Author Topic: VCE Specialist 3/4 Question Thread! (Read 2555355 times) Tweet Share

sin0001 · « **Reply #2280 on:** September 22, 2013, 10:52:24 pm »

Thanks nliu & Alwin; both the explanations made sense

Jaswinder · « **Reply #2281 on:** September 23, 2013, 03:04:48 pm »

an intensible string passes over a smooth pulley. Objects of mass m kg and 3kg are attached to each end of the string. The objects are released from rest. The tension in the string in newtons, in newtons, i equal to:
a) 3mg/2
b)0
c)3mg
d)mg/4
e)6mg

i get only mg :s

b^3 · « **Reply #2282 on:** September 23, 2013, 03:27:11 pm »

We have the situation below

Since both are going to be moving in the same plane, and the block on the left has a larger weight force acting, we know that the left block will move downwards and the right block will move upwards.
If we look at the $3m$ , which is moving downwards we have:
$m_{1}g-T=m_{1}a$ ...[1]
Looking at the block on the right, which is moving upwards we have
$T-m_{2}g=m_{2}a$ ...[2]
Rearranging equ [1]
$a=\frac{m_{1}g-t}{m_{1}}$
Substituting [1] into [2]
$\begin{alignedat}{1}T-m_{2}g & =\frac{m_{2}}{m_{1}}\left(m_{1}g-T\right) \\ T-m_{1}m_{2}g & =m_{1}m_{2}g-m_{2}T \\ T\left(m_{1}+m_{2}\right) & =2m_{1}m_{2}g \\ T & =\frac{2m_{1}m_{2}g}{m_{1}+m_{2}} \\ m_{1}=3m,\: m_{2}=m \\ T & =\frac{2\times3m\times m\times g}{3m+m} \\ & =\frac{3mg}{2}\: N \end{alignedat}$

Which is option A.

zvezda · « **Reply #2283 on:** September 23, 2013, 07:54:29 pm »

Hey, ive come across a neap trial paper where i need to antidifferentiate sprt(9-t^2)

Never needed to do something like this.
Help is much appreciated

Ronald Zhang · « **Reply #2284 on:** September 23, 2013, 08:16:25 pm »

Whenever you get a square root of an expression with a square and then a constant, it's usually an integration by substitution. Here's the solution, apologies for the plain text - don't quite know how to use LaTeX just yet! Gonna use B in place of theta

Let t = 3sin(B)

Hence:
dt/dB = 3cos(B)
dt = 3cos(B) dB

So, bringing that back into the question:

int [sqrt(9 - t^2)] dt = int [sqrt(9 - (3sin(B))^2)] dt = dt = int [sqrt(9 - (3sin(B))^2) 3cos(B)] dB = int [sqrt(9 - 9sin^2(B)) 3cos(B)] dB = int [3 sqrt(1 - sin^2(B)) 3cos(B)] dB = int [3 sqrt(cos^2(B)) 3cos(B)] dB = int [3cos(B) 3cos(B)] dB = int [9cos^2(B)] dB = int [9cos^2(B)] dB = 9 int [cos^2(B)] dB = 9 int [cos^2(B)] dB

Then you use the identity cos(2B) = 1 + 2cos^2(B) and you're good to go

EDIT: As SocialRhubarb rightly points out below, the identity is not cos(2B) = 1 - 2cos^2(B) as I originally wrote.

SocialRhubarb · « **Reply #2285 on:** September 23, 2013, 08:25:52 pm »

$\int{\sqrt{9-t^2}}\ dt$

$\text{Let t=3 sin(u)}, u\in [-\frac{\pi}{2},\ \frac{\pi}{2}]$

$\frac{dt}{du}=3\cos(u)$

$1=3\frac{du}{dt}\cos(u)$

$\int{\sqrt{9-t^2}}\ dt$

$=\int{\sqrt{9-9\sin^2(u)}(3\frac{du}{dt}\cos(u))}\ dt$

$=\int{3\sqrt{1-\sin^2(u)}(3\cos(u))}\ du$

$=9\int{\cos(u)\cos(u)}\ du$

$=9\int{\cos^2(u)}\ du$

$=9\int{(\frac{1}{2}+\frac{1}{2}\cos(2u))}\ du$

$=9(\frac{1}{2}u+\frac{1}{4}\sin(2u))+C$

$=\frac{9}{2}\sin^{-1}(\frac{t}{3})+\frac{9}{4}\sin(2\sin^{-1}(\frac{t}{3}))+C$

$=\frac{9}{2}\sin^{-1}(\frac{t}{3})+\frac{9}{4}2\sin(\sin^{-1}(\frac{t}{3}))\cos(\sin^{-1}(\frac{t}{3}))+C$

