# What is the arc length of f(x) = (x^2-1)^(3/2)  on x in [1,3] ?

May 31, 2018

A first-order approximation gives $L = 23 - \frac{5}{24 \sqrt{2}} \ln \left(\frac{5 + 2 \sqrt{2}}{5 - 2 \sqrt{2}}\right)$ units.

#### Explanation:

$f \left(x\right) = {\left({x}^{2} - 1\right)}^{\frac{3}{2}}$

$f ' \left(x\right) = 3 x \sqrt{{x}^{2} - 1}$

Arc length is given by:

$L = {\int}_{1}^{3} \sqrt{1 + 9 {x}^{2} \left({x}^{2} - 1\right)} \mathrm{dx}$

Expand:

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

Complete the square:

$L = \frac{1}{2} {\int}_{1}^{3} \sqrt{9 {\left(2 {x}^{2} - 1\right)}^{2} - 5} \mathrm{dx}$

Factorize:

$L = \frac{3}{2} {\int}_{1}^{3} \left(2 {x}^{2} - 1\right) \sqrt{1 - \frac{5}{9 {\left(2 {x}^{2} - 1\right)}^{2}}} \mathrm{dx}$

For $x \in \left[1 , 3\right]$, 5/(9(2x^2-1)^2<1. Take the series expansion of the square root:

$L = \frac{3}{2} {\int}_{1}^{3} \left(2 {x}^{2} - 1\right) \left\{{\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{5}{9 {\left(2 {x}^{2} - 1\right)}^{2}}\right)}^{n}\right\} \mathrm{dx}$

Isolate the $n = 0$ term and simplify:

$L = \frac{3}{2} {\int}_{1}^{3} \left(2 {x}^{2} - 1\right) \mathrm{dx} + \frac{3}{2} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{5}{9}\right)}^{n} {\int}_{1}^{3} {\left(\frac{1}{2 {x}^{2} - 1}\right)}^{2 n - 1} \mathrm{dx}$

Apply the difference of squares:

$L = \frac{3}{2} {\left[\frac{2}{3} {x}^{3} - x\right]}_{1}^{3} + \frac{3}{2} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{5}{9}\right)}^{n} {\int}_{1}^{3} {\left(\frac{1}{\left(\sqrt{2} x - 1\right) \left(\sqrt{2} x + 1\right)}\right)}^{2 n - 1} \mathrm{dx}$

Apply partial fraction decomposition:

$L = 23 + 3 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{5}{36}\right)}^{n} {\int}_{1}^{3} {\left(\frac{1}{\sqrt{2} x - 1} - \frac{1}{\sqrt{2} x + 1}\right)}^{2 n - 1} \mathrm{dx}$

Isolate the $n = 1$ term and simplify:

$L = 23 - \frac{5}{24} {\int}_{1}^{3} \left(\frac{1}{\sqrt{2} x - 1} - \frac{1}{\sqrt{2} x + 1}\right) \mathrm{dx} + 3 {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{5}{36}\right)}^{n} {\int}_{1}^{3} {\left(\frac{1}{\sqrt{2} x - 1} - \frac{1}{\sqrt{2} x + 1}\right)}^{2 n - 1} \mathrm{dx}$

Hence

$L = 23 - \frac{5}{24 \sqrt{2}} {\left[\ln | \sqrt{2} x - 1 | - \ln | \sqrt{2} x + 1 |\right]}_{1}^{3} + 3 {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{5}{36}\right)}^{n} {\int}_{1}^{3} {\left(\frac{1}{\sqrt{2} x - 1} - \frac{1}{\sqrt{2} x + 1}\right)}^{2 n - 1} \mathrm{dx}$

Giving:

$L = 23 - \frac{5}{24 \sqrt{2}} \ln \left(\frac{5 + 2 \sqrt{2}}{5 - 2 \sqrt{2}}\right) + 3 {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{5}{36}\right)}^{n} {\int}_{1}^{3} {\left(\frac{1}{\sqrt{2} x - 1} - \frac{1}{\sqrt{2} x + 1}\right)}^{2 n - 1} \mathrm{dx}$