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

Feb 16, 2018

$y = \frac{2}{5 \pi} x + 1$

Explanation:

We want to figure out two things: the slope of the curve and the position of the point. Therefore, we can find a line that crosses perpendicular to that point.

The point is easy. Plugging in $\theta = \frac{\pi}{2}$, we get $f \left(\frac{\pi}{2}\right) = \cos \left(0\right) + \frac{\pi}{2} \sin \left(2 \pi\right) = 1$, i.e. a point at a radius of 1 and an angle of $\frac{\pi}{2}$, which is the point (0, 1). Therefore, this is our y-intercept.

Next, we find the slope. We can do this by calculating $\frac{\mathrm{df}}{d \theta}$ and using it in the following:
$y = r \sin \theta \to \mathrm{dy} = \sin \theta \mathrm{dr} + r \cos \theta d \theta = \frac{y}{r} \mathrm{dr} + x d \theta$
$x = r \cos \theta \to \mathrm{dx} = \cos \theta \mathrm{dr} - r \sin \theta d \theta = \frac{x}{r} \mathrm{dr} - y d \theta$

In other words,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{d \theta} / \frac{\mathrm{dx}}{d \theta} = \frac{\frac{y}{r} \cdot \frac{\mathrm{dr}}{d \theta} - x}{\frac{x}{r} \cdot \frac{\mathrm{dr}}{d \theta} - y}$

Since our point is at (0, 1), this simplifies significantly to
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta}}{- 1} = - \frac{\mathrm{dr}}{d \theta}$

Now what is $\frac{\mathrm{dr}}{d \theta}$? We calculate using the formula given:

$f \left(\theta\right) = \cos \left(\pi - 2 \theta\right) + \theta \sin \left(5 \theta - \frac{\pi}{2}\right)$
$f ' \left(\theta\right) = 2 \sin \left(\pi - 2 \theta\right) + \sin \left(5 \theta - \frac{\pi}{2}\right) + 5 \theta \cos \left(5 \theta - \frac{\pi}{2}\right)$
$f ' \left(\frac{\pi}{2}\right) = 2 \sin \left(0\right) + \sin \left(2 \pi\right) + 5 \cdot \frac{\pi}{2} \cdot \cos \left(2 \pi\right) = \frac{5 \pi}{2}$

Therefore, the slope of the curve (in Cartesian coordinates) is $- \frac{5 \pi}{2}$.

Orthogonal slopes are determined using the negative reciprocal, i.e. the slope for the line is $\frac{2}{5 \pi}$. From the slope and the intercept, we get the equation

$y = \frac{2}{5 \pi} x + 1$