What is the arclength of the polar curve f(theta) = 4sin(2theta)-2sec^2theta over theta in [0,pi/8] ?

1 Answer
May 2, 2017

"Arclength " approx 2.509

Explanation:

Arclength on the polar plane has the following formula:
L=int_(theta_1)^(theta_2)sqrt(r^2+((dr)/(d theta))^2)" "d theta

Given the equation f(theta)=r=4sin(2theta)-2sec^2theta, we can differentiate to find
(dr)/(d theta)=4cos(2theta)(2)-4sectheta(sec theta tan theta)

(dr)/(d theta)=8cos(2theta)-4sec^2thetatantheta

Plug in the expressions for r and (dr)/(d theta) into the polar arclength formula to get:
"Arclength "= int_0^(pi/8)sqrt((4sin(2theta)-2sec^2theta)^2+(8cos(2theta)-4sec^2theta tantheta)^2)" "d theta

Use a graphing calculator to evaluate:

approx 2.509258859