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

Question

1593 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-25T07:49:04

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

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

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