# Orthogonal Trajectories/Cardioids

## Theorem

Consider the one-parameter family of curves of cardioids given in polar form as:

$(1): \quad r = c \paren {1 + \cos \theta}$

Its family of orthogonal trajectories is given by the equation:

$r = c \paren {1 - \cos \theta}$

## Proof

Differentiating $(1)$ with respect to $r$ gives:

 $(2):\quad$ $\displaystyle \frac {\d r} {\d \theta}$ $=$ $\displaystyle - c \sin \theta$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d r} {\d \theta}$ $=$ $\displaystyle - \frac {r \sin \theta} {1 + \cos \theta}$ eliminating $c$ between $(1)$ and $(2)$ $\displaystyle \leadsto \ \$ $\displaystyle r \frac {\d \theta} {\d r}$ $=$ $\displaystyle - \frac {1 + \cos \theta} {\sin \theta}$

Thus from Orthogonal Trajectories of One-Parameter Family of Curves, the family of orthogonal trajectories is given by:

$r \dfrac {\d \theta} {\d r} = \dfrac {\sin \theta} {1 + \cos \theta}$

So:

 $\displaystyle r \dfrac {\d \theta} {\d r}$ $=$ $\displaystyle \frac {\sin \theta} {1 + \cos \theta}$ $\displaystyle \leadsto \ \$ $\displaystyle \int \frac {\d r} r$ $=$ $\displaystyle \int \frac {1 + \cos \theta} {\sin \theta} \rd \theta$ Separation of Variables $\displaystyle$ $=$ $\displaystyle \int \paren {\csc \theta + \cot \theta} \rd \theta$ $\displaystyle \leadsto \ \$ $\displaystyle \ln r$ $=$ $\displaystyle \ln \size {\csc \theta - \cot \theta} + \ln \size {\sin \theta} + c$ $\displaystyle$ $=$ $\displaystyle \ln \size {\paren {\csc \theta - \cot \theta} \sin \theta} + c$ $\displaystyle$ $=$ $\displaystyle \ln \size {1 - \cos \theta} + c$ $\displaystyle \leadsto \ \$ $\displaystyle r$ $=$ $\displaystyle c \paren {1 - \cos \theta}$

Hence the result.

$\blacksquare$