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) :}#