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

Apr 24, 2017

The slope, $m \approx - 1.72$

#### Explanation:

To find the slope, m, of the tangent line, we must compute $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of $\theta$ and then evaluate it at $\theta = \frac{- 7 \pi}{4}$.

Here is a reference Tangents with Polar Coordinates that will give us the general equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of $\theta$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta} \sin \left(\theta\right) + r \cos \left(\theta\right)}{\frac{\mathrm{dr}}{d \theta} \cos \left(\theta\right) - r \sin \left(\theta\right)} \text{ }$

We are given r:

$r = 2 \theta - \cos \left(5 \theta - \frac{2 \pi}{3}\right) \text{ }$

We must compute $\frac{\mathrm{dr}}{d \theta}$:

$\frac{\mathrm{dr}}{d \theta} = 2 - 5 \sin \left(5 \theta - \frac{2 \pi}{3}\right) \text{ }$

Substitute the right sides of equations  and  into equation :

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(2 - 5 \sin \left(5 \theta - \frac{2 \pi}{3}\right)\right) \sin \left(\theta\right) + \left(2 \theta - \cos \left(5 \theta - \frac{2 \pi}{3}\right)\right) \cos \left(\theta\right)}{\left(2 - 5 \sin \left(5 \theta - \frac{2 \pi}{3}\right)\right) \cos \left(\theta\right) - \left(2 \theta - \cos \left(5 \theta - \frac{2 \pi}{3}\right)\right) \sin \left(\theta\right)} \text{ }$

When we evaluate equation  at $\theta = \frac{- 7 \pi}{4}$ the left side becomes the slope, m:

$m = \frac{\left(2 - 5 \sin \left(5 \frac{- 7 \pi}{4} - \frac{2 \pi}{3}\right)\right) \sin \left(\frac{- 7 \pi}{4}\right) + \left(2 \frac{- 7 \pi}{4} - \cos \left(5 \frac{- 7 \pi}{4} - \frac{2 \pi}{3}\right)\right) \cos \left(\frac{- 7 \pi}{4}\right)}{\left(2 - 5 \sin \left(5 \frac{- 7 \pi}{4} - \frac{2 \pi}{3}\right)\right) \cos \left(\frac{- 7 \pi}{4}\right) - \left(2 \frac{- 7 \pi}{4} - \cos \left(5 \frac{- 7 \pi}{4} - \frac{2 \pi}{3}\right)\right) \sin \left(\frac{- 7 \pi}{4}\right)} \text{ }$

$m \approx - 1.72$