What the is the polar form of $y = \frac{y - x}{x^{3} + y}$ $y = (y-x)/(x^3+y)$ ?

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What the is the polar form of $y = \frac{y - x}{x^{3} + y}$ $y = (y-x)/(x^3+y)$ ?

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Answer 1 · 2016-07-03T07:10:43

$r^{3} {cos}^{3} θ sin θ + r {sin}^{2} θ = (sin θ - cos θ)$ $r^3cos^3thetasintheta+rsin^2theta=(sintheta-costheta)$

Explanation:

When polar coordinate are $(r, θ)$ $(r,theta)$ and corresponding Cartesian coordinates are $(x, y)$ $(x,y)$ , the relation between them is $x = r c \otimes θ$ $x=rcoxtheta$ and $y = r sin θ$ $y=rsintheta$ and $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ .

Hence $y = \frac{y - x}{x^{3} + y}$ $y=(y-x)/(x^3+y)$ can be written as $x^{3} y + y^{2} = y - x$ $x^3y+y^2=y-x$ or

$r^{4} {cos}^{3} θ sin θ + r^{2} {sin}^{2} θ = r (sin θ - cos θ)$ $r^4cos^3thetasintheta+r^2sin^2theta=r(sintheta-costheta)$

or $r^{3} {cos}^{3} θ sin θ + r {sin}^{2} θ = (sin θ - cos θ)$ $r^3cos^3thetasintheta+rsin^2theta=(sintheta-costheta)$

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What the is the polar form of $y = \frac{y - x}{x^{3} + y}$ $y = (y-x)/(x^3+y)$ ?

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What the is the polar form of y = (y-x)/(x^3+y) y=y−xx3+yy = (y-x)/(x^3+y) ?

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What the is the polar form of $y = \frac{y - x}{x^{3} + y}$ $y = (y-x)/(x^3+y)$ ?