Author Topic: VCE Specialist 3/4 Question Thread! (Read 2193703 times) Tweet Share

Hutchoo · « **Reply #780 on:** October 09, 2012, 06:37:56 pm »

I have a quick question:
Question 10: VCAA 2007. I really suck with these types of questions:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths1.pdf

I did something funky D= It's wrong, I gave myself 0/3 for it, but I was hoping someone could help me understand what I did wrong =)
{SEE ATTACHED IMAGE}

@b^3 I'd like to publicly thank you for all of your efforts in regards to helping the VN community. You're an absolute legend. I MEAN, LOOK AT THOSE DIAGRAMS!

Jenny_2108 · « **Reply #781 on:** October 09, 2012, 07:26:55 pm »

Quote from: Hutchoo on October 09, 2012, 06:37:56 pm

I have a quick question:
Question 10: VCAA 2007. I really suck with these types of questions:
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths1.pdf

I did something funky D= It's wrong, I gave myself 0/3 for it, but I was hoping someone could help me understand what I did wrong =)
{SEE ATTACHED IMAGE}

@b^3 I'd like to publicly thank you for all of your efforts in regards to helping the VN community. You're an absolute legend. I MEAN, LOOK AT THOSE DIAGRAMS!

You did wrong at line 3. It should be

$2sin(x)=\frac{4\sqrt{2}}{7} \times \frac{1-2sin^2(x)}{cos(x)}$

BubbleWrapMan · « **Reply #782 on:** October 09, 2012, 07:36:21 pm »

$\cos(2x)=1-2\sin^{2}(x)=\frac{7}{\sqrt{(4\sqrt{2})^{2}+7^{2}}}=\frac{7}{9}$ (use triangle)

$\Rightarrow 1-\frac{7}{9}=\frac{2}{9}=2\sin^{2}(x)$

$\Rightarrow \sin^{2}(x)=\frac{1}{9}$

$\Rightarrow \sin(x)=\frac{1}{3}$ as $x\in[0,\frac{\pi}{4})$

soccerboi · « **Reply #783 on:** October 09, 2012, 08:33:42 pm »

A particle moves along a curved path in such a way that its position vector r(t) at any time t is given by r(t)=acos(bt) i + asin(bt) j, where a,b E R+

How would i "verify that the acceleration vector is always directed towards the origin" ?

b^3 · « **Reply #784 on:** October 09, 2012, 08:56:15 pm »

If a particle is moving in a circle, then it's velocity is tangential to it's path.
Now if we find the acceleration vector.
$\begin{alignedat}{1}\underset{\sim}{r}(t) & =a\cos(bt)\underset{\sim}{i}+a\sin(bt)\underset{\sim}{j} \\ \underset{\sim}{\dot{r}}(t) & =-ab\sin(bt)\underset{\sim}{i}+ab\cos(bt)\underset{\sim}{j} \\ \underset{\sim}{\ddot{r}}(t) & =-ab^{2}\cos(bt)\underset{\sim}{i}-ab^{2}\sin(bt)\underset{\sim}{j} \\ & =-b^{2}\left(a\cos(bt)\underset{\sim}{i}+a\sin(bt)\underset{\sim}{j}\right) \\ & =-b^{2}\underset{\sim}{r}(t) \end{alignedat}$

So this means that the acceleration vector is the displacement vector mu,tipled by the constant -b². I.e. The acceleration vector will be in the opposite direction to the displacement vector, and as the displacement vector starts at the origin, this means that the acceleration vector is always directed towards the origin.

NOTE: I feel there is something I've missed above, maybe even missed the point of the question completely, if there is someone pick me up on it (I went on the path of finding that the velocity vectors and acceleration vectors are perpendicular then realised that wasn't needed either).... I might just be going crazy a bit..

Anyway, hope that helps

Jenny_2108 · « **Reply #785 on:** October 09, 2012, 09:00:44 pm »

^ OMG, you type so quickly b^3
I was typing and now delete LOL Hate LaTex now

soccerboi · « **Reply #786 on:** October 10, 2012, 05:00:04 pm »

I sketched the graph of y=2sin^-1(x/3) and noticed that the domain is [-3,3]. However, i always thought that the domain of a sin^-1 graph is always between [-1,1]. Can someone please explain where i'm lost at?
Thanks

b^3 · « **Reply #787 on:** October 10, 2012, 05:05:44 pm »

The factor in front of the x in the sine is $\frac{1}{3}$
$\begin{alignedat}{1}x & =\frac{1}{3}x' \\ x' & =3x \\ (x,y)\rightarrow & (3x,y) \end{alignedat}$

That is we are dilating the graph of $sin^{-1}(x)$ by a factor of 3 from the y-axis, this changes the domain of the function we have. (there is also a dilation from the x-axis involved in this one too which I haven't considered in the above working, this dilation doesn't change the domain, but changes the range).

