.
Post by: hansolo9 on July 07, 2017, 02:22:42 am
Can someone show how to do part B?
My teacher didn't explain the Simpson/Trapezoidal rotation very well and I'm confused.
Post by: RuiAce on July 07, 2017, 10:06:09 am
Hey guys,It is all formula work. You can choose to apply Simpson's rule twice separately or use the generalised Simpson's rule. Which one do you want to use?
Can someone show how to do part B?
My teacher didn't explain the Simpson/Trapezoidal rotation very well and I'm confused.
Post by: katnisschung on July 09, 2017, 01:32:37 pm
e.g. A scientific researcher was studying the population growth of insects
He knew the population grew at a rate of 1% compounding daily.
The insect population was 1000.
how many insects would be in the colony after 100 days? you may assume the mortality rate is
zero over this period
so they answer i got by applying the compound interest formula was
2704.8138...
so do u just round up as normal for these questions. kinda counter-intuitive :(
Post by: Sine on July 09, 2017, 01:35:40 pm
for series application questions how do i know how to round up or round down.for any sort of question involving an organism I always round down and feel that is the correct method to do so. (answering VCAA questions I have found this to be correct). However I am not sure about HSC stuff.
e.g. A scientific researcher was studying the population growth of insects
He knew the population grew at a rate of 1% compounding daily.
The insect population was 1000.
how many insects would be in the colony after 100 days? you may assume the mortality rate is
zero over this period
so they answer i got by applying the compound interest formula was
2704.8138...
so do u just round up as normal for these questions. kinda counter-intuitive :(
So I would say 2704.81 insects is 2704 since you can't have .81 of an insect
Post by: RuiAce on July 09, 2017, 01:39:29 pm
for any sort of question involving an organism I always round down and feel that is the correct method to do so. (answering VCAA questions I have found this to be correct). However I am not sure about HSC stuff.For this particular question yes.
So I would say 2704.81 insects is 2704 since you can't have .81 of an insect
In general, you will need some common sense for all of these questions.
Post by: Sine on July 09, 2017, 01:53:22 pm
For this particular question yes.yup definitely a case by case thing :) :)
In general, you will need some common sense for all of these questions.
Post by: 12070 on July 10, 2017, 06:24:26 pm
Post by: jakesilove on July 10, 2017, 06:35:09 pm
I'm baffled. I have expanded to get: ax^2-4ax+4a+bx-2b+c
You'll get used to questions like this; they're always attacked in exactly the same way.
Now, we factorise out such that we get a coefficient of each power of x.
Finally, we EQUATE the coefficients of each power of x on the 'right hand side' with that of the original function. So, in this case, we get three equations
The first one is already solved, and thus we can use it to solve the second
Subbing in these values into the third equation yields
Post by: vandanac55555 on July 11, 2017, 07:06:33 pm
I was wondering if anyone could please help me with this extension 1 maths q i have attached.
I wanted to integrate the acceleration formula to get the velocity but is there an easy way to expand the formula for acceleration? Or do i use binomial theorem? Not sure:(
Does anyone know ? :)
Thankyou:)
Post by: Shadowxo on July 11, 2017, 07:32:01 pm
Hi:)
I was wondering if anyone could please help me with this extension 1 maths q i have attached.
I wanted to integrate the acceleration formula to get the velocity but is there an easy way to expand the formula for acceleration? Or do i use binomial theorem? Not sure:(
Does anyone know ? :)
Thankyou:)
See the trick here is the x out the front. The derivative of (x2-3) is 2x. If you've learned substitution use that trying u=x2-3.
If you still need help let me know :)
Edit: Believe the formula you have to use is a = v * dv/dx
Post by: MisterNeo on July 11, 2017, 08:30:23 pm
Hi:)
I was wondering if anyone could please help me with this extension 1 maths q i have attached.
I wanted to integrate the acceleration formula to get the velocity but is there an easy way to expand the formula for acceleration? Or do i use binomial theorem? Not sure:(
Does anyone know ? :)
Thankyou:)
Remember these formulas for derivatives/integrals?
Just sub into the equation.
Post by: RuiAce on July 11, 2017, 08:35:21 pm
See the trick here is the x out the front. The derivative of (x2-3) is 2x. If you've learned substitution use that trying u=x2-3.Might want to note that v.dv/dx isn't required in 3U. They are allowed to use it but they are only expected to know d/dx v^2/2
If you still need help let me know :)
Edit: Believe the formula you have to use is a = v * dv/dx
Also, in the future, please put Ext 1 questions in the ext 1 section of the forum.
Hi:)
I was wondering if anyone could please help me with this extension 1 maths q i have attached.
I wanted to integrate the acceleration formula to get the velocity but is there an easy way to expand the formula for acceleration? Or do i use binomial theorem? Not sure:(
Does anyone know ? :)
Thankyou:)
Post by: RuiAce on July 11, 2017, 08:37:38 pm
Remember these formulas for derivatives/integrals?Carefully note that you actually did not use these formulae as you're dealing with a squared term. Those formulae are for linear expressions only, NOT quadratics.
Just sub into the equation.
Shadowxo's strategy of integration by substitution should be employed, which is 3U.
But I will remind that this question was 3U posted incorrectly into 2U. 2U deal with accelerations in terms of \(t\), not \(x\)
Post by: katnisschung on July 12, 2017, 10:42:55 pm
eg. 1/3 x^3 - 3x^2 + 11x-9=0
also need to justify my answer?
Post by: RuiAce on July 12, 2017, 10:51:23 pm
x-intercepts for cubic curves???You are not required to know how to deal with these in 2U.
eg. 1/3 x^3 - 3x^2 + 11x-9=0
also need to justify my answer?
Post by: Raymond_Cen on July 12, 2017, 10:59:14 pm
Find the dimensions of he largest rectangle that can be inscribed in the semicircle y=root(4-x^2)
Post by: RuiAce on July 12, 2017, 11:06:32 pm
Hello! Can you please help me with this question, thanks.
Find the dimensions of he largest rectangle that can be inscribed in the semicircle y=root(4-x^2)
Post by: katnisschung on July 13, 2017, 06:21:57 pm
You are not required to know how to deal with these in 2U.cries why was it in a 2 unit trial test!!! is there any way of solving this with 2u knowledge thanks rui
Post by: josephmalicdem on July 13, 2017, 07:21:10 pm
Thanks.
Post by: RuiAce on July 13, 2017, 08:24:48 pm
cries why was it in a 2 unit trial test!!! is there any way of solving this with 2u knowledge thanks ruiNothing appeared obvious to me and when I plugged it into WoiframAlpha it gave me really bizarre results for that one I doubt it.
Whilst some cubic equations are solvable, most aren't. That one in particular requires stuff beyond even 4U unless guided extensively along the way. What trial did it come from?
Post by: RuiAce on July 13, 2017, 08:28:19 pm
Can anyone please help me with this question? (Trigonometric Functions Topic)
Thanks.
Post by: katnisschung on July 13, 2017, 10:53:35 pm
Nothing appeared obvious to me and when I plugged it into WoiframAlpha it gave me really bizarre results for that one I doubt it.
Whilst some cubic equations are solvable, most aren't. That one in particular requires stuff beyond even 4U unless guided extensively along the way. What trial did it come from?
sydney grammar boys 2015
q15 a i) show that the curve has no stationary points
ii) show that there is a point of inflexion at (3,6)
iii) how many x intercepts does the curve have? justify your answer
Post by: RuiAce on July 13, 2017, 11:02:15 pm
sydney grammar boys 2015The question never asked you to find the intercepts themselves. It just asked how many there were.
q15 a i) show that the curve has no stationary points
ii) show that there is a point of inflexion at (3,6)
iii) how many x intercepts does the curve have? justify your answer
And the answer is 1. Because the curve has no stationary points, it's monotonic increasing. If it's monotonic increasing, carefully visualising it graphically it will only have one x-intercept.
Post by: josephmalicdem on July 14, 2017, 03:07:58 pm
Thanks a lot
Post by: josephmalicdem on July 15, 2017, 01:09:27 am
Post by: RuiAce on July 15, 2017, 09:07:55 am
This is another question I struggle with a bit.Maths in focus and it's dumb wording... not mentioning the fact that the interest is compounded monthly.
Post by: _____ on July 15, 2017, 05:16:57 pm
For shading the area, once I have x <= 3 what do I do? Test points for the 4 regions?
2. (http://i.imgur.com/Ime2UxD.png)
Am I supposed to recognise this as involving the area of a cone and have the formula memorised or is there a better way? The answer uses the volume of a cone (with h = SQRT(2)) and then adds the top part using a regular integration.
Thanks!
Post by: RuiAce on July 15, 2017, 05:25:41 pm
1. (http://i.imgur.com/Krr5c6v.png)
For shading the area, once I have x <= 3 what do I do? Test points for the 4 regions?
2. (http://i.imgur.com/Ime2UxD.png)
Am I supposed to recognise this as involving the area of a cone and have the formula memorised or is there a better way? The answer uses the volume of a cone (with h = SQRT(2)) and then adds the top part using a regular integration.
Thanks!
WolframAlpha input
______________________________________
You could've used integration for both if you wanted to. Their method is also acceptable.
Post by: jaskirat on July 15, 2017, 06:34:50 pm
Post by: RuiAce on July 15, 2017, 06:48:29 pm
Hey guys, i'm currently prepping for trials and just wanted some tips on what the best way to study for the maths exams isIt is a known fact that the best way to study is to continuously do past papers and to do as many as you can. You can only get better and more well versed against the exam by doing questions that came from exams in the first place.
After that, everything becomes entirely relative. What will work for someone else may not necessarily work for you.
Post by: _____ on July 15, 2017, 07:25:46 pm
WolframAlpha input
______________________________________
You could've used integration for both if you wanted to. Their method is also acceptable.
So just to clarify the procedure for 1 (non 3U algebra method):
- Interpret from the information given that the region occurs when y >= 1 for the original equation
- Express the original equation in terms of x to find x <= 3
- Test a point on either side of the asymptote (x=2) with the original equation considering what's been obtained
- Shade the region
For 2 I tried integration initially but got the wrong answer. Don't think we covered volumes generated by an area between two intersecting functions. Are any of these on the right track? Should I be trying to split up the volumes and subtracting/adding them with separate integrations or does it work like the area between two intersecting functions?
(http://i.imgur.com/PK4Pkl4.png)
Post by: RuiAce on July 15, 2017, 07:40:37 pm
So just to clarify the procedure for 1 (non 3U algebra method):Testing is up to you and perfectly safe. I prefer using my intuition or doing it in my head but there's nothing wrong with testing if you know what you're doing.
- Interpret from the information given that the region occurs when y >= 1 for the original equation
- Express the original equation in terms of x to find x <= 3
- Test a point on either side of the asymptote (x=2) with the original equation considering what's been obtained
- Shade the region
For 2 I tried integration initially but got the wrong answer. Don't think we covered volumes generated by an area between two intersecting functions. Are any of these on the right track? Should I be trying to split up the volumes and subtracting/adding them with separate integrations or does it work like the area between two intersecting functions?
(http://i.imgur.com/PK4Pkl4.png)
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If you were rotating about the x-axis, you'd have a volume between two curves similar to the first one.
But you're rotating about the y-axis. If you treated this is an area question, you would have a compound region and be adding the area integrals. Hence, you should be adding the volume integrals here as in equation 3.
Post by: jaskirat on July 15, 2017, 08:05:38 pm
Post by: junzhang on July 15, 2017, 09:29:56 pm
Could you help me with this question?
I'm not sure how to find the focal length and I can't get the required proof. Thank you!
Post by: _____ on July 15, 2017, 09:44:21 pm
Testing is up to you and perfectly safe. I prefer using my intuition or doing it in my head but there's nothing wrong with testing if you know what you're doing.
________________________
If you were rotating about the x-axis, you'd have a volume between two curves similar to the first one.
But you're rotating about the y-axis. If you treated this is an area question, you would have a compound region and be adding the area integrals. Hence, you should be adding the volume integrals here as in equation 3.
Thanks a bunch m8
Post by: kiwiberry on July 15, 2017, 09:47:24 pm
Hey,
Could you help me with this question?
I'm not sure how to find the focal length and I can't get the required proof. Thank you!
Hello!
h=40 and k=0 because the vertex is at (40,0). Because A(30,50) lies on the curve, we can sub it into the equation to find the focal length
Post by: hansolo9 on July 15, 2017, 10:33:51 pm
I haven't seen these types of radian questions yet
Post by: RuiAce on July 15, 2017, 11:00:01 pm
Hello, can I have help with the attached question?
I haven't seen these types of radian questions yet
Post by: hansolo9 on July 15, 2017, 11:11:25 pm
Omg how did i not see that...
Thanks Rui :)
Post by: sophiegmaher on July 16, 2017, 11:51:08 am
Post by: ellipse on July 16, 2017, 11:57:06 am
So I'm really confused as to how the x-intercept of the function y=6ln(x-1) is (2,0)? I understand you let the equation equal to 0, but I don't get how that will equal to 2?
0=6ln(x-1)
ln(x-1)=0
x-1=e^0
x-1=1
x=2
Post by: sophiegmaher on July 16, 2017, 12:00:21 pm
0=6ln(x-1)
ln(x-1)=0
x-1=e^0
x-1=1
x=2
Thank you!!
Post by: Kekemato_BAP on July 16, 2017, 07:41:12 pm
Thanks :)
Post by: RuiAce on July 16, 2017, 07:58:11 pm
Hi, may I have help on question 14b?Important warning: The Extension questions in Cambridge are designed purely for the sake of self-interest. They are only for you to build superior techniques when you feel necessary. In general, I will not do questions from this section, as they very often go well beyond the scope of the course in question.
Thanks :)
Post by: itssona on July 17, 2017, 06:28:31 pm
Do we see tan? And how do we use our knowledge of quadrants
Thank you
Post by: RuiAce on July 17, 2017, 06:36:24 pm
This is rlly dumb but.. how would you exactly know cot (-135)Assuming degrees, although not explicitly stated.
Do we see tan? And how do we use our knowledge of quadrants
Thank you
Post by: georgiia on July 17, 2017, 07:16:14 pm
And this one
Moderator action: Posts merged. At times like these, please resort to the Modify button at the top right corner of a post.
Post by: RuiAce on July 17, 2017, 07:28:02 pm
And this one
Please provide progress on the previous parts as they are necessary for part d).
Post by: 1937jk on July 18, 2017, 08:21:59 am
A particle is accelerating according to the equation a = (3t + 1)2 . if the particle is initially at rest 2 m to the left of 0, find its displacement after 4 secs.
Post by: RuiAce on July 18, 2017, 08:30:37 am
Just stuck with this particle question, any help would be appreciated :)
A particle is accelerating according to the equation a = (3t + 1)2 . if the particle is initially at rest 2 m to the left of 0, find its displacement after 4 secs.
Post by: itssona on July 18, 2017, 08:58:33 am
Rewrite the following as a positive acute angle:
Cos 195
I know its -cos(theta) but how can we get a positive answer? :/
Post by: RuiAce on July 18, 2017, 09:04:02 am
Pls help :)
Rewrite the following as a positive acute angle:
Cos 195
I know its -cos(theta) but how can we get a positive answer? :/
Post by: itssona on July 18, 2017, 09:11:53 am
Ah thank you! Glad it wasn't a problem with my lack of understanding xD
Post by: jaysun on July 18, 2017, 03:21:46 pm
Thank youu!!!!
Post by: sudodds on July 18, 2017, 03:28:26 pm
Hey, could I please get some help with question b) both parts... I'm really confused how to go about this; my teacher taught this really badly and Im super scared for trials!!!
Thank youu!!!!
Hi Jaysun,
so for (i) I was thinking you would differentiate (xe^x) and get xe^x + e^x which would be your first mark.
For (ii):
d/dx(xe^x) = e^x+xe^x
d/dx(xe^x) - e^x = xe^x
then 'integral' d/dx (xe^x) - e^x.dx = 'integral' xe^x.dx
finally: xe^x - e^x + C = 'integral' xe^x.dx
For this just try to make the second part resemble the first bit - so you see how part (i) had an extra e^x, you just manipulate the equation in part (ii) to resemble the earlier one.
Hope this helps!
Susie
Post by: Shadowxo on July 18, 2017, 03:35:21 pm
So
b) So we know
Basically, you rearrange it, and also the integral of the derivative of something is just the thing itself. You rearrange the equation so that what you're integrating is the subject, and usually you end up with regular integrals or integral of d/dx(something). Hard to explain but examples help :)
Also sorry reply took a while, wanted to make it easy to read :)
Post by: RuiAce on July 18, 2017, 04:41:56 pm
Hi Jaysun,Thanks Isaac/Jamon
so for (i) I was thinking you would differentiate (xe^x) and get xe^x + e^x which would be your first mark.
For (ii):
d/dx(xe^x) = e^x+xe^x
d/dx(xe^x) - e^x = xe^x
then 'integral' d/dx (xe^x) - e^x.dx = 'integral' xe^x.dx
finally: xe^x - e^x + C = 'integral' xe^x.dx
For this just try to make the second part resemble the first bit - so you see how part (i) had an extra e^x, you just manipulate the equation in part (ii) to resemble the earlier one.
Hope this helps!
Susie
Post by: sudodds on July 18, 2017, 04:46:12 pm
Thanks Isaac/JamonYou spelt Susie wrong
Post by: itssona on July 18, 2017, 05:33:27 pm
Parabola:
Domain - all real x
Range- all real y
Log x
Domain - all positivr x
Range- all real y
Absolute value:
Range: all real y
Cubic:
Domain: all real x??
Range: all real y
Exponential :
Range: all positive y
Domain: all real x
Post by: RuiAce on July 18, 2017, 05:44:13 pm
Heey can u pls check if these are right:Not sure what's going on with the red ones.
Parabola:
Domain - all real x
Range- all real y
Log x
Domain - all positivr x
Range- all real y
Absolute value:
Range: all real y
Cubic:
Domain: all real x??
Range: all real y
Exponential :
Range: all positive y
Domain: all real x
Post by: itssona on July 18, 2017, 05:48:39 pm
Thank u for explaining that part :))
Post by: RuiAce on July 18, 2017, 05:53:55 pm
Post by: Shadowxo on July 18, 2017, 05:55:44 pm
Ohh wait can u correct me on the red ones
Thank u for explaining that part :))
x2 (basic parabola)
Domain: all real x
Range: [0,infinity)
logex (basic log)
Domain: All positive x (0,infinity)
Range: All real y
|x| (basic modulus graph)
Domain: All real x
Range: [0,infinity)
x3 (basic cubic)
Domain: All real x
Range: All real y
ex (basic exponential)
Domain: All real x
Range: All positive y (0,infinity)
Post by: itssona on July 18, 2017, 05:58:07 pm
Post by: itssona on July 18, 2017, 05:59:08 pm
x2 (basic parabola)Ah thank u!!
Domain: all real x
Range: [0,infinity)
logex (basic log)
Domain: All positive x (0,infinity)
Range: All real y
|x| (basic modulus graph)
Domain: All real x
Range: [0,infinity)
x3 (basic cubic)
Domain: All real x
Range: All real y
ex (basic exponential)
Domain: All real x
Range: All positive y (0,infinity)
Post by: RuiAce on July 18, 2017, 06:03:45 pm
x2 (basic parabola)They don't use interval notation in the HSC.
Domain: all real x
Range: [0,infinity)
logex (basic log)
Domain: All positive x (0,infinity)
Range: All real y
|x| (basic modulus graph)
Domain: All real x
Range: [0,infinity)
x3 (basic cubic)
Domain: All real x
Range: All real y
ex (basic exponential)
Domain: All real x
Range: All positive y (0,infinity)
So for log (x+1) is the domain x>-1 ?Yes
Post by: bigsweetpotato2000 on July 18, 2017, 09:26:05 pm
I'm still rote-learning it and I just can't seem to find the logic in it!
Unsatisfied,
Bigsweetpotato Farm
Post by: RuiAce on July 18, 2017, 09:48:19 pm
What is the best way of understanding Motion ?
I'm still rote-learning it and I just can't seem to find the logic in it!
Unsatisfied,
Bigsweetpotato Farm
Post by: bigsweetpotato2000 on July 18, 2017, 09:50:45 pm
Apply to real life situation logic?
Post by: RuiAce on July 18, 2017, 10:12:02 pm
Apply to real life situation logic?Can you please provide an example to supplement your question? It's very ambiguous as to what you want right now.
2U scenarios are pretty much immediate applications of the above results.
Post by: bigsweetpotato2000 on July 18, 2017, 10:28:05 pm
Can you please provide an example to supplement your question? It's very ambiguous as to what you want right now.
2U scenarios are pretty much immediate applications of the above results.
Graphing?
Post by: RuiAce on July 18, 2017, 10:32:19 pm
Alternatively, just recognise that it's decreasing without sketching the tangent by inspection.
Post by: Shadowxo on July 18, 2017, 10:33:55 pm
Graphing?v=dx/dt. As the gradient is negative, v is negative at P
a=dv/dt. As the gradient is increasing (becoming less negative) a is positive at P
Edit: Rui beat me to it but hope this helps anyway
Post by: bigsweetpotato2000 on July 19, 2017, 12:03:12 am
Alternatively, just recognise that it's decreasing without sketching the tangent by inspection.
Isn't it A?
;)
Thanks btw :D
Post by: bigsweetpotato2000 on July 19, 2017, 12:05:02 am
v=dx/dt. As the gradient is negative, v is negative at P
a=dv/dt. As the gradient is increasing (becoming less negative) a is positive at P
Edit: Rui beat me to it but hope this helps anyway
Heheh
Repeated trials with consistent results lead to reliability :P
Many thanks!
Post by: RuiAce on July 19, 2017, 12:06:51 am
Post by: bigsweetpotato2000 on July 19, 2017, 12:10:02 am
I can't read. Yes it is A (fixing in original post now)
Need to double the glasses power :D
Post by: itssona on July 19, 2017, 09:46:15 pm
Whats domain of underroot (x-2)
Our school says x greater than and equal to 2 but in another exam a similar question had the answer of GREATER than only
Thank you
Post by: RuiAce on July 19, 2017, 09:48:13 pm
Um so i was doing past papers for my school exam andddd i need clarificationIt's greater than or equal to. If we sub in x=2 we don't run into problems.
Whats domain of underroot (x-2)
Our school says x greater than and equal to 2 but in another exam a similar question had the answer of GREATER than only
Thank you
Post by: itssona on July 19, 2017, 09:49:13 pm
It's greater than or equal to. If we sub in x=2 we don't run into problems.Ohhh THANK YOU RUI :)
Post by: Youssk on July 19, 2017, 09:57:20 pm
The acceleration of a particle is given by
a=1-2t,
where a is measured in cm/second and t is measured in seconds.
a) At what time is the particle at rest?
b) Where is the particle at this time?
Post by: RuiAce on July 19, 2017, 09:59:09 pm
Hi, Could I have some help with this question.This question lacks information and cannot be done. Information needs to be given about initial conditions.
The acceleration of a particle is given by
a=1-2t,
where a is measured in cm/second and t is measured in seconds.
a) At what time is the particle at rest?
b) Where is the particle at this time?
Post by: Youssk on July 19, 2017, 10:02:12 pm
Post by: RuiAce on July 19, 2017, 10:05:49 pm
Post by: georgiia on July 20, 2017, 02:47:15 pm
Thanks!
Post by: jamonwindeyer on July 20, 2017, 02:52:47 pm
Could someone please help me for this? Its probably very simple but I have no clue??
Thanks!
Hey! This is a quotient rule problem, definitely not simple! Helpful to differentiate \(\ln{x^2}\) separately so we'll do \(u\) and \(v\) as well as their derivatives:
Note you can also do that last one with log laws ;D put this into the quotient rule:
Post by: georgiia on July 20, 2017, 02:58:01 pm
Post by: RuiAce on July 20, 2017, 03:01:06 pm
Could someone please help me for this? Its probably very simple but I have no clue??Presumably the attachment changed.
Thanks!
Post by: kylesara on July 20, 2017, 04:03:58 pm
The cubic y=ax^3+bx^2+cx+d has a point of inflexion at x=p.
Show that p = − b . 3a
Thanks:)
Post by: RuiAce on July 20, 2017, 04:08:31 pm
Hi could i please have help with this question.A typo was assumed.
The cubic y=ax^3+bx^2+cx+d has a point of inflexion at x=p.
Show that p = − b . 3a
Thanks:)
Post by: georgiia on July 20, 2017, 04:25:01 pm
Post by: jaskirat on July 20, 2017, 04:59:28 pm
Post by: RuiAce on July 20, 2017, 05:03:29 pm
Need help with this please! Forgot this topic 😐
Literally sub in t=0 to do part a).
______________________
______________________
______________________
Post by: georgiia on July 21, 2017, 12:29:30 pm
Can someone please explain this?
Post by: RuiAce on July 21, 2017, 12:39:32 pm
I ve solved the integral and get that K needs to be 0??
Can someone please explain this?
Post by: georgiia on July 21, 2017, 01:45:29 pm
Thanks!!
Post by: jaskirat on July 21, 2017, 04:52:14 pm
Post by: RuiAce on July 21, 2017, 04:57:14 pm
Need help with these two questions, thanks :)
_____
This second one involved nothing challenging. What was causing the struggle?
Post by: jaskirat on July 21, 2017, 05:04:39 pm
_____
This second one involved nothing challenging. What was causing the struggle?
Right after i posted about needing help with the second question, i remembered it lol! Thanks bro!
Post by: 12070 on July 22, 2017, 02:05:42 pm
Post by: RuiAce on July 22, 2017, 02:06:52 pm
Hey, was just wondering how if 3e^t=12/(e^2t), e^3t=4
Post by: winstondarmawan on July 23, 2017, 02:29:51 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20270047_1538514876198918_1314381077_n.jpg?oh=3ce907b341fda76bc69f71d6d59fe730&oe=5975F432
TIA
Post by: RuiAce on July 23, 2017, 02:34:33 pm
Post by: winstondarmawan on July 23, 2017, 02:57:22 pm
I was able to do that part, but theres a 2nd part to the QSubject to computational inaccuracy
Post by: RuiAce on July 23, 2017, 03:11:14 pm
I won't be able to get to it for a while though; I can't do that without a diagram
That being said, please make it clearer what you required for future benefit
Post by: Thebarman on July 23, 2017, 03:19:27 pm
Post by: Shadowxo on July 23, 2017, 04:01:09 pm
How would I go about answering this question? ThanksHi :)
I don't know what method they'd like you to use for this, so have you been taught either
1) integration by parts or
2) using a function like e^x to find the area?
Post by: georgiia on July 23, 2017, 04:05:41 pm
I am slightly confused by this question, is it asking for growth in rate or growth in population because if it's population growth, there isn't enough information provided so ???
Thank you
Post by: Shadowxo on July 23, 2017, 04:11:02 pm
Post by: RuiAce on July 23, 2017, 04:11:21 pm
Hi :)Show them the latter please; IBP is for MX2
I don't know what method they'd like you to use for this, so have you been taught either
1) integration by parts or
2) using a function like e^x to find the area?
Post by: georgiia on July 23, 2017, 04:12:26 pm
I believe it's asking for the change in population. Since you're given the rate of change of population (ie dP/dt where P is population) you should be able to find the change in population by finding the integral between 1 and 3 of R(t) :)Ahh! Didn't think of attaching limits to the integral :)
I was stuck with a C I couldn't find
Thx
Post by: georgiia on July 23, 2017, 04:21:31 pm
Thanks
Post by: georgiia on July 23, 2017, 04:27:28 pm
Thanks
Post by: Shadowxo on July 23, 2017, 04:28:25 pm
Show them the latter please; IBP is for MX2Thought so but wasn't sure :)
How would I go about answering this question? ThanksSo finding the integral of ln(x) is difficult, but finding the integral of ex is easy. ex and ln(x) are inverse functions of each other (if you don't know this or are given a more complicated question, you can figure it out by finding the inverse function like usual).
In this kind of question, a diagram is useful :) I'll attach my working/solution(http://uploads.tapatalk-cdn.com/20170723/bd6ece3a886f410803ecdc984ac60505.jpg)
My working is in black and the answer provided is pink. Would I still get the mark for failing to see if the quicker way of answering the question?I think your method is valid but since it asks how you know the population is decreasing, rather than asking what will happen to the population, it would probably be safer to find dp/dt and hence show it's less than 0, ie decreasing
Thanks
Post by: georgiia on July 23, 2017, 04:49:52 pm
Thought so but wasn't sure :)Thanks, yeah that makes sense
So finding the integral of ln(x) is difficult, but finding the integral of ex is easy. ex and ln(x) are inverse functions of each other (if you don't know this or are given a more complicated question, you can figure it out by finding the inverse function like usual).
In this kind of question, a diagram is useful :) I'll attach my working/solution(http://uploads.tapatalk-cdn.com/20170723/bd6ece3a886f410803ecdc984ac60505.jpg)
I think your method is valid but since it asks how you know the population is decreasing, rather than asking what will happen to the population, it would probably be safer to find dp/dt and hence show it's less than 0, ie decreasing
Post by: georgiia on July 23, 2017, 04:53:15 pm
(For iii.)
Because I get two answers, is it safe to just ignore t=75 because of the fact that it takes him t=50 to empty the contents thus I can assume that t=75 is inadmissible even if they didn't supply a domain in the Q. ?
Post by: Shadowxo on July 23, 2017, 04:57:41 pm
I just have a clarification question here:Can't see the question, did you attach it properly? :)
(For iii.)
Because I get two answers, is it safe to just ignore t=75 because of the fact that it takes him t=50 to empty the contents thus I can assume that t=75 is inadmissible even if they didn't supply a domain in the Q. ?
Post by: mlarsson on July 23, 2017, 04:58:00 pm
Post by: georgiia on July 23, 2017, 05:03:29 pm
Hey just wanted to confirm a few things, because my teacher hasnt been very clear about this topic. How do you know when to use logs in an equation ? (besides the questions where it is obviously asking about logs?)When the unknown is to the power of e
Like this:
e^t=5
Ln(e)^t=ln(5)
t=ln(5)
Post by: georgiia on July 23, 2017, 05:03:53 pm
Can't see the question, did you attach it properly? :)Fixed it
Post by: Shadowxo on July 23, 2017, 05:13:11 pm
I just have a clarification question here:Correct, the implied domain is 0≤t≤50 as this is the time he's drinking the water and the volume is decreasing :)
(For iii.)
Because I get two answers, is it safe to just ignore t=75 because of the fact that it takes him t=50 to empty the contents thus I can assume that t=75 is inadmissible even if they didn't supply a domain in the Q. ?(https://uploads.tapatalk-cdn.com/20170723/89f6bfbb6af23218833c051ca7dd3400.jpg)
Also, in questions like this you usually take the earliest answer as that's when it first occurs
Post by: georgiia on July 23, 2017, 05:16:42 pm
Correct, the implied domain is 0≤t≤50 as this is the time he's drinking the water and the volume is decreasing :)
Also, in questions like this you usually take the earliest answer as that's when it first occurs
Thanks!
Post by: RuiAce on July 23, 2017, 05:40:55 pm
Post by: Shadowxo on July 23, 2017, 05:43:13 pm
Anything still unattended to? Just quickly checking I wasn't sure what wasn't addressed but I'll be able to handle a few things soon - just not sure if I missed anyThink everything's been answered :)
Post by: RuiAce on July 23, 2017, 05:43:44 pm
Think everything's been answered :)Awesome. Even the locus one? - that one appeared in my head just now
Post by: Shadowxo on July 23, 2017, 05:47:07 pm
Hey just wanted to confirm a few things, because my teacher hasnt been very clear about this topic. How do you know when to use logs in an equation ? (besides the questions where it is obviously asking about logs?)Also just in addition to what georgiia said,
You use logs when what you want to find is in the power.
ab=c
logac=b
You can use this for any power and any base, doesn't have to be e (but for this a>0 and c>0)
Awesome. Even the locus one? - that one appeared in my head just nowEverything but that question I mean :P forgot about it too but I'm not familiar with solving those kinds of problems, I'll leave that to you :)
Post by: RuiAce on July 23, 2017, 06:49:21 pm
I was able to do that part, but theres a 2nd part to the QI ended up finding something (in what little time I ended up having) but it involves a 3U circle geometry theorem. Before I jump into it, what was the source of this question?
Post by: winstondarmawan on July 23, 2017, 08:50:28 pm
I ended up finding something (in what little time I ended up having) but it involves a 3U circle geometry theorem. Before I jump into it, what was the source of this question?Fitzpatrick Textbook, not sure if 2U or 3U, because my friend sent me that photo, sorry.
I just assumed it was 2U because it was a locus Q.
Post by: Youssk on July 23, 2017, 09:10:57 pm
The acceleration of a moving body is given by a=√2t+1 ms^-2. If the body starts from rest, find its velocity after 4 seconds.
Post by: MisterNeo on July 23, 2017, 10:12:06 pm
Hello, could I please have some help with this question.
The acceleration of a moving body is given by a=√2t+1 ms^-2. If the body starts from rest, find its velocity after 4 seconds.
Hi! For this question, you would integrate the acceleration to find the velocity function with a +C.
It says that the body starts from rest, so velocity is 0 when t=0. (You have to solve for C)
Then sub in t=4 into the full velocity function to find the velocity after 4 seconds! ;D
Post by: RuiAce on July 24, 2017, 01:03:36 pm
Hello! Would appreciate help with Q3:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20270047_1538514876198918_1314381077_n.jpg?oh=3ce907b341fda76bc69f71d6d59fe730&oe=5975F432
TIA
(http://i.imgur.com/IerlQXZ.png)
Post by: georgiia on July 24, 2017, 02:48:20 pm
Thanks!
Post by: Shadowxo on July 24, 2017, 07:59:22 pm
Can someone please walk me through drawing the primitive function? I haven't done these for ages and Im ok drawing the gradient function but i can remember what turns into what for primitive :(Hi :)
Thanks!
So, while I haven't done primitive functions, I have done anti-differentiation / integration. So sorry if my terminology is wrong or I don't address everything etc :)
Where y=0 on the graph / original function, the gradient of the primitive function is 0 (basically opposite of finding the gradient function), which means either a turning point or a point of inflection. As on the graph there is no turning point at y=0 x=2, (just goes straight through 0, one side is negative one side is positive), this means x=2 on the primitive function must be a turning point, and it's a local minimum as the original function goes from negative to positive.
Where there's a turning point on the original function, there is a point of inflection on the primitive function. In this case the original function becomes more positive, then more negative, then more positive again, meaning the gradient on the primitive function will do the same.
So for this primitive function, it will start off high and be decreasing, have a couple points of inflections approx x=-1 and x=1 (not a stationary point of inflection though, and it will continue decreasing this whole time as the original graph has y as negative until x=2). Then, at x=2 there will be a turning point and it will go upwards.
Hope this helps, let me know if you're still having trouble with it or if you'd like a diagram :D
Post by: georgiia on July 24, 2017, 08:20:58 pm
Hi :)
So, while I haven't done primitive functions, I have done anti-differentiation / integration. So sorry if my terminology is wrong or I don't address everything etc :)
Where y=0 on the graph / original function, the gradient of the primitive function is 0 (basically opposite of finding the gradient function), which means either a turning point or a point of inflection. As on the graph there is no turning point at y=0 x=2, (just goes straight through 0, one side is negative one side is positive), this means x=2 on the primitive function must be a turning point, and it's a local minimum as the original function goes from negative to positive.
Where there's a turning point on the original function, there is a point of inflection on the primitive function. In this case the original function becomes more positive, then more negative, then more positive again, meaning the gradient on the primitive function will do the same.
So for this primitive function, it will start off high and be decreasing, have a couple points of inflections approx x=-1 and x=1 (not a stationary point of inflection though, and it will continue decreasing this whole time as the original graph has y as negative until x=2). Then, at x=2 there will be a turning point and it will go upwards.
Hope this helps, let me know if you're still having trouble with it or if you'd like a diagram :D
Thank You!! I just needed clarification as to what turns into what when you integrate and you've refreshed my memory now, Thanks!!
Post by: Shadowxo on July 24, 2017, 08:25:25 pm
Thank You!! I just needed clarification as to what turns into what when you integrate and you've refreshed my memory now, Thanks!!No problem :D
Post by: hansolo9 on July 26, 2017, 01:52:15 pm
Can someone help with these 4 trial questions?
TIA
Post by: RuiAce on July 26, 2017, 01:54:31 pm
Hi :)I don't know what's confusing about the first question and all I'd be doing is regurgitating the answer.
Can someone help with these 4 trial questions?
TIA
Post by: RuiAce on July 26, 2017, 01:59:51 pm
Hi :)
Can someone help with these 4 trial questions?
TIA
__________________________
Post by: katnisschung on July 26, 2017, 06:37:34 pm
needing help with this question
its ii) (also its from 2010 paper q 5)
I don't understand how I show that the graph has a minimum value.
should I find the stationary points?
(how do i do that for graphs in general? generally only know how to do it with a parabola and differentiate)
helppp :'( :'( :'(
Post by: Shadowxo on July 26, 2017, 06:51:06 pm
Bonjour! :)First you differentiate, then you find the stationary point/s. To find out whether it's a max/min, find whether the second derivative is positive/neg for that value and that should show it's a min (*in VCE we use different methods usually). If it's a minimum, the gradient should be increasing at that point (going from neg to 0 to positive) meaning the rate of change of gradient is positive at that value ie second derivative is positive
needing help with this question
its ii) (also its from 2010 paper q 5)
I don't understand how I show that the graph has a minimum value.
should I find the stationary points?
(how do i do that for graphs in general? generally only know how to do it with a parabola and differentiate)
helppp :'( :'( :'(
Post by: RuiAce on July 26, 2017, 07:36:13 pm
Bonjour! :)As far as 2U goes, the "existence" of the minimum value is verified so long as you may find a local minimum.
needing help with this question
its ii) (also its from 2010 paper q 5)
I don't understand how I show that the graph has a minimum value.
should I find the stationary points?
(how do i do that for graphs in general? generally only know how to do it with a parabola and differentiate)
helppp :'( :'( :'(
That being said, this type of question is very standard and you have seen many instances of it in the geometrical applications of differentiation topic.
Post by: Mymy409 on July 27, 2017, 01:07:52 pm
A particle is moving along the x-axis. The displacement of the particle at time t seconds
is x metres.
At a certain time, v = -3 m/s and a = 2m/s^2.
Which statement describes the motion of the particle at that time?
(A) The particle is moving to the right with increasing speed.
(B) The particle is moving to the left with increasing speed.
(C) The particle is moving to the right with decreasing speed.
(D) The particle is moving to the left with decreasing speed.
The answer is D.
Post by: Shadowxo on July 27, 2017, 01:20:51 pm
Hi, I've got a question that's confusing me quite a bit.The velocity is negative so it's moving to the left
A particle is moving along the x-axis. The displacement of the particle at time t seconds
is x metres.
At a certain time, v = -3 m/s and a = 2m/s^2.
Which statement describes the motion of the particle at that time?
(A) The particle is moving to the right with increasing speed.
(B) The particle is moving to the left with increasing speed.
(C) The particle is moving to the right with decreasing speed.
(D) The particle is moving to the left with decreasing speed.
The answer is D.
The acceleration is positive so the 'velocity' is increasing but it's just becoming less negative. The 'speed' however is decreasing (as the initial speed is 3 m/s, speed doesn't have a direction). So it's slowing down (speed decreasing). It's just a terminology thing, does this help?
Post by: 1937jk on July 27, 2017, 03:50:00 pm
A coin is tossed n times. Find the probability in terms of n of tossing,
a) all heads
b) no tails
c) at least one tails
Thanks!
Post by: RuiAce on July 27, 2017, 03:58:45 pm
Hey! Just need some help in understanding how to go about this question, like I think I get how to get the answer but I don't quite think I get the concept behind it,Non-mathematical, only explanation:
A coin is tossed n times. Find the probability in terms of n of tossing,
a) all heads
b) no tails
c) at least one tails
Thanks!
Of course, the probability of tossing a head in the first turn does not impact the probability of tossing a head on the second turn, and so on. i.e. all the tosses are independent.
a) The probability of tossing a head on the first turn is 1/2. The probability of tossing a head on the second turn is still 1/2. And we keep going, and the probability of tossing a head on the n-th turn is still 1/2. So we just multiply them together.
b) If exactly zero of your tosses are tails, then all of them are heads. This is the same as above.
c) At least one tail implies exactly one + exactly two + exactly three + so on and that takes work. So we consider the complement, which is the probability of tossing no tails. This is essentially the opposite event, which you should recognise fairly easily.
The probability of no tails from part b) is (1/2)^n, so the probability of at least one tail is 1 minus (1/2)^n
Post by: winstondarmawan on July 27, 2017, 05:10:31 pm
Mainly not sure on how the n from n-sided polygon gets incorporated into the angle.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20427880_1282466078548868_1386597183_n.png?oh=a54f49ae1c9642fa30bdf54ebbf9cd6d&oe=597C71D1
Post by: jaskirat on July 27, 2017, 05:41:02 pm
Post by: Natasha.97 on July 27, 2017, 06:02:10 pm
Need help with this question :)
Hi!
I've attached the solution below:
(http://i.imgur.com/tvyQRuK.jpg)
Hope this helps!
Note: Not 100% sure that this is correct, maybe someone below could check :D
Post by: jaskirat on July 27, 2017, 06:07:52 pm
Hi!
I've attached the solution below:
(http://i.imgur.com/tvyQRuK.jpg)
Hope this helps!
Note: Not 100% sure that this is correct, maybe someone below could check :D
Post by: Shadowxo on July 27, 2017, 06:45:45 pm
Hi!Looks good :)
I've attached the solution below:
(http://i.imgur.com/tvyQRuK.jpg)
Hope this helps!
Note: Not 100% sure that this is correct, maybe someone below could check :D
Post by: RuiAce on July 27, 2017, 07:05:43 pm
Hello, would appreciate help with the following, TIA.This one was asked a while back
Mainly not sure on how the n from n-sided polygon gets incorporated into the angle.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20427880_1282466078548868_1386597183_n.png?oh=a54f49ae1c9642fa30bdf54ebbf9cd6d&oe=597C71D1
Post by: katnisschung on July 27, 2017, 07:10:08 pm
Find the area between the curve y=2ln(x+1), the x axis and the line x=2
How do I rearrange this for x.
Post by: RuiAce on July 27, 2017, 07:12:58 pm
Hi need help with this thanks ins advance!
Find the area between the curve y=2ln(x+1), the x axis and the line x=2
How do I rearrange this for x.
I'll let you have a go at the remainder first. Comment again if you need further help
Post by: katnisschung on July 27, 2017, 08:27:39 pm
Sin x and 1/2 tanx I tried it solving it graphically but that didn't work out so well
Post by: RuiAce on July 27, 2017, 08:29:41 pm
How do I find the points of intersections between the 2 trig graphs
Sin x and 1/2 tanx I tried it solving it graphically but that didn't work out so well
The relevant related angle for whatever purpose is \(\frac\pi3\)
Post by: VanillaRice on July 27, 2017, 08:47:39 pm
How do I find the points of intersections between the 2 trig graphsWe can use an algebraic method to solve for x, and then substitute those values into either equation to obtain the y value:
Sin x and 1/2 tanx I tried it solving it graphically but that didn't work out so well
Spoiler
and
Spoiler
After finding a general solution for both cases, substitute a specific x-value from each general solution (should be four different cases) to obtain the y-values.
EDIT: Oops got beaten to it :-X
EDIT2:
I think you've accidentally deleted a solution by dividing both sides by sin(x) :P
The relevant related angle for whatever purpose is \(\frac\pi3\)
Post by: RuiAce on July 27, 2017, 09:02:36 pm
Hmm. Walked into my own trap it seems; fixing.
EDIT2: I think you've accidentally deleted a solution by dividing both sides by sin(x) :P
Post by: kiiaaa on July 27, 2017, 10:17:38 pm
could someones please help me how to tackle this question please? I couldnt find the answers and nor do i know where to start from?
Thanks soo much guys!!:)))
Post by: RuiAce on July 27, 2017, 10:22:00 pm
hey
could someones please help me how to tackle this question please? I couldnt find the answers and nor do i know where to start from?
Thanks soo much guys!!:)))
I will let you have a go at ii).
Post by: Mymy409 on July 28, 2017, 08:43:43 am
The velocity is negative so it's moving to the left
The acceleration is positive so the 'velocity' is increasing but it's just becoming less negative. The 'speed' however is decreasing (as the initial speed is 3 m/s, speed doesn't have a direction). So it's slowing down (speed decreasing). It's just a terminology thing, does this help?
Thanks for your reply! Why is the speed decreasing, though? I still don't quite get it.
Post by: RuiAce on July 28, 2017, 09:25:19 am
Thanks for your reply! Why is the speed decreasing, though? I still don't quite get it.We have stated that the particle is moving to the left (because v is negative).
We have also stated that the particle is accelerating to the right (a is positive). Consider what this means.
If the particle is moving to the left, AND accelerating to the left, it will move to the left more quickly. I.e. When the acceleration and velocity work in the SAME direction, the particle speeds up.
But we have the opposite. If the particle is moving to the left, BUT it's being accelerated towards the other direction (the right), something is causing the particle to actually WANT to move to the right, instead of to the left. This means that the particle slows down when the velocity and acceleration work against each other.
Post by: Mymy409 on July 28, 2017, 10:25:23 am
We have stated that the particle is moving to the left (because v is negative).
We have also stated that the particle is accelerating to the right (a is positive). Consider what this means.
If the particle is moving to the left, AND accelerating to the left, it will move to the left more quickly. I.e. When the acceleration and velocity work in the SAME direction, the particle speeds up.
But we have the opposite. If the particle is moving to the left, BUT it's being accelerated towards the other direction (the right), something is causing the particle to actually WANT to move to the right, instead of to the left. This means that the particle slows down when the velocity and acceleration work against each other.
That makes sense, thank you!
Post by: Mounica on July 29, 2017, 06:58:49 pm
Can someone pls help me with this question.
thanks
Post by: RuiAce on July 29, 2017, 07:07:05 pm
Hey Guys,
Can someone pls help me with this question.
thanks
____________________________________
____________________________________
Post by: Mounica on July 29, 2017, 07:15:34 pm
Thank You :D
____________________________________
____________________________________
Post by: ekhan_01 on July 30, 2017, 06:44:36 pm
Can someone pls help, I have no idea what to do // the answer is A
Thank you
Post by: RuiAce on July 30, 2017, 06:51:00 pm
Hi!
Can someone pls help, I have no idea what to do // the answer is A
Thank you
Validity check: After 7 white marbles are added we have 10 green marbles and 15 white marbles. Pr(White) = 3/5 as expected.
Post by: 12070 on August 01, 2017, 12:18:41 pm
I was wondering how I should study for Mathematics on Thursday. I have gone through about 7 CSSA trials over the past two weeks and I still feel weak with some areas like Max/Min but I could spend the next two days doing that topic only to not understand the question in the test. I want my study to be really efficient but am unsure how to achieve this.
Post by: RuiAce on August 01, 2017, 12:29:46 pm
Hey,Potentially something like this:
I was wondering how I should study for Mathematics on Thursday. I have gone through about 7 CSSA trials over the past two weeks and I still feel weak with some areas like Max/Min but I could spend the next two days doing that topic only to not understand the question in the test. I want my study to be really efficient but am unsure how to achieve this.
Do one more past paper or something once you've recovered from exam fatigue today. Then focus on it tomorrow, because you only have one day left?
Post by: 12070 on August 01, 2017, 12:43:28 pm
Potentially something like this:
Do one more past paper or something once you've recovered from exam fatigue today. Then focus on it tomorrow, because you only have one day left?
Okay, I'll do the 2016 one again today and tomorrow should I do more past papers?
Post by: RuiAce on August 01, 2017, 12:46:51 pm
Okay, I'll do the 2016 one again today and tomorrow should I do more past papers?Only if both:
a) You still have enough energy in you to
b) It's not too late in the day (you really shouldn't be doing a whole past paper the day before an exam after 6:30 PM)
Post by: gh971 on August 01, 2017, 03:15:21 pm
My math trial exam is on thursday and I was wondering how best I should structure my time management because as I've been doing practice papers sometimes I do the test super quickly and then other times I don't finish.
Thanks!!
Post by: RuiAce on August 01, 2017, 03:26:22 pm
Hey ATAR Notes,A question to think about. Does anything (or things) in particular cause you to finish some early and struggle with others? E.g. The year of the paper, the company/school who writes it, a bunch of topics that you dislike, time of day you do the paper?
My math trial exam is on thursday and I was wondering how best I should structure my time management because as I've been doing practice papers sometimes I do the test super quickly and then other times I don't finish.
Thanks!!
Post by: winstondarmawan on August 01, 2017, 04:45:38 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20615428_1287430571382463_536652419_o.png?oh=5e7214d5505fd8305c65d29000ccd6f7&oe=5981D56E
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20616187_1287430424715811_1281406970_o.png?oh=3b04c1f9e87c952043520a56af576ba7&oe=5981C3A3 Parts (ii), (iii)
Post by: RuiAce on August 01, 2017, 07:42:12 pm
Hello! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20615428_1287430571382463_536652419_o.png?oh=5e7214d5505fd8305c65d29000ccd6f7&oe=5981D56E
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/20616187_1287430424715811_1281406970_o.png?oh=3b04c1f9e87c952043520a56af576ba7&oe=5981C3A3 Parts (ii), (iii)
The other question is ambiguous, but I believe it's saying that 5 is certainly attained and must be the highest number.
Recall that the sum of two odd numbers is even but the sum of an odd and never number is odd. So effectively we need the second number to be even. So we want (4,5), (2,5), (5,2) or (5,4), where this bracket notation denotes (number on first roll, number on second roll).
Post by: gilliesb18 on August 01, 2017, 08:41:40 pm
Find the derivative of x^3 using first principles.
I know the answer, being 3x, but I don't know how to get this using first principles. Can someone help me?
Thanks soo much:):):)
B.
Post by: EEEEEEP on August 01, 2017, 08:46:40 pm
Hello, can someone please help me with this question:You sure it's 3x?
Find the derivative of x^3 using first principles.
I know the answer, being 3x, but I don't know how to get this using first principles. Can someone help me?
Thanks soo much:):):)
B.
Shouldn't the derivative be 3x^2?
The result from using first principles and standard differentiation should be the same.
Post by: RuiAce on August 01, 2017, 08:48:21 pm
Hello, can someone please help me with this question:Not only is it 3x^2, note that you have to expand (x+h)^3 by expanding (x^2+2xh+h^2)(x+h)
Find the derivative of x^3 using first principles.
I know the answer, being 3x, but I don't know how to get this using first principles. Can someone help me?
Thanks soo much:):):)
B.
Post by: gilliesb18 on August 01, 2017, 08:50:14 pm
But once I have expanded that out, which is what I did in the first place but got no where, what do I do?
Mod Edit: Merge :)
Post by: RuiAce on August 01, 2017, 09:01:29 pm
But once I have expanded that out, which is what I did in the first place but got no where, what do I do?
Edit: Typo
Post by: Claudiaa on August 01, 2017, 09:03:15 pm
Post by: gilliesb18 on August 01, 2017, 09:14:50 pm
Yes I agree with you Claudia!!!! Definitely definitely definitely, thankyou sooo much!!!
Post by: winstondarmawan on August 01, 2017, 10:17:27 pm
How about part ii?
Post by: RuiAce on August 01, 2017, 10:29:48 pm
How about part ii?What was the answer to (i)?
Post by: winstondarmawan on August 01, 2017, 10:36:23 pm
What was the answer to (i)?
I got 2 seconds
Post by: RuiAce on August 01, 2017, 10:47:12 pm
It may be hard understanding how the absolute value brackets vanished - comment if that's an issue
Post by: smshs2017 on August 01, 2017, 10:54:10 pm
Thank you
Post by: RuiAce on August 01, 2017, 10:55:47 pm
Hey, can you please help me with this question? integration and differenciation always confuse me.
Thank you
Post by: winstondarmawan on August 01, 2017, 10:56:04 pm
(http://i.imgur.com/sRVTXw4.jpg)
It may be hard understanding how the absolute value brackets vanished - comment if that's an issue
Thank you! How did you go from 2nd last to last line?
Post by: smshs2017 on August 01, 2017, 11:05:19 pm
Ah!!! Thank you!!! and sorry!!!!
Post by: RuiAce on August 01, 2017, 11:05:49 pm
Thank you! How did you go from 2nd last to last line?There's an error in the last line it seems; there should be a coefficient of 3 in front of the log
Post by: asd987 on August 01, 2017, 11:40:34 pm
Post by: jamonwindeyer on August 01, 2017, 11:57:17 pm
hi how would find the integral of sec3x tan3x dx neither of them are on the formula sheet
Hey! Don't stress about this, this exact integral used to be on the old reference sheet (table of integrals) but it isn't on there anymore, not assessable now ;D
Post by: 12070 on August 02, 2017, 10:03:05 am
Post by: RuiAce on August 02, 2017, 10:14:27 am
I don't understand this law. If you had e^x^2 and substituted 1 in for x. If you times the powers together you get e^2 however if you bring the 2 to the front you get 2e and e^2 does not equal 2e.Please add bracketing. It's ambiguous as to whether you mean \(e^{x^2}\) or \( \left(e^x\right)^2\)
Assuming you meant the latter though, I don't see how you can bring the 2 in front.
Post by: 12070 on August 02, 2017, 10:23:45 am
The answer just switches between the two.
Post by: RuiAce on August 02, 2017, 10:26:43 am
I mean the former.
The answer just switches between the two.
Also don't forget that a 2 automatically appears from the trapezoidal rule formula, not from actually computing f((a+b)/2)
Post by: 12070 on August 02, 2017, 10:38:10 am
[/quote]
Ahhh. I completely missed the two. Thank you. So only when it's a log you can bring a power to the front?
Post by: Opengangs on August 02, 2017, 10:40:32 am
Also don't forget that a 2 automatically appears from the trapezoidal rule formula, not from actually computing f((a+b)/2)If the power is inside, then you can bring it down.
Ahhh. I completely missed the two. Thank you. So only when it's a log you can bring a power to the front?
For example, log(x^2) is the same as 2logx, but not the same as [log(x)]^2
Post by: RuiAce on August 02, 2017, 10:42:04 am
Also don't forget that a 2 automatically appears from the trapezoidal rule formula, not from actually computing f((a+b)/2)If you're referring to the log law \( \log M^n = n \log M\) then yes
Ahhh. I completely missed the two. Thank you. So only when it's a log you can bring a power to the front?
Post by: biffi023 on August 02, 2017, 02:05:57 pm
im warning u now that this might not make sense.. but hopefully i can kinda explain it! :o
this is a question from 2016 hsc paper.. i finished working it out and got it right, but on looking back at it.. shouldn't you only have to rotate it 'half' of the way around, seeing as you are given the whole shape? like if you were given this same question but with only the 0<y<2 portion, wouldnt you use the same working and rotate it the whole way?
im just thinking that seeing as you have 2 symmetrical 'parts' of the same curve, both rotating that you would only have to rotate it half way around the x-axis for them to 'meet up' and form the whole volume?? btw.. im only really talking abt the 'C2' part of it.. C1 i just found using volume of a sphere formula..
but maybe my whole thinking is wrong!?? if someone can make that clear to me wld be awesome!!
thank u!
Post by: gilliesb18 on August 02, 2017, 02:18:21 pm
f(x)= 2x^3 +5x
for some reason i keep getting 6x^3 +5, but the answer is 6x^2 +5 so i don't know how to get it.
Can someone help me once again???
Thanks...
Post by: RuiAce on August 02, 2017, 02:23:20 pm
Hey :)
im warning u now that this might not make sense.. but hopefully i can kinda explain it! :o
this is a question from 2016 hsc paper.. i finished working it out and got it right, but on looking back at it.. shouldn't you only have to rotate it 'half' of the way around, seeing as you are given the whole shape? like if you were given this same question but with only the 0<y<2 portion, wouldnt you use the same working and rotate it the whole way?
im just thinking that seeing as you have 2 symmetrical 'parts' of the same curve, both rotating that you would only have to rotate it half way around the x-axis for them to 'meet up' and form the whole volume?? btw.. im only really talking abt the 'C2' part of it.. C1 i just found using volume of a sphere formula..
but maybe my whole thinking is wrong!?? if someone can make that clear to me wld be awesome!!
thank u!
(http://i.imgur.com/ZZsoELd.png)
Subtleties are left to second year uni.
Post by: RuiAce on August 02, 2017, 02:26:00 pm
Another one on derivatives using first principles (sorry :-[)
f(x)= 2x^3 +5x
for some reason i keep getting 6x^3 +5, but the answer is 6x^2 +5 so i don't know how to get it.
Can someone help me once again???
Thanks...
If you're having trouble, please post relevant working out.
Post by: biffi023 on August 02, 2017, 02:29:31 pm
(http://i.imgur.com/ZZsoELd.png)
Subtleties are left to second year uni.
ahh!! thanks so much ;D ;D
Post by: Mymy409 on August 02, 2017, 03:35:46 pm
Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% per annum paid monthly.
I've seen superannuation questions where the investment is done at the beginning of the period, but not at the end.
Post by: RuiAce on August 02, 2017, 03:39:16 pm
How would you do this question?
Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% per annum paid monthly.
I've seen superannuation questions where the investment is done at the beginning of the period, but not at the end.
Post by: Mymy409 on August 02, 2017, 04:29:32 pm
Thanks, Rui
Post by: Opengangs on August 02, 2017, 04:31:52 pm
Hopefully, you're able to solve it from here.
(:
Post by: RuiAce on August 02, 2017, 04:34:33 pm
You will need to form a generalised equation for An, using geometric progression.I presume they already knew that since they could handle deposits at the beginning of the year. I just illustrated the other one for means of contrast (because it was at the end of the year, i.e. we're considering an annuity in arrears as opposed to annuity in advance).
Hopefully, you're able to solve it from here.
(:
Post by: jaskirat on August 02, 2017, 04:45:26 pm
Post by: RuiAce on August 02, 2017, 04:55:06 pm
Need help with this question :)
Post by: ekhan_01 on August 02, 2017, 06:41:56 pm
there's no solutions and I have noooooo idea on what to do
Thank you
Post by: 12070 on August 02, 2017, 07:48:08 pm
Hi,
there's no solutions and I have noooooo idea on what to do
Thank you
I wouldn't be worried about questions like at the moment. The first two are some what manageable but the last one is designed to sort out extension students. For i you have to use right angle trig to find UO and then minus TO thus giving you UT. You'll then use UTxST for ii but honestly, unless you have absolutely everything else nailed in the course I wouldn't spend any time worrying about it.
Post by: jamonwindeyer on August 02, 2017, 07:59:19 pm
Next, we just multiply the width of the rectangle from Part (i), with the height, which we established was \(r\sin{\theta}\). You'll get:
The last bit requires you to differentiate this function of area and find the maximum, but the method is a little different than normal. The hardest bit is the derivative itself, which requires the product rule and the chain rule as well. You should get:
Now to prove that \(\theta=\frac{\pi}{12}\) corresponds to a maximum, we actually need to just substitute into this expression to prove it puts the derivative equal to zero. You'll also need to substitute values slightly smaller and larger to prove it is a max and not a min :) this is the atypical bit, but we do it because finding the exact value of \(\frac{\pi}{12}\) is, if memory serves, otherwise impossible with 2U methods - So we just use their indicator to guide us ;D a brief rundown but hopefully it helps!
Post by: Mymy409 on August 02, 2017, 08:12:04 pm
Need help with this. Thanks.
Post by: Ali_Abbas on August 02, 2017, 08:23:15 pm
Now to prove that \(\theta=\frac{\pi}{12}\) corresponds to a maximum, we actually need to just substitute into this expression to prove it puts the derivative equal to zero. You'll also need to substitute values slightly smaller and larger to prove it is a max and not a min :) this is the atypical bit, but we do it because finding the exact value of \(\frac{\pi}{12}\) is, if memory serves, otherwise impossible with 2U methods - So we just use their indicator to guide us ;D a brief rundown but hopefully it helps!
Or you can do this:
Post by: winstondarmawan on August 02, 2017, 08:25:31 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20623991_1288382307953956_1340814897_n.png?oh=19ee69f9b0f7024aa8306b170b649109&oe=59841D89 (Part iii)
I got alpha = 0.64 radians and beta = 0.93 radians, for part ii if that helps. TIA
Post by: Shadowxo on August 02, 2017, 08:38:39 pm
Hello! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20623991_1288382307953956_1340814897_n.png?oh=19ee69f9b0f7024aa8306b170b649109&oe=59841D89 (Part iii)
I got alpha = 0.64 radians and beta = 0.93 radians, for part ii if that helps. TIA
What I would do is get the area of each sector and add them, then subtract the area of the rectangle. This should give half of the area enclosed.
It would be something along the lines of
P.S. Way I solved the earlier question by ekhan was to find the maximum area was using sin(2x) and cos(2x) double angle formulae, is this taught in 2U?
Post by: jamonwindeyer on August 02, 2017, 08:42:01 pm
Very true! But 2U students don't learn the method to get the exact answer there, only a decimal approximation with their calculator - Which seems really strange in honesty because it means you HAVE to prove it by substitution, but that's what it was :P
A man contributes $1200 each year into a superannuation for the first 30 years of his working life. For the next 15 years, until retirement, He decides to increase this and invests a total of $5000 each year. If the investment earns 8% p.a. paid yearly over the whole year period, how much will his investment be worth upon retirement?
Need help with this. Thanks.
Hey Mymy! Have you made a start at all? Any progress you can share for us? Only because this is a question students normally can start but not finish and I'd love to take it from where you left off ;D
Remember to build up the pattern!
P.S. Way I solved the earlier question to find the maximum area was using sin(2x) and cos(2x) double angle formulae, is this taught in 2U?
Nah, unfortunately not!
Post by: RuiAce on August 02, 2017, 09:27:29 pm
I wouldn't be worried about questions like at the moment. The first two are some what manageable but the last one is designed to sort out extension students. For i you have to use right angle trig to find UO and then minus TO thus giving you UT. You'll then use UTxST for ii but honestly, unless you have absolutely everything else nailed in the course I wouldn't spend any time worrying about it.No this was certainly in my 2013 2U CSSA paper. I think it's fair to worry about it.
Post by: Disney on August 02, 2017, 09:45:49 pm
Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a. paid monthly?
Also how is best to prepare the night before for a 2u exam? Kinda stressing that I don't know my stuff
Post by: 12070 on August 02, 2017, 09:59:38 pm
No this was certainly in my 2013 2U CSSA paper. I think it's fair to worry about it.
I'm just saying that there probably won't be a max/min question involving right trig tomorrow. I did do that question a while ago but I could have spent the whole day studying questions like that only to walk into the exam and not know how to do one involving some other piece of theory. Then my whole day would have basically been wasted.
Post by: RuiAce on August 02, 2017, 10:02:14 pm
I'm just saying that there probably won't be a max/min question involving right trig tomorrow. I did do that question a while ago but I could have spent the whole day studying questions like that only to walk into the exam and not know how to do one involving some other piece of theory. Then my whole day would have basically been wasted.The way I see it, it's a max/min question. Maybe it won't involve trig, but it will be in there. The previous parts were trig, and that may or may not appear, but it doesn't change that the last part is a max/min question and I think it's fair enough to worry about max/min problems if they cause trouble. (I highly doubt they would've neglected all other kinds of problems.) The fact that trig was combined it with is a less important point.
I also don't think 2U maths is a good subject to be predicting questions.
Post by: RuiAce on August 02, 2017, 10:04:52 pm
Hey guys! I'm very confused how to answer questions on superannuation with series! :( Particularly something like:This question was started off at the start of this page. Please give more insight as to where the problems are.
Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a. paid monthly?
Also how is best to prepare the night before for a 2u exam? Kinda stressing that I don't know my stuff
How much studying have you been doing despite the stress? Because even people who have studied lots stress, but sometimes they just don't realise what they know.
Post by: 12070 on August 02, 2017, 10:22:37 pm
The way I see it, it's a max/min question. Maybe it won't involve trig, but it will be in there. The previous parts were trig, and that may or may not appear, but it doesn't change that the last part is a max/min question and I think it's fair enough to worry about max/min problems if they cause trouble. (I highly doubt they would've neglected all other kinds of problems.) The fact that trig was combined it with is a less important point.
I also don't think 2U maths is a good subject to be predicting questions.
I totally agree. I feel like I'm being misinterpreted here. If you are learning an instrument such as the piano, you don't start by trying to learn a Rachmaninoff piece. You build technique over time. Today I went through topics I felt I was weak at and did some easier questions so I at least knew the fundamentals of each topic and hopefully if it comes up at the back of the paper tomorrow I will know how to start and can get a few marks. I'm not at all saying to not worry about certain topics.
Post by: jamonwindeyer on August 02, 2017, 10:31:42 pm
Hey guys! I'm very confused how to answer questions on superannuation with series! :( Particularly something like:
Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a. paid monthly?
Also how is best to prepare the night before for a 2u exam? Kinda stressing that I don't know my stuff
Hey! As Rui said the idea here is to build up a series, for you:
That interest rate comes from dividing 8.2% by 12 to get a monthly rate. What you'll notice is a geometric series will build up:
This is always the approach with these questions - Step through what happens in the account, spot the series, and from there you can use your sum of a geometric series formula to make it look simpler and rearrange from there! ;D for you you'd be looking at the value after 3 years (36 months) so:
I've fudged this question a bit because it doesn't quite explain it properly to know 100% how it would structure, but this is the rough idea - Hope it helps :)
Post by: smshs2017 on August 02, 2017, 10:39:01 pm
Thanks.
Post by: RuiAce on August 02, 2017, 10:42:10 pm
Hey, can i please get help for this question, especially when it gets to part three.What was the value of k in part one?
Thanks.
Post by: jamonwindeyer on August 02, 2017, 10:44:51 pm
Hey, can i please get help for this question, especially when it gets to part three.
Thanks.
Hey! So Part III asks for a rate of change, which is just a derivative:
Substitute \(t=1\) to get your answer there! For Part IV, we set \(N=3000\) and find how long it takes to get there:
Hope this helps! :)
Post by: Claudiaa on August 02, 2017, 10:50:48 pm
Because walking into my first 3u exam, I scored a shocking 65%! But I had solely relied on Maths in Focus...:/
So any suggestions? Are there textbooks that are particularly good for certain topics etc.
Thanks :)
Post by: smshs2017 on August 02, 2017, 10:56:31 pm
What was the value of k in part one?the value of k is 0.050953563
Post by: fantasticbeasts3 on August 02, 2017, 10:57:34 pm
Hi! Could any recommend what they found was the best textbook for Ext 1 math? (or combination of textbooks?)
Because walking into my first 3u exam, I scored a shocking 65%! But I had solely relied on Maths in Focus...:/
So any suggestions? Are there textbooks that are particularly good for certain topics etc.
Thanks :)
oh, maths in focus... my favourite book! (not) i only do 2u, but i can recommend cambridge, which is pretty good, but past papers are the best way to study for exams :-)
Post by: RuiAce on August 02, 2017, 11:00:26 pm
Hey, can i please get help for this question, especially when it gets to part three.
Thanks.
the value of k is 0.050953563
Post by: smshs2017 on August 02, 2017, 11:07:05 pm
Hey! So Part III asks for a rate of change, which is just a derivative:hey, i did that but i got 1.12595933 but the answer is 3.6.
Substitute \(t=1\) to get your answer there! For Part IV, we set \(N=3000\) and find how long it takes to get there:
Hope this helps! :)
Post by: smshs2017 on August 02, 2017, 11:11:58 pm
thanks but i am not getting the same answer for the rate. is t = 1?
Post by: RuiAce on August 02, 2017, 11:13:23 pm
hey, i did that but i got 1.12595933 but the answer is 3.6.They tricked you.
t is measured in hours, not days. Have to sub in t=24.
Post by: smshs2017 on August 02, 2017, 11:21:38 pm
They tricked you.oh! but my answer still isnt the same. i'll attact the solution here. they multiplied k with N....
t is measured in hours, not days. Have to sub in t=24.
Post by: RuiAce on August 02, 2017, 11:43:54 pm
WolframAlpha input
Post by: MegatronMaximus on August 03, 2017, 05:43:24 pm
Post by: Mounica on August 03, 2017, 07:21:53 pm
can someone help me with this question
thanks :D
Post by: jamonwindeyer on August 03, 2017, 08:00:59 pm
Can someone please explain how gradient graphs work? Like how to draw from f(x) to f'(x) and vice versa. I get confused when there needs to be an inflection point when you do it backwards.
Welcome to the forums!! So a few tips here...
When going from \(f(x)\) to \(f'(x)\), mark the sign of your gradient at every point on the graph with a '+' or '-'. This will help in your sketch, because anywhere there is a plus, the new graph should be above the axis. Minus, it is below. Also keep in mind:
- Turning points become x-intercepts
- Points of inflexion become turning points
When going from \(f'(x)\) to \(f(x)\), the opposite applies. The value of the function corresponds to the gradient: If it is above the axis your new graph should slope up. If it is below your new graph should slope down.
- x-intercepts become turning points
- Turning points become points of inflexion
Feel free to post an example we can help with ;D
Post by: jamonwindeyer on August 03, 2017, 08:16:50 pm
hey
can someone help me with this question
thanks :D
Hey! Let me run you through the processes for each. The first bit, we know two coordinates, \((0,-6)\) and \((5,-1)\). So we can substitute to form two equations:
You can solve those simultaneously ;D for Part (ii), we know the x-coordinate is \(\alpha\), so let's substitute that into the equation (with the h and k we now have) and we'll get a y-coordinate:
That y-coordinate we just found is actually the HEIGHT of the rectangle (but we ignore a negative if there is one) - The width is \(2\alpha\). Why? The function we are examining is even, so if Q is at \(\alpha\), then P is at \(-\alpha\). PQ is therefore \(2\alpha\)!
So the area is:
From this, I reckon \(h=30\) and \(k=5\) to match the statement in the question ;)
Right, so we are working with the function they give in Part (iii).
To find a maximum, we differentiate (quotient rule):
Now put it equal to zero:
Then you test that with a point either side check as usual - Then pop it back into the area formula :)
Note: Did this quickly, subject to computational errors.
Post by: RuiAce on August 03, 2017, 08:19:11 pm
Welcome to the forums!! So a few tips here...Hey by the way Jamon, I haven't checked if this stuff is in your guides yet but if not, we really need to add it there so we can just link it...:P
When going from \(f(x)\) to \(f'(x)\), mark the sign of your gradient at every point on the graph with a '+' or '-'. This will help in your sketch, because anywhere there is a plus, the new graph should be above the axis. Minus, it is below. Also keep in mind:
- Turning points become x-intercepts
- Points of inflexion become turning points
When going from \(f'(x)\) to \(f(x)\), the opposite applies. The value of the function corresponds to the gradient: If it is above the axis your new graph should slope up. If it is below your new graph should slope down.
- x-intercepts become turning points
- Turning points become points of inflexion
Feel free to post an example we can help with ;D
Post by: Kekemato_BAP on August 03, 2017, 09:58:43 pm
Solve for x. Thanks :)
Post by: kiwiberry on August 03, 2017, 10:06:36 pm
Hi I need help with this question^-^
Solve for x. Thanks :)
Post by: good_stuff on August 03, 2017, 10:33:59 pm
Studying for trials and I keep getting confused by the [roots] kind of questions, with the alphas + betas,,, then the alphaxbetas,,,, (Im fine with these ones) AND THEN OUT OF THE BLUE
ALPHA + 1/ALPHA
ALPHA^3 + BETA ^3
and so on. I'm always get caught out and end up spending forever on them but I just can't get the hang of them.
Wondering if anyone has lil hacks?
Also, is there a specific list of common questions for the types where they make you figure the formula out after the initial a+b, ab kind of questions? (Like the ones i get stuck on)
Post by: RuiAce on August 03, 2017, 11:29:39 pm
Hey mathematicians!!
Studying for trials and I keep getting confused by the [roots] kind of questions, with the alphas + betas,,, then the alphaxbetas,,,, (Im fine with these ones) AND THEN OUT OF THE BLUE
ALPHA + 1/ALPHA
ALPHA^3 + BETA ^3
and so on. I'm always get caught out and end up spending forever on them but I just can't get the hang of them.
Wondering if anyone has lil hacks?
Also, is there a specific list of common questions for the types where they make you figure the formula out after the initial a+b, ab kind of questions? (Like the ones i get stuck on)
The other one cannot be asked as using only one of the root (as opposed to both of them) runs into severe ambiguity, unless you explicitly find what alpha is.
Post by: Natasha.97 on August 04, 2017, 12:06:32 am
How would the graph esinx be graphed, with stat points and an indication of what the value is when x = 0, pi, and 2pi?
Post by: RuiAce on August 04, 2017, 12:19:54 am
Hi :)
How would the graph esinx be graphed, with stat points and an indication of what the value is when x = 0, pi, and 2pi?
If you put it into graphing software, you will see that it looks like a distorted sine curve.
Post by: Natasha.97 on August 04, 2017, 01:55:45 am
If you put it into graphing software, you will see that it looks like a distorted sine curve.
:O are you serious about it being 4U? I ask because I got this question yesterday for my 2U trial...
Post by: RuiAce on August 04, 2017, 10:02:20 am
:O are you serious about it being 4U? I ask because I got this question yesterday for my 2U trial...Which company/school?
Post by: caitlinlddouglas on August 04, 2017, 11:18:49 am
Post by: RuiAce on August 04, 2017, 11:24:10 am
Hey i was wondering how to find the total distance a particle travels in a given time frame? Thanks!Recall that the distance differs from the displacement in that we only consider the magnitude. If the particle turns around, its displacement may start going the opposite way whereas its distance increases.
Hence, you should consider where the particle is at rest (to address the fact it may turn around), and then use the equation of the displacement to figure out the distance.
Please provide an example if you need more assistance. (There was an already answered example a few posts back but I'm not sure if you want to go scrolling for it.)
Post by: caitlinlddouglas on August 04, 2017, 12:46:45 pm
Recall that the distance differs from the displacement in that we only consider the magnitude. If the particle turns around, its displacement may start going the opposite way whereas its distance increases.Thanks heaps,
Hence, you should consider where the particle is at rest (to address the fact it may turn around), and then use the equation of the displacement to figure out the distance.
Please provide an example if you need more assistance. (There was an already answered example a few posts back but I'm not sure if you want to go scrolling for it.)
THe question was: acceleration of a particle is given by x:=4sin2t. Initially particle is 1m to left of origin and has a velocity of 2m/s. Find the distance travelled in the first 4 seconds.
Post by: RuiAce on August 04, 2017, 01:52:34 pm
Thanks heaps,Are you sure there isn't a minus sign missing? Because the question is uglier but easier without it.
THe question was: acceleration of a particle is given by x:=4sin2t. Initially particle is 1m to left of origin and has a velocity of 2m/s. Find the distance travelled in the first 4 seconds.
Post by: anotherworld2b on August 05, 2017, 11:35:48 pm
I had a feeling that I did the question wrong because I was confused about how to find delta x
I believe I interpreted the responses to my question incorrectly.
I thought that when you are given information about the question having 4 strips/trapeziums/subsections n becomes 4 +1 when you want to find delta x. Or is n simply 4?
So I assumed to find delta x I had to divide by 5.
I will post future questions in the mathematics (2U) thread. I'm not quite sure how I migrated to this thread :o
Post by: Opengangs on August 05, 2017, 11:45:04 pm
I hope I am in the right thread this time :)I answered in the 3U section, but for everyone's sake, I'll answer it again.
I had a feeling that I did the question wrong because I was confused about how to find delta x
I believe I interpreted the responses to my question incorrectly.
I thought that when you are given information about the question having 4 strips/trapeziums/subsections n becomes 4 +1 when you want to find delta x. Or is n simply 4?
So I assumed to find delta x I had to divide by 5.
I will post future questions in the mathematics (2U) thread. I'm not quite sure how I migrated to this thread :o
When we say "number of subsections or strips", we are merely talking about the interval between the two ending function values. For n strips, there will be n + 1 function values. However, both Simpson's rule and Trapezoidal Rule merely want the interval between two such function values, which is denoted as the sum of the two end points divided by the number of subsections or strips or applications.
This tells us the relative position between two function values, which is the interval they are after.
So, if they ask four function values between 0 and 9, they simply want the change in x (delta x) to be: (0 + 9)/3, since for there to be four such function values, there are only three strips.
Post by: pikachu975 on August 06, 2017, 12:12:32 am
Hi :)
How would the graph esinx be graphed, with stat points and an indication of what the value is when x = 0, pi, and 2pi?
Should be doable in 2 unit with knowledge of stationary points. Find their nature, find x and y intercepts and plot the points at 0, pi and 2pi then sketch.
Post by: anotherworld2b on August 06, 2017, 10:03:32 am
I'm not sure what to do
Post by: RuiAce on August 06, 2017, 10:13:53 am
Can I have help with this question please?What you wrote out so far is correct. Now, replace \(x_0, x_1, \dots, x_4\) with \(0, \frac\pi4, \dots, \pi\) as specified.
I'm not sure what to do
Post by: anotherworld2b on August 06, 2017, 10:56:57 am
I was wondering for this question what [O;5] is. Does that mean the upper limit is five and the lower limit 0?
What you wrote out so far is correct. Now, replace \(x_0, x_1, \dots, x_4\) with \(0, \frac\pi4, \dots, \pi\) as specified.
Post by: jakesilove on August 06, 2017, 11:09:55 am
I see :D
I was wondering for this question what [O;5] is. Does that mean the upper limit is five and the lower limit 0?
Yep, it does!
Post by: Jenny_trn on August 06, 2017, 08:41:55 pm
1. Why isn't the y=3 included in the volume function thing . like V= pie integral [3 - log2x] (as shown in the question?)
2. also when you re-arrange y= log2x, in terms of x, when you square it, what are the steps to reach the final answer --> e^yln4 as shown?
Post by: RuiAce on August 06, 2017, 08:46:22 pm
Hi, currently stuck on this question!1 - Is unnecessary, as your rotation means that you're integrating with respect to \(y\) and not \(x\). Also note that the \(y=3\) actually appears as the upper boundary of your integral: \(V=\pi\int_0^{\boxed{\textbf{3}}}...\). Of course, if we rotated about the x-axis then we'd have \( V=\pi \int_1^{\log_2 3}(3-\log_2 x)^2\,dx \)
1. Why isn't the y=3 included in the volume function thing . like V= pie integral [3 - log2x] (as shown in the question?)
2. also when you re-arrange y= log2x, in terms of x, when you square it, what are the steps to reach the final answer --> e^yln4 as shown?
Post by: Jenny_trn on August 06, 2017, 09:01:01 pm
stuck on part ii) of this question
answer to part i) was (square root n + 1) - ( square root n)
I'm not sure how part i is used to find out the value of the sum from part i)
Thanks in advance!
Post by: RuiAce on August 06, 2017, 09:14:59 pm
Hi!
stuck on part ii) of this question
answer to part i) was (square root n + 1) - ( square root n)
I'm not sure how part i is used to find out the value of the sum from part i)
Thanks in advance!
Post by: anotherworld2b on August 08, 2017, 12:57:04 am
I also had a question about whether or not my answer 18.9 (using trapezoidal rule) would be considered off? I used my class pad and the answer was 19.63?
Post by: jamonwindeyer on August 08, 2017, 08:49:10 am
Hi I attempted to do a question using the trapezoidal rule. I was wondering if someone could check my working please.
I also had a question about whether or not my answer 18.9 (using trapezoidal rule) would be considered off? I used my class pad and the answer was 19.63?
Looks good to me assuming there is no calculator error! Your answer is within 5% of the actual, that's a good estimate!
Post by: anotherworld2b on August 08, 2017, 09:17:02 am
I thought the answer had to be exact :3
I was wondering why do you need 'n' to be even to use Simpson's rule? Why can't n be odd like in the trapezoidal rule?
Looks good to me assuming there is no calculator error! Your answer is within 5% of the actual, that's a good estimate!
Post by: kiiaaa on August 08, 2017, 11:15:35 am
I was doing this paper from another school and question ii) and iii) confused me as i dont know how did they find 'a' and the common ratio. could someone please help me and explain that to me?
Thank you so much :)
Post by: RuiAce on August 08, 2017, 11:26:23 am
Hey guys,
I was doing this paper from another school and question ii) and iii) confused me as i dont know how did they find 'a' and the common ratio. could someone please help me and explain that to me?
Thank you so much :)
Post by: Natasha.97 on August 08, 2017, 11:38:59 am
Just adding onto the above, the proof is as follows:
(http://i.imgur.com/YHH7RkI.png)
Hope this helps :)
Post by: kiiaaa on August 08, 2017, 11:39:23 am
OMG i completely didint relize i) was part of ii) the graph completely threw me off. Thank you sooo much! :) :) :) =P
Just adding onto the above, the proof is as follows:
(http://i.imgur.com/YHH7RkI.png)
Hope this helps :)
thank you so much. this question makes soo much sense now :)) :) :) :) :)
Post by: katnisschung on August 08, 2017, 06:47:42 pm
y=lnx/x
how the flip do i find the inflection point
i've done double differentiation (attached) but need help factorising for x
thanks!
Post by: katnisschung on August 08, 2017, 06:53:15 pm
derived it and got... (attached also capture 1)
answer in book is 2x+(1/1-x)
Post by: pikachu975 on August 08, 2017, 07:17:57 pm
trying to sketch this graph
y=lnx/x
how the flip do i find the inflection point
i've done double differentiation (attached) but need help factorising for x
thanks!
-x - 2x + 2xlnx = 0
-3x + 2xlnx = 0
x (2lnx - 3) = 0
x = 0, lnx = 3/2 so x = e^(3/2)
But we know x = 0 is not possible as y = lnx / x so x = e^(3/2) and just test that with the points on both sides.
Post by: RuiAce on August 08, 2017, 07:25:11 pm
trying to sketch this graphGiven that the second derivative is bizarre, it is highly recommend that you test a bit on both sides instead of apply the second derivative test.
y=lnx/x
how the flip do i find the inflection point
i've done double differentiation (attached) but need help factorising for x
thanks!
Post by: katnisschung on August 08, 2017, 07:46:46 pm
Given that the second derivative is bizarre, it is highly recommend that you test a bit on both sides instead of apply the second derivative test.
[/quote
yeah i did, did a table of values yet the answers still called for a max at (e,1/e)
and inflexion pt at e^3/4 , -3/(2e^3/2)
weirdest q i've seen
Post by: katnisschung on August 08, 2017, 07:48:10 pm
step by step differentiation for this q pls (capture)this one got lost thanks :)
derived it and got... (attached also capture 1)
answer in book is 2x+(1/1-x)
Post by: RuiAce on August 08, 2017, 10:01:57 pm
Given that the second derivative is bizarre, it is highly recommend that you test a bit on both sides instead of apply the second derivative test.This was the output Wolfram gave me, and the root of the second derivative is where the inflexion point is.
[/quote
yeah i did, did a table of values yet the answers still called for a max at (e,1/e)
and inflexion pt at e^3/4 , -3/(2e^3/2)
weirdest q i've seenstep by step differentiation for this q pls (capture)Where did this come from? This seems like a pointlessly long derivative
derived it and got... (attached also capture 1)
answer in book is 2x+(1/1-x)
Post by: anotherworld2b on August 08, 2017, 10:14:17 pm
I was wondering why do you need 'n' to be even to use Simpson's rule? Why can't n be odd like in the trapezoidal rule?
Post by: RuiAce on August 08, 2017, 10:24:55 pm
I was wondering why do you need 'n' to be even to use Simpson's rule? Why can't n be odd like in the trapezoidal rule?It's more a consequence of the fact that n+1 has to be odd, which consequently implies n is even.
The reason n+1, i.e. the number of function values needs to be odd, has to do with how Simpson's rule actually works in the first place.
Simpson's rule doesn't use straight lines to estimate areas under curves, but rather uses parabolas. There is a theorem (which is provable, albeit a bit messy) that any three points UNIQUELY define a parabola. This serves as a basis.
Because we need three points to uniquely define the parabola, as opposed to just two points, we need to group points in groups of three instead. So we'd group \(x_0, x_1, x_2\), then \(x_2, x_3, x_4\), then keep going up until \(x_{n-2}, x_{n-1}, x_n\). And these are the points of interest. The number of points we have is of course the number of function values we must therefore consider.
\(x_0\) is the first point. With \(x_0\), comes \(x_1\) and \(x_2\), so that's two more points. With \(x_2\), comes \(x_3\) and \(x_4\), and that's another two more points. This will keep going until we reach \(x_n\).
Notice how we started off with just one single point, but then when we added we always added two more points. When you add 2 to an odd number, you get an odd number.
Visualising it, when you add 2 to 1 you get 3, which is odd. When you add 2 to 3 you get 5 which is still odd. And you're only gonna be stuck with odd terms forever.
Hence, the odd number of function values, and thus the even number of sub-intervals.
Post by: kiiaaa on August 09, 2017, 05:22:48 pm
I was wondering if you could tell me how you would solve this. Like I know it isn't A or D due to the process of elimination but I have absolutely no idea how to solve it or to eliminate another option. Could you also please tell me the process you used?
Thank you sooo much! :) :) :) :)
Post by: RuiAce on August 09, 2017, 05:24:42 pm
hello, guys could you please help me in this question?
I was wondering if you could tell me how you would solve this. Like I know it isn't A or D due to the process of elimination but I have absolutely no idea how to solve it or to eliminate another option. Could you also please tell me the process you used?
Thank you sooo much! :) :) :) :)
Post by: kiiaaa on August 09, 2017, 07:13:02 pm
Thank you! that really clears it up
I also had another question- may be a sort of stupid one but:
for what type of questions do we change our calculator into radians mode and which types of questions should it be in degrees?
I've confused myself a lot and would like some clarification please so during the exam in don't waste a minute or two contemplating which mode it should be in hahaha =P =P =P =P
Thank you :)
Post by: Natasha.97 on August 09, 2017, 07:33:07 pm
Thank you! that really clears it up
I also had another question- may be a sort of stupid one but:
for what type of questions do we change our calculator into radians mode and which types of questions should it be in degrees?
I've confused myself a lot and would like some clarification please so during the exam in don't waste a minute or two contemplating which mode it should be in hahaha =P =P =P =P
Thank you :)
Hi!
- Radians usually always involve pi, e.g. find tan theta = 1 over sqroot of 2 from 0<x<2pi
- Degrees is the exact opposite, e.g. find tan theta = sqroot 3 from 0<x<360 degrees
Hope this helps! :)
Post by: Shadowxo on August 09, 2017, 07:53:30 pm
Thank you! that really clears it upJust adding, almost everything is in radians (eg differentiation, integration, etc) while degrees is often used for angles in trig, such as "find this angle in degrees". Usually the default is radians unless they ask for it in degrees or give you a value in degrees.
I also had another question- may be a sort of stupid one but:
for what type of questions do we change our calculator into radians mode and which types of questions should it be in degrees?
I've confused myself a lot and would like some clarification please so during the exam in don't waste a minute or two contemplating which mode it should be in hahaha =P =P =P =P
Thank you :)
You should also be able to tell when something is in degrees and radians - radians will be a smaller number often including Pi (typically 0-2pi ie 0-6.28), while degrees will be much larger, eg 30-360°
Also converting degrees to radians just requires *pi/180 while radians to degrees is *180/pi
Post by: RuiAce on August 09, 2017, 09:09:51 pm
Thank you! that really clears it upEssentially, the above are the main guidelines you should need. However, the following are the only two necessary and sufficient checks:
I also had another question- may be a sort of stupid one but:
for what type of questions do we change our calculator into radians mode and which types of questions should it be in degrees?
I've confused myself a lot and would like some clarification please so during the exam in don't waste a minute or two contemplating which mode it should be in hahaha =P =P =P =P
Thank you :)
1. If an angle is involved IN the question, look out for the circle. Because if I told you to compute \(\sin 45\), without that \(^\circ\) degree symbol there, I mean radians.
2. If an angle is NOT involved in the question, ordinarily you should always be assuming radians for the exact reason Shadow said at the start.
Post by: kiiaaa on August 10, 2017, 05:33:02 pm
Hi!
- Radians usually always involve pi, e.g. find tan theta = 1 over sqroot of 2 from 0<x<2pi
- Degrees is the exact opposite, e.g. find tan theta = sqroot 3 from 0<x<360 degrees
Hope this helps! :)
Just adding, almost everything is in radians (eg differentiation, integration, etc) while degrees is often used for angles in trig, such as "find this angle in degrees". Usually the default is radians unless they ask for it in degrees or give you a value in degrees.
You should also be able to tell when something is in degrees and radians - radians will be a smaller number often including Pi (typically 0-2pi ie 0-6.28), while degrees will be much larger, eg 30-360°
Also converting degrees to radians just requires *pi/180 while radians to degrees is *180/pi
Essentially, the above are the main guidelines you should need. However, the following are the only two necessary and sufficient checks:
1. If an angle is involved IN the question, look out for the circle. Because if I told you to compute \(\sin 45\), without that \(^\circ\) degree symbol there, I mean radians.
2. If an angle is NOT involved in the question, ordinarily you should always be assuming radians for the exact reason Shadow said at the start.
THANK YOU SO MUCH GUYS FOR YOUR HELP!
also, i had a few more questions and was wondering if you could help me please.
- how do you integrate for chain rule type questions? like for like you know how for chain rule for differentiation you bring down the power of the brackets then times by what's in the bracket. how do you integrate a chain rule type question? ( have attached an example) [ INTEGRATE: (5+3x)2
- also if they ask to integrate something which has the denominator of 'x' how do you integrate it as if i convert it to be 'x^-1' and integrate it will be '-1^0' which is basically 1 which is why I'm confused how to tackle such questions and was wondering if you could sort of teach me how to solve these questions please? ( i also attached an example of what I meant in case my blabber didn't make sense) [INTEGRATE: ( x2 +3x)/x]
Thank you soo much and i hope all of that makes sense? i really appreciate your help :D :D :D :D :D
EDIT: so the picture that has the example is too big of a file to upload so ill try my best to do write the question to show what i mean. i hope it makes sense. if not please let me know :)
Post by: pikachu975 on August 10, 2017, 06:04:25 pm
THANK YOU SO MUCH GUYS FOR YOUR HELP!
also, i had a few more questions and was wondering if you could help me please.
- how do you integrate for chain rule type questions? like for like you know how for chain rule for differentiation you bring down the power of the brackets then times by what's in the bracket. how do you integrate a chain rule type question? ( have attached an example) [ INTEGRATE: (5+3x)2
- also if they ask to integrate something which has the denominator of 'x' how do you integrate it as if i convert it to be 'x^-1' and integrate it will be '-1^0' which is basically 1 which is why I'm confused how to tackle such questions and was wondering if you could sort of teach me how to solve these questions please? ( i also attached an example of what I meant in case my blabber didn't make sense) [INTEGRATE: ( x2 +3x)/x]
Thank you soo much and i hope all of that makes sense? i really appreciate your help :D :D :D :D :D
EDIT: so the picture that has the example is too big of a file to upload so ill try my best to do write the question to show what i mean. i hope it makes sense. if not please let me know :)
For the first example since the terms in the bracket are linear then you just do reverse of chain rule: add one to the power, and DIVIDE by the derivative of the brackets.
For the second example you split up the fraction into x + 3 by dividing through by x. Sometimes when x is on the bottom you use logs, or you split up the fraction.
Post by: kiiaaa on August 10, 2017, 11:53:59 pm
For the first example since the terms in the bracket are linear then you just do reverse of chain rule: add one to the power, and DIVIDE by the derivative of the brackets.
For the second example you split up the fraction into x + 3 by dividing through by x. Sometimes when x is on the bottom you use logs, or you split up the fraction.
Absolute legend! thank you sooo much :) :) :)
Post by: anotherworld2b on August 11, 2017, 12:42:40 am
Post by: Shadowxo on August 11, 2017, 07:06:40 am
Can I have help with part d please?
If you're doing it by hand you may want to expand the brackets :)
Post by: RuiAce on August 11, 2017, 09:58:22 am
That symbol is just \geq or \ge=0.6 \\ \int_q^3 k(1-x)(x-3) dx=0.6 \text{ and solve from there})
If you're doing it by hand you may want to expand the brackets :)
Post by: katnisschung on August 11, 2017, 09:49:08 pm
If one particle starts at the pt x=0 and travels with a velocity of v=2t^3 -7t+5
And a second particle starts at Tue point x=7 and travels with a velocity of v=8+2t^3. Will these particles collide?
Integrated to find x=1/2 t^4 -7/2t^2 +5t
And x=8t+2/4t^4+7
Equated these and answers suggest using discriminant?? Pls explain
Post by: Opengangs on August 11, 2017, 10:11:59 pm
How and why can I use the discriminant for this q.Let's forget about particle movement for the time being, and recall what the discriminant stands for.
If one particle starts at the pt x=0 and travels with a velocity of v=2t^3 -7t+5
And a second particle starts at Tue point x=7 and travels with a velocity of v=8+2t^3. Will these particles collide?
Integrated to find x=1/2 t^4 -7/2t^2 +5t
And x=8t+2/4t^4+7
Equated these and answers suggest using discriminant?? Pls explain
If we're trying to find the number of roots of a polynomial (such as a quadratic), we use the discriminant, because that will indicate how many real roots the function is going to yield.
For instance: x2 + 3x + 2 = 0 is going to yield two roots, since the discriminant, 9 - 8, is > 0.
In this way, we can also find whether or not two lines will intersect.
To further illustrate this, let's assume two quadratics: y = x2 - 2 and y = -2x2 + 3
To find whether or not they will intersect (or collide), we need to find the points of intersection (by equating).
x2 - 2 = -2x2 + 3
Move everything to one side to form a third polynomial. This quadratic is the equation of the quadratic when the two previous quadratic functions are subtracted from each other (so, we simply find its root to find the intersection point).
3x2 - 5 = 0
To determine the no. of roots within this new quadratic, we need to find the x-values for this statement to be true (using the quadratic formula). If the discriminant is less than zero, then the two original functions will not meet, because there are no real x-values to yield the new quadratic to be true.
We can relate this concept back to our question with velocity and displacement.
The first particle moves with a velocity of: v = 2t3 - 7t + 5
So, integrating this, we get: x1 = (t4)/2 - (7/2)t2 + 5t + C
Substituting in x = 0, when t = 0, C = 0.
x1 = (t4)/2 - (7/2)t2 + 5
The second particle moves with a velocity of: v = 8 + 2t3
So, integrating this, we get: x2 = 8t + (t4/2) + C
Substituting in x = 7, when t = 0, C = 7
x2 = 8t + (t4/2) + 7
Equating: (t4)/2 - (7/2)t2 + 5 = 8t + (t4/2) + 7
From here, we move every term to one side:
(7/2)t2 + 8t + 2 = 0
7t2 + 16t + 4 = 0
From here, we get a simple quadratic. To determine the number of roots that this quadratic has, we take the discriminant.
This new quadratic is simply the first particle subtracted with the second particle, and is just as relevant to finding the point of intersection.
Thus, to determine whether or not the two particles collide, we use the discriminant.
Post by: RuiAce on August 11, 2017, 10:17:11 pm
How and why can I use the discriminant for this q.
If one particle starts at the pt x=0 and travels with a velocity of v=2t^3 -7t+5
And a second particle starts at Tue point x=7 and travels with a velocity of v=8+2t^3. Will these particles collide?
Integrated to find x=1/2 t^4 -7/2t^2 +5t
And x=8t+2/4t^4+7
Equated these and answers suggest using discriminant?? Pls explain
Which is related to the fact that \(\Delta \ge 0\) implies solutions exist. Whereas \(\Delta < 0\) implies that they do not.
____________________________
Post by: katnisschung on August 12, 2017, 07:20:11 am
Post by: anotherworld2b on August 12, 2017, 05:46:50 pm
So far I gave this for part a
E(X) = 3(12) = 36
Post by: Sine on August 12, 2017, 05:55:06 pm
I was wondering how to do this question.(c) Y = 2X + 5
So far I gave this for part a
E(X) = 3(12) = 36
E(X) = 12
So
E(Y) = E(2X+5) = 2E(X) + 5 = 2 * 12 + 5 = 29
Sd(X) = 3
So
Sd(Y) = Sd(2x+5) = 2 * Sd(X) = 2 * 3 = 6
In general terms
E(aX+b) = aE(X) + b
Sd(aX+b) = |a|Sd(X)
and for completeness
Var(aX+b) = a2Var(X)
Hopefully you can do the other questions now :)
Post by: Adammurad on August 12, 2017, 08:28:48 pm
possible answers. At least three correct answers are required to pass the examination.
Suppose that a student guesses the answer to each question.
a What is the probability the student guesses every question correctly?
b What is the probability the student will pass the examination?
Would anyone help and show the working out and solution thanks
Post by: RuiAce on August 12, 2017, 08:32:06 pm
An examination consists of six multiple-choice questions. Each question has four
possible answers. At least three correct answers are required to pass the examination.
Suppose that a student guesses the answer to each question.
a What is the probability the student guesses every question correctly?
b What is the probability the student will pass the examination?
Would anyone help and show the working out and solution thanks
The second one is a binomial-probability, which is 3U. What is the source of this question?
Post by: Adammurad on August 12, 2017, 08:36:42 pm
Post by: RuiAce on August 12, 2017, 08:42:39 pm
Cambridge Senior maths ex14a unit 3/4By any chance, are you after the VCE section of the forum?
Post by: Adammurad on August 12, 2017, 08:50:36 pm
Post by: RuiAce on August 12, 2017, 09:01:06 pm
Post by: anotherworld2b on August 13, 2017, 11:36:07 am
Post by: RuiAce on August 13, 2017, 11:43:19 am
I am not sure how to find he standard deviation for q12. I am not quite sure what to do for q13The standard deviation is just the square root of the variance. Once you have Var(Y), you should only be taking the square root of it to find Stddev(Y).
Note that the properties of the standard deviation are NOT similar to those of the expected value. The standard deviation is the square root of the variance and nothing else.
Hint for 13: If X is the random variable measuring the length in centimetres, then Y = 0.01X is the random variable measuring it in metres.
Post by: 12070 on August 13, 2017, 03:31:23 pm
Post by: RuiAce on August 13, 2017, 03:32:34 pm
Hey just wondering if I can post Uni questions as my brother needs help with engineering and if so, where should I post them? This is the question:Can you please post this not in the HSC section but down here?
Post by: caitlinlddouglas on August 13, 2017, 05:44:10 pm
Find the domain of the graph for y=log|2x-3|
Thanks!
Hi I was wondering if someone could also help me with this question?
Find the equation of the line passing through the lint of intersection of 2x-3y-5=0 and x+2y=7 and the point (2,4). The solution says you need to put them together with a k in front of the second line but I don't get why? Thanks heaps :)
Mod edit: Posts merged. At times like this, please resort to the modify function at the top right corner of a post to refrain from double posting.
Post by: RuiAce on August 13, 2017, 05:56:32 pm
Hey I was wondering why the answeR is x>0 and not all real x, x>3/2?
Find the domain of the graph for y=log|2x-3|
Thanks!
For the other question, have you been taught the k-method in class yet?
Post by: caitlinlddouglas on August 13, 2017, 06:20:04 pm
Thanks heaps!
For the other question, have you been taught the k-method in class yet?
No I haven't been taught that but would you mind explaining it please?
Post by: RuiAce on August 13, 2017, 06:24:40 pm
If you don't use it, the alternate approach is this, which is for granted:
1. Use simultaneous equations to find the point of intersection of the lines (call the point of intersection, say, P)
2. Find the gradient of the line between P and the other point, in your case being (2,4)
3. Use the point-gradient formula to find the equation of your required line.
Post by: jamonwindeyer on August 13, 2017, 08:29:01 pm
To be honest, in the HSC that's either a method you just accept and take for granted, or don't use altogether. I didn't understand it until first year uni.
If you don't use it, the alternate approach is this, which is for granted:
1. Use simultaneous equations to find the point of intersection of the lines (call the point of intersection, say, P)
2. Find the gradient of the line between P and the other point, in your case being (2,4)
3. Use the point-gradient formula to find the equation of your required line.
Strongly advocate for this method over the k-method, it isn't that long and I don't think many schools teach the k-method anyway - Makes it easier on everyone ;D
Post by: beatroot on August 15, 2017, 03:13:05 pm
Post by: RuiAce on August 15, 2017, 03:18:51 pm
This isn't an actual Maths question, but a question regarding my rank and my marks in Maths. So I got back my mark back for my trial paper (I did the CSSA one) and I got 56% (lowest in a class of ten). My average for my internal mark is now 69%. Is there any hope for me left? Will I ever make it into the Band 5 range? If I get around 80-90% during the HSC, would I still make it into the Band 5 range. Last time I checked my ranking, it was 9 out 10. I feel like I'm dragging everyone down.Your performance in the final HSC exam always has the greatest impact on your final mark. How you perform in the final exam will determine whether or not you can enter the band 5 range.
This is all thanks to how moderation works (you can ask Jamon for the guide if you need it). At the end of the day, regardless of the actual marks you got in the internals, your final mark can go up depending on how you (AND ALSO you cohort) performs on the day.
Now that you've received your marks, you should now be focused on what you did do and did not do during the trial preparation block, and work to fixing all of these and improving for the finals. Despite your ranking, you're not gonna drag the cohort down provided you pull yourself up. It's now ALL about mark maximisation and using whatever resources you have to assist you.
If you fix it all up, you have no reason not to be able to get it.
Post by: _____ on August 16, 2017, 08:49:26 am
Post by: RuiAce on August 16, 2017, 08:56:03 am
For the past HSC papers what years am I safe in doing? I'm just worried that doing some of the earlier ones might be too easy/too hard due to changes in the syllabus or the introduction of the reference sheet (not sure what year that was). Is it a good idea to skip some or should I start at 2001?Papers that are too hard will only be papers that are prior to 2001. Anything from 2001 to 2016 inclusive is perfectly fine DESPITE the fact that 2016 was the first year they had the formula sheet.
Not more than 20% of the paper typically even requires the formula sheet (you should know the required formulas off by heart, everything else is just a bunch of methods). So there's no major problem at all.
In fact, you should push yourself to do all of them (unless of course, time constraints on other subjects) so that you can expose yourself to as much material and problem solving skills as possible.
Post by: anotherworld2b on August 18, 2017, 12:07:57 am
Post by: RuiAce on August 18, 2017, 12:14:26 am
I am not quite sure what I did wrong for this question. Can I have help please?At the start, you wrote 0.5 as 5/100 instead of 5/10
Post by: 12070 on August 18, 2017, 02:55:08 pm
Thank you!
Post by: RuiAce on August 18, 2017, 06:44:14 pm
Hey, I was just wondering how to find x find this information. It's nowhere near to scale as I don't have the actual question. The 9 in the middle is degrees by the way.This question lacks information; with only the given information any (positive) value of x could work.
Thank you!
Whilst the upper triangle was uniquely defined, the lower was not.
Post by: 12070 on August 18, 2017, 10:26:40 pm
This question lacks information; with only the given information any (positive) value of x could work.
Whilst the upper triangle was uniquely defined, the lower was not.
Sorry, I drew this on Microsoft Word as I don't have the question with me. The top triangle is meant to have one verticle side meaning the angle opposite x in the bottom triangle is 4.5. Also, the very bottom length of the bottom triangle is 100.
Edit: Just realised this question is incredibly easy with this information.
Post by: good_stuff on August 19, 2017, 11:19:06 pm
This isn't a specific math question but I'm confused about when to use quotient and product rules. There seem to be a lot more ways to solve equations in answer sections of the booklet including moving fractions out to the front of the swirly integrating sign (sorry I don't know what it's called :/) and changing the numberator's coefficient to the denominator's indice power...
Please help?
Thank you
Post by: RuiAce on August 20, 2017, 12:51:51 am
Hi all,The fact that you've indicated the quotient and product rules (which are for derivatives), but also mention the word "solve" (which is used for equations) and the integral symbol (which is for integration, not differentiation) makes me cautious as to if you juggled the methods. We must not get differentiation and integration mixed up.
This isn't a specific math question but I'm confused about when to use quotient and product rules. There seem to be a lot more ways to solve equations in answer sections of the booklet including moving fractions out to the front of the swirly integrating sign (sorry I don't know what it's called :/) and changing the numberator's coefficient to the denominator's indice power...
Please help?
Thank you
Please provided example questions.
(Also, for reference, that is called the integral symbol.)
Post by: anotherworld2b on August 20, 2017, 12:59:44 am
Post by: jamonwindeyer on August 20, 2017, 01:55:35 am
Can I have help with part d and e
For D, we know how likely it is a pilot can make it to 20 hours from part (b). We can do a similar calculation to find how likely it is they make it 15 hours. So you'd then be doing what is essentially a conditional probability calculation. I believe it would be as below, but I'm moving quickly, I await corrections if they are needed ;)
For E, you know the probability of totalling at least 20 hours before being shot down, from part (b). Use this in the binomial distribution with \(n=5\). So, if we let the number of pilots who satisfy the condition be the random variable \(Y\):
Where \(q\) is the probability you don't get 20 hours, and \(p\) is the probability that you do ;D
Post by: anotherworld2b on August 20, 2017, 10:04:17 am
Post by: jamonwindeyer on August 20, 2017, 11:24:17 am
Can I have help with this question too?
Warning: Not 2U content
Hey! First step is to standardise the distribution by defining a new random variable, Y:
This new random variable follows a standard normal distirbution, that is, \(Y\sim\mathcal{N}\left(0,1\right)\). For this we can now use a table of standard normal probabilities. Now what we need corresponds to:
You should be able to use the table to find the answer from there :)
PS - Is there any reason you stopped posting your questions in this thread? We'd be happy to help you out there, that way we aren't constantly reminding NSW students that your stuff isn't assessable in their course. Would be easier if you posted there from now on if you could :)
Post by: katnisschung on August 20, 2017, 02:21:22 pm
Post by: RuiAce on August 20, 2017, 02:27:28 pm
can somebody help me integrate this? thanks
For extra clarity, the index law \( a^m a^n = a^{m+n}\) was used. This was not necessary, and the same working can be done without it.
______________________________
Side remark: It may have been easier to notice how we have \(\int e^{x+2} dx = e^{x+2}+C \) as an example. Essentially, all we do is replace 2, which is a constant, with -2p, which is also a constant. \(\int e^{x-2p}dx = e^{x-2p}+C\)
Post by: anotherworld2b on August 20, 2017, 06:31:40 pm
Warning: Not 2U content
Hey! First step is to standardise the distribution by defining a new random variable, Y:
This new random variable follows a standard normal distirbution, that is, \(Y\sim\mathcal{N}\left(0,1\right)\). For this we can now use a table of standard normal probabilities. Now what we need corresponds to:
You should be able to use the table to find the answer from there :)
PS - Is there any reason you stopped posting your questions in this thread? We'd be happy to help you out there, that way we aren't constantly reminding NSW students that your stuff isn't assessable in their course. Would be easier if you posted there from now on if you could :)
Post by: jamonwindeyer on August 20, 2017, 06:34:26 pm
I was advised to post questions here by RuiAce. Im a bit confused where I should post questions
Post them in the other section, you are WA after all - Rui probably asked you to post here so he can see your questions. In honesty, I'm not overly fussed and if you really want to post here you can, but between Rui, Shadow and myself, one of us will spot your questions in the other board.
Post by: winstondarmawan on August 20, 2017, 06:37:26 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20939088_2364480123777043_876034297_n.jpg?oh=6f825f073179892ac96bbca1fe01300e&oe=599AE038
TIA!
Post by: RuiAce on August 20, 2017, 07:04:57 pm
Would appreciate help with c)(i).
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/20939088_2364480123777043_876034297_n.jpg?oh=6f825f073179892ac96bbca1fe01300e&oe=599AE038
TIA!
Post by: RuiAce on August 20, 2017, 07:36:51 pm
Hi...this is only simple yr 11 stuff...but could I have help with this?You can use brackets to emphasise the √(5−x)
Differentiate 2x√5−x
That square root is meant to continue all the way over the x...
Thanks heaps
Post by: gilliesb18 on August 20, 2017, 11:21:19 pm
1/[x^4 - 3x^3 + 3x]
Thanks...
Post by: Natasha.97 on August 20, 2017, 11:31:36 pm
I've attached the solution below, hope this helps ;D
Post by: EEEEEEP on August 20, 2017, 11:44:34 pm
Hello, can someoneplease help me to differentiate the following question??
1/[x^4 - 3x^3 + 3x]
Thanks...
WHen you get a differentiation.. (express it terms of something that you recognise)
- Then differentiate the aspects of it using the rules =)
To differentiate an entire bracket (with x's in it), do the outside then the inside =)
Post by: gilliesb18 on August 21, 2017, 08:45:39 am
Hi :)Thats awesome, makes heaps of sense now thanks!!!
I've attached the solution below, hope this helps ;D
Post by: RuiAce on August 21, 2017, 12:07:34 pm
Hi :)Slight remark - You should have \(\frac{d}{dx} \) in front of your first line. Because the way you wrote it implies that you're saying \( (x^4-3x^3+3x)^{-1}\) actually EQUALS its derivative
I've attached the solution below, hope this helps ;D
Post by: phebsh on August 21, 2017, 02:44:31 pm
Sorry if this sounds like I'm having a sook, I'm just honestly losing hope... Thanks :)
Post by: RuiAce on August 21, 2017, 03:22:50 pm
Hey, so I have my math trial on Wednesday and I've been practising past trial papers and I'm constantly failing them even though I'm always practising and asking for help in areas that I'm getting wrong and I just don't know what else to do to improve.When you say that you're "constantly failing them", in your scenario what does that encompass? Some things to think about:
Sorry if this sounds like I'm having a sook, I'm just honestly losing hope... Thanks :)
- Not understanding the question
- Not associating the question with relevant topics
- Simple getting lost in the wording
- Silly mistakes such as misreading numbers and valuse
- Not choosing correct formulae/methods
- Unable to make a choice on formulae/methods
- Sample solutions make no sense
- Inspiration to use said formula/method confusing
- Flow of logic confusing
- Spending too long on a question and not moving on
- Grasping probably one half of the content but not understanding the other half
- Difficulty in retaining all the methods
Think about all of these, then clarify your issue further (preferably also with your own areas of concern, not just the list above)
Post by: phebsh on August 21, 2017, 04:04:02 pm
When you say that you're "constantly failing them", in your scenario what does that encompass? Some things to think about:
- Not understanding the question
- Not associating the question with relevant topics
- Simple getting lost in the wording
- Silly mistakes such as misreading numbers and valuse
- Not choosing correct formulae/methods
- Unable to make a choice on formulae/methods
- Sample solutions make no sense
- Inspiration to use said formula/method confusing
- Flow of logic confusing
- Spending too long on a question and not moving on
- Grasping probably one half of the content but not understanding the other half
- Difficulty in retaining all the methods
Think about all of these, then clarify your issue further (preferably also with your own areas of concern, not just the list above)
Not knowing how to answer questions and the answers were so far from what I would have guessed. And by failing I literally mean when I count up my marks, I fail the test.
Post by: RuiAce on August 21, 2017, 04:15:55 pm
Not knowing how to answer questions and the answers were so far from what I would have guessed. And by failing I literally mean when I count up my marks, I fail the test.Then may the thought process continue.
When you attempt to answer a question, what do you do? Why is it potentially unlike anything you guessed?
- Have you thought about what topic the question might come from?
- Have you spent enough time thinking about a method/formula (with OR without your reference sheet)?
- If what you guessed is far off, explain a thought process with examples of a question, and how you unintentionally derailed from the intended approach
- How do you break down a question?
- Are you able to take the time out to compare a question you've done/seen, to another you're now encountering?
- Are you proficient with the standard methods for each topic, before adapting to weirder questions like those on past papers?
- Extra: Have you considered the possibility of multiple approaches arriving at the correct answer?
There's no point in just telling you what to do without sufficient context. Everyone has their own struggle, and it's about being able to clearly identify where your own are, before you can get some properly beneficial advice.
Post by: phebsh on August 21, 2017, 04:24:45 pm
Then may the thought process continue.
When you attempt to answer a question, what do you do? Why is it potentially unlike anything you guessed?
- Have you thought about what topic the question might come from?
- Have you spent enough time thinking about a method/formula (with OR without your reference sheet)?
- If what you guessed is far off, explain a thought process with examples of a question, and how you unintentionally derailed from the intended approach
- How do you break down a question?
- Are you able to take the time out to compare a question you've done/seen, to another you're now encountering?
- Are you proficient with the standard methods for each topic, before adapting to weirder questions like those on past papers?
- Extra: Have you considered the possibility of multiple approaches arriving at the correct answer?
There's no point in just telling you what to do without sufficient context. Everyone has their own struggle, and it's about being able to clearly identify where your own are, before you can get some properly beneficial advice.
So when I see a question I think about the method associated with the topic and any relevant formulas. If the question is out of the ordinary or of a higher difficulty level, this sounds silly but it scares me and I can't think of how to approach it.
Post by: pikachu975 on August 21, 2017, 04:53:25 pm
So when I see a question I think about the method associated with the topic and any relevant formulas. If the question is out of the ordinary or of a higher difficulty level, this sounds silly but it scares me and I can't think of how to approach it.
Sounds like a problem that you don't know the content and instead are rote learning textbook/past paper questions. Instead, try to go and understand the content FULLY, like where formulas come from and what they mean, then try and see if you can apply it to questions.
Post by: RuiAce on August 22, 2017, 07:57:19 pm
So when I see a question I think about the method associated with the topic and any relevant formulas. If the question is out of the ordinary or of a higher difficulty level, this sounds silly but it scares me and I can't think of how to approach it.For the most part, what pikachu said.
On top of that, honestly being scared isn't abnormal. I study a maths degree and even then I still get intimidated by some 4U questions every now and then.
But if you let yourself stay scared, that's going to cause problems. It's just like anything in life; you have to control that fear. Instead of worrying about how weird the question is, take the time to start breaking it down into a sequence of steps, and slowly piece every bit of the puzzle together. Not only do you need to know what you know, but you need to anticipate the possibility of adapting it to a foreign scenario, and trying out ideas in the hope it'll actually get there.
Post by: gilliesb18 on August 22, 2017, 10:13:22 pm
And this is a yr 11 question btw...
(x^3)/(x^2+ 1)
Thanks!!
Post by: EEEEEEP on August 22, 2017, 10:16:40 pm
Another one on derivatives sorry :-[Quotient Rule
And this is a yr 11 question btw...
(x^3)/(x^2+ 1)
Thanks!!
f/g
(f’ g − g’ f )/g2
Post by: RuiAce on August 22, 2017, 10:18:30 pm
Another one on derivatives sorry :-[
And this is a yr 11 question btw...
(x^3)/(x^2+ 1)
Thanks!!
Post by: Leah_Mer on August 23, 2017, 07:54:42 pm
Post by: Opengangs on August 23, 2017, 08:08:38 pm
Hey there! Just some HSC internal/external mark confusion here... I have read the really great article on this site about the way the HSC mark is calculated, but I'd love to get someones opinion on my personal situation! In 2U, I'm currently ranked about 4th in large cohort, but the average marks for our cohort are pretty low, and will probably be this way in the HSC as well! My question is, if I'm sitting on a band 4 after the trial exam for my internal mark (due to lack of study, so room for improvement in the hsc exam), how much potential is there for me to increase my final mark? Hypothetically, say I score over 90% the HSC, will my poor internal rank/mark and poor performance of my cohort in the exam mean that this significantly better score doesn't make a different to my final band? Hope this makes sense!Hey, Leah_Mer!
Your internal mark doesn't matter per se, rather your ranking and mark difference is. So, as you probably do know, the person who places first in the course will take the highest external mark to be their internal mark, and everyone will get scaled according to the gap between these respective ranks.
So, if you and the top 3 marks are relatively close, you'll receive a relative 'close' internal mark (maybe 2 - 4 mark difference). So, even if you didn't perform well internally, you can still make up for it with your external mark. Even if you have a weak cohort, placing fourth should still put you in a good position for a band 6 HSC mark.
Post by: jamonwindeyer on August 23, 2017, 09:56:17 pm
Hey there! Just some HSC internal/external mark confusion here... I have read the really great article on this site about the way the HSC mark is calculated, but I'd love to get someones opinion on my personal situation! In 2U, I'm currently ranked about 4th in large cohort, but the average marks for our cohort are pretty low, and will probably be this way in the HSC as well! My question is, if I'm sitting on a band 4 after the trial exam for my internal mark (due to lack of study, so room for improvement in the hsc exam), how much potential is there for me to increase my final mark? Hypothetically, say I score over 90% the HSC, will my poor internal rank/mark and poor performance of my cohort in the exam mean that this significantly better score doesn't make a different to my final band? Hope this makes sense!