$=\frac{9}{2}\sin^{-1}(\frac{t}{3})+\frac{9}{2}(\frac{t}{3})\(\sqrt{1-\sin^2(\sin^{-1}(\frac{t}{3}))}+C$

$=\frac{9}{2}\sin^{-1}(\frac{t}{3})+\frac{3}{2}(t)\(\sqrt{\frac{9-t^2}{9}}+C$

$=\frac{9}{2}\sin^{-1}(\frac{t}{3})+\frac{t\(\sqrt{9-t^2}}{2}+C$

Quote from: Ronald Zhang on September 23, 2013, 08:16:25 pm

Then you use the identity cos(2B) = 1 - 2cos^2(B) and you're good to go

Hahaha, oh Ronald. Did you really get above 45 in specialist last year? Also, who uses their real name as their username? And I think your methods score is in the wrong year of your signature.

Alwin · « **Reply #2286 on:** September 23, 2013, 08:31:45 pm »

Quote from: stankovic123 on September 23, 2013, 07:54:29 pm

Hey, ive come across a neap trial paper where i need to antidifferentiate sqrt(9-t^2)
Never needed to do something like this.
Help is much appreciated

Hmm, tirg substitution isn't on the course iirc

it's a cas question I think, but the "by hand" method is this:

$\begin{alignedat}{1}\text{let } t&= 3\sin\theta \rightarrow \frac{dt}{d\theta}=3\cos\theta \rightarrow dt = 3\cos\theta d\theta \\ \int{\sqrt{9-t^2}} dt &= \int{\sqrt{9-(3\sin\theta)^2} \cdot 3\cos\theta } d\theta \\ &=\int{\sqrt{9-(3\sin\theta)^2} \cdot 3\cos\theta } d\theta \\ &=\int{3\sqrt{1-(\sin\theta)^2} \cdot 3\cos\theta } d\theta \\ &=\int{3\cos\theta \cdot 3\cos\theta } d\theta \\ &=\int{9 \cos^{2}\theta } d\theta \\ &=9\int{\frac{1}{2} (1+\cos{2\theta}} d\theta \\ &=\frac{9}{2} \int{1+\cos{2\theta}} d\theta \\ \text{my latex messed up somewhere in here, so just deleting it all} &- \text{you should be able to do it from here - anyways!} \\ &= \frac{9}{2} \sin^{-1}(\frac{t}{3})+\frac{3}{2}(t)\(\sqrt{\frac{9-t^2}{9}}+C \end{alignedat}$

EDIT: beaten, severely

Quote from: SocialRhubarb on September 23, 2013, 08:25:52 pm

Hahaha, oh Ronald. Did you really get above 45 in specialist last year? Also, who uses their real name as their username? And I think your methods score is in the wrong year of your signature.

What's wrong with using real names huh. I think out of the last 3 posts you're in the minority

lzxnl · « **Reply #2287 on:** September 23, 2013, 08:48:10 pm »

Here's a different way of doing it:

$\int{\sqrt{a^2-x^2}dx}$
$=\int{\frac{a^2-x^2}{\sqrt{a^2-x^2}}dx}$
Decomposing the fraction, the first term is just $a^2\arcsin\frac{x}{a}$

The second term is $=\int{\frac{x^2}{\sqrt{a^2-x^2}}dx}$

Integrating by parts, letting $u=x, du=dx$ and $dv=\frac{x}{\sqrt{a^2-x^2}}$
You can see that $v=-\sqrt{a^2-x^2}$
So this term becomes $-x\sqrt{a^2-x^2} -\int{-\sqrt{a^2-x^2}dx}$
Now...remember that the second term had a negative sign in front of it?
Putting everything together, we have
$\int{\sqrt{a^2-x^2}dx} = a^2\arcsin\frac{x}{a} + x\sqrt{a^2-x^2} -\int{\sqrt{a^2-x^2}dx}$
Rearranging the $\int{\sqrt{a^2-x^2}dx}$ terms on the one side and halving
$\int{\sqrt{a^2-x^2}dx} = \frac12( a^2\arcsin\frac{x}{a} + x\sqrt{a^2-x^2} )$

zvezda · « **Reply #2288 on:** September 23, 2013, 09:16:42 pm »

Cheers guys, all quality replies.
One thing however: was looking through socialrhubarb's working; whats the concept behind sin^-1(2x) equaling 2sin^-1(x)??
Also, would any of you be able to evaluate the definite integral of that same expression? Somehow im getting (9/2)sin^-1(2/3) + 3 rather than (9/2)sin^-1(2/3) + sqrt(5)
Thanks guys!

SocialRhubarb · « **Reply #2289 on:** September 23, 2013, 09:28:00 pm »

It's actually just the double angle formula for sine.

