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

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Answer 1 · 2016-11-21T18:35:09

The arclength is infinite.

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

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.

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