100% what Opengangs said - So like, if you and a handful of other students can perform really well (most courses have "those students" that do quite well!) then the effect is likely to be quite low! 4th isn't a bad rank at all, so provided your cohort performs decently and you smash it out of the park, no reason you can't get a Band 6 ;D
Post by: asd987 on August 24, 2017, 04:07:47 pm
Post by: RuiAce on August 24, 2017, 04:22:32 pm
hi can i get some help for part 2 of this question. Its confusing ??? thanksThis was a past HSC question and was already addressed in the compilation.
Post by: _____ on August 24, 2017, 04:44:19 pm
Example from HSC 2003 Q10 b (looking at iii):
(http://i.imgur.com/LlpkwCP.png)
I can find the required result (http://i.imgur.com/H7Wy4u4.png) but I struggle to test both sides (<x and >x) when some other variable is involved as is the case here. What's the best way to do this?
Thanks
Post by: RuiAce on August 24, 2017, 09:19:28 pm
When doing those practical GAD questions, how do I prove that a stationary point is a maximum/minimum when some other variable, say r, is involved?
Example from HSC 2003 Q10 b (looking at iii):
(http://i.imgur.com/LlpkwCP.png)
I can find the required result (http://i.imgur.com/H7Wy4u4.png) but I struggle to test both sides (<x and >x) when some other variable is involved as is the case here. What's the best way to do this?
Thanks
That second derivative is FAR too ridiculous. So we will consider testing both sides instead.
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___________________________
(http://i.imgur.com/IBFrQQP.png)
Note that in general, there may not always be a nice approach. It varies from question to question how you could approach them. These ones are some of the worst types in 2U.
This will be added to the compilation thread.
Post by: _____ on August 24, 2017, 09:51:43 pm
That second derivative is FAR too ridiculous. So we will consider testing both sides instead.
___________________________
___________________________
(http://i.imgur.com/IBFrQQP.png)
Note that in general, there may not always be a nice approach. It varies from question to question how you could approach them. These ones are some of the worst types in 2U.
This will be added to the compilation thread.
Thanks a bunch.
I think I understand what you've done but there's no way I would have thought of doing that.
Just so I can understand your process for approaching different types of questions, may I ask how you'd prove a min TP with the question below (2003 HSC 9c). It's probably more typical of the difficulty level they tend to go for as the value to prove is simpler (V = 3/2U).
(http://i.imgur.com/Ivr2ChF.png)
Post by: RuiAce on August 24, 2017, 10:05:09 pm
Thanks a bunch.
I think I understand what you've done but there's no way I would have thought of doing that.
Just so I can understand your process for approaching different types of questions, may I ask how you'd prove a min TP with the question below (2003 HSC 9c). It's probably more typical of the difficulty level they tend to go for as the value to prove is simpler (V = 3/2U).
(http://i.imgur.com/Ivr2ChF.png)
In the future, please provide relevant progress. Here, the derivative could've helped
Post by: Leah_Mer on August 27, 2017, 06:25:57 pm
Hey, Leah_Mer!
Your internal mark doesn't matter per se, rather your ranking and mark difference is. So, as you probably do know, the person who places first in the course will take the highest external mark to be their internal mark, and everyone will get scaled according to the gap between these respective ranks.
So, if you and the top 3 marks are relatively close, you'll receive a relative 'close' internal mark (maybe 2 - 4 mark difference). So, even if you didn't perform well internally, you can still make up for it with your external mark. Even if you have a weak cohort, placing fourth should still put you in a good position for a band 6 HSC mark.
Thankyou! That's great news to hear! Also a big thankyou to atar notes in general, forum advice and the notes I've purchased at lectures throughout the year have really made all the difference to my marks and taken so much stress away when dealing with exams! The maths topic tests have been a big help too, so thanks for everything you guys do :)
Post by: jakesilove on August 28, 2017, 10:49:38 am
Thankyou! That's great news to hear! Also a big thankyou to atar notes in general, forum advice and the notes I've purchased at lectures throughout the year have really made all the difference to my marks and taken so much stress away when dealing with exams! The maths topic tests have been a big help too, so thanks for everything you guys do :)
So glad that you've benefited from the forums and the notes! We'll be here for you for the rest of the year (and beyond!) so make sure to ask as many questions as you'd like :)
Post by: Daniyahasan on August 28, 2017, 12:02:26 pm
Can someone help me with this question
Post by: RuiAce on August 28, 2017, 12:59:14 pm
Hey guys
Can someone help me with this question
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Post by: daniel.hu3 on August 28, 2017, 04:53:11 pm
A box of rectangular cross-section sits on a train luggage rack as shown with the point C touching the wall. P, the point in contact with the rack, is the midpoint of AD. If D is 8 cm from the wall and P, the edge of the rack, is 20 cm from the wall, find how far A is from the wall.
Hint: Draw parallels to the wall through A, P and D.
Post by: RuiAce on August 28, 2017, 06:56:44 pm
Hey Guys, can someone please help me with this question?
A box of rectangular cross-section sits on a train luggage rack as shown with the point C touching the wall. P, the point in contact with the rack, is the midpoint of AD. If D is 8 cm from the wall and P, the edge of the rack, is 20 cm from the wall, find how far A is from the wall.
Hint: Draw parallels to the wall through A, P and D.
(http://i.imgur.com/5VaToQl.png)
Extra intuition: As everything is linear, an arithmetic progression is unsurprising.
Post by: _____ on August 29, 2017, 03:09:07 pm
(http://i.imgur.com/N6SBhQN.png)
How do I prove theta = <DBE here for (i)? I understand how to do the rest of it from there (finding DE in terms of theta and then FE using DE).
Thanks
Post by: Opengangs on August 29, 2017, 03:33:32 pm
Quick question from 2005 HSC:
Post by: Shadowxo on August 29, 2017, 04:22:41 pm
Quick question from 2005 HSC:Or, even easier:
(http://i.imgur.com/N6SBhQN.png)
How do I prove theta = <DBE here for (i)? I understand how to do the rest of it from there (finding DE in terms of theta and then FE using DE).
Thanks
Post by: fantasticbeasts3 on August 31, 2017, 10:17:39 pm
Post by: RuiAce on August 31, 2017, 10:34:11 pm
don't know if it's me who can't read, or the question isn't worded nicely
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Post by: winstondarmawan on September 01, 2017, 06:53:45 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21269494_1314379005354286_1994981189_n.jpg?oh=f219784caf6eb433c41fbdf2e1531b4a&oe=59ABAA26
TIA!
Post by: RuiAce on September 01, 2017, 07:02:38 pm
Hey! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21269494_1314379005354286_1994981189_n.jpg?oh=f219784caf6eb433c41fbdf2e1531b4a&oe=59ABAA26
TIA!
Post by: Dragomistress on September 02, 2017, 04:49:47 pm
I need help with all questions except for question 12 a and b.
Post by: RuiAce on September 02, 2017, 04:54:46 pm
This will probably be a pain for you to answer but...
I need help with all questions except for question 12 a and b.
____________________________
Post by: RuiAce on September 03, 2017, 11:13:58 am
This will probably be a pain for you to answer but...
I need help with all questions except for question 12 a and b.
_________________
_________________
(Because consecutive numbers are numbers one after the other.)
As it so stands, this coincides with c).
15 follows a similar procedure to 14 and you should now give it a go using similar ideas, and post any progress if you remain stuck.
Post by: Kevinl123 on September 03, 2017, 11:07:27 pm
determine the age of ancient objects. A Babylonian cloth fragment now has 40% of the
carbon-14 that it contained originally. How old is the fragment of cloth?
Post by: RuiAce on September 03, 2017, 11:12:09 pm
Carbon-14 is a radioactive substance with a half-life of 5730 years. It is used to
determine the age of ancient objects. A Babylonian cloth fragment now has 40% of the
carbon-14 that it contained originally. How old is the fragment of cloth?
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Post by: amberberrie on September 05, 2017, 08:51:16 pm
Much appreciated,
Amber
Post by: RuiAce on September 05, 2017, 09:12:45 pm
Please help with this question!! No one I ask can seem to figure it out!
Much appreciated,
Amber
Remark that this is also the probability of two tails and double six.
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Favourable outcomes: (1,1), (2,2), (3,3), (4,4), (5,5)
Total possible outcomes: 36 of them
Remark that this is also the probability of two heads and a double that's not double six
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It was also clear that this question came out of Maths in Focus (so therefore I went to check after I was done typing). Their answers for parts c) and d) look completely unrealistic.
Post by: arunasva on September 06, 2017, 01:07:45 am
Post by: RuiAce on September 06, 2017, 08:47:53 am
I'm like really bad in Probability. I thought practicing would solve it like anything else in life, however turns out I'm not even good with the concepts. Like what to multiply what to add, why multiply these fractions what kinda scenarios exist etc. Is there a really good guide on probability or advice for that topic ? ThanksAnd means Multiply
Or means Plus
i.e. When you're considering one single case and there's more than one factor involved, you're multiplying. But when you're regrouping seperate cases, you add.
Post by: arunasva on September 06, 2017, 10:16:08 pm
And means Multiply
Or means Plus
i.e. When you're considering one single case and there's more than one factor involved, you're multiplying. But when you're regrouping seperate cases, you add.
makes more sense than anything ive read on that topic T.Thanks
Post by: hansolo9 on September 07, 2017, 12:55:06 pm
May I have help with this question? (Part a)
Excuse the scribbling :) thanks
Post by: mary123987 on September 07, 2017, 01:28:34 pm
regards Mary
Post by: pikachu975 on September 07, 2017, 01:35:33 pm
Hey guys I was just wondering if anyone knows what to do if their is a horizontal assymptote and the question asks you to sketch the derivative just for reference i'm referring to question 9c 2011 HSC 2 unit paper
regards Mary
At x = 1 f'(x) intercepts the x axis as f(x) has a stationary point
At x = 3 f'(x) has a stationary point since f(x) has a point of inflection
I think on f'(x) the horizontal asymptote would be y = 0 since the gradient approaches 0, and it would approach from the negative side
Post by: mary123987 on September 07, 2017, 02:03:17 pm
Post by: left right gn on September 07, 2017, 05:06:04 pm
I'm usually stumped when questions like these show up.
Post by: jamonwindeyer on September 07, 2017, 08:40:17 pm
Hello
May I have help with this question? (Part a)
Excuse the scribbling :) thanks
Hey!! So we can deduce fairly easily that \(\triangle ATD\) and \(\triangle FTB\) are congruent (use an AAS proof), so you'd start there. This lets you prove that \(AT=FT\). Now, assign a variable to \(AT\), let it be \(x\), and label \(BT=7-x\) accordingly. From there, use pythagoras theorem in \(\triangle BFT\):
Post by: jamonwindeyer on September 07, 2017, 08:50:36 pm
Hey! can someone explain how to do this question??
I'm usually stumped when questions like these show up.
Hey! So when going from a derivative to an original function, remember that the given graph represents the gradient of the graph you are drawing. So, for this line, the gradient starts negative, and gradually increases to a positive value. The line turns into a parabola, that decreases initially (negative gradient), turns around where the line intercepts the axis, then increases from there. That's the red curve.
Doing the same with the red curve, we simply mark where the red curve is positive - Our blue graph slopes UPWARDS there. When the red curve is negative, our blue graph slopes DOWNWARDS. Again, the intercepts of the red curve are where the blue curve turns around! :)
Note also that \(f'(0)=a, a>0\) means the first derivative needs to have a y-intercept above the x-axis. My red curve does this (accidentally to be honest, I've only just read that bit ;))
(https://i.imgur.com/S8GR37J.png)
Post by: gilliesb18 on September 08, 2017, 09:58:16 am
Can someone please help me with these questions- g, h, i, and j of question 1??
Lmk if you can't read it!!
Thanks 😀😀
Post by: Natasha.97 on September 08, 2017, 10:30:40 am
Hello-
Can someone please help me with these questions- g, h, i, and j of question 1??
Lmk if you can't read it!!
Thanks 😀😀
Hi!
I've attached my solution to part g
Hope this helps
Edit: Attached the rest of the solutions ;D
Post by: Shadowxo on September 08, 2017, 10:35:54 am
Hello-Let's say the angle inside the brackets is a*x (x is easier to type than theta)
Can someone please help me with these questions- g, h, i, and j of question 1??
Lmk if you can't read it!!
Thanks 😀😀
First find a*x (like you'd do if it were just x by itself)
Then divide all the solutions by a to get x by itself
eg g)
Tan(2x) = sqrt(3)
2x = 60°, 240°, 420°, 600°
(Note 420 = 60+360, 600=240+360)
x = 30°, 120°, 210°, 300°
Also you should know the table of values off by heart (or the triangle method for memorising them).
Post by: gilliesb18 on September 08, 2017, 10:38:34 am
Hi!
I've attached my solution to part g (not enough time for the rest)
Hope this helps
Thanks sooooo sooo much!!!!! Very kind:):)
Post by: _____ on September 09, 2017, 06:24:28 pm
If they give us the rate of change (kP), can I just say P = P0e^kt or do I need to do some proof to get the population equation? No need to solve this for me I'm just wondering if I need to show an integration of kP and if so, how I'd do that.
(http://i.imgur.com/7u2tqml.png)
Stuck with IV on this one, answers I've seen don't make much sense. I tried to set 1 - 2cosΘ >= 0 as a and y need to be positive but I wasn't able to get the answer.
Thanks.
Post by: RuiAce on September 09, 2017, 06:40:55 pm
(http://i.imgur.com/2lbk1q9.png)If you want to be safe you can prove it anyway. But personally I'd just assume it, because in 2U you aren't taught how to properly solve that differential equation. This is also because you don't have enough tools to solve it in 2U anyway.
If they give us the rate of change (kP), can I just say P = P0e^kt or do I need to do some proof to get the population equation? No need to solve this for me I'm just wondering if I need to show an integration of kP and if so, how I'd do that.
(http://i.imgur.com/7u2tqml.png)
Stuck with IV on this one, answers I've seen don't make much sense. I tried to set 1 - 2cosΘ >= 0 as a and y need to be positive but I wasn't able to get the answer.
Thanks.
In 3U, you develop a useful tool to solving them. If you wish to see the method, you can just post your question again in the 3U thread.
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Solving an equation is not the intended approach here because of the fact they wanted this 'upper bound'. It's not impossible to use it, but it would get incredibly hand-wavy.
Post by: katnisschung on September 10, 2017, 11:12:02 am
from the top of a vertical cliff the angle of depression to a marker at the base of the cliff is
39 degrees. On the other side of the cliff is another marker, and the angle of depression from
the top of the cliff to this marker is 43 degrees. The markers are 300m apart. Find the height of
the cliff to the nearest metre. (answer 130m)
thanks :)
Post by: MisterNeo on September 10, 2017, 05:15:18 pm
i need help visualising this...could someone pls draw a diagram thanks
from the top of a vertical cliff the angle of depression to a marker at the base of the cliff is
39 degrees. On the other side of the cliff is another marker, and the angle of depression from
the top of the cliff to this marker is 43 degrees. The markers are 300m apart. Find the height of
the cliff to the nearest metre. (answer 130m)
thanks :)
The cliff is like a telegraph pole where it has virtually no thickness and has two faces. :)
(https://i.imgur.com/gk0xKkN.jpg)
So my way of working is basically:
-Find the two distances from the markers at each side using Sine Rule.
-Find the area of the triangle diagram using that formula on the Ref Sheet.
-Convert it to height using the easy triangle formula, which has a height variable.
Also, you can't just halve the triangles and use Sin Rule because the height is unknown and the base isn't necessarily going to be a perfect 150-150 metres.
Hope this helps :D
Post by: itssona on September 10, 2017, 06:46:20 pm
do i take seperate cases
|4x +3| + |x-1| <11
Post by: Natasha.97 on September 10, 2017, 06:51:44 pm
hiii how do i do this
do i take seperate cases
|4x +3| + |x-1| <11
Hi!
Yeah I would do the four cases:
- Pos/Pos
- Pos/Neg
- Neg/Pos
- Neg/Neg
After finding x, test it with the equation to ensure that it holds true (that the value is smaller than 11).
Hope this helps
My solution below:
(https://i.imgur.com/uCn2iDn.png)
Post by: RuiAce on September 10, 2017, 07:19:36 pm
Hi!
Yeah I would do the four cases:
- Pos/Pos
- Pos/Neg
- Neg/Pos
- Neg/Neg
After finding x, test it with the equation to ensure that it holds true (that the value is smaller than 11).
Hope this helps
hiii how do i do this
do i take seperate cases
|4x +3| + |x-1| <11
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__________________
Post by: itssona on September 10, 2017, 07:25:29 pm
Post by: georgiia on September 11, 2017, 09:42:13 am
Post by: RuiAce on September 11, 2017, 09:44:03 am
(http://uploads.tapatalk-cdn.com/20170910/1bc7195a98affa433fd75f4a21446520.jpg)
Post by: georgiia on September 11, 2017, 05:17:28 pm
Post by: itssona on September 12, 2017, 10:29:47 am
how would the graph of y=1-cosx look?
domain x between -180 and 180
Post by: RuiAce on September 12, 2017, 10:36:32 am
hiii kinda stuck
how would the graph of y=1-cosx look?
domain x between -180 and 180
Post by: itssona on September 12, 2017, 10:42:01 am
ahh thank you Rui :)
Post by: _____ on September 13, 2017, 06:59:57 pm
(https://i.imgur.com/0an84If.png)
Looking at v. Got P > 2r therefore r < P/2 but couldn't progress from there.
Post by: mary123987 on September 13, 2017, 07:21:39 pm
StupidHey in regards to this you must consider the fact that :sexyhard inequality questions always throw me off.
(https://i.imgur.com/0an84If.png)
Looking at v. Got P > 2r therefore r < P/2 but couldn't progress from there.
θ is ≤ 2π(as it is a circle)
now from part (i) we know that p=r( θ+2)
so p/r =2 + θ
and p/r -2= θ
however we said that θ<2π
so it can be wriiten as : p/r -2<2π
now just rearrange so that p/r <2(π+1)
p<2(π+1)r
p/2(π+1)<r
this is the left side of the inequality
Now for the left p=2r +rθ
since θ>0
p>2r
p/2>r
now just combine to get
p/2(π+1)<r <p/2
Hope this helps
Post by: _____ on September 13, 2017, 09:47:54 pm
Hey in regards to this you must consider the fact that :
θ is ≤ 2π(as it is a circle)
now from part (i) we know that p=r( θ+2)
so p/r =2 + θ
and p/r -2= θ
however we said that θ<2π
so it can be wriiten as : p/r -2<2π
now just rearrange so that p/r <2(π+1)
p<2(π+1)r
p/2(π+1)<r
this is the left side of the inequality
Now for the left p=2r +rθ
since θ>0
p>2r
p/2>r
now just combine to get
p/2(π+1)<r <p/2
Hope this helps
Thanks!
Post by: _____ on September 16, 2017, 03:22:01 pm
(https://i.imgur.com/q9hfSkF.png)
Post by: RuiAce on September 16, 2017, 03:25:08 pm
How do I find that T is (cosx, sinx) here? Apart from that I can complete the question.This question was already addressed in the compilation.
(https://i.imgur.com/q9hfSkF.png)
Post by: _____ on September 16, 2017, 07:16:11 pm
This question was already addressed in the compilation.
Didn't consider it a particularly hard one so I didn't look, mb.
Unfortunately that solution didn't help me much, as it says to use trig to find T which I what I'm struggling with. If I do this I get sinx = y/TO and cosx = x/TO (drawing a perpendicular line as suggested). How do I remove the denominator?
Post by: RuiAce on September 16, 2017, 07:19:03 pm
Didn't consider it a particularly hard one so I didn't look, mb.TO is the radius of the circle which has length 1
Unfortunately that solution didn't help me much, as it says to use trig to find T which I what I'm struggling with. If I do this I get sinx = y/TO and cosx = x/TO (drawing a perpendicular line as suggested). How do I remove the denominator?
Also, I believe you meant sin(theta).
Post by: _____ on September 16, 2017, 10:15:52 pm
TO is the radius of the circle which has length 1
Also, I believe you meant sin(theta).
Thanks. My own inability to see what's right in front of me is quite amazing ???
Post by: winstondarmawan on September 17, 2017, 12:11:17 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21733200_1324623537666455_1924185311_n.png?oh=755e3ab03cb1b287d4ad471e3535d5c2&oe=59C00DE2
Apparently my friend got it from a 2U Trial paper. If this is not a 2U/3U question, then don't worry about it.
Thanks in advance.
Post by: georgiia on September 17, 2017, 12:32:58 pm
Post by: RuiAce on September 17, 2017, 12:58:00 pm
Hello! Would appreciate help with the following:I personally don't believe your friend at all. The \( w^{n+1}=-1 \) thing and proving that it's real hints that this is actually a 4U question. If anything, given the typeface it looks as though it'd be a question from brilliant.org
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21733200_1324623537666455_1924185311_n.png?oh=755e3ab03cb1b287d4ad471e3535d5c2&oe=59C00DE2
Apparently my friend got it from a 2U Trial paper. If this is not a 2U/3U question, then don't worry about it.
Thanks in advance.
I'm happy to not worry about it or take more time if it gets shoved into the 4U thread (up to you if you want to do this), but unless your friend cannot clearly identify what paper this is from I will certainly not do it here. It's potentially possible that 2U knowledge is sufficient enough to actually do the question, but it goes beyond the level of understanding required.
My first guess: Geometric series on the top.
Edit: Just had a go at it. I managed to get it to work but I could only finish it off with complex numbers.
(http://uploads.tapatalk-cdn.com/20170917/b15160969ca4c463c1db4b862b7d0656.jpg)could I pls have help with both parts of the question? Thxx
So the information we have is:
Radius of cone = 2sqrt(3/pi)
Height of cone = 6
Radius of liquid = sqrt(3/pi)
Height of cone = 3
And part 2 is just simple subtraction: 24 - 6 = 18.
Post by: hansolo9 on September 18, 2017, 09:36:40 am
How do you do this? :)
Post by: RuiAce on September 18, 2017, 09:43:13 am
HiNote that csc is how \( \LaTeX \) denotes 'cosec', and thus is what is displayed
How do you do this? :)
\[ \text{From continued use of Pythagorean identities} \]
\begin{align*}\sqrt{\frac{1-\sin^2\theta}{\csc^2\theta -\cot^2\theta -\cos^2\theta}} &= \sqrt{\frac{\cos^2\theta}{(\csc^2\theta -\cot^2\theta) -\cos^2\theta}}\\ &= \sqrt{\frac{\cos^2\theta}{1-\cos^2\theta}}\\ &= \sqrt{\frac{\cos^2\theta}{\sin^2\theta}}\\ &= \frac{\cos\theta}{\sin\theta}\\ &= \cot \theta\end{align*}
Post by: davidss on September 19, 2017, 06:10:39 pm
(https://i.imgur.com/MF05vVg.png)
Post by: Mathew587 on September 19, 2017, 06:22:25 pm
Hi can someone help with this pls thanks <3whcich q
(https://i.imgur.com/MF05vVg.png)
Post by: JuliaPascale123 on September 19, 2017, 07:05:32 pm
Could I please have working out and answer for this question as I have found others like it and need to practice the skill.
(http://i67.tinypic.com/jkbyn6.png)
Thanks <3,
Julia
Post by: persistent_insomniac on September 19, 2017, 07:28:22 pm
If logax=4 and logay=5, use log laws to find the exact values of:
logax^2y
logaaxy
logaxy
I literally don't know who to ask since theres nothing on my textbook or even on google. Thanks so much in advance!
Post by: RuiAce on September 19, 2017, 07:31:47 pm
Hey guys I don't even know if this question belongs here so sorry if it doesn't but how do you solve:If you want to avoid hitting the subscript button, can you please use more brackets to make it clear what we're talking log-base-a of?
If logax=4 and logay=5, use log laws to find the exact values of:
logax^2y
logaaxy
logaxy
I literally don't know who to ask since theres nothing on my textbook or even on google. Thanks so much in advance!
Post by: Shadowxo on September 19, 2017, 07:48:45 pm
Hey guys I don't even know if this question belongs here so sorry if it doesn't but how do you solve:As Rui said it's a bit difficult to determine what you want to find but the basic log laws should make this question easy :)
If logax=4 and logay=5, use log laws to find the exact values of:
logax^2y
logaaxy
logaxy
I literally don't know who to ask since theres nothing on my textbook or even on google. Thanks so much in advance!
Try using these, in particular the first one, for your problems :)
Post by: jamonwindeyer on September 19, 2017, 11:46:31 pm
Hi Guys,
Could I please have working out and answer for this question as I have found others like it and need to practice the skill.
Thanks <3,
Julia
Hi Julia - I think you'll have trouble getting an answer to this here, it doesn't look like anything from a HSC course to me! Where did you find it? :)
Post by: davidss on September 20, 2017, 07:23:59 am
whcich q
the first bit pls xd
Post by: Shadowxo on September 20, 2017, 09:07:51 am
Hi can someone help with this pls thanks <3Let CAD be alpha
(https://i.imgur.com/MF05vVg.png)
So ACD = alpha (isosceles)
2alpha + theta = 180°
CBA = alpha (isosceles)
Hence ACB = theta (angles add to 180°)
Hence they're similar as they have the same three angles (alpha alpha and theta)
There are many ways to approach this problem, I just thought angles was the easiest way. Their solution may be different, and you may need to be more specific than I've been :)
Post by: RuiAce on September 20, 2017, 09:37:09 am
Let CAD be alphaIt suffices to only prove that two angles are equal for the equiangular test. The third can be assumed to equal by the angle sum
So ACD = alpha (isosceles)
2alpha + theta = 180°
CBA = alpha (isosceles)
Hence ACB = theta (angles add to 180°)
Hence they're similar as they have the same three angles (alpha alpha and theta)
There are many ways to approach this problem, I just thought angles was the easiest way. Their solution may be different, and you may need to be more specific than I've been :)
Post by: caitlinlddouglas on September 20, 2017, 03:34:46 pm
for part ii) it says just to use parts from 1) and 2) but i'm unsure how they've manipulated it! thanks :)
http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
solutions: http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-marking-guide-maths.pdf
Post by: RuiAce on September 20, 2017, 04:20:13 pm
hi i didn't understand the proof for question 16 c) from the 2013 mathematics paper.
for part ii) it says just to use parts from 1) and 2) but i'm unsure how they've manipulated it! thanks :)
http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
solutions: http://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-marking-guide-maths.pdf
________________________________
Observation: AB appears in the denominator twice. Also, AX + BX = AB, which is extremely convenient. This second observation was deduced by looking at the diagram, because clearly, AB is literally just AX and BX joined together.
And hence the intuition to add the equations
Note that clearly XY was just factorised.
Post by: Dragomistress on September 22, 2017, 06:57:59 am
Post by: RuiAce on September 22, 2017, 09:52:20 am
May I have these questions answered?In order, you would start with something along the lines of:
\( PR = 2\times \text{"distance from }y=1\text{"} \) so \( (x-3)^2+(y-2)^2 =4[ (x-x)^2 + (y-1)^2] \)
\( \text{"distance from }x=0\text{"} = \text{"distance from }y=0\text{"} \) so \( (x-0)^2+(y-y)^2 = (x-x)^2+(y-0)^2 \)
\( \frac{2}{5} PA = \frac{3}{5} PB \) so \( \frac{4}{25}[(x+6)^2+(y-5)^2] = \frac{9}{25}[(x-3)^2+(y+1)^2]\)
These questions follow the exact same working and are repetitive so I will not clarify or complete the question any further. If you still have troubles, please indicate clearly where the issue is and/or post relevant working-out progress.
Post by: katnisschung on September 22, 2017, 12:01:57 pm
for q 9 a) ii
of this paper my answer was 174.1431...... so money will run out on the 175th month therefore the answer is 174 because until that month there will be money remaining in the account correct? but i'm wrong which is annoying considering i'll lose a mark. can someone pls explain :)
Post by: EEEEEEP on September 22, 2017, 12:14:53 pm
http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9TIn HSC maths, when it comes to exponentials, financial maths and all... they usually round up the END answer..,
for q 9 a) ii
of this paper my answer was 174.1431...... so money will run out on the 175th month therefore the answer is 174 because until that month there will be money remaining in the account correct? but i'm wrong which is annoying considering i'll lose a mark. can someone pls explain :)
So if its.. 1.56 days, they say 2 days.
IF its 174.1431 days (in this case), they round up to 175.
Post by: katnisschung on September 22, 2017, 12:22:03 pm
for 9 b) iv) i can't quite tell from the marking guidelines if the graphs at x>4 is a straight line graph...considering the gradient from the given graph is a constant -3
Post by: RuiAce on September 22, 2017, 12:25:56 pm
http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9TIt should most certainly be a straight line over the interval 4 < x < 6. The gradient of this line should be -3.
for 9 b) iv) i can't quite tell from the marking guidelines if the graphs at x>4 is a straight line graph...considering the gradient from the given graph is a constant -3
http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9TTechnically speaking, during the 175th month there is still money. It is only at the end of the 175th month where there won't be money anymore.
for q 9 a) ii
of this paper my answer was 174.1431...... so money will run out on the 175th month therefore the answer is 174 because until that month there will be money remaining in the account correct? but i'm wrong which is annoying considering i'll lose a mark. can someone pls explain :)
Post by: mary123987 on September 22, 2017, 12:31:39 pm
http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T
for q 9 a) ii
of this paper my answer was 174.1431...... so money will run out on the 175th month therefore the answer is 174 because until that month there will be money remaining in the account correct? but i'm wrong which is annoying considering i'll lose a mark. can someone pls explain :)
Hey , with respect to this question it is not right to assume that , what you rather should do is sub in the value (i.e n months ) in to the equation which you worked out in part (ii) 1 which is this formula : An = (P − 400 000) × 1.005n + 400 000.
now the question asks
For how many months after retirement will there be money left in
the account?
so what you need to do is sub in the value if you sub in 174.1431...... you get exactly 0
now consider rounding:
when tyou sub in 174 you will see you get $285.4307496
when you sub in 175 you will see you get $-1713.142097
so if you look at the question it states:Chris withdraws $2000 from the account at the end
of each month, without making any further deposits......
thus considering it is withdrawn at the end of 175 months not the beginning 175 is the answer.
Hope this helps
Post by: georgiia on September 23, 2017, 04:10:50 pm
Post by: RuiAce on September 23, 2017, 04:12:52 pm
Could I please have help for these two questions? Thanks!!(http://uploads.tapatalk-cdn.com/20170923/c7b5d07bcd9116e3dded8d75808858ec.jpg)(http://uploads.tapatalk-cdn.com/20170923/21c623c4234a3257d7d34fe79a290e91.jpg)You have already asked the first one
Post by: georgiia on September 23, 2017, 04:25:21 pm
You have already asked the first onethank you!
Could I please also have help for this? (http://uploads.tapatalk-cdn.com/20170923/bf429800ae132d8360897329857d5018.jpg)
Post by: RuiAce on September 23, 2017, 04:54:04 pm
thank you!Double check your graph. Over the domain \(-\frac{\pi}{2} < x < \frac{\pi}{2} \), \(y=\tan x\) only intersects \(y=x\) at 0; never again.
Could I please also have help for this? (http://uploads.tapatalk-cdn.com/20170923/bf429800ae132d8360897329857d5018.jpg)
Post by: georgiia on September 23, 2017, 04:58:49 pm
Double check your graph. Over the domain \(-\frac{\pi}{2} < x < \frac{\pi}{2} \), \(y=\tan x\) only intersects \(y=x\) at 0; never again.
Im having trouble with graphing it. I know i can search it online but I want the process of finding it explained
Post by: RuiAce on September 23, 2017, 05:05:15 pm
Im having trouble with graphing it. I know i can search it online but I want the process of finding it explainedI'm going to assume that you understand how to draw the shapes of the graphs separately with ease.
___________________________________
Post by: Nuwanda_g on September 23, 2017, 08:50:12 pm
I need help with 'iii.'
I dont get how the DF = FE-DE, became equal to EB-AE.
Thanks in advance :)
Post by: Mathew587 on September 23, 2017, 08:59:08 pm
I dont get how the DF = FE-DE, became equal to EB-AE.
Thanks in advance :)
we know AE = DE = X (isoceles t)
we know EF = EB = Y (isos)
chuck that into your q;
DF = FE-DE, became equal to EB-AE...
DF = Y - X, ... Y-X
therefore EB-AE = FE-DE
Post by: Dragomistress on September 24, 2017, 08:38:52 pm
The point M divides the interval AB internally in the ratio 3:1. The point N divides the interval AB externally in the ratio 3:1. What is the value of NB:MB.
Post by: RuiAce on September 24, 2017, 08:41:38 pm
How does one solve this?This is an Extension 1 level question. If you need help with it, please post it in the more relevant thread.
The point M divides the interval AB internally in the ratio 3:1. The point N divides the interval AB externally in the ratio 3:1. What is the value of NB:MB.
Post by: Nuwanda_g on September 25, 2017, 02:04:22 pm
I need help for HSC 2016 Mathematics paper, question 16 (b) of i.
I dont get how to differentiate the inside bracket of (1+19e^-0.5t).
Thanks in advance.
Post by: Shadowxo on September 25, 2017, 03:02:09 pm
Hey guys,
I need help for HSC 2016 Mathematics paper, question 16 (b) of i.
I dont get how to differentiate the inside bracket of (1+19e^-0.5t).
Thanks in advance.
Remember, the derivative of eax=a*eax and the derivative of a constant is 0
Let me know if you're still confused :)
Post by: _____ on September 26, 2017, 02:39:42 pm
I don't want it solved, I'm just wondering what's the reasoning behind the working for iv? Why is (where x is theta) sinx/(cosx+1) > 0 and tan2x < 0?
Thanks!
Post by: RuiAce on September 26, 2017, 02:53:20 pm
I've attached a question and its solution as a word document because it's too cumbersome for images/text.
I don't want it solved, I'm just wondering what's the reasoning behind the working for iv? Why is (where x is theta) sinx/(cosx+1) > 0 and tan2x < 0?
Thanks!
__________________________________________________
Post by: Becky234 on September 27, 2017, 05:10:03 pm
can someone pls help me with these 2 questions:
1) Solve 2^(2x + 1) = 32.
2) Differentiate ln(5x + 2) with respect to x.
Post by: sophiegmaher on September 27, 2017, 05:13:27 pm
Post by: Natasha.97 on September 27, 2017, 05:18:32 pm
hey guys
can someone pls help me with these 2 questions:
1) Solve 2^(2x + 1) = 32.
2) Differentiate ln(5x + 2) with respect to x.
Hi!
(https://i.imgur.com/6SNe0KP.png)
Hope this helps
Post by: RuiAce on September 27, 2017, 05:22:54 pm
How do you complete the attached question??
Post by: sophiegmaher on September 27, 2017, 05:25:50 pm
ohhhh okay awesome thank you!
Post by: Becky234 on September 27, 2017, 05:33:34 pm
Hi!for part 1) can u pls explain how u got 5
(https://i.imgur.com/6SNe0KP.png)
Hope this helps
Post by: Mathew587 on September 27, 2017, 05:38:27 pm
Post by: Mathew587 on September 27, 2017, 05:40:58 pm
for part 1) can u pls explain how u got 5
for qs where its ln(f(x)), dy/dx = f'(x)/f(x)
so for ln (5x+2)
its
5/5x+2 + c
:)
Post by: RuiAce on September 27, 2017, 05:41:51 pm
How do we integrate y=ln(x-2) ?You don't have any techniques for doing this in 2U. Please post the full question.
for part 1) can u pls explain how u got 5From basic arithmetic/the calculator, 25 = 32.
Post by: pikachu975 on September 27, 2017, 05:47:57 pm
How do we integrate y=ln(x-2) ?
This is 4 unit so there must be a previous part
set u = ln(x-2) and dv = dx
du = 1/(x-2) and v = x
so the integral = xln(x-2) - integral of x/(x-2)
= xln(x-2) - integral of (1 + 2/(x-2) )
= xln(x-2) - x - 2ln(x-2) + C
Post by: Mathew587 on September 27, 2017, 05:56:46 pm
You don't have any techniques for doing this in 2U. Please post the full question.oh its hsc 2008 q10 a)
From basic arithmetic/the calculator, 25 = 32.
why can't we just use dy/dx = f'x / fx?
This is 4 unit so there must be a previous part
set u = ln(x-2) and dv = dx
du = 1/(x-2) and v = x
so the integral = xln(x-2) - integral of x/(x-2)
= xln(x-2) - integral of (1 + 2/(x-2) )
= xln(x-2) - x - 2ln(x-2) + C
ahah i was gonna ask you personally XD
Post by: Becky234 on September 27, 2017, 05:59:21 pm
You don't have any techniques for doing this in 2U. Please post the full question.ok thanks
From basic arithmetic/the calculator, 25 = 32.
Hi!thanks for ur help
(https://i.imgur.com/6SNe0KP.png)
Hope this helps
Mod action: Posts merged. At times like this, please resort to the "modify" function at the top right corner of a post, to refrain from multiple-posting.
Post by: RuiAce on September 27, 2017, 06:08:24 pm
This is 4 unit so there must be a previous partNot going to lie; I had already suspected what the actual question was. But without an example, I wasn't going to go into definite integral territory with finding areas.
set u = ln(x-2) and dv = dx
du = 1/(x-2) and v = x
so the integral = xln(x-2) - integral of x/(x-2)
= xln(x-2) - integral of (1 + 2/(x-2) )
= xln(x-2) - x - 2ln(x-2) + C
oh its hsc 2008 q10 a)
why can't we just use dy/dx = f'x / fx?
ahah i was gonna ask you personally XD
(https://i.imgur.com/eyRSrTg.png)
___________________________
Post by: Mathew587 on September 27, 2017, 06:41:15 pm
Not going to lie; I had already suspected what the actual question was. But without an example, I wasn't going to go into definite integral territory with finding areas.
(https://i.imgur.com/eyRSrTg.png)
___________________________
well...
that was dumb by me :/
thanks for the help :)
Post by: Becky234 on September 27, 2017, 07:08:39 pm
The number of members of a new social networking site doubles every day. On
Day 1 there were 27 members and on Day 2 there were 54 members.
(i) How many members were there on Day 12? 1
(ii) On which day was the number of members first greater than 10 million? 2
(iii) The site earns 0.5 cents per member per day. How much money did the
site earn in the first 12 days? Give your answer to the nearest dollar
Post by: Natasha.97 on September 27, 2017, 08:01:24 pm
can someome pls solve this question
The number of members of a new social networking site doubles every day. On Day 1 there were 27 members and on Day 2 there were 54 members.
(i) How many members were there on Day 12? 1
(ii) On which day was the number of members first greater than 10 million? 2
(iii) The site earns 0.5 cents per member per day. How much money did the
site earn in the first 12 days? Give your answer to the nearest dollar
Hi!
(https://i.imgur.com/RUI0kBS.png)
Hope this helps
Edit: Fixed part ii and iii
Post by: RuiAce on September 27, 2017, 08:06:07 pm
Hi!Fairly sure ii) is asking for \(T_n > 10^7\), not \(S_n\). Instead, iii) should be the one considering \( S_n\)
(https://i.imgur.com/KPFO6v2.png)
Hope this helps
Post by: Becky234 on September 27, 2017, 08:27:14 pm
Fairly sure ii) is asking for \(T_n > 10^7\), not \(S_n\). Instead, iii) should be the one considering \( S_n\)Thank you
Hi!thanks
(https://i.imgur.com/RUI0kBS.png)
Hope this helps
Edit: Fixed part ii and iii
Mod edit: Posts merged. You're welcome for the help, but you will need to resort to the modify button as chain-posting is against the forum rules. Further cases of this may result in a formal warning being issued.
Post by: Becky234 on September 27, 2017, 08:40:00 pm
Kim has three red shirts and two yellow shirts. On each of the three days,
Monday, Tuesday and Wednesday, she selects one shirt at random to wear. Kim
wears each shirt that she selects only once.
(i) What is the probability that Kim wears a red shirt on Monday? 1
(ii) What is the probability that Kim wears a shirt of the same colour on all
three days?
1
(iii) What is the probability that Kim does not wear a shirt of the same colour
on consecutive days?
Post by: Opengangs on September 27, 2017, 08:48:49 pm
Can someone pls solve this question(i) 3 red shirts, 5 shirts altogether. Thus, probability is 3/5
Kim has three red shirts and two yellow shirts. On each of the three days,
Monday, Tuesday and Wednesday, she selects one shirt at random to wear. Kim
wears each shirt that she selects only once.
(i) What is the probability that Kim wears a red shirt on Monday? 1
(ii) What is the probability that Kim wears a shirt of the same colour on all
three days?
1
(iii) What is the probability that Kim does not wear a shirt of the same colour
on consecutive days?
(ii) If she were to wear a red shirt on Monday, she'd have to wear a red shirt on Tuesday and Wednesday.
(3/5)(2/4)(1/3)
She can't wear a yellow shirt on the three consecutive days.
Thus, the probability is simply (3/5)(2/4)(1/3) = 1/10
(iii) P(not wear a shirt of the same colour) = P(RYR) + P(YRY) = (3/5)(2/4)(2/3) + (2/5)(3/4)(1/3) = 18/60 = 3/10
Post by: Becky234 on September 27, 2017, 09:28:35 pm
(i) 3 red shirts, 5 shirts altogether. Thus, probability is 3/5
(ii) If she were to wear a red shirt on Monday, she'd have to wear a red shirt on Tuesday and Wednesday.
(3/5)(2/4)(1/3)
She can't wear a yellow shirt on the three consecutive days.
Thus, the probability is simply (3/5)(2/4)(1/3) = 1/10
(iii) P(not wear a shirt of the same colour) = P(RYR) + P(YRY) = (3/5)(2/4)(2/3) + (2/5)(3/4)(1/3) = 18/60 = 3/10
Thank you :D
Can someone please help me with these two questions
Mod Edit: Post merge
Post by: Opengangs on September 27, 2017, 09:49:48 pm
Can someone please help me with these two questions
Post by: Fahim486 on September 28, 2017, 03:34:48 pm
Post by: RuiAce on September 28, 2017, 03:37:02 pm
Hey can someone help me out with these two questions. Thanks!
The other question is already addressed in the compilation.
Post by: Fahim486 on September 28, 2017, 04:05:41 pm
So when you sub in x = 0, does the cos and sec squared disappear? because when i put cos(0) into my calculator, it gives me 1 so i thought cos3x = 3
The other question is already addressed in the compilation.
Post by: RuiAce on September 28, 2017, 04:08:56 pm
So when you sub in x = 0, does the cos and sec squared disappear? because when i put cos(0) into my calculator, it gives me 1 so i thought cos3x = 3
Post by: MisterNeo on September 28, 2017, 08:33:53 pm
Post by: Opengangs on September 28, 2017, 09:01:13 pm
Hey guys, this may be easy, but how do you do part ii ?
Post by: RuiAce on September 28, 2017, 09:06:56 pm
Hey guys, this may be easy, but how do you do part ii ?
To observe why, consider a few examples.
30 has 2 digits. Additionally, \(\log_10 30\approx 1.47\) and 1.47 rounded up gives you 2.
987 has 3 digits. Additionally, \(\log_10 987 \approx 2.99\) and 2.99 rounded up gives you 3.
The simple reason behind this is just the observation that log is a monotonic increasing function, and furthermore,
\( \log_{10} 10 = 1, \log_{10} 100 = 2, \log_{10} 1000 = 3, \log_{10} 10000 = 4 \dots\)
So one of infinitely many consequences is that if \(10 < x < 100 \), then \( 1\le \log_{10} x < 2\), so if you round up \( \log_{10}x\) where x is a two digit number that's not 10 itself, you get back to 2.
Post by: Mounica on September 29, 2017, 07:28:47 pm
Post by: RuiAce on September 29, 2017, 07:42:23 pm
Can someone pls help me with this question
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Post by: Mounica on September 29, 2017, 09:25:29 pm
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Thanks you
Post by: Becky234 on September 29, 2017, 09:28:02 pm
Can someone pls help me with this question
Post by: Shadowxo on September 29, 2017, 09:49:14 pm
HeyWhich part are you stuck on?
Can someone pls help me with this question
Post by: sidzeman on September 30, 2017, 04:17:06 pm
Post by: RuiAce on September 30, 2017, 04:21:38 pm
I'm at a complete loss as how to do this question, even from part iPart i has already been addressed in the compilation.
Go back and think about the remaining parts again. They are standard applications of coordinate geometry, followed by a max/min problem (albeit with the wording overcomplicated). As a hint, for part ii, recall that BQ is a horizontal line. Also, Q is the point where the lines \( x \cos \theta + y\sin\theta = 1\) and \( y = 1\), which should now be all the information you need.
(You may want to anticipate a messy quotient rule for part iv.)
Please post your thought process if you need further help. Part i) is the hardest part of this question.
Post by: _____ on September 30, 2017, 10:01:27 pm
Is integration of tanx a part of the 2 unit syllabus or is this trial just being hard? Wondering if it is because I haven't encountered it before (I know now to use sinx/cosx).
(https://i.imgur.com/jUc3IKi.png)
Seems straightforward, a + B = -B + B therefore b/a = 0. But why is (4k+1) = b here? Shouldn't it be (4k+1)x? Solution:
(https://i.imgur.com/xfKYo8U.png)
Thanks
Post by: RuiAce on October 01, 2017, 12:07:59 am
(https://i.imgur.com/5MmSHYr.png)It really isn't if you ask me. As you've just seen, the trick was to use sin(x)/cos(x), but personally I find it unfair how a 2U student should be told to do that without any guidance whatsoever.
Is integration of tanx a part of the 2 unit syllabus or is this trial just being hard? Wondering if it is because I haven't encountered it before (I know now to use sinx/cosx).
(https://i.imgur.com/jUc3IKi.png)
Seems straightforward, a + B = -B + B therefore b/a = 0. But why is (4k+1) = b here? Shouldn't it be (4k+1)x? Solution:
(https://i.imgur.com/xfKYo8U.png)
Thanks
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Post by: _____ on October 01, 2017, 01:05:03 pm
It really isn't if you ask me. As you've just seen, the trick was to use sin(x)/cos(x), but personally I find it unfair how a 2U student should be told to do that without any guidance whatsoever.
_____________________________________
Thanks.
For the second one, the equation is given as 2x^2 - (4k+1) + 2k^2 - 1 = 0. Shouldn't there be an x next to the -(4k+1) (so the original equation is 2x^2 - (4k+1)x + 2k^2 - 1 = 0) so that I know that -(4k+1) is b? If not, how do I know what's b and what's c? I should have explained this better earlier.
Post by: Shadowxo on October 01, 2017, 01:36:07 pm
Thanks.It looks like there was supposed to be an x there, they just didn't put it in, so b would be -(4k+1). If there was no 'x' term then b would be 0.
For the second one, the equation is given as 2x^2 - (4k+1) + 2k^2 - 1 = 0. Shouldn't there be an x next to the -(4k+1) (so the original equation is 2x^2 - (4k+1)x + 2k^2 - 1 = 0) so that I know that -(4k+1) is b? If not, how do I know what's b and what's c? I should have explained this better earlier.
Post by: RuiAce on October 01, 2017, 01:37:21 pm
Thanks.Ah. Whilst technically that should be a part of the constant, I believe whoever wrote that paper made an extremely clumsy typo and forgot the x.
For the second one, the equation is given as 2x^2 - (4k+1) + 2k^2 - 1 = 0. Shouldn't there be an x next to the -(4k+1) (so the original equation is 2x^2 - (4k+1)x + 2k^2 - 1 = 0) so that I know that -(4k+1) is b? If not, how do I know what's b and what's c? I should have explained this better earlier.
Post by: _____ on October 01, 2017, 02:23:09 pm
Ah. Whilst technically that should be a part of the constant, I believe whoever wrote that paper made an extremely clumsy typo and forgot the x.
Whew, glad I wasn't being dumb. Didn't think James Ruse (2015 paper) of all schools would make a mistake like that. They didn't even note it in the marker's comments lol
Post by: RuiAce on October 01, 2017, 02:28:47 pm
Whew, glad I wasn't being dumb. Didn't think James Ruse (2015 paper) of all schools would make a mistake like that. They didn't even note it in the marker's comments lolYou shouldn't discard that possibility just because they're the #1 school. These mistakes happen all the time.
Post by: Thebarman on October 01, 2017, 02:59:12 pm
Post by: RuiAce on October 01, 2017, 03:22:02 pm
Any tips on how to integrate functions? I'm completely lost when it comes to it.There's only so many functions to integrate in 2U.
Please provide examples indicating where the difficulty lies.
Post by: Thebarman on October 01, 2017, 03:33:44 pm
For a while, I tried using the quotient rule here before realising that the rule didn't apply for integration.
Post by: RuiAce on October 01, 2017, 03:44:20 pm
Say t /(t^2 + 3). How would I integrate that?
For a while, I tried using the quotient rule here before realising that the rule didn't apply for integration.
Usually, what constitutes "almost the same" is when you're out by a factor of a constant.
Note that we fudge by multiplying by 2, but then multiplying by 1/2 to cancel it out. This is effectively equivalent to multiplying by 1, so we have not changed the question.
Post by: Thebarman on October 01, 2017, 05:06:53 pm
I'm still ending up with answers that are slightly off when it comes to these kind of questions. With this question, I'm off by a few decimals, but it screws up the following answers.
A particle's acceleration is given by a=2/(t+3)^2 with the particle initially resting at 2 Ln(3) m to the left of the origin. Evaluate the velocity after 2 seconds.
I integrated it and reached ln(t+3) + C, but everything went fuzzy afterwards. Where would I go from here?
Post by: RuiAce on October 01, 2017, 05:09:38 pm
Ahhhh. Yeah, that makes sense.
I'm still ending up with answers that are slightly off when it comes to these kind of questions. With this question, I'm off by a few decimals, but it screws up the following answers.
A particle's acceleration is given by a=2/(t+3)^2 with the particle initially resting at 2 Ln(3) m to the left of the origin. Evaluate the velocity after 2 seconds.
I integrated it and reached ln(t+3) + C, but everything went fuzzy afterwards. Where would I go from here?
Post by: sophiegmaher on October 01, 2017, 07:13:20 pm
Post by: RuiAce on October 01, 2017, 07:15:16 pm
Hey so I'm confused on part (ii) of the attached question. I've looked through the answers but I still don't get it!Already addressed by Jake in the compilation.
Post by: jaskirat on October 01, 2017, 09:10:39 pm
Post by: RuiAce on October 01, 2017, 09:12:36 pm
why does my question keep getting removed?Because I have already moved it elsewhere. Like stated, please refrain from spam posting.
Post by: jaskirat on October 01, 2017, 09:27:10 pm
Post by: sweetiepi on October 01, 2017, 09:39:46 pm
where did you move it to?Hey there! I believe Rui moved it to here: https://atarnotes.com/forum/index.php?topic=173885.msg985661#new :)
Post by: bdobrin on October 02, 2017, 05:16:54 pm
I was just wondering if you could help me with this HSC question from the 2004 exam.
1) The limiting sum of the geometric series: '1-Tan2 theta + Tan4 theta' is Cos2 theta. For what values of theta in the interval -π/2 < theta < π/2 does the limiting sum of the series exist?
Thanks,
Ben
Post by: RuiAce on October 02, 2017, 05:22:26 pm
Hey jake,
I was just wondering if you could help me with this HSC question from the 2004 exam.
1) The limiting sum of the geometric series: '1-Tan2 theta + Tan4 theta' is Cos2 theta. For what values of theta in the interval -π/2 < theta < π/2 does the limiting sum of the series exist?
Thanks,
Ben
________________________
The square rooting was justified as all quantities are positive.
Post by: Piza on October 02, 2017, 06:13:20 pm
1. A cubic curve has a minimum turning point at the point (-2, -15), a maximum turning point at (1,12) and it cuts the y axis at 5. What is the equation of the cubic?
I know that the constant will be 5 and that the curve will be negative, given the location of the minimum and maximum turning points. I tried subbing in the points (differentiated curve equation) and doing simultaneous equations but it didn't work out for me.
Spoiler
Answer: y = -2x3 - 3x2 + 12x + 5
2. A sheet of paper for a poster has an area of 6m2. The margins at the top and bottom are 0.75m and at the sides 0.5m. Show that the dimensions for maximum printed area are 2m in length and 3m in width.
(I saw an answer to a similar question on this forum but I had trouble applying it)
3. Find the smallest distance between the graphs of y = x2 - 4x + 12 and y = 2x +1
Spoiler
2/5 √5
Post by: RuiAce on October 02, 2017, 06:27:14 pm
Hi guys, I am having quite some trouble with the following questions:
1. A cubic curve has a minimum turning point at the point (-2, -15), a maximum turning point at (1,12) and it cuts the y axis at 5. What is the equation of the cubic?
I know that the constant will be 5 and that the curve will be negative, given the location of the minimum and maximum turning points. I tried subbing in the points (differentiated curve equation) and doing simultaneous equations but it didn't work out for me.SpoilerAnswer: y = -2x3 - 3x2 + 12x + 5
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Note that this question could've been done a whole variety of ways due to the 4 simultaneous equations involving a,b,c
Post by: RuiAce on October 02, 2017, 06:45:40 pm
3. Find the smallest distance between the graphs of y = x2 - 4x + 12 and y = 2x +1Spoiler2/5 √5
The other question is VERY ambiguous. Please link to the question that looked "similar".
Post by: pokemonlv10 on October 02, 2017, 06:53:03 pm
Post by: RuiAce on October 02, 2017, 06:57:03 pm
MY friend sent me this question and i'm stumped. Any ideas for this?This question has gone well into Extension 1 territory and I will not address it in this thread. Please post it in the more relevant thread if you wish for an answer.
Post by: pokemonlv10 on October 02, 2017, 06:59:38 pm
This question has gone well into Extension 1 territory and I will not address it in this thread. Please post it in the more relevant thread if you wish for an answer.
Ah alright, i'll move it over to there
Post by: Piza on October 02, 2017, 07:53:51 pm
The other question is VERY ambiguous. Please link to the question that looked "similar".
Hi RuiAce, thank you so much for answering my questions :D
Here is the question which looked similar: https://atarnotes.com/forum/index.php?topic=164548.msg867873#msg867873
Post by: RuiAce on October 02, 2017, 09:14:24 pm
Hi RuiAce, thank you so much for answering my questions :DHm, I apologise about overlooking something earlier. I had done something which had actually gotten the right answer, but was beyond the scope of the HSC
Here is the question which looked similar: https://atarnotes.com/forum/index.php?topic=164548.msg867873#msg867873
However, I remain my stance that this question is ambiguous. They relate the dimensions of the "maximum-printed-area" to that of the entire poster, which was quite annoyingly confusing. The way I read the question, it had appeared as though it was just the dimensions of the maximum printed area that required, not the poster.
(https://i.imgur.com/7QRgwDe.png)
Post by: bdobrin on October 03, 2017, 04:19:41 pm
just wondering if you could help me on this little integration question.
1.) Find the integral of 'sec2(x) + sec(x)•tan(x)
Cheers
Post by: Eric11267 on October 03, 2017, 04:28:03 pm
Hi Jake,I do VCE so I'm not sure if this applies to you but I would split it up into two different integrals and integrate sec2 as normal. For the other half I would put it in terms of sin and cos and use the substition u=cos(x)
just wondering if you could help me on this little integration question.
1.) Find the integral of 'sec2(x) + sec(x)•tan(x)
Cheers
Post by: RuiAce on October 03, 2017, 04:31:26 pm
Hi Jake,
just wondering if you could help me on this little integration question.
1.) Find the integral of 'sec2(x) + sec(x)•tan(x)
Cheers
I do VCE so I'm not sure if this applies to you but I would split it up into two different integrals and integrate sec2 as normal. For the other half I would put it in terms of sin and cos and use the substition u=cos(x)If you look at the syllabus (when you are free), integration by substitution is only expected at the Extension 1 level. (Usually the substitution is also provided.)
Post by: Mary_a on October 03, 2017, 04:34:28 pm
I know that the answer to studying for maths is practice, practice, practice!
It's just I've done about 3 hsc papers in the last week and I've gotten around 60% every time, and I don't know how to improve that?
Should i just keep going with hsc papers?
I guess I'm just a little bit upset because maths is the subject I put the most work into and it's my worst subject.
Thanks guys,
Mary
Post by: RuiAce on October 03, 2017, 04:39:39 pm
Hey,If you're only doing past papers in exam conditions, relax and do not be scared to do open-book papers and out of those exam conditions.
I know that the answer to studying for maths is practice, practice, practice!
It's just I've done about 3 hsc papers in the last week and I've gotten around 60% every time, and I don't know how to improve that?
Should i just keep going with hsc papers?
I guess I'm just a little bit upset because maths is the subject I put the most work into and it's my worst subject.
Thanks guys,
Mary
Consider where the marks are being lost. How is it occurring? Some starting points to consider:
1. Silly mistakes
2. Not understanding the question
3. Not using key bits of information
4. Approaching it from the wrong method
5. Calculator problems
6. Inability to deduce a method
and more.
Carefully consider where, why and how the marks were lost, and make a resolution from there. Of course, you may choose to point to us how the marks are getting lost if you want further insight on those specific areas of concern.
Post by: Fahim486 on October 03, 2017, 08:16:39 pm
Post by: justwannawish on October 03, 2017, 08:22:59 pm
Just a quick question on sig figs. Does 2800 count as 2 sf or 4? And is the HSC consistent with significant figures across science and maths, because my chemistry teacher seems to think that NESA is wrong about how they do it :(
Post by: RuiAce on October 03, 2017, 08:25:06 pm
Hey could someone pls explain to me how x was made the subject. Thanks!
Hey guys,It counts as two, three or four, and without further information we are not really able to deduce this.
Just a quick question on sig figs. Does 2800 count as 2 sf or 4? And is the HSC consistent with significant figures across science and maths, because my chemistry teacher seems to think that NESA is wrong about how they do it :(
However, in mathematics, whilst you are expected to know what significant figures are, you will not be given the same types of question like in science. You should probably ask this question again in the physics thread for appropriate treatment in physics (or chemistry if applicable).
Post by: sidzeman on October 03, 2017, 09:37:26 pm
Post by: RuiAce on October 03, 2017, 09:48:57 pm
Hey for this, shouldn't the expansion be 4n + 2?They miscounted the number of terms. There's only 1+n terms; not 1+2n, as a consequence of the common ratio being \(a^2\) and not \(a\).
Post by: justwannawish on October 03, 2017, 10:41:27 pm
It counts as two, three or four, and without further information we are not really able to deduce this.
However, in mathematics, whilst you are expected to know what significant figures are, you will not be given the same types of question like in science. You should probably ask this question again in the physics thread for appropriate treatment in physics (or chemistry if applicable).
Sorry about that! Will post in the correct forum now :)
Post by: JuliaPascale123 on October 04, 2017, 09:07:38 am
Post by: sidzeman on October 04, 2017, 11:24:13 am
(http://i63.tinypic.com/2ntv5g9.png)Not 100% sure on this at all but I'll give it a shot
a. a/q is ordering while bq is storage. We know that its cheaper to place large orders - therefore as "q" (ordered amount) goes up, total price C must go down - which is given by a/q (inverse relationship)
b. I think is just simple differentiation. I'll upload a picture showing what I did - I don't think theres any other way of doing this question?
Post by: Mary_a on October 04, 2017, 11:43:58 am
If you're only doing past papers in exam conditions, relax and do not be scared to do open-book papers and out of those exam conditions.
Consider where the marks are being lost. How is it occurring? Some starting points to consider:
1. Silly mistakes
2. Not understanding the question
3. Not using key bits of information
4. Approaching it from the wrong method
5. Calculator problems
6. Inability to deduce a method
and more.
Carefully consider where, why and how the marks were lost, and make a resolution from there. Of course, you may choose to point to us how the marks are getting lost if you want further insight on those specific areas of concern.
Okay, thanks so much, I'll definitely try to work out what's going wrong.
Cheers,
Mary
Post by: _____ on October 04, 2017, 03:05:15 pm
(https://i.imgur.com/3HY6qXX.png)
Answer is D. Is this because the positive speed means that the particle is moving towards a min TP/point of inflexion on the displacement graph (bottoming out)? If it were a negative speed, would the graph instead be dropping off like the example below? Haven't quite got the x v a relationship figured out yet.
(https://i.imgur.com/RoNaP3R.png)
Post by: RuiAce on October 04, 2017, 03:11:46 pm
2013 paper:
(https://i.imgur.com/3HY6qXX.png)
Answer is D. Is this because the positive speed means that the particle is moving towards a min TP/point of inflexion on the displacement graph (bottoming out)? If it were a negative speed, would the graph instead be dropping off like the example below? Haven't quite got the x v a relationship figured out yet.
(https://i.imgur.com/RoNaP3R.png)
__________________
So here, it slows down as it moves towards the left.
Post by: arunasva on October 05, 2017, 06:10:44 am
Post by: RuiAce on October 05, 2017, 09:09:53 am
Can I get some help with part b ? TY famz
And here's the part where math "clashes" with what really happens. Technically here, you could argue that after an infinite amount of time has passed, the solvent has fully evaporated. But infinity is not a number, so that never happens. So technically it never fully evaporates; it just gets asymptotically closer to 0 volume but never quite gets there.
When the questions are worded this way, it's quite poor. Usually, they would specifically ask for the "limiting behaviour", or in this case the "limiting volume of the solvent". Not bluntly whether or not it will evaporate.
Post by: davidss on October 05, 2017, 09:48:13 am
(https://i.imgur.com/6zpahWq.png)
Post by: Eric11267 on October 05, 2017, 10:13:59 am
hey can someone help with part iii please. thx :DAn easier way of approaching it is first looking at the probability of getting 45 or higher. To get this you would need either (20,25), (25,20) or (25,25).
(https://i.imgur.com/6zpahWq.png)
The probability of (20,25) is 2/36x1/36
The probability of (25,20) is 1/36x2/36
The probability of (25,25) is 1/36x1/36
So the probability of getting 45 or above is 5/1296
Thus the probability of getting less than 45 is 1-5/1296= 1291/1296
Edit: added working
Post by: EEEEEEP on October 05, 2017, 11:23:51 am
(https://i.imgur.com/UrPsNt6.png)
Post by: bdobrin on October 05, 2017, 11:54:11 am
Could someone please explain how to do part two of this question for me - i am really confused.
Thanks,
Ben
Post by: Shadowxo on October 05, 2017, 01:17:46 pm
Hi,In part i) You've found the area of the cross section, which you'll use here.
Could someone please explain how to do part two of this question for me - i am really confused.
Thanks,
Ben
So you want to find the volume of water that passes through the cross section. It's easiest to imagine this, water is travelling at 0.4 m/s through this cross sectional area (which I'll call A). The volume passing through it in one second is simply area * velocity = A*0.4.
The amount of water passing through it in 10 seconds is volume per second * time = 0.4*A*10 = 4*A
Post by: fantasticbeasts3 on October 05, 2017, 01:24:45 pm
Post by: Shadowxo on October 05, 2017, 01:31:35 pm
i'm terrible at everything to do with logs - can someone explain why the answer's D?
All the answers have it as a single term in base 2. So, you want to convert everything to log2. Remember logaa=1
Which is D :)
Post by: inescelic on October 05, 2017, 02:56:34 pm
I'm really struggling with these derivative graph questions, does anyone have some general tips for them or know where I can practice questions like these (I've really only seen questions like these in the hsc and not in any textbooks)
Much appreciated :)
Post by: JeffChiang on October 05, 2017, 05:47:22 pm
Post by: Thebarman on October 05, 2017, 06:06:19 pm
Post by: RuiAce on October 05, 2017, 07:28:13 pm
Hey can someone please help me with this question part (iii) and (iv)?Please provide the answers to the previous parts. Also, note that there are some exercises in maths in focus in regards to (iv), just not at the same difficulty.
I'm really struggling with these derivative graph questions, does anyone have some general tips for them or know where I can practice questions like these (I've really only seen questions like these in the hsc and not in any textbooks)
Much appreciated :)
This seemingly simple question has made quite a kerfuffle for part ii. The answer says top graph minus bottom graph but then that disregards the negative area below the axis. Please explain, thanks!
For now, I will not provide an extensive explanation as to why this is the case, but only a rough sketch. That rough sketch is that we know the integral (without absolute values) measures a signed area, and when we subtract something negative we get back to something positive.
Post by: RuiAce on October 05, 2017, 07:33:14 pm
I'm watching the max/min HSC revision video right now, and I'm a bit confused regarding the differentiation. Originally y' = 2H(pi)x - (3/R)H(pi)x^2, where H and R are constants. Later on in the video, it was written as y' = 2x - (3x^2)/R. Why so? Shouldn't the pi and H be left in the equation?(https://i.imgur.com/9OIi4sM.png)
Judging by the red lines, Jamon has clearly divided both sides by \(\pi\) and \(H\)[/tex]. Of course, \( \frac{0}{\pi H} = 0\) .
There is no reason to leave \( \pi \) and \( H \) in there if they can be divided out without harm.
Post by: Opengangs on October 05, 2017, 08:23:46 pm
Hey can someone please help me with this question part (iii) and (iv)?Assuming that you've done the parts before it, you'll find that the distance travelled for the first 4 seconds is 6 metres.
I'm really struggling with these derivative graph questions, does anyone have some general tips for them or know where I can practice questions like these (I've really only seen questions like these in the hsc and not in any textbooks)
Much appreciated :)
This is important to note, but we'll save it for a little bit later.
The next thing to note is that the area under the curve between 4 and 5 seconds is exactly the same as the area under the curve between 5 and 6 seconds. This implies it takes as much time going vertically upwards and coming back down to 6 metres; this means we need to find the time it takes to reach 6 metres vertically down.
The final thing we need to note is that the graph given is the velocity graph, or its speed with a vector quantity. It is the velocity at which the particle travels at any point in time. So, once we understand that it's the dx/dt graph, we can use it to find the time. Note that the time is given by the distance over the speed at a particular time.
Thus, the time it takes for the particle to reach 6 metres is given by: distance (6m) / speed (5ms^-1) = 1.2 seconds.
This means, it takes 1.2 seconds to reach a displacement of 6 metres, meaning that for the particle to have no displacement, it will take: 6 + 1.2 = 7.2 seconds.
Graphing the displacement graph shouldn't come at a surprise to you.
You're given that it starts at the origin; it will keep increasing until 5 seconds, where there is a maximum turning point. It will curve until about 6 seconds, where the displacement graph then drops at a constant rate until it reaches the origin again at 7.2 seconds.
Post by: inescelic on October 05, 2017, 10:30:08 pm
Please provide the answers to the previous parts. Also, note that there are some exercises in maths in focus in regards to (iv), just not at the same difficulty.
Sure, solution is attached- I think my my main issue is how does a velocity graph tell us when a particle returns to the origin? Is it when the area of the top half of the axis cancels the bottom half of the x axis to = 0? Does that mean if area under the curve of velocity graph is 0, the displacement is also 0?
I have seen a similar question in the 2013 hsc (attached with solutions) which would also be great if you could explain :)
Thank you :)
Post by: RuiAce on October 05, 2017, 10:43:58 pm
Please provide the answers to the previous parts. Also, note that there are some exercises in maths in focus in regards to (iv), just not at the same difficulty.
Sure, solution is attached- I think my my main issue is how does a velocity graph tell us when a particle returns to the origin? Is it when the area of the top half of the axis cancels the bottom half of the x axis to = 0? Does that mean if area under the curve of velocity graph is 0, the displacement is also 0?
I have seen a similar question in the 2013 hsc (attached with solutions) which would also be great if you could explain :)
Thank you :)
Some information on tackling that 2013 question was provided in the compilation thread already. But if it's not enough, you may come back and further your question.
tagged: 2007
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where f(t) is any random function that we plucked out of thin air because maths is magic.
________________________________________
The point: We got back to the same answer, and we didn't have to do that useless finding +C stuff
________________________________________
A key point is that this measured the change in displacement, SPECIFICALLY between times t=0 and t=4. This is significant, as otherwise the boundaries of our integral would've been different numbers.
________________________________________
________________________________________
Reason being, this value \(t_1\) will ensure that the change in displacement, from 0, to \(t_1\), will be zero. Because the particle starts at the origin, the time when it returns to the origin is the time when the change in displacement is zero.
________________________________________
Post by: winstondarmawan on October 05, 2017, 10:53:01 pm
I was just wondering for Series Application questions, particularly time repayments and the like, which steps of working to we ALWAYS have to include? I have lost marks for omitting steps in the past and I want to make sure it doesn't happen again.
Thanks in advance.
Post by: RuiAce on October 05, 2017, 10:57:26 pm
Hello!1. Establishing the very first case \(A_1\)
I was just wondering for Series Application questions, particularly time repayments and the like, which steps of working to we ALWAYS have to include? I have lost marks for omitting steps in the past and I want to make sure it doesn't happen again.
Thanks in advance.
2. Explicitly showing a pattern beginning to form in \(A_2\). Some computations should be kept there - do not assume the pattern, prove it.
3. Show how a rearrangement of the terms demonstrates the same pattern in \(A_3\). (Recommended: Keep the explicit computations, but I'm not sure how markers care about this)
4. Only after you've made the pattern explicit in the first three cases, can you jump straight to the n-th case.
5. Mention the first term and common ratio of your geometric series
6. Fair game
Also step 0: Make clear what your appropriate interest rate is, based off the compounding frequency.
Post by: inescelic on October 05, 2017, 11:05:36 pm
Some information on tackling that 2013 question was provided in the compilation thread already. But if it's not enough, you may come back and further your question.
tagged: 2007
________________________________________
where f(t) is any random function that we plucked out of thin air because maths is magic.
________________________________________
The point: We got back to the same answer, and we didn't have to do that useless finding +C stuff
________________________________________
Ok, I'm a little confused by all that, I understand i) and ii) but how does that help us with part 3?
Also how can I find that 2013 q when this thread has 180 pages haha
Appreciate the help :)
Post by: RuiAce on October 05, 2017, 11:06:45 pm
Ok, I'm a little confused by all that, I understand i) and ii) but how does that help us with part 3?I had added in more information.
Also how can I find that 2013 q when this thread has 180 pages haha
Appreciate the help :)
The compilation thread is a separate thread.
Post by: pokemonlv10 on October 05, 2017, 11:42:37 pm
Post by: RuiAce on October 06, 2017, 12:01:18 am
Unsure how to do the 2nd part of this question? I got 3 points of intercepts for part a - (-2,0) (-1,0) and (3,0)
Remember, you can read off the graph they actually gave you to deduce that f(-2) = -3, f(-1) = -1, f(3) = 5.
Post by: davidss on October 06, 2017, 12:07:22 pm
(https://i.imgur.com/zIHirsQ.png)
Post by: Tempestuous on October 06, 2017, 12:56:35 pm
From part ii we know that angle BDC = angle BCD (assuming you've done the first 2 parts)
Thus BC = BD (equal sides opposite equal angles in triangle BCD)
Then, AE:ED = AB:BC (a line parallel to one side of a triangle divides the other two sides in the same ratio)
Therefore AE:ED = AB:BD
Post by: jaskirat on October 06, 2017, 01:23:55 pm
Post by: RuiAce on October 06, 2017, 02:17:10 pm
Need help with this question
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The last part is the challenging part as it involves doing battle with algebra and manipulation
Substitute back in and you are done.
Post by: sidzeman on October 06, 2017, 03:18:06 pm
Sorry I think I'm being stupid but I don't understand the last part. It says to find the min length of the fence, which is ST or "z" right?. So why did you use the cosine rule to find the angle A?
___________________________
___________________________
___________________________
The last part is the challenging part as it involves doing battle with algebra and manipulation
Substitute back in and you are done.
Post by: davidss on October 06, 2017, 03:31:29 pm
Hey davidss,
From part ii we know that angle BDC = angle BCD (assuming you've done the first 2 parts)
Thus BC = BD (equal sides opposite equal angles in triangle BCD)
Then, AE:ED = AB:BC (a line parallel to one side of a triangle divides the other two sides in the same ratio)
Therefore AE:ED = AB:BD
Thanksss!!! Didn't know about that rule :ooo
Post by: _____ on October 06, 2017, 03:52:38 pm
What's the technique to solve this?
I got y' as a concave up parabola with vertex at x = (the point where the y'' graph touches the x axis), and y as a cubic with a point of inflexion at this point, which is correct. But how do I determine elevation (y value) at this point for y' and y?
---------
When I'm asked to graph say "y = SQRT(16-x^2)" do I assume this just means the positive square root and not the negative square root because it is a function? Do I only sketch the full circle if the question specifically states that it's a relation or a full circle?
Thanks
Post by: RuiAce on October 06, 2017, 04:19:28 pm
Sorry I think I'm being stupid but I don't understand the last part. It says to find the min length of the fence, which is ST or "z" right?. So why did you use the cosine rule to find the angle A?I always was solving for z.
But z was written in terms of cos A. So I needed to get rid of it somehow
Thanksss!!! Didn't know about that rule :oooIt relies on similar triangles. Quite risky to use, even though I would happily give marks for it.
Post by: bdobrin on October 06, 2017, 04:45:58 pm
Can someone please explain to me how the answer to this question is D.
I thought it would be 'b' because the velocity is negative meaning it moves in a negative direction and the acceleration is positive so it is picking up speed.
Thanks,
Ben :)
Post by: Eric11267 on October 06, 2017, 04:50:41 pm
HiIn this case the positive and negative signs refer to a direction, with negative being left and positive being right. So in this case it is moving to the left but is accelerating to the right. So it is moving to the left with decreasing speed.
Can someone please explain to me how the answer to this question is D.
I thought it would be 'b' because the velocity is negative meaning it moves in a negative direction and the acceleration is positive so it is picking up speed.
Thanks,
Ben :)
Post by: RuiAce on October 06, 2017, 04:55:08 pm
HiThis was already addressed here
Can someone please explain to me how the answer to this question is D.
I thought it would be 'b' because the velocity is negative meaning it moves in a negative direction and the acceleration is positive so it is picking up speed.
Thanks,
Ben :)
Post by: kiiaaa on October 06, 2017, 06:51:38 pm
i need help in q14a iv / q14b and q14 iii/iv
so for q14a iv - was wondering how to exactly do it as i was really lost
for q14b i dort of tried and got stuck at a dead end having no idea where I was going
and finally, for q14 iii/iv I'm really lost how to find the rate of change =/
thank you very very much and sorry for bombarding you with sooo many questions
Post by: RuiAce on October 06, 2017, 06:57:38 pm
hi, guys, i was doing the 2012 HSC paper and was having some difficulties and was wondering if you could help me out explain and teach me how to tackle these please so I know in the future? I looked at the answers but I didn't really understand what they were doing exactly
i need help in q14a iv / q14b and q14 iii/iv
so for q14a iv - was wondering how to exactly do it as i was really lost
for q14b i dort of tried and got stuck at a dead end having no idea where I was going
and finally, for q14 iii/iv I'm really lost how to find the rate of change =/
thank you very very much and sorry for bombarding you with sooo many questions
_________________________
_________________________
Post by: RuiAce on October 06, 2017, 07:05:30 pm
hi, guys, i was doing the 2012 HSC paper and was having some difficulties and was wondering if you could help me out explain and teach me how to tackle these please so I know in the future? I looked at the answers but I didn't really understand what they were doing exactly
i need help in q14a iv / q14b and q14 iii/iv
so for q14a iv - was wondering how to exactly do it as i was really lost
for q14b i dort of tried and got stuck at a dead end having no idea where I was going
and finally, for q14 iii/iv I'm really lost how to find the rate of change =/
thank you very very much and sorry for bombarding you with sooo many questions
e.g. y=sin(x) + 2 is just y = sin(x), but shifted up by 2 units.
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Post by: Tempestuous on October 06, 2017, 08:59:22 pm
I always was solving for z.
But z was written in terms of cos A. So I needed to get rid of it somehowIt relies on similar triangles. Quite risky to use, even though I would happily give marks for it.
Would it be safe to use in the HSC then?
Post by: RuiAce on October 06, 2017, 09:15:48 pm
Would it be safe to use in the HSC then?They’d have to be pretty evil to deduct you marks for it. But if you want to be absolutely safe, don’t
Post by: _____ on October 06, 2017, 10:08:28 pm
Thanks for monitoring the thread so closely Rui, it's a real help.
Post by: RuiAce on October 06, 2017, 10:18:16 pm
(https://i.imgur.com/RGBxRI7.png)Ah yes all good, unfortunately it was lost indeed and your bump was rightly justified. I've had a busy week and my mind's been all over the place.
What's the technique to solve this?
I got y' as a concave up parabola with vertex at x = (the point where the y'' graph touches the x axis), and y as a cubic with a point of inflexion at this point, which is correct. But how do I determine elevation (y value) at this point for y' and y?
---------
When I'm asked to graph say "y = SQRT(16-x^2)" do I assume this just means the positive square root and not the negative square root because it is a function? Do I only sketch the full circle if the question specifically states that it's a relation or a full circle?
Thanks
First bit - Seeing as though they give you no real means of handling the +C...
Whereas for y' you must draw it as you correctly did, you can draw it however high you want for y and you will still be marked correct. Your answer can only be marked on what you've been given, so your approach was fine.
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An easy way to see this is really just a table of values. You will only get positive y-values for one of them and negative y-values for the other.
Post by: _____ on October 06, 2017, 10:50:30 pm
Ah yes all good, unfortunately it was lost indeed and your bump was rightly justified. I've had a busy week and my mind's been all over the place.
First bit - Seeing as though they give you no real means of handling the +C...
Whereas for y' you must draw it as you correctly did, you can draw it however high you want for y and you will still be marked correct. Your answer can only be marked on what you've been given, so your approach was fine.
__________________________
An easy way to see this is really just a table of values. You will only get positive y-values for one of them and negative y-values for the other.
Thanks a bunch.
For the second one, I look at SQRT(16-x^2) and think that it can refer to the upper or lower half. This is probably because I've been doing questions (like rotation volume questions where you must find say x^2 = SQRT(y)) where you find the square root of something and then must test the positive and negative possibilities and choose the correct one.
Thinking about this more, the correct notation for both possibilities must always be +-SQRT(y) even though my calculator just produces SQRT(y), right? So I need to write it like this. If I (or the examiners) write SQRT(y) I'm/they're saying +SQRT(y), not +-SQRT(y). I think I understand now.
Post by: sophiegmaher on October 07, 2017, 01:47:43 pm
(i) In the following series, for what values of x will a sum to infinity exist?
(1-x)^1/2 + (1-x) + (1-x)^3/2+...
(ii) Find the value of x, if the series in part (i) has a limiting sum of 2root3 + 3
Post by: RuiAce on October 07, 2017, 02:24:31 pm
How do you solve part (ii) of this question? I have confused on the simplification part of the equation you eventually get.
(i) In the following series, for what values of x will a sum to infinity exist?
(1-x)^1/2 + (1-x) + (1-x)^3/2+...
(ii) Find the value of x, if the series in part (i) has a limiting sum of 2root3 + 3
Edit: Mistake in sign
Post by: sophiegmaher on October 07, 2017, 02:53:17 pm
Post by: Shadowxo on October 07, 2017, 03:10:49 pm
I worked it out by expanding the brackets, but I'm not sure how to simplify it like the answers I have do, like how they go from line 3 of the answers to line 4 of the answers:They just skipped a few steps. All they did was move all the sqrt(1-x) terms to one side
Post by: sophiegmaher on October 07, 2017, 04:24:52 pm
They just skipped a few steps. All they did was move all the sqrt(1-x) terms to one side
Thank you!! That makes so much more sense, I was just getting confused with the factorising :)
Post by: hobocop on October 08, 2017, 05:42:20 pm
Post by: Shadowxo on October 08, 2017, 08:24:58 pm
Hi! Could I get some help with part iii) please?
Post by: hansolo9 on October 09, 2017, 12:18:38 am
Is it possible to get a question in the HSC about approximating a solid of revolution using the trapezoidal or Simpson rule? Our teacher taught us this in our very last lesson but not wasn’t sure if it was actually in the syllabus
Post by: RuiAce on October 09, 2017, 07:02:48 am
Hey guysSure, why not?
Is it possible to get a question in the HSC about approximating a solid of revolution using the trapezoidal or Simpson rule? Our teacher taught us this in our very last lesson but not wasn’t sure if it was actually in the syllabus
Combining concepts was never outside the scope of the course
Post by: av-angie-er on October 09, 2017, 12:13:36 pm
Post by: RuiAce on October 09, 2017, 12:30:20 pm
Hi! Could I get some help with part (i)? The solutions explain that you're meant to use the formula x/a + y/b = 1 where a = OP and b = OR to just use trigonometric ratios, but I'm not sure where this formula came from. Thanks! :)This was already addressed in the compilation.
Note, however, that the form x/a + y/b = 1 is referred to as the "two-intercept" form. This is a rarely used formula and usually unnecessary.
Post by: sidzeman on October 09, 2017, 03:05:21 pm
Post by: Zaye on October 09, 2017, 03:26:08 pm
thanks
Post by: EEEEEEP on October 09, 2017, 03:56:29 pm
Hey could someone please explain how to do part ii to me pleaseHint: ln (e^x) = x
1. Sub the Intensity in the equation,
2. divide both sides by 10^-12
3. Times both sides by ln
The answer should be 90.
Post by: RuiAce on October 09, 2017, 07:03:24 pm
Hi, I really need help with similarity- basically its whole concept and rules and approach to questions i.e the fundamental steps.Start by recognising what similarity is. Two triangles are similar if one's basically another, stretched.
thanks
Then, know your tests.
1. Equiangular (two angles equal)
2. All three sides in proportion
3. Two sides in proportion, included angle equal
Of course, 1 appears the most, and you should always watch out for two angles being equal. But 3 also appears occasionally. (2 is, however, rarer.)
Know the tests, because they are self explanatory - they describe what they require FOR you (unlike congruency, which involves abbreviations). Then, think about which is appropriate.
Those are some general tips, but if you need serious help you should provide examples of questions you're struggling on, for a clearer thought process to be illustrated.
Post by: _____ on October 09, 2017, 09:10:49 pm
This is an example:
(https://i.imgur.com/TX79Bh1.png)
Can I just express SQRT(A1) and SQRT(A2) and add them to show they equal SQRT(A)? This isn't exactly what the answer does (see below) so I'd like to make sure I'd get the marks.
(https://i.imgur.com/e3wtdxt.png)
Thanks
Edit:
Hey guys
Is it possible to get a question in the HSC about approximating a solid of revolution using the trapezoidal or Simpson rule? Our teacher taught us this in our very last lesson but not wasn’t sure if it was actually in the syllabus
Just to make sure I know how to do this, are these the correct steps?
1. approximate the area
2. square this result
3. multiply it by pi
Post by: av-angie-er on October 09, 2017, 09:23:48 pm
This was already addressed in the compilation.That makes it much clearer, thanks so much! :D
Note, however, that the form x/a + y/b = 1 is referred to as the "two-intercept" form. This is a rarely used formula and usually unnecessary.
Post by: RuiAce on October 09, 2017, 09:27:05 pm
When they tell us to "deduce" does this imply any restrictions on getting to the result, or can we just show it in any way possible?"Deduce" is pretty much another (but actually a stronger) way of saying hence.
This is an example:
(https://i.imgur.com/TX79Bh1.png)
Can I just express SQRT(A1) and SQRT(A2) and add them to show they equal SQRT(A)? This isn't exactly what the answer does (see below) so I'd like to make sure I'd get the marks.
(https://i.imgur.com/e3wtdxt.png)
Thanks
Edit:
Just to make sure I know how to do this, are these the correct steps?
1. approximate the area
2. square this result
3. multiply it by pi
Also, the word similar was given in the question, so if you find that some result follows the exact same lines of working (in which here, it does, just in a different triangle), you are allowed to use the word similarly.
That "deduce" is there for a reason. Trying to add \(\sqrt{A_1}\) and \(\sqrt{A_2}\) by brute force (i.e. without your expression for \( \sqrt{\frac{A_1}{A_2}}\) ) will be
________________________________
Post by: sidzeman on October 10, 2017, 12:18:25 pm
For the similar triangles question Im not sure how to do part ii - Im never quite sure on how to approach plane geo questions
c) How are we meant to show part iii?
The working out for 10 is weird could someone also tell me what their answer was with working out?
Sorry for the spam thanks so much in advance guys
Post by: RuiAce on October 10, 2017, 12:56:21 pm
For the geometric series question, I got part 1 correct (cos squared theta if im remembering correctly), but cant do part ii.The first one was already addressed (and will potentially be added to the compilation)
For the similar triangles question Im not sure how to do part ii - Im never quite sure on how to approach plane geo questions
c) How are we meant to show part iii?
The working out for 10 is weird could someone also tell me what their answer was with working out?
Sorry for the spam thanks so much in advance guys
You should either copy out the diagram and annotate this, or print off the diagram and then annotate it. You should also annotate further knowing that you now have a pair of congruent triangles, therefore all sides on the triangle are equal.
To what extent was it obvious - I do not know. But the diagram really looked suspicious - the three congruent triangles were just staring at me after I looked hard enough.
_______________________
Label some more information on your diagram.
Post by: RuiAce on October 10, 2017, 03:57:25 pm
For the geometric series question, I got part 1 correct (cos squared theta if im remembering correctly), but cant do part ii.
For the similar triangles question Im not sure how to do part ii - Im never quite sure on how to approach plane geo questions
c) How are we meant to show part iii?
The working out for 10 is weird could someone also tell me what their answer was with working out?
Sorry for the spam thanks so much in advance guys
Post by: Aaron12038488 on October 10, 2017, 06:39:05 pm
how many weeks will it take for the topic Series and Sequences AND geomterical applications of calculus?
Post by: Savas_P on October 10, 2017, 06:56:11 pm
i have my first 2 unit test in week 7 i believe...
how many weeks will it take for the topic Series and Sequences AND geomterical applications of calculus?
lol at school it took a few weeks for each topic, you could probably cover at home in a few hours. I wouldn't stress too much.
Post by: RuiAce on October 10, 2017, 07:42:51 pm
The working out for 10 is weird could someone also tell me what their answer was with working out?
Sorry for the spam thanks so much in advance guys
Nothing looks overly confusing here. Please mention which line you feel uncomfortable with.
i have my first 2 unit test in week 7 i believe...Depends for school to school. Probably a fair call in your concern in that those are two of the three largest topics, but it isn't an impossibility.
how many weeks will it take for the topic Series and Sequences AND geomterical applications of calculus?
Usually, if your school messes up with the schedule, they will work their way around it, possibly changing their mind and only examine half of one of the topics. They are (as alluded to above) "learn-able" in a few hours (spread over probably two or three days though), but your school will deal with things as they come if anything.
Post by: nattynatman on October 11, 2017, 10:24:07 am
Usually I struggle during practice questions and doubt my answer because it looks so absurd. I feel like it's easy to second guess myself in maths. This has been the downfall of my marks (average 60%) and once I check the solutions it all makes sense - but I still can't seem to apply it in the process of finding the answers. D:
What are some ways I can fix this before my 2u exam? (I've been doing past papers too)
Thanks!!
Post by: RuiAce on October 11, 2017, 11:33:33 am
Helloo I have a question regarding more so how to think about questions/apply your knowledge/formulas:Second guessing can be quite normal, and sometimes put you back on track. When you should act on it, however, is usually not when you actually start doubting it. Unless you're fully confident that you know the correct method, you should leave that question for now and complete what's left on the paper, because it's all about mark maximisation. Keep on trying and attempting all the remaining questions, and then once you feel as though you can't do any more (paper too hard by then, time etc.) go back to what the question you were concerned about.
Usually I struggle during practice questions and doubt my answer because it looks so absurd. I feel like it's easy to second guess myself in maths. This has been the downfall of my marks (average 60%) and once I check the solutions it all makes sense - but I still can't seem to apply it in the process of finding the answers. D:
What are some ways I can fix this before my 2u exam? (I've been doing past papers too)
Thanks!!
It's easy to look at a question and realise "how" it's done, but it can be far more difficult understanding WHY that approach was taken. When you first started doing past papers, this was not problematic, as you didn't need to develop any sense on intuition on what method to use. Now that you have done a fair few, you need to consider why. Instead of "how is this question done?", be more focused on "how was this one of the many correct approaches, if not the only correct approach to tackling it?". Also, remember that there can be multiple ways of doing the same question.
Because the questions themselves tend to differ but involve the exact same methods/formulae/techniques, look out for any key characteristics as well. E.g. in geometry, use all key details but also try to "see" the theorem. With quadratics, the discriminant can appear at some really weird places. If you need guidance on this, provide further examples of questions with their solutions and explain where the confusion lies.
Post by: gilliesb18 on October 11, 2017, 12:07:23 pm
I just need help on a locus question. I am new to the topic so I need heaps of help!!!
Question: Find the equation of the locus of point P(x,y) that moves so that it is equidistant from the points (3,2) and (-1,5).
I started to work out the line perpendicular to the line joining the two points. Is that right? Cause I then looked at the answers and I was completly wrong... At least it looked that way!
Thanks heaps in advance....
Post by: Shadowxo on October 11, 2017, 12:22:51 pm
Hello,Drawing it can help you visualise it better
I just need help on a locus question. I am new to the topic so I need heaps of help!!!
Question: Find the equation of the locus of point P(x,y) that moves so that it is equidistant from the points (3,2) and (-1,5).
I started to work out the line perpendicular to the line joining the two points. Is that right? Cause I then looked at the answers and I was completly wrong... At least it looked that way!
Thanks heaps in advance....
So this results in a line that is perpendicular to the line joining them (find the gradient of the line joining them and use m1*m2= -1 to find the gradient of the normal / perpendicular line) and passes through the midpoint
I suggest using y-y1=m(x-x1) to find the equation, with (x1,y1) being the midpoint
Post your working if you're still stuck :) Good luck!
Post by: Natasha.97 on October 11, 2017, 12:29:20 pm
Hello,
I just need help on a locus question. I am new to the topic so I need heaps of help!!!
Question: Find the equation of the locus of point P(x,y) that moves so that it is equidistant from the points (3,2) and (-1,5).
I started to work out the line perpendicular to the line joining the two points. Is that right? Cause I then looked at the answers and I was completly wrong... At least it looked that way!
Thanks heaps in advance....
Hi!
Agree with Shadowxo in terms of drawing it out :)
An alternative method would be getting an arbitrary point on the line e.g. P = (x, y), use the distance formula to get PA and PB, then apply the condition in the question (equidistant: PA = PB) to solve :)
Edit: Added solution in case you're stuck :)
(https://i.imgur.com/dhanc98.png)
Post by: RuiAce on October 11, 2017, 12:38:02 pm
Drawing it can help you visualise it betterThis approach is perfectly valid (and I don’t think you can be marked wrong for it), but in the HSC the one they teach is Jess’s version. An arbitrary point P(x,y) is made to satisfy the equation PA=PB, and then distance formulae are applied. It’s usually the recommended approach because it relaxes the need to assume the shape of the locus (I.e. a line), which is only really expected in 4u
So this results in a line that is perpendicular to the line joining them (find the gradient of the line joining them and use m1*m2= -1 to find the gradient of the normal / perpendicular line) and passes through the midpoint
I suggest using y-y1=m(x-x1) to find the equation, with (x1,y1) being the midpoint
Post your working if you're still stuck :) Good luck!
(I don’t have enough time right now to actually write up the solution using that method right now)
Post by: kiiaaa on October 11, 2017, 09:19:29 pm
I'm struggling with all of question b and when I looked at the answer i got even more confused on how they did it.lol Could you please explain to me how to solve this please? as like I found the answers sort of vague so i don't know what they did exactly and why
it is q14 from the 2013 hsc paper incase you are wondering
Thank you very much! :))
Post by: RuiAce on October 11, 2017, 09:39:06 pm
Hi guys
I'm struggling with all of question b and when I looked at the answer i got even more confused on how they did it.lol Could you please explain to me how to solve this please? as like I found the answers sort of vague so i don't know what they did exactly and why
it is q14 from the 2013 hsc paper incase you are wondering
Thank you very much! :))
because distance = speed*time and we have speed=80, time=t.
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But part ii) is just your average max/min problem. Please expand on the problem regarding that part.
Post by: gilliesb18 on October 12, 2017, 09:41:48 am
This approach is perfectly valid (and I don’t think you can be marked wrong for it), but in the HSC the one they teach is Jess’s version. An arbitrary point P(x,y) is made to satisfy the equation PA=PB, and then distance formulae are applied. It’s usually the recommended approach because it relaxes the need to assume the shape of the locus (I.e. a line), which is only really expected in 4uThanks heaps!! I will work on that then...
(I don’t have enough time right now to actually write up the solution using that method right now)
Post by: sidzeman on October 12, 2017, 11:32:21 am
The first one was already addressed (and will potentially be added to the compilation)Sorry, I'm still failing to understand what you did for the similar triangle question. I was able to prove congruency for all 3 triangles. I then calculated angle YCM using the angle sum of a triangle (as angle MYC must be 60). However I get lost up to that point - when you say using trig and find those ratios, what exactly are you doing? Sin rule??
You should either copy out the diagram and annotate this, or print off the diagram and then annotate it. You should also annotate further knowing that you now have a pair of congruent triangles, therefore all sides on the triangle are equal.
To what extent was it obvious - I do not know. But the diagram really looked suspicious - the three congruent triangles were just staring at me after I looked hard enough.
_______________________
Label some more information on your diagram.
Post by: RuiAce on October 12, 2017, 11:40:07 am
Sorry, I'm still failing to understand what you did for the similar triangle question. I was able to prove congruency for all 3 triangles. I then calculated angle YCM using the angle sum of a triangle (as angle MYC must be 60). However I get lost up to that point - when you say using trig and find those ratios, what exactly are you doing? Sin rule??Right angled trigonometry. Always be on the look-out for right-angled triangles, because that allows you to use the very fundamental formulae.
Post by: sidzeman on October 12, 2017, 11:52:55 am
Right angled trigonometry. Always be on the look-out for right-angled triangles, because that allows you to use the very fundamental formulae.
Hmmm, for your 2nd trig use then (tan(angle YCM)) - is it not meant to be MY over MC? How did you get AC instead?
Post by: RuiAce on October 12, 2017, 11:54:11 am
Post by: winstondarmawan on October 12, 2017, 12:45:30 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22450384_1859562304360226_1131010543_o.jpg?oh=c9539facf7d280f9852385e9ac2652e1&oe=59E0BC01
BOS solutions: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22414623_1348343865291133_172365985_n.png?oh=4e05da4594845a032459785e2fe7ee0a&oe=59E12EB3
I'm not sure why they let discriminant = 0 when there are two solutions. Also an explanation for the solution to (ii) would be appreciated.
TIA
Post by: RuiAce on October 12, 2017, 03:42:23 pm
For this question (2012 HSC):
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22450384_1859562304360226_1131010543_o.jpg?oh=c9539facf7d280f9852385e9ac2652e1&oe=59E0BC01
BOS solutions: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22414623_1348343865291133_172365985_n.png?oh=4e05da4594845a032459785e2fe7ee0a&oe=59E12EB3
I'm not sure why they let discriminant = 0 when there are two solutions. Also an explanation for the solution to (ii) would be appreciated.
TIA
Two solutions for x, yes, but only one solution for y. Focus on the y-coordinate of the points of intersection, not the x-coordinates.
__________________________________________
I will do things a bit differently. I'm not fully sure how they were able to assume stuff so easily.
Post by: itssona on October 12, 2017, 05:25:03 pm
find the equation of the parabola passing through (-2,1) with vertex (0,0) and axis x=0
Post by: RuiAce on October 12, 2017, 05:38:29 pm
heya pls help with how to do this
find the equation of the parabola passing through (-2,1) with vertex (0,0) and axis x=0
Post by: itssona on October 12, 2017, 06:22:50 pm
also
does the line 3x + 4y=5 represent a focal chord o n the parabola x^2=5y
whats the best method? just subbing im? or could i equate gradients..
Post by: RuiAce on October 12, 2017, 07:13:15 pm
thank you^A focal chord passes through (0,a). In your case, a=5/4 so you just need to check if the point (0,5/4) lies on the line.
also
does the line 3x + 4y=5 represent a focal chord o n the parabola x^2=5y
whats the best method? just subbing im? or could i equate gradients..
Post by: hobocop on October 12, 2017, 07:56:07 pm
Would this be a good idea, in regards for our answer being able to be scanned. I'm just afraid it would be too time consuming.
Post by: bdobrin on October 12, 2017, 08:00:40 pm
I was just wondering if someone could tell me for a distance travelled question how I do this.
| 2-4ln2 | (there are absolute values on the outside)
Do i just treat it as normal and make it -(2-4ln2) and +(2-4ln2)?
Thanks,
Ben
Post by: RuiAce on October 12, 2017, 08:01:00 pm
Hey, so when answering graph questions, would you draw it out in pencil first then when finalised you would go over it in pen?By the time I was in Year 12, I just drew all of my graphs in pen on the go, and I had no problems with doing so.
Would this be a good idea, in regards for our answer being able to be scanned. I'm just afraid it would be too time consuming.
If you want to use this approach, it's viable only if it won't eat into valuable time for other questions at all. You should draw your x and y axes in pen immediately, and then draw it in pencil as fast as you can. Then, don't come back to it until you have time to, and that's when you go over it in pen.
Post by: RuiAce on October 12, 2017, 08:02:26 pm
Hi,
I was just wondering if someone could tell me for a distance travelled question how I do this.
| 2-4ln2 | (there are absolute values on the outside)
Do i just treat it as normal and make it -(2-4ln2) and +(2-4ln2)?
Thanks,
Ben
Post by: hobocop on October 12, 2017, 08:51:08 pm
By the time I was in Year 12, I just drew all of my graphs in pen on the go, and I had no problems with doing so.
If you want to use this approach, it's viable only if it won't eat into valuable time for other questions at all. You should draw your x and y axes in pen immediately, and then draw it in pencil as fast as you can. Then, don't come back to it until you have time to, and that's when you go over it in pen.
Thanks for your answer.
But would it be detrimental for me if I leave it in pencil? I use 2B lead.
Post by: RuiAce on October 12, 2017, 10:26:06 pm
Post by: mitchello on October 13, 2017, 10:55:46 am
'Helen has just retired with $500000 in her account and wishes to start an annuity. She intends to withdraw $M at the end of each year for 30 years. Reducible interest is calculated at 6% p.a., compounded annually just before each withdrawal.
Calculate $M, the amount of each withdrawal.' (3 marks)
Post by: Natasha.97 on October 13, 2017, 11:55:00 am
Heyo, could I please have help with this question:
'Helen has just retired with $500000 in her account and wishes to start an annuity. She intends to withdraw $M at the end of each year for 30 years. Reducible interest is calculated at 6% p.a., compounded annually just before each withdrawal.
Calculate $M, the amount of each withdrawal.' (3 marks)
Hi!
Financial maths is not my strongest topic so I'm not 100% sure if this is correct:
(https://i.imgur.com/1nn4Poc.png)
Hope this helps
Post by: inescelic on October 13, 2017, 01:22:19 pm
Post by: itssona on October 13, 2017, 02:00:39 pm
a parabola with vertical axis has its vertex at origin. if the line 8x-y-4=0 is a tangent to the parabola fine yhe value of k
Post by: RuiAce on October 13, 2017, 02:17:35 pm
Please help with these questions from 2004 HSC, no solutions from nesa :(Some of these were previously answered.
more
pls help<br>What's k?
a parabola with vertical axis has its vertex at origin. if the line 8x-y-4=0 is a tangent to the parabola fine yhe value of k
Post by: Natasha.97 on October 13, 2017, 02:33:49 pm
Please help with these questions from 2004 HSC, no solutions from nesa :(
2004 6bii)
Hi!
Edit: Added LaTeX
Post by: av-angie-er on October 13, 2017, 11:47:34 pm
Post by: RuiAce on October 14, 2017, 12:12:08 am
Hi! Can I get some help for part (ii)? I've got the rationalised denominators down, but I'm not quite sure how to approach applying AP/GP sums. Thanks! :)Already addressed in the compilation. It is not related to arithmetic or geometric sums.
Post by: georgiia on October 14, 2017, 12:33:15 pm
Could I please have help with this? Thanks!
Post by: RuiAce on October 14, 2017, 04:41:05 pm
(http://uploads.tapatalk-cdn.com/20171014/6498cde8f6c91e29ef28d02b891e34b0.jpg)
Could I please have help with this? Thanks!
Inspiration: If you try graphing \(x_A\) and \(x_B\), you will find that the graph of \(x_B\) will always be above \(x_A\). Because one graph is always above the other, there will not be any points of intersection, and hence the particles will not meet. But we will pretend we didn't know that.
_________________________________________
(https://i.imgur.com/P3ew0iZ.png)
Post by: Mathew587 on October 14, 2017, 07:50:18 pm
Couldn't we have just graphed both the equations of the same graph and show that they never meet?
Inspiration: If you try graphing \(x_A\) and \(x_B\), you will find that the graph of \(x_B\) will always be above \(x_A\). Because one graph is always above the other, there will not be any points of intersection, and hence the particles will not meet. But we will pretend we didn't know that.
_________________________________________
(https://i.imgur.com/P3ew0iZ.png)
Post by: RuiAce on October 14, 2017, 07:54:21 pm
Couldn't we have just graphed both the equations of the same graph and show that they never meet?How would you have known that the graphs don't intersect?
I've seen students not introduce points of intersection when they should be there. I've also seen students who introduce points of intersection when they should not have as well.
Post by: bdobrin on October 15, 2017, 11:47:42 am
I was just wondering if someone could explain for the answer to this question, why they did the 'democracy rule' where they share the denominator instead of integrating it normally? Dont you just check if the derivative of the bottom is equal to the numerator (which it is in this case).
Thanks,
Ben
Post by: Natasha.97 on October 15, 2017, 12:12:15 pm
Hi there,
I was just wondering if someone could explain for the answer to this question, why they did the 'democracy rule' where they share the denominator instead of integrating it normally? Dont you just check if the derivative of the bottom is equal to the numerator (which it is in this case).
Thanks,
Ben
Hi Ben!
In this case, the "democracy rule" would have to be used, as the derivative of the denominator is \(2x\). If the denominator was \(x^2+x\), then the answer would be \(ln(x^2+x) + C\).
Hope this helps
Post by: caitlinlddouglas on October 17, 2017, 04:42:28 pm
Thanks!:)
Post by: pokemonlv10 on October 17, 2017, 06:29:20 pm
Post by: RuiAce on October 17, 2017, 07:56:51 pm
Hey i was just wondering how to work out which side of the triangle went with which for this question? I can't work it outBecause the HSC always does things correctly (as opposed to some certain textbooks), the ordering goes with the order they are listed in.
Thanks!:)
Hey, was wondering if c and d are both the correct answer in the 2nd screenshot? apparently the answer is c but i think d is plausible as well. Also, is my drawing for 1st picture correct? (the point of inflection is poorly drawn sorry)I don't know what you mean by c and d. Your second screenshot involves a graph.
I assume that in your first screenshot you were given the black curve and needed to draw the blue one, which was its primitive. If that were the case, the only problem I have with it is that as you go further to the left \( (x < -4) \), your graph becomes horizontal. That definitely should not happen.
Other than that it looks fine.
Post by: pokemonlv10 on October 17, 2017, 08:02:01 pm
Post by: RuiAce on October 17, 2017, 08:06:51 pm
Post by: pokemonlv10 on October 17, 2017, 08:10:02 pm
EDIT: Thank you for the response below!
Post by: RuiAce on October 17, 2017, 08:11:03 pm
Your sketch implied that there would be a horizontal asymptote. But actually the graph can just keep going on without a problem.
Post by: jaskirat on October 17, 2017, 08:27:01 pm
Post by: RuiAce on October 17, 2017, 11:41:25 pm
Need help with questions iii and iv. checked the answers but still a little confused :/Here is a breakdown of the BoS answers.
Post by: bsdfjnlkasn on October 18, 2017, 11:36:27 am
Am so confused about the first part of the last (ii)...
Would really appreciate an explanation
Post by: gilliesb18 on October 18, 2017, 02:15:49 pm
Just wondering if someone could help me on the following question:
Find the equation of the locus of a point that moves so that it's distance from the line 3x + 4y +5= 0 is always 4 units.
I hardly know where to start so I would appreciate any help!!
Thanks heaps and heaps...
Post by: angelahchan on October 18, 2017, 02:37:57 pm
[img]http://i.imgur.com/tLOFDnR.png[/img]
Post by: Natasha.97 on October 18, 2017, 02:55:18 pm
Hello...
Just wondering if someone could help me on the following question:
Find the equation of the locus of a point that moves so that its distance from the line 3x + 4y +5= 0 is always 4 units.
I hardly know where to start so I would appreciate any help!!
Thanks heaps and heaps...
Hi!
- Draw a diagram to help visualise, using the condition given in the question (If it is always 4 units away from the line, that must mean that the locus is parallel to the equation: the perpendicular distance between the equation and the locus is 4)
- Substitute the values of the equation and an arbitrary point P(x,y) into the perpendicular distance formula (there should be two equations for the locus: one above and one below 3x+4y+5=0)
I have attached my working below, hope this helps
Post by: sidzeman on October 18, 2017, 03:33:09 pm
I'm not sure how to do part 1 and 3
For part ii, i got the Di/Dx as positive on both sides of x = 0, so I'm not sure what I'm doing wrong.
Post by: jamonwindeyer on October 18, 2017, 03:37:53 pm
Hey there,
Am so confused about the first part of the last (ii)...
Would really appreciate an explanation
So we need to spot that this situation models the volume we found previously. When \(\theta=0\), the volume would have been a full hemisphere (tilted). The height of the water is actually \(AB\), where the original depth is \(OB\). You need to find the value of \(\theta\) that sets \(A\) halfway between \(O\) and \(B\). That is, \(OA=\frac{r}{2}\). You should be able to do some right angled trigonometry to find the angle:
Post by: jamonwindeyer on October 18, 2017, 03:45:36 pm
Hi, could someone please help explain part (ii)? I don't understand why C is 0. The answers say when t=0, v=0 therefore C=0, but doesn't part (iii) contradict with (ii) since (iii) seems to say when t=0, v=1500?
[img]http://i.imgur.com/tLOFDnR.png[/img]
\(C=0\) because we don't care about the initial quantity in Part (ii)! We only want how much has flowed in, so the initial amount is irrelevant to us. So, we can effectively disregard the constant! ;D
Post by: Mary_a on October 18, 2017, 05:47:28 pm
No matter how much work I'm doing, it's not paying off!
Where do I go from here?
Post by: jaskirat on October 18, 2017, 07:34:10 pm
Post by: hansolo9 on October 18, 2017, 07:35:41 pm
Need help with the last part because I got 11 whereas the 2016 guidelines said 12
Post by: Mymy409 on October 18, 2017, 08:46:13 pm
(c) There’s been a cane beetle infestation at Mortdale! To combat this 5 cane toads
are introduced to start eating the cane beetles. Consequently, the number of
cane beetles decreases according to
B = 5000-10e^(0.01t)
where t is the time in months after the toads are introduced.
The population of cane toads, C, increases according to
dC/dt = 0.01C.
(i) How many beetles were in the town when the toads were introduced. 1
(ii) When will the population of beetles be one tenth of the population at t = 0. 1
(iii) Find an expression for the number of toads C in terms of t. 1
(iv) When is the rate of increase of toads equal to the rate of decrease of beetles? 3
Post by: RuiAce on October 18, 2017, 08:58:13 pm
Hey could someone help me with the final question from the 2002 hsc?
I'm not sure how to do part 1 and 3
For part ii, i got the Di/Dx as positive on both sides of x = 0, so I'm not sure what I'm doing wrong.
_____________________________
_____________________________
Note from the calculator, \(\sqrt{60}\approx 7.74 \).
tagged: 2002
(Will most likely be added to the compilation at some point.)
Post by: RuiAce on October 18, 2017, 09:04:40 pm
Okay, so I thought averaging 62 in all the past papers I did was bad, but I just sat the 2016 paper and I got 42.Following from your previous post, what have you done? What are the causes for the lost marks and how have you been trying to improve
No matter how much work I'm doing, it's not paying off!
Where do I go from here?
need help with this question :)
Remark: We also have a fourth equation \( -19 = a + b + c + 7 \) from just substituting the point in. This becomes \( -26 = a + b + c\), and can be convenient to use.
Post by: allmyfranzaredead on October 18, 2017, 09:21:20 pm
Post by: RuiAce on October 18, 2017, 09:25:37 pm
Hi, this is probs easy as,but I am awful at trig. In the 2014 multiple choice there was this, and it stumped me.The compilation briefly mentions why the answer is not 3, but rather 2. If you need any help beyond that, please provide more details.
Post by: RuiAce on October 18, 2017, 09:28:34 pm
Hey :)
Need help with the last part because I got 11 whereas the 2016 guidelines said 12
If you solve this the same way you tried solving it originally, you will now get n=12. That is to say, solve the equation
with that n-1, instead of n.
Post by: RuiAce on October 18, 2017, 09:41:13 pm
Can someone help me with part iv)
(c) There’s been a cane beetle infestation at Mortdale! To combat this 5 cane toads
are introduced to start eating the cane beetles. Consequently, the number of
cane beetles decreases according to
B = 5000-10e^(0.01t)
where t is the time in months after the toads are introduced.
The population of cane toads, C, increases according to
dC/dt = 0.01C.
(i) How many beetles were in the town when the toads were introduced. 1
(ii) When will the population of beetles be one tenth of the population at t = 0. 1
(iii) Find an expression for the number of toads C in terms of t. 1
(iv) When is the rate of increase of toads equal to the rate of decrease of beetles? 3
____________________________________
____________________________________
All of these three parts are fairly standard and you need to get used to doing them.
____________________________________
Which makes me suspect that you made a typo somewhere.
Post by: Mymy409 on October 18, 2017, 09:53:11 pm
____________________________________
____________________________________
All of these three parts are fairly standard and you need to get used to doing them.
____________________________________
Which makes me believe that you made a typo somewhere.
I only needed help with part iv) :P
Nope, that's how the question is exactly. Error in the question maybe?
Post by: RuiAce on October 18, 2017, 10:03:31 pm
I only needed help with part iv) :PLol, sorry. Probably went blind after already typing a lot of others before I got around to your's.
Nope, that's how the question is exactly. Error in the question maybe?
But in that case, the correct answer is definitely "never".
Post by: caitlinlddouglas on October 19, 2017, 07:06:35 am
Post by: RuiAce on October 19, 2017, 07:14:08 am
Hey i a was wondering why for this question you couldn't just do the top curve minus the bottom curve? Was there a reason they had to split it at y=4 and then add them? Thanks!Because you are rotating it about the y-axis.
If you were rotating about the x-axis, then you would have an area between two curves, or rather volume between two curves problem. Since you're rotating about the y-axis, the analogy is different.
Post by: gilliesb18 on October 19, 2017, 09:17:59 am
Hi!Wow! Thanks so much... Very helpful.
- Draw a diagram to help visualise, using the condition given in the question (If it is always 4 units away from the line, that must mean that the locus is parallel to the equation: the perpendicular distance between the equation and the locus is 4)
- Substitute the values of the equation and an arbitrary point P(x,y) into the perpendicular distance formula (there should be two equations for the locus: one above and one below 3x+4y+5=0)
I have attached my working below, hope this helps
Post by: gilliesb18 on October 19, 2017, 09:41:03 am
I just need help on the following question...
Find the equation of the locus of a point that moves so that it is equidistant from the line 4x - 3y + 2 =0 and the line 3x + 4y - 7 =0
Thanks once again!!
Post by: RuiAce on October 19, 2017, 09:48:10 am
Hello again,
I just need help on the following question...
Find the equation of the locus of a point that moves so that it is equidistant from the line 4x - 3y + 2 =0 and the line 3x + 4y - 7 =0
Thanks once again!!
______________________________
\begin{align*}PA^2&=PB^2\\ (x-x)^2+\left(y - \frac{1}{3}(2+4x)\right)^2&= (x-x)^2 + \left(y-\frac14(7-3x)\right)^2\end{align*}
\[ \text{These will give the same equations as above.} \]
Post by: gilliesb18 on October 19, 2017, 10:00:18 am
Post by: caitlinlddouglas on October 19, 2017, 11:14:35 am
Hi was looking at this question and I was wondering why there only has to be 2 cases? Wouldn't there be 4 as you'd need to have one case positive, the other negative
______________________________\\ \text{but }A\text{ and }B\text{ are now }\left(x, \frac13(2+4x)\right)\text{ and }\left(x, \frac14 (7-3x)\right))
\begin{align*}PA^2&=PB^2\\ (x-x)^2+\left(y - \frac{1}{3}(2+4x)\right)^2&= (x-x)^2 + \left(y-\frac14(7-3x)\right)^2\end{align*}
\[ \text{These will give the same equations as above.} \]
Then both cases positive and both cases negative?
Thanks!
Post by: RuiAce on October 19, 2017, 11:23:07 am
Hi was looking at this question and I was wondering why there only has to be 2 cases? Wouldn't there be 4 as you'd need to have one case positive, the other negativeBecause when you take all four cases, you will realise that some of the cases coincide.
Then both cases positive and both cases negative?
Thanks!
Post by: jamonwindeyer on October 19, 2017, 11:34:38 am
Because when you take all four cases, you will realise that some of the cases coincide.
To expand on that, consider the case:
Multiply by \(-1\) on both sides:
Similarly, we can turn \(A=-B\), into \(-A=B\) if we need to - So it is actually only two cases, though it can be written in four ways ;D
Post by: georgiia on October 19, 2017, 03:02:11 pm
Also if it says "copy into your writing book" can I use pen?
Thanks!! :)
Post by: RuiAce on October 19, 2017, 03:02:59 pm
Am I allowed to use a pen to dram a set of axes on the exams paper and then graph in pencil so that If i needed to rub out the axes wouldn't rub out? Or is this not allowed :/At the end of the day everything should be in pen. If you used pencil, you're ultimately going to have to trace over it in pen.
Also if it says "copy into your writing book" can I use pen?
Thanks!! :)
Post by: georgiia on October 19, 2017, 03:07:34 pm
At the end of the day everything should be in pen. If you used pencil, you're ultimately going to have to trace over it in pen.
Arn't graphs meant to be in pencil?
Post by: RuiAce on October 19, 2017, 03:20:19 pm
Arn't graphs meant to be in pencil?Nope.
(Graphs used to be allowed in pencil. But then for some reason they changed their scanning system and it's now terrible. Pencil will most likely not get picked up by it unless it's a very dark one (4B+).)
Post by: georgiia on October 19, 2017, 04:12:53 pm
Nope.
(Graphs used to be allowed in pencil. But then for some reason they changed their scanning system and it's now terrible. Pencil will most likely not get picked up by it unless it's a very dark one (4B+).)
Oh wow ok lucky i asked bc in all my internals marks were deducted if you used pen
Post by: winstondarmawan on October 19, 2017, 04:28:09 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22643316_1354494134676106_1428485746_o.png?oh=a4b3b3b16bd432cdaa3c24ae1adb3b22&oe=59EB0985
Would it be better to use the rounded value in the next steps or store the unrounded and use that?
Post by: georgiia on October 19, 2017, 05:33:13 pm
For a question like the one below where the question asks you to round in an earlier step:I usually store it in the calculator because when you need to find the time later then you sub the calc value in and gove the time to the nearest second, min, hour whatever is asked.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22643316_1354494134676106_1428485746_o.png?oh=a4b3b3b16bd432cdaa3c24ae1adb3b22&oe=59EB0985
Would it be better to use the rounded value in the next steps or store the unrounded and use that?
Hope that helps :)
Post by: Mymy409 on October 19, 2017, 05:40:08 pm
Sorry, the part that's cut off says 'Claire's path' and the corner that's cut off is labelled F and is the farmhouse.
Post by: pokemonlv10 on October 19, 2017, 06:05:48 pm
Post by: asahyoun3 on October 19, 2017, 06:19:35 pm
I know one method for this is to graph it and calculate the area of the two triangles. Is there another method through integrating the absolute values? Do i consider cases?
For this question, it is best to solve it graphically, as its more likely to make mistakes doing it algebraically
Post by: pokemonlv10 on October 19, 2017, 06:38:27 pm
For this question, it is best to solve it graphically, as its more likely to make mistakes doing it algebraically
Is there a chance of them asking to integrate absolute values in the exam (Not in MC)? Because i would like to know the algebraic method
Post by: Opengangs on October 19, 2017, 07:00:37 pm
Is there a chance of them asking to integrate absolute values in the exam (Not in MC)? Because i would like to know the algebraic methodWe define the absolute value to be x + 1 for x >= - 1 and -(x+1) for x < -1
Perhaps, this helps
Post by: asahyoun3 on October 19, 2017, 07:02:20 pm
Is there a chance of them asking to integrate absolute values in the exam (Not in MC)? Because i would like to know the algebraic method
Yes they could, most likely as a question 11 or 12
I have sent the solution to the question you asked, solving it algebraiclly
Post by: winstondarmawan on October 19, 2017, 07:10:06 pm
I usually store it in the calculator because when you need to find the time later then you sub the calc value in and gove the time to the nearest second, min, hour whatever is asked.
Hope that helps :)
This is what I usually always do, but in the answers they used the rounded off value.
