What are $r(s)$ and $\theta(s)$ for an ellipse?

Question

Book:
https://www.amazon.com/Vector-Calculus-Dover-Books-Mathematics/dp/0486466205

We have this problem here on page 89.

I think I was able to solve a) and c).

I know that b) is an ellipse. And I think we can also assume that the parameters $a,b$ are positive.

But for b) I don't know how to present the parametrization in the form

$$\boldsymbol{r}(s) = r(s) \cdot (\cos{(\theta(s))}, \sin{(\theta(s))}) \tag{*}$$

where $r,\theta$ are scalar functions.

This presentation is needed in order to apply the formulas given in the text.

I think this makes b) much harder than a) and c).
I think also that there might be some theory here which I am missing.
Or maybe the problem was meant to be just for the case $a=b$?!

Any ideas or hints?

EDIT: Radial and transverse components refer to the representations of $\boldsymbol{\dot{r}(s)}$ and $\boldsymbol{\ddot{r}(s)}$ as linear combinations of these two vectors.

$\boldsymbol{\hat{r}}(s) = (\cos{(\theta(s))}, \sin{(\theta(s))})$
$\boldsymbol{\hat{s}}(s) = (-\sin{(\theta(s))}, \cos{(\theta(s))})$

Also, $s$ is just some parameter, not necessarily path-length.

More details here:

Answer 1

• To avoid ambiguity, try using polar coordinates:

\begin{align*} \mathbf{r} &= r \, \hat{r} \\ \mathbf{v} &= \dot{r} \, \hat{r}+r\dot{\theta} \, \hat{\theta} \\ \mathbf{a} &= (\ddot{r}-r\dot{\theta}^2) \, \hat{r}+ (2\dot{r}\dot{\theta}+r\ddot{\theta}) \, \hat{\theta} \end{align*}

• In the language of differential geometry, $s$ is the arclength, $\kappa$ is the curvature and $\rho$ is the radius of curvature. Acceleration will include the tangential component (along $\mathbf{T}$) and normal component (along $\mathbf{N}$) which is the centripetal acceleration but is given by radius of curvature instead.

\begin{align} \mathbf{v} &= \dot{s} \, \mathbf{T} \\ &= v \, \mathbf{T} \\ \mathbf{a} &= \ddot{s} \, \mathbf{T}+ \kappa \, \dot{s}^2\mathbf{N} \\ &= \frac{d^2s}{dt^2} \mathbf{T} + \frac{v^2}{\rho} \mathbf{N} \\ &= v\frac{dv}{ds} \mathbf{T} + \frac{v^2}{\rho} \mathbf{N} \\ \end{align}

• Please also refer to my answer in Physics SE here.

• Back to your problem, it's just a simple exercise of calculus

\begin{align} \mathbf{r} &= (a\cos \omega t,b\sin \omega t) \\ \mathbf{v} &= \omega \, (-a\sin \omega t,b\cos \omega t) \\ v &= \omega \sqrt{a^2\sin^2 \omega t+b^2\cos^2 \omega t} \\ \mathbf{a} &= -\omega^2 \mathbf{r} \end{align}