M408M Learning Module Pages
Main page Chapter 10: Parametric Equations and Polar CoordinatesLearning module LM 10.1: Parametrized Curves:Learning module LM 10.2: Calculus with Parametrized Curves:Learning module LM 10.3: Polar Coordinates:Learning module LM 10.4: Areas and Lengths of Polar Curves:Area inside a polar curveArea between polar curves Arc lengths of polar curves Learning module LM 10.5: Conic Sections:Learning module LM 10.6: Conic Sections in Polar Coordinates:Chapter 12: Vectors and the Geometry of SpaceChapter 13: Vector FunctionsChapter 14: Partial DerivativesChapter 15: Multiple Integrals |
Arc lengths of polar curvesThe 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 $$\left (\frac{dx}{d\theta}\right )^2 + \left ( \frac{dy}{d\theta} \right )^2 = r^2 + \left ( \frac{dr}{d\theta}\right)^2, \qquad \hbox{so}$$
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