Author Topic: VCE Specialist 3/4 Question Thread! (Read 2543314 times) Tweet Share

Alwin · « **Reply #2235 on:** September 07, 2013, 05:22:22 pm »

Quote from: nliu1995 on September 07, 2013, 05:20:16 pm

nek minnit...integrate e^x*sin x by hand yeah? =D

AHAHAHAHAHA IMAGINE THE COMPLAINTS VCAA WOULD GET BUT IT WOULD BE SO WORTH IT ^^

lzxnl · « **Reply #2236 on:** September 08, 2013, 02:21:29 am »

They can do that; use the derivatives of e^x sin x and e^x cos x to work this one out.

barydos · « **Reply #2237 on:** September 08, 2013, 12:44:37 pm »

How does one work out this question?

answers:

Spoiler

[/img]http://i.imgur.com/XgQbcZK.png[/img]

Thanks in advance

b^3 · « **Reply #2238 on:** September 08, 2013, 01:08:18 pm »

The components of the tangent vector are given by the derivatives of the components of the position vector. I.e. we will end up with a velocity vector. This is because, if we were to draw a tangent to the position curve, we get the direction of the velocity vector, as this is the direction that the object is moving in.
$\begin{alignedat}{1}\underset{\sim}{r}\left(t\right) & =\left(\cos\left(t\right)+t\sin\left(t\right)\right)\underset{\sim}{i}+\left(\sin\left(t\right)-t\cos\left(t\right)\right)\underset{\sim}{j} \\ \underset{\sim}{r'}\left(t\right) & =\frac{d}{dt}\left(\cos\left(t\right)+t\sin\left(t\right)\right)\underset{\sim}{i}+\frac{d}{dt}\left(\sin\left(t\right)-t\cos\left(t\right)\right)\underset{\sim}{j} \\ \underset{\sim}{r'}\left(t\right) & =\left(-\sin\left(t\right)+t\left(\cos\left(t\right)\right)+\sin\left(t\right)\left(1\right)\right)\underset{\sim}{i}+\left(\cos\left(t\right)-t\left(-\sin\left(t\right)\right)-\cos\left(t\right)\left(1\right)\right)\underset{\sim}{j} \\ \underset{\sim}{r'}\left(t\right) & =t\cos\left(t\right)\underset{\sim}{i}+t\sin\left(t\right)\underset{\sim}{j} \end{alignedat}$

But we want the unit tangent vector, so to find the unit vector, we divide by the magnitude of the tangent vector.
$\begin{alignedat}{1}|\underset{\sim}{r'}\left(t\right)| & =\sqrt{t^{2}\cos^{2}\left(t\right)+t^{2}\sin^{2}\left(t\right)} \\ & =\sqrt{t^{2}\left(\cos^{2}\left(t\right)+\sin^{2}\left(t\right)\right)} \\ \cos^{2}\left(t\right)+\sin^{2}\left(t\right)=1 \\ |\underset{\sim}{r'}\left(t\right)| & =\sqrt{t^{2}} \\ & =t \\ \underset{\sim}{\hat{r}'}\left(t\right) & =\frac{t\cos\left(t\right)}{t}\underset{\sim}{i}+\frac{t\sin\left(t\right)}{t}\underset{\sim}{j} \\ & =\cos\left(t\right)\underset{\sim}{i}+\sin\left(t\right)\underset{\sim}{j} \end{alignedat}$

barydos · « **Reply #2239 on:** September 08, 2013, 01:18:58 pm »

Quote from: b^3 on September 08, 2013, 01:08:18 pm

The components of the tangent vector are given by the derivatives of the components of the position vector. I.e. we will end up with a velocity vector. This is because, if we were to draw a tangent to the position curve, we get the direction of the velocity vector, as this is the direction that the object is moving in.
$\begin{alignedat}{1}\underset{\sim}{r}\left(t\right) & =\left(\cos\left(t\right)+t\sin\left(t\right)\right)\underset{\sim}{i}+\left(\sin\left(t\right)-t\cos\left(t\right)\right)\underset{\sim}{j} \\ \underset{\sim}{r'}\left(t\right) & =\frac{d}{dt}\left(\cos\left(t\right)+t\sin\left(t\right)\right)\underset{\sim}{i}+\frac{d}{dt}\left(\sin\left(t\right)-t\cos\left(t\right)\right)\underset{\sim}{j} \\ \underset{\sim}{r'}\left(t\right) & =\left(-\sin\left(t\right)+t\left(\cos\left(t\right)\right)+\sin\left(t\right)\left(1\right)\right)\underset{\sim}{i}+\left(\cos\left(t\right)-t\left(-\sin\left(t\right)\right)-\cos\left(t\right)\left(1\right)\right)\underset{\sim}{j} \\ \underset{\sim}{r'}\left(t\right) & =t\cos\left(t\right)\underset{\sim}{i}+t\sin\left(t\right)\underset{\sim}{j} \end{alignedat}$

