Question

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

Answer 1

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

Explanation:

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`