Author Topic: VCE Specialist 3/4 Question Thread! (Read 2172874 times) Tweet Share

Jenny_2108 · « **Reply #660 on:** September 01, 2012, 09:05:53 pm »

Quote from: ClimbTooHigh on September 01, 2012, 08:26:36 pm

Quote from: studynotes on September 01, 2012, 06:20:33 pm
can i please have help with this definite integral

0 to pi/4 tan^3(x) dx
$\int_{0}^{\frac{\pi }{4}}\tan ^{3}(x)\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}\tan (x)\tan ^{2}(x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\tan (x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x$

Let $u = \tan (x)$ and $y = \cos(x)$

Then $\frac{\mathrm{d} u}{\mathrm{d} x}=\sec^{2}(x)$ and $\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (x)\Leftrightarrow -\frac{\mathrm{d} y}{\mathrm{d} x}=\sin (x)$

$\therefore \int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}u\frac{\mathrm{d} u}{\mathrm{d} x}\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{1}{y}(-\frac{\mathrm{d} y}{\mathrm{d} x})\mathrm{d} x$

$=\int_{0}^{1}u\mathrm{d} u+\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{y}\mathrm{d} y$

$=(\frac{1}{2}(1)^{2}-\frac{1}{2}(0)^{2})+(\ln |\frac{1}{\sqrt{2}}|-\ln |1|)$

$=\frac{1}{2}(1-\ln (2))$

Wow, I didnt think of the method let y=cos (x)

What I do is

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$let u=tan(x) => \frac{du}{dx}=sec^2(x) => dx=\frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1) \frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(1- \frac{1}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{\frac{\pi }{4}}\tan (x) - \frac{tan(x)}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{1} u - \frac{u}{1+u^2} \mathrm{d}u$

$= [\frac{u^2}{2} - \frac{1}{2} ln(u^2+1)]_ 0^1$

$= \frac{1}{2}-\frac{1}{2}ln(2)+\frac{1}{2}ln(1)$

$= \frac{1}{2}(1-ln(2))$

Edit: correct LaTex

BubbleWrapMan · « **Reply #661 on:** September 01, 2012, 09:55:07 pm »

^I didn't think of that either, haha. Seems quicker.

paulsterio · « **Reply #662 on:** September 01, 2012, 10:22:37 pm »

Quote from: LovesPhysics on September 01, 2012, 08:32:46 pm

I'm confused by this :/ Surely if k is negative the vectors are still parallel and the proof Paul is suggesting is valid.

Yeah, I'm confused as well. Some things I want to point out:

1) Jenny, when you say scalar multiple, it is automatically assumed that you mean scalar multiple of a vector - like why would you want the scalar multiple of the magnitude - that doesn't make sense.

2) Don't worry about parallel or anti-parallel, you're getting too caught up in words - just stick to the basic concept, parallel is parallel, it means the same gradient, the same slope, the same angle made with a particular axis.

3) Why do you keep making things more complicated, just stick to the basic rules and you'll be fine, just note that when a vector is scaled by a constant amount (i.e. a scalar) it will be parallel to the original vector, i.e. a = kb, then a//b - don't worry about parallel, anti-parallel and all that other bs, just stick to the basic definition that is used in VCE.

BlueSky_3 · « **Reply #663 on:** September 01, 2012, 11:20:58 pm »

Quote from: ClimbTooHigh on September 01, 2012, 09:55:07 pm

^I didn't think of that either, haha. Seems quicker.

Want to see something even quicker?

Well: S tan(x) x ( sec^2(x) - 1)dx, and let sec^2(x) -1 = u
Then dx= 2tan(x)sec^2(x) du
Next, I = .5S u/(u+1) du
= .5S (u+1-1)/(u+1) du
= .5S 1 -1/(u+1) du
= same answer

TrueTears · « **Reply #664 on:** September 02, 2012, 01:09:47 am »

Quote from: Jenny_2108 on September 01, 2012, 09:05:53 pm

Quote from: ClimbTooHigh on September 01, 2012, 08:26:36 pm

Quote from: studynotes on September 01, 2012, 06:20:33 pm
can i please have help with this definite integral

0 to pi/4 tan^3(x) dx
$\int_{0}^{\frac{\pi }{4}}\tan ^{3}(x)\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}\tan (x)\tan ^{2}(x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\tan (x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x$

