What is the Cartesian form of -2+r^2 = -6theta+sin(theta)cos(theta) −2+r2=−6θ+sin(θ)cos(θ)?

1 Answer
Shwetank Mauria
May 23, 2016

x^2+y^2=6tan^(-1)(y/x)+(xy)/(x^2+y^2)+2x2+y2=6tan−1(yx)+xyx2+y2+2

Explanation:

A Cartesian point (x,y)(x,y) in polar form is (r,theta)(r,θ), where

x=rcosthetax=rcosθ and y=rsinthetay=rsinθ and hence

x^2+y^2=r^2cos^2theta+r^2sin^2theta=r^2x2+y2=r2cos2θ+r2sin2θ=r2 and theta=tan^(-1)(y/x)θ=tan−1(yx)

Hence -2+r^2=-6theta+sinthetacostheta−2+r2=−6θ+sinθcosθ is

-2+x^2+y^2=6tan^(-1)(y/x)+(xy)/(x^2+y^2)−2+x2+y2=6tan−1(yx)+xyx2+y2

or x^2+y^2=6tan^(-1)(y/x)+(xy)/(x^2+y^2)+2x2+y2=6tan−1(yx)+xyx2+y2+2

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