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

1 Answer
A. S. Adikesavan
Aug 4, 2018

See answer in the explanation.

Explanation:

Using

#0 <= r = sqrt ( x^2 + y^2 ), r ( cos theta, sin theta ) = ( x, y )#.

#theta = arctan (y/x ), theta in Q_1# or #Q_4# and

#theta = pi + arctan (y/x ), theta in Q_2# or #Q_3#

#r = - sin^2theta -theta csc^2theta# converts to

#sqrt ( x^2 + y^2 ) = -y^2/( x^2 + y^2) - phi/y^2( x^2 + y^2 )#,

where

#phi = arctan (y/x ), theta in Q_1# or #Q_4# and

#= pi + arctan (y/x ), theta in Q_2# or #Q_3#

See graph.
graph{y^2(x^2+y^2)^1.5+y^4 + (x^2+y^2)^2 arctan (y/x )=0}
Graph for the wholesome inverse.
graph{tan((y^2(x^2+y^2)^1.5+y^4)/(x^2+y^2))-1=0}

Related questions
See all questions in Converting Between Systems
Impact of this question
1389 views around the world
You can reuse this answer
Creative Commons License