# What is the arc length of the curve given by r(t)= (e^-t,e^t,1) on  t in [1, 2]?

May 29, 2018

$L = {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{{e}^{2 \left(1 - 4 n\right)} - {e}^{1 - 4 n}}{1 - 4 n}$ units.

#### Explanation:

$r \left(t\right) = \left({e}^{-} t , {e}^{t} , 1\right)$

$r ' \left(t\right) = \left(- {e}^{-} t , {e}^{t} , 0\right)$

Arc length is given by:

$L = {\int}_{1}^{2} \sqrt{{e}^{- 2 t} + {e}^{2 t} + 0} \mathrm{dt}$

Rearrange:

$L = {\int}_{1}^{2} {e}^{t} \sqrt{1 + {e}^{- 4 t}} \mathrm{dt}$

For $t \in \left[1 , 2\right]$, ${e}^{- 4 t} < 1$. Take the series expansion of the square root:

$L = {\int}_{1}^{2} {e}^{t} \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {e}^{- 4 n t}\right\} \mathrm{dt}$

Simplify:

$L = {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{1}^{2} {e}^{\left(1 - 4 n\right) t} \mathrm{dt}$

Integrate directly:

$L = {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left[{e}^{\left(1 - 4 n\right) t}\right]}_{1}^{2} / \left(1 - 4 n\right)$

Hence

$L = {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{{e}^{2 \left(1 - 4 n\right)} - {e}^{1 - 4 n}}{1 - 4 n}$