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

Jun 20, 2017

$y = \sqrt{3} x$

#### Explanation:

From the reference Tangents with Polar Coordinates , we obtain the equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

The slope, m, of the normal the polar curve will be the negative of the reciprocal of equation [1] evaluated at $\theta = \frac{\pi}{3}$:

$m = \frac{r \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{3}\right) - \frac{\mathrm{dr} \left(\frac{\pi}{3}\right)}{d \theta} \cos \left(\frac{\pi}{3}\right)}{\frac{\mathrm{dr} \left(\frac{\pi}{3}\right)}{d \theta} \sin \left(\frac{\pi}{3}\right) + r \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right)} \text{ [2]}$

I need to backtrack a bit and explain that:

$r \left(\theta\right) = f \left(\theta\right) = \sin \left(\theta\right) - \frac{\theta}{2}$

We must compute $\frac{\mathrm{dr} \left(\theta\right)}{d \theta}$ and evaluate at $\theta = \frac{\pi}{3}$

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

$\frac{\mathrm{dr} \left(\frac{\pi}{3}\right)}{d \theta} = \cos \left(\frac{\pi}{3}\right) - \frac{1}{2} = 0$

This greatly simplifies equation [2]:

$m = \frac{r \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{3}\right) - \left(0\right) \cos \left(\frac{\pi}{3}\right)}{\left(0\right) \sin \left(\frac{\pi}{3}\right) + r \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right)} \text{ [2.1]}$

$m = \frac{r \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{3}\right)}{r \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}\right)} \text{ [2.2]}$

$m = \frac{\sin \left(\frac{\pi}{3}\right)}{\cos \left(\frac{\pi}{3}\right)} \text{ [2.3]}$

$m = \tan \left(\frac{\pi}{3}\right) \text{ [2.4]}$

$m = \sqrt{3} \text{ [2.5]}$

Compute the Cartesian point for $\left(r \left(\frac{\pi}{3}\right) , \frac{\pi}{3}\right)$:

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

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

$x = \frac{3 \sqrt{3} - \pi}{12}$

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

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

$y = \frac{9 - \sqrt{3} \pi}{12}$

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

$y = \sqrt{3} \left(x - \frac{3 \sqrt{3} - \pi}{12}\right) + \frac{9 - \sqrt{3} \pi}{12}$

This simplifies to:

$y = \sqrt{3} x$