# What is the slope of the polar curve f(theta) = theta^2-theta - cos^3theta + tan^2theta at theta = pi/4?

Sep 4, 2017

The slope is: $\frac{{\pi}^{2} + 4 \pi + 8 \sqrt{2} + 64}{{\pi}^{2} + 12 \pi + 16 \sqrt{2} + 32} \approx 0.9564910$

#### Explanation:

The slope will be given by $\frac{\mathrm{dy}}{\mathrm{dx}}$ of the curve evaluated at $\theta = \frac{\pi}{4}$.

To get the equation in terms of $x$ and $y$, we should use the polar-to-rectangular identities: $\left\{\begin{matrix}x = r \cos \theta \\ y = r \sin \theta\end{matrix}\right.$

So here, where $r = f \left(\theta\right)$, we have:

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

$\textcolor{w h i t e}{x} = {\theta}^{2} \cos \theta - \theta \cos \theta - {\cos}^{4} \theta + \tan \theta \sin \theta$

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

$\textcolor{w h i t e}{y} = {\theta}^{2} \sin \theta - \theta \sin \theta - {\cos}^{3} \theta \sin \theta + \sin \theta {\tan}^{2} \theta$

To find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we first need to find $\frac{\mathrm{dx}}{d \theta}$ and $\frac{\mathrm{dy}}{d \theta}$. This will take a fair amount of product and chain rule, so be careful.

$\frac{\mathrm{dx}}{d \theta} = 2 \theta \cos \theta - {\theta}^{2} \sin \theta - \cos \theta + \theta \sin \theta + 4 {\cos}^{3} \theta \sin \theta + {\sec}^{2} \theta \sin \theta + \tan \theta \cos \theta$

$\textcolor{w h i t e}{\frac{\mathrm{dx}}{d \theta}} = 2 \theta \cos \theta - {\theta}^{2} \sin \theta - \cos \theta + \theta \sin \theta + 4 {\cos}^{3} \theta \sin \theta + \tan \theta \sec \theta + \sin \theta$

$\frac{\mathrm{dy}}{d \theta} = 2 \theta \sin \theta + {\theta}^{2} \cos \theta - \sin \theta - \theta \cos \theta + 3 {\cos}^{2} \theta {\sin}^{2} \theta - {\cos}^{4} \theta + \cos \theta {\tan}^{2} \theta + 2 \sin \theta \tan \theta {\sec}^{2} \theta$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{d \theta}} = 2 \theta \sin \theta + {\theta}^{2} \cos \theta - \sin \theta - \theta \cos \theta + 3 {\cos}^{2} \theta {\sin}^{2} \theta - {\cos}^{4} \theta + \sin \theta \tan \theta + 2 {\tan}^{2} \theta \sec \theta$

Now we can find $\frac{\mathrm{dy}}{\mathrm{dx}}$ using $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / d \theta}{\mathrm{dx} / d \theta}$.

Just so we don't have to write out a giant fraction, let's evaluate $\frac{\mathrm{dx}}{d \theta}$ and $\frac{\mathrm{dy}}{d \theta}$ at $\theta = \frac{\pi}{4}$ separately.

$\frac{\mathrm{dx}}{d \theta} = 2 \frac{\pi}{4} \frac{1}{\sqrt{2}} + {\pi}^{2} / 16 \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \frac{\pi}{4} \frac{1}{\sqrt{2}} + 4 \frac{1}{2 \sqrt{2}} \frac{1}{\sqrt{2}} + 2 \frac{1}{\sqrt{2}} + 1 \frac{1}{\sqrt{2}}$

$\textcolor{w h i t e}{\frac{\mathrm{dx}}{d \theta}} = \frac{\pi}{2 \sqrt{2}} + {\pi}^{2} / \left(16 \sqrt{2}\right) + \frac{\pi}{4 \sqrt{2}} + 1 + \sqrt{2}$

$\textcolor{w h i t e}{\frac{\mathrm{dx}}{d \theta}} = \frac{1}{16 \sqrt{2}} \left({\pi}^{2} + 12 \pi + 16 \sqrt{2} + 32\right)$

$\frac{\mathrm{dy}}{d \theta} = 2 \frac{\pi}{4} \frac{1}{\sqrt{2}} + {\pi}^{2} / 16 \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} - \frac{\pi}{4} \frac{1}{\sqrt{2}} + 3 \frac{1}{2} \frac{1}{2} - \frac{1}{4} + \frac{1}{\sqrt{2}} 1 + 2 \cdot 1 \sqrt{2}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{d \theta}} = \frac{\pi}{2 \sqrt{2}} + {\pi}^{2} / \left(16 \sqrt{2}\right) - \frac{\pi}{4 \sqrt{2}} + \frac{1}{2} + 2 \sqrt{2}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{d \theta}} = \frac{1}{16 \sqrt{2}} \left({\pi}^{2} + 4 \pi + 8 \sqrt{2} + 64\right)$

Thus, at $\theta = \frac{\pi}{4}$, the slope is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / d \theta}{\mathrm{dx} / d \theta} = \frac{{\pi}^{2} + 4 \pi + 8 \sqrt{2} + 64}{{\pi}^{2} + 12 \pi + 16 \sqrt{2} + 32} \approx 0.9564910$