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

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Answer 1 · 2016-05-23T09:25:02

$x^2+y^2=6tan^(-1)(y/x)+(xy)/(x^2+y^2)+2$

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

$x=rcostheta$ and $y=rsintheta$ and hence

$x^2+y^2=r^2cos^2theta+r^2sin^2theta=r^2$ and $theta=tan^(-1)(y/x)$

Hence $-2+r^2=-6theta+sinthetacostheta$ is

$-2+x^2+y^2=6tan^(-1)(y/x)+(xy)/(x^2+y^2)$

or $x^2+y^2=6tan^(-1)(y/x)+(xy)/(x^2+y^2)+2$

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