Author Topic: VCE Methods Question Thread! (Read 6058553 times) Tweet Share

IndefatigableLover · « **Reply #9390 on:** March 23, 2015, 09:53:44 pm »

Quote from: Cosec on March 23, 2015, 09:44:11 pm

Healpppppp

Probably best to go over the quick method of chain rule

Just a heads up but a^2 is a constant so won't do anything!

a). $\frac{d}{dx}(\sqrt{x^2\pm a^2})$

$\therefore \frac{d}{dx}((x^2\pm a^2)^\frac{1}{2})$

$= \frac{1}{2} \cdot (x^2 \pm a^2)^{-\frac{1}{2}} \cdot 2x$

$=\frac{2x}{2\sqrt{x^2\pm a^2}}$

$=\frac{x}{\sqrt{x^2\pm a^2}}$

From there I think you can work out b). since it's literally the same thing but change the sign of x^2 and there's no plus minus sign (not that it should have an effect on the end result as shown in a)!

Cosec · « **Reply #9391 on:** March 23, 2015, 10:05:53 pm »

Quote from: IndefatigableLover on March 23, 2015, 09:53:44 pm

Probably best to go over the quick method of chain rule
Just a heads up but a^2 is a constant so won't do anything!

a). $\frac{d}{dx}(\sqrt{x^2\pm a^2})$

$\therefore \frac{d}{dx}((x^2\pm a^2)^\frac{1}{2})$

$= \frac{1}{2} \cdot (x^2 \pm a^2)^{-\frac{1}{2}} \cdot 2x$

$=\frac{2x}{2\sqrt{x^2\pm a^2}}$

$=\frac{x}{\sqrt{x^2\pm a^2}}$

From there I think you can work out b). since it's literally the same thing but change the sign of x^2 and there's no plus minus sign (not that it should have an effect on the end result as shown in a)!

Hahaha, for some reason i was trying to expand the (x^2+- a^2)^1/2
Thanks buddy.

Cosec · « **Reply #9392 on:** March 23, 2015, 10:36:10 pm »

Heallppp, again. Dont know where im screwing up.

I_I · « **Reply #9393 on:** March 24, 2015, 08:40:40 am »

Hi guys,
I have a lot of questions that I need to ask you,
I would greatly appreciate it if you could answer it!!!

I've attached the document with the questions
Thank you so much

Apink! · « **Reply #9394 on:** March 24, 2015, 10:12:09 am »

Hello,
Could someone please tell how to solve this question?

A supermarket has four check outs in operation. A customer is in hurry and leaves without making a purchase if all the checkouts are busy. At that time of day the probability of each checkout being free is 0.25. Assume that whether or not a checkout is busy is independent of any other checkout. What is the probability that the customer will make a purchase?

Thank you!

wunderkind52 · « **Reply #9395 on:** March 24, 2015, 10:37:50 am »

Quote from: I_I on March 24, 2015, 08:40:40 am

Hi guys,
I have a lot of questions that I need to ask you,
I would greatly appreciate it if you could answer it!!!
I've attached the document with the questions
Thank you so much

