Author Topic: 3U Maths Question Thread (Read 1553156 times) Tweet Share

jamonwindeyer · « **Reply #3615 on:** August 21, 2018, 06:42:57 pm »

Quote from: jazcstuart on August 21, 2018, 06:00:10 pm

Hi, I'm fairly slow at extension maths and struggle with time management in the exam. For trials there were quite a few questions I couldn't complete simply because of time (I think almost 10 marks worth). I usually get most things that I manage to complete correct, but I think that's because I take more time to do the questions. Especially for multiple choice, I find they take a long time to work out for just one mark. Does anyone else struggle with this, or have any tips to improve?

If you are spending a lot of time on MC and struggling to find time for other stuff, my first bit of advice would be to do the MC last! 10 marks, you are statistically likely to get at least 3 of them, and with educated guesses you could easily double those odds. Far better to come back with 10 minutes left and finish them off in a rush, get a few extra marks, than miss stuff altogether at the back end of the exam!

In terms of getting faster, it is just practice. You need to have done a question type enough times that you don't think about how it is done. You just do it. It becomes muscle memory almost! This gives you more time to devote to the curveballs which typically start getting thrown at you in Q13. Practice makes perfect

ravgill17 · « **Reply #3616 on:** August 23, 2018, 10:05:13 am »

I had a question in which it was not specified to prove my mathematical induction, however I did use this method. Instead it required finding sum of LHS and then differentiating both sides and subbing n=2. If I got the same result using mathematical induction, is this wrong?

RuiAce · « **Reply #3617 on:** August 23, 2018, 10:59:35 am »

Quote from: ravgill17 on August 23, 2018, 10:05:13 am

I had a question in which it was not specified to prove my mathematical induction, however I did use this method. Instead it required finding sum of LHS and then differentiating both sides and subbing n=2. If I got the same result using mathematical induction, is this wrong?

This is a very tough call and depends very much on the question itself.

Firstly, induction is reserved for proofs that involve only the integers, as opposed to all real numbers. Depending on the scenario, if your question was a binomial theorem proof it might be safe, but I can’t assume that it was binomial theorem given what you’ve stated.

Secondly, there’s also the issue of “hence”. If it specifies hence, as opposed to nothing or “hence or otherwise”, that’s also dangerous

envisagator · « **Reply #3618 on:** August 27, 2018, 09:05:40 pm »

Hei Rui, for the following S.H.M question how do I determine direction (+ or -)

The period of a particle moving in SHM is 6s and its amplitude is 8cm. Calculate it's velocity and acceleration (1 d.p) when the displacement is 5cm from the centre of the motion.

My working out: $\text{ I assumed acceleration as -ve since using} a=-n^2x \text{ substituting in 5 i got 'a' as negative } a\approx-5.5 \\ \text{ for v taking the square root of} \approx6.539 \text{ taking 2 solutions, +ve or -ve but I'm not sure how to determine which one it is}$

Thanks in Advance!!

RuiAce · « **Reply #3619 on:** August 27, 2018, 09:26:08 pm »

Quote from: envisagator on August 27, 2018, 09:05:40 pm

Hei Rui, for the following S.H.M question how do I determine direction (+ or -)

The period of a particle moving in SHM is 6s and its amplitude is 8cm. Calculate it's velocity and acceleration (1 d.p) when the displacement is 5cm from the centre of the motion.

My working out: $\text{ I assumed acceleration as -ve since using} a=-n^2x \text{ substituting in 5 i got 'a' as negative } a\approx-5.5 \\ \text{ for v taking the square root of} \approx6.539 \text{ taking 2 solutions, +ve or -ve but I'm not sure how to determine which one it is}$

Thanks in Advance!!

Did this come from a textbook by any chance? Because your concern is pretty legit and the correct answer is that there is no real way of determining which direction to use.

(Or at least from all the information provided.)

envisagator · « **Reply #3620 on:** August 27, 2018, 09:28:05 pm »

Quote from: RuiAce on August 27, 2018, 09:26:08 pm

Did this come from a textbook by any chance? Because your concern is pretty legit and the correct answer is that there is no real way of determining which direction to use.

good old margaret, thats wat I was thinking aswell (unable to determine direction), just needed to confirm. Thank You!!

Never.Give.Up · « **Reply #3621 on:** August 27, 2018, 09:49:22 pm »

Hi!

