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

Mar 9, 2016

$L = {\int}_{1}^{3} \sqrt{{\left[\left(2 t + 1\right) {e}^{2 t} - {e}^{t} - 3\right]}^{2} + 16 {t}^{2}} \mathrm{dt}$
After setting the above Integral I used an online Integration Calculator for the numerical approximation. No closed integral exist, but the numerical approximation of L is: $L = 1179.856725422904$

#### Explanation:

Given $f \left(t\right) = \left(t {e}^{2 t} - {e}^{t} - 3 t , - 2 {t}^{2}\right)$ on $t \in \left[1 , 3\right]$
Find the Arclength, L:
$L = {\int}_{a}^{b} \sqrt{{\left(f ' \left(t\right)\right)}^{2} + {\left(g ' \left(t\right)\right)}^{2}} \mathrm{dt}$

$f \left(t\right) = t {e}^{2 t} - {e}^{t} - 3 t$
$f ' \left(t\right) = \left(2 t + 1\right) {e}^{2 t} - {e}^{t} - 3$
${\left(f ' \left(t\right)\right)}^{2} = {\left[\left(2 t + 1\right) {e}^{2 t} - {e}^{t} - 3\right]}^{2}$

$g \left(t\right) = - 2 {t}^{2}$
$g ' \left(t\right) = - 4 t$
${\left(g ' \left(t\right)\right)}^{2} = 16 {t}^{2}$

$L = {\int}_{1}^{3} \sqrt{{\left[\left(2 t + 1\right) {e}^{2 t} - {e}^{t} - 3\right]}^{2} + 16 {t}^{2}} \mathrm{dt}$
No closed integral, but numerically you can Approximate, L:
$L = 1179.856725422904$