Question

What is the arclength of `f(t) = (tsqrt(lnt),t^3/(4-t))` on `t in [1,e]`?

Answer 1

`\approx 16.0871`

Explanation:

We get
`x'(t)=1/(2sqrt(log(t)))+sqrt(log(t))`
`y'(t)=3t^2/(4-t)+t^3/(4-t)^2`
so we have the integral
`int_{0}^{e}sqrt((1/(2*sqrt(log(t)))+sqrt(log(t)))^2+(3*t^2/(4-t)+t^3/(4-t)^2)^2)dt`
Good luck!