# What is the slope of the polar curve f(theta) = theta^2 - thetasintheta  at theta = (11pi)/8?

Feb 21, 2017

11.2206#

#### Explanation:

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

$f ' \left(\theta\right) = 2 \theta - \left(\theta \cos \theta + \sin \theta\right)$

at $\theta = \frac{11 \pi}{8}$,

$f ' \left(\theta\right) = 2 \cdot \frac{11 \pi}{8} - \left[\frac{11 \pi}{8} \cos \left(\frac{11 \pi}{8}\right) + \sin \left(\frac{11 \pi}{8}\right)\right]$

$f ' \left(\theta\right) = \frac{11 \pi}{4} - \left[\frac{11 \pi}{8} \left(- 0.3827\right) + \left(- 0.9239\right)\right]$

$f ' \left(\theta\right) = 8.6429 - \left(- 1.6538 - 0.9239\right)$

$f ' \left(\theta\right) = 8.6429 - \left(- 2.5777\right)$

$f ' \left(\theta\right) = 11.2206$