# What is the arclength of r=-10sin(theta/4+(5pi)/16)  on theta in [(-5pi)/16,(9pi)/16]?

May 27, 2018

$L = \frac{133 \pi}{20} - \frac{48}{5} \sin \left(\frac{7}{16} \pi\right) \cos \left(\frac{11}{8} \pi\right) + 40 {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{6}{25}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{7 \pi}{32} + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{7 \pi}{16}\right) \cos \left(\left(n - k\right) \frac{11 \pi}{8}\right)}{n - k}\right\}$ units.

#### Explanation:

r=−10sin(θ/4+(5pi)/16)
r^2=100sin^2(θ/4+(5pi)/16)

r'=−10/4cos(θ/4+(5pi)/16)
(r')^2=100/25cos^2(θ/4+(5pi)/16)

Arclength is given by:

L=int_((-5pi)/16)^((9pi)/16)sqrt(100sin^2(θ/4+(5pi)/16)+100/25cos^2(θ/4+(5pi)/16))d theta

Apply the substitution θ/4+(5pi)/16=phi and rearrange:

$L = 40 {\int}_{\frac{15 \pi}{64}}^{\frac{29 \pi}{64}} \sqrt{1 - \frac{24}{25} {\cos}^{2} \phi} \mathrm{dp} h i$

For $\phi \in \left[\frac{15 \pi}{64} , \frac{29 \pi}{64}\right]$, $\frac{24}{25} {\cos}^{2} \phi < 1$. Take the series expansion of the square root:

$L = 40 {\int}_{\frac{15 \pi}{64}}^{\frac{29 \pi}{64}} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{24}{25} {\cos}^{2} \phi\right)}^{n} \mathrm{dp} h i$

Isolate the $n = 0$ term and rearrange:

$L = 40 {\int}_{\frac{15 \pi}{64}}^{\frac{29 \pi}{64}} \mathrm{dp} h i + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{24}{25}\right)}^{n} {\int}_{\frac{15 \pi}{64}}^{\frac{29 \pi}{64}} {\cos}^{2 n} \phi \mathrm{dp} h i$

Apply the trigonometric power-reduction formula:

$L = \frac{35 \pi}{4} + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{24}{25}\right)}^{n} {\int}_{\frac{15 \pi}{64}}^{\frac{29 \pi}{64}} \left\{\frac{1}{4} ^ n \left(\begin{matrix}2 n \\ n\end{matrix}\right) + \frac{2}{4} ^ n {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \cos \left(\left(2 n - 2 k\right) \phi\right)\right\} \mathrm{dp} h i$

Integrate term by term:

$L = \frac{35 \pi}{4} + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{6}{25}\right)}^{n} {\left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) \phi + {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \sin \frac{\left(2 n - 2 k\right) \phi}{n - k}\right]}_{\frac{15 \pi}{64}}^{\frac{29 \pi}{64}}$

Insert the limits of integration:

$L = \frac{35 \pi}{4} + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{6}{25}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{7 \pi}{32} + {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{29 \pi}{32}\right) - \sin \left(\left(n - k\right) \frac{15 \pi}{32}\right)}{n - k}\right\}$

Apply the trigonometric sum-to-product formula:

$L = \frac{35 \pi}{4} + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{6}{25}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{7 \pi}{32} + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{7 \pi}{16}\right) \cos \left(\left(n - k\right) \frac{11 \pi}{8}\right)}{n - k}\right\}$

Isolate the $n = 1$ terms:

$L = \frac{35 \pi}{4} - \frac{24}{5} \left(\frac{7 \pi}{16} + 2 \sin \left(\frac{7}{16} \pi\right) \cos \left(\frac{11}{8} \pi\right)\right) + 40 {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{6}{25}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{7 \pi}{32} + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{7 \pi}{16}\right) \cos \left(\left(n - k\right) \frac{11 \pi}{8}\right)}{n - k}\right\}$

Simplify:

$L = \frac{133 \pi}{20} - \frac{48}{5} \sin \left(\frac{7}{16} \pi\right) \cos \left(\frac{11}{8} \pi\right) + 40 {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{6}{25}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{7 \pi}{32} + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{7 \pi}{16}\right) \cos \left(\left(n - k\right) \frac{11 \pi}{8}\right)}{n - k}\right\}$