How do you convert $r = 4 θ - sin θ + {cos}^{2} θ$ $r=4 theta - sin theta +cos^2theta$ to Cartesian form?

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Answer 1 · 2016-10-09T20:36:49

Here is 1 of the 3 equations:

$(x² + y²)^(3/2) = 4(x² + y²)(tan^-1(y/x)) - ysqrt(x² + y²) + x²; x > 0, y >= 0$

Please see the explanation for the other two.

Let's begin by multiplying both sides of the equation by $r²$

$r^3 = 4r²theta - r²sin(theta) + r²cos²(theta)$

Because $x = rcos(theta)$ , we can replace $r²cos²(theta)$ with $x²$ :

$r^3 = 4r²theta - r²sin(theta) + x²$

Because $y = rsin(theta)$ and $r = sqrt(x² + y²)$ , we can replace $- r²sin(theta)$ with $-ysqrt(x² + y²)$ :

$r^3 = 4r²theta - ysqrt(x² + y²) + x²$

The $4r²theta$ term makes the one equation become 3 equations with 3 domain restrictions:

$r^3 = 4(x² + y²)(tan^-1(y/x)) - ysqrt(x² + y²) + x²; x > 0, y >= 0$

$r^3 = 4(x² + y²)(tan^-1(y/x) + pi) - ysqrt(x² + y²) + x²; x < 0$

$r^3 = 4(x² + y²)(tan^-1(y/x) + 2pi) - ysqrt(x² + y²) + x²; x > 0, y < 0$

Replace $r³$ with $(x² + y²)^(3/2)$

$(x² + y²)^(3/2) = 4(x² + y²)(tan^-1(y/x)) - ysqrt(x² + y²) + x²; x > 0, y >= 0$

$(x² + y²)^(3/2)= 4(x² + y²)(tan^-1(y/x) + pi) - ysqrt(x² + y²) + x²; x < 0$

$(x² + y²)^(3/2) = 4(x² + y²)(tan^-1(y/x) + 2pi) - ysqrt(x² + y²) + x²; x > 0, y < 0$

How do you convert r=4 theta - sin theta +cos^2thetar=4θ−sinθ+cos2θ r=4 theta - sin theta +cos^2theta to Cartesian form?