Question

What is the arclength of `r=5/2sin(theta/8+(3pi)/16) -theta/4` on `theta in [(-3pi)/16,(7pi)/16]`?

Answer 1

`s = 2.72416`

Explanation:

In polar coordinates `ds^2 = r(theta)^2+((d r(theta))/(d theta))^2`
here `r(theta)=5/5sin(theta/8+(3pi)/16)-theta/4`
`(dr(theta))/(d theta) =-1/4+5/16cos((3pi)/16+theta/8)` doing
`int_((-3pi)/16)^((7pi)/16)sqrt(r(theta)^2+((d r(theta))/(d theta))^2)d theta = 2.72416`