# What is the slope of the tangent line of r=theta-3cos((2theta)/3+(pi)/2) at theta=(-5pi)/6?

Aug 15, 2017

The slope $m \approx - 1.34$

#### Explanation:

The reference Tangents with Polar Coordinates gives us the equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta} \sin \left(\theta\right) + r \cos \left(\theta\right)}{\frac{\mathrm{dr}}{d \theta} \cos \left(\theta\right) - r \sin \left(\theta\right)} \text{ [1]}$

Compute $\frac{\mathrm{dr}}{d \theta}$

$\frac{\mathrm{dr}}{d \theta} = 1 + 2 \sin \left(\frac{2 \theta}{3} + \frac{\pi}{2}\right)$

$\frac{\mathrm{dr}}{d \theta} = 1 + 2 \cos \left(\frac{2 \theta}{3}\right)$

The substitute the equation for r and $\frac{\mathrm{dr}}{d \theta}$ into equation [1]:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 + 2 \cos \left(\frac{2 \theta}{3}\right)\right) \sin \left(\theta\right) + \left(\theta - 3 \cos \left(\frac{2 \theta}{3} + \frac{\pi}{2}\right)\right) \cos \left(\theta\right)}{\left(1 + 2 \cos \left(\frac{2 \theta}{3}\right)\right) \cos \left(\theta\right) - \left(\theta - 3 \cos \left(\frac{2 \theta}{3} + \frac{\pi}{2}\right)\right) \sin \left(\theta\right)}$

The slope, m, is the above equation evaluated at $\theta = \frac{- 5 \pi}{6}$

$m = \frac{\left(1 + 2 \cos \left(\frac{2 \left(\frac{- 5 \pi}{6}\right)}{3}\right)\right) \sin \left(\frac{- 5 \pi}{6}\right) + \left(\frac{- 5 \pi}{6} - 3 \cos \left(\frac{2 \left(\frac{- 5 \pi}{6}\right)}{3} + \frac{\pi}{2}\right)\right) \cos \left(\frac{- 5 \pi}{6}\right)}{\left(1 + 2 \cos \left(\frac{2 \left(\frac{- 5 \pi}{6}\right)}{3}\right)\right) \cos \left(\frac{- 5 \pi}{6}\right) - \left(\frac{- 5 \pi}{6} - 3 \cos \left(\frac{2 \left(\frac{- 5 \pi}{6}\right)}{3} + \frac{\pi}{2}\right)\right) \sin \left(\frac{- 5 \pi}{6}\right)}$

$m \approx - 1.34$