Question

What is the arclength of `r=costheta-3sin2theta` on `theta in [-pi/4,pi]`?

Answer 1

`approx 16.7916`

Explanation:

From
`r(theta)=cos(theta)-3sin(3theta)`
we get by differentiating

`r'(theta)=-sin(theta)-6cos(2theta)`
so we get

`int_(-pi/4)^pisqrt((cos(theta)-3*sin(2theta))^2+(-sin(theta)-6cos(2theta))^2)d theta`

By a numerical method we obtain

`16.7916`