Question

How do you graph `r^2=2sin(3θ)`?

Answer 1

Thanks to Sam's updating, it is a 3-petal curve, without using non-negative r, For the three loops, choose the ranges `theta in [0, pi/3] , [(2pi)/3, pi] and [(4pi)/3, (5pi)/3]`..

Explanation:

Upon insertion of the graph by Sam, I have duly revised my answer,

to make the answer almost perfect..

`r^2 = 2 sin 3theta >=0`. Maximum r = 2.

Choose the range `theta in [0, pi/3]`,

so that `3 theta in [0, pi]` is in the first two quadrants, where sine is

positive. It is similar, for the the other two petals, in the respective

ranges for `theta`....

For every `sin 3theta, r=+-sqrt(2 sin 3theta)`

(If negative r is allowed, there would be six loops altogether, keeping `2 sin 3theta>=0`. For the range `theta in [0, pi/3]`, one loop traced by the (+r)-hand, is in the first quadrant, and its mirror image about the pole, traced with the (-r)-hand, is in the third quadrant. If the convention is to treat r as modulus (length) of a vector, r`>=` 0, the mirror images are not done.)

As the period of `2 sin 3theta` is `(2pi)/3` the petals are retraced

when the ends of the intervals for `theta` are increased by integer

multiple `n((2pi)/3)` of `(2pi)/3`...