Let $u = \tan (x)$ and $y = \cos(x)$

Then $\frac{\mathrm{d} u}{\mathrm{d} x}=\sec^{2}(x)$ and $\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (x)\Leftrightarrow -\frac{\mathrm{d} y}{\mathrm{d} x}=\sin (x)$

$\therefore \int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}u\frac{\mathrm{d} u}{\mathrm{d} x}\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{1}{y}(-\frac{\mathrm{d} y}{\mathrm{d} x})\mathrm{d} x$

$=\int_{0}^{1}u\mathrm{d} u+\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{y}\mathrm{d} y$

$=(\frac{1}{2}(1)^{2}-\frac{1}{2}(0)^{2})+(\ln |\frac{1}{\sqrt{2}}|-\ln |1|)$

$=\frac{1}{2}(1-\ln (2))$

Wow, I didnt think of the method let y=cos (x)

What I do is

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$let u=tan(x) => \frac{du}{dx}=sec^2(x) => dx=\frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1) \frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(1- \frac{1}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{\frac{\pi }{4}}\tan (x) - \frac{tan(x)}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{1} u - \frac{u}{1+u^2} \mathrm{d}u$

$= [\frac{u^2}{2} - \frac{1}{2} ln(u^2+1)]_ 0^1$

$= \frac{1}{2}-\frac{1}{2}ln(2)+\frac{1}{2}ln(1)$

$= \frac{1}{2}(1-ln(2))$

Edit: correct LaTex

someone's enjoying the use of latex

Jenny_2108 · « **Reply #665 on:** September 02, 2012, 08:50:32 pm »

Quote from: laseredd on September 01, 2012, 08:38:02 pm

Quote from: Jenny_2108 on September 01, 2012, 08:06:21 pm
Dont make it so complicated. They are both parallel, anti-parallel just gives extra info that they are in opposite direction
Oh my bad, I misinterpreted what you meant then. I thought you were implying that 'anti-parallel' wasn't parallel at all.

Nah, it was my fault actually, I didnt expain clearly and actually, anti-parallel means they are in opposite sense, not direction

Quote from: paulsterio on September 01, 2012, 10:22:37 pm

Some things I want to point out:

1) Jenny, when you say scalar multiple, it is automatically assumed that you mean scalar multiple of a vector - like why would you want the scalar multiple of the magnitude - that doesn't make sense.

In my understanding, scalar multiplies just means "times with a number" (of a quantity, having only magnitude, no direction. In this situation, this number is k. Thus, why cant I say |a|=k|b|?

Anyway, its just due to my misunderstanding the question. I read "regardless i,j,k" in the question and I thought without i,j,k => implies no direction, no sense at all => regardless vectors

Quote from: paulsterio on September 01, 2012, 10:22:37 pm

2) Don't worry about parallel or anti-parallel, you're getting too caught up in words - just stick to the basic concept, parallel is parallel, it means the same gradient, the same slope, the same angle made with a particular axis.

I just wanna be specific a bit, vectors have direction. Without direction, its just a line
"Parallel means same gradient, same slope, same angle made with a particular axis" is applied for gradient of lines or parallel vectors in the same direction, same sense

Gradients of parallel vectors= $(f_x, f_y,f_z)$
The tangent plane to the surface given by f(x,y,z) while with an antiparallel vector given by f(-x,-y,-z)
If they are anti-parallel, they will have opposite slope and gradient, NOT same

Quote from: paulsterio on September 01, 2012, 10:22:37 pm

3) Why do you keep making things more complicated, just stick to the basic rules and you'll be fine, just note that when a vector is scaled by a constant amount (i.e. a scalar) it will be parallel to the original vector, i.e. a = kb, then a//b - don't worry about parallel, anti-parallel and all that other bs, just stick to the basic definition that is used in VCE.

I dont wanna make things complicated but in VCE, parallel vectors are not defined accurately

Parallel vectors should be considered when they have: same direction + same sense
Antiparallel vectors have: same direction +opposite sense

Direction just means a horizontal line
Sense: can be to the right or left

What I said before about k, I meant: +if k is positive, they are parallel (same direction, same sense)
+if k is negative, they are antiparallel (same direction, opposite sense)

Anyway, as you said, just stick with VCE level, so just ignore what I mentioned earlier about antiparallel.
Its too much and unecessary for the exams

Quote from: TrueTears on September 02, 2012, 01:09:47 am

someone's enjoying the use of latex

Haha, actually I like LaTex now

soccerboi · « **Reply #666 on:** September 03, 2012, 08:51:31 pm »

Can somebody help me with this:
A train is moving with uniform acceleration is observed to take 20s and 30s to travel successive half kilometres. How much farther will it travel before coming to rest if the acceleration remains constant?

