Question

What is the arclength of `f(t) = (sint*cos2t,sint)` on `t in [-pi,0]`?

Answer 1

`approx 3.943805190587053`

Explanation:

We have
`x(t)=sin(t)cos(2*t)`
`x'(t)=cos(t)cos(2t)-2sin(t)sin(2t)`

`y(t)=sin(t)`
`y'(t)=cos(t)`
so we have to integrate

`int_(-pi)^0sqrt((cos(t)cos(2t)-2sin(t)sin(2t))^2+cos^2(t))dt`
i got only a numerical result

`approx 3.943805190587053`