Question

How do you graph `r = 1 + 2 sin(3theta)`?

Answer 1

See the graphs and the explanation.

Explanation:

The period for `r(theta)` is `(2pi)/3`. So, in one period

`r>=1+2sin 3theta >=0 to sin 3theta >=-1/2 to 3theta >=-pi/6 to`

`theta >=-pi/18`

Max r = 3 and min r = 0.

In half period #theta -n [-pi/18, 5/18pi], the whole loop is drawn. In the

other half `theta in (5/18pi, 11/18pi), r < 0`

The first graph is locally zoomed at the pole to reveal the dimple

therein. The second reveals the loop representing the whole

periodic curve, redrawn periodically, with period `2/3pi`

graph{(x^2+y^2)(x^2+y^2-sqrt(x^2+y^2)-6y)-8y^3=0 [-2.5, 2.5, -1.25, 1.25]}
graph{(x^2+y^2)(x^2+y^2-sqrt(x^2+y^2)-6y)-8y^3=0 [-80, 80, -40, 40]}