# What is the slope of the polar curve f(theta) = theta - sec^3theta+thetasin^3theta  at theta = (5pi)/8?

Apr 20, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 0.54$

#### Explanation:

For a polar function $f \left(\theta\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(\theta\right) \sin \theta + f \left(\theta\right) \cos \theta}{f ' \left(\theta\right) \cos \theta - f \left(\theta\right) \sin \theta}$

$f \left(\theta\right) = \theta - {\sec}^{3} \theta + \theta {\sin}^{3} \theta$

$f ' \left(\theta\right) = 1 - 3 \left({\sec}^{2} \theta\right) \left(\frac{d}{\mathrm{dx}} \left[\sec \theta\right]\right) - {\sin}^{3} \theta + 3 \theta {\sin}^{2} \theta \left(\frac{d}{\mathrm{dx}} \left[\sin \theta\right]\right)$

$f ' \left(\theta\right) = 1 - 3 {\sec}^{3} \theta \tan \theta - {\sin}^{3} \theta + 3 \theta {\sin}^{2} \theta \cos \theta$

$f ' \left(\frac{5 \pi}{3}\right) = 1 - 3 {\sec}^{3} \left(\frac{5 \pi}{3}\right) \tan \left(\frac{5 \pi}{3}\right) - {\sin}^{3} \left(\frac{5 \pi}{3}\right) + 3 \left(\frac{5 \pi}{3}\right) {\sin}^{2} \left(\frac{5 \pi}{3}\right) \cos \left(\frac{5 \pi}{3}\right) \approx - 9.98$

$f \left(\frac{5 \pi}{3}\right) = \left(\frac{5 \pi}{3}\right) - {\sec}^{3} \left(\frac{5 \pi}{3}\right) + \left(\frac{5 \pi}{3}\right) {\sin}^{3} \left(\frac{5 \pi}{3}\right) \approx - 6.16$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 9.98 \sin \left(\frac{5 \pi}{3}\right) - 6.16 \cos \left(\frac{5 \pi}{3}\right)}{- 9.98 \cos \left(\frac{5 \pi}{3}\right) + 6.16 \sin \left(\frac{5 \pi}{3}\right)} = - 0.54$