Question

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

Answer 1

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Explanation:

Answer 2

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`