What the is the polar form of #y = y^3-xy+x^2/y #?

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Answer 1 · 2018-07-23T09:16:46

See graph and details.

Using #( x, y ) = r ( cos theta, sin theta )#. conversion gives

#r = 0# and

#r^2sin^4theta - rsin^2thetacos theta+ cos^2theta - sin^2theta =0# So,

#r^2 -r csc theta cot theta +csc^2theta ( cot^2 theta - 1 )#.

And so,

#0 <= r = 1/2 csc theta ( cot theta +- sqrt( 4 - 3 cot^2 theta )) #
.
graph{(y*4-xy^2+x^2-y^2)=0[-10 10-10.10]}

Slide the graph #uarr larr darr rarr # to see more of the graph.