Author Topic: VCE Specialist 3/4 Question Thread! (Read 2578331 times) Tweet Share

Henreezy · « **Reply #1515 on:** March 27, 2013, 07:52:51 pm »

e^1, I can't actually check if you're wrong, my teacher just gave us this exam prep. paper to get us used to the style of questions we could be asked.

I assumed that it would be where it was undefined but reciting that logic in my head only made it hurt

I'm just going to proceed as if it were correct and I'll ask tomorrow.
Thanks for the help!

sin0001 · « **Reply #1516 on:** March 28, 2013, 05:25:10 pm »

Are questions asking for the same thing:
1) find the vector with the same magnitude as b and with the same direction as a
2) find the vector resolute of b in the direction of a

b^3 · « **Reply #1517 on:** March 28, 2013, 05:33:18 pm »

Not quite, think about it this way. The vector resolute of b in the direction of a is projecting b onto a, the length of the vector that you obtain will be smaller than the original vector, since for a right angled triangle the hypotenuse will be longer than the adjacent side. But for 1), you want a vector that has the same magnitude as b.

To find 1), you can find a unit vector in the direction of a, and then multiply by the magnitude of b (the unit vector will ensure the vector has the same direction as a while multiply by the magnitude of b, will result in the vector having the same magnitude of b, as the unit vector's length is 1).

EDIT: Adding an example.

Let
$\begin{alignedat}{1}\underset{\sim}{a} & =6\underset{\sim}{i}+8\underset{\sim}{k} \\ \underset{\sim}{b} & =2\underset{\sim}{i}-\underset{\sim}{j}-2\underset{\sim}{k} \end{alignedat} $

Now for (1), we want a vector in the same direction as $\underset{\sim}{a}$ , so we first need to find the unit vector in the direction of $\underset{\sim}{a}$
$\begin{alignedat}{1}\underset{\sim}{\hat{a}} & =\frac{1}{\sqrt{6^{2}+0^{2}+8^{2}}}\left(6\underset{\sim}{i}+8\underset{\sim}{k}\right) \\ & =\frac{1}{\sqrt{100}}\left(6\underset{\sim}{i}+8\underset{\sim}{k}\right) \\ & =\frac{1}{10}\left(6\underset{\sim}{i}+8\underset{\sim}{k}\right) \\ & =\frac{1}{5}\left(3\underset{\sim}{i}+4\underset{\sim}{k}\right) \end{alignedat}$

So now we have a vector of magnitude 1 in the direction of $\underset{\sim}{a}$ , so we need to now find the magnitude of $\underset{\sim}{b}$ .
$\begin{alignedat}{1}|\underset{\sim}{b}| & =\sqrt{2^{2}+\left(-1\right)^{2}+\left(-2\right)^{2}} \\ & =\sqrt{9} \\ & =3 \end{alignedat}$

So our final vector will be
$\begin{alignedat}{1}\underset{\sim}{c} & =3\times\frac{1}{5}\left(3\underset{\sim}{i}+4\underset{\sim}{k}\right) \\ & =\frac{9}{5}\underset{\sim}{i}+\frac{12}{5}\underset{\sim}{k} \end{alignedat}$

(2) For the vector resolute of $\underset{\sim}{b}$ in the direction of $\underset{\sim}{a}$ we will obtain.
$\begin{alignedat}{1}\frac{\underset{\sim}{b}.\underset{\sim}{a}}{|\underset{\sim}{a}|^{2}}\underset{\sim}{a} & =\frac{\left(2\underset{\sim}{i}-\underset{\sim}{j}-2\underset{\sim}{k}\right).\left(6\underset{\sim}{i}+8\underset{\sim}{k}\right)}{\sqrt{6^{2}+0^{2}+8^{2}}}\left(6\underset{\sim}{i}+8\underset{\sim}{k}\right) \\ & =\frac{12-16}{\sqrt{100}}\left(6\underset{\sim}{i}+8\underset{\sim}{k}\right) \\ & =-\frac{2}{5}\left(6\underset{\sim}{i}+8\underset{\sim}{k}\right) \\ & =-\frac{12}{5}\underset{\sim}{i}-\frac{16}{5}\underset{\sim}{k} \end{alignedat}$

EDIT2: I hope that the above all right, took a bit to get it all typed together properly. Anyways, hope it helps.

sin0001 · « **Reply #1518 on:** March 28, 2013, 06:25:22 pm »

Thanks b^3, you rule!!!
Thanks for going to the trouble of including the example

lzxnl · « **Reply #1519 on:** March 28, 2013, 10:50:54 pm »

Quote from: Henreezy on March 27, 2013, 07:34:31 pm

I'm stuck on this question:

find dy/dx in terms of both x and y for the equation 3x^2+18x-y^2+4y+11=0,
then find the coordinates for which the tangent is parallel to the y axis.

