What is the arclength of r=-3cos(theta/16+(pi)/16)  on theta in [(-5pi)/16,(9pi)/16]?

May 30, 2018

$L = \frac{32 \pi}{128} + 3 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{255}{1024}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{7 \pi}{128} + {\sum}_{k = 0}^{n - 1} {\left(- 1\right)}^{n - k} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{25 \pi}{128}\right) - \sin \left(\left(n - k\right) \frac{11 \pi}{128}\right)}{n - k}\right\}$ units.

Explanation:

$r = - 3 \cos \left(\frac{\theta}{16} + \frac{\pi}{16}\right)$
${r}^{2} = 9 {\cos}^{2} \left(\frac{\theta}{16} + \frac{\pi}{16}\right)$

$r ' = \frac{3}{16} \sin \left(\frac{\theta}{16} + \frac{\pi}{16}\right)$
${\left(r '\right)}^{2} = \frac{9}{256} {\sin}^{2} \left(\frac{\theta}{16} + \frac{\pi}{16}\right)$

Arclength is given by:

$L = {\int}_{\frac{- 5 \pi}{16}}^{\frac{9 \pi}{16}} \sqrt{9 {\cos}^{2} \left(\frac{\theta}{16} + \frac{\pi}{16}\right) + \frac{9}{256} {\sin}^{2} \left(\frac{\theta}{16} + \frac{\pi}{16}\right)} d \theta$

Apply the substitution $\frac{\theta}{16} + \frac{\pi}{16} = \phi$:

$L = 3 {\int}_{\frac{11 \pi}{256}}^{\frac{25 \pi}{256}} \sqrt{{\cos}^{2} \phi + \frac{1}{256} {\sin}^{2} \phi} \mathrm{dp} h i$

Rearrange:

$L = 3 {\int}_{\frac{11 \pi}{256}}^{\frac{25 \pi}{256}} \sqrt{1 - \frac{255}{256} {\sin}^{2} \phi} \mathrm{dp} h i$

From here we can use the small-angle approximation for an easy answer.

Alternatively take the series expansion of the square root:

$L = 3 {\int}_{\frac{11 \pi}{256}}^{\frac{25 \pi}{256}} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{255}{256} {\sin}^{2} \phi\right)}^{n} \mathrm{dp} h i$

Isolate the $n = 0$ term and simplify:

$L = 3 {\int}_{\frac{11 \pi}{256}}^{\frac{25 \pi}{256}} \mathrm{dp} h i + 3 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{255}{256}\right)}^{n} {\int}_{\frac{11 \pi}{256}}^{\frac{25 \pi}{256}} {\sin}^{2 n} \phi \mathrm{dp} h i$

Apply the trigonometric power reduction formula:

$L = \frac{32 \pi}{128} + 3 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{255}{256}\right)}^{n} {\int}_{\frac{11 \pi}{256}}^{\frac{25 \pi}{256}} \left\{\frac{1}{4} ^ n \left(\begin{matrix}2 n \\ n\end{matrix}\right) + \frac{2}{4} ^ n {\sum}_{k = 0}^{n - 1} {\left(- 1\right)}^{n - k} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \cos \left(\left(2 n - 2 k\right) \phi\right)\right\} \mathrm{dp} h i$

Integrate directly:

$L = \frac{32 \pi}{128} + 3 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{255}{1024}\right)}^{n} {\left[\left(\begin{matrix}2 n \\ n\end{matrix}\right) \phi + {\sum}_{k = 0}^{n - 1} {\left(- 1\right)}^{n - k} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \sin \frac{\left(2 n - 2 k\right) \phi}{n - k}\right]}_{\frac{11 \pi}{256}}^{\frac{25 \pi}{256}}$

Hence:

$L = \frac{32 \pi}{128} + 3 {\sum}_{n = 1}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) {\left(- \frac{255}{1024}\right)}^{n} \left\{\left(\begin{matrix}2 n \\ n\end{matrix}\right) \frac{7 \pi}{128} + {\sum}_{k = 0}^{n - 1} {\left(- 1\right)}^{n - k} \left(\begin{matrix}2 n \\ k\end{matrix}\right) \frac{\sin \left(\left(n - k\right) \frac{25 \pi}{128}\right) - \sin \left(\left(n - k\right) \frac{11 \pi}{128}\right)}{n - k}\right\}$