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

## (parametric equations)

Jun 14, 2018

We need to start by deriving a speed function.

Recall that $s = \sqrt{{\left(x ' \left(t\right)\right)}^{2} + {\left(y ' \left(t\right)\right)}^{2}}$

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

$x ' \left(t\right) = 6 \cos \left(2 t\right)$
$y ' \left(t\right) = - 8 \sin \left(2 t\right)$

Therefore

$s = \sqrt{36 {\cos}^{2} \left(2 t\right) + 64 {\sin}^{2} \left(2 t\right)}$

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

$s ' = \frac{28 \sin \left(4 x\right)}{\sqrt{64 {\sin}^{2} \left(2 x\right) + 36 {\cos}^{2} \left(2 x\right)}}$

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

$4 x = \pi \mathmr{and} 0 \mathmr{and} 2 \pi$
$x = \frac{\pi}{4} \mathmr{and} 0 \mathmr{and} \frac{\pi}{2} , \frac{3 \pi}{4}$

At $\frac{\pi}{3}$ the derivative is negative. Therefore $x = \frac{\pi}{4} + \frac{\pi}{2} n$ will always be a maximum.

We can confirm graphically

Hopefully this helps!

Jun 14, 2018

$t = \left(2 k + 1\right) \frac{\pi}{4} q \quad k = 0 , 1 , 2 , \ldots$

#### Explanation:

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

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

Speed $s$:

${s}^{2} = {\left\mid \boldsymbol{v} \right\mid}^{2} = 36 {\cos}^{2} \left(2 t\right) + 64 {\sin}^{2} \left(2 t\right)$

$= 36 + 28 {\sin}^{2} \left(2 t\right) q \quad \square$

You don't need calculus to solve this, ${s}^{2} \in \left[36 , 64\right]$, but that's the section it is in.

$\frac{d \left({s}^{2}\right)}{\mathrm{dt}} = 112 \sin \left(2 t\right) \cos \left(2 t\right)$

$= 56 \sin \left(4 t\right)$

$\therefore \frac{d \left({s}^{2}\right)}{\mathrm{dt}} = 0$ for:

• $4 t = 0 , \pi , 2 \pi , \ldots$

• $t = 0 , \frac{\pi}{4} , \frac{\pi}{2} , \ldots$

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

• $\left\{\begin{matrix}{s}^{2} \left(0\right) = 36 \\ \boldsymbol{{s}^{2} \left(\frac{\pi}{4}\right) = 64} \\ {s}^{2} \left(\frac{\pi}{2}\right) = 36\end{matrix}\right.$

So the particle is travelling fastest at:

$t = \frac{\pi}{4} + k \frac{\pi}{2} = \left(2 k + 1\right) \frac{\pi}{4} q \quad k = 0 , 1 , 2 , \ldots$