Question

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

Answer 1

Cartesian form:
`x^2+y^2+sqrt(x^2+y^2)=2*arctan(y/x)-((x^2+y^2)/x^2)+(4*sqrt(x^2+y^2))/y`

Explanation:

From the given:
`r^2+r=2theta-sec^2 theta+4*csc theta`
We make use of the equivalent of
`r=sqrt(x^2+y^2)`
`theta=arctan (y/x)`

Direct substitution
`r^2+r=2theta-sec^2 theta+4*csc theta`
`(sqrt(x^2+y^2))^2+sqrt(x^2+y^2)=2*arctan(y/x)-(sqrt(x^2+y^2)/x)^2+(4*sqrt(x^2+y^2))/y`

The graph of `x^2+y^2+sqrt(x^2+y^2)=2*arctan(y/x)-((x^2+y^2)/x^2)+(4*sqrt(x^2+y^2))/y`
graph{x^2+y^2+sqrt(x^2+y^2)=2arctan(y/x)-((x^2+y^2)/x^2)+(4sqrt(x^2+y^2))/y [-20, 20, -10, 10]}

God bless....I hope the explanation is useful.