But we want the unit tangent vector, so to find the unit vector, we divide by the magnitude of the tangent vector.
$\begin{alignedat}{1}|\underset{\sim}{r'}\left(t\right)| & =\sqrt{t^{2}\cos^{2}\left(t\right)+t^{2}\sin^{2}\left(t\right)} \\ & =\sqrt{t^{2}\left(\cos^{2}\left(t\right)+\sin^{2}\left(t\right)\right)} \\ \cos^{2}\left(t\right)+\sin^{2}\left(t\right)=1 \\ |\underset{\sim}{r'}\left(t\right)| & =\sqrt{t^{2}} \\ & =t \\ \underset{\sim}{\hat{r}'}\left(t\right) & =\frac{t\cos\left(t\right)}{t}\underset{\sim}{i}+\frac{t\sin\left(t\right)}{t}\underset{\sim}{j} \\ & =\cos\left(t\right)\underset{\sim}{i}+\sin\left(t\right)\underset{\sim}{j} \end{alignedat}$

Ahh I see where I went wrong, I was treating the 't' coefficient of sin and cos, as a constant rather than a variable.
Thanks a lot b^3!

lzxnl · « **Reply #2240 on:** September 08, 2013, 01:41:10 pm »

I'm going to be picky and say...be careful of signs. Here, the working is valid because t>=0, but generally you cannot just assume sqrt(t^2) = t.

b^3 · « **Reply #2241 on:** September 08, 2013, 01:45:20 pm »

Yeah, but for problems like these we generally assume $t\geq0$ unless otherwise stated (well for vector functions in VCE anyways). I probably should have stated $t\geq0$ , but anyways.

If you really want to be picky, every time you introduce a new constant you should state what it can be, e.g. $C\in \mathbb{R}$ e.t.c While it's good to be picky (to some extent), and you will be expected to be later at uni, it's not always necessary during VCE.

lzxnl · « **Reply #2242 on:** September 08, 2013, 03:20:11 pm »

Well, I'm sure we can't lose a mark for being overly cautious.

barydos · « **Reply #2243 on:** September 09, 2013, 08:50:44 pm »

Easy one here; what's the best way to answer this question:

The position of a particle at time t seconds is given by the vector r(t) = ( $t^2 - 6t)$ (i+2j-2k), measured in metres form a fixed point. The distance in metres travelled in the first 5 seconds is:
A) 15
B) 13
C) 5
D) 20
E) 39

b^3 · « **Reply #2244 on:** September 09, 2013, 08:58:33 pm »

Since we want distance and not displacement, break it up into sections, sections where it is moving in the positive direction or the negative direction. To find out when it changes direction, we need to find when the velocity becomes zero, so firstly you would differentiate each component to find the velocity vector equation, and then equation that to zero (normally you equate each of the three components to zero and find a value of $t$ for which they are all zero, but since they all vary with the same function of $t$ , it's a little bit easier here) to find value(s) of $t$ . Then you can find the displacement at the points that it stops, relative to the point it started at. The distance will be the sum of these differences.

i.e. It stops at $t=3$ , checking that at $t=0$ our displacement is zero, we can just subtitute $t=3$ into the displacement vector equation, and then substitute $t=5$ in. The distance travelled between $t=0$ and $t=3$ will be given by $|\underset{\sim}{r}\left(3\right)-\underset{\sim}{r}\left(0\right)|$ . Then the distance travelled between $t=3$ and our endpoint $t=5$ will be $|\underset{\sim}{r}\left(5\right)-\underset{\sim}{r}\left(3\right)|$ . Add those two together and you get the total distance traveled in that interval.

The reason it works is because when you break it down into sections where the direction of motion does not change, then the change in displacement will be equal to the distance travelled in that interval.

Probably took more space to explain it than to actually do the question but oh well, hope that helps and makes sense.

Now back to not procrastinating from midsem exam studying...

EDIT: Added modulus to difference in displacements, as we want the distances to be positive.

barydos · « **Reply #2245 on:** September 09, 2013, 09:03:05 pm »

Quote from: b^3 on September 09, 2013, 08:58:33 pm

Since we want distance and not displacement, break it up into sections, sections where it is moving in the positive direction or the negative direction. To find out when it changes direction, we need to find when the velocity becomes zero, so firstly you would differentiate each component to find the velocity vector equation, and then equation that to zero (normally you equate each of the three components to zero and find a value of $t$ for which they are all zero, but since they all vary with the same function of $t$ , it's a little bit easier here) to find value(s) of $t$ . Then you can find the displacement at the points that it stops, relative to the point it started at. The distance will be the sum of these differences.

i.e. It stops at $t=3$ , checking that at $t=0$ our displacement is zero, we can just subtitute $t=3$ into the displacement vector equation, and then substitute $t=5$ in. The distance travelled between $t=0$ and $t=3$ will be given by $\underset{\sim}{r}\left(3\right)-\underset{\sim}{r}\left(0\right)$ . Then the distance travelled between $t=3$ and our endpoint $t=5$ will be $\underset{\sim}{r}\left(5\right)-\underset{\sim}{r}\left(3\right)$ . Add those two together and you get the total distance traveled in that interval.

