Question

What is the arclength of the polar curve `f(theta) = 2thetasin(5theta)-thetacot2theta` over `theta in [pi/12,pi/2]`?

Answer 1

`infty`

Explanation:

For this solution, I am assuming that your `f` is actually referring to the radial component.

If we travel an infinitely small angle, the distance will be f(`theta`) d`theta`. Therefore, we can integrate that function across that whole range to get the value, i.e. `int_(pi/12)^(pi/2) f(theta)d theta`

However, since cotangent goes to infinity very quickly at `pi/2`, this integral does not converge to a finite value, hence the length is infinite.