Question

What is the arclength of the polar curve `f(theta) = -3sin(theta)-2sec^2theta` over `theta in [0,pi/8]`?

Answer 1

`approx 1.827918426331`

Explanation:

We are using the formula

`s=int_alpha^betasqrt(r^2+((d r)/(d theta))^2)d theta`
`r(theta)=-3sin(theta)-2sec^2(theta)`

`r'(theta)=-3cos(theta)-4sec^2(theta)tan(theta)`

so we have to integrate

`int_0^(pi/8)sqrt((-3sin(theta)-2sec^2(theta))^2+(-3cos(theta)-4sec^2(theta)tan(theta))^2)d theta`

by a numerical method we get `approx 1.827918426331`