Polar coordinates and polar equations provide an alternative way of representing points and curves in a plane, different from the usual Cartesian (x, y) system. For electronics engineering technicians in Kenya, understanding polar coordinates is crucial, especially when dealing with phasor diagrams, antenna radiation patterns, and signal processing where angles and magnitudes are more natural descriptors than Cartesian coordinates. This chapter focuses on defining polar coordinates, formulating polar equations, and representing curves using these equations, preparing technicians to analyze and visualize complex electronic systems effectively.

4.1 Polar equations

4.1.1 Definition of polar coordinates

Polar coordinates represent a point in a plane by its distance from a fixed origin and the angle it makes with a fixed reference direction, usually the positive x-axis. The two parameters used are \(r\), the radial distance, and \(\theta\), the angle measured in radians or degrees. The relationship between polar coordinates \((r, \theta)\) and Cartesian coordinates \((x, y)\) is given by:

$$ x = r \cos \theta $$

$$ y = r \sin \theta $$

Conversely,

$$ r = \sqrt{x^2 + y^2} $$

$$ \theta = \tan^{-1} (\frac{y}{x}) $$

Worked Examples

Example 1: Convert the point with Cartesian coordinates (3 m, 4 m) to polar coordinates.

Given: \(x=3\,m\), \(y=4\,m\)

$$ r = \sqrt{x^2 + y^2} $$

$$ r = \sqrt{3^2 + 4^2} $$

$$ r = \sqrt{9 + 16} = \sqrt{25} $$

$$ r = 5\,m $$

$$ \theta = \tan^{-1} (\frac{y}{x}) $$

$$ \theta = \tan^{-1} (\frac{4}{3}) $$

$$ \theta = 53.13^\circ $$

Answer: \(r=5\,m\), \(\theta=53.13^\circ\)

Example 2: Find the Cartesian coordinates of a point given in polar coordinates as \(r = 7\,m\), \(\theta = 45^\circ\).

Given: \(r=7\,m\), \(\theta=45^\circ\)

$$ x = r \cos \theta $$

$$ x = 7 \times \cos 45^\circ $$

$$ x = 7 \times 0.7071 = 4.95\,m $$

$$ y = r \sin \theta $$

$$ y = 7 \times \sin 45^\circ $$

$$ y = 7 \times 0.7071 = 4.95\,m $$

Answer: \(x=4.95\,m\), \(y=4.95\,m\)

Example 3: An antenna is located at Cartesian coordinates \((-5\,m, 5\,m)\). Find its polar coordinates.

Given: \(x = -5\,m\), \(y=5\,m\)

$$ r = \sqrt{(-5)^2 + 5^2} $$

$$ r = \sqrt{25 + 25} = \sqrt{50} $$

$$ r = 7.07\,m $$

$$ \theta = \tan^{-1} (\frac{5}{-5}) $$

$$ \theta = \tan^{-1} (-1) = -45^\circ $$

Since the point is in the second quadrant,

$$ \theta = 180^\circ - 45^\circ = 135^\circ $$

Answer: \(r=7.07\,m\), \(\theta=135^\circ\)

4.1.2 Definition and examples of polar equations

Polar equations express \(r\) as a function of \(\theta\), often written as \(r = f(\theta)\). These equations define curves in the plane by specifying the distance from the origin for each angle. Common polar equations include circles, spirals, and cardioids, which appear frequently in electronics engineering, such as in plotting antenna radiation patterns or impedance loci.

A general polar equation:

$$ r = f(\theta) $$

Worked Examples

Example 1: Consider the circle defined by \(r = 4\). Find the Cartesian coordinates of the point where \(\theta = 60^\circ\).

Given: \(r=4\), \(\theta=60^\circ\)

$$ x = r \cos \theta = 4 \times \cos 60^\circ = 4 \times 0.5 = 2\,m $$

$$ y = r \sin \theta = 4 \times \sin 60^\circ = 4 \times 0.866 = 3.464\,m $$

Answer: \(x=2\,m\), \(y=3.464\,m\)

Example 2: For the spiral defined by \(r = 2\theta\) (with \(\theta\) in radians), find the Cartesian coordinates at \(\theta = \pi/3\).

Given: \(r = 2 \times \frac{\pi}{3} = \frac{2\pi}{3} \approx 2.094\,m\), \(\theta = \pi/3\)

$$ x = r \cos \theta = 2.094 \times \cos 60^\circ = 2.094 \times 0.5 = 1.047\,m $$

$$ y = r \sin \theta = 2.094 \times \sin 60^\circ = 2.094 \times 0.866 = 1.813\,m $$

Answer: \(x=1.047\,m\), \(y=1.813\,m\)

Example 3: For the cardioid given by \(r = 3(1 + \cos \theta)\), find \(r\) and Cartesian coordinates at \(\theta = 90^\circ\).

Given: \(\theta=90^\circ\)

$$ r = 3(1 + \cos 90^\circ) = 3(1 + 0) = 3\,m $$

$$ x = r \cos \theta = 3 \times \cos 90^\circ = 3 \times 0 = 0\,m $$

$$ y = r \sin \theta = 3 \times \sin 90^\circ = 3 \times 1 = 3\,m $$

Answer: \(r=3\,m\), \(x=0\,m\), \(y=3\,m\)

4.1.3 Representing curves using polar equations

Curves in the polar coordinate system are plotted by calculating \(r\) for various values of \(\theta\) and then converting to Cartesian coordinates for graphing or directly plotting in polar form. This method is useful in electronics for visualizing impedance loci on Smith charts or radiation patterns of antennas. Understanding how to represent and interpret these curves supports circuit analysis and antenna design.

