What is the Cartesian form of #r^2+r = 2theta-2sectheta-csctheta #?

Question

What is the Cartesian form of #r^2+r = 2theta-2sectheta-csctheta  #?

Nolan Alexander · Accepted Answer

The conversion maps to 3 equations. Please see the explanation.

Before we begin the conversion, please observe that the secant function has a division by zero issue at #theta = pi/2 + npi#. The same is true for the cosecant function and #theta = npi#. This translates into the Cartesian as a restriction of #x !=0 and y != 0# For #csc(theta)#, begin with: #y = rsin(theta)# #1/sin(theta) = r/y# #1/sin(theta) = sqrt(x^2 + y^2)/y# #csc(theta) = sqrt(x^2 + y^2)/y# A similar substitution exists for the secant function: #sec(theta) = sqrt(x^2 + y^2)/x# Substitute #x^2 + y^2# for #r^2# and #sqrt(x^2 + y^2)# for r: #x^2 + y^2 + sqrt(x^2 + y^2) = 2theta - 2sqrt(x^2 + y^2)/x - sqrt(x^2 + y^2)/y; x != 0 and y!=0# The substitution for #theta# breaks the equation into 3 equations: #x^2 + y^2 + sqrt(x^2 + y^2) = 2tan^-1(y/x) - 2sqrt(x^2 + y^2)/x - sqrt(x^2 + y^2)/y; x > 0 and y>0# #x^2 + y^2 + sqrt(x^2 + y^2) = 2(tan^-1(y/x) + pi) - 2sqrt(x^2 + y^2)/x - sqrt(x^2 + y^2)/y; x < 0 and y!=0# #x^2 + y^2 + sqrt(x^2 + y^2) = 2(tan^-1(y/x) + 2pi) - 2sqrt(x^2 + y^2)/x - sqrt(x^2 + y^2)/y; x > 0 and y<0# Undefined elsewhere.

Aaliyah Abdullah · Answer

undefined The Cartesian form of the equation $r^2 + r = 2\theta - 2\sec(\theta) - \csc(\theta)$ is $x^2 + y^2 + x = 2\arctan\left(\frac{y}{x}\right) - 2\sec\left(\arctan\left(\frac{y}{x}\right)\right) - \csc\left(\arctan\left(\frac{y}{x}\right)\right)$.