Author Topic: VCE Specialist 3/4 Question Thread! (Read 2545839 times) Tweet Share

Phy124 · « **Reply #435 on:** April 28, 2012, 07:52:40 pm »

$d=vt$ (or d = tv, that's how I'll set it up)

For the purposes of this question I'll be taking d in km, t in h and v in km/h

$d_1=v_1t_1=2v_1$

$d_2=v_2t_2=1.5(v_1+2) = 1.5v_1+3$

$d_t = d_1 + d_2 = 2v_1 + 1.5v_1 + 3 = 3.5v_1 +3 = 26$

$3.5v_1+3=26$

$3.5v_1=23$

$\frac{7}{2}v_1=23$

$v_1=\frac{23\times2}{7}=\frac{46}{7} \ km/h$

If you need me to explain each line, just ask

(should be right

)

rife168 · « **Reply #436 on:** April 30, 2012, 06:37:04 pm »

show that:

$\int{\frac{dy}{y(1-y)}=\ln \left(\frac{y}{1-y}\right)+c$

for some constant $c$ , where $0<y<1$

edit: Haha, I got it, didn't realise it was a partial fractions question -.-

b^3 · « **Reply #437 on:** April 30, 2012, 07:37:11 pm »

Firstly split $\frac{1}{y(1-y)}$ into partial fractions.
$\begin{alignedat}{1}\frac{1}{y(1-y)} & =\frac{A}{y}+\frac{B}{1-y} \\ 1 & =A(1-y)+By \\ 1 & =y(B-A)+A \end{alignedat}$
Equating coefficients
$A=1$
$\begin{alignedat}{1}B-A & =0 \\ B-1 & =0 \\ B & =1 \end{alignedat}$
So now we have
$\begin{alignedat}{1}\intop\frac{dy}{y(1-y)} & =\intop\frac{1}{y}+\frac{1}{1-y}dy \\ & =\ln(|y|)-\ln(|1-y|)+C \\ & =\ln\left(\frac{|y|}{|1-y|}\right)+C \\ & =\ln\left(\frac{y}{1-y}\right)+C\:(\mathrm{Remove\: the\: modulus\: as\: y>0\: and\:1-y>0\: for\:0<y<1)} \end{alignedat}$

Hope that helps

EDIT: Just thought I'd add this, it might help.
For $\int\frac{1}{1-y}dy$
Let $u=1-y$
then $\frac{du}{dy}=-1$
So $dy=-du$
So we have
$\begin{alignedat}{1}\int\frac{1}{1-y}dy & =\int\frac{1}{u}\times-du \\ & =-\int\frac{1}{u}du \\ & =-\ln(|1-y|)+C \end{alignedat}$

Fishyiscool · « **Reply #438 on:** May 07, 2012, 05:34:38 pm »

Idk if people still post in this thread but with vector proof questions, generally i stumble around and find myself at the answer accidentally. Anyone know about mark allocation for these questions or is there a particular thought process i should follow for each question? :s

nisha · « **Reply #439 on:** May 07, 2012, 05:57:09 pm »

Basically write down what you have to find, and what you are given. Then manipulate...somehow. In generally all the questions I have done so far, vector proofs can range from 2-4 mark questions.

Hutchoo · « **Reply #440 on:** May 07, 2012, 09:29:08 pm »

Show that cos 3Ɵ = 4 cos^3 Ɵ - 3 cos Ɵ. Similar express sin3Ɵ as a polynomial in sin Ɵ.

This would be an epic help!
I understand that you have to use the double angle formulas (and perhaps compound formulas), but I'm confused!

THANKS GUYS!

b^3 · « **Reply #441 on:** May 07, 2012, 09:50:06 pm »

Firstly we know our Compound Angle Formula $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$
So..
$\begin{alignedat}{1}\cos(3\theta) & =\cos(\theta+2\theta) \\ & =\cos(\theta)\cos(2\theta)-\sin(\theta)\sin(2\theta) \end{alignedat}$
Now we also know our Double Angle Formula $cos(2\theta)=2\cos^{2}(\theta)-1=1-2\sin^{2}(\theta)=\cos^{2}(\theta)-\sin^{2}(\theta)$ and our one for sin $\sin(2\theta)=2\cos(\theta)\sin(\theta)$
So..
$\begin{alignedat}{1}\cos(3\theta) & =\cos(\theta)\left(2\cos^{2}(\theta)-1\right)-\sin(\theta)\times2\cos(\theta)\sin(\theta) \\ & =2\cos^{3}(\theta)-\cos(\theta)-2\cos(\theta)\sin^{2}(\theta) \\ & =2\cos^{3}(\theta)-\cos(\theta)-2\cos(\theta)\left(1-\cos^{2}(\theta)\right)\:\left(\mathrm{as\:}\cos^{2}(\theta)+\sin^{2}(\theta)=1\right) \\ & =2\cos^{3}(\theta)-\cos(\theta)-2\cos(\theta)+2\cos^{3}(\theta) \\ & =4\cos^{3}(\theta)-3\cos(\theta) \end{alignedat}$

