# What is the slope of the polar curve r(theta) = theta^2 - sec(theta)+cos(theta)*sin^3(theta)  at theta = (2pi)/3?

Dec 30, 2015

$\frac{64 {\pi}^{2} - 192 \pi \sqrt{3} - 27 \sqrt{3} + 1152}{64 {\pi}^{2} \sqrt{3} + 192 \pi - 81} \approx 0.42824$

#### Explanation:

First convert the Polar equation to Parametric form.

$y = r \sin \left(\theta\right)$
$= {\theta}^{2} \sin \left(\theta\right) - \tan \left(\theta\right) + \cos \left(\theta\right) \cdot {\sin}^{4} \left(\theta\right)$

$x = r \cos \left(\theta\right)$
$= {\theta}^{2} \cos \left(\theta\right) - 1 + {\cos}^{2} \left(\theta\right) \cdot {\sin}^{3} \left(\theta\right)$

To find the slope, we need to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, which we can get by using the Chain Rule.

Since, from the Chain Rule, $\frac{\mathrm{dy}}{d \theta} = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{d \theta}$,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}}$.

We find $\frac{\mathrm{dy}}{d \theta}$ and $\frac{\mathrm{dx}}{d \theta}$ separately.

$\frac{\mathrm{dy}}{d \theta} = \frac{d}{d \theta} \left({\theta}^{2} \sin \left(\theta\right) - \tan \left(\theta\right) + \cos \left(\theta\right) \cdot {\sin}^{4} \left(\theta\right)\right)$

$= \left[{\theta}^{2} \cos \left(\theta\right) + \sin \left(\theta\right) \left(2 \theta\right)\right] - {\sec}^{2} \left(\theta\right) + \left[\cos \left(\theta\right) \cdot \left(4 {\sin}^{3} \left(\theta\right) \cos \left(\theta\right)\right) + {\sin}^{4} \left(\theta\right) \left(- \sin \left(\theta\right)\right)\right]$

$= {\theta}^{2} \cos \left(\theta\right) + 2 \theta \sin \left(\theta\right) - {\sec}^{2} \left(\theta\right) + 4 {\cos}^{2} \left(\theta\right) \cdot {\sin}^{3} \left(\theta\right) - {\sin}^{5} \left(\theta\right)$

$= {\theta}^{2} \cos \left(\theta\right) + 2 \theta \sin \left(\theta\right) - {\sec}^{2} \left(\theta\right) + 4 \left(1 - {\sin}^{2} \left(\theta\right)\right) \cdot {\sin}^{3} \left(\theta\right) - {\sin}^{5} \left(\theta\right)$

$= {\theta}^{2} \cos \left(\theta\right) + 2 \theta \sin \left(\theta\right) - {\sec}^{2} \left(\theta\right) + 4 {\sin}^{3} \left(\theta\right) - 4 {\sin}^{5} \left(\theta\right) - {\sin}^{5} \left(\theta\right)$

$= {\theta}^{2} \cos \left(\theta\right) + 2 \theta \sin \left(\theta\right) - {\sec}^{2} \left(\theta\right) + 4 {\sin}^{3} \left(\theta\right) - 5 {\sin}^{5} \left(\theta\right)$

$\frac{\mathrm{dy}}{d \theta} {|}_{\theta = \frac{2 \pi}{3}} = {\left(\frac{2 \pi}{3}\right)}^{2} \cos \left(\frac{2 \pi}{3}\right) + 2 \left(\frac{2 \pi}{3}\right) \sin \left(\frac{2 \pi}{3}\right) - {\sec}^{2} \left(\frac{2 \pi}{3}\right) + 4 {\sin}^{3} \left(\frac{2 \pi}{3}\right) - 5 {\sin}^{5} \left(\frac{2 \pi}{3}\right)$

$= - \frac{2 {\pi}^{2}}{9} + \frac{2 \pi \sqrt{3}}{3} + \frac{3 \sqrt{3}}{32} - 4$

$\frac{\mathrm{dx}}{d \theta} = \frac{d}{d \theta} \left({\theta}^{2} \cos \left(\theta\right) - 1 + {\cos}^{2} \left(\theta\right) \cdot {\sin}^{3} \left(\theta\right)\right)$

$= \left[{\theta}^{2} \left(- \sin \left(\theta\right)\right) + \cos \left(\theta\right) \left(2 \theta\right)\right] - 0 + \left[{\cos}^{2} \left(\theta\right) \left(3 {\sin}^{2} \left(\theta\right) \cos \left(\theta\right)\right) + {\sin}^{3} \left(\theta\right) \left(2 \cos \left(\theta\right) \left(- \sin \left(\theta\right)\right)\right)\right]$

$= - {\theta}^{2} \sin \left(\theta\right) + 2 \theta \cos \left(\theta\right) + 3 {\sin}^{2} \left(\theta\right) {\cos}^{3} \left(\theta\right) - 2 {\sin}^{4} \left(\theta\right) \cos \left(\theta\right)$

$\frac{\mathrm{dx}}{d \theta} {|}_{\theta = \frac{2 \pi}{3}} = - {\left(\frac{2 \pi}{3}\right)}^{2} \sin \left(\frac{2 \pi}{3}\right) + 2 \left(\frac{2 \pi}{3}\right) \cos \left(\frac{2 \pi}{3}\right) + 3 {\sin}^{2} \left(\frac{2 \pi}{3}\right) {\cos}^{3} \left(\frac{2 \pi}{3}\right) - 2 {\sin}^{4} \left(\frac{2 \pi}{3}\right) \cos \left(\frac{2 \pi}{3}\right)$

$= - \frac{2 {\pi}^{2} \sqrt{3}}{9} - \frac{2 \pi}{3} + \frac{9}{32}$

Now, to calculate $\frac{\mathrm{dy}}{\mathrm{dx}}$.

${\frac{\mathrm{dy}}{\mathrm{dx}}}_{\theta = \frac{2 \pi}{3}} = \frac{- \frac{2 {\pi}^{2}}{9} + \frac{2 \pi \sqrt{3}}{3} + \frac{3 \sqrt{3}}{32} - 4}{- \frac{2 {\pi}^{2} \sqrt{3}}{9} - \frac{2 \pi}{3} + \frac{9}{32}}$

$= \frac{64 {\pi}^{2} - 192 \pi \sqrt{3} - 27 \sqrt{3} + 1152}{64 {\pi}^{2} \sqrt{3} + 192 \pi - 81}$

$\approx 0.42824$