Question

How do you graph `r^2 = 3 sin 2θ`?

Answer 1

One loop, in the first quadrant, is for `r =sqrt(3 sin 2theta`. The mirror image, in the third quadrant, with respect to `theta =3pi/4`, is from `r =-sqrt(3 sin 2theta`.

Explanation:

The curve is called Limacon. The standard form is `r^2 = a^2 cos 2theta`. Here, this is rotated clockwise through `pi/4`, about the pole r = 0.

`r^2` is non-negative. So, ir is undefined when `sin 2theta < 0`. This happens when,`pi/2 < theta < pi and 3pi/2 < theta < 2pi`.

The loops meet at r = 0 and the axis of symmetry is `theta = pi/4` for the first-quadrant loop, and the reverse line `theta = 5pi/4`, for the other.