Question

What is the arclength of `(t^2-t-1,9t^2-2/t)` on `t in [1,4]`?

Answer 1

`approx 137.033`

Explanation:

Let `x(t)=t^2-t-1` then

`x'(t)=2t-1`

`y(t)=9t^2-2/t`

`y'(t)=18t+2/t^2`
then we have to calculate

`int_1^4sqrt((2t-1)^2+(18t+2/t^2)^2)dt`

by a numerical Methode we get `approx 137.033`