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

1 Answer
Jan 17, 2018

Answer:

#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.