Question

How do you graph `r=6sin2theta`?

Answer 1

Graph for the cartesian form `(x^2+y^2)sqrt(x^2+y^2)=12xy` is inserted.

Explanation:

`r >= 0 to 2theta in Q_1 or Q_2 to theta in Q_1`

Mini/Max r: 0/6

Period for graph : `(2pi)/2 = pi`.

So, for `theta in [0, 2pi],` two loops are created.

Note that for `theta in Q_2 or Q_4, r < 0`.

graph{(x^2+y^2)^1.5-12xy=0 [-6, 6, -6, 6]}