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

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Answer 1 · 2018-02-19T14:15:36

#r-theta=-2sin^3theta+sec^2theta# in cartesian form is

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

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

#sintheta=y/sqrt(x^2+y^2)#

#-2sin^3theta=-2(sintheta)^3=-2(y/sqrt(x^2+y^2))^3#

#sec^2theta=1/cos^2theta=1/(x^2/(x^2+y^2))=(x^2+y^2)/x^2#

Thus,
#r-theta=-2sin^3theta+sec^2theta# in cartesian form is

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