What the is the polar form of $y = \frac{1}{y} - x y + \frac{x^{2}}{y}$ $y = 1/y-xy+x^2/y$ ?

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What the is the polar form of $y = \frac{1}{y} - x y + \frac{x^{2}}{y}$ $y = 1/y-xy+x^2/y$ ?

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Answer 1 · 2018-06-27T01:05:34

$r sin (θ) = \frac{1}{r sin (θ)} - r^{2} sin (θ) cos (θ) + \frac{r {cos}^{2} (θ)}{sin (θ)}$ $rsin(theta)=1/(rsin(theta))-r^2sin(theta)cos(theta)+(rcos^2(theta))/(sin(theta))$

Simplified: $r^{2} (cos (θ) - cos (3 θ)) - 4 = {(2 r)}^{2} cos (2 θ)$ $r^2(cos(theta)-cos(3theta))-4=(2r)^2cos(2theta)$

Explanation:

$x = r cos (θ), y = r sin (θ)$ $x=rcos(theta), y=rsin(theta)$

Substitute and Simplify

$r sin (θ) = \frac{1}{r sin (θ)} - r^{2} sin (θ) cos (θ) + \frac{r {cos}^{2} (θ)}{sin (θ)}$ $rsin(theta)=1/(rsin(theta))-r^2sin(theta)cos(theta)+(rcos^2(theta))/sin(theta)$

Alternate: $r^{2} sin (θ) = csc (θ) - r^{3} sin (θ) cos (θ) + r^{2} cot (θ) cos (θ)$ $r^2sin(theta)=csc(theta)-r^3sin(theta)cos(theta)+r^2cot(theta)cos(theta)$

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What the is the polar form of $y = \frac{1}{y} - x y + \frac{x^{2}}{y}$ $y = 1/y-xy+x^2/y$ ?

1 Answer

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What the is the polar form of y = 1/y-xy+x^2/y y=1y−xy+x2yy = 1/y-xy+x^2/y ?

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What the is the polar form of $y = \frac{1}{y} - x y + \frac{x^{2}}{y}$ $y = 1/y-xy+x^2/y$ ?