Aeroplane question - In this instance, it's a good idea to draw a diagram - (well the question already has one). You know that the height of the aeroplane is always 1000m. Here, it is a good idea to work backwards, and see where you need to go. The question is asking for instantaneous rate of angle decreasing, or $\frac{d\theta}{dt}$ This is a related rates question. You already know your speed ( $\frac{dx}{dt}$ ), so the last bit you need to find is $\frac{d\theta}{dx}$ . Your final equation will become $\frac{d\theta}{dx} \times \frac{dx}{dt}=\frac{d\theta}{dt}$
Now you need to find the relationship between theta and x, the horizontal distance from the plane. You may recall the ever-so-handy SOH CAH TOA. $tan^{-1}(\frac{1000}{x})$ is the relationship between your angle and x, however you need to find the rate of change. YOu have this weird tan in front of your theta, so you can't immediately differentiate that. So, you can rearrange the equation to $x=\frac{1000}{tan(\theta)}$ and then find $\frac{dx}{d\theta}$ . If you plug that into your calculator, $\frac{dx}{d\theta}=\frac{-1000}{sin^2(\theta)}$
However, you want $\frac{d\theta}{dx}$ , so you can just flip that around to become $\frac{d\theta}{dx}=\frac{sin^2(\theta)}{-1000}$
Looking at $\frac{dx}{dt}$ , we have an issue, because the unit is in km/hr, so we need to make 480km/hr m/s, so divide 480 by 3.6 giving you 400/3.
Now, you can multiply $\frac{d\theta}{dx}$
$\frac{d\theta}{dx} \times \frac{dx}{dt}=\frac{sin^2(\theta)}{-1000} \times \frac{400}{3}$ You are told the angle is 30 degrees, so you can sub that into the equation as x.
$\frac{sin^2(pi/6)}{-1000} \times \frac{400}{3}=\frac{-1}{30}$ radians/second. TO convert to degrees, simply multiply by 180/pi, giving you -6/pi degrees/second. So in your answer you would put decreasing by 6/pi degrees/second. I have a slight feeling I MAY have mixed up my negative and positive signs somewhere, but I don't have time to check.
Part 2 - You want to find the straight line distance with respect to time, so let that be $\frac{dd}{dt}$ . Once again, you know $\frac{dx}{dt}$ (given previously as 400/3), you just need to find ${dd}{dx}$ . So you will need something that links your distance and x. This should shout "distance formula" to you. So it is $d=\sqrt{x^2+y^2}$ but as you know that the vertical distance is 1000m, this is $\sqrt{x^2+1000^2}$ .
To find $\frac{dd}{dx}$ , simply differentiate, which will give you $\frac{x}{\sqrt{x^2+1000^2}}$ .
Thus, $\frac{dd}{dt}=\frac{x}{\sqrt{x^2+1000^2}} \times \frac{400}{3}$
However, you know that when the angle is 30degrees, tan(30degrees)=1000/x, and x= $1000\sqrt{3}$ .
So subbing that in, you get $\frac{dd}{dt}=\frac{200\sqrt{3}}{3}m/s$

Question 4 -
It is exactly the same - if you consider $|2x-5|$ and $|5-2x|$ they are the same. The calculator just likes to make it pretty.
Question 5 - Your answer is right I think - not sure what's the problem here.

Question 6 - this is a question that comes from experience, and here you need to know the rule $\int\frac{f'(x)}{f(x)}}=ln(f(x)}$ . Consider it the opposite way - when you differentiate a ln, you get the derivative of the function divided by the function itself. So, you can rewrite $\int{\frac{x}{x^2+1}}$ as $\frac{1}{2} \int{\frac{2x}{x^2+1}}$ Now, you have the derivative on the top, and the function on the bottom, so you can write it as $\frac{1}{2} \int{\frac{2x}{x^2+1}}=\frac{1}{2}ln(x^2+1)$

Question 7 -
I have a feeling this isn't part of the methods course, and can't be done with normal "methods" methods. But I'm not sure.

Question 8 - Once again, the calculator has just made things pretty - sin(3pi-x)=sin(pi-x) is the same as sin(x)

wunderkind52 · « **Reply #9396 on:** March 24, 2015, 11:33:22 am »

Quote from: Apink! on March 24, 2015, 10:12:09 am

Hello,
Could someone please tell how to solve this question?

A supermarket has four check outs in operation. A customer is in hurry and leaves without making a purchase if all the checkouts are busy. At that time of day the probability of each checkout being free is 0.25. Assume that whether or not a checkout is busy is independent of any other checkout. What is the probability that the customer will make a purchase?

Thank you!

So, the question is saying, if all the checkouts are busy, the customer won't make a purchase. I think here you need to find the probability that there will be AT LEAST 1 checkout free. You could do this manually, ie add up the proabilities of 1 checkout free, 2 checkout free etc., but it's much easier to take 1-Pr(no checkouts free). Pr(checkout not free)=1-Pr(checout free)=1-0.25=0.75. Pr(no checkouts free)=Pr(checkout not free)^4=81/256.
Answer should be 1-0.75^4=175/256.
Haven't done prob in a while so I MAY be wrong.

