Author Topic: VCE Specialist 3/4 Question Thread! (Read 2549139 times) Tweet Share

507 · « **Reply #1260 on:** February 08, 2013, 07:47:03 pm »

b) I'm getting $\frac{3\sqrt{26}}{26}(3\mathbf{i}+4\mathbf{j}-\mathbf{k})$

Answer from the book says $\frac{\sqrt{78}}{26}(3\mathbf{i}+4\mathbf{j}-\mathbf{k})$

Am I doing something wrong, or did the authour just accidentally multiply $\hat{a}$ by $\sqrt{3}$ rather than $3$ .

Thanks

polar · « **Reply #1261 on:** February 08, 2013, 07:56:05 pm »

Quote from: 507 on February 08, 2013, 07:47:03 pm

b) I'm getting $\frac{3\sqrt{26}}{26}(3\mathbf{i}+4\mathbf{j}-\mathbf{k})$

Answer from the book says $\frac{\sqrt{78}}{26}(3\mathbf{i}+4\mathbf{j}-\mathbf{k})$

Am I doing something wrong, or did the authour just accidentally multiply $\hat{a}$ by $\sqrt{3}$ rather than $3$ .

Thanks

so, you know that the vector you want is $|\mathbf{b}|\mathbf{\hat{a}}$
first, lets work out the unit vector of a: $\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{\sqrt{26}}{26}(3\mathbf{i} + 4\mathbf{j} - \mathbf{k})$

next, work out the magnitude of b: $|\mathbf{b}| = \sqrt{(1)^2+(-1)^2+(-1)^2} = \sqrt{3}$

now, multiplying these gives $\sqrt{3}\times\frac{\sqrt{26}}{26}(3\mathbf{i} + 4\mathbf{j} - \mathbf{k}) = \frac{\sqrt{78}}{26}(3\mathbf{i} + 4\mathbf{j} - \mathbf{k})$

507 · « **Reply #1262 on:** February 08, 2013, 07:59:31 pm »

Quote from: polar on February 08, 2013, 07:56:05 pm

so, you know that the vector you want is $|\mathbf{b}|\mathbf{\hat{a}}$
first, lets work out the unit vector of a: $\mathbf{\hat{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{\sqrt{26}}{26}(3\mathbf{i} + 4\mathbf{j} - \mathbf{k})$

next, work out the magnitude of b: $|\mathbf{b}| = \sqrt{(1)^2+(-1)^2+(-1)^2} = \sqrt{3}$

now, multiplying these gives $\sqrt{3}\times\frac{\sqrt{26}}{26}(3\mathbf{i} + 4\mathbf{j} - \mathbf{k}) = \frac{\sqrt{78}}{26}(3\mathbf{i} + 4\mathbf{j} - \mathbf{k})$

Oh that was dumb lol. Accidentally found $|\mathbf{b}|$ from the previous question

Thanks

saba.ay · « **Reply #1263 on:** February 09, 2013, 03:36:16 pm »

This is gonna be a very stupid question, but how would I do the following:
Find the Cartesian equation of the circle:
x=3 cos (2t) , y = 3 sin (2t) , given that 't' belongs to [0, pi]

I don't know how to deal with the '2t' seeing as we've only been shown questions with only t? :/

Homer · « **Reply #1264 on:** February 09, 2013, 03:46:33 pm »

How would I differentiate 1\(1+x^2)

polar · « **Reply #1265 on:** February 09, 2013, 03:53:45 pm »

Quote from: shemurr on February 09, 2013, 03:36:16 pm

This is gonna be a very stupid question, but how would I do the following:
Find the Cartesian equation of the circle:
x=3 cos (2t) , y = 3 sin (2t) , given that 't' belongs to [0, pi]

I don't know how to deal with the '2t' seeing as we've only been shown questions with only t? :/

$\begin{aligned} \sin^2(x) + \cos^2(x) &=1, <br />\\ \sin^2(1000) + \cos^2(1000) &=1, <br />\\ \sin^2(x^2) + \cos^2(x^2) &=1\end{aligned}$ ,

it doesn't matter what's inside the brackets (provided it's defined of course)

- always check the domain as well to see how much of the path you need to draw $t \in [0,\pi] \implies 2t \in [0,2\pi]$ , hence this gives a full circle
- you can't really see what effect the 2t has, but it makes travelling along the path a bit 'faster', but doesn't have any effect on the shape of the path (t takes $2\pi$ to make a whole circle, 2t takes $\pi$ )
- you still get $x^2+y^2=9, x,y \in [-3,3]$
$x=3\cos(2t) \implies \cos(2t) = \frac{x}{3}$ and $y=3\sin(2t) \implies \sin(2t) = \frac{y}{3}$

since $\cos^2(2t) + \sin^2(2t)=1, \left(\frac{x}{3}\right)^2 + \left(\frac{y}{3}\right)^2 =1, x^2+y^2 =9$

