# What is the equation of the line that is normal to the polar curve f(theta)=sin(2theta+pi) -theta at theta = pi/4?

May 19, 2017

$\left(y - \frac{- 4 \sqrt{2} - \pi \sqrt{2}}{8}\right) = \left(\frac{\pi}{8 + \pi}\right) \left(x - \frac{- 4 \sqrt{2} - \pi \sqrt{2}}{8}\right)$

#### Explanation:

Find $r$ when $\theta = \frac{\pi}{4}$
$r = \sin \left(2 \left(\frac{\pi}{4}\right) + \pi\right) - \theta = \frac{- 4 - \pi}{4}$

Find $\frac{\mathrm{dr}}{d \theta} :$
$r = \sin \left(2 \theta + \pi\right) - \theta$
$\frac{\mathrm{dr}}{d \theta} = \cos \left(2 \theta + \pi\right) \left(2\right) - 1 = 2 \cos \left(2 \theta + \pi\right) - 1$

Substitute $\theta = \frac{\pi}{4}$:
$= 2 \cos \left(2 \left(\frac{\pi}{4}\right) + \pi\right) - 1 = - 1$

Find the derivative:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(- 1\right) \sin \left(\frac{\pi}{4}\right) + \left(\frac{- 4 - \pi}{4}\right) \cos \left(\frac{\pi}{4}\right)}{\left(- 1\right) \cos \left(\frac{\pi}{4}\right) - \left(\frac{- 4 - \pi}{4}\right) \sin \left(\frac{\pi}{4}\right)} = - \frac{8}{\pi} - 1$

Find the respective $x$ and $y$.
$x = r \cos \left(\theta\right) = \left(\frac{- 4 - \pi}{4}\right) \left(\frac{\sqrt{2}}{2}\right) = \frac{- 4 \sqrt{2} - \pi \sqrt{2}}{8}$

$y = r \sin \left(\theta\right) = \left(\frac{- 4 - \pi}{4}\right) \left(\frac{\sqrt{2}}{2}\right) = \frac{- 4 \sqrt{2} - \pi \sqrt{2}}{8}$

The slope of a normal line is the negative reciprocal of the derivative.
$- \frac{1}{- \frac{8}{\pi} - 1} = \frac{\pi}{8 + \pi}$

Putting it all together:
$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$
$\left(y - \frac{- 4 \sqrt{2} - \pi \sqrt{2}}{8}\right) = \left(\frac{\pi}{8 + \pi}\right) \left(x - \frac{- 4 \sqrt{2} - \pi \sqrt{2}}{8}\right)$