# What is the slope of the polar curve f(theta) = -4theta - thetacos^2theta  at theta = (15pi)/8?

Jul 5, 2018

Slope of the polar curve at $\theta = \left(\frac{15 \pi}{8}\right)$ is $- 9.02$

#### Explanation:

$f \left(\theta\right) = - 4 \theta - \theta {\cos}^{2} \theta$

$\theta = \frac{15 \pi}{8} = {337.5}^{0}$

${f}^{'} \left(\theta\right) = - 4 - \left(- \theta \cdot 2 \cos \theta \sin \theta + {\cos}^{2} \theta\right)$

${f}^{'} \left(\theta\right) = - 4 + \theta \cdot 2 \cos \theta \sin \theta - {\cos}^{2} \theta$

${f}^{'} \left(\frac{15 \pi}{8}\right) = - 4 + \theta \cdot \sin 2 \theta - {\cos}^{2} \theta$

${f}^{'} \left(\frac{15 \pi}{8}\right) = - 4 + \frac{15 \pi}{8} \cdot \sin \left(\frac{15 \pi}{4}\right) - {\cos}^{2} \left(\frac{15 \pi}{8}\right)$

${f}^{'} \left(\frac{15 \pi}{8}\right) = - 9.02 \left(2 \mathrm{dp}\right)$

Slope of the polar curve at $\theta = \left(\frac{15 \pi}{8}\right)$ is $- 9.02$ [Ans]