# What is the arc length of f(t)=(3te^t,t-e^t)  over t in [2,4]?

May 15, 2017

$612.530$ (3dp)

#### Explanation:

We have:

$f \left(t\right) = \left(3 t {e}^{t} , t - {e}^{t}\right)$ where $t \in \left[2 , 4\right]$

The parametric arc-length is given by:

$L = {\int}_{\alpha}^{\beta} \setminus \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \setminus \mathrm{dt}$

We can differentiate the parameters:

$x \left(t\right) = 3 t {e}^{t} \implies \frac{\mathrm{dx}}{\mathrm{dt}} = 3 t {e}^{t} + 3 {e}^{t}$

$y \left(t\right) = t - {e}^{t} \implies \frac{\mathrm{dy}}{\mathrm{dt}} = 1 - {e}^{t}$

Then the arc-length is given by:

$L = {\int}_{2}^{4} \setminus \sqrt{{\left(3 t {e}^{t} + 3 {e}^{t}\right)}^{2} + {\left(1 - {e}^{t}\right)}^{2}} \setminus \mathrm{dt}$

This integral dos not have a trivial anti-derivative, and so is evacuated using numerical methods to give:

$L = 612.530$ (3dp)