Quote from: SocialRhubarb on September 23, 2013, 08:25:52 pm

$=\frac{9}{2}\sin^{-1}(\frac{t}{3})+\frac{9}{4}\sin(2\sin^{-1}(\frac{t}{3}))+C$

$=\frac{9}{2}\sin^{-1}(\frac{t}{3})+\frac{9}{4}2\sin(\sin^{-1}(\frac{t}{3}))\cos(\sin^{-1}(\frac{t}{3}))+C$

It's just a bit confusing because instead of a simple pronumeral like 'x' as the argument of the trigonometric term, we have $\sin^{-1}(\frac{t}{3})$ .

BubbleWrapMan · « **Reply #2290 on:** September 28, 2013, 11:39:09 am »

Quote from: stankovic123 on September 23, 2013, 09:16:42 pm

Cheers guys, all quality replies.
One thing however: was looking through socialrhubarb's working; whats the concept behind sin^-1(2x) equaling 2sin^-1(x)??
Also, would any of you be able to evaluate the definite integral of that same expression? Somehow im getting (9/2)sin^-1(2/3) + 3 rather than (9/2)sin^-1(2/3) + sqrt(5)
Thanks guys!

What terminals are you using?

Quote from: SocialRhubarb on September 23, 2013, 08:25:52 pm

Also, who uses their real name as their username?

saba.ay · « **Reply #2291 on:** September 28, 2013, 04:00:19 pm »

Could someone please explain how to do the following.
Question 1:

A particle moves from rest at the origin with an acceleration of v^3 + (pi^2)*v m/s^2 where v is the velocity in m/s. Find the velocity of the particle when it is 0.25 m to the right of the origin

Also Question 2:
Find the square root of -2+2 root(3)i in Cartesian form. I found this to be 1 + root (3)i.
Hence, find the solution to {z: z^2 + (root(3) - i)z + (1-root(3)i) = 0}.

Please and thank you.

09Ti08 · « **Reply #2292 on:** September 28, 2013, 04:50:23 pm »

Quote from: sabzie. on September 28, 2013, 04:00:19 pm

Could someone please explain how to do the following.
Question 1:

A particle moves from rest at the origin with an acceleration of v^3 + (pi^2)*v m/s^2 where v is the velocity in m/s. Find the velocity of the particle when it is 0.25 m to the right of the origin

Also Question 2:
Find the square root of -2+2 root(3)i in Cartesian form. I found this to be 1 + root (3)i.
Hence, find the solution to {z: z^2 + (root(3) - i)z + (1-root(3)i) = 0}.

Please and thank you.

Question 1:
a=v*dv/dx=>dv/dx=v^2+pi^2
Then you integrate to get a relation between x and v, then use the initial condition v=0 and x=0 to work out c. When you get your final equation, just plug in x=0.25 to figure out v.

Question 2:
Delta=-2+2*sqrt(3)*i
Now notice that you have already figured out sqrt(delta)
Now just apply the quadratic formula.

saba.ay · « **Reply #2293 on:** September 28, 2013, 05:53:01 pm »

Quote from: 09Ti08 on September 28, 2013, 04:50:23 pm

Question 1:
a=v*dv/dx=>dv/dx=v^2+pi^2
Then you integrate to get a relation between x and v, then use the initial condition v=0 and x=0 to work out c. When you get your final equation, just plug in x=0.25 to figure out v.

Question 2:
Delta=-2+2*sqrt(3)*i
Now notice that you have already figured out sqrt(delta)
Now just apply the quadratic formula.

Thanks- I didn't think to try the quadratic formula. :/ And I've been trying the same method for question 1, but I guess I'm going wrong somewhere in my working. :/

Also, could some please explain how to go about vector questions which require you to show a geometrical relationship. I've attached a question. I never know how to do these so could someone please explain it from step 1.

Please and thank you again

SocialRhubarb · « **Reply #2294 on:** September 28, 2013, 06:05:01 pm »

Find $\vec{OM}$ in terms of a and c.

Then find $\vec{OP}$ in terms of a and c using the fact that $\vec{OP}=\frac{2}{3}\vec{OM}$ .

Find $\vec{PC}$ in terms of a and c using the fact that $\vec{PC}=\vec{PO}+\vec{OC}$ .

Find $\vec{AP}$ in terms of a and c using the fact that $\vec{AP}=\vec{AO}+\vec{OP}$ .

Answer.

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Author Topic: VCE Specialist 3/4 Question Thread! (Read 2555355 times) Tweet Share

sin0001

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Jaswinder

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b^3

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zvezda

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Ronald Zhang

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SocialRhubarb

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Alwin

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lzxnl

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zvezda

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SocialRhubarb

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BubbleWrapMan

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saba.ay

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09Ti08

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SocialRhubarb

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