Thanks

InsaneMcFries · « **Reply #667 on:** September 03, 2012, 09:10:59 pm »

At the moment Dynamics is doing my head in. My first problem is a simple pulley problem with two particles.

The mass of one particle is 1.5kg and the other is 2kg. They hang vertically around a pulley. There are two Tension forces, one on each side of the pulley rope.

The first question is to find the value of the tension.

My main issue is I don't understand how to set out the vector equations for this setup, as I set them up as T-1.5g=1.5a and T-2g=2a.

The answer is T=16.8N.

Jenny_2108 · « **Reply #668 on:** September 03, 2012, 09:19:51 pm »

Quote from: soccerboi on September 03, 2012, 08:51:31 pm

Can somebody help me with this:
A train is moving with uniform acceleration is observed to take 20s and 30s to travel successive half kilometres. How much farther will it travel before coming to rest if the acceleration remains constant?

Thanks

This question ClimbTooHigh answered already. Have a look

Quote from: ClimbTooHigh on August 30, 2012, 10:28:31 pm

Let's say after 20 s, the train has travelled 0.5 km = 500 m, then after 20 + 30 = 50 s, it has travelled 1000 m. Consider the initial time and distance to both be 0. Use the equation s = ut + 1/2at^2. Your unknowns are u and a, which you can find since you have two sets of values for s and t (i.e. two equations). Once you have u and a, use v^2 = u^2 + 2as to find s, then subtract 1000 m.

Quote from: InsaneMcFries on September 03, 2012, 09:10:59 pm

At the moment Dynamics is doing my head in. My first problem is a simple pulley problem with two particles.

The mass of one particle is 1.5kg and the other is 2kg. They hang vertically around a pulley. There are two Tension forces, one on each side of the pulley rope.

The first question is to find the value of the tension.

My main issue is I don't understand how to set out the vector equations for this setup, as I set them up as T-1.5g=1.5a and T-2g=2a.

The answer is T=16.8N.

You have to decide which direction is positive.
Then the equations should be: T-1.5g=1.5a and 2g-T=2a
Or: 1.5g-T=1.5a and T-2g=2a

Solve it, you get T=16.8N

InsaneMcFries · « **Reply #669 on:** September 03, 2012, 10:21:59 pm »

Oh right! I didn't know I had to decide a direction. Thanks!

Mr. Study · « **Reply #670 on:** September 05, 2012, 08:54:31 am »

The solutions to this question seem quite ... vague ... (Either that or I am quite stupid

)

I got two marks, out of 4, for this question. (Kilbaha 2009 Specialist Exam 1)

I converted each .... form (Can't think of correct terminology), into polar form.

$z=2cis(\pi/6)$ and $z=2cis(-\pi/6)$

Hence, $(2cis(\pi/6))^m-(2cis(-\pi/6))^m=0$

'Expanded' by De Moivre's Theorem.

$2^mcis(m\pi/6)-2^m(cis(-m\pi/6)=0$

I really do not understand what the solutions did after 'expanding'.

Thanks for any help.

tony3272 · « **Reply #671 on:** September 05, 2012, 10:05:14 am »

Quote from: Mr. Study on September 05, 2012, 08:54:31 am

The solutions to this question seem quite ... vague ... (Either that or I am quite stupid )

I got two marks, out of 4, for this question. (Kilbaha 2009 Specialist Exam 1)

I converted each .... form (Can't think of correct terminology), into polar form.

$z=2cis(\pi/6)$ and $z=2cis(-\pi/6)$

Hence, $(2cis(\pi/6))^m-(2cis(-\pi/6))^m=0$

'Expanded' by De Moivre's Theorem.

$2^mcis(m\pi/6)-2^m(cis(-m\pi/6)=0$

I really do not understand what the solutions did after 'expanding'.

Thanks for any help.

