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

1 Answer
Dec 1, 2016

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]}