Convert Equations from Rectangular to Polar Form

This page presents problems in which equations written in rectangular form are converted to polar form using the standard relationships between rectangular and polar coordinates. Each problem is accompanied by a complete and detailed solution.

In what follows, the polar coordinates of a point are written as \((R, t)\), where \(R\) is the radial coordinate and \(t\) is the angular coordinate.

The relationships between rectangular coordinates \((x, y)\) and polar coordinates \((R, t)\) are given by

\[ R^2 = x^2 + y^2, \qquad x = R \cos t, \qquad y = R \sin t. \]

Problems on Converting Rectangular Equations to Polar Form

Problem 1

Convert the equation

\[ 2x^2 + 2y^2 - x + y = 0 \]

to polar form.

Solution to Problem 1

Rewrite the equation by grouping the squared terms: \[ 2(x^2 + y^2) - x + y = 0. \]
Use the relationships \(R^2 = x^2 + y^2\), \(x = R \cos t\), and \(y = R \sin t\): \[ 2R^2 - R \cos t + R \sin t = 0. \]
Factor out \(R\): \[ R(2R - \cos t + \sin t) = 0. \]
This equation gives \[ R = 0 \quad \text{or} \quad 2R - \cos t + \sin t = 0. \]
The equation \(R = 0\) represents the pole. The pole is included in the graph of the second equation because, for example, when \(t = \pi/4\), we obtain \(R = 0\). Therefore, we keep only the second equation.
Solving for \(R\), we obtain the polar form of the equation: \[ 2R - \cos t + \sin t = 0 \quad \Rightarrow \quad R = \frac{1}{2}(\cos t - \sin t). \]

Problem 2