Now you can do a similar thing for $\sin(3\theta)$ , using $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$ , and some of the other compound angle and double angle formulas above.

EDIT: Just an extra note, the reason I subbed in the $2\cos^{2}(\theta)-1$ instead of any of the other compound angle formulas, was because the RHS is all in terms of cos(theta) to the power of something, so naturally subbing in the formula with the most cos terms and the least sine terms would be the better option.

Hutchoo · « **Reply #442 on:** May 07, 2012, 09:59:39 pm »

Thanks so much man! Yeah, i know why you subbed in 2cos^2Ɵ -1, i did that as well, but I failed hardcore on the other steps

THANKS B^3, you're mah main man.

Hutchoo · « **Reply #443 on:** May 08, 2012, 12:34:31 pm »

I have a SAC tomorrow and on the revision sheet given to us, the teacher talks about "sketching complicated circular functions". What would a complicated circ. function look like :/? Can someone give me some ideas as to what is considered "complicated" and worthy for an analysis task? Thanks

Dominatorrr · « **Reply #444 on:** May 08, 2012, 06:49:45 pm »

Quote from: Hutchoo on May 08, 2012, 12:34:31 pm

I have a SAC tomorrow and on the revision sheet given to us, the teacher talks about "sketching complicated circular functions". What would a complicated circ. function look like :/? Can someone give me some ideas as to what is considered "complicated" and worthy for an analysis task? Thanks

I'm assuming graphs with a lot of transformations.

VECTORS) If AC = AD + DC and BD = BA + AD then how would we figure out the dot product of AC and BD? There are no values given, the letters are just points on a rhombus, it is part of a vector proof question and vectors AC and BD are obviously expressed as a sum of 2 vectors.

generalkorn12 · « **Reply #445 on:** May 09, 2012, 04:55:32 pm »

For Vector Calculus, when we have found the velocity vector and the question asks for the maximum and minimum speeds, do we equate the i or j components to 0?

I keep getting the i and j's mixed up when determining the minimum/maximum speeds and minimum/maximum heights..

Fishyiscool · « **Reply #446 on:** May 09, 2012, 09:49:44 pm »

A particle's position as a function of time is given by u = cos (2t)i + sin (2t)j.
a) Find the equation of the path
b) State the period of the motion of this particle

(P.S. Would this be a tech-assumed question?)

Mr. Study · « **Reply #447 on:** May 09, 2012, 10:01:47 pm »

Hmm.

If I assume is correct, For a), you want the cartesian form.

Let $x=cos(2t)$ and $y=sin(2t).$

$2t=cos^-^1(x)$
$t=cos^-^1(x)/2$

$2t=sin^-^1(y)$
$t=sin^-^1(y)/2$

$sin^-^1(y)/2=cos^-^1(x)/2$
$sin^-^1(y)=cos^-^1(x)$
$y=\sqrt1-x^2$

Someone check If I'm correct as I haven't done too many problems on vector calculus.

Hmm, I am not too sure about b). Sorry.

(Square root over the entire answer, I have no idea how to do that for LaTeX >.<)

Lasercookie · « **Reply #448 on:** May 09, 2012, 11:13:55 pm »

Quote from: Mr. Study on May 09, 2012, 10:01:47 pm

(Square root over the entire answer, I have no idea how to do that for LaTeX >.<)

You just stick it around curly brackets

Code: [Select]

[tex]\sqrt{moooooooooooooooooo + roooooar}[/tex]

$\sqrt{moooooooooooooooooo + roooooar}$

The curly bracket thing goes for most things in LaTeX when you want it to whatever you're doing to apply to the entire block, e.g. $X_{hello}$

#1procrastinator · « **Reply #449 on:** May 25, 2012, 01:13:54 am »

Express (1-i)^3 in cartesian form. isn't that already in cartesian form? if not, what is that form called? the solution just expanded and simplified it

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Author Topic: VCE Specialist 3/4 Question Thread! (Read 2545839 times) Tweet Share

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