Question

How do you convert `(r+1)^2= - sin theta costheta` to Cartesian form?

Answer 1

`(x^2+y^2)^2+2(x^2+y^2)^(3/2)+x^2+y^2+xy=0`

Explanation:

The relation between polar coordinates `(r,theta)` and Cartesian coordinates `(x,y)` is given by

`x=rcostheta`, `y=rsintheta` and `x^2+y^2=r^2`

Hence we can write `(r+1)^2=-sinthetacostheta` in Cartesian form as

`x^2+y^2+2sqrt(x^2+y^2)+1=-y/rxx x/r`

or `x^2+y^2+2sqrt(x^2+y^2)+1=-(xy)/(x^2+y^2)`

or `(x^2+y^2)^2+2(x^2+y^2)^(3/2)+x^2+y^2+xy=0`