Question

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

Answer 1

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

Explanation:

From the pass equations
`x = r cos(theta)`
`y = r sin(theta)`
We find out that
`y/x=sin(theta)/cos(theta) = tan(theta)->theta = arctan(y/x)`
`cot(theta) = 1/tan(theta) = x/y`
`r = x/cos(theta)`