How do you graph r^2= - cos theta?

2 Answers
May 12, 2016

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Explanation:

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May 12, 2016

The pass equations are:
x = r*cos(theta), y = r*sin(theta)
squaring them, x^2=r^2*cos(theta)^2,y^2=r^2*sin(theta). Substituting now, x^2=-cos(theta)*cos(theta)^2,y^2=-cos(theta)*sin(theta)^2. Those equations make sense only for cos(theta) le 0 so for pi le theta le 2pi. Having this in mind we can draw: x = pm cos(theta)*sqrt(-cos(theta)),y = pm sin(theta)*sqrt(-cos(theta)) for pi le theta le 2pi