# What is the arclength of r=-10sin(theta/4+(9pi)/8)  on theta in [(pi)/4,(7pi)/4]?

Mar 19, 2018

$L = 15 \pi + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{15}{64}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{3 \pi}{8} + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{3 \pi}{4}\right) \cos \left(\left(n - k\right) \frac{11 \pi}{2}\right)}{n - k}\right\}$ units.

#### Explanation:

$r = - 10 \sin \left(\frac{\theta}{4} + \frac{9 \pi}{8}\right)$
${r}^{2} = 100 {\sin}^{2} \left(\frac{\theta}{4} + \frac{9 \pi}{8}\right)$

$r ' = - \frac{10}{4} \cos \left(\frac{\theta}{4} + \frac{9 \pi}{8}\right)$
${\left(r '\right)}^{2} = \frac{100}{16} {\cos}^{2} \left(\frac{\theta}{4} + \frac{9 \pi}{8}\right)$

Arclength is given by:

$L = {\int}_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} \sqrt{100 {\sin}^{2} \left(\frac{\theta}{4} + \frac{9 \pi}{8}\right) + \frac{100}{16} {\cos}^{2} \left(\frac{\theta}{4} + \frac{9 \pi}{8}\right)} d \theta$

Apply the substitution $\frac{\theta}{4} + \frac{9 \pi}{8} = \phi$

$L = 40 {\int}_{\frac{19 \pi}{16}}^{\frac{25 \pi}{16}} \sqrt{{\sin}^{2} \phi + \frac{1}{16} {\cos}^{2} \phi} \mathrm{dp} h i$

Apply the Trigonometric identity ${\sin}^{2} x + {\cos}^{2} x = 1$:

$L = 40 {\int}_{\frac{19 \pi}{16}}^{\frac{25 \pi}{16}} \sqrt{1 - \frac{15}{16} {\cos}^{2} \phi} \mathrm{dp} h i$

Since $\frac{15}{16} {\cos}^{2} \phi < 1$, take the series expansion of the square root:

$L = 40 {\int}_{\frac{19 \pi}{16}}^{\frac{25 \pi}{16}} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{15}{16} {\cos}^{2} \phi\right)}^{n} \mathrm{dp} h i$

Isolate the $n = 0$ term and simplify:

$L = 40 {\int}_{\frac{19 \pi}{16}}^{\frac{25 \pi}{16}} \mathrm{dp} h i + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{15}{16}\right)}^{n} {\int}_{\frac{19 \pi}{16}}^{\frac{25 \pi}{16}} {\cos}^{2 n} \phi \mathrm{dp} h i$

Apply the Trigonometric power-reduction formula:

$L = 40 \left(\frac{25 \pi}{16} - \frac{19 \pi}{16}\right) + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{15}{16}\right)}^{n} {\int}_{\frac{19 \pi}{16}}^{\frac{25 \pi}{16}} \left\{\frac{1}{4} ^ n \left(\begin{matrix}2 n \\ n\end{matrix}\right) + \frac{2}{4} ^ n {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \cos \left(\left(2 n - 2 k\right) \phi\right)\right\} \mathrm{dp} h i$

Integrate directly:

$L = 15 \pi + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{15}{64}\right)}^{n} {\left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) \phi + {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \sin \frac{\left(2 n - 2 k\right) \phi}{n - k}\right]}_{\frac{19 \pi}{16}}^{\frac{25 \pi}{16}}$

Insert the limits of integration:

$L = 15 \pi + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{15}{64}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{3 \pi}{8} + {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{25 \pi}{8}\right) - \sin \left(\left(n - k\right) \frac{19 \pi}{8}\right)}{n - k}\right\}$

Apply the Trigonometric sum-to-product identity:

$L = 15 \pi + 40 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{15}{64}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{3 \pi}{8} + 2 {\sum}_{k = 0}^{n - 1} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{3 \pi}{4}\right) \cos \left(\left(n - k\right) \frac{11 \pi}{2}\right)}{n - k}\right\}$