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

Jun 10, 2018

$\approx 10.37224180$

#### Explanation:

We have

$x \left(t\right) = {e}^{- 2 t} - {t}^{2}$
then
$x ' \left(t\right) = - {2}^{- 2 t} - 2 t$+

$y \left(t\right) = t - \frac{t}{e} ^ \left(t - 1\right)$
then

$y ' \left(t\right) = 1 - {e}^{1 - t} + {e}^{1 - t} t$
so we get the integral
${\int}_{- 1}^{1} \sqrt{{\left(- 2 {e}^{- 2 t} - 2 t\right)}^{2} + {\left(1 - {e}^{1 - t} + {e}^{1 - t} t\right)}^{2}} \mathrm{dt}$
By a numerical method we get

$\approx 10.37224180$