Question

What the is the polar form of `y^2 = 1/y-1/x+x^2/y`?

Answer 1

`r^3 sin^3 theta cos theta -r^2 cos^3 theta = cos theta -sin theta`

Explanation:

Using the relation `x = r cos theta , quad y= r sin theta` between Cartesian and polar coordinates, we have

`r^2 sin^2 theta = 1/{r sin theta}-1/{r cos theta}+{r^2 cos ^2 theta}/{r sin theta}`

Multiplying both sides by `r sin theta cos theta` lets us simplify this a bit to

`r^3 sin^3 theta cos theta = cos theta -sin theta +r^2 cos^3 theta`
or
`r^3 sin^3 theta cos theta -r^2 cos^3 theta = cos theta -sin theta`