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

Jan 6, 2017

0.48799, nearly.

#### Explanation:

In polar form. slope

$\frac{\mathrm{dy}}{\mathrm{dx}}$

$= \frac{\frac{d}{d \theta} \left(r \sin \theta\right)}{\frac{d}{d \theta} \left(r \cos \theta\right)}$

$= \frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$,

where$r ' = \frac{\mathrm{dr}}{d \theta}$.

Here,

$r = 2 \theta + {\tan}^{2} \theta - \cos \theta = \frac{\pi}{4} + {\tan}^{2} \left(\frac{\pi}{8}\right) - \cos n \left(\frac{\pi}{8}\right)$

at $\theta = \frac{\pi}{8} = {22.5}^{o}$

$= 0.20466$, nearly.

$r ' = 2 + 2 \tan \theta {\sec}^{2} \theta + \sin t h e a$

$= 2 + 2 \tan \left(\frac{\pi}{8}\right) {\sec}^{2} \left(\frac{\pi}{8}\right) + \sin \left(\frac{\pi}{8}\right)$

#=3.335325, nearly.

Now the slope at $\theta = \frac{\pi}{8}$ is

$= \frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$,

where $r ' = \frac{\mathrm{dr}}{d \theta} \mathmr{and} \theta = \frac{\pi}{8}$

Upon substitutions and simplification,

slope = 0.48799, nearly.