# What is the slope of the tangent line of r=thetacos(theta/4-(5pi)/3) at theta=(pi)/3?

Nov 14, 2016

$r = r ' \left(\frac{\pi}{3}\right) \left(\theta - \frac{\pi}{3}\right) + \frac{\pi}{3} \cos \left(- 19 \frac{\pi}{12}\right)$
with
$r ' \left(\frac{\pi}{3}\right) = \cos \left(- 19 \frac{\pi}{12}\right) - \frac{\pi}{12} \sin \left(- 19 \frac{\pi}{12}\right)$

#### Explanation:

$\frac{\mathrm{dr}}{d \theta} = r ' \left(\theta\right) = \cos \left(\frac{\theta}{4} - 5 \frac{\pi}{3}\right) + \theta \left(- \sin \left(\frac{\theta}{4} - 5 \frac{\pi}{3}\right) \cdot \frac{1}{4}\right)$ that

evaluated in $\theta = \frac{\pi}{3}$ becomes

$r ' \left(\frac{\pi}{3}\right) = \cos \left(\frac{\pi}{12} - 5 \frac{\pi}{3}\right) - \frac{\pi}{12} \sin \left(\frac{\pi}{12} - 5 \frac{\pi}{3}\right)$

$r ' \left(\frac{\pi}{3}\right) = \cos \left(- 19 \frac{\pi}{12}\right) - \frac{\pi}{12} \sin \left(- 19 \frac{\pi}{12}\right)$

the point of tangence is $\left(\frac{\pi}{3} , \frac{\pi}{3} \cos \left(- 19 \frac{\pi}{3}\right)\right)$ and the wished line is

$r = r ' \left(\frac{\pi}{3}\right) \left(\theta - \frac{\pi}{3}\right) + \frac{\pi}{3} \cos \left(- 19 \frac{\pi}{12}\right)$