The key to computing the length of a polar curve is to think of it as a parametrized curve with parameter $\theta$. (When computing the slope of a polar curve, we called the parameter $t$ and set $\theta=t$. Calling the parameter $\theta$ is equivalent and saves a step.) Then \begin{eqnarray*}x&=& r \cos(\theta); \cr y &=& r \sin(\theta). \end{eqnarray*}Taking derivatives we get\begin{eqnarray*}\frac{dx}{d\theta} &=& r' \cos(\theta) - r \sin(\theta); \cr \frac{dy}{d\theta} &=& r' \sin(\theta) + r \cos(\theta), \end{eqnarray*} where $r'$ is shorthand for $dr/d\theta$. Squaring gives \begin{eqnarray*}\left ( \frac{dx}{d\theta}\right)^2 &=& (r')^2 \cos^2(\theta) + r^2 \sin^2(\theta) - 2 r r' \sin(\theta) \cos(\theta); \cr \left ( \frac{dy}{d\theta}\right)^2 &=& (r')^2 \sin^2(\theta) + r^2 \cos^2(\theta) + 2 r r' \sin(\theta) \cos(\theta). \end{eqnarray*} Adding then gives $$\left (\frac{dx}{d\theta}\right )^2 + \left ( \frac{dy}{d\theta} \right )^2 = r^2 + \left ( \frac{dr}{d\theta}\right)^2, \qquad \hbox{so}$$