Question

What is the Cartesian form of `(r+3)^2 = sin^3theta+csctheta`?

Answer 1

We know the relations

`x=rcostheta and y =rsintheta`

Again `x^2+y^2=r^2`

where r and `theta` are the polar coordinate of a point having rectangular coordinate `(x,y)`

The given equation in polar form is

`(r+3)^2=sin^3theta+csctheta`

`=>r^2+6r+9=(r^3sin^3theta)/r^3+r/(rsintheta)`

`=>r^2+6r+9=y^3/r^3+r/y`

`=>yr^5+6yr^4+9yr^3=y^4+r^4`

`=>yr^5+(6y-1)r^4+9yr^3=y^4`

`=>y(x^2+y^2)^(5/2)+(6y-1)(x^2+y^2)^2+9y(x^2+y^2)^(3/2)=y^4`

This is the Cartesian form of the given polar equation.