What is the Cartesian form of $r^{2} + θ = - {tan}^{2} θ + 4 {cot}^{3} θ$ $r^2+theta = -tan^2theta+4cot^3theta$ ?

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What is the Cartesian form of $r^{2} + θ = - {tan}^{2} θ + 4 {cot}^{3} θ$ $r^2+theta = -tan^2theta+4cot^3theta$ ?

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Answer 1 · 2016-05-15T13:37:47

$x^{2} - \frac{4 x^{3}}{y^{3}} + y^{2} + \frac{y^{2}}{x^{2}} + arctan (\frac{y}{x}) = 0$ $x^2 - (4 x^3)/y^3 + y^2 + y^2/x^2 + arctan(y/x) = 0$

Explanation:

From pass equations
$x = r cos (θ)$ $x = r cos(theta)$
$y = r sin (θ)$ $y=r sin(theta)$
dividing we get
$\frac{y}{x} = tan (θ), \frac{x}{y} = cot (θ)$ $y/x=tan(theta), x/y=cot(theta)$
and also
$θ = arctan (\frac{y}{x})$ $\theta=arctan(y/x)$
After substituting $r = \frac{x}{cos (θ)}$ $r = x/cos(theta)$ we substitute for $θ$ $\theta$
obtaining the result shown

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What is the Cartesian form of $r^{2} + θ = - {tan}^{2} θ + 4 {cot}^{3} θ$ $r^2+theta = -tan^2theta+4cot^3theta$ ?

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Explanation:

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What is the Cartesian form of r^2+theta = -tan^2theta+4cot^3theta r2+θ=−tan2θ+4cot3θr^2+theta = -tan^2theta+4cot^3theta ?

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Explanation:

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What is the Cartesian form of $r^{2} + θ = - {tan}^{2} θ + 4 {cot}^{3} θ$ $r^2+theta = -tan^2theta+4cot^3theta$ ?