Question

What is the arclength of `(t/(1+t)^2,-1/t)` on `t in [2,4]`?

Answer 1

`approx 0.258144`

Explanation:

We have
`x(t)=t/(1+t)^2`
so
`x'(t)=(t-1)^2/(1+t)^2`

`y(t)=-1/t`
so
`y'(t)=1/t^4`
and we get the integral

`int_2^4sqrt(1/t^4+(t-1)^2/(1+t)^6)dt`
by a numerical method we get

`approx 0.258144`