What is the Cartesian form of r^2-rtheta = -cottheta-sintheta ?

1 Answer
Jan 17, 2017

The presence of theta causes the equation to become 3 equations. Please see the explanation.

Explanation:

Given: r^2 - rtheta = -cot(theta) - sin(theta)

Because of the cotangent function, the restriction that theta cannot be an integer multiple of pi must be added but this translates into the Cartesian restriction y != 0 and we can replace cot(theta) with x/y

r^2 - rtheta = -x/y - sin(theta); y !=0

We can replace r^2 with x^2 + y^2:

x^2 + y^2 - rtheta = -x/y - sin(theta); y !=0

Because y = rsin(theta), we can replace sin(theta) with y/r

x^2 + y^2 - rtheta = -x/y - y/r; y !=0

We can replace r with sqrt(x^2 + y^2)

x^2 + y^2 - sqrt(x^2 + y^2)theta = -x/y - y/sqrt(x^2 + y^2); y !=0

Converting the theta into Cartesian means that we break this into 3 equations:

x^2 + y^2 = { (sqrt(x^2 + y^2)tan^-1(y/x) -x/y - y/sqrt(x^2 + y^2); y >0", "x > 0), (sqrt(x^2 + y^2)(pi + tan^-1(y/x)) -x/y - y/sqrt(x^2 + y^2); y != 0", "x < 0), (sqrt(x^2 + y^2)(2pi + tan^-1(y/x)) -x/y - y/sqrt(x^2 + y^2); y < 0", "x > 0) :}