I am pretty worried about ext 1 maths hsc and was wondering if i could get some help from those here...
I reckon i failed my trials (we havent got our marks back yet), but we still have a lot of content to cover and I have only been getting average marks all year for this subject.. around the mid 70s- mid 80s mark...
I would love to get an E4 in this subject but I just don't know where to start (we missed a lot of yr 11 content because our teacher had an accident- so I dont know a lot about parametrics/locus, permutations/combination etc. and we haven't yet learnt extension: application of calculus to physical world, probability or binomials... so yes I am freaking out about the HSC and really want to do better

any tips would be super appreciated as well as any ways to understand the content... when you don't have a face-to-face teacher

Thanks so much!

envisagator · « **Reply #3622 on:** August 27, 2018, 10:20:22 pm »

Quote from: Never.Give.Up on August 27, 2018, 09:49:22 pm

Hi!
I am pretty worried about ext 1 maths hsc and was wondering if i could get some help from those here...
I reckon i failed my trials (we havent got our marks back yet), but we still have a lot of content to cover and I have only been getting average marks all year for this subject.. around the mid 70s- mid 80s mark...
I would love to get an E4 in this subject but I just don't know where to start (we missed a lot of yr 11 content because our teacher had an accident- so I dont know a lot about parametrics/locus, permutations/combination etc. and we haven't yet learnt extension: application of calculus to physical world, probability or binomials... so yes I am freaking out about the HSC and really want to do better
any tips would be super appreciated as well as any ways to understand the content... when you don't have a face-to-face teacher
Thanks so much!

Hi mate, I'm really sorry about the unfortunate circumstance with your teacher. Our school did our trials last term, and we hadnt covered applications of calculus nor perms and combs. We just finished these 2 topics.

You can still achieve that E4 even if you think failed, you just have to smash the externals, which is what actually counts. Trials as the name implies are a trial and are supposed to be harder than HSC.

My tips for you would be:
1. Watch Eddie Woo, hes got fantastic explanation of every topic you need to learn, I pretty much learned most of the course like this.
2. Keep your head high and always maintain a postive attitude, without this, watever you do wont have 100% effect.
3. Post questions you need help with, mathematical geniuses like Rui will help you out in no time.
4. With regard to not knowing where to start, I would suggest making a list of descending order of difficulty or lack of understanding. Work your way through this.
5. Make sure to take a break every now and then from doing a lot of maths, your brain can get clouded and your critical thinking and probleming solving ability drops.

Keep pushing ON!!!, your chance to get an E4 is not over until you finish that HSC mxt 1 exam

envisagator · « **Reply #3623 on:** August 29, 2018, 02:53:08 pm »

Hei Rui, can I please have some help with the following three questions:

$\text{1. Given that } \frac{d^2y}{dx^2}=\frac{2}{x^2} + 2e^{2x} \text{and that, when x=1,} \frac{dy}{dx}=e^2 \text{ and } y=\frac{e^2}{2} \text{ find an expression for y in terms of x, with no other unknon values} \\ \text{2. The sketch shows the curve } y^2+x^4=16 \text{ The area enclosed within the curve is rotated about the x-axis. Find the volume of the solid of revolution so formed.} \\ \text{3. Given } \log_m{p}=1.75 \text{ and } \log_m{q}=2.25 \text{ find } \sqrt[5]{pq^2} \text{ in terms of m}$

For question 2 what I did was 4 times the intergral from 0 to 2, but I dont quite understand how and why its wrong.

Thanks in advance !!!!!!

3.14159265359 · « **Reply #3624 on:** August 29, 2018, 03:18:34 pm »

Quote from: envisagator on August 29, 2018, 02:53:08 pm

Hei Rui, can I please have some help with the following three questions:

$\text{1. Given that } \frac{d^2y}{dx^2}=\frac{2}{x^2} + 2e^{2x} \text{and that, when x=1,} \frac{dy}{dx}=e^2 \text{ and } y=\frac{e^2}{2} \text{ find an expression for y in terms of x, with no other unknon values} \\ \text{2. The sketch shows the curve } y^2+x^4=16 \text{ The area enclosed within the curve is rotated about the x-axis. Find the volume of the solid of revolution so formed.} \\ \text{3. Given } \log_m{p}=1.75 \text{ and } \log_m{q}=2.25 \text{ find } \sqrt[5]{pq^2} \text{ in terms of m}$

For question 2 what I did was 4 times the intergral from 0 to 2, but I dont quite understand how and why its wrong.

Thanks in advance !!!!!!

hey Im not Rui but I tried q1 but I got a really weird answer. is it
y= -2lnx + 0.5e^(2x) + xe^2 -ex +2x -e^2 +e -2
or in simplified
y= -2lnx + 0.5e^(2x) + (x-1)e^2 + x(2-e) + e - 2

if yes, ill send my working out through

RuiAce · « **Reply #3625 on:** August 29, 2018, 03:24:39 pm »

Quote from: envisagator on August 29, 2018, 02:53:08 pm

Hei Rui, can I please have some help with the following three questions:

$\text{1. Given that } \frac{d^2y}{dx^2}=\frac{2}{x^2} + 2e^{2x} \text{and that, when x=1,} \frac{dy}{dx}=e^2 \text{ and } y=\frac{e^2}{2} \text{ find an expression for y in terms of x, with no other unknon values} \\ \text{2. The sketch shows the curve } y^2+x^4=16 \text{ The area enclosed within the curve is rotated about the x-axis. Find the volume of the solid of revolution so formed.} \\ \text{3. Given } \log_m{p}=1.75 \text{ and } \log_m{q}=2.25 \text{ find } \sqrt[5]{pq^2} \text{ in terms of m}$

For question 2 what I did was 4 times the intergral from 0 to 2, but I dont quite understand how and why its wrong.

