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

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

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Answer 1 · 2016-10-20T17:49:52

Please see the explanation, because this breaks into 3 equations.

Explanation:

Multiply both sides by $r^{5} {sin}^{3} (θ)$ $r^5sin^3(theta)$ :

$r^{7} {sin}^{3} (θ) + (θ) r^{5} {sin}^{3} (θ) = 4 r^{5} {cos}^{3} (θ) - r^{5} {sin}^{5} (θ)$ $r^7sin^3(theta) + (theta)r^5sin^3(theta) = 4r^5cos^3(theta) - r^5sin^5(theta)$

Do some regrouping:

$r^{4} {(r sin (θ))}^{3} + (θ) r^{2} {(r sin (θ))}^{3} = 4 r^{2} {(r cos (θ))}^{3} - {(r sin (θ))}^{5}$ $r^4(rsin(theta))^3 + (theta)r^2(rsin(theta))^3 = 4r^2(rcos(theta))^3 - (rsin(theta))^5$

Substitute x for $(r cos (θ))$ $(rcos(theta))$ and y for $(r sin (θ))$ $(rsin(theta))$ :

$r^{4} y^{3} + (θ) r^{2} y^{3} = 4 r^{2} x^{3} - y^{5}$ $r^4y^3 + (theta)r^2y^3 = 4r^2x^3 - y^5$

Substitute $(x^{2} + y^{2})$ $(x^2 + y^2)$ for $r^{2}$ $r^2$

${(x^{2} + y^{2})}^{2} y^{3} + (θ) (x^{2} + y^{2}) y^{3} = 4 (x^{2} + y^{2}) x^{3} - y^{5}$ $(x^2 + y^2)^2y^3 + (theta)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5$

We cannot merely substitute ${tan}^{- 1} (\frac{y}{x})$ $tan^-1(y/x)$ for $θ$ $theta$ ; we must specify that $x \neq 0 and y \neq 0$ $x!=0 and y!=0$ and then handle special cases for $x < 0$ $x < 0$ , $x > 0 and y \geq 0$ $x > 0 and y>=0$ , and $x > 0 and y < 0$ $x > 0 and y < 0$

For $x > 0 and y > 0$ $x > 0 and y>0$ :

${(x^{2} + y^{2})}^{2} y^{3} + {tan}^{- 1} (\frac{y}{x}) (x^{2} + y^{2}) y^{3} = 4 (x^{2} + y^{2}) x^{3} - y^{5}$ $(x^2 + y^2)^2y^3 + tan^-1(y/x)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5$

For $x < 0$ $x < 0$ :

${(x^{2} + y^{2})}^{2} y^{3} + ({tan}^{- 1} (\frac{y}{x}) + π) (x^{2} + y^{2}) y^{3} = 4 (x^{2} + y^{2}) x^{3} - y^{5}$ $(x^2 + y^2)^2y^3 + (tan^-1(y/x) + pi)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5$

For $x > 0 and y < 0$ $x > 0 and y < 0$

${(x^{2} + y^{2})}^{2} y^{3} + ({tan}^{- 1} (\frac{y}{x}) + 2 π) (x^{2} + y^{2}) y^{3} = 4 (x^{2} + y^{2}) x^{3} - y^{5}$ $(x^2 + y^2)^2y^3 + (tan^-1(y/x) + 2pi)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5$

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

1 Answer

Explanation:

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What is the Cartesian form of r^2+theta = -sin^2theta+4cot^3theta r2+θ=−sin2θ+4cot3θr^2+theta = -sin^2theta+4cot^3theta ?

1 Answer

Explanation:

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