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

Question

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

Violet Anderson · Accepted Answer

Please see the 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.

Adam Aguirre · Answer

undefined To convert the polar equation $r^2 \sin(\theta) = 2\theta - 4\tan(\theta) - \csc(\theta)$ to Cartesian form, we can use the relationships between polar and Cartesian coordinates:

$$x = r \cos(\theta)$$
$$y = r \sin(\theta)$$

First, we need to express $r$ in terms of $x$ and $y$. Since $r^2 = x^2 + y^2$, we have $r = \sqrt{x^2 + y^2}$.

Now, substitute $r$ in the given equation with $\sqrt{x^2 + y^2}$:

$$\sqrt{x^2 + y^2} \cdot \sin(\theta) = 2\theta - 4\tan(\theta) - \csc(\theta)$$

Then, replace $\sin(\theta)$ and $\csc(\theta)$ with their Cartesian equivalents:

$$\frac{y}{\sqrt{x^2 + y^2}} = 2\theta - 4\left(\frac{y}{x}\right) - \frac{1}{\sin(\theta)}$$

Since $\sin(\theta) = \frac{y}{r}$ and $r = \sqrt{x^2 + y^2}$, we can substitute $\sin(\theta)$ with $\frac{y}{\sqrt{x^2 + y^2}}$:

$$\frac{y}{\sqrt{x^2 + y^2}} = 2\theta - 4\left(\frac{y}{x}\right) - \frac{\sqrt{x^2 + y^2}}{y}$$

To eliminate $\theta$, we need to express it in terms of $x$ and $y$. We know that $\tan(\theta) = \frac{y}{x}$, so $\theta = \arctan\left(\frac{y}{x}\right)$.

Substitute $\theta$ with $\arctan\left(\frac{y}{x}\right)$:

$$\frac{y}{\sqrt{x^2 + y^2}} = 2\arctan\left(\frac{y}{x}\right) - 4\left(\frac{y}{x}\right) - \frac{\sqrt{x^2 + y^2}}{y}$$

This is the Cartesian form of the given polar equation.