Post by: RuiAce on October 19, 2017, 07:16:25 pm
A little help with part 2 of this, please. Thanks in advance.I will get back to this one. This was the dreaded 2001 question that was so hard, apparently the committee got fired for it.
Sorry, the part that's cut off says 'Claire's path' and the corner that's cut off is labelled F and is the farmhouse.
This will take considerable time to explain. I also want to type up a solution for the first part as well, so that it can be added to the compilation in entirety.
For this question, it is best to solve it graphically, as its more likely to make mistakes doing it algebraically"Best" is debatable - I've seen students accidentally sketch these wrong. "Most efficient" - yes.
Is there a chance of them asking to integrate absolute values in the exam (Not in MC)? Because i would like to know the algebraic methodAs Opengangs said, you must take cases.
This is what I usually always do, but in the answers they used the rounded off value.This is my recommendation.
You are safe to use an unrounded value if the question actually told you to compute an approximate value, like the one you've shown. Since they've stated that you must show \(k \approx 0.0347 \), you can keep using it if you wish.
If they just ask you to bluntly find the value of k, as opposed to showing an approximation holds, then you should use the exact value.
Note, however, that the one you've shown is the ONLY time I've seen a rounded value get used without problems.
Post by: gilliesb18 on October 19, 2017, 07:31:34 pm
If R is the fixed point(2,3) and P is a movable point(x,y), find the equation of the locus of P if the distance PR is twice the distance from P to the line y=
-1.
Thanks heaps :D :)
Post by: RuiAce on October 19, 2017, 07:36:12 pm
Could someone help me with the following question? Any help whatsoever is appreciated!!
If R is the fixed point(2,3) and P is a movable point(x,y), find the equation of the locus of P if the distance PR is twice the distance from P to the line y=
-1.
Thanks heaps :D :)
Aside: It can actually be shown that this will be a hyperbola, but oriented differently to the usual \( y=\frac1x \)
Post by: hansolo9 on October 19, 2017, 07:39:07 pm
Post by: Sukakadonkadonk on October 19, 2017, 07:51:18 pm
Or should I do more past HSC papers?
Post by: RuiAce on October 19, 2017, 07:53:34 pm
Need help with this one, thank you
for some constant \( a\).
Post by: RuiAce on October 19, 2017, 07:55:13 pm
Curious, I've been doing trial 2U past papers from 'good' schools like selective ones and what not, is that a good idea?You should be the judge of what papers you are doing.
Or should I do more past HSC papers?
Personally, however, I never bothered with any trial papers during the lead-up to the HSC and focused exclusively on past HSC papers. I also feel that some "good" selective schools don't have the "best" selective papers either; i felt many of James Ruse's papers were easier than Sydney Grammar's by far.
Post by: Sukakadonkadonk on October 19, 2017, 07:59:00 pm
You should be the judge of what papers you are doing.
Personally, however, I never bothered with any trial papers during the lead-up to the HSC and focused exclusively on past HSC papers. I also feel that some "good" selective schools don't have the "best" selective papers either; i felt many of James Ruse's papers were easier than Sydney Grammar's by far.
Thats very true. I will probably focus on HSC past papers then. Or maybe do some Sydney Grammar??
Post by: RuiAce on October 19, 2017, 08:24:26 pm
A little help with part 2 of this, please. Thanks in advance.
Sorry, the part that's cut off says 'Claire's path' and the corner that's cut off is labelled F and is the farmhouse.
____________________________________________________________________
Lemma: The derivative of sec
\begin{align*}\frac{d}{dx}\sec x &= \frac{d}{dx} (\cos x)^{-1}\\ &= - (-\sin x)(\cos x)^{-2}\tag{chain rule}\\ &= \sec x \tan x\tag{using some identities}\end{align*}
(Example: if t = 24, then the bus arrives at 8:24 AM. So Claire must be there by 8:24AM as well.)
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Post by: julia_warren13 on October 19, 2017, 08:25:29 pm
Thank you :)
Post by: RuiAce on October 19, 2017, 08:29:27 pm
Could someone please explain how to do part ii) of this question (it's from the 2009 HSC paper)
Thank you :)
This is because if we try to minimise \(\sin ax\), the minimum value will be -1. This can be proven via calculus, but is more easily deduced just because the range of \( \sin ax \) is \( -1 \le y \le 1\).
Post by: Mymy409 on October 19, 2017, 08:31:37 pm
Thanks very much, Rui :D
____________________________________________________________________Lemma: The derivative of sec\begin{align*}\frac{d}{dx}\sec x &= \frac{d}{dx} (\cos x)^{-1}\\ &= - (-\sin x)(\cos x)^{-2}\tag{chain rule}\\ &= \sec x \tan x\tag{using some identities}\end{align*}
(Example: if t = 24, then the bus arrives at 8:24 AM. So Claire must be there by 8:24AM as well.)
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Post by: julia_warren13 on October 19, 2017, 08:40:05 pm
This is because if we try to minimise \(\sin ax\), the minimum value will be -1. This can be proven via calculus, but is more easily deduced just because the range of \( \sin ax \) is \( -1 \le y \le 1\).\begin{align*}1+0.7\sin \frac{\pi t}{6}&=0.3\\ \sin \frac{\pi t}{6}&=-0.7\\ \frac{\pi t}{6} &=\frac{3\pi}{2}\\ t&= 9\end{align*}
Thanks Rui but I still don't quite understand how to get h=0.3 at low tide
Post by: RuiAce on October 19, 2017, 08:45:22 pm
Thanks Rui but I still don't quite understand how to get h=0.3 at low tide
This is because h is the height of the tide itself. Logically speaking, the low tide should be when the height of the tide is minimised.
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Post by: sidzeman on October 19, 2017, 11:31:06 pm
Why is the rationalisation of this root (n+1)− root n and not root n - root (n+1)?
Post by: RuiAce on October 19, 2017, 11:33:31 pm
Hey I know this has been covered before but,If you use that other one it will still work. They just chose to replace \( \sqrt{n}+\sqrt{n+1} \) with \( \sqrt{n+1} + \sqrt{n} \) first instead, which is still perfectly allowed.
Why is the rationalisation of this root (n+1)− root n and not root n - root (n+1)?
Post by: biffi023 on October 20, 2017, 09:47:41 am
sorry very basic but hmm.. sadly i can hardly even do the basic stuff ??? :o ;)
thanks!
Post by: RuiAce on October 20, 2017, 09:52:58 am
hey! bit of a dumb one sorry.. but how do you calculate something that has 'cos^2(theta)' or something like that?? as in to put it in the calculator i am guessing you have to do something with it cos u cant put a cos^2 or sin^2 etc in??
sorry very basic but hmm.. sadly i can hardly even do the basic stuff ??? :o ;)
thanks!
So you type cos of whatever number you have, and then put that in brackets before putting a square around it.
Post by: hansolo9 on October 20, 2017, 12:30:00 pm
I'm stuck on part 2 and 3
thank you :D
Post by: RuiAce on October 20, 2017, 12:50:39 pm
Hello
I'm stuck on part 2 and 3
thank you :D
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\begin{align*}\frac94 CX^2 &= \frac14 AB^2 + AB^2\\ \frac94 CX^2 &= \frac54 AB^2\\ 9 CX^2 &= 5 AB^2\end{align*}
Post by: caitlinlddouglas on October 20, 2017, 01:22:33 pm
iii) what does it mean by doing d/V? I get that the displacement is 6units to the right (it cancels out between t=4 and t=6) but how do you then go on to workout when it returns to the origin?
Iv) what happens to the graph of the displacement if there is a point of inflexion on the velocity graph? The solution is indicated that it was just concave up?
Thanks heaps!!:)
Post by: sidzeman on October 20, 2017, 01:22:59 pm
Post by: hobocop on October 20, 2017, 01:45:46 pm
Post by: RuiAce on October 20, 2017, 01:58:18 pm
Hey could someone please help me with this question? ForSome comprehensive discussion was given somewhere around here for part iii). Maybe the later bit of it will be useful to you.
iii) what does it mean by doing d/V? I get that the displacement is 6units to the right (it cancels out between t=4 and t=6) but how do you then go on to workout when it returns to the origin?
Iv) what happens to the graph of the displacement if there is a point of inflexion on the velocity graph? The solution is indicated that it was just concave up?
Thanks heaps!!:)
As for iv), If the derivative graph has a point of inflexion then you actually DO NOT KNOW what happens to the original graph. So they probably just drew it concave up because it is meant to concave up.
(Remember: The second derivative of the derivative is the third derivative of the original thing. And you've never touched on applications of the third derivative at all.)
Post by: RuiAce on October 20, 2017, 01:59:33 pm
For part iii, you get a result of n = 178.37 to repay the 346 095. The answers then round this down to 178, but shouldn't it be rounded up to 179? As is takes 178 AND a bit to repay the loan, surely just 178 payments would leave some amount still owingThe HSC has left quite some contradictory evidence with this stuff. Technically, you should round up like you said so.
Post by: RuiAce on October 20, 2017, 02:01:59 pm
Hi, could I get some help with these two pleasePlease consult the compilation for the first one. You may expand on your question if the advice given was too brief though.
Post by: caitlinlddouglas on October 20, 2017, 02:09:31 pm
Some comprehensive discussion was given somewhere around here for part iii). Maybe the later bit of it will be useful to you.Hi thanks for that I can't see the discussion for iii) though?
As for iv), If the derivative graph has a point of inflexion then you actually DO NOT KNOW what happens to the original graph. So they probably just drew it concave up because it is meant to concave up.
(Remember: The second derivative of the derivative is the third derivative of the original thing. And you've never touched on applications of the third derivative at all.)
Post by: sidzeman on October 20, 2017, 02:27:35 pm
Post by: Fahim486 on October 20, 2017, 03:05:11 pm
Post by: julia_warren13 on October 20, 2017, 03:15:05 pm
Hey so I thought the answer to this was A but it is in fact B. I have a feeling it is something to do with the All Stations To Central rule but I can't seem to understand it so could someone pls explain why it is B. Thanks!
When using the m = tan x rule, x is the angle the line makes with the POSITIVE X AXIS
m = tan (180 - 60)
m = tan 120
m = - srt 3 = B
m = tan 120
m = - srt 3 = B
Post by: Mymy409 on October 20, 2017, 04:01:22 pm
Post by: kemi on October 20, 2017, 04:15:55 pm
For part 1 of this, my solution for k was - ln(1/2) / 1600. Would that still be correct?
Yes
Remember the log rule, log (M^k) = klog (M)
So if you raise (1/2) to the power (-1), you will get k = (ln2)/1600
Post by: Shadowxo on October 20, 2017, 04:22:11 pm
Little help with this qn, please. I don't understand the solution. TIA. ;DIt's rotated about the y axis so the volume is
If you're still confused, let me know where :) Also the reason why they did the whole "e to a power" thing was so it's easier to find the derivative.
Hope this helps
Hey so I thought the answer to this was A but it is in fact B. I have a feeling it is something to do with the All Stations To Central rule but I can't seem to understand it so could someone pls explain why it is B. Thanks!You should also be able to tell since the line is sloped down, the gradient will be negative. That's one of the reasons you can eliminate A :)
Post by: sophiegmaher on October 20, 2017, 04:37:27 pm
Post by: sophiegmaher on October 20, 2017, 04:56:58 pm
Post by: Mymy409 on October 20, 2017, 05:17:37 pm
It's rotated about the y axis so the volume is
If you're still confused, let me know where :) Also the reason why they did the whole "e to a power" thing was so it's easier to find the derivative.
Hope this helps
Thanks, I'm just confused about how we get x=3^y
Post by: RuiAce on October 20, 2017, 05:19:29 pm
Thanks, I'm just confused about how we get x=3^y
Post by: Mymy409 on October 20, 2017, 05:19:50 pm
Post by: RuiAce on October 20, 2017, 05:21:14 pm
Also, how do we solve sin2x=cosx?You cannot do this via 2U methods. The best you can do is sketch both of them and approximate a solution.
Post by: RuiAce on October 20, 2017, 05:23:16 pm
Hi thanks for that I can't see the discussion for iii) though?When I click the hyperlink, it links to a huge amount of text related to part iii).
Some information on tackling that 2013 question was provided in the compilation thread already. But if it's not enough, you may come back and further your question.
tagged: 2007
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where f(t) is any random function that we plucked out of thin air because maths is magic.
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The point: We got back to the same answer, and we didn't have to do that useless finding +C stuff
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A key point is that this measured the change in displacement, SPECIFICALLY between times t=0 and t=4. This is significant, as otherwise the boundaries of our integral would've been different numbers.
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Reason being, this value \(t_1\) will ensure that the change in displacement, from 0, to \(t_1\), will be zero. Because the particle starts at the origin, the time when it returns to the origin is the time when the change in displacement is zero.
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Post by: RuiAce on October 20, 2017, 05:32:59 pm
I don't know how to do question (iii) from the attached question, it's from the CSSA trial :)
I strongly believe that this question was unfair and I'm actually appalled that they put that in a 2U paper. Please be advised the limit they made you evaluate was actually a 3U limit.
Post by: RuiAce on October 20, 2017, 05:42:35 pm
Also not sure how to answer part (ii) of this attached question from the CSSA trial(http://uploads.tapatalk-cdn.com/20171020/a349d0bf7bb16102ecea4f70f5da99bc.jpg)
Post by: sophiegmaher on October 20, 2017, 06:52:47 pm
I strongly believe that this question was unfair and I'm actually appalled that they put that in a 2U paper. Please be advised the limit they made you evaluate was actually a 3U limit.
Thank you!
Post by: sophiegmaher on October 20, 2017, 06:54:28 pm
Post by: RuiAce on October 20, 2017, 06:59:43 pm
I'm also stuck on the question 8 multiple choice question and question 15a(ii) attached from the CSSA trial :/
The other one just looks like an area between curves problem.
For 0≤x≤π/8, the upper curve is y=1 and the lower curve is y=cos(2x).
For π/8≤x≤π/4, the upper curve is y=1 and the lower curve is y=sin(2x).
Post by: sophiegmaher on October 20, 2017, 07:08:59 pm
The other one just looks like an area between curves problem.
For 0≤x≤π/8, the upper curve is y=1 and the lower curve is y=cos(x).
For π/8≤x≤π/4, the upper curve is y=1 and the lower curve is y=sin(x).
Yeah, but the answers say the integral has 1-cos2x dx to begin with and I don't understand how :/
Post by: clarence.harre on October 20, 2017, 07:10:41 pm
Answers claim 6am to 11am. Everytime I do it I get 6am to 10am. Figured since it's BOSTES they aren't going to mess up the marking guidelines for a paper, but I still fail to see where I'm going wrong.
Question is attached as a screenshot.
Post by: sidzeman on October 20, 2017, 07:10:52 pm
Yes
Remember the log rule, log (M^k) = klog (M)
So if you raise (1/2) to the power (-1), you will get k = (ln2)/1600
Hmmmm I see. Could you explain how to do part ii then please
Post by: RuiAce on October 20, 2017, 07:11:57 pm
Yeah, but the answers say the integral has 1-cos2x dx to begin with and I don't understand how :/Fixed up a typo in what I wrote earlier - forgot about the 2.
Symmetry can be exploited wherever possible, but it is not mandatory that it is used.
Post by: RuiAce on October 20, 2017, 07:15:32 pm
Question 7.b.(iii) from the 2009 HSC Paper.
Answers claim 6am to 11am. Everytime I do it I get 6am to 10am. Figured since it's BOSTES they aren't going to mess up the marking guidelines for a paper, but I still fail to see where I'm going wrong.
Question is attached as a screenshot.
\begin{align*}\sin \frac{\pi t}{6} &= \frac12\\ \frac{\pi t}{6} &= \frac\pi6, \frac{5\pi}6\\ t &= 1, 5\end{align*}
There weren't any traps here.
Post by: sophiegmaher on October 20, 2017, 07:15:52 pm
Fixed up a typo in what I wrote earlier - forgot about the 2.
Symmetry can be exploited wherever possible, but it is not mandatory that it is used.
Sorry I'm still not sure how that makes it equal that :/
Post by: roygbivmagic on October 20, 2017, 07:18:24 pm
'Find all values of m for which the equation |2x-3|=mx+1 has exactly one solution.'
Post by: RuiAce on October 20, 2017, 07:21:04 pm
Sorry I'm still not sure how that makes it equal that :/The symmetry means the area between 0 and π/4, and the area between 0 and π/8, are the exact same. That is to say, the integrals actually equal each other: \( \int_0^{\frac\pi 4}(1-\cos 2x)\,dx = \int_{\frac\pi4}^{\frac\pi 8}(1-\sin 2x)\,dx \)
It's like with \( y=x^2 \). The curve is symmetric about the line \( y=0\). So if you compute the area between -1 and 0, and also the area between 0 and 1, by symmetry they will be the same. That is to say, \( \int_{-1}^0 x^2\,dx = \int_0^1 x^2\,dx \)
What happens when you add two of the exact same things? You just get double that thing.
Hi, can you please explain how to solve this?Please find the solution in the compilation.
'Find all values of m for which the equation |2x-3|=mx+1 has exactly one solution.'
Post by: RuiAce on October 20, 2017, 07:27:27 pm
Hmmmm I see. Could you explain how to do part ii then please
Post by: roygbivmagic on October 20, 2017, 07:55:04 pm
Wait but isn't 5 hours past 5AM = 10AM not 11AM? (I keep getting the answer as 10AM as well)
\begin{align*}\sin \frac{\pi t}{6} &= \frac12\\ \frac{\pi t}{6} &= \frac\pi6, \frac{5\pi}6\\ t &= 1, 5\end{align*}
There weren't any traps here.
Post by: kemi on October 20, 2017, 08:21:09 pm
Wait but isn't 5 hours past 5AM = 10AM not 11AM? (I keep getting the answer as 10AM as well)
Hey!
Did this question and got the same answer as you. Lo and behold, Success One answers are incorrect. Here is the true BOSTES HSC sample solution.
Post by: Justinhales on October 20, 2017, 08:33:40 pm
Sorry i'm quite stuck on a certain question! A textbook question mind u!!! :-\
Q13
The half-life of radium is 1600 years.
(a) Find the percentage of radium that will be decayed after 500 years.
(b) Find the number of years that it will take for 75% of the radium to decay.
Thanks in advance!!
Post by: RuiAce on October 20, 2017, 08:35:45 pm
Wait but isn't 5 hours past 5AM = 10AM not 11AM? (I keep getting the answer as 10AM as well)For some reason I could not count. Or I interpreted 5 as 6.
My apologies. Yes, they (and I) are wrong. kemi's remark on the mistake in Excel Success One could prove useful.
Post by: RuiAce on October 20, 2017, 08:42:43 pm
Hey Guys!!
Sorry i'm quite stuck on a certain question! A textbook question mind u!!! :-\
Q13
The half-life of radium is 1600 years.
(a) Find the percentage of radium that will be decayed after 500 years.
(b) Find the number of years that it will take for 75% of the radium to decay.
Thanks in advance!!
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Post by: Justinhales on October 20, 2017, 08:46:23 pm
Makes sense now!!
Post by: sidzeman on October 20, 2017, 10:33:46 pm
Post by: RuiAce on October 20, 2017, 10:38:40 pm
Could someone please help me with part iii, iv and v
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This is just the formula \(A = \frac{1}{2}r^2\theta\). Just remember to use the correct value for \(\theta\), i.e. the one for the major sector. Not the minor sector.
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Find the area of the major sector AOB using the same formula as above, and then just add all the areas you had from iii and iv with that one.
Post by: sidzeman on October 20, 2017, 10:53:31 pm
For part iii, I proved the triangles were congruent as well, but then used 1/2absinC (times 2) to calculate the area, and got 2root3.
________________________________
This is just the formula \(A = \frac{1}{2}r^2\theta\). Just remember to use the correct value for \(\theta\), i.e. the one for the major sector. Not the minor sector.
________________________________
Find the area of the major sector AOB using the same formula as above, and then just add all the areas you had from iii and iv with that one.
My working out was A= 2(0.5)(sin60)(2)(1) - using the absolute value for sin (root3 /2) - shouldnt the answer be 2root3 or am I doing something wrong?
For part iv - i used the angle pi/3 is this not correct?
Post by: RuiAce on October 20, 2017, 11:29:42 pm
For part iii, I proved the triangles were congruent as well, but then used 1/2absinC (times 2) to calculate the area, and got 2root3.
My working out was A= 2(0.5)(sin60)(2)(1) - using the absolute value for sin (root3 /2) - shouldnt the answer be 2root3 or am I doing something wrong?
For part iv - i used the angle pi/3 is this not correct?
Your expression "A= 2(0.5)(sin60)(2)(1)" already counts the two triangles together due to that extra '2' in front. Else, you fudged something in when you should not have.
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Post by: clarence.harre on October 21, 2017, 05:59:06 am
\begin{align*}\sin \frac{\pi t}{6} &= \frac12\\ \frac{\pi t}{6} &= \frac\pi6, \frac{5\pi}6\\ t &= 1, 5\end{align*}
There weren't any traps here.
Ok Great!! So I guess I didn't make any mistakes in the working but I still don't understand how if 5 + 1 = 6 but 5 + 5 = 11. I still swear it's 10; even plugged it into a time difference calculator (see attachtment). Thank you RuiAce for the prompt response though :)
Post by: RuiAce on October 21, 2017, 08:26:38 am
Ok Great!! So I guess I didn't make any mistakes in the working but I still don't understand how if 5 + 1 = 6 but 5 + 5 = 11. I still swear it's 10; even plugged it into a time difference calculator (see attachtment). Thank you RuiAce for the prompt response though :)I apologised about this one at the top of this page aha
Post by: nattynatman on October 21, 2017, 09:30:06 am
but which questions require you to use the discriminant to prove it exists?
I always seem to miss using it mostly cos I'm not sure when to apply it
Post by: RuiAce on October 21, 2017, 09:34:46 am
this might be a stupid question...When you're interested in the number of solutions, not the actual solution itself.
but which questions require you to use the discriminant to prove it exists?
I always seem to miss using it mostly cos I'm not sure when to apply it
Sure, there are a whole bunch of applications of the discriminant, and they generally target band 6 as well. A common one can be in the number of points of intersections. But ultimately they still rely on the number of solutions to a quadratic equation you get given.
Post by: K9810 on October 21, 2017, 10:55:46 am
Post by: RuiAce on October 21, 2017, 10:58:17 am
Hi, I was wondering if you can please explain why the first line of the answers for part iii) says angle CBD=36Because if you replicate the exact same proof for how you found \(\angle BAC\), but just replaced some of the sides, you would get the exact same answer for \(\angle CBD\).
"Similarly" can only be used when you try to find something else, but the steps in the proof goes the exact same way.
(Alternatively by inspection \( \triangle BAC \equiv \triangle CBD \) (SAS) )
Post by: K9810 on October 21, 2017, 11:04:30 am
Because if you replicate the exact same proof for how you found \(\angle BAC\), but just replaced some of the sides, you would get the exact same answer for \(\angle CBD\).
"Similarly" can only be used when you try to find something else, but the steps in the proof goes the exact same way.
(Alternatively by inspection \( \triangle BAC \equiv \triangle CBD \) (SAS) )
Thank you! That makes so much more sense
Post by: asd987 on October 21, 2017, 11:18:09 am
Post by: K9810 on October 21, 2017, 11:24:31 am
Post by: RuiAce on October 21, 2017, 11:28:02 am
Hi can i get help with ii) please. thanksAssuming that the point of intersection was (2,4)
Post by: _____ on October 21, 2017, 12:04:41 pm
Post by: RuiAce on October 21, 2017, 12:08:54 pm
Can you please help me with part iii) and iv)Some comprehensive discussion on part iii can be found here.
Is it OK to use a pencil for diagrams? Front of 2016 paper just says black pen.You really should be using pen. From previous posts, the general advice is to draw the axes in pen, then if you want draw the curve itself in pencil before tracing it. At least, that's if you insist on using the pencil.
Post by: sidzeman on October 21, 2017, 12:09:26 pm
Is it OK to use a pencil for diagrams? Front of 2016 paper just says black pen.^expanding on that, will we have marks deducted for sketching lets say number plane axis freehand rather than with a ruler?
Also, could someone help me with part iv of this question please, as well as part ii from b
Post by: RuiAce on October 21, 2017, 12:10:44 pm
^expanding on that, will we have marks deducted for sketching lets say number plane axis freehand rather than with a ruler?The HSC has never given any instruction on this. So if you want to do this, you're gambling on chance (a nice/mean marker).
Post by: winstondarmawan on October 21, 2017, 12:16:06 pm
Do they have to be completely blank with just a brand name and measurements?
Post by: RuiAce on October 21, 2017, 12:17:28 pm
Adding onto these technicality questions, what are the specifications for rulersOf course you can’t have a ruler with a bunch of formulas written over it.
Do they have to be completely blank with just a brand name and measurements?
They can’t do anything about your name or just the brand name though as far as I’m aware of.
Post by: bsdfjnlkasn on October 21, 2017, 12:22:27 pm
I've got a bunch of questions from the 2001 but i'll stagger them as I post :)
Thanks so much!
These are the specifics for the questions I need help with:
Finance - ii and iii
Geometry - ii and iii
Post by: bsdfjnlkasn on October 21, 2017, 12:23:20 pm
Here's the second round :)
Thanks so much!
These are the specifics for the questions I need help with:
Trapezoidal rule - legit don't think I applied it right but don't know how to correct it?
My working: 2(1.7 + 2(1.3))
Answer is 12 m/s ??
Post by: bsdfjnlkasn on October 21, 2017, 12:35:41 pm
Here's the third round :)
Thanks so much!
These are the specifics for the questions I need help with:
Plane geometry: v) - I thought to find where BC intersects the x-axis and the x-coordinate would be the perpendicular distance but I don't know the coordinates of C...
Range: not sure how to find the range..
Post by: bsdfjnlkasn on October 21, 2017, 12:40:17 pm
Here's the final round :)
Thanks so much!
Just need help with the first part, then I'll attempt the second
Post by: RuiAce on October 21, 2017, 02:03:07 pm
Hey there!1.3 and 1.7 are both "inbetweens" for the trapezoidal rule. It should be \( 2 \left[0 + 0 + 2(1.7+1.3) \right] \) which is equal to 12.
Here's the second round :)
Thanks so much!
These are the specifics for the questions I need help with:
Trapezoidal rule - legit don't think I applied it right but don't know how to correct it?
My working: 2(1.7 + 2(1.3))
Answer is 12 m/s ??
Basically, the "first" and the "last" were both 0.
Hey there!This one is in the compilation. It's ridiculous.
Here's the final round :)
Thanks so much!
Just need help with the first part, then I'll attempt the second
Post by: RuiAce on October 21, 2017, 02:09:58 pm
Hey there!
Here's the third round :)
Thanks so much!
These are the specifics for the questions I need help with:
Plane geometry: v) - I thought to find where BC intersects the x-axis and the x-coordinate would be the perpendicular distance but I don't know the coordinates of C...
Range: not sure how to find the range..
This is basically because it's just a "stretch" of the semicircle up and down. (But there is no down, because this is the upper semicircle. So it's only stretched up.)
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Post by: RuiAce on October 21, 2017, 02:17:29 pm
Hey there!
I've got a bunch of questions from the 2001 but i'll stagger them as I post :)
Thanks so much!
These are the specifics for the questions I need help with:
Finance - ii and iii
Geometry - ii and iii
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Post by: winstondarmawan on October 21, 2017, 02:29:18 pm
Can someone please explain why for this question:
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22689757_1356165251175661_1010246494_o.png?oh=9c15550ed02a06f9bd76c168e6a75e59&oe=59ECDE84
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22685022_1356184464507073_13089599_n.png?oh=04eab45a00570206fe870ce096865e5c&oe=59EDC439
Why the graph tends towards zero?
TIA.
Post by: RuiAce on October 21, 2017, 02:38:08 pm
Hello!Because think about what the derivative is and what happens to the original graph as well.
Can someone please explain why for this question:
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22689757_1356165251175661_1010246494_o.png?oh=9c15550ed02a06f9bd76c168e6a75e59&oe=59ECDE84
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22685022_1356184464507073_13089599_n.png?oh=04eab45a00570206fe870ce096865e5c&oe=59EDC439
Why the graph tends towards zero?
TIA.
As the original graph flattens out and tends towards its horizontal asymptote, the gradient of the tangent to the curve is tending towards 0.
This is what the derivative is; the gradient of the tangent. So on the sketch of the derivative, it must tend to 0.
Post by: winstondarmawan on October 21, 2017, 02:46:22 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22712062_1356191744506345_1426088660_o.png?oh=a041315ba9f538f3e83a8837421517c6&oe=59ECE7F5
Part (iii).
Would it be enough for 2 marks to state that for t>0, v>0 and a>0 (after finding the acceleration).
Post by: RuiAce on October 21, 2017, 02:53:20 pm
For this question:Yeah.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22712062_1356191744506345_1426088660_o.png?oh=a041315ba9f538f3e83a8837421517c6&oe=59ECE7F5
Part (iii).
Would it be enough for 2 marks to state that for t>0, v>0 and a>0 (after finding the acceleration).
Although you probably should explain why for v. It's obvious that a>0 but not so obvious that v>0.
Post by: RuiAce on October 21, 2017, 03:25:24 pm
^expanding on that, will we have marks deducted for sketching lets say number plane axis freehand rather than with a ruler?
Also, could someone help me with part iv of this question please, as well as part ii from b
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Post by: RuiAce on October 21, 2017, 03:31:03 pm
Can you please help me with part iii) and iv)Guidelines:
The point of inflexion at A (on the velocity graph) isn't too useful. Because there is a local maximum at B, the displacement will increase (at an increasing rate) until it reaches t=4. Then, it will start increasing at a decreasing rate.
(Effectively, for the displacement graph, t=4 will "almost" exhibit the properties of a point of inflexion)
Until you reach t=5, where the particle is at rest. Then, the displacement graph starts turning around (i.e. decreases) until you reach t=6.
(It may be worth mentioning that between the times t=4 and t=6, the graph will appear to be quite parabolic.)
And then at t=6, because the velocity is now constant, the displacement is now linear (straight line). So the displacement decreases in the same way a straight line with gradient m=-5 does.
Post by: RuiAce on October 21, 2017, 03:54:42 pm
Hey there!
I've got a bunch of questions from the 2001 but i'll stagger them as I post :)
Thanks so much!
These are the specifics for the questions I need help with:
Finance - ii and iii
Geometry - ii and iii
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You really shouldn't be doing this, but for convenience in typing I will round to 841.83.
Post by: ~BK~ on October 21, 2017, 04:14:12 pm
i would really appreciate some help on the attached question...
i always struggle on these sorta ones and they are worth 6 marks (at least)
TIA
Post by: RuiAce on October 21, 2017, 04:32:22 pm
hey RuiAce (or anyone for that matter ;))
i would really appreciate some help on the attached question...
i always struggle on these sorta ones and they are worth 6 marks (at least)
TIA
although since this part said "write down" you didn't need to show working if you didn't want to.
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Post by: hansolo9 on October 21, 2017, 04:34:16 pm
I also need help for part 2, 3, and 4. Not that good at these types of questions
Post by: ~BK~ on October 21, 2017, 04:41:03 pm
thanks heaps, clears it up, just wish i could think like that during an exam!! :o
that last section, as in the bit above, you've done differently to succus one and got the same answer.... personally, your method makes way more sense...
tvm btw- can't believe you're sacrificing your time on the w/e!?
Post by: gabi.c on October 21, 2017, 04:49:06 pm
Post by: RuiAce on October 21, 2017, 04:50:57 pm
Do we get marked down for giving more answers than the marking criteria asks for? In the past papers I've done, I keep going past the answer (simplifying more, or factorising)Not unless you write something wrong.
A whole bunch of correctness - ok. Add a mistake in there - not ok.
Post by: RuiAce on October 21, 2017, 04:51:28 pm
Thank you for the previous answers, I understand now. :)
I also need help for part 2, 3, and 4. Not that good at these types of questions
This can be proven. \( \angle ADG = \frac\pi2 - \theta\) from the angle sum of \( \triangle ADG\).
This part was more or less the hardest part because of that tiny triangle down the corner, which can easily throw you off.
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Which, I find funny.
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I feel as though parts iii and iv were easier than ii. This is an example of a question where it helps to skip previous parts if you are stuck, and then come back to them.
Post by: RuiAce on October 21, 2017, 04:52:33 pm
thanks heaps, clears it up, just wish i could think like that during an exam!! :oThere's multiple methods, but personally I find the recursive one makes more sense all the time. The other method is more or less only if you need it.
that last section, as in the bit above, you've done differently to succus one and got the same answer.... personally, your method makes way more sense...
tvm btw- can't believe you're sacrificing your time on the w/e!?
Yeah, I knew that these 4 days would be the period where I get the most questions :P
Post by: pokemonlv10 on October 21, 2017, 04:57:13 pm
Post by: RuiAce on October 21, 2017, 05:00:34 pm
If a question has multiple parts to it, and one question asks you to find a solution to 2dp, and the next question requires you to use that answer, do you use the rounded solution or the exact value?You should be using the exact value wherever possible.
BoS answers often chose not do this. My general advice is that if they tell you to find a rounded value, you can use the rounded value. If they just say to find k or something, use the exact value.
Post by: ~BK~ on October 21, 2017, 05:12:23 pm
i am really struggling on the attached and the succes one answers aren't clearing anything up either!!
thanks again
Post by: RuiAce on October 21, 2017, 05:16:29 pm
another q... JIC you haven't got enough to do ;)That one's in the compilation (luckily)
i am really struggling on the attached and the succes one answers aren't clearing anything up either!!
thanks again
Post by: sidzeman on October 21, 2017, 05:38:49 pm
Post by: ~BK~ on October 21, 2017, 05:43:50 pm
That one's in the compilation (luckily)
??? may i ask where?
Post by: RuiAce on October 21, 2017, 05:45:07 pm
??? may i ask where?It's stickied in the main board
Post by: RuiAce on October 21, 2017, 05:46:37 pm
Hey could someone help me with this question? For some reason the answers have the domain of x as being in between - pi/3 and pi/3 so I'm not sure they are right
Post by: ~BK~ on October 21, 2017, 05:54:09 pm
It's stickied in the main boardthanks... but woooooah!! that is such a confusing question!!
i really hope they're a little nicer this year *high hopes* <doubt it>
Post by: RuiAce on October 21, 2017, 05:56:37 pm
thanks... but woooooah!! that is such a confusing question!!Yeah part (i) was the hardest part about that question. Every now and then something is just weird (and if you just don't know, skip it and come back to it later).
i really hope they're a little nicer this year *high hopes* <doubt it>
Post by: magnesium4 on October 21, 2017, 06:14:38 pm
I do not understand financial math questions. Can someone please explain to me how to tackle them with regards to this question attached.
Thanks
Post by: RuiAce on October 21, 2017, 06:18:25 pm
Hello,This particular one was addressed just a few moments ago
I do not understand financial math questions. Can someone please explain to me how to tackle them with regards to this question attached.
Thanks
Post by: ~BK~ on October 21, 2017, 06:31:50 pm
Yeah part (i) was the hardest part about that question. Every now and then something is just weird (and if you just don't know, skip it and come back to it later).
yeah, it was really strange that part (i) and for some reason i had never heard of m=tan(theta)?? where exactly does this apply!? thanks for the advice... trouble is, i end up with more than half the paper skipped.. hahahaha ???
Post by: RuiAce on October 21, 2017, 06:35:08 pm
yeah, it was really strange that part (i) and for some reason i had never heard of m=tan(theta)?? where exactly does this apply!? thanks for the advice... trouble is, i end up with more than half the paper skipped.. hahahaha ???
For some examples:
If y = x, then theta = 45 degrees or pi/4.
If y = 2x, then theta = whatever tan^-1(2) is
If y = -x, then theta = -45 degrees (or 135 degrees; both work)
If y = sqrt(3)x, then theta = 60 degrees or pi/3.
If y = 0, then theta = 0.
Post by: sidzeman on October 21, 2017, 06:35:41 pm
Post by: RuiAce on October 21, 2017, 06:39:26 pm
Hey could someone help me out with this please
Post by: sidzeman on October 21, 2017, 07:01:29 pm
ahhhh I see thank you so much!
Post by: winstondarmawan on October 21, 2017, 07:51:39 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22690507_1356345717824281_1978276958_o.png?oh=1e9877b120499b8eb20278f75eb94eb5&oe=59EDA08E
Post by: mxrylyn on October 21, 2017, 07:55:03 pm
This a some prelim revision but, How do you prove that and absolute value of a + b is greater than or equal to the absolute value of a + the absolute value of b?
Post by: RuiAce on October 21, 2017, 07:57:10 pm
Hello, just a little confused as to why Simple Harmonic Motion is in the 2009 HSC..It's weird and it most certainly is SHM (and I said the same thing when I did 2U), but nah it's fine. It's only problematic once they start throwing stuff like \( \ddot{x} = -n^2 x \) in there, which they didn't.
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22690507_1356345717824281_1978276958_o.png?oh=1e9877b120499b8eb20278f75eb94eb5&oe=59EDA08E
So long as everything remains a function of time in the whole question (as opposed to a function of displacement), it will be fine. Period and max/min (range of trig functions) is technically still 2U material.
Post by: RuiAce on October 21, 2017, 07:59:42 pm
Hey.It's not. It's less than.
This a some prelim revision but, How do you prove that and absolute value of a + b is greater than or equal to the absolute value of a + the absolute value of b?
Post by: mxrylyn on October 21, 2017, 08:03:52 pm
How do you prove that it is less than?
Post by: RuiAce on October 21, 2017, 08:15:06 pm
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Because this is not a 4U course, the only approach we can take is dealing with cases
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Post by: mxrylyn on October 21, 2017, 08:53:47 pm
If this as a 4Unit question, would I have to answer it differently?
Post by: RuiAce on October 21, 2017, 09:05:00 pm
THANK YOU!The 4U method is actually in the 4U thread right now.
If this as a 4Unit question, would I have to answer it differently?
It relies on something you can't really do in 2U, because the number line is what's called "degenerate".
Post by: sidzeman on October 21, 2017, 09:37:58 pm
Post by: hobocop on October 21, 2017, 09:53:23 pm
This is Q9 b of 2011 HSC paper.
Thanks
Post by: RuiAce on October 21, 2017, 09:56:31 pm
Can someone please help with part ii and iii
Note that 6% is also the interest rate. So some really nice things happen in the computations and we dodge the need for a geometric series.
The payment at the start of year 1 is 1000
The payment at the start of year 2 is 1000(1.06)
The payment at the start of year 3 is 1000(1.06)^2
...so...
\begin{align*}A_1 &= 1000(1.06)\\ A_2& = [A_1 + 1000(1.06)](1.06)\\ &=1000(1.06)^2 + 1000(1.06)^2 \\ A_3 &=[A_2 + 1000(1.06)^2](1.06)\\ &= 1000(1.06)^3+1000(1.06)^3+1000(1.06)^3 \end{align*}
Post by: RuiAce on October 21, 2017, 10:01:50 pm
Hi, could I get an explanation on why you need to integrate 't' for part ii of this question?
This is Q9 b of 2011 HSC paper.
Thanks
Of course, you could've just integrated the rates independently. But good luck trying to integrate \(2 +\frac{t^2}{t+1} \)
Post by: blasonduo on October 21, 2017, 10:53:46 pm
For questions such as these, I always normally answer the question with decimal places, however, the answers normally give it in exact form.
My question is if they don't specify the exact value, and I put an approximate answer (to 3 d.p for example) will I lose a mark?
Post by: RuiAce on October 21, 2017, 10:55:51 pm
Hey! Just a quick question :)The examiner would be extremely rude to deduct you marks for it, but personally I believe that whenever it can be given in exact form it should be. You're taking a gamble if you just obtain a rounded answer without having an exact answer before-hand.
For questions such as these, I always normally answer the question with decimal places, however, the answers normally give it in exact form.
My question is if they don't specify the exact value, and I put an approximate answer (to 3 d.p for example) will I lose a mark?
Post by: asd987 on October 21, 2017, 11:47:17 pm
Post by: kemi on October 21, 2017, 11:58:26 pm
Hey so for number iv) the answer is 45.433... does that mean this is the 46th year and so the year will be 1991+46 which is the year 2037, or will it be rounded down to 45 then 1991+46 which is 2036. the answers/markers notes say its 2036, but i always thought you had to round it up. Can anyone please clarify, thanks.
Hey :)
They way I'd think about it is, the count is from the beginning of 1991, and the question says during which year? So if you got 45.433... that would be from the beginning of 1991, then as it is only partially through the 45th year, it must be 2036, not 2037 because well, it really hasn't reached 2037 at all. It's still part way through the 45th year - so add 45 to 1991.
Post by: RuiAce on October 22, 2017, 12:05:29 am
Hey so for number iv) the answer is 45.433... does that mean this is the 46th year and so the year will be 1991+46 which is the year 2037, or will it be rounded down to 45 then 1991+46 which is 2036. the answers/markers notes say its 2036, but i always thought you had to round it up. Can anyone please clarify, thanks.
i.e. in 2036.
Post by: bsdfjnlkasn on October 22, 2017, 08:43:24 am
Can I get some help as to how they derived the equation for the total income in the last part?
I don't get why it's (n-1) :)
Post by: RuiAce on October 22, 2017, 09:07:15 am
Hey!These dodgy questions are part of the reason why I generally don't vouch for questions before 2001 unless you really need it. The method they provide is correct but trying to interpret why can be hard. Here's the reasoning:
Can I get some help as to how they derived the equation for the total income in the last part?
I don't get why it's (n-1) :)
That was basically the whole point of part iii), to find when this happens.
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Post by: bsdfjnlkasn on October 22, 2017, 10:05:08 am
Just need some help with the last probability, thought it was 1/3 just by reading off the branch because that's what I thought the probability was... answer is 1/5
Post by: Nuwanda_g on October 22, 2017, 10:09:28 am
Post by: RuiAce on October 22, 2017, 10:15:58 am
Are we ought to mentally find the equation of velocity to integrate then find maximum displacement. Or is there an easier way? Thanks in advanceAlready addressed in the compilation. The equation of the velocity isn't actually too hard to find here, because we have a straight line.
Mental computations are one possible way as well but that requires quite some good memory
Post by: K9810 on October 22, 2017, 10:16:55 am
Thanks
Post by: RuiAce on October 22, 2017, 10:17:11 am
Hey, could you please help me with part iv)Also already addressed in the compilation
Thanks
Post by: RuiAce on October 22, 2017, 10:22:00 am
Hey there!
Just need some help with the last probability, thought it was 1/3 just by reading off the branch because that's what I thought the probability was... answer is 1/5
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Post by: bsdfjnlkasn on October 22, 2017, 11:47:02 am
Post by: teapancakes08 on October 22, 2017, 11:53:34 am
...help.
Post by: teapancakes08 on October 22, 2017, 11:55:25 am
Post by: angelahchan on October 22, 2017, 12:07:36 pm
https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
Post by: EEEEEEP on October 22, 2017, 12:14:48 pm
Hi, could someone please help with Q16 b(iii) of 2013? I don't get how we're meant to work out it's shape - is it by plotting points?
https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
Hint hint: It's an exponential. ALso consider the previous answers.
(i) How many trout were in the lake when the carp were introduced? This tells you the value of N when T is 0.
(ii) When will the population of trout be zero? This tells you the value of T (when N = 0).
Given these 2, you can graph it.
YOu only need 2 or 3 points and you know the general shape of the graph =)
Post by: sidzeman on October 22, 2017, 12:25:25 pm
Post by: adelaidecruz on October 22, 2017, 12:34:08 pm
Halp pls
Q7:
ln (6-x) = 2lnx
loge (6-x) = 2logex
loge (6-x) = loge (x^2)
6-x = x^2 ..... x^2 + x - 6 =0 .... (x+3) (x-2)
x = -3, 2
Post by: EEEEEEP on October 22, 2017, 12:36:37 pm
Halp pls
q7
The 2 rules that are applicable here are...
ln(x ^ y) = y ∙ ln(x) AND e^ ln(x) = x
1. 2 ln x = X^2
ln (6-x) = ln (X^2)
2. Make both sides the superscipt for e
(6-x) = (X^2)
Take the LHS to the RHS and you get...
X^2 + x - 6
Solve for x.
Post by: RuiAce on October 22, 2017, 12:37:56 pm
How often do we have to use the product of roots in questions? Like is it something we should keep in mind when dealing with intersection points?The product and sum of roots are useful when you aren't interested in intersection points themselves, but any relationships between the intersection points. They're a bit less likely to appear than the discriminant, but probably the next thing as obscure to it.
Post by: asahyoun3 on October 22, 2017, 12:38:16 pm
Halp pls
Hi I have attached the solution to Q10, If you have any problems understanding just let me know
Post by: RuiAce on October 22, 2017, 12:41:17 pm
Halp pls
Post by: asahyoun3 on October 22, 2017, 12:42:36 pm
Q7:
ln (6-x) = 2lnx
loge (6-x) = 2logex
loge (6-x) = loge (x^2)
6-x = x^2 ..... x^2 + x - 6 =0 .... (x+3) (x-2)
x = -3, 2
Solution is only x=2 as x>0 for logs
Post by: RuiAce on October 22, 2017, 12:42:50 pm
Halp pls
(https://i.imgur.com/ZCLAJfi.png)
You can plug this into the calculator
Post by: RuiAce on October 22, 2017, 12:50:16 pm
Hi, could someone please help with Q16 b(iii) of 2013? I don't get how we're meant to work out it's shape - is it by plotting points?The procedure is to build it up.
https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
1. Sketch \(e^{0.04t}\), because you know how to sketch the ordinary exponential.
2. Sketch \(-e^{0.04t}\) by reflecting about the \(t\)-axis.
3. Sketch \(375-e^{0.04t}\) by translating the graph up by 375
You were taught all of these techniques; the hardest bit is about doing them in the correct order.
Post by: RuiAce on October 22, 2017, 12:53:58 pm
Part 1/2 List of SOS 2010 2U HSC Asks – 5(b)(iii), 8(d), 9(b)
...help.
Post by: RuiAce on October 22, 2017, 12:56:59 pm
Part 1/2 List of SOS 2010 2U HSC Asks – 5(b)(iii), 8(d), 9(b)
...help.
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Post by: RuiAce on October 22, 2017, 01:00:59 pm
Part 1/2 List of SOS 2010 2U HSC Asks – 5(b)(iii), 8(d), 9(b)
...help.
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Note: You can actually check that in fact, \(f(4) = 0\) as well.
The graph should look something like a sine curve between 0 and 4, and then a straight line with gradient m=-3, from 4 to 6.
Post by: sidzeman on October 22, 2017, 01:07:43 pm
Sorry for the spam but, for part iv of this question, I found P and so found the rate of increase of carp. When I go to find the rate of decrease of trout so as to equate the two and find t, I get -0.04e^0.04t - leaving me unable to solve for t as you cannot log a negative. The answers has it as postive 0.04... - Am i differentiating wrong or what I'm not sure what I'm missing
Post by: RuiAce on October 22, 2017, 01:17:59 pm
Part 2/2 2010 2U HSC Asks – 10(a)(iv), 10(b)(ii)First one was answered here
Post by: RuiAce on October 22, 2017, 01:21:31 pm
Part 2/2 2010 2U HSC Asks – 10(a)(iv), 10(b)(ii)
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You can now just plug π/3 in to find how much volume is left, and thus the required fraction.
Post by: Thebarman on October 22, 2017, 01:58:39 pm
Thanks
Post by: RuiAce on October 22, 2017, 02:02:40 pm
Hey, I'm constantly getting confused when it comes to Simpson's rule and the trapezoidal rule in regards to the number of subintervals. When working out h = (b - a)/n, how many subintervals are there? Because for some questions, it will be one more than the number provided (say there are 5 function values provided, then the number of sub intervals will be 6), while other times it will be one less. Sorry if that's a bit confusing, but I;m a bit confused myself.n is the number of intervals but n+1 is the number of function values.
Thanks
For Simpson's rule, there will always be an even number of subintervals, and an odd number of function values.
It never goes the other way. If somehow you ended up with n-1 function values it means you have made a mistake somewhere.
Sometimes, that mistake happens because you forget that one of the function values can actually be 0.
Post by: _____ on October 22, 2017, 02:03:35 pm
(https://i.imgur.com/NP51jlb.png)
Thanks
Post by: MisterNeo on October 22, 2017, 02:08:44 pm
How do I find the required angle here? I know to use two segments but apart from that I'm stuck. Solution says to use equilateral triangles but I don't understand how that would work.(https://i.imgur.com/wqWVpA2.png)
(https://i.imgur.com/NP51jlb.png)
Thanks
There are 2 equiliateral triangles that make up the segments, so the angle would be 120o because each angle is 60o. :)
Post by: RuiAce on October 22, 2017, 02:11:57 pm
How do I find the required angle here? I know to use two segments but apart from that I'm stuck. Solution says to use equilateral triangles but I don't understand how that would work.(https://i.imgur.com/WWq6jv1.png)
(https://i.imgur.com/NP51jlb.png)
Thanks
Post by: cxmplete on October 22, 2017, 03:42:33 pm
Post by: jamonwindeyer on October 22, 2017, 03:52:16 pm
How do you do part i
Hey! This is actually a 3U question (purely because of that factorial notation at the bottom), did you still want us to tackle it for you? We can move it to the MX1 section and help you out if it is something you need help with, but if it is just 2U you can definitely ignore ;D
Post by: cxmplete on October 22, 2017, 03:55:46 pm
Hey! This is actually a 3U question (purely because of that factorial notation at the bottom), did you still want us to tackle it for you? We can move it to the MX1 section and help you out if it is something you need help with, but if it is just 2U you can definitely ignore ;DHi. Yah, I'll move it to the MX1 section
Post by: georgiia on October 22, 2017, 03:58:57 pm
Thx !
Post by: jamonwindeyer on October 22, 2017, 04:02:18 pm
What am I doing wrong with this probability question? (http://uploads.tapatalk-cdn.com/20171022/e9f97b57ea9739ca9379074221ce727e.jpg)(http://uploads.tapatalk-cdn.com/20171022/69bcb88b61ef43f28691116004e25f85.jpg)
Thx !
Hey! For your tree diagram, instead of a branch for each person, do a stage for each person. So three stages, each with two branches of support (70%) and not support (30%) ;D
Post by: georgiia on October 22, 2017, 04:04:02 pm
Hey! For your tree diagram, instead of a branch for each person, do a stage for each person. So three stages, each with two branches of support (70%) and not support (30%) ;Dcool thanks I’ll try that, but how come this method doesn’t work?
Thank you
Post by: jamonwindeyer on October 22, 2017, 04:07:44 pm
cool thanks I’ll try that, but how come this method doesn’t work?
Thank you
Basically, it is because you are assigning a probability to which person you pick, when really that isn't something that matters - Because all of their opinions are counted.
The best way to think of it is to think, well, if all three are for it, then
Person A is for it
AND
Person B is for it
AND
Person C is for it
So you can turn that into an equation (which a tree diagram helps you visualise) -
Remember, the word AND means you multiply, the word OR means you add the respective probabilities ;D
Post by: RuiAce on October 22, 2017, 04:10:38 pm
What am I doing wrong with this probability question? (http://uploads.tapatalk-cdn.com/20171022/e9f97b57ea9739ca9379074221ce727e.jpg)(http://uploads.tapatalk-cdn.com/20171022/69bcb88b61ef43f28691116004e25f85.jpg)You can label the F's and A's as appropriate.
Thx !
The idea is that your probability tree diagram assumed that you had to pick one out of three, then ask for only THAT person's opinion, when in reality you're interested in all three of their opinions.
Post by: georgiia on October 22, 2017, 04:11:09 pm
Basically, it is because you are assigning a probability to which person you pick, when really that isn't something that matters - Because all of their opinions are counted.
The best way to think of it is to think, well, if all three are for it, then
Person A is for it
AND
Person B is for it
AND
Person C is for it
So you can turn that into an equation (which a tree diagram helps you visualise) -
Remember, the word AND means you multiply, the word OR means you add the respective probabilities ;D
Ok thanks, now I’m struggling with the second part
Post by: RuiAce on October 22, 2017, 04:13:16 pm
Ok thanks, now I’m struggling with the second part
Post by: jamonwindeyer on October 22, 2017, 04:14:37 pm
Ok thanks, now I’m struggling with the second part
So the majority would mean 2 OR 3 people are supportive of the change! So, find the branches of your tree diagram that reflect this (there should be four of them) and add up their probabilities! Essentially you are doing:
- A, B AND C are all supportive (same as your answer from Part (i))
OR (meaning, plus)
- A AND B are supportive but C isn't
OR
- A AND C are supportive but B isn't
OR
- B AND C are supportive but A isn't
Which is why we take four branches ;D
Post by: georgiia on October 22, 2017, 04:14:57 pm
I get it now with this type of diagram ! Thanks
Post by: emmy_b on October 22, 2017, 04:34:13 pm
http://educationstandards.nsw.edu.au/wps/wcm/connect/2cdf8537-92dc-4328-ab39-74a1f1dd489f/maths-hsc-exam-2013.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-2cdf8537-92dc-4328-ab39-74a1f1dd489f-lGhNsb6
Post by: K9810 on October 22, 2017, 04:58:33 pm
A superannuation fund pays an interest rate of 8.75% per annum which compounds annually. Stephanie decides to invest $5000 in the fund at the beginning of each year, commencing on 1 January 2003.
What will be the value of Stephanie’s superannuation when she retires on 31 December 2023?
Thanks
Post by: ~BK~ on October 22, 2017, 05:05:15 pm
could you pls help me out with the attached q from the 2011 paper...
don't even know where to start ???still kinda just staring at it blankly
thanks in advance
Post by: RuiAce on October 22, 2017, 05:22:23 pm
Hi can someone make worked solutions for 2013 HSC Q6? Super stuck!Back when I did this paper, I just subbed in points and looked out for which one worked.
http://educationstandards.nsw.edu.au/wps/wcm/connect/2cdf8537-92dc-4328-ab39-74a1f1dd489f/maths-hsc-exam-2013.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-2cdf8537-92dc-4328-ab39-74a1f1dd489f-lGhNsb6
The only one that intercepts the x-axis at -π/6 is D.
Post by: MisterNeo on October 22, 2017, 05:26:32 pm
hey...
could you pls help me out with the attached q from the 2011 paper...
don't even know where to start ???still kinda just staring at it blankly
thanks in advance
Post by: jamonwindeyer on October 22, 2017, 05:27:35 pm
Can you please help me with this question:
A superannuation fund pays an interest rate of 8.75% per annum which compounds annually. Stephanie decides to invest $5000 in the fund at the beginning of each year, commencing on 1 January 2003.
What will be the value of Stephanie’s superannuation when she retires on 31 December 2023?
Thanks
Hey! What was your approach to starting the question? Did you have a go at forming the required series? So something like this would be your approach:
Post by: ~BK~ on October 22, 2017, 05:34:40 pm
wow, that's awesome thanks so much ;DAnd you can find the centre/radius from this equation (1,0) r=4. ;)
makes heaps more sense now, it just seemed to be worded strangely (as BOSTES love to do with any maths q)
Post by: Natasha.97 on October 22, 2017, 05:34:56 pm
hey...
could you pls help me out with the attached q from the 2011 paper...
don't even know where to start ???still kinda just staring at it blankly
thanks in advance
Hi!
(https://i.imgur.com/RMz8zFJ.png)
Hope this helps
Post by: _____ on October 22, 2017, 06:18:28 pm
(https://i.imgur.com/SgvaBlf.png)
Can I just sub in y into the equation given or do I need to "show" how the equation is found by 'going the other way' if you get what I mean? If so, how would this be done?
Thanks!
Post by: blasonduo on October 22, 2017, 06:21:08 pm
(I totally didn't accidentally put this in 4u :P)
For questions like this (ii) how do I adequately prove that the value is always positive?
I have lost a mark in trials for not doing this, but I'm unsure how to do this apart from subbing in values and going "look its still positive!"
Post by: Natasha.97 on October 22, 2017, 06:26:14 pm
Hey!
(I totally didn't accidentally put this in 4u :P)
For questions like this (ii) how do I adequately prove that the value is always positive?
I have lost a mark in trials for not doing this, but I'm unsure how to do this apart from subbing in values and going "look its still positive!"
Hi
(Don't worry, no one saw :P)
For 1 mark, I don't think you need that much proof to go with it. I would derive the acceleration formula (16e-2t) and give the statement that e-2t > 0 for all values of t, which makes acceleration always positive.
Hope this helps
Post by: RuiAce on October 22, 2017, 06:46:50 pm
For this (iii):Yeah if you subbed it in you would be assuming the result. You have to show it.
(https://i.imgur.com/SgvaBlf.png)
Can I just sub in y into the equation given or do I need to "show" how the equation is found by 'going the other way' if you get what I mean? If so, how would this be done?
Thanks!
Post by: jamonwindeyer on October 22, 2017, 06:55:10 pm
For this (iii):
(https://i.imgur.com/SgvaBlf.png)
Can I just sub in y into the equation given or do I need to "show" how the equation is found by 'going the other way' if you get what I mean? If so, how would this be done?
Thanks!
Will just say that, while Rui's method is definitely more mathematically correct, in general you can definitely just substitute, especially for a 1 marker. NESA's solutions do it ;D
Post by: kiiaaa on October 22, 2017, 07:03:17 pm
would you be able to please explain to me what is ' small angles and limits' and where/what types of questions should i beware of them?
i was away that day in class and every time i look at the notes i copied off my friend they make little sense =/
thank you :)
Post by: RuiAce on October 22, 2017, 07:07:50 pm
hi guys
would you be able to please explain to me what is ' small angles and limits' and where/what types of questions should i beware of them?
i was away that day in class and every time i look at the notes i copied off my friend they make little sense =/
thank you :)
Post by: pokemonlv10 on October 22, 2017, 07:11:01 pm
Post by: kiiaaa on October 22, 2017, 07:11:39 pm
oooh thats much more clearer than what i had down lool
what type of question would they ask such in? would it be in MC by any chance?
Post by: RuiAce on October 22, 2017, 07:12:44 pm
Hey, for this question i just assigned a value and tested all the answers, but would like to know the proper method of doing this?
Post by: RuiAce on October 22, 2017, 07:14:33 pm
oooh thats much more clearer than what i had down loolThey seem to appear extremely often in 3U multiple choice questions. Usually it involves you to manipulate your way around it through the limit \[\lim_{x\to 0}\frac{\sin (ax)}{ax} = 1 .\]
what type of question would they ask such in? would it be in MC by any chance?
It's a 2U concept but seems to appear more often in 3U exams than 2U exams for some reason.
Post by: MisterNeo on October 22, 2017, 07:15:40 pm
Hey, for this question i just assigned a value and tested all the answers, but would like to know the proper method of doing this?
Post by: 3.14159265359 on October 22, 2017, 07:46:11 pm
Post by: _____ on October 22, 2017, 07:59:57 pm
Will just say that, while Rui's method is definitely more mathematically correct, in general you can definitely just substitute, especially for a 1 marker. NESA's solutions do it ;D
Thanks for pointing that out, I'll just do that unless it's 3+ marks or a relatively easy manipulation
Post by: roygbivmagic on October 22, 2017, 08:02:42 pm
Find the exact length of the tangent from (4,-5) to the circle x^2 + 4x + y^2 - 2y - 11 = 0.
(Answer = square root of 56)
Post by: wulverine on October 22, 2017, 08:18:05 pm
(i) Sketch the graph y = I 2x − 3 I
(ii) Using the graph from part (i), or otherwise, find all values of m for which
the equation I 2x − 3 I = mx + 1 has exactly one solution.
Also, question 9 from 2014 mc. These pop up every year and are the only ones I get wrong in mc
Post by: kiiaaa on October 22, 2017, 08:28:52 pm
i was wodnering if you could please help me tackle this question from the 2014 hsc paper. i dont know why i just found it confusing and i legit dont know what to do
thank you very much
Post by: RuiAce on October 22, 2017, 08:37:36 pm
Hi, please can you explain how to do this question?
Find the exact length of the tangent from (4,-5) to the circle x^2 + 4x + y^2 - 2y - 11 = 0.
(Answer = square root of 56)
(https://i.imgur.com/Z5eBHip.png)
Yeah. Technically not a 2U question, because the fact that \(\angle ABC = 90^\circ \) relies on circle geometry.. But I've seen the Cambridge textbook throw it at 2U students anyway.
Post by: jamonwindeyer on October 22, 2017, 08:38:06 pm
hey! I wanted to ask about part iv of this 2006 hsc paper q6b. when I completed this question I did the equation less than 200 (i.e. P<200) and solved but in the answers they had solved P=200 and I was confused as to why they did that since the question says "when the population falls below 200". with my working out, is the way I answered the question wrong? please clarify. thanks!
Hey! Both are fine - They are just taking a little shortcut, since they know it is decaying (getting smaller) then the time it hits 200 (=) is also the moment it falls below 200 (<), so solving the equality and the inequality are equivalent in this case! You can almost always do this for exponential growth and decay ;D
Post by: Infinitex on October 22, 2017, 08:38:39 pm
Hi, please can you explain how to do this question?
Find the exact length of the tangent from (4,-5) to the circle x^2 + 4x + y^2 - 2y - 11 = 0.
(Answer = square root of 56)
Try form a triangle with the centre of the circle and the point using the tangent. It might become more clear.
Edit: I was too slow typing
Post by: 12070 on October 22, 2017, 08:40:49 pm
hi guys!
i was wodnering if you could please help me tackle this question from the 2014 hsc paper. i dont know why i just found it confusing and i legit dont know what to do
thank you very much
For part (i)
A1= 500(1.003)
A2= [A1+500(1.01)]x(1.003)
= [500(1.003)+500(1.01]x(1.003)
Therefore A2= 500(1.003)^2 + 500(1.01)(1.003)
I'll write out the solution for part (ii) before explaining it because I will probably get something wrong
Post by: RuiAce on October 22, 2017, 08:42:24 pm
Hi, could anyone explain 15c part (ii) from the 2013 paper? The solutions make no sense to me...The 2013 question is already addressed in the compilation.
(i) Sketch the graph y = 2x − 3 .
(ii) Using the graph from part (i), or otherwise, find all values of m for which
the equation 2x − 3 = mx + 1 has exactly one solution.
Also, question 9 from 2014 mc. These pop up every year and are the only ones I get wrong in mc
Post by: 3.14159265359 on October 22, 2017, 09:04:33 pm
Hey! Both are fine - They are just taking a little shortcut, since they know it is decaying (getting smaller) then the time it hits 200 (=) is also the moment it falls below 200 (<), so solving the equality and the inequality are equivalent in this case! You can almost always do this for exponential growth and decay ;D
thank youuu!!!!!
Post by: 3.14159265359 on October 22, 2017, 09:21:15 pm
Post by: jamonwindeyer on October 22, 2017, 09:22:54 pm
hi guys!
i was wodnering if you could please help me tackle this question from the 2014 hsc paper. i dont know why i just found it confusing and i legit dont know what to do
thank you very much
Hey! You want to start by building up the series. Let the balance at the end of the month be \(A_n\). We step through what happens to the account balance in each month:
- Add $500 at the start
- Apply interest at the end
So at the end of the first month we have:
Now the next month, we don't add $500. We add 1% more, \(500(1.01)\). So now we have this:
This is Part (i) done!
Next month, we don't add \(500(1.01)\). We add 1% greater AGAIN, so, \(500(1.01)^2\). So now we have:
By now we should notice the pattern. Jump to \(A_n\):
Now what is inside the brackets is actually a very messy geometric series. We could define it in various ways by pulling out different terms, but I'll define it as a series with first term \(a=1.003^n\), and common ratio:
We get this by looking at what happens to the powers of (1.003) and (1.01) - The powers of the former go down, so we are dividing by it. The powers of the latter go up, so we are multiplying by it, hence this value of \(r\)!
Besides this, this is just a standard question. Use the formula for the sum of a geometric series:
Now thankfully, we don't need to rearrange this monstrosity - Just set \(n=60\) and calculator:
To the nearest cent ;D
Post by: jamonwindeyer on October 22, 2017, 09:50:40 pm
hey Jamon with the revision videos that you had posted for the applications of series question you done, with the last part, do you have to reserve that formula or can you use the formula from part ii but change the value from 100000 to 48500?
Hey! I'm pretty sure I just reuse the formula right, changing to 48500? You can definitely do that ;D
Post by: liya1234 on October 22, 2017, 09:57:27 pm
Post by: RuiAce on October 22, 2017, 10:00:08 pm
Hi I just had a general question concerning graphs of displacement/velocity/acceleration. If we get a graph of one of these (e.g. velocity over time) and are asked to determine a different aspect (e.g. acceleration at a certain point in time), how would we do that? I feel like i'm missing something because I never know how to read those graphs and always get confused with those three? Any tips would be greatly appreciated! Thanks!!
Recall that the first derivative measures the gradient of the tangent (and thus the slope, i.e. whether the curve is increasing or decreasing). Also, the second derivative measures the concavity of the curve (up or down).
__________________
Post by: 3.14159265359 on October 22, 2017, 10:01:01 pm
Hey! I'm pretty sure I just reuse the formula right, changing to 48500? You can definitely do that ;D
sorry I meant re-derive not reserve. but yh you just reused it. so if I do that I wont lose any marks? coz I was told you had to re-derive.
Post by: jamonwindeyer on October 22, 2017, 10:03:09 pm
sorry I meant re-derive not reserve. but yh you just reused it. so if I do that I wont lose any marks? coz I was told you had to re-derive.
Provided you make specific reference (like I did aurally) to the fact that we have the same scenario, but with a different starting amount, I am almost certain you would still be awarded the marks. But of course if you wanted to be safe you could re-derive - It wouldn't take more than two minutes so that might be worth the peace of mind ;D
Post by: winstondarmawan on October 22, 2017, 10:03:25 pm
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22773454_1357349967723856_102862891_n.png?oh=32a8ea99a0286365b829dc824bd07616&oe=59EF071C
Part (ii), would it be acceptable to prove EDB is similar to ABE, and since ABE is similar to BCD (from part i) then EDB is similar to BCD?
Also:
In the answer for this max/min question, would it be acceptable to leave my answer as this:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22711487_1357352164390303_1734180122_n.jpg?oh=721aa87626b816afee1cb391e694f9b2&oe=59EF0CE2
And not simplify fully?
TIA.
Post by: RuiAce on October 22, 2017, 10:05:11 pm
For this question:Yes to your first question.
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22773454_1357349967723856_102862891_n.png?oh=32a8ea99a0286365b829dc824bd07616&oe=59EF071C
Part (ii), would it be acceptable to prove EDB is similar to ABE, and since ABE is similar to BCD (from part i) then EDB is similar to BCD?
Also:
In the answer for this max/min question, would it be acceptable to leave my answer as this:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22711487_1357352164390303_1734180122_n.jpg?oh=721aa87626b816afee1cb391e694f9b2&oe=59EF0CE2
And not simplify fully?
TIA.
That already looks simplified to me. Not sure how you want to simplify it further.
Post by: winstondarmawan on October 22, 2017, 10:07:33 pm
Yes to your first question.
That already looks simplified to me. Not sure how you want to simplify it further.
This is what the answers did:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22711377_1357354501056736_913648934_n.png?oh=f241001e14e4563ed5eb50030ccd6925&oe=59EF1A1F
Post by: liya1234 on October 22, 2017, 10:08:46 pm
Oh okay that makes sense thank you so much! Just a few more things I wanted to clarify - so if I was given a graph on acceleration, could I find velocity by finding the area under the graph then as well? Also I understand that from a displacement graph, velocity is given by the slope so I would find the gradient. But for acceleration, which is given by concavity, how exactly would I find the value for it?
Post by: RuiAce on October 22, 2017, 10:10:39 pm
Oh okay that makes sense thank you so much! Just a few more things I wanted to clarify - so if I was given a graph on acceleration, could I find velocity by finding the area under the graph then as well? Also I understand that from a displacement graph, velocity is given by the slope so I would find the gradient. But for acceleration, which is given by concavity, how exactly would I find the value for it?Yes to the first question.
You wouldn't be able to find the actual value. You can only find the sign of the acceleration.
Post by: bimberfairy on October 22, 2017, 10:15:37 pm
Post by: RuiAce on October 22, 2017, 10:18:17 pm
Hi! I was doing a paper and this question suddenly popped in my head. I'm only up to 2012 so far and haven't checked the more recent ones, but do the questions go from easy to hard ?? Or are all the questions jumbled up in the exam.Yes.
Math papers are specifically designed so that the difficulty goes in ascending order (with the exception of the multiple choice).
Post by: bimberfairy on October 22, 2017, 10:34:47 pm
Yes.
Math papers are specifically designed so that the difficulty goes in ascending order (with the exception of the multiple choice).
Thank you !! I'll most likely use my reading time to flip to those first haha
Post by: wulverine on October 22, 2017, 10:55:09 pm
And if given the graph of f'(x), how do you draw f(x)?
Post by: pokemonlv10 on October 22, 2017, 11:03:53 pm
Post by: RuiAce on October 22, 2017, 11:08:01 pm
Hello, would like to know how to solve this type of equation?I'm not lying when I say this.
You need to sketch a graph, drawn to scale. Because there IS no algebraic method of solving that equation.
Post by: pokemonlv10 on October 22, 2017, 11:09:05 pm
Post by: RuiAce on October 22, 2017, 11:10:08 pm
For questions asking to draw the derivative of a graph f(x), how would you approach this? Is it stationary points of f(x) drawn as x-intercepts of f'(x) and inflexion points of f(x) drawn as max/min points of f'(x) and whatever is under the curve as a negative slope vice versa for above curve?
And if given the graph of f'(x), how do you draw f(x)?
So yes, a S.P. on y=f(x) corresponds to an x-intercept on y=f'(x).
The same can be deduced from a similar analysis for inflexion points as you stated.
Post by: RuiAce on October 22, 2017, 11:13:24 pm
And these two not sure about, i subbed in a, but didn't get any of the mc. Also not sure about the last question.
Note: This question could've been done differently if you chose to raise everything to the power of 3/2, before taking logs of both sides.
Post by: RuiAce on October 22, 2017, 11:20:24 pm
And these two not sure about, i subbed in a, but didn't get any of the mc. Also not sure about the last question.
\[ \text{The common trick to these questions is the identity }\boxed{e^{\ln a} = a}\text{ for all }a > 0. \]
\begin{align*}\int_3^{10}3^x\,dx &= \int_3^{10} e^{\ln (3^x)}dx\\ &= \int_3^{10} e^{x\ln 3} \tag{log law}\\ &= \left[\frac{e^{x\ln 3}}{\ln 3}\right]_3^{10} \tag{ln 3 is a constant}\\ &= \left[\frac{3^x}{\ln 3}\right]_3^{10}\tag{reverse engineering some steps}\\ &= \frac{3^{10} - 3^3}{\ln 3}\end{align*}
Post by: sidzeman on October 22, 2017, 11:35:52 pm
Post by: pokemonlv10 on October 22, 2017, 11:39:53 pm
Post by: RuiAce on October 22, 2017, 11:46:31 pm
For question 16 part i does my answer (1/tantheta) satisfy the question. The solutions said there are many forms x could take i just wanted to make sure mine was correct. Note: I used to proportional division theorem of a triangle, I just didnt write it out (hence why AD also equals x)I'm not sure how you got that answer. You assumed that C is the midpoint of AB and D is the midpoint of AE, but I don't see anything that permits this.
Post by: RuiAce on October 22, 2017, 11:47:27 pm
Hello, how would i factorise this completely: x^2 -9y^2 -x - 3y. Apparently completing the square would lose marks here\begin{align*}x^2-9y^2-x-3y&= (x-3y)(x+3y)-1(x+3y)\\ &= (x+3y)\left[(x-3y)+1\right]\\ &= (x+3y)(x-3y+1)\end{align*}
Post by: sidzeman on October 22, 2017, 11:51:19 pm
I'm not sure how you got that answer. You assumed that C is the midpoint of AB and D is the midpoint of AE, but I don't see anything that permits this.I was attempting to use the Proportionality theorem - "A line drawn parallel to one side of a triangle divides the other two sides into parts of equal proportion". CD should be parallel to BE right?
Edit: I just realised I've been mistaking equal proportional to mean bisect nvm then
Post by: RuiAce on October 22, 2017, 11:54:03 pm
I was attempting to use the Proportionality theorem - "A line drawn parallel to one side of a triangle divides the other two sides into parts of equal proportion". CD should be parallel to BE right?Yeah, it's really just similar triangles (proportional sides). Bisection would be too good to be true I'm afriad
Edit: I just realised I've been mistaking equal proportional to mean bisect nvm then
Post by: pokemonlv10 on October 22, 2017, 11:56:55 pm
\begin{align*}x^2-9y^2-x-3y&= (x-3y)(x+3y)-1(x-3y)\\ &= (x-3y)\left[(x+3y)+1\right]\\ &= (x-3y)(x+3y+1)\end{align*}
Where did the last bit come from? the -1(x-3y) part
Post by: RuiAce on October 22, 2017, 11:59:50 pm
Where did the last bit come from? the -1(x-3y) partThere's a mistake, sorry fixing it now.
The idea was to factorise -1, but it was factorised incorrectly.
Post by: RuiAce on October 23, 2017, 12:06:13 am
There should be people online to take last minute questions tomorrow morning so don't fuss too much about that. Just try not making them too last minute.
Post by: winstondarmawan on October 23, 2017, 12:07:43 am
Seems like the link to the farmhouse question is broken, so I'm just gonna repost the Q here:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22713025_1357409827717870_1319962237_o.png?oh=f5adcb772fbeb84dc95709f06ccfc5ae&oe=59EF0120
TIA.
Post by: RuiAce on October 23, 2017, 12:36:47 am
Hello!Yep, should be fixed now
Seems like the link to the farmhouse question is broken, so I'm just gonna repost the Q here:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22713025_1357409827717870_1319962237_o.png?oh=f5adcb772fbeb84dc95709f06ccfc5ae&oe=59EF0120
TIA.
Post by: sidzeman on October 23, 2017, 12:44:01 am
Post by: RuiAce on October 23, 2017, 12:47:13 am
Hey can someone tell me where i went wrong in part iii of this question - I got an answer of 16 while the solutions says its -6
Post by: nattynatman on October 23, 2017, 01:19:27 am
Post by: itssona on October 23, 2017, 02:08:18 am
kinda confused. if a parabola is in the form x^2=4ay then is directrix always y=k-a
where a is focal length
because according to direction I get different answers and idk what to do since my topic test is afternoon
Post by: Opengangs on October 23, 2017, 07:10:42 am
hiiiIf the parabola is in the form:
kinda confused. if a parabola is in the form x^2=4ay then is directrix always y=k-a
where a is focal length
because according to direction I get different answers and idk what to do since my topic test is afternoon
If, however, the parabola is of the form:
Post by: caitlinlddouglas on October 23, 2017, 07:25:25 am
Post by: RuiAce on October 23, 2017, 07:31:03 am
hey I was wondering how to find when a particle was furthest from the origin of they were to give you the velocity or acceleration graph? Thanks!
You always need to make some kind of deduction based off any information they give you.
You need to go back to the velocity first, or else you're stuck
Rui, genuine question: how did you get so damn good at maths?Uhhhh, idk probably just too much practice.. Thanks? :P
Post by: Justinhales on October 23, 2017, 08:41:58 am
Sorry real quick question..!!
A parabola has focus (5, 0) and directrix x = 1.
What is the equation of the parabola?
Can u pls do full working out too!!
thanks!
Post by: RuiAce on October 23, 2017, 08:47:16 am
Hey!!
Sorry real quick question..!!
A parabola has focus (5, 0) and directrix x = 1.
What is the equation of the parabola?
Can u pls do full working out too!!
thanks!
Nobody really cares if the sketch is rough or not, but you need to sketch it.
(https://i.imgur.com/CpsPLUL.png)
There's not much working out at all for these questions.
Post by: Justinhales on October 23, 2017, 08:53:25 am
Post by: angelahchan on October 23, 2017, 09:30:44 am
http://educationstandards.nsw.edu.au/wps/wcm/connect/47b7a5d2-8729-4d55-825a-7e840444d93f/2016-hsc-maths.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-47b7a5d2-8729-4d55-825a-7e840444d93f-lCmfMeF
Post by: winstondarmawan on October 23, 2017, 09:32:53 am
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22809677_1357774924348027_1303197799_n.jpg?oh=76d0e9125ddfa8d3f76ff4dbe96fe3ed&oe=59EF9CD2
Even though it's not in the syllabus?
Post by: Opengangs on October 23, 2017, 09:36:15 am
Hi, can someone please help with Q9 of 2016? I'm not sure how to integrate with absolute valuesRecall that the integral is merely the area under the curve.
http://educationstandards.nsw.edu.au/wps/wcm/connect/47b7a5d2-8729-4d55-825a-7e840444d93f/2016-hsc-maths.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-47b7a5d2-8729-4d55-825a-7e840444d93f-lCmfMeF
Sketch the absolute function to find the area of the two triangles formed between the two bounds.
If you're still unsure, let me know!!
Are we permitted to use this formula:I'm pretty sure then you have to derive it, if it's not in the syllabus
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22809677_1357774924348027_1303197799_n.jpg?oh=76d0e9125ddfa8d3f76ff4dbe96fe3ed&oe=59EF9CD2
Even though it's not in the syllabus?
Post by: pokemonlv10 on October 23, 2017, 09:39:22 am
Post by: mbdtHSC on October 23, 2017, 10:13:56 am
how would you do this question?I think you split it into an AP with 7, 14, 21... and GP with 3, 6, 12 ... then sum each
Post by: Opengangs on October 23, 2017, 10:15:49 am
how would you do this question?
If we split up the series, we can form something like:
Post by: pokemonlv10 on October 23, 2017, 10:39:16 am
Post by: RuiAce on October 23, 2017, 10:43:19 am
Not sure about this trig
Post by: winstondarmawan on October 23, 2017, 10:59:56 am
Just a little confused on how they got 5pi4 out.
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22790869_816333725195103_370366407_o.png?oh=65e3b77173cc2771d0f8b9c4ab7bbf0a&oe=59EF5BC6
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22809895_816336158528193_1950803996_n.jpg?oh=c5927f4ef4917a4b2cbce9df04d4e20f&oe=59EF96A1
TIA.
Post by: RuiAce on October 23, 2017, 11:02:57 am
Hello!They didn't really adjust the domains like they should have. Which is fair enough, because adjusting the domains would be dodgy (since we need to overlap the domains, and THEN work out the relevant quadrants). But I don't really like that method because it can be too confusing to explain.
Just a little confused on how they got 5pi4 out.
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22790869_816333725195103_370366407_o.png?oh=65e3b77173cc2771d0f8b9c4ab7bbf0a&oe=59EF5BC6
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22809895_816336158528193_1950803996_n.jpg?oh=c5927f4ef4917a4b2cbce9df04d4e20f&oe=59EF96A1
TIA.
Just use my method above.
Post by: caitlinlddouglas on October 23, 2017, 11:04:34 am
Post by: RuiAce on October 23, 2017, 11:05:25 am
They could someone please explain the solution to 15c) ii please? Especially why there are the less than and greater than/equal to signs Thanks!Already addressed in the compilation.
This question isn't easy because algebraic techniques are insufficient.
Post by: Justinhales on October 23, 2017, 11:16:45 am
just a little confused about 13d) iii 2013 hsc pp! Can someone pls explain the step in the answers from
370 000(1.005)^n = 599 600 (1.005^n − 1) to
599 600 = (599 600 − 370 000)(1.005^n)
thanks!!
https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-marking-guide-maths.pdf
Post by: RuiAce on October 23, 2017, 11:19:49 am
Hey!\begin{align*}370000(1.005)^n &= 599600\left(1.005^n - 1\right)\\ 370000(1.005)^n &= 599600\left(1.005\right)^n - 599600 \tag{expanding brackets}\\ 599600 &= 599600(1.005)^n - 370000(1.005)^n \tag{rearranging}\\ 599600 &= \left(599600-370000\right)(1.005)^n\tag{factorising}\end{align*}
just a little confused about 13d) iii 2013 hsc pp! Can someone pls explain the step in the answers from
370 000(1.005)^n = 599 600 (1.005^n − 1) to
599 600 = (599 600 − 370 000)(1.005^n)
thanks!!
https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-hsc-maths.pdf
https://www.boardofstudies.nsw.edu.au/hsc_exams/2013/pdf_doc/2013-marking-guide-maths.pdf
Post by: Justinhales on October 23, 2017, 11:22:11 am
Post by: taylorlucy on October 23, 2017, 12:00:58 pm
http://educationstandards.nsw.edu.au/wps/wcm/connect/47b7a5d2-8729-4d55-825a-7e840444d93f/2016-hsc-maths.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-47b7a5d2-8729-4d55-825a-7e840444d93f-lCmfMeF
Thanks in advance
Post by: RuiAce on October 23, 2017, 12:02:16 pm
A basic question, but would anyone be able to explain why the answer to 2016 HSC Q10 is D?
http://educationstandards.nsw.edu.au/wps/wcm/connect/47b7a5d2-8729-4d55-825a-7e840444d93f/2016-hsc-maths.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-47b7a5d2-8729-4d55-825a-7e840444d93f-lCmfMeF
Thanks in advance
\begin{align*}4+\log_2 x &= \log_2 16 + \log_2 x \\ &= \log_2 (16x)\end{align*}
Post by: asd987 on October 23, 2017, 12:10:16 pm
the equation ax^2-x+b=0 has roots -2 and 4. find the value of a and b
Post by: RuiAce on October 23, 2017, 12:12:46 pm
hey could someone help me with this:
the equation ax^2-x+b=0 has roots -2 and 4. find the value of a and b
Post by: gilliesb18 on October 23, 2017, 04:18:19 pm
I just need a small explanation on how to do this question... I bet its real simple for those of u out there but i'm new to this and just need help.
Question: The focal chord that cuts the parabola x^2= -6y at (6, -6) cuts the parabola again at X. Find the coordinates of X.
Thanks heaps in advance!!
Post by: MisterNeo on October 23, 2017, 10:18:30 pm
Hello!
I just need a small explanation on how to do this question... I bet its real simple for those of u out there but i'm new to this and just need help.
Question: The focal chord that cuts the parabola x^2= -6y at (6, -6) cuts the parabola again at X. Find the coordinates of X.
Thanks heaps in advance!!
Post by: gilliesb18 on October 25, 2017, 01:51:26 pm
Also, just on a side note, I have just bought the ATARNotes HSC Mathematics complete course notes book and it is awesome!! I seriously recommend this book to anyone out there finding this course hard like I have been. It covers everything you need to know!!
Heres the link to it: https://atarnotes.com/product/hsc-mathematics-notes/
Everyone on here asking questions should really get one :)
(In fact anyone just doing this course should get one ;))
Post by: itssona on October 25, 2017, 02:50:25 pm
a woman is employed at $1600 a month but at the end of each month she receives an increase of $60.
what will be her annual income?
Post by: RuiAce on October 25, 2017, 06:06:43 pm
been stuck on how to do this in a clever way using arithmetic sequenceThe question lacks information unless we assume she's employed in January. When that happens it's just an AP with a = 1600 and d = 60 (and you'd be calculation S12)
a woman is employed at $1600 a month but at the end of each month she receives an increase of $60.
what will be her annual income?
Post by: itssona on October 25, 2017, 06:21:55 pm
The question lacks information unless we assume she's employed in January. When that happens it's just an AP with a = 1600 and d = 60 (and you'd be calculation S12)ohh yes thats right, thank you!!
Post by: Mate2425 on October 28, 2017, 06:58:01 pm
The curve y=ax*3 + bx*2 - x + 5 has a point of inflexion at (1,-2). Find the values of 'a' and 'b'.
Thanks.
Post by: RuiAce on October 28, 2017, 07:34:17 pm
Hi could i please get some help on this question...
The curve y=ax*3 + bx*2 - x + 5 has a point of inflexion at (1,-2). Find the values of 'a' and 'b'.
Thanks.
Post by: Mate2425 on October 28, 2017, 09:03:16 pm
Post by: Aaron12038488 on October 29, 2017, 02:25:24 pm
Bill thinks he can afford a
mortgage payment of $800
each month. How much can he
borrow
, to the nearest $100, over
25 years at 11.5% p.a.?
Post by: itssona on October 29, 2017, 08:05:43 pm
cosx=1/root 2 and sinx=-1/root 2
and my domain is x is between 0 and pi/2
what answer fits both equations??
would I just say that sinx has no solutions in our domain?
Post by: jamonwindeyer on October 29, 2017, 08:11:50 pm
help!
Bill thinks he can afford a
mortgage payment of $800
each month. How much can he
borrow
, to the nearest $100, over
25 years at 11.5% p.a.?
Hey! Form a series of the amount owed after each month, with \(P\) being the principal. Note that \(1+r=1+\frac{0.115}{12}\approx1.0096\). Your first few terms would be:
This would look the same as usual for this sort of question. Generalising:
Then just set \(n=25\times12=300\), and evaluate! ;D
Post by: jamonwindeyer on October 29, 2017, 08:13:56 pm
heey, suppose I have the equations
cosx=1/root 2 and sinx=-1/root 2
and my domain is x is between 0 and pi/2
what answer fits both equations??
would I just say that sinx has no solutions in our domain?
You are looking for the quadrant where sine is negative and cosine is positive, that's the fourth quadrant! We can see the related angle is 45 degrees using the exact ratio triangles on our reference sheet, so that means the common solution is \(x=360-45=315\) degrees, using the angles of any magnitude 'system' (angles in the fourth quadrant are expressed as \(360-\theta\) ;D
Post by: RuiAce on October 29, 2017, 10:13:05 pm
You are looking for the quadrant where sine is negative and cosine is positive, that's the fourth quadrant! We can see the related angle is 45 degrees using the exact ratio triangles on our reference sheet, so that means the common solution is \(x=360-45=315\) degrees, using the angles of any magnitude 'system' (angles in the fourth quadrant are expressed as \(360-\theta\) ;DAlthough it specified that x was between 0 and pi/2 instead of 0 and 2pi. So therefore there'd be no solutions?
(Of course, there's always the possibility that pi/2 was a typo.)
Post by: jamonwindeyer on October 29, 2017, 10:44:08 pm
Although it specified that x was between 0 and pi/2 instead of 0 and 2pi. So therefore there'd be no solutions?
(Of course, there's always the possibility that pi/2 was a typo.)
Oh I totally read it as \(2\pi\), so yeah, whichever sssona ;D
Post by: itssona on October 31, 2017, 05:38:40 pm
alsoo so if I was summing a geometric sequence - how many terms of -16+8-4+... must be taken to give a sum of -43/4
Post by: Opengangs on October 31, 2017, 05:44:26 pm
Thanks Jamon and Rui :)Use the sum to n terms of a GP, and equate it to -43/4. Your goal now is to find the value of n.
alsoo so if I was summing a geometric sequence - how many terms of -16+8-4+... must be taken to give a sum of -43/4
Post by: itssona on October 31, 2017, 05:52:42 pm
Use the sum to n terms of a GP, and equate it to -43/4. Your goal now is to find the value of n.oh thanks!!
Post by: taliakaragiannis8 on October 31, 2017, 06:50:43 pm
Hi there, could I have some help with this. Find the x-intercept of the tangent to the curve y = (4x - 3)/(2x+1) at the point where
x=1.
Post by: RuiAce on October 31, 2017, 06:57:38 pm
Hi there, could I have some help with this. Find the x-intercept of the tangent to the curve y = (4x - 3)/(2x+1) at the point where
x=1.
Post by: itssona on November 05, 2017, 02:44:18 pm
Post by: jamonwindeyer on November 05, 2017, 04:36:18 pm
for limiting sum, is it always a/1-r
Yep! Provided the limiting sum actually exists, that is the formula you use to calculate it ;D
Post by: itssona on November 05, 2017, 04:47:12 pm
Yep! Provided the limiting sum actually exists, that is the formula you use to calculate it ;Dooh awesomeee! :)
Post by: itssona on November 05, 2017, 09:33:51 pm
18+6+2+...
Post by: RuiAce on November 05, 2017, 09:36:39 pm
why will this series have a limiting sum?The common ratio is 1/3, so its absolute value is 1/3 which is less than 1. Therefore it has a limiting sum.
18+6+2+...
Post by: itssona on November 05, 2017, 09:47:18 pm
The common ratio is 1/3, so its absolute value is 1/3 which is less than 1. Therefore it has a limiting sum.ohh thank you!! :)
also i have this question- for what values of x does 1+(x-2)+(x-2)^2... converge and whats the limiting sum
so do i find r (which is x-2) and say that -1<x-2<1 so those are the values
and then to find limiting sum, i plug in those values? or am i off?
Post by: RuiAce on November 05, 2017, 09:55:22 pm
ohh thank you!! :)You really should simplify \( -1 < x-2 < 1\) to \( 1 < x < 3\) first, but then just plug straight into the limiting sum formula. You'll get \( \frac{1}{1-(x-2)} \)
also i have this question- for what values of x does 1+(x-2)+(x-2)^2... converge and whats the limiting sum
so do i find r (which is x-2) and say that -1<x-2<1 so those are the values
and then to find limiting sum, i plug in those values? or am i off?
Post by: itssona on November 05, 2017, 10:09:32 pm
You really should simplify \( -1 < x-2 < 1\) to \( 1 < x < 3\) first, but then just plug straight into the limiting sum formula. You'll get \( \frac{1}{1-(x-2)} \)oh so i leave it in the form 1/3-x ?
Post by: jamonwindeyer on November 05, 2017, 11:39:28 pm
oh so i leave it in the form 1/3-x ?
Yep! :)
Post by: itssona on November 06, 2017, 07:29:17 am
Yep! :)thank you :))
Post by: gilliesb18 on November 06, 2017, 06:21:07 pm
This is probably a regularly asked question, but How do I know how to write geometry proofs quickly and on the spot in an exam?
Anyone else struggle with this??
Thanks once again...
Post by: jamonwindeyer on November 06, 2017, 08:27:56 pm
Hey!
This is probably a regularly asked question, but How do I know how to write geometry proofs quickly and on the spot in an exam?
Anyone else struggle with this??
Thanks once again...
Hey! This is very much a practice makes perfect thing, you need to just do enough questions that it flows naturally! However, memorising the rules properly will help you write them succinctly, so make sure you do that. You are allowed to use symbols too. Don't be afraid to scribble out the proof using pro-numerals and labels on the exam sheet before actually writing out the proof, to give you a better idea of the path to the solution before you start :)
Post by: gilliesb18 on November 06, 2017, 09:14:41 pm
What I've been struggling with recently is writing out the proof, but then checking the answers and it has something completely different!! Probably a very very dumb question, but is there always more than one answer?
Thanks again :D
Post by: RuiAce on November 06, 2017, 09:16:32 pm
Ok, thanks.Of course there is.
What I've been struggling with recently is writing out the proof, but then checking the answers and it has something completely different!! Probably a very very dumb question, but is there always more than one answer?
Thanks again :D
Why wouldn't there be? A question can be done in thousands of different ways and so long as the logic follows, each of those thousands of ways is a correct answer. The only time you can be wrong is if your logic and reasoning was incorrect
The "best" set of answers just demonstrates the "most optimal" method.
Post by: gilliesb18 on November 06, 2017, 09:31:21 pm
Thanks!
Post by: Mate2425 on November 11, 2017, 11:14:29 am
《 = integral, 1 = b , 0=a
《^1 0 x^0.5 dx using 5 subintervals
Post by: RuiAce on November 11, 2017, 11:35:31 am
Hi could i please get some help with this question (trapezoidal rule) thanks 😃
《 = integral, 1 = b , 0=a
《^1 0 x^0.5 dx using 5 subintervals
Post by: owidjaja on November 11, 2017, 09:21:54 pm
I need help with the following question:
Find any stationary points on the curve y=(4x^2-1)^4 and determine their nature.
Thanks!
Post by: RuiAce on November 11, 2017, 10:21:02 pm
Hey guys,
I need help with the following question:
Find any stationary points on the curve y=(4x^2-1)^4 and determine their nature.
Thanks!
Post by: owidjaja on November 11, 2017, 10:39:33 pm
Wait, where did the x= -1/4, 1/4 come from?
Post by: RuiAce on November 11, 2017, 10:59:38 pm
Wait, where did the x= -1/4, 1/4 come from?From the fact that we needed to test both sides of 0 to determine the nature of the stationary point.
Note that choices such as -1/3 could've replaced -1/4, and 2/5 could've replaced 1/4. The only important thing was that they didn't go too far and exceeded 1/2 (or went below -1/2), which were the other stationary points.
Post by: owidjaja on November 11, 2017, 11:04:58 pm
The curve y=ax^3+bx^2-x+5 has a point of inflexion at (1,-2). Find the values of a and b.
Thanks!
Post by: jamonwindeyer on November 12, 2017, 10:48:07 am
Hey, not entirely sure how to do this question as well:
The curve y=ax^3+bx^2-x+5 has a point of inflexion at (1,-2). Find the values of a and b.
Thanks!
Hey! The second derivative of that function is:
Now we know there is a point of inflexion at (1,-2), so substituting \(x=1\) should give a second derivative of zero:
And we also know (1,-2) must lie on the curve, so we can substitute those into the original curve:
Solve (1) and (2) simultaneously ;D
Post by: taliakaragiannis8 on November 16, 2017, 03:59:51 pm
(a) $6311.48
how do i do this ?
Post by: RuiAce on November 16, 2017, 05:29:49 pm
Jude invested $4500 five years ago at x% p.a. Evaluate x if Jude now has an amount of
(a) $6311.48
how do i do this ?
Post by: taliakaragiannis8 on November 19, 2017, 01:50:00 pm
3 months, while Sydney Bank offers loans at 7% p.a. Compare these loans on an amount of $5000 over 3 years and state which bank offers the best loan and why.
could I please get some help with this question ? thank you
Post by: RuiAce on November 19, 2017, 04:04:26 pm
But it's maths in focus.
Remark: I just used 7/1200 to avoid errors in rounding.
_________________________________________________________
Now, maths in focus uses 'interest free' period in the question, which is not terminology you're expected to know (and thus very unfair). An interest free-period does NOT mean that interest is not charged for the first three months, but rather that you don't PAY in the first three months. i.e. you're still being charged interest; but it sits there for three months and that's it.
So upon comparison, the second choice is better.
Post by: gilliesb18 on November 22, 2017, 12:27:52 pm
I just need help on a (probably fairly easy) question;
The curve y= x^5 + mx^3 - 7x +5 has a stationary point at x= -2. Find the value of m.
Thanks!
Post by: RuiAce on November 22, 2017, 12:35:09 pm
Hello!!
I just need help on a (probably fairly easy) question;
The curve y= x^5 + mx^3 - 7x +5 has a stationary point at x= -2. Find the value of m.
Thanks!
Post by: Dragomistress on November 23, 2017, 09:51:03 pm
I would like some assistance with this question.
Post by: Eric11267 on November 23, 2017, 10:05:58 pm
Hello!i) you have two equations in the form of y=
I would like some assistance with this question.
so to find the point of intersection equate the two equations so that
x2+4=x+k
then x2-x+4+k=0
ii) The equation from part 1 is given in the form of a quadratic, so to get one point of intersection the discriminant needs to be 0
In this case the discriminant is
1-4(4-k)=
4k-15=0
so k=15/4
iii) Now the equation is
x2-x+4-15/4=0
putting this into the quadratic formula gives you the x coordinate of P as 1/2
then putting 1/2 into either equation will give the y value as 17/4
so the coordinates of P are (1/2,17/4)
Post by: RuiAce on November 23, 2017, 10:08:28 pm
Hello!
I would like some assistance with this question.
_____________________________________________________________
_____________________________________________________________
Post by: owidjaja on November 28, 2017, 08:16:54 pm
Not so much of me needing help with a specific question, but tomorrow's my 2U exam (topics covering quadratic function, locus and geometric application). Any last minute tips or anything I should do before my exam?
Thanks!
Post by: mary123987 on November 28, 2017, 08:43:13 pm
Hey guys,Hey firstly all the best for tomorrow ! i think at this point just chill familiarise yourself with the formula sheet and any formulas that mightnt be on it . Next thing is at this point you should really be focussing your weakness if you think hey i am actually great at quadratic function then skip over to the other topics and vice versa.
Not so much of me needing help with a specific question, but tomorrow's my 2U exam (topics covering quadratic function, locus and geometric application). Any last minute tips or anything I should do before my exam?
Thanks!
If your mind is wondering and you feel like oh i dont feel ready even though you are do some harder questions , if your up for it past HSC questions.
But overall honstly just chill now the most important thing before a maths exam is SLEEP!
All the best , your gonna smash it :)
Post by: nhlj143 on November 29, 2017, 03:37:40 pm
Post by: RuiAce on November 29, 2017, 06:18:15 pm
If you have any time to spare please help our dying brain. If you can please help us with all but If not just the ones circled in red please. if you thank you so so so so bery much :-[ :'(There is no limit to the amount of questions you can ask but when you have multiple questions please consider spreading them out in the future. It can be very off-putting seeing a whole pile of questions all at once, as opposed to having them appear one at a time.
Q5 of the first paper is quite unsuitable to the 2U course, as the algebraic approach requires far too much handwork. The geometric approach would be to write down the answer by inspection.
i) The perpendicular bisector of the interval joining the points A and B (which is a line)
ii) The line basically halfway between them
Due to how similar they are, and the fact similar questions should've been covered in class, please provide any thought process for the leftovers.
________________________________________________________________
Note: The question was clearly wrong to have dropped \(AM^2\), because we don't know that \(AM = 1\).
Post by: nhlj143 on November 30, 2017, 12:33:57 am
Post by: jamonwindeyer on November 30, 2017, 12:44:36 am
Thank you, and sorry If it looked so confronting. We panicked because there is a test today
Good luck with the test!! Fingers crossed you smash it ;D
Post by: Never.Give.Up on December 05, 2017, 07:21:14 am
i was just wondering whether anyone could help me with how to use desmos...
we have to graph the following equation:
y= -1/3125 (x-250)^2 +50
What other points do I have to find/how do I find other points to make this graph work??
Thanks so much for your help :) :D
Post by: Opengangs on December 05, 2017, 07:44:36 am
hi,Find the x and y intercepts.
i was just wondering whether anyone could help me with how to use desmos...
we have to graph the following equation:
y= -1/3125 (x-250)^2 +50
What other points do I have to find/how do I find other points to make this graph work??
Thanks so much for your help :) :D
x-intercept occurs when y = 0 (that is, -1/3125 (x - 250)^2 + 50 = 0
-x^2 + 500x - 62500 + 156250 = 0
x^2 - 500x - 93750 = 0
Use the quadratic formula to find where the x intercepts are.
y-intercept occurs when x = 0
y = -1/3125 (-250)^2+50 = 30
Find stationary points.
Take the first derivative.
dy/dx = -2/3125 * (x - 250) = (500 - 2x)/3125
Let dy/dx = 0. 500 - 2x = 0
2x = 500
x = 250
Since it's a concave down parabola, we know at (250, 50), the parabola will be a maximum value. Thus, the range becomes: y <= 50
From these points, we can graph the parabola.
You can see this graph in action here.
Post by: itssona on December 07, 2017, 05:27:47 pm
2. the sum of two numbers is 6. find the numbers if the sum of their cubes is a minimum
apparently the approach is to differentiate, for maxima/minima problems? how do I do this?
thank youu ;D
Post by: RuiAce on December 07, 2017, 06:24:30 pm
Just do the same old thing from here; differentiate, set derivative = 0 and test that it's a minimum. You'll find that \(x=3\) which consequently also implies \(y=3\).
Post by: itssona on December 07, 2017, 09:03:40 pm
A rectangular block, the length whose base is twice the width, has a total surface area of 300cm^2. Find the dimensions of the block if it is of maximum value
Post by: RuiAce on December 07, 2017, 09:17:10 pm
only got one equation for this but dunno how to get the other :/Did you mean maximum
A rectangular block, the length whose base is twice the width, has a total surface area of 300cm^2. Find the dimensions of the block if it is of maximum value
(Hints though: \(b = 2l\) and \(lb+bh+lh=300\))
Post by: itssona on December 07, 2017, 09:24:28 pm
Did you mean maximum value?ah forgot about using b=2l thanks ruii :)
(Hints though: \(b = 2l\) and \(lb+bh+lh=300\)) (Edited)
Post by: Dragomistress on December 09, 2017, 09:09:18 pm
Post by: Shadowxo on December 09, 2017, 09:35:02 pm
Hello hello!Hey :)
So we want to find the minimum cost for the trip, ie find Total Cost in terms of V (average speed) then find where d(Total Cost)/dV =0
c is the cost in cents per hour, so TC (Total Cost) = c*t (cost per hour multiplied by the number of hours)
We know distance travelled = velocity * time, so
1000=V*t
t=1000/V
Substitute that in and we get
From there you should be able to find where the derivative = 0 (you should find V = 86.6 rounded to 87)
Post by: itssona on December 11, 2017, 07:59:02 pm
the answer is 17/42
the question is:
students studying at least one of the languages, french and Japanese, attend a meeting. Of the 28 students present, 18 study french and 22 study Japanese.
(i)whats the probability that two randomly chosen students both study french?
(ii) what is the probability that a randomly chosen student studies both languages?
thank you :)
Post by: RuiAce on December 11, 2017, 08:07:01 pm
wondering why the answer to this isn't (9/14 x 9/14)
the answer is 17/42
the question is:
students studying at least one of the languages, french and Japanese, attend a meeting. Of the 28 students present, 18 study french and 22 study Japanese.
(i)whats the probability that two randomly chosen students both study french?
(ii) what is the probability that a randomly chosen student studies both languages?
thank you :)
Post by: itssona on December 11, 2017, 08:58:20 pm
david has invented a game in which he throws two die repeatedly until the sum of the two numbers shown is either 7 or 9. If the sum is 9, david wins. If the sum is 7, he loses. If it's another number, he continues to throw until it's a 7 or 9.
(I) probability he wins on first throw is 1/9 (I found out)
(ii) prob that a second throw is needed is 26/36 (I found out)
(iii) what is the probability tgat david wins on his first, second or third throw? leave answer in unsimplified form
(iv) probability that david wins the game?
Post by: RuiAce on December 11, 2017, 09:08:26 pm
got another
david has invented a game in which he throws two die repeatedly until the sum of the two numbers shown is either 7 or 9. If the sum is 9, david wins. If the sum is 7, he loses. If it's another number, he continues to throw until it's a 7 or 9.
(I) probability he wins on first throw is 1/9 (I found out)
(ii) prob that a second throw is needed is 26/36 (I found out)
(iii) what is the probability tgat david wins on his first, second or third throw? leave answer in unsimplified form
(iv) probability that david wins the game?
____________________________________________________________________________________
Post by: itssona on December 11, 2017, 10:33:45 pm
A box contains
twelve chocolates all of exactly the same appearance. Four of the chocolates are hard and eight are soft. Kim eats three chocolates chosen randomly from the box.
Find probability that:
I) the first chocolate kim eats is hard
ii) kim eats three hard chocolates
iii) kim eats exactly one hard chocolate
thank youuu
Post by: RuiAce on December 11, 2017, 10:37:32 pm
stumped for part iii in hereHint: There's more than one way in which he can eat exactly 1 hard chocolate, since he could eat the hard chocolate first, second or third.
A box contains
twelve chocolates all of exactly the same appearance. Four of the chocolates are hard and eight are soft. Kim eats three chocolates chosen randomly from the box.
Find probability that:
I) the first chocolate kim eats is hard
ii) kim eats three hard chocolates
iii) kim eats exactly one hard chocolate
thank youuu
Post by: Calley123 on December 14, 2017, 06:56:31 pm
showing any stationary points and inflexions.
I know how to do the question but i'm not getting all the points
Answers: its a graph answer -Something at ( 1,2/e) )-got this one AND ( 3, 10/e^3) ---not getting this answer
-thanks
Post by: RuiAce on December 14, 2017, 07:15:09 pm
Sketch the curve y= (x^2+1)/e^2
showing any stationary points and inflexions.
I know how to do the question but i'm not getting all the points
Answers: its a graph answer -Something at ( 1,2/e) )-got this one AND ( 3, 10/e^3) ---not getting this answer
-thanks
Post by: Calley123 on December 14, 2017, 07:37:45 pm
Sorry !!
- meant to say e^x at the bottom.
Post by: RuiAce on December 14, 2017, 07:40:41 pm
Sorry !!I had a quick look on GeoGebra and for \(y=\frac{x^2+1}{e^x} \) there appears to be a point of inflexion at \(\left(3,\frac{10}{e^3}\right)\). This comes out from setting \( \frac{d^2y}{dx^2}=0\).
- meant to say e^x at the bottom.
Personally, I reckon using the quotient rule on that is disgusting and it'd be better to differentiate with \( y=e^{-x}(x^2+1)\). (At least, using the quotient rule the second time will be disgusting.) If that doesn't work, tell me what your \(y^{\prime\prime}\) was and I'll investigate
Note: The one at \( \left(1,\frac{2}{e}\right) \) is a horizontal point of inflexion, and thus satisfies both \(\frac{dy}{dx}=0\) and \(\frac{d^2y}{dx^2}=0\).
Post by: Calley123 on December 14, 2017, 08:39:10 pm
I don't understand why I couldn't have gotten both points through the first derivative ( only gave me x=1) though.
- Thanks :)
Post by: RuiAce on December 14, 2017, 09:25:43 pm
I got both the points using the 2nd derivative !!! So both points are inflexions and there is no stationary points ??
I don't understand why I couldn't have gotten both points through the first derivative ( only gave me x=1) though.
- Thanks :)
__________________________________________________
Of course, in saying all that stuff about points of inflexion you do have to test both sides of \(\frac{d^2y}{dx^2}\). But I'll just assume you already knew that.
Post by: Calley123 on December 14, 2017, 10:15:11 pm
__________________________________________________
Of course, in saying all that stuff about points of inflexion you do have to test both sides of \(\frac{d^2y}{dx^2}\). But I'll just assume you already knew that.
Thank you!!!
Post by: gilliesb18 on December 15, 2017, 09:55:40 am
Just need help on maxima/minima, stationary points question:
The formula for the surface area of a cylinder is given by S=2πr(r+h). Show that if the cylinder holds a volume of 54πm^3, the surface area is given by the equation S=2πr^2 + 108π/r. Hence find the radius that gives the minimum surface area.
Thanks heaps!
Post by: Natasha.97 on December 15, 2017, 10:43:03 am
Helloo,
Just need help on maxima/minima, stationary points question:
The formula for the surface area of a cylinder is given by S=2πr(r+h). Show that if the cylinder holds a volume of 54πm^3, the surface area is given by the equation S=2πr^2 + 108π/r. Hence find the radius that gives the minimum surface area.
Thanks heaps!
Hi!
a) We know that V = πr2h. Making h the subject will give you 54/r2. Substituting h back into S=2πr(r+h) will give you S=2πr2 + 108π/r.
Not 100% sure on this:
b) We've already taken into account the given volume in part a. Deriving S will give 4πr - 108π/r2. Equating it to 0 and simplifying will result in a radius of 3m.
Hope this helps :)
Post by: gilliesb18 on December 15, 2017, 10:47:46 am
Yes that makes sense. And I just checked the back of the book, and 3m is the right answer.
Thanks heaps!!
Post by: Dragomistress on December 16, 2017, 09:28:46 pm
Many thanks.
Post by: RuiAce on December 16, 2017, 09:42:32 pm
Hello!
Many thanks.
(https://i.imgur.com/ZEvz2Em.png)
Post by: Mate2425 on December 20, 2017, 09:02:44 am
Find the Exact area enclosed between the curve y =e^2x and the lines y=1 and x=2
Answer: 0.5 (e^4 -5) units squared
Thank you :)
Post by: Natasha.97 on December 20, 2017, 09:46:25 am
Hi could i please get some help with this question i don't know how to get the five in the final answer.
Find the Exact area enclosed between the curve y =e^2x and the lines y=1 and x=2
Answer: 0.5 (e^4 -5) units squared
Thank you :)
Hi!
Hope this helps :)
Post by: mxrylyn on December 21, 2017, 12:14:37 pm
On stationary point. On this worksheet, I am fine doing Q1 I, II, III, and V.
but for Q1 IV , i tried to differentiate Y and ended up with f'(x) = 3x^2 + 6x. and i equated that to 0, but i have no idea what to do with it next.
On Q1 VI, I ended up with X=12 and x=-6, which I know is wrong because the values are in none of the possible answers.
What should I do next for Q1 IV, and how do i know where i am going wrong in Q1 VI?
Post by: Opengangs on December 21, 2017, 12:21:23 pm
Hello.Q1 (iv)
On stationary point. On this worksheet, I am fine doing Q1 I, II, III, and V.
but for Q1 IV , i tried to differentiate Y and ended up with f'(x) = 3x^2 + 6x. and i equated that to 0, but i have no idea what to do with it next.
On Q1 VI, I ended up with X=12 and x=-6, which I know is wrong because the values are in none of the possible answers.
What should I do next for Q1 IV, and how do i know where i am going wrong in Q1 VI?
Q1 (vi)
Rinse and repeat from above.
Post by: mxrylyn on December 21, 2017, 01:48:02 pm
Q1 (iv)
Q1 (vi)
Rinse and repeat from above.
THANK YOU!
It all makes sense now
Post by: mxrylyn on December 21, 2017, 03:12:14 pm
Post by: Opengangs on December 21, 2017, 03:20:01 pm
Hey, not quite sure what method to use to do Q2, I tried to use substitution, but I confused myself and kept going in circles.Hello, mxrylyn.
If a curve has a stationary point, as in Q1, we need to find when the tangent line is equal to zero. This simply means that we set the derivative to zero.
So, at x = 1, the derivative is equal to zero.
We also know that the cubic curve passes through the point (1, -4). We use these two equations to form two equations which can then be solved simultaneously.
Thus, a is 4 and b is -12.
See if you can do (ii) based on (i).
If you have any more troubles, feel free to reply on this thread.
Post by: Dragomistress on December 30, 2017, 03:51:46 pm
Post by: RuiAce on December 30, 2017, 03:59:34 pm
How do I do f?
\begin{align*}\log_7 14 &= \log_7 2 + \log_7 7\\ &= 0.36 + 1\\ &= 1.36\end{align*}
Post by: swordkillz on January 02, 2018, 03:39:57 pm
Hey Guys, can someone please help me with this question?
A box of rectangular cross-section sits on a train luggage rack as shown with the point C touching the wall. P, the point in contact with the rack, is the midpoint of AD. If D is 8 cm from the wall and P, the edge of the rack, is 20 cm from the wall, find how far A is from the wall.
Hint: Draw parallels to the wall through A, P and D.
Isn't that a question from our 2017 trial paper?
Post by: RuiAce on January 02, 2018, 11:58:13 pm
Isn't that a question from our 2017 trial paper?P.S. It is possible, given the date it was posted
Post by: Dragomistress on January 05, 2018, 03:10:13 pm
Post by: brightsky on January 05, 2018, 03:44:00 pm
Not too sure what exactly is meant by 'minor segment cut off by the angle' in part b.
If they are asking for area of minor segment cut off by the chord (i.e. the region bounded by the chord and the minor arc intercepted by the chord), then all you need to do is figure out the area of the minor sector and subtract the area of the triangle:
Area of Minor Segment = 45/360*pi*[8/sqrt(2-sqrt(2))]^2 - 1/2*[8/sqrt(2-sqrt(2))][8/sqrt(2-sqrt(2))] sin(45 degrees) = 4.3 units^2 (to 1 decimal place)
If they are asking for area of minor sector cut off by the angle (i.e. the region bounded by the two radii and the minor arc), then all you need to do is find the area of the minor sector by taking the appropriate fraction of entire circle:
Area of Minor Sector = 45/360*pi*[8/sqrt(2-sqrt(2))]^2 = 42.9 units^2 (to 1 decimal place)
Post by: RuiAce on January 17, 2018, 07:39:40 pm
Hey could anyone answer these questions? (any help appreciated)
1) 1) Rotate about the y-axis the region bounded by the curve y = loge(x-2), y=0 and y=h to create a bowl. Find the exact volume of the bowl.
2) The portion of the curve y = ex -1 from x=0 to x=1 is rotated about the x=axis. Find the volume of revolution generated
3) Calculate the exact area of the region bounded by the curve y = e2x, y axis and the line y=e4
Before we dive into questions 1 and 3, have you drawn a diagram for both of them? You will not see what to do by purely relying on the given information itself.
Post by: hsotuhsa on January 17, 2018, 07:56:35 pm
And yes, I have drawn up graphs for question 1 and 3
Post by: RuiAce on January 17, 2018, 08:03:09 pm
Possible method
But of course, you don't know how to integrate log in 2U. So you need an alternate approach.
Note: The point \((2,e^4)\) actually lies on the curve \(y=e^{2x}\).
Note: \(e^4\) is a constant. Therefore its primitive is \(e^4 x\)
_____________________________________________________________________________
Post by: beeangkah on January 19, 2018, 05:44:47 pm
Any help would be appreciated!
Post by: RuiAce on January 19, 2018, 05:48:33 pm
Post by: beeangkah on January 19, 2018, 05:57:21 pm
Thanks! But the answers at the back say
Post by: RuiAce on January 19, 2018, 06:03:53 pm
Other than that it's the same. They just factorised: \(e^6-e^2 = e^2(e^4-1)\)
Post by: brooksykait on January 21, 2018, 02:30:59 pm
A circle has centre C(-1,3) and radius 5 units
a) Find the equation of the circle
b) The line 3x - y +1 = 0 meets the circle at 2 points, find their coordinates
c) Let the coordinates be X and Y, where Y is the coordinate directly below centre C. Find the coordinates of point Z, where YZ is a diameter of the circle
d) Hence show ZXY = 90 degrees
Specifcally parts c) and d), a) and b) I'm ok with.
Thank youuuu
Post by: RuiAce on January 21, 2018, 02:58:11 pm
Hey Jake, I have a question.The wording of this question is extremely ambiguous, as the way part c) is written does not make sense. Following some backtracking, it was presumed that X and Y are the coordinates of the points of intersection found in part b), with the remainder information suggesting that Y must be the point (-1,-2) (because it is directly below the circle's center, C).
A circle has centre C(-1,3) and radius 5 units
a) Find the equation of the circle
b) The line 3x - y +1 = 0 meets the circle at 2 points, find their coordinates
c) Let the coordinates be X and Y, where Y is the coordinate directly below centre C. Find the coordinates of point Z, where YZ is a diameter of the circle
d) Hence show ZXY = 90 degrees
Specifcally parts c) and d), a) and b) I'm ok with.
Thank youuuu
Assuming no typo was placed, please state the source of this question for future reference.
Making the assumption I suggested above
Essentially, because \(Y\) is directly below the circle's centre, the diameter of the circle through \(Y\) must pass through the point on the circle, but directly above the circle's centre. This can be found by just noting that the circle's radius has length \(5\), so its diameter has length \(10\).
Post by: as111 on January 21, 2018, 06:37:21 pm
Post by: RuiAce on January 21, 2018, 06:45:34 pm
Hi, Could somebody please help me with this question?
Post by: as111 on January 21, 2018, 06:58:19 pm
Post by: angela23 on January 21, 2018, 07:48:52 pm
A point P(x, y) moves so that the sum of the squares of its distance from each of the points A(-1, 0) and B(3, 0) is equal to 40.
Show that the locus of P(x, y) is a circle, and state its radius and centre.
Thanks!
Post by: RuiAce on January 21, 2018, 08:04:59 pm
Hi! I'm struggling with this locus question:
A point P(x, y) moves so that the sum of the squares of its distance from each of the points A(-1, 0) and B(3, 0) is equal to 40.
Show that the locus of P(x, y) is a circle, and state its radius and centre.
Thanks!
Post by: gumscape on January 22, 2018, 02:47:47 pm
Post by: RuiAce on January 22, 2018, 02:59:31 pm
hellooooooo! can anyone give me a general guideline/tips & tricks when it comes to sketching the primitive function? I'm having a hard time with it :(Think about how you sketched the derivative first. You converted the gradient of the tangent, into the y-coordinate.
For the primitive, you're doing the exact opposite. So if the curve is above the x-axis, then its primitive will be increasing. If the curve is below the x-axis, then its primitive will be decreasing. (And of course if you have an x-intercept then the primitive has a stationary point).
Remember that in general, a primitive function can be as high or as low as you want it to be (thanks to the +C you get from integrating). So unless they specify something about a point that the primitive passes through, you have flexibility over this.
Post by: BridgitteH on January 23, 2018, 06:08:50 pm
"The sum of the first 5 terms of an arithmetic series is 100 and the sum of the first 10 terms is 320. Find the first term and the common difference."
Post by: RuiAce on January 23, 2018, 06:14:36 pm
Hi! Could someone please help with this question? I haven't touched maths since the end of last term and forgot how to do it (whoops!)
