What is the Cartesian form of $r = - {sin}^{2} θ - r^{2} {csc}^{2} θ$ $r = -sin^2theta-r^2csc^2theta$ ?

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What is the Cartesian form of $r = - {sin}^{2} θ - r^{2} {csc}^{2} θ$ $r = -sin^2theta-r^2csc^2theta$ ?

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Answer 1 · 2017-01-03T09:08:58

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

Explanation:

The relation between polar coordinates $(r, θ)$ $(r,theta)$ and Cartesian coordinates $(x, y)$ $(x,y)$ is

$x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ i.e. $r^{2} = x^{2} + y^{2}$ $r^2=x^2+y^2$ and $\frac{y}{x} = tan θ$ $y/x=tantheta$

Hence $r = - {sin}^{2} θ - r^{2} {csc}^{2} θ$ $r=-sin^2theta-r^2csc^2theta$ can be written as

$r = - \frac{y^{2}}{r^{2}} - r^{2} \times \frac{r^{2}}{y^{2}} = \frac{y^{2}}{r^{2}} - \frac{r^{4}}{y^{2}}$ $r=-y^2/r^2-r^2xxr^2/y^2=y^2/r^2-r^4/y^2$ and multiplying each term by $r^{2} y^{2}$ $r^2y^2$ , this can be written as

$r^{3} y^{2} = - y^{4} - r^{6}$ $r^3y^2=-y^4-r^6$

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

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What is the Cartesian form of $r = - {sin}^{2} θ - r^{2} {csc}^{2} θ$ $r = -sin^2theta-r^2csc^2theta$ ?

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Explanation:

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What is the Cartesian form of r = -sin^2theta-r^2csc^2theta r=−sin2θ−r2csc2θr = -sin^2theta-r^2csc^2theta ?

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Explanation:

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What is the Cartesian form of $r = - {sin}^{2} θ - r^{2} {csc}^{2} θ$ $r = -sin^2theta-r^2csc^2theta$ ?