Where abouts did you stop in your working? I'll just start at the last line you posted

You have $2^m(cis(\frac{m\pi}{6})-cis(-\frac{m\pi}{6}))=0$
And since we know $2^m\neq 0$ we can say
$cis(\frac{m\pi}{6})-cis(-\frac{m\pi}{6})=0$

$\therefore cos(\frac{m\pi}{6})+isin(\frac{m\pi}{6})-cos(-\frac{m\pi}{6})-isin(-\frac{m\pi}{6})=0$

$cos(\frac{m\pi}{6})+isin(\frac{m\pi}{6})-(cos(\frac{m\pi}{6}))-(-isin(\frac{m\pi}{6}))=0$

$\therefore 2isin(\frac{m\pi}{6})=0$

$\therefore sin(\frac{m\pi}{6})=0$ solve to find m

Is this what you were looking for?

Mr. Study · « **Reply #672 on:** September 05, 2012, 03:51:05 pm »

Yep, that's exactly it!

Thanks tony.

Sweetpotato · « **Reply #673 on:** September 07, 2012, 12:48:40 pm »

Quote from: TrueTears on September 02, 2012, 01:09:47 am

Quote from: Jenny_2108 on September 01, 2012, 09:05:53 pm
Quote from: ClimbTooHigh on September 01, 2012, 08:26:36 pm

Quote from: studynotes on September 01, 2012, 06:20:33 pm
can i please have help with this definite integral

0 to pi/4 tan^3(x) dx
$\int_{0}^{\frac{\pi }{4}}\tan ^{3}(x)\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}\tan (x)\tan ^{2}(x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\tan (x)\mathrm{d} x$

$=\int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x$

Let $u = \tan (x)$ and $y = \cos(x)$

Then $\frac{\mathrm{d} u}{\mathrm{d} x}=\sec^{2}(x)$ and $\frac{\mathrm{d} y}{\mathrm{d} x}=-\sin (x)\Leftrightarrow -\frac{\mathrm{d} y}{\mathrm{d} x}=\sin (x)$

$\therefore \int_{0}^{\frac{\pi }{4}}\tan (x)\sec ^{2}(x)\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{\sin(x)}{\cos(x)}\mathrm{d} x=\int_{0}^{\frac{\pi }{4}}u\frac{\mathrm{d} u}{\mathrm{d} x}\mathrm{d} x-\int_{0}^{\frac{\pi }{4}}\frac{1}{y}(-\frac{\mathrm{d} y}{\mathrm{d} x})\mathrm{d} x$

$=\int_{0}^{1}u\mathrm{d} u+\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{y}\mathrm{d} y$

$=(\frac{1}{2}(1)^{2}-\frac{1}{2}(0)^{2})+(\ln |\frac{1}{\sqrt{2}}|-\ln |1|)$

$=\frac{1}{2}(1-\ln (2))$

Wow, I didnt think of the method let y=cos (x)

What I do is

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1)\mathrm{d} x$

$let u=tan(x) => \frac{du}{dx}=sec^2(x) => dx=\frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(\sec ^{2}(x)-1) \frac{du}{sec^2(x)}$

$\int_{0}^{\frac{\pi }{4}}\tan (x)(1- \frac{1}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{\frac{\pi }{4}}\tan (x) - \frac{tan(x)}{sec ^2(x)}) \mathrm{d}u$

$\int_{0}^{1} u - \frac{u}{1+u^2} \mathrm{d}u$

$= [\frac{u^2}{2} - \frac{1}{2} ln(u^2+1)]_ 0^1$

$= \frac{1}{2}-\frac{1}{2}ln(2)+\frac{1}{2}ln(1)$

$= \frac{1}{2}(1-ln(2))$

Edit: correct LaTex

someone's enjoying the use of latex

Hey help with this qs please!!!!!!!!!

A particle P of unit mass moves on the positive x axis. At time t, the velocity of the
particle is v and the force F acting on the particle is given by
F= 50/(25+v) for 0<t< and including 50
F= −v^2/1000 for t>50
Initially the particle is at rest at the origin O.
a Show that v=50 when t=50.
b Find the distance of P from O when v = 50.
c Find the distance of P from O when v=25 and t>50.

monkeywantsabanana · « **Reply #674 on:** September 07, 2012, 08:59:37 pm »

When you have these types of questions:

A particle moves in a straight line. When its displacement from a fixed origin is x m, its velocity is v m/s and its acceleration is a m/s^2.

Given that a=16x and that v = -5 when x = 0, the relation between v and x is?

I got two answers for v but I used the positive square root instead of the negative square root. How does one work out if it's + or -?

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Jenny_2108

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