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

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Answer 1 · 2016-05-03T08:20:55

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

If $(r,theta)$ is in polar form and $(x,y)$ in Cartesian form the relation between them is as follows:

$x=rcostheta$ , $y=rsintheta$ , $r^2=x^2+y^2$ and $tantheta=y/x$

Or, $costheta=x/r$ , $sintheta=y/r$ , $theta=tan^(-1)(y/x)$ and $cottheta=x/y$ .

Hence, $r^2+theta=-sin^2theta-cot^2theta$ can be written as

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

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

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