# What is the slope of the tangent line of r=5theta+cos(-theta/3-(pi)/2) at theta=(-5pi)/6?

Dec 20, 2015

slope would be $\left[\left(- \frac{1}{2}\right) \left(5 - \frac{1}{3} \cos \left(\frac{5 \pi}{18}\right)\right) + \left\{\frac{- 25 \pi}{6} + \sin \left(\frac{5 \pi}{18}\right)\right\} \left(\frac{- \sqrt{3}}{2}\right)\right]$ / $\left[\left\{5 - \frac{1}{3} \cos \left(\frac{5 \pi}{18}\right)\right\} \left(\frac{- \sqrt{3}}{2}\right) - \left\{\frac{- 25 \pi}{6} + \sin \left(\frac{5 \pi}{18}\right)\right\} \left(- \frac{1}{2}\right)\right]$

#### Explanation:

write r= $5 \theta + \cos \left(\frac{\theta}{3} + \frac{\pi}{2}\right) = 5 \theta - \sin \left(\frac{\theta}{3}\right)$

For $\theta = \frac{- 5 \pi}{6}$, $r = \frac{- 25 \pi}{6} + \sin \left(\frac{5 \pi}{18}\right)$

$\frac{\mathrm{dr}}{d \left(\theta\right)} = 5 - \frac{1}{3} \cos \left(\frac{\theta}{3}\right)$

For $\theta = \frac{- 5 \pi}{6}$

$\frac{\mathrm{dr}}{d \left(\theta\right)} = 5 - \frac{1}{3} \cos \left(\frac{5 \pi}{18}\right)$

$\left[\cos \left(- \theta\right) = \cos \theta\right]$

Formula for slope in polar coordinates is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \left(\theta\right)} \sin \theta + r \cos \theta}{\frac{\mathrm{dr}}{d \left(\theta\right)} \cos \theta - r \sin \theta}$

Also $\sin \left(\frac{- 5 \pi}{6}\right) = - \frac{1}{2}$
$\cos \left(\frac{- 5 \pi}{6}\right) = - \frac{\sqrt{3}}{2}$

slope would be $\left[\left(- \frac{1}{2}\right) \left(5 - \frac{1}{3} \cos \left(\frac{5 \pi}{18}\right)\right) + \left\{\frac{- 25 \pi}{6} + \sin \left(\frac{5 \pi}{18}\right)\right\} \left(\frac{- \sqrt{3}}{2}\right)\right]$ / $\left[\left\{5 - \frac{1}{3} \cos \left(\frac{5 \pi}{18}\right)\right\} \left(\frac{- \sqrt{3}}{2}\right) - \left\{\frac{- 25 \pi}{6} + \sin \left(\frac{5 \pi}{18}\right)\right\} \left(- \frac{1}{2}\right)\right]$