Question

A particle is travelling an elliptical path described by `(3sin(2t)), 4cos(2t))`. Find the points at which it is travelling fastest?

(parametric equations)

Answer 1

We need to start by deriving a speed function.

Recall that `s = sqrt((x'(t))^2 + (y'(t))^2)`

Therefore we need to find the derivative of the position functions.

`x'(t) = 6cos(2t)`
`y'(t) = -8sin(2t)`

Therefore

`s = sqrt(36cos^2(2t) + 64sin^2(2t))`

We need to differentiate this in order to find the maximum.

`s' = (28sin(4x))/sqrt(64sin^2(2x) + 36cos^2(2x))`

Critical points occur when the derivative equals `0`, in other words when `28sin(4x) = 0`.

`4x = pi or 0 or 2pi`
`x= pi/4 or 0 or pi/2, (3pi)/4`

At `pi/3` the derivative is negative. Therefore `x = pi/4 +pi/2n` will always be a maximum.

We can confirm graphically

Hopefully this helps!

Answer 2

`t =(2k+1) pi/4 qquad k = 0,1,2,...`

Explanation:

`bbr = (3sin(2t)), 4cos(2t))`

`bbv = (6cos(2t)), -8sin(2t))`

Speed `s`:

`s^2 = abs( bbv)^2 = 36 cos^2(2t) + 64 sin^2(2t)`

`= 36 + 28 sin^2(2t) qquad square`

You don't need calculus to solve this, `s^2 in [36,64]`, but that's the section it is in.

`(d(s^2))/(dt) = 112 sin (2t) cos (2t)`

`= 56 sin (4t)`

`:. (d(s^2))/(dt) = 0` for:

• `4t = 0, pi , 2pi , ...`

• `t = 0, pi/4, pi/2 , ...`

Given the symmetry of the orbit, you only really need to look at the first 3 solutions, as they will repeat:

• `{(s^2(0) = 36 ),(bb(s^2(pi/4) = 64)),(s^2(pi/2) = 36) :}`

So the particle is travelling fastest at:

`t = pi/4 + k pi/2 = (2k+1) pi/4 qquad k = 0,1,2,...`