Question

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

Answer 1

`"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`