How do you convert # 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#