# What is the arc length of the curve given by r(t)= (e^tsqrt(2t),e^t,e^(-2t)) on  t in [3,4]?

Aug 25, 2016

$\approx 110.75$

#### Explanation:

$\vec{r} \left(t\right) = \left(\begin{matrix}{e}^{t} \sqrt{2 t} \\ {e}^{t} \\ {e}^{- 2 t}\end{matrix}\right)$

for arclength: $s = {\int}_{3}^{4} \dot{s} \setminus \mathrm{dt}$

$= {\int}_{3}^{4} \sqrt{\vec{v} \cdot \vec{v}} \setminus \mathrm{dt}$

$\vec{v} = \frac{d \vec{r}}{\mathrm{dt}} = \left(\begin{matrix}{e}^{t} \sqrt{2 t} + \frac{{e}^{t}}{\sqrt{2 t}} \\ {e}^{t} \\ - 2 {e}^{- 2 t}\end{matrix}\right)$

$\implies {\int}_{3}^{4} \sqrt{{\left({e}^{t} \sqrt{2 t} + \frac{{e}^{t}}{\sqrt{2 t}}\right)}^{2} + {\left({e}^{t}\right)}^{2} + {\left(- 2 {e}^{- 2 t}\right)}^{2}} \setminus \mathrm{dt}$

$= {\int}_{3}^{4} \sqrt{{e}^{2 t} \left(2 t\right) + 2 {e}^{2 t} + \frac{{e}^{2 t}}{2 t} + {e}^{2 t} + 4 {e}^{- 4 t}} \setminus \mathrm{dt}$

$= {\int}_{3}^{4} \sqrt{{e}^{2 t} \left(2 t + 3 + \frac{1}{2 t}\right) + 4 {e}^{- 4 t}} \setminus \mathrm{dt}$

$\approx 110.75$ by computer