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#