How do you graph #r=2cos3theta#?

Question

Violet Anderson · Accepted Answer

See Socratic graph for the Cartesian form of the equation

#r=2cos3theta>=0,# giving #3theta in Q_1 and Q_4#. So, for one loop of this 3-petal rose, #theta in [-pi/6, pi/6]# and this range is #1/2(2/3pi)# = 1/2(period of #cos 3theta)#) For the other half of one period r is negative and, if you choose to admit negative r, the third loop would be redrawn. For #r = 2sin3theta#, r_ loops would add three more. I reiterate that #r = sqrt(x^2+y^2) >= 0# I have used the Cartesian form of the equation, using #cos 3theta=4cos^3theta-3costheta# graph{(x^2+y^2)^2+6x(x^2+y^2)=8x^3 [-4, 4, -2, 2]} I introduce here a 2-cosine-combined 10-petal rose for my viewers of this answer. Observe that four of the petals are much smaller. graph{0.01((x^2+y^2)^2.5-x^4+6x^2y^2-y^4)(0.25(x^2+y^2)^3.5-x^6+15x^4y^2-15x^2y^4+y^6)(x^2+y^2-.04)=0}

Daniel Acosta · Answer

undefined To graph $ r = 2\cos(3\theta) $, follow these steps:

1. Plot points for various values of $ \theta $ ranging from $ 0 $ to $ 2\pi $.
2. Substitute each value of $ \theta $ into the equation $ r = 2\cos(3\theta) $ to find the corresponding value of $ r $.
3. Use polar coordinates to plot each point $ (r, \theta) $ on the graph.
4. Connect the points to form the graph of the equation.

Keep in mind that $ r $ represents the distance from the origin (the pole), and $ \theta $ represents the angle measured counterclockwise from the positive x-axis.