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

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Answer 1 · 2017-05-25T07:49:04

${(x^{2} + y^{2})}^{2} + 2 {(x^{2} + y^{2})}^{\frac{3}{2}} + x^{2} + y^{2} + x y = 0$ $(x^2+y^2)^2+2(x^2+y^2)^(3/2)+x^2+y^2+xy=0$

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

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ and $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Hence we can write ${(r + 1)}^{2} = - sin θ cos θ$ $(r+1)^2=-sinthetacostheta$ in Cartesian form as

$x^{2} + y^{2} + 2 \sqrt{x^{2} + y^{2}} + 1 = - \frac{y}{r} \times \frac{x}{r}$ $x^2+y^2+2sqrt(x^2+y^2)+1=-y/rxx x/r$

or $x^{2} + y^{2} + 2 \sqrt{x^{2} + y^{2}} + 1 = - \frac{x y}{x^{2} + y^{2}}$ $x^2+y^2+2sqrt(x^2+y^2)+1=-(xy)/(x^2+y^2)$

or ${(x^{2} + y^{2})}^{2} + 2 {(x^{2} + y^{2})}^{\frac{3}{2}} + x^{2} + y^{2} + x y = 0$ $(x^2+y^2)^2+2(x^2+y^2)^(3/2)+x^2+y^2+xy=0$

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