Author Topic: VCE Specialist 3/4 Question Thread! (Read 2602337 times) Tweet Share

Jeggz · « **Reply #2625 on:** November 07, 2013, 07:44:03 pm »

Quote from: Stevensmay on November 07, 2013, 07:32:49 pm

Does anyone mind explaining Question 4 part a for me?

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012specmath1-w.pdf
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/specmaths1_assessrep12.pdf

Thank you.

Is it the actual labelling of forces that you needed help with, or was it part b) ?

b^3 · « **Reply #2626 on:** November 07, 2013, 07:50:15 pm »

Just be careful with the difference between the subscript notation and what you're subbing in. I.e. $y_{2}$ is at $x+h=1.2$ , so using $x=1.1$ .
$\begin{alignedat}{1}y_{1} & =y_{0}+hy'\left(1\right) \\ & =1+0.1\times\frac{1}{1} \\ & =1.1 \\ y_{2} & =y_{1}+hy'\left(1.1\right) \\ & =\frac{11}{10}+\frac{1}{10}\times\frac{1}{\frac{11}{10}} \\ & =\frac{11}{10}+\frac{1}{10}\times\frac{10}{11} \\ & =\frac{11}{10}+\frac{1}{11} \\ & =\frac{121+10}{110} \\ & =\frac{131}{110} \end{alignedat}$

ahat · « **Reply #2627 on:** November 07, 2013, 08:05:17 pm »

Quote from: SocialRhubarb on November 07, 2013, 07:26:29 pm

(Image removed from quote.)

I'm honestly clueless, is there a solution?

xenial · « **Reply #2628 on:** November 07, 2013, 08:07:06 pm »

Hey, just covering all my weakpoints - these babies are one of them.

Arg(z-1-4i)= 7pi/12

I got y= - rt(3)x - 2x + rt(3) + 6

Could somebody post how they would work out this question? I've got one method (might've screwed up somewhere) but I'd like to see how other people go about it.

Stevensmay · « **Reply #2629 on:** November 07, 2013, 08:07:24 pm »

Quote from: Jeggz on November 07, 2013, 07:44:03 pm

Is it the actual labelling of forces that you needed help with, or was it part b) ?

Ah yes, part b sorry.

xenial · « **Reply #2630 on:** November 07, 2013, 08:16:12 pm »

Quote from: SocialRhubarb on November 07, 2013, 07:26:29 pm

(Image removed from quote.)

I did it! Wooo. But that was hard :\

Hint guys: z=acis(angle1) and w=bcis(angle2)

hence area of triangle= 1/2 ab sin(angle1-angle2)

another hint: don't use a double angle formula, just show left side = right side

Edit: oop, looks like you can use double angle formula, you can do it without though.

SocialRhubarb · « **Reply #2631 on:** November 07, 2013, 08:18:01 pm »

Quote from: ahat on November 07, 2013, 08:05:17 pm

I'm honestly clueless, is there a solution?

Haha, be funny if it was unsolvable, but here's my solution.

Solution:

Spoiler

$\text{Let z=a+bi and w=x+yi}$

$A=\frac{1}{2}(|z|\sin(\theta-\phi))(|w|)$ Area of a triangle

$2A=|z||w|(\sin(\theta)\cos(\phi)-\sin(\phi)\cos(\theta))$ sin subtraction of angles formula

$=|z||w|\sin(\theta)\cos(\phi)-|z||w|\sin(\phi)\cos(\theta)$

$=(|z|\sin(\theta))(|w|\cos(\phi))-(|z|\cos(\theta))(|w|\sin(\phi))$

$=bx-ay$

$LHS=z\bar{w}-w\bar{z}$

$=(a+bi)(x-yi)-(x+yi)(a-bi)$

$=(ax+bxi-ayi-byi^2)-(ax+ayi-bxi-byi^2)$

$=2bxi-2ayi$

$=2i(bx-ay)$

$=2i(2A)$

$=4iA=RHS$

I feel like there's a more elegant solution, but I haven't found it, please post one if you guys find something.

Daenerys Targaryen · « **Reply #2632 on:** November 07, 2013, 09:06:50 pm »

How else can you show:

$\text{Show } \frac{2x^2+3}{x^2+1} \text{ as }2+\frac{1}{x^2+1}$

Without long division

ahat · « **Reply #2633 on:** November 07, 2013, 09:12:48 pm »

Quote from: The xx on November 07, 2013, 09:06:50 pm

How else can you show:

$\text{Show } \frac{2x^2+3}{x^2+1} \text{ as }2+\frac{1}{x^2+1}$

Without long division

$\frac{(2x^2 + 2) -2 +3}{x^2 + 1}$

$\frac{2(x^2+1)}{x^2 + 1} + \frac{3-2}{x^2 + 1}$

$2+\frac{1}{x^2+1}$ as required

$\int{\frac{x^2}{9 - x^2}}$ anyone?

*do I simplify it to $-1 + \frac{9}{9-x^2}$ first?* Seems legit
Edit: Got it, shaweeet

BubbleWrapMan · « **Reply #2634 on:** November 07, 2013, 09:16:14 pm »

Yeah then partial fractions

Stevensmay · « **Reply #2635 on:** November 07, 2013, 09:18:36 pm »

Quote from: ahat on November 07, 2013, 09:12:48 pm

$\int{\frac{x^2}{9 - x^2}}$ anyone?

Partial fractions I believe.
Edit: I suck.

ahat · « **Reply #2636 on:** November 07, 2013, 09:23:25 pm »

Quote from: Stevensmay on November 07, 2013, 09:18:36 pm

Partial fractions I believe.
Spoiler
$-\int{\frac{x^2}{-9 + x^2}}$

$\frac{x^2}{-9 + x^2} = \frac{x^2}{(x-3)(x+3)}=\frac{a}{x-3} + \frac{b}{x+3}$

$x^2 = a(x+3) + b(x-3)$ Let x = 3

$9= 6a$ Let x = -3

$9= -6b$

$\therefore a = \frac{3}{2} and b =\frac{-3}{2}$

$\frac{x^2}{-9 + x^2} =\frac{1}{2}(\frac{3}{x-3} + \frac{-3}{x+3})$

Normal stuff now.

[/spoiler]

You have to simplify the expression first because the numerator is to the same degree as the denominator (remembered when I was doing it) so if you did your method, you would miss out on an x term in the end. Have to long divide first.

Alright, one more question

I've asked this before, but I need some closure.
What method do we use to show a point of inflexion. Not a stationary point of inflexion. i.e. A point of inflexion when we let f''(x) = 0. How do we prove what it is? Change in concavity?
Thanks.

zvezda · « **Reply #2637 on:** November 07, 2013, 09:24:39 pm »

Has anybody looked at the neap 2011 exam 1?
For q6,
They give a=(3-x)/x^3, with v=0 when x=1.
When you find v in terms of x, you get v^2=(x^2+2x-3)/x^2. So when you square root both sides, neap says that you take the positive as a>0 when x=1?
Is there a better explanation for why we take the positive?
Cheers

ahat · « **Reply #2638 on:** November 07, 2013, 09:26:49 pm »

Quote from: Stevensmay on November 07, 2013, 08:07:24 pm

Ah yes, part b sorry.

The maximum value of T will occur as N --> 0 (as N gets smaller). So the crate will leave the floor if Tsin(30^o) > 50g. So solve for T_max by doing Tsin(30^o) = 50g

zvezda · « **Reply #2639 on:** November 07, 2013, 09:44:55 pm »

Hey,
If something is an element of Q, what does this mean?
Cheers

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