Differentiating implictly, we end up with 6x+18-2y*dy/dx+4*dy/dx=0
dy/dx(2y-4)=6x+18
dy/dx=(6x+18)/(2y-4)=(3x+9)/(y-2)

The tangent is parallel to the y axis when we have a vertical tangent, i.e. dy/dx=+-infinity
This happens when the denominator=0, or y=2
Subbing in the original equation, we find that 3x^2+18x-4+8+11=0
3x^2+18x+15=0
x^2+6x+5=0
(x+1)*(x+5)=0 => x=-1 or x=-5
so (-1,2) and (-5,2) are your coordinates

TimmyC · « **Reply #1520 on:** March 30, 2013, 03:21:43 pm »

No idea how to do this!!
This is adv gen, but still ties in with spesh:

Vector a=3i+2j-k b=4i+3k
vector resolute of a in the direction of b= 9/25(4i+3k)

Find the vector resolute of a perpendicular to b

Thanks in advance!!

Lasercookie · « **Reply #1521 on:** March 30, 2013, 03:34:18 pm »

Quote from: TrentCotchin on March 30, 2013, 03:21:43 pm

Vector a=3i+2j-k b=4i+3k
vector resolute of a in the direction of b= 9/25(4i+3k)

Find the vector resolute of a perpendicular to b

So you've found the vector resolute of a in the direction of b. That's the vector a1 there. You could also consider this to be the 'horizontal' component (though not necessarily horizontal)

We want the vector that's perpendicular to that (the 'vertical' component), a2.

It's just a bit of vector addition to find that vector: a2 = a - a1

This is a recent post by polar which is related, might be worth reading Vectors

P0ppinfr3sh · « **Reply #1522 on:** April 01, 2013, 06:29:27 pm »

Hey guys,

So this question has been giving me a bit of trouble:
(Just pretend @=theta, unfortunately i don't know latex $:-\$ )
ABCDE is a pentagon inscribed in a circle.
AB=BC=CD=DE=1 and angle BOA=2@
O is the centre of the circle
Let AE=p
Show p=sin(4@)/sin(@)
Refer to the attached file for the diagram.

If it helps, using circle theorems, angle BEA=@.

Thanks for your help.

Homer · « **Reply #1523 on:** April 01, 2013, 06:57:13 pm »

a particle moves in a straight line, left and right of an origin O. The position, x(t) of the particle at any time, t (seconds) is given by:

$x(t)=3t^2-12$

what is the average velocity during the fourth second? ANS: 21cm/s how?

b^3 · « **Reply #1524 on:** April 01, 2013, 07:04:38 pm »

The average velocity of the particle will be given by the change in position of the particle (during that time interval), divided by the time interval. The first second is from t=0 s to t=1 s, so the fourth second will be from t=3 s to t=4 s. So during the fourth second we are looking for
$\begin{alignedat}{1}v_{avg} & =\frac{\Delta x}{\Delta t} \\ & =\frac{x_{2}-x_{1}}{t_{2}-t_{1}} \\ & =\frac{x\left(4\right)-x\left(3\right)}{4-3} \end{alignedat} $

So we need to find the displacement of the particle at t=3 s and t=4 s.
$\begin{alignedat}{1}x\left(4\right) & =3\left(4\right)^{2}-12 \\ & =36 \\ x\left(3\right) & =3\left(3{}^{2}\right)-12 \\ & =15 \\ v_{avg} & =\frac{x\left(4\right)-x\left(3\right)}{4-3} \\ & =36-15 \\ & =21\:\text{cm/s} \end{alignedat} $

Phy124 · « **Reply #1525 on:** April 01, 2013, 09:17:50 pm »

Quote from: Deleted User on April 01, 2013, 07:54:05 pm

How do I prove using complex exponential to prove that
antiderivative(e^(-t)*(sin(t))^2 dt) from 0 to pi = 0.4*(1 - e^(-pi))

One arduous way of going about it is as follows;

$\int^\pi_0e^{-t}\sin^2(t)dt$

Using the idendity $\sin^2(x) = \frac{1}{2}(1-\cos(2x))$ :

$= \frac{1}{2}\int^\pi_0e^{-t}(1-\cos(2t))dt$

$= \frac{1}{2}\left[\int^\pi_0e^{-t}dt-\int^\pi_0e^{-t}\cos(2t)dt\right]$

$= \frac{1}{2}\left[\int^\pi_0e^{-t}dt-Re\int^\pi_0e^{-t}e^{2it}dt\right]$

$= \frac{1}{2}\left[\int^\pi_0e^{-t}dt-Re\int^\pi_0e^{(2i-1)t}dt\right]$

$= \frac{1}{2}\left[-e^{-t}-Re \left\{\frac{1}{2i-1}e^{(2i-1)t}\right\}\right]^{\pi}_0$

$= \frac{1}{2}\left[-e^{-t}-Re \left\{-\frac{2i+1}{5}e^{(2i-1)t}\right\}\right]^{\pi}_0$

$= \frac{1}{2}\left[-e^{-t}+Re \left\{\frac{2i+1}{5}e^{-t}(\cos(2t)+i\sin(2t))\right\}\right]^{\pi}_0$

