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

2 Answers
Sam
May 12, 2016

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Explanation:

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Cesareo R.
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#

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