# What is the arc length of the curve given by r(t)= (ln(1/t),t^2,t) on  t in [1, 10]?

Mar 8, 2018

The arc length is $99 + \ln 10 - \frac{3}{4} \left(\ln 10 + {\sum}_{m = 1}^{\infty} \left(\begin{matrix}- 1 \\ m\end{matrix}\right) {\left(\frac{1}{2}\right)}^{m} \left(\frac{1}{2 m}\right) \left(1 - \frac{1}{100} ^ m\right)\right) + {\sum}_{n = 2}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}1 - 2 n \\ m\end{matrix}\right) {\left(- \frac{3}{4}\right)}^{n} {\left(\frac{1}{2}\right)}^{m} \left(\frac{1}{n + m}\right) \left(1 - \frac{1}{100} ^ \left(n + m\right)\right)$ units.

#### Explanation:

$r \left(t\right) = \left(\ln \left(\frac{1}{t}\right) , {t}^{2} , t\right) = \left(- \ln t , {t}^{2} , t\right)$

$r ' \left(t\right) = \left(- \frac{1}{t} , 2 t , 1\right)$

Arc length is given by:

$L = {\int}_{1}^{10} \sqrt{\frac{1}{t} ^ 2 + 4 {t}^{2} + 1} \mathrm{dt}$

Complete the square in the square root:

$L = {\int}_{1}^{10} \sqrt{{\left(2 t + \frac{1}{t}\right)}^{2} - 3} \mathrm{dt}$

Factor out the larger piece:

$L = {\int}_{1}^{10} \left(2 t + \frac{1}{t}\right) \sqrt{1 - \frac{3}{2 t + \frac{1}{t}} ^ 2} \mathrm{dt}$

For $t \in \left[1 , 10\right]$, $\frac{3}{2 t + \frac{1}{t}} ^ 2 < 1$. Take the series expansion of the square root:

$L = {\int}_{1}^{10} \left(2 t + \frac{1}{t}\right) \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{3}{2 t + \frac{1}{t}} ^ 2\right)}^{n}\right\} \mathrm{dt}$

Isolate the $n = 0$ term and simplify:

$L = {\int}_{1}^{10} \left(2 t + \frac{1}{t}\right) \mathrm{dt} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- 3\right)}^{n} {\int}_{1}^{10} {\left(2 t + \frac{1}{t}\right)}^{1 - 2 n} \mathrm{dt}$

Rearrange:

$L = {\left[{t}^{2} + \ln t\right]}_{1}^{10} + 2 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{3}{4}\right)}^{n} {\int}_{1}^{10} {t}^{1 - 2 n} {\left(1 + \frac{1}{2 {t}^{2}}\right)}^{1 - 2 n} \mathrm{dt}$

Take a second series expansion:

$L = 99 + \ln 10 + 2 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{3}{4}\right)}^{n} {\int}_{1}^{10} {t}^{1 - 2 n} \left\{{\sum}_{m = 0}^{\infty} \left(\begin{matrix}1 - 2 n \\ m\end{matrix}\right) {\left(\frac{1}{2 {t}^{2}}\right)}^{m}\right\} \mathrm{dt}$

Simplify:

$L = 99 + \ln 10 + 2 {\sum}_{n = 1}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}1 - 2 n \\ m\end{matrix}\right) {\left(- \frac{3}{4}\right)}^{n} {\left(\frac{1}{2}\right)}^{m} {\int}_{1}^{10} {t}^{1 - 2 n - 2 m} \mathrm{dt}$

Isolate the $n = 1$ case:

$L = 99 + \ln 10 - \frac{3}{4} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}- 1 \\ m\end{matrix}\right) {\left(\frac{1}{2}\right)}^{m} {\int}_{1}^{10} {t}^{- 1 - 2 m} \mathrm{dt} + 2 {\sum}_{n = 2}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}1 - 2 n \\ m\end{matrix}\right) {\left(- \frac{3}{4}\right)}^{n} {\left(\frac{1}{2}\right)}^{m} {\int}_{1}^{10} {t}^{1 - 2 n - 2 m} \mathrm{dt}$

Isolate the $n = 1 , m = 0$ case:

$L = 99 + \ln 10 - \frac{3}{4} \left({\int}_{1}^{10} \frac{1}{t} \mathrm{dt} + {\sum}_{m = 1}^{\infty} \left(\begin{matrix}- 1 \\ m\end{matrix}\right) {\left(\frac{1}{2}\right)}^{m} {\int}_{1}^{10} {t}^{- 1 - 2 m} \mathrm{dt}\right) + 2 {\sum}_{n = 2}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}1 - 2 n \\ m\end{matrix}\right) {\left(- \frac{3}{4}\right)}^{n} {\left(\frac{1}{2}\right)}^{m} {\int}_{1}^{10} {t}^{1 - 2 n - 2 m} \mathrm{dt}$

The remaining integrals are trivial:

$L = 99 + \ln 10 - \frac{3}{4} \left({\left[\ln t\right]}_{1}^{10} + {\sum}_{m = 1}^{\infty} \left(\begin{matrix}- 1 \\ m\end{matrix}\right) {\left(\frac{1}{2}\right)}^{m} {\left[{t}^{- 2 m}\right]}_{1}^{10} / \left(- 2 m\right)\right) + 2 {\sum}_{n = 2}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}1 - 2 n \\ m\end{matrix}\right) {\left(- \frac{3}{4}\right)}^{n} {\left(\frac{1}{2}\right)}^{m} {\left[{t}^{- 2 n - 2 m}\right]}_{1}^{10} / \left(- 2 n - 2 m\right)$

Insert the limits of integration and simplify:

$L = 99 + \ln 10 - \frac{3}{4} \left(\ln 10 + {\sum}_{m = 1}^{\infty} \left(\begin{matrix}- 1 \\ m\end{matrix}\right) {\left(\frac{1}{2}\right)}^{m} \left(\frac{1}{2 m}\right) \left(1 - \frac{1}{100} ^ m\right)\right) + {\sum}_{n = 2}^{\infty} {\sum}_{m = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \left(\begin{matrix}1 - 2 n \\ m\end{matrix}\right) {\left(- \frac{3}{4}\right)}^{n} {\left(\frac{1}{2}\right)}^{m} \left(\frac{1}{n + m}\right) \left(1 - \frac{1}{100} ^ \left(n + m\right)\right)$