"The sum of the first 5 terms of an arithmetic series is 100 and the sum of the first 10 terms is 320. Find the first term and the common difference."
The rest should be easy
Post by: BridgitteH on January 23, 2018, 06:23:26 pm
The rest should be easy
Thank you! Now I get it :))))
Post by: angela23 on January 24, 2018, 10:11:32 am
For which angles in the domain 0 ≤ x ≤ 2π does the series 1 + cos²x + cos⁴x +... not converge?
Thanks heaps!
Post by: RuiAce on January 24, 2018, 10:14:00 am
Hi! Could someone help me with this sequences and series question:
For which angles in the domain 0 ≤ x ≤ 2π does the series 1 + cos²x + cos⁴x +... not converge?
Thanks heaps!
Edit: Mistake halfway through
____________________________________________________
This is essentially because the function \( f(x) = \cos x\) has range \(-1 \le y \le 1\), which is the same thing as \( y\le 1\).
Post by: kauac on January 24, 2018, 10:52:11 am
My holidays have turned my maths brain to mush... :-[
I keep getting stuck on questions like this:
Find the points of inflexion for f(x) = 3x^5 - 10x^3 + 7
I can't quite work out how to manipulate the equation after finding the second derivative... So far I haven't had any luck with factorisation or anything else...
Any help greatly appreciated! ;D
Post by: RuiAce on January 24, 2018, 10:57:08 am
Hi...
My holidays have turned my maths brain to mush... :-[
I keep getting stuck on questions like this:
Find the points of inflexion for f(x) = 3x^5 - 10x^3 + 7
I can't quite work out how to manipulate the equation after finding the second derivative... So far I haven't had any luck with factorisation or anything else...
Any help greatly appreciated! ;D
Testing a bit to both sides we see that all of them are points of inflexion.
Post by: kauac on January 24, 2018, 11:02:29 am
Testing a bit to both sides we see that all of them are points of inflexion.
Thanks so much!
I think I was on the right track with the factorisation, but now I can understand how to approach these kinds of questions. Thanks again. :D
Post by: vic321 on January 26, 2018, 06:36:13 pm
Simplify:
sin(pi -x)
Thanks
Post by: RuiAce on January 26, 2018, 06:38:51 pm
How do you do thisThis is just \( \sin x \). Recall your ASTC identities from prelim: \( \sin (180^\circ - \theta) = \sin \theta\).
Simplify:
sin(pi -x)
Thanks
Post by: vic321 on January 26, 2018, 06:40:54 pm
This is just \( \sin x \). Recall your ASTC identities from prelim: \( \sin (180^\circ - \theta) = \sin \theta\).
ohh ok, thanks
Post by: Calley123 on January 28, 2018, 06:49:59 pm
I keep getting the wrong answer for this question !
Find the area enclosed by the curve y=x+e^-x, the x-axis and the lines x=o and y=2, correct to 2 decimal places
Thanks :)
Post by: RuiAce on January 28, 2018, 07:11:36 pm
Hey,This question lacks information. Ignoring that the sketch of \( y = x + e^{-x} \) is certainly not 2U material, there are two regions that could be implied by this information. One in the first quadrant, and one in the second quadrant.
I keep getting the wrong answer for this question !
Find the area enclosed by the curve y=x+e^-x, the x-axis and the lines x=o and y=2, correct to 2 decimal places
Thanks :)
Post by: Calley123 on January 28, 2018, 08:02:16 pm
This question lacks information. Ignoring that the sketch of \( y = x + e^{-x} \) is certainly not 2U material, there are two regions that could be implied by this information. One in the first quadrant, and one in the second quadrant.
This question is from the log and Exponential Functions, a two unit topic but I think its for 3U students...
Sorry
Post by: RuiAce on January 28, 2018, 08:05:47 pm
This question is from the log and Exponential Functions, a two unit topic but I think its for 3U students...Even if it's for 3U students that doesn't address the main problem.
Sorry
there are two regions that could be implied by this information.
Post by: gumscape on January 31, 2018, 01:21:09 pm
what are the equations of the vertical asymptotes of y=1/2sex(2x-1)
Thank you!!!
Post by: RuiAce on January 31, 2018, 01:29:48 pm
You might wanna fix your typo.
Note that the 1/2 in front does not change the asymptotes.
Post by: owidjaja on January 31, 2018, 08:52:15 pm
I've been kinda stuck with this question because I keep getting the same answer but it's different to the answers. Is there something I did wrong in my working out? (Still haven't figured out how to embed images so I'll try via Google Drive Link)
Thanks in advance!
Post by: RuiAce on January 31, 2018, 09:06:46 pm
Hey,Presumably everything else was just an error carry through, because I think the factorisation might be wrong. I’m on my phone right now but when I plugged into Wolfram I got -2x^2-5x+3=-(2x-1)(x+3)
I've been kinda stuck with this question because I keep getting the same answer but it's different to the answers. Is there something I did wrong in my working out? (Still haven't figured out how to embed images so I'll try via Google Drive Link)
Thanks in advance!
Post by: owidjaja on February 03, 2018, 10:58:36 pm
I need help with the following question:
Consider the function x^2=8y. Tangents are drawn at the points A(4,2) and B(-4,2) and intersect on the y-axis. Find the area bounded by the curve and the tangents.
Thanks in advance.
Post by: jamonwindeyer on February 04, 2018, 09:01:09 am
Hey guys,
I need help with the following question:
Consider the function x^2=8y. Tangents are drawn at the points A(4,2) and B(-4,2( and intersect on the y-axis. Find the area bounded by the curve and the tangents.
Thanks in advance.
Hey hey!! This is a big question, the first few steps laid out are:
- Rearrange the equation to make \(y\) the subject
- Differentiate the function
- Determine the gradients of the function at the two points of interest using the derivative
- Use these gradients and the point-gradient formula to find the equations of the two tangents
Equations are
At this point, you have the equation of the parabola and two lines. The area enclosed between them can be split in half, one from \(x=-4\) to \(x=0\) (the y-axis), and one from \(x=0\) to \(x=4\). You need to then find each of those using an integral! The first half, for example, would be:
And the second half:
Calculate those and add them up! The tricky bit about this is really just sequencing your actions, and then visualising how to divide that area you get as a result - Hopefully you are able to form a sketch because that helps heaps! I wanted to put one but Imgur is failing on me for some reason :P anyways, hopefully this helps add some clarity, but definitely skipped the bulk of it. If you let me know where you are having trouble specifically I can help some more! :)
Post by: Aaron12038488 on February 04, 2018, 08:02:03 pm
Cheers
Post by: owidjaja on February 05, 2018, 07:23:57 pm
So the question I need help in is: The parabola y= (x+2)^2 is rotated about the x-axis from x=0 to x=2. Find the volume of the solid formed.
I attached my working out because I'm not sure where I went wrong. The answer is supposed to be 992/5 π u^3.
Thanks in advance.
Post by: Sine on February 05, 2018, 07:30:52 pm
Hey guys,Hey owidjaja,
So the question I need help in is: The parabola y= (x+2)^2 is rotated about the x-axis from x=0 to x=2. Find the volume of the solid formed.
I attached my working out because I'm not sure where I went wrong. The answer is supposed to be 992/5 π u^3.
Thanks in advance.
you made an error i've done before ahah ;D. You didn't use the 0 when evaluting the intergral .maybe thinking it cancels down to 0? Anyway you should have a line of working (3rd line) saying pi[(4)^5/5 - (2)^5/5].
Hopefully this makes sense :D
Post by: jamonwindeyer on February 05, 2018, 07:37:18 pm
The limiting sum of a geometric series is 20. what series might this be? (Give 2 possible answers)
Cheers
Hey mate! If the limiting sum is 20:
Choose values of \(a\) and \(r\) that make this true! For example, \(a=10\) and \(r=0.5\) :)
Post by: RuiAce on February 08, 2018, 09:22:22 am
The limiting sum of a geometric series is 20. what series might this be? (Give 2 possible answers)Will just add a comment: This question is open ended, because there's actually an infinite number of possible answers.
Cheers
Perhaps the most convenient way of doing this is to use the formula (as stated above), and just put any random values of \(r\) into it (such that \( |r|<1 \) of course), and then solve for \(a\) in both cases.
Post by: Mate2425 on February 08, 2018, 09:11:26 pm
b) Sketch, showing any stationary points and inflexions:
Y = (log e x - 1 )^3
Post by: Opengangs on February 08, 2018, 09:28:19 pm
Hi could i please get some help with the question below regarding exponentials and logs and also with the setting out of the x and y intercepts thank you!Hey, Mate2425!
b) Sketch, showing any stationary points and inflexions:
Y = (log e x - 1 )^3
----
Post by: cnimm2000 on February 09, 2018, 04:02:25 pm
i was doing this series and applications hsc question:
2007 Q3
Can you please let me know if i have the correct answers pls.
thank you :)
i) Tn = 750 + 100 (n-1)
ii) 1650 m
iii) 12 000 m
iv) 20 days
Post by: RuiAce on February 09, 2018, 04:05:26 pm
Hi guys,Quick check on WolframAlpha shows that you're correct
i was doing this series and applications hsc question:
2007 Q3
Can you please let me know if i have the correct answers pls.
thank you :)
i) Tn = 750 + 100 (n-1)
ii) 1650 m
iii) 12 000 m
iv) 20 days
Post by: vic321 on February 09, 2018, 06:16:07 pm
Post by: RuiAce on February 09, 2018, 06:17:15 pm
Hi, how do you differentiate tan(x^0.5)? Thanks
Post by: Asereta on February 10, 2018, 09:52:30 am
Why is 0.65 * 3 different to 0.65*0.65*0.65?
Q12) Alan buys 4 tickets in a raffle in which 100 tickets are sold altogether. There are two prizes in the raffle. Find the probability that Alan will win
(c) 1 prize
For this question, I used replacement since that's have raffle usually work, where they take out the winning raffle. So my equation was (4/100*97/99) +(96/100*4/99) =193/2475 but the answer is 64/825.
(d) no prizes
I used replacement as usual and I got the equation 96/100*96/99 but the answer's 152/165
(e) at least 1 prize.
Since, I couldn't do d), I can't exactly do e)
13) Mary buys 20 tickets in a lottery that has 5000 tickets altogether. Find the probability that Mary will win
(c) neither first nor second prize.
With replacement since it's a lottery, I did (4980/5000 * 4980/4999) but the answer's 1239771/1249750.
Thank you!
Post by: RuiAce on February 10, 2018, 10:20:24 am
I have a few maths questions I don't get. Can you help? :-\\(0.65 \times 3\) is \( 0.65+0.65+0.65\)
Why is 0.65 * 3 different to 0.65*0.65*0.65?
Q12) Alan buys 4 tickets in a raffle in which 100 tickets are sold altogether. There are two prizes in the raffle. Find the probability that Alan will win
(c) 1 prize
For this question, I used replacement since that's have raffle usually work, where they take out the winning raffle. So my equation was (4/100*97/99) +(96/100*4/99) =193/2475 but the answer is 64/825.
(d) no prizes
I used replacement as usual and I got the equation 96/100*96/99 but the answer's 152/165
(e) at least 1 prize.
Since, I couldn't do d), I can't exactly do e)
13) Mary buys 20 tickets in a lottery that has 5000 tickets altogether. Find the probability that Mary will win
(c) neither first nor second prize.
With replacement since it's a lottery, I did (4980/5000 * 4980/4999) but the answer's 1239771/1249750.
Thank you!
\(0.65\times0.65\times0.65\) is \(0.65^3\)
_________________________________________________________
Be careful to make sure you know what replacement means. Replacement is when you draw something, and then put it back in there. If you do things with replacement, you denominator should not change.
Since your denominator changed, what you've done is that after you've drawn something, you don't put it back in there. This is without replacement. Which is correct, because it is a raffle, but even then you need to make sure you do not mix up your terminology.
You will find that this same issue is what caused your incorrect answers in 12c and 12d. For 12c, it would be \( \frac{4}{100}\times \frac{96}{99} + \frac{96}{100}\times \frac{4}{99} \)
Post by: owidjaja on February 10, 2018, 06:41:09 pm
I need help with question 15c i). Just a bit unsure with what limits to use and which pronumeral to integrate.
Thanks in advance.
Post by: RuiAce on February 10, 2018, 07:14:29 pm
Hey guys,After you sketch the semi-circle, you'll see that the limits should be \( -r \) and \(r\). But try to figure out why that's the case.
I need help with question 15c i). Just a bit unsure with what limits to use and which pronumeral to integrate.
Thanks in advance.
Here, \(r\) is a constant. Therefore you integrate w.r.t. \(x\)
Post by: owidjaja on February 10, 2018, 09:03:30 pm
I'm not entirely sure how to do question 16 a).
Thanks in advance.
Post by: Sine on February 10, 2018, 09:07:26 pm
Hey,It's the same methods as you would do otherwise for a volume of revolution. First sketch the graph in terms of "a" and "b". Then you can find the terminals required in terms of "a" and "b". Then use the relation in terms of "a" and "b" in the "volume of revolution formula" nothing changes from a question where you would have numbers in place of "a" and "b" you will just get an answer in terms of "a" and "b".
I'm not entirely sure how to do question 16 a).
Thanks in advance.
Post by: RuiAce on February 10, 2018, 09:08:08 pm
It's the same methods as you would do otherwise for a volume of revolution. First sketch the graph in terms of "a" and "b". Then you can find the terminals required. Then use the relation in terms of "a" and "b" in the "volume of revolution formula" nothing changes from a question where you would have numbers in place of "a" and "b" you will just get an answer in terms of "a" and "b".I think a part of the problem is that in 2U, they aren't taught the graph of an ellipse. (Nor 3U for that matter)
Post by: kauac on February 12, 2018, 06:24:17 pm
I have spent wayyyyy too long on question 24 and haven't really got anywhere... Any help greatly appreciated. ;D
Edit: I think the image is attached now :)
Post by: RuiAce on February 13, 2018, 08:31:38 pm
Hi...The image would actually be very helpful here. I have a feeling I know what's going on in this question but I'm not fully sure, so I don't want to dive straight into it.
I have spent wayyyyy too long on this question and haven't really got anywhere... (There was a diagram to go with it, but I haven't yet worked out how to attach an image to my posts). Any help greatly appreciated. ;D
Q: Grant is at a point A on one side of a 20m wide river and needs to get to point B on the other side 80m along the bank as shown. Grant swims to any point on the other bank and then runs along the side of the river to point B. If he can swim at 7km/h an run at 11km/h, find the distance he swims (x) to minimise the time taken to reach point B. (Answer to nearest metre).
If you have trouble with the direct method, then make an account on imgur, then upload a screenshot, click on it, and then copy and paste the BBC code that looks something like this
Code: [Select]
[img]some random url[/img]
Post by: gilliesb18 on February 14, 2018, 04:07:37 pm
Using the trapezoidal rule, find an approximation for;
\int_1^2 x^(2)dx using 1 subinterval
Is it possible to make sense of that integral??
Thanks!!
Post by: RuiAce on February 14, 2018, 04:12:32 pm
Hello! Just wondering if someone could help me with this question... its probably the most simple integration question but i cant find anyone to explain it to me!!!
Using the trapezoidal rule, find an approximation for;
\int_1^2 x^(2)dx using 1 subinterval
Is it possible to make sense of that integral??
Thanks!!
Post by: gilliesb18 on February 14, 2018, 04:46:35 pm
Post by: cocopops201 on February 14, 2018, 05:09:47 pm
I totally forgot how to.
Thank you for anyone who replies!
Post by: kauac on February 14, 2018, 05:33:03 pm
I need help with finding the radius of a circle: x^2+y^2-4x+8y+11=0
I totally forgot how to.
Thank you for anyone who replies!
Hi...
I think this might be how you do it...
Rearrange formula:
Then complete the square:
Plug into circle formula:
Therefore:
Hope this is helpful (and correct, of course)! :D
Post by: StupidProdigy on February 14, 2018, 05:47:10 pm
Hi...Don't forget to add 4 and 16 to the RHS when completing the square. R=sqrt(-11) doesn't sound right for this question. Instead we should have R=sqrt(-11+20)=3 :)
I think this might be how you do it...
Rearrange formula:
x^2 -4x + y^2 + 8y = -11
Then complete the square:
x^2 - 4x + 4 +y^2 +8y +16 = -11
Plug into circle formula:
(x - a)^2 + (y - b)^2 = r^2
(x - 2)^2 + (y + 4)^2 = -11
Therefore r = the square root of -11
Hope this is helpful (and correct, of course)! :D
Post by: kauac on February 14, 2018, 06:23:40 pm
Don't forget to add 4 and 16 to the RHS when completing the square. R=sqrt(-11) doesn't sound right for this question. Instead we should have R=sqrt(-11+20)=3 :)
Sorry, it's all fixed now. I had a feeling that I did something wrong, but couldn't work out what. Thanks for reminding me! :D
Post by: RuiAce on February 15, 2018, 10:02:13 am
Hi...Your thing's definitely up indeed :) sorry for the delay, I've only had enough time to properly concentrate on this question just now.
I have spent wayyyyy too long on question 24 and haven't really got anywhere... Any help greatly appreciated. ;D
Edit: I think the image is attached now :)
Deleted materials: Spotted the problem - the textbook rounded (gross) (again thanks Shadowxo)
What I've found is that ultimately you're trying to minimise \( t = \frac{x}{7} + \frac{80 - \sqrt{x^2-400}}{11} \) (thanks Shadowxo for confirmation). But when I checked the textbook, the final answer was completely different to what WolframAlpha gave me. Wolfram claims that the minimum is at \(x = \frac{55\sqrt2}{3}\) with a minimum value of \( \frac{40}{77}(14+3\sqrt2) \)
If you set this equal to 0 you'll get \(x = \frac{55\sqrt2}{3}\) as stated in the spoiler. Which rounds to 26.
Post by: gilliesb18 on February 15, 2018, 10:22:44 am
\int_1^3 x^(4)dx using 3 function values
Thanks heaps..
Post by: RuiAce on February 15, 2018, 10:35:42 am
Hello again... just need help on another integration one... this time using simpson's rule.\begin{align*}\int_1^3 x^4\,dx &\approx \frac{3-1}{6} \left(1^4 + 4(2^4) + 3^4 \right)\\ &= 38\end{align*}
\int_1^3 x^(4)dx using 3 function values
Thanks heaps..
Post by: gilliesb18 on February 15, 2018, 10:42:11 am
Post by: kauac on February 15, 2018, 04:32:26 pm
Your thing's definitely up indeed :) sorry for the delay, I've only had enough time to properly concentrate on this question just now.
If you set this equal to 0 you'll get \(x = \frac{55\sqrt2}{3}\) as stated in the spoiler. Which rounds to 26.
Thanks so much!! The wording of the question had me stumped, but your explanation is really clear. :D
Post by: Calley123 on February 15, 2018, 06:21:32 pm
How do I show that y=x^3-3x^2+27x-3 is monotonically increasing for all values of x?
THanks
Post by: Opengangs on February 15, 2018, 06:55:48 pm
HEy,Hey, Calley123!
How do I show that y=x^3-3x^2+27x-3 is monotonically increasing for all values of x?
THanks
To show that a function is monotonically increasing, we need to show that the derivative at any point for x is positive.
So, we need to show that f'(x) > 0 for all x
Since our derivative is always positive, then we have a monotonically increasing function. :)
Post by: Calley123 on February 15, 2018, 07:38:47 pm
Hey, Calley123!
To show that a function is monotonically increasing, we need to show that the derivative at any point for x is positive.
So, we need to show that f'(x) > 0 for all x
Since our derivative is always positive, then we have a monotonically increasing function. :)
Thank youu!
Post by: owidjaja on February 16, 2018, 10:17:30 pm
I need help with the following question.
Thanks in advance.
Post by: RuiAce on February 16, 2018, 10:38:41 pm
Hey guys,
I need help with the following question.
Thanks in advance.
Post by: brooksykait on February 19, 2018, 12:25:57 pm
Grant is at point A on one side
of a 20 m wide river and needs to
get to point B on the other side
80 m along the bank.
Grant swims to any point on the
other bank and then runs along
the side of the river to point B. If
he can swim at 7 km/h and run
at 11 km/h, find the distance he
swims (x) to minimise the time
taken to reach point B. Answer to
the nearest metre.
If somebody could please help me with this solution that would be great! thank you :)
Post by: jazzycab on February 19, 2018, 02:30:59 pm
Hey :). This question is from the Magaret Grove Extension Math Book, (chapter 2). I've been stuck on in for ages.
Grant is at point A on one side
of a 20 m wide river and needs to
get to point B on the other side
80 m along the bank.
Grant swims to any point on the
other bank and then runs along
the side of the river to point B. If
he can swim at 7 km/h and run
at 11 km/h, find the distance he
swims (x) to minimise the time
taken to reach point B. Answer to
the nearest metre.
If somebody could please help me with this solution that would be great! thank you :)
Note that I used speeds of 7000 m/hr and 11000 m/hr to ensure I was working with appropriate units, however, due to the fact that the thousandth is a factor of the derivative, it should work even if you use the units given.
Tip: Don't try and do this by specifying x as the run distance, use x as the swim distance as given in the question. It can be done, but the resultant quadratic is horrendous to solve algebraically.
Post by: kauac on February 19, 2018, 04:36:49 pm
When I was trying to work out how to find it, when the question gave one point where it intersects the parabola, I found the distance between this point and the focus, then multiplied it by the focal length (a) and got the right answer. Will this method work for all parabolas, or was this just a coincidence?
Post by: cocopops201 on February 19, 2018, 05:46:09 pm
So I'm currently stuck on this question.
I've figured out the x-coordinates: x=5 or -1
but i've totally forgotten how to find k the constant.
Thank you for anyone who replies!!
Post by: jazzycab on February 19, 2018, 06:10:07 pm
Hi guys,
So I'm currently stuck on this question.
I've figured out the x-coordinates: x=5 or -1
but i've totally forgotten how to find k the constant.
Thank you for anyone who replies!!
This is a discriminant question. The resultant quadratic obtained in the previous part of the question will have a discriminant equal to zero, because there is one solution to its equation
Post by: RuiAce on February 19, 2018, 06:40:00 pm
Is there a solid formula for finding a focal chord's length in a parabola (not a latus rectum)?Please send a screenshot, I got a bit lost reading this aha
When I was trying to work out how to find it, when the question gave one point where it intersects the parabola, I found the distance between this point and the focus, then multiplied it by the focal length (a) and got the right answer. Will this method work for all parabolas, or was this just a coincidence?
Note that I used speeds of 7000 m/hr and 11000 m/hr to ensure I was working with appropriate units, however, due to the fact that the thousandth is a factor of the derivative, it should work even if you use the units given.You don't need to worry about units in HSC maths. (Same cannot be said for HSC physics though.)
Tip: Don't try and do this by specifying x as the run distance, use x as the swim distance as given in the question. It can be done, but the resultant quadratic is horrendous to solve algebraically.
This question was done previously; the textbook answers just use the units provided.
Post by: jamonwindeyer on February 19, 2018, 06:40:25 pm
Is there a solid formula for finding a focal chord's length in a parabola (not a latus rectum)?
When I was trying to work out how to find it, when the question gave one point where it intersects the parabola, I found the distance between this point and the focus, then multiplied it by the focal length (a) and got the right answer. Will this method work for all parabolas, or was this just a coincidence?
Hey! As far as I know this would be a coincidence, I'm not aware of any theorem relating the distance from a point to a focus to the length of that focal chord! Would love to be corrected though - Do keep in mind that you'd likely need to show working in an exam situation anyway since such a formula isn't taught in the HSC :)
Post by: RuiAce on February 19, 2018, 06:42:30 pm
Hey! As far as I know this would be a coincidence, I'm not aware of any theorem relating the distance from a point to a focus to the length of that focal chord! Would love to be corrected though - Do keep in mind that you'd likely need to show working in an exam situation anyway since such a formula isn't taught in the HSC :)Oh, decided to draw it out after your comment. Ya, you're right, it's a coincidence.
Only thing that can be quoted is the focus-directrix definition \(PS = PM\). Nothing can really be said about the length of a chord
Post by: kauac on February 19, 2018, 07:12:50 pm
Post by: RuiAce on February 19, 2018, 07:28:42 pm
With that in mind, what is the best way in approaching questions such as part c?Tbh I would've just brute forced this. A clever way is possible with 3U techniques but I don't even think that the clever method is worthwhile.
Step 1. Find the equation of the line though \( (0,-3) \) and \( \left(2, \frac13\right) \)
Step 2. Determine where that line re-intersects with the parabola through simultaneous equations
Step 3. Plug into the distance formula.
Post by: kauac on February 19, 2018, 07:50:52 pm
Step 1. Find the equation of the line though \( (0,-3) \) and \( \left(2, \frac13\right) \)
Step 2. Determine where that line re-intersects with the parabola through simultaneous equations
Step 3. Plug into the distance formula.
Great! Thanks!! ;D
Post by: SophiePalmer26 on February 19, 2018, 08:40:26 pm
If anyone has an easy way to find the solution for these types of questions I would be grateful! ;D
Post by: slinkybench on February 19, 2018, 08:41:42 pm
Simplify:
1. a^3 + b^3 + a + b
2. x^3 - 1 / x^2 - 1
3. (3 / a - 2) - (3a / a^2 + 2a + 4)
Thanks in advance!
Post by: RuiAce on February 19, 2018, 08:46:59 pm
I'm struggling with this question: For what value of n is the sum of the arithmetic series 5+9+13+... equal to 152?Presumably \(n\) is the number of terms, or else this question makes no sense.
If anyone has an easy way to find the solution for these types of questions I would be grateful! ;D
\begin{align*}\frac{n}{2}[10 + 4(n-1)]&=152\\ n[5 + 2(n-1)]&=152\\ 2n^2 + 3n - 152 &= 0\\ n &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-152)}}{2(2)}\\ &= -\frac{19}{2}, 8\end{align*}
Post by: SophiePalmer26 on February 19, 2018, 08:50:50 pm
Presumably \(n\) is the number of terms, or else this question makes no sense.
\begin{align*}\frac{n}{2}[10 + 4(n-1)]&=152\\ n[5 + 2(n-1)]&=152\\ 2n^2 + 3n - 152 &= 0\\ n &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-152)}}{2(2)}\\ &= -\frac{19}{2}, 8\end{align*}
Thank you so much, that actually helps so much considering that i've never been good at maths :)
Post by: RuiAce on February 20, 2018, 09:23:58 am
A few difference of 2 cubes questions I was stuck on.Presumably you meant to factorise, as Q1 and Q2 are already fully simplified
Simplify:
1. a^3 + b^3 + a + b
2. x^3 - 1 / x^2 - 1
3. (3 / a - 2) - (3a / a^2 + 2a + 4)
Thanks in advance!
\begin{align*}a^3+b^3 + a+b &= (a+b)(a^2-ab+b^2)+(a+b)\\ &= (a+b)\left[(a^2-ab+b^2)+1 \right]\\ &= (a+b)(a^2-ab+b^2+1)\end{align*}
If the transition was not clear enough from line 1 to line 2
\begin{align*}(a+b)(a^2-ab+b^2)+(a+b) &= uv + u\\ &= u(v+1)\\ &= (a+b)(a^2-ab+b^2+1)\end{align*}
Post by: Calley123 on February 22, 2018, 07:28:24 pm
How do I find the shaded area of a parabola with x^2=4ay with a focus (0,5) and a line y=a ? ( answer 8a^2/3) )
Also how do volume when it is rotated around y-axis ? (Answer: 2 times pi times a^3 )
Thanks !!
Post by: gilliesb18 on February 22, 2018, 08:03:56 pm
I'm needing help(again) on an integration one...
this time its the sum of areas etc...
Qu. Find the area bounded by the curve y= 9-x^2 and the line y=5
Thanks so much ;D
Post by: jamonwindeyer on February 22, 2018, 08:43:47 pm
Hello:)
I'm needing help(again) on an integration one...
this time its the sum of areas etc...
Qu. Find the area bounded by the curve y= 9-x^2 and the line y=5
Thanks so much ;D
Hey! So you must draw a sketch for stuff like this, it makes things loads easier! If you sketch the parabola and the line, you'll see that the area we need is enclosed by the parabola on top, and the line on the bottom.
Remember, whenever we have an area between two functions, we just integrate the top curve minus the bottom curve:
But what are the limits? They tell us where the area starts and stops - We need the points of intersection of the parabola and the line. Equate them:
So, we need to evaluate the integral:
Reckon you might be able to take it from there? ;D this is a super important question to understand btw so if anything is confusing, ask! :)
Post by: jamonwindeyer on February 22, 2018, 08:48:55 pm
Hey,
How do I find the shaded area of a parabola with x^2=4ay with a focus (0,5) and a line y=a ? ( answer 8a^2/3) )
Also how do volume when it is rotated around y-axis ? (Answer: 2 times pi times a^3 )
Thanks !!
Hey! Do you mean the area enclosed between the two? Take a look at the post I've just made above, because it's the same logic!! In your case we have an unknown \(a\), but it's the same method, integrate top minus bottom. In your case, the line is on top! Rearrange the parabola equation to make \(y\) the subject and:
The limits for you are where the parabola cuts the line, same as above!
So those are your limits, you're evaluating:
Edit: Actually this question is a tad confusing, because we can evaluate \(a\) using the known focus, \(a=5\), so I'm not sure why \(a\) even appears in the answer!
Post by: Calley123 on February 23, 2018, 07:12:15 am
Hey! Do you mean the area enclosed between the two? Take a look at the post I've just made above, because it's the same logic!! In your case we have an unknown \(a\), but it's the same method, integrate top minus bottom. In your case, the line is on top! Rearrange the parabola equation to make \(y\) the subject and:
The limits for you are where the parabola cuts the line, same as above!
So those are your limits, you're evaluating:
Edit: Actually this question is a tad confusing, because we can evaluate \(a\) using the known focus, \(a=5\), so I'm not sure why \(a\) even appears in the answer!
Thank youuu :)
Post by: slinkybench on February 23, 2018, 05:44:48 pm
x^4 - x^2 - 2x -1
Post by: arii on February 23, 2018, 06:41:47 pm
I need some help with this factorisation question:
x^4 - x^2 - 2x -1
Hey there,
Please refer to the attachment for the solutions. I'm pretty sure that's all you can factorise (unless you are an Extension 2 Mathematics student).
Let me know if you don't understand anything.
Post by: gilliesb18 on February 26, 2018, 06:22:22 pm
Hey! So you must draw a sketch for stuff like this, it makes things loads easier! If you sketch the parabola and the line, you'll see that the area we need is enclosed by the parabola on top, and the line on the bottom.
Remember, whenever we have an area between two functions, we just integrate the top curve minus the bottom curve:
But what are the limits? They tell us where the area starts and stops - We need the points of intersection of the parabola and the line. Equate them:
So, we need to evaluate the integral:
Reckon you might be able to take it from there? ;D this is a super important question to understand btw so if anything is confusing, ask! :)
Ok great thanks heaps!!!
Post by: mirakhiralla on March 02, 2018, 08:07:33 pm
I have attempted this question but I dont have answers and I am pretty sure it is not correct, if it isn't, can you please show me how to do it?
Terry borrowed $20000 on 1 January 2008. He agreed that on 1 January in each succeeding year he would pay back $3000 and add 6%p.a. interest on the amount owing during the year just completed. Find:
a) the amount still owing after ! January 2013
b) the number of years needed to pay off the debt
My answers: a) $22181.51 b) 120.7 years
Post by: jamonwindeyer on March 03, 2018, 12:24:38 am
Hey! (:
I have attempted this question but I dont have answers and I am pretty sure it is not correct, if it isn't, can you please show me how to do it?
Terry borrowed $20000 on 1 January 2008. He agreed that on 1 January in each succeeding year he would pay back $3000 and add 6%p.a. interest on the amount owing during the year just completed. Find:
a) the amount still owing after ! January 2013
b) the number of years needed to pay off the debt
My answers: a) $22181.51 b) 120.7 years
Hey! Yeah not quite, but that's all good, let me show you!
So we start with $20,000, let that be \(A_0\). After 1 year (so, January 2009), we add 6% interest and then pay back $3000. That looks like this:
The next year, we take that amount, and do the same thing. Add 6%, subtract $3000:
If you expand, you'll get what I've got above! And if you do it again, you should get:
See the pattern? If you're just starting this topic it might look strange, but do a few of these and this is what they all look like, more or less. After 5 years (2013), we have:
Pop that in your calculator, I get $9853.23!! For Part (b), you need to instead consider a general version of the expression, after \(n\) years:
We need \(A_n=0\) -> See if you can manipulate that expression to find \(n\)! If you've never done a question like this before let me know and I'd be happy to show you, or perhaps read this guide which steps through it for you! Happy to help if anything above was confusing as well - I'm assuming you've seen something similar to it before but can definitely go slower if you haven't :)
Post by: kauac on March 03, 2018, 08:23:25 pm
Post by: RuiAce on March 03, 2018, 08:27:07 pm
How do you solve simultaneous equations when one of the equations is a cubic? Such as:You don't.and
What's the full question?
Post by: kauac on March 03, 2018, 08:29:30 pm
Post by: RuiAce on March 03, 2018, 08:34:53 pm
Find the area enclosed between y=x^3, x-axis and y= -3x+4?If that was it, then that is actually a very unfair question. You are NOT expected to solve general cubic equations in the HSC course, and it is unfair for them to make you guess that \(x=1\) is the solution (and hence \( (1,1) \) is the point of intersection.
Post by: kauac on March 03, 2018, 08:39:29 pm
If that was it, then that is actually a very unfair question. You are NOT expected to solve general cubic equations in the HSC course, and it is unfair for them to make you guess that \(x=1\) is the solution (and hence \( (1,1) \) is the point of intersection.
Intriguing... there are a few similar questions with cubic equations in the textbook exercise that I am working on...Is there a certain way to approach these types of questions?
Post by: RuiAce on March 03, 2018, 08:40:51 pm
Intriguing... there are a few similar questions with cubic equations in the textbook exercise that I am working on...Is there a certain way to approach these types of questions?Unless you wanna try guessing (which is fine), there really isn't. Which textbook is this?
Post by: kauac on March 03, 2018, 08:42:18 pm
Which textbook is this?
Maths in Focus - Margaret Grove
Post by: RuiAce on March 03, 2018, 08:47:20 pm
Maths in Focus - Margaret GroveQuestions like Q13 and Q6 in that exercise look fine to me. Occasionally, they can be solved because the expression should be easy to factorise.
However, that particular one disgusts me and just further lowered my opinion of the Maths in Focus textbook. It is definitely not expected out of a 2U student and it's beyond unfair to throw it at them.
(There are 3U cheats around it. But a 2U student is not expected to know of these.)
Post by: kauac on March 03, 2018, 08:50:10 pm
Questions like Q13 and Q6 in that exercise look fine to me. Occasionally, they can be solved because the expression should be easy to factorise.
However, that particular one disgusts me and just further lowered my opinion of the Maths in Focus textbook. It is definitely not expected out of a 2U student and it's beyond unfair to throw it at them.
(There are 3U cheats around it. But a 2U student is not expected to know of these.)
Alright... Thanks for looking into it! :)
Post by: jamonwindeyer on March 03, 2018, 09:16:51 pm
However, that particular one disgusts me
Rui is disgusted! :o
Post by: brooksykait on March 05, 2018, 11:49:23 am
Find the sum of all integers between 1 and 200 that are not multiples of 9
(from Magaret Grove textbook - Challenge Exercise 8)
i know this is supposed to be an easy question but i cannot think of a way to do this without literally counting the numbers up. any help would be appreciated :)
Post by: EEEEEEP on March 05, 2018, 11:53:26 am
helloooo, i’m having some difficulties, could someone please help?Hint:
Find the sum of all integers between 1 and 200 that are not multiples of 9
(from Magaret Grove textbook - Challenge Exercise 8)
i know this is supposed to be an easy question but i cannot think of a way to do this without literally counting the numbers up. any help would be appreciated :)
Try using pairs. For numbers between 1 and 200, 2 numbers at each ends will go together to make 200, e.g.
(I excluded 199 for now)
2, 198
3, 197
4, 196
5, 195
Once you get the right amount of pairs , you can do 200 * X (which denotes the number of pairs), minus and plus some other things =)
Post by: RuiAce on March 05, 2018, 12:00:36 pm
Hint:That's a bit overkill, especially with the whole "not multiples of 9" issue
Try using pairs. For numbers between 1 and 200, 2 numbers at each ends will go together to make 200, e.g.
(I excluded 199 for now)
2, 198
3, 197
4, 196
5, 195
Once you get the right amount of pairs , you can do 200 * X (which denotes the number of pairs), minus and plus some other things =)
helloooo, i’m having some difficulties, could someone please help?
Find the sum of all integers between 1 and 200 that are not multiples of 9
(from Magaret Grove textbook - Challenge Exercise 8)
i know this is supposed to be an easy question but i cannot think of a way to do this without literally counting the numbers up. any help would be appreciated :)
Post by: brooksykait on March 05, 2018, 12:13:21 pm
[/quote]
Thank you, that worked. Except I think your calculation was wrong. My answer was 17823, and so was the textbook. Doesn't matter but thank you ;D
Post by: RuiAce on March 05, 2018, 12:15:09 pm
I fixed it. I accidentally punched a + into the calculator instead of a -. Check updated post
Thank you, that worked. Except I think your calculation was wrong. My answer was 17823, and so was the textbook. Doesn't matter but thank you ;D
Post by: mirakhiralla on March 05, 2018, 12:15:09 pm
Hey! Yeah not quite, but that's all good, let me show you!
So we start with $20,000, let that be \(A_0\). After 1 year (so, January 2009), we add 6% interest and then pay back $3000. That looks like this:
The next year, we take that amount, and do the same thing. Add 6%, subtract $3000:
If you expand, you'll get what I've got above! And if you do it again, you should get:
See the pattern? If you're just starting this topic it might look strange, but do a few of these and this is what they all look like, more or less. After 5 years (2013), we have:
Pop that in your calculator, I get $9853.23!! For Part (b), you need to instead consider a general version of the expression, after \(n\) years:
We need \(A_n=0\) -> See if you can manipulate that expression to find \(n\)! If you've never done a question like this before let me know and I'd be happy to show you, or perhaps read this guide which steps through it for you! Happy to help if anything above was confusing as well - I'm assuming you've seen something similar to it before but can definitely go slower if you haven't :)
So I got a right after doing it again, thank you.
For b, do I use the sum formula and then log? I get the same answer but I feel like my sum is wrong...
Post by: RuiAce on March 05, 2018, 12:18:44 pm
So I got a right after doing it again, thank you.
For b, do I use the sum formula and then log? I get the same answer but I feel like my sum is wrong...
\begin{align*}20000(1.06)^n - 3000\left(\frac{1.06^n - 1}{0.06} \right)&= 0\\ 20000(1.06)^n -50000(1.06^n - 1)&=0 \tag{from calculator}\\ 20000(1.06)^n - 50000(1.06)^n + 50000 &= 0\\ 30000(1.06)^n &= 50000\\ 1.06^n &= \frac{5}{3}\\ n \log 1.06 &= \log \frac{5}{3}\\ n &= \frac{\log \frac53}{\log 1.06}\\ &\approx 8.767\end{align*}
Post by: gilliesb18 on March 06, 2018, 11:55:45 am
Just need help on a question; Find the equation of the normal to the curve, y=e^x at the point where x=3, in exact form.
I just can't remember how to do this!!
Any help is appreciated.....
Thanks heaps:):)
Post by: jazzycab on March 06, 2018, 01:51:27 pm
Hello,
Just need help on a question; Find the equation of the normal to the curve, y=e^x at the point where x=3, in exact form.
I just can't remember how to do this!!
Any help is appreciated.....
Thanks heaps:):)
The normal is perpendicular to the curve, so you can find it by first finding the gradient of the tangent at that point (by differentiating), then evaluating the gradient of the normal (the product of perpendicular gradients is -1).
You will also need the coordinates of a point on the line, but the line passes through \(y=e^x\) at \(x=3\), so it passes through \(\left(3,e^3\right)\)
Post by: gilliesb18 on March 06, 2018, 02:58:47 pm
Post by: mirakhiralla on March 12, 2018, 06:51:06 pm
Can someone help me out with this question
Thank you (:
Post by: RuiAce on March 12, 2018, 06:53:56 pm
Heyyy!The attachment is kinda small and becomes blurry when zoomed in. What was the source of the question?
Can someone help me out with this question
Thank you (:
(So I can locate it myself, to save you from reuploading)
Post by: mirakhiralla on March 12, 2018, 07:02:37 pm
I reuploaded the wrong question haahah
Post by: RuiAce on March 12, 2018, 07:08:21 pm
Note that we could've used \( y - y_1 = m(x-x_1) \) with either the point \( (0,4) \) or \( \left( -\frac23, 0 \right) \). But that was overkill for this question, since they gave us the \(y\)-intercept.
From here, just do the same thing again. Integrate to get an expression for \( f(x) \), and to solve for the second constant \( C_2\) just use the fact that the curve \(y=f(x)\) passes through (2,4)
Post by: Mate2425 on March 12, 2018, 11:36:46 pm
Q) Find the exact area bounded by the curves y = sinx and y=cos x in the domain 0 < = X <= 2 pie. (Where 'arrows' are signs greater than/equal and less than/equal.
Thank you ;)
Post by: Sine on March 13, 2018, 12:20:22 am
Hi Could i please have some help with the following question:
Q) Find the exact area bounded by the curves y = sinx and y=cos x in the domain 0 < = X <= 2 pie. (Where 'arrows' are signs greater than/equal and less than/equal.
Thank you ;)
Post by: Mate2425 on March 13, 2018, 07:31:11 am
Thank you :) :)
Post by: gilliesb18 on March 21, 2018, 02:13:26 pm
Ok so I have a few questions re logs n exponentials...
1. what is the derivative of log_(10)x? and..
2. Find the equation of the tangent to the curve y= log_(e)x at the point (2, log_e(2))
I've been working on derivatives of logs etc. but these have just confused me a tad!!
Thanks heaps...
Post by: Opengangs on March 21, 2018, 03:20:16 pm
Hello,
Ok so I have a few questions re logs n exponentials...
1. what is the derivative of log_(10)x? and..
2. Find the equation of the tangent to the curve y= log_(e)x at the point (2, log_e(2))
I've been working on derivatives of logs etc. but these have just confused me a tad!!
Thanks heaps...
Hopefully, this helps with the intuition behind logarithmic differentiation.
If you're still stuck on part (2), post all relevant working out and ask specifically where you're confused.
Post by: gilliesb18 on March 21, 2018, 03:54:09 pm
Ohhh ok rightio thanks.... I just forgot that change of base thingo....
Thanks heaps!!
Post by: Mate2425 on March 23, 2018, 08:20:41 am
A) Evaluate (2x^3 -x^2 +5x +3) ÷ x with definite integral between 1 and 2
B) Use trapezoidal rule with 4 strips to find the area bounded by the curve y=In (x^2 -1), x-axis and the lines x =3 and x=5.
Thank you very much :)
Post by: RuiAce on March 23, 2018, 08:45:58 am
Hi could please have some help with the following questions:I'll just start you off.
A) Evaluate (2x^3 -x^2 +5x +3) ÷ x with definite integral between 1 and 2
B) Use trapezoidal rule with 4 strips to find the area bounded by the curve y=In (x^2 -1), x-axis and the lines x =3 and x=5.
Thank you very much :)
__________________
where \(f(x) = \ln (x^2-1)\).
If you have further problems you should post up any progress you've made thus far.
Post by: LaraC on March 23, 2018, 11:37:01 am
Could someone please help me to differentiate loge(2x+4)(3x-1)??
I know I should be using function of a function rule, but have completely confused myself!!
Thanks!
Post by: jazzycab on March 23, 2018, 12:26:20 pm
Hello :)
Could someone please help me to differentiate loge(2x+4)(3x-1)??
I know I should be using function of a function rule, but have completely confused myself!!
Thanks!
Is your question to differentiate \(\log_e{\left(\left(2x+4\right)\left(3x-1\right)\right)}=\log_e{\left(6x^2+10x-4\right)}\) or \(\log_e{\left(2x+4\right)}\left(3x-1\right)=\left(3x-1\right)\times\log_e{\left(2x+4\right)}\)?
Given your comment at the end, assuming the former:
From here, just evaluate each derivative, multiply them together, then substitute \(u=6x^2+10x-4\) back in.
If it is the latter expression, you would need to use the product and chain rules.
Post by: LaraC on March 23, 2018, 12:30:29 pm
Post by: jazzycab on March 23, 2018, 03:09:30 pm
Oh sorry....how stupid of me! :-[ No it must be the product rule not the function of a function......the brackets are multiplied by each other and by loge.....they are not all in one big lot of brackets if that makes sense?
So is it \(\log_e{\left(2x+4\right)}\times\left(3x-1\right)\)?
If so, you'll need to apply the product rule where \(y=\left(3x-1\right)\log_e{\left(2x+4\right)}=u\times v\) where \(u=3x-1\) and \(v=\log_e{\left(2x+4\right)}\).
That is:
Post by: LaraC on March 23, 2018, 05:24:07 pm
Post by: LaraC on March 24, 2018, 07:14:49 pm
Umm yeah...sorry...another question from me!!
Can someone help me how to find the point of inflexion on the curve y=xlog_ex-x^2?
Thanks!
Post by: RuiAce on March 24, 2018, 07:31:05 pm
Hello :)
Umm yeah...sorry...another question from me!!
Can someone help me how to find the point of inflexion on the curve y=xlog_ex-x^2?
Thanks!
Post by: LaraC on March 24, 2018, 09:44:16 pm
I was thinking it was the first derivative =0 at point of inflexion.....but oops! Makes sense now...thanks!
Post by: Jada03 on March 26, 2018, 08:33:58 pm
I'm struggling a little with the question:
Find the equation of the tangent to the curve y=4^(x+1) (as in 4 to the power of x+1) at the point (0,4).
I'm stuck with how to find the differential of a constant raised to a more complex index such as (x+1) as opposed to just x by itself.
Thanks in advance!! :D
Post by: RuiAce on March 26, 2018, 09:47:37 pm
Hello :)
I'm struggling a little with the question:
Find the equation of the tangent to the curve y=4^(x+1) (as in 4 to the power of x+1) at the point (0,4).
I'm stuck with how to find the differential of a constant raised to a more complex index such as (x+1) as opposed to just x by itself.
Thanks in advance!! :D
(Although we don't quote it - this is how we do it)
\begin{align*}\frac{d}{dx} 4^{x+1} &= \frac{d}{dx} e^{\ln 4^{x+1}}\tag{formula above}\\ &= \frac{d}{dx} e^{(x+1) \ln 4} \tag{log law}\\ &= e^{(x+1)\ln 4} \ln 4\tag{chain rule}\\ &= 4^{x+1}\ln 4\tag{reversing everything}\end{align*}
Post by: gilliesb18 on March 27, 2018, 12:23:57 pm
Find the area enclosed between the curve y=x^3, the x-axis and the line y=-3x +4.
Me and my friend just can't seem to find where the two points intercept...
Any help??
thanks heaps
Post by: Opengangs on March 27, 2018, 12:32:22 pm
Hello, just need help on this question....Just a quick observation:
Find the area enclosed between the curve y=x^3, the x-axis and the line y=-3x +4.
Me and my friend just can't seem to find where the two points intercept...
Any help??
thanks heaps
Try substituting x = 1 and you'll see the point of intersection.
Post by: RuiAce on March 27, 2018, 12:46:30 pm
Hello, just need help on this question....Will make a small remark because I do remember seeing this in maths in focus. The fact they're expecting you to solve that question in 2U is unrealistic - this should not be a 2U question. So don't worry too much about it (or just take Opengangs' suggestion for granted to prove that there's an intercept at \(x=1\)).
Find the area enclosed between the curve y=x^3, the x-axis and the line y=-3x +4.
Me and my friend just can't seem to find where the two points intercept...
Any help??
thanks heaps
Post by: gilliesb18 on March 27, 2018, 01:32:27 pm
If you don't think it's necessary for 2U, I will just leave it....
Just btw- do you feel that 'Maths in Focus' covers everything? I just get a bit worried when I look at past papers that it doesn't!
Post by: RuiAce on March 27, 2018, 02:15:10 pm
Textbooks like Terry Lee’s are also good but involve generally much harder questions.
Post by: skisso on March 27, 2018, 07:02:33 pm
I'm trying to do this question in a HSC paper (1994), and I have the answer but I dont get whats happening...
Post by: RuiAce on March 27, 2018, 07:20:05 pm
Hey :)They simply used the fact that integrals and derivatives undo each other to go from \( xe^{x^2} = \frac12 \, \frac{d}{dx} e^{x^2} \) to \( \int_0^1 xe^{x^2}\,dx = \frac12 \left[e^{x^2} \right]_0^1 \). The whole point is that because \( x e^{x^2} \) is a derivative of \( \frac12 e^{x^2}\), consequently we must have \( \frac12 e^{x^2} \) being an antiderivative of \( x e^{x^2} \)
I'm trying to do this question in a HSC paper (1994), and I have the answer but I dont get whats happening...
Post by: whitney.dent on March 28, 2018, 02:08:06 pm
Find the volume of the solid formed when the line x+3y-1=0 is rotated around the x-axis from x=0 and x=8.
Thanks!!
Post by: NowYouTseMe on March 28, 2018, 03:17:37 pm
Quote
Hey, just needing some help on this question:Rearrange your line equation into the form y2=1/9(1-x)2 , then sub into the formula V= π∫ab y2dx to get the volume of the solid
Find the volume of the solid formed when the line x+3y-1=0 is rotated around the x-axis from x=0 and x=8.
Thanks!!
Post by: Alex Molloy on March 29, 2018, 09:11:07 am
(a) 2^x (b) 5^-x
Q2. Find a primitive of: (6x^2 - 8x + 6x).e^(x^3 -2x^2 +3x -5)
Q3. How do you sketch a primitive function given a picture of a graph? Help
Post by: jazzycab on March 29, 2018, 09:58:25 am
Q1. Use the identity a=e^logea, to express the following as a power of e, and hence find its primitive.
(a) 2^x (b) 5^-x
Q2. Find a primitive of: (6x^2 - 8x + 6x).e^(x^3 -2x^2 +3x -5)
Q3. How do you sketch a primitive function given a picture of a graph? Help
In Question 1, start with \(2=e^{\log_e{\left(2\right)}}\Rightarrow 2^x=\left(e^{\log_e{\left(2\right)}}\right)^x=e^{x\log_e{\left(2\right)}}\) then use the rule for the primitive of an exponential, \(\int{e^{ax}}dx=\frac{1}{a}e^{ax}+c,c\in\mathbb{R}.\)
For Question 2, note that \(\frac{d}{dx}\left(e^{x^3-2x^2+3x-5}\right)=\left(3x^2-4x+3\right)e^{x^3-2x^2+3x-5}=\frac{1}{2}\left(6x^2-8x+6\right)e^{x^3-2x^2+3x-5}\)
For your final question, it is often easier to try and think in reverse (i.e. the graph given looks as it does due to the fact that it is the graph of the derivative of the graph that you are trying to draw).
This means that:
- when the given graph is at zero (i.e. an x-intercept), the gradient of the primitive graph is zero (so a stationary point at the same x-value)
- when the given graph is greater than zero (i.e. above the x-axis), the gradient of the primitive graph is positive at the same x-values
- when the given graph is less than zero (i.e. below the x-axis), the gradient of the primitive graph is negative at the same x-values
- when the given graph is increasing, the gradient of the primitive graph is increasing
- when the given graph is decreasing, the gradient of the primitive graph is decreasing
Due to each of these properties, the y-values of the primitive graph can't be explicitly stated, unless more information is given (there are infinitely many possibilities, each of which is a vertical translation of every other)
Post by: gilliesb18 on March 29, 2018, 01:35:57 pm
Can someone please help me with the attached HSC question??
I'm very confused about it...
Thanks heaps.
Post by: RuiAce on March 29, 2018, 01:42:31 pm
Hello,Addressed in the compilation.
Can someone please help me with the attached HSC question??
I'm very confused about it...
Thanks heaps.
Post by: username2000 on March 29, 2018, 08:43:15 pm
i know this must be any easy question :o...but forgetten how to do differentiation :-[ :-\
its attached cos thats the best way to do it...
thanks for yah help :D
Post by: username2000 on March 29, 2018, 08:47:35 pm
and fyi anyone who needs help with maths, mathsonline.com.au is very helpful, it has videos and worksheets for the whole syllabus, every year, all levels 8)
Post by: Opengangs on March 29, 2018, 08:51:47 pm
hey whoever can help me...This is just a simple case of the "chain rule".
i know this must be any easy question :o...but forgetten how to do differentiation :-[ :-\
its attached cos thats the best way to do it...
thanks for yah help :D
Suppose that: \( u = x^2 \), then by the chain rule, we can see that:
Post by: minnak on March 30, 2018, 02:22:02 pm
I was just wondering in 2unit exams, are we allowed to use extension 1 content? such as integration by substitution. Will we still get full marks if the answer is correct or will we lose them because we haven't learnt it in 2 unit?
Post by: RuiAce on March 30, 2018, 02:41:07 pm
Hi!Whilst I'd call the marker "excessively harsh" if they did, there's still a risk of getting penalised. Every 2U problem in the exam is designed to be doable via only 2U techniques.
I was just wondering in 2unit exams, are we allowed to use extension 1 content? such as integration by substitution. Will we still get full marks if the answer is correct or will we lose them because we haven't learnt it in 2 unit?
Although having said that, the gamble can be worth it if you're running low on time and genuinely can't figure out the 2U method.
Post by: jamonwindeyer on March 30, 2018, 11:42:45 pm
Hi!
I was just wondering in 2unit exams, are we allowed to use extension 1 content? such as integration by substitution. Will we still get full marks if the answer is correct or will we lose them because we haven't learnt it in 2 unit?
To add to above, my understanding has always been that you can only get marks for 2U working. But if you get the correct answer then the 3U method will give you full marks. I have absolutely no source for this, but it's how I tended to operate - It's always better to use the 2U methods imo!! ;D
Post by: username2000 on March 31, 2018, 12:19:17 pm
This is just a simple case of the "chain rule".thanks a lot opengangs 8) ;D :D
Suppose that: \( u = x^2 \), then by the chain rule, we can see that:
could someone also please help me with finding the equation of normal/tangent to the curve please?
i know its easter weekend...but my exams on tuesday :'(
Post by: jamonwindeyer on March 31, 2018, 12:46:39 pm
thanks a lot opengangs 8) ;D :D
could someone also please help me with finding the equation of normal/tangent to the curve please?
i know its easter weekend...but my exams on tuesday :'(
Hey! This is the process (roughly) in steps:
- Differentiate your curve function to get the first derivative.
- Substitute the point you are given (where you want the tangent/normal) to find the gradient of the tangent to the curve at that point.
- If you are finding the tangent, you are done with the gradient. If you are finding the normal, adjust the gradient by taking the negative reciprocal (EG - \(m_t=5\) would be \(m_n=-\frac{1}{5}\).
- If you are not given it, find the y-coordinate by substituting the x-coordinate into the curve.
- With your coordinates and gradient, substitute into the point gradient formula:
Then just rearrange to make it look good! Usually in general form ;D if you have a particular question you are struggling with we can help you with it if you like!
Post by: username2000 on March 31, 2018, 01:46:46 pm
Hey! This is the process (roughly) in steps:thanks a lot, that explains it well..just needed a refresh ;D
- Differentiate your curve function to get the first derivative.
- Substitute the point you are given (where you want the tangent/normal) to find the gradient of the tangent to the curve at that point.
- If you are finding the tangent, you are done with the gradient. If you are finding the normal, adjust the gradient by taking the negative reciprocal (EG - \(m_t=5\) would be \(m_n=-\frac{1}{5}\).
- If you are not given it, find the y-coordinate by substituting the x-coordinate into the curve.
- With your coordinates and gradient, substitute into the point gradient formula:
Then just rearrange to make it look good! Usually in general form ;D if you have a particular question you are struggling with we can help you with it if you like!
Post by: username2000 on March 31, 2018, 02:23:48 pm
Hey! This is the process (roughly) in steps:Hey Jamon,
- Differentiate your curve function to get the first derivative.
- Substitute the point you are given (where you want the tangent/normal) to find the gradient of the tangent to the curve at that point.
- If you are finding the tangent, you are done with the gradient. If you are finding the normal, adjust the gradient by taking the negative reciprocal (EG - \(m_t=5\) would be \(m_n=-\frac{1}{5}\).
- If you are not given it, find the y-coordinate by substituting the x-coordinate into the curve.
- With your coordinates and gradient, substitute into the point gradient formula:
Then just rearrange to make it look good! Usually in general form ;D if you have a particular question you are struggling with we can help you with it if you like!
more specifically...ive attached some questions i need help with, just get a bit confused with the e's, and whether to leave it as decimal places or what.
thanks heaps!
cheers
Post by: RuiAce on March 31, 2018, 02:34:56 pm
Hey Jamon,
more specifically...ive attached some questions i need help with, just get a bit confused with the e's, and whether to leave it as decimal places or what.
thanks heaps!
cheers
Also note that when differentiating in Q4, you should use log laws to rewrite \( \ln \sqrt{x} \) as \( \frac12\ln x\) before differentiating. Note that you should never round unless explicitly told to do so.
If you have further problems, you should post any relevant progress.
Post by: Mate2425 on March 31, 2018, 03:11:51 pm
Q. Solve 7Sin3x = 2x-1 for .... 0 <= x <= 2Pi
Where: < = is x is greater than/ equal to
<= is X is less than/ equal to
Thanks heaps ;)
Post by: RuiAce on March 31, 2018, 03:14:37 pm
Hey could someone please help me with this question.You don't. Either you approximate the solution by drawing two graphs or there's a mistake.
Q. Solve 7Sin3x = 2x-1 for .... 0 <= x <= 2Pi
Where: < = is x is greater than/ equal to
<= is X is less than/ equal to
Thanks heaps ;)
Post by: jamonwindeyer on March 31, 2018, 03:14:41 pm
Hey could someone please help me with this question.
Q. Solve 7Sin3x = 2x-1 for .... 0 <= x <= 2Pi
Where: < = is x is greater than/ equal to
<= is X is less than/ equal to
Thanks heaps ;)
Hey! There's no way to solve this precisely, so what you'll want to do is draw graphs of both \(y=7\sin{3x}\) and \(y=2x-1\) to scale, on the same axes. Then, use the x-coordinates of the points of intersection as estimates for your answer (since those x-coordinates are where the LHS equals the RHS, where the two curves cross) :)
Post by: username2000 on March 31, 2018, 05:26:06 pm
thanks, v helpful RuiAce 8)
Also note that when differentiating in Q4, you should use log laws to rewrite \( \ln \sqrt{x} \) as \( \frac12\ln x\) before differentiating. Note that you should never round unless explicitly told to do so.
If you have further problems, you should post any relevant progress.
but i cant seem to find that answer on the sheet...well it is P when i chnaged it to general form...but when i put in in it says its wrong. its from mathsonline
Post by: RuiAce on March 31, 2018, 05:58:26 pm
thanks, v helpful RuiAce 8)Erm, not sure what you're trying to tell me here. Are these from mathsonline or are they a worksheet but without answers, or something else?
but i cant seem to find that answer on the sheet...well it is P when i chnaged it to general form...but when i put in in it says its wrong. its from mathsonline
Post by: username2000 on March 31, 2018, 06:49:44 pm
its a worksheet from mathsonline and the thing i sent before was just a snip..
thanks again for your help...especially when its easter ;D atar notes is sooo good
Post by: RuiAce on March 31, 2018, 06:54:15 pm
Post by: username2000 on March 31, 2018, 07:02:30 pm
I'm looking at their "answers" and my answer to Q3 looks like the same as their answer K)is that the same as p or am i seeing things...i might be doing something wrong but it still says thats wrong...unless their answers are incorrect ???
unless it can be factorised or simplified further...maybe im just wasting ur time sorry :-\
Post by: RuiAce on March 31, 2018, 07:17:20 pm
is that the same as p or am i seeing things...i might be doing something wrong but it still says thats wrong...unless their answers are incorrect ???Nope you're right. That's definitely the same as P) there.
unless it can be factorised or simplified further...maybe im just wasting ur time sorry :-\
(Nor would I know why, rip)
Post by: username2000 on March 31, 2018, 07:19:47 pm
Nope you're right. That's definitely the same as P) there.so do u agreee your answers right
(Nor would I know why, rip)
Post by: RuiAce on March 31, 2018, 07:22:49 pm
Post by: sady123 on April 01, 2018, 01:20:29 pm
Post by: RuiAce on April 01, 2018, 01:25:46 pm
Hi can someone help me with this question?? Thanks!
Post by: Never.Give.Up on April 01, 2018, 10:27:28 pm
Find the exact area bounded by the curve y=e^2x, the x-axis and the lines x=2 and x=5.
sorry, i know this is really easy but i've had a mind blank :( :o
thanks heaps
Post by: EEEEEEP on April 01, 2018, 11:13:25 pm
Find the exact area bounded by the curve y=e^2x, the x-axis and the lines x=2 and x=5.To find the exact area, we need to integrate y=e^2x and then apply the values of x =2, and x =5.
sorry, i know these are really easy but i've had a mind blank :( :o
thanks heaps
The integration of y=e^2x = 1/2 * e^2x (We can test this using differentiation, but it should be right =)
The actual application of the integral bounds equals to:
((e^10)/2) - ((e^4)/2) ... note, this isn't in the form of the answer with a lot of decimals.
Remember that the integral of e^x = e^x.
If you do a calculation with a calculator, unless it's a whole number, you will get a sequence of about 9 or 10 digits on screen. This is not an exact answer BUT an approximation.
Post by: Never.Give.Up on April 02, 2018, 08:10:02 am
Also, Find the exact gradient of the normal to the curve
and,
thanks heaps!
Post by: RuiAce on April 02, 2018, 08:15:14 am
Thanks for that EEEEEEP ;DPutting \(x=2\) into \( \frac{dy}{dx} = 1 + e^{-x} \) just gives \( m = 1 + e^{-2} \).
Also, Find the exact gradient of the normal to the curve y=x-e^-x at the point where x=2.
and,
thanks heaps!
(After subbing in, note that \( \ln 1 = 0\).)
Post by: Never.Give.Up on April 02, 2018, 09:09:43 am
Putting \(x=2\) into \( \frac{dy}{dx} = 1 + e^{-x} \) just gives \( m = 1 + e^{-2} \).thanks RuiAce!!v kind of u :D
(After subbing in, note that \( \ln 1 = 0\).)
im getting one wrong stationary point here....
Find the stationary points on the curve
thankyou!!
Post by: RuiAce on April 02, 2018, 10:04:31 am
thanks RuiAce!!v kind of u :D
im getting one wrong stationary point here....
Find the stationary points on the curveand determine their nature. for some reason i keep getting x=0,3, when its x=0,-3...???
thankyou!!
Which gives 0 and -3
Post by: LaraC on April 02, 2018, 10:56:41 am
Could someone please help me with two general rules. I'm a tad confused sorry! :o
How to integrate a to the function of x??
And how to differentiate a to the function of x??
E.g, the question 3^(2x-1).
Could someone please explain how to both differentiate and integrate this as well as tell me the rules involved??
Thanks so much!!!!
Post by: Dragomistress on April 02, 2018, 11:35:28 am
I was wondering how are state-rankers given state-ranks if a vast majority of people are able to obtain 100 raw in Mathematics.
Post by: RuiAce on April 02, 2018, 12:28:06 pm
Hey,They go into the school's internal results as well.
I was wondering how are state-rankers given state-ranks if a vast majority of people are able to obtain 100 raw in Mathematics.
Hey!
Could someone please help me with two general rules. I'm a tad confused sorry! :o
How to integrate a to the function of x??
And how to differentiate a to the function of x??
E.g, the question 3^(2x-1).
Could someone please explain how to both differentiate and integrate this as well as tell me the rules involved??
Thanks so much!!!!
Post by: Dragomistress on April 02, 2018, 01:00:31 pm
Post by: jamonwindeyer on April 02, 2018, 01:12:23 pm
What is their criteria for internals?
This is quoted from NESA:
If students are equal on the highest HSC marks in a course, then the following process is used to determine the recipient(s):
1. take an average of each student's exam mark and assessment mark after alignment to performance bands, each to one decimal place
2. take an average of each student's exam mark and assessment mark before alignment to performance bands, each to two decimal places
3. if an extension course, use the marks awarded for other courses in the subject area.
Post by: skisso on April 02, 2018, 03:20:53 pm
I was wondering if there was a quicker way of doing probability questions without drawing tree diagrams because they are really time consuming and messy.
Thanks heaps :)
Post by: RuiAce on April 02, 2018, 03:26:29 pm
Helloo!Personally, I've somewhat developed the ability to visualise the important branches of a tree diagram in my head.
I was wondering if there was a quicker way of doing probability questions without drawing tree diagrams because they are really time consuming and messy.
Thanks heaps :)
If they start throwing you off with those "with or without replacement" questions then you might not have much of a choice. Having said that, you don't lose marks for omitting the tree diagram, so you can always just draw a draft tree diagram (or even just 1/4 of it) on the question booklet to use as a reference. You're also quite limited in 2U because you don't have access to perms and combs.
If there's any specific problems causing you grief, please provide them.
Post by: skisso on April 02, 2018, 04:12:01 pm
Personally, I've somewhat developed the ability to visualise the important branches of a tree diagram in my head.
If they start throwing you off with those "with or without replacement" questions then you might not have much of a choice. Having said that, you don't lose marks for omitting the tree diagram, so you can always just draw a draft tree diagram (or even just 1/4 of it) on the question booklet to use as a reference. You're also quite limited in 2U because you don't have access to perms and combs.
If there's any specific problems causing you grief, please provide them.
Okk thank you :)
I was doing this question when I wrote this:
Post by: RuiAce on April 02, 2018, 04:23:18 pm
Okk thank you :)Yeah, I'm afraid that one is a bit more of a tough one.
I was doing this question when I wrote this:
With perms and combs at your side, this would be much easier. But because we cannot do that in 2U, the only 'shorter' way I can think of would be this.
Note, of course, 'shorter' is debatable here. It's up to you if you'd take this suggestion on board.
Post by: skisso on April 02, 2018, 07:48:01 pm
Yeah, I'm afraid that one is a bit more of a tough one.
With perms and combs at your side, this would be much easier. But because we cannot do that in 2U, the only 'shorter' way I can think of would be this.
Note, of course, 'shorter' is debatable here. It's up to you if you'd take this suggestion on board.
Yeah I guess its easier to do the combinations like that with more practice
Thank you :)
Post by: LaraC on April 02, 2018, 10:04:54 pm
Post by: Never.Give.Up on April 02, 2018, 10:21:33 pm
thanks!!
Which gives 0 and -3
(I had forgotten to use product rule :-[ :-[ :-[ ::))
Post by: Alex Molloy on April 03, 2018, 10:13:52 am
Post by: RuiAce on April 03, 2018, 10:21:11 am
1. Find the area under the curve y=loge2x, bounded by x= 4 and the x-axis.Have you drawn a diagram to help you do this question?
Post by: Alex Molloy on April 03, 2018, 10:49:58 am
Have you drawn a diagram to help you do this question?
just did, not sure if correct.
Post by: RuiAce on April 03, 2018, 11:00:41 am
(https://i.imgur.com/pe3mTVb.png)
Post by: Mate2425 on April 06, 2018, 06:15:16 pm
Thank you 😃
Post by: RuiAce on April 06, 2018, 07:29:49 pm
Hi could someone please help me understand why in Q6. 2007 Advanced Math paper they were able to get x = ln0.5. What is the process from lne^x to x.Not really worried this time because the issue is clear. In the future, please link to the paper.
Thank you 😃
The formal reasoning relies on inverse functions, but of course that's not 2U material.
Post by: Opengangs on April 06, 2018, 09:18:47 pm
Hi could someone please help me understand why in Q6. 2007 Advanced Math paper they were able to get x = ln0.5. What is the process from lne^x to x.Just going to expand on Rui's answer.
Thank you 😃
Post by: Calley123 on April 07, 2018, 09:36:25 pm
How do I do this question from trig functions?
Thanks :)
Post by: Calley123 on April 07, 2018, 09:38:44 pm
Hey,
How do I do this question from trig functions?
Thanks :)
The question was:
A sector of a circle with radius 5cm and an angle of pi/ subtended at the centre is cut out of the cardboard. It is then curved around to form a cone . Find its exact area and volume.
:)
Post by: Mate2425 on April 08, 2018, 10:49:40 am
Thanks ;D
Post by: kauac on April 08, 2018, 11:35:13 am
Hey,could someone please help understand when do i use the trapezoidal rule / Simpson's rule on the formula sheet vs. the trapezoidal rule/ Simpson's rule otherwise used such as i.e Trap = h/2 and Simp = h/3. I always get confused :o.
Thanks ;D
The Trap and Simpson's rules on the formula sheet are used when you are finding the estimate for only one interval- which uses two function values.
Anytime you are asked to find multiple intervals, you will need to use the more complex version of the formulas (not on the sheet).
Post by: kauac on April 08, 2018, 12:58:27 pm
Hey,
How do I do this question from trig functions?
Thanks :)
Hi...
I have no idea if this is the right answer, but here is how I would solve it logically...
Post by: Calley123 on April 08, 2018, 04:12:28 pm
Hi...
I have no idea if this is the right answer, but here is how I would solve it logically...
Both answers are correct !
Thank you so much !
Post by: Calley123 on April 08, 2018, 04:51:56 pm
How do I solve this trig equation between o and 2pi
sin3x+ sinx=0
Thank you :)
Post by: StupidProdigy on April 08, 2018, 06:25:20 pm
Hey again,Basically just apply the relevant compound angle formula first (sin(3x) becomes sin(2x+x)), then apply the sin double angle formula and cosine double angle formula. The overall goal is to get the argument in any of the trig functions to all be the same and in this case be 'x' by itself. See the pic attached for working. I didn't do the final step because I'm trusting you know where to go from there, hope it helps :)
How do I solve this trig equation between o and 2pi
sin3x+ sinx=0
Thank you :)
Post by: RuiAce on April 08, 2018, 08:45:34 pm
Hey again,Answer provided above. The solutions will then just be \(x=0,\frac\pi2,\pi,\frac{3\pi}2,2\pi \).
How do I solve this trig equation between o and 2pi
sin3x+ sinx=0
Thank you :)
However please note that this is a 3U question, and for these situations in the future I will move it to the relevant thread with only a small warning.
Post by: Calley123 on April 09, 2018, 09:01:33 am
Basically just apply the relevant compound angle formula first (sin(3x) becomes sin(2x+x)), then apply the sin double angle formula and cosine double angle formula. The overall goal is to get the argument in any of the trig functions to all be the same and in this case be 'x' by itself. See the pic attached for working. I didn't do the final step because I'm trusting you know where to go from there, hope it helps :)Thank you !!
Post by: Mate2425 on April 10, 2018, 03:53:24 pm
f(x) = 2x^3 +9x^2 +12x +1
Q. By halving the interval twice find approximation to this. (Answer = 0)
Thank you!!
Also when they ask to use the method of halving the interval twice to find an approximation / estimate to cube root of 12 ........; why do they have the answer as only one of the numbers of where the root lies e.g After halving interval twice gives me 2.25< Cube root of 12 < 2.5 and Answer is only 2.25.
Any help in understanding why, would be of much appreciation!! :)
Mod Edit: Post merge, use the 'Modify' button to add to your last post if no one has responded yet ;D
Post by: RuiAce on April 10, 2018, 10:23:42 pm
Hi could someone please help me with this question:
f(x) = 2x^3 +9x^2 +12x +1
Q. By halving the interval twice find approximation to this. (Answer = 0)
Thank you!!
Also when they ask to use the method of halving the interval twice to find an approximation / estimate to cube root of 12 ........; why do they have the answer as only one of the numbers of where the root lies e.g After halving interval twice gives me 2.25< Cube root of 12 < 2.5 and Answer is only 2.25.
Any help in understanding why, would be of much appreciation!! :)
Mod Edit: Post merge, use the 'Modify' button to add to your last post if no one has responded yet ;D
This is actually a 3U concept! We can help with that in our 3U thread if you like, just so everything is in one spot when people read it back later!
Click here!
Mod Edit: Added link :)
Post by: LaraC on April 18, 2018, 10:26:36 am
Just not quite sure how to tackle this question:
The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the first 20 terms.
So I've done part 1: as in, got an equation for the sum of the first 4 terms = 42 (my equation is 2a + d = 21.....is that right?!), and I know how to find the sum of the first 20 terms once I have the middle bit....I'm just a little confused about the 2nd step.....how do I work out an equation for the 3rd and 7th term adding to 46?
Thanks!
Post by: Opengangs on April 18, 2018, 11:34:17 am
Hello :)Hey, so we begin with the sum of the first 4 terms being 42.
Just not quite sure how to tackle this question:
The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the first 20 terms.
So I've done part 1: as in, got an equation for the sum of the first 4 terms = 42 (my equation is 2a + d = 21.....is that right?!), and I know how to find the sum of the first 20 terms once I have the middle bit....I'm just a little confused about the 2nd step.....how do I work out an equation for the 3rd and 7th term adding to 46?
Thanks!
Post by: LaraC on April 18, 2018, 12:33:47 pm
Dur! Just saw where I went wrong with my first equation for that question! :-[ :-[
Sorry, could I also ask about this qu....(I'm v dumb! ::)) :
The 20th term of an arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. Find the sum of the first 20 terms.
Its kind of similar, but do I need to form an equation that goes something like (sum of 10 terms - sum of 5 terms = 235)....or how do I go about it?
Post by: Opengangs on April 18, 2018, 12:41:17 pm
Thanks opengangs!Hey, LaraC!
Dur! Just saw where I went wrong with my first equation for that question! :-[ :-[
Sorry, could I also ask about this qu....(I'm v dumb! ::)) :
The 20th term of an arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. Find the sum of the first 20 terms.
Its kind of similar, but do I need to form an equation that goes something like (sum of 10 terms - sum of 5 terms = 235)....or how do I go about it?
That's a very good question - don't feel dumb! ;D
So, basically yep! You will need to form an equation where \( S_{10} - S_5 = 235 \) - that'd be the right approach!
[Hint: Use the fact that \( a_n = a_1 + (n - 1)d \) and use the formula for the sum to n terms]
Let me know if you get stuck!
Post by: Calley123 on April 20, 2018, 12:10:13 pm
It would be great if someone could help me with these questions. They are a bit tricky.
Thanks :)
Post by: RuiAce on April 20, 2018, 01:08:32 pm
Hey,
It would be great if someone could help me with these questions. They are a bit tricky.
Thanks :)
___________________________________________
\begin{align*}\therefore \int_1^3 2x(1+2\log_ex)\,dx &= \left[2x^2\log_e x\right]_1^3\\ &= 18\log_e 3 - 0 \\ &= 18\log_e 3\end{align*}
Post by: Calley123 on April 20, 2018, 03:33:17 pm
___________________________________________
\begin{align*}\therefore \int_1^3 2x(1+2\log_ex)\,dx &= \left[2x^2\log_e x\right]_1^3\\ &= 18\log_e 3 - 0 \\ &= 18\log_e 3\end{align*}
Thanks heaps
Post by: Calley123 on April 20, 2018, 03:36:22 pm
I thought I attached this question before but obviously not.
Help please...
Cheers
Post by: RuiAce on April 20, 2018, 06:08:48 pm
Hey again,
I thought I attached this question before but obviously not.
Help please...
Cheers
Post by: Calley123 on April 20, 2018, 09:41:11 pm
Much appreciated !
Post by: Calley123 on April 21, 2018, 08:06:36 am
I keep getting the wrong answers for these ! - Question 10 & 12
Thank you in advance :)
Post by: RuiAce on April 21, 2018, 09:59:50 am
Lol its me again,Q12 is just \( \pi \int_0^{\pi/2}\cos x\,dx =\pi [\sin x]_0^{\pi/2}\)
I keep getting the wrong answers for these ! - Question 10 & 12
Thank you in advance :)
(https://i.imgur.com/esDDQcS.png)
Alternatively, treat it as the area under \( y = \sin x\), minus a rectangle of height \( \frac12 \) and length \( \frac{5\pi}{6} - \frac\pi6\).
Post by: Calley123 on April 21, 2018, 05:36:22 pm
Q12 is just \( \pi \int_0^{\pi/2}\cos x\,dx =\pi [\sin x]_0^{\pi/2}\)(https://i.imgur.com/esDDQcS.png)
Alternatively, treat it as the area under \( y = \sin x\), minus a rectangle of height \( \frac12 \) and length \( \frac{5\pi}{6} - \frac\pi6\).
Thank you !
Post by: Calley123 on April 25, 2018, 12:46:34 pm
Help please..
Thanks in advance :)
Post by: RuiAce on April 25, 2018, 01:10:18 pm
Hey,\[ m_{PA} = \frac{y+2}{x-3}\text{ and }m_{PB} = \frac{y-7}{x+1} \]
Help please..
Thanks in advance :)
Post by: aryan.gupta1 on April 25, 2018, 05:45:59 pm
in a litter of seven pups it is known that the first three born are all male. What is the probability that the next pup born will also be male?
Thanks in advance :))
Post by: RuiAce on April 25, 2018, 05:47:24 pm
Hey I have a probability question that im really struggling with !!!The intuitive answer is just 1/2 assuming that both genders are equally likely. What are the answers saying?
in a litter of seven pups it is known that the first three born are all male. What is the probability that the next pup born will also be male?
Thanks in advance :))
Post by: aryan.gupta1 on April 25, 2018, 05:48:43 pm
Post by: jamonwindeyer on April 25, 2018, 05:50:55 pm
Lol, my teacher just got these questions and pasted them on a word doc, there aren't any answers :/
This is a logic problem more than a maths problem ;D
Definitely 1/2. Once we know the first three pups are male then we've sort of moved along the probability tree - So the next stage just becomes another 50/50 shot ;D it would be different if we didn't know for sure about the first three!
Post by: RuiAce on April 25, 2018, 05:52:02 pm
So if we assume that both genders are equally likely (which is just the natural assumption), then it'd just be 1/2. Otherwise, the question lacks information by a lot.
(Assuming that the probability that a puppy is male is 3/7 just using the information given is invalid. This is because we have no information about the fifth, sixth or seventh puppy to base our assumption on. If your teacher does that, I'd be questioning it a lot.)
Post by: Calley123 on May 03, 2018, 05:41:22 pm
\[ m_{PA} = \frac{y+2}{x-3}\text{ and }m_{PB} = \frac{y-7}{x+1} \]
THANK YOUUUUUUUU!
Please help!
Mod Edit: Post merge :)
Post by: Calley123 on May 03, 2018, 06:14:19 pm
Hey RuiAce,
In the second line, did you mean ln(y)^2 dy rather than ln(x)^2 dx ( Since it rotates around y-axis )
Mod Edit: Post merge, use the "Modify" button to add to your previous post ;D
Post by: owidjaja on May 03, 2018, 07:03:08 pm
So my teacher basically spent the last 5 minutes of the period to explain geometric progression to us and set us homework. Considering he's the type of teacher to just push on and I'm a slow learner, can someone please explain how geometric progression works?
Thanks in advance :)
Post by: RuiAce on May 03, 2018, 07:14:45 pm
This one too please :)Yeah my bad, I'll edit that later. The final answer remains the same but for the working out I should've used \(y\)'s.
Hey RuiAce,
In the second line, did you mean ln(y)^2 dy rather than ln(x)^2 dx ( Since it rotates around y-axis )
Mod Edit: Post merge, use the "Modify" button to add to your previous post ;D
Hey guys,
So my teacher basically spent the last 5 minutes of the period to explain geometric progression to us and set us homework. Considering he's the type of teacher to just push on and I'm a slow learner, can someone please explain how geometric progression works?
Thanks in advance :)
(As opposed to arithmetic progressions, where you add.)
Anything else - provide more details.
Post by: owidjaja on May 03, 2018, 07:41:30 pm
I'm not entirely sure where I went wrong in my working out- could someone help me find my mistake?
Here's the question (the answer as well) and here's my working out.
Post by: RuiAce on May 03, 2018, 07:45:50 pm
Hey guys,\[ T_n = a r^{n-1} =4 \left(\frac12\right)^{n-1}.\\ \text{You can't throw the 4 under the bracket here.} \]
I'm not entirely sure where I went wrong in my working out- could someone help me find my mistake?
Here's the question (the answer as well) and here's my working out.
\begin{align*}4 \times \frac{1}{2^{n-1}} &= \frac1{128}\\ \frac{1}{2^{n-1}}&=\frac1{512}\end{align*}
Post by: owidjaja on May 03, 2018, 08:07:14 pm
Post by: RuiAce on May 03, 2018, 08:30:57 pm
I'm also not sure where I went wrong in this other question. Here is the question and here is my working out.Your working out is fine. You just haven't finished the question
Post by: owidjaja on May 07, 2018, 06:59:43 pm
I'm not entirely sure how to figure out this question, so it'll be great to get some help.
Thanks in advance :)
Post by: RuiAce on May 07, 2018, 07:07:32 pm
Hey guys,
I'm not entirely sure how to figure out this question, so it'll be great to get some help.
Thanks in advance :)
Post by: owidjaja on May 09, 2018, 08:53:51 pm
I'm not entirely sure how to do the following question, so help is appreciated.
Thanks in advance :)
Post by: RuiAce on May 09, 2018, 09:04:46 pm
Hey guys,(Shows how shit MIF is as a textbook with that horrendous typo...)
I'm not entirely sure how to do the following question, so help is appreciated.
Thanks in advance :)
Post by: Mate2425 on May 11, 2018, 11:00:21 am
Q. Find the sum of all integers between 1 and 100 that are not multiples of 6.
Post by: RuiAce on May 11, 2018, 12:04:18 pm
Hi could i please have some help with the following AP Question.
Q. Find the sum of all integers between 1 and 100 that are not multiples of 6.
Post by: Mate2425 on May 16, 2018, 07:40:52 pm
Thanks RuiAce
Post by: LaraC on May 22, 2018, 12:14:09 pm
Could someone please help me with the following question...I'm a little confused as I used the limiting sum formula with a= 1.5m and r=2/5....but that gives me 2.5m and the answer says 3.5m?!
Mary bounces a ball, dropping it from 1.5m on its first bounce. It then rises up to 2/5 of its height on each bounce. Find the distance through which the ball travels.
Thanks! ;) :D
Post by: RuiAce on May 22, 2018, 12:32:50 pm
Hello :)Some similar questions were asked here and here. Take a look at the first.
Could someone please help me with the following question...I'm a little confused as I used the limiting sum formula with a= 1.5m and r=2/5....but that gives me 2.5m and the answer says 3.5m?!
Mary bounces a ball, dropping it from 1.5m on its first bounce. It then rises up to 2/5 of its height on each bounce. Find the distance through which the ball travels.
Thanks! ;) :D
The idea is that you can’t treat these as just one geometric series; you actually have two geometric series added together. In your case, you’ll have \((1.5 + 1.5(0.4) + 1.5(0.4)^2 + \dots) + (1.5(0.4) + 1.5(0.4)^2 + 1.5(0.4)^3 + \dots)\) which will give \( \frac{1.5}{1-0.4} + \frac{1.5(0.4)}{1 - 0.4} \).
Post by: LaraC on May 22, 2018, 12:40:49 pm
Post by: LaraC on May 23, 2018, 10:13:19 am
This isn't exactly a maths question, but more a calculator qu :-[ :o
I have a series of questions that include things like "find correct to 2 decimal place cos 0.589", and " find correct to 2 decimal places tan 0.056" etc. They are all along the same lines. Punching that into my calculator is easy enough, but it says it has to be done using the radian mode on my calculator (clearly giving a different answer).....how do I convert my calculator into radian mode?
It is a Casio fx-82AU PLUS.
Sorry I've looked up online manuals but still a bit confused!! ::) :-[
Thanks!
Post by: RuiAce on May 23, 2018, 11:04:43 am
Hello again,
This isn't exactly a maths question, but more a calculator qu :-[ :o
I have a series of questions that include things like "find correct to 2 decimal place cos 0.589", and " find correct to 2 decimal places tan 0.056" etc. They are all along the same lines. Punching that into my calculator is easy enough, but it says it has to be done using the radian mode on my calculator (clearly giving a different answer).....how do I convert my calculator into radian mode?
It is a Casio fx-82AU PLUS.
Sorry I've looked up online manuals but still a bit confused!! ::) :-[
Thanks!
Code: [Select]
0. Calculator turned on and ready
1. Press "SHIFT", followed by <MODE>.
- By doing so you go into <SETUP>, instead of <MODE>. That's how the "SHIFT" button works.
2. Choose "Rad"
You can also use this method to go back to degrees.
Post by: mxrylyn on May 28, 2018, 12:37:50 pm
I am not quite sure how to differentiate
Ln (2x + 4) (3x - 1)
I have looked at a solution but I dont understand the steps.
Post by: RuiAce on May 28, 2018, 01:32:56 pm
Hello.Did you mean \( [\ln (2x+4) ] \times (3x-1) \) or \( \ln [(2x+4)(3x-1)] \)?
I am not quite sure how to differentiate
Ln (2x + 4) (3x - 1)
I have looked at a solution but I dont understand the steps.
Post by: mxrylyn on May 28, 2018, 01:38:24 pm
But I wasn't sure if the × (3x - 1) was implied
Post by: RuiAce on May 28, 2018, 02:19:03 pm
Therefore \( \frac{d}{dx} \ln [(2x+4)(3x-1)] = \frac{2}{2x+4} + \frac{3}{3x-1} \)
Post by: gilliesb18 on May 29, 2018, 05:00:17 pm
15. A wedge is cut so that its cross-sectional area is a sector of a circle, radius 15cm and subtending an angle pi/6 at the centre. Find
a) the volume of the wedge
b) the surface area of the wedge
Any help is appreciated... I just cant think how to start to work it out..
Thanks heaps:):)
Post by: RuiAce on May 29, 2018, 07:36:37 pm
Hello, just need help on a trig question...The question doesn't provide any assumption on the shape that the wedge originally came from. If it's cylindrical then we're missing its height; if it's spherical or something else then we're lacking in heaps of information.
15. A wedge is cut so that its cross-sectional area is a sector of a circle, radius 15cm and subtending an angle pi/6 at the centre. Find
a) the volume of the wedge
b) the surface area of the wedge
Any help is appreciated... I just cant think how to start to work it out..
Thanks heaps:):)
Post by: Mate2425 on May 30, 2018, 01:56:00 pm
Kate has $4,000 in a bank account that pays 5% p.a. with interest paid annually and Rachel has a different account $4,000 paying 4% quarterly. Which person will receive the most interest over 5 yrs and by how much?
ANS: Kate $224.37
Thanks :)
Post by: RuiAce on May 30, 2018, 02:29:51 pm
Hi could i please have help with the working out for this compound interest Question.
Kate has $4,000 in a bank account that pays 5% p.a. with interest paid annually and Rachel has a different account $4,000 paying 4% quarterly. Which person will receive the most interest over 5 yrs and by how much?
ANS: Kate $224.37
Thanks :)
Note that the rates described are all annual. So 4% p.a. turns into 1% per quarter for Rachel. It also means that Rachel receives interest 4 times a year, not just once at the end.
Note that there is no annuity here. This is just a straightforward application of compound interest by itself.
Post by: Dragomistress on May 31, 2018, 08:54:44 pm
Post by: RuiAce on May 31, 2018, 08:59:50 pm
How am I meant to do this question?You know that in degrees, the interior angle sum is \( 180^\circ (n-2) \). So just replace \(180^\circ\) with \( \pi\) to get \(3\pi\).
(Essentially speaking, the smallest angle becomes the first term of your arithmetic progression.)
Post by: Dragomistress on June 02, 2018, 11:44:35 am
Post by: Opengangs on June 02, 2018, 02:04:00 pm
I am wondering how I am meant to do e.
So because \( n \) is a solution to the equation, \( 2\cos x = 1 - \frac{1}{2}x \) then any \( x \) that follows must strictly be in between these two bounds. Hence, it follows that the solution \( n \) must also lie in between the two.
Post by: kauac on June 03, 2018, 01:10:11 pm
In Simpson's rule, my teacher has a bit of a mnemonic to remember it, which is: First + last, 4 x odd and 2 x even.
What would the 'odd' and 'even' be referring to? Like, if you had a table of values, is it whether the x value is an odd/even number? Or is it if you count across from the first and last values, every second one is odd, and the others are even?
Any clarification would be super helpful! :)
Post by: RuiAce on June 03, 2018, 01:18:43 pm
This confuses me every single time:
In Simpson's rule, my teacher has a bit of a mnemonic to remember it, which is: First + last, 4 x odd and 2 x even.
What would the 'odd' and 'even' be referring to? Like, if you had a table of values, is it whether the x value is an odd/even number? Or is it if you count across from the first and last values, every second one is odd, and the others are even?
Any clarification would be super helpful! :)
\begin{align*} f(3) &\to \text{first}\\ f(3.2) &\to \text{odd}\\ f(3.4) &\to \text{even}\\ f(3.6)&\to \text{odd}\\ f(3.8 ) &\to \text{even}\\ f(4) &\to \text{odd}\\ f(4.2) &\to \text{even}\\ f(4.4) &\to \text{odd}\\ f(4.6) &\to \text{even}\\ f(4.8 )&\to \text{odd}\\ f(5)&\to \text{last} \end{align*}
That "mnemonic" is a very well known way of memorising the generalised Simpson's rule. Essentially your "first" input value is \(x_0 = 3\). The next input value, \(x_1 = 3.2\), is called "odd" in this context because the index is \(1\). The subsequent input value, \(x_2 = 3.4\), is called "even" in this context because the index is \(2\). And so on.
Essentially, you start counting after the "first". That one, will be the first "odd".
You should always end up with exactly one more "odd" than "even"
Post by: kauac on June 03, 2018, 01:21:28 pm
\begin{align*} f(3) &\to \text{first}\\ f(3.2) &\to \text{odd}\\ f(3.4) &\to \text{even}\\ f(3.6)&\to \text{odd}\\ f(3.8 ) &\to \text{even}\\ f(4) &\to \text{odd}\\ f(4.2) &\to \text{even}\\ f(4.4) &\to \text{odd}\\ f(4.6) &\to \text{even}\\ f(4.8 )&\to \text{odd}\\ f(5)&\to \text{last} \end{align*}
That "mnemonic" is a very well known way of memorising the generalised Simpson's rule. Essentially your first function value is \(x_0 = 3\). The next function value, \(x_1 = 3.2\), is called "odd" in this context because the index is \(1\). The subsequent function value, \(x_2 = 3.4\), is called "even" in this context because the index is \(2\). And so on.
Essentially, you start counting after the "first". That one, will be the first "odd".
You should always end up with exactly one more "odd" than "even"
Thanks so much for your explanation! Everything is a lot more clear now. :)
Post by: Dragomistress on June 05, 2018, 06:42:41 pm
sin(pi-2x)=sin(2x)?
Post by: RuiAce on June 05, 2018, 06:43:52 pm
I am confused on how I can convert the LHS into the RHS.
sin(pi-2x)=sin(2x)?
Second quadrant angle.
Post by: LaraC on June 07, 2018, 09:06:52 pm
Could I please have help with this question:
Find lim (theta approaches 0) ((tan theta/3)/theta)
Hopefully you can decipher that! :P
Thanks in advance! :D
Post by: clovvy on June 10, 2018, 06:19:37 pm
Hello :)
Could I please have help with this question:
Find lim (theta approaches 0) ((tan theta/3)/theta)
Hopefully you can decipher that! :P
Thanks in advance! :D
Post by: mirakhiralla on June 10, 2018, 06:23:01 pm
could someone please help/// I think it was from the 2007 HSC:
Mrs Cain decided to invest some money each year to help pay for her son's uni education.
She contributes $1000 on the day of her son's birth and increases her annual contribution by 6% each year.
Her investment also earns 6% compound interest p/a.
Find the total value of Mrs Cain's investment on her son's birthday (just before her fourth contribution)
Post by: clovvy on June 10, 2018, 06:45:50 pm
Hey I have attempted this question multiple times but couldn't do it,
could someone please help/// I think it was from the 2007 HSC:
Mrs Cain decided to invest some money each year to help pay for her son's uni education.
She contributes $1000 on the day of her son's birth and increases her annual contribution by 6% each year.
Her investment also earns 6% compound interest p/a.
Find the total value of Mrs Cain's investment on her son's birthday (just before her fourth contribution)
Post by: RuiAce on June 10, 2018, 06:54:13 pm
Hey I have attempted this question multiple times but couldn't do it,I found the question - I think you meant her son's third birthday.
could someone please help/// I think it was from the 2007 HSC:
Mrs Cain decided to invest some money each year to help pay for her son's uni education.
She contributes $1000 on the day of her son's birth and increases her annual contribution by 6% each year.
Her investment also earns 6% compound interest p/a.
Find the total value of Mrs Cain's investment on her son's birthday (just before her fourth contribution)
Note that, because of what the question wants, we've set \(A_n\) to be the value right before we make the next deposit. So we need to ensure that after we compute \(A_1\), we deposit immediately after, before smacking on the interest factor of 1.06
Which you can put into your calculator.
The reason why this annuity turned out so surprisingly nicely is because the deposits were increasing alongside the interest accrued. This cleans up a lot of the mess for us.
Post by: clovvy on June 10, 2018, 07:01:49 pm
That was embarrasing
Post by: mirakhiralla on June 10, 2018, 07:38:01 pm
Oh lol I just realised I did the wrong question when I had a look at it lol, so as Rui says just said just sub them in to the calculatorAhahah I wrote the question wrong, all good.
That was embarrasing
Thank you both!
Post by: ilikeapples on June 14, 2018, 07:50:32 pm
A bandicoot population is decreasing by 5% each year. a) what % of population is left after 5 years. b) after how many years will the population only be 50%. c) how many years will I take for the population to decrease by 80%.
Thankyou
Post by: jamonwindeyer on June 14, 2018, 08:01:31 pm
Hi I am having difficulty with this series question:
A bandicoot population is decreasing by 5% each year. a) what % of population is left after 5 years. b) after how many years will the population only be 50%. c) how many years will I take for the population to decrease by 80%.
Thankyou
Welcome to the forums!! Let me give you a few hints!
- For (a), a decrease of 5% is the same as there being 95% remaining. So, every year we are multiplying by 0.95. The number left after the first year is \(N=N_0(0.95)\), where \(N_0\) is the initial. The next year will be \(N=N_0(0.95)^2\). And so on!
- For (b), you'll need to generalise the formula like this:
You want the value of N that leaves 50% of the initial, or \(0.5N_0\), generate a formula! And c) is the same, just you want it equal to \(0.2N_0\) (20% left since there is 80% gone!).
Hope this helps ;D
Post by: talitha_h on June 17, 2018, 05:47:02 pm
I have a cited task coming up for math - basically just a bunch of hsc past papers- and I was wondering what would be the best way to study for this (other than actually doing the papers)?
Thanks :)
Post by: RuiAce on June 17, 2018, 05:49:55 pm
Hi,Still not sure what you mean by "cited task". Is it just something like an in-class test?
I have a cited task coming up for math - basically just a bunch of hsc past papers- and I was wondering what would be the best way to study for this (other than actually doing the papers)?
Thanks :)
Post by: talitha_h on June 17, 2018, 08:56:38 pm
Still not sure what you mean by "cited task". Is it just something like an in-class test?
sorry, I meant 'sighted' task, we get given a bunch of past papers to do before the exam and they select a number of them to be in the assessment
Post by: RuiAce on June 17, 2018, 09:21:04 pm
sorry, I meant 'sighted' task, we get given a bunch of past papers to do before the exam and they select a number of them to be in the assessmentNot fully sure what you mean but I'm interpreting that you're essentially being given past papers as a question bank and the questions in the exam can only be pooled from there.
If you get to see a question bank beforehand, and they can't pick anything that's not in what they give you, then your job really depends on how large the question bank is.
If the question bank is reasonably small (i.e. not more than 2 full HSC papers worth, i.e. <200 marks in total), then your job is to prepare solutions for them and work on being able to write down them fast. That'll involve essentially doing the papers over and over again, but almost like essay writing in the sense you're trying to speed up as well. Strategy here would be to do the easier questions enough times so that you're confident that your answer is correct. For computational questions, you can confirm your answer by checking on Wolfram, but for proof-based questions you will need to be able to convince yourself that your proof makes sense.
If the question bank is big then essentially you'll only have enough time to do the easy stuff properly once. It should not be the only time you do the easy questions, but you're gonna have fewer opportunities to do them again (so be careful about that). But you'll need to both constantly revise/study the hard ones as well as write out the answer over and over again. Reason being you never know which of the hard ones will show up, and you want to be prepared for all of them without being hammered by the pressure of time. You should also try mixing up the order of the questions, and attempt to do them closed book even after the first time you succeed in doing them.
(On the other hand, if things can be asked outside of the question bank, you really should be treating it like studying for a usual test; albeit with a bit more focus on the ones that were given.)
Essentially, if you know what the questions are, then you know whether or not you'll be right before you walk in. So studying the concepts is useless. However you should, of course, understand how the concepts work to create the solutions that you've prepared for to begin with.
Post by: cocopops201 on June 23, 2018, 07:37:17 pm
Thanks :)
Post by: RuiAce on June 23, 2018, 07:41:24 pm
Hey can someone help me with this question?Resisted motion is not covered in 2U (nor 3U). What is the source of the question?
Thanks :)
Post by: cocopops201 on June 23, 2018, 08:56:27 pm
The relationship I only saw was dat dp/dt = kAe^kt where dp/dt can equal the velocity, but I just don't know I could get k
Post by: cocopops201 on June 23, 2018, 08:58:56 pm
Post by: RuiAce on June 23, 2018, 09:01:58 pm
Note that there should not be an extra \(k\) in front. This is because what we've given relates the velocity to the acceleration. At no point do we actually need to consider the displacement of the particle.
Edit: Cool - won't remove this though for anyone else's reference
Post by: cocopops201 on June 24, 2018, 11:29:00 am
I've got a question regarding superannuation and loan repayments in the applications of series topic.
For supperannuation what are the formulas that's needed and the process of finding how much an amount has grown from one year to another year and the total investment overall.
For loan repayments what are the formulas thats needed and the process to it?
I just find these two a really long process that I kind of have trouble revising.
Here's an example of questions
Thanks :)
Post by: RuiAce on June 24, 2018, 04:01:25 pm
Hi,
I've got a question regarding superannuation and loan repayments in the applications of series topic.
For supperannuation what are the formulas that's needed and the process of finding how much an amount has grown from one year to another year and the total investment overall.
For loan repayments what are the formulas thats needed and the process to it?
I just find these two a really long process that I kind of have trouble revising.
Here's an example of questions
Thanks :)
This is in contrast to the usual recursive method, where we have a running total of what goes on but don't really care about what happens to each individual investment.
The other part is just the old usual recursive approach.
__________________________________________________________________
Personal opinion: In the rare case this happens, I like to do the usual thing with whichever is more frequent. Here, the compounding frequency is just monthly, whereas the payments are made annually, so I analyse the whole thing on a monthly basis.
The recursive approach should be understood by trying to read out loud what is going on. Here, between the jump from the beginning to the end of the first month, the only thing that occurs is that it picks up one month's worth of interest. So at this stage, we still require only the compound interest formula.
\begin{align*}A_1 &= A_0 (1.01)\\ \therefore A_1 &= 50000(1.01) \end{align*}
Between the end of the first month and the end of the second month, the same thing happens.
\begin{align*}A_2 &= A_1(1.01)\\ \therefore A_2 &= 50000(1.01)^2\end{align*}
And it should be clear that similarly, \(A_3 =50000(1.01)^3\). This repeats itself all the way up to \(A_11\).
\[ A_{11} = 50000(1.01)^{11} \]
And then after the 12th month, after the money picks up interest yet again, we make our first payment. Recall that the payment we make is \(M\). So we have two things going on here now; the interest, followed by payment.
\begin{align*}A_{12} &= A_{11}(1.01) - M\\ &= 50000(1.01)^{12} - M. \end{align*}
Remark: iv) is nothing fancy; they just want you to calculate what the "average interest across the two years" was, which you can do by comparing how much you had to pay off (i.e. \(2M\)) to the initial amount.
Post by: owidjaja on June 25, 2018, 09:52:54 pm
I'm a bit unsure how to do this question.
Thanks in advance :)
Post by: EEEEEEP on June 25, 2018, 10:06:33 pm
Hey there,Hi!
I'm a bit unsure how to do this question.
Thanks in advance :)
lets assume that the pollution level at time 0 is 100, a 20% increase of that is 110, a 50% increase is 150.
:) don't forget that the exponential function formula is A = Pe^rt, where A = result, P = Principal (what you start off with), r =rate and t = time
Sidenote: Since the question is asking for an increase in the value exponentially and not a decrease, we use a positive sign for the power!
110 = 100e^2r (t =2 , r = we dont know)
1.1 = e^2r
We now take the ln of both sides! (REMEMBER, ln (e) = 1
ln (1.1) = 2r, (you can find out r from here.. which should be 0.047655
.........
To find out how many years it would take for the level to exceed 150... (50%)
150 = 100e ^ 0.047655t
1.5 = e^ 0.047655t
You can do the rest from here ..
t should be 8.508343...
BUT... they may be asking for the answer in years, so it would be 9 years since, 100e ^(0.047655*8) is too low !
Post by: secretweapon on June 26, 2018, 04:45:38 pm
Post by: clovvy on June 26, 2018, 05:47:40 pm
how would (0.5e^((log(e,64))/2)) be evaluated with no calculator? the e is meant to be a suffix, don't know how to type latex, hope it's still understood what i typed?
Post by: secretweapon on June 26, 2018, 06:27:40 pm
How do you get from log(e,64)/2 to 3*log(e,2)?
the e is meant to be suffix for both btw
Post by: clovvy on June 26, 2018, 06:30:11 pm
How do you get from log(e,64)/2 to 3*log(e,2)?log laws, I'll show you:
the e is meant to be suffix for both btw
Post by: secretweapon on June 26, 2018, 06:55:53 pm
log laws, I'll show you:Legend, thanks 8)
Post by: owidjaja on June 27, 2018, 07:18:48 pm
I'm a bit unsure how to do question d).
Thanks in advance :)
Post by: Opengangs on June 27, 2018, 08:25:45 pm
Hey guys,Note that \( x = 2 - t - t^2 \) can be factored nicely to give: \( x(t) = (t + 2)(1 - t) \).
I'm a bit unsure how to do question d).
Thanks in advance :)
What's neat about this is that we can graph this and see what's happening around the interval \( [0, 2] \), which is what we're interested in.
(https://i.imgur.com/pu1VZNQ.png)
Now, we notice that the particle starts at \( t = 0 \) and finishes at \( t = 2 \). So it must travel first from \( t = 0 \) to \( t = 1 \), and then \( t = 1 \) to \( t = 2 \).
Now, between \( t = 0 \) and \( t = 1 \), we see that it would have travelled \( x(0) = 2 \) units because it would have taken the particle 2 units downwards to get to \( x = 0 \). Now, it will continue to travel downwards until it hits \( x(2) = -4 \). Thus, it travelled 4 units (just in the downwards direction). So if we consider the path the particle takes between \( [0, 2] \), we see that it 2 + 4 = 6 units altogether.
Whenever you're not sure where to start, I recommend visualising it or draw a diagram to aid with your working. :)
Post by: Yiruma on July 02, 2018, 11:14:08 pm
find the exact area bounded by the curves y=sin x and y=cos x in the domain 0<x<2pi
Post by: Opengangs on July 03, 2018, 03:05:18 pm
Hey guys, I have a question for 2 unit maths integration!Hey, Yiruma!
find the exact area bounded by the curves y=sin x and y=cos x in the domain 0<x<2pi
Have you drawn a diagram? :) If not, that would be a good place to start.
Let me know if you have any problems
Post by: jamonwindeyer on July 03, 2018, 06:44:43 pm
Click here! :)
Post by: kauac on July 10, 2018, 09:30:46 am
My textbook works through three different formulas for the same example...So, I was wondering whether it just preference, or if some formulas suit particular questions more than others??
Post by: RuiAce on July 10, 2018, 09:59:44 am
When asked to differentiate using first principles, which formula should I use?Ignore the extra stuff maths in focus puts there for no good reason and just use \( \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \)
My textbook works through three different formulas for the same example...So, I was wondering whether it just preference, or if some formulas suit particular questions more than others??
(Remark: The formula \( \lim_{x\to c} \frac{f(x)-f(c)}{x-c} \) is good for mathematicians to establish where differentiation comes from in the first place. But to actually use it is a waste - it overcomplicates calculations.)
Post by: terassy on July 10, 2018, 03:50:05 pm
Post by: jamonwindeyer on July 10, 2018, 09:41:46 pm
Am I correct? When using the Simpson's rule, one application uses three function values and two applications uses 5 function values. And when using the Trapezoidal rule, one application uses two function values, and two applications uses three function values and so on.
You are indeed correct ;D
Post by: LaraC on July 12, 2018, 09:46:04 am
Just trying to do a bit of revision and I got a little stuck on this question. I have the first derivative and a point it passes through and I have to find the equation of the curve.
dy/dx = 6x^2 + 12x -5
If the curve passes through the point (2,-3), find the equation of the curve.
Thanks in advance! :D :D
Post by: Opengangs on July 12, 2018, 09:49:06 am
Hello :)Hey, LaraC.
Just trying to do a bit of revision and I got a little stuck on this question. I have the first derivative and a point it passes through and I have to find the equation of the curve.
dy/dx = 6x^2 + 12x -5
If the curve passes through the point (2,-3), find the equation of the curve.
Thanks in advance! :D :D
So what you'd want to do is to integrate it to find the equation if the curve (don't forget your constant of integration).
And using the fact that it passes through (2, -3), are you able to finish it off?
Let me know if you have any problems. :)
Post by: LaraC on July 13, 2018, 09:21:27 am
A rectangular prism with a square base is to have a surface area of 250cm^2. Its volume is given by V=(125x - x^3) / 2
Find the dimensions that will give the maximum volume.
I started by differentiating the volume formula....is this the right way to go or not? I can't remember sorry!! :-\ :-[
Thankks!!
Post by: RuiAce on July 13, 2018, 09:40:48 am
I have another question sorry guys!! :-[ ::) :PYeah.
A rectangular prism with a square base is to have a surface area of 250cm^2. Its volume is given by V=(125x - x^3) / 2
Find the dimensions that will give the maximum volume.
I started by differentiating the volume formula....is this the right way to go or not? I can't remember sorry!! :-\ :-[
Thankks!!
Now set \( \frac{dV}{dx}=0 \) to find the value of \(x\) that maximises the volume. And remember to test that it is a max, as with all problems of this nature.
Post by: LaraC on July 13, 2018, 09:55:49 am
So in this case it is a cube with dimensions 6.4cm x 6.4cm x 6.4cm....
But what about in the case it wasn't a cube, and had three different dimensions? Would there be three different stationery points and the three dimensions would be those three x values?
Sorry I don't have any specific question here, I'm just wondering in general as this type of question seems to crop up a bit ;) :D
Post by: RuiAce on July 13, 2018, 10:05:19 am
Thanks a million RuiAce!! #manyrespectpoints!!From what I've seen when there's multiple stationary points in 2U, most of the time the other ones are all for negative values of \(x\). Which we can typically discard for various reasons (e.g. side length cannot be negative).
So in this case it is a cube with dimensions 6.4cm x 6.4cm x 6.4cm....
But what about in the case it wasn't a cube, and had three different dimensions? Would there be three different stationery points and the three dimensions would be those three x values?
Sorry I don't have any specific question here, I'm just wondering in general as this type of question seems to crop up a bit ;) :D
But that's exactly why we must test for a max/min. Usually if both stationary points end up being at positive values of \(x\), one of them gives a min whilst the other gives a max.
(The techniques for 3+ stationary points are more advanced and require more work, but I've never seen that appear in 2U as of yet)
Post by: LaraC on July 13, 2018, 04:27:28 pm
E.g integral of (x/ (x^2 - 9))
Or integral of (1/(x+4))
I know how to for exponentials or logarithmic equations, but can't seem to make it work for these ones!
Thankss!! ;D
Post by: RuiAce on July 13, 2018, 06:08:20 pm
Accidently uploaded 2 of the same pictures, it wouldnt let me delete it :(Integration by substitution is an MX1 concept and not examinable here. They're required to use the rule \( \int \frac{f^\prime(x)}{f(x)}\,dx = \ln [f(x)]+C \).
@Lara.C, basically the goal is recognise the numerator is the differential of the denominator, which when you integrate gives a log.
Just wondering if there is a general rule for integrating fractions? It doesn't seem to be in my textbook and I seem to hit a deadend every time I run into a similar sort of problem.For the second one, integrals of the form \( \int \frac{1}{ax+b}dx \) you're just expected to know that turns into \( \frac{1}{a}\ln(ax+b)+C \)
E.g integral of (x/ (x^2 - 9))
Or integral of (1/(x+4))
I know how to for exponentials or logarithmic equations, but can't seem to make it work for these ones!
Thankss!! ;D
Whereas for the first one, you need to get used to using that rule. Which relies essentially on you recognising that what's on the top, looks very similar to what's on the bottom, except it's out by a constant. In particular, the derivative of the bottom here is \( \frac{d}{dx}(x^2-9)=2x\), but on the numerator we just have \(x\) by itself.
...So we multiply some fudge factors to force the numerator into being \(2x\).
\begin{align*}\int \frac{x}{x^2-9}\,dx &= \frac12\int \frac{2x}{x^2-9}\,dx\\ &= \frac12\ln (x^2-9)+C\end{align*}
Post by: prakash.ram on July 18, 2018, 09:45:36 am
Post by: key to success on July 18, 2018, 10:56:19 pm
hi, can you please recommend me a maths study textbook. I want a textbook that will help me understand hard questions through examples.Hey!
For me past paper questions and the worked solutions that come with that work the best!
Or else try Coroneous, they have separate worked solutions books that come with them!
Hope it helps- Good luck!
Post by: NowYouTseMe on July 18, 2018, 11:17:11 pm
hi, can you please recommend me a maths study textbook. I want a textbook that will help me understand hard questions through examples.
Honestly at this point, you're better off doing past papers and trial papers from other schools. If you can find them, selective school papers tend to be more difficult and with worked solutions, so they're super useful. That being said, Cambridge 2U has examples before each exercise so that's not bad and I believe New Senior Mathematics, or the Fitzpatrick 2U textbook has separate worked solutions as well.
Post by: skisso on July 21, 2018, 03:58:27 pm
This question is frm CSSA 2017, last question. I dont understand part iii. Like in the answers i dont get how they get to q=+- root 3/2 and how the third quadrant makes it a minus since there are no sin, cos etc.
Thank you :)
Post by: jamonwindeyer on July 21, 2018, 04:45:06 pm
Hello :))
This question is frm CSSA 2017, last question. I dont understand part iii. Like in the answers i dont get how they get to q=+- root 3/2 and how the third quadrant makes it a minus since there are no sin, cos etc.
Thank you :)
Hey hey! So they conclude that \(q\) must be negative, because the point Q is to the left of the origin. So, it's \(x\) coordinate must be negative.
So:
Since Q is left of the origin, then \(q\) must be a negative number ;D
Post by: amelia20181 on July 21, 2018, 09:10:05 pm
Mary buys 20 tickets in a lottery that has 5000 tickets altogether. Find the probability that Mary will win
(a) first and second prize (b) second prize only
(c) neither first nor second prize.
also this question
i just need help with a question from my textbook
The two machines in a workshop each have a probability of 1
of breaking down. Find the probability that at any one time (a) neither machine will be broken down
(b) 1 machine will be broken down.
Mod edit: merged multiple posts
Post by: envisagator on July 21, 2018, 09:39:19 pm
how do you do this question?a) This is a basic application of the product rule. First, recognise the fact that since its a lottery each time someone wins a prize a ticket is 'used up', that is, the total number of tickets decreases after each prize is given out.
Mary buys 20 tickets in a lottery that has 5000 tickets altogether. Find the probability that Mary will win
(a) first and second prize (b) second prize only
(c) neither first nor second prize.
b) To win the second prize only implies she doesnt win the first prize, the chance of not winning the first prize is 5000-20 = 4980, that is mary has 4980/5000 chance of losing.
c) Using the same logic in b)
Hope this helps :)
Post by: envisagator on July 21, 2018, 09:42:38 pm
helloIs there a typo with this question?? , since if there is a probability of 1 of breaking down then its certain it will breakdown regardless of what the situation is.
i just need help with a question from my textbook
The two machines in a workshop each have a probability of 1
of breaking down. Find the probability that at any one time (a) neither machine will be broken down
(b) 1 machine will be broken down.
Post by: amelia20181 on July 21, 2018, 10:28:25 pm
Is there a typo with this question?? , since if there is a probability of 1 of breaking down then its certain it will breakdown regardless of what the situation is.
oh sorry its 1/45
Post by: amelia20181 on July 21, 2018, 10:36:32 pm
Post by: envisagator on July 21, 2018, 10:44:53 pm
helloa)
i just need help with a question from my textbook
The two machines in a workshop each have a probability of 1
of breaking down. Find the probability that at any one time (a) neither machine will be broken down
(b) 1 machine will be broken down.
b) As the question doesnt state a specific order in which the either one of the machine will break down, there are two possibilities: Machine 1 breaks down and machine 2 doesnt, OR machine 1 doesnt break down and machine 2 breaks, so:
Post by: amelia20181 on July 21, 2018, 10:55:11 pm
a)=\frac { 44 }{ 45 } \times \frac { 44 }{ 45 } \\ )
b) As the question doesnt state a specific order in which the either one of the machine will break down, there are two possibilities: Machine 1 breaks down and machine 2 doesnt, OR machine 1 doesnt break down and machine 2 breaks, so:
how would you draw a tree diagram for that question
Post by: envisagator on July 21, 2018, 11:05:15 pm
how do you do this
Post by: amelia20181 on July 21, 2018, 11:16:20 pm
Post by: amelia20181 on July 21, 2018, 11:22:03 pm
Post by: envisagator on July 21, 2018, 11:48:55 pm
how do you do this
Post by: skisso on July 22, 2018, 04:38:48 pm
Hey hey! So they conclude that \(q\) must be negative, because the point Q is to the left of the origin. So, it's \(x\) coordinate must be negative.
So:
Since Q is left of the origin, then \(q\) must be a negative number ;D
Ohh i see, thank you :))
btw your lecture was superrr helpful so thank you!
Post by: jamonwindeyer on July 22, 2018, 04:42:49 pm
also how do you differentiate this
Hi! Chain rule says that you differentiate the inside of the function:
Then this goes out the front of the sine function, which you then turn into a cosine function because of differentiation:
I hope this helps!! Also, try not to post a heap of questions as separate posts - Put them all in one post! You can actually go back and add to your last post by clicking 'Modify' just above the post you made ;D
Post by: jamonwindeyer on July 22, 2018, 04:44:49 pm
Ohh i see, thank you :))
btw your lecture was superrr helpful so thank you!
You are so welcome!! Really glad it helped ;D
Post by: infectmarshroom on July 22, 2018, 10:26:34 pm
Came across this exciting (not! :-[) log and exp question and was stumped for part 3 and hence was not able to complete part 4. I know it probably has something to do with the features of y=x like gradients etc but can't get any conclusive answer with my maths of a coordinate purely in terms of 'a' without x's. Help is greatly appreciated!
Q16. part b
For some a>0 two curves f(x)=a^x and g(x)=loga(x) are drawn on the same axes so that they touch on y=x.
i) Write down expressions for f'(x) and g'(x) (2mks)
ii) Write down an equation involving natural logarithms whose solution is the x value at their point of contact. (1mk)
iii) Find the coordinate of the point of contact (2mks)
iv) What is the value of a? (1mk)
Btw Jamon your 2u and 3U lectures were amazing, love what you guys do and what you stand for. ;D
Post by: RuiAce on July 23, 2018, 08:28:31 am
Hey,
Came across this exciting (not! :-[) log and exp question and was stumped for part 3 and hence was not able to complete part 4. I know it probably has something to do with the features of y=x like gradients etc but can't get any conclusive answer with my maths of a coordinate purely in terms of 'a' without x's. Help is greatly appreciated!
Q16. part b
For some a>0 two curves f(x)=a^x and g(x)=loga(x) are drawn on the same axes so that they touch on y=x.
i) Write down expressions for f'(x) and g'(x) (2mks)
ii) Write down an equation involving natural logarithms whose solution is the x value at their point of contact. (1mk)
iii) Find the coordinate of the point of contact (2mks)
iv) What is the value of a? (1mk)
Btw Jamon your 2u and 3U lectures were amazing, love what you guys do and what you stand for. ;D
______________________________________________________
______________________________________________________
Note that \(a\neq 1\) for the sake of logarithms.
or equivalently, \(x = \left(\ln\left( a^{-1}\right)\right)^{-1} \).
