# What is the arclength of (1/(t+te^t),-t) on t in [3,4]?

Aug 7, 2017

Approximate arc length via a numerical method is:

$1.00$ (2dp)

#### Explanation:

The arc length of a curve:

$\boldsymbol{\vec{r}} \left(t\right) = \left\langlex \left(t\right) , y \left(t\right)\right\rangle$

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

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

So, for the given curve:

$\boldsymbol{\vec{r}} \left(t\right) = \left\langle\frac{1}{t + t {e}^{t}} , - t\right\rangle \setminus \setminus \setminus t \in \left[3 , 4\right]$

Differentiating the components wrt $t$ we get:

$x ' \left(t\right) = - {\left(t + t {e}^{t}\right)}^{- 2} \frac{d}{\mathrm{dt}} \left(t + t {e}^{t}\right)$
$\text{ } = - {\left(t + t {e}^{t}\right)}^{- 2} \left(1 + \left(t\right) \left({e}^{t}\right) + \left(1\right) \left({e}^{t}\right)\right)$
$\text{ } = - \frac{\left(1 + {e}^{t} + t {e}^{t}\right)}{t + t {e}^{t}} ^ \left(2\right)$

$y ' \left(t\right) = - 1$

So, the arc length is given by:

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

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

$L = 1.000072796710095 \ldots$