I_I · « **Reply #9397 on:** March 24, 2015, 12:53:15 pm »

Quote from: wunderkind52 on March 24, 2015, 10:37:50 am

Aeroplane question - In this instance, it's a good idea to draw a diagram - (well the question already has one). You know that the height of the aeroplane is always 1000m. Here, it is a good idea to work backwards, and see where you need to go. The question is asking for instantaneous rate of angle decreasing, or $\frac{d\theta}{dt}$ This is a related rates question. You already know your speed ( $\frac{dx}{dt}$ ), so the last bit you need to find is $\frac{d\theta}{dx}$ . Your final equation will become $\frac{d\theta}{dx} \times \frac{dx}{dt}=\frac{d\theta}{dt}$
Now you need to find the relationship between theta and x, the horizontal distance from the plane. You may recall the ever-so-handy SOH CAH TOA. $tan^{-1}(\frac{1000}{x})$ is the relationship between your angle and x, however you need to find the rate of change. YOu have this weird tan in front of your theta, so you can't immediately differentiate that. So, you can rearrange the equation to $x=\frac{1000}{tan(\theta)}$ and then find $\frac{dx}{d\theta}$ . If you plug that into your calculator, $\frac{dx}{d\theta}=\frac{-1000}{sin^2(\theta)}$
However, you want $\frac{d\theta}{dx}$ , so you can just flip that around to become $\frac{d\theta}{dx}=\frac{sin^2(\theta)}{-1000}$
Looking at $\frac{dx}{dt}$ , we have an issue, because the unit is in km/hr, so we need to make 480km/hr m/s, so divide 480 by 3.6 giving you 400/3.
Now, you can multiply $\frac{d\theta}{dx}$
$\frac{d\theta}{dx} \times \frac{dx}{dt}=\frac{sin^2(\theta)}{-1000} \times \frac{400}{3}$ You are told the angle is 30 degrees, so you can sub that into the equation as x.
$\frac{sin^2(pi/6)}{-1000} \times \frac{400}{3}=\frac{-1}{30}$ radians/second. TO convert to degrees, simply multiply by 180/pi, giving you -6/pi degrees/second. So in your answer you would put decreasing by 6/pi degrees/second. I have a slight feeling I MAY have mixed up my negative and positive signs somewhere, but I don't have time to check.
Part 2 - You want to find the straight line distance with respect to time, so let that be $\frac{dd}{dt}$ . Once again, you know $\frac{dx}{dt}$ (given previously as 400/3), you just need to find ${dd}{dx}$ . So you will need something that links your distance and x. This should shout "distance formula" to you. So it is $d=\sqrt{x^2+y^2}$ but as you know that the vertical distance is 1000m, this is $\sqrt{x^2+1000^2}$ .
To find $\frac{dd}{dx}$ , simply differentiate, which will give you $\frac{x}{\sqrt{x^2+1000^2}}$ .
Thus, $\frac{dd}{dt}=\frac{x}{\sqrt{x^2+1000^2}} \times \frac{400}{3}$
However, you know that when the angle is 30degrees, tan(30degrees)=1000/x, and x= $1000\sqrt{3}$ .
So subbing that in, you get $\frac{dd}{dt}=\frac{200\sqrt{3}}{3}m/s$

Question 4 -
It is exactly the same - if you consider $|2x-5|$ and $|5-2x|$ they are the same. The calculator just likes to make it pretty.
Question 5 - Your answer is right I think - not sure what's the problem here.

Question 6 - this is a question that comes from experience, and here you need to know the rule $\int\frac{f'(x)}{f(x)}}=ln(f(x)}$ . Consider it the opposite way - when you differentiate a ln, you get the derivative of the function divided by the function itself. So, you can rewrite $\int{\frac{x}{x^2+1}}$ as $\frac{1}{2} \int{\frac{2x}{x^2+1}}$ Now, you have the derivative on the top, and the function on the bottom, so you can write it as $\frac{1}{2} \int{\frac{2x}{x^2+1}}=\frac{1}{2}ln(x^2+1)$

Question 7 -
I have a feeling this isn't part of the methods course, and can't be done with normal "methods" methods. But I'm not sure.

Question 8 - Once again, the calculator has just made things pretty - sin(3pi-x)=sin(pi-x) is the same as sin(x)

Thank you so much, wunderkind52!!!!!!!

