What is the Cartesian form of #r^2sintheta = 2theta-4tantheta-csctheta #?

1 Answer
Oct 27, 2016

Please see the explanation

Explanation:

Before we begin the conversion, please observe that the cosecant function and the tangent function have division by zero issues at integer multiples of #pi# offset by #0# and #pi/2#, respectively; this translates into the Cartesian restrictions #x !=0 and y!=0#.

The derivation for the substitution for #csc(theta)# is as follows:

#y = rsin(theta)#

#1/sin(theta) = r/y#

#csc(theta) = r/y#

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

Allowing y to equal zero would be trouble.

The derivation for the substitution for #tan(theta)# is as follows:

#y = rsin(theta)# and #x = rcos(theta)#

#y/x = (rsin(theta))/(rcos(theta)) = sin(theta)/cos(theta) = tan(theta)#

Allowing x to equal zero is a division by zero issue.

The substitution for #r^2sin(theta)#:

#r^2sin(theta) = (rsin(theta))r = ysqrt(x^2 + y^2)#

I am saving the best for last so let's write the equation with these 3 substitutions:

#ysqrt(x^2 + y^2) = 2theta - 4y/x -sqrt(x^2+y^2)/y#

The substitution for #theta# is:

#theta = tan^-1(y/x); x > 0 and y > 0#

#theta = tan^-1(y/x) + pi; x < 0, and y !=0#

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

This creates the 3 following equations:

#ysqrt(x^2 + y^2) = 2tan^-1(y/x) - 4y/x -sqrt(x^2+y^2)/y; x > 0 and y > 0#

#ysqrt(x^2 + y^2) = 2(tan^-1(y/x) + pi) - 4y/x -sqrt(x^2+y^2)/y; x < 0 and y != 0#

#ysqrt(x^2 + y^2) = 2(tan^-1(y/x) + 2pi) - 4y/x -sqrt(x^2+y^2)/y; x > 0 and y < 0#

Undefined elsewhere.