# What is the arc length of r(t)=(te^(t^2),t^2e^t,1/t) on tin [1,ln2]?

Jan 30, 2018

Arc Length ~~ −2.42533 \ \  (5dp)

The arc length is negative due to the lower bound $1$ being greater than the upper bound of $\ln 2$

#### Explanation:

We have a parametric vector function, given by:

$\boldsymbol{\underline{r}} \left(t\right) = \left\langlet {e}^{{t}^{2}} , {t}^{2} {e}^{t} , \frac{1}{t}\right\rangle$

In order to calculate the arc-length we will require the vector derivative, which we can compute using the product rule:

$\boldsymbol{\underline{r}} ' \left(t\right) = \left\langle\left(t\right) \left(2 t {e}^{{t}^{2}}\right) + \left(1\right) \left({e}^{{t}^{2}}\right) , \left({t}^{2}\right) \left({e}^{t}\right) + \left(2 t\right) \left({e}^{t}\right) , - \frac{1}{t} ^ 2\right\rangle$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left\langle2 {t}^{2} {e}^{{t}^{2}} + {e}^{{t}^{2}} , {t}^{2} {e}^{t} + 2 t {e}^{t} , - \frac{1}{t} ^ 2\right\rangle$

Then we compute the magnitude of the derivative vector:

 | bb ul r'(t) | = sqrt( (2t^2e^(t^2) + e^(t^2))^2 + (t^2e^t + 2te^t)^2 + (-1/t^2)^2) )

$\text{ } = \sqrt{{e}^{2 t} {t}^{4} + \frac{1}{t} ^ 4 + 4 {e}^{2 t} {t}^{3} + 4 {e}^{2 t} {t}^{2} + 4 {e}^{2 {t}^{2}} {t}^{2} + {e}^{2 {t}^{2}} + 4 {e}^{2 {t}^{2}} {t}^{4}}$

Then we can compute the arc-length using:

$L = {\int}_{1}^{\ln 2} \setminus | \boldsymbol{\underline{r}} ' \left(t\right) | \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{1}^{\ln 2} \setminus \sqrt{{e}^{2 t} {t}^{4} + \frac{1}{t} ^ 4 + 4 {e}^{2 t} {t}^{3} + 4 {e}^{2 t} {t}^{2} + 4 {e}^{2 {t}^{2}} {t}^{2} + {e}^{2 {t}^{2}} + 4 {e}^{2 {t}^{2}} {t}^{4}} \setminus \mathrm{dt}$

It is unlikely we can compute this integral using analytical technique, so instead using Numerical Methods we obtain an approximation:

 L ~~ −2.42533 \ \  (5dp)

The arc length is negative due to the lower bound $1$ being greater than the upper bound of $\ln 2$