How do you graph r=6sin2theta?

1 Answer
Dec 18, 2016

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]}