What is the Cartesian form of rtheta+r = 2costheta-csctheta ?

1 Answer
Oct 25, 2016

There are 5 conditions that yield 4 equations and a division by 0. Please see the explanation.

Explanation:

Before we begin. Please notice that theta must not be a 2npi multiple of 0 or pi, because the cosecant function is undefined at the these points. This converts to the Cartesian restriction y != 0

Eventually, we will substitute tan^-1(y/x) for theta, therefore, we must look at the pristine function, r = (2cos(theta) - csc(theta))/(theta + 1), and see what happens at theta = pi/2 and (3pi)/2

r = -2/(pi + 2) and r = -2/(3pi + 2)

These are the y coordinates at x = 0.

y = -2/(pi + 2) and y = -2/(3pi + 2)

Now, we may proceed with the conversion.

For csc(theta):

y = rsin(theta)

1/sin(theta) = r/y

csc(theta) = r/y

csc(theta) = sqrt(x^2 + y^2)/y

r = (2cos(theta) - sqrt(x^2 + y^2)/y)/(theta + 1)

Multiply both sides by r:

r^2 = (2rcos(theta) - (x^2 + y^2)/y)/(theta + 1)

Substitute x for rcos(theta)

r^2 = (2x - (x^2 + y^2)/y)/(theta + 1)

Substitute x^2 + y^2 for r^2

(x^2 + y^2) = (2x - (x^2 + y^2)/y)/((theta + 1)

Multiply the right side by y/y:

(x^2 + y^2) = (2xy - x^2 - y^2)/(y(theta + 1))

The substitution for theta breaks the equation into 3 equations plus the 2 for x = 0 and y = 0:

Undefined for y = 0

y = -2/(pi + 2) and y = -2/(3pi + 2); x = 0

(x^2 + y^2) = (2xy - x^2 - y^2)/(y(tan^-1(y/x) + 1)); x > 0 and y > 0

(x^2 + y^2) = (2xy - x^2 - y^2)/(y(tan^-1(y/x) + pi + 1)); x < 0 and y != 0

(x^2 + y^2) = (2xy - x^2 - y^2)/(y(tan^-1(y/x) + 2pi + 1)); x > 0 and y < 0