______________________________________________________
Post by: jamonwindeyer on July 23, 2018, 10:39:26 am
Btw Jamon your 2u and 3U lectures were amazing, love what you guys do and what you stand for. ;D
Hey! Welcome to the forums! Super glad you found the lectures helpful! ;D
Was coming back to sit down and work through this this morning, but now I've seen Rui's working, maybe I'm glad he beat me to it ;) will just note that this question (while it is a fantastic question) is really beyond what would be asked in 2U Exams. Especially don't like how much emphasis is placed on the word 'touch,' if they were to ask anything like this they would want to be more specific about exactly what information you need to take from that.
Super cool challenge though - Be sure to post any others you need help with ;D
Post by: priyanka.sharma13 on July 23, 2018, 12:48:26 pm
“Abby has set up her superannuation fund and after 10 years she has accumulated $134,000. However due to an accident she is no longer able to work and make further contributions to the fund. Abby is leaving the money in the superannuation fund to accumulate interest at 8% p.a. compounded annually. However she needs to withdraw $24,000 at the end of each year for normal living expenses. Write an expression for the amount in the fund after 3 years.”
Thanks!
Post by: RuiAce on July 23, 2018, 12:58:56 pm
Hi! I need help in this superannuation question:
“Abby has set up her superannuation fund and after 10 years she has accumulated $134,000. However due to an accident she is no longer able to work and make further contributions to the fund. Abby is leaving the money in the superannuation fund to accumulate interest at 8% p.a. compounded annually. However she needs to withdraw $24,000 at the end of each year for normal living expenses. Write an expression for the amount in the fund after 3 years.”
Thanks!
We don't care about the other pieces given for the sake of the question.
Post by: amelia20181 on July 23, 2018, 03:10:25 pm
13. The number of cats to dogs at a pet hotel is in the ratio of 4 to 7. If 3 pets are chosen at random, find the probability that
(a) they are all dogs (b) just one is a dog
(c) at least one is a cat.
15. At City Heights School it was found that 75% of students in year 12 study 13 units, 21% study 12 units and 4% study 11 units. If 2 students are selected at random from year 12, find the probability that
(a) 1 student will study 12 units (b) at least 1 student will study 13 units.
Post by: Mate2425 on July 23, 2018, 03:11:50 pm
Thanks.
Post by: RuiAce on July 23, 2018, 03:19:18 pm
Hi when the questions aks to e.g. use osine rule to find..., what is the directive meaning behind deduce?Deduce is just another way of saying "hence, prove (with the aid of ...)". The cosine rule has nothing to do with it directly; if the cosine rule has been mentioned, it means you need to use it TO prove something.
Thanks.
(In practice, deduce means to draw a conclusion out of something. In the world of maths, conclusions are of course, drawn from proofs.)
Post by: fun_jirachi on July 23, 2018, 04:25:00 pm
13. The number of cats to dogs at a pet hotel is in the ratio of 4 to 7. If 3 pets are chosen at random, find the probability that
(a) they are all dogs (b) just one is a dog
(c) at least one is a cat.
i think you just draw a tree diagram with cats and dogs, and since it doesnt matter what order you pick them in, its kinda like a three step tree diagram with cats and dogs at every step i think?
so for a), the only option is dog dog dog, so that's (7/11)^3
b) one dog means two cats and there are three with that option so the answer i think is 3*((4/11)^2)*(7/11)
c) at least one is a cat means no all dogs, so that equals 1 minus all dogs = 1 - (7/11)^3
can someone pls confirm
Post by: fun_jirachi on July 23, 2018, 04:40:07 pm
15. At City Heights School it was found that 75% of students in year 12 study 13 units, 21% study 12 units and 4% study 11 units. If 2 students are selected at random from year 12, find the probability that
(a) 1 student will study 12 units (b) at least 1 student will study 13 units.
you do sorta the same thing, where you pick two students, order doesnt matter, someone pls correct me if i'm wrong
a) only one student does 12 units, so there are four possibilities with that, by probability tree = 2*(0.21*0.04)+2*(0.21*0.75) = roughly 33.18%
b) at least one student studies 13 units
so i think here as long as you see 13 units in one of the options its a valid option that you can count
so i think by that logic and the probability tree, there are five options, the whole 13 line [13, 12 and 13, 11 and 13, 13] as well as 12, 13 and 11, 13
adding the probabilities up you get 0.75+0.21*0.75+0.04*0.75 which roughly equals 93.75%
i didnt really get your q14 cos i think the formatting of the question made it kinda hard for me to get all the info from the question
so yeah i think that's how you do it, but PLEASE correct me if im wrong so i can learn from my mistakes and not screw up in an exam situation
Post by: amelia20181 on July 23, 2018, 04:51:42 pm
Post by: S200 on July 23, 2018, 05:02:26 pm
also this
Post by: jamonwindeyer on July 23, 2018, 06:25:34 pm
can someone pls confirm
so yeah i think that's how you do it, but PLEASE correct me if im wrong so i can learn from my mistakes and not screw up in an exam situation
Nicely done fun_jirachi, just confirming your answers look all sweet to me - Thanks for helping out! Legend! ;D
Post by: priyanka.sharma13 on July 23, 2018, 10:18:41 pm
What is the greatest value of the function y = 4-2cosx?
I’m struggling with trig functions, need help
What is the solution to the equation (sinx+2)(2sinx+1)=0?
Mod Edit: Post merge, modify your previous post using 'Modify' instead of posting twice in a row :)
Post by: amelia20181 on July 23, 2018, 10:37:16 pm
Post by: envisagator on July 23, 2018, 10:44:50 pm
I’m struggling with trig functions, need helpfor some reason the my response wasnt showing the whole LaTex script so I posted the answer as an image
What is the solution to the equation (sinx+2)(2sinx+1)=0?
EDIT: I forgot to state that: therefore, sinx=-2, has no real solutions
Post by: envisagator on July 23, 2018, 10:52:11 pm
how do you do b?So you have found the points of inflexion from the first derivative, now find the second derivative.
Solve the second derivative equal to zero, this should give you an x-coordinate identical to one of the x-coordinate of the point of inflexion, so at that point there is a horizontal point of inflexion, sub the x-value into the original equation to find the corresponding y-value :)
Hope this helps, if you need further assistance just ask :)
Post by: amelia20181 on July 23, 2018, 11:00:38 pm
how do you do this?
If 4 dice are thrown, find the probability that the dice will have
(a) four 6’s
(b) only one 6 (c) at least one 6.
Post by: amelia20181 on July 23, 2018, 11:09:08 pm
So you have found the points of inflexion from the first derivative, now find the second derivative.
Solve the second derivative equal to zero, this should give you an x-coordinate identical to one of the x-coordinate of the point of inflexion, so at that point there is a horizontal point of inflexion, sub the x-value into the original equation to find the corresponding y-value :)
Hope this helps, if you need further assistance just ask :)
i though you find the points of inflexion from the second derivative , but then how do you know if its a horizontal point of inflexion
Post by: jamonwindeyer on July 23, 2018, 11:14:38 pm
i though you find the points of inflexion from the second derivative , but then how do you know if its a horizontal point of inflexion
Indeed you do find points of inflexion with the second derivative! If any of these are also stationary points from the first derivative, then they are horizontal points of inflexion ;D
Post by: amelia20181 on July 23, 2018, 11:18:48 pm
Indeed you do find points of inflexion with the second derivative! If any of these are also stationary points from the first derivative, then they are horizontal points of inflexion ;D
ohh ok thanks!
Post by: jamonwindeyer on July 23, 2018, 11:26:02 pm
Hi! Need help, I don’t understand the question
What is the greatest value of the function y = 4-2cosx?
Hi! The question is asking how high the curve gets at the peak of its oscillation. So the \(y=4\) part of the equation shifts the curve upwards to oscillate about \(y=4\), and the function will go 2 above and 2 below that (since it is \(2\cos{x}\). 2 above 4 is 6, so \(y=6\) is the greatest value of the function :)
how do you do this?
If 4 dice are thrown, find the probability that the dice will have
(a) four 6’s
(b) only one 6 (c) at least one 6.
Hi! So the trick with these is to always draw tree diagrams. Like, anytime you get a probability question, try drawing a tree diagram as the default response. It will help you visualise the problem on top of helping you calculate the answer itself.
For Part (a), each dice has a probability of \(\frac{1}{6}\) of being a six. So the probability of all four being a six:
For (b), the tree diagram comes in handy. You'll see four branches, each with three \(\frac{5}{6}\) and one \(\frac{1}{6}\), but in different orders. Either way, the answer is:
For Part C, we go to the complementary event. This is usually the case when we see at least in the question. In this case, the opposite of having at least one six is having no sixes. So, calculate the probability of no sixes then subtract that from 1 (or 100%, same thing). This is easier than individually counting every possible way to get at least one six.
Post by: amelia20181 on July 23, 2018, 11:29:56 pm
The probability of a pair of small parrots breeding an albino bird
is 2 . If they lay three eggs, find 33
the probability of the pair
(a) not breeding any albinos (b) having all three albinos
(c) breeding at least one albino.
Post by: jamonwindeyer on July 23, 2018, 11:35:10 pm
need help with this
The probability of a pair of small parrots breeding an albino bird
is 2 . If they lay three eggs, find 33
the probability of the pair
(a) not breeding any albinos (b) having all three albinos
(c) breeding at least one albino.
Hey! Have you had a read of my reply above? This question is almost identical. For example, not breeding any albinos should be a single branch on your tree diagram - The formatting doesn't look right for us to provide any worked solutions, the probability of breeding an albino bird can't be 2 because nothing can happen more than 100% of the time!
Whatever the numbers actually are, give a tree diagram a go, see how my working from above might be able to parallel to the question here :)
Post by: amelia20181 on July 24, 2018, 08:24:29 pm
use simpson's rule to find an approximation for
Post by: amelia20181 on July 24, 2018, 08:25:51 pm
use the trapezoidal rule to find an approximation for
Post by: emily_p on July 25, 2018, 12:09:39 am
how do you do this
use simpson's rule to find an approximation for
need help with an integration question
use the trapezoidal rule to find an approximation for
Attached two photos of calculations.
First image (Simpson’s Rule Q)
6 subintervals/strips means 7 function values
- work out width of each strip using h = (b-a)/n
- work out each x value for the function values
- work out each y value for the corresponding x value
- substitute into the Simpson’s Rule formula
For Trapezoidal rule question you’re given the function values, so:
- find out width of strips
- sub into Trapezoidal Rule formula
Hope this helps! :) If you aren’t sure about anything let me know and I’ll try to clarify.
(http://i67.tinypic.com/102ro7m.jpg) (http://i67.tinypic.com/mudftu.jpg)
Post by: amelia20181 on July 25, 2018, 03:59:42 pm
Post by: S200 on July 25, 2018, 04:01:36 pm
how do you do
For question 14
This graph is the inverse to an exponential graph.
Take the graph of y=e^x, and reflect it across the line y=x
The domain of the first becomes the range of the second and vice versa...
Take the graph of y=e^x, and reflect it across the line y=x
The domain of the first becomes the range of the second and vice versa...
Post by: envisagator on July 25, 2018, 04:05:44 pm
how do you doBy this stagye in hsc maths you should be quite familiar with the basic graph of y=lnx
For a recap, Domain refers to what values of x the curve exists
Range: refers to what y values exists for the curve.
So, the domain for y=lnx = x>0, and range is all real y
If you are unsure how this came about, use Geogebra - its a graphing calculator, helps with determining domain and range if you cant picture what the curve looks like.
https://www.geogebra.org/graphing
Post by: envisagator on July 25, 2018, 04:06:59 pm
It is a good way to figure out the domain and range, but, inverse function is a mathematics extension 1 course.For question 14This graph is the inverse to an exponential graph.
Take the graph of y=e^x, and reflect it across the line y=x
The domain of the first becomes the range of the second and vice versa...
Post by: S200 on July 25, 2018, 04:08:21 pm
It is a good way to figure out the domain and range, but, inverse function is a mathematics extension 1 course.Yeah Ok. I don't know my way around HSC maths that well sorry... :)
Post by: amelia20181 on July 25, 2018, 04:15:32 pm
which textbook do you think its best to use for 2unit maths? Is grove the easiest textbook
Mod Edit: Post merge, use Modify to avoid posting twice in a row!
Post by: vic321 on July 26, 2018, 11:39:34 pm
how do you sketch curves with exponentials
which textbook do you think its best to use for 2unit maths? Is grove the easiest textbook
Mod Edit: Post merge, use Modify to avoid posting twice in a row!
Hey!
If you're stuck, you could just sub in x values into your calculator and plot them on the number plane. However, it is good to know the basic form of an exponential. You can use desmos, a graphing website to help you if you aren't sure.
Hope that helps
Post by: jamonwindeyer on July 27, 2018, 12:00:07 am
how do you sketch curves with exponentials
which textbook do you think its best to use for 2unit maths? Is grove the easiest textbook
Mod Edit: Post merge, use Modify to avoid posting twice in a row!
I'd also suggest checking out this cheat sheet for sketching functions, including exponentials, it might help a bit! ;D
As for textbook - I would say Grove is the easiest one out there. Cambridge tends to be the most difficult. Fitzpatrick is in the middle - At least that is my view. The "best" is a matter of preference I'd say!! :)
Post by: amelia20181 on July 27, 2018, 12:44:00 pm
Post by: jamonwindeyer on July 27, 2018, 01:02:22 pm
how do you do this
Hey! So for the first bit, the total cost is given by the the cost per hour, multiplied by the number of hours spent travelling:
Now \(h\) can be found, because we know it is travelling 1500km at a speed of \(s\) kilometres per hour (we assume the units match, at least). So:
So we sub that in above:
The rest of the question is your usual process:
- Differentiate the function
- Find where that derivative is equal to zero to find a maxima/minima for speed
- Prove it is a minimum using second derivative check (or otherwise)
- Sub this speed back into the cost function to get a value
This is something you should have been doing a lot of in this topic, but if you've had trouble with it feel free to post some working and we can try and guide you! :)
Post by: amelia20181 on July 28, 2018, 09:21:57 pm
Mod Edit: Post merge, use 'Modify' to add to your previous posts.
Post by: S200 on July 28, 2018, 09:30:35 pm
also
Shouldn't be that hard to solve from there...
Mod Edit: Fixed LaTex, think the site glitched for a second! - Jamon
Post by: moq418 on July 29, 2018, 04:48:18 am
Post by: fun_jirachi on July 29, 2018, 12:41:26 pm
send help pls
(https://i.imgur.com/a9rcwcp.jpg)
Post by: RuiAce on July 29, 2018, 01:18:54 pm
I'm struggling with Max/Min problems and this one is the worst(https://i.imgur.com/VPzrA67.png)
send help pls
(https://i.imgur.com/a9rcwcp.jpg)
(may need to click on image to resize).
And then you should be able to check the second derivative yourself. So subbing back in for the length and the breadth the appropriate dimensions are \( \frac{2a}{3} \times \frac{4a \sqrt{3}}{3} \).
Post by: jamonwindeyer on July 29, 2018, 08:36:53 pm
how to prepare for 2unit math trial exam i only have 1 week left what should i do ive been doing some past papers trails is there more i should do
Hi! Past papers is an excellent way to go, keep doing them! Make sure you are marking them and actually revisiting stuff you are getting wrong. Reteach yourself using textbooks, online videos, get help from peers/teachers, etc ;D
Post by: owidjaja on July 29, 2018, 09:52:16 pm
I'm a bit unsure on how to do this question so it would be great if you could help me :)
Post by: RuiAce on July 29, 2018, 10:42:09 pm
Hey guys,
I'm a bit unsure on how to do this question so it would be great if you could help me :)
Post by: vic321 on July 30, 2018, 12:50:39 am
Hey guys,
I'm a bit unsure on how to do this question so it would be great if you could help me :)
What topic is this may i ask? :)
Post by: itssona on July 30, 2018, 01:18:40 am
In a kitchen where the temp is 20 degrees C, Mary takes a boiling water kettle off the stove at time zero. Five minutes later the temp of the boiling water is 70 degrees C.
(a) show that T=20 +80e^-kt satisfies the cooling equation dT/dt = -k (T-20) and gives the correct value of 100 at t=0
Post by: emily_p on July 30, 2018, 09:18:16 pm
whats the proper way to prove this?
In a kitchen where the temp is 20 degrees C, Mary takes a boiling water kettle off the stove at time zero. Five minutes later the temp of the boiling water is 70 degrees C.
(a) show that T=20 +80e^-kt satisfies the cooling equation dT/dt = -k (T-20) and gives the correct value of 100 at t=0
- differentiate the equation and sub in T-20 instead of 80e^-kt
- sub t=0 into the T equation
Hope this helps!
(http://i64.tinypic.com/1pytf4.jpg)
Post by: dermite on July 30, 2018, 09:23:35 pm
Hi! Past papers is an excellent way to go, keep doing them! Make sure you are marking them and actually revisiting stuff you are getting wrong. Reteach yourself using textbooks, online videos, get help from peers/teachers, etc ;D
wootube is the best
Post by: amelia20181 on July 30, 2018, 10:09:48 pm
Post by: fun_jirachi on July 30, 2018, 11:01:02 pm
how do you do
oooooh triangle inequality
(https://i.imgur.com/f29eAOZ.jpg)
Post by: RuiAce on July 31, 2018, 09:07:38 am
oooooh triangle inequalityBe careful in the jump from line 1 to line 2. The way you've written it, it looks as though you've assumed what you were trying to prove, and then squared it.
(https://i.imgur.com/f29eAOZ.jpg)
A workaround could be to add a few words and say that proving \( |a|+|b| \leq |a+b| \) is equivalent to proving \( ( |a|+|b| ) ^2 \leq |a+b|^2\), due to the fact that all terms are non negative.
Working out is otherwise fine though; basically the right idea
Post by: amelia20181 on July 31, 2018, 09:33:41 am
Post by: fun_jirachi on July 31, 2018, 10:47:45 am
Be careful in the jump from line 1 to line 2. The way you've written it, it looks as though you've assumed what you were trying to prove, and then squared it.
A workaround could be to add a few words and say that proving \( |a|+|b| \leq |a+b| \) is equivalent to proving \( ( |a|+|b| ) ^2 \leq |a+b|^2\), due to the fact that all terms are non negative.
Working out is otherwise fine though; basically the right idea
ok thanks! i was kinda implying that, but i didnt state it explicitly should i do that next time
also answer to question is attached
Post by: RuiAce on July 31, 2018, 10:53:38 am
ok thanks! i was kinda implying that, but i didnt state it explicitly should i do that next timeYeah it's highly recommended because assuming what you're trying to prove is technically classified as a mistake. Whilst in high school maths I would probably just ignore it, some harsh markers can pick you out for it. Best be safe than sorry :P
also answer to question is attached
Post by: amelia20181 on July 31, 2018, 11:16:37 pm
Post by: Ali_Abbas on August 01, 2018, 12:19:17 am
how do you do
Recall that \(\sqrt{a}\) = \(a^{1/2}\) for all non-negative real numbers \(a\). Simply make this substitution into the integrand and multiply the powers, giving \(5/2\) as the new exponent. You now have something of the form \(\int(ax+b)^{n}dx\) which will be equal to \(\frac{1}{a(n+1)}(ax+b)^{n+1} + C\).
Post by: amelia20181 on August 01, 2018, 12:54:22 pm
Post by: Fergus6748 on August 01, 2018, 03:57:01 pm
how do you doHey, so the way I would do the question would be to first convert it to index form, so: 2(x-3)^-2, and from there integrate to get -2(x-3)^-1 or -2/(x-3). That would be when you use your limits, x=0,1.
f(x)=(-2/(1-3))-(-2/(0-3))
=(1-2/3)
so the answer would be:
=1/3 units^2
im pretty sure
Post by: LaraC on August 02, 2018, 09:55:39 am
Could someone please help me with the following question:
In a set of 30 cards, each one has a number on it from 1 to 30. If 1 card is drawn out, then replaced and another drawn out, find the probability of getting:
a) a 3 on the first card and an 18 on the second card (I have this answer --> 1/30 x 1/30 = 1/900
but the second part is:
b) a 3 on one card and an 18 on the other card.
The answer to b) is 1/450. I'm a little confused how it is different to the first part and how they got the different answer!?
Thanks :D
Post by: fun_jirachi on August 02, 2018, 11:23:08 am
Hello!
Could someone please help me with the following question:
In a set of 30 cards, each one has a number on it from 1 to 30. If 1 card is drawn out, then replaced and another drawn out, find the probability of getting:
a) a 3 on the first card and an 18 on the second card (I have this answer --> 1/30 x 1/30 = 1/900
but the second part is:
b) a 3 on one card and an 18 on the other card.
The answer to b) is 1/450. I'm a little confused how it is different to the first part and how they got the different answer!?
Thanks :D
ok the answer is correct
this is because 3 on one card and 3 on another means they can be in any order ie. 3 first, 18 second or 18 first 3 second.
part a) you had the specific scenario of 3 first 18 second, so the answer was 1/900, but since you have a second identical scenario, its 2/900, or 1/450
Post by: Mate2425 on August 02, 2018, 01:57:43 pm
i) What is the probability that the name of a girl and then a boy is picked?
ii) What is the probability of getting a boy and a girl?
Thank you.
Post by: Fergus6748 on August 02, 2018, 08:13:03 pm
Hey guys how would i approach this question. Names of 10 boys and 8 girls are placed in a hat. Another set of different names of 8 boys and 5 girls are placed in another hat. One name is selected from each hat.Heya, honestly the easiest way to approach this would be to make a tree diagram with two events. So the first branches is the first hat, while the next section is the second hat. So:
i) What is the probability that the name of a girl and then a boy is picked?
ii) What is the probability of getting a boy and a girl?
Thank you.
i) Probability of event 1 being a girl times the proability of event 2 being a boy
(8/18) x (8/13) = 64/234
ii) Simply, the above answer times two. If you draw out the tree diagram there are four possibilites; Boy/Boy, Boy/Girl, Girl/Boy, and Girl/Girl
(64/234) + (64/234) = 128/234
Post by: fun_jirachi on August 02, 2018, 08:31:52 pm
ii) Simply, the above answer times two. If you draw out the tree diagram there are four possibilites; Boy/Boy, Boy/Girl, Girl/Boy, and Girl/Girl
(64/234) + (64/234) = 128/234
isnt the answer a boy in the first girl in the second + girl in the first and boy in the second
im pretty sure, correct me if im wrong, B1G2 doesnt equal G1B2
so the answer is (8/18) x (8/13) + (10/18) x (5/13) = 114/234 = 19/39
Post by: Fergus6748 on August 02, 2018, 09:23:32 pm
isnt the answer a boy in the first girl in the second + girl in the first and boy in the secondOh right, yeah you're right, I just assumed they were the same
im pretty sure, correct me if im wrong, B1G2 doesnt equal G1B2
so the answer is (8/18) x (8/13) + (10/18) x (5/13) = 114/234 = 19/39
Post by: amelia20181 on August 03, 2018, 12:44:54 pm
Post by: fun_jirachi on August 03, 2018, 01:56:22 pm
y=1-x^2
=(1+x)(1-x)
ie. it cuts the x-axis at -1 and 1
as such we integrate y=1-x^2 between 1 and -1, or we can just integrate y=1-x^2 from 1 to 0 and multiply the result by two, seeing as it is an even function
integrating 1-x^2 between 1 and 0 gets you 2 x ((1-(1^3)/3)-(0-(0^3)/3))
= 2 x 2/3 = 4/3
Post by: amelia20181 on August 03, 2018, 03:39:59 pm
Post by: fun_jirachi on August 03, 2018, 04:50:22 pm
Post by: S200 on August 03, 2018, 04:57:16 pm
how do you doBy this time in the year, you should know what the graph of
The +5 is just translation to the left - As shown by @fun_jirachi
Post by: amelia20181 on August 03, 2018, 08:22:05 pm
Post by: jamonwindeyer on August 03, 2018, 10:59:40 pm
is it possible to get a band 5/6 if you fail everything but do well in trials and the hsc?
Absolutely! You can read the specifics here, but the HSC exam is really the thing you want to do well in. A super strong performance there can largely make up for less than ideal results earlier in the year.
Like, to go from failing your exams to getting a Band 5/6 is a big jump, and you'd need to put in an insane amount of work. But it is definitely possible :)
Post by: gumscape on August 04, 2018, 01:54:22 pm
Post by: dermite on August 04, 2018, 02:01:16 pm
Hello! Trials is going to get me k i l l e d. I'm absolutely stuck when it comes to Simpson's and Trapezoidal rule for more than one application. I just cannot get my head around it. Can someone give me a thorough answer for Q12 d) i. 2012 HSC please? Thank you!
well h = 3 as the interval lengths are 3.
then you have 1(0.5) + 4(2.3) + 2(2.9) + 4(3.8) + 1(2.1)
so
A ~ 3/3 [1(0.5) + 4(2.3) + 2(2.9) + 4(3.8) + 1(2.1)]
= 32.8m2
Post by: LaraC on August 04, 2018, 05:25:29 pm
Could I have help with this probability question please and show me how it would be done?
Two musicians are selected at random to lead their band. One person is chosen from Band A, which has 8 females and 7 males, and the other is chosen from band B, which has 6 females and 9 males. Find the probability of choosing 1 female and 1 male?
thx
Post by: RuiAce on August 04, 2018, 05:33:50 pm
Hello!
Could I have help with this probability question please and show me how it would be done?
Two musicians are selected at random to lead their band. One person is chosen from Band A, which has 8 females and 7 males, and the other is chosen from band B, which has 6 females and 9 males. Find the probability of choosing 1 female and 1 male?
thx
Post by: amelia20181 on August 04, 2018, 10:38:58 pm
Post by: fun_jirachi on August 04, 2018, 10:42:50 pm
I don't think you can solve that without other information
Post by: amelia20181 on August 04, 2018, 10:54:47 pm
is there any other information from that question?
I don't think you can solve that without other information
theres no other information
Post by: fun_jirachi on August 04, 2018, 10:57:59 pm
*smacks face*
so R is the rate of change, or dy/dx
d/dx(e^4x) = 4e^4x
you notice that e^4x is equal to y, as given in y=e^4x
you substitute that back into the derivative of the function
giving R=4y
Post by: S200 on August 04, 2018, 11:00:31 pm
__________________________
Oops. I'll stick to VCE Maths... :D
Post by: amelia20181 on August 05, 2018, 02:59:17 pm
Post by: jamonwindeyer on August 05, 2018, 03:00:51 pm
how do you integrate this
Hey! Like this:
This is just using the rule on your reference sheet for integrating exponentials! :)
Post by: amelia20181 on August 05, 2018, 03:19:01 pm
Post by: fun_jirachi on August 05, 2018, 04:25:30 pm
EDIT: I just realised I forgot the dx when writing down the integral of 3x^2 - 2x + 1, but make sure you put that in when writing the integral! :)
Post by: amelia20181 on August 05, 2018, 05:18:48 pm
Post by: amelia20181 on August 05, 2018, 05:27:20 pm
:)
EDIT: I just realised I forgot the dx when writing down the integral of 3x^2 - 2x + 1, but make sure you put that in when writing the integral! :)
Thanks!
Post by: fun_jirachi on August 05, 2018, 06:34:18 pm
Post by: amelia20181 on August 06, 2018, 02:10:09 pm
Post by: dermite on August 06, 2018, 02:13:47 pm
how do you doform a GP:
a = 0
r = 10%
n = ?
then use sum of a go
s = a(rn-1)/r-1
i dont hve a calculator on me currently, soz, but thatsthe method
Post by: amelia20181 on August 06, 2018, 02:37:20 pm
Post by: Fergus6748 on August 06, 2018, 03:21:58 pm
how do you do this questionHeya,
So the first question is about primitive functions, first you want to integrate to get the first derivative:
second derivative = 12x +6
first derivative = 6x^2 + 6x + C
So now you have to find out what C is by equating the first derivative with 1 and subbing in x (-1) from the point:
first derivative = 1
first derivative also = 6x^2 + 6x + C
1 = 6(-1)^2 +6(-1) + C
1 = 6 - 6 + C
1 = C
Therefore, first derivative = 6x^2 + 6x + 1
Then find the equation of the curve:
First derivative = 6x^2 + 6x +1
Equation = 2x^3 + 3x^2 + x + C
Then find C again:
-2=2(-1)^3 + 3(-1)^2 + (-1) + C
-2=-2 + 3 - 1 + C
0=2 + C
C=-2
y = 2x^3 + 3x^2 + x - 2
Post by: Fergus6748 on August 06, 2018, 03:30:06 pm
how do you do this questionWith the second question, you have to integrate the acceleration equation to find the velocity equation:
Acceleration = 25 e^5t
Velocity = 5e^5t
(a)
Then sub in t = 9:
v = 5e^5(9)
= 5e^45
(b) Then integrate the velocity equation, to find the displacement equation:
Velocity = 5e^5t
Displacement = e^5t
Then sub in t = 6
x = e^5(6)
= e^30
(c)
a = 25e^5t
= 25 (e^5t)
= 25x
As x = e^5t
(d)
Sub in x = 2 to the acceleration equation
a = 25(2)
= 50 ms^-2
And that's it
Post by: amelia20181 on August 06, 2018, 09:20:59 pm
Post by: Opengangs on August 06, 2018, 09:29:48 pm
\[ P = Ae^{-kt}\]
We can easily notice that \(A\) is just 150 since at \(t = 0\), the blood alcohol level is \(150\). Could you do the rest from here? :)
Post by: fun_jirachi on August 06, 2018, 09:42:18 pm
a) so after three hours, his BAC will have decayed to approximately 51.2%
to get this answer, you simply take 0.8^3 and convert it to a percentage
the amount that is actually in his bloodstream is about 76.8 mg/dL
if you actually wanted to use calculus/any graphing technique you could look at the graph y=0.8^x, where y = %alcohol still in system and x = time
you'll still get the same answer
b) 20/150= 0.133333333333...
so you calculate when y=0.8^x cuts y=0.133333333...
0.8^x=0.133333333333...
x=log 0.1333333333.../log0.8 (using log laws)
x= approximately 9.030 hrs
you don't actually need calculus to solve this question
hope this helps :)
Post by: fergo on August 07, 2018, 10:08:02 am
Can someone help me with the sum and products of roots? I know the two main formulas are on the formula sheet but I always forget how to do the ones when they manipulate the equations, like when you have to square or times the basic formulas to get the new ones.
I'm trying to just find the other formulas but am having no luck, can anyone help me?!
Thanks!
Post by: fun_jirachi on August 07, 2018, 10:20:07 am
Usually, you try and simplify the ones where they manipulate the roots into forms like (a-b)^2 into a+b and ab. Also, it's very common to find that when there is a more complex expression with the only variables being the roots, that there are prior questions that you answer to make the simplification easier. There are no other formulas, its just learning how to simplify algebraic expressions into simpler expressions. If you have any specific examples that could help me better explain this to you, feel free to post them up. :)
Hope this helps! :)
Post by: amelia20181 on August 07, 2018, 10:27:01 am
Post by: fun_jirachi on August 07, 2018, 10:45:19 am
The second derivative is -6, showing the that x has a maximum. The maximum is at the point where the derivative = 0, ie. where 15-6t=0
At the point t=15/6, the ball is 18.75m up the slope.
b) With the current formula of distance, you're looking at the ball going up the slope, so when you differentiate the velocity will be negative, as the ball is travelling down the slope, not up.
You need to first find the time t when it reaches the bottom of the slope. 15t-3t^2 = 3t(5-t) , so it will reach the bottom of the slope at t=5s. Subbing into the derivative 15-6t, the ball is travelling 15m/s down the slope.
c) I may be wrong for this one as the wording is a bit weird. I assume that the whole motion means the time it takes to go all the way up and all the way down. That's just 5s as I figured it part b)
Hope this helps! :)
Post by: amelia20181 on August 07, 2018, 05:13:43 pm
Post by: dermite on August 07, 2018, 05:42:18 pm
if this is the graph for displacement how would you graph velocity and acceleration
take the gradients of each subsequent graph
taking the gradient of a displacement graph gives velocity as dx/dt = v = x(dot)
taking the gradient of a velocitry/time graph will give the acceleration as dv/dt = a = x(double_dot)
Post by: amelia20181 on August 07, 2018, 05:45:13 pm
Post by: amelia20181 on August 07, 2018, 05:48:56 pm
Post by: fun_jirachi on August 07, 2018, 05:55:02 pm
for velocity should the graph be a straight line below the x axis
yes for that graph the velocity should be a horizontal line below the x-axis, as the gradient of the 'curve' is constant, and negative
for this question how would you find the total area
The garden is 40x25. The path surrounding it is 2m wide, which means it adds 2m to each side of the garden's dimensions. The yard as such is 44x29m.
The percentage of the yard that is garden I'm pretty sure you can figure out for yourself.
Hope this helps :)
Post by: amelia20181 on August 07, 2018, 07:17:42 pm
Post by: jamonwindeyer on August 07, 2018, 07:21:02 pm
how do you describe the motion of the particle at T1 for this displacement graph
Hey! So to describe the motion you need three things:
- Where is it (position)
- How fast is it moving and in what direction (velocity)
- What about acceleration? Is It speeding up or slowing down?
We can see from the graph that it is passing through \(x=0\) at that time, so we're at the origin. As for speed, well we're moving to the right, since the slope of the graph is positive (we're going from negative x to positive x). For acceleration, I throw to you, do you think it is getting faster? Slowing down? About the same? :)
Post by: amelia20181 on August 07, 2018, 07:40:48 pm
Post by: jamonwindeyer on August 07, 2018, 07:46:09 pm
is it getting faster
Yep, awesome stuff! If the line were straight it would be going a constant speed, but it is curving upwards, so it is getting faster/accelerating! :)
Post by: StephTol on August 07, 2018, 08:07:08 pm
I was just wondering how to do this question, would I need to draw it out?
Thanks
Post by: Opengangs on August 07, 2018, 09:27:03 pm
Hey!Hey, StephTol.
I was just wondering how to do this question, would I need to draw it out?
Thanks
So for these types of questions, note that solutions occur when \(\tan x = 2\) OR \(\sin x = 1/2\).
In knowing this, we note that there is exactly one solution for \(\tan x = 2\) in between 0 and \(\pi\), and that occurs at \(x = \tan^{-1}(2)\).
How many solutions are there for \(\sin x = 1/2\)?
Well, we know that \(\frac{\pi}{6}\) is another unique solution. And finally, there lies another solution: \(\frac{5\pi}{6}\). We can find this by determining the nature of the sine curve.
(https://i.imgur.com/0g6mFFA.png)
Notice that there is a symmetry between \(\frac{\pi}{2}\) and \(\frac{\pi}{6}\)! So, to find where it hits the sine curve, we simply note that:
\[ \begin{align*}x &= \left(\frac{\pi}{2} - \frac{\pi}{6}\right) + \frac{\pi}{2} \\ &= 1 - \frac{\pi}{6} \\ &= \frac{5\pi}{6} \\ &\leq \pi\end{align*}\]
In all, we know there are 3 unique solutions!
Post by: amelia20181 on August 07, 2018, 11:28:51 pm
Post by: fun_jirachi on August 08, 2018, 05:17:56 am
how do you do this question
Integrating the second derivative gets you the derivative. As such, integrating 8x gives you the derivative of the function which is 4x^2 + C, since we aren't given any upper or lower limits. however, at (-2, 5) the derivative y = 4x^2 + C = 1 because it states that the tangent makes an angle with the x axis of 45 degrees. It's important to remember tangent means gradient ie. first derivative at a point and also that gradient = tan theta. As tan 45 = 1, we know the gradient is 1. so you sub in x = -2 and you get C = -15.
So we now know that y' = 4x^2 - 15
integrating we get y = 2x^3 - 15x + C
now because it has a point (-2, 5) on its curve, we substitute those values in
5 = -16 + 30 + C
so C = -9
so the function is y = 2x^3 - 15x -9
Hope this helps! :)
Post by: isabella104 on August 08, 2018, 08:13:58 am
I'm pretty new here, this is my first post! Hope I'm doing it properly haha.
I've been having trouble with these questions I have attached below from the 2017 CSSA Trial Paper. It provides solutions but I don't know how they got there!
In 13.a.i, I don't understand how they got the answer (I thought your were meant to use the product rule?)
In 14.a.ii, I don't understand where the π comes from for (5-π)/2
I'm sure there are other people out there that have been struggling with these questions because my maths class have been talking about them non-stop!
Post by: jamonwindeyer on August 08, 2018, 08:27:26 am
Hi everyone!
I'm pretty new here, this is my first post! Hope I'm doing it properly haha.
I've been having trouble with these questions I have attached below from the 2017 CSSA Trial Paper. It provides solutions but I don't know how they got there!
In 13.a.i, I don't understand how they got the answer (I thought your were meant to use the product rule?)
In 14.a.ii, I don't understand where the π comes from for (5-π)/2
I'm sure there are other people out there that have been struggling with these questions because my maths class have been talking about them non-stop!
Hi Isabella! You are indeed doing it right, welcome! ;D
- That first one is a trick - That first exponential isn't a function. It is just a number! Like, put \(e^\pi\) in your calculator, it has a value! So this is just a case like \(y=2x\) or \(y=\frac{5}{4}x\). It just so happens the number out front is an exponential. Just like those other cases, the derivative of a number out the front of \(x\) is just the number itself.
- In that second one, the angle they are talking about is the angle in that little right angled triangle. To get that, we need the answer we had from Part (i), the whole angle, the 5/2. Then we subtract 90 degrees (or \(\frac{\pi}{2}\)) to just leave the triangle.
I hope this helps ;D
Post by: SanaBanana on August 08, 2018, 12:01:30 pm
say you do a math question and it has parts to it, like (i),(ii), and so on, and you figure out how to solve the latter parts before the former, and then you USE the answers from the latter parts to solve the former parts. can you be marked down or not get the marks for the former parts?
(do i make any sense at all??)
Post by: RuiAce on August 08, 2018, 12:03:24 pm
hiya, so ihave a question.Yes. Whilst you're always allowed to use the former parts to solve the latter, in general you cannot use the latter to solve the former.
say you do a math question and it has parts to it, like (i),(ii), and so on, and you figure out how to solve the latter parts before the former, and then you USE the answers from the latter parts to solve the former parts. can you be marked down or not get the marks for the former parts?
(do i make any sense at all??)
The only exception is the oddball case where you actually solved both the former and the latter at the same time, possibly without realising it.
(If I were an examiner, I probably would've just said whatever. But questions are designed so that each part is supposed to somehow follow from the previous ones unless explicitly stated otherwise. So other examiners do have every right to penalise you for it.)
Post by: billy.ohlmeyer on August 08, 2018, 04:42:33 pm
Given that the wingspan of an aeroplane is 30m, find the plane's altitude to the nearest metre if the wingspan subtends an angle of 14' when it is directly overhead.
how do i do this?
thanks
Post by: isabella104 on August 08, 2018, 04:50:27 pm
Thank you so much Jamon for your help with the previous questions I posted on this thread!
These multiple choice questions are the death of me...without the worked solutions I am lost!!! Does anyone know how to solve any of these questions I have attached below?
The answers are:
5. C
7. D
8. C
9. A
Thanks!
Also, just a question in terms of how to approach mathematics exams...I've overheard some people with tutors recommend finishing the 3 hr exam in 2 hrs so that there is 1 hr remaining to go through and double check your work. I tend to be very slow in maths exams, with 5-10 mins (if any) to go back through my work. Should I be concerned?? Or is it unnecessary to finish the exam 1 hour early?? I try to go slowly so that I don't make mistakes the first time round, and in the time remaining I tend to quickly go through the questions I couldn't complete previously because they were too hard. Any thoughts?
Post by: isabella104 on August 08, 2018, 04:54:13 pm
hiya, so ihave a question.
say you do a math question and it has parts to it, like (i),(ii), and so on, and you figure out how to solve the latter parts before the former, and then you USE the answers from the latter parts to solve the former parts. can you be marked down or not get the marks for the former parts?
(do i make any sense at all??)
In my experience, if you get the wrong answer in (i) but use the correct processes in (ii), (iii) and so on, you can still get the marks for (ii), (iii), etc. because you understood the process. But in my experience this only tends to happen if you accidentally copy down the question wrong or something in (i). Hope this helps :)
Post by: Never.Give.Up on August 08, 2018, 05:16:31 pm
Hi again everyone!Hey ;D
Thank you so much Jamon for your help with the previous questions I posted on this thread!
These multiple choice questions are the death of me...without the worked solutions I am lost!!! Does anyone know how to solve any of these questions I have attached below?
The answers are:
5. C
7. D
8. C
9. A
Thanks!
Also, just a question in terms of how to approach mathematics exams...I've overheard some people with tutors recommend finishing the 3 hr exam in 2 hrs so that there is 1 hr remaining to go through and double check your work. I tend to be very slow in maths exams, with 5-10 mins (if any) to go back through my work. Should I be concerned?? Or is it unnecessary to finish the exam 1 hour early?? I try to go slowly so that I don't make mistakes the first time round, and in the time remaining I tend to quickly go through the questions I couldn't complete previously because they were too hard. Any thoughts?
For q 5
You can use null factor law between o and pi
so,
tanx-2=0 and 2sinx-1=0
therefore tanx= 2 between 0 and pi (gives one solution as tan is only positive in this domain in first quadrant)
and 2sinx=1, sinx=1/2 (gives 2 solutions as sin is positive in both 1st and 2nd quadrants which is in domain of 0 and pi)
therefore, there are 3 solutions.
for question 7
this is a geometric series where n= p-q+1, which is 15-3+1, n=13
t1= 2^3 =8
t2 = 2^4 = 16
t3 = 2^5 = 32 (and so on...)
from this a=8, r=2
therefore terms of geometric series --> a(r^n -1)
so, 8 (2^13 -1)
= 2^3 (2^13 -1)
= 2^16 -8 (with index law m^a (m^b) = m^a+b
therefore it is equal to d.
for question 8
you can find the area of the trapezium a= 1/2h(a+b) --> you must find area of trapezium above x-axis first so, 1/2(10)(2+5) = 35 (you need this later)
therefore because it is between -6 and k, we know it is the trapezium below the x-axis
so, a=1/2 (k-4) (4+6) h=k-a a=6 and b=4 (these are positive because they are a distance)
therefore a= 1/2 (k-4) (10)
=5(k-4)
=5k-20 =35 (because it is equal to zero you need to cancel them out by making them equal to each other)
5k= 55
k=11
therefore, C.
for question 9
I was also confused by this but i found if you sub in points for x, you can solve the equation (it must equal 0)
hope this makes sense ;D
Post by: isabella104 on August 08, 2018, 06:23:01 pm
Hey ;D
For q 5
You can use null factor law between o and pi
so,
tanx-2=0 and 2sinx-1=0
therefore tanx= 2 between 0 and pi (gives one solution as tan is only positive in this domain in first quadrant)
and 2sinx=1, sinx=1/2 (gives 2 solutions as sin is positive in both 1st and 2nd quadrants which is in domain of 0 and pi)
therefore, there are 3 solutions.
for question 7
this is a geometric series where n= p-q+1, which is 15-3+1, n=13
t1= 2^3 =8
t2 = 2^4 = 16
t3 = 2^5 = 32 (and so on...)
from this a=8, r=2
therefore terms of geometric series --> a(r^n -1)
so, 8 (2^13 -1)
= 2^3 (2^13 -1)
= 2^16 -8 (with index law m^a (m^b) = m^a+b
therefore it is equal to d.
for question 8
you can find the area of the trapezium a= 1/2h(a+b) --> you must find area of trapezium above x-axis first so, 1/2(10)(2+5) = 35 (you need this later)
therefore because it is between -6 and k, we know it is the trapezium below the x-axis
so, a=1/2 (k-4) (4+6) h=k-a a=6 and b=4 (these are positive because they are a distance)
therefore a= 1/2 (k-4) (10)
=5(k-4)
=5k-20 =35 (because it is equal to zero you need to cancel them out by making them equal to each other)
5k= 55
k=11
therefore, C.
for question 9
I was also confused by this but i found if you sub in points for x, you can solve the equation (it must equal 0)
hope this makes sense ;D
Thank you so much! You are a legend!
I'm still a bit confused on Q7,
where you wrote:
therefore terms of geometric series --> a(r^n -1)
so, 8 (2^13 -1)
= 2^3 (2^13 -1)
= 2^16 -8 (with index law m^a (m^b) = m^a+b
therefore it is equal to d.
I don't quite understand the those three lines. I think I'm misreading it, I keep interpreting it as:
8 x (2^[13-1])
=2^3 x 2^12
=2^15 ????
I don't understand where the 2^16 and the -8 comes from?? I thought they were all positive???
If you could clarify that would be great, if not, you were a MASSIVE help anyway. Thank you!!!
Post by: fun_jirachi on August 08, 2018, 06:32:21 pm
Hi again everyone!
Thank you so much Jamon for your help with the previous questions I posted on this thread!
These multiple choice questions are the death of me...without the worked solutions I am lost!!! Does anyone know how to solve any of these questions I have attached below?
The answers are:
5. C
7. D
8. C
9. A
Thanks!
Also, just a question in terms of how to approach mathematics exams...I've overheard some people with tutors recommend finishing the 3 hr exam in 2 hrs so that there is 1 hr remaining to go through and double check your work. I tend to be very slow in maths exams, with 5-10 mins (if any) to go back through my work. Should I be concerned?? Or is it unnecessary to finish the exam 1 hour early?? I try to go slowly so that I don't make mistakes the first time round, and in the time remaining I tend to quickly go through the questions I couldn't complete previously because they were too hard. Any thoughts?
so for your q9,
i guess maybe its expected for 2u that you know your basic trigonometric functions as graphed on a cartesian plane, but anyway
the regular y = tan x tends to positive infinity at pi/2 and negative infinity at -pi/2, and this continues at intervals of pi.
From knowing our tan graph, we know the answer isnt b
seeing that it cuts the x-axis at pi/4 also rules out C
now we notice that the asymptotes are in increments of pi/2, not pi, so we go to A, which is the only option left with a transformation that allows for this change -- you could look inside the brackets and simplify it down from y = tan (2x-pi/2) to y = tan (2(x-pi/4))
Hope this helps :)
just gonna edit in an explanation for q7 so you fully understand :)
the expression is essentially 2^3+2^4+2^5+...+2^14+2^15
from the sum of a geometric series we get the sum equal to (8(2^13-1))/(2-1) from a(r^n-1)/(r-1)
because there is a 1 on the denominator, the sum is equal to 8(2^13-1) = 2^3(2^13-1)=2^16-1 as previously stated
i think you're getting confused because you're inserting brackets from nowhere; you're combining the indice 13 and the number 1 into a single expression, like so
8(2^13-1) --> 8(2^(13-1))
which is wrong because BIMA
the 2^16 and the -8 come from just expanding the brackets and using indice laws
im gonna insert a paint picture just in case it helps, the not thing is what you're doing and that's how you're getting the 2^15, while the correct answer is derived from the thing on the left
:) Hope this helps
Post by: isabella104 on August 08, 2018, 07:50:35 pm
so for your q9,
i guess maybe its expected for 2u that you know your basic trigonometric functions as graphed on a cartesian plane, but anyway
the regular y = tan x tends to positive infinity at pi/2 and negative infinity at -pi/2, and this continues at intervals of pi.
From knowing our tan graph, we know the answer isnt b
seeing that it cuts the x-axis at pi/4 also rules out C
now we notice that the asymptotes are in increments of pi/2, not pi, so we go to A, which is the only option left with a transformation that allows for this change -- you could look inside the brackets and simplify it down from y = tan (2x-pi/2) to y = tan (2(x-pi/4))
Hope this helps :)
just gonna edit in an explanation for q7 so you fully understand :)
the expression is essentially 2^3+2^4+2^5+...+2^14+2^15
from the sum of a geometric series we get the sum equal to (8(2^13-1))/(2-1) from a(r^n-1)/(r-1)
because there is a 1 on the denominator, the sum is equal to 8(2^13-1) = 2^3(2^13-1)=2^16-1 as previously stated
i think you're getting confused because you're inserting brackets from nowhere; you're combining the indice 13 and the number 1 into a single expression, like so
8(2^13-1) --> 8(2^(13-1))
which is wrong because BIMA
the 2^16 and the -8 come from just expanding the brackets and using indice laws
im gonna insert a paint picture just in case it helps, the not thing is what you're doing and that's how you're getting the 2^15, while the correct answer is derived from the thing on the left
:) Hope this helps
Aaaaaaaaah, got it! Thanks so much!!!!!
Post by: amelia20181 on August 08, 2018, 08:26:39 pm
The combination lock on a safe has three concentric circular discs, each showing the digits 0 to 9. Only one combination of digits will open the safe. what is the probability of opening the safe at my first attempt if i do not know the combination?
Post by: fun_jirachi on August 08, 2018, 08:34:22 pm
how do you do this question
The combination lock on a safe has three concentric circular discs, each showing the digits 0 to 9. Only one combination of digits will open the safe. what is the probability of opening the safe at my first attempt if i do not know the combination?
I'm not too sure on this one, but I think this is how it works;
Because there are three concentric circular discs each with digits 0 to 9, you're essentially looking at the chance of guessing correctly a number for 000 to 999, in which case the answer is 1 in 1000. :)
Post by: jamonwindeyer on August 08, 2018, 09:58:30 pm
I'm not too sure on this one, but I think this is how it works;
Because there are three concentric circular discs each with digits 0 to 9, you're essentially looking at the chance of guessing correctly a number for 000 to 999, in which case the answer is 1 in 1000. :)
Perfect example of breaking down a wordy problem!! Love your work ;D
Post by: darlo69 on August 09, 2018, 06:56:47 pm
Post by: jamonwindeyer on August 09, 2018, 10:50:03 pm
Is it too late to drop to General? Trials is in less than a week and I can't even do the questions from the year 11 course ahhhh
You'd be relearning an entire course in three months! Definitely not a good call - Stick with it, pick a topic at a time and work really hard to understand it. I promise it will get easier! You just need to be willing to really invest a heap of time to make it happen ;D
Post by: amelia20181 on August 11, 2018, 06:18:06 pm
Post by: S200 on August 11, 2018, 06:40:54 pm
what should you do if you have no motivation to study or do anythingI'd say join Me and Techno in procrastinating on the NSW vs VIC thread, but the competition might be to much for us..
On the flip side, maybe dedicate half an hour to just chill, and then try to come back to work?
Post by: Lollzza on August 11, 2018, 07:17:10 pm
Thanks!
Post by: RuiAce on August 11, 2018, 07:18:43 pm
hi I had my trials recently (not CSSA so not security issues dw) and we had a question that asked us to integrate lnx and i was so confused because I thought 2U wasn't taught integration by parts? Isnt that a low blow??? Is there any other way to integrate it? I didnt actually get this in exam, it was an awful epiphany five minutes after the fact.There is if it's a definite integral.
Thanks!
Post by: amelia20181 on August 11, 2018, 07:42:32 pm
1- (0.85x0.85x0.85)
a certain type of plant has a probability of 0.85 of producing a variegated leaf. if i grow 3 of these plants, find the probability of getting a variegated leaf in
a) 2 of the plants
b) none of the plants
c) at least 1 plant
when you're integrating 4/x can you integrate 4x-1 or do you have to In
Post by: fun_jirachi on August 11, 2018, 08:13:03 pm
for this question for b can you do 1- probability of all of the plants
1- (0.85x0.85x0.85)
a certain type of plant has a probability of 0.85 of producing a variegated leaf. if i grow 3 of these plants, find the probability of getting a variegated leaf in
a) 2 of the plants
b) none of the plants
c) at least 1 plant
when you're integrating 4/x can you integrate 4x-1 or do you have to In
for part b no you can't because the part inside the brackets is only considering if all of them have variegated leaves. you're forgetting that two of them can have different leaves or one of them can have different leaves
the way to approach it would be (1-0.85)^3, using the chance of it being non-variegated on all three
also when integrating 4/x no you can't do the 4x^-1 thing because when you integrate it in the latter form, you have to add one to the power then divide by the new power, which causes you to divide by zero. the integral will always be 4 ln|x| + C.
Post by: ilikeapples on August 11, 2018, 08:59:57 pm
Post by: RuiAce on August 11, 2018, 09:06:33 pm
Hey I am having trouble with this question: There are 10 green marbles and 'W' white marbles in a bag. The probability of selecting a white marble is 4/9. How many more white marbles need to be added to the bag so that the probability of selecting a white marble from the bag is 3/5?
Post by: ilikeapples on August 11, 2018, 09:19:56 pm
Post by: Lollzza on August 11, 2018, 11:07:15 pm
There is if it's a definite integral.
Oh really?! I haven't been able to find any other methods for it online. If it's alright could you please tell me the name of the method? We were never taught how to integrate lnx so the exam stooped most of us ahah. :) Would love to know for next time.
Post by: RuiAce on August 11, 2018, 11:13:21 pm
Oh really?! I haven't been able to find any other methods for it online. If it's alright could you please tell me the name of the method? We were never taught how to integrate lnx so the exam stooped most of us ahah. :) Would love to know for next time.It's not really a method per se but more of a technique. Basically you draw out the graph, noting that \(y=\ln x\) converts to \(x = e^y\) and then do a difference of areas between a rectangle and something of the form \( \int_c^d e^y\,dy \), that of which you know how to integrate.
For example, \( \int_1^e \ln x\,dx = 1\times e - \int_0^1 e^y\,dy \)
Typically in the actual HSC it gets hinted, but for trials it's hard to say.
Post by: Lollzza on August 11, 2018, 11:33:02 pm
It's not really a method per se but more of a technique. Basically you draw out the graph, noting that \(y=\ln x\) converts to \(x = e^y\) and then do a difference of areas between a rectangle and something of the form \( \int_c^d e^y\,dy \), that of which you know how to integrate.
For example, \( \int_1^e \ln x\,dx = 1\times e - \int_0^1 e^y\,dy \)
Typically in the actual HSC it gets hinted, but for trials it's hard to say.
Oh wow okay! I'd never seen a question like this before so I'll be sure to remember for next time. Thanks for answering so well so late ahah. I think I overthought the question and didnt think on the graph too much. Thanks again, math really isn't my strong suit ;D
Post by: amelia20181 on August 12, 2018, 02:50:47 pm
Post by: jamonwindeyer on August 12, 2018, 04:01:53 pm
can someone pls explain simpsons and trapezoidal rule. I'm so confused rn
Hey! They are both essentially ways of estimating the value of definite integrals/approximating the area under a curve between two points. The trapezoidal rule does so by forming multiple trapeziums and summing their areas, and the Simpson's Rule uses (you can think of it this way at least) a parabola to make the approximation.
To use either rule, you just need the value of the curve/function in question at equally spaced points along the interval you care about! Once you have that, it is just substituting into the formula. Although you can understand how the results are derived (particularly for Trapezoidal), you don't need to, you can just substitute and go! ;D
Were there specific questions that were giving you trouble? Would be happy to lend a hand :)
Post by: amelia20181 on August 13, 2018, 03:41:50 pm
Post by: jamonwindeyer on August 13, 2018, 11:11:36 pm
what would happen if you didn't write your number on the exam paper
They would catch it picking it up, usually! Otherwise they cross reference the student list with the scripts to see who is missing, and other similar logistical checks ;D either way, no reason to be concerned about it!
Post by: S200 on August 14, 2018, 10:24:49 am
my textbook has this question:Hey there!
Given that the wingspan of an aeroplane is 30m, find the plane's altitude to the nearest metre if the wingspan subtends an angle of 14' when it is directly overhead.
how do i do this?
thanks
Just stalked your posts.. (
) and realised that no-one had answered this...Only 6 days late, but here goes.
Basically, a diagram of this would be an isosceles triangle, with a base of thirty and an angle of 14' opposite the base. By cutting it in two we can make a right-angled triangle, and with the base of this triangle (i.e: 15) and the angle given halved, we can find the height of the triangle...
So, by going
Therefore, the plane is 7367 Meters above the ground...
Post by: darlo69 on August 15, 2018, 03:53:17 pm
Find the coordinates of the center of a circle that passes (7,2) and (2,3) and (-4,-1)
Post by: fun_jirachi on August 15, 2018, 04:42:59 pm
How do you do this?
Find the coordinates of the center of a circle that passes (7,2) and (2,3) and (-4,-1)
all explained on the sheet :)
Post by: amelia20181 on August 18, 2018, 04:52:39 pm
Post by: fun_jirachi on August 18, 2018, 07:06:27 pm
how do you do
Cheers S200, saved me a bit of time on the last q there :)
Hey there!
Just stalked your posts.. (
Only 6 days late, but here goes.
Basically, a diagram of this would be an isosceles triangle, with a base of thirty and an angle of 14' opposite the base. By cutting it in two we can make a right-angled triangle, and with the base of this triangle (i.e: 15) and the angle given halved, we can find the height of the triangle...
So, by going
Therefore, the plane is 7367 Meters above the ground...
Answers to the other two attached :)
Post by: fun_jirachi on August 18, 2018, 07:31:25 pm
Just wondering where you guys (Jamon, Rui, S200) get that really cool font that looks like it came off Symbolab so I can stop using photos :)
Post by: S200 on August 18, 2018, 07:35:17 pm
Also I just realised my photos are really unclear, because I take them on my phone, then send them over Messenger to myself on my laptop, then post them. I'm guessing my laptop does this weird laptop rendering thing????It's called LaTex. :D
Just wondering where you guys (Jamon, Rui, S200) get that really cool font that looks like it came off Symbolab so I can stop using photos :)
The board has it inbuilt. Link in my sig is kinda exhaustive, and not all of it works. But if you just search the entire forum for Latex, you will get a few hits on tutorial threads
Post by: RuiAce on August 18, 2018, 07:37:51 pm
Also I just realised my photos are really unclear, because I take them on my phone, then send them over Messenger to myself on my laptop, then post them. I'm guessing my laptop does this weird laptop rendering thing????If you wanted to get started with \( \LaTeX \)
Just wondering where you guys (Jamon, Rui, S200) get that really cool font that looks like it came off Symbolab so I can stop using photos :)
Symbolab uses it, Wolfram uses it, a lot of things use it to display mathematics nowadays
I need to update it though. Two years or so later and I've become more proficient with it. Whilst not all of it, some of the stuff on there is a bit overkill
Post by: S200 on August 18, 2018, 07:43:36 pm
I've been wanting to do inline for ages! :D
Post by: fun_jirachi on August 18, 2018, 08:04:09 pm
Post by: amelia20181 on August 18, 2018, 08:22:30 pm
Post by: S200 on August 18, 2018, 08:24:10 pm
how do you doLiterally
This time you are solving for base, not height.
Edited working from last question
Basically, a diagram of this would be an isosceles triangle, with a
So, by going
Therefore, the sun is \( 2 \times \alpha\) km in diameter
Post by: amelia20181 on August 18, 2018, 09:14:02 pm
Thanks!
Post by: fun_jirachi on August 18, 2018, 10:10:46 pm
Looks a little messy rn, but I'm getting there with the code! :)
Hope this helps!
Post by: amelia20181 on August 19, 2018, 04:28:45 pm
Post by: fun_jirachi on August 19, 2018, 04:40:02 pm
how do you do this
I don't know if this is a legit way of doing it, but this is the way my teacher taught me.
So with evaluating limits, you just sub in a small number that is close to the number x is approaching, so for this example, sub in 0.001 or something, and you get 0.2499999583. If you keep subbing in even smaller values, the calculator just spits out a definitive value of 0.25, which is your answer.
:)
Post by: amelia20181 on August 19, 2018, 04:47:41 pm
Post by: RuiAce on August 19, 2018, 04:57:34 pm
I don't know if this is a legit way of doing it, but this is the way my teacher taught me.You can't sadly; that technique more or less exists to help make students' lives easier when graphing things with asymptotes.
So with evaluating limits, you just sub in a small number that is close to the number x is approaching, so for this example, sub in 0.001 or something, and you get 0.2499999583. If you keep subbing in even smaller values, the calculator just spits out a definitive value of 0.25, which is your answer.
:)
That particular question requires the limit \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) which is a known result. Although you'd sub \(\theta = 4x\) to obtain \( \lim_{x\to 0} \frac{\sin 4x}{4x} = 1\). It's a limit that's taught in the 2U course but never examined in 2U; only in MX1.
Post by: fun_jirachi on August 19, 2018, 05:08:21 pm
also this questionFirst one is similar to integrating sin x, but in degrees.
Second one
Post by: amelia20181 on August 19, 2018, 07:32:43 pm
the ratio of boys to girls in Redwell high school is 3:5. The ratio of boys to girls in Southcourt high school is 2:3. A student is chosen from redwell high school and then another student is chosen from Southcourt high school. Find the probability that at least one girl is chosen.
claudia is one of 10 employees in a company. This month the company will commence holding a charity fundraising raffle at the end of each month. There is only one winner each month. what is the probability that claudia will win the raffle exactly once within the first four months?
for this question would it be 2.8
Post by: fun_jirachi on August 19, 2018, 08:42:37 pm
would the answer for this question be 1.88
the ratio of boys to girls in Redwell high school is 3:5. The ratio of boys to girls in Southcourt high school is 2:3. A student is chosen from redwell high school and then another student is chosen from Southcourt high school. Find the probability that at least one girl is chosen.
claudia is one of 10 employees in a company. This month the company will commence holding a charity fundraising raffle at the end of each month. There is only one winner each month. what is the probability that claudia will win the raffle exactly once within the first four months?
for this question would it be 2.8
Ok firstly, just by looking at your answers, they're both wrong because the probability of something happening is always greater than or equal to 0 and less than or equal to 1. Also, in general its just better to express probability as a fraction, not a decimal.
For the first question,
For the second question, its her winning once, losing three times, multiplied by 4 for each month
Post by: amelia20181 on August 19, 2018, 08:59:27 pm
Post by: jamonwindeyer on August 19, 2018, 10:31:36 pm
how do you do this
Hey! So if you break it down you have:
- 13 married men
- 21 non-married men
- 19 married women
- 13 non-married women
For the first one, there are 34 men and we are picking two, so:
In general, the easiest way to approach any question like this involving multiple selections is to use a tree diagram. So I'd be drawing one of those for this question. It will have two stages, and each stage will have four branches coming off it to match the four groups. Once you put that together you can use it to make the rest of the questions much easier!! It helps visualise how the product/addition rules can be used :)
Post by: amelia20181 on August 19, 2018, 10:56:54 pm
For this question for a can you do 1/5 x 2/9 + 1/5 x 2/9
Mod Edit: Post merge :)
Post by: jamonwindeyer on August 19, 2018, 11:17:00 pm
Thanks!
For this question for a can you do 1/5 x 2/9 + 1/5 x 2/9
You are welcome!! That's not how I'd approach it, where are you getting that \(\frac{2}{9}\) from? :)
Post by: amelia20181 on August 19, 2018, 11:27:53 pm
Post by: jamonwindeyer on August 20, 2018, 06:51:11 am
wait how do you do the question
Another tree diagram is what I would do! Tree diagram is default response for when you are doing one thing, then another thing, etc. All the probabilities will be \(\frac{1}{5}\) at the first stage, as there are five cards, then \(\frac{1}{4}\) at the second stage, because there are four cards.
Part (a) for D and E, for example, would be:
Post by: amelia20181 on August 20, 2018, 09:15:10 pm
Post by: S200 on August 20, 2018, 09:23:47 pm
Each ticket has \(\left(\frac {1} {50} \right ) \) chance of winning a prize.
Anna has 4 tickets.
P= \(\frac {4} {50} \)
Post by: amelia20181 on August 20, 2018, 09:32:11 pm
The probability of an arrow hitting a target is 85%. If 3 arrows are shot, find the probability as a percentage, correct to 2 decimal places, of a) all arrows hitting the target b) no arrows hitting the target c) at least one arrow hitting the target
Post by: jamonwindeyer on August 20, 2018, 09:34:24 pm
for part c in the this question why cant you do 1- probability of no arrows hitting the target
The probability of an arrow hitting a target is 85%. If 3 arrows are shot, find the probability as a percentage, correct to 2 decimal places, of a) all arrows hitting the target b) no arrows hitting the target c) at least one arrow hitting the target
You can do that! 1 - the answer from (b) should be correct ;D
Post by: fun_jirachi on August 20, 2018, 09:43:11 pm
100 tickets, 2 prizes.
Each ticket has \(\left(\frac {1} {50} \right ) \) chance of winning a prize.
Anna has 4 tickets.
P= \(\frac {4} {50} \)
I dunno about this, because part a) saying first prize made me think there's a second prize, so like maybe there isnt an equal chance of winning them ie. second prize is drawn after the first prize.
If Alan wins first prize only its 4/100x96/99 and second prize only is 96/100*4/99, so its 768/9900? idk, might be wrong
EDIT: in bold, i previously wrote 792 because i can't type properly on a calculator :'( the new answer cancels down to the 64/825 you want
Post by: amelia20181 on August 20, 2018, 09:48:15 pm
Post by: S200 on August 20, 2018, 10:01:37 pm
The answer is 64/825 but i don't know how to get it
I dunno about this, because part a) saying first prize made me think there's a second prize, so like maybe there isnt an equal chance of winning them ie. second prize is drawn after the first prize.
If Alan wins first prize only its 4/100x96/99 and second prize only is 96/100*4/99, so its 768/9900? idk, might be wrong
EDIT: in bold, i previously wrote 792 because i can't type properly on a calculator :'( the new answer cancels down to the 64/825 you want
Hmm. I need to rework my maths... :-[ sorry...
Lucky New south Welshmen are smarter! ;)
Post by: amelia20181 on August 20, 2018, 10:05:34 pm
for real roots is the discriminant greater than or equal to zero
do the cambridge and fitzpatrick textbooks have worked solutions for all the answers
would this question be a band 6 type
Post by: jamonwindeyer on August 20, 2018, 10:41:21 pm
for real roots is the discriminant greater than or equal to zero
Yep! Otherwise you have the square root of a negative, which for 2U is total nonsense ;D
Quote
do the cambridge and fitzpatrick textbooks have worked solutions for all the answers
Pretty sure it is just solutions, no working, for both :)
Quote
would this question be a band 6 type
Yeah for sure, testing your ability to adapt to a new situation of the same type! ;D
Post by: amelia20181 on August 21, 2018, 05:18:20 pm
Post by: fun_jirachi on August 21, 2018, 05:52:09 pm
For part c in this question can you do 1- probability of not breeding any albinosYes you can! :)
Post by: amelia20181 on August 21, 2018, 05:53:58 pm
Post by: fun_jirachi on August 21, 2018, 06:01:35 pm
Post by: amelia20181 on August 21, 2018, 06:08:01 pm
Post by: fun_jirachi on August 21, 2018, 06:19:16 pm
Sometimes, if the answer doesnt make sense after an hour or so, you're probably right.
Post by: amelia20181 on August 21, 2018, 06:45:14 pm
Post by: jamonwindeyer on August 21, 2018, 06:50:18 pm
how would you do this question
You really don't have the information you'd need to know for sure, but the answer would be 'incorrect.' Bill is assuming they would actively not choose numbers 1 through 6, when in reality, even that highly coincidental combo has a chance of appearing :)
Post by: amelia20181 on August 22, 2018, 03:33:36 pm
Post by: fun_jirachi on August 22, 2018, 04:29:02 pm
First throw is a branch with 6 options {1,2,3,4,5,6}, and from each of those branch out another 6 options {1,2,3,4,5,6} and from each of THOSE options branch out another 6 options {1,2,3,4,5,6}.
Post by: gilliesb18 on August 23, 2018, 11:37:33 am
30. A set of 20 cards is numbered 1 to 20 with replacement. Find the probability of selecting:
a) all three 10's
b) no 10s
c) at least one 10.
I have done part a) but just can't seem to work out how to do parts b and c...
Any help is appreciated:)
Thanks...
Post by: RuiAce on August 23, 2018, 05:19:11 pm
Hello, just needing help on a question re Probability...I cannot figure out what this question is asking. What do you mean by the cards being numbered 1 to 20 “with replacement”? You haven’t specified that we’re drawing the cards, let alone how many cards are being drawn.
30. A set of 20 cards is numbered 1 to 20 with replacement. Find the probability of selecting:
a) all three 10's
b) no 10s
c) at least one 10.
I have done part a) but just can't seem to work out how to do parts b and c...
Any help is appreciated:)
Thanks...
However, judging by question a), were you meant to have said that “3 cards were being draw with replacement” in there somewhere?
Post by: amelia20181 on August 23, 2018, 07:02:58 pm
Post by: clovvy on August 23, 2018, 07:28:27 pm
how do you do b for this questionTo get any number in each dice is 1/6... in this case you roll 3 dices yea?
The probability of rolling 2 dices is
Post by: amelia20181 on August 23, 2018, 07:36:14 pm
Post by: amelia20181 on August 23, 2018, 07:43:43 pm
Post by: clovvy on August 23, 2018, 07:45:37 pm
The answer is 5/72Plug it in the calculator it is 5/72... I don't really know how to do it the 2U way... You basically chose 2 dice to be the value you want out of the 3.. hence 3 choose 2 (look for nCr button in the calculator, press shift then divide to get the c, 3C2)... you power the number of possibility that is 1/6 by two and the remaining that is NOT 2 by 1 in this case
Post by: amelia20181 on August 23, 2018, 08:11:40 pm
Post by: RuiAce on August 23, 2018, 08:40:58 pm
Plug it in the calculator it is 5/72... I don't really know how to do it the 2U way... You basically chose 2 dice to be the value you want out of the 3.. hence 3 choose 2 (look for nCr button in the calculator, press shift then divide to get the c, 3C2)... you power the number of possibility that is 1/6 by two and the remaining that is NOT 2 by 1 in this caseThey don't do nCr in 2U mathematics.
Post by: fun_jirachi on August 23, 2018, 09:09:26 pm
how do you do b for this question
Plug it in the calculator it is 5/72... I don't really know how to do it the 2U way... You basically chose 2 dice to be the value you want out of the 3.. hence 3 choose 2 (look for nCr button in the calculator, press shift then divide to get the c, 3C2)... you power the number of possibility that is 1/6 by two and the remaining that is NOT 2 by 1 in this case
The 2U way is to just analyse the chances indivdually.
So you have the chance of rolling a 6 as 1 in 6, the chance of not rolling a 6 is 5 in 6. You need two sixes, and one non-six, so you have
But the dice that isn't a six can be the first, second or third die, so you have 3 different forms of that, so you multiply that by 3 and you get your 5/72.
for this question is the answer for c 3/5
No, the answer is not 3/5 because you have restrictions on the selections you can place.
If you picked one card, yeah the probability would be 3/5 because you can't pick A & D, but when you pick two cards without replacement, restricting two cards the second selection has a probability of 2/4. tbh you could list out all the possibilities, since there are 20 (10, and their reverses, see which don't have A and/or D) and you'd still get the same answer. FYI the answer is for that is 3/10.
How do you do b for this questionIf you've drawn your Venn diagram correctly, you'll have 12 doing music only, 7 doing music and sports, 2 doing nothing and 9 doing sports only.
Since you're picking two different people, its without replacement ie your answer will be
Post by: amelia20181 on August 23, 2018, 09:18:10 pm
also how do you do this
Post by: fun_jirachi on August 23, 2018, 10:02:15 pm
b) Same sort of thing.
c) Is just asking how likely it is you will score exactly 13.
There are a only a few ways to do this, namely 5,5,3 (or some variant of that) 5,4,4 (or some variant of that). Each of those has 3 variants, including the listed variant.
Essentially it's just this, using the results from before.
d) Is just asking how likely it is you will score 14 or 15.
There are only 2 ways to do this, namely 5,5,4 (or some variant of that) and 5,5,5.
Hope this helps! :)
Post by: amelia20181 on August 23, 2018, 10:39:42 pm
Post by: shellmekler on August 25, 2018, 02:46:39 pm
Can someone please help me with Q13d from the 2016 HSC exam?
To find the shaded area - I thought to find the area of the curve between 2 and 0 and then deduct the area of the triangle using 1/2bh but I might of integrated wrong? I am not sure.
Thanks in advance!
Post by: envisagator on August 25, 2018, 03:30:37 pm
I think you can take it from here, hope this helps :) :)
Post by: Calley123 on September 02, 2018, 01:42:52 pm
Please help with question Q4) b and c). This is part of the applications for sequence and series.
Thanks :)
Post by: Calley123 on September 02, 2018, 02:30:17 pm
Still sequence and series questions.
Post by: fun_jirachi on September 02, 2018, 03:46:06 pm
Hey,For this one, b) is essentially the same as a), except you're plugging in 11 years instead of 1, 2 or 3 years.
Please help with question Q4) b and c). This is part of the applications for sequence and series.
Thanks :)
c) for me is a little dodgy, might mean time, days years whatever, but im going to go ahead and assume it means how many years does it take to get 50k
so it takes 7 years to get to 50k, but if you really want to know how long it takes to get to 50k exactly, its just that 6.17 whatever whatever in years, around 2255 days, 54118 hours 3.247m minutes, and so on. just the question is really non specific, but i think that really the answer should be 7 ie. it takes 7 years to get to 50000 dollars
Please help again!and i dont think i can't help with this one, because you've just posted a statement with some information! what is the question? post it, and ill help you out ;D
Still sequence and series questions.
Post by: Calley123 on September 02, 2018, 04:26:01 pm
Hey there!For this one, b) is essentially the same as a), except you're plugging in 11 years instead of 1, 2 or 3 years.
c) for me is a little dodgy, might mean time, days years whatever, but im going to go ahead and assume it means how many years does it take to get 50k
so it takes 7 years to get to 50k, but if you really want to know how long it takes to get to 50k exactly, its just that 6.17 whatever whatever in years, around 2255 days, 54118 hours 3.247m minutes, and so on. just the question is really non specific, but i think that really the answer should be 7 ie. it takes 7 years to get to 50000 dollarsand i dont think i can't help with this one, because you've just posted a statement with some information! what is the question? post it, and ill help you out ;D
Thank you so much for doing part a! Sorry I thought I posted the question. Here it is now. Also please help me with question 10 !
Thanks in advance
Post by: S200 on September 02, 2018, 05:26:13 pm
compound interest formula?
(https://www.thecalculatorsite.com/images/compound-interest-formula-diagram.png)
EDIT:
You have to minus the principle. so \(\overbrace {{A-1400}}^ \text {Equals the interest...} \therefore ~$377\)
Post by: fun_jirachi on September 02, 2018, 05:43:02 pm
Thank you so much for doing part a! Sorry I thought I posted the question. Here it is now. Also please help me with question 10 !
Thanks in advance
So hopefully you can see that for the nth apple where n is an integer, she runs 6n metres to get it ie. for the first one, she runs 6, 3m there and 3m back. Just note that for now.
a) To actually get to the kth apple, she only needs to run 3k metres since the question doesnt mention anything about coming back.
b) And this is where we get into the fun stuff because its an arithmetic series, you're looking at the sum of the trips from the first apple to the kth apple
c) now you just set Sk to equal 270.
Hope this helps! :D
Post by: Calley123 on September 03, 2018, 05:43:58 pm
To start off I'd like to say cheers S200, saved me a bit of time there
So hopefully you can see that for the nth apple where n is an integer, she runs 6n metres to get it ie. for the first one, she runs 6, 3m there and 3m back. Just note that for now.
a) To actually get to the kth apple, she only needs to run 3k metres since the question doesnt mention anything about coming back.
b) And this is where we get into the fun stuff because its an arithmetic series, you're looking at the sum of the trips from the first apple to the kth apple
c) now you just set Sk to equal 270.
Hope this helps! :D
YASSSS Thanks
Post by: Calley123 on September 07, 2018, 09:25:21 am
Please help with Q11 d) and e).
Thanks :)
Post by: RuiAce on September 07, 2018, 09:47:36 am
HI,This is an Extension 1 level question. Please post it in the relevant thread if you require help with it.
Please help with Q11 d) and e).
Thanks :)
Post by: ilikeapples on September 17, 2018, 09:11:31 pm
Post by: Opengangs on September 17, 2018, 09:24:39 pm
Could someone please help me with this question: solve (3x^2 -2x)^0.5 = x^1.5We can simplify the left side if we recall the following property:
\[ (a^m)^n = a^{mn} \]
So if we raise both sides to the 1/0.5th power, we end up with:
\[ ((3x^2 - 2x)^{0.5})^2 = (x^{1.5})^2 \]
which simplifies down to:
\[ 3x^2 - 2x = x^3 \]
The rest should be fairly easy to simplify. If you have any questions, feel free to clarify :)
Post by: S200 on September 17, 2018, 09:25:21 pm
Could someone please help me with this question: solve (3x^2 -2x)^0.5 = x^1.5OG has given the index laws, so I'll just throw this up here... ;)
This is equal to; \(\sqrt{3x^2 -2x} = \sqrt{x^3} \)
square both sides and you have \(3x^2-2x=x^3\)
Swing the \(x^3\) across and make a cubic...
Post by: ilikeapples on September 21, 2018, 02:50:41 pm
Post by: S200 on September 21, 2018, 05:07:08 pm
Post by: RuiAce on September 23, 2018, 02:47:42 pm
isn't the distance \(\sqrt {(p^2,q^2))}\)?What you mean by the square root of a point?
This seems like a really simple mc question but I keep getting the wrong answer: What is the distance from the point (p,q) to the line px+qy=0, could anyone help me out? (:
Post by: Calley123 on September 23, 2018, 06:09:10 pm
Post by: S200 on September 23, 2018, 06:13:33 pm
What you mean by the square root of a point?Sorry, I was a little gone when I posted that. I meant \(\sqrt{p^2+q^2}\) (from distance between two points, (0,0) and (p,q))...
Please help! Loan repayments under series and sequence. How do I take the 3 months of no interest into consideration?Could you attach the question instead of (or as well as) the answer?
Post by: jamonwindeyer on September 23, 2018, 09:51:28 pm
Please help! Loan repayments under series and sequence. How do I take the 3 months of no interest into consideration?
At a guess (as S200 says, the question might be helpful!) but I think all you would do is just take the monthly repayments off without adding any interest for the first three months of your derivation! So:
Or something similar ;D
Post by: kauac on October 02, 2018, 12:07:07 pm
HSC 2013 13 d) iii:
After 20 years the family borrows an extra amount, so that the family
then owes a total of $370 000. The monthly repayment remains $2998,
and the interest rate remains the same.
How long will it take to repay the $370 000?
Thanks!
Post by: lexi24216 on October 02, 2018, 03:28:14 pm
If y = a^x find dy/dx when x=1:
Solution says:
When x=1;
In step 2, I don’t understand why lna was placed in front of a^x
Thanks :)
Post by: fun_jirachi on October 02, 2018, 03:33:45 pm
Hi, would love some working out for this question (sample answer seems to skip a few steps).
HSC 2013 13 d) iii:
After 20 years the family borrows an extra amount, so that the family
then owes a total of $370 000. The monthly repayment remains $2998,
and the interest rate remains the same.
How long will it take to repay the $370 000?
Thanks!
Yessssss this thread is alive again!!!!
Ok so, essentially you're looking at 370000 being your new starting amount ie. P, and you're already given M as 2998 and r as 1.005. Since you want to know how long it takes to repay the whole thing, you set up the equation to be equal to zero, and your only variable left to find is n, the number of years, and that's what you want.
Just manipulate the equation, moving the 2998 by whatever to the right side, and go from there to get to:
And from there you manipulate again to get:
Use your log laws, and you should get the 192.46. Round it up, as they won't have finished in 192 months, and you have your answer.
Hope this helps!
EDIT: just saw the new question, gonna edit the solution in so i don't double post :P
The thing is, when you differentiate an exponential, the derivative is only the same as the original function if the base of the exponential is e. In this case, it isn't and its an arbritary constant a.
So basically you convert it to base e like so:
Using log laws, so you get to:
and then differentiating the expontential like usual, bring down the constant in the power, in this case lna, then keeping the top, which is
and yeah just sub in 1 at the end to get your answer
Post by: kauac on October 04, 2018, 12:48:37 pm
Is there a set method to solving questions like 'find the distance the particle travels in x seconds' ?
Post by: S200 on October 04, 2018, 12:58:19 pm
For the graph f(x)=displacement of x
Post by: kauac on October 04, 2018, 04:14:04 pm
Does area under the graph work?
For the graph f(x)=displacement of x
Sometimes it seems to work, other times, it doesn't?
Post by: S200 on October 04, 2018, 04:29:23 pm
Gotta be a shorter way though. :-\
Sometimes it seems to work, other times, it doesn't?When doesn't it work?
Post by: RuiAce on October 04, 2018, 04:45:57 pm
If \(f(x)\)=displacement, \(f^{\prime}(x)\)=velocity. You could get the final velocity add all of the velocities before?When the displacement is negative.
Gotta be a shorter way though. :-\
When doesn't it work?
He asked for the distance, so we need to consider the unsigned area. This means that if we want to use your approach, we must treat any areas below the t-axis as positive as well, which can be done by putting absolute value brackets around them.
For motion questions in app. of calculus:Essentially, you do need to watch out for the particle turns around. (Of course, if it does turn around, then we expect that \(v = 0\) there.)
Is there a set method to solving questions like 'find the distance the particle travels in x seconds' ?
Once you figure out where the particle turns around, divide your region into those intervals and consider the distance travelled between those intervals separately. For example, if a particle follows the equation \( x = 2(t-1)(t-3)\), then consider the distance travelled between 0 and 1, separate to the distance travelled between 1 and 3, separate to any distance travelled after 3.
Having said that, that is a very crude example. A proper question would be appreciated if you require more detail.
Post by: kauac on October 04, 2018, 08:01:43 pm
Once you figure out where the particle turns around, divide your region into those intervals and consider the distance travelled between those intervals separately. For example, if a particle follows the equation \( x = 2(t-1)(t-3)\), then consider the distance travelled between 0 and 1, separate to the distance travelled between 1 and 3, separate to any distance travelled after 3.
So work out the separate distances according to where the particle turns around, and then add them all up?
Post by: S200 on October 04, 2018, 08:09:59 pm
If it wants distance from origin, you would probably have to minus one from the other.
Post by: ilikeapples on October 05, 2018, 09:07:29 am
Post by: RuiAce on October 05, 2018, 09:38:53 am
Hey could someone explain something to me in regards to the last question from the 2017 exam... why in the solutions attached for part (i) does BD=DE? Is this some ancient rule from year ten that I have forgotten? Could someone just explain what it means by equal intercept. thanks!It's one of those obscure theorems that some schools teach whilst other schools take for granted. I feel it's slightly too rushed, but the following explanation is still a bit rushed.
Post by: Mate2425 on October 05, 2018, 08:07:23 pm
http://educationstandards.nsw.edu.au/wps/wcm/connect/f718de94-0e72-4393-bb72-714be7283934/maths-hsc-exam-2011.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-f718de94-0e72-4393-bb72-714be7283934-lGhOrSw
Thanks all!!
Post by: kauac on October 05, 2018, 08:17:09 pm
Hi guys this question is from HSC 2011, was wondering if someone could please help me understand how to work out part Q9(ii).
Which part of question 9 was it?
Post by: Mate2425 on October 05, 2018, 09:02:56 pm
Which part of question 9 was it?
Q9a (ii) please
Post by: RuiAce on October 05, 2018, 09:16:47 pm
Q9a (ii) pleaseBasically the similar triangles in part i) were proven using "two sides in proportion, included angle equal". But now that we have them, we can then use the other property, i.e. equiangular.
Of course, could also use the other ones.
_________________________________________
Post by: Mate2425 on October 05, 2018, 09:51:51 pm
Basically the similar triangles in part i) were proven using "two sides in proportion, included angle equal". But now that we have them, we can then use the other property, i.e. equiangular.
Of course, could also use the other ones.
_________________________________________
Thanks RuiAce :)
With the numbers 1:2 are these just figures that represent they are similar but different in size?
Could you please also explain why in question HSC 2010 Q6b (iii) and (iv) it was a necessity to change angle to degrees rather than keep it RAD. I tried my method and still don't understand why it is necessary and how to recognise when to do this.
Thankkkksss!!! :) :)
http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T
Post by: RuiAce on October 05, 2018, 10:13:26 pm
Thanks RuiAce :)I'm looking at your iv) and your "area of sector" is definitely correct. Also your angle in iii) makes sense.
With the numbers 1:2 are these just figures that represent they are similar but different in size?
Could you please also explain why in question HSC 2010 Q6b (iii) and (iv) it was a necessity to change angle to degrees rather than keep it RAD. I tried my method and still don't understand why it is necessary and how to recognise when to do this.
Thankkkksss!!! :) :)
http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T
On the other hand, your "area of total" though - I'm not too sure how you got that? Also I'm not sure why you used Pythagoras on 9^2 when 9 is the length of the arc? Have a look again, because I believe it's the other stuff giving you a wrong answer.
Post by: Mate2425 on October 05, 2018, 11:39:53 pm
I'm looking at your iv) and your "area of sector" is definitely correct. Also your angle in iii) makes sense.
On the other hand, your "area of total" though - I'm not too sure how you got that? Also I'm not sure why you used Pythagoras on 9^2 when 9 is the length of the arc? Have a look again, because I believe it's the other stuff giving you a wrong answer.
Ahhhhhhh, yes that makes sense thank you! :)
Just in regards to previous question - With the numbers 1:2 are these just figures that represent they are similar but different in size? (where do they come from? how do we know it is 1:2 specifically?)
Thanks 8)
Post by: RuiAce on October 06, 2018, 09:57:37 am
Ahhhhhhh, yes that makes sense thank you! :)
Just in regards to previous question - With the numbers 1:2 are these just figures that represent they are similar but different in size? (where do they come from? how do we know it is 1:2 specifically?)
Thanks 8)
Post by: Mate2425 on October 06, 2018, 12:42:44 pm
Many thanks Rui, true Ace!! :D
Post by: kauac on October 06, 2018, 01:16:34 pm
Could someone please explain the answer for HSC 2015 MC Q9?
Post by: RuiAce on October 06, 2018, 01:17:25 pm
Hi...Already addressed in the compilation.
Could someone please explain the answer for HSC 2015 MC Q9?
Post by: kauac on October 06, 2018, 01:19:03 pm
Already addressed in the compilation.
Sweet, thanks for that. :)
Post by: Mate2425 on October 06, 2018, 03:45:25 pm
And also could someone please provide a really simple explanation to the topic of prob. when do you add and when do you times, what do each represent?
Thanks. :)
Post by: RuiAce on October 07, 2018, 05:33:21 pm
Hi is there a process in order to derive the formulas for surface area or volume in an array of shapes in case you forget them in the test!In general, for the non circular prisms and pyramids, for the surface area you can always just literally compute the area of each face by hand. For the volume, for the prism it is is always of the form \(V = Ah\), where \(A\) is the area of the base. For a pyramid, it is always 1/3 that of the volume of a prism.
And also could someone please provide a really simple explanation to the topic of prob. when do you add and when do you times, what do each represent?
Thanks. :)
The surface area of the cone and sphere are pretty hard to derive. For the cylinder, you can be careful, so long as you know that the circumference of a circle is \(2\pi r\) to figure out the area of the rectangle.
The volume of the cylinder and cone can be done the same way as above: basically just let \(A = \pi r^2\). The volume of the sphere is again pretty hard to derive.
Post by: jamonwindeyer on October 07, 2018, 09:53:19 pm
The volume of the sphere is again pretty hard to derive.
100% an example in one of my Tutesmart classes or exams ;D
Post by: RuiAce on October 07, 2018, 10:10:07 pm
100% an example in one of my Tutesmart classes or exams ;DBut like why \(\pi \int_{-r}^r \left(\sqrt{r^2-x^2}\right)^2 \mathrm{d}x \) when one can \( \int_0^{2\pi} \int_0^\pi \int_0^r \rho^2 \sin\phi \,\mathrm{d}\rho\mathrm{d}\phi\mathrm{d}\theta \) :'(
Post by: Mate2425 on October 08, 2018, 06:43:43 pm
If you write answer as 2Kg ( nearest whole number) as A was originally given as 20kg?
Thanks :)
Post by: clovvy on October 08, 2018, 07:13:56 pm
Hey guys for Exponential growth and decay questions if they do not specify how many numbers to round to/sig fig to, would you be marked correct e.g in HSC 2014 - http://educationstandards.nsw.edu.au/wps/wcm/connect/aea5d47c-525d-4202-9d8c-72f76575a1fc/maths-hsc-mg-2014.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-aea5d47c-525d-4202-9d8c-72f76575a1fc-lGhCSoBIf they don't signify how many sig figs required, you just need to state it at the end of your answer I believe... Because my teacher told me that for HSC maths, what markers look for it the process you take to obtain your answer
If you write answer as 2Kg ( nearest whole number) as A was originally given as 20kg?
Thanks :)
Post by: 8veFable on October 11, 2018, 03:36:18 pm
I am struggling with Q16 ii) from the 2012 2U exam.
I tried to look at the solutions and manage to get that point Q is ([1 - sinθ]/cosθ) by using part i) to find the point of intersection with y = 1 but I am massively confuzzled as to how you are meant to use that information to find BQ.
Any help would be appreciated! :)
Edit: I did try and use the point intercept formula y - y1 = m(x - x1) but that didn't work out :O
Post by: RuiAce on October 11, 2018, 03:40:28 pm
Hello!If you know that \(Q\) is the point \( \left( \frac{1-\sin\theta}{\cos\theta}, 1 \right) \), then by the distance formula the length of \(BQ\) is just \( \sqrt{\left( \frac{1-\sin\theta}{\cos\theta} - 0\right)^2 + (1-1)^2} \), and note that the 1 - 1 cancels out to 0.
I am struggling with Q16 ii) from the 2012 2U exam.
I tried to look at the solutions and manage to get that point Q is ([1 - sinθ]/cosθ) by using part i) to find the point of intersection with y = 1 but I am massively confuzzled as to how you are meant to use that information to find BQ.
Any help would be appreciated! :)
Edit: I did try and use the point intercept formula y - y1 = m(x - x1) but that didn't work out :O
(Alternatively, it's just the difference between their \(x\)-coordinates, because \(BQ\) is a horizontal line segment. And \(B\) obviously has \(x\)-coordinate 0 since it lies on the \(y\)-axis.)
Post by: S200 on October 11, 2018, 03:53:01 pm
Hello!Hmm. I'm missing something here I think, but yeah.
I am struggling with Q16 ii) from the 2012 2U exam.
I tried to look at the solutions and manage to get that point Q is ([1 - sinθ]/cosθ) by using part i) to find the point of intersection with y = 1 but I am massively confuzzled as to how you are meant to use that information to find BQ.
Any help would be appreciated! :)
Edit: I did try and use the point intercept formula y - y1 = m(x - x1) but that didn't work out :O
The only way I can think of doing it is to find the length of PO using \(\sin{\theta}\) and the length PT.
Then make a right angled triangle with Q, a point vertically down from Q (say, Y), and P.
You can use tan to find length PY and minus that from PO to give you BQ.
The angle QPY would be \(\frac{-1}{\tan {\theta}}\)
If you know that \(Q\) is the point \( \left( \frac{1-\sin\theta}{\cos\theta}, 1 \right) \), then by the distance formula the length of \(BQ\) is just \( \sqrt{\left( \frac{1-\sin\theta}{\cos\theta} - 0\right)^2 + (1-1)^2} \), and note that the 1 - 1 cancels out to 0.Edit - Rui's way is a lot easier.
(Alternatively, it's just the difference between their \(x\)-coordinates, because \(BQ\) is a horizontal line segment. And \(B\) obviously has \(x\)-coordinate 0 since it lies on the \(y\)-axis.)
Post by: terassy on October 11, 2018, 06:40:51 pm
(https://i.gyazo.com/3e78a27954d79e5dd00250f6020f2101.png)
Part (ii) states 'during which month and year'. So does that mean the month it will reach 4000 million (i.e. October) or like the month it will enter, with more than 4000 million (i.e. November).
Thanks
Final answer to part (i)
help ;(
Post by: jamonwindeyer on October 11, 2018, 07:55:31 pm
Help for part(ii). This is from HSC 2U Maths 2007
(https://i.gyazo.com/3e78a27954d79e5dd00250f6020f2101.png)
Part (ii) states 'during which month and year'. So does that mean the month it will reach 4000 million (i.e. October) or like the month it will enter, with more than 4000 million (i.e. November).
Thanks
Final answer to part (i)
help ;(
Hey! I read it as during the month, so October should be the answer ;D
Post by: 8veFable on October 11, 2018, 10:20:38 pm
I also have another question if you anyone could please help me with it.
I don't know how on earth I forgot this all of a sudden but could someone please clarify if calculus of trigonometric functions is done in degrees or radians? Like for example if you were given a motion question and it told you to sub in 2 seconds (At t = 2), then would the calculator be set in degrees or radians?
Is using the wrong mode a mistake that is usually done in motion questions with trigonometric functions? :o
Much appreciated!
Post by: RuiAce on October 11, 2018, 10:23:52 pm
Hi! Thank you Rui and S200 for answering my question! After your responses I realised the question asked for length' instead of finding the equation, so that made me much less confused ;DAlways radians with the calculus of trig functions.
I also have another question if you anyone could please help me with it.
I don't know how on earth I forgot this all of a sudden but could someone please clarify if calculus of trigonometric functions is done in degrees or radians? Like for example if you were given a motion question and it told you to sub in 2 seconds (At t = 2), then would the calculator be set in degrees or radians?
Is using the wrong mode a mistake that is usually done in motion questions with trigonometric functions? :o
Much appreciated!
You should never use degrees unless you explicitly see the little circle hinting at degrees
Post by: markkhodair on October 12, 2018, 12:13:50 pm
Post by: fun_jirachi on October 12, 2018, 12:19:44 pm
You should never use degrees unless you explicitly see the little circle hinting at degrees
Pretty much this.
Post by: RuiAce on October 12, 2018, 02:25:38 pm
Hey! I know that this question has probably been asked and answered so many times in the past, so I apologise for repeating, but I just wanted to clarify when to use radians mode in the calculator as opposed to using degrees mode? Thank you!Just watch out for any posts that are made on the same page as well in the future. Your question was answered in the post immediately above yours.
(That rule of thumb (as fun_jirachi bumped) applies not just to calculus, but in any generic contexts as well.)
Post by: msimo4 on October 13, 2018, 10:07:42 pm
Post by: RuiAce on October 13, 2018, 10:16:04 pm
That's what "limiting" means in the applications topic. It means to consider the eventual behaviour of your particle, which is what happens as your \(t\) gets very large.
Post by: jamonwindeyer on October 14, 2018, 09:52:28 am
Post by: RuiAce on October 14, 2018, 10:08:51 am
Ily Rui <3This is a miracle. Did Jamon just make my life complete <3
Post by: jamonwindeyer on October 14, 2018, 01:29:37 pm
This is a miracle. Did Jamon just make my life complete <3
I was demoing the forums, but if it completes you I am happy to have played a part ;)
Post by: saige_abigail on October 14, 2018, 03:51:59 pm
The correct answer was 48.2% ? In the textbook
Permit k to be negative in your working out.
This is the question that I'm struggling with. Why 0.9 instead of 0.1? And what happens to the M0? These questions might sound stupid but I'm just trying to understand.
Post by: RuiAce on October 14, 2018, 04:21:15 pm
This is the question that I'm struggling with. Why 0.9 instead of 0.1? And what happens to the M0? These questions might sound stupid but I'm just trying to understand.10% is just how much of it has decayed. But that’s not gonna help us measure anything easily.
Because how else can we measure what percentage decays in the next year? Or the year after that? (Note that in fact, in the second year what’s decayed is 90% * 10%, i.e. 9%. And in fact, another 8.1% decays in the third year.)
This is because the amount that decays per year is not dependent on how much has decayed last year, but rather how much from last year has not decayed. The problem is then that to figure out the total decay, we need to keep a running record of how much decays only in the n-th year, and then do the sum of a GP.
It turns out it is much easier to just resort to keeping track of how much is there still left, as opposed to what’s already decayed. This simplified the computations a lot more.
The M0 simply gets cancelled out on both sides of the equation. We don’t need to know what M0 is to complete the question
Also note that the post had a mistake, which was addressed a bit further down
Post by: saige_abigail on October 14, 2018, 04:32:40 pm
10% is just how much of it has decayed. But that’s not gonna help us measure anything easily.
Because how else can we measure what percentage decays in the next year? Or the year after that? (Note that in fact, in the second year what’s decayed is 90% * 10%, i.e. 9%. And in fact, another 8.1% decays in the third year.)
This is because the amount that decays per year is not dependent on how much has decayed last year, but rather how much from last year has not decayed. The problem is then that to figure out the total decay, we need to keep a running record of how much decays only in the n-th year, and then do the sum of a GP.
It turns out it is much easier to just resort to keeping track of how much is there still left, as opposed to what’s already decayed. This simplified the computations a lot more.
The M0 simply gets cancelled out on both sides of the equation. We don’t need to know what M0 is to complete the question
Also note that the post had a mistake, which was addressed a bit further down
So to clarify: if the substance decays by 10% - 90% is what's left?
Post by: RuiAce on October 14, 2018, 07:06:11 pm
So to clarify: if the substance decays by 10% - 90% is what's left?Yeah.
Because 100% minus 10% is 90%.
Post by: skisso on October 15, 2018, 06:25:49 pm
I dont understand the answers online or the answers in the excel book i have.
Post by: clovvy on October 15, 2018, 06:34:57 pm
Helloo :) having trouble with HSC paper 2009, Question 10 e).Hey, what they did is simply using product rule for (1+x)ln(1+x).... If you still don't understand feel free to reply and I'll try to explain further as best as I can..
I dont understand the answers online or the answers in the excel book i have.
Post by: fun_jirachi on October 15, 2018, 06:37:42 pm
Post by: skisso on October 15, 2018, 06:53:56 pm
Hey, what they did is simply using product rule for (1+x)ln(1+x).... If you still don't understand feel free to reply and I'll try to explain further as best as I can..
Ohh i see. I had interpreted the question wrong and though that you had to integrate or something, but I got in now. Thanks!!
Post by: SebastianHabibi on October 17, 2018, 06:02:05 pm
The focal chord that cuts the parabola x^2 = -6y at (6,-6) cuts the parabola again at X. Find the coordinates of X
Post by: fun_jirachi on October 17, 2018, 06:11:03 pm
Hope I helped :)
Post by: jamesrandom on October 19, 2018, 03:54:10 pm
Anyway, im reviewing product rule and forgot how to do this.
Differentiate:
(x + 1)(2x + 5)^4
I know its probably easy im just confused as to how my textbook got the answer
(10x + 13)(2x + 5)^3
Post by: RuiAce on October 19, 2018, 04:14:48 pm
Hey, Im just starting Year 12 now so dont think im doing my HSC next week and dont know how to do this haha.
Anyway, im reviewing product rule and forgot how to do this.
Differentiate:
(x + 1)(2x + 5)^4
I know its probably easy im just confused as to how my textbook got the answer
(10x + 13)(2x + 5)^3
\begin{align*} \frac{d}{dx} (x+1)(2x+5)^4 &= 1 \times (2x+5)^4 + (x+1) \times 8(2x+5)^3\\ &= (2x+5)^4 + (8x+8)(2x+5)^3\\ &= (2x+5)^3 \left[ (2x+5) + (8x+8) \right] \\ &= (10x+13)(2x+5)^3\end{align*}
Post by: Mate2425 on October 19, 2018, 05:55:01 pm
I thought that you consider where sin is pos in ASTC which would be in the first and second quadrant but my answers do not relate.
Thank you.
Post by: RuiAce on October 19, 2018, 06:00:03 pm
Hey Rui, could you please explain why the answer is pi/3 and 5pi/3 in HSC 2016 Advanced Math, rather than pi/3 and 2pi/3.Which question of the 2016 paper was this?
I thought that you consider where sin is pos in ASTC which would be in the first and second quadrant but my answers do not relate.
Thank you.
Post by: fun_jirachi on October 19, 2018, 06:17:25 pm
Hey Rui, could you please explain why the answer is pi/3 and 5pi/3 in HSC 2016 Advanced Math, rather than pi/3 and 2pi/3.
I thought that you consider where sin is pos in ASTC which would be in the first and second quadrant but my answers do not relate.
Thank you.
You should probably provide the question with your question, so we know what we're looking at! I found it anyway, but maybe do that next time, it really helps us help you! :)
I'm assuming this is 11(g) in 2016?
Assuming I'm correct, the answer is right. You solve
Hope this helps :)
Post by: raghav_singh on October 19, 2018, 08:07:31 pm
Post by: RuiAce on October 19, 2018, 08:11:14 pm
Hi so I was wondering whether the 2U Maths work done in Year 11 will be tested in the HSC exams. OR, will it it just come back in the form of harder HSC topics, and be somewhat of a foundation to complete questions in these topics.For all mathematics courses, preliminary content IS examinable in the HSC and makes up 20% of the marks.
Post by: Mate2425 on October 19, 2018, 10:34:43 pm
http://educationstandards.nsw.edu.au/wps/wcm/connect/2cdf8537-92dc-4328-ab39-74a1f1dd489f/maths-hsc-exam-2013.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-2cdf8537-92dc-4328-ab39-74a1f1dd489f-lGhNsb6
Many thanks, in advance! :)
Post by: jamesrandom on October 19, 2018, 10:37:07 pm
Post by: S200 on October 19, 2018, 11:05:18 pm
Can someone help me with the attached question.Find the derivative, sub in the point to get the gradient..
Then use the gradient-point formula (\(y-y_1 =m(x-x_1)\)), where \(m\) is the gradient, to get the full equation.
Post by: RuiAce on October 20, 2018, 12:04:57 am
Hi guys for this question HSC 2013 Q14d, the working out does not resonate a complete understanding, could you please explain it in another way please.
http://educationstandards.nsw.edu.au/wps/wcm/connect/2cdf8537-92dc-4328-ab39-74a1f1dd489f/maths-hsc-exam-2013.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-2cdf8537-92dc-4328-ab39-74a1f1dd489f-lGhNsb6
Many thanks, in advance! :)
The last two come from noticing symmetry - if we have the same area appearing above and below the x-axis, they cancel each other out. Whereas the first one occurs because the region between \(x=-1\) and \(x=2\) is just a rectangle with length 3 and breadth 1.
(https://i.imgur.com/urTSPp2l.jpg)
Post by: Sr-1425 on October 20, 2018, 07:52:52 am
Any help would be much appreciated.
Post by: fun_jirachi on October 20, 2018, 08:33:20 am
Post by: clovvy on October 20, 2018, 08:37:30 am
You have from rearranging the question that 4.8t=0. Therefore, t=0. :)
it says ln(0.2t)=5t... I tried regular rearranging and I am not getting anywhere so I check wolfram and it seems that it is way beyond HSC unless I misread the question or the person mistyped
Post by: fun_jirachi on October 20, 2018, 08:43:11 am
Well thinking about it graphically, you have y=ln0.2x and y=5x. The graphs of both those functions never cross, so there is no solution, which is why you're struggling to find it. Sometimes you've gotta look at it unconventionally and think about the graphical method, which many people seem to forget :)
Post by: RuiAce on October 20, 2018, 08:50:52 am
Hi, I'm struggling to find a value for t in which ln 0.2t = 5t.It is essentially as Clovvy has stated. In general, there is no algebraic method whatsoever of solving equations of this form. (In fact, \(bx = f(x)\) can not be solved algebraically at all, where \(f(x)\) is any exponential, logarithmic or trigonometric function.)
Any help would be much appreciated.
Therefore, I am not sure where you got this question from.
Note that there is also some difficulty using the graphical method for these types of problems. This is because \( ax = \ln bx\) will have solutions for some values of \(a\) and \(b\), but not for all possible values. Therefore, it is very hard to tell just by staring at the equation of the graph, whether or not there will be points of intersection between the two.
Post by: kauac on October 20, 2018, 12:43:41 pm
More of an exam technique question:
Is it generally better to try and do the paper as fast as possible, to maximize time spent checking over, or do it a bit slower and check as you go?
When I do past papers with the first approach, I generally have about 20-30min check-over time, but then I don't always pick up on the silly errors I make. But I'm worried if I try and do it slower and check as I go, that I will run out of time.
Post by: RuiAce on October 20, 2018, 01:18:14 pm
Hi...The most optimal balance varies from person to person. It requires balancing out factors including (but not necessarily limited to) how error prone you are, and your time management skills in the exam.
More of an exam technique question:
Is it generally better to try and do the paper as fast as possible, to maximize time spent checking over, or do it a bit slower and check as you go?
When I do past papers with the first approach, I generally have about 20-30min check-over time, but then I don't always pick up on the silly errors I make. But I'm worried if I try and do it slower and check as I go, that I will run out of time.
For me, I try to avoid checking immediately after I do a question and taking it too slow. More or less because I know I'll have preconceptions about what I think is right, and then not spot it so easily in comparison to when I check it later. But I do choose to check the paper the instant I hit that point where all the remaining marks are no longer so obvious to me. That way, at least I'm checking the paper with reasonable amount of time left, but also at a point where I'm comfortable (for now) with the amount of marks I had already scored.
Post by: Sr-1425 on October 20, 2018, 03:45:49 pm
I got ln (0.2/t) = -1.1t, but I'm not sure how to continue from there.
Post by: RuiAce on October 20, 2018, 04:04:23 pm
Oops, my bad. I miscalculated and got stuck.Still cannot be solved algebraically. As stated, no equation with logs on one side and linear terms on the other can be solved algebraically.
I got ln (0.2/t) = -1.1t, but I'm not sure how to continue from there.
Please provide the original question in its entirety.
Post by: Mate2425 on October 20, 2018, 04:18:24 pm
Is there a standardised process to 'Find the greatest' questions?
Thank you :)
Post by: RuiAce on October 20, 2018, 04:24:24 pm
Hey guys for rates questions, when they ask what is the 'greatest' e.g. in HSC 2015 Q15c , is it always when it hits the x-intercept as for this question it seemed to be the case but in HSC 2016 Q16b they took the halfway distance between the two roots (= maximum peak of a parabola).They are asking for slightly different things.
Is there a standardised process to 'Find the greatest' questions?
Thank you :)
In the 2015 paper, they want you to maximise the quantity, i.e. the volume. We know from doing classical maxima/minima problems that \(V\) can be maximised by setting \( \frac{dV}{dt} = 0\), and solving for the time \(t\). This is the usual method.
But in the 2016 paper, they don't want you to maximise the quantity, i.e. the population itself. They want you to maximise the rate of growth of the quantity, i.e. the rate of growth of the population. To maximise \( \frac{dy}{dt} \), in theory what we would want is to set \( \frac{d^2y}{dt^2} = 0\). But because we don't know how to obtain that (because we can't differentiate with respect to \(t\) here anymore), we can just try maximising \( \frac{dy}{dt} \) by literally plotting the graph of \( \frac{dy}{dt} \) v.s. \(y\). From there, we can read the value of \(y\) (NOT \(t\)) that maximises \( \frac{dy}{dt} \).
Post by: Mate2425 on October 20, 2018, 04:33:13 pm
They are asking for slightly different things.
In the 2015 paper, they want you to maximise the quantity, i.e. the volume. We know from doing classical maxima/minima problems that \(V\) can be maximised by setting \( \frac{dV}{dt} = 0\), and solving for the time \(t\). This is the usual method.
But in the 2016 paper, they don't want you to maximise the quantity, i.e. the population itself. They want you to maximise the rate of growth of the quantity, i.e. the rate of growth of the population. To maximise \( \frac{dy}{dt} \), in theory what we would want is to set \( \frac{d^2y}{dt^2} = 0\). But because we don't know how to obtain that (because we can't differentiate with respect to \(t\) here anymore), we can just try maximising \( \frac{dy}{dt} \) by literally plotting the graph of \( \frac{dy}{dt} \) v.s. \(y\). From there, we can read the value of \(y\) (NOT \(t\)) that maximises \( \frac{dy}{dt} \).
Thanks Rui, 8) 8)
Post by: Sr-1425 on October 20, 2018, 04:42:47 pm
The concentration of a certain drug in the blood at a time t hours after taking a dose is x units, where x=0.3t*(e)^(-1.1t)
A) Determine the maximum concentration and the time at which this is reached.
B) Plot the function for t=0, 0.1, 0.5, 1, 2, 3.
C) This drug kills germs only while its concentration is at least 0.06 units. From the graph, find the length of time during which the drug will kill germs.
It's part c) where I can't seem to find a solution
Post by: RuiAce on October 20, 2018, 04:54:25 pm
This is the question:Because it says "from the graph", you aren't being asked to determine the value of \(t\) algebraically. You're being asked to read off your graph and interpret an approximate value.
The concentration of a certain drug in the blood at a time t hours after taking a dose is x units, where x=0.3t*(e)^(-1.1t)
A) Determine the maximum concentration and the time at which this is reached.
B) Plot the function for t=0, 0.1, 0.5, 1, 2, 3.
C) This drug kills germs only while its concentration is at least 0.06 units. From the graph, find the length of time during which the drug will kill germs.
It's part c) where I can't seem to find a solution
In fact, given how informal part b) is, only asking you to plot some dots, it's really up to you to infer the shape of the graph itself. In the HSC you would not be asked a question as vague as this - they will run you through the usual steps of stationary points if you had to do a plot.
But otherwise, you can draw the curve however you want to draw it. From Desmos output, I approximate that \(t \approx 0.3\) and \(t \approx 2.2\), so that would be a length of 1.9 hours.
(https://i.imgur.com/d4kWeFf.png)
Post by: Sr-1425 on October 20, 2018, 05:03:26 pm
Because it says "from the graph", you aren't being asked to determine the value of \(t\) algebraically. You're being asked to read off your graph and interpret an approximate value.
In fact, given how informal part b) is, only asking you to plot some dots, it's really up to you to infer the shape of the graph itself. In the HSC you would not be asked a question as vague as this - they will run you through the usual steps of stationary points if you had to do a plot.
But otherwise, you can draw the curve however you want to draw it. From Desmos output, I approximate that \(t \approx 0.3\) and \(t \approx 2.2\), so that would be a length of 1.9 hours.
(https://i.imgur.com/d4kWeFf.png)
Thank you.
Post by: LaraC on October 20, 2018, 05:32:40 pm
I have a really simple question that I can't work out.....its only at the start of the paper :( is there just a rule I'm forgetting?
Solve the equation lnx=2. Give your answer correct to four decimal places.
Thanks :)
Post by: RuiAce on October 20, 2018, 05:34:17 pm
Hello,\[ \ln x = 2\text{ becomes }\boxed{x = e^2}\\ \text{so just plug }e^2\text{ in your calculator to obtain }7.3891\]
I have a really simple question that I can't work out.....its only at the start of the paper :( is there just a rule I'm forgetting?
Solve the equation lnx=2. Give your answer correct to four decimal places.
Thanks :)
(rounded up)
Post by: LaraC on October 20, 2018, 05:38:25 pm
Sorry....i should know :-[ :-[
Post by: fun_jirachi on October 20, 2018, 05:40:57 pm
When
it's the same as
Just put in the numbers and you get your x=e2
Post by: RuiAce on October 20, 2018, 05:41:48 pm
But how do u just 'know' that x = e^2?From the definition of the logarithm.
Sorry....i should know :-[ :-[
\( y = \log_a x \) is defined to be equivalent to \( x = a^y\). This is something that we've essentially declared to be the case as mathematicians.
In your case, also recall that \( \ln 2\) is a shorthand for \(\log_e 2\)
Post by: LaraC on October 20, 2018, 05:48:15 pm
Post by: Mate2425 on October 20, 2018, 07:12:20 pm
Thanks. :D
Post by: RuiAce on October 20, 2018, 07:14:51 pm
Hey, i don't understand why in HSC 2013 Answers they drew all those lines and were somehow able to deduce the answer from that, any help would be appreciated.Already addressed in the compilation. Check out the GeoGebra simulation for more details
Thanks. :D
Post by: Mate2425 on October 20, 2018, 07:56:21 pm
Already addressed in the compilation. Check out the GeoGebra simulation for more detailsHey Rui, i understand it down to the last part. Is the reason why m = -2/3 is because or perpendicular gradient or something?
Post by: RuiAce on October 20, 2018, 08:06:13 pm
Hey Rui, i understand it down to the last part. Is the reason why m = -2/3 is because or perpendicular gradient or something?Fair, I'll add some detail to it. Basically, I just computed the gradient of the line that passes through the points \( (0,1) \) and \( \left(\frac32, 0\right) \) using the usual gradient formula \( m = \frac{y_2-y_1}{x_2-x_1} \)
Post by: chayek on October 20, 2018, 09:29:55 pm
Thanks
A worker's salary was $10500 at the beginning of 1977 and it increased by 5% at the beginning of each year thereafter. Find the worker's salary at the beginning of 1994.
Post by: RuiAce on October 20, 2018, 09:32:42 pm
Hi, I am having difficulties with this question if I could please get some help.
Thanks
A worker's salary was $10500 at the beginning of 1977 and it increased by 5% at the beginning of each year thereafter. Find the worker's salary at the beginning of 1994.
Or one could've done all that working out in one line by quoting the compound interest formula.
Post by: askaerdf on October 21, 2018, 10:37:00 am
Post by: RuiAce on October 21, 2018, 10:42:18 am
Hey sorry to be a bother but from 2013 q 11(f) i got to a point where 1/3 (ln 2 - ln1) and it supposedly equals 1/3 ln 2 and i dont really understand itYou can plug \( \ln 1 \) into your calculator and check that it equals to 0.
Post by: askaerdf on October 21, 2018, 10:45:48 am
Post by: kauac on October 21, 2018, 11:11:34 am
(Sorry I couldn't type it out).
Post by: Em1313 on October 21, 2018, 12:50:11 pm
f(x) = x^3 - 3x^2 - 9x -2
f'(x) = 3x^2 - 6x - 9
= 3(x-3)(x+1)
By sketching y = f'(x) show that f(x) is increasing when x > 3 or x < -1
Post by: ee1233 on October 21, 2018, 01:11:36 pm
Hi, could someone please explain HSC 2016 MC Q10?
(Sorry I couldn't type it out).
Hey kauac,
What I did was write 4 as logbase216.
So the expression would look like logbase216 + logbase2x. Using log laws (loga + logb = logab), the expression would then look like logbase216x (D).
Hope that helps/makes sense!
Post by: imogen.b on October 21, 2018, 01:21:16 pm
I'm having a bit of trouble with part 2 of this question, I can find the time in terms of t = d/s, but not sure how to then find theta.
Thanks! :) :)
Post by: fun_jirachi on October 21, 2018, 01:22:47 pm
Sorry if i post this in the wrong place this is my first time on the site. Can someone please help me with this question.
f(x) = x^3 - 3x^2 - 9x -2
f'(x) = 3x^2 - 6x - 9
= 3(x-3)(x+1)
By sketching y = f'(x) show that f(x) is increasing when x > 3 or x < -1
You can see that f'(x) has zeroes at 3 and -1, and is a parabola that has a minimum. So you draw a parabola cutting at those points, like in the picture (bad picture, i know). And you can see it's above the x-axis when x>3 and x<-1, so since the gradient is >0 at those points, by definition f(x) is increasing when x>3 and x<-1.
Hope this helps :)
EDIT:
Hey guys!
I'm having a bit of trouble with part 2 of this question, I can find the time in terms of t = d/s, but not sure how to then find theta.
Thanks! :) :)
Just saw this, gonna edit something in :)
Ah yes, the dastardly farmhouse question :'( This has already been addressed in the HARD 2U QUESTIONS COMPILATION (soz for caps, im pretty sure the topic is in caps :)) so go and check it out there :) !
Post by: Mate2425 on October 22, 2018, 08:36:23 am
Thank you!
Post by: RuiAce on October 22, 2018, 08:38:32 am
I guys i was wondering if you could please help me with understanding Q16bi from the 2012 HSC paper, i am finding it hard to understand their worked solutions.Already addressed in the compilation
Thank you!
Post by: Mate2425 on October 22, 2018, 09:20:06 pm
Also for probability questions could you please tell me what process i follow when they say 'exactly' e.g HSC 2017 Q12e(iv) exactly one of three spins
Thank you, :)
Post by: RuiAce on October 22, 2018, 09:46:37 pm
Hey, RuiAce could you please tell me a really easy way to understand when you either + probabilities or x.Simple:
Also for probability questions could you please tell me what process i follow when they say 'exactly' e.g HSC 2017 Q12e(iv) exactly one of three spins
Thank you, :)
- Times means 'and'
- Plus means 'or' (under the assumption they're mutually exclusive)
The crucial thing about that part is that there's an extra step, in contrast to the part above it. Note that the previous part specifically asked for an even number on the first spin and an odd number on the second and third spins. Whereas this part doesn't cater for that, so you need to consider the possibility where the even number occurs on the second spin instead, or on the third spin instead.
This is why your answer will be similar to the one above it, but you have to multiply by 3 to cater for how there's multiple possibilities.
(Alternatively, you could treat it like an OR scenario, and do \( \frac{18}{125} + \frac{18}{125} + \frac{18}{125} \) instead.
