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

2 Answers
Sam
May 12, 2016

..

Explanation:

enter image source here

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#

Related questions
See all questions in Graphing Basic Polar Equations
Impact of this question
11249 views around the world
You can reuse this answer
Creative Commons License