Question

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

Answer 1

`~4.298`

Explanation:

If we have a really small change in `t`, we will travel according to the following equation:
`ds^2 = dx^2 + dy^2 = [((dx)/(dt))^2 +((dy)/(dt))^2]dt^2`
i.e.
`ds = dt sqrt(((dx)/(dt))^2 +((dy)/(dt))^2)`

The total length is therefore
`int_(t_0)^(t_f) ds = int_(t_0)^(t_f) sqrt(((dx)/(dt))^2 +((dy)/(dt))^2) dt`

Calculating the values,
`(dx)/(dt) = 1 - e^t`
`(dy)/(dt) = 1 + 2/t^2`

Therefore, plugging in the numbers, we have the equation
`int_(1)^(2) sqrt((( 1 - e^t )^2 +(1 + 2/t^2)^2) dt`

We can't solve this exactly (or at least not in any easy way if it is technically possible), so we plug this into a calculator such as WolframAlpha, getting the value around `4.298`.