Post by: Opengangs on October 22, 2018, 09:47:56 pm
Hey, RuiAce could you please tell me a really easy way to understand when you either + probabilities or x.Hey, I'm not Rui but hopefully I can provide the same insights as Rui :)
Also for probability questions could you please tell me what process i follow when they say 'exactly' e.g HSC 2017 Q12e(iv) exactly one of three spins
Thank you, :)
With probability, you want to see whether your two events are simultaneous or not. If you want two events together, then you use the \(\times\) rule. For example, if you tossed a coin AND rolled a die, then you want these events together. So you'd use the \(\times\) rule. If the question was if you tossed a coin OR rolled a die, then you could have option a or option b. So we use the \(+\) rule.
Exactly one means you could have the following events:
1) Even-odd-odd,
2) Odd-even-odd,
3) Odd-odd-even.
Either one of these have exactly one of the three spins being an even number. So you'd have to use both, addition and multiplication rule. :)
As an example, to complete the first event, you would want even AND odd AND odd. So you would have \(P(\text{even}) \times P(\text{odd}) \times P(\text{odd})\), since you want these events happening together.
If you have any more questions, feel free to send a reply! :)
oh, he beat me to it but I'll happily have my explanation up in case you're still confused :)
Post by: saige_abigail on October 22, 2018, 10:08:09 pm
[This is from Eddie Woo's YouTube video on Superannuation (2 of 3) from his playlist HSC Series and Sequences]
Johnny regularly invests $750 at an interest rate of 8% p.a. and would like to end up with $50, 000. How long will it take him to reach this amount?
The working out starts off like this...
$50 000 = $750 x (1.08 [1.08^n -1]/0.08)
but it gets to this...
50 000 x (0.08 / 750 x 1.08) = 1.08^n -1
...which I don't get. How did he get the 1.08^n-1 on the right hand side please?
Cheers
Post by: RuiAce on October 22, 2018, 10:13:37 pm
Hey - can I please have some help if you don't mind? :)
[This is from Eddie Woo's YouTube video on Superannuation (2 of 3) from his playlist HSC Series and Sequences]
Johnny regularly invests $750 at an interest rate of 8% p.a. and would like to end up with $50, 000. How long will it take him to reach this amount?
The working out starts off like this...
$50 000 = $750 x (1.08 [1.08^n -1]/0.08)
but it gets to this...
50 000 x (0.08 / 750 x 1.08) = 1.08^n -1
...which I don't get. How did he get the 1.08^n-1 on the right hand side please?
Cheers
\begin{align*} 50000 &= 750 \times \left( \frac{1.08(1.08^n - 1)}{0.08} \right)\\ \frac{50000}{750} &= \frac{1.08(1.08^n - 1)}{0.08}\\ \frac{50000}{750} \times 0.08 &= 1.08(1.08^n - 1)\\ \frac{50000}{750} \times 0.08 \times \frac{1}{1.08} &= 1.08^n - 1\end{align*}
Here, I do it step by step so the order in which things got moved over has been shuffled. But it is clear that the 750 and the 1.08 both land on the bottom, whilst the 0.08 lands on the top.
Post by: emily_p on October 23, 2018, 05:28:38 pm
How many strips/function values are used in one, two and three applications of the simpsons and trapezoidal rule? I’m not too familiar with the reference sheet methods and prefer to use the short cut.
Thanks :) :)
Post by: kauac on October 23, 2018, 05:32:33 pm
Hi! Quick question-
How many strips/function values are used in one, two and three applications of the simpsons and trapezoidal rule? I’m not too familiar with the reference sheet methods and prefer to use the short cut.
Thanks :) :)
Hi...
For both Simpson's and trapezoidal's, the rule for the number of function values = number sub-intervals + 1.
E.g. the formulas on the formula sheet use 2 function values for 1 sub-interval.
Post by: fun_jirachi on October 23, 2018, 05:44:17 pm
In general, trapezoidal rule uses n function values for n-1 applications while Simpson's rule does the same for n-2 applications.
Hope this helps :D
Post by: emily_p on October 23, 2018, 05:48:21 pm
Pretty sure Simpson's Rule uses three function values, while trapezoidal uses two. Reason being from the reference sheet for simpsons rule you have f(a), f(b) and f((a+b)/2). THis is for one application.
In general, trapezoidal rule uses n function values for n-1 applications while Simpson's rule does the same for n-2 applications.
Hope this helps :D
Ah yep that definitely helps. So just clarifying if one application of Simpson’s is 2 strips/3 f values, does that mean two applications is 4 strips/5 f values, three applications is 6 strips etc? I’m not sure if the parabolic arcs overlap.
Post by: RuiAce on October 23, 2018, 05:49:22 pm
Ah yep that definitely helps. So just clarifying if one application of Simpson’s is 2 strips/3 f values, does that mean two applications is 4 strips/5 f values, three applications is 6 strips etc? I’m not sure if the parabolic arcs overlap.For Simpson's rule, yes you always increment by 2 at a time. So for two applications of Simpson's rule, as you correctly stated you'd require 5 function values.
Post by: emily_p on October 23, 2018, 05:51:08 pm
For Simpson's rule, yes you always increment by 2 at a time. So for two applications of Simpson's rule, as you correctly stated you'd require 5 function values.
Ohhh right got it, thanks Rui!! :D :D
Post by: fun_jirachi on October 23, 2018, 05:56:28 pm
For Simpson's rule, yes you always increment by 2 at a time. So for two applications of Simpson's rule, as you correctly stated you'd require 5 function values.
Oops, my mistake. Was meant to write for n applications you need 2n+1 function values, but oh well.
Post by: bahramnilu on October 23, 2018, 07:33:43 pm
How would you do this question and why?
Post by: 8veFable on October 23, 2018, 07:54:38 pm
Find the equation of the locus of a point that moves so that it is equidistant from the line 4x-3y+2=0 and the line 3x+4y-7.
How would you do this question and why?
I’ll try my best but I’m not sure if I got the last part of this correct.
I would let P be point (x,y), and would use the perpendicular distance formula between P and each of those two lines, as the perpendicular formula allows you to find the shortest distance from a point to a line.
d = |a(x1)+b(y1)+c| / (a2 + b2)1/2
For P(x,y) and 4x-3y+2=0
|4(x) - 3(y) + 2|/√25 — (1)
For P(x,y) and 3x+4y-7=0
|3(x) + 4(y) - 7|/√25 — (2)
And since equations (1) and (2) give the distance from point P(x,y) to each line, and we want our point to be equistant (equal distance) from both of them, we will make them equal to each other.
|4x - 3y + 2|/√25 = |3x+4y-7|/√25
Cancel out the √25
|4x - 3y +2| = |3x + 4y - 7|
And then this is the dodgy part that I am not sure about. I would cancel out the absolute values but I have no idea if this is even allowed.
I got: x - 7y + 9 = 0, and graphing it on Desmos, it seems okay, but I would like to know if I broke mathematics with making the absolute values equal to each other and if it’s even allowed
Post by: fun_jirachi on October 23, 2018, 08:00:20 pm
So we have two cases.
Case 1
Case 2
Notice that the two solutions are perpendicular to each other.
Hope this helps :)
EDIT: I saw your solution above, but you've got to note you can't just cancel the absolute values. When equating two absolute values, you have to consider both the negative case and the positive case. In a sense there are four cases, but you can narrow it down to two (which is what I've done) by considering that two pairs of cases are the same, where they are opposite in sign or have the same sign.
Post by: 8veFable on October 23, 2018, 08:06:35 pm
Thanks for fixing it up fun_jirachi :D
I also have a small question that I wanted to get clarified about the 2016 HSC 2U Math question b) iii) --> Attached
For part iii) when you are expected to 'show' that dy/dt = y/400(200-y) I thought that you were not allowed to go 'reverse' and sub y into that equation and turn it into dy/dt but that is how the sample answers did it.
Is this a case with all math questions like this, where you are allowed to go 'reverse' and start with the equation that you are provided with to show that something is equal to something?
Thanks :)
Post by: RuiAce on October 23, 2018, 08:12:50 pm
Hi!You aren't working backwards there. You were given that \(y = \frac{200}{1+19e^{-0.5t}}\) at the start to begin with, and hence are not assuming anything you want to prove. Because you've already proven that, you can then sub it straight into \( \frac{y}{400}(200-y)\) and see what falls out.
Thanks for fixing it up fun_jirachi :D
I also have a small question that I wanted to get clarified about the 2016 HSC 2U Math question b) iii) --> Attached
For part iii) when you are expected to 'show' that dy/dt = y/400(200-y) I thought that you were not allowed to go 'reverse' and sub y into that equation and turn it into dy/dt but that is how the sample answers did it.
Is this a case with all math questions like this, where you are allowed to go 'reverse' and start with the equation that you are provided with to show that something is equal to something?
Thanks :)
What you want to prove is when you simplify that expression, coincidentally you get to the same answer as in (i). Which proves that in fact, \( \frac{y}{400}(200-y)\) is an alternate way of writing \( \frac{dy}{dt} \).
A case of working backwards would be assuming that \( \frac{dy}{dt} = \frac{y}{400}(200-y) \), and then subbing in your answer from part i), and rearranging the gibberish until you get something that's obviously true such as \(1 = 1\). Now that, would be unacceptable. Again, what you're doing here is manually computing \( \frac{y}{400}(200-y)\) given what you already know, and checking it actually equals \( \frac{dy}{dt} \).
Edit: Perhaps the resolve the confusion, a case of where you can't just sub \(y\) in without good reason would be if you weren't given the formula for \(y\) until that part. But here, we were given the formula for \(y\) at the very beginning.
Post by: jamesrandom on October 23, 2018, 09:44:21 pm
Spoiler
( x4 -2x3 -4x2 -1 ) / ( x2 - x - 1 )
Post by: RuiAce on October 23, 2018, 10:04:41 pm
Hi, can someone please show me how to do this? I cant find which line of working i screwed up on. I can apply the quotient rule but I cant get the answer in the textbook which isThe denominator should have a square around it.Spoiler( x4 -2x3 -4x2 -1 ) / ( x2 - x - 1 )
\begin{align*} \frac{d}{dx} \left( \frac{x^3+x}{x^2-x-1} \right)&= \frac{(3x^2+1)(x^2-x-1) - (2x-1)(x^3+x)}{(x^2-x-1)^2}\\ &= \frac{(3x^4-3x^3-3x^2+x^2-x-1)-(2x^4+2x^2-x^3-x)}{(x^2-x-1)^2}\\ &= \frac{x^4-2x^3-4x^2-1}{(x^2-x-1)^2}\end{align*}
Post by: jamesrandom on October 23, 2018, 10:11:29 pm
The denominator should have a square around it.
\begin{align*} \frac{d}{dx} \left( \frac{x^3+x}{x^2-x-1} \right)&= \frac{(3x^2+1)(x^2-x-1) - (2x-1)(x^3+x)}{(x^2-x-1)^2}\\ &= \frac{(3x^4-3x^3-3x^2+x^2-x-1)-(2x^4+2x^2-x^3-x)}{(x^2-x-1)^2}\\ &= \frac{x^4-2x^3-4x^2-1}{(x^2-x-1)^2}\end{align*}
Thanks mate, as I thought, I messed up on one line... and yes the denominator should be squared, i forgot to type that in my question.
Post by: steviemay2000 on October 24, 2018, 10:56:04 am
I am confused by this, as how did they jump from finding the points of intersection to then using something to do with he discriminant?? I though the discriminant was to do with where the parabola cuts on the axis... Could someone please clarify my confusion. Thanks!
Post by: RuiAce on October 24, 2018, 11:02:06 am
Could someone help clarify something about the attached question from the 2012 hsc exam. So I had a look at the success workbook solutions. And they substituted y into the formula and such to find the points of intersection and put the equation into a neat quadratic form. But they then went to say to let the discriminant be equal to zero since there are two equal roots...Actually, the discriminant being equal to zero is by definition what happens when there are two equal roots.
I am confused by this, as how did they jump from finding the points of intersection to then using something to do with he discriminant?? I though the discriminant was to do with where the parabola cuts on the axis... Could someone please clarify my confusion. Thanks!
- \( \Delta > 0\) implies two distinct roots
- \( \Delta = 0\) implies two equal roots
- \( \Delta < 0\) implies no (real) roots
What you've stated is a follow-up result, and not a part of the definition of the discriminant. The idea is that if we then plot the graph of the quadratic (i.e. the new one we obtained from doing that subbing and rearranging), we get a (new) parabola. But, if the parabola has two equal roots, then the parabola barely touches the axis. Similarly, if the parabola has two distinct roots, then the parabola cuts the axis twice.
So, whilst the discriminant is related to where that new parabola cuts the axis, it's only related because it is a consequence of the three points I've stated above.
Post by: parallaxd on October 24, 2018, 03:55:05 pm
Thanks.
Post by: RuiAce on October 24, 2018, 04:15:49 pm
In 10aiv from the 2010 hsc maths exam I have no clue how to prove that result??Simply use the range of the cosine function.
Thanks.
\( \cos \theta\) has range \(-1\leq \cos\theta \leq 1\).
Therefore \(-2\cos \theta\) has range \(-2\leq -2\cos\theta \leq 2\).
Therefore \(1-2\cos\theta\) has range \(-1\leq1-2\cos\theta\leq3\).
From here, we can just extract off the right bit to conclude that \(1-2\cos\theta\leq 3\). Multiplying both sides of the equation by \(a\) gives the desired result.
Post by: Bells_123 on October 24, 2018, 04:22:34 pm
Post by: dazza2020 on October 24, 2018, 04:40:01 pm
Hello! I was wondering with this multiple choice question, how to find all of the solutions of what x could be. I know the answer is C (6 solutions) but am not sure how to get all of them. Thank you!
You'll have two equations to solve, sin3x = 1/2 & sin3x = -1/2, due to the nature of the absolute value
Then, as you're finding solutions for 3x, triple your range, and find all solutions for both equations for 0 < 3x < 3pi
FInally, divide all your solutions by 3 and you'll find 6 solutions betweem 0 < x < pi
The solutions are pi/18, 5pi/18, 13pi/18, 17pi/18, 7pi/18, 11pi/18
Post by: 8veFable on October 24, 2018, 06:33:49 pm
I'm having trouble with a line in the 2017 HSC Sample Answers for Q16 c) i) if anyone could please help me with it :) - Attached
They get 'then BD = DE (equal intercept)', but I am unsure how they got this?
Much appreciated!! :D
Post by: RuiAce on October 24, 2018, 06:36:38 pm
Hi!
I'm having trouble with a line in the 2017 HSC Sample Answers for Q16 c) i) if anyone could please help me with it :) - Attached
They get 'then BD = DE (equal intercept)', but I am unsure how they got this?
Much appreciated!! :D
Post by: michael.shin on October 24, 2018, 07:18:55 pm
Post by: RuiAce on October 24, 2018, 07:25:32 pm
qs probability
Post by: alisoneom on October 24, 2018, 09:58:05 pm
1) I don’t understand what that statement means :(((
2) I’m not sure why they equated the discriminant to 0 (implying that there is only one solution) - shouldn’t there be two solutions, since the circle touches the parabola twice?
(It would also be great to have more guided answers to the remainder of the question)
Thank you in advance and I’d appreciate anyone’s help! :D
Post by: alisoneom on October 24, 2018, 10:01:12 pm
Post by: fun_jirachi on October 24, 2018, 10:23:54 pm
Just two things to start off with:
1. Please don't double post! There's an edit button on your original post if you want to append anything.
2. As for your first question, Rui already addressed this on the previous page! You might want to check it out. :)
Differentiating from first principles is essentially what you started out with when before you did all your product rule and your chain rule and your f'(x)=nxn-1 jazz. Remember from the definition of the function that f(x)= whatever the function is is essentially the value of the function when you sub in x for x. A similar thing happens for f(x+h), sub in (x+h) every time you see an x, that's a great way to think about it. An example would be f(x)=x2, you get that f(x+h)=(x+h)2.
Differentiating from first principles is usually easy from here. You ideally want to factorise out the h and try to remove the f(x) by relating it to f(x+h) in some shape or form, so you get an expression independent of h (in the denominator) since the denominator can't equal zero.
If you ave any more questions, feel free to ask! Hope I helped :)
Post by: Opengangs on October 24, 2018, 10:28:09 pm
Hey guys! I need help with both parts of this HSC 2012 question attached. I’m most confused about the solution, which states that “The circle is a tangent if there is precisely one solution, so the discriminant has to vanish”, and then equates the discriminant to 0.Hey, alisoneom!
1) I don’t understand what that statement means :(((
2) I’m not sure why they equated the discriminant to 0 (implying that there is only one solution) - shouldn’t there be two solutions, since the circle touches the parabola twice?
(It would also be great to have more guided answers to the remainder of the question)
Thank you in advance and I’d appreciate anyone’s help! :D
(I've avoided looking at the solution to provide a full explanation into how to approach this type of question!)
First, notice that the parabola and the circle touch at some point. So we solve the two curves simultaneously. Substituting \(y = x^2\) into the circle, we get:
\[ y + (y - c)^2 = r^2.\]
Upon expanding, you'll end up with a quadratic in terms of \(y\). The question tells us that the point of intersection is symmetric for \(y\). So, while we do have two points of intersection, these points of intersection occur on the same \(y\) value! This tells us that you have one distinct root. And hence, you have the discriminant being zero, which answers your first and second question!
Setting the discriminant equal to 0 will give you the required result!
Post by: jamesrandom on October 28, 2018, 04:32:45 pm
Post by: RuiAce on October 28, 2018, 04:41:43 pm
Hey can someone help me out with this question. Have no clue where im going wrong....
Or if you chose the product rule approach to differentiate it you'd obtain \( \frac{dy}{dx} = 2(x+3)^2 + 4x(x+3) \), but expanding that out still gives \(6x^2+24x+18\).
Post by: jamesrandom on November 01, 2018, 11:04:52 pm
Post by: RuiAce on November 01, 2018, 11:23:10 pm
Can someone just quickly explain why x > +- 3 becomes x > 3 and x < -3. Why does the inequality sign switch.It doesn't.
Are you solving a quadratic inequality? Because note that for quadratic inequalities such as \(x^2-9 > 0\), you should not be treating the same way as though you were solving linear inequalities.
Post by: Bilbo on November 14, 2018, 08:58:26 pm
Post by: jamonwindeyer on November 14, 2018, 11:04:08 pm
Hi, found the lecture super helpful, will these come out each month.
Welcome to the forums!! Our Maths lectures typically happen in the holiday periods, so every few months :)
Post by: jamesrandom on November 19, 2018, 01:02:18 pm
Post by: S200 on November 19, 2018, 04:21:06 pm
Plz help this locus chapter is doing my head in :)
Question 8
\(Eq_1: y=2x-4\) (re-arranged)
\(Eq_2: y=\frac{1}{4}x^2\) (also re-arranged)
So, we know that the derivative at a point will equal the gradient of the tangent at that point (\(2\)), so lets go ahead and do that.
\(2=2\times \frac{1}{4}x \quad \therefore \quad x=4\)
Then, sub this x-value back into either of the given equations to get the y-coordinate, and hence the full co-ordinates of the point.
\(Eq_2: y=\frac{1}{4}x^2\) (also re-arranged)
So, we know that the derivative at a point will equal the gradient of the tangent at that point (\(2\)), so lets go ahead and do that.
\(2=2\times \frac{1}{4}x \quad \therefore \quad x=4\)
Then, sub this x-value back into either of the given equations to get the y-coordinate, and hence the full co-ordinates of the point.
Question 10
I had no clue what a focus or a latus rectum was before this question, so thanks for posting this question! :D
This gives a nifty formula to help here.
So, we re-arrange into the "vertex form"...
\(x^2=-8y \quad \therefore \quad y=\frac{-1}{8}(x)^2\)
We can clearly see that this parabola has the vertex at \((0,0)\), because \(h\) and \(k\) are not present (i.e; They are zero).
So, we move on to the focus, using the formula given... \((h,k+\frac{1}{4a})\)
\(h=0\), so no worries there.
\((k+\frac{1}{4a}) = \left (0+\frac{1}{4\times \frac{-1}{8}}\right )\) which then gives us the \(y\) (or \(k\)) position of \(-2\)
Now, we just equate the parabola to the line of \(y=-2\) to get the x-points of the intersection. The length of the latus rectum can be found by taking the length of the semi-latus rectum (i.e; the positive x-coordinate of the intersection points you just found) and doubling it.
This gives a nifty formula to help here.
Quote
If you have the equation of a parabola in vertex form \(y=a(x−h)^2+k\), then the vertex is at \((h,k)\) and the focus is \((h,k+\frac{1}{4a})\).
So, we re-arrange into the "vertex form"...
\(x^2=-8y \quad \therefore \quad y=\frac{-1}{8}(x)^2\)
We can clearly see that this parabola has the vertex at \((0,0)\), because \(h\) and \(k\) are not present (i.e; They are zero).
So, we move on to the focus, using the formula given... \((h,k+\frac{1}{4a})\)
\(h=0\), so no worries there.
\((k+\frac{1}{4a}) = \left (0+\frac{1}{4\times \frac{-1}{8}}\right )\) which then gives us the \(y\) (or \(k\)) position of \(-2\)
Now, we just equate the parabola to the line of \(y=-2\) to get the x-points of the intersection. The length of the latus rectum can be found by taking the length of the semi-latus rectum (i.e; the positive x-coordinate of the intersection points you just found) and doubling it.
Did you try to work it out?
Yes
\(2=\frac {x^2} {8}\) as the negatives cancelled.
\(x=\sqrt {2\times 8}\)
\(x=\pm 4 \quad \therefore \quad 4\times 2=8\), so the Latus Rectum is equal to \(8\) units!
\(x=\sqrt {2\times 8}\)
\(x=\pm 4 \quad \therefore \quad 4\times 2=8\), so the Latus Rectum is equal to \(8\) units!
No
Go try it please, 'cause just following my answers most likely won't help unless you try similar questions yourself and develop your problem solving abilities (and that's what we're all about) ;) :D
Post by: iktimal on November 20, 2018, 08:39:44 pm
Can someone please assist me in finding Tn of this series?
5-2+10-8+15-32
Post by: RuiAce on November 20, 2018, 08:46:52 pm
Hello,From what it looks like, this looks like two sequences clumped together into one. For each odd value, we have \( T_{2n-1} = 5n \) and \(T_{2n} = -2 \times 4^{n-1} \).
Can someone please assist me in finding Tn of this series?
5-2+10-8+15-32
(Note that this is because the odd terms form an A.P. with common difference 5, and the even terms form a G.P. with common ratio 4.)
Note that writing out \(S_n\) would therefore be a bit awkward.
Post by: iktimal on November 20, 2018, 08:51:51 pm
From what it looks like, this looks like two sequences clumped together into one. For each odd value, we have \( T_{2n-1} = 5n \) and \(T_{2n} = -2 \times 4^{n-1} \).
(Note that this is because the odd terms form an A.P. with common difference 5, and the even terms form a G.P. with common ratio 4.)
Note that writing out \(S_n\) would therefore be a bit awkward.
Hi Rui,
Well actually the next part of the question asks for the sum of the first 20 terms of the series. How would I do it using these equations?
Would I look for the sum of the first 20 terms for the AP and GP then add them together?
Post by: RuiAce on November 20, 2018, 08:54:02 pm
Hi Rui,You can think of it like this. 10 of those terms will be "odd terms", namely position 1, 3, 5, ..., 19. The other terms will be the even terms.
Well actually the next part of the question asks for the sum of the first 20 terms of the series. How would I do it using these equations?
So you can add: {sum of 10 terms in an AP with first term 5 and common difference 5} + {sum of 10 terms in a GP with first term -2 and common ratio 4}. This would be equivalent to \(\boxed{ \frac{10}{2} \left[2(5) + (10-1)(5) \right]+ \frac{-2\left(4^{10}-1\right)}{4-1}}\).
Post by: iktimal on November 20, 2018, 08:58:25 pm
You can think of it like this. 10 of those terms will be "odd terms", namely position 1, 3, 5, ..., 19. The other terms will be the even terms.
So you can add: {sum of 10 terms in an AP with first term 5 and common difference 5} + {sum of 10 terms in a GP with first term -2 and common ratio 4}. This would be equivalent to \(\boxed{ \frac{10}{2} \left[2(5) + (10-1)(5) \right]+ \frac{-2\left(4^{10}-1\right)}{4-1}}\).
Oh I see now, thank you very much :)
Post by: jamonwindeyer on January 12, 2019, 09:51:30 am
I love you! <3
Post by: jamonwindeyer on January 12, 2019, 09:52:00 am
Help me plz...
I love you! <3
weird...
Post by: emilyyyyyyy on January 12, 2019, 07:12:09 pm
Just a general question; how much of the prelim course do we have to know? and should I be actively studying it?
Post by: r1ckworthy on January 12, 2019, 07:55:00 pm
Hi!
Just a general question; how much of the prelim course do we have to know? and should I be actively studying it?
Hi Emily,
I assume you will be doing the HSC this year, so you're still part of the old syllabus. Yes, you will have to know ALL of the preliminary course. As mentioned by RUIace below, you will still have to know all of the preliminary course for the new syllabus as well. In the 2-Unit maths HSC, you will be tested on topics such as plane geometry, trig functions, real functions etc, which you have learned last year. These types of questions typically come in the earlier questions of the HSC 2-Unit exam, although they do have a slight chance to come up in more difficult questions.
In terms of studying actively, the best thing you can do is to go through HSC papers. When I accelerated the course last year, we haven't covered that much of Yr12 topics when it was time for our half yearlies. Unlike others, I went through and answered the HSC papers from 2012-17, skipping the questions that I didn't know. This advice came from a teacher, and it really helped me. This is the best way you can learn. Attempt a paper from start to finish and check the answers if you are having a hard time and see if you have learned that topic yet. If not, just keep going. If you are struggling with a topic (eg. trig functions), go back to your textbook and do more questions. That is the best way to study for mathematics (or what I found really helpful).
Hope that helped!!
Post by: emilyyyyyyy on January 12, 2019, 10:33:37 pm
Post by: RuiAce on January 12, 2019, 10:36:51 pm
Hi Emily,Remark: For both the old and new syllabuses, you're expected to know all of the preliminary/Year 11 course.
I assume you will be doing the HSC this year, so you're still part of the old syllabus. Yes, you will have to know the most important bits of the preliminary course. x
Post by: r1ckworthy on January 13, 2019, 09:53:24 am
Remark: For both the old and new syllabuses, you're expected to know all of the preliminary/Year 11 course.Thanks, I'll correct my answer soon.
Post by: shaynec19 on January 14, 2019, 01:48:21 pm
Just struggling with this question atm. Keep seeming to get the wrong answer.
"Find the equation of the tangent to the curve y=lnx at the point (2,ln2)."
VVVV simple question, I just can't see what I'm doing wrong.
Cheers
Post by: MB_ on January 14, 2019, 01:56:03 pm
Hi anyone that can answer my question...Would you be able to post your working so we can see where you're getting stuck?
Just struggling with this question atm. Keep seeming to get the wrong answer.
"Find the equation of the tangent to the curve y=lnx at the point (2,ln2)."
VVVV simple question, I just can't see what I'm doing wrong.
Cheers
Post by: shaynec19 on January 14, 2019, 02:09:57 pm
I have differentiated lnx to get 1/x. This is my gradient right?
I then used the point gradient formula with the x&y values.
Post by: MB_ on January 14, 2019, 02:12:36 pm
Yep Sure,Once you differentiate, you have to sub in the x value of the point specified to find the gradient
I have differentiated lnx to get 1/x. This is my gradient right?
I then used the point gradient formula with the x&y values.
Post by: shaynec19 on January 14, 2019, 02:17:19 pm
What is wrong with me....???
#holidaymaths
Thankyou very much!
Post by: emilyyyyyyy on January 14, 2019, 03:15:00 pm
Post by: fun_jirachi on January 14, 2019, 06:48:53 pm
Can someone please help me with what values or whatever i put into this trapezoidal rule formula (attached)? i've watched yt vids and looked in multiple textbooks but i don't really get how i'm meant to use this formula. Thanks!
If you look at the formula, you see that the first and terms are only multiplied by one while every other term is multiplied by two. Think about what the trapezoidal rule actually does over multiple applications; with two trapeziums side by side ie for the integral
when applying the trapezoidal rule with two applications you use
Notice that since the 'gaps' between the numbers are the same you can just reduce that to h/2 in the formula, where h is the distance between applications. Another thing to notice is that you count the first term (ie. y0, or f(x) of the lower bound) and last term (ie. yn, or f(x) of the upper bound) only once as they intuitively are on the 'outside' of the many trapeziums (ie. they are not included by two trapeziums). On the other hand, anything in between is counted twice because the function value is used twice by two different trapeziums. hence the 2(y1+y2+y3+...+yn-1).
Basically for n applications of the trapezoidal rule over the curve f(x), subtract the lower bound from the upper bound, divide n into that number to get another number, say h. h+the lower bound becomes your next point of reference, 2h+lowerbound is the next and so on until you get to the upper bound. Find f(x) of these values to get what you need to sub in. f(x) of the lower bound and upper bound will get counted once only (first and last) and everything else that you've figured out will get counted twice. h is also your height that you sub into that formula.
If you need a better/quick way to memorise it, I remembered it like this; height/2 x (first + last + 2(everything else)) which is the same thing as what you've given, but much easier to remember. :)
Hope this helps! ")
Post by: shaynec19 on January 15, 2019, 02:56:34 pm
Can someone please explain why when differentiating this exponential, that ln2 becomes part of the coefficient? I thought it would just be the 3, as the 3x-4 is the term with the pronumeral.
Thank You for anyone's help.
Post by: RuiAce on January 15, 2019, 03:00:33 pm
Hi,\(\ln 2\) is a constant.
Can someone please explain why when differentiating this exponential, that ln2 becomes part of the coefficient? I thought it would just be the 3, as the 3x-4 is the term with the pronumeral.^{3x-4})
 )
Thank You for anyone's help.
That last step was redundant but I only did it to illustrate that what comes next is we expand the \(\ln 2\) term into the bracket.
\[ e^{\ln 2 \times (3x-4)} = e^{3x\ln 2 - 4\ln 2} \]
Therefore the coefficient on \(x\) is \(3\ln 2\), justifying why that's what comes down after differentiating
Post by: shaynec19 on January 15, 2019, 03:12:16 pm
\(\ln 2\) is a constant.
That last step was redundant but I only did it to illustrate that what comes next is we expand the \(\ln 2\) term into the bracket.
\[ e^{\ln 2 \times (3x-4)} = e^{3x\ln 2 - 4\ln 2} \]
Therefore the coefficient on \(x\) is \(3\ln 2\), justifying why that's what comes down after differentiating
Aaah Thanks Rui!
Didn't think about expanding the brackets! Makes perfect sense now.
Cheers
Shayne
Post by: shaynec19 on January 15, 2019, 04:30:39 pm
Is there any tips and and tricks to finding it? I know its 3/2 in this case, and I can understand why it's there but can't seem to be able to reverse engineer it.
Post by: RuiAce on January 15, 2019, 04:45:15 pm
Could Someone please help me with finding the fraction that leads to finding primitive function of:\[ \text{At the end of the day, somehow you're trying to invoke that}\\ \int \frac{f^\prime(x)}{f(x)}\,dx = \ln [f(x)]+C.\\ \text{It's your job to actually make }f^\prime(x)\text{ appear.} \]
Is there any tips and and tricks to finding it? I know its 3/2 in this case, and I can understand why it's there but can't seem to be able to reverse engineer it.
\[ \text{That manipulation in line 1 where we force }2x\text{ to appear}\\ \text{is the crucial step.} \]
Post by: shaynec19 on January 16, 2019, 11:54:40 am
Post by: RuiAce on January 16, 2019, 04:36:21 pm
Could someone please help with this question;
\begin{align*}y&= \log_e x\\ x&= e^y\\ x^2 &= e^{2y} \end{align*}
So your volume integral will be \( \pi\int_1^3 e^{2y}\,dy \).
Post by: shaynec19 on January 16, 2019, 05:28:40 pm
Thanks Rui!
\begin{align*}y&= \log_e x\\ x&= e^y\\ x^2 &= e^{2y} \end{align*}
So your volume integral will be \( \pi\int_1^3 e^{2y}\,dy \).
Also, thank you for your advice with the exam!
Post by: emmajb37 on January 17, 2019, 11:26:01 am
I have attached a photo.
Why does geometry exist >:(
Post by: RuiAce on January 17, 2019, 02:06:02 pm
Thanks Rui!Haha no worries. Your exam was certainly a weird case. I feel a bit bad for you given the circumstances around it but there was still a very subtle mistake anyhow
Also, thank you for your advice with the exam!
Hi, I am struggling with this past trial question please help! Western Region Trial Exam 2005 Question 5 a) ii)Well there's a bit of trigonometry involved on top of the geometry here.
I have attached a photo.
Why does geometry exist >:(
\[ \text{Part i) should be obvious, given that }AB \parallel CD\\ \text{so clearly the two alternate angles will help complete the equiangular test.} \]
\[ \text{Now as a consequence of the similar triangles, we have the proportional sides}\\ \frac{AM}{DM} = \frac{MB}{MC} = \frac{AB}{DC}.\\ \text{But we're also told that }\frac{AB}{CD} = \frac25.\\ \text{This consequently implies that }\frac{AM}{DM} = \frac25 \implies \boxed{DM = \frac52 AM}\\ \text{and also implies that }\frac{MB}{MC} = \frac25 \implies \boxed{CM = \frac52 BM}.\]
\[ \text{To invoke areas, let }\angle AMB = \theta.\text{ Then from vertically opposite angles, }\angle CMD = \theta\\ \text{and from angles on straight angles, }\angle AMC = \angle BMD = 180^\circ - \theta.\]
\[ \text{Consequently }\sin \angle AMB = \sin \angle CMD = \sin \theta\\ \text{and }\sin \angle AMC = \sin \angle BMD = \sin (180^\circ - \theta).\\ \text{However using our ASTC identities, we know that }\boxed{\sin(180^\circ - \theta) = \sin \theta}\\ \text{so very conveniently, we have}\\ \boxed{\sin \angle AMB = \sin \angle AMC = \sin \angle BMD = \sin \angle CMD = \sin \theta} \]
\[ \text{This set-up was used because since}\operatorname{Area}_{\triangle AMB} = 10,\\ \text{we have }\boxed{\frac12 \, AM \, BM\sin \theta = 10 }.\\ \text{Note that the area we require can be thought of as}\\ \boxed{\operatorname{Area}_{ABCD} = \operatorname{Area}_{\triangle AMB} + \operatorname{Area}_{\triangle AMC} + \operatorname{Area}_{\triangle BMD} + \operatorname{Area}_{\triangle CMD}}\\ \text{and we now have the ingredients to find the areas of all four triangles.} \]
\begin{align*} \operatorname{Area}_{\triangle AMB} &= \frac12 \, AM\, BM \sin \theta = 10\\ \operatorname{Area}_{\triangle AMC} &= \frac12\, AM\, CM\sin\theta = \frac12\,AM \left(\frac52BM\right)\sin\theta = 25\\ \operatorname{Area}_{\triangle BMD} &= \frac12\, BM\, DM\sin\theta = \frac12\,BM \left(\frac52AM\right)\sin\theta = 25\\ \operatorname{Area}_{\triangle CMD} &= \frac12\, CM\, DM\sin\theta = \frac12\left( \frac52 AM\right) \left(\frac52BM\right)\sin\theta =62.5 \end{align*}
The final answer is hence just the sum of these, which is 122.5 units2.
Post by: emmajb37 on January 17, 2019, 03:10:19 pm
Haha no worries. Your exam was certainly a weird case. I feel a bit bad for you given the circumstances around it but there was still a very subtle mistake anyhowWell there's a bit of trigonometry involved on top of the geometry here.
\[ \text{Part i) should be obvious, given that }AB \parallel CD\\ \text{so clearly the two alternate angles will help complete the equiangular test.} \]
\[ \text{Now as a consequence of the similar triangles, we have the proportional sides}\\ \frac{AM}{DM} = \frac{MB}{MC} = \frac{AB}{DC}.\\ \text{But we're also told that }\frac{AB}{CD} = \frac25.\\ \text{This consequently implies that }\frac{AM}{DM} = \frac25 \implies \boxed{DM = \frac52 AM}\\ \text{and also implies that }\frac{MB}{MC} = \frac25 \implies \boxed{CM = \frac52 BM}.\]
\[ \text{To invoke areas, let }\angle AMB = \theta.\text{ Then from vertically opposite angles, }\angle CMD = \theta\\ \text{and from angles on straight angles, }\angle AMC = \angle BMD = 180^\circ - \theta.\]
\[ \text{Consequently }\sin \angle AMB = \sin \angle CMD = \sin \theta\\ \text{and }\sin \angle AMC = \sin \angle BMD = \sin (180^\circ - \theta).\\ \text{However using our ASTC identities, we know that }\boxed{\sin(180^\circ - \theta) = \sin \theta}\\ \text{so very conveniently, we have}\\ \boxed{\sin \angle AMB = \sin \angle AMC = \sin \angle BMD = \sin \angle CMD = \sin \theta} \]
\[ \text{This set-up was used because since}\operatorname{Area}_{\triangle AMB} = 10,\\ \text{we have }\boxed{\frac12 \, AM \, BM\sin \theta = 10 }.\\ \text{Note that the area we require can be thought of as}\\ \boxed{\operatorname{Area}_{ABCD} = \operatorname{Area}_{\triangle AMB} + \operatorname{Area}_{\triangle AMC} + \operatorname{Area}_{\triangle BMD} + \operatorname{Area}_{\triangle CMD}}\\ \text{and we now have the ingredients to find the areas of all four triangles.} \]
\begin{align*} \operatorname{Area}_{\triangle AMB} &= \frac12 \, AM\, BM \sin \theta = 10\\ \operatorname{Area}_{\triangle AMC} &= \frac12\, AM\, CM\sin\theta = \frac12\,AM \left(\frac52BM\right)\sin\theta = 25\\ \operatorname{Area}_{\triangle BMD} &= \frac12\, BM\, DM\sin\theta = \frac12\,BM \left(\frac52AM\right)\sin\theta = 25\\ \operatorname{Area}_{\triangle CMD} &= \frac12\, CM\, DM\sin\theta = \frac12\left( \frac52 AM\right) \left(\frac52BM\right)\sin\theta =62.5 \end{align*}
The final answer is hence just the sum of these, which is 122.5 units2.
Thanks so much Rui!
But the answers say that it is 72.5u^2. Would hat then be wrong?
Post by: RuiAce on January 17, 2019, 03:21:04 pm
Thanks so much Rui!Well, taking a look at the diagram and the computations, 72.5u^2 would just be the combined area of just \( \triangle AMB\) and \(\triangle CMD\), i.e. the two triangles originally drawn and excluding the ones you have in pencil. However I feel as though the wording of the question is ambiguous then because these "self-intersecting quadrilaterals" are not a part of the course, so I interpreted the quadrilateral to actually include the new triangles you've drawn in.
But the answers say that it is 72.5u^2. Would hat then be wrong?
So I blame the exam writers on this one. In the HSC, this kind of ambiguity would not occur.
Post by: emmajb37 on January 18, 2019, 12:43:40 pm
Well, taking a look at the diagram and the computations, 72.5u^2 would just be the combined area of just \( \triangle AMB\) and \(\triangle CMD\), i.e. the two triangles originally drawn and excluding the ones you have in pencil. However I feel as though the wording of the question is ambiguous then because these "self-intersecting quadrilaterals" are not a part of the course, so I interpreted the quadrilateral to actually include the new triangles you've drawn in.
So I blame the exam writers on this one. In the HSC, this kind of ambiguity would not occur.
Awesome thanks again so much!!
Post by: shaynec19 on January 22, 2019, 04:34:46 pm
I have found the x-coordinate, how do I find the y-coordinate.
Post by: RuiAce on January 22, 2019, 04:42:14 pm
Could someone please help me with the 1997 2u HSC exam, Question 10. b. iv.At a quick glance, you could probably use the fact that \(P\) lies on the circle \(x^2+y^2=1\)?
I have found the x-coordinate, how do I find the y-coordinate.
Post by: shaynec19 on January 22, 2019, 04:46:19 pm
At a quick glance, you could probably use the fact that \(P\) lies on the circle \(x^2+y^2=1\)?Hi Rui,
Thankyou! I will try again, I already tried that and got nowhere but figured that because it was so long and messy, there must have been a better way...
Post by: RuiAce on January 22, 2019, 04:53:28 pm
Hi Rui,
Thankyou! I will try again, I already tried that and got nowhere but figured that because it was so long and messy, there must have been a better way...
Well the computations can be made cleaner if we're super clever about it
By doing this, we've completely avoided the quadratic formula
\[ \text{Now using }x^2+y^2=1\text{ we have}\\ \begin{align*}y^2 &= 1- x^2\\ &= 1 - \left(\frac{1-m^2}{1+m^2}\right)^2\\ &= \frac{(1+m^2)^2 - (1-m^2)^2}{(1+m^2)^2}\\ &= \frac{4m^2}{(1+m^2)^2}\\ \therefore y &= \frac{2m}{1+m^2} \end{align*} \]
Post by: Rabi on January 23, 2019, 07:40:56 pm
"A water trough is 200 cm long and has a cross section of a right angled isosceles triangle. Show that when the depth of the water is x cm, the volume of water in the tank is 200x^2 cm^3. Water is poured in at a constant rate of 5 litres per minute. Find the rate at which the water level is rising when the depth is 30cm.
Post by: RuiAce on January 23, 2019, 07:54:50 pm
Hey I really need help with this question plz plz help.This is an MX1 related rates question. Please consult the MX1 thread if you require help with these concepts.
"A water trough is 200 cm long and has a cross section of a right angled isosceles triangle. Show that when the depth of the water is x cm, the volume of water in the tank is 200x^2 cm^3. Water is poured in at a constant rate of 5 litres per minute. Find the rate at which the water level is rising when the depth is 30cm.
Having said that, this question was asked recently. You should also check the hints that were recently provided on addressing this question before indicating more of the problem.
Post by: Georgakopoulou on January 26, 2019, 06:27:23 pm
Post by: RuiAce on January 26, 2019, 07:29:42 pm
Hi! I am having trouble solving the attached question regarding the geometrical applications of calculus, could you please solve it for me? :D\[ \text{Let }C\text{ be the total cost of the trip. Then}\\ C = ct\\ \text{where }t\text{ is the time taken for the trip.} \]
\[ \text{Using the speed-distance-time formula we have }v = \frac{500}{t}\\ \text{so therefore }\boxed{C = \left( 150+\frac{v^2}{80} \right) \frac{500}{v}} \]
\begin{align*} C &=\frac{25}{4} \left( \frac{12000}{v}+v\right) \\ \frac{dC}{dv} &= \frac{25}{4} \left( -\frac{12000}{v^2}+1 \right)\end{align*}
You should be able to continue from here following the usual process. The minimum occurs at \(v = 20\sqrt{30} \) which is approximately 110.
Post by: Georgakopoulou on January 26, 2019, 07:35:41 pm
\[ \text{Let }C\text{ be the total cost of the trip. Then}\\ C = ct\\ \text{where }t\text{ is the time taken for the trip.} \]
\[ \text{Using the speed-distance-time formula we have }v = \frac{500}{t}\\ \text{so therefore }\boxed{C = \left( 150+\frac{v^2}{80} \right) \frac{500}{v}} \]
\begin{align*} C &=\frac{25}{4} \left( \frac{12000}{v}+v\right) \\ \frac{dC}{dv} &= \frac{25}{4} \left( -\frac{12000}{v^2}+1 \right)\end{align*}
You should be able to continue from here following the usual process. The minimum occurs at \(v = 20\sqrt{30} \) which is approximately 110.
Thank you so much!!!!!!!
Post by: alexnero7 on January 29, 2019, 03:24:17 pm
Would someone be able to help me with a question? I am having trouble solving the attached question regarding the geometrical applications of calculus.
Thanks :)
Post by: meerae on January 29, 2019, 03:34:52 pm
Hi everyone,
Would someone be able to help me with a question? I am having trouble solving the attached question regarding the geometrical applications of calculus.
Thanks :)
Hi!
So what we know directly from the question is that y’=0 when x= -1 or 2
So all you would need to do is differentiate and do simultaneous equations.
Hope this helped!
meerae :)
Post by: alexnero7 on January 29, 2019, 05:52:16 pm
Thanks
Post by: meerae on January 29, 2019, 06:09:27 pm
Could someone please work this out? :)
Thanks
Hey alexnero7!
I've attached the working out below, if you need any clarification please let me know as I tend to skip steps without even realising.
P.S for some reason its showing dark on my laptop - it should still be readable, if not please let me know!
Hope this helped!
Post by: alexnero7 on January 29, 2019, 07:04:20 pm
Post by: meerae on January 29, 2019, 07:07:38 pm
Anyone know how to do this?
Hey alexnero7!
You can differentiate this by using the quotient rule, you would differentiate it twice to find first and second derivative.
Hope this helped!
meerae :)
Post by: jamonwindeyer on January 29, 2019, 07:16:53 pm
Anyone know how to do this?
Here's the first one!
Have a shot at the second one, it is really similar! :)
Post by: abdulrahmanb8 on January 29, 2019, 07:28:03 pm
Post by: shaynec19 on January 30, 2019, 12:33:59 pm
Post by: RuiAce on January 30, 2019, 12:37:49 pm
Can someone please help me to solve this definite integral.\begin{align*}\int_3^4 \frac{x^2+x+3}{3x^5}\,dx &= \frac13 \int_3^4 \left( \frac{x^2}{x^5}+\frac{x}{x^5}+\frac{3}{x^5} \right)dx\tag{split fraction up}\\ &= \frac13 \int_3^4 \left( x^{-3}+x^{-4}+3x^{-5}\right)dx \end{align*}
You should be able to take it from here.
Post by: shaynec19 on January 30, 2019, 12:42:11 pm
Post by: alexnero7 on January 30, 2019, 12:45:09 pm
Post by: RuiAce on January 30, 2019, 12:51:23 pm
Hi can someone please help me solve this? For some reason I'm getting an inflexion point, but it should be getting a minimum. Thanks. :)It is true that if we have a point of inflexion, the second derivative is equal to 0.
However, it is not always true that if the second derivative equals to 0, we have a point of inflexion.
For this reason, once we find \( \frac{d^2y}{dx^2} = 0 \) (which is true at your stationary point \(x=0\)), we must always test both sides of the equation to see if there is a concavity change. That forms the distinction between a horizontal point of inflexion and a turning point.
Here, testing a bit to the left, say \(x=-1\) we have \( \frac{d^2y}{dx^2} = 1 > 0\). Testing a bit to the right, say \(x = 1\) we have \( \frac{d^2y}{dx^2} = 1 > 0 \). Hence there is no concavity change, and thus we do not have a point of inflexion.
In fact, because the concavity remains concave up, we deduce it is a local minimum.
Post by: alexnero7 on January 30, 2019, 01:00:14 pm
It is true that if we have a point of inflexion, the second derivative is equal to 0.
However, it is not always true that if the second derivative equals to 0, we have a point of inflexion.
For this reason, once we find \( \frac{d^2y}{dx^2} = 0 \) (which is true at your stationary point \(x=0\)), we must always test both sides of the equation to see if there is a concavity change. That forms the distinction between a horizontal point of inflexion and a turning point.
Here, testing a bit to the left, say \(x=-1\) we have \( \frac{d^2y}{dx^2} = 1 > 0\). Testing a bit to the right, say \(x = 1\) we have \( \frac{d^2y}{dx^2} = 1 > 0 \). Hence there is no concavity change, and thus we do not have a point of inflexion.
In fact, because the concavity remains concave up, we deduce it is a local minimum.
Post by: alexnero7 on January 30, 2019, 03:58:17 pm
Post by: jamonwindeyer on January 30, 2019, 06:17:36 pm
.
Hey! Curve sketches like this are a really common HSC question, so common that I actually wrote a guide on them! Read it here, it might be enough to get you through! Otherwise feel free to let us know which step you get stuck on ;D
Post by: emilyyyyyyy on January 30, 2019, 07:40:00 pm
Thanks :)
Post by: fun_jirachi on January 30, 2019, 08:05:46 pm
All you have to do for this question is find the second derivative Since you're given that x=2 and x=-1 are points of inflexion, you have that f''(2) and f''(-1)=0. Solve simultaneously to find a and b. :) Answers in the spoiler :)
Spoiler
a=3, b=-3
Post by: alexnero7 on January 31, 2019, 05:38:42 pm
Can somebody show me their working out to the attached question? I want to compare it to mine. The answer claims that the maximum is -2.
Post by: RuiAce on January 31, 2019, 05:51:46 pm
Hey everyone,It should be. The only stationary point is at \(x=0\), so we only need to check the value of \(y\) when \(x=0\), and those at the endpoints \(-2\) and \(1\). The candidate values are therefore \(-3\), \(-35\) and \(-2\), so the global maximum value is \(-2\).
Can somebody show me their working out to the attached question? I want to compare it to mine. The answer claims that the maximum is -2.
Note that global maxima/minima do not necessarily coincide with local maxima/minima. They can (and sometimes do in fact) occur on the endpoints of the domain.
Post by: alexnero7 on January 31, 2019, 05:53:43 pm
It should be. The only stationary point is at \(x=0\), so we only need to check the value of \(y\) when \(x=0\), and those at the endpoints \(-2\) and \(1\). The candidate values are therefore \(-3\), \(-35\) and \(-2\), so the global maximum value is \(-2\).
Note that global maxima/minima do not necessarily coincide with local maxima/minima. They can (and sometimes do in fact) occur on the endpoints of the domain.
Post by: meerae on January 31, 2019, 07:20:09 pm
Hey everyone,
Can somebody show me their working out to the attached question? I want to compare it to mine. The answer claims that the maximum is -2.
Hey!
I've attached my working out - if you need any clarification please let me know!
Hope this helped!
meerae :)
Post by: fun_jirachi on January 31, 2019, 08:42:37 pm
The simplest way is probably to take a geometrical approach. By rearranging the second equation to y=x+2, and equating (or recognising) that they intersect at (0, 2) and (-2, 0). The area enclosed by the curve is a quarter of a circle (top right quadrant of the circle x^2+y^2=4), plus a triangle enclosed by the coordinate axes and the line y=x+2, that cuts at (0, 2) and (-2, 0). This triangle has base 2 and height 2 (thinking geometrically) and likewise the semicircle given gives you the dimensions of the quadrant that you need to calculate the area for. I got pi+2 using this method. :)
Hope this helps :D
Post by: eeshab7 on February 02, 2019, 05:57:50 pm
Thanks :)
Post by: meerae on February 02, 2019, 06:00:50 pm
Hi guys, I've had a complete mind blank and can't remember how to do this question. Could someone please help me.
Thanks :)
Hey!
You would use the product rule to differentiate this. just remember for exponentials that if y= e^2x then y'= 2e^2x
You would then find the x and y intercepts of the graph and possibly inflection points (achieved by differentiating again and making y''=0) and then utilise this information to sketch the graph. Does that make sense?
Hope this helped!
meerae :)
Post by: eeshab7 on February 02, 2019, 06:23:15 pm
Hey!
You would use the product rule to differentiate this. just remember for exponentials that if y= e^2x then y'= 2e^2x
You would then find the x and y intercepts of the graph and possibly inflection points (achieved by differentiating again and making y''=0) and then utilise this information to sketch the graph. Does that make sense?
Hope this helped!
meerae :)
Hi,
Thanks for that and yeah it makes sense. This might be a bit messy but am I on the right track?
Post by: RuiAce on February 02, 2019, 06:48:44 pm
Hi,\[ \text{Your product rule computation should be}\\ \begin{align*}y&=x^2e^{2x}\\ \frac{dy}{dx} &= (2x)(e^{2x}) + (x^2)(2e^{2x})\\ &= 2e^{2x} x(x+1) \end{align*} \]
Thanks for that and yeah it makes sense. This might be a bit messy but am I on the right track?
Also, are you sure you meant to have two dashes there on your \(f^{\prime\prime}(x)\)?
Post by: eeshab7 on February 02, 2019, 07:41:12 pm
\[ \text{Your product rule computation should be}\\ \begin{align*}y&=x^2e^{2x}\\ \frac{dy}{dx} &= (2x)(e^{2x}) + (x^2)(2e^{2x})\\ &= 2e^{2x} x(x+1) \end{align*} \]
Also, are you sure you meant to have two dashes there on your \(f^{\prime\prime}(x)\)?
yeah sorry I forgot to include the previous steps in the picture - that was just using the double derivative
yeah sorry I forgot to include the previous steps in the picture - that was just using the double derivative
is that right for the next step?
Post by: RuiAce on February 02, 2019, 07:50:25 pm
is that right for the next step?Still looks off for the second derivative. Just off by a +1 term.
\begin{align*} \frac{dy}{dx} &= (2x^2+2x)e^{2x}\\ \frac{d^2y}{dx^2} &= (4x+2)e^{2x} + (2x^2+2x)(2e^{2x}) \\ &= (4x+2+4x^2+4x)e^{2x}\\ &= 2(2x^2+4x+1)e^{2x} \end{align*}
Although, if we just want to test the nature of the stationary points, leaving it as \( \frac{d^2y}{dx^2} = (4x+2)e^{2x} + 2(2x^2+2x)e^{2x} \) is fine - no need to simplify.
Post by: alexnero7 on February 02, 2019, 08:09:18 pm
Post by: RuiAce on February 02, 2019, 08:12:40 pm
I cant seem to get the right answer for this... it's SO frustrating!! Please HELP!!You have \( \frac{d^2y}{dx^2} = 12x - 14\) so all you're trying to solve is \(12x - 14 > 0\), which becomes \(x > \frac76\).
Post by: alexnero7 on February 02, 2019, 08:16:06 pm
You have \( \frac{d^2y}{dx^2} = 12x - 14\) so all you're trying to solve is \(12x - 14 > 0\), which becomes \(x > \frac76\).
Post by: dhaider on February 03, 2019, 11:26:56 am
Q) A certain type of bird has a probability of 1/12 of hatching a bird with white feathers. If a bird lays three eggs, find the probability of hatching:
i) exactly one bird with white feathers
iI) at least one bird with white feathers.
Any help or or solution will be appreciated guys. Thanks
Post by: darkz on February 03, 2019, 11:38:02 am
Hey everyone, I need help with a probability question from HSC course.
Q) A certain type of bird has a probability of 1/12 of hatching a bird with white feathers. If a bird lays three eggs, find the probability of hatching:
i) exactly one bird with white feathers
iI) at least one bird with white feathers.
Any help or or solution will be appreciated guys. Thanks
Try using binomial probability i.e.
Let X be the number of birds with white feathers
Then, for i) Pr(X=1)
Then, for ii) Pr(X≥1)
(Success; p=1/12)
(Sample size; n = 3)
Post by: RuiAce on February 03, 2019, 11:45:27 am
Try using binomial probability i.e.Please keep in mind that for the current syllabus they're not introduced to random variable notation. They simply refer to a "binomial probability" by the formula \( \binom{n}{k}p^k(1-p)^{n-k} \). That and it is only examined in Extension 1/2 in NSW.
Let X be the number of birds with white feathers
Then, for i) Pr(X=1)
Then, for ii) Pr(X≥1)
(Success; p=1/12)
(Sample size; n = 3)
Hey everyone, I need help with a probability question from HSC course.\[ \text{Note that there are three different ways this can happen.}\\ \text{Either the first bird has white feathers, or the second does, or the third does.} \]
Q) A certain type of bird has a probability of 1/12 of hatching a bird with white feathers. If a bird lays three eggs, find the probability of hatching:
i) exactly one bird with white feathers
iI) at least one bird with white feathers.
Any help or or solution will be appreciated guys. Thanks
Denoting W - white feathers and N - no white feathers here:
\[ \text{Thus we have}\\ \begin{align*} P(\text{Exactly 1 with white}) &= P(WNN) + P(NWN) + P(NNW)\\ &= \left( \frac1{12} \right)\left( \frac{11}{12}\right)^2 + \left( \frac1{12} \right)\left( \frac{11}{12}\right)^2 + \left( \frac1{12} \right)\left( \frac{11}{12}\right)^2\\&= \frac{121}{576} \end{align*} \]
\[ \text{For the other one, we can just consider the complement.}\\ \begin{align*}P(\text{At least 1 white}) &= 1 - P(\text{No white})\\ &= 1 - P(NNN)\\ &= 1 - \left( \frac{11}{12}\right)^3\\ &= \frac{397}{1728} \end{align*}\]
Post by: alexnero7 on February 03, 2019, 02:47:00 pm
x^2-\frac{x^3}{3}
to...
x^2\left(1-\frac{x}{3}\right)
If you don't understand the above... please let me know, and ill send a photo instead.. thanks
Post by: Opengangs on February 03, 2019, 03:11:11 pm
Can somebody please explain to me how we go from:Hey there!
x^2-\frac{x^3}{3}
to...
x^2\left(1-\frac{x}{3}\right)
If you don't understand the above... please let me know, and ill send a photo instead.. thanks
Notice that:
So, factorising \(x^2\) gives us:
Post by: alexnero7 on February 03, 2019, 03:19:21 pm
Hey there!
Notice that:
So, factorising \(x^2\) gives us:as required.
Post by: alexnero7 on February 03, 2019, 07:00:26 pm
Post by: dream chaser on February 03, 2019, 07:17:54 pm
Hey can someone please solve these? Thanks
Hi alexnero7,
I will try and answer your questions.
Q9. First of all, lets establish what maximum and minimum values are. They correspond to the y values. For instance, the graph y=x^2 +3 would have a minimum value of 3 and y=-x^ -3 would have a maximum value of -3.
Now to find these values. We have to differentiate the equation. Once you do that, you equal the differentiated equation to zero and solve for x. Once you solve for x, sub the x value into the original equation to obtain the maximum and minimum values. Note, make sure the x values fall under the implied domain.
Q10. For this question, they state their is no stationary points. Therefore, normally the maximum and minimum values will be at the endpoints. To make sure of this, sketch the graph.
Hope this helps.
Post by: meerae on February 03, 2019, 07:45:09 pm
Hey can someone please solve these? Thanks
Hey!
I thought I'd give insight on how I'd solve them, but dream chaser's way should get you to the correct answer.
Q9: I'd first sketch the curve, so you'd differentiate (keeping in mind that y= (x+5)^1/2 and then using chain rule) make y'=0 and all the curve sketching
So, now when you attempt to find the maximum/minimum values you can cross reference with your graph to ensure you haven't made a silly algebraic mistake.
To find the max/min values, you would test your value when you made y'=0 as well as the end points (-4 and 4) by putting them into the y equation to find the max/min.
Q10: As the question doesn't ask to sketch, I wouldn't waste my time (especially in an exam). I would differentiate the equation, make it equal 0 and then conclude there is no stat. point.
Like in the previous question, you would have to test the stat. points and the end points (in this case, -3 and 3), but you do not have any stat points which means you would just sub in the end points and determine which is max and min.
Hope this helped!
meerae :)
Post by: alexnero7 on February 03, 2019, 09:48:21 pm
Post by: RuiAce on February 03, 2019, 09:50:32 pm
Hi everyone.. can someone please help me understand what is the significance of the 45 degree angle in this question? thank you. :)\[ \text{We have }\frac{dy}{dx} = 4x^2+C\\ \text{but that extra piece of information says that}\\ \text{at the point }(-2,5)\text{, we have }\frac{dy}{dx}=\tan 45^\circ = 1. \]
This is because \( \frac{dy}{dx} \) is just the gradient of the tangent. And we also saw the formula \(m = \tan \theta\) in coordinate geometry, where \(\theta\) is the angle made between a line and the \(x\)-axis. The significance is that here, that \(45^\circ\) is the \(\theta\).
Post by: UStoleMyBike on February 04, 2019, 11:52:57 pm
In lotto, a machine holds 45 balls, each with a number between 1 and 45 on it. the machine draws out one ball at a time at random. Find the probability that the first ball drawn out will be
(a) less than 10 or an even
(b)between 1 and 15 inclusive
or divisible by 6
(c)greater than 30 or an odd
number.
Post by: jamonwindeyer on February 05, 2019, 12:02:21 am
Hi, these stupid probability questions are triggering me, help me find out what I'm doing wrong in this question:
Hey! All of these look at the probabilities of mutually non-exclusive events. The formula you need:
That is, the probability of one thing OR another thing happening is the sum of each individual probability, minus the probability that they BOTH happen at once. Let's take a)
Numbers Less than 10: 1,2,3,4,5,6,7,8,9
Evens: 2,4,6,8...,38,40,42,44
Numbers Less than 10 THAT ARE ALSO Evens: 2,4,6,8
So to calculate this probability:
Try and apply this same idea to the other two! :)
Post by: jamonwindeyer on February 05, 2019, 12:09:02 am
Hey everyone!
Can anyone help with this question including worked solutions?
Find the area bounded by the curve y=(x+3)[squared], the y axis and the lines y=9 and y=16 in the first quadrant.
Find the volume of the solid formed if this area is rotated about the y axis.
Many thanks!
Hello! So drawing a bit of a sketch might help you visualise this, but essentially we are looking at the area between the right arm of that parabola and the y-axis, between the two y-values given. We need to rearrange to make \(x\) the subject:
So the area would be:
To swap to the volume, you use \(x^2\) instead of just \(x\), and add a \(\pi\) out front:
Both of these are just definite integrals using the regular rules, let us know how you go with them! :)
Post by: alexnero7 on February 12, 2019, 06:02:01 pm
Post by: myopic_owl22 on February 12, 2019, 06:40:33 pm
Hey everyone, hope you all well. Can somebody please help me find the primitive function of this attached question? Thanks. :)
Hi there,
To integrate functions like these, I'd recommend moving the (1/2) bit to the left side of the integral, as it's a constant multiplier. Dealing with the rest...
1. Convert into index notation
2. Integrate as usual. Treat the bracket as a term - power up by one, divide by new power.
We divide through by 4 as well to essentially 'reverse' the effects of the function of a function rule if the answer was to be diffed. General formula:
Note that this will only work for functions whose leading power of x is 1. Quadratics, etc. do not follow this rule.
Thankfully, this is about as tricky as 2U integration will get. Hopefully this helps!
Oh and don't forget +c :)
Post by: myopic_owl22 on February 12, 2019, 07:09:16 pm
Can anyone explain the working out for this? Is there a method that you are supposed to use?
Question: Write this series in sigma notation.
1+1/2+1/4+...+1/512
Also in some examples I have seen like 3+6+12+...+3×2(power n) you are supposed to change the 'n' to a 'k'. Why is that?
Hi there,
Noticing that the terms in your series are all 1 divided by increasing powers of 2 (i.e. 2^0, 2^1, 2^2, up to 2^9), we'd write something like this:
The bottom is where n begins (0 in this case for our first term of 1) and the top number is when we finish. You can use trial and error without too much fuss to see what power of 2 that 512 is raised to. Alternatively, you could use logarithms, which is a later topic in the maths course. Here, log2512 = 9. We use n as our only variable as this is a finite series, which will stop when we finish adding 1/512 to our sum.
On the right, is what we're summing over and over again, with different values of n for each time we do it. Hopefully I'm making some sense.
We generally use k if the sum's final term is in general form (i.e. all the terms in this series follow this rule, where k is replaced by the term number). It will either denote some value that we either need to find, or don't need to worry about. It's essentially the math's conventional placeholder, which isn't a major focus of the course anyway, but to see an example of it in action:
So yeah, it's more focusing on the process of summation rather than the answer. Let me know if I'm not making sense :)
Post by: emilyyyyyyy on February 15, 2019, 07:43:55 pm
thanks!
Post by: meerae on February 15, 2019, 08:28:33 pm
hello can someone please help me how to find the enclosed area between the y-axis and the curve x=y(y-2)?? I'm just not sure what the y-values im meant to be integrating are.
thanks!
Hey!
You'd be integrating between y=0 and y=2
Hope this helps!
meerae :)
Post by: jamonwindeyer on February 15, 2019, 08:49:51 pm
Can someone help with this question? Particularly stuck on how to write the equation for the second bit.
The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the first 20 terms.
Hey! So your first equation is:
The next one is tougher, but it is really just the sum of \(T_3\) and \(T_7\):
Solving those simultaneously will give you your first term and common difference! And that can be used to find \(S_{20}\) - Is that enough to get you rolling? :)
Post by: jamonwindeyer on February 15, 2019, 09:00:38 pm
hello can someone please help me how to find the enclosed area between the y-axis and the curve x=y(y-2)?? I'm just not sure what the y-values im meant to be integrating are.
thanks!
Hey! As meerae says, the values are \(y=0\) and \(y=2\). This is because these are the two points where the curve cuts the y-axis (sideways parabola!), so this defines the start and stop point for the area. The graph is here:
(https://i.imgur.com/a17TDqw.png)
Also notice the area is to the left of the y-axis, meaning you'll need to compensate for the negative as well! Hope this helps :)
Post by: goodluck on February 15, 2019, 09:22:06 pm
|(x-1)/(x+1)| <1
Post by: jamonwindeyer on February 15, 2019, 09:44:30 pm
Ah yep thank you so much! :)
Though I think you wrote T4 instead of T3 by accident??
I did indeed, sorry!! Just tidied it up but the general idea is the same :)
Post by: RuiAce on February 15, 2019, 09:50:39 pm
Hey how would I solve this? (it's absolute values if the writing doesn't make it clear!)\[ \text{The solution is equivalent to that of}\\ |x-1| < |x+1| \]
|(x-1)/(x+1)| <1
(However, note that \(x \neq -1\) because otherwise in the original equation we have dividing by 0 issues.)
Recall
The definition of the absolute value states that
\[ |x| = \begin{cases} x & \text{if }x\geq 0\\ -x & \text{if }x < 0\end{cases} \]
Therefore we have
\begin{align*} |x-1| &= \begin{cases} x-1 & \text{if }x-1 \geq 0\\ -(x-1) & \text{if }x-1 < 0\end{cases} \\ &= \begin{cases} x-1 & \text{if }x \geq 1\\ -x+1 &\text{if }x < 1\end{cases} \end{align*}
And similarly
\[ |x+1| = \begin{cases} x+1 &\text{ if }x \geq -1\\ -x-1 &\text{if }x < -1\end{cases} \]
\[ |x| = \begin{cases} x & \text{if }x\geq 0\\ -x & \text{if }x < 0\end{cases} \]
Therefore we have
\begin{align*} |x-1| &= \begin{cases} x-1 & \text{if }x-1 \geq 0\\ -(x-1) & \text{if }x-1 < 0\end{cases} \\ &= \begin{cases} x-1 & \text{if }x \geq 1\\ -x+1 &\text{if }x < 1\end{cases} \end{align*}
And similarly
\[ |x+1| = \begin{cases} x+1 &\text{ if }x \geq -1\\ -x-1 &\text{if }x < -1\end{cases} \]
\[ \text{Case 1: } x < -1.\\ \text{For this case the inequality becomes}\\ \begin{align*} -x+1 &< -x-1\\ 1 &< -1\end{align*}\\ \text{This obviously has no solution}\\ \text{so no values of }x\text{ in this case belong in the solution.} \]
\[ \text{Case 2: }-1\leq x < 1.\\ \text{For this case the inequality becomes}\\ \begin{align*}-x+1 &< x+1\\ 0 &< 2x\\ x &> 0 \end{align*}\]
\[ \text{So overlapping }x > 0\text{ with }-1\leq x < 1\\ \text{we see that }\boxed{0 < x < 1}\text{ is a part of the solution.}\]
\[ \text{Case 3: }x\geq 1.\\ \text{For this case the inequality becomes}\\ \begin{align*}x-1 &< x+1\\ -1 &< 1 \end{align*}\]
\[ \text{This is obviously true for every value of }x.\\ \text{So all values of }x\text{ in this case, i.e. }\\\boxed{ x\geq 1}\text{, is a part of the solution.}\]
\[ \text{Combining all parts, our solution is thus}\\ \boxed{x > 0}.\]
Post by: goodluck on February 16, 2019, 06:11:59 pm
Post by: RuiAce on February 16, 2019, 06:41:31 pm
The groundwork
\[ \text{Essentially we are solving }1-2\sin x \geq 0\\ \text{which rearranges to }\sin x \leq \frac12. \]
\[ \text{First note that the solutions to }\sin x = \frac12 \text{ for }0\leq x \leq 2\pi\\ \text{are }x=\frac\pi6\text{ or } \frac{5\pi}6.\]
\[ \text{Following the periodicity of the trigonometric functions, the other solutions will be at}\\ x = 2k\pi+\frac\pi6\text{ or }x=2k\pi + \frac{5\pi}{6}\\ \text{where }k\text{ is an integer.}\]
_____________________________
\[\text{Now for the inequality, we can sketch the curve }y=\sin x\\ \text{along with the line }y=\frac12.\\ \text{The above computations track down each point of intersection for us.}\]
\[ \text{As we are interested in }\sin x \leq \frac12,\\ \text{we're interested in the }x\text{ values for which}\\ y=\sin x\text{ lies }\textbf{below}\text{ the line }y=\frac12.\]
\[ \text{We see that we'll have }-\frac{7\pi}{6} \leq x \leq \frac\pi6, \, \frac{5\pi}{6}\leq x \leq \frac{13\pi}6\\ \text{and so on.}\\ \text{All of these intervals will form our answer.} \]
\[ \text{First note that the solutions to }\sin x = \frac12 \text{ for }0\leq x \leq 2\pi\\ \text{are }x=\frac\pi6\text{ or } \frac{5\pi}6.\]
\[ \text{Following the periodicity of the trigonometric functions, the other solutions will be at}\\ x = 2k\pi+\frac\pi6\text{ or }x=2k\pi + \frac{5\pi}{6}\\ \text{where }k\text{ is an integer.}\]
_____________________________
\[\text{Now for the inequality, we can sketch the curve }y=\sin x\\ \text{along with the line }y=\frac12.\\ \text{The above computations track down each point of intersection for us.}\]
\[ \text{As we are interested in }\sin x \leq \frac12,\\ \text{we're interested in the }x\text{ values for which}\\ y=\sin x\text{ lies }\textbf{below}\text{ the line }y=\frac12.\]
\[ \text{We see that we'll have }-\frac{7\pi}{6} \leq x \leq \frac\pi6, \, \frac{5\pi}{6}\leq x \leq \frac{13\pi}6\\ \text{and so on.}\\ \text{All of these intervals will form our answer.} \]
Note that I purposely avoided "general solutions" as this was posted in the 2U section.
Edit: fixed typos
Post by: goodluck on February 16, 2019, 07:01:58 pm
but how does the possible answer work if k is like 10 for example since it will be bigger than pi/6? would we just do general solution there?
So expressing it in a nice way is reasonably hard, because we have to express an infinite number of intervals. But we can always work around this by just setting \(k\) to be an arbitrary integer. One possible answer is \( -\frac{5\pi}{6}+2k\pi \leq x \leq \frac\pi6 \), where \(k\) is an integer. Having this arbitrary integer placeholder lets us get around this issue, because \(k\) can then be substituted for every integer possible.The groundwork\[ \text{Essentially we are solving }1-2\sin x \geq 0\\ \text{which rearranges to }\sin x \leq \frac12. \]
\[ \text{First note that the solutions to }\sin x = \frac12 \text{ for }0\leq x \leq 2\pi\\ \text{are }x=\frac\pi6\text{ or } \frac{5\pi}6.\]
\[ \text{Following the periodicity of the trigonometric functions, the other solutions will be at}\\ x = 2k\pi+\frac\pi6\text{ or }x=2k\pi + \frac{5\pi}{6}\\ \text{where }k\text{ is an integer.}\]
_____________________________
\[\text{Now for the inequality, we can sketch the curve }y=\sin x\\ \text{along with the line }y=\frac12.\\ \text{The above computations track down each point of intersection for us.}\]
\[ \text{As we are interested in }\sin x \leq \frac12,\\ \text{we're interested in the }x\text{ values for which}\\ y=\sin x\text{ lies }\textbf{below}\text{ the line }y=\frac12.\]
\[ \text{We see that we'll have }-\frac{7\pi}{6} \leq x \leq \frac\pi6, \, \frac{5\pi}{6}\leq x \leq \frac{13\pi}6\\ \text{and so on.}\\ \text{All of these intervals will form our answer.} \]
Note that I purposely avoided "general solutions" as this was posted in the 2U section.
Post by: AlphaZero on February 16, 2019, 07:11:06 pm
Oh you can express it via general solutions? (maybe should have posted it in the 3U section then!)
but how does the possible answer work if k is like 10 for example since it will be bigger than pi/6? would we just do general solution there?
I think he meant to write \[\frac{-7\pi}{6}+2\pi k\leq x\leq \frac{\pi}{6}+2\pi k,\quad \text{where }k\ \text{is an integer.}\]
Post by: Youssefh_ on February 18, 2019, 07:48:57 pm
Post by: meerae on February 18, 2019, 08:05:11 pm
can someone please explain these two questions for me because i am getting confused at what to do, thank you
Hey!
For the first question, the formula to find arc length is
Luckily, they gave you the circumference, so you would use the circumference formula and solve for r
For the second question, you are required to find the angle. The question begins with telling you the area of the circle, so it would make sense to note this formula down as
Hope this helps!
meerae :)
Post by: Thankunext on February 18, 2019, 09:34:10 pm
1. Which term of the series 8 - 4 + 2 - ... is 1/128?
2. Which term of 54 + 18 + 6 + ... is 2/243?
3. Find the value of n if the nth term of the series -2 + 3/2 - 9/8 + ... is -81/128.
Post by: fun_jirachi on February 18, 2019, 09:49:03 pm
You can also use other log laws here to help you out, but in this case by knowing powers of two you can technically do it by inspection without logs and calculators.
A similar process follows for the second and third question, try them out!
Post by: AlphaZero on February 18, 2019, 09:50:29 pm
Can someone please help with the working out of these? I tried using log but I think I'm doing something wrong :'( :'(
1. Which term of the series 8 - 4 + 2 - ... is 1/128?
2. Which term of 54 + 18 + 6 + ... is 2/243?
3. Find the value of n if the nth term of the series -2 + 3/2 - 9/8 + ... is -81/128.
Indeed, using logarithms may bring a little trouble in Q1 and Q3 since the common ratio of the sequences are negative. But, we can avoid these if we just make some clever observations.
Question 1\begin{align*}\frac{1}{128}&=8\times \left(\frac{-1}{2}\right)^{n-1}\\
\implies \left(\frac{-1}{2}\right)^{n-1}&=\frac{1}{1024}\\
\implies \frac{(-1)^{n-1}}{2^{n-1}}&=\frac{1}{2^{10}}\tag{1}\end{align*} Now, let's look back at the original sequence. Clearly, the \(n\)th element of the sequence is positive only if \(n\) is odd. Notice that, if \(n\) is odd, then \(n-1\) is even, and so equation \((1)\) becomes \[\frac{1}{2^{n-1}}=\frac{1}{2^{10}}.\]Thus, \(n=11\).
I'll let you try Q2 and Q3 on your own. If you're still having trouble, don't hesitate to ask again :)
Post by: emilyyyyyyy on February 20, 2019, 12:35:20 pm
Thanks!
Post by: RuiAce on February 20, 2019, 01:41:06 pm
Hi, with this question, i've drawn the graph and found the point of intersection. I've integrated what I think I have to, but I keep getting the wrong answer! How do I solve: Find the area enclosed between the curve y = x^3, the x-axis and the line y = -3x + 4.Upon re-inspecting the question, because the region is bound also by the \(x\)-axis, it is the small triangle-like shape below both curves. This is a compound region, whose area is described by \( \int_0^1 x^3\,dx + \int_1^{4/3} (-3x+4)\,dx \).
Thanks!
Post by: emilyyyyyyy on February 20, 2019, 02:08:56 pm
I get that the intercept for the line y=-3x+4 is 4/3, but why do I use that instead of 1 when integrating the line, seeing as 1 is where the line and curve intersect?
Post by: AlphaZero on February 20, 2019, 02:36:24 pm
thank you!
I get that the intercept for the line y=-3x+4 is 4/3, but why do I use that instead of 1 when integrating the line, seeing as 1 is where the line and curve intersect?
Not too sure what you mean by using 4/3 instead of 1. We require both values. This image may help though.
(https://i.imgur.com/Dl7kKrc.png)
We are trying to find the area of the red and blue regions. \[\text{The area of the red region is given by }\int_0^1 x^3\,dx.\] \[\text{The area of the blue region is given by }\int_1^{4/3} (-3x+4)\,dx.\]To find the total area, just add the two together.
Post by: Kalliope.m on February 20, 2019, 04:13:37 pm
The first term of a geometric series is 3 and the common ratio is 4. Find the first term of the series that exceeds 300 000.
I got the inequality to find n, but I think I might be solving for it wrong. Could someone please help out? Thank you :)
Post by: AlphaZero on February 20, 2019, 04:49:56 pm
Hi, I’m struggling a bit with this question:
The first term of a geometric series is 3 and the common ratio is 4. Find the first term of the series that exceeds 300 000.
I got the inequality to find n, but I think I might be solving for it wrong. Could someone please help out? Thank you :)
So, this question asks us essentially to find the minimal value of \(n\) for which \(3\times 4^{n-1}\geq300\,000\). There are a couple of ways to go about this. I've presented two methods below.
Method 1: (probably the better method)
First note that the sequence given by \(T_n=3\times 4^{n-1}\) is clearly strictly increasing. So, one way we could go about this problem is by solving \(3\times 4^{n-1}=300\,000\) and rounding our answer up to the nearest integer: \begin{align*}4^{n-1}&=300\,000\\
n-1&=\log_4(300\,000)\\
n&=1+\log_4(300\,000)\approx 9.30\ \ (\text{2DP})\end{align*} Thus, the first element to exceed \(300\,000\) is the \(10\)th.
Method 2: (trial and error)
You would find that \begin{align*}T_8&=49\,152<300\,000\\
T_9&=196\,608<300\,000\\
T_{10}&=786\,432>300\,000.\end{align*}Thus, our answer is the \(10\)th element.
Post by: AlphaZero on February 20, 2019, 10:42:25 pm
Can someone please help with these series questions? Sorry for asking a lot but I'm really stuck :'(
1. Lucia currently earns $25000. Her wage increases by 5% each year. Find her total earnings (before tax) in 6 years.
2. Write down an expression for the series 2 - 10 + 50 - ... + 2(-5[power k - 1]) in sigma notation and as a sum of n terms.
3. Find the sum of the first 10 terms of the series 3 + 7 + 13 + ... + [2(power n) + (2n - 1)] + ...
There is nothing wrong with asking a lot of questions :)
Question 1
Edit: I missed the word "total" in the question. Apologies.
\[S_6=\frac{\$25\,000(1.05^6-1)}{1.05-1}=\$170\,048.82\]
Question 2
This is a geometric series whose first term is \(2\) and common ratio is \(-5\).
So, the \(k\)th term of the series is given by \(T_k=2\times(-5)^{n-1}\).
Therefore, the sum of the first \(n\) terms, \(S_n=T_1+T_2+\dots+T_n\) in sigma notation is just given by \[\sum_{k=1}^nT_k=\sum_{k=1}^n 2\!\times\!(-5)^{k-1}.\]
Question 3
In this question, \begin{align*}S=\sum_{k=1}^{10}\Big[2^k+2k-1\Big]&=\sum_{k=1}^{10}2^k+\sum_{k=1}^{10}(2k-1)\\
&=\underbrace{\sum_{k=1}^{10}2\!\times\! 2^{k-1}}_{\text{geometric series}}+\underbrace{\sum_{k=1}^{10}\big[1+2(k-1)\big]}_{\text{arithmetic series}}\end{align*}
The first series is just a geometric series with first term \(a=2\) and common ratio \(r=2\).
The second series is just an arithmetic series with first term \(a=1\) and common difference \(d=2\). Therefore, \begin{align*}S&=\frac{2(2^{10}-1)}{2-1}+\frac{10}{2}\big[2\!\times\! 1+2(10-1)\big]\\
&=2146\end{align*}
Post by: RuiAce on February 20, 2019, 11:40:03 pm
Thank you so much for taking the time to answer my questions. :)\[ \text{His solution is slightly off for Q1.}\\ \text{We're interested in the }\textbf{total}\text{ earnings accumulated.}\\ \text{Not just the sixth period itself.} \]
In regards to question 1, I kept getting $170047.82 (2dp) as my answer??
In regards to question 2, how do you know that k=1 and how come you need to change from k to n? I don't know when you have to do that and which part you need to change.
Apologies, I just need some clarification.
\[ \text{Because the earnings in the }n\text{-year can be modelled by}\\ \text{a geometric sequence, with }a=25000\text{ and }r=1.05,\\ \text{what we're interested in is }S_6 = 25000\left( \frac{1.05^6-1}{0.05} \right) \]
Which evaluates to the given answer.
With Q2, that one has more than one correct answer in regards to what \(k\) we can start from. We can start from whatever value of \(k\) we like, so long as it works. Starting with \(k=1\), we have \( \sum_{k=1}^n 2(5)^{k-1} \) like he wrote. But we could, for example, start from \(k=0\) to obtain \( \sum_{k=0}^{n-1}2(-5)^k \). Or we could start from say \(k=2\), and obtain \( \sum_{k=2}^{n+1} 2(5)^{k-2}\).
Convention is typically to start with \(k=1\) though as far as the HSC course is concerned.
\[ \sum_{j=1}^k 2(5)^{j-1} \]
Post by: RuiAce on February 21, 2019, 07:43:17 pm
Can someone help with this question? I do not know how to write the equation or what formula to use. It is a Fibonacci problem which is that 1, 1, 2, 3, 5, 8, ... thing.Where is this question from? Through recurrence relations I obtain an answer of 382 (which seemed to work when I backtracked as well), but apart from annuity like applications recurrences are not a part of the current HSC course.
A man entered an orchard through 7 guarded gates and gathered a certain number of apples. As he left the orchard he gave the guard at the first gate half the apples he had and 1 more. He repeated this process for each of the remaining 6 guards and eventually left the orchard with 1 apple. How many apples did he gather? (He did not give away any half apples.)
Thank you to whoever will answer this question.
Post by: RuiAce on February 21, 2019, 08:07:52 pm
What really? The Fibonacci thing is not in the HSC course? It is in the Year 12 2u Grove Textbook in Chapter 7 - Series and my teacher showed us the pattern but not a lot of examples.Ok I tracked down the question in the textbook and it looks like it's a puzzle. (And it got tagged with a hilarious "This is a hard one!" in the MX1 version at least.)
May I still have the working out because my teacher assigned that bit for homework. I will clarify with my teacher whether we will be tested on that.
I would be extremely concerned if this appeared in any HSC exam. Nevertheless, here's the full recursive approach.
\[ \text{Let }T_n\text{ be the amount of apples the man has left}\\ \textbf{after }\text{giving the }n\text{-th guard his apples.} \]
Note that the question can be done using 'before' instead of 'after'. You may like to experiment using 'before' instead.
\[ \text{We're essentially given that }T_7 = 1.\\ \text{Our aim is to backtrack to find }T_0\\ \text{which is the amount before even the 1st guard got his apples.} \]
\[ \text{The recurrence of interest is}\\ \boxed{T_n = \frac12 T_{n-1} - 1}.\\ \text{This results from the take-half, then give one more apple scenario}\\ \text{the question presented.} \]
\[ \text{The recurrence relation can be rearranged to}\\ \boxed{T_{n-1} = 2(T_n + 1)}.\\ \text{Once we're here, we can start backtracking.}\\ \begin{align*} T_6 &= 2(1 + 1) = 4\\ T_5 &= 2(4+1) = 10\\ T_4 &= 2(10+1) = 22\\ T_3 &= 2(22+1) = 46\\ T_2 &= 2(46+1) = 94\\ T_1 &= 2(94+1) = 190\\ T_0 &= 2(190+1) = 382\end{align*} \]
\[ \text{So he originally had 382 apples.} \]
Note that the problem was put in the geometric series section, so it had to somehow be related to geometric series. It turns out that one can actually prove that \(T_n = 384(2^{-n}) - 2\). This looks almost like a geometric series, except there is an extra \(-2\) hanging on the end interfering with the pattern. The formal derivation more or less uses first year uni techniques and is hence avoided, but there may be an intuition behind this formula that I haven't had the time to investigate yet.
Post by: RuiAce on February 21, 2019, 08:51:00 pm
Wow thank you so much for taking the time to type the working out for me. :D"_______ of interest" is just a fancy way of saying "the __________ that we are interested in". So in English, it's basically saying it's the correct recurrence relation for this particular question.
I understand most of what is going on apart from how to get the "recurrence of interest" bit: Tn = 1/2 Tn -1 -1.
But as for why it works, it's simply just using what they gave us. We firstly give them half the number of apples we currently have, so therefore we still have the other half left - this is the \( \frac12 T_{n-1}\) bit. But then we need to give them one more apple. This leaves us with one more apple less that we have, and hence we have the trailing \(-1\) term in \( \frac12 T_{n-1} - 1 \).
Post by: Cherre Ho on February 21, 2019, 09:20:34 pm
"_______ of interest" is just a fancy way of saying "the __________ that we are interested in". So in English, it's basically saying it's the correct recurrence relation for this particular question.Ahhhh I see. Yay I understand!! Thank you very much. You explain it very well and in great detail (thumbs up) :D
But as for why it works, it's simply just using what they gave us. We firstly give them half the number of apples we currently have, so therefore we still have the other half left - this is the \( \frac12 T_{n-1}\) bit. But then we need to give them one more apple. This leaves us with one more apple less that we have, and hence we have the trailing \(-1\) term in \( \frac12 T_{n-1} - 1 \).
Post by: annabeljxde on February 22, 2019, 10:29:12 pm
A sector of a circle with a radius 5cm and an angle of π/3 subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find it’s exact surface area and volume.
Is the question asking us to use the SA and V formula for a cone or is the resulting cone empty (where the SA includes both the interior and exterior of the cone)?
Can anyone please help me? Thank you!
Post by: RuiAce on February 22, 2019, 10:34:39 pm
I’ve been stuck on this question for days now! I just can’t wrap my head around what the question is asking...That is a fair question, since they don't seem to tell you whether we should treat it as a 'sealed-off' cone or an empty cone. So why not try out both methods and see which one works?
A sector of a circle with a radius 5cm and an angle of π/3 subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find it’s exact surface area and volume.
Is the question asking us to use the SA and V formula for a cone or is the resulting cone empty (where the SA includes both the interior and exterior of the cone)?
Can anyone please help me? Thank you!
(However, it is a bit tricky. Recall that the net of a cone looks like what's shown on this website. Note that the sector is wrapped around to form the apex of the cone. Therefore, that radius of 5 becomes the slant height of the cone. Furthermore, the arc length \( \ell = 5\times \frac\pi3\) becomes the circumference of the attached circle.)
Although having said that, this kind of ambiguity would not appear in the HSC. Where did this question originally come from?
Post by: fun_jirachi on February 22, 2019, 11:13:48 pm
EDIT: to be fair though, if you have a closed cone it's just one extra step using the radius 5/6 to find the area of the base. Regardless of this, the volume remains the same.
Post by: annabeljxde on February 24, 2019, 11:39:29 am
That is a fair question, since they don't seem to tell you whether we should treat it as a 'sealed-off' cone or an empty cone. So why not try out both methods and see which one works?
(However, it is a bit tricky. Recall that the net of a cone looks like what's shown on this website. Note that the sector is wrapped around to form the apex of the cone. Therefore, that radius of 5 becomes the slant height of the cone. Furthermore, the arc length \( \ell = 5\times \frac\pi3\) becomes the circumference of the attached circle.)
Although having said that, this kind of ambiguity would not appear in the HSC. Where did this question originally come from?
It was a question from my textbook, HSC Maths in Focus as one of the harder questions of the 'arc length' exercise.
The answers are:
SA= 175π/36 cm²
V= 125√35/648 cm^3
Post by: annabeljxde on February 24, 2019, 11:42:29 am
Already been answered, but here's my two cents. Pretty sure that given you slice it out of a circle, the bottom is empty since you can't have a circle appear out of nowhere to form the base (that's my interpretation at least). The area of the sector should be the surface area of the cone, so subbing into half x r^2 x theta, you get 25/6 pi. For the volume, you need the height of the cone, which by Pythagoras' theorem (subbing in slant height and radius calculated from arc length) is 5root35/6, the radius of the cone is just 5/6, so sub that into the formula for volume of a cone and you get 125/648 x sqrt 35 x pi. Hope this helps :)
EDIT: to be fair though, if you have a closed cone it's just one extra step using the radius 5/6 to find the area of the base. Regardless of this, the volume remains the same.
You got the Volume answer correct (thank you so much for the working :) ) but I also got the same answer for the surface area but apparently it says otherwise in the answers in the textbook... why is that??
Post by: annabeljxde on February 24, 2019, 02:15:31 pm
So apparently, it is assumed that the cone is closed off and you have to use the formula for surface area (SA= pi x r(r+l))
I figured out the radius to be 5/6 from the circumference of the cone (which is the arc length of the cut-off sector) and I substituted this value to r in the SA formula.
The radius of the sector becomes the slant height of the cone (which is 5)
My substitution looked like:
SA = pi x 5/6(5/6 + 5)
My answer was correct (according to the textbook) = 175pi/36 cm^2
I really hope these questions don't pop up in future trial or HSC exams. It took me a whole week (and your guys' help!) to finally figure it out!!
Anyways, thanks for the help you guys :) ;D ;D
Post by: RuiAce on February 24, 2019, 02:24:38 pm
I figured it out RuiAce and fun_jirachi !Typical maths in focus and its lack of clarity.
So apparently, it is assumed that the cone is closed off and you have to use the formula for surface area (SA= pi x r(r+l))
I figured out the radius to be 5/6 from the circumference of the cone (which is the arc length of the cut-off sector) and I substituted this value to r in the SA formula.
The radius of the sector becomes the slant height of the cone (which is 5)
My substitution looked like:
SA = pi x 5/6(5/6 + 5)
My answer was correct (according to the textbook) = 175pi/36 cm^2
I really hope these questions don't pop up in future trial or HSC exams. It took me a whole week (and your guys' help!) to finally figure it out!!
Anyways, thanks for the help you guys :) ;D ;D
Glad you solved it :) but yeah, in the HSC exam this kind of ambiguity is guaranteed to not appear. (Else the examiners get fired.)
Post by: alexnero7 on February 27, 2019, 08:09:43 pm
Post by: RuiAce on February 27, 2019, 08:12:13 pm
Can somebody please help me with these. I can't seem to get the correct answer. Thanks.Both of them rely on the formula \( \int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} +C \). Please provide working out showing usage of those formulas for further assistance. Or alternatively you may ask about how that formula provided works if necessary.
Post by: alexnero7 on February 27, 2019, 08:19:38 pm
Both of them rely on the formula \( \int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} +C \). Please provide working out showing usage of those formulas for further assistance. Or alternatively you may ask about how that formula provided works if necessary.
Thank you Rui!! Really helped, got both answers right this time. :)
Post by: alexnero7 on March 04, 2019, 06:48:57 pm
Post by: jamonwindeyer on March 04, 2019, 06:54:23 pm
What you really need to do is sketch these curves and see the area they are talking about. These are all standard curves, parabolas and semicircles, so you should be able to do a rough sketch of them and find the relevant intercepts (if you can’t, it is worth revising the Function sketches part of your Prelim textbook!)
Post by: jamonwindeyer on March 04, 2019, 10:06:43 pm
Help please!
What is the formula for solving this? Please provide working out!
Bill thinks he can afford a mortgage payment of $800 each month. How much can he borrow, to the nearest $100, over 25 years at 11.5% p.a.?
Hey there! So this is actually a tricky question because it is our standard question type, but the unknown is actually the principal! So the setup looks much the same as normal, except \(P\) is now an unknown. I'm considering \(1+r=1.0098...\approx1.01\) to make it easier to type, but you'll probably want to be more precise:
We know \(A_{300}=0\) if the loan is repaid in 25 years, so now just substitute \(n=12\times25=300\) and then solve for \(P\)! :)
Post by: eeshab7 on March 06, 2019, 05:35:47 pm
Post by: david.wang28 on March 06, 2019, 05:42:41 pm
Hi could I get some help with this question. I thought I had it but the answers say something completely different.You need to integrate this to find the total volume of the container at a point of time. After integrating, let t = 0, so the C becomes 15000. Then let t = 10, and you should be there. Hope this helps :)
Post by: eeshab7 on March 06, 2019, 05:52:08 pm
Post by: david.wang28 on March 06, 2019, 05:53:42 pm
I'm not doing 2u right now but I stumbled across this random challenging 2u question, and I have no idea how to do it. Can anyone please help me out? Thanks :)
Post by: david.wang28 on March 06, 2019, 05:55:38 pm
okay I got it, thanks for the help :) :)No problem :)
Post by: RuiAce on March 06, 2019, 06:03:53 pm
Hi there,That question certainly has a typo. It should be \( S_n = \frac12 (a+\ell) \left(1+ \frac{\ell-a}{d} \right) \). Note the 1.
I'm not doing 2u right now but I stumbled across this random challenging 2u question, and I have no idea how to do it. Can anyone please help me out? Thanks :)
Other than that though, it can be fudged. Recall that \(S_n = \frac{n}{2}(a+\ell) \) where \(\ell = T_n = a+ (n-1)d\). Literally rearranging \(\ell = a+(n-1)d\) gives \(n = 1+\frac{\ell-a}{d}\), which you can sub into there. The second part is then easy - just sub \(a=550\), \(\ell=990\) and \(d=11\) in, noting that the last multiple of 11 between 550 and 1000 is actually 990.
Post by: david.wang28 on March 06, 2019, 06:06:06 pm
That question certainly has a typo. It should be \( S_n = \frac12 (a+\ell) \left(1+ \frac{\ell-a}{d} \right) \). Note the 1.Thanks for the help! :)
Other than that though, it can be fudged. Recall that \(S_n = \frac{n}{2}(a+\ell) \) where \(\ell = T_n = a+ (n-1)d\). Literally rearranging \(\ell = a+(n-1)d\) gives \(n = 1+\frac{\ell-a}{d}\), which you can sub into there. The second part is then easy - just sub \(a=550\), \(\ell=990\) and \(d=11\) in, noting that the last multiple of 11 between 550 and 1000 is actually 990.
Post by: shaynec19 on March 08, 2019, 10:39:12 am
Post by: jamonwindeyer on March 08, 2019, 11:34:34 am
Could someone please help me with this.
Hey! The trick with the sum of the 6th-10th terms is to take \(S_{10}\) and subtract \(S_5\)! :) the equations you'll form are:
Simplify and solve these simultaneously!! Let us know if you need help with that ;D
Post by: shaynec19 on March 08, 2019, 05:14:14 pm
They are the equations I already derived, but I cannot for the life of me solve them to get an integer of a or d.
I have tried a couple of methods..
Cheers
Post by: RuiAce on March 08, 2019, 05:23:26 pm
Thanks Jamon!\begin{align*}a+19d &= 131\tag{1}\\ \frac{10}{2}(2a+9d) - \frac{5}{2}(2a+4d)&=235\\ 5(2a+9d) - 5(a+2d)&=235\\ a+7d&=47\tag{2}\end{align*}
They are the equations I already derived, but I cannot for the life of me solve them to get an integer of a or d.
I have tried a couple of methods..
Cheers
\[ (2)-(1)\text{ gives }12d = 84 \implies \boxed{d=7}. \]
Post by: shaynec19 on March 08, 2019, 05:43:24 pm
\begin{align*}a+19d &= 131\tag{1}\\ \frac{10}{2}(2a+9d) - \frac{5}{2}(2a+4d)&=235\\ 5(2a+9d) - 5(a+2d)&=235\\ a+7d&=47\tag{2}\end{align*}
\[ (2)-(1)\text{ gives }12d = 84 \implies \boxed{d=7}. \]
Thanks a mill Rui!
Post by: Youssefh_ on March 08, 2019, 10:22:06 pm
Post by: david.wang28 on March 08, 2019, 11:06:35 pm
Can someone please help me understand how to differentiate these sorts of questionsJust remember to take the power down, multiply the coefficient, and then minus the power by one. Also, can you take a clearer screenshot please? Thanks :)
Post by: fun_jirachi on March 08, 2019, 11:23:06 pm
Just remember to take the power down, multiply the coefficient, and then minus the power by one. Also, can you take a clearer screenshot please? Thanks :)
Note that you can't actually do this. Think a) is asking for
Basically you want to be rearranging in into the form e^(ln (whatever you have)) which is true by log laws (if you don't understand, ask again :) ). Then you follow your regular thing for differentiating the natural exponential (basically the chain rule, and it's on your reference sheet), and simplify where necessary.
Hope this helps :)
Post by: spnmox on March 11, 2019, 04:49:04 pm
Post by: fun_jirachi on March 11, 2019, 06:56:14 pm
From this we can see we have a stationary point at (1, 2/e), and that the function is decreasing for all real x.
Note also that the domain is all real x, and that since x^2+1>=1 for all real x and e^x>0 for all real x, then f(x) > 0 for all real x. Also note that this will result in a horizontal asymptote at y=0. Do your usual thing with limits as x approaches -infinity and nature around the stationary point to finish up, then sketch. You should get something relatively similar to e^-x for the most part.
Also try and show some working next time, so instead of me telling you basically everything, I can properly help out with where you're going wrong in particular!
Hope this helps :)
Post by: spnmox on March 11, 2019, 10:27:43 pm
From this we can see we have a stationary point at (1, 2/e), and that the function is decreasing for all real x.
Note also that the domain is all real x, and that since x^2+1>=1 for all real x and e^x>0 for all real x, then f(x) > 0 for all real x. Also note that this will result in a horizontal asymptote at y=0. Do your usual thing with limits as x approaches -infinity and nature around the stationary point to finish up, then sketch. You should get something relatively similar to e^-x for the most part.
Also try and show some working next time, so instead of me telling you basically everything, I can properly help out with where you're going wrong in particular!
Hope this helps :)
Hi! Sorry, I'll be more specific.
I found the stat pt and tried to find slope around it, except I got that the gradient is the same ? so is it an inflexion point?
as for limits, I got as x--> infinity, y--> - infinity and as x--> - infinity, y--> infinity. is there a way to figure these out without subbing in very large and very small numbers?
Post by: spnmox on March 11, 2019, 10:34:06 pm
so i differentiated and got (1/e, -1/e) as a min TP, and no inflexion points; x--> infinity and y--> infinity, and x can't approach negative infinity because x can't be equal to 0 therefore there's an asymptote at x=0, is that right? I have looked at the answers and i just don't get the open circle thing at (0,0). There's an asymptote at x=0 and also an x-intercept at 0, why
Post by: fun_jirachi on March 11, 2019, 10:57:56 pm
Hi! Sorry, I'll be more specific.
I found the stat pt and tried to find slope around it, except I got that the gradient is the same ? so is it an inflexion point?
as for limits, I got as x--> infinity, y--> - infinity and as x--> - infinity, y--> infinity. is there a way to figure these out without subbing in very large and very small numbers?
Out of curiosity, which points did you test? Because it's highly unlikely they were the same. But yes, it is an inflexion point as you can find by finding the second derivative (which I haven't bothered to add in, seems unnecessary.) Your limits are a bit off, as x tends to infinity, y should tend to zero (remember to sub into the original function, not the derivative). Subbing in big numbers is the easiest way, and I guess that's all you need to use in 2U. It's the easiest and quickest way to do so. :)
Sketch y=xlnx.
so i differentiated and got (1/e, -1/e) as a min TP, and no inflexion points; x--> infinity and y--> infinity, and x can't approach negative infinity because x can't be equal to 0 therefore there's an asymptote at x=0, is that right? I have looked at the answers and i just don't get the open circle thing at (0,0). There's an asymptote at x=0 and also an x-intercept at 0, why
There isn't an asymptote at x=0; there is an open endpoint. Remember that lnx is defined only for x>0. Because of this, there is an open circle at (0,0) because lnx isn't defined there. You assess the limiting behaviour of xlnx as x->0 and it because it approaches zero, you just let the 'endpoint' be (0,0). Also, xlnx only has negative values for 0<x<1 because over this domain y=x is positive, but y=lnx is negative --> you will never approach negative infinity. For all other x, lnx is positive as is x over the same domain, and thus it approaches infinity. Important to note there is not technically an asymptote or a x-intercept at x=0. (explained above due to graph being undefined there - an asymptote is a line of discontinuity: think y=1/x). Also note that having an x-intercept and a asymptote at the same point is impossible, they are two mutually exclusive things for any function. In fact, you can't have any point with an x-value corresponding to a vertical asymptote.
Hope this helps :)
Post by: david.wang28 on March 12, 2019, 09:01:40 am
I am stuck on a challenging 2 unit question in the attachment below (I have no idea how to do it). Can anyone please help me out? Thanks :)
Post by: jamonwindeyer on March 12, 2019, 07:55:08 pm
Hello,
I am stuck on a challenging 2 unit question in the attachment below (I have no idea how to do it). Can anyone please help me out? Thanks :)
This looks impossible because there is an \(a\) in the answer but not in the given terms!! ;D
Post by: julz_roha on March 13, 2019, 02:00:52 pm
How do we use the first derivative to check for points of inflexion?
Please Reply.
- Juliana.
Post by: Bri MT on March 13, 2019, 03:16:47 pm
Edit: I misinterpreted something due to differences across state's curriculms. Just go read Rui's response.
my original response
Hey Juliana,
To find a point of inflection for a function f(x):
1. Solve f'(x) = 0 for x
This gives you the stationary points (point of inflection, local maximums, and/or local minimums)
2. Now we need to figure out what type of stationary point it is. Around a local minimum, the graph goes down then up. So we have: f'(x)<0 on the left, f'(x) = 0 at the centre, f'(x)>0 on the right.
Around a point of inflection we have f'(x)>0 on both sides of f'(x) = 0 or f'(x)<0 on both sides of f'(x)
I'll leave it to you to see what the "rule" is for a local max but if you don't get it feel free to ask :)
The above info means that we can test the nature of a stationary point by finding f'(x) a little to the left & a little to the right of the x value you found in part 1.
Let me know if this is unclear :)
(Note: I did VCE not HSC so I'm unsure if this method is part of your curriculm)
To find a point of inflection for a function f(x):
1. Solve f'(x) = 0 for x
This gives you the stationary points (point of inflection, local maximums, and/or local minimums)
2. Now we need to figure out what type of stationary point it is. Around a local minimum, the graph goes down then up. So we have: f'(x)<0 on the left, f'(x) = 0 at the centre, f'(x)>0 on the right.
Around a point of inflection we have f'(x)>0 on both sides of f'(x) = 0 or f'(x)<0 on both sides of f'(x)
I'll leave it to you to see what the "rule" is for a local max but if you don't get it feel free to ask :)
The above info means that we can test the nature of a stationary point by finding f'(x) a little to the left & a little to the right of the x value you found in part 1.
Let me know if this is unclear :)
(Note: I did VCE not HSC so I'm unsure if this method is part of your curriculm)
Post by: RuiAce on March 13, 2019, 04:08:20 pm
Hello Atar Notes.
How do we use the first derivative to check for points of inflexion?
Please Reply.
- Juliana.
Hey Juliana,Hold. I would like to interfere here.
To find a point of inflection for a function f(x):
1. Solve f'(x) = 0 for x
This gives you the stationary points (point of inflection, local maximums, and/or local minimums)
2. Now we need to figure out what type of stationary point it is. Around a local minimum, the graph goes down then up. So we have: f'(x)<0 on the left, f'(x) = 0 at the centre, f'(x)>0 on the right.
Around a point of inflection we have f'(x)>0 on both sides of f'(x) = 0 or f'(x)<0 on both sides of f'(x)
I'll leave it to you to see what the "rule" is for a local max but if you don't get it feel free to ask :)
The above info means that we can test the nature of a stationary point by finding f'(x) a little to the left & a little to the right of the x value you found in part 1.
Let me know if this is unclear :)
(Note: I did VCE not HSC so I'm unsure if this method is part of your curriculm)
A point of inflexion is not at all related to the first derivative. The question provided does not flow logically because it contradicts basic principles.
Points of inflexion do not refer to points of horizontal tangents in general (although occasionally they can, see below). They refer to points where the concavity of the curve changes. By itself, the first derivative is useless for this. We require the second derivative to deduce possible points of inflexion.
In general, setting the second derivative equal to 0 gathers all such points for ya. THEN, we apply a change in concavity test (sub a point close to both sides of the possible inflexion point) to determine if it actually is a point of inflexion.
The first derivative only comes into play for the horizontal point of inflexion. This is a special case where both the first and second derivatives are zero. But again, the second derivative is necessary.
Post by: julz_roha on March 13, 2019, 08:44:39 pm
Hey Juliana,
To find a point of inflection for a function f(x):
1. Solve f'(x) = 0 for x
This gives you the stationary points (point of inflection, local maximums, and/or local minimums)
2. Now we need to figure out what type of stationary point it is. Around a local minimum, the graph goes down then up. So we have: f'(x)<0 on the left, f'(x) = 0 at the centre, f'(x)>0 on the right.
Around a point of inflection we have f'(x)>0 on both sides of f'(x) = 0 or f'(x)<0 on both sides of f'(x)
I'll leave it to you to see what the "rule" is for a local max but if you don't get it feel free to ask :)
The above info means that we can test the nature of a stationary point by finding f'(x) a little to the left & a little to the right of the x value you found in part 1.
Let me know if this is unclear :)
(Note: I did VCE not HSC so I'm unsure if this method is part of your curriculm)
Hold. I would like to interfere here.
A point of inflexion is not at all related to the first derivative. The question provided does not flow logically because it contradicts basic principles.
Points of inflexion do not refer to points of horizontal tangents in general (although occasionally they can, see below). They refer to points where the concavity of the curve changes. By itself, the first derivative is useless for this. We require the second derivative to deduce possible points of inflexion.
In general, setting the second derivative equal to 0 gathers all such points for ya. THEN, we apply a change in concavity test (sub a point close to both sides of the possible inflexion point) to determine if it actually is a point of inflexion.
The first derivative only comes into play for the horizontal point of inflexion. This is a special case where both the first and second derivatives are zero. But again, the second derivative is necessary.
Thank you \ ^ o ^ /
I understand now.
- Juliana.
Post by: shaynec19 on March 19, 2019, 05:23:25 pm
A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.
a) How much is each instalment?
b) How much is paid back altogether?
Post by: david.wang28 on March 19, 2019, 05:44:33 pm
Could someone please help me here.For a), you want to start off with A1 = 6000[1+(0.15/12)] - M and then write a few similar lines to A1 until A60 (60 months in 5 years). Let A60 = 0, then do some simple rearranging. Then, you want to use the sum of GP formula, so you get M[(1.0125^60 - 1)/0.0125] = 6000(1.0125)^60. Then find M, then times 12 to get the answer.
A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.
a) How much is each instalment?
b) How much is paid back altogether?
For b), times the answer of a) by 5, and you should be there.
Hope this helps :)
Post by: shaynec19 on March 19, 2019, 06:41:20 pm
For a), you want to start off with A1 = 6000[1+(0.15/12)] - M and then write a few similar lines to A1 until A60 (60 months in 5 years). Let A60 = 0, then do some simple rearranging. Then, you want to use the sum of GP formula, so you get M[(1.0125^60 - 1)/0.0125] = 6000(1.0125)^60. Then find M, then times 12 to get the answer.That's what i did, and the answer is apparently wrong. It has got to do with the interest being charged monthly, but each installment being paid annually. From what I can work out, the r value of the series is 1.0125^12, but I can't seem to get the right answer.
For b), times the answer of a) by 5, and you should be there.
Hope this helps :)
Part a answer is $1835.68
Post by: david.wang28 on March 19, 2019, 08:05:56 pm
That's what i did, and the answer is apparently wrong. It has got to do with the interest being charged monthly, but each installment being paid annually. From what I can work out, the r value of the series is 1.0125^12, but I can't seem to get the right answer.If the power is 12, then you should be writing 1.15^12. Try using that and let me know if you have any problems :)
Part a answer is $1835.68
Post by: spnmox on March 19, 2019, 08:07:24 pm
So I'm not sure if this is right: area=37/12, and if I were to evaluate the integral, it would be 5/12 - 8/3 = -9/4??
Post by: david.wang28 on March 19, 2019, 08:35:01 pm
Find the area bounded by the graph of f(x) = x(x-2)(x+1) and the x-axisYou got the final answer right. But you evaluated the integral erroneously. the -8/3 part should be positive, by putting that as an absolute value. Drawing the graph out will help you out. Hope this helps :)
So I'm not sure if this is right: area=37/12, and if I were to evaluate the integral, it would be 5/12 - 8/3 = -9/4??
Post by: spnmox on March 19, 2019, 09:05:17 pm
You got the final answer right. But you evaluated the integral erroneously. the -8/3 part should be positive, by putting that as an absolute value. Drawing the graph out will help you out. Hope this helps :)
Wait but isn't evaluating the integral and finding the area different? The integral can be negative but the area has to have the absolute value signs? If I were to put abs value signs around the 8/3, then the integral would be the same as the area ?
Post by: david.wang28 on March 19, 2019, 09:32:00 pm
Wait but isn't evaluating the integral and finding the area different? The integral can be negative but the area has to have the absolute value signs? If I were to put abs value signs around the 8/3, then the integral would be the same as the area ?Yes, evaluating the integral is finding a certain region of the area using the limits given to you. The reason why you should put absolute value signs is for the area below the x-axis. Then the integral you find will be the correct area.
Post by: szetotess on March 21, 2019, 06:19:09 pm
I was wondering why Simpson's Rule is more accurate in the equation -
f(x)= 0.00446x^4 - 0.01508x^3 - 0.17816x^2 + 0.73131x +1
(i know its a stupid equation but its my given one...)
Ive written a page on why trapezoidal rule introduces errors with calculations to the original graph but Simpson's rule is very hard...
Thanks
Post by: jamonwindeyer on March 21, 2019, 11:46:18 pm
Hello :)
I was wondering why Simpson's Rule is more accurate in the equation -
f(x)= 0.00446x^4 - 0.01508x^3 - 0.17816x^2 + 0.73131x +1
(i know its a stupid equation but its my given one...)
Ive written a page on why trapezoidal rule introduces errors with calculations to the original graph but Simpson's rule is very hard...
Thanks
Hey there! The answer to this one is actually fairly simple - Simpson's Rule uses parabolas to estimate the area. Thus, Simpson's Rule will be more accurate for polynomials with curvature similar to that of a parabola. That's the case for the curve you've presented (graph it on Desmos to see if you like!
Post by: emmajb37 on March 23, 2019, 02:26:28 pm
Find the exact gradient of the normal to the curve y = x-e-x at the point where x = 2.
I am getting some really yucky numbers and don't know why :-\
Thanks!
Post by: AlphaZero on March 23, 2019, 02:52:50 pm
Hey, I can't get this question out please help!!
Find the exact gradient of the normal to the curve y = x-e-x at the point where x = 2.
I am getting some really yucky numbers and don't know why :-\
Thanks!
The numbers are a little 'yucky'. \[y=x-e^{-x}\implies\frac{dy}{dx}=1+e^{-x},\] and so at \(x=2\), \[\text{gradient of normal}=\frac{-1}{1+e^{-2}}\]
Post by: emmajb37 on March 23, 2019, 03:10:03 pm
Use simpson’s rule with 3 function values to find the area bounded by the curve y = ln x , the x-axis and the lines x = 2 and x = 4 .
Post by: jamonwindeyer on March 23, 2019, 03:19:23 pm
Please help!
Use simpson’s rule with 3 function values to find the area bounded by the curve y = ln x , the x-axis and the lines x = 2 and x = 4 .
Hey! So the rule is on your reference sheet, should look like this:
It's just finding the three function values (at x=2,3,4) and subbing them in to the formula! ;D
Post by: not a mystery mark on March 24, 2019, 04:27:17 pm
Find the area bounded by the curve y=ln(x), the x-axis, and the lines x=2 and x=4.
Thank you!
EDIT: Funnily enough, I just noticed the question above is apart of the same set of questions this is from.
Post by: fun_jirachi on March 24, 2019, 04:43:32 pm
I know that the integral of ln x isn't on the reference sheet, but for this, consider the derivative of xlnx.
Have a go, it should be a lot easier to integrate ln x now you have this information. (Hint: manipulate the integral so it appears in the form 1 + lnx, then work from there.)
Hope this helps :)
Post by: not a mystery mark on March 24, 2019, 05:20:04 pm
Hope this helps :)
Hey!! Thanks heaps. Just curious if this is a 2U method? Our teacher is so strict on only using advanced methods.
Thanks man.
Post by: jamonwindeyer on March 24, 2019, 05:39:27 pm
Hey!! Thanks heaps. Just curious if this is a 2U method? Our teacher is so strict on only using advanced methods.
Thanks man.
The more "2U" way could be to find the area with respect to the y-axis in that range, then cleverly use it to find the x-referenced area. That is, do:
That will give you an area with respect to the y-axis. If you draw a diagram, you might be able to see how that might help you - You'll need the area of two other rectangles to make it work! ;D
Post by: RuiAce on March 24, 2019, 05:44:30 pm
(https://i.imgur.com/tU1nESn.jpg)
The bounding by \(x=2\) wrecks things up a bit. I had to do the usual method twice to get the answer.
Edit: Actually I think I was tired. It could be done more quickly with \( \int_{\ln 2}^{\ln 4}e^y\,dy\). Just need to consider the \(2\ln 2\) rectangle differently.
I think it's something like \( 2\ln 2 + \int_{\ln 2}^{\ln 4}e^y\,dy = 4\ln 4\)?
Post by: jamonwindeyer on March 24, 2019, 05:54:17 pm
Post by: not a mystery mark on March 24, 2019, 06:29:39 pm
Yep, so if the integral I referred to above is \(I\), the area we want is:
Ahhh, sneaky. Haha. Thanks man, I'll pass the working on :)
Post by: fun_jirachi on March 24, 2019, 07:31:11 pm
Hey!! Thanks heaps. Just curious if this is a 2U method? Our teacher is so strict on only using advanced methods.
Thanks man.
Probably not a 2U method. If your teacher is like that, definitely use Jamon/Rui's method. However, this method is still pretty useful for finding unfamiliar integrals because it's based off the intuition that every derivative gives us another integral ie. you find something with a derivative pretty close to the thing in the integral, then use that information to find the integral. Basically it's still pretty useful for scoping out some integrals, but otherwise stick to some more familiar methods :)
Post by: spnmox on March 25, 2019, 08:06:34 pm
The curve y=2^x is rotated about the x-axis between x=1 and x=2. Use Simpson's value with 3 function values to find an approximation of the volume of the solid formed to 2 s.f.
So I calculated the area and squared it and multiplied by pi to get 26.2, but answers say 27.2 Typo or ?
and: Find the exact area bounded by the parabola y=x^2 and the line y=4-x.
I found the x-intercepts to be (-1+-sqr17)/2. So when I sub into the integral, I have to expand those big fractions?? Am I doing this right/is there an easier method
Post by: emmajb37 on March 27, 2019, 08:13:36 am
Find the exact gradient of the tangent to the curve y = ex+lnx at the point where x = 1.
Thank you!!
Post by: emmajb37 on March 27, 2019, 08:28:17 am
I got 0.18 somehow >:(
Post by: MB_ on March 27, 2019, 08:39:05 am
Hey, I got the answer 1 + e for this question but the answer is actually 2e. I have no idea how to get that though, please help!
Find the exact gradient of the tangent to the curve y = ex+lnx at the point where x = 1.
Thank you!!
Please Help!!Can you show your working so we can identify where you may have made a mistake?
I got 0.18 somehow >:(
Post by: jamonwindeyer on March 27, 2019, 10:13:18 am
I need some help!
The curve y=2^x is rotated about the x-axis between x=1 and x=2. Use Simpson's value with 3 function values to find an approximation of the volume of the solid formed to 2 s.f.
So I calculated the area and squared it and multiplied by pi to get 26.2, but answers say 27.2 Typo or ?
and: Find the exact area bounded by the parabola y=x^2 and the line y=4-x.
I found the x-intercepts to be (-1+-sqr17)/2. So when I sub into the integral, I have to expand those big fractions?? Am I doing this right/is there an easier method
Hey! Sorry for the late reply - But yep you are over-complicating it a tad! They've told you where the bounds are for the volume, so no need to calculate x-intercepts. For 3 function values, you'll use \(x=1\), \(x=1.5\), and \(x=2\). The Simpson's Rule for volume is:
Notice it is just the normal rule with a \(\pi\) out the front and all function values squared, give that a go! :)
Post by: jamonwindeyer on March 27, 2019, 10:25:58 am
Hey, I got the answer 1 + e for this question but the answer is actually 2e. I have no idea how to get that though, please help!
Find the exact gradient of the tangent to the curve y = ex+lnx at the point where x = 1.
Thank you!!
Hey! As a quick check, is your derivative:
Please Help!!
I got 0.18 somehow >:(
And equally, is your integral here...
Always good to check the Calculus bits first! If these match, then as MB_ said I reckon we could spot the mistake if you post your working! ;D
Post by: emmajb37 on March 30, 2019, 01:15:09 pm
Find the exact area enclosed by the curve y = sinx and the line y = 1/2 for 0 ≤ x ≤ 2pi
Post by: fun_jirachi on March 30, 2019, 02:58:32 pm
Notice that when you do draw the graph, it results in only one area. You can find the intersection of the two curves by solving sinx=1/2, which should yield two solutions ie. pi/6 and 5pi/6. These solutions will provide your lower bound and upper bound respectively. From here, use the fact that the area is equal to the integral of the top curve minus the integral of the bottom curve ie. sinx-1/2. Work from there to get your answer! You should get sqrt3-pi/3. :)
Hope this helps :)
Post by: emmajb37 on March 30, 2019, 03:20:21 pm
Have you tried drawing out the graphs of these two functions over the domain provided? It really helps you visualise what you're actually looking for. If you haven't that's okay, draw it up as you answer the question :)Thank you so much!!
Notice that when you do draw the graph, it results in only one area. You can find the intersection of the two curves by solving sinx=1/2, which should yield two solutions ie. pi/6 and 5pi/6. These solutions will provide your lower bound and upper bound respectively. From here, use the fact that the area is equal to the integral of the top curve minus the integral of the bottom curve ie. sinx-1/2. Work from there to get your answer! You should get sqrt3-pi/3. :)
Hope this helps :)
Post by: emmajb37 on March 30, 2019, 03:22:19 pm
Thanks
Find the exact area bounded by the curves y = sinx and y = cos x in the domain 0 < x < 2pi
Post by: fun_jirachi on March 30, 2019, 03:47:58 pm
Hope this helps :)
Post by: emmajb37 on April 01, 2019, 04:09:20 pm
Solve cos2x = 2x/3 for 0<x<2pi
Post by: benneale on April 01, 2019, 05:10:03 pm
find the equation of the normal to the curve y=e^x at the point where x=3, in exact form ...?
thanks!
Post by: RuiAce on April 01, 2019, 05:22:00 pm
Please Help!!You cannot do this algebraically. (As in, you physically cannot - there is no mathematical method in existence to do it.)
Solve cos2x = 2x/3 for 0<x<2pi
If the question was given to you like that without any more information, it is a faulty question. If you were told some other information (e.g. use a graph), please provide it.
heyo could you please help with this:This looks like the usual method. You should be able to check that the gradient of the tangent at 3 is just \(e^3\), so the normal has gradient \( -e^{-3}\), and then sub into \(y-y_1=m(x-x_1)\).
find the equation of the normal to the curve y=e^x at the point where x=3, in exact form ...?
thanks!
Post by: benneale on April 01, 2019, 06:09:56 pm
You cannot do this algebraically. (As in, you physically cannot - there is no mathematical method in existence to do it.)
If the question was given to you like that without any more information, it is a faulty question. If you were told some other information (e.g. use a graph), please provide it.
This looks like the usual method. You should be able to check that the gradient of the tangent at 3 is just \(e^3\), so the normal has gradient \( -e^{-3}\), and then sub into \(y-y_1=m(x-x_1)\).
did all this and still couldn't get to the answer. The book says the answer is x+e^3 y-3-e^6 = 0... how would I be able to get to that? thanks :)
Post by: RuiAce on April 01, 2019, 06:16:27 pm
did all this and still couldn't get to the answer. The book says the answer is x+e^3 y-3-e^6 = 0... how would I be able to get to that? thanks :)\begin{align*} y - e^3 &=- \frac{1}{e^3} (x-3)\\ e^3(y - e^3) &= -(x - 3)\\ e^3 y - e^6 &= -x + 3\\ x + e^3 y - 3 - e^6 &= 0 \end{align*}
Post by: georgebanis on April 02, 2019, 07:26:44 pm
Thanks
Post by: RuiAce on April 02, 2019, 08:06:02 pm
Hey guys, just need a hand with part B of both of these applications of calculus questions.
Thanks
These questions involve related rates which are only a part of the MX1 course. It would be great if you could repost this to our MX1 Question Thread, just so we can do the proper solution and it isn't confusing for any 2U students looking through this later. Here is the link!
Jamon Edit: Added link!
Post by: benneale on April 04, 2019, 10:49:17 am
Having trouble working out the indefinite integral of e^x^2...
Thanks!
Post by: jamonwindeyer on April 04, 2019, 10:59:40 am
Hey!
Having trouble working out the indefinite integral of e^x^2...
Thanks!
Hey! That isn't an integral you can perform, at least not nicely ;) so I don't blame you for having trouble!! You don't need to be able to do it :)
Post by: benneale on April 04, 2019, 11:10:37 am
Find the exact area enclosed between the curve y=e^2x and the lines y=1 and x=2
Post by: fun_jirachi on April 04, 2019, 12:21:01 pm
Noting this, graph everything. It'll make it easier to visualise. Essentially now, we have an enclosed area between the line y=1 and y=e^2x, so this becomes a top curve minus bottom curve sort of question.
Your area should be something similar to the below, work from there to get your answer!
Hope this helps :)
Post by: emmajb37 on April 05, 2019, 01:10:01 pm
Post by: RuiAce on April 05, 2019, 01:18:16 pm
Having a struggle please help!This is just plugging into the Simpson's rule formula the usual way. Noting that \( \frac\pi8 = \frac{4\pi}{32}\) and \(\frac\pi4 = \frac{8\pi}{32}\), the required function values are \( \frac{4\pi}{32}\), \(\frac{5\pi}{32}\), \( \frac{6\pi}{32}\), \(\frac{7\pi}{32}\), \( \frac{8\pi}{32}\). If you have any further queries, please specify exactly where they are.
Post by: emmajb37 on April 05, 2019, 01:23:19 pm
This is just plugging into the Simpson's rule formula the usual way. Noting that \( \frac\pi8 = \frac{4\pi}{32}\) and \(\frac\pi4 = \frac{8\pi}{32}\), the required function values are \( \frac{4\pi}{32}\), \(\frac{5\pi}{32}\), \( \frac{6\pi}{32}\), \(\frac{7\pi}{32}\), \( \frac{8\pi}{32}\). If you have any further queries, please specify exactly where they are.Thanks so much
Post by: emmajb37 on April 05, 2019, 01:25:05 pm
The area of the sector of a circle is 4π untis2 and the length of the arc bounded by this sector is π/8 units. Find the radius of the circle and the angle that is subtended at the centre.
Post by: fun_jirachi on April 05, 2019, 02:42:09 pm
Work from there to find your radius, and then find your angle by substituting all the given information back into one of the two formulas (easier to do it with the length of an arc (less computations)).
Hope this helps :)
Post by: emmajb37 on April 06, 2019, 02:58:04 pm
Use the trapezoidal rule with 4 subintervals to find, correct to 3 decimal places, an approximation to the volume of the solid formed by rotating the curve y = sin x about the x-axis from x=0.2 to x = 0.6.
Post by: emmajb37 on April 06, 2019, 03:30:39 pm
Find all the points of inflexion on the curve y = 3cos(2x +π/4) for 0 ≤ x ≤ 2π
Post by: fun_jirachi on April 06, 2019, 03:55:23 pm
Somehow I got 2.18 when the answer is 2.04 >:(
Use the trapezoidal rule with 4 subintervals to find, correct to 3 decimal places, an approximation to the volume of the solid formed by rotating the curve y = sin x about the x-axis from x=0.2 to x = 0.6.
That's odd, the answer you've given is wrong? It's a magnitude of 10 off; I got roughly 0.204.
Basically, remember that for volumes you use the curve squared instead of the curve itself. ie.
When using the trapezoidal rule, you're approximating the integral of y squared, so you need to sub your values into y squared, not y. Given your boundaries are 0.2 and 0.6, with four sections you have the x values 0.2, 0.3, 0.4, 0.5 and 0.6. Sub these into (sin x)^2 to find your y values (your side lengths of the trapezium if you will). Remember that the trapezoidal rule is height/2 multiplied by (first + last + 2(everything else)), ie.
Tried this algebraically and graphically but still getting the wong values.
Find all the points of inflexion on the curve y = 3cos(2x +π/4) for 0 ≤ x ≤ 2π
For this one, note that the cosine graph isn't actually shifted up or down. If you know your trigonometric graphs well enough, you know that the sine and cosine waves change concavity on the x-axis, and since the graph isn't shifted up or down, the inflexion points should still lie on the x-axis. You're essentially solving for 3cos(2x +π/4) = 0, if that helps.
Hope this helps :)
Post by: emmajb37 on April 06, 2019, 04:12:00 pm
That's odd, the answer you've given is wrong? It's a magnitude of 10 off; I got roughly 0.204.Unfrtunately I am still getting the wrong values for that second question. I am getting 0, π/2, π, 3π/2 and 2π
Basically, remember that for volumes you use the curve squared instead of the curve itself. ie.
When using the trapezoidal rule, you're approximating the integral of y squared, so you need to sub your values into y squared, not y. Given your boundaries are 0.2 and 0.6, with four sections you have the x values 0.2, 0.3, 0.4, 0.5 and 0.6. Sub these into (sin x)^2 to find your y values (your side lengths of the trapezium if you will). Remember that the trapezoidal rule is height/2 multiplied by (first + last + 2(everything else)), ie.
For this one, note that the cosine graph isn't actually shifted up or down. If you know your trigonometric graphs well enough, you know that the sine and cosine waves change concavity on the x-axis, and since the graph isn't shifted up or down, the inflexion points should still lie on the x-axis. You're essentially solving for 3cos(2x +π/4) = 0, if that helps.
Hope this helps :)
However the answers are π/8, 5π/8, 9π/8 and 13π/8
Post by: fun_jirachi on April 06, 2019, 05:32:48 pm
And you can see only the first four solutions fit the domain, so that's where the answer comes from. I'm kind of curious as to how you got your answer, since it looks as if your answers are solving for sin 2x = 0, not 3cos(2x +π/4) = 0. Maybe share your working out, and we can see where you went wrong specifically! :)
Post by: emilyyyyyyy on April 06, 2019, 06:16:46 pm
Can someone please help me with the two questions attached?
thanksssss
Post by: benneale on April 06, 2019, 06:21:46 pm
evaluate:
bounds 2 and 1 for 2x^3-x^2+5x+3 all over x, dx
Post by: RuiAce on April 06, 2019, 06:25:46 pm
Hi,\[ \text{Since you have the sketch of the cubic given you know your integration boundaries.}\\ \text{To do the integral, you can just expand the brackets to get}\\ y=x (-x^2+5x-6) \implies \boxed{y = -x^3+5x^2-6x} \]
Can someone please help me with the two questions attached?
thanksssss
\begin{align*}\therefore A&= \left| \int_0^2 -x^3+5x^2-6x\, dx \right| + \int_2^3 -x^3+5x^2-6x\,dx \end{align*}
\[ \text{The rest is left as your exercise to compute.} \]
With regards to the second question, it technically is not a part of the 2U course. However some textbooks teach the rule \( \int f^\prime(x) e^{f(x)}dx = e^{f(x)}+C\) anyway. If your textbook does this, you should consider the fact that \( \int (x^2+1)e^{x^3+3x}dx = \frac13 \int (3x^2+3)e^{x^3+3x}dx \), and try applying the formula from there.
hey! please help...I don't see where the main difficulty is in this question. Please post relevant working out or add specifically where the problem is.
evaluate:
bounds 2 and 1 for 2x^3-x^2+5x+3 all over x, dx
Post by: emilyyyyyyy on April 06, 2019, 07:15:53 pm
\[ \text{Since you have the sketch of the cubic given you know your integration boundaries.}\\ \text{To do the integral, you can just expand the brackets to get}\\ y=x (-x^2+5x-6) \implies \boxed{y = -x^3+5x^2-6x} \]
\begin{align*}\therefore A&= \left| \int_0^2 -x^3+5x^2-6x\, dx \right| + \int_2^3 -x^3+5x^2-6x\,dx \end{align*}
\[ \text{The rest is left as your exercise to compute.} \]
With regards to the second question, it technically is not a part of the 2U course. However some textbooks teach the rule \( \int f^\prime(x) e^{f(x)}dx = e^{f(x)}+C\) anyway. If your textbook does this, you should consider the fact that \( \int (x^2+1)e^{x^3+3x}dx = \frac13 \int (3x^2+3)e^{x^3+3x}dx \), and try applying the formula from there.
thank you!
I've attached my working for the area question and i cant seem to get the right answer! What have I done wrong in my working out?
Post by: RuiAce on April 06, 2019, 07:34:57 pm
thank you!Not too sure why you only put on absolute values at the end, instead of around only the first integral like what I wrote down?
I've attached my working for the area question and i cant seem to get the right answer! What have I done wrong in my working out?
Recall that absolute values are only used for regions below the \(x\)-axis in order to focus on the magnitude of the area only. By essentially ignoring the absolute values, you end up calculating the sum of the signed areas instead, i.e. in the process of doing so you treat the area under the \(x\)-axis as negative.
Post by: emilyyyyyyy on April 06, 2019, 07:35:21 pm
I'm having trouble solving questions like the one attached with x's in the numerator!
How do I solve it?
thanks :)
Post by: emilyyyyyyy on April 06, 2019, 07:39:14 pm
Not too sure why you only put on absolute values at the end, instead of around only the first integral like what I wrote down?
Recall that absolute values are only used for regions below the \(x\)-axis in order to focus on the magnitude of the area only. By essentially ignoring the absolute values, you end up calculating the sum of the signed areas instead, i.e. in the process of doing so you treat the area under the \(x\)-axis as negative.
my bad! i put them around the first one instead and my answer is 37/12, but the I'm still not getting the answer.
I've attached the solution given to me, but I don't really understand how they did it.
Post by: fun_jirachi on April 06, 2019, 07:41:35 pm
Hi I need help again!
I'm having trouble solving questions like the one attached with x's in the numerator!
How do I solve it?
thanks :)
In general, you want to be splitting up the integral and manipulating it into a better form to integrate. :)
Post by: RuiAce on April 06, 2019, 07:44:29 pm
my bad! i put them around the first one instead and my answer is 37/12, but the I'm still not getting the answer.That looks like the solution to a completely different question to me.
I've attached the solution given to me, but I don't really understand how they did it.
I checked on Wolfram and the answer to the one you posted is definitely \( \frac{37}{12} \).
Post by: emilyyyyyyy on April 06, 2019, 08:10:51 pm
How do I integrate logs without fractions?
For e.g. y=ln(x-2) ?
thanks!
Post by: slinkybench on April 06, 2019, 11:04:10 pm
Thanks in advance.
Post by: RuiAce on April 06, 2019, 11:16:12 pm
Hi,Short answer: You don't (in 2U).
How do I integrate logs without fractions?
For e.g. y=ln(x-2) ?
thanks!
Long answer: You still don't, but if you're given a definite integral in 2U you can potentially draw a picture and consider something involving \( \int_a^b e^y\,dy\). Depends on the question at hand.
I can't figure out how to do this question: Consider the two simultaneous equations mx + y = 24 & 6x + my = m. Find the exact values of m for which there is no solution to the pair of simultaneous equations.\[ \text{For a purely algebraic approach, we can attempt to start solving.}\\ \text{The first equation rearranges to }y=-mx+24.\text{ So subbing into the second gives} \\ \begin{align*} 6x+m(24-mx) &= m\\ 6x + 24m - m^2x &= m\\ x(6-m^2) &= -23m\\ x &= \frac{-23m}{6-m^2}.\end{align*} \]
Thanks in advance.
\[ \text{From here, we see that we will not have any solutions if }6-m^2=0\\ \text{which solves to give }\boxed{m=\pm \sqrt{6}}. \]
Note that this essentially boiled down to when do we have "divide by zero" problems. i.e. when the denominator is zero. Because then the result is undefined.
Post by: emilyyyyyyy on April 07, 2019, 05:34:49 pm
With this question, when I'm working it out, what am I meant to do to know whether the square root of y is + or - in the graph?
Thanks!
Post by: benneale on April 08, 2019, 06:21:20 pm
Find the arc length, correct to 2 decimal places, given radius is 5.9cm and angle subtended is 23degrees 12minutes...
I understand the process of l=r.theta , but when I punch in the degrees bit into my calculator, the answer is way off!
Post by: RuiAce on April 08, 2019, 06:30:57 pm
Hi,If you had to go from \(x = y^2\) to \(y = \pm \sqrt{x}\), you'd identify that the top branch is \(y=\sqrt{x}\) and the bottom branch is \(y=-\sqrt{x}\). This is because above the \(x\)-axis, the \(y\)-coordinates are positive, and below the \(x\)-axis, the \(y\)-coordinates are negative.
With this question, when I'm working it out, what am I meant to do to know whether the square root of y is + or - in the graph?
Thanks!
Similarly, if you had to go from \(y=x^2\) to \(x=\pm \sqrt{y}\), you'd identify that the left branch is \(x=\sqrt{y}\) and the right branch is \(x=-\sqrt{y}\). This is because to the right of the \(y\)-axis, the \(x\)-coordinates are positive, and to the left of the \(x\)-axis, the \(y\)-coordinates are negative.
Same goes here. You would take \(\boxed{x-2 = -\sqrt{y}}\) and hence \(x = 2-\sqrt{y} \), because the expression \(x-2\) will be negative for values to the left of the line \(x=2\).
hey, having a bit of a 'moment'...That formula only works when \(\theta\) is radians. Write the angle as \( \left( 23 + \frac{12}{60} \right) \) degrees, and hence convert it to radians by writing it as \( \left( 23 + \frac{12}{60} \right) \times \frac\pi{180} \).
Find the arc length, correct to 2 decimal places, given radius is 5.9cm and angle subtended is 23degrees 12minutes...
I understand the process of l=r.theta , but when I punch in the degrees bit into my calculator, the answer is way off!
Post by: emilyyyyyyy on April 08, 2019, 07:37:46 pm
How do I do part iii?
thanks!
Post by: fun_jirachi on April 08, 2019, 07:47:54 pm
There are two cases; getting a 45 and getting a 50.
There's only one way of getting a 50; rolling 5 all four times. The chance of that happening is (1/6)^4, or 1/1296.
There's four ways of getting a 45; if you think about it, a 45 is made of a 4 and 5, and a 5 and 5. What you should notice is that the four appears once and once only, and can appear anywhere, as long as all the other rolls are fives ie. 5 and 4 + 5 and 5 is different to 4 and 5 + 5 and 5. ie. the probability of this happening is 4 x (1/6)^4, or 4/1296.
Add these up, and subtract them from 1, and you get 1291/1296.
Hope this helps :)
Post by: emilyyyyyyy on April 08, 2019, 07:54:11 pm
You could think about the probability of getting 45 or more in the first two sums, then subtracting that probability to obtain the answer. This basically uses the idea of the complement to make your life a lot easier.
There are two cases; getting a 45 and getting a 50.
There's only one way of getting a 50; rolling 5 all four times. The chance of that happening is (1/6)^4, or 1/1296.
There's four ways of getting a 45; if you think about it, a 45 is made of a 4 and 5, and a 5 and 5. What you should notice is that the four appears once and once only, and can appear anywhere, as long as all the other rolls are fives ie. 5 and 4 + 5 and 5 is different to 4 and 5 + 5 and 5. ie. the probability of this happening is 4 x (1/6)^4, or 4/1296.
Add these up, and subtract them from 1, and you get 1291/1296.
Hope this helps :)
thank you! i understand the part about the 50 and how there are 4 chances of that happening - so to get a 50 it's (1/6)^4
But I'm still a little lost on the part about getting 45, is there another way you might be able to explain it to m (if not don't worry about it hahha)
Post by: emilyyyyyyy on April 08, 2019, 09:46:15 pm
Is someone able to tell me where I've gone wrong in working out the area of this graph? I cant get the right answer
thanks :))
Post by: jamonwindeyer on April 08, 2019, 09:59:17 pm
thank you! i understand the part about the 50 and how there are 4 chances of that happening - so to get a 50 it's (1/6)^4
But I'm still a little lost on the part about getting 45, is there another way you might be able to explain it to m (if not don't worry about it hahha)
I always revert to listing the possible outcomes when it gets confusing! Here are the four outcomes Fun_Jirachi is referencing:
- (4x5) + (5x5)
- (5x4) + (5x5)
- (5x5) + (4x5)
- (5x5) + (5x4)
All of these give an answer of 45. Notice it is just all fives with a four in position #1, #2, #3 or #4 - Hence the four possible ways it can be done ;D
Post by: jamonwindeyer on April 08, 2019, 10:01:44 pm
Hi,
Is someone able to tell me where I've gone wrong in working out the area of this graph? I cant get the right answer
thanks :))
Hi! If you just want the blue shaded area, then you only need the \(A_1\) from your working! That gives you the area between the curve and the x-axis for x=0 to x=1, which is exactly what you are looking for ;D the dotted line above is just the asymptote of the graph but it actually doesn't matter because the area never touches it!
Post by: emilyyyyyyy on April 08, 2019, 10:06:37 pm
Hi! If you just want the blue shaded area, then you only need the \(A_1\) from your working! That gives you the area between the curve and the x-axis for x=0 to x=1, which is exactly what you are looking for ;D the dotted line above is just the asymptote of the graph but it actually doesn't matter because the area never touches it!
okay thank you! I think I was getting a bit confused bc when I was doing this question attached, I had to find the area under the curve and also of the rectangular shape? are you able to possibly explain why?
thanks!
Post by: jamonwindeyer on April 08, 2019, 10:12:45 pm
okay thank you! I think I was getting a bit confused bc when I was doing this question attached, I had to find the area under the curve and also of the rectangular shape? are you able to possibly explain why?
thanks!
So in this question, what you want (just straight up) is:
Problem is, we can't find that directly because we can't integrate logarithms. So, we instead rearrange it to \(x=2+e^y\) and find the area with respect to the y-axis, that's the area to the left of the curve.
That's not the one we want - The one we want is the area of the rectangle MINUS that area! And that's why we do it that way!
In summary, we need to use the rectangle for this question because we can't directly integrate a logarithm. So, we need to use tricks - We don't need to do that for exponentials ;D
Post by: emilyyyyyyy on April 09, 2019, 07:31:36 pm
How do I find the area in this question? I've tried splitting it up into positive and negative areas, but my equations or something aren't working because I can't seem to get the answer.
Thanks! :)
Post by: r1ckworthy on April 09, 2019, 08:50:25 pm
Hi!Hey,
How do I find the area in this question? I've tried splitting it up into positive and negative areas, but my equations or something aren't working because I can't seem to get the answer.
Thanks! :)
Here is my solution of the problem. I don't think you need to split the curve up into positive and negative areas, as you are only working out the area between two curves. I think the time you do split the curve into positive and negative areas is when working out the area under a curve in a negative region. Hopefully someone can back this up, as I am not too sure on whether this is correct or not.
(https://i.imgur.com/yo8afhv.jpg)
Post by: emilyyyyyyy on April 09, 2019, 08:55:39 pm
Hey,
Here is my solution of the problem. I don't think you need to split the curve up into positive and negative areas, as you are only working out the area between two curves. I think the time you do split the curve into positive and negative areas is when working out the area under a curve in a negative region. Hopefully someone can back this up, as I am not too sure on whether this is correct or not.
(https://i.imgur.com/yo8afhv.jpg)
Thanks! what you've done is actually what I did the first time, but for some reason I guess I put the numbers in my calculator wrong :)
I think I was getting confused with working out two separate areas on a curve - i.e. one that is above the x-axis and one that is below.
Post by: r1ckworthy on April 09, 2019, 09:32:22 pm
Thanks! what you've done is actually what I did the first time, but for some reason I guess I put the numbers in my calculator wrong :)
I think I was getting confused with working out two separate areas on a curve - i.e. one that is above the x-axis and one that is below.
Cool, no worries! Yeah, I get what you mean, I got my numbers all wrong too until I checked it on an integration calculator online (where they calculate the answer to any integral- so useful).
Post by: JkbC on April 10, 2019, 07:07:23 pm
Post by: RuiAce on April 10, 2019, 07:27:53 pm
To clarify on the above question, the graph shown is y = f(x) and the question wants you to sketch y=f^-1(x)In saying that, this is an MX1 question, so you should post in the appropriate MX1 thread if you require assistance.
Post by: emilyyyyyyy on April 10, 2019, 10:32:11 pm
I have two questions.
In the integration question, my answer is x^4/4 -2lnx + C. The answer says it should be x^2/3 + 2lnx + C, but I'm unsure how they got that.
And can someone help me with iii and iv of the probability question? Thanks!
Post by: fun_jirachi on April 10, 2019, 11:20:44 pm
Hi!
I have two questions.
In the integration question, my answer is x^4/4 -2lnx + C. The answer says it should be x^2/3 + 2lnx + C, but I'm unsure how they got that.
And can someone help me with iii and iv of the probability question? Thanks!
Are you sure you're looking at the right question and the right solution? You're right, but possibly the solutions/questions are jumbled up.
For the second one, A wins if he draws green, B wins if he draws red. For nothing to happen in a round (thus proceeding to the next round), A must draw red, then B must draw green. ie, 3/5 x 2/5, or 6/25. For iii), do it casewise; what's the probability that A wins in one turn? Two? Three? Add these up and you should have the answer. For iv) consider what you had in iii). Notice, that for the win in each increasing number of turns, you need an extra term of nothing happening ie 6/25. This forms a geometric series, see if you can work it out from there!
Hope this helps :)
Post by: emilyyyyyyy on April 11, 2019, 05:07:25 pm
How do I do question ii? I'm unsure how to do it seeing as the question wants me to integrate with x-values, even though the curve is enclosed by the y-axis?
Thanks!
Post by: RuiAce on April 11, 2019, 05:10:17 pm
Hi Im back again!By just treating it as an area between two curves. Note that the upper curve is \(y = 5e^{-x}\), and the lower curve is \( y=e^{\frac{x}{2}}\). That is where they get the integrand (expression to be integrated) from.
How do I do question ii? I'm unsure how to do it seeing as the question wants me to integrate with x-values, even though the curve is enclosed by the y-axis?
Thanks!
Note that for the upper limit in integration, you will need to calculate the point of intersection of the two curves.
Post by: emilyyyyyyy on April 11, 2019, 05:14:28 pm
By just treating it as an area between two curves. Note that the upper curve is \(y = 5e^{-x}\), and the lower curve is \( y=e^{\frac{x}{2}}\). That is where they get the integrand (expression to be integrated) from.
Note that for the upper limit in integration, you will need to calculate the point of intersection of the two curves.
thanks!
If the question didn't specify how to solve it, if I used y values, would I still get the same answer in finding the area? Or is it just easier to use the x values?
Post by: RuiAce on April 11, 2019, 05:31:58 pm
thanks!In theory you would, but you wouldn't want to. You'd have to deal with \( \int \ln x\,dx\), which is an MX2 integral.
If the question didn't specify how to solve it, if I used y values, would I still get the same answer in finding the area? Or is it just easier to use the x values?
Post by: emilyyyyyyy on April 11, 2019, 08:22:14 pm
I get that I'm integrating it, but I'm unsure which numbers to sub in each time?
thanks!!
Post by: jamonwindeyer on April 11, 2019, 09:54:27 pm
How do I do this question?
I get that I'm integrating it, but I'm unsure which numbers to sub in each time?
thanks!!
Howdy! Yep, sounds like you are close, hopefully these hints are just the right amount of help ;D
- Integrate once to get \(f'(x)\). You know there is a stationary point at \(x=0\). So, when \(x=0\), \(f'(x)=0\) for the stationary point! That's how you evaluate the first constant.
- Integrate again to get \(f(x)\), and substitute the given coordinate to evaluate the second constant.
Let us know how you go :)
Post by: emilyyyyyyy on April 11, 2019, 10:00:14 pm
Howdy! Yep, sounds like you are close, hopefully these hints are just the right amount of help ;D
- Integrate once to get \(f'(x)\). You know there is a stationary point at \(x=0\). So, when \(x=0\), \(f'(x)=0\) for the stationary point! That's how you evaluate the first constant.
- Integrate again to get \(f(x)\), and substitute the given coordinate to evaluate the second constant.
Let us know how you go :)
thank you! I got the right answer now :)
Post by: emilyyyyyyy on April 11, 2019, 10:29:49 pm
I eventually got the answers by a bit of trial and error, but I don't understand why for a) I have to multiply by 5 and for b) I only have to multiply by 4 not 5 (bc the person is dealt 5 cards). Hopefully that makes sense?
thanks :)
Post by: fun_jirachi on April 11, 2019, 11:12:48 pm
For b), you can first pick the suit, (of which there are 4), then pick any 5 cards from that suit. Similar to above, to pick 5 from 13 we have 13x12x11x10x9 over 5! which eliminates order. We then have all that over the total number of ways, ie the answer will be
Here, we multiply by four for the four suits :)
Hopefully this makes sense, and hope this helps :)
Post by: RuiAce on April 12, 2019, 09:13:47 am
Hello!\[ \text{There isn't, however the following }\textbf{limit}\text{ holds:}\\ \boxed{\lim_{x\to 0^+ }x\ln x = 0} \]
I just want to ask for the graph, y=xInx, why is there a point at x=0? Because when I put in 0In0 in the calculator it says MATHS ERROR.
Thank you!
Note that if you just naively sub 0 into \(x \ln x\), you get \(0 \times \ln 0\), which can be thought of as \(0\times -\infty\) (due to the asymptote of \(y=\ln x\), which is a math error.
Graphing software like Desmos and GeoGebra would not be able to show you this, but the idea is that in a hand-drawn sketch, a discontinuity (i.e. a hole in the graph) should be located at the point (0,0).
Note that, of course, this limit is not meant to be doable with HSC methods. And that this graph should not be asked for a 2U student.
Post by: Thankunext on April 15, 2019, 02:27:46 pm
(A) Use Simpson's rule with 3 function values to find the area bounded by the curve y=Inx, the x-axis and the lines x=2 and x=4.
(B) Change the subject of y=Inx to x.
(C) Hence find the exact area in part (A).
:'( :'( :'( Please answer me
Post by: RuiAce on April 17, 2019, 09:30:48 pm
Can someone help with part (C) plz.This particular question was discussed here.
(A) Use Simpson's rule with 3 function values to find the area bounded by the curve y=Inx, the x-axis and the lines x=2 and x=4.
(B) Change the subject of y=Inx to x.
(C) Hence find the exact area in part (A).
:'( :'( :'( Please answer me
Post by: spnmox on April 19, 2019, 02:25:56 pm
So I found the A with x=0,pi/8,pi/4,3pi/8,pi/2 and it was equal to approx 1.4 (maybe someone could confirm this?). I just don't know how to find the volume from this area. Do I square it and then multiply by pi??
Post by: RuiAce on April 19, 2019, 03:41:52 pm
Use the trapezoidal rules with 4 sub intervals to find an approximation to the volume of the solid formed when rotating y=sqrcosx about the x-axis from x=0 to x=pi/2.\[ \text{Recall that the volume is given by}\\ V =\pi \int_a^b y^2\,dx. \]
So I found the A with x=0,pi/8,pi/4,3pi/8,pi/2 and it was equal to approx 1.4 (maybe someone could confirm this?). I just don't know how to find the volume from this area. Do I square it and then multiply by pi??
\[ \text{So here, }y = \sqrt{\cos x}\implies \boxed{y^2 = \cos x}\text{. Hence your volume is}\\ \boxed{V = \pi \int_0^{\pi/2} \cos x\,dx}.\\ \text{This is the integral you would like to use on the trapezoidal rule.} \]
The area is not helpful in finding the volume of solid of revolution because the \(y^2\) is entirely under the integral. As opposed to something like \( \pi \left( \int_a^b y\,dx \right)^2\).
Post by: spnmox on April 19, 2019, 04:55:30 pm
Post by: therese07 on April 20, 2019, 06:58:47 pm
So I was doing some revision on curve sketching, and I came across this question (attached below).
I understand you have to do the produce rule first (I could be wrong while doing this working out, so sorry in advance)
u = 2x +1 v = (x-2)^4
u' = 2 v' = 4(x-2)^3
dy/dx = (2x +1) x 4(x-2)^3 + (x-2)^4 x 2
= 4(2x +1)(x-2)^3 + 2(x-2)^4
I don't know what do to, or what to factor out to get to the next step and to find stationary points, and sketch it.
Thank you for your help!! ;D
Post by: jamonwindeyer on April 20, 2019, 07:11:20 pm
Hi!
So I was doing some revision on curve sketching, and I came across this question (attached below).
I understand you have to do the produce rule first (I could be wrong while doing this working out, so sorry in advance)
u = 2x +1 v = (x-2)^4
u' = 2 v' = 4(x-2)^3
dy/dx = (2x +1) x 4(x-2)^3 + (x-2)^4 x 2
= 4(2x +1)(x-2)^3 + 2(x-2)^4
I don't know what do to, or what to factor out to get to the next step and to find stationary points, and sketch it.
Thank you for your help!! ;D
Smashed the first bit! If you pull out a factor of \(\left(x-2\right)^3\), you'll get:
From there you can solve for the \(x=2\) and \(x=0\) stationary points, and determine their nature using a table (the second derivative test is probably a bit cumbersome here) ;D
Post by: Thankunext on April 22, 2019, 10:14:21 pm
For what value of k will the equation 6x*2 + (k+15)x +(4+k)=0 have roots which are
I. Equal but opposite in sign?
2. Reciprocals?
Post by: RuiAce on April 22, 2019, 10:42:47 pm
Help!\[ \text{You can just use the sum and product of roots.}\\ \text{In the first case, suppose that the roots are negatives of each other, i.e. }\boxed{\beta = -\alpha}. \]
For what value of k will the equation 6x*2 + (k+15)x +(4+k)=0 have roots which are
I. Equal but opposite in sign?
2. Reciprocals?
\[ \text{But since }\alpha+\beta = -(k+15),\text{ we therefore have}\\ \alpha-\alpha = -k-15 \implies \boxed{0 = -k-15}. \]
You can do something similar for the next question.
Post by: spnmox on April 23, 2019, 09:29:44 pm
How many solutions of the equation (sinx-1)(tanx+2)=0 lie between 0 and 2pi?
Post by: jamonwindeyer on April 23, 2019, 09:35:02 pm
2014 HSC Q7 Multiple Choice
How many solutions of the equation (sinx-1)(tanx+2)=0 lie between 0 and 2pi?
Hey! So all you are doing here is solving each part of the equation separately, like a quadratic!
And I don't have a calculator handy, but you'll get two answers for that second bit - So it would appear that the answer is three! However, \(x=\frac{\pi}{2}\) isn't a solution! Why? If you substitute \(x=\frac{\pi}{2}\) into \(\tan{x}\), it breaks! Math error! The tangent ratio doesn't allow odd multiples of 90 degrees :)
Really tricky, but you have to ignore that solution. Which means only the two solutions from the \(\tan\) part of the equation work - The answer is two ;D
Post by: spnmox on April 23, 2019, 10:09:00 pm
Hey! So all you are doing here is solving each part of the equation separately, like a quadratic!
And I don't have a calculator handy, but you'll get two answers for that second bit - So it would appear that the answer is three! However, \(x=\frac{\pi}{2}\) isn't a solution! Why? If you substitute \(x=\frac{\pi}{2}\) into \(\tan{x}\), it breaks! Math error! The tangent ratio doesn't allow odd multiples of 90 degrees :)
Really tricky, but you have to ignore that solution. Which means only the two solutions from the \(\tan\) part of the equation work - The answer is two ;D
oh, thank you so much! haha now it seems obvious
Post by: Karlamineeeee on April 24, 2019, 01:01:59 pm
For the first question, why do the answers calculate for the 15th term instead of the 30th?
Post by: RuiAce on April 24, 2019, 01:40:48 pm
I hope this question is allowed here, I'm trying out the first series and sequence test in the ATAR notes' HSC mathematics topic tests.I had a look. That's definitely some accidental error. Should certainly sub in \(n=30\) in \(T_n = 6(2)^{n-1}\)
For the first question, why do the answers calculate for the 15th term instead of the 30th?
Post by: Karlamineeeee on April 24, 2019, 07:30:00 pm
I had a look. That's definitely some accidental error. Should certainly sub in \(n=30\) in \(T_n = 6(2)^{n-1}\)
Thank you so much!!
Also, for the final question of that test, where did the 500 value come from? Wasn't $1000 deposited? Sorry for asking so late!! DX
Post by: RuiAce on April 24, 2019, 10:31:58 pm
Thank you so much!!Only just arrived back at home and yea, that's another accidental error I think. Should be 1000.
Also, for the final question of that test, where did the 500 value come from? Wasn't $1000 deposited? Sorry for asking so late!! DX
Post by: jamonwindeyer on April 25, 2019, 12:45:36 am
Apologies for the confusion caused - It's been fixed for future prints, but we must still be working through existing stock! ;D
Post by: Karlamineeeee on April 25, 2019, 01:41:17 am
Confirming both of those are errors, and both have corrections at this link: https://atarnotes.com/product-updates/. There's a full fixed solution to that second one too!
Apologies for the confusion caused - It's been fixed for future prints, but we must still be working through existing stock! ;D
Thank you so much!!
Post by: therese07 on May 07, 2019, 08:01:07 pm
I was doing some homework tasks for math, and I got stuck on 10ii, proving that the stationary point is a maximum. I was wondering how would I be able to do this question. Would I need to second derivative at all?
Thank you!!!
Post by: RuiAce on May 07, 2019, 08:28:17 pm
Hi there!The question wants you to avoid using the second derivative but rather "consider the gradient on each side of \(x=e\)".
I was doing some homework tasks for math, and I got stuck on 10ii, proving that the stationary point is a maximum. I was wondering how would I be able to do this question. Would I need to second derivative at all?
Thank you!!!
\[ \text{So we just sub some numeric values in.}\\ \text{For a point on the left, we can try }x=2.\\ \text{When }x=2\text{, the calculator says that }\frac{dy}{dx} = 0.0767\dots > 0 \]
Therefore on the left of \(x=e\), the curve is increasing.
\[ \text{For a point on the right, we can try }x=3.\\ \text{When }x=3\text{, the calculator says that }\frac{dy}{dx} = -0.0109\dots < 0\]
Therefore on the right of \(x=e\), the curve is decreasing.
\[ \text{So the curve increases, until it reaches }x=e.\\ \text{Then after }x=e\text{, it decreases.}\\ \text{That's all that is required to conclude that }x=e\\ \text{must therefore be a local maximum.} \]
This method (which is really just a table of values) should be noted, because sometimes computing the second derivative is just nastily exhausting effort. The whole point of this method is to get you out of that trouble.
Remark: Second derivative
\[ \text{If you insist on using the second derivative}\\ \text{you more or less need the quotient rule }\textit{again}. \]
\begin{align*}\frac{dy}{dx} &= \frac{x\cdot \frac1x - 1 \cdot \ln x}{x^2}\\ &= \frac{1-\ln x}{x^2}\\ \therefore \frac{d^2y}{dx^2} &= \frac{x^2\cdot -\frac1x - 2x(1-\ln x)}{x^4}\end{align*}
\[ \text{At this point, you can now sub }x=e\text{ in straight away.}\]
This can optionally be simplified to \( \frac{d^2y}{dx^2} = \frac{-3x+2x\ln x}{x^4} = \frac{-3+2\ln x}{x^3}\) but you don't really need to. You can just sub \(x=e\) into the unsimplified expression.
For this question though, in theory you should not do this. This is because they told you to use the other method.
\begin{align*}\frac{dy}{dx} &= \frac{x\cdot \frac1x - 1 \cdot \ln x}{x^2}\\ &= \frac{1-\ln x}{x^2}\\ \therefore \frac{d^2y}{dx^2} &= \frac{x^2\cdot -\frac1x - 2x(1-\ln x)}{x^4}\end{align*}
\[ \text{At this point, you can now sub }x=e\text{ in straight away.}\]
This can optionally be simplified to \( \frac{d^2y}{dx^2} = \frac{-3x+2x\ln x}{x^4} = \frac{-3+2\ln x}{x^3}\) but you don't really need to. You can just sub \(x=e\) into the unsimplified expression.
For this question though, in theory you should not do this. This is because they told you to use the other method.
Post by: Kombmail on May 12, 2019, 07:25:29 pm
'The rate of percentage of leakage of water out of a container is proportional to the amount of water in the container at any one time. If the container is 60% empty after 5 minutes, find how long it will take for the container to be 90% empty'
I got an answer of 6.82 minutes but the actual answer is 12.6 minutes. Please assist in telling me how the answer 12.6 is found.
thanks!
Post by: RuiAce on May 12, 2019, 07:46:19 pm
So I was doing 6.2 ( exponential growth and decay) for mathing and came across this:\[ \text{If }V_0\text{ is the initial quantity}\\ \text{then }V = V_0 e^{-kt}. \]
'The rate of percentage of leakage of water out of a container is proportional to the amount of water in the container at any one time. If the container is 60% empty after 5 minutes, find how long it will take for the container to be 90% empty'
I got an answer of 6.82 minutes but the actual answer is 12.6 minutes. Please assist in telling me how the answer 12.6 is found.
thanks!
\[ \text{When }t=5, \quad V = 0.4V_0\\ \text{Therefore }0.4V_0 = V_0 e^{-5k} \implies \boxed{k = -\frac15 \ln 0.4} \]
\[\text{Therefore }V = V_0 e^{\frac{t}5\ln 0.4}\\ \text{We want when }V = 0.1 V_0.\\ \text{Solving }0.1V_0 = V_0 e^{\frac{t}{5}\ln 0.4}\text{ gives }\boxed{t = \frac{\ln 0.1}{\frac15 \ln 0.4} \approx 12.6}. \]
There is a slight trap in this question worth mentioning. It is in that the analysis must somewhat be done in reverse. The question gives what percentage the tank is empty at, but actually measuring how much of the initial quantity we have means that our calculations must be based on what percentage the tank is full at.
Aside from that, if you have any further questions you should post relevant working out, or ask for key areas to clarify.
Post by: Kombmail on May 17, 2019, 09:57:13 am
\[ \text{If }V_0\text{ is the initial quantity}\\ \text{then }V = V_0 e^{-kt}. \]
\[ \text{When }t=5, \quad V = 0.4V_0\\ \text{Therefore }0.4V_0 = V_0 e^{-5k} \implies \boxed{k = -\frac15 \ln 0.4} \]
\[\text{Therefore }V = V_0 e^{\frac{t}5\ln 0.4}\\ \text{We want when }V = 0.1 V_0.\\ \text{Solving }0.1V_0 = V_0 e^{\frac{t}{5}\ln 0.4}\text{ gives }\boxed{t = \frac{\ln 0.1}{\frac15 \ln 0.4} \approx 12.6}. \]
There is a slight trap in this question worth mentioning. It is in that the analysis must somewhat be done in reverse. The question gives what percentage the tank is empty at, but actually measuring how much of the initial quantity we have means that our calculations must be based on what percentage the tank is full at.
Aside from that, if you have any further questions you should post relevant working out, or ask for key areas to clarify.
Thanks!
Post by: spnmox on May 19, 2019, 07:10:57 pm
PS. please ignore the scribbles down the side of the page!
Post by: RuiAce on May 19, 2019, 08:42:35 pm
Hey guys, I need a bit of help with this question. I understand part i) but not part ii). I don't know how to figure it out and don't understand the solutions either :( Any help appreciated!This will be added to the compilation. This is still NESA's solution but with more added depth (no pun intended)
PS. please ignore the scribbles down the side of the page!
\[ \textbf{2010 HSC Mathematics - Q10 b) ii)} \]
\[ \text{In part (1), a subtle thing to note is how they mention that the}\\ \text{hemispherical container is initially laid out }\textbf{horizontal.}\\ \text{That, and it is }\textbf{full.} \]
\[ \text{So initially it must've looked something like this.} \]
For reference I intentionally plot the centre of the circle.
(https://i.imgur.com/dWqglHj.png)
\[ \text{Hence the initial depth is literally just the distance between the}\\ \text{centre of a circle, and any point on its circumference.}\\ \text{Which is of course, the radius, and hence equal to }r. \]Note therefore that half of the initial depth must be \( \boxed{\frac{r}{2}} \).
\[ \text{Now after doing that rotation, the situation is as follows.}\\ \text{We wish to understand why.} \]
(https://i.imgur.com/25VhMIl.png)
\[ \text{The }r\text{ towards the right should be reasonably clear.}\\ \text{However the red line segment }\textbf{also}\text{ has length }r. \]\[ \text{This is because }\textbf{any}\text{ line from the centre of the circle to its circumference}\\ \text{has length }r\text{, where }r\text{ is the radius.}\\ \text{Note that the red line segment }\textbf{also}\text{ satisfies this.}\]
It just so happens that the red line is special, in that it's perpendicular to the ground (or whatever surface the container is lying on. And consequently the red line is vertical.
\[ \text{Again, the blue region shaded reflects the water in the bowl.}\\ \text{Note that the depth of the water is then just from the point where the bowl is at the ground}\\ \text{up to the black line, i.e. where the water goes up to.} \]
\[ \text{Since the question instructs us to make this depth equal to }\frac{r}{2}\\ \text{we do so. But then because the red line segment}\\ \text{has length }r\text{, everything above the water to the centre of the circle}\\ \text{must }\textbf{also}\text{ have length }\frac{r}{2}. \]
\[ \text{The last thing to justify is where the }\theta\text{ is placed.}\\ \text{This relies on the fact that }\theta\text{ measures simply what angle the bowl is }\textbf{tilted at}\\ \text{and therefore must be the angle drawn from the horizontal, as required.}\]
Or alternatively, if the diagram they gave was clear enough, you should be able to automatically see why \(\theta\) is there. I literally just copied \(\theta\) across - it was the \(r\) that required more deduction.
\[ \text{With the set-up done, the trigonometry is now easy.}\\ \text{Because we have opposite/hypotenuse here, we just consider}\\ \sin \theta = \frac{\frac{r}{2}}{r} \implies \sin \theta = \frac12 \implies \boxed{\theta = \frac\pi6}. \]
__________________________________________________________
\[ \text{Put simply, part (2) now wishes us to simplify the following fraction:}\\ \frac{\text{Remaining volume left}}{\text{Initial volume}} \]
Because we're given that the water initially fills the entire hemisphere, we just need to compute the volume of said hemisphere. Which is of course half that of the volume of the entire sphere, i.e. \(\frac12 \times \frac{4\pi r^3}{3}\), which simplifies to \( \boxed{\frac{2\pi r^3}{3}} \).
\[ \text{The remaining bit is to make use of part i).}\\ \text{It was more or less up to you to recognise that the 'solid' that the water manifests in}\\ \text{is basically the same as what you do in that volumes question above.} \]
By simply rotating NESA's own diagram by \(90^\circ\) anticlockwise this may be clearer. Focus on the grey shaded area, because that reflects the volume of the water.
(https://i.imgur.com/E5MBCfo.png?1)
Note that in the diagram in part i), only the top half is shaded. But recall that when you rotate to form your solid, you go through a full \(360^\circ\) rotation about the \(x\)-axis. By doing so, the bottom half of the minor segment does get included as well.\[ \text{So in our case, the volume of the water is just the answer in part i)}\\ \text{except thanks to (1), we know now that }\theta = \frac\pi6.\]
\[ \text{Thus we get}\\ \begin{align*}V&= \frac{\pi r^3}{3} \left(2-3\sin \frac\pi6 +\sin^3\frac\pi6 \right)\\ &= \frac{\pi r^3}{3} \times \frac{5}{8}\\ &= \frac{5\pi r^3}{24}. \end{align*} \]
\[ \text{Finally subbing back, the required fraction is}\\ \frac{ \frac{5\pi r^3}{24}}{\frac{2\pi r^3}{3}} = \frac{5}{16}. \]
Post by: spnmox on May 19, 2019, 09:31:42 pm
This will be added to the compilation. This is still NESA's solution but with more added depth (no pun intended)
\[ \textbf{2010 HSC Mathematics - Q10 b) ii)} \]
\[ \text{In part (1), a subtle thing to note is how they mention that the}\\ \text{hemispherical container is initially laid out }\textbf{horizontal.}\\ \text{That, and it is }\textbf{full.} \]
\[ \text{So initially it must've looked something like this.} \]
For reference I intentionally plot the centre of the circle.(https://i.imgur.com/dWqglHj.png)\[ \text{Hence the initial depth is literally just the distance between the}\\ \text{centre of a circle, and any point on its circumference.}\\ \text{Which is of course, the radius, and hence equal to }r. \]
Note therefore that half of the initial depth must be \( \boxed{\frac{r}{2}} \).
\[ \text{Now after doing that rotation, the situation is as follows.}\\ \text{We wish to understand why.} \](https://i.imgur.com/25VhMIl.png)\[ \text{The }r\text{ towards the right should be reasonably clear.}\\ \text{However the red line segment }\textbf{also}\text{ has length }r. \]
\[ \text{This is because }\textbf{any}\text{ line from the centre of the circle to its circumference}\\ \text{has length }r\text{, where }r\text{ is the radius.}\\ \text{Note that the red line segment }\textbf{also}\text{ satisfies this.}\]
It just so happens that the red line is special, in that it's perpendicular to the ground (or whatever surface the container is lying on. And consequently the red line is vertical.
\[ \text{Again, the blue region shaded reflects the water in the bowl.}\\ \text{Note that the depth of the water is then just from the point where the bowl is at the ground}\\ \text{up to the black line, i.e. where the water goes up to.} \]
\[ \text{Since the question instructs us to make this depth equal to }\frac{r}{2}\\ \text{we do so. But then because the red line segment}\\ \text{has length }r\text{, everything above the water to the centre of the circle}\\ \text{must }\textbf{also}\text{ have length }\frac{r}{2}. \]
\[ \text{The last thing to justify is where the }\theta\text{ is placed.}\\ \text{This relies on the fact that }\theta\text{ measures simply what angle the bowl is }\textbf{tilted at}\\ \text{and therefore must be the angle drawn from the horizontal, as required.}\]
Or alternatively, if the diagram they gave was clear enough, you should be able to automatically see why \(\theta\) is there. I literally just copied \(\theta\) across - it was the \(r\) that required more deduction.
\[ \text{With the set-up done, the trigonometry is now easy.}\\ \text{Because we have opposite/hypotenuse here, we just consider}\\ \sin \theta = \frac{\frac{r}{2}}{r} \implies \sin \theta = \frac12 \implies \boxed{\theta = \frac\pi6}. \]
__________________________________________________________
\[ \text{Put simply, part (2) now wishes us to simplify the following fraction:}\\ \frac{\text{Remaining volume left}}{\text{Initial volume}} \]
Because we're given that the water initially fills the entire hemisphere, we just need to compute the volume of said hemisphere. Which is of course half that of the volume of the entire sphere, i.e. \(\frac12 \times \frac{4\pi r^3}{3}\), which simplifies to \( \boxed{\frac{2\pi r^3}{3}} \).
\[ \text{The remaining bit is to make use of part i).}\\ \text{It was more or less up to you to recognise that the 'solid' that the water manifests in}\\ \text{is basically the same as what you do in that volumes question above.} \]
By simply rotating NESA's own diagram by \(90^\circ\) anticlockwise this may be clearer. Focus on the grey shaded area, because that reflects the volume of the water.(https://i.imgur.com/E5MBCfo.png?1)Note that in the diagram in part i), only the top half is shaded. But recall that when you rotate to form your solid, you go through a full \(360^\circ\) rotation about the \(x\)-axis. By doing so, the bottom half of the minor segment does get included as well.
\[ \text{So in our case, the volume of the water is just the answer in part i)}\\ \text{except thanks to (1), we know now that }\theta = \frac\pi6.\]
\[ \text{Thus we get}\\ \begin{align*}V&= \frac{\pi r^3}{3} \left(2-3\sin \frac\pi6 +\sin^3\frac\pi6 \right)\\ &= \frac{\pi r^3}{3} \times \frac{5}{8}\\ &= \frac{5\pi r^3}{24}. \end{align*} \]
\[ \text{Finally subbing back, the required fraction is}\\ \frac{ \frac{5\pi r^3}{24}}{\frac{2\pi r^3}{3}} = \frac{5}{16}. \]
Thank you so much for this lengthy explanation, and especially for rotating the diagram! It made things a lot clearer. I just don't understand the solutions provided:
if depth of water remaining = half original
r-rsintheta = 1/2 r
where did the r-rsintheta come from?
Post by: RuiAce on May 19, 2019, 10:21:11 pm
Thank you so much for this lengthy explanation, and especially for rotating the diagram! It made things a lot clearer. I just don't understand the solutions provided:That's what you get if you don't do what I did and sub \( \frac{r}{2} \) straight in. The length of that red line segment will still be \(r\), but if you choose to work backwards, then you first conclude using \( \sin = \frac{opp}{hyp} \) that the opposite side of that right angle triangle has length \(r\sin \theta\). But this is also just the top part of the red line, so the bottom part of the red line (which corresponds to the water's depth) has length \( r - r\sin\theta\).
if depth of water remaining = half original
r-rsintheta = 1/2 r
where did the r-rsintheta come from?
Post by: Thankunext on May 21, 2019, 12:38:45 pm
Use the trapezoidal rule with 4 subintervals to find an approximation to the volume of the solid formed by rotating the curve y=square root cosx about the x axis from x=0 to x= pi/2.
Answer : 3.1
Post by: Thankunext on May 21, 2019, 12:49:20 pm
A curve has double derivative 18sin3x and a stationary point at (pi/6, -2). Find the equation of the curve.
Answer: y=-2sin3x
Post by: Madison22 on May 21, 2019, 03:11:52 pm
My friend's and I are really stuck on this rates question.
A tap releases liquid A into a tank at the rate of (2+t^2/t+1) litres per minute, where t is time in minutes.
A second tap releases liquid B into the same tank at the rate of (1+1/t+1) litres per minute.
The tapes are opened at the same time and release the liquids into an empty tank.
ii) The taps are closed after 4 minutes. By how many litres is the volume of liquid A greater than the volume of liquid B in the tank when the taps are closed?
Post by: fun_jirachi on May 21, 2019, 05:12:09 pm
Just a friendly reminder to please post any working you have so we can target the areas you're struggling with (unless you have no idea, then at least tell us what you've learned!) It really helps us help you :)
Please help with full working work.
Use the trapezoidal rule with 4 subintervals to find an approximation to the volume of the solid formed by rotating the curve y=square root cosx about the x axis from x=0 to x= pi/2.
Answer : 3.1
With this question, remember that the trapezoidal rule for one application is (f(a)+f((a+b)/2)+f(b)) * (b-a)/2, and with multiple applications it's (b-a)/2 * first + last + 2(everything in between). Here, since your function f(x)=cos1/2 x, your function that you're using for the trapezoidal rule will just be cos x (remembering to square the function prior to doing any work with volumes).
From here, you're given four subintervals ie. the boundaries should be 0, pi/8, pi/4, 3pi/8, pi/2.
Using that, you have the area being roughly equal to:
And also this:
A curve has double derivative 18sin3x and a stationary point at (pi/6, -2). Find the equation of the curve.
Answer: y=-2sin3x
Integrating once, then twice like so:
Substitute the x value given into the equation ie. pi/6 to find that C1 and C2 are zero, since f(x) at that point is equal to -2.
Heyy!
My friend's and I are really stuck on this rates question.
A tap releases liquid A into a tank at the rate of (2+t^2/t+1) litres per minute, where t is time in minutes.
A second tap releases liquid B into the same tank at the rate of (1+1/t+1) litres per minute.
The tapes are opened at the same time and release the liquids into an empty tank.
ii) The taps are closed after 4 minutes. By how many litres is the volume of liquid A greater than the volume of liquid B in the tank when the taps are closed?
Welcome to atarnotes :)
Firstly, I'm not actually sure I'm solving this correctly because I might not have the question down right :(
Right now, they're reading like (given no intuition and worst case scenario)
Please make sure you use parentheses properly so it's clear what the fractions are :) If I've got it down wrong, point it out and I'll happily do it again, but I'm going to be making a few assumptions as I solve this due to that (the process should still be the same though)
Essentially they've made this question a bit too wordy. All they want you to do is integrate the functions given for A and B with respect to t, with bounds 4 minutes and 0 minutes, then find the difference. So for a), then b)
So basically they differ by 2ln5, which you can verify is roughly 3.2L.
Hope this helps :)
Post by: Thankunext on May 21, 2019, 06:50:45 pm
Hey there!
Just a friendly reminder to please post any working you have so we can target the areas you're struggling with (unless you have no idea, then at least tell us what you've learned!) It really helps us help you :)
Integrating once, then twice like so:
Substitute the x value given into the equation ie. pi/6 to find that C1 and C2 are zero, since f(x) at that point is equal to -2.
I wish I could post my working but I dont know how to use the italic text with the integration signs and the like.
I dont really understand the +c part and substituting into where. I got the answer just by integrating twice but dont know what to do with the points given.
Post by: fun_jirachi on May 21, 2019, 08:28:37 pm
With the +C, you have to have a constant when evaluating an indefinite integral. This is because say 4x-4 has the same derivative as 4x, and thus the primitive of 4 'differs by a constant', given as the +C. Since the C is just a constant, we can then integrate that term to get a linear term as seen in the second line (it's like integrating 1, pi, e, 3/2, sqrt2 or any other constant!) That's how you get to the integral in the second line :) You shouldn't in any case be 'just getting the answer' by integrating twice, because you're omitting constants you don't know exist or not without new information. It's so important that you remember +C (not only because of the memes about it) but because of the marks you get docked for forgetting it in the HSC.
As for the substitution, the fact that the point (pi/6, 2) lies on the curve tells you that f(pi/6)=-2. In this case, your f(x)becomes -2sin3x+C1x+C2, and you sub in x=pi/6. By already being given that f(pi/6)=-2, and that result becoming -2+pi/6 * C1+C2, none of the constants should exist to maintain equality with the already known fact. Therefore, C1 and C2 are both zero and the answer is just -2sin3x.
Hope thois helps :)
Post by: kector on May 22, 2019, 07:32:16 pm
Jane’s mother puts $300 into an account
at the beginning of each year to pay for
Jane’s education in 5 years’ time. If 6%
p.a. interest is paid quarterly, how much
money will Jane’s mother have at the
end of the 5 years?
Post by: RuiAce on May 22, 2019, 07:35:06 pm
Hi, not sure what I'm doing wrong, i did P=300, I=0.06 , n=20 , A1= 300(1.06) ---> A20 = 300(1.06)^20Have you covered annuities in class? This question involves repeated deposits that each earn interest, not just a single sum, and thus has to be computed the usual way.
Jane’s mother puts $300 into an account
at the beginning of each year to pay for
Jane’s education in 5 years’ time. If 6%
p.a. interest is paid quarterly, how much
money will Jane’s mother have at the
end of the 5 years?
\begin{align*}A_1 &= 300(1.06)\\ A_2 &= A_1(1.06) + 300(1.06)\\ &= 300(1.06)^2 + 300(1.06)\\ A_3 &= A_2 (1.06) + 300(1.06)\\ &= 300(1.06)^3 + 300(1.06)^2 + 300(1.06)\\ &\vdots\\ A_{20} &= 300(1.06)^{20} + \dots + 300(1.06)^2 + 300(1.06) \end{align*}
Note that the question says "at the beginning of each year".
Post by: kector on May 22, 2019, 07:36:13 pm
Hi, not sure what I'm doing wrong, i did P=300, I=0.06 , n=20 , A1= 300(1.06) ---> A20 = 300(1.06)^20Sorry I am also struggling with this question - thank you!
Jane’s mother puts $300 into an account
at the beginning of each year to pay for
Jane’s education in 5 years’ time. If 6%
p.a. interest is paid quarterly, how much
money will Jane’s mother have at the
end of the 5 years?
27. Scott borrows $200 000 to buy a house.
If the interest is 6% p.a. and the loan is
over 20 years,
(a) how much is each monthly
repayment?
(b) how much does Scott pay altogether?
A farmer borrows $50 000 for farm
machinery at 18% p.a. over 5 years and
makes equal yearly repayments on the
loan at the end of each year.
(a) How much does he owe at the end
of the first year, just before he makes the
first repayment?
(b) How much is each yearly repayment?
Post by: kector on May 22, 2019, 07:38:33 pm
Have you covered annuities in class? This question involves repeated deposits that each earn interest, not just a single sum, and thus has to be computed the usual way.Ah, i see my mistake, thank you, i've covered it in class and finished recently but need revising it again
\begin{align*}A_1 &= 300(1.06)\\ A_2 &= A_1(1.06) + 300(1.06)\\ &= 300(1.06)^2 + 300(1.06)\\ A_3 &= A_2 (1.06) + 300(1.06)\\ &= 300(1.06)^3 + 300(1.06)^2 + 300(1.06)\\ &\vdots\\ A_{20} &= 300(1.06)^{20} + \dots + 300(1.06)^2 + 300(1.06) \end{align*}
Note that the question says "at the beginning of each year".
Post by: fun_jirachi on May 22, 2019, 08:07:55 pm
Part b) is actually a lot easier, once you have the above value, multiply it by 60 for the number of repayments he makes to find the amount he pays.
Try the second one yourself with the same approach, and if you need further help please ask :)
Post by: therese07 on May 23, 2019, 07:20:50 pm
This isn't a question I'm stuck on but for my maths assessment, I had make an exam full of past HSC questions, only allocating 12 marks. My topic I had to do was geometrical applications of differentiation, and I was wondering if you could take a look, and judge whether these questions have a range of bands (2-6). If they don't, please tell me so I can fix it up :)
Thank you!
Post by: RuiAce on May 23, 2019, 07:55:55 pm
Hi There!I would say so. Part a) looks like 2-4 and part b) looks like 3-6
This isn't a question I'm stuck on but for my maths assessment, I had make an exam full of past HSC questions, only allocating 12 marks. My topic I had to do was geometrical applications of differentiation, and I was wondering if you could take a look, and judge whether these questions have a range of bands (2-6). If they don't, please tell me so I can fix it up :)
Thank you!
Post by: kiwii on May 27, 2019, 04:35:15 pm
Just a quick question about probability. When do you do with or without replacement? There were two questions that didn't specify anything, yet one was with replacement and one was without replacement. Does it depend on the context of the question?
Post by: RuiAce on May 27, 2019, 04:45:01 pm
Hello!Usually you can infer it. So yes, depends on context. If you’re still unsure please post the actual relevant questions.
Just a quick question about probability. When do you do with or without replacement? There were two questions that didn't specify anything, yet one was with replacement and one was without replacement. Does it depend on the context of the question?
Post by: kiwii on May 27, 2019, 09:48:29 pm
Usually you can infer it. So yes, depends on context. If you’re still unsure please post the actual relevant questions.Thank you!
Post by: Kombmail on May 27, 2019, 09:48:50 pm
Does this mean I draw a graph of the previous data collected from (b) ?
(b)= find the values of x, x differentiated and x second dfiferentiated after 1 s.
Post by: fun_jirachi on May 27, 2019, 10:10:06 pm
I just did a motion and differentiation question and I came across this: (c) Describe the motion of the body after 1 s.
Does this mean I draw a graph of the previous data collected from (b) ?
(b)= find the values of x, x differentiated and x second dfiferentiated after 1 s.
Not necessarily. What the information gives you is an indication of where the body is (displacement), what direction it's moving in, and how fast it's moving (velocity) and if it's speeding up or slowing down (acceleration). This should be enough info to describe the motion of the particle.
For example if at t=1, x=1m, v=2m/s, a=4m/s/s, the particle is 1m to the right of the origin, travelling in the positive direction at 2m/s, and is speeding up at 4m/s/s in the positive direction. You shouldn't really need a graph if you have that information readily available.
Hope this helps :)
Post by: Kombmail on May 27, 2019, 10:25:27 pm
Not necessarily. What the information gives you is an indication of where the body is (displacement), what direction it's moving in, and how fast it's moving (velocity) and if it's speeding up or slowing down (acceleration). This should be enough info to describe the motion of the particle.
For example if at t=1, x=1m, v=2m/s, a=4m/s/s, the particle is 1m to the right of the origin, travelling in the positive direction at 2m/s, and is speeding up at 4m/s/s in the positive direction. You shouldn't really need a graph if you have that information readily available.
Hope this helps :)
thanks mate!
Post by: Kombmail on May 29, 2019, 09:09:59 pm
Does anyone now how this equals to four positive negative three root five?
Post by: fun_jirachi on May 29, 2019, 09:27:12 pm
I'm not sure where you got these values from, since remembering the quadratic formula is in the form (and the above doesn't really match any quadratic on inspection):
Other than that though, the expansion and simplification of the surd involves the 'identity' that
Technically you should have |a| for the result moving from the second thing to the third, but ignoring that that's basically how the expansion of surds works :)
Post by: Thankunext on May 31, 2019, 08:09:18 pm
How do you draw a tree diagram for this question and solve:
If 4 dice are thrown, find the probability that the dice will have only one 6.
I can work this out manually but not with a tree diagram because I don't know how to draw it.
Post by: RuiAce on May 31, 2019, 08:44:33 pm
Hello, can someone help with this. Many thanks!(https://i.imgur.com/Wc7qZpAl.png)
How do you draw a tree diagram for this question and solve:
If 4 dice are thrown, find the probability that the dice will have only one 6.
I can work this out manually but not with a tree diagram because I don't know how to draw it.
Trying to list out all the options {1,2,3,4,5,6} will probably take forever. At times like these, because we know that only the number 6 is important, we can just squash all the other cases into some dummy variable.
Post by: Kombmail on June 01, 2019, 02:51:16 pm
I did a few factorization questions but I did not understand how you get the answers!
Q1) (x-3)^2 + 5(x-3) = (x-3)(x+2)
Q2) a(a+1)-a(a+1)^2= -(a+1)
Please help me in understanding these someone?
Post by: fun_jirachi on June 01, 2019, 03:37:59 pm
Post by: Kombmail on June 01, 2019, 08:13:23 pm
For Q1 (x-3) is a common factor. Hence, you can write this as (x-3)(x-3+5), which is just (x-3)(x+2). Similarly, for Q2, a(a+1) is a common factor, and thus it can be factorised to become a(a+1)(1-(a+1)), which should be -a^2(a+1), not what you've got there :)
Thanks again!
Post by: Kombmail on June 01, 2019, 10:02:14 pm
Post by: Thankunext on June 02, 2019, 01:53:18 pm
(https://i.imgur.com/Wc7qZpAl.png)Ah I see. Thank you for your diagram!
Trying to list out all the options {1,2,3,4,5,6} will probably take forever. At times like these, because we know that only the number 6 is important, we can just squash all the other cases into some dummy variable.
Post by: Kombmail on June 02, 2019, 03:08:13 pm
Please someone help mehhh!!!
Here is question 7: write the definite integral for the area of the region under the curve y= root 9-x^2 . By using appropriate area formulae, find the value of this area.
Post by: fun_jirachi on June 02, 2019, 03:26:06 pm
Note that this is in fact a semicircle with radius 3, centred at the origin. You can write this as:
But you should probably evaluate the area using the area of a semicircle:
Note: You can't integrate quadratics like that using the reverse chain rule. This only works for linear equations ie. ax+b. Refer to the reference sheet for the appropriate formula! You also shouldn't need to be finding stationary points when integrating!
Hope this helps :)
Post by: Thankunext on June 02, 2019, 03:27:13 pm
There are 34 men and 32 women at a party. Of these, 13 men and 19 women are married. If 2 people are chosen at random, find the probability that
A) both will be men
B) 1 will be a married woman and the other an unmarried man
C) both will be married.
Thank you.
Post by: Kombmail on June 02, 2019, 07:28:29 pm
Hey there!
Note that this is in fact a semicircle with radius 3, centred at the origin. You can write this as:
But you should probably evaluate the area using the area of a semicircle:
Note: You can't integrate quadratics like that using the reverse chain rule. This only works for linear equations ie. ax+b. Refer to the reference sheet for the appropriate formula! You also shouldn't need to be finding stationary points when integrating!
Hope this helps :)
Thanks and yes it does:)
Post by: Kombmail on June 02, 2019, 07:30:47 pm
Post by: RuiAce on June 02, 2019, 07:56:45 pm
Can someone help with drawing a tree diagram for this question? I tried many attempts but dont think I drew it correctly.Note that there is no single correct tree diagram. This tree diagram is quite generic, but is completely overkill for parts A and C. Yet simpler tree diagrams would probably be insufficient for part B.
There are 34 men and 32 women at a party. Of these, 13 men and 19 women are married. If 2 people are chosen at random, find the probability that
A) both will be men
B) 1 will be a married woman and the other an unmarried man
C) both will be married.
Thank you.
(https://i.imgur.com/Pz1kPwm.png)
Can someone explain what it means by limiting velocity pls?It means the velocity as \(t\to \infty\).
Post by: Thankunext on June 03, 2019, 06:00:22 pm
Note that there is no single correct tree diagram. This tree diagram is quite generic, but is completely overkill for parts A and C. Yet simpler tree diagrams would probably be insufficient for part B.Thank you for the tree diagram! Will the HSC ask us to draw a complex one like that? Because it takes quite some time to get the answers.
(https://i.imgur.com/Pz1kPwm.png)It means the velocity as \(t\to \infty\).
Post by: stella_atarnotes on June 03, 2019, 06:03:26 pm
Thank you for the tree diagram! Will the HSC ask us to draw a complex one like that? Because it takes quite some time to get the answers.
They usually won't ask you to draw a tree diagram, and if you do it shouldn't be that hard. Tree diagrams should mainly just be used as part of your working out, and for most questions there would be another way to go about it if you have to draw a tree with too many branches.
Post by: RuiAce on June 03, 2019, 06:30:47 pm
Thank you for the tree diagram! Will the HSC ask us to draw a complex one like that? Because it takes quite some time to get the answers.Will comment that I've never seen a tree diagram be demanded ever. It's purely there as a reference tool (and can grab you marks for working out when done appropriately).
It's heavily advised at the start (because otherwise people get lost too easily), but every probability question can ultimately be done without a tree diagram. But as a general rule of thumb, the more complicated the questions get, the nastier tree diagrams become to draw. Whilst there are some ways to overcome it:
- at some point you just get the idea and cannot be bothered to draw everything else, so you only draw like half of the tree diagram (or some incomplete tree diagram)
- be more selective of what your labels are, constantly adjusting them as appropriate, rather than relying on one generic tree diagram
at the end of the day, they are not necessary for any question regardless of context.
If your teacher enforced the tree diagram for internal assessment, I would say that it's for your own benefit since they're giving you good habits and you should do it. I just would not happen to agree with that teaching philosophy because I know many bright students can just do without. For the final exam, this won't be something you need to worry about.
Post by: therese07 on June 03, 2019, 09:04:23 pm
I'm doing some revision for integration trig functions, and I'm really struggling to do this question!
All help is appreciated!
Post by: fun_jirachi on June 03, 2019, 11:22:54 pm
Remember that the volume when rotated around the x-axis is defined by
You should be able to take it from here :)
Post by: Thankunext on June 04, 2019, 10:51:54 am
The ratio of girls to boys at a school is four to five. Two students are surveyed at random from the school. Find the probability that the students are
A) both boys
B) a girl and a boy
C) at least one girl.
The answers provided were
A) 25/81
B) 40/81
C) 56/81
I had the same problem with most of the questions from this exercise as by common sense I think it is without replacement but maybe I am reading the question wrong??
Post by: RuiAce on June 04, 2019, 11:31:12 am
Hello! I need help with the interpretation of this question as I believe this is without replacement, however the answers suggest that it is with replacement.These questions have an extra subtlety behind them. CSSA papers love doing this but they try to make it a bit clearer.
The ratio of girls to boys at a school is four to five. Two students are surveyed at random from the school. Find the probability that the students are
A) both boys
B) a girl and a boy
C) at least one girl.
The answers provided were
A) 25/81
B) 40/81
C) 56/81
I had the same problem with most of the questions from this exercise as by common sense I think it is without replacement but maybe I am reading the question wrong??
\[ \text{The hidden assumption is that we're considering}\\ \text{a }\textbf{very large}\text{ sample.}\\ \text{Just for the sake of example, let's assume that the school}\\ \text{has 900 students.} \]
\[ \text{Then there would be 400 girls and 500 boys.}\\ \text{Suppose we want the probability of them both being boys.}\\ \text{Then the answer would be }\frac{500}{900}\times \frac{499}{899}. \]
In a similar way, let's say it has 1800 students (an arguably giant school). Then the probability would be \( \frac{1000}{1800} \times \frac{999}{1799}\).
\[ \text{Observe how technically speaking, the probability actually changes}\\ \text{a little when we change the number of students at the school.} \]
\[ \text{However if you plug into the calculator, you'll see that }\frac{499}{899} \approx \frac{500}{900} = 5/9\text{ anyway.}\\ \text{Similarly, }\frac{999}{1799}\approx \frac{1000}{1800} = \frac59. \]
Here's where everyone gets confused. People tend to think that because the ratio is \(4:5\), there should only be \(9\) students. Thus they're lead to think that the required probability should be \( \frac{5}{9} \times \frac{4}{8} \).
When written in English, this makes no sense at all. But it's easy to miss this subtlety in the question because a lot of people forget that a ratio tells us nothing about how big the original sample is!
\[ \text{This is why we usually assume that the sample is very large.}\\ \text{When we do so, we see that after surveying one student}\\ \text{the ratio of remaining students isn't necessarily equal to }4:5\text{ anymore},\\ \text{but is still }\textbf{approximately}\text{ equal to }4:5. \]
\[\text{So we }\textbf{estimate}\text{ our required probability under this assumption.}\\ \text{This is why we arrive at }\frac59 \times \frac59\text{ instead.} \]
Note that these probabilities can never be computed to be exact. But at some point the error in the approximation is just negligible so we assume it doesn't matter. This question wasn't the best in my opinion because I feel uncomfortable assuming schools have 10000+ people, but it still illustrate the idea.
Post by: Kombmail on June 04, 2019, 08:19:38 pm
What should I note rather than revising exponentials and log seperately since I need time to do past papers since that it what I will be tested upon in my exam.
Any suggestions?
Post by: Georgakopoulou on June 04, 2019, 08:57:52 pm
Post by: MB_ on June 05, 2019, 10:17:17 am
Hey! Attached I have a question I came across and had a bit of trouble with. It would be greatly appreciated if someone could help me!\[\begin{align*}\text{The revenue is } R(t) &= 1000000 \cdot 0.5 \cdot (1-e^{-0.04t})\\
&= 500000(1-e^{-0.04t})\\
\text{The cost is } C(t) &= 1000t\\~\\
\therefore \text{ The profit is } P(t) &= 500000(1-e^{-0.04t})-1000t
\end{align*}\]
Now you have to find \(t\) that maximizes profit by differentiating the profit function, setting it equal to \(0\) and solving for \(t\).
Post by: Georgakopoulou on June 05, 2019, 02:17:03 pm
\begin{align*}\text{The revenue is } R(t) &= 1000000 \cdot 0.5 \cdot (1-e^{-0.04t})\\
&= 500000(1-e^{-0.04t})\\
\text{The cost is } C(t) &= 1000t\\~\\
\therefore \text{ The profit is } P(t) &= 500000(1-e^{-0.04t})-1000t
\end{align*}
Now you have to find \(t\) that maximizes profit by differentiating the profit function, setting it equal to \(0\) and solving for \(t\).
Post by: mirakhiralla on June 11, 2019, 10:22:08 pm
Post by: RuiAce on June 11, 2019, 10:24:18 pm
When a q says "using one application of Simpsons rule, find the area", how many function values is that?3.
If \(N\) is the number of applications of Simpson's rule, you'll be using \(2N+1\) function values.
Post by: mirakhiralla on June 11, 2019, 10:25:26 pm
3.
If \(N\) is the number of applications of Simpson's rule, you'll be using \(2N+1\) function values.
is that the same for trapezoidal?
Post by: RuiAce on June 11, 2019, 10:28:00 pm
I haven't seen that wording be used for trapezoidal rule before. It's quite ambiguous, because one application of the trapezoidal rule usually means only one trapezium (i.e. one sub-interval). Yet some may argue it actually requires two sub-intervals as well (i.e. 3 function values) because the formula is written as such.
is that the same for trapezoidal?
Usually, for the trapezoidal rule, the number of sub-intervals is specified instead to make it abundantly clear what is required.
Post by: mirakhiralla on June 11, 2019, 10:31:32 pm
I haven't seen that wording be used for trapezoidal rule before. It's quite ambiguous, because one application of the trapezoidal rule usually means only one trapezium (i.e. one sub-interval). Yet some may argue it actually requires two sub-intervals as well (i.e. 3 function values) because the formula is written as such.
Usually, for the trapezoidal rule, the number of sub-intervals is specified instead to make it abundantly clear what is required.
okay thank you so much
Post by: Kombmail on June 12, 2019, 06:33:20 pm
Post by: MB_ on June 12, 2019, 06:37:48 pm
Guys help pls! If the question is given f(t)=sin2t, fins the rate of change when t=pi/6, I end up with 11pi/6 but the answer is one in the book?Rate of change is \(f'(t)=2\cos(2t)\), at \(t=\frac{\pi}{6}\) then \(f'(\frac{\pi}{6})=2\cos(\frac{\pi}{3})=1\)
Post by: emilyyyyyyy on June 12, 2019, 06:49:46 pm
I need help in where the values of the x-axis line up to the graph in this question (i hope that makes sense):
Sketch the curve y = 4 + 3sin2x for 0<x<2p
Like I know what the graph should look like, just not sure where the x-values line up
Thanks!!
Post by: fun_jirachi on June 12, 2019, 11:39:16 pm
If ever in doubt, always plot some points. With sine curves (and trigonometric curves in general) multiples of pi/2 are always good x-values to test out, as they're usually the x-intercepts or turning points of the curve.
eg.
The curve is also the original sine curve compressed by a factor of two, stretched vertically by a factor of 3, then shifted up by 4 units. If you like, the amplitude gets tripled, the period gets halved and it gets shifted up 4 units. ie. x-values will line up with triple the y-value on the curve sin2x shifted up 4.
I hope this makes some sort of sense :)
Post by: kector on June 17, 2019, 10:01:38 pm
Post by: Abhiram on June 18, 2019, 02:03:10 pm
Sarah borrows $450000 from a bank. The loan is to be repaid in 20 years. The interest rate is 6%p.a. compounded monthly. There is no repayment for the first three months. Let A[n] be the amount owing after n months and M be the monthly repayments.
(i) Find an expression for A[4]
(ii) Show that A[5] = 450000(1.005)^5 - M(1+1.005)
(iii) Find the monthly repayments if the loan is to be repaid in 20 years.
Post by: InnererSchweinehund on June 18, 2019, 02:42:49 pm
Sarah borrows $450000 from a bank. The loan is to be repaid in 20 years. The interest rate is 6%p.a. compounded monthly. There is no repayment for the first three months. Let A[n] be the amount owing after n months and M be the monthly repayments.
(i) Find an expression for A[4]
(ii) Show that A[5] = 450000(1.005)^5 - M(1+1.005)
(iii) Find the monthly repayments if the loan is to be repaid in 20 years.
Hi!
Unfortunately I'm not Jake... but I did do the HSC last year so hopefully my memory is still up to scratch.
I have attached my working and answer to your question below.
Hopefully I got the correct answer ( ??? ) and this can help you out a bit.
If it doesn't, or it's not right, let me know!! :)
Ps. Sorry if the image is bad quality - I had to resize it so it would attach. Also let me know if you need a better one!!
Post by: fun_jirachi on June 18, 2019, 03:34:35 pm
Hi I'm really confused on doing Part III and sketching in general... Thanks for any help. 2004 HSC 7(b) **hopefully the screenshot went thru**
Remember that when you're sketching the acceleration, you're essentially sketching the derivative of velocity ie. f'(x) for the graph you already have. As a start, try looking at the graph between x=0 and x=1 + x=3 and x=5: in these restricted domains the derivative is zero since the velocity is constant. As the velocity becomes more negative, you want to make sure the acceleration is somewhere below the x-axis, and at any points of inflexion there must be a max/min for the acceleration :)
Have a go, and if you need more help ask away :)
Post by: Abhiram on June 18, 2019, 06:30:34 pm
Hi!
Unfortunately I'm not Jake... but I did do the HSC last year so hopefully my memory is still up to scratch.
I have attached my working and answer to your question below.
Hopefully I got the correct answer ( ??? ) and this can help you out a bit.
If it doesn't, or it's not right, let me know!! :)
Ps. Sorry if the image is bad quality - I had to resize it so it would attach. Also let me know if you need a better one!!
thanks for the reply the image is perfect.
Post by: Thankunext on June 19, 2019, 06:21:03 pm
Tim plays a video game 3 times and the probability that he wins at least once is 37/64. What is Tim's probability of winning one game?
Post by: fun_jirachi on June 19, 2019, 06:48:53 pm
Hope this helps :)
EDIT: If P(lose) = x, then P(losing all n games) = xn. Conversely, if P(losing all n games) = x, then P(lose) is the nth root of x. In this case, we take the cube root of 27/64.
Post by: Thankunext on June 19, 2019, 07:17:24 pm
Post by: RuiAce on June 19, 2019, 09:04:05 pm
Sorry but how did you get the 3/4?\[ P(\text{No wins in 3 games}) = [P(\text{Does not win one game})]^3,\\ \text{therefore }[P(\text{Does not win one game})]^3 = \frac{27}{64}. \]
Hence he took cube roots.
Post by: Thankunext on July 08, 2019, 07:20:15 pm
The population of a city is P(t) at any one time. The rate of decline in population is proportional to the population P(t), that is, dP(t)/dt = -kP(t). What will the percentage rate of decline in population be after 10 years?
Post by: Thankunext on July 08, 2019, 09:12:15 pm
If dQ/kQ, prove that Q=Ae^kt satisfies this equation by integrating dQ/dt=kQ.
Post by: Kombmail on July 08, 2019, 11:43:40 pm
Can someone help with this one please. Thanks.I think it’s one of those were you differentiate the first given formula and break it down to the given answer ie. Ae^kt = kAe^kt = k(Ae^kt) = kQ since Q = Ae^kt
If dQ/kQ, prove that Q=Ae^kt satisfies this equation by integrating dQ/dt=kQ.
Post by: LoneWolf on July 11, 2019, 05:49:56 am
Can you help me understand why the second derivative of dy/dx is d^2y/dx^2 not dy^2/dx^2
Im in year 11 and doing a journal assessment! :D
Post by: blyatman on July 11, 2019, 10:51:27 am
Post by: Thankunext on July 11, 2019, 02:39:05 pm
A particle is moving in a straight line so that its displacement x cm over time t seconds is given by x = t [square root (49-t^2)].
A) For how many seconds does the particle travel? Answer: 7s
B) How far does the particle move altogether? Answer: 49cm
Post by: LoneWolf on July 11, 2019, 08:34:43 pm
It's just mathematical convention for the notation of the second derivative. dx^2 is literally the square of the differential. So dy^2/dx^2 would actually be (dy/dx)^2, which is the square of the first derivative. This is not the same as the second derivative, which is completely different. As a result, the notation d^2y/dx^2 is used to avoid confusion.
Thank you Blyatman
Appreciate your help
Post by: Minivasili on July 12, 2019, 11:32:26 am
Post by: fun_jirachi on July 12, 2019, 01:27:42 pm
Can someone please help with this motion question?
A particle is moving in a straight line so that its displacement x cm over time t seconds is given by x = t [square root (49-t^2)].
A) For how many seconds does the particle travel? Answer: 7s
B) How far does the particle move altogether? Answer: 49cm
It's pretty clear from the square root function that t must be strictly less than or equal to 7, and since t cannot be negative, the particle is in motion between t=0 and t=7, hence it is in motion for 7 seconds.
For the second question, we find that:
Sub that value of t back into the original equation, and you'll find the particle travels there and back to the origin for a total of 49cm.
Hey guys, this multi has me stuck i cant work it out. Help is much appreciated. Thanks :)
As soon as you see an at least question for probability, your mind should instantly think 'how can I use the complement?' ie. (1-P(opposite of situation)).
Here, the complement of at least one movie ticket is going to be no movie tickets. The only way Darren can get no movie tickets is if he doesn't win at all ie. from both prizes, he doesn't win. The chance of this happening is 15/20 * 14/19, since the tickets are implied to not be replaced (and they never are in prize draws anyway.) Therefore, the chance of him winning at least one ticket is 1 minus this value, which is 17/38, hence the answer should be D.
Post by: fiafishsauce on July 13, 2019, 01:31:09 pm
Rachel starts working for a business at the beginning of 2005. If she retires at the end of 2034, how much superannuation will she have if $1000 is invested at the beginning of each year at 9.5% p.a.?
Post by: r1ckworthy on July 13, 2019, 01:43:27 pm
Hi! For this question, I keep receiving 148 687.50 when the answers say $163 907.81. Where am I going wrong?
Rachel starts working for a business at the beginning of 2005. If she retires at the end of 2034, how much superannuation will she have if $1000 is invested at the beginning of each year at 9.5% p.a.?
Hey! This question has a tiny little trick to it! Great job on finding $148 687.50. Everything you have done is correct except the total number of years. Look at the bit where the questions mentions the number of years:
Quote
If she retires at the end of 2034...That means that we must account for an extra year as well. This would mean that n=30, not n=29. We would get the answer $163 907.81 if we use n=30. I thought of this first but then dismissed it, and then got the same answer as you. When I used n=30, I got the correct answer.
Hopefully that helps!
Post by: Shiv3n on July 16, 2019, 07:58:43 pm
Post by: redpanda83 on July 16, 2019, 08:07:27 pm
Hi guys, I'm having trouble with answering this question 16, quite unsure where to start even(a) pretty much try substituting those two quardinates into the equation - make two simultaneous equation and then solve. Try doing this for now!
Post by: Shiv3n on July 16, 2019, 09:11:35 pm
(a) pretty much try substituting those two quardinates into the equation - make two simultaneous equation and then solve. Try doing this for now!
Ahhh, yup after doing that, I was able to work thorugh the rest of the question with ease. Thank you!
Post by: redpanda83 on July 16, 2019, 09:23:05 pm
Ahhh, yup after doing that, I was able to work thorugh the rest of the question with ease. Thank you!Your Welcome
Post by: therese07 on July 17, 2019, 01:44:44 pm
I was just wondering to anyone who's doing CSSA or has done CSSA in the past, what trial papers would be the best to study for? I'm currently doing the James Ruse and the Baulkham Hills trial papers, but I was wondering is there any schools trial papers that may seem redundant to study for trials? as in its level of difficulty would never be in the CSSA?
Thank you!!
Post by: emilyyyyyyy on July 19, 2019, 11:16:27 am
A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.
(a) What is the amount of each instalment?
The Smith family buys a car for $38 000, paying a 10% deposit and taking a loan out for the balance. if the loan is over 5yrs with interest of 1.5% monthly, find amount of each monthly repayment.
--> with this question, I took away the amount paid for the deposit and then answered how I normally would, but I still couldn't get the answer.
thanks! :)
Post by: david.wang28 on July 19, 2019, 01:11:53 pm
Can someone please help me w these 2 loan repayment questions:For a), make sure you divide 0.15 by 12, as the question states 'monthly' (12 months in 1 year). Let P = 6000, let M be instalments, and let An be amount owing (equals zero if all is paid back). Now An = P*1.0125^n - M(1.0125^n - 1)/0.0125, then do some simple algebraic manipulation and sub in An = 0, n = 60 months, P = 6000. You should get $142.74.
A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.
(a) What is the amount of each instalment?
The Smith family buys a car for $38 000, paying a 10% deposit and taking a loan out for the balance. if the loan is over 5yrs with interest of 1.5% monthly, find amount of each monthly repayment.
--> with this question, I took away the amount paid for the deposit and then answered how I normally would, but I still couldn't get the answer.
thanks! :)
For b), don't take away the amount paid for the deposit. Take P = 38000 - (38000*0.1) = 34200, and repeat the same process as for a), except for the fact that you are finding M. Have a go, and ask for more help if you need it :)
Post by: RuiAce on July 19, 2019, 04:21:21 pm
Hi there!CSSA past papers tend to have considerable difficulty compared to most past papers. Among selective papers, it's generally regarded that Sydney Grammar offers the hardest of the lot, so you can probably focus in that direction.
I was just wondering to anyone who's doing CSSA or has done CSSA in the past, what trial papers would be the best to study for? I'm currently doing the James Ruse and the Baulkham Hills trial papers, but I was wondering is there any schools trial papers that may seem redundant to study for trials? as in its level of difficulty would never be in the CSSA?
Thank you!!
I doubt any paper would be "redundant" necessarily speaking, but it is interesting to note that despite having reasonable difficulty, James Ruse papers aren't actually the hardest
(Note that you can always start by asking your school if they have past papers they can offer you first. But presumably you've already done that)
Post by: boulos on July 19, 2019, 09:54:21 pm
Post by: Sine on July 19, 2019, 10:00:44 pm
Hey there, i am unsure on what i'm doing wrong in the second part of this question, and the working out is confusing me. I keep getting the final answer as -1 but it's not. Hope you can helpWhat particular aspect of this question is troubling you?
The "hence" in the second part pretty much directly tells you that you need to use something from the first part to solve it. Hopefully, you can recognise it is integration by recognition.
Post by: Kombmail on July 19, 2019, 11:00:32 pm
I just did a question from the 2006 paper 7a and didnt understand how the answer was three?
The question was to let 'a' and' b' be the solutions of x squared minus 3x +1= 0 and find a + 1/a
this was a stepwise question with part a being find the product of the roots which equals to 1 when i solved it.
I tried solving it a rearranged it to a plus b /ab divided by 1/b.
This further equals 3/1 divided by 1/b and b becomes equal to three which was the correct answer.
Is this the right process?
Post by: fun_jirachi on July 20, 2019, 12:24:00 am
Hey Guys!
I just did a question from the 2006 paper 7a and didnt understand how the answer was three?
The question was to let 'a' and' b' be the solutions of x squared minus 3x +1= 0 and find a + 1/a
this was a stepwise question with part a being find the product of the roots which equals to 1 when i solved it.
I tried solving it a rearranged it to a plus b /ab divided by 1/b.
This further equals 3/1 divided by 1/b and b becomes equal to three which was the correct answer.
Is this the right process?
I'm not quite sure what you've done there, seems a tad dodgy on inspection (and tbh got me a bit confused).
An easier method (by easier I mean less convoluted) would be to note that the roots a and b satisfy the following set of equations (by sum and product of roots):
a+b=3 (1)
ab=1 (2)
From (2), note that b=1/a. Substituting this into (1), we get that a+1/a=3.
The fact that the first part forced you to find the product of roots indicates you should use that result in some way, which should be the case for all 'step-wise questions'.
Hope this helps :)
Post by: boulos on July 20, 2019, 10:50:04 am
What particular aspect of this question is troubling you?
The "hence" in the second part pretty much directly tells you that you need to use something from the first part to solve it. Hopefully, you can recognise it is integration by recognition.
Yea i recognised it was integration, I rearranged to get the integration of xe^-x, but i am some how not getting the answer. And the working out is pretty confusing.
Post by: fun_jirachi on July 20, 2019, 11:11:20 am
From part a),
Post by: boulos on July 20, 2019, 11:52:00 am
As with many previous posts, the key here is that 'every derivative gives us another integral' and vice versa. If a question asks us to differentiate something, then asks us to integrate a similar function to the derivative, it's pretty much asking us to manipulate the integral and use the result in some way.
From part a),
Got it, saw where i went wrong. Thanks mate
Post by: LoneWolf on July 22, 2019, 07:32:45 am
Obscure question!!!
Year 11 student posting 'ere seeking a brief explanation from a 'senior' on the chain rule and why it is integral to mathematics, is applicable to real life and why it is worth learning!*
*For a mathematics journal i am keeping as part of an assessment!
Post by: RuiAce on July 22, 2019, 09:58:19 am
Hi PPlWithout spoiling your assignment too much, in my opinion the main benefit of the chain rule lies in a certain concept that happens to be taught in MX1. It is the concept of analysing rates of change, among variables that are somehow related.
Obscure question!!!
Year 11 student posting 'ere seeking a brief explanation from a 'senior' on the chain rule and why it is integral to mathematics, is applicable to real life and why it is worth learning!*
*For a mathematics journal i am keeping as part of an assessment!
The chain rule says that \( \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\). So if you want to analyse how \(y\) changes with respect to \(x\), but you know stuff about \(u\)....
Post by: LoneWolf on July 22, 2019, 10:17:17 am
I would be glad of any other contributions from anyone!!!!
It all helps!
Post by: blyatman on July 22, 2019, 10:25:08 am
Hi PPl
Obscure question!!!
Year 11 student posting 'ere seeking a brief explanation from a 'senior' on the chain rule and why it is integral to mathematics, is applicable to real life and why it is worth learning!*
*For a mathematics journal i am keeping as part of an assessment!
Hi there, engineer here.
In the real world, everything is measured in rates of change, and we can make various predictions based on these rates. For example, suppose you drop a ball from a cliff, and you want to know where the ball will be after a certain time. To calculate this, you need to know the forces are acting on the ball (which will tell you how the speed of the ball is changing). This is, of course, it's weight, which is the force due to gravity. Now, the force acting on an objective is equal to it's RATE OF CHANGE of momentum with respect to time. The fact that it's a rate of change allows us to use elementary calculus to determine the position of the ball as a function of time.
Basic differentiation provides us with the tools to calculate these rates of change. However, unlike the idealised problems you deal with in high school calculus, problems in the real world are multivariable, meaning that they depend on a multitude of variables. In the above example, other factors that come into play are air resistance, which depends on the density of the air as well as the geometry of the ball. If you were to drop the ball from a significant altitude, the density of the air would change as the ball falls through the earth's atmosphere, which would in turn change the air resistance. Thus, you need to take into account the rate of change of numerous variables. The chain rule allows us to calculate the rate of change of a function that depends on numerous variables.
Post by: KingTings on July 22, 2019, 01:01:24 pm
Answer is C.
Post by: annabeljxde on July 22, 2019, 01:23:11 pm
I have a question from the 2018 CSSA Trials!
Answer is C.
I'm not too sure if this is the proper way to go about the question, but this is how I would end up with the answer C:
Since f(-x) = -f(x), this means that f(x) has odd symmetry. On the other hand, if f(x)=f(-x), the graph would have even symmetry.
I would recommend doing a rough sketch if you're still unsure, but basically, since (3,7) is a maximum turning point, there will be a minimum turning point at (-3,-7), which is symmetrical to the maximum but negative. (If the condition was even symmetry, there would be a maximum turning point again at (-3,7) )
Hopefully this makes some sort of sense... If not, let me know and I'll demonstrate with a diagram :)
Post by: KingTings on July 22, 2019, 04:41:43 pm
Thanks so much!
And if it was f(x) = f(-x) there would be a max turning point at (-3,7)??
Post by: RuiAce on July 22, 2019, 04:43:39 pm
Ahhhh I totally get it.Correct, because even function symmetry is about the \(y\)-axis, so it's just a reflection.
Thanks so much!
And if it was f(x) = f(-x) there would be a max turning point at (-3,7)??
Post by: LoneWolf on July 23, 2019, 12:28:07 pm
I am learning the product rule atm however i cannot solve this question:
-2(x^2+x+1)^3 x.
Post by: RuiAce on July 23, 2019, 12:37:32 pm
Hey People,\begin{align*}
I am learning the product rule atm however i cannot solve this question:
-2(x^2+x+1)^3 x.
u&=x^2+x+1\\
u^\prime &= 2x+1\\
v &= x\\
v^\prime &= 1
\end{align*}
\begin{align*}
\therefore \frac{d}{dx} -2(x^2+x+1)^3 x &= -2 \frac{d}{dx} (x^2+x+1)^3 x\\
&= -2 \left[x (2x+1) + (x^2+x+1) \right]
\end{align*}
You could probably simplify this further.
Post by: LoneWolf on July 23, 2019, 01:40:36 pm
Post by: benneale on August 03, 2019, 03:03:48 pm
A little confused as to how sinxcosx all over cos^2x is equal to tanx... can u show me like the proof or something?? thanks!
Post by: Jakeybaby on August 03, 2019, 03:22:52 pm
Hey guys!
A little confused as to how sinxcosx all over cos^2x is equal to tanx... can u show me like the proof or something?? thanks!
Some useful things to keep in mind!
Use these 2 pieces of information to progress through the proof - let me know how you go!
Post by: Kombmail on August 05, 2019, 06:56:15 pm
Integrate 2/(5-x)^4dx ?
Ps any tips for index law help?
Post by: fun_jirachi on August 05, 2019, 07:58:26 pm
All there is to the index law is on your reference sheet.
For reference, this is what it has:
Hope this helps :)
Post by: mirakhiralla on August 06, 2019, 03:35:41 pm
Without finding the point of intersection, find the equation of the line which passes through the point of intersection of 7x+3y-13=0 and 3x+2y-12=0, and also passes through D(1.5, -1)
Post by: emmajb37 on August 07, 2019, 11:06:33 am
Find the value of p, where x=p is a vertical line that divides the area between y=√x, x=16 and the x-axis into 2 equal parts.
Thanks in advance :)
Post by: DrDusk on August 07, 2019, 11:58:07 am
Hi, with this question I understand the concept but I cannot seem to simplify the equation and the working of the answers isn't helping so i would love an explanation.
Find the value of p, where x=p is a vertical line that divides the area between y=√x, x=16 and the x-axis into 2 equal parts.
Thanks in advance :)
Solve for p from there because you know what the total area is from the first integral
Post by: Coolmate on August 07, 2019, 08:00:15 pm
I am having a bit of trouble with these exponential equations (see attached) and was wondering if anyone could please step me through how to do them?
Also, do you times the front and the end numbers (outside of the brackets) together? Or do you just leave them?
Thankyou!
Coolmate :D👍
Post by: RuiAce on August 09, 2019, 09:30:06 am
please help with this question:If you're not allowed to do it the classic way, then you more or less have no choice but to use the mythical 'k'-method. Incredibly disgusting, but again, no choice.
Without finding the point of intersection, find the equation of the line which passes through the point of intersection of 7x+3y-13=0 and 3x+2y-12=0, and also passes through D(1.5, -1)
\[ \text{The equation will take the form}\\ 7x+3y-13 + k(3x+2y-12) = 0. \]
\[\text{If the point }D\left( \frac32, -1\right) \text{lies on the line, then upon substituting in we obtain}\\ 7(1.5)+3(-1)-13 + k(3(1.5)+2(-1)-12) = 0 \implies \boxed{k=-\frac{11}{19}}.\]
\[ \text{Hence the equation of the line will be}\\ 7x+3y-13 - \frac{11}{19}(3x+2y-12)=0\\ \text{which you can multiply by 19, then expand and simplify.} \]
Post by: RuiAce on August 09, 2019, 09:35:41 am
Hello everyone,You haven't specified what you want me to do for question j.
I am having a bit of trouble with these exponential equations (see attached) and was wondering if anyone could please step me through how to do them?
Also, do you times the front and the end numbers (outside of the brackets) together? Or do you just leave them?
Thankyou!
Coolmate :D👍
With that sketch in question a, you should first note that \( f(x)=a^x\) has asymptote \(y=0\). So in your case, \(f(x) =4(3^x) + 1\) has asymptote \(y=1\). (Which can be properly verified - as \(x\to \infty\), it is clear that there is no asymptote. But as \(x\to -\infty\), we have \(4(3^x)+1 \to 0 + 1 = 1\).
Now the \(4\) gets multiplied to the entire exponential term, i.e. the \(3^x\) bit. So it forms some kind of a stretch (dilation) to just the exponential - it doesn't move the asymptote around. As an example, your \(y\)-intercept is now at \(y=4(1)+1 = 5\), instead of at \(y=2\).
In general, graphically showing the extra factor in front isn't exactly possible. Some points need to be plotted (and the graph drawn to scale) for it to have a noticeable impact. But if you try plotting on Desmos/GeoGebra, it will be 100% clear that some kind of stretching is happening.
Post by: Coolmate on August 09, 2019, 01:05:34 pm
I have re looked over the question and it makes much more sense now, thanks for the thorough explanation! ;D
With regards to Question j I have attached the original question of what to do (Sorry)
Thanks Again!
Coolmate 8)
Post by: fun_jirachi on August 09, 2019, 05:30:44 pm
- Asymptotes, limiting value as x approaches positive and negative infinity
In this case, the horizontal asymptote is at x=4, and as x approaches positive infinity y approaches 4 from below.
- Dilation/Shift to the right/left, orientation
The dilation is likewise pretty hard to see, but basically, note that from the two negative signs (or by subbing in numbers) that the exponential will tend towards negative infinity as y approaches negative infinity.
Hope this helps :)
Post by: Coolmate on August 10, 2019, 01:19:34 pm
You should try following a similar process to what Rui previously outlined:
- Asymptotes, limiting value as x approaches positive and negative infinity
In this case, the horizontal asymptote is at x=4, and as x approaches positive infinity y approaches 4 from below.
- Dilation/Shift to the right/left, orientation
The dilation is likewise pretty hard to see, but basically, note that from the two negative signs (or by subbing in numbers) that the exponential will tend towards negative infinity as y approaches negative infinity.
Hope this helps :)
Hi fun_jirachi!
This was extremely helpful and I understood the question perfectly, thankyou for your help!
Coolmate 8)
Post by: LoneWolf on August 13, 2019, 11:44:50 am
Bit confused over logs atm (in year 11)
the question says:
"Find which two integers each expression lies between"
Log2(50) and the answers say 5 & 6 can someone explain this please!
Post by: LoneWolf on August 13, 2019, 11:49:54 am
Post by: Coolmate on August 23, 2019, 10:16:22 pm
Would someone please be able to help me with these questions based on Logarithms, b,e, and h?(Attached) I have no idea about how to go about answering them :-\ --> and also explain a bit about natural logs compared against normal logs?
Cheers,
Coolmate 8)
Post by: RuiAce on August 23, 2019, 10:31:40 pm
Hello Everyone!Your calculator should have a button that can compute all of those for you. And then you just round it.
Would someone please be able to help me with these questions based on Logarithms, b,e, and h?(Attached) I have no idea about how to go about answering them :-\ --> and also explain a bit about natural logs compared against normal logs?
Cheers,
Coolmate 8)
The "natural" logarithm is just a special name we give to the base \(e\) logarithm, where \(e\) is this fancy number (Euler's number) that behaves like \(\pi\) with its weird decimals and \(e\approx 2.718281828459045\).
Saying \(y = \ln x\) is the exact same thing as saying \(y = \log_e x\).
Post by: Coolmate on August 23, 2019, 11:00:02 pm
Your calculator should have a button that can compute all of those for you. And then you just round it.
The "natural" logarithm is just a special name we give to the base \(e\) logarithm, where \(e\) is this fancy number (Euler's number) that behaves like \(\pi\) with its weird decimals and \(e\approx 2.718281828459045\).
Saying \(y = \ln x\) is the exact same thing as saying \(y = \log_e x\).
Hey Rui!
Thankyou! I just checked with the calculator and it does have the button! ;D
Also, just clarifying; so ln is essentially just exactly written as loge? :)
Cheers,
Coolmate 8)
Post by: RuiAce on August 23, 2019, 11:01:08 pm
Hey Rui!Awesome :)
Thankyou! I just checked with the calculator and it does have the button! ;D
Also, just clarifying; so ln is essentially just exactly written as loge? :)
Cheers,
Coolmate 8)
And yep
Post by: Coolmate on August 23, 2019, 11:04:35 pm
Cheers,
Coolmate 8)
Post by: mirakhiralla on August 29, 2019, 10:47:30 pm
In the Jackpot Lottery, the probability of the Jackpot prize being won in any draw is approximately 1 in 50.
i) What is the probability that the jackpot prize will be won in each of the three consecutive draws?
ii) How many consecutive draws must be made for it to be 99% certain that a Jackpot prize will have been won?
Post by: DrDusk on August 29, 2019, 11:36:48 pm
Hey sorry I need help in ii
In the Jackpot Lottery, the probability of the Jackpot prize being won in any draw is approximately 1 in 50.
i) What is the probability that the jackpot prize will be won in each of the three consecutive draws?
ii) How many consecutive draws must be made for it to be 99% certain that a Jackpot prize will have been won?
Post by: RuiAce on August 30, 2019, 10:14:21 am
Hey sorry I need help in iiBasically see above for the answer. Just want to provide a small remark for extra intuition.
In the Jackpot Lottery, the probability of the Jackpot prize being won in any draw is approximately 1 in 50.
i) What is the probability that the jackpot prize will be won in each of the three consecutive draws?
ii) How many consecutive draws must be made for it to be 99% certain that a Jackpot prize will have been won?
The question is probably doable directly, but it would probably be messy. Because you only require the probability that it is won at least one out of the first \(n\) draws, you get different results depending on if it's won exactly once, twice, three times, all the way up to \(n\) times.
And in 2U maths, of course we learn that the complement is a natural way to navigate around the "at least" issue wherever possible. The complement is when you don't win it at all, which you know can only happen one possible way. (Namely, it is never won.)
So this probability will be \( \left( \frac{49}{50} \right)^n \), and hence what we require is \( 1 - \left( \frac{49}{50} \right)^n = 0.99 \). Which of course, becomes what was computed above.
-snip-You can use \times for \(\times\) and you should use \ln for \(\ln\) for better LaTeX in the future
Post by: Kombmail on September 09, 2019, 06:17:46 pm
becomes d=e?
Post by: RuiAce on September 09, 2019, 06:19:45 pm
does anyone know how Ln(1)= Ln(d)Except it doesn’t, so somewhere in there is a mistake.
becomes d=e?
\( \ln d=1\) becomes \(d=e\).
Post by: Coolmate on September 11, 2019, 09:58:09 pm
I am in Year 11 revising Calculus for my prelims and was wondering if someone would be able to step through how to do the Product Rule with these questions below (attached). I am confused with 'a', 'h', and 'i'.
Thanks in advance!
Coolmate 8)
Post by: RuiAce on September 11, 2019, 10:04:28 pm
Hi Everyone! ;DI'll only do i) for now. For the rest, if you have further trouble please post up any progress.
I am in Year 11 revising Calculus for my prelims and was wondering if someone would be able to step through how to do the Product Rule with these questions below (attached). I am confused with 'a', 'h', and 'i'.
Thanks in advance!
Coolmate 8)
\begin{align*}
u &= x+1 &&& v &= (2x+5)^4\\
u^\prime &= 1 &&& v^\prime &= 8(2x+5)^3
\end{align*}
Note that the chain rule was required to obtain \( v^\prime\). The "inner" function was \(2x+5\) and the outer function was \((...)^4\).
\begin{align*}
\frac{dy}{dx} &= vu^\prime + u v^\prime\\
&= (2x+5)^4 + 8(2x+5)^3(x+1)\\
&= (2x+5)^3\left[ (2x+5) + 8(x+1) \right] \\
&= (2x+5)^3 (10x+13)
\end{align*}
Post by: fun_jirachi on September 11, 2019, 10:08:23 pm
(Alternatively, expand and then differentiate by the power rule (which is what is simpler in an exam), but that's clearly not the point of the exercise :) )
Hope this helps :）
Post by: Coolmate on September 11, 2019, 10:16:43 pm
Your explanation was very clear and I understood it!
Thanks also, fun_jirachi! ;D ;D That is a good point!
Post by: annabeljxde on September 11, 2019, 11:02:37 pm
Post by: fun_jirachi on September 11, 2019, 11:10:09 pm
Hope this helps :)
Post by: RuiAce on September 11, 2019, 11:35:48 pm
Need help with working out this question:Here's a visual aid to go with fun_jirachi's explanation.
(https://i.imgur.com/I8EpE5y.png)
The idea is that ultimately, no matter what it's going to be a parabola. So it's easier to just rotate the paper until it's oriented the same way as an \(x^2=4ay\) parabola. (Note that the vertex was chosen to be \((0,0)\) for the sake of convenience - you could try a different vertex and find that you'll get the same answer, but it's a needless trek.)
Post by: dani01 on October 01, 2019, 11:13:52 am
The concentration of salt in Liquid A is 20g per 100 mL and the concentration of salt in Liquid B is 40g per 100 mL. What is the concentration of salt when 250mL of Liquid A is mixed with 250mL of liquid B.
Thanks!
Post by: RuiAce on October 01, 2019, 11:26:04 am
I know this is the 2u thread but moderator hasn't posted on standard thread for a quite a bit so I'll just ask on here.(Note: Answers from a non-moderator don't mean they are not credible :). I just don't like providing answers when they're about matters I am not confident on.)
The concentration of salt in Liquid A is 20g per 100 mL and the concentration of salt in Liquid B is 40g per 100 mL. What is the concentration of salt when 250mL of Liquid A is mixed with 250mL of liquid B.
Thanks!
New volume: 250mL + 250mL = 500mL
Original masses:
250mL of Liquid A will have (20/100) * 250 = 50g of salt.
250mL of Liquid B will have (40/100) * 250 = 100g of salt.
New mass of salt = 50g + 100g = 150g.
New concentration: 150/500 = 30/100. Hence 30g per 100mL.
Post by: Raissa on October 01, 2019, 10:09:48 pm
Post by: RuiAce on October 01, 2019, 10:13:25 pm
Hey I'm really stuck on question 15. c) (ii) HSC 2013. I looked at the solutions but still didn't understand how they got the answer.You can find a solution for this in the compilation.
(You're welcome to ask for more clarification in any further areas you have trouble with.)
Post by: therese07 on October 02, 2019, 11:00:25 pm
As hsc maths is coming up, I wanted to ask, besides doing past hsc papers, would it be worth doing trial questions?? I know trials are harder than hsc, but has there ever been instances where a maths trial question and it’s difficulty been on a hsc paper? Especially in question 16?
Thank you!!
Post by: InnererSchweinehund on October 04, 2019, 07:55:13 pm
Hey,
As hsc maths is coming up, I wanted to ask, besides doing past hsc papers, would it be worth doing trial questions?? I know trials are harder than hsc, but has there ever been instances where a maths trial question and it’s difficulty been on a hsc paper? Especially in question 16?
Thank you!!
Hi therese07!
It's definitely worth continuing to do trial papers in the lead up to the HSC.
I sat the Mathematics HSC last year and definitely found that the trial papers were really beneficial for preparing me for the actual HSC, especially the later questions (Q14 onwards).
I honestly believe that if I hadn't done so many trial papers, I wouldn't have done as well as I did.
Good luck!!
8)
Post by: spnmox on October 07, 2019, 12:48:56 am
The line 3x − 4y + 3 = 0 is a tangent to a circle with centre (3, −2).
What is the equation of the circle?
A runner has four different pairs of shoes.
If two shoes are selected at random, what is the probability that they will be a matching pair?
A radio telescope has a parabolic dish. The width of the opening is 24 m and the distance
along the axis from the vertex to the opening is 4 m, as shown in the diagram. (attached)
What is the focal length of the parabola?
Post by: fun_jirachi on October 07, 2019, 12:15:08 pm
hey guys, a few questions from 2018 mc:
The line 3x − 4y + 3 = 0 is a tangent to a circle with centre (3, −2).
What is the equation of the circle?
A runner has four different pairs of shoes.
If two shoes are selected at random, what is the probability that they will be a matching pair?
A radio telescope has a parabolic dish. The width of the opening is 24 m and the distance
along the axis from the vertex to the opening is 4 m, as shown in the diagram. (attached)
What is the focal length of the parabola?
Hey there!
If a line is tangent to a circle, we know that the distance from the centre to the point of contact is the radius, which meets the tangent at a right angle. Basically, we substitute (3, -2) into the perpendicular distance formula, then use the equation of a circle (x-h)2+(y-k)2=r2 to find the answer (given the centre (h, k)).
When picking the shoes, note that when we pick the first shoe we don't actually care what shoe it is; we just care that the second shoe we pick matches the first one. Hence, once we pick the first shoe, there's only one shoe that matches out of the remaining seven ie. the answer is going to be 1/7.
The last one was addressed already further up the page! (about 5 messages or so up :) )
Hope this helps :)
Post by: spnmox on October 07, 2019, 12:51:28 pm
Hey there!
If a line is tangent to a circle, we know that the distance from the centre to the point of contact is the radius, which meets the tangent at a right angle. Basically, we substitute (3, -2) into the perpendicular distance formula, then use the equation of a circle (x-h)2+(y-k)2=r2 to find the answer (given the centre (h, k)).
When picking the shoes, note that when we pick the first shoe we don't actually care what shoe it is; we just care that the second shoe we pick matches the first one. Hence, once we pick the first shoe, there's only one shoe that matches out of the remaining seven ie. the answer is going to be 1/7.
The last one was addressed already further up the page! (about 5 messages or so up :) )
Hope this helps :)
oh, i see haha. thanks very much!
Post by: spnmox on October 08, 2019, 02:50:42 am
John has 10 marbles in a bag, 3 of which are green and the rest are yellow. He randomly draws 1 out of
the bag, notes its colour and then puts it back before drawing another out of the bag. What is the
probability that both marbles drawn are the same colour?
is the second one 0.58?? haha i'm so unsure with probability questions
Post by: InnererSchweinehund on October 08, 2019, 08:23:14 am
The quadrilateral formed by (1, 2), (3, 3), (1, 7), (-1, 3) is ?
John has 10 marbles in a bag, 3 of which are green and the rest are yellow. He randomly draws 1 out of
the bag, notes its colour and then puts it back before drawing another out of the bag. What is the
probability that both marbles drawn are the same colour?
is the second one 0.58?? haha i'm so unsure with probability questions
Hi!
For the quadrilateral question, I would start by drawing a cartesian plane and plotting the points to create the shape. see first image attached
From this I got the impression that the quadrilateral is a kite, but you would look to prove that by showing that the diagonals of the kite meet at a right angle (I'm not 100% sure but I think you can do this by showing that the gradients of the two diagonals, when multiplied together equal -1)
For the probability question, I used a tree diagram to clearly step it out and show everything I needed to include
It's important to remember that this question uses replacement
see the second image attached
Hope this helps!!
Post by: spnmox on October 08, 2019, 11:53:40 am
Hi!
For the quadrilateral question, I would start by drawing a cartesian plane and plotting the points to create the shape. see first image attached
From this I got the impression that the quadrilateral is a kite, but you would look to prove that by showing that the diagonals of the kite meet at a right angle (I'm not 100% sure but I think you can do this by showing that the gradients of the two diagonals, when multiplied together equal -1)
For the probability question, I used a tree diagram to clearly step it out and show everything I needed to include
It's important to remember that this question uses replacement
see the second image attached
Hope this helps!!
thank you so much!!
Post by: spnmox on October 08, 2019, 01:11:46 pm
with 4 suits (hearts, diamonds, spades
and clubs). Each suit has 13 cards,
consisting of an ace, cards numbered
from 2 to 10, a jack, queen and king.
If a person is dealt 5 cards find the
probability of getting a flush (all cards the same suit)
Sorry for weird formatting!
A squad of 8 is chosen at random from 3 baseball teams with 10 players in each team. Find the probability that Joe from the B team and Dan from the A team will be chosen.
I saw that the answer is 8/30 * 7/29. Isn't that the probability of choosing any 2 people - should we specify that it is Joe and Dan, or does it not matter?
Post by: fun_jirachi on October 08, 2019, 02:36:23 pm
For the first question, first note that the total number of ways of getting a 5 card hand from 52 cards is equal to (52x51x50x49x48)/5!. The numerator should make sense (tell me if it doesn't!) and dividing by 5! removes order ie. picking an Ace, 2, 3, 4, 5 of diamonds is the same as picking a 2, Ace, 3, 4, 5 of diamonds in that order. Getting a flush means you're limiting yourself to a certain suit, and then picking 5 cards in a similar way; ie. the number of ways will be (13x12x11x10x9)/5!, for much the same reasons as the total number of ways. Since there are four suits, we multiply this value by four, and put it all over the total number of ways
For the second question, that's not the probability of choosing any two people; rather, it's the probability of choosing any two PARTICULAR people. What this means is that we pick the people first, then choose the team, not choose the team and check if those people are there. By choosing Joe and Dan first, then finding the probability they're in the team, we simply make sure they're picked ie. 8/30 x 7/29, then disregard the rest of the team, since it doesn't matter who the rest of the team is, as long as Joe and Dan are in it. Essentially, the 8/30 x 7/29 have to be Joe and Dan.
Hope this helps :)
Post by: spnmox on October 08, 2019, 04:04:23 pm
Hey there!
For the first question, first note that the total number of ways of getting a 5 card hand from 52 cards is equal to (52x51x50x49x48)/5!. The numerator should make sense (tell me if it doesn't!) and dividing by 5! removes order ie. picking an Ace, 2, 3, 4, 5 of diamonds is the same as picking a 2, Ace, 3, 4, 5 of diamonds in that order. Getting a flush means you're limiting yourself to a certain suit, and then picking 5 cards in a similar way; ie. the number of ways will be (13x12x11x10x9)/5!, for much the same reasons as the total number of ways. Since there are four suits, we multiply this value by four, and put it all over the total number of ways
For the second question, that's not the probability of choosing any two people; rather, it's the probability of choosing any two PARTICULAR people. What this means is that we pick the people first, then choose the team, not choose the team and check if those people are there. By choosing Joe and Dan first, then finding the probability they're in the team, we simply make sure they're picked ie. 8/30 x 7/29, then disregard the rest of the team, since it doesn't matter who the rest of the team is, as long as Joe and Dan are in it. Essentially, the 8/30 x 7/29 have to be Joe and Dan.