Quote from: Homer on February 09, 2013, 03:46:33 pm

How would I differentiate 1\(1+x^2)

use the quotient rule
$\frac{d}{dx}\left(\frac{1}{1+x^2}\right) = \frac{0\times(1+x^2) - 1\times(2x)}{(1+x^2)^2}=-\frac{2x}{(x^2+1)^2}$

Homer · « **Reply #1266 on:** February 09, 2013, 04:13:07 pm »

Oh yeah haha what a dummy! Must have had a mental blank :-/

nspire · « **Reply #1267 on:** February 10, 2013, 05:26:21 pm »

Can somebody help me with the following vectors question from the Essentials textbook?

ABC is a right-angled triangle with the right angle at B. If AC = 2i + 4j and AB is parallel to i + j, find AB.

507 · « **Reply #1268 on:** February 10, 2013, 05:48:36 pm »

Quote from: AimingHigh on February 10, 2013, 05:26:21 pm

Can somebody help me with the following vectors question from the Essentials textbook?

ABC is a right-angled triangle with the right angle at B. If AC = 2i + 4j and AB is parallel to i + j, find AB.

Since $\vec{AB}$ is parallel to $\bf{i}+\bf{j}$ , $\vec{AB}=k(\bf{i}+\bf{j})$

$\vec{AB}=k{\bf{i}}+k{\bf{j}}$

Also, $\vec{BC}=\vec{AC}-\vec{AB}$

$=(2-k){\bf{i}}+(4-k){\bf{j}}$

$\vec{AB}.\vec{BC}=0$ (perpendicular to one another)

Therefore, $k(2-k)+k(4-k)=0$

$-2k^{2}+6k=0$

$k=0, k=3$

$k=0$ Isn't applicable as the dot product only applies to non-zero vectors.

So $\vec{AB}=3(\bf{i}+\bf{j})$

I'll try to fix the latex up a bit lol.

Stick · « **Reply #1269 on:** February 10, 2013, 05:52:14 pm »

Hmm... I was going to suggest generating three equations and solving them simultaneously, but that method is much more efficient.

Bad Student · « **Reply #1270 on:** February 10, 2013, 06:20:28 pm »

Quote from: AimingHigh on February 10, 2013, 05:26:21 pm

Can somebody help me with the following vectors question from the Essentials textbook?

ABC is a right-angled triangle with the right angle at B. If AC = 2i + 4j and AB is parallel to i + j, find AB.

I think you can also use vector resolutes to solve this.

Let b = i + j

$\vec{AB} = \vec{AC}.\hat{b}\hat{b}$

$\hat{b} = \frac{1}{\sqrt{2}}(i + j)$

$\vec{AC}.\hat{b} = 2 * \frac{1}{\sqrt{2}} + 4 * \frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}}$

$\vec{AC}.\hat{b}\hat{b} = \frac{6}{\sqrt{2}} * \frac{1}{\sqrt{2}}(i + j) = \frac{6}{2}(i + j) = 3(i + j)$

Stick · « **Reply #1271 on:** February 11, 2013, 06:14:39 pm »

Just stuck on part e.

Mao · « **Reply #1272 on:** February 11, 2013, 06:58:55 pm »

Hint: if a circle with centre Q goes through E, D, B and F, that means QE, QD, QB and QF are all radii, and must have the same magnitude.

507 · « **Reply #1273 on:** February 11, 2013, 07:05:22 pm »

Think of the coefficient of $\bf{i}$ to be like an x co-ordinate, and the coefficient of $\bf{j}$ to be a y co-ordinate.

So $\bf{e}$ = the point $(-4,6)$
$\bf{d}$ = $(-2,8)$
$\bf{b}$ = $(2,8)$

Sub these values into $(x-h)^{2}+(y-k)^{2}=r^{2}$ and solve simultaneously.
Giving you the equation $x^{2}+(y-4)^{2}=20$
CP is $(0,4)$
Which can be expressed as the vector $4{\bf{j}}$

bucklr · « **Reply #1274 on:** February 11, 2013, 07:13:18 pm »

Quote from: Stick on February 11, 2013, 06:14:39 pm

(Image removed from quote.)

Just stuck on part e.

or think that the angles EDF and EBF are right angles which means EF is the diameter of the circle. The midpoint of EF is C.

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