Question

What is the arclength of `r=2sin(theta-(3pi)/16) + 9theta` on `theta in [(-5pi)/16,(9pi)/16]`?

Answer 1

Store `r=2sin(theta-(3pi)/16)+9theta` in your calculator.

Find the derivative `(dr)/(d theta)=2cos(theta-(3pi)/16)+9` and store this separately.

Plug these into the integral that gives the arc length for polar equations:

`s=int_((-5pi)/16)^((9pi)/16)sqrt(r^2+((dr)/(d theta))^2)d theta=37.1478...`