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.