# How do you graph the function r^2 = 9cos(2θ)?

Nov 17, 2016

I have inserted the graph of this Lemniscate, using the Socratic facility for the cartesian frame.

#### Explanation:

The period of $\cos k \theta$ is $\frac{2}{k} \pi$. Here, k = 2. So, the period is $\pi$.

${r}^{2} = 9 \cos 2 \theta \ge 0 \to \cos 2 \theta \ge 0$. So, conveniently, one

period can be chosen as $\theta \in \left[0. \pi\right]$.

wherein $\cos 2 \theta \ge 0$.

As $\cos \left(2 \left(- \theta\right)\right) = \cos 2 \theta$, the graph is symmetrical about the

initial line.

Also, as cos (2(pi-theta) = cos 2theta, the graph is symmetrical

about the vertical $\theta = \frac{\pi}{2}$.

A Table for half period $\left[0 , \frac{\pi}{4}\right]$ is sufficient for the part in ${Q}_{1}$..

Use symmetry. for the other three quarters..

$\left(r , \theta\right) : \left(0 , 3\right) \left(\frac{3}{\sqrt{\sqrt{2}}} , \frac{\pi}{8}\right) \left(\frac{3}{2} \sqrt{2} , \frac{\pi}{6}\right) \left(0 , \frac{\pi}{4}\right)$

graph{(x^2+y^2)^2=9(x^2-y^2) [-10, 10, -5, 5]}