How do you convert $y = {(x - y)}^{2} - x^{2} y + x y^{2}$ $y=(x-y)^2-x^2y +xy^2$ into a polar equation?

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How do you convert $y = {(x - y)}^{2} - x^{2} y + x y^{2}$ $y=(x-y)^2-x^2y +xy^2$ into a polar equation?

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Answer 1 · 2016-04-15T21:49:30

$0 = r^{2} - r^{2} sin (2 θ) - \frac{1}{2} r^{3} cos θ sin (2 θ) + \frac{1}{2} r^{3} sin θ sin (2 θ) - r sin θ$ $0=r^2-r^2sin(2theta)-1/2 r^3costhetasin(2theta)+1/2r^3sinthetasin(2theta)-rsintheta$

Explanation:

$0 = x^{2} - 2 x y + y^{2} - x^{2} y + x y^{2} - y$ $0=x^2-2xy+y^2-x^2y+xy^2-y$

$0 = x^{2} + y^{2} - 2 x y - x^{2} y + x y^{2} - y$ $0=x^2+y^2-2xy-x^2y+xy^2-y$

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

$0 = r^{2} - 2 r cos θ \times r sin θ - r^{2} {cos}^{2} θ \times r sin θ + r cos θ \times r^{2} {sin}^{2} θ - r sin θ$ $0=r^2-2rcosthetaxxrsintheta-r^2cos^2thetaxxrsintheta+rcos thetaxxr^2sin^2theta-rsintheta$

$0 = r^{2} - 2 r^{2} sin θ cos θ - r^{3} sin θ {cos}^{2} θ + r^{3} {sin}^{2} θ cos θ - r sin θ$ $0=r^2-2r^2sinthetacostheta-r^3sinthetacos^2theta+r^3sin^2thetacostheta-rsintheta$

$0 = r^{2} - r^{2} (2 sin θ cos θ) - \frac{1}{2} r^{3} cos θ (2 sin θ cos θ) + \frac{1}{2} r^{3} sin θ (2 sin θ cos θ) - r sin θ$ $0=r^2-r^2(2sinthetacostheta)-1/2 r^3costheta(2sinthetacostheta)+1/2r^3 sintheta(2sinthetacostheta)-rsintheta$

$0 = r^{2} - r^{2} sin (2 θ) - \frac{1}{2} r^{3} cos θ sin (2 θ) + \frac{1}{2} r^{3} sin θ sin (2 θ) - r sin θ$ $0=r^2-r^2sin(2theta)-1/2 r^3costhetasin(2theta)+1/2r^3sinthetasin(2theta)-rsintheta$

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How do you convert $y = {(x - y)}^{2} - x^{2} y + x y^{2}$ $y=(x-y)^2-x^2y +xy^2$ into a polar equation?

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How do you convert y=(x-y)^2-x^2y +xy^2 y=(x−y)2−x2y+xy2y=(x-y)^2-x^2y +xy^2 into a polar equation?

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How do you convert $y = {(x - y)}^{2} - x^{2} y + x y^{2}$ $y=(x-y)^2-x^2y +xy^2$ into a polar equation?