Question

What is the arclength of `f(t) = (sin^2t-cos2t,t/pi)` on `t in [-pi/4,pi/4]`?

Answer 1

`approx 3.06967`

Explanation:

we have
`x(t)=sin^2(t)-cos(2t)`
so
`x'(t)=2sin(t)cos(t)+2sin(t)`
`y(t)=t/pi`
`y'(t)=1/pi`
and we have the integral
`int_(-pi/4)^(pi/4)sqrt(1/pi^2+(2sin(t)cos(t)+2sin(2t))^2)dt`
This leads to an elliptic integral.