How do you graph #r = theta, theta in [ 0, 2pi ]#, using its Cartesian equivalent?

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Answer 1 · 2018-08-11T12:16:08

See explanation and graphs.

Range-restricted #abs r <= pi/2 # combined graph, for #theta uarr#

in both clockwise and anticlockwise sense, using

#sqrt(x^2+y^2) = arctan(y/x)#

graph{sqrt(x^2+y^2) - arctan(y/x) = 0 }

Range-unrestricted combined graph, for #theta uarr#

in both clockwise and anticlockwise sense, using the wholesome

inverse

y = x tan (sqrt ( x^2 + y^2 ))

graph{(x tan ((x^2+y^2)^0.5)-y)(x-2pi+0.0001y)(x+2pi+0.0001y)=0}

It appears that the algorithm for the Cartesian

frame has to be improved, to separate the conjoined twin graphs.