# What is the arclength of the polar curve f(theta) = sin(3theta)-4cot6theta  over theta in [0,pi/4] ?

Nov 21, 2016

The arclength is infinite.

#### Explanation:

From the reference Arc Length with Polar Coordinates

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

Given: $r \left(\theta\right) = \sin \left(3 \theta\right) - 4 \cot \left(6 \theta\right) , \alpha = 0 \mathmr{and} \beta = \frac{\pi}{4}$

(dr(theta))/(d"theta) = 3cos(3theta) + 24csc^2(6theta)

Substituting into the integral:

$L = {\int}_{0}^{\frac{\pi}{4}} \sqrt{{\left(\sin \left(3 \theta\right) - 4 \cot \left(6 \theta\right)\right)}^{2} + {\left(3 \cos \left(3 \theta\right) + 24 {\csc}^{2} \left(6 \theta\right)\right)}^{2}} d \theta$

This integral does not converge.