We could also look at it this way.
$\sin^{-1}(x)$ can 'take' values between -1 and 1 as input values.
So if we have $\frac{x}{3}$ , then this can be between -1 and 1.
$-1\leq\frac{x}{3}\leq1$
So solving for x, our domain will become
$-3\leq x\leq3$

soccerboi · « **Reply #788 on:** October 10, 2012, 06:19:26 pm »

Thanks b^3!

Could someone explain to me why Friction is pointing the way it is in the diagram below?

EDIT: Nevermind, i think i got it. It's because it says "Albert is on the verge of moving down the slide."

soccerboi · « **Reply #789 on:** October 10, 2012, 06:50:59 pm »

The info in the above digram relates to this question that i'm stuck on.

Take the bottom end of the slide as the origin,i as the unit vector horizontally to the right and jas the unit vector vertically up.
d. Find the velocity vector v(t) which represents the velocity of Albert t seconds after he leaves the slide.

Note: in part c) i found that the speed of Albert as he reaches the end of the slide is 7m/s.
The answer to part d) is v (t) = (3.5 root3) i − (3.5+ gt) j

b^3 · « **Reply #790 on:** October 10, 2012, 07:09:19 pm »

Assuming that Albert was travelling with a constant velocity once he reached the end of the slide, then once he leaves the slide the only force that will be acting on him is gravity.
Now since the slide is at an angle, we can make the velocity vector at the end of the slide into it's x (well i) and y (well j) components.

So our initial velocity vector will be $\underset{\sim}{u}(t)&=V_{x}\underset{\sim}{i}+V_{y}\underset{\sim}{j}$ . (you will need to find these

)
Now the as we said before the only force acting on the object is gravity at this point, so that means $a=-g\underset{\sim}{j}$
So
$\begin{alignedat}{1}\frac{d\underset{\sim}{v}(t)}{dt} & =-g\underset{\sim}{j} \\ \underset{\sim}{v}(t) & =\int-g\underset{\sim}{j}dt+\underset{\sim}{u}(t) \end{alignedat}$

So from that I think you should be able to get it

soccerboi · « **Reply #791 on:** October 10, 2012, 07:51:54 pm »

What i initially did was just subbed in t=0, v(t)=7 into v(t)=-gt i+c to solve for c, but why is this wrong? Why does the initial velocity have to be changed into i,j from?

b^3 · « **Reply #792 on:** October 10, 2012, 08:01:40 pm »

Remember $\underset{\sim}{v}(t)$ is a vector, not a scalar. 7 is just the magnitude of the vector, so we need to make it into it's components. Also be careful with your coordinate system, the gt should be on j not i as gravity acts downwards not sidewards

martin1106 · « **Reply #793 on:** October 10, 2012, 10:11:46 pm »

Q: why do we need to add the unit vectors of a and b?

THANKS

Jenny_2108 · « **Reply #794 on:** October 10, 2012, 11:21:25 pm »

Quote from: martin1106 on October 10, 2012, 10:11:46 pm

Q: why do we need to add the unit vectors of a and b?

THANKS

Its the formula of unit vector bisects angle of 2 vectors. It takes long time to type LaTex and I'm doing homework but if you want ~~I can post the proof tomorrow~~

I post now

They add $\underset {\sim}{\hat a} and \underset {\sim}{\hat b}$ by using parallelogram rule

Before that, they find unit vectors so that their magnitudes are equal because the diagonals of a rhombus bisects the angle at vertex.

$\underset {\sim}{c}= \underset{\sim}{\hat a} + \underset {\sim}{\hat b}$ is vector bisects the angle of vector $\underset {\sim}{a}$ and $\underset {\sim}{b}$

$\Rightarrow \underset {\sim}{\hat c}= \frac{\underset{\sim}{\hat a} + \underset {\sim}{\hat b}}{ |\underset{\sim}{\hat a} + \underset {\sim}{\hat b}|}$

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