The reason it works is because when you break it down into sections where the direction of motion does not change, then the change in displacement will be equal to the distance travelled in that interval.

Probably took more space to explain it than to actually do the question but oh well, hope that helps and makes sense.

Now back to not procrastinating from midsem exam studying...

Thanks for the quick response

G'luck on your exam haha

b^3 · « **Reply #2246 on:** September 09, 2013, 09:06:23 pm »

I should have had a modulus on those differences in displacement. My bad, will add in now (since we only want the distance).

lzxnl · « **Reply #2247 on:** September 09, 2013, 09:32:28 pm »

Quote from: b^3 on September 09, 2013, 08:58:33 pm

Since we want distance and not displacement, break it up into sections, sections where it is moving in the positive direction or the negative direction. To find out when it changes direction, we need to find when the velocity becomes zero, so firstly you would differentiate each component to find the velocity vector equation, and then equation that to zero (normally you equate each of the three components to zero and find a value of $t$ for which they are all zero, but since they all vary with the same function of $t$ , it's a little bit easier here) to find value(s) of $t$ . Then you can find the displacement at the points that it stops, relative to the point it started at. The distance will be the sum of these differences.

i.e. It stops at $t=3$ , checking that at $t=0$ our displacement is zero, we can just subtitute $t=3$ into the displacement vector equation, and then substitute $t=5$ in. The distance travelled between $t=0$ and $t=3$ will be given by $|\underset{\sim}{r}\left(3\right)-\underset{\sim}{r}\left(0\right)|$ . Then the distance travelled between $t=3$ and our endpoint $t=5$ will be $|\underset{\sim}{r}\left(5\right)-\underset{\sim}{r}\left(3\right)|$ . Add those two together and you get the total distance traveled in that interval.

The reason it works is because when you break it down into sections where the direction of motion does not change, then the change in displacement will be equal to the distance travelled in that interval.

Probably took more space to explain it than to actually do the question but oh well, hope that helps and makes sense.

Now back to not procrastinating from midsem exam studying...

EDIT: Added modulus to difference in displacements, as we want the distances to be positive.

Quote from: Anonymiza on September 09, 2013, 08:50:44 pm

Easy one here; what's the best way to answer this question:

The position of a particle at time t seconds is given by the vector r(t) = ( $t^2 - 6t)$ (i+2j-2k), measured in metres form a fixed point. The distance in metres travelled in the first 5 seconds is:
A) 15
B) 13
C) 5
D) 20
E) 39

Erm...the particle travels along a curved path. At least from what I remember, an integral is needed for this one.
The integral is often given in VCAA exams, but if you're not familiar with it, distance = integral of speed dt.
Velocity vector is (2t-6)*(i+2j-2k)
Speed = integral of 3*abs(2t-6) dt from 0 to 5
calculator says 39
b^3's method is for when the particle is moving in a straight line. It's not here.
Although somehow her/his approach yields 39 as well.

SocialRhubarb · « **Reply #2248 on:** September 09, 2013, 09:43:26 pm »

Quote from: nliu1995 on September 09, 2013, 09:32:28 pm

Erm...the particle travels along a curved path.

... isn't it moving in a straight line in the same direction as the vector i+2j-2k?

Alwin · « **Reply #2249 on:** September 09, 2013, 09:49:17 pm »

Quote from: nliu1995 on September 09, 2013, 09:32:28 pm

Erm...the particle travels along a curved path. At least from what I remember, an integral is needed for this one.
The integral is often given in VCAA exams, but if you're not familiar with it, distance = integral of speed dt.
Velocity vector is (2t-6)*(i+2j-2k)
Speed = integral of 3*abs(2t-6) dt from 0 to 5
calculator says 39
b^3's method is for when the particle is moving in a straight line. It's not here.
Although somehow her/his approach yields 39 as well.

Firstly, b^3 is guy nliu

look that the little symbol next to his gender, Gender: Male

Secondly, aren't you doing the exact same thing?

The particle is travelling along a straight line in the direction of the vector (i+2j-2k). The particle is always at some scalar value of it, so it is travelling along that line (ill obmit the uni maths representation)

Thirdly,
r(t) = (t^2 - 6t)(i+2j-2k)
v(t) = (2t-6)*(i+2j-2k)
Since we want only positives, we positify it

ie:
$\left|\int_0^3 v(t) dt\right| + \left|\int_3^5 v(t) dt\right|=|r(3)-r(0)| + |r(5)-r(3)|$
Which is exactly the same as b^3, no wonder you got the same answer

If it was a curved path, the you have to use ds = sqrt(dx^2 + dy^2) which can be written as s= int_a^b (sqrt(1+[f'(x)]^2)) for those of you who where interested. And no, this is not in the spesh course

EDIT: beaten

Im such a slowpoke tonight haha

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