# What is the arclength of (t/(t+5),t) on t in [-1,1]?

May 25, 2018

$L = 2 + 4 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4 n - 1} {\left(\frac{5}{16}\right)}^{2 n} \left(1 - {\left(\frac{2}{3}\right)}^{4 n - 1}\right)$ units.

#### Explanation:

$f \left(t\right) = \left(\frac{t}{t + 5} , t\right) = \left(1 - \frac{5}{t + 5} , t\right)$

$f ' \left(t\right) = \left(\frac{5}{t + 5} ^ 2 , 1\right)$

Arclength is given by:

$L = {\int}_{-} {1}^{1} \sqrt{\frac{25}{t + 5} ^ 4 + 1} \mathrm{dt}$

Apply the substitution $t + 5 = u$:

$L = {\int}_{4}^{6} \sqrt{1 + \frac{25}{u} ^ 4} \mathrm{du}$

For $u \in \left[4 , 6\right]$, $\frac{25}{u} ^ 4 < 1$. Take the series expansion of the square root:

$L = {\int}_{4}^{6} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(\frac{25}{u} ^ 4\right)}^{n} \mathrm{du}$

Isolate the $n = 0$ term:

$L = {\int}_{4}^{6} \mathrm{du} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {25}^{n} {\int}_{4}^{6} {u}^{- 4 n} \mathrm{du}$

Integrate directly:

$L = 2 + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {25}^{n} / \left(1 - 4 n\right) {\left[{u}^{1 - 4 n}\right]}_{4}^{6}$

Hence

$L = 2 + 4 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4 n - 1} {\left(\frac{5}{16}\right)}^{2 n} \left(1 - {\left(\frac{2}{3}\right)}^{4 n - 1}\right)$