# What is the equation of the line that is normal to the polar curve f(theta)=2sin(3theta+pi/3) - 2thetacostheta  at theta = pi/4?

Jul 27, 2018

$y = \frac{- 8 - 8 \sqrt{3}}{- 12 - 4 \sqrt{3} + 2 \pi} \left(x - \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)\right) + \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)$

#### Explanation:

To find the equation of the line, we need a $\textcolor{red}{p o i n t}$, and we need a $\textcolor{b l u e}{s l o p e}$.

First, let's find the $\textcolor{red}{p o i n t}$.

Plug in $\frac{\pi}{4}$ into $f \left(\theta\right)$.

$f \left(\frac{\pi}{4}\right) = 2 \sin \left(3 \left(\frac{\pi}{4}\right) + \frac{\pi}{3}\right) - 2 \left(\frac{\pi}{4}\right) \left(\cos \left(\frac{\pi}{4}\right)\right)$
$f \left(\frac{\pi}{4}\right) = \frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}$

We have the polar coordinate $\left(\frac{\pi}{4} , \frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}\right)$.
We have to convert this to Cartesian form. Using the formulas:
$x = r \cos \theta$
$y = r \cos \theta$

We find that the Cartesian coordinate is color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4).

Good! We have found the $\textcolor{red}{p o i n t}$.

Now, we need to find the $\textcolor{b l u e}{s l o p e}$.

The formula for the tangent line of a polar function is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta} \sin \theta + r \cos \theta}{\frac{\mathrm{dr}}{d \theta} \cos \theta - r \sin \theta}$

If you notice, $\sin \theta$ and $\cos \theta$ will be the same in this case, because $\sin \left(\frac{\pi}{4}\right)$ and $\cos \left(\frac{\pi}{4}\right)$ are the same, so for this case, our $\frac{\mathrm{dy}}{\mathrm{dx}}$ is simply:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta} + r}{\frac{\mathrm{dr}}{d \theta} - r}$

We solved $r$ already previously, which was $\frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}$.

Now, we need to find $\frac{\mathrm{dr}}{d \theta}$.

$f ' \left(\theta\right) = \frac{\mathrm{dr}}{d \theta} = 6 \cos \left(3 \theta + \frac{\pi}{3}\right) + 2 \theta \sin \theta - 2 \cos \theta$

Now, we plug in $\frac{\pi}{4}$ and get:

$f ' \left(\frac{\pi}{4}\right) = \frac{\mathrm{dr}}{d \theta} = \frac{\sqrt{2} \left(- 10 - 6 \sqrt{3} + \pi\right)}{4}$

Going back to the tangent line formula:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\sqrt{2} \left(- 10 - 6 \sqrt{3} + \pi\right)}{4} + \frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}}{\frac{\sqrt{2} \left(- 10 - 6 \sqrt{3} + \pi\right)}{4} - \frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}}$

After some tedious simplification, we find this reduces to color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)

We have our $\textcolor{red}{p o i n t}$:
color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4)

We have our $\textcolor{b l u e}{s l o p e}$:
color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)

All that is left is to plug in these values into the point slope formula:
$\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$

We get:

$y - \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right) = \frac{- 8 - 8 \sqrt{3}}{- 12 - 4 \sqrt{3} + 2 \pi} \left(x - \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)\right)$

So the equation is:

$y = \frac{- 8 - 8 \sqrt{3}}{- 12 - 4 \sqrt{3} + 2 \pi} \left(x - \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)\right) + \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)$