How do you graph #r=4+6costheta#?

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Answer 1 · 2016-11-15T07:59:28

For making this cardioid-like graph, see explanation. The graph that is inserted is for the equivalent cartesian form #x^2+y^2-6x-4sqrt(x^2+y^2)=0#

#r=4+6 cos theta>=0 to cos theta >=-2/3 to#

#theta in (-131.81^o, 131.81^o)#, nearly.

As #cos(-theta)=costheta#, the graph is symmetrical with respect to

the initial line #theta = 0#.

The period of #r(theta)# is #2pi#.

So, a Table for #theta in (1. 131.61^o)# is sufficient, for making the

graph..

#(r, theta): (10, 0^o) (7, 60^o) (4, 90^o) (1, 120^o) (0, 131.81^o)#

The other half can be made using symmetry about the axis

#theta=0#.
graph{x^2 + y^2- 6x - 4sqrt (x^2+y^2) =0 [-10, 10, -5, 5]}