# What is the arc length of f(x)=sqrt(1+64x^2) on x in [1,5]?

Jun 1, 2018

A first-order approximation gives $L = 4 \sqrt{65} - \frac{4}{\sqrt{65}} {\tan}^{- 1} \left(\frac{32}{321}\right)$ units.

#### Explanation:

$f \left(x\right) = \sqrt{1 + 64 {x}^{2}}$

$f ' \left(x\right) = \frac{64 x}{\sqrt{1 + 64 {x}^{2}}}$

Arc length is given by:

$L = {\int}_{1}^{5} \sqrt{1 + \frac{4096 {x}^{2}}{1 + 64 {x}^{2}}} \mathrm{dx}$

Apply the substitution $8 x = u$ and rearrange:

$L = \frac{\sqrt{65}}{8} {\int}_{8}^{40} \sqrt{1 - \frac{64}{65} \frac{1}{1 + {u}^{2}}} \mathrm{du}$

For $u \in \left[8 , 40\right]$, $\frac{64}{65} \frac{1}{1 + {u}^{2}} < 1$. Take the series expansion of the square root:

$L = \frac{\sqrt{65}}{8} {\int}_{8}^{40} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{64}{65} \frac{1}{1 + {u}^{2}}\right)}^{n} \mathrm{du}$

Isolate the $n = 0$ term and simplify:

$L = \frac{\sqrt{65}}{8} {\int}_{8}^{40} \mathrm{du} + \frac{\sqrt{65}}{8} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{64}{65}\right)}^{n} {\int}_{8}^{40} {\left(\frac{1}{1 + {u}^{2}}\right)}^{n} \mathrm{du}$

Apply the substitution $u = \tan \theta$:

$L = 4 \sqrt{65} + \frac{\sqrt{65}}{8} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{64}{65}\right)}^{n} {\int}_{{\tan}^{- 1} \left(8\right)}^{{\tan}^{- 1} \left(40\right)} {\cos}^{2 n - 2} \theta d \theta$

Isolate the $n = 1$ term:

$L = 4 \sqrt{65} - \frac{4}{\sqrt{65}} {\int}_{{\tan}^{- 1} \left(8\right)}^{{\tan}^{- 1} \left(40\right)} d \theta + \frac{\sqrt{65}}{8} {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{64}{65}\right)}^{n} {\int}_{{\tan}^{- 1} \left(8\right)}^{{\tan}^{- 1} \left(40\right)} {\cos}^{2 n - 2} \theta d \theta$

Rescale $n$:

$L = 4 \sqrt{65} - \frac{4}{\sqrt{65}} \left({\tan}^{- 1} \left(40\right) - {\tan}^{- 1} \left(8\right)\right) + \frac{\sqrt{65}}{8} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{64}{65}\right)}^{n + 1} {\int}_{{\tan}^{- 1} \left(8\right)}^{{\tan}^{- 1} \left(40\right)} {\cos}^{2 n} \theta d \theta$

Apply the Trigonometric power-reduction formula:

$L = 4 \sqrt{65} - \frac{4}{\sqrt{65}} {\tan}^{- 1} \left(\frac{32}{321}\right) - \frac{8}{\sqrt{65}} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{64}{65}\right)}^{n} {\int}_{{\tan}^{- 1} \left(8\right)}^{{\tan}^{- 1} \left(40\right)} \left\{\frac{1}{4} ^ n \left(\begin{matrix}2 n \\ n\end{matrix}\right) + \frac{2}{4} ^ n {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \cos \left(\left(2 n - 2 k\right) \theta\right)\right\} d \theta$

Integrate term by term:

$L = 4 \sqrt{65} - \frac{4}{\sqrt{65}} {\tan}^{- 1} \left(\frac{32}{321}\right) - \frac{8}{\sqrt{65}} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{16}{65}\right)}^{n} {\left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) \theta + {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \sin \frac{\left(2 n - 2 k\right) \theta}{n - k}\right]}_{{\tan}^{- 1} \left(8\right)}^{{\tan}^{- 1} \left(40\right)}$

Hence:

$L = 4 \sqrt{65} - \frac{4}{\sqrt{65}} {\tan}^{- 1} \left(\frac{32}{321}\right) - \frac{8}{\sqrt{65}} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{16}{65}\right)}^{n} \left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) {\tan}^{- 1} \left(\frac{32}{321}\right) + {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(2 n - 2 k\right) {\tan}^{- 1} 40\right) - \sin \left(\left(2 n - 2 k\right) {\tan}^{- 1} 8\right)}{n - k}\right]$

Apply the sum-to-product formula for the sine function:

$L = 4 \sqrt{65} - \frac{4}{\sqrt{65}} {\tan}^{- 1} \left(\frac{32}{321}\right) - \frac{8}{\sqrt{65}} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{16}{65}\right)}^{n} \left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) {\tan}^{- 1} \left(\frac{32}{321}\right) + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left\{\left(n - k\right) \left({\tan}^{- 1} 40 - {\tan}^{- 1} 8\right)\right\} \cos \left\{\left(n - k\right) \left({\tan}^{- 1} 40 + {\tan}^{- 1} 8\right)\right\}}{n - k}\right]$

Apply the angle-sum identities for the tangent function:

$L = 4 \sqrt{65} - \frac{4}{\sqrt{65}} {\tan}^{- 1} \left(\frac{32}{321}\right) - \frac{8}{\sqrt{65}} {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n + 1\end{matrix}\right) {\left(- \frac{16}{65}\right)}^{n} \left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) {\tan}^{- 1} \left(\frac{32}{321}\right) - 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left\{\left(n - k\right) {\tan}^{- 1} \left(\frac{32}{321}\right)\right\} \cos \left\{\left(n - k\right) {\tan}^{- 1} \left(\frac{48}{319}\right)\right\}}{n - k}\right]$