# What is the arclength of f(t) = (t^3-t+55,t^2-1) on t in [2,3]?

Jun 20, 2018

$L = \frac{56}{3} + \frac{2}{9} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {2}^{n} {\int}_{6}^{9} {\left(\frac{1}{u - 1} - \frac{1}{u + 1}\right)}^{2 n - 1} \mathrm{du}$ units.

#### Explanation:

$f \left(t\right) = \left({t}^{3} - t + 55 , {t}^{2} - 1\right)$

$f ' \left(t\right) = \left(3 {t}^{2} - 1 , 2 t\right)$

Arclength is given by:

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

Expand the squares:

$L = {\int}_{2}^{3} \sqrt{9 {t}^{4} - 2 {t}^{2} + 1} \mathrm{dt}$

Complete the square in the square root:

$L = \frac{1}{3} {\int}_{2}^{3} \sqrt{{\left(9 {t}^{2} - 1\right)}^{2} + 8} \mathrm{dt}$

Rearrange:

$L = \frac{1}{3} {\int}_{2}^{3} \left(9 {t}^{2} - 1\right) \sqrt{1 + \frac{8}{9 {t}^{2} - 1} ^ 2} \mathrm{dt}$

For $t \in \left[2 , 3\right]$, $\frac{8}{9 {t}^{2} - 1} ^ 2 < 1$. Take the series expansion of the square root:

$L = \frac{1}{3} {\int}_{2}^{3} \left(9 {t}^{2} - 1\right) \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(\frac{8}{9 {t}^{2} - 1} ^ 2\right)}^{n}\right\} \mathrm{dt}$

Isolate the $n = 0$ term and simplify:

$L = \frac{1}{3} {\int}_{2}^{3} \left(9 {t}^{2} - 1\right) \mathrm{dt} + \frac{1}{3} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {8}^{n} {\int}_{2}^{3} \frac{1}{9 {t}^{2} - 1} ^ \left(2 n - 1\right) \mathrm{dt}$

Apply partial fraction decomposition:

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

For ease of reading, apply the substitution $3 t = u$:

$L = \frac{56}{3} + \frac{2}{9} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {2}^{n} {\int}_{6}^{9} {\left(\frac{1}{u - 1} - \frac{1}{u + 1}\right)}^{2 n - 1} \mathrm{du}$

The $n = 1$ case is trivial.