path in polar coordinates

Question

Instead of a path $γ(t) = (x(t), y(t))$, we could have a path in polar coordinates, $γ(t) = (r(t), θ(t))$. How would you calculate $ds$ in polar coordinates? I know $ds$ is just another way of saying a "bit of length"

Answer 1

We have $x(t) = r\cos\theta$ and $y(t)=r\sin\theta$, so using the chain rule, \begin{equation*} \frac{dx}{dt} = \cos\theta\frac{dr}{dt} - r\sin\theta\frac{d\theta}{dt} \end{equation*} and \begin{equation*} \frac{dy}{dt} = \sin\theta\frac{dr}{dt} + r\cos\theta\frac{d\theta}{dt}. \end{equation*} Thus, \begin{equation*} \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \left(\frac{dr}{dt}\right)^2 + r^2\left(\frac{d\theta}{dt}\right)^2. \end{equation*} This yields \begin{equation*} \int_a^b \sqrt{\left(\frac{dr}{dt}\right)^2 + r^2\left(\frac{d\theta}{dt}\right)^2} \ dt = \int_a^b \sqrt{r^2 + \frac{dr}{d\theta}}\frac{d\theta}{dt} \ dt = \int_a^b \sqrt{r^2 + \frac{dr}{d\theta}} \ d\theta. \end{equation*}