# What is the equation of the tangent line to the polar curve f(theta)=theta- sin((3theta)/2-pi/2)+tan((2theta)/3-pi/2)  at theta = pi?

Nov 9, 2016

$y = - \frac{18}{61} \left(\pi + \frac{\sqrt{3}}{3}\right) \left(x + \pi + \frac{\sqrt{3}}{3}\right)$

#### Explanation:

Let's begin by determining the $\left(x , y\right)$ point through which the line must pass. Because $\theta = \pi$, we know that $y = r \sin \left(\pi\right) = 0$

$x = r \cos \left(\pi\right)$

$x = - r$

$r = f \left(\pi\right) = \pi - \sin \left(\frac{3 \pi}{2} - \frac{\pi}{2}\right) + \tan \left(\frac{2 \pi}{3} - \frac{\pi}{2}\right)$

$r = \pi + \frac{\sqrt{3}}{3}$

The point through which the line must pass is:

$\left(- \pi - \frac{\sqrt{3}}{3} , 0\right)$

This reference Tangents to polar curves gives us an equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

To obtain the slope of the tangent line, m, we evaluate the above at $\theta = \pi$ so the above equation becomes:

m = r/(((dr)/(d theta))

Therefore, we need to evaluate $\frac{\mathrm{dr}}{d \theta}$ at $\pi$. I am going to use WolframAlpha to do the computation:

$\frac{\mathrm{dr}}{d \theta} = \frac{61}{18}$

$m = - \frac{18}{61} \left(\pi + \frac{\sqrt{3}}{3}\right)$

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

$y = - \frac{18}{61} \left(\pi + \frac{\sqrt{3}}{3}\right) \left(x + \pi + \frac{\sqrt{3}}{3}\right)$