Thanks in advance !!!!!!

Isn't Q1 just integrating it out twice? You first have $ \frac{dy}{dx} = -\frac{2}{x} + e^{2x} + c_1$ and the initial conditions give you $c_1 = 2$. And then you have $ y = -2 \ln x + \frac12 e^{2x} + 2x + c_2$...
$\text{Q2 should just be }\pi \int_{-2}^2 (16-x^4)\mathrm{d}x\\ \text{or if you want to exploit symmetry, }2\pi \int_0^{2} (16-x^4)\mathrm{d}x.$
$\text{This is because the formula }V = \pi \int_a^b y^2\mathrm{d}x\text{ assumes}\\ \text{that we have performed a }360^\circ\text{ rotation to obtain our solid.}$
$\text{However, if we keep both the positive }\textbf{and}\text{ negative branches of the curve,}\\ \text{without addressing this issue, then what we've really done}\\ \text{is rotated the region upon itself }\textbf{twice}.$
Picture an animation of what happens when you rotate just the upper branch. If you rotate by $360^\circ$ you trace out the solid you require. Then picture what happens when you rotate the entire shape. If you rotate it by $180^\circ$, you trace out the solid you require. But if you rotate it by another $180^\circ$, which is effectively speaking rotating by $360^\circ$, you've traced out the solid a second time.
$\text{Thus, given the curve's symmetry, for the sake of a volume of solid of revolution,}\\ \text{we should ignore the negative branch of the curve and only focus on the positive.}$
(Or ignore the positive and only focus on the negative, but that'd give you the same answer.)
A more formal analysis involves thinking about it in terms of the actual slices formed, as per the usual 4U approach.
$\text{For Q3, observe that}\\ \begin{align*}\log_m \sqrt[5]{pq^2} &= \frac15 \left( \log_m p + 2 \log_m q \right)\\ &= \frac15 \left(1.75 + 2\times 2.25 \right)\\ &=1.25\end{align*}\\ \text{so therefore }\sqrt[5]{pq^2} = m^{1.25}.$

Quote from: 3.14159265359 on August 29, 2018, 03:18:34 pm

hey Im not Rui but I tried q1 but I got a really weird answer. is it
y= -2lnx + 0.5e^(2x) + xe^2 -ex +2x -e^2 +e -2
or in simplified
y= -2lnx + 0.5e^(2x) + (x-1)e^2 + x(2-e) + e - 2

if yes, ill send my working out through

You posted your reply as I was typing - it looks like some bits match up, but probably compare it more carefully in case I made a mistake myself

3.14159265359 · « **Reply #3626 on:** August 29, 2018, 03:33:39 pm »

Quote from: RuiAce on August 29, 2018, 03:24:39 pm

$\text{For Q3, observe that}\\ \begin{align*}\log_m \sqrt[5]{pq^2} &= \frac15 \left( \log_m p + 2 \log_m q \right)\\ &= \frac15 \left(1.75 + 2\times 2.25 \right)\\ &=1.25\end{align*}\\ \text{so therefore }\sqrt[5]{pq^2} = m^{1.25}.$ You posted your reply as I was typing - it looks like some bits match up, but probably compare it more carefully in case I made a mistake myself

LOOL IS THAT ALL YOU DO FOR Q3 HAHAHA

ive attached my working out for q1 below.

3.14159265359 · « **Reply #3627 on:** August 29, 2018, 04:02:57 pm »

https://scontent.xx.fbcdn.net/v/t1.15752-0/p280x280/40421289_2208929539205207_8962370860006506496_n.jpg?_nc_cat=0&_nc_ad=z-m&_nc_cid=0&oh=3b38ca57b231ddcb8752074a146d667a&oe=5BEE3891

I tried my best to upload. not sure how it turned out

EDIT: I checked my working out and I had done a mistake with subbing in. Rui is right.
the equation should be
y= -2lnx + 0.5e^(2x) +2x -2

S200 · « **Reply #3628 on:** August 29, 2018, 04:05:30 pm »

Just use the

Code: [Select]

[img]...[/img] tags...

Like this....

3.14159265359 · « **Reply #3629 on:** August 29, 2018, 04:10:01 pm »

Quote from: S200 on August 29, 2018, 04:05:30 pm

Just use the
Code: [Select]
[img]...[/img] tags...

Like this....
(Image removed from quote.)

thank you so much!!!

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