Cosec · « **Reply #9398 on:** March 24, 2015, 06:41:00 pm »

Cant find where im screwing up.

wunderkind52 · « **Reply #9399 on:** March 24, 2015, 07:36:17 pm »

Quote from: Cosec on March 24, 2015, 06:41:00 pm

Cant find where im screwing up.

So you have $(x+\sqrt{x^2+1})^2$
I won't go through the whole u substitution, but let me know if you need that.
$\frac{dy}{dx}=2\times (1+\frac{x}{\sqrt{x^2+1})})\times(x+\sqrt{x^2+1})$
Multiplying the 1 in the first bracket by root (x^2+1) to give a common denominator, you get
$2 \times (\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1})})\times(x+\sqrt{x^2+1})$
Simplifying, it becomes
$2\times (\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}})\times(x+\sqrt{x^2+1})$
And because your you have the x+squaroot (x^2+1) on both fractions and you are multiplying, it becomes
$2\times (\frac{(\sqrt{x^2+1}+x)^2}{\sqrt{x^2+1}})$
And thus you get $\frac{2y}{\sqrt{x^2+1}}$

garytheasian · « **Reply #9400 on:** March 26, 2015, 12:21:05 am »

I'm having trouble answering this question:

Q7.A piece of string 10 metres long is cut into two pieces to form two squares.
a) If one piece of string has length x show that the combined area of the two squares is given A=1/8(x^2-10x+50).

-What I got when I attempted to answer this question was x^2/32 can someone please explain how this is wrong?

Thanks!

Apink! · « **Reply #9401 on:** March 26, 2015, 09:16:00 am »

Hi!
How could I solve this probability question?

Rex is shooting at a target. His probability of hitting the target is 0.6. What is the minimum number of shots needed for the probability of Rex hitting the target exactly five times to be more than 25%?

I can solve questions similar to this where they ask you to find at least 2, 3, 4, etc shots but not when it is exactly 2,3,4 et.c shots
Please help!

Apink! · « **Reply #9402 on:** March 26, 2015, 09:48:57 am »

Hi!

Sorry, I have another question

Monique is practising goaling for netball. She knows from past experience that her changes of making any one shot is about 70%. Her coach has asked her to keep practising until she score 50 goals. How many shots would she need to attempt to ensure that the probability of making at least 50 shots is more than 0.99?

I had gotten until: Pr (X < or = 49) <0.01

I know I'm meant to use the spreadsheet on the CAS but I don't know what I need to do exactly!
Thank you so much!!

I_I · « **Reply #9403 on:** March 26, 2015, 11:57:30 am »

Hi guys,
I have a page (and a little bit more) questions I want to ask you guys

Sorry for being such a bother

Thank you so much (and happy holidays! yoohoo!!!

)
Catherine

Apink! · « **Reply #9404 on:** March 26, 2015, 04:34:34 pm »

Hello,
I feel really bad asking yet another question but probability is really torturing me. It's so confusing! $:-\$

Suppose that it is known that the probability of the Liberal party winning the election if they are already in power is 0.6, and the probability of the Labor party winning the election if they are already in power is 0.65

If the Liberal party is already in power, what is the probability of the Liberal party winning the next two elections?

I solved this like this:
1. First I let A = event where labor wins and I = event where Liberal wins
2. Synthesise transition matrix [0.65; 0.4; 0.35 ; 0.6] ----> (top-left, top-right, bottom left, bottom right)
3. Initial state would be [0; 1] ----> 0 is the top, 1 is the bottom

4. 1st election would be:
[0.65; 0.4; 0.35 ; 0.6] * [0; 1] = [0.4;0.6] ----> (0.4 is the top showing the probability of labor winning and 0.6, bottom, liberal)

then on the 2nd election:
[0.65; 0.4; 0.35 ; 0.6]² * [0; 1] = [0.5;0.5]

so since it's both election 1 and 2 liberal needs to win, I multiplied them together to get 0.5 * 0.6 = 0.3

But the correct answer is 0.36 where all I had to do was 0.6 * 0.6 *face palm*

I do understand how I could get the correct answer, but I don't see what is wrong with my methods above. Could someone please point it out for me? It would be really appreciated

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