# What is the arclength of the polar curve f(theta) = cos^2theta-3sin^2theta  over theta in [pi/3,pi/2] ?

Mar 14, 2018

$L = \sqrt{\frac{13}{3}} \left(\frac{\pi}{3}\right) + \sqrt{\frac{52}{3}} {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{2 n} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) {6}^{m} / {\left(- 52\right)}^{n} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} {\cos}^{m} 2 \theta d \theta$ units.

#### Explanation:

$f \left(\theta\right) = {\cos}^{2} \theta - 3 {\sin}^{2} \theta = 2 \cos 2 \theta - 1$

$f ' \left(\theta\right) = 4 \sin 2 \theta$

Arclength is given by:

$L = {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sqrt{{\left(2 \cos 2 \theta - 1\right)}^{2} + 16 {\sin}^{2} 2 \theta} d \theta$

Expand the square and combine terms:

$L = {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sqrt{17 - 4 \cos 2 \theta - 12 {\cos}^{2} 2 \theta} d \theta$

Complete the square in the square root:

$L = \frac{1}{\sqrt{3}} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sqrt{52 - {\left(6 \cos 2 \theta + 1\right)}^{2}} d \theta$

Factor out the larger piece:

$L = \sqrt{\frac{52}{3}} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sqrt{1 - \frac{1}{52} {\left(6 \cos 2 \theta + 1\right)}^{2}} d \theta$

For $\theta \in \left[\frac{\pi}{3} , \frac{\pi}{2}\right]$, $\frac{1}{52} {\left(6 \cos 2 \theta + 1\right)}^{2} < 1$. Take the series expansion of the square root:

$L = \sqrt{\frac{52}{3}} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{1}{52} {\left(6 \cos 2 \theta + 1\right)}^{2}\right)}^{n} d \theta$

Isolate the $n = 0$ term:

$L = \sqrt{\frac{52}{3}} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} d \theta + \sqrt{\frac{52}{3}} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{1}{52}\right)}^{n} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} {\left(6 \cos 2 \theta + 1\right)}^{2 n} d \theta$

Apply binominal expansion:

$L = \sqrt{\frac{13}{3}} \left(\frac{\pi}{3}\right) + \sqrt{\frac{52}{3}} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{1}{52}\right)}^{n} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} {\sum}_{m = 0}^{2 n} \left(\begin{matrix}2 n \\ m\end{matrix}\right) {\left(6 \cos 2 \theta\right)}^{m} d \theta$

Simplify:

$L = \sqrt{\frac{13}{3}} \left(\frac{\pi}{3}\right) + \sqrt{\frac{52}{3}} {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{2 n} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}2 n \\ m\end{matrix}\right) {6}^{m} / {\left(- 52\right)}^{n} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} {\cos}^{m} 2 \theta d \theta$