Question

What is the Cartesian form of `r^2+theta = -tan^2theta+4cot^3theta`?

Answer 1

`x^2 - (4 x^3)/y^3 + y^2 + y^2/x^2 + arctan(y/x) = 0`

Explanation:

From pass equations
`x = r cos(theta)`
`y=r sin(theta)`
dividing we get
`y/x=tan(theta), x/y=cot(theta)`
and also
`\theta=arctan(y/x)`
After substituting `r = x/cos(theta)` we substitute for `\theta`
obtaining the result shown