# What is the slope of the tangent line of r=2theta-cos(5theta+(5pi)/3) at theta=(-7pi)/4?

Dec 27, 2017

$\frac{2 + \frac{\pi}{2} + 5 \sin \left(\frac{11 \pi}{12}\right) - \cos \left(\frac{11 \pi}{12}\right)}{2 - \frac{\pi}{2} + 5 \sin \left(\frac{11 \pi}{12}\right) + \cos \left(\frac{11 \pi}{12}\right)}$

#### Explanation:

Hand written answer is now replaced by typed answer, with an error in calculating(dr)/(d(theta) rectified:

First of all $5 \theta + \frac{5 \pi}{3}$ is equivalent to $5 \theta + 2 \pi - \frac{\pi}{3}$ which is equivalent to $5 \theta - \frac{\pi}{3}$. Hence $r = 2 \theta - \cos \left(5 \theta - \frac{\pi}{3}\right) \to \frac{\mathrm{dr}}{d \left(\theta\right)} = 2 + 5 \sin \left(5 \theta - \frac{\pi}{3}\right)$

Formula for the slope of a polar curve is $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \left(\theta\right)} \sin \theta + r \cos \theta}{\frac{\mathrm{dr}}{d \left(\theta\right)} \cos \theta - r \sin \theta}$

Slope is required at $\theta = \frac{- 7 \pi}{4}$ which is equivalent to $\frac{\pi}{4} - 2 \pi \to \frac{\pi}{4}$. For this $\theta$, sine and cosine values would be both $\frac{1}{\sqrt{2}}$. Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \left(\theta\right)} + r}{\frac{\mathrm{dr}}{d \left(\theta\right)} - r}$

Now at $\theta = \frac{\pi}{4}$, $r = \frac{\pi}{2} - \cos \left(\frac{5 \pi}{4} - \frac{\pi}{3}\right) = \frac{\pi}{2} - \cos \left(\frac{11 \pi}{12}\right)$ and $\frac{\mathrm{dr}}{d \left(\theta\right)} = 2 + 5 \sin \left(\frac{11 \pi}{12}\right)$

Substituting the values of $\frac{\mathrm{dr}}{d \left(\theta\right)}$ and r, slope would be

$\frac{2 + \frac{\pi}{2} + 5 \sin \left(\frac{11 \pi}{12}\right) - \cos \left(\frac{11 \pi}{12}\right)}{2 - \frac{\pi}{2} + 5 \sin \left(\frac{11 \pi}{12}\right) + \cos \left(\frac{11 \pi}{12}\right)}$