$= \frac{1}{2}\left[-e^{-t}+\frac{1}{5}e^{-t}(\cos(2t)-2\sin(2t))\right]^{\pi}_0$

$= \frac{1}{2}\left(\left[-e^{-\pi}+\frac{1}{5}e^{-\pi}(\cos(2\pi)-2\sin(2\pi))\right]-\left[-e^{0}+\frac{1}{5}e^{0}(\cos(0)-2\sin(0))\right]\right)$

$= \frac{2}{5}(1-e^{-\pi})$

Additionally, sorry about splitting the integrals then combining them, but I couldn't be bothered doing the separately after that

Also, for the right side integral it can be done easily by knowing that:

$\int e^{ax}\cos(bx) \ dx = \frac{e^{ax}(a\cos(bx)+b\sin(bx))}{a^2+b^2}$

b^3 · « **Reply #1526 on:** April 01, 2013, 09:23:11 pm »

Just going to note that you shouldn't be expected to do that by hand in Specialist Maths^

bapp · « **Reply #1527 on:** April 02, 2013, 01:48:07 pm »

Quote from: P0ppinfr3sh on April 01, 2013, 06:29:27 pm

Hey guys,

So this question has been giving me a bit of trouble:
(Just pretend @=theta, unfortunately i don't know latex $:-\$ )
ABCDE is a pentagon inscribed in a circle.
AB=BC=CD=DE=1 and angle BOA=2@
O is the centre of the circle
Let AE=p
Show p=sin(4@)/sin(@)
Refer to the attached file for the diagram.

If it helps, using circle theorems, angle BEA=@.

Thanks for your help.

Alright I did this question a while back, first you have to realise that this is a circle so the angle AOE will equal 2pi-8@ then you can sort of makeout a triangle. Call the midpoint of AE M. Which will mean the angle AOM is pi-4@ Now you know AB is equal to 1 so you can draw another triangle to work out the length AO in terms of sin@ you end up getting 1/2sin@ now you can write the equation sin(pi-4@)=AE/2 x 2sin@ expand the LHS and you get -cos(pi)sin4@=AEsin@ rearrange AE and you get your answer I attached my working out as well. It's a little messy!!!

Kanye East · « **Reply #1528 on:** April 03, 2013, 01:10:53 pm »

Consider the periodic funtion F(x)=cos(2x)-sqrt(3) sin(2x)
Express this as a single sine function. Please show working out!
I keep getting 2sin(2x-pi/6) when the answer is supposed to be 2sin(pi/6-2x)
Thanks in advance.

b^3 · « **Reply #1529 on:** April 03, 2013, 01:36:31 pm »

You might want to check your value of A and the base angle corresponding to that

$\begin{alignedat}{1}\cos\left(2x\right)-\sqrt{3}\sin\left(2x\right) & =A\sin\left(Bx+C\right)\end{alignedat}$
Using the compound angle formula for sine
$\begin{alignedat}{1}\sin\left(x+y\right) & =\sin\left(x\right)\cos\left(y\right)+\cos\left(x\right)\sin\left(y\right) \\ \cos\left(2x\right)-\sqrt{3}\sin\left(2x\right) & =A\sin\left(Bx\right)\cos\left(C\right)+A\cos\left(Bx\right)\sin\left(C\right) \end{alignedat}$

Equating $A\cos\left(Bx\right)\sin\left(C\right)$ with $\cos\left(2x\right)$ , (we equate these two since these are the terms with the $x$ inside the cosine.
$\begin{alignedat}{1}\cos\left(2x\right) & =A\cos\left(Bx\right)\sin\left(C\right) \\ \implies B & =2 \\ A\sin\left(C\right) & =1 \\ A & =\frac{1}{\sin\left(C\right)}....[1] \end{alignedat}$

Equating the other terms
$\begin{alignedat}{1}-\sqrt{3}\sin\left(2x\right) & =A\sin\left(2x\right)\cos\left(C\right) \\ A\cos\left(C\right) & =-\sqrt{3}....[2] \\ \text{Substitute [1] into [2]} \\ \frac{\cos\left(C\right)}{\sin\left(C\right)} & =-\sqrt{3} \\ \tan\left(C\right) & =-\frac{1}{\sqrt{3}} \\ \implies C & =-\frac{\pi}{6} \\ A & =\frac{1}{\sin\left(-\frac{\pi}{6}\right)} \\ \implies A & =-2 \\ \cos\left(2x\right)-\sqrt{3}\sin\left(2x\right) & =-2\sin\left(2x-\frac{\pi}{6}\right) \end{alignedat}$

Sine is an odd function, i.e. $f\left(-x\right)=-f\left(x\right)$
$\begin{alignedat}{1}\therefore & \cos\left(2x\right)-\sqrt{3}\sin\left(2x\right)=2\sin\left(\frac{\pi}{6}-2x\right)\end{alignedat}$

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Author Topic: VCE Specialist 3/4 Question Thread! (Read 2578331 times) Tweet Share

Henreezy

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sin0001

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