What is the arc length of f(x) = sinx  on x in [pi/12,(5pi)/12] ?

1 Answer
Mar 13, 2018

$L = \frac{\pi}{3} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{\pi}{3} + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{\pi}{3}\right) \cos \left(\left(n - k\right) \frac{\pi}{2}\right)}{n - k}\right\}$

Explanation:

$f \left(x\right) = \sin x$

$f ' \left(x\right) = \cos x$

Arc length is given by:

$L = {\int}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} \sqrt{1 + {\cos}^{2} x} \mathrm{dx}$

For $x \in \left[\frac{\pi}{12} , \frac{5 \pi}{12}\right]$, ${\cos}^{2} x < 1$. Take the series expansion of the square root:

$L = {\int}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\cos}^{2 n} x \mathrm{dx}$

Isolate the $n = 0$ case:

$L = {\int}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} \mathrm{dx} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} {\cos}^{2 n} x \mathrm{dx}$

Apply the Trigonometric power-reduction formula:

$L = {\left[x\right]}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\int}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} \left\{\frac{1}{4} ^ n \left(\begin{matrix}2 n \\ n\end{matrix}\right) + \frac{2}{4} ^ n {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}n \\ k\end{matrix}\right) \cos \left(2 \left(n - k\right) x\right)\right\} \mathrm{dx}$

Simplify:

$L = \frac{\pi}{3} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n {\int}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}n \\ k\end{matrix}\right) \cos \left(2 \left(n - k\right) x\right)\right\} \mathrm{dx}$

Integrate directly:

$L = \frac{\pi}{3} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n {\left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) x + {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}n \\ k\end{matrix}\right) \sin \frac{2 \left(n - k\right) x}{n - k}\right]}_{\frac{\pi}{12}}^{\frac{5 \pi}{12}}$

Simplify:

$L = \frac{\pi}{3} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{\pi}{3} + {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{5 \pi}{6}\right) - \sin \left(\left(n - k\right) \frac{\pi}{6}\right)}{n - k}\right\}$

Apply the Trigonometric sum-to-product identity $\sin a - \sin b = 2 \sin \left(\frac{a - b}{2}\right) \cos \left(\frac{a + b}{2}\right)$:

$L = \frac{\pi}{3} + {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{\pi}{3} + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{\pi}{3}\right) \cos \left(\left(n - k\right) \frac{\pi}{2}\right)}{n - k}\right\}$

Isolate the $n = 1$ case:

$L = \frac{\pi}{3} + \frac{\pi}{12} + {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{\pi}{3} + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{\pi}{3}\right) \cos \left(\left(n - k\right) \frac{\pi}{2}\right)}{n - k}\right\}$