What the is the polar form of $y = \frac{y}{x} + x^{2} {(2 - y)}^{2}$ $y = y/x+x^2(2-y)^2$ ?

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What the is the polar form of $y = \frac{y}{x} + x^{2} {(2 - y)}^{2}$ $y = y/x+x^2(2-y)^2$ ?

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Answer 1 · 2017-04-24T08:44:01

Use the conversion:
$x = r cos (θ)$ $x=r cos(theta)$
$y = r sin (θ)$ $y=r sin(theta)$

Now substitute:
$r sin (θ) = \frac{r sin (θ)}{r cos (θ)} + {(r cos (θ))}^{2} {(2 - r sin (θ))}^{2}$ $r sin(theta)=(rsin(theta))/(rcos(theta))+(rcos(theta))^2(2-rsin(theta))^2$

Expand the final term:
$r sin (θ) = \frac{sin (θ)}{cos (θ)} + r^{2} {cos}^{2} (θ) (4 - 2 r sin (θ) + r^{2} {sin}^{2} (θ))$ $r sin(theta)=(sin(theta))/(cos(theta))+r^2cos^2(theta)(4-2rsin(theta)+r^2sin^2(theta))$

Multiply by $cos (θ)$ $cos(theta)$ :
$r sin (θ) cos (θ) = sin (θ) + r^{2} {cos}^{3} (θ) (4 - 2 r sin (θ) + r^{2} {sin}^{2} (θ))$ $r sin(theta)cos(theta)=sin(theta)+r^2cos^3(theta)(4-2rsin(theta)+r^2sin^2(theta))$

At this point it is not looking like the solution can be represented 'nicely'. You could try converting everything into $sin$ $sin$

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What the is the polar form of $y = \frac{y}{x} + x^{2} {(2 - y)}^{2}$ $y = y/x+x^2(2-y)^2$ ?

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What the is the polar form of y = y/x+x^2(2-y)^2 y=yx+x2(2−y)2y = y/x+x^2(2-y)^2 ?

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What the is the polar form of $y = \frac{y}{x} + x^{2} {(2 - y)}^{2}$ $y = y/x+x^2(2-y)^2$ ?