Question

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

Answer 1

Set:

`x=rcosθ`

`y=rsinθ`

Answer is:

`sqrt(x^2+y^2)-arccos(x/sqrt(x^2+y^2))=-2x^2/(x^2+y^2)-x^3/y^3`

Explanation:

According to the following picture:

![https://www.mathsisfun.com/]()

Set:

`x=rcosθ`

`y=rsinθ`

So we have:

`cosθ=x/r`

`sinθ=y/r`

`θ=arccos(x/r)=arcsin(y/r)`

`r=sqrt(x^2+y^2)`

The equation becomes:

`r-θ=-2sin^2θ-cot^3θ`

`r-θ=-2sin^2θ-cos^3θ/sin^3θ`

`sqrt(x^2+y^2)-arccos(x/r)=-2x^2/r^2-(x^3/r^3)/(y^3/r^3)`

`sqrt(x^2+y^2)-arccos(x/r)=-2x^2/r^2-x^3/y^3`

`sqrt(x^2+y^2)-arccos(x/sqrt(x^2+y^2))=-2x^2/sqrt(x^2+y^2)^2-x^3/y^3`

`sqrt(x^2+y^2)-arccos(x/sqrt(x^2+y^2))=-2x^2/(x^2+y^2)-x^3/y^3`