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

1 Answer
Nov 17, 2016

Answer:

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]}