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

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

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Answer 1 · 2018-05-02T10:38:32

$r^{2} (r {cos}^{2} θ + r cos θ {sin}^{2} θ - sin θ) = cot θ csc θ$ $r^2(rcos^2theta+rcosthetasin^2theta-sintheta)=cotthetacsctheta$

Explanation:

For this we will use:
$x = r cos θ$ $x=rcostheta$
$y = r sin t h e t r a$ $y=rsinthetra$

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

$r sin θ = r^{2} {cos}^{2} θ - \frac{cot θ csc θ}{r} + r^{2} cos θ {sin}^{2} θ$ $rsintheta=r^2cos^2theta-(cotthetacsctheta)/r+r^2costhetasin^2theta$

$r^{2} sin θ = r^{3} {cos}^{2} θ - cot θ csc θ + r^{3} cos θ {sin}^{2} θ$ $r^2sintheta=r^3cos^2theta-cotthetacsctheta+r^3costhetasin^2theta$

$r^{3} {cos}^{2} θ + r^{3} cos θ {sin}^{2} θ - r^{2} sin θ = cot θ csc θ$ $r^3cos^2theta+r^3costhetasin^2theta-r^2sintheta=cotthetacsctheta$

$r^{2} (r {cos}^{2} θ + r cos θ {sin}^{2} θ - sin θ) = cot θ csc θ$ $r^2(rcos^2theta+rcosthetasin^2theta-sintheta)=cotthetacsctheta$

This cannot be simplified further and has to be left as an implicit equation.

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

1 Answer

Explanation:

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