Question

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

Answer 1

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`

Explanation:

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`