1. Speed Problem

Question: When a particle moves in a plane so that x=2cos3t and y=sin3t, then the speed on the interval 0 =< t =< $\displaystyle \frac{\pi}{2}$ is a maximum when t equals ...

A. 0

B. $\displaystyle \frac{\pi}{6}$

C. $\displaystyle \frac{\pi}{4}$

D. $\displaystyle \frac{\pi}{3}$

E. $\displaystyle \frac{\pi}{2}$

I took the derivatives of both x and y equations and put them in the speed formula. But now that I have the derivatives in the speed formula, I'm not sure what the next step is algebraically.

I have, all in the square root,

(36sin3t)^2 + (9cos3t)^2

So how do I expand that out and what are the next steps?

2. Hi

You mean $\displaystyle 36 \sin^2(3t) + 9 \cos^2 (3t)$ I suppose ?

Use $\displaystyle \cos^2(3t) = 1 - \sin^2(3t)$

3. Originally Posted by Lord Darkin
Question: When a particle moves in a plane so that x=2cos3t and y=sin3t, then the speed on the interval 0 =< t =< $\displaystyle \frac{\pi}{2}$ is a maximum when t equals ...

A. 0

B. $\displaystyle \frac{\pi}{6}$

C. $\displaystyle \frac{\pi}{4}$

D. $\displaystyle \frac{\pi}{3}$

E. $\displaystyle \frac{\pi}{2}$

I took the derivatives of both x and y equations and put them in the speed formula. But now that I have the derivatives in the speed formula, I'm not sure what the next step is algebraically.

I have, all in the square root,

(36sin3t)^2 + (9cos3t)^2

So how do I expand that out and what are the next steps?
Wouldn't an application of the chain rule be easier?

$\displaystyle \frac{dx}{dt} = -6sin(3t)$

$\displaystyle \frac{dy}{dt} = 3cos(3t)$

$\displaystyle \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = -\frac{3cos(3t)}{6sin(3t)} = -\frac{1}{2}cot(3t)$

IIRC a maximum occurs when $\displaystyle \frac{dy}{dx} = 0$

$\displaystyle -\frac{cos(3t)}{2sin(3t)} = 0$

$\displaystyle cos(3t) = 0$

$\displaystyle 3t = \frac{\pi}{2}$

$\displaystyle t = \frac{\pi}{6}$

4. I'm still a little shaky on this. My teacher, on the answer key he gave me, gave me this all under the square root:

$\displaystyle 36 \sin^2(3t) + 9 \cos^2 (3t)$

Then he wrote:

$\displaystyle 72sin(3t)cos(t)*3 + 18cos(3t)(-sin(3t))*3 = 0$

Simplifies to:

$\displaystyle 81sin(6t)$

I'm not sure how to get to the part starting with 72 sin 3 etc. I think I understand the concept of this problem, I just want to know the algebra on the answer key.

5. $\displaystyle \sqrt{36 \sin^2(3t) + 9 \cos^2 (3t)}$ is maximum when $\displaystyle 36 \sin^2(3t) + 9 \cos^2 (3t)$ is maximum which means when the derivative of $\displaystyle 36 \sin^2(3t) + 9 \cos^2 (3t)$ is equal to 0

Derivative of
$\displaystyle \sin^2(3t)$
The derivative of $\displaystyle u^2(t)$ is $\displaystyle 2 \:u(t)\: u'(t)$
The derivative of $\displaystyle u(t) = \sin(3t)$ is $\displaystyle 3 \:\cos(3t)$
Therefore the derivative of
$\displaystyle \sin^2(3t)$ is $\displaystyle 2 \:sin(3t)\: 3 \:\cos(3t) = 6 \:sin(3t) \:cos(3t)$

In the same way the derivative of $\displaystyle \cos^2(3t)$ is $\displaystyle 2 \:cos(3t)\: (-3) \:\sin(3t) = -6 \:sin(3t) \:cos(3t)$

Therefore the derivative of $\displaystyle 36 \sin^2(3t) + 9 \cos^2 (3t)$ is $\displaystyle 36 \times 6 \:sin(3t) \:cos(3t) - 9 \times 6 \:sin(3t) \:cos(3t) = 162 \:sin(3t) \:cos(3t)$

Using $\displaystyle \sin(2x) = 2 \:\sin x \: \cos x$ the derivative of $\displaystyle 36 \sin^2(3t) + 9 \cos^2 (3t)$ is $\displaystyle 81 \: \sin(6t)$ which is equal to 0 when $\displaystyle t = k\:\frac{\pi}{6}$

6. Thanks for the help. I think I get it.

1 last thing.

If I have

$\displaystyle sin3(\frac{\pi}{3})$

does that simplify to

$\displaystyle sin(\pi)$?

7. Originally Posted by Lord Darkin
Thanks for the help. I think I get it.

1 last thing.

If I have

$\displaystyle sin3(\frac{\pi}{3})$

does that simplify to

$\displaystyle sin(\pi)$?
No, it won't.

$\displaystyle \sin \left[3\left(\frac{\pi}{3}\right)\right]$ is the graph of $\displaystyle sin \left(\frac{\pi}{3}\right)$ but 'squashed' by a factor of 3 such that a period is now $\displaystyle \frac{2\pi}{3}$

8. I get it. Thanks.