Question

What is the arclength of `f(t) = (t/sqrt(t-1),t/(t^2-1))` on `t in [2,3]`?

Answer 1

`0.326586`

Explanation:

We have
`x(t)=t/sqrt(t-1)`
`x'(t)=1/sqrt(t-1)-t/(2*(t-1)^(3/2)`
`y(t)=t/(t^2-1)`
then
`y'(t)=1/(t^2-1)-2*t^2/(t^2-1)^2`
and our integral will be
`int_2^3sqrt((t-2)^2/(4*(t-1)^3)+(t^2+1)^2/((t-1)^4*(t+1)^4))dt`
which i can only solve by a numerical method.