What is the Cartesian form of #r^2-rtheta = -cottheta+3tantheta #?

1 Answer
Feb 18, 2018

Cartesian form of
#r^2-rtheta=-cottheta+3tantheta# is

#sqrt(x^2+y^2)(sqrt(x^2+y^2)-tan^-1(y/x))=-x/y+(3y)/x#

Explanation:

Given:
#r^2-rtheta=-cottheta+3tantheta#

#x=rcostheta#
#y=rsintheta#
#r=sqrt(x^2+y^2)#

#theta=tan^-1(y/a)#
#tantheta=y/x#

#cottheta=x/y#

With this transformations,

#r^2-rtheta=r(r-theta)#
#sqrt(x^2+y^2)(sqrt(x^2+y^2)-tan^-1(y/x))#

#-cottheta+3tantheta=-x/y+3y/x#

Thus,

#sqrt(x^2+y^2)(sqrt(x^2+y^2)-tan^-1(y/x))=-x/y+(3y)/x#