Question

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

Answer 1

See graph and details.

Explanation:

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.