Question

What is the arclength of `f(t) = (t^2-5,e^t)` on `t in [1,3]`?

Answer 1

`approx 19.2253`

Explanation:

We get
`x'(t)=2t` so `x'(t)^2=4t^2`

`y'(t)=e^t` so `y'(t)^2=e^(2t)`
and the integral is given by
`int_1^3sqrt(4t^2+e^(2t))dt`
With a numerical method we get
`approx 19.2253`