What is the Cartesian form of #r = -sin^2theta-r^2csc^2theta #?

1 Answer
Shwetank Mauria
Jan 3, 2017

#(x^2+y^2)^(3/2)y^2=-y^4-(x^2+y^2)^3#

Explanation:

The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is

#x=rcostheta# and #y=rsintheta# i.e. #r^2=x^2+y^2# and #y/x=tantheta#

Hence #r=-sin^2theta-r^2csc^2theta# can be written as

#r=-y^2/r^2-r^2xxr^2/y^2=y^2/r^2-r^4/y^2# and multiplying each term by #r^2y^2#, this can be written as

#r^3y^2=-y^4-r^6#

or #(x^2+y^2)^(3/2)y^2=-y^4-(x^2+y^2)^3#

Related questions
See all questions in Converting Between Systems
Impact of this question
1176 views around the world
You can reuse this answer
Creative Commons License