Question

How do you graph `r^2=cos(2theta)`?

Answer 1

Called Lemniscate, the graph looks like `oo`, symmetrical about the initial line `theta=0` and the pole r = 0.

Explanation:

In the 1st and 4th quadrants, `|theta|<=pi/4` In the 2nd and 3rd,

`|pi-theta|<=pi/4`.

I used infinity symbol `oo` to depict the shape of this double-

loop, looking like a fallen 8. For getting 8-erect, the equation is

`r^2=-cos 2theta`

Use a table

`{(r, theta)}.= {(0, -pi/4) (1/sqrt2, -pi/8) (1, 0) (1/sqrt2, pi/8) (0,`

`pi/4)}`, for one loop. Its mirror image with respect to `theta=pi/2`

is the other loop.

Strictly, `r =sqrt(x^2 + y^2)>=0`.

Graphs of both `r^2 =+- cos 2theta` are combined.
graph{((x^2+y^2)^2-x^2+y^2)((x^2+y^2)^2+x^2-y^2)=0[-2 2 -1 1]}.

Interestingly, an easy rotation of this graph through `pi/4`

produces a grand 8-petal flower.

graph{((x^2+y^2)^2-x^2+y^2)((x^2+y^2)^2+x^2-y^2)((x^2+y^2)^2-2xy)((x^2+y^2)^2+2xy)=0[-2 2 -1 1]}.