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

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Answer 1 · 2018-02-18T09:15:02

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

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

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,

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

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