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

Sep 3, 2016

$\approx 0.4029$

#### Explanation:

$\vec{r} = {\theta}^{2} - \theta - 3 {\sin}^{2} \theta + {\tan}^{2} \theta = r \setminus \hat{r}$

the velocity vector is the tangent vector and is....

$\vec{v} = \dot{r} \setminus \hat{r} + \dot{\theta} r \setminus \hat{\theta}$

$= \left(\begin{matrix}2 \theta \dot{\theta} - \dot{\theta} - 6 \sin \theta \cos \theta \dot{\theta} + 2 \tan \theta {\sec}^{2} \theta \dot{\theta} \\ \dot{\theta} \left({\theta}^{2} - \theta - 3 {\sin}^{2} \theta + {\tan}^{2} \theta\right)\end{matrix}\right) \left(\begin{matrix}\hat{r} \\ \hat{\theta}\end{matrix}\right)$

$= \dot{\theta} \left(\begin{matrix}2 \theta - 1 - 3 \sin 2 \theta + 2 \tan \theta {\sec}^{2} \theta \\ {\theta}^{2} - \theta - 3 {\sin}^{2} \theta + {\tan}^{2} \theta\end{matrix}\right) \left(\begin{matrix}\hat{r} \\ \hat{\theta}\end{matrix}\right)$

$= \dot{\theta} \left(\begin{matrix}\frac{\pi}{2} \\ \frac{1}{16} \left(- 8 - 4 \pi + {\pi}^{2}\right)\end{matrix}\right) \left(\begin{matrix}\hat{r} \\ \hat{\theta}\end{matrix}\right)$

We can discard the $\dot{\theta}$ term because it is merely a scalar multiple. It is the direction of this vector that we are interested in

To convert to Cartesian, we know that

$\hat{r} = \cos \theta \hat{x} + \sin \theta \hat{y}$ and $\hat{\theta} = - \sin \theta \hat{x} + \cos \theta \hat{y}$

or $\left(\begin{matrix}\hat{r} \\ \hat{\theta}\end{matrix}\right) = \left(\begin{matrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{matrix}\right) \left(\begin{matrix}\hat{x} \\ \hat{y}\end{matrix}\right)$

So $\left(\begin{matrix}\hat{x} \\ \hat{y}\end{matrix}\right) = \left(\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right) \left(\begin{matrix}\hat{r} \\ \hat{\theta}\end{matrix}\right)$

$\vec{\overline{v}} = \frac{1}{\sqrt{2}} \left(\begin{matrix}1 & - 1 \\ 1 & 1\end{matrix}\right) \left(\begin{matrix}\frac{\pi}{2} \\ \frac{1}{16} \left(- 8 - 4 \pi + {\pi}^{2}\right)\end{matrix}\right)$

$= \frac{1}{\sqrt{2}} \left(\begin{matrix}\frac{\pi}{2} - \frac{1}{16} \left(- 8 - 4 \pi + {\pi}^{2}\right) \\ \frac{\pi}{2} + \frac{1}{16} \left(- 8 - 4 \pi + {\pi}^{2}\right)\end{matrix}\right)$

Again we can drop the $\frac{1}{\sqrt{2}}$ and so the slope is

$\frac{\frac{\pi}{2} + \frac{1}{16} \left(- 8 - 4 \pi + {\pi}^{2}\right)}{\frac{\pi}{2} - \frac{1}{16} \left(- 8 - 4 \pi + {\pi}^{2}\right)}$

$\approx 0.4029$