The procedure to represent a curve defined by \(r = f(\theta)\) is:

1. Select a range of \(\theta\) values (e.g., 0 to \(2\pi\) radians).
2. Calculate corresponding \(r\) values using the polar equation.
3. Convert each \((r, \theta)\) pair to Cartesian coordinates if needed using \(x = r \cos \theta\), \(y = r \sin \theta\).
4. Plot the points on polar or Cartesian axes.
5. Connect the points smoothly to reveal the curve shape.
6. Analyze curve properties such as symmetry and intercepts.

Worked Examples

Example 1: Plot points for the circle \(r=5\) at \(\theta = 0^\circ, 90^\circ, 180^\circ, 270^\circ\).

Calculate \(r\) at each \(\theta\):

• \(\theta=0^\circ\), \(r=5\)
• \(\theta=90^\circ\), \(r=5\)
• \(\theta=180^\circ\), \(r=5\)
• \(\theta=270^\circ\), \(r=5\)

Convert to Cartesian:

At \(0^\circ\):

$$ x = 5 \times \cos 0^\circ = 5 \times 1 = 5 $$

$$ y = 5 \times \sin 0^\circ = 5 \times 0 = 0 $$

At \(90^\circ\):

$$ x = 5 \times \cos 90^\circ = 5 \times 0 = 0 $$

$$ y = 5 \times \sin 90^\circ = 5 \times 1 = 5 $$

At \(180^\circ\):

$$ x = 5 \times \cos 180^\circ = 5 \times (-1) = -5 $$

$$ y = 5 \times \sin 180^\circ = 5 \times 0 = 0 $$

At \(270^\circ\):

$$ x = 5 \times \cos 270^\circ = 5 \times 0 = 0 $$

$$ y = 5 \times \sin 270^\circ = 5 \times (-1) = -5 $$

Answer: Points are (5,0), (0,5), (-5,0), (0,-5), forming a circle centered at origin with radius 5 m.

Example 2: For the rose curve \(r = 4 \cos 3\theta\), calculate \(r\) and Cartesian coordinates at \(\theta = 0^\circ, 30^\circ, 60^\circ, 90^\circ\).

\(\theta=0^\circ\):

$$ r = 4 \cos 0^\circ = 4 \times 1 = 4 $$

$$ x = 4 \times \cos 0^\circ = 4 \times 1 = 4 $$

$$ y = 4 \times \sin 0^\circ = 4 \times 0 = 0 $$

\(\theta=30^\circ\):

$$ r = 4 \cos 90^\circ = 4 \times 0 = 0 $$

$$ x = 0, \quad y=0 $$

\(\theta=60^\circ\):

$$ r = 4 \cos 180^\circ = 4 \times (-1) = -4 $$

$$ x = -4 \times \cos 60^\circ = -4 \times 0.5 = -2 $$

$$ y = -4 \times \sin 60^\circ = -4 \times 0.866 = -3.464 $$

\(\theta=90^\circ\):

$$ r = 4 \cos 270^\circ = 4 \times 0 = 0 $$

$$ x=0, \quad y=0 $$

Answer: Points plotted show petal-like shapes characteristic of rose curves, useful in antenna pattern analysis.

Example 3: Represent the lemniscate defined by \(r^2 = 16 \cos 2\theta\). Find \(r\) at \(\theta=0^\circ, 45^\circ, 90^\circ\).

At \(\theta=0^\circ\):

$$ r^2 = 16 \cos 0^\circ = 16 \times 1 = 16 $$

$$ r = \pm 4 $$

At \(\theta=45^\circ\):

$$ r^2 = 16 \cos 90^\circ = 16 \times 0 = 0 $$

$$ r = 0 $$

At \(\theta=90^\circ\):

$$ r^2 = 16 \cos 180^\circ = 16 \times (-1) = -16 $$

Since \(r^2\) cannot be negative, no real points exist at \(\theta=90^\circ\).

Answer: Points at \(\theta=0^\circ\) are \(r = \pm 4\), at \(45^\circ\), \(r=0\); no points at \(90^\circ\). The curve loops around the origin, useful in complex impedance plots.

Chapter Summary

This chapter introduced the concept of polar coordinates and polar equations, explaining how points and curves can be represented in the polar coordinate system. It then covered Cartesian coordinates, defining points and equations in this system and exploring the relationship and conversion between Cartesian and polar forms. The chapter detailed methods for plotting polar equations, including the use of symmetry and periodicity to analyze key features such as intercepts and extrema. It also examined tangents and normals to curves, providing techniques for determining their slopes and equations in both Cartesian and polar contexts. The final section focused on loci, defining the locus of points and emphasizing its relevance in engineering applications, particularly in relation to circles. Practical uses of loci were discussed, including tracing paths and optimizing mechanical systems, demonstrating their importance in real-world engineering problems. Overall, the chapter provided a comprehensive foundation for understanding and applying coordinate geometry in engineering contexts.