# Determining the Length of a Polar Curve

## Key Questions

• The Arc Length in Polar Coordinates is given bu:

$L = \int \setminus \mathrm{dS}$ where $\mathrm{dS} = \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} \setminus d \theta$

Use the chain rule.

#### Explanation:

By Chain Rule,

$\frac{\mathrm{dr}}{d \theta} = 3 {\cos}^{2} \left(\frac{\theta}{3}\right) \cdot \left[- \sin \left(\frac{\theta}{3}\right)\right] \cdot \frac{1}{3}$

by cleaning up a bit,

$= - {\cos}^{2} \left(\frac{\theta}{3}\right) \sin \left(\frac{\theta}{3}\right)$

Let us first look at the curve $r = {\cos}^{3} \left(\frac{\theta}{3}\right)$, which looks like this:

Note that $\theta$ goes from $0$ to $3 \pi$ to complete the loop once.

Let us now find the length $L$ of the curve.

$L = {\int}_{0}^{3 \pi} \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$

$= {\int}_{0}^{3 \pi} \sqrt{{\cos}^{6} \left(\frac{\theta}{3}\right) + {\cos}^{4} \left(\frac{\theta}{3}\right) {\sin}^{2} \left(\frac{\theta}{3}\right)} d \theta$

by pulling ${\cos}^{2} \left(\frac{\theta}{3}\right)$ out of the square-root,

$= {\int}_{0}^{3 \pi} {\cos}^{2} \left(\frac{\theta}{3}\right) \sqrt{{\cos}^{2} \left(\frac{\theta}{3}\right) + {\sin}^{2} \left(\frac{\theta}{3}\right)} d \theta$

by ${\cos}^{2} \theta = \frac{1}{2} \left(1 + \cos 2 \theta\right)$ and ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$,

$= \frac{1}{2} {\int}_{0}^{3 \pi} \left[1 + \cos \left(\frac{2 \theta}{3}\right)\right] d \theta$

$= \frac{1}{2} {\left[\theta + \frac{3}{2} \sin \left(\frac{2 \theta}{3}\right)\right]}_{0}^{3 \pi}$

$= \frac{1}{2} \left[3 \pi + 0 - \left(0 + 0\right)\right] = \frac{3 \pi}{2}$

I hope that this was helpful.

• We can find the arc length $L$ of a polar curve $r = r \left(\theta\right)$ from $\theta = a$ to $\theta = b$ by

$L = {\int}_{a}^{b} \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$