# What is the equation of the tangent line of r=-2sin(6theta-(4pi)/3)  at theta=(2pi)/3?

Aug 15, 2017

$y = - 4 \left(x - \frac{\sqrt{3}}{2}\right) - 1.5$

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

$\frac{\mathrm{dr}}{d \theta} = - 12 \cos \left(6 \theta - \frac{4 \pi}{3}\right)$

$\frac{\mathrm{dr}}{d \theta} = 12 \sin \left(6 \theta + \frac{\pi}{6}\right)$

Substitute right sides of the equation for r and $\frac{\mathrm{dr}}{d \theta}$ into equation [1]:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(12 \sin \left(6 \theta + \frac{\pi}{6}\right)\right) \sin \left(\theta\right) + \left(- 2 \sin \left(6 \theta - \frac{4 \pi}{3}\right)\right) \cos \left(\theta\right)}{\left(12 \sin \left(6 \theta + \frac{\pi}{6}\right)\right) \cos \left(\theta\right) - \left(- 2 \sin \left(6 \theta - \frac{4 \pi}{3}\right)\right) \sin \left(\theta\right)}$

The slope, m, of the tangent line is the above equation evaluated at $\theta = \frac{2 \pi}{3}$

$m = \frac{\left(12 \sin \left(6 \frac{2 \pi}{3} + \frac{\pi}{6}\right)\right) \sin \left(\frac{2 \pi}{3}\right) + \left(- 2 \sin \left(6 \frac{2 \pi}{3} - \frac{4 \pi}{3}\right)\right) \cos \left(\frac{2 \pi}{3}\right)}{\left(12 \sin \left(6 \frac{2 \pi}{3} + \frac{\pi}{6}\right)\right) \cos \left(\frac{2 \pi}{3}\right) - \left(- 2 \sin \left(6 \frac{2 \pi}{3} - \frac{4 \pi}{3}\right)\right) \sin \left(\frac{2 \pi}{3}\right)}$

$m \approx - 4$

The x coordinate, ${x}_{1}$, of the point of tangency is:

${x}_{1} = - 2 \sin \left(6 \frac{2 \pi}{3} - \frac{4 \pi}{3}\right) \cos \left(\frac{2 \pi}{3}\right)$

${x}_{1} = \frac{\sqrt{3}}{2}$

The y coordinate, ${y}_{1}$, of the point of tangency is:

${y}_{1} = - 2 \sin \left(6 \frac{2 \pi}{3} - \frac{4 \pi}{3}\right) \sin \left(\frac{2 \pi}{3}\right)$

${y}_{1} = - 1.5$

Using the point-slope form of the equation of a line:

$y = m \left(x - {x}_{1}\right) + {y}_{1}$

We obtain the following equation:

$y = - 4 \left(x - \frac{\sqrt{3}}{2}\right) - 1.5$

Here is a graph of the function, the point of tangency, and the tangent line: