Question

What is the arclength of the polar curve `f(theta) = 2cos(9theta)-3sintheta` over `theta in [0,pi/3]`?

Answer 1

`approx 13.20973`

Explanation:

We use the formula

`s=int_alpha^beta\sqrt(r^2+((dr)/(d theta))^2)d theta`
where

`r(theta)=2cos(9theta)-3sin(theta)`
`r'(theta)=-18sin(9theta)-3cos(theta)`

so we have to integrate

`int_0^(pi/3)sqrt(173-160cos(18 theta)+60sin(8theta)+48sin(10theta))d theta`
by a numerical method we get

`approx 13.20973`