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

Jan 17, 2018

~4.298

#### Explanation:

If we have a really small change in $t$, we will travel according to the following equation:
${\mathrm{ds}}^{2} = {\mathrm{dx}}^{2} + {\mathrm{dy}}^{2} = \left[{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}\right] {\mathrm{dt}}^{2}$
i.e.
$\mathrm{ds} = \mathrm{dt} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}}$

The total length is therefore
${\int}_{{t}_{0}}^{{t}_{f}} \mathrm{ds} = {\int}_{{t}_{0}}^{{t}_{f}} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

Calculating the values,
$\frac{\mathrm{dx}}{\mathrm{dt}} = 1 - {e}^{t}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = 1 + \frac{2}{t} ^ 2$

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 $4.298$.