How do you convert $r = 3 θ - tan θ$ $r=3theta - tan theta$ to Cartesian form?

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Answer 1 · 2016-10-02T01:46:26

$x² + y² = (3tan^-1(y/x) - y/x)²; x > 0, y > 0$
Please see the explanation for the other two equations

$r = 3theta - tan(theta)$

Substitute $sqrt(x² + y²)$ for r:

$sqrt(x² + y²) = 3theta - tan(theta)$

Square both sides:

$x² + y² = (3theta - tan(theta))²$

Substitute $y/x$ for $tan(theta)$ :

$x² + y² = (3theta - y/x)²; x !=0$

Substitute $tan^-1(y/x)$ for $theta$ . NOTE: We must adjust for the $theta$ returned by the inverse tangent function based on the quadrant:

First quadrant:

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

Second and Third quadrant:

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

Fourth quadrant:

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

How do you convert r=3theta - tan theta r=3θ−tanθ r=3theta - tan theta to Cartesian form?