What is the Cartesian form of #r^2+theta = -sin^2theta-cot^3theta #?

1 Answer
Shwetank Mauria
May 3, 2016

#x^2+y^2+tan^(-1)(y/x)= -y^2/(x^2+y^2)-x^2/y^2#

Explanation:

If #(r,theta)# is in polar form and #(x,y)# in Cartesian form the relation between them is as follows:

#x=rcostheta#, #y=rsintheta#, #r^2=x^2+y^2# and #tantheta=y/x#

Or, #costheta=x/r#, #sintheta=y/r#, #theta=tan^(-1)(y/x)# and #cottheta=x/y#.

Hence, #r^2+theta=-sin^2theta-cot^2theta# can be written as

#x^2+y^2+tan^(-1)(y/x)=-y^2/r^2-x^2/y^2# or

#x^2+y^2+tan^(-1)(y/x)= -y^2/(x^2+y^2)-x^2/y^2#

Related questions
See all questions in Converting Between Systems
Impact of this question
1098 views around the world
You can reuse this answer
Creative Commons License