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

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

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Answer 1 · 2018-05-21T16:06:17

$r = \frac{cos θ - sin θ}{(2 - cos θ) sin θ cos θ}$ $r=(costheta-sintheta)/((2-costheta)sinthetacostheta)$

Explanation:

$y = x - 2 y + x y^{2}$ $y=x-2y+xy^2$
$x = r cos θ$ $x=rcostheta$
$y = r sin θ$ $y=rsintheta$
Thus,
$r sin θ = r cos θ - 2 r cos θ \times r sin θ + r cos θ \times {(r sin θ)}^{2}$ $rsintheta=rcostheta-2rcosthetaxxrsintheta+rcosthetaxx(rsintheta)^2$

$r sin θ = r cos θ - 2 r^{2} sin θ cos θ + r^{2} {sin}^{2} θ cos θ$ $rsintheta=rcostheta-2r^2sinthetacostheta+r^2sin^2thetacostheta$

$r sin θ - r cos θ = r^{2} (- 2 sin θ cos θ + {sin}^{2} θ cos θ)$ $rsintheta-rcostheta=r^2(-2sinthetacostheta+sin^2thetacostheta)$

$r (sin θ - cos θ) = r^{2} sin θ cos θ (- 2 + cos θ)$ $r(sintheta-costheta)=r^2sinthetacostheta(-2+costheta)$

$r (sin θ - cos θ) + r^{2} sin θ cos θ (2 - cos θ)$ $r(sintheta-costheta)+r^2sinthetacostheta(2-costheta)$

$r (sin θ - cos θ + r sin θ cos θ (2 - cos θ)) = 0$ $r(sintheta-costheta+rsinthetacostheta(2-costheta))=0$

$r = 0$ $r=0$

$sin θ - cos θ + r sin θ cos θ (2 - cos θ) = 0$ $sintheta-costheta+rsinthetacostheta(2-costheta)=0$

$r sin θ cos θ (2 - cos θ) = cos θ - sin θ$ $rsinthetacostheta(2-costheta)=costheta-sintheta$
$r = \frac{cos θ - sin θ}{(2 - cos θ) sin θ cos θ}$ $r=(costheta-sintheta)/((2-costheta)sinthetacostheta)$

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

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