# What is the equation of the line that is normal to the polar curve f(theta)=-5theta- sin((3theta)/2-pi/3)+tan((theta)/2-pi/3)  at theta = pi?

Jul 13, 2016

The line is $y = \frac{6 - 60 \pi + 4 \sqrt{3}}{9 \sqrt{3} - 52} x + \frac{{\left(\sqrt{3} \left(1 - 10 \pi\right) + 2\right)}^{2}}{9 \sqrt{3} - 52}$

#### Explanation:

This behemoth of an equation is derived through a somewhat lengthy process. I will first outline the steps by which the derivation will proceed and then perform those steps.

We are given a function in polar coordinates, $f \left(\theta\right)$. We can take the derivative, $f ' \left(\theta\right)$, but in order to actually find a line in cartesian coordinates, we will need $\frac{\mathrm{dy}}{\mathrm{dx}}$.

We can find $\frac{\mathrm{dy}}{\mathrm{dx}}$ by using the following equation:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(\theta\right) \sin \left(\theta\right) + f \left(\theta\right) \cos \left(\theta\right)}{f ' \left(\theta\right) \cos \left(\theta\right) - f \left(\theta\right) \sin \left(\theta\right)}$

Then we'll plug that slope into the standard cartesian line form:
$y = m x + b$
And insert the cartesian converted polar coordinates of our point of interest:
$x = f \left(\theta\right) \cos \left(\theta\right)$
$y = f \left(\theta\right) \sin \left(\theta\right)$

A few things that should be immediately obvious and will save us time down the line. We are taking a line tangent to the point $\theta = \pi$. This means that $\sin \left(\theta\right) = 0$ so...
1) Our equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$ will actually be:
$\frac{\mathrm{dy}}{\mathrm{dx}} = f \frac{\pi}{f ' \left(\pi\right)}$

2) Our equations for the cartesian coordinates of our point will become:
$x = - f \left(\theta\right)$
$y = 0$

Starting to actually solve the problem, then, our first order of business is finding $f ' \left(\theta\right)$. It isn't hard, just three easy derivatives with chain rule applied to two:
$f ' \left(\theta\right) = - 5 - \frac{3}{2} \cos \left(\frac{3 \pi}{2} - \frac{\pi}{3}\right) + \frac{1}{2} {\sec}^{2} \left(\frac{\theta}{2} - \frac{\pi}{3}\right)$

Now we want to know $f \left(\pi\right)$:
$f \left(\pi\right) = - 5 \pi - \sin \left(\frac{7 \pi}{6}\right) + \tan \left(\frac{\pi}{6}\right)$
$= - 5 \pi - \frac{1}{2} + \frac{1}{\sqrt{3}}$
$= \frac{\sqrt{3} \left(1 - 10 \pi\right) + 2}{2 \sqrt{3}}$

And $f ' \left(\pi\right)$...
$f ' \left(\pi\right) = - 5 - \frac{3}{2} \cos \left(\frac{7 \pi}{6}\right) + \frac{1}{2} {\sec}^{2} \left(\frac{\pi}{6}\right)$
$= - 5 + \frac{3 \sqrt{3}}{4} + \frac{2}{3}$
$= \frac{9 \sqrt{3} - 52}{12}$

With these in hand, we are ready to determine our slope:
$\frac{\mathrm{dy}}{\mathrm{dx}} = f \frac{\pi}{f ' \left(\pi\right)}$
$= \frac{\sqrt{3} \left(1 - 10 \pi\right) + 2}{2 \sqrt{3}} \cdot \frac{12}{9 \sqrt{3} - 52}$
$= \frac{6 \left(1 - 10 \pi\right) + 4 \sqrt{3}}{9 \sqrt{3} - 52}$

We can plug this in as $m$ in $y = m x + b$. Recall that we previously determined that $y = 0$ and $x = - f \left(\theta\right)$:
$0 = - \left(\frac{6 \left(1 - 10 \pi\right) + 4 \sqrt{3}}{9 \sqrt{3} - 52}\right) \left(\frac{\sqrt{3} \left(1 - 10 \pi\right) + 2}{2 \sqrt{3}}\right) + b$
$0 = - \left(\frac{3 \left(1 - 10 \pi\right) + 2 \sqrt{3}}{9 \sqrt{3} - 52}\right) \left(\frac{\sqrt{3} \left(1 - 10 \pi\right) + 2}{\sqrt{3}}\right) + b$
$0 = - \left(\frac{\sqrt{3} \left(1 - 10 \pi\right) + 2}{9 \sqrt{3} - 52}\right) \left(\sqrt{3} \left(1 - 10 \pi\right) + 2\right) + b$
$b = \frac{{\left(\sqrt{3} \left(1 - 10 \pi\right) + 2\right)}^{2}}{9 \sqrt{3} - 52}$

We can combine our previously determined $m$ with our newly determined $b$ to give the equation for the line:
$y = \frac{6 - 60 \pi + 4 \sqrt{3}}{9 \sqrt{3} - 52} x + \frac{{\left(\sqrt{3} \left(1 - 10 \pi\right) + 2\right)}^{2}}{9 \sqrt{3} - 52}$