Paul's Online Notes
Home / Calculus II / Parametric Equations and Polar Coordinates / Arc Length with Polar Coordinates
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 9.9 : Arc Length with Polar Coordinates

3. Set up, but do not evaluate, and integral that gives the length of the following polar curve. You may assume that the curve traces out exactly once for the given range of $$\theta$$.

$r = \cos \left( {2\theta } \right) + \sin \left( {3\theta } \right), \,\, 0 \le \theta \le 2\pi$

Show All Steps Hide All Steps

Start Solution

The first thing we’ll need here is the following derivative.

$\frac{{dr}}{{d\theta }} = - 2\sin \left( {2\theta } \right) + 3\cos \left( {3\theta } \right)$ Show Step 2

We’ll need the $$ds$$ for this problem.

$ds = \sqrt {{{\left[ {\cos \left( {2\theta } \right) + \sin \left( {3\theta } \right)} \right]}^2} + {{\left[ { - 2\sin \left( {2\theta } \right) + 3\cos \left( {3\theta } \right)} \right]}^2}} \,d\theta$ Show Step 3

The integral for the arc length is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{L = \int_{0}^{{2\pi }}{{\sqrt {{{\left[ {\cos \left( {2\theta } \right) + \sin \left( {3\theta } \right)} \right]}^2} + {{\left[ { - 2\sin \left( {2\theta } \right) + 3\cos \left( {3\theta } \right)} \right]}^2}} \,d\theta }}}}$