Question

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

Answer 1

The arclength is infinite.

Explanation:

From the reference Arc Length with Polar Coordinates

`L = int_alpha^beta sqrt{(r(theta))^2 + ((dr(theta))/(d theta))^2}d theta`

Given: `r(theta) = sin(3theta) - 4cot(6theta), alpha = 0 and beta = pi/4`

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

Substituting into the integral:

`L = int_0^(pi/4) sqrt{(sin(3theta) - 4cot(6theta))^2 + (3cos(3theta) + 24csc^2(6theta))^2}d theta`

This integral does not converge.