How do you convert #r^2 theta=1# into cartesian form?

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Answer 1 · 2016-10-01T05:37:25

#y=x tan(1/(x^2+y^2))#, representing a spiral from the origin that turns forever, to reach the origin..

As #r^2>0#, its reciprocal #theta > 0#

Using the conversion formula #r(cos theta, sin theta) = (x, y),

#theta=tan^(-1)(y/x)=1/(x^2+y^2) to y = x tan (1/(x^2+y^2))#.

I want variation of #theta in (0, oo)#.

The graph is a spiral that approaches origin only in the limit,

as #x, y to oo#.. .

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