Question

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

Answer 1

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

Explanation:

The period of `cos ktheta` is `2/kpi`. Here, k = 2. So, the period is `pi`.

`r^2 = 9 cos 2theta >=0 to cos 2theta >=0`. So, conveniently, one

period can be chosen as `theta in [0.pi]`.

wherein `cos 2theta >=0`.

As `cos(2(-theta)) = cos 2theta`, the graph is symmetrical about the

initial line.

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

about the vertical `theta=pi/2`.

A Table for half period `[0, pi/4]` is sufficient for the part in `Q_1`..

Use symmetry. for the other three quarters..

`(r, theta): (0, 3) (3/sqrt(sqrt2), pi/8) (3/2sqrt2, pi/6) (0, pi/4)`

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