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

Jul 21, 2017

Approximate arc length via a numerical method is:

$577.06$ (2dp)

#### Explanation:

The arc length of a curve:

 vec(r) (t) = << x(t), y(t)) >>

Over an interval $\left[a , b\right]$ is given by:

$L = {\int}_{a}^{b} \setminus | | \vec{r} \left(t\right) | | \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{a}^{b} \setminus \sqrt{x ' {\left(t\right)}^{2} + y ' {\left(t\right)}^{2}} \setminus \mathrm{dt}$

So, for the given curve:

$\vec{r} \left(t\right) = \left\langle\left(4 - {t}^{2}\right) {e}^{t} , t {e}^{t}\right\rangle \setminus \setminus \setminus t \in \left[3 , 4\right]$

Differentiating the components wrt $t$ we get:

$x ' \left(t\right) = \left(4 - {t}^{2}\right) \left(\frac{d}{\mathrm{dt}} {e}^{t}\right) + \left(\frac{d}{\mathrm{dt}} \left(4 - {t}^{2}\right)\right) \left({e}^{t}\right)$
$\text{ } = \left(4 - {t}^{2}\right) \left({e}^{t}\right) + \left(- 2 t\right) \left({e}^{t}\right)$
$\text{ } = - \left({t}^{2} + 2 t - 4\right) {e}^{t}$

$y ' \left(t\right) = \left(t\right) \left(\frac{d}{\mathrm{dt}} {e}^{t}\right) + \left(\frac{d}{\mathrm{dt}} t\right) \left({e}^{t}\right)$
$\text{ } = t {e}^{t} + {e}^{t}$
$\text{ } = \left(t + 1\right) {e}^{t}$

So, the arc length is given by:

$L = {\int}_{3}^{4} \setminus \sqrt{x ' {\left(t\right)}^{2} + y ' {\left(t\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{3}^{4} \setminus \sqrt{{\left(- \left({t}^{2} + 2 t - 4\right) {e}^{t}\right)}^{2} + {\left(\left(t + 1\right) {e}^{t}\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{3}^{4} \setminus \sqrt{{\left({t}^{2} + 2 t - 4\right)}^{2} {e}^{2 t} + {\left(t + 1\right)}^{2} {e}^{2 t}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{3}^{4} \setminus \sqrt{\left\{{\left({t}^{2} + 2 t - 4\right)}^{2} + {\left(t + 1\right)}^{2}\right\} {e}^{2 t}} \setminus \mathrm{dt}$

The integral does not have an elementary antiderivative,and so we evaluate the definite integral al numerically:

$L = 577.055737868 \ldots$