Hope this helps :)
wait, so is 8/30 * 7/29 picking the other people after assuming that joe and dan are already picked?
Post by: Grace0702 on October 09, 2019, 04:48:20 pm
Show that
Hence find expressions for the velocity and the acceleration in terms of t
Post by: fun_jirachi on October 09, 2019, 05:25:57 pm
wait, so is 8/30 * 7/29 picking the other people after assuming that joe and dan are already picked?No, this is the probability that Dan and Joe are picked. This probability ensures they are picked in the group; note that the probability of picking the other 6 people is just 1, since we don't care who they are. I mustn't have made it clear, thanks for bringing this up!
A particle moves in a straight line so that its displacement, in meters, is given by
Show that
Hence find expressions for the velocity and the acceleration in terms of t
Hey there!
This question is just asking you to differentiate the equation in a pretty convoluted way :)
Hope this helps :)
Post by: spnmox on October 09, 2019, 06:26:11 pm
Post by: fun_jirachi on October 09, 2019, 06:38:13 pm
Pat goes first. If Pat wins, the game ends. Pat wins on the first throw.
If he loses, it's now Chandra's turn. If Chandra wins, the game ends. If Chandra loses, it's Pat's turn again, and he has the chance to win on his second throw.
Note that Pat must lose the first round for it to reach any further round. In general, each player must lose in order for the game to progress to the next round. Hence, we can't just disregard the result of Pat's first go; if he wins, the game is already won on the first round, and he only gets to win the second round if and only if both Pat and Chandra lose on their first throws.
Post by: spnmox on October 09, 2019, 06:57:14 pm
There have to be two losses; consider what happens as the game is played out.
Pat goes first. If Pat wins, the game ends. Pat wins on the first throw.
If he loses, it's now Chandra's turn. If Chandra wins, the game ends. If Chandra loses, it's Pat's turn again, and he has the chance to win on his second throw.
Note that Pat must lose the first round for it to reach any further round. In general, each player must lose in order for the game to progress to the next round. Hence, we can't just disregard the result of Pat's first go; if he wins, the game is already won on the first round, and he only gets to win the second round if and only if both Pat and Chandra lose on their first throws.
oh, haha yeah i completely forgot about pat losing first.thanks!!!
Post by: Hawraa on October 09, 2019, 09:19:14 pm
I need some help with a past hsc question. It's from 2008 (Q10-b-ii) for math advance. My problem is with the differentiation part, when we have to differentiate the area A. I looked through the answers and the marking centre feedback but I still don't get it.
If S, L and Sin are constants, why they are not cancelled in the derivative?
And why they cancelled one L inside the brackets but left the other one squared?
I hope my question makes sense. If anyone can help me with this, it would be much appreciated. Thanks.
Post by: RuiAce on October 09, 2019, 09:38:23 pm
Hi guys,I'm afraid I'm not 100% certain about the issue, but it sounds like the differentiation itself is what you're asking about?
I need some help with a past hsc question. It's from 2008 (Q10-b-ii) for math advance. My problem is with the differentiation part, when we have to differentiate the area A. I looked through the answers and the marking centre feedback but I still don't get it.
If S, L and Sin are constants, why they are not cancelled in the derivative?
And why they cancelled one L inside the brackets but left the other one squared?
I hope my question makes sense. If anyone can help me with this, it would be much appreciated. Thanks.
\[ \text{As }s \text{ and }\sin \alpha\text{ are constants, we can move them outside the derivative}\\ \text{using the rule }\frac{d}{dx}(cf(x)) = c\frac{d}{dx}f(x) \]
\begin{align*}
\frac{dA}{dx} &= \frac{d}{dx} \left[s\left(x-\ell +\frac{\ell^2}{2x} \right)\sin\alpha\right]\\
&= s\sin \alpha \times \frac{d}{dx}\left(x-\ell + \frac{\ell^2}{2x} \right)
\end{align*}
\[ \text{Now recall that the derivative is with respect to }x.\\ \text{As always, when there's a sum of terms, we differentiate term by term.}\\ \text{Firstly, }\frac{d}{dx} x = 1. \]
\[ \text{Then, }\frac{d}{dx}\ell = 0\text{, because }\ell\text{ is a constant.}\\ \text{So differentiating }\ell\text{ has the same effect as differentiating a number, like }2. \]
\[ \text{With the last term, the }\frac{\ell^2}{2}\text{ bit is also a constant.}\\ \text{So we move it out in front.}\\ \frac{d}{dx} \left( \frac{\ell^2}{2x} \right) = \frac{\ell^2}{2} \times \frac{d}{dx} \frac1x \]
\[ \text{And since }\frac1x\text{ is all we're left with, we just use the power rule to deduce that}\\ \frac{d}{dx} \frac1x = -\frac1{x^2}. \]
Which gives the final answer \( \frac{dA}{dx} = s\sin \alpha \times \left(1-\frac{\ell^2}{2x^2} \right) \)
Make sure you understand that the derivative is with respect to \(x\). Not with respect to something else like \(\ell\). The \(\ell\) inside was not cancelled because of the fact it was an \(\ell\), but rather because it was a constant when as opposed to \(x\) and hence its derivative is zero. Whereas that \( \frac{\ell^2}{2x}\) term actually has an \(x\) attached to it, and hence requires use of laws of derivatives.
Post by: Hawraa on October 09, 2019, 09:40:55 pm
A strip of wire 96 cm in length is used to build a square prism. Supposing that the length of the square side is x cm, show that the surface area of the square prism is given by S= 6x(16 - x). Hence, find the volume of the prism with the maximum surface area.
Thanks guys.
Post by: Hawraa on October 09, 2019, 09:46:26 pm
I'm afraid I'm not 100% certain about the issue, but it sounds like the differentiation itself is what you're asking about?
\[ \text{As }s \text{ and }\sin \alpha\text{ are constants, we can move them outside the derivative}\\ \text{using the rule }\frac{d}{dx}(cf(x)) = c\frac{d}{dx}f(x) \]
\begin{align*}
\frac{dA}{dx} &= \frac{d}{dx} \left[s\left(x-\ell +\frac{\ell^2}{2x} \right)\sin\alpha\right]\\
&= s\sin \alpha \times \frac{d}{dx}\left(x-\ell + \frac{\ell^2}{2x} \right)
\end{align*}
\[ \text{Now recall that the derivative is with respect to }x.\\ \text{As always, when there's a sum of terms, we differentiate term by term.}\\ \text{Firstly, }\frac{d}{dx} x = 1. \]
\[ \text{Then, }\frac{d}{dx}\ell = 0\text{, because }\ell\text{ is a constant.}\\ \text{So differentiating }\ell\text{ has the same effect as differentiating a number, like }2. \]
\[ \text{With the last term, the }\frac{\ell^2}{2}\text{ bit is also a constant.}\\ \text{So we move it out in front.}\\ \frac{d}{dx} \left( \frac{\ell^2}{2x} \right) = \frac{\ell^2}{2} \times \frac{d}{dx} \frac1x \]
\[ \text{And since }\frac1x\text{ is all we're left with, we just use the power rule to deduce that}\\ \frac{d}{dx} \frac1x = -\frac1{x^2}. \]
Which gives the final answer \( \frac{dA}{dx} = s\sin \alpha \times \left(1-\frac{\ell^2}{2x^2} \right) \)
Make sure you understand that the derivative is with respect to \(x\). Not with respect to something else like \(\ell\). The \(\ell\) inside was not cancelled because of the fact it was an \(\ell\), but rather because it was a constant when as opposed to \(x\) and hence its derivative is zero. Whereas that \( \frac{\ell^2}{2x}\) term actually has an \(x\) attached to it, and hence requires use of laws of derivatives.
Thanks so much. This actually makes sense now. ☺️🌷
Post by: Hawraa on October 10, 2019, 09:20:16 pm
Can someone please explain this question for me, I'm so confused about it.
2U math advance hsc 2010 Q9-b
Thanks🌷
Post by: fun_jirachi on October 10, 2019, 10:20:15 pm
Remember what the function y=f'(x) actually denotes; it's just the slope of the function y=f(x). If f'(x)>0 over some domain, then f(x) is increasing over that domain. Similarly, if f'(x)<0 over some domain, then f(x) is decreasing over that domain. And if f'(x)=0, we have a stationary point.
Also, since
We have
ie. the area under the curve of y=f'(x) denotes the function value of f(x) at a specific x value.
From my first point, we see that f'(x)>0 for 0≤x<2, hence it is for the same domain that f(x) is increasing.
From my second point, we see that the maximum signed area under the curve is 4 units. Hence, the maximum value of f(x) is also 4.
From my second point, we note that f(6) will be the total signed area from 0 to 6. Seeing that A1 and A2 cancel, we just note the area between x=4 and x=6 (which is just a rectangle!) to be 6 units squared; hence f(6)=-6 (the area is below the x-axis!).
The graph is rather more difficult.
Key points should be noted on the graph, and can be deduced from the signed area! For example, (0, 0), (2, 4), (4, 0) and (6, -6) should all be on your graph. Since f'(2)=0, there should be a stationary point there. Note that also the gradient of f'(x) is negative at x=2, this indicates that f''(2)<0 and thus we have a maximum at x=2. Also, ensure that the gradient at x=0 looks convincingly like a gradient of 3, and similarly for x=4, make sure the gradient looks convincingly like a gradient of -3. Since between x=4 and x=6 f'(x) is a constant, between x=4 and x=6 f(x) should also be a straight line that has a constant gradient of -3.
If you need an actual picture of the graph, let me know and I'll attach one.
Hope this helps :)
Post by: Hawraa on October 10, 2019, 10:38:59 pm
Hey there! :)
Remember what the function y=f'(x) actually denotes; it's just the slope of the function y=f(x). If f'(x)>0 over some domain, then f(x) is increasing over that domain. Similarly, if f'(x)<0 over some domain, then f(x) is decreasing over that domain. And if f'(x)=0, we have a stationary point.
Also, since
We have
ie. the area under the curve of y=f'(x) denotes the function value of f(x) at a specific x value.
From my first point, we see that f'(x)>0 for 0≤x<2, hence it is for the same domain that f(x) is increasing.
From my second point, we see that the maximum signed area under the curve is 4 units. Hence, the maximum value of f(x) is also 4.
From my second point, we note that f(6) will be the total signed area from 0 to 6. Seeing that A1 and A2 cancel, we just note the area between x=4 and x=6 (which is just a rectangle!) to be 6 units squared; hence f(6)=-6 (the area is below the x-axis!).
The graph is rather more difficult.
Key points should be noted on the graph, and can be deduced from the signed area! For example, (0, 0), (2, 4), (4, 0) and (6, -6) should all be on your graph. Since f'(2)=0, there should be a stationary point there. Note that also the gradient of f'(x) is negative at x=2, this indicates that f''(2)<0 and thus we have a maximum at x=2. Also, ensure that the gradient at x=0 looks convincingly like a gradient of 3, and similarly for x=4, make sure the gradient looks convincingly like a gradient of -3. Since between x=4 and x=6 f'(x) is a constant, between x=4 and x=6 f(x) should also be a straight line that has a constant gradient of -3.
If you need an actual picture of the graph, let me know and I'll attach one.
Hope this helps :)
Thanks a lot 🌷 that actually clarifies the question.
Post by: fun_jirachi on October 10, 2019, 11:48:59 pm
And one more question please. This is from the (ATAR Notes topic tests HSC mathematics Edition 1 2017-2019. Page 35 Q9.) I think they accidentally forgot to put the answer for this one. The question is :
A strip of wire 96 cm in length is used to build a square prism. Supposing that the length of the square side is x cm, show that the surface area of the square prism is given by S= 6x(16 - x). Hence, find the volume of the prism with the maximum surface area.
Thanks guys.
Sorry for missing this one!
The question's wording is actually quite ambiguous! However, if you initially thought that the object was a cube, and the surface area should've been 6x2 (like I did!), you would've noticed there would have been no maximum. Hence, the object is just a square-sided rectangular prism with at least one set of opposite sides being squares.
Let the square sides have length x, and the other edge have length y.
Since the total length of wire is 96cm, we have that 8x+4y=96 ie. y=24-2x.
Now, we have the surface area being 2(x2+2xy).
ie. S=2x(x+2y)
S=2x(x+2(24-2x))
S=2x(48-3x)
S=6x(16-x)
From here, we find that \(\frac{dS}{dx} = 96-12x\) ie. there's a maximum when x=8 (you can test for a maximum in whichever way you like) (and therefore also when y=8! basically the object was a cube the whole time, which makes sense.) And thus the volume which maximises the surface area is just 8x8x8, or 512cm3.
Hope this helps :)
Post by: Hawraa on October 11, 2019, 01:01:20 am
Sorry for missing this one!
The question's wording is actually quite ambiguous! However, if you initially thought that the object was a cube, and the surface area should've been 6x2 (like I did!), you would've noticed there would have been no maximum. Hence, the object is just a square-sided rectangular prism with at least one set of opposite sides being squares.
Let the square sides have length x, and the other edge have length y.
Since the total length of wire is 96cm, we have that 8x+4y=96 ie. y=24-2x.
Now, we have the surface area being 2(x2+2xy).
ie. S=2x(x+2y)
S=2x(x+2(24-2x))
S=2x(48-3x)
S=6x(16-x)
From here, we find that \(\frac{dS}{dx} = 96-12x\) ie. there's a maximum when x=8 (you can test for a maximum in whichever way you like) (and therefore also when y=8! basically the object was a cube the whole time, which makes sense.) And thus the volume which maximises the surface area is just 8x8x8, or 512cm3.
Hope this helps :)
OMG! This question was actually killing me. But the way you explained it is just awesome! Many thanks 🌷
Post by: Hawraa on October 11, 2019, 01:34:11 am
I'm back with another question 😅. This time it's trig functions.
Q: find the area of the minor segment which subtends an angle of 3/4 pi at the centre of a circle with radius 20m.
Can you please explain how to work it out?
Post by: Grace0702 on October 11, 2019, 10:08:35 am
Hi again,
I'm back with another question 😅. This time it's trig functions.
Q: find the area of the minor segment which subtends an angle of 3/4 pi at the centre of a circle with radius 20m.
Can you please explain how to work it out?
Hey!
So there is actually a formula to work this out which is:
(https://www.mathsisfun.com/geometry/images/circle-segment-area.svg)
Take a look at the diagram. Basically the formula is finding the area of the sector using:
and the subtracting the area of the triangle with angle theta using:
If you sub in the radius and angle theta to the area of a triangle formula you get
So the Area of sector - Area of triangle = Area of minor segment
So then we sub in what we know
FYI this formula is not on the formula sheet so you will need to remember it or how to derive it
Hope this helps! :)
Post by: Hawraa on October 11, 2019, 11:14:28 am
Hey!
So there is actually a formula to work this out which is:
(https://www.mathsisfun.com/geometry/images/circle-segment-area.svg)
Take a look at the diagram. Basically the formula is finding the area of the sector using:
and the subtracting the area of the triangle with angle theta using:
If you sub in the radius and angle theta to the area of a triangle formula you get
So the Area of sector - Area of triangle = Area of minor segment
So then we sub in what we know
FYI this formula is not on the formula sheet so you will need to remember it or how to derive it
Hope this helps! :)
That's great, thanks a lot🌹. I actually came across this formula when I was looking for an answer, but I don't remember using it in school, so I thought maybe there is another way to solve the question. . But it makes sense now, especially knowing how to derive it.
Post by: Estermont on October 17, 2019, 07:54:11 pm
Post by: DrDusk on October 17, 2019, 08:28:49 pm
I've just started year 12 and I was wondering whether it is worth keeping Advanced Mathematics, I feel that I have the ability to get 80-84 for my Advanced exams and 90-93 if I dropped down to General Mathematics. Is it worthwhile dropping to General maths or should I stick with Advanced? Thanks :)I would honestly advise anyone to stay with advanced. Even though advanced forces you to learn much more advanced concepts, the actual ratio of knowledge/ability vs difficulty is basically the same as Standard Maths. Standard math students know much less Maths than advanced students but to them the exam tests their abilities. In advanced you have much more knowledge and it tests your abilities just as much as Standard does for its students. It's rather unlike the Extension 1 and 2 courses where the difficulty of the exam gets much more than the amount of knowledge/ability that most students will have.
Post by: RuiAce on October 18, 2019, 10:45:08 am
Standard math students know much less Maths than advanced studentsPlease watch your wording. When you say that it becomes a needlessly derogatory remark to standard maths students instead.
I've just started year 12 and I was wondering whether it is worth keeping Advanced Mathematics, I feel that I have the ability to get 80-84 for my Advanced exams and 90-93 if I dropped down to General Mathematics. Is it worthwhile dropping to General maths or should I stick with Advanced? Thanks :)Just some food for thought: If you compare the results in the HSC raw marks database you'll find that the marks in Advanced you're earning also has good chances of attaining a B6. Of course, it may not always be the case - the alignment algorithm does vary from year to year.
Otherwise, perhaps just consider whether or not you'd need Advanced for whatever it is you want to pursue at university. You don't want to fall into the trap of relying on Standard for reasonably maths-intensive disciplines. Also, keep in mind that you may need to relearn a few concepts from Year 11 Standard along the way; note that Year 11 content is examinable in the HSC for mathematics subjects.
But of course, you should also weigh out if Advanced is actually stressing you. (Would be a bit surprised if it did with 80+ raw marks, but can't fault you if it does anyway!)
Post by: Estermont on October 18, 2019, 06:15:58 pm
Post by: Kombmail on October 19, 2019, 11:17:25 pm
Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.
(ii) what is the probability that Pat wins the game on the first or on the second throw?
(iii) fins the probability that pat eventually wins the game
Post by: fun_jirachi on October 19, 2019, 11:53:16 pm
Does anyone now how to do this question?
Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.
(ii) what is the probability that Pat wins the game on the first or on the second throw?
(iii) fins the probability that pat eventually wins the game
Hey there!
This question was previously asked here. There's also a small explanation below that post to have a read as well if you don't understand part ii). If there's any other part you don't understand, have a query and we'll answer it :)
Hope this helps :)
Post by: boulos on October 20, 2019, 12:22:21 pm
Post by: fun_jirachi on October 20, 2019, 12:56:16 pm
Essentially, when we have f(x)-k = 0, we're looking for the solutions as a horizontal line cuts through f(x) (we're just considering where y=f(x) and y=k intersect!). Since you have the sketch of the graph from part ii), you'll notice that above the local maximum, y=k cuts the graph only once, while a similar occurs below the local minimum. Between the local maximum and the local minimum, you'll also notice that the line y=k cuts f(x) three times. Hence, the only values where the line y=k will cut the graph at two distinct places is when k is equal to the y-value of the local max and local min ie. when the horizontal line is tangential to the graph! Hence, k would equal either 7 or -20.
Hope this helps :)
Post by: Hawraa on October 20, 2019, 12:56:53 pm
Can someone please help me with this probability question.
Gerry has a standard deck of 52 cards in his hands, consisting of 4 different suits, each suit has 13 cards. What is the probability of Gerry picking a queen of spades then picking up any card from the hearts suit?
I know that the P of queen of spades would be 1/52 (which is also correct in the answers given) but for the other part of P of any card of the heart suit, I think it should be 13/51 since he already has one card in his hand but the answers say it's 13/52? So that's why I'm confused.
Post by: fun_jirachi on October 20, 2019, 01:01:36 pm
I think it might be an issue with interpreting the question.
I agree with you, when the question is worded like that, the answer for picking a heart would be 13/51 given Gerry picks the queen of spades then picks any heart (thus implying no replacement). However, if we try to justify the question's answer, there would be two possibilities I can think of (which are basically one possibility)
a) There's replacement
b) They're two separate events ie What is the probability of Gerry picking a queen of spades as opposed to a heart
Just make sure you check the wording of the question very carefully, because one or two words alter what the question implies entirely.
Hope this helps :)
Post by: Hawraa on October 20, 2019, 01:10:15 pm
Hey there!
I think it might be an issue with interpreting the question.
I agree with you, when the question is worded like that, the answer for picking a heart would be 13/51 given Gerry picks the queen of spades then picks any heart (thus implying no replacement). However, if we try to justify the question's answer, there would be two possibilities I can think of (which are basically one possibility)
a) There's replacement
b) They're two separate events ie What is the probability of Gerry picking a queen of spades as opposed to a heart
Just make sure you check the wording of the question very carefully, because one or two words alter what the question implies entirely.
Hope this helps :)
Thanks for your help 🌷
But which one do you think U should go with if I get a question like this in the exam?
Post by: spnmox on October 20, 2019, 02:30:46 pm
Also, for v=2- 4/(t+1). Find the exact distance travelled by the particle in the first 7 seconds.
I got 10-4ln2, but answers say 10-4log2. These give different answers, right? How do I get log instead of ln?
Thanks so much :) 5 more days !!
Post by: fun_jirachi on October 20, 2019, 03:48:54 pm
Thanks for your help 🌷
But which one do you think U should go with if I get a question like this in the exam?
Always go with what the question implies :) But there's no way the HSC will be this ambiguous ever, so you shouldn't have to worry about it at all.
Hey guys, need help with attached MC. I integrated as normal and just put some abs value signs around it, I got A but answer is C
Also, for v=2- 4/(t+1). Find the exact distance travelled by the particle in the first 7 seconds.
I got 10-4ln2, but answers say 10-4log2. These give different answers, right? How do I get log instead of ln?
Thanks so much :) 5 more days !!
Hey there!
You can't just 'integrate as normal' here and then stick some absolute value signs; that's always very risky and nearly always wrong. A rule of thumb is to never integrate absolute values conventionally :)
There's a few better ways to consider this:
a) Draw up the graph, then calculate the area using triangles
b) Split up the integral using the definition of the absolute value like so
For your second question, log always denotes ln ie. they are the same thing except on your calculator, where log is log base 10, while ln is log base e. They are the same answer :)
Post by: violet123 on October 20, 2019, 07:20:17 pm
why does sin (x+
Post by: DrDusk on October 20, 2019, 07:26:33 pm
Hi,
why does sin (x+
Post by: Kombmail on October 21, 2019, 10:36:54 am
Consider the function f(x)= xe^-x it says show that f’’(x) = e^-x(x-2) would you use product rule on the xe^-x?
Ps. Can anyone help me with understanding how to sketch this? Indicating stationary points and nature..
Post by: fun_jirachi on October 21, 2019, 12:02:37 pm
Yes, you would have to use the product rule.
In sketching this:
a) You would've found f'(x) by finding f''(x) --> Note that there is a singular stationary point at x=1, where f'(x)=0, and since f''(1) is negative, this indicates that the stationary point is a local maximum
b) Similarly, there is a point of inflexion at x=2
c) Have a look at the values of f(x) as x approaches positive and negative infinity to find any asymptotes
d) Clearly also, f(0)=0, so put that point on your graph, as well as the inflexion point (2, 2/e2) and the local max (1, 1/e)
Hope this helps :)
EDIT: Typo as kindly pointed out by Hawraa and Rui :)
Post by: Hawraa on October 21, 2019, 12:44:30 pm
Hey there!
Yes, you would have to use the product rule.
In sketching this:
a) You would've found f'(x) by finding f''(x) --> Note that there is a singular stationary point at x=1, where f'(x)=0, and since f''(1) is positive, this indicates that the stationary point is a local maximum
b) Similarly, there is a point of inflexion at x=2
c) Have a look at the values of f(x) as x approaches positive and negative infinity to find any asymptotes
d) Clearly also, f(0)=0, so put that point on your graph, as well as the inflexion point (2, 2/e2) and the local max (1, 1/e)
Hope this helps :)
Isn't like when the second derivative is positive then it is a minimum, not maximum?
Post by: RuiAce on October 21, 2019, 12:48:41 pm
Isn't like when the second derivative is positive then it is a minimum, not maximum?Yeah. Think he just made a typo. The second derivative should be negative when \(x=1\), and hence we have a local maximum.
This formula is not taught in the 2U course. It's a valid cheat from a 3U context, but we cannot assume 2U students know it.
Hi,Your LaTeX isn't showing up, but if like DrDusk said you meant \(\sin(x+\pi)\) on the left, you basically need to think back to ASTC.
why does sin (x+
Throughout the course, you've done things like \( \sin \left( \frac{5\pi}{4} \right) = -\sin \frac\pi4\), and \( \sin \frac{3\pi}{4} = \sin \frac\pi4\). These were done identifying things like \( \frac{5\pi}4\) is in the 3rd quadrant and \( \frac{3\pi}{4}\) is in the 2nd quadrant. But you may have learnt it as nothing more but a trick, without thinking about why they work.
In reality, the identity \( \boxed{\sin(\pi + x) = -\sin x} \) for the 3rd quadrant, along with other identities for the 2nd and 4th quadrants, are the formulas that let you do this in the first place. In general, depending on which quadrant you're in, you can use the `ASTC' trick to figure out whether the trig ratio is positive or negative. But the actual formulae being used are:
\begin{align*}
\sin(\pi - x) &= \sin x\tag{+'ve in 2nd quad}\\
\sin(\pi + x) &= -\sin x\\
\sin (2\pi - x) &= -\sin x
\end{align*}
\begin{align*}
\cos(\pi - x) &= -\cos x\\
\cos(\pi + x) &= -\cos x\\
\cos(2\pi - x) &= \cos x\tag{+'ve in 4th quad}
\end{align*}
\begin{align*}
\tan(\pi - x) &= -\tan x\\
\tan (\pi + x) &= \tan x\tag{+'ve in 3rd quad}\\
\tan(2\pi - x) &= -\tan x
\end{align*}
If you're genuinely interested in derivations of such results, you have to go back to the unit circle definition of the trigonometric ratios. This video can help get you started.
Post by: Youssefh_ on October 21, 2019, 03:20:19 pm
Post by: fun_jirachi on October 21, 2019, 03:25:48 pm
If a line is tangent to a circle, we know that the distance from the centre to the point of contact is the radius, which meets the tangent at a right angle. Basically, we substitute (3, -2) into the perpendicular distance formula, then use the equation of a circle (x-h)2+(y-k)2=r2 to find the answer (given the centre (h, k)).
Hope this helps :)
Hey there!
This was actually asked a while back! If there's anything else that needs clarification please ask :D
Hope this helps :)
Post by: annabeljxde on October 21, 2019, 08:39:34 pm
I'm stuck on this 2010 HSC Question (8d)
I got the required k value, which was 3, but the answer said the inequality was k≥3, while I got k≤3.
The marker's notes states that many students were unable to recognise the necessary sign change. Can someone please explain why the signs need to change?
Thank you.
Post by: DrDusk on October 21, 2019, 08:58:41 pm
Hi!
I'm stuck on this 2010 HSC Question (8d)
I got the required k value, which was 3, but the answer said the inequality was k≥3, while I got k≤3.
The marker's notes states that many students were unable to recognise the necessary sign change. Can someone please explain why the signs need to change?
Thank you.
Something my tutor always told me was always consider what everything means physically. Like what does discriminant mean and how does it affect the roots of the polynomial? What is it's physical significance?
That way of thinking will help with more conceptual questions or this one for example where you have to link the fact that the discriminant implies the nature of the roots, i.e. are they real or not which will allow you to visualize the polynomial.
Post by: Kombmail on October 21, 2019, 10:53:42 pm
A)3072x^10
B)-3072x^10
C)3072x^11
D)-3072x^11
But how would you do this question?
Post by: not a mystery mark on October 22, 2019, 11:19:09 am
If you get a 3 mark question correct but provide no working - will you get the 3 marks or will points be deducted?
Thanks heaps to the og who answers this <3
Post by: RuiAce on October 22, 2019, 11:26:14 am
Not a calculation question, but a HSC maths marking question.For a 3 marker, working out is definitely mandatory.
If you get a 3 mark question correct but provide no working - will you get the 3 marks or will points be deducted?
Thanks heaps to the og who answers this <3
If you get the final answer correct with no working, the most likely scenario is that you'd get 1/3. (2/3 and 0/3 would be in exceptional circumstances in my opinion.)
Post by: not a mystery mark on October 22, 2019, 01:35:35 pm
For a 3 marker, working out is definitely mandatory.
If you get the final answer correct with no working, the most likely scenario is that you'd get 1/3. (2/3 and 0/3 would be in exceptional circumstances in my opinion.)
Legendary!! Thank you heaps. You have legendary OG status.
Post by: s.jay on October 22, 2019, 09:19:49 pm
I know this may sound silly but with days before my maths exams, and after prepping extensively on the more difficult topics, I realised today that I forgot how to do questions with bearings (as in prelim trig). I know that I need to use sine and cosine rule etc. but I have forgotten how to actually find the angles!
If anyone could give me a quick refresher that would be great!!
Also here is a question that I am stuck with:
Sally starts on J and swims on a bearing of 105 deg for 5km out to L. She then changes direction and swims on a bearing of 045 deg for a further 16km.
(i) find angle JLM
(ii) find the distance JM that Sally will swim back to J
Thanks so much ;D
Post by: fun_jirachi on October 22, 2019, 10:51:01 pm
Always a good idea to draw a diagram; that should always be your first step. I've put it in the spoiler below so this post takes up less space, but please take a look at it, it will go a long way to helping you understand any bearing questions that come your way :)
Spoiler
(https://i.imgur.com/2u9lwm0.png)
I've actually forgotten to put in the 45 degree angle, but from the information given in the question and angle sum of a triangle, you should be able to deduce that angle JLM = 120 degrees.
And for part b), it's just one application of the cosine rule, with sides 5 and 16, with enclosed angle 120 degrees as found in part a). Have a go at more questions - starting with drawing a diagram! While you might be able to visualise it well, exams aren't really the places to hedge bets, it's better to draw them out always :)
Hope this helps :)
Post by: Kombmail on October 22, 2019, 11:08:18 pm
Post by: Hawraa on October 23, 2019, 01:35:50 pm
Guys how do you find solutions to tan x =-2?
Hi,
I'd do it this way. Can u confirm if the answer is correct ☺️?
Post by: Hawraa on October 23, 2019, 01:46:06 pm
This is a multiple choice question from 2015 : which expression is a term of the geometric series 3x-6x^2+12x^3-...?
A)3072x^10
B)-3072x^10
C)3072x^11
D)-3072x^11
But how would you do this question?
Hi there,
Post by: RuiAce on October 23, 2019, 01:47:52 pm
Hi,Their question was unclear because they didn't specify the domain for \(\theta\).
I'd do it this way. Can u confirm if the answer is correct ☺️?
If we assume that \(0^\circ \leq \theta \leq 360^\circ\), then your method is certainly correct. (Of course, in the exam, they could ask for radians instead.)
Post by: mani.s_ on October 23, 2019, 05:14:11 pm
Thanks for your time to answer my question :)
Post by: RuiAce on October 23, 2019, 05:47:35 pm
Hi, I just wanted to know the difference between standard deviation and variance? I know that standard deviation measures the spread of data from the mean, so 68% of the data will be in 1 standard deviation and 95% of the data will be in 2 standard deviations. I was also told that variance is the same thing as standard deviation except its squared. So what's the point of having variance and how is it actually different to standard deviation and how is it useful in maths?The standard deviation is nothing more but the square root of the variance.
Thanks for your time to answer my question :)
\[SD(X) = \sqrt{\operatorname{Var}(X)} \]
Why do we have both? Because in the context of random variables, the absence of the square root makes the formula look nicer.
\[ \operatorname{Var}(X) = E\left[(x-\mu)^2\right] \]
That, and it's also more cleaner to use in more advanced (university level) mathematical statistics proofs.
But once the results are derived, it's usually more of interest to report standard deviations to the public. The variance rescales things according to square of quantities, whilst the standard deviation is on the same scale.
Edit: Upon looking at the question again, I guess the main answer is that the variance is more commonly used for proofs in mathematical statistics at university. The variance was the first convention used by mathematicians, and a lot of the theory of statistics has been developed based off it. For many famous distributions, the variance generally takes a nicer form - the standard deviation introduces a square root out of nowhere. Sure, a square root isn't overly harmful or anything, but it's just nicer to not have it there altogether. Also, computers may run into precision errors when dealing with square roots, as opposed to just positive integer powers.
But for summary statistics, I'd probably say one would be crazy to only use the variance instead of the SD.
Post by: mani.s_ on October 23, 2019, 06:04:28 pm
The standard deviation is nothing more but the square root of the variance.So pretty much the variance and standard deviation are the exact same thing, except that variance is used to make the formulas look nicer and
\[SD(X) = \sqrt{\operatorname{Var}(X)} \]
Why do we have both? Because in the context of random variables, the absence of the square root makes the formula look nicer.
\[ \operatorname{Var}(X) = E\left[(x-\mu)^2\right] \]
That, and it's also more cleaner to use in more advanced (university level) mathematical statistics proofs.
But once the results are derived, it's usually more of interest to report standard deviations to the public. The variance rescales things according to square of quantities, whilst the standard deviation is on the same scale.
working out more cleaner.
Post by: mani.s_ on October 23, 2019, 06:08:59 pm
Thanks :)
Post by: RuiAce on October 23, 2019, 06:30:29 pm
So pretty much the variance and standard deviation are the exact same thing, except that variance is used to make the formulas look nicer andYep.
working out more cleaner.
Rui one more thing, for the Expected value of binomial distribution E(X) = np and Variance Var(X) = np(1-p), do we learn how to derive these formulas in school, cause I havent really seen anything in the books and if not, can u please show me how to derive these formulas???These proofs are actually heavily involved. I had trouble doing them in first year; wasn't until second year university when I started finding it easy. Here's the proof for the expected value at least just for viewing pleasure.
Thanks :)
(Note: So therefore you definitely won't need it at the high school level.)
\begin{align*}
E(X) &= \sum_x x P(X=x)\\
&= \sum_{x=0}^n x \binom{n}{x}p^x (1-p)^{n-x}\\
&= \sum_{x=0}^n x \frac{n!}{x!(n-x)!}p^x (1-p)^{n-x}.
\end{align*}
Now, we first evaluate the sum at \(x=0\). We pull the value when \(x=0\) outside of the sum.
\begin{align*}
E(X) &= 0 \binom{n}{0}p^0 (1-p)^{n-0} + \sum_{x=1}^n x\binom{n}{x}p^x (1-p)^{n-x}\\
&= \sum_{x=1}^n x\binom{n}{x}p^x (1-p)^{n-x}
\end{align*}
because that first term actually evaluates to 0. Now we note that the terms in the sum are indexed by \(x=1, 2, 3, \dots, n\). So every term in the sum now always has the property \(x=0\), and hence we can cancel any \(x\)'s in the numerator and denominator.
In addition to this, we need the trick \(N! = N(N-1)!\). (You can try to convince yourself that this formula is true by simply expanding both of the factorials.) Here, I first use that \(x! = x(x-1)!\). (Note: The assumption for this question is that \(x\) is an integer; not any real number.)
\begin{align*}
E(X)& = \sum_{x=1}^n x\binom{n}{x}p^x (1-p)^{n-x}\\
&= \sum_{x=1}^n x \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x}\\
&= \sum_{x=1}^n \frac{n!}{(x-1)(n-x)!} p^x (1-p)^{n-x}
\end{align*}
Observe that we had to tear apart the binomial coefficient (i.e. convert it to its factorial notation) to cancel out the \(x\) in front. We now have to rebuild a new binomial coefficient using the remaining factorials. It turns out that we now need to manipulate it to obtain \( \binom{n-1}{x-1} \).
This requires that we use \(n! = n(n-1)!\) in the numerator. Once we do that, observe that the sum is in terms of \(x\), and not in terms of \(n\). Hence the extra \(n\) can now be moved in front.
\begin{align*}
E(X) &= \sum_{x=1}^n \frac{n(n-1)!}{(x-1)!(n-x)!}p^x (1-p)^{n-x}\\
&= n \sum_{x=1}^n \frac{(n-1)!}{(x-1)!(n-x)!} p^x (1-p)^{n-x}\\
&= n \sum_{x=1}^n \frac{(n-1)!}{(x-1)![(n-1)-(x-1)!]} p^x (1-p)^{n-x}\\
&= n \sum_{x=1}^n \binom{n-1}{x-1}p^x (1-p)^{n-x}
\end{align*}
Now I'll decompose \(p^x\) into \(p\times p^{x-1}\), which allows me to pull out a factor of \(p\) in front as well.
\begin{align*}
E(X) &= n \sum_{x=1}^n\binom{n-1}{x-1} p\, p^{x-1}(1-p)^{n-x}\\
&= np \sum_{x=1}^n \binom{n-1}{x-1} p^{x-1}(1-p)^{n-x}
\end{align*}
__________________________________________________________________
Observe that our end goal \(np\) has now showed up. It remains to prove that the ugly sum actually equals to 1. To do this, I will now tear apart the sum, by subbing every value of \(x\) from \(1\) to \(n\) in. I'll focus only on the sum here.
\begin{align*}
&\quad \sum_{x=1}^n \binom{n-1}{x-1} p^{x-1}(1-p)^{n-x}\\
&= \binom{n-1}{1-1} p^{1-1} (1-p)^{n-1} + \binom{n-1}{2-1} p^{2-1}(1-p)^{n-2} + \binom{n-1}{3-1} p^{3-1}(1-p)^{n-3} + \cdots + \binom{n-1}{n-1} p^{n-1} (1-p)^{n-n}\\
&= \binom{n-1}{0} (1-p)^{n-1} + \binom{n-1}{1} p(1-p)^{(n-1)-1} + \binom{n-1}{2} p^2 (1-p)^{(n-1)-2} + \cdots + \binom{n-1}{n-1} p^{n-1}
\end{align*}
It may or may not be obvious here, but this now looks disturbing like the statement of the binomial theorem!
Recall: Statement of the binomial theorem
\[ (x+y)^n = \binom{n}{0} x^n + \binom{n}{1} x^{n-1}y + \binom{n}{2} x^{n-2} y^2 + \cdots + \binom{n}{n}y^n. \]
So from using the binomial theorem, that expression now simplifies to
\[ \left[ p + (1-p)\right]^{n-1}. \]
Hopefully it is clear that this expression indeed evaluates to 1. Thus upon substituting into what we had earlier,
\[ E(X) = np \times 1 = np \]
as required.
For the variance, one uses a similar strategy to this, but they start by computing \( E(X(X-1))\) instead. Since \( E(X(X-1)) = E(X^2 - X) = E(X^2) - E(X)\), this then allows us to find the second moment \(E(X^2)\). And as usual, conclude with \(\operatorname{Var}(X) = E(X^2) - [E(X)]^2\).
Post by: Hawraa on October 24, 2019, 08:50:48 pm
Just a question from 2011 Q5-iii
After they find the number of members using Sn, shouldn't they multiply the answer by (0.5)? They multiplied it by (0.005) and I don't understand why?
Post by: fun_jirachi on October 24, 2019, 09:03:53 pm
They earn half a cent per member per day, not half a dollar. This means they are correct in multiplying by $0.005 as opposed to the $0.50 that half a dollar would get you.
Hope this helps :)
Post by: Hawraa on October 24, 2019, 09:04:48 pm
Can someone please explain how did they get the answer as 18 litres? And like how do you know that you have to integrate (t)? Because I tried to sub 4 into both equations of liquid A and B and then subtract to get the answer as 4 litres? Help please!
Post by: Hawraa on October 24, 2019, 09:08:02 pm
Hey there!
They earn half a cent per member per day, not half a dollar. This means they are correct in multiplying by $0.005 as opposed to the $0.50 that half a dollar would get you.
Hope this helps :)
But then would you do 0.5x1/2? But that would equal 0.25? Sorry I know it sounds stupid but I'm actually confused how to get half a cent?
Post by: fun_jirachi on October 24, 2019, 09:46:43 pm
But then would you do 0.5x1/2? But that would equal 0.25? Sorry I know it sounds stupid but I'm actually confused how to get half a cent?
There are never stupid questions, ever. You can't know what you don't know, so don't be afraid to ask!
Recall that 100 cents make a dollar; this means that 1 cent is $0.01. Then, we halve it to get half a cent ie. $0.005 :) Does this make more sense?
2011-Q9-b-ii
Can someone please explain how did they get the answer as 18 litres? And like how do you know that you have to integrate (t)? Because I tried to sub 4 into both equations of liquid A and B and then subtract to get the answer as 4 litres? Help please!
You have to integrate here to find the volumes; you can't subtract the rates of change then multiply by the time taken. The rate of change given in the question is essentially just telling you how fast the water is coming out; it doesn't tell us anything about how much water is in the actual tank. And the answer is wrong; it's actually supposed to be 8 litres, not 18. There's a load of typos in the various sample answers, so just be wary of that :)
Post by: Hawraa on October 24, 2019, 09:55:03 pm
There are never stupid questions, ever. You can't know what you don't know, so don't be afraid to ask!
Recall that 100 cents make a dollar; this means that 1 cent is $0.01. Then, we halve it to get half a cent ie. $0.005 :) Does this make more sense?
You have to integrate here to find the volumes; you can't subtract the rates of change then multiply by the time taken. The rate of change given in the question is essentially just telling you how fast the water is coming out; it doesn't tell us anything about how much water is in the actual tank. And the answer is wrong; it's actually supposed to be 8 litres, not 18. There's a load of typos in the various sample answers, so just be wary of that :)
Yup, got it now. Thanks a lot 🌷
Post by: Coolmate on November 05, 2019, 07:09:34 pm
I am currently studying the topic: "Series and Sequences" and have come across both of these questions (questions are attached) and have no idea about how to go about solving them. Could someone please help me with them? --> btw, the question states:
"Find the value of the pronumeral in each arithmetic sequence"
Thanks in advance,
Coolmate 8)
Post by: r1ckworthy on November 05, 2019, 07:30:04 pm
Hi Everyone! :D
I am currently studying the topic: "Series and Sequences" and have come across both of these questions (questions are attached) and have no idea about how to go about solving them. Could someone please help me with them? --> btw, the question states:
"Find the value of the pronumeral in each arithmetic sequence"
Thanks in advance,
Coolmate 8)
Hey!
Arithmetic series have the following property (due to the common difference):
\[ T_2 - T_1 = T_3 - T_2 \]
We can use this to find the pro-numeral:
\[ (5k + 2) - 3 = 21 - (5k +2) \\ 5k + 2 -3 = 21 -5k -2 \\ 5k -1 = 19 - 5k \\ 10k = 20 \\ k = 2 \]
I'll let you have a go with the other question :)
Let me know if you need any more help!
Post by: Coolmate on November 05, 2019, 07:46:52 pm
Hey!
Arithmetic series have the following property (due to the common difference):
\[ T_2 - T_1 = T_3 - T_2 \]
We can use this to find the pro-numeral:
\[ (5k + 2) - 3 = 21 - (5k +2) \\ 5k + 2 -3 = 21 -5k -2 \\ 5k -1 = 19 - 5k \\ 10k = 20 \\ k = 2 \]
I'll let you have a go with the other question :)
Let me know if you need any more help!
Hi r1ckworthy,
Thankyou for your reply, I do have a couple of questions though. For the first one why didn't you times (5k+2) by -3? and for the second question (h) I wrote this as my working out and got a wrong answer of x = 11:
(h):
\[ (x+3) - x = (2x +5) - (x+3) \\ 3 = 2x + 5 - x - 3 \\ 6 = x + 5 \\ x = 11 \]
Could you please suggest what I have done wrong with my working out please?
Thanks again for your help!
Coolmate 8)
Post by: r1ckworthy on November 05, 2019, 07:58:56 pm
Hi r1ckworthy,
Thankyou for your reply, I do have a couple of questions though. For the first one why didn't you times (5k+2) by -3? and for the second question (h) I wrote this as my working out and got a wrong answer of x = 11:
(h):
\[ (x+3) - x = (2x +5) - (x+3) \\ 3 = 2x + 5 - x - 3 \\ 6 = x + 5 \\ x = 11 \]
Could you please suggest what I have done wrong with my working out please?
Thanks again for your help!
Coolmate 8)
Hey Coolmate!
As for your first question, I don't really need to multiply by -3, just subtract 3 from (5k + 2) as per the formula. As for your second question, you made an error on the third line of your working out. You should subtract 5 from 6 instead of add :), so it will look like this:
\[ (x+3) - x = (2x +5) - (x+3) \\ 3 = 2x + 5 - x - 3 \\ 6 = x + 5 \\ 6 \color{red}{- 5} = x \\ x = 1 \]
Hope that helps! Let me know if you need further help :)
Post by: Coolmate on November 05, 2019, 08:04:52 pm
Thankyou so much, this makes a whole lot more sense and I am getting the questions right now! I like how you previously did the formula as:
T2 - T1 = T3 - T2
And for my second question...... that was such a silly mistake😂😂, thanks for clarifying this for me
Thanks again!
Coolmate 8)
Post by: sharlt on November 28, 2019, 03:44:40 pm
I was wondering if I could have help with this question - thank you!
A bag contains 6 white, 7 red and 8 blue balls. Create a probability function table for the number of white balls selected when drawing two balls from the bag:
A) With replacement
B) Without replacement
Post by: RuiAce on November 28, 2019, 10:26:40 pm
Hi there,Can you please be more specific on where it is you're having trouble? Or alternatively provide some insight on what you tried thus far?
I was wondering if I could have help with this question - thank you!
A bag contains 6 white, 7 red and 8 blue balls. Create a probability function table for the number of white balls selected when drawing two balls from the bag:
A) With replacement
B) Without replacement
Post by: sharlt on November 29, 2019, 02:04:37 pm
Post by: Coolmate on January 20, 2020, 08:57:32 pm
I have a test coming up regarding the syllabus topics of:
- ✔Functions
✔Trigonometric Functions
I was just wondering whether anyone would have any tips for tests regarding these topics
Thanks in advance,
Coolmate 8)
Post by: kauac on January 20, 2020, 09:51:59 pm
Hi Everyone, ;D
I have a test coming up regarding the syllabus topics of:
- ✔Functions
✔Trigonometric Functions
I was just wondering whether anyone would have any tips for tests regarding these topics
Thanks in advance,
Coolmate 8)
Hi Coolmate!
Not too familiar with the new maths advanced syllabus, but I had a look at my 2U notes for the same topics and have some (but definitely not exhaustive) general things that could help:
Trigonometric Functions
• Take note of whether the question is asking for the answer in degrees or radians (and double check your calculator is in this mode) – easy marks can be lost there.
• On a similar note: know how to convert radians to degrees and vice versa.
• Know well and be able to draw your basic y= sin x, cos & tan graphs (and their inverse).
Functions:
• Know the difference between domain & range
• With a region question, check you have drawn it correctly by plugging a set of coordinates inside the region and making sure the inequality is true for this.
• Be familiar with the format/wording of locus questions so that you can easily work out what it’s meant to look like (e.g. a circle or a straight line).
I’m sure there are plenty more tips that others could add as well! :)
Post by: Coolmate on January 25, 2020, 09:35:57 am
Hi Coolmate!
Not too familiar with the new maths advanced syllabus, but I had a look at my 2U notes for the same topics and have some (but definitely not exhaustive) general things that could help:
Trigonometric Functions
• Take note of whether the question is asking for the answer in degrees or radians (and double check your calculator is in this mode) – easy marks can be lost there.
• On a similar note: know how to convert radians to degrees and vice versa.
• Know well and be able to draw your basic y= sin x, cos & tan graphs (and their inverse).
Functions:
• Know the difference between domain & range
• With a region question, check you have drawn it correctly by plugging a set of coordinates inside the region and making sure the inequality is true for this.
• Be familiar with the format/wording of locus questions so that you can easily work out what it’s meant to look like (e.g. a circle or a straight line).
I’m sure there are plenty more tips that others could add as well! :)
Thankyou kauac for the advice! :D
Post by: shekhar.patel on February 09, 2020, 09:56:33 am
Post by: fun_jirachi on February 09, 2020, 01:03:27 pm
Given that the sums of three different sections need to be the same, we should automatically be considering a sort of inverse to that statement ie. that the three different sections need to each have a sum a third of the total, which is 78. Therefore, each section should have a total of 26. Also, note that each number can be paired up with one other to form 13, ie. two pairs form 26. It's a similar sort of idea to summing arithmetic series quickly by adding the first and last terms, the second and second last, etc. Therefore, any two pairs will work, but given that two lines need to be drawn as opposed to randomly picking two pairs, the following would be a good solution.
Spoiler
(https://i.imgur.com/jWNg0G7.png)
Post by: shekhar.patel on February 10, 2020, 04:16:43 pm
Hey there!
Given that the sums of three different sections need to be the same, we should automatically be considering a sort of inverse to that statement ie. that the three different sections need to each have a sum a third of the total, which is 78. Therefore, each section should have a total of 26. Also, note that each number can be paired up with one other to form 13, ie. two pairs form 26. It's a similar sort of idea to summing arithmetic series quickly by adding the first and last terms, the second and second last, etc. Therefore, any two pairs will work, but given that two lines need to be drawn as opposed to randomly picking two pairs, the following would be a good solution.Hope this helps!Spoiler(https://i.imgur.com/jWNg0G7.png)
Oh right!
That makes sense. Thanks a lot for your help.
Post by: Tierney_P on February 25, 2020, 02:50:12 pm
like for example y=
?????
Post by: RuiAce on February 25, 2020, 04:06:03 pm
Hey guys this is probably a dumb question......how do I integrate a square root???This reads as \( 4\sqrt{2} - x\). Did you mean this or did you mean \(4\sqrt{2-x}\)?
like for example y=
?????
Post by: Tierney_P on February 25, 2020, 04:48:52 pm
This reads as \( 4\sqrt{2} - x\). Did you mean this or did you mean \(4\sqrt{2-x}\)?Sorry :-[ the second one!
Post by: RuiAce on February 25, 2020, 05:08:09 pm
\[ \text{Now apply the reverse chain rule result}\\ \int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)}+c \]
Post by: Tierney_P on February 25, 2020, 05:16:26 pm
\[ \int 4\sqrt{2-x}\,dx = 4\int (2-x)^{\frac12}\,dx \]
\[ \text{Now apply the reverse chain rule result}\\ \int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)}+c \]
Ok awesome! Thank you!
Post by: sunflowah on February 29, 2020, 10:13:40 pm
How would I work out this question:
In an arithmetic series, T3 = -2 and T9 = 28. How many terms of this series are required to give a sum of 1092?
Thanks in advance!
Post by: fun_jirachi on February 29, 2020, 10:23:55 pm
Hello :)
How would I work out this question:
In an arithmetic series, T3 = -2 and T9 = 28. How many terms of this series are required to give a sum of 1092?
Thanks in advance!
Hey there!
There are a few key steps to doing this question:
a) Consider that there are 9 - 3 = 6 'jumps' between T3 and T9, all with the same value. ie. \(T_i - T_j = (i-j)d, \ [i, j \in \mathbb{Z}^+, \ i > j]\). This step will enable you to find the common difference.
b) Use this common difference to find the first term in the sequence, then substitute the relevant values into the sum of an arithmetic sequence formula, to then solve for Sn = 1092.
Hope this helps :)
Post by: Einstein_Reborn_97 on March 11, 2020, 09:51:11 pm
My textbook has lots of questions that involve sketching graphs first and solving an equation graphically. This seems unreasonable to me. Isn't the idea of a sketch that it doesn't have to be accurate? Especially in exam scenarios. For example, how can we be expected to use our sketch of y=e^{x-1} - 2 to solve for e^{x-1} - 2 = 8, without doing is algebraically? Would we actually get such questions in an exam? Because so far, I've just sketched the graph and solved the given equation algebraically. I'd imagine they'd give us an accurately drawn graph in an exam and ask us to solve an equation or inequality using that.
Post by: fun_jirachi on March 12, 2020, 07:41:46 am
Hi!
My textbook has lots of questions that involve sketching graphs first and solving an equation graphically. This seems unreasonable to me. Isn't the idea of a sketch that it doesn't have to be accurate? Especially in exam scenarios. For example, how can we be expected to use our sketch of y=e^{x-1} - 2 to solve for e^{x-1} - 2 = 8, without doing is algebraically? Would we actually get such questions in an exam? Because so far, I've just sketched the graph and solved the given equation algebraically. I'd imagine they'd give us an accurately drawn graph in an exam and ask us to solve an equation or inequality using that.
Usually in exams, graphs are used more for detecting if points of intersection exist, and for visualising inequalities. While textbooks do this as an exercise, its important to note it down as a decent technique to see if you algebraic answer is 'in the ballpark' of where the answer should be - does it make sense? Solving for the exact values of x would of course be done algebraically :) No exam will ever ask you to take values from your graph, though some more informal exams might ask you to take approximations.
Hope this helps :)
Post by: Einstein_Reborn_97 on March 12, 2020, 01:20:53 pm
Usually in exams, graphs are used more for detecting if points of intersection exist, and for visualising inequalities. While textbooks do this as an exercise, its important to note it down as a decent technique to see if you algebraic answer is 'in the ballpark' of where the answer should be - does it make sense? Solving for the exact values of x would of course be done algebraically :) No exam will ever ask you to take values from your graph, though some more informal exams might ask you to take approximations.
Hope this helps :)
Ok, thanks!
Post by: Coolmate on April 06, 2020, 11:18:30 pm
I have two questions regarding the topic of the differentiation of exponential functions that I am confused with and was wondering is someone could please help me.
I have attached the questions
Thanks in advance! ;D
Coolmate 8)
Post by: fun_jirachi on April 06, 2020, 11:31:47 pm
Recall that for some exponential denoted \(e^{f(x)}\), its derivative is going to be \(f'(x) \times e^{f(x)}\). Around this, we can apply things like the product rule and quotient rule - which you'll have to use for the first question!
For the second question, recall that you can find the equation of a line using \(y-y_1 = m(x-x_1)\). For some curve \(f(x)\), its tangent and normal at some point \((\alpha, f(\alpha))\) are \(y-f(\alpha) = f'(\alpha) \times (x-\alpha)\) and \(y-f(\alpha) = -\frac{1}{f'(\alpha)} \times (x-\alpha)\). Consider why this is the case, and apply it to the question!
Hope this helps :)
Post by: Coolmate on April 09, 2020, 08:39:02 am
Hey there!
Recall that for some exponential denoted \(e^{f(x)}\), its derivative is going to be \(f(x) \times e^{f'(x)}\). Around this, we can apply things like the product rule and quotient rule - which you'll have to use for the first question!
For the second question, recall that you can find the equation of a line using \(y-y_1 = m(x-x_1)\). For some curve \(f(x)\), its tangent and normal at some point \((\alpha, f(\alpha))\) are \(y-f(\alpha) = f'(\alpha) \times (x-\alpha)\) and \(y-f(\alpha) = -\frac{1}{f'(\alpha)} \times (x-\alpha)\). Consider why this is the case, and apply it to the question!
Hope this helps :)
Thankyou for the help fun_jirachi! ;D
Post by: jasminerulez9 on April 09, 2020, 11:16:08 am
A company finds that the function f(x)= x^3 - 96x^2 +2880x provides a good approximation for their profit f(x) in dollars, where x is the advertising expenditure in thousands of dollars
(i) What expenditure of advertising would produce the maximum profit (4 marks)
(ii) What is the maximum profit? (1 mark)
Post by: Grace0702 on April 09, 2020, 12:06:35 pm
Could someone PLEASE help me with this question, i have been working on it for a while but just do not understand how to solve it. Help will be really appreciated, as maths is my weakest subject :(
A company finds that the function f(x)= x^3 - 96x^2 +2880x provides a good approximation for their profit f(x) in dollars, where x is the advertising expenditure in thousands of dollars
(i) What expenditure of advertising would produce the maximum profit (4 marks)
(ii) What is the maximum profit? (1 mark)
So basically this question is asking us to find the maximum turning point on the function f(x)= x^3 - 96x^2 +2880x. So we would do this how we would normally. Differentiate the curve, find the stationary points and test to see whether they are a maximum or minimum, the maximum turning point here is the answer to (i). Part (ii) then asks us to sub in the stationary point/answer we got in part (i) into f(x) and this will give us the answer for maximum profit.
Follow these steps and it should produce the right answer! Hope this helps :)
Post by: jasminerulez9 on April 09, 2020, 12:57:32 pm
Post by: jasminerulez9 on April 09, 2020, 01:38:28 pm
During a survey the area of an irregular headland was to be found. The surveyor used a base line divided into ten equal sub-intervals, each of width 5m. Measurements f(x) were taken at each interval across the base line and tabulated below:
x metres 0 5 10 15 20 25 30 35 40 45
f(x) 0 10.2 13.2 16.3 13.8 16.0 17.0 18.6 10.8 0
Use the trapezoidal rule to find the approximate area of the headland ( 5 marks)
THANK YOU!!
Post by: fun_jirachi on April 09, 2020, 03:47:25 pm
With multiple applications of the trapezoidal rule on some function \(f(x)\) over some domain \(x \in [a, b]\) partitioned by x-values \(x_1, x_2 ... x_n\) such that \(a < x_1 < x_2 < ... < x_n < b\), we have that the area obtained by using the trapezoidal rule is \(\frac{h}{2}\left(f(a) + f(b) + 2(f(x_1) + f(x_2)+...+ f(x_n))\right)\), assuming, of course, that each of the partitions are equally spaced, represented by the value h \((\text{ie. }x_1 - a = x_2 - x_1 = x_3 - x_2 = ... = b - x_n = h\)) . Essentially what this means is that we have the space between each given value of x, divided by 2, multiplied by the total of the sum of the first and last terms of the function output and the sum of twice all the middle terms.
On this application, we have that \(h = 5\), and that \(f(a) = 0, f(b) = 0\). Thus, we have that the approx. area is roughly \(\frac{5}{2} (0+ 0 +2(10.2 +13.2 + 16.3 + 13.8+16.0+17.0+18.6+10.8 )) \\ = 5(115.9) = 579.5 \text{ square metres}\). Try this with any other trapezoidal rule questions you might have!
Hope this helps :)
Post by: jasminerulez9 on April 09, 2020, 08:02:43 pm
Post by: Coolmate on April 14, 2020, 07:12:43 pm
I am confused with 2 questions and how to get to the answer, from the topic of differentiating logs.
1. The first attached image; The answer says that it is -1/x, but where did the 3 go?
2. I am generally confused with this question
Thanks in advance,
Coolmate 8)
Post by: RuiAce on April 14, 2020, 08:25:21 pm
Hey Everyone,\begin{align*}
I am confused with 2 questions and how to get to the answer, from the topic of differentiating logs.
1. The first attached image; The answer says that it is -1/x, but where did the 3 go?
2. I am generally confused with this question
Thanks in advance,
Coolmate 8)
\frac{d}{dx} (1- \ln 3x) &= -\frac{3}{3x}\\
&= -\frac1x.
\end{align*}
Simply put, the 3's cancel each other out.
For the second one, you should apply the logarithm law \( \log \frac{A}{B} = \log A - \log B\) first, to make the differentiation procedure easier.
Post by: Coolmate on April 14, 2020, 10:06:10 pm
\begin{align*}
\frac{d}{dx} (1- \ln 3x) &= -\frac{3}{3x}\\
&= -\frac1x.
\end{align*}
Simply put, the 3's cancel each other out.
For the second one, you should apply the logarithm law \( \log \frac{A}{B} = \log A - \log B\) first, to make the differentiation procedure easier.
Thankyou so much Rui! This makes heaps more sense now! ;D ;D
Post by: 2020hsc on April 23, 2020, 04:09:39 pm
Could I get a hand on this question please....
Find the side a in triangleABC, where <C=60, b=24cm and the area is 30cm^2
Post by: fun_jirachi on April 23, 2020, 10:56:56 pm
If we construct a perpendicular through the side AC (ie. b = 24cm) and vertex B, we have that the perpendicular is 2.5cm (0.5(24 x 2.5) = 30cm^2) - from the area of a triangle! Now, we have that 2.5/(sin 60) = BC (side a!), which is going to be \(\frac{5\sqrt{3}}{3}\) cm.
Hope this helps :)
Post by: 2020hsc on April 24, 2020, 06:06:29 am
Hey there!
If we construct a perpendicular through the side AC (ie. b = 24cm) and vertex B, we have that the perpendicular is 2.5cm (0.5(24 x 2.5) = 30cm^2) - from the area of a triangle! Now, we have that 2.5/(sin 60) = BC (side a!), which is going to be \(\frac{5\sqrt{3}}{3}\) cm.
Hope this helps :)
Ah! Is that all :o :D That's great...thanks so much fun_jirachi...!
Post by: 2020hsc on April 28, 2020, 07:57:55 pm
could someone give me a hand on the attached question please?
solving for 0 < x < 360...
Post by: Einstein_Reborn_97 on April 28, 2020, 09:35:05 pm
hey!
could someone give me a hand on the attached question please?
solving for 0 < x < 360...
Hi. (Edit: Ignore this, it's incorrect)
Since
Hence, the two possible values of x between 0 and 360 degrees are 135 and 375. Solve the equation using either value to check if it's correct.
Let me know if you need me to clarify anything. :)
Post by: 2020hsc on April 29, 2020, 04:53:01 am
Hi.
Since(from the original equation), the possible values of (x-75) must lie in the 1st and 4th quadrant. 60 degrees is the reference angle in the first quadrant. To find the value from the 4th quadrant, we use 360 - θ. So...
Hence, the two possible values of x between 0 and 360 degrees are 135 and 375. Solve the equation using either value to check if it's correct.
Let me know if you need me to clarify anything. :)
hey! thats cool, thanks for doing that for me..
is pretty much what i had come up with myself, however, and yet the answers tell me it should be x = 15 or 135.
any ideas why that would be?
sorry! thanks anyway
Post by: Einstein_Reborn_97 on April 29, 2020, 09:50:32 am
hey! thats cool, thanks for doing that for me..
is pretty much what i had come up with myself, however, and yet the answers tell me it should be x = 15 or 135.
any ideas why that would be?
sorry! thanks anyway
That would be because I wasn't thinking properly last night. :(
You have to change the domain first...
So the new domain is (-75, 285)
Post by: 2020hsc on April 29, 2020, 10:43:41 am
That would be because I wasn't thinking properly last night. :(
You have to change the domain first...
So the new domain is (-75, 285):)
ah yes! thats the step i must have been missing with a few trig questions i was doing :-[
thanks so much Einstein_Reborn_97 ;D
Post by: Coolmate on May 09, 2020, 07:47:55 pm
Could someone please help me with this question (attached) on finding more than one stationary point please?
Thanks in advance! 8)
Post by: Einstein_Reborn_97 on May 09, 2020, 10:55:36 pm
Hi Everyone,
Could someone please help me with this question (attached) on finding more than one stationary point please?
Thanks in advance! 8)
For stationary points:
When x=0, y=5 and
Check that concavity changes by choosing values on the LHS and RHS, e.g. x=±1. When x=-1, the second derivative is equal to -6 (concave downwards). When x=1, the second derivative is equal to 6 (concave upwards).
Since concavity changes, (0,5) is a horizontal point of inflection.
When x=2, y=1 and
>0 (concave upwards)
So (2,1) is a minimum turning point.
Hope that makes sense. If you need more help, feel free to ask! ;)
Post by: Coolmate on May 14, 2020, 09:07:26 pm
Post by: svnflower on May 15, 2020, 04:43:44 pm
When finding the area bounded by the y-axis (this is for integration), is the first step always to write the equation as a function of y (i.e. make x the subject)?
Also, is it necessary to always find the point of intersection for each question that requires us to find the area between 2 curves?
Post by: Opengangs on May 15, 2020, 05:02:03 pm
Hello,Hi there!
When finding the area bounded by the y-axis (this is for integration), is the first step always to write the equation as a function of y (i.e. make x the subject)?
Also, is it necessary to always find the point of intersection for each question that requires us to find the area between 2 curves?
To answer your first question, generally yes. This is typically because when we have a function \(f(x)\), the area underneath \(f(x)\) is the area bounded by the \(x\) axis and not the \(y\) axis. So if we're considering the area that is bounded by the \(y\) axis, we require \(g(y)\) instead which can be found by making \(x\) the subject. The bounds will also change so to really capture this, I suggest you draw a diagram!
To answer your second question, yes! That is because the area bounded by two curves is the entire area that is squeezed by these two curves. These two curves bound by 2 points and they are the points of intersection! Again, draw a diagram and you should be able to see this geometrically! :)
Post by: svnflower on May 15, 2020, 05:11:36 pm
Would you mind showing how to solve part (b) of this question? I understand how to find the points of intersection but am slightly confused with how to find the area of the region
Post by: Opengangs on May 15, 2020, 05:31:56 pm
Thankyou so much, I understand better now!Sure, so the first thing to note is that our region of interest is bounded by these three curves \(y = 3x\), \(y = x^2 - 4\) and \(x = 0\). We can see that we actually have two separate areas! One is bounded above by the \(x\) axis, I labelled it as \(A_1\) and the other is bounded below by the \(x\) axis, I labelled it as \(A_2\).
Would you mind showing how to solve part (b) of this question? I understand how to find the points of intersection but am slightly confused with how to find the area of the region
Now, the bounds of \(A_1\) is from 0 to the point of intersection between \(y = 3x\) and \(y = x^2 - 4\) which you can find to be \(x = 4\). So \(A_1\) is just \(\displaystyle \int_0^4 [\text{Top } - \text{Bottom}]\,dx\), with the top curve being \(y = 3x\) and the bottom curve being \(y = x^2 - 4\). That gives us
\begin{align*}
A_1 &= \int_0^4 \left(3x - x^2 + 4\right)\,dx \\
&= \frac{3}{2}x^2 - \frac{1}{3}x^3 + 4x \bigg|_0^4 \\
&= 24 - \frac{64}{3} + 16 \\
&= 40 - \frac{64}{3}.
\end{align*}
On the other hand, \(A_2\) is a little bit harder to find. It's unclear how we could rearrange this region. But notice how this is just \(y = x^2 - 4\) bounded by the \(y\) axis! So we need to write \(x\) in terms of \(y\) which gives us \(x = \sqrt{y + 4}\). Our bounds for \(y\) go from \(y = -4\) to \(y = 0\) (hopefully you can see that on the diagram.
So \(A_2\) is just \(\int_{-4}^0 \sqrt{y + 4}\,dy\) which turns out to be \(\displaystyle \frac{16}{3}\). So the entire area is just
\begin{align*}
A &= A_1 + A_2 \\
&= 40 - \frac{64}{3} + \frac{16}{3} \\
&= 40 - \frac{48}{3} \\
&= 40 - 16 \\
&= 24\text{u}^2.
\end{align*}
If that's incorrect, pls let me know and I will try to correct it :))
Post by: svnflower on May 15, 2020, 05:53:05 pm
I had a look and it states the answer as 18 and 2/3 units2
Post by: Opengangs on May 15, 2020, 06:02:30 pm
I see! Soo if the shaded area spans above and below the x-axis we always work it out as two separate areas?Normally yes! It depends on the graph but generally you'd break it up into separate areas and calculate them independently!
I had a look and it states the answer as 18 and 2/3 units2
Ah, the question seemed to have only calculated \(A_1\). I think the intended question was an area bounded by two curves above the \(x\) axis.
Post by: svnflower on May 15, 2020, 11:37:36 pm
I've also tried doing this question a few times but keep getting the wrong answer? Would you mind letting me know where I went downhill?
Thanks in advance!!
Post by: Opengangs on May 15, 2020, 11:57:59 pm
Ahh thanks heaps, this topic is becoming much more clearer to me now :)Hi there!
I've also tried doing this question a few times but keep getting the wrong answer? Would you mind letting me know where I went downhill?
Thanks in advance!!
Check your bounds :-)) They should be the points of intersection! To find your points of intersection, set \(x^2 = 2x + 3\). This is just a quadratic in \(x\) that you can solve! This gives you
\begin{align*}
x^2 - 2x - 3 &= 0 \\
(x - 3)(x + 1) &= 0 \\
x = 3, &\quad x = -1.
\end{align*}
So these are your bounds! The next step is to figure out which curve is the top curve! This can be visualised using a diagram and it's pretty easy to see that \(y = 2x + 3\) is the top curve as you successfully worked out! So all we need to do is to evaluate the following integral \(\displaystyle \int_{-1}^3 (2x + 3 - x^2)\,dx\), which becomes
\begin{align*}
\int_{-1}^3 (2x + 3 - x^2)\,dx &= \left(x^2 + 3x - \frac{1}{3}x^3\right)\bigg|_{-1}^3 \\
&= \left(9 + 9 - 9\right) - \left(1 - 3 + \frac{1}{3}\right) \\
&= 9 + 2 - \frac{1}{3} \\
&= 11 - \frac{1}{3} \\
&= \frac{32}{3} \\
&= 10 \frac{2}{3}.
\end{align*}
I think the mistake here is really figuring out what the bounds are, which are the points of intersection between the two curves. :-))
Post by: svnflower on May 16, 2020, 01:02:34 am
Post by: 2020hsc on June 01, 2020, 03:44:25 pm
could i get some help on the question below...please and thanks :) its in the year 12 unit on sequences and series, looking at solving problems with arithmetic sequences and geometric sequences.
'find a and b if a,b,1 forms a GP and b,a,10 forms and AP'
thanks!
Post by: fun_jirachi on June 01, 2020, 05:07:32 pm
Think about what an AP and a GP define for any three consecutive terms. For \(n \in \mathbb{Z}^+, n > 2\), the former states that \(T_n - T_{n-1} = T_{n+1} - T_n\), while the latter states that \(\frac{T_n}{T_{n-1}} = \frac{T_{n+1}}{T_n}\).
Now, considering this, we have that \(\frac{b}{a}= \frac{1}{b} \implies a=b^2\). We also have that \(a-b = 10-a \implies 2a = b+10\).
See how you go from here :)
Post by: 2020hsc on June 01, 2020, 05:22:05 pm
considering the GP is a,b,1 and the AP is b,a,10...then should it not be b/a = 1/b and a-b = 10-a ??
Post by: fun_jirachi on June 01, 2020, 05:34:52 pm
thanks....but should it not be the other way?
considering the GP is a,b,1 and the AP is b,a,10...then should it not be b/a = 1/b and a-b = 10-a ??
Yes, it should! I misread it ~ the point still stands, however - you'll still have a quadratic. I'll edit that now so it's correct. In general, it is handy to know what makes a GP a GP, and what makes an AP an AP, even though these questions are relatively uncommon - the theory is so important to reasoning out further applications :)
Post by: 2020hsc on June 01, 2020, 05:52:40 pm
Yes, it should! I misread it ~ the point still stands, however - you'll still have a quadratic. I'll edit that now so it's correct. In general, it is handy to know what makes a GP a GP, and what makes an AP an AP, even though these questions are relatively uncommon - the theory is so important to reasoning out further applications :)
ok, sweet. thanks a lot :) from there then do you solve it as a simultaneous equation or how would be best?
Post by: Einstein_Reborn_97 on June 01, 2020, 07:15:16 pm
ok, sweet. thanks a lot :) from there then do you solve it as a simultaneous equation or how would be best?Yes, solve it using a simultaneous equation. Let me know what answers you get ;)
Post by: 2020hsc on June 01, 2020, 07:27:11 pm
Yes, solve it using a simultaneous equation. Let me know what answers you get ;)
thanks!!
coming out with b= -2, a = 4 or b= 5/2, a = 25/4
sound about right?
Post by: Einstein_Reborn_97 on June 01, 2020, 08:00:28 pm
thanks!!Yep, correct. Both solutions work for both the GP and the AP.
coming out with b= -2, a = 4 or b= 5/2, a = 25/4
sound about right?
Post by: RuskiBrah on June 02, 2020, 09:04:26 pm
Post by: Einstein_Reborn_97 on June 02, 2020, 09:30:37 pm
hey guys, was wondering what formula to use for this question and why. cheersHey RuskiBrah,
For (a): Use the formula for the sum of a geometric series:
a=45, n=6, r=0.4 (so make sure to use to the correct formula; -1<r<1).
Why? Initial height (first term) is 45 cm. Common ratio, r is 0.4: "grows by 2/5 of its previous growth each month". n in this case is the number of months so is equal to 6.
Your answer should be 74.7 cm.
For (b): Use the formula for limiting sum of an infinite geometric series:
a=45 and r=0.4
Why? The question is asking for the lamb's "final height" and for geometric series where -1<r<1, you can find its limiting sum because the series converge.
Your answer should be 75 cm.
Hope that helps! Let me know if you need any further explanations ;)
Post by: RuskiBrah on June 02, 2020, 10:11:00 pm
Post by: RuskiBrah on June 02, 2020, 10:18:04 pm
Hey RuskiBrah,
For (a): Use the formula for the sum of a geometric series:
a=45, n=6, r=0.4 (so make sure to use to the correct formula; -1<r<1).
Why? Initial height (first term) is 45 cm. Common ratio, r is 0.4: "grows by 2/5 of its previous growth each month". n in this case is the number of months so is equal to 6.
Your answer should be 74.7 cm.
For (b): Use the formula for limiting sum of an infinite geometric series:
a=45 and r=0.4
Why? The question is asking for the lamb's "final height" and for geometric series where -1<r<1, you can find its limiting sum because the series converge.
Your answer should be 75 cm.
Hope that helps! Let me know if you need any further explanations ;)
so in other words, its a limiting sum because its height could go on infinitely?
Post by: Einstein_Reborn_97 on June 02, 2020, 10:35:26 pm
so in other words, its a limiting sum because its height could go on infinitely?Not exactly. When -1<r<1, the sum of a geometric series does not continue increasing greatly after a few times, instead it approaches some constant value - the limiting sum. E.g. 8+4+2+1+..., this series is said to converge, you can see that as the series continues, each term gets smaller and smaller (approaching zero) and eventually any changes to the sum of the series will be infinitesimal.
Contrast this with the series 2+4+8+16+32+..., the sum will become very large as n increases and we say these series diverge (their sum is infinite).
Hope that clarifies everything ;)
Post by: RuiAce on June 05, 2020, 11:25:37 am
If a question says "give an answer correct to ... decimal places" how do you know if you should round? From what I've seen, some answers are rounded and some aren't.If they're not rounding correctly then that's incorrect. You should always be rounding.
Post by: Einstein_Reborn_97 on June 05, 2020, 11:26:59 am
If a question says "give an answer correct to ... decimal places" how do you know if you should round? From what I've seen, some answers are rounded and some aren't.Always round off to whatever number of decimal places the question specifies...that's the correct way to present your answer. Don't worry if the answers from a textbook are different from yours by a single digit at the end because of different rounding. Just check your working out and re-calculate the answer. If your process is correct, you wouldn't lose marks for that.
Post by: Coolmate on June 10, 2020, 11:09:23 pm
Could I please have some help with this Definite Integral question:
I get the answer of:
But the answer is meant to be:
I have attached my full working, could someone please identify where I am going wrong?
Thanks in advance!
Coolmate 8)
Post by: wsdm on June 11, 2020, 12:12:31 am
I have attached my full working, could someone please identify where I am going wrong?There doesn't seem to be any error in your working out. I also attained \(\frac{65}{3}\) as my final answer. Are you completely sure there wasn't any typo in the question? If not, the textbook (if you're getting this off a textbook) might have the wrong answer, but don't worry, that happens a lot.
Post by: fun_jirachi on June 11, 2020, 12:44:02 pm
Hey Everyone!
Could I please have some help with this Definite Integral question:
I get the answer of:
But the answer is meant to be:
I have attached my full working, could someone please identify where I am going wrong?
Thanks in advance!
Coolmate 8)
Your working is totally fine - the only way I think \(\frac{2}{3}\) could be a reasonable answer is if we had the integral \(\int_{-1}^1 x^2 \ dx\). Perhaps you copied the bounds wrong (highly unlikely, but possible) or more likely the answers are just plain wrong.
Note that it's often redundant to rewrite your answer as a mixed fraction/decimal as the improper fraction is sufficient - leave it as an improper fraction (it'll save you time and computation!). Depending on how pedantic your teachers are, you may want to add \(u^2\) as units as well.
Post by: Coolmate on June 11, 2020, 06:13:04 pm
There doesn't seem to be any error in your working out. I also attained \(\frac{65}{3}\) as my final answer. Are you completely sure there wasn't any typo in the question? If not, the textbook (if you're getting this off a textbook) might have the wrong answer, but don't worry, that happens a lot.
Your working is totally fine - the only way I think \(\frac{2}{3}\) could be a reasonable answer is if we had the integral \(\int_{-1}^1 x^2 \ dx\). Perhaps you copied the bounds wrong (highly unlikely, but possible) or more likely the answers are just plain wrong.
Note that it's often redundant to rewrite your answer as a mixed fraction/decimal as the improper fraction is sufficient - leave it as an improper fraction (it'll save you time and computation!). Depending on how pedantic your teachers are, you may want to add \(u^2\) as units as well.
Thankyou wsdm and fun_jirachi, for the assistance!
So, funny story, I asked my teacher about it, then she looked at the question and what I wrote, and she showed me how I copied the bounds from another question, but wrote the "x" value from the right question..... ::) ::). So you were correct fun_jirachi, I did copy the bounds wrong. ;D
Thankyou again for the help!
Coolmate 8)
Post by: twelftholmes on June 22, 2020, 04:16:39 pm
(the question is attached as a jpg if you don't mind downloading it, I'm not sure how you insert a picture properly into a post oop)
I understand proving parts (i) and (ii) but I don't understand how to connect it to part (iii). I looked at the worked solutions but I can't make sense of it, could someone please explain it to me like I'm 5? haha thanks in advance!
Post by: fun_jirachi on June 22, 2020, 04:45:09 pm
You can use an image hosting service like imgur to include photos - if you want more information or are unsure how to do it, there's plenty of info around the forum/ you can ask again!
It's important to note that the question is a hence question. This implies you have to use the previous parts at some point or other. While this is usually as early as possible in the question, it very occasionally isn't - but luckily, here it definitely is. Thus, we sub in what we just proved in part (ii) to conclude that
From here, it's important to recognise that \(\int \sec ^2x \ dx\) is a standard integral that appears on your reference sheet - and that \(\int \sec x \tan x \ dx\) is given to you as part of the question. Hence, we find that
Hope this helps :)
Post by: twelftholmes on June 22, 2020, 06:17:10 pm
Also thanks for telling me I could use Imagur, I went and found out how to do it under New Users Lounge and its much better
If you don't mind there's this other question that I tried to work out myself but I have no idea where to start, could you please help me understand it? It's not from any HSC paper so I can't find worked solutions online, but I think it's from a bank where teachers have access to HSC style questions.
https://imgur.com/a/nFPINIq
Thank you so much in advance!
Post by: fun_jirachi on June 22, 2020, 06:40:25 pm
From here, you can integrate to find \(A(a)\) in simpler terms. Answer in the spoiler, but try and avoid using it where possible - I think it's important to get as much practice as possible with questions of this sort :)
Spoiler
Post by: Sine on July 17, 2020, 01:39:30 pm
Can someone please provide me the solution to the attached question (HSC 2007, Q7), I'm really finding it hardi) Solve the equation and find the x value solutions that are between 0 and 2pi. Sub the solutions for the x values back into either equation to find the y-values of the intersection thus you find the coordinates of A and B
Thank you soo much!
ii) Using the coordinates A and B use integration to find the shaded area between the two curves
Post by: jasminerulez9 on July 17, 2020, 04:33:58 pm
Post by: jasminerulez9 on July 17, 2020, 04:42:38 pm
Thank you soo much in advance!! :)
Post by: fun_jirachi on July 17, 2020, 05:40:26 pm
For the last part of the question, consider the fact that we can rewrite \(v\) as \(\frac{ds}{dt}\), where \(s\) is the displacement of the particle. We also have the initial condition that at t=0, s=0. Letting s=S at t=4, we have that:
Try evaluating this :)
For earlier parts of the question, here are my solutions for you to compare:
Spoiler
(a) = \frac{2t}{16+t^2} \\ v(0) = 0)
(b) = \frac{d}{dt} \frac{2t}{16+t^2} \\ a(t) = \frac{2(16+t^2)-4t^2}{(16+t^2)^2} = \frac{32-2t^2}{(16+t^2)^2})
(c)=0 \implies \frac{32-2t^2}{(16+t^2)^2} = 0 \\ 2t^2 = 32 \implies t = 4 \text{ (time can't be negative)} )
(b)
(c)
Hope this helps :)
Post by: jasminerulez9 on July 17, 2020, 06:30:19 pm
Post by: Coolmate on September 05, 2020, 04:16:25 pm
Could someone please explain why a fraction comes out the front with this integral (attached), the reasoning behind it and how to do this with any other integral?
Also, when doing revision for the HSC exam for maths, is it best to just work through the NESA past papers? Or is there anything else that would benefit my studies?
Thanks in advance!
Coolmate 8)
Post by: Justin_L on September 05, 2020, 04:39:06 pm
Hi Everyone! :D
Could someone please explain why a fraction comes out the front with this integral (attached), the reasoning behind it and how to do this with any other integral?
Also, when doing revision for the HSC exam for maths, is it best to just work through the NESA past papers? Or is there anything else that would benefit my studies?
Thanks in advance!
Coolmate 8)
Hey Coolmate,
-1/3 in this case is a constant, which can be removed from the integral. This can be done with any integral as the constant doesn't affect the end result since it's applied at the end anyways.
Not sure if I explained properly, hopefully this makes sense!
EDIT: With doing revision, I'm wondering that myself too. Based on my trials, my teacher actually discouraged me from doing past papers and advised that I work through revision first as to build my way up to exam level questions. Not sure if this is useful to you, hopefully someone else in this thread will have some good tips.
Post by: Justin_L on September 05, 2020, 05:40:59 pm
Simplify sinθcosθ + cos³θ/sinθ into a single trigonometric ratio
I've gotten to sinθcosθ + cotθcos²θ, but I'm not quite sure how to go further. Any help would be appreciated, as well as any tips on solving these types of problems!
Post by: FlammaZ on September 05, 2020, 05:58:37 pm
a-If c=12, calculate the probability that a type A cell will be misclassified, and the probability that a type B cell will be misclassified.
b-Find the value of c for which the two probabilities of misclassification are equal.
Post by: jamonwindeyer on September 05, 2020, 06:57:00 pm
Heyo, would appreciate help with this question:
Simplify sinθcosθ + cos³θ/sinθ into a single trigonometric ratio
I've gotten to sinθcosθ + cotθcos²θ, but I'm not quite sure how to go further. Any help would be appreciated, as well as any tips on solving these types of problems!
Hey! I'll try and give you my thought process. So I actually don't necessarily think your first step was a good move - Why? Because I see sines and cosines, I like this because it lets us use that great \(\sin^2+\cos^2=1\) rule. It takes a bit, but you get a bit of a gut feel to keep cosines and sines around as much as possible.
New first step - It wants it as a single ratio, so let's get a common denominator asap. Pull a \(\frac{1}{\sin{\theta}}\) out of the first term, and hopefully you can follow my steps:
These questions take practice, and my only advice is to try and get closer to what they want as soon as you can. They want a single ratio? Get a single fraction first. And in general, keep cosines and sines around as long as you can because they generally are more flexible to work with :)
Post by: jamonwindeyer on September 05, 2020, 07:15:30 pm
The amount of a certain chemical in a type A cell is normally distributed with a mean of 10 and a standard deviation of 1. The amount in a type B cell is normally distributed with a mean of 14 and a standard deviation of 2. To determine whether a cell is type A or type B, the amount of chemical in the cell is measured. The cell is classified as type A if the amount is less than a specified value c, and as type B otherwise.
a-If c=12, calculate the probability that a type A cell will be misclassified, and the probability that a type B cell will be misclassified.
b-Find the value of c for which the two probabilities of misclassification are equal.
Hey! Have you tried drawing the two distributions on a graph to visualise the problem at all? I think that would be a good start. Here's the diagram I pulled up to help myself with this:
(https://upload.wikimedia.org/wikipedia/commons/thumb/8/8c/Standard_deviation_diagram.svg/1200px-Standard_deviation_diagram.svg.png)
Draw yourself the two distributions for Type A and Type B, draw a dotted line at 12, and hopefully these will guide you:
- Consider just Type A for a bit. It is mean 10, std-dev of 1. 12 is two standard deviations above the mean. This means that only about 2.2% of Type A cells will have more than 12 units of this chemical. So there is a 2.2% chance that a Type A cell is misclassified.
- For Type B, it has a larger standard deviation (there is more spread in the distribution). If you had a to scale drawing provided, you'd see that the Type B curve spreads below the line at c=12 more than the Type A one spreads above it. 12 is only standard deviation below the mean of Type B cells. Therefore, you see a 15.8% chance of being misclassified.
For the last bit, you need to find where to draw the line such that the percentage of Type A cells above the line, is equal to the percentage of Type B cells below the line. Conceptually, you are looking for the value which is equally 'far' from the centre of the two distributions, when you consider that one standard deviation is larger than the other. Do you know of a numerical measure which takes into account mean and standard deviation in this way?
Have a think and if you're stuck...
CLICK ME IF STUCK
Use z scores! The misclassification rate will be the same when the z-score is the same (except one will be positive, one will be negative, because \(c\) sits on opposite sides of the mean for each distribution).
Solve for your answer :)
Solve for your answer :)
Hopefully this helps!
Post by: fun_jirachi on September 05, 2020, 09:28:49 pm
Also, when doing revision for the HSC exam for maths, is it best to just work through the NESA past papers? Or is there anything else that would benefit my studies?
EDIT: With doing revision, I'm wondering that myself too. Based on my trials, my teacher actually discouraged me from doing past papers and advised that I work through revision first as to build my way up to exam level questions. Not sure if this is useful to you, hopefully someone else in this thread will have some good tips.
At this point, it's probably best to be doing papers since that is what you'll be doing 6 weeks or so down the track. However, there are still very good reasons not to be doing them as well, especially if there are holes in your understanding (based on trial results, which are a good indicator of what you need to work on in the next month or two). In general, give yourself a break and tackle light stuff (you'll have graduated real soon - enjoy yourself!) as a bare minimum (the break part is super important!) - and depending on how confident you are, try papers/extra targeted revision. Just remember not to leave doing papers until too late - there really is no substitute for those :)
Post by: Coolmate on September 05, 2020, 10:02:36 pm
Hey Coolmate,
-1/3 in this case is a constant, which can be removed from the integral. This can be done with any integral as the constant doesn't affect the end result since it's applied at the end anyways.
Not sure if I explained properly, hopefully this makes sense!
EDIT: With doing revision, I'm wondering that myself too. Based on my trials, my teacher actually discouraged me from doing past papers and advised that I work through revision first as to build my way up to exam level questions. Not sure if this is useful to you, hopefully someone else in this thread will have some good tips.
Thankyou Justin_L for the clarification! :)
At this point, it's probably best to be doing papers since that is what you'll be doing 6 weeks or so down the track. However, there are still very good reasons not to be doing them as well, especially if there are holes in your understanding (based on trial results, which are a good indicator of what you need to work on in the next month or two). In general, give yourself a break and tackle light stuff (you'll have graduated real soon - enjoy yourself!) as a bare minimum (the break part is super important!) - and depending on how confident you are, try papers/extra targeted revision. Just remember not to leave doing papers until too late - there really is no substitute for those :)Thanks fun_jirachi for the advice, I really appreciate it! ;D
Coolmate 8)
Post by: FlammaZ on September 23, 2020, 09:56:18 pm
Find the initial rate of flow of water into the tank.
Find the value of t for which dV/dt=0.
Find the time, to the nearest second, when the rate is 1 litre per minute.
Find the first time, to the nearest second, when dV/dt<0.1.
4-Find the amount of water in the tank when t=5.
5-Find the time, to the nearest second, when there are 10 litres of water in the tank.
the last two is all i am unclear with(4,5), because when you integrate dv/dt, you get the v (the amount of water in the tank) and t( the time at which this volume is), why do you need to find the area? note, dv/dt is 10 times e to the power of (-t-1) times (5-t)
Post by: fun_jirachi on September 24, 2020, 01:39:33 pm
Reasonably sure that integral is outside the scope of the syllabus as well (unless they gave you some other prelude that would point you in that direction). Where did you get this question from? It would really help if a) you showed us what you've done already and/or b) made your request a little clearer because it'd really help us help you - hope this makes sense :)
Post by: FlammaZ on September 24, 2020, 05:53:46 pm
What i am trying to say is, once you integrate dv/dt, you get v(t), v representing the amount of water in the tank, and t the time in minutes. Now, number 4 is asking when t=5, what is the amount in the tank, it seems simple to just sub 5 into the equation and get the amount. Why do you have to find the area under the curve instead ?
Post by: fun_jirachi on September 24, 2020, 06:15:21 pm
You can find the area under the curve, but you don't have to - that is the point I'm making. Your method is entirely valid. It is also valid to find the area under the curve because the time was constrained to \(0 \leq t \leq 5\). Since we're given the initial condition that the tank was empty, it's logical that \(v(5)\) is equal to the area under the curve anyway., since an upper limit was already placed on t = 5. Well within your rights to go for whichever way you choose - as long as you have valid working and the correct answer, no one will dock you marks, even if your method is different. This applies for any other question (you said: 'Why do you have to find the area under the curve instead ? ') - you NEVER have to do x method to solve y question. You can use any valid method to get to the one correct answer. Fundamentally everything runs under the same or similar principles, or they 'coincidentally' align.
Hope this helps :)
Post by: FlammaZ on September 24, 2020, 07:00:00 pm
Post by: keltingmeith on September 24, 2020, 07:06:17 pm
I really appreciate your help, here are some photos of solution. Here V(5) is different to the correct answer, but the question says the tank is initially empty, so shouldn't v(0)=0, ? confused. :-\
V(0)=0 is a correct assumption - the problem is there seems to be an issue with your integration. Here's a hint - just because V(0)=0, does not mean that C=0. (also, have you written x's instead of t's? This is a bit of an issue, make sure to write for the pronumeral given in the question!)
Also, your 1s look a lot like 7s - if I were you, I'd consider trying to write 1s as just a straight line, and draw 7s with a line through the middle, like so:
(https://upload.wikimedia.org/wikipedia/commons/thumb/5/5c/Hand_Written_7.svg/800px-Hand_Written_7.svg.png)
I feel this is just a little more obviously different than the way you're doing 1 is.
Post by: FlammaZ on September 24, 2020, 07:16:06 pm
Post by: keltingmeith on September 24, 2020, 07:38:42 pm
apologies, not so easy to write on a computer with your finger ;D . could you explain what you mean by the c=0 part. fyi, V(0) = -14.72, however this is not logical. (But it does not = 0)
Wait, HSC uses CAS calculators now? I thought they weren't allowed
Okay, so, if we take:
And integrate with respect to t, we get:
Now, this ONLY matches what you wrote if C=0. Watch what happens if I say V(0)=0, though:
Which is very different to what you wrote.
---
After writing this, I think I've realised the confusion - you do know that after every indefinite integral, you MUST include the constant of integrate (often referred to as, "always remember your +C"), right?
Post by: FlammaZ on September 24, 2020, 07:49:23 pm
Also im not doing HSC im in a different state, i am really new to the site and don't know how to really use it thank you for helping me out though.
Post by: fun_jirachi on September 24, 2020, 07:50:01 pm
Wait, HSC uses CAS calculators now? I thought they weren't allowed
We don't - evaluating this by hand requires IBP, which is outside the scope of Maths Advanced. Must be either the wrong board or some textbook exercise :)
apologies, not so easy to write on a computer with your finger ;D . could you explain what you mean by the c=0 part. fyi, V(0) = -14.72, however this is not logical. (But it does not = 0)
Don't like answering questions that have already been answered, but I feel like this is important enough to warrant an additional response :)
This sort of working is sort of what we require to help you adequately - in hindsight the confusion wasn't really about methodology as previously thought (I am no longer confused :D). Assuming the initial condition isn't a requirement since it's given in the question (and I quote from the question 'Initially the tank is empty.') - you've got to do what keltingmeith has already said you should do, as this implies that V(0) = 0, not -14.72 as you've said. Watch out for these sorts of statements as these give you the initial condition that is required to get the proper solution.
You either have two choices when finding V(t) - a) integrate as an indefinite integral, then substitute the initial condition in as shown by keltingmeith (don't forget +C), or b) integrate as a definite integral with v and t as upper bounds and the initial condition as a lower bound (or vice versa). If you find yourself forgetting the constant of integration, definitely go with the second method - this is the reason you have negative volume as an initial condition (ideally something this weird should tip you off that's something's gone wrong in your calculation rather than your methodology - since methods are something people tend to get correct in my experience).
Hope this helps :)
Post by: Ookei on September 30, 2020, 06:32:03 pm
Post by: fun_jirachi on October 01, 2020, 03:50:39 pm
You could probably get away with it if using a sign table. If you're using the second derivative, definitely write the evaluated value of the second derivative at the used point. That being said, it's always safer to write the value down.
For school assessments, regardless of what's said here or anywhere else, go by your teacher's directives. Otherwise the above is a good rule of thumb.
Post by: Coolmate on October 11, 2020, 11:55:58 pm
I am currently working through the NESA maths past papers and have come across some questions and was wondering if someone could please let me know if these questions are applicable to the new syllabus?
- Q15c (iii) and (iv) (2018)
- Q16a (i), (ii) and (iii) (2018)
- Q16c (2017)
- Q12 (2016)
- Q15b (2015)
Thanks in advance!
Coolmate 8)
Post by: Coolmate on October 22, 2020, 02:24:29 pm
Could someone please step through on how to do the following questions as I am unsure how to do them:
- 12a (2019)
- 12d (2019)
- 15bi and ii (2017)
Thanks in advance!
Coolmate 8)
Post by: fun_jirachi on October 22, 2020, 05:20:39 pm
12a)i)
- Find the gradient for the line AC. Recall that for lines \(l_1\) and \(l_2\), the respective gradients \(m_{l_1}\) and \(m_{l_2}\) must have product -1.
- Use the point C (which has coordinates given), and substitute the previous result into the point-gradient form of the line.
12a)ii).
- You can find the x-intercept of the line calculated above, then find the length of BC (reasonably easy since they both have y coordinate 0), then use formula \(A = \frac{bh}{2}\) (think about where you can find the height!)
12d)
Basically asking you to evaluate \(\int_0^3 \frac{3x}{x^2+1} \ dx\). Recall that \(\int \frac{f'(x)}{f(x)} \ dx = f(x) + C\) - try manipulating this integral into a better form so you can use this formula.
15b)i)
Note that \(M_0 = 0\), and that \(M_1 = X(1+\frac{\frac{4.2}{12}}{100})-2500 = (1.0035)X - 2500\). Always start with \(M_0\), and find \(M_1\) after, executing each successive action in order. Things to note here are that \(M_0 = 0\) because the withdrawal happens the day before the deposit (withdrawal is on the last day, deposit on the first day of the next month), and \(M_1\) is that expression because the month's deposit has been placed, interest has been tacked on, then the withdrawal occurs.
Executing instructions in this order on \(M_1\) means that \(M_2 = 1.0035(M_1) - 2500\). You can work it out from there - take away is the setup is so important and order matters.
ii)
After this, you can generalise the formula for \(M_n\) based on the pattern. Try working out \(M_3\) and even \(M_4\) if you have to, so you can see what it is. This can be applied for any question.
here, they want you to use n = 4, and \(M_4 = 80000\). After this, you have an equation in X that you can rearrange and solve.
Hope this helps :)
Post by: HS26 on October 23, 2020, 06:44:37 pm
Post by: keltingmeith on October 23, 2020, 07:11:58 pm
Hi, can someone please help me with this question? I don't understand the highlighted part... Thank you so much :D
Note that I'm going to be answering this for discrete random variables, because that's what's in this question - however, all of this also applies for continuous random variables, just replace the summation with an integration. I also don't know which part in there you're confused by, so forgive me if I cover something you already know. In general:
You are most familiar with this in terms of the mean, or the expected value of X. That is:
But I could put any function in there. Here's some examples:
So that's what the E[X^2] is. As for the formula:
This one is on your formula sheet, and is just a different way of calculating the variance. Let me know if I didn't answer what you were confused by
Post by: Coolmate on October 23, 2020, 07:54:56 pm
Hey :)
12a)i)
- Find the gradient for the line AC. Recall that for lines \(l_1\) and \(l_2\), the respective gradients \(m_{l_1}\) and \(m_{l_2}\) must have product -1.
- Use the point C (which has coordinates given), and substitute the previous result into the point-gradient form of the line.
12a)ii).
- You can find the x-intercept of the line calculated above, then find the length of BC (reasonably easy since they both have y coordinate 0), then use formula \(A = \frac{bh}{2}\) (think about where you can find the height!)
12d)
Basically asking you to evaluate \(\int_0^3 \frac{3x}{x^2+1} \ dx\). Recall that \(\int \frac{f'(x)}{f(x)} \ dx = f(x) + C\) - try manipulating this integral into a better form so you can use this formula.
15b)i)
Note that \(M_0 = 0\), and that \(M_1 = X(1+\frac{\frac{4.2}{12}}{100})-2500 = (1.0035)X - 2500\). Always start with \(M_0\), and find \(M_1\) after, executing each successive action in order. Things to note here are that \(M_0 = 0\) because the withdrawal happens the day before the deposit (withdrawal is on the last day, deposit on the first day of the next month), and \(M_1\) is that expression because the month's deposit has been placed, interest has been tacked on, then the withdrawal occurs.
Executing instructions in this order on \(M_1\) means that \(M_2 = 1.0035(M_1) - 2500\). You can work it out from there - take away is the setup is so important and order matters.
ii)
After this, you can generalise the formula for \(M_n\) based on the pattern. Try working out \(M_3\) and even \(M_4\) if you have to, so you can see what it is. This can be applied for any question.
here, they want you to use n = 4, and \(M_4 = 80000\). After this, you have an equation in X that you can rearrange and solve.
Hope this helps :)
Hey fun_jirachi! ;D
Thankyou so much for the help, it now makes sense!
Coolmate 8)
Post by: rirerire on October 23, 2020, 11:28:02 pm
Equation: 2x^2 +8x +k=0
the roots are: a and B (alpha beta signs)
if a + B= -4
And a^2B + aB^2=6
Find k
Answer is: k/2(-4)=6
-2k=6
k= 6
I don’t get the working out at all... why it’s k/2, shouldn’t it be 2k?
thanks!
Post by: keltingmeith on October 23, 2020, 11:58:15 pm
could someone pls help with HSC 2014 Question 14 b ii?
Equation: 2x^2 +8x +k=0
the roots are: a and B (alpha beta signs)
if a + B= -4
And a^2B + aB^2=6
Find k
Answer is: k/2(-4)=6
-2k=6
k= 6
I don’t get the working out at all... why it’s k/2, shouldn’t it be 2k?
thanks!
Nope. We want to solve:
We also know that alpha and beta are roots. So you'd think that should be equivalent to saying:
But if you expand the RHS, you'll notice that the x^2 doesn't have a two out the front! So instead, you need to get the coefficient of the x^2 to be 1. To do that, divide everything by 2:
Next, we can do the equality as you were thinking of:
And substituting into the third equation you gave:
Which is exactly what is given by the supplied answer.
Tl;dr, you might think it's 2k because the coefficient in front of the x^2 is 2, but remember that you want to undo that coefficient, as you can see by my working from first principles - hence why it's k/2, not 2k.
Post by: Coolmate on October 24, 2020, 08:32:34 pm
Could I please have help with working out the correlation coefficient (r) and part ii, in the attached question?
Thanks in advance,
Coolmate 8)
Post by: svnflower on October 24, 2020, 10:25:33 pm
Hi Everyone!
Could I please have help with working out the correlation coefficient (r) and part ii, in the attached question?
Thanks in advance,
Coolmate 8)
Hello :)
To find r, input your values into a table. To do this, click MODE > 2 > 2 then plug in your values. After that, hit the AC button > SHIFT > 1 > 5 > 3. And there's your correlation coefficient number!
Spoiler
p.s. i got an answer of 0.959 for that question
Post by: Coolmate on October 24, 2020, 10:47:07 pm
Hello :)
To find r, input your values into a table. To do this, click MODE > 2 > 2 then plug in your values. After that, hit the AC button > SHIFT > 1 > 5 > 3. And there's your correlation coefficient number!Spoilerp.s. i got an answer of 0.959 for that question
Wow! Thanks, svnflower ;D This really helped me, I was so confused, but now it is so clear ;D. Also, I got the same answer as you
Coolmate 8)
Post by: svnflower on October 24, 2020, 11:00:21 pm
Wow! Thanks, svnflower ;D This really helped me, I was so confused, but now it is so clear ;D. Also, I got the same answer as you
Coolmate 8)
YAY happy to help, all the best for Monday! :)
Post by: BakerDad12 on October 25, 2020, 11:14:50 am
Post by: fun_jirachi on October 25, 2020, 01:38:37 pm
If the graph is concave down over an interval [a, b], then the trapezoidal rule will evaluate a value less than the actual integrated value over that same interval, and vice versa - you can think about this by drawing it out and seeing that the trapezoidal rule effectively truncates area/adds area on. If the slope of the curve is constant, then the trapezoidal rule will evaluate a value equal to the actual integrated value.
Post by: BakerDad12 on October 25, 2020, 02:11:54 pm
Post by: fun_jirachi on October 25, 2020, 02:51:50 pm
Consider what happens if we have a function \(f(x)\) bounded by the interval \([a, c]\) and \(b = \frac{a+c}{2}\). By the trapezoidal rule, \(\int_a^c f(x) \ dx \approx \frac{b-a}{2}(f(a) + 2f(b) + f(c))\), if we use two intervals. However, we can rewrite the right hand side as \(\frac{b-a}{2}(f(a)+f(b)) + \frac{c-b}{2}(f(b)+f(c))\) since b is the midpoint of a and c. Note that this also approximates the integral and is also the application of the trapezoidal rule on one interval twice!
Hope this helps :)
Post by: Coolmate on October 25, 2020, 05:20:15 pm
YAY happy to help, all the best for Monday! :)
Thanks, you too! :D
Post by: Coolmate on October 25, 2020, 08:19:52 pm
Just a quick question before tomorrows exam:
Using the attached question as an example, would someone please show me how to determine the equation of the least-squares regression line?
Thanks in advance!
Coolmate 8)
Post by: needhelp101 on February 03, 2021, 10:17:03 pm
I'm in Year 11 and I need help on this Question:
The questions asks me to prove the following inequality is true by considering the reciprocals of both sides:
Thanks in advance.
Post by: fun_jirachi on February 03, 2021, 10:27:18 pm
They're basically asking you to take the reciprocals then rationalise the denominator without explicitly telling you the second step.
This is the true statement you're supposed to find. Note that you can't actually present this as a proof because you're starting with what you're trying to prove, but working backwards from the last line of working is an acceptable proof of the inequality in the question (and this method of 'discovering a proof' is very handy in a lot of HS maths, especially compared to blindly using brute force). Working backwards is fine because you're starting with a statement of fact, then proving (or maybe in some other case disproving) the statement presented in the question.
Hope this helps
Post by: Maroon and Gold Never Fold on May 31, 2021, 11:18:56 pm
Post by: RuiAce on June 01, 2021, 12:17:31 am
Attached a question I am unable to do.For \(y>0\),
\begin{align*}
F_Y(y) &= P(Y\leq y)\\
&= P(\sqrt{X} \leq y)\\
&= P(X \leq y^2) \tag{square both sides}\\
&= F_X(y^2) \tag{because we now have $X$}\\
&= 1-e^{-y^2}.
\end{align*}
You can then find the density function through the usual way (taking derivative).
For any future questions, you should post your attempts at solving them, no matter how right or wrong they are. That way we can be sure you've given them a go first, and can also understand your thought process to give you actual valuable help.
Post by: Maroon and Gold Never Fold on June 01, 2021, 08:10:45 am
For any future questions, you should post your attempts at solving them, no matter how right or wrong they are. That way we can be sure you've given them a go first, and can also understand your thought process to give you actual valuable help.
Thanks for your help. I was really stuck on the wording of the question and I'm still a bit lost on what I'm sort of meant to do.
Post by: RuiAce on June 01, 2021, 12:32:36 pm
Thanks for your help. I was really stuck on the wording of the question and I'm still a bit lost on what I'm sort of meant to do.You started with a random variable \(X\). You were given the CDF of \(X\).
You then pull out a new random variable \(Y\). You are told that this new random variable is related to the old one, by the equation \( Y = \sqrt{X}\). (So you see that \(Y\) is in fact a function of another random variable, and hence also a random variable.)
Since \(Y\) is a new random variable, it should also have a CDF. You are asked to find this CDF in the first part. (And then using it, you can find the PDF in the second part.)
(But how would you be able to find the CDF of \(Y\) to begin with? Well, you're given the relationship \( Y = \sqrt{X}\), and the CDF of \(X\). So somehow you had to puzzle the two together.)
The actual technique used is understandably one you might've not seen before. But as for what the question was asking, you needed to realise most of the above.
Post by: ethan.lozevski on June 03, 2021, 03:53:50 pm
I have attached a question that has confused me a little. I used Ali's model and put all my solutions over 19, to sum up to 1 as a pdf requires. Please let me know if I am doing this question correctly. Thanks, guys.
Post by: fun_jirachi on June 03, 2021, 06:44:43 pm
A few questions in return:
- What are the chances you roll a 1 (and subsequently score 4 points) on the new distribution? It is still a fair die.
- Do some values on the original die yield duplicate numbers of points when rolled? What does this do for the probability of scoring a particular y value?
Hope this helps :)
Post by: RuiAce on June 03, 2021, 09:47:11 pm
Hey guys,Building onto fun_jirachi's answer here. I would strongly advise considering the second point that he mentioned.
I have attached a question that has confused me a little. I used Ali's model and put all my solutions over 19, to sum up to 1 as a pdf requires. Please let me know if I am doing this question correctly. Thanks, guys.
Your table of values is good up until the second row. Then I believe you are confusing yourself with the third row. You want to find the probability distribution for \(Y\). Just using the actual values that \(Y\) takes on, is not the correct approach here.
Hint: Firstly, what are the values that \(Y\) can actually be? (Refer to the second row of your table of values. There are 3 distinct values that \(Y\) can take on.) Secondly, for each of those three values of \(y\), what is \(P(Y=y)\)?
Post by: ethan.lozevski on June 04, 2021, 09:17:11 am
Thank you.
Post by: fun_jirachi on June 04, 2021, 12:34:29 pm
The second question still stands however.
Hint: Firstly, what are the values that \(Y\) can actually be? (Refer to the second row of your table of values. There are 3 distinct values that \(Y\) can take on.) Secondly, for each of those three values of \(y\), what is \(P(Y=y)\)?
Should such a distribution have duplicate values for \(Y\)? If so, how would we amend our probability \(P(Y = y)\)? Definitely getting closer though, give it another shot :)
Post by: Maroon and Gold Never Fold on June 06, 2021, 08:45:07 pm
Thanks
Post by: fun_jirachi on June 06, 2021, 08:52:03 pm
Post by: Maroon and Gold Never Fold on June 07, 2021, 06:13:01 pm
Seems about right :) - just be careful with the first line, you haven't subscripted the i, it looks like it's being multiplied onto the numerator as opposed to indexing the score
What do you mean by I haven't subscripted the 'i'. In my picture it is being multiplied on my numerator, is that wrong?
Post by: fun_jirachi on June 07, 2021, 06:24:52 pm
Post by: Maroon and Gold Never Fold on June 07, 2021, 09:50:32 pm
Yes - \(\sum_{i=1}^{n} \frac{(a+x)i}{n} = \frac{(a+x)(n - 1)}{2}\), which will lead you to an incorrect conclusion. If you notice in the formula that is given for the mean, i is a subscript denoting index (ie. \(x_i\) denotes the \(i^{\text{th}}\) score). Your first line should be something like \(\sum_{i=1}^{n} \frac{(a+x_i)}{n}\).
thanks, should I leave the subscript 'i' on 'a' in my second line of working out then.
Post by: fun_jirachi on June 07, 2021, 10:34:36 pm
Post by: neha.singh4 on July 03, 2021, 09:42:23 pm
As apart of a maths assessment task I was wondering what were some examples of bivariate data where one pair of variables shows a
positive correlation and the other showing a negative correlation. For example, some that I have are: Arm Length vs Foot Length, Height and Running Speed but am looking for some other ones.
And also what kinds of predictions can you gather from data when interpolating and extrapolating values?
Thank you! Any help would be greatly appreciated!!!
Post by: fun_jirachi on July 03, 2021, 10:20:32 pm
Hello!
As apart of a maths assessment task I was wondering what were some examples of bivariate data where one pair of variables shows a
positive correlation and the other showing a negative correlation. For example, some that I have are: Arm Length vs Foot Length, Height and Running Speed but am looking for some other ones.
And also what kinds of predictions can you gather from data when interpolating and extrapolating values?
These are good examples of positive correlation; was there anything in particular you were looking for for your examples? The question is pretty vague :(
Extrapolating is generally risky, since you can't always guarantee trends and correlation extend to a range values not within your measured range. Interpolation of bivariate data given one variable can allow you to predict with reasonable accuracy the other variable given a large enough sample size ie. P(Y|X) or P(X|Y).
Post by: neha.singh4 on July 04, 2021, 11:37:29 pm
These are good examples of positive correlation; was there anything in particular you were looking for for your examples? The question is pretty vague :(
Extrapolating is generally risky, since you can't always guarantee trends and correlation extend to a range values not within your measured range. Interpolation of bivariate data given one variable can allow you to predict with reasonable accuracy the other variable given a large enough sample size ie. P(Y|X) or P(X|Y).
Oh yes that makes sense! Thank you! : ))))
Yes I agree the question is pretty vague too. In terms of what my teacher had explained, she'd like for us to show one set of data that reveals a positive correlation and another set of data that reveals a negative correlation. I was looking for some examples where two sets of variables could fulfill this requirement as well as allow me to analyse data in terms of the Pearson's correlation. Any recommendations?
Post by: fun_jirachi on July 05, 2021, 10:32:11 pm
Post by: Maroon and Gold Never Fold on July 20, 2021, 09:02:54 pm
I initially attempted it by making it into this like I would with any absolute function
This did not work, I looked at the solutions and they revomed the absolute value bars, why is this? is anyone able to explain to me how to do this inequality.
Thanks
Post by: fun_jirachi on July 20, 2021, 09:27:45 pm
As for why they removed the absolute value bars, it's because they can; you can always construct a solution that is logically/mathematically identical but without them.
Please post your working, I'd really like to see it :) (if required I'll post an actual solution afterwards)
Post by: 1729 on July 20, 2021, 10:09:14 pm
For this inequalityAdding onto fun_jirachi's advice (and yes please tell us what you have tried to do before asking a question - helps us outline the errors you have made to better help you)^2|<1)
I initially attempted it by making it into this like I would with any absolute function
This did not work, I looked at the solutions and they revomed the absolute value bars, why is this? is anyone able to explain to me how to do this inequality.
Thanks
If we have function \(f(x)\), and make it \(\left|f\left(x\right)\right|\) it essentially flips any cordinates of \(f\left(x\right)<0\) (cordinates that lie under the \(y\) axis) over the \(x\) axis since absolute values make negative numbers positive by definition, ie. \(\left|-p\right|=p\) and \(\left|p\right|=p\)
Notice how in the function you sent there is nothing to flip over the \(x\) axis or there isn't anything below \(y=0\)? In this case, your function is equal to the absolute value of it since it's always positive similarly, if p is positive, the absolute value of p is just itself \(\left|p\right|=p\) ie. \(\frac{x^2}{\left(1-x\right)^2}>\:0\) is always more than 0 (positive)
To get more of a visual understanding click here
Post by: Maroon and Gold Never Fold on July 21, 2021, 10:23:06 am
Adding onto fun_jirachi's advice (and yes please tell us what you have tried to do before asking a question - helps us outline the errors you have made to better help you)
If we have function \(f(x)\), and make it \(\left|f\left(x\right)\right|\) it essentially flips any cordinates of \(f\left(x\right)<0\) (cordinates that lie under the \(y\) axis) over the \(x\) axis since absolute values make negative numbers positive by definition, ie. \(\left|-p\right|=p\) and \(\left|p\right|=p\)
Notice how in the function you sent there is nothing to flip over the \(x\) axis or there isn't anything below \(y=0\)? In this case, your function is equal to the absolute value of it since it's always positive similarly, if p is positive, the absolute value of p is just itself \(\left|p\right|=p\) ie. \(\frac{x^2}{\left(1-x\right)^2}>\:0\) is always more than 0 (positive)
To get more of a visual understanding click here
So because the function is positive for all values of x, we can remove the absolute value bars because they make no difference?
Post by: 1729 on July 21, 2021, 10:28:12 am
So because the function is positive for all values of x, we can remove the absolute value bars because they make no difference?Yep that is correct.
Think about it, if you take the absolute value of a positive number, you are left with the same number! This rule applies to functions as well (:
Post by: crippledbyenglish on November 11, 2021, 05:24:23 pm
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I want to let you have a go at (iii) first. Use a graph similar to the graph in (ii). Come back with whatever progress you have.
Pleaseee, what is the answer to the last part
Post by: crippledbyenglish on November 12, 2021, 10:40:09 am
Thanks!
Post by: crippledbyenglish on November 12, 2021, 10:45:16 am
Post by: crippledbyenglish on November 12, 2021, 10:51:54 am
Post by: SnekiSnek on November 12, 2021, 02:10:13 pm
When I looked at the solutions for the question below it said that you should dilate and then translate which is weird because aren't you supposed to do what is in the parenthesis first? Is this a misconception?
I am in VCE not HSC but for methods it was to follow DRT
Dilations first, then reflections and then translations
Post by: fun_jirachi on November 12, 2021, 03:14:58 pm
Pleaseee, what is the answer to the last part
Have you had a go at the last part? That's the advice given in the original response, and I'm reluctant to deviate from it because I think it's 100% correct - you should have a go first.
Hi everyone! Just wondering if someone could let me know how they would have thought about the question below, needing to find tan from sin. I understand how they got the answer but didn't think to simplify the denominator by factorising. What would you look for/what hints would you take from the question to do this?
Thanks!
This sort of factorisation should be relatively standard as you do look into completing the square when you study solutions to quadratics. It's in a different form so it may not be immediately obvious, but you should be on the lookout for things like this regardless. More of a practice thing to get your eye in on these sorts of tricks. It's not strictly necessary to employ this trick in this particular instance, in any case.
When I looked at the solutions for the question below it said that you should dilate and then translate which is weird because aren't you supposed to do what is in the parenthesis first? Is this a misconception?
You're thinking of computing equations. SnekiSnek is correct
Someone please help I don't think I know how to use a calculator :'(
\(z = \frac{x - \mu}{\sigma}\). If the numbers in the table were all correct, \(\mu, \sigma\) would be equal in a system of linear equations. Solve two simultaneously to find values of \(\mu, \sigma\) then sub into the other result to see what adjustment needs to be made (and if it matches one of the other options given).
I'm not giving you the answer because I don't really see an attempt ('someone please help' is not really helpful to me because I want to help you target the bits you're struggling with, I'm only able to give you general advice in this case).
In future, please add more working and refrain from doing a double post, let alone a quadruple
Post by: Albertenouttaten on March 10, 2022, 06:03:32 pm
Post by: 1729 on March 11, 2022, 01:24:12 pm
Normal Distribution question that is absolutely melting my mind. Even the wording of the question is so tuff. Any tips and guidance on how to do this would be greatly appreciated.Hi - here are some tips on how you would go about this problem
For the purpose of my explanation let \(X\) be the normally distributed random variable such that \(X\)~\(N\left(\mu ,3^2\right)\). We want to find out what \(\mu\) is given that \(\text{Pr}\left(X\le 29\right)=0.16\).
I want you to consider that \(\text{Pr}\left(X<a\right)=\text{Pr}\left(Z<\frac{a-\mu }{\sigma }\right)\) where \(Z\) is the standard normal random variable.
Now solve \(\text{Pr}\left(Z\le \frac{29-\mu }{6}\right)=0.16\) for \(\mu\).
However, this method assumes access to technology to get a more accurate approximation.
If you did not have access to CAS, you could simply use the fact that approximately \(68\)% of values lie within 1 standard deviation of the mean, that is \(\mu -\sigma <x<\mu +\sigma \). This means that \(32\)% lie everywhere else, which means that \(16\)% lie \(x<\mu -\sigma \).
This essentially means that \(\text{Pr}\left(X<\mu -\sigma \right)\approx \text{Pr}\left(X<29\right)\approx 0.16\). Now solve \(29=\mu -\sigma \) or \(29=\mu -6\).
Post by: Sagarc on May 30, 2022, 11:31:42 pm
A flat verandah roof 1.8 m deep is 2.6 m up from the ground. At a certain time of day, the sun makes an angle of elevation of 72°25′. How much shade is provided on the ground by the verandah roof at that time, to 1 decimal place?
Post by: fun_jirachi on June 01, 2022, 10:55:42 am