Coordinate geometry is essential for industrial control and installation professionals, particularly in tasks such as sensor placement, robotic arm path planning, and control panel layout. Polar equations allow engineers to model circular and spiral paths efficiently, which is common in rotating machinery, conveyors, and automated systems. Mastery of polar coordinates ensures accurate measurements, optimal layouts, and precise control in a variety of industrial environments across Kenya.

4.1 Polar Equations

In engineering mathematics, polar equations describe curves using a radius and angle rather than Cartesian \(x\) and \(y\) coordinates. This makes them ideal for representing circular motion, rotating components, and locations around a central point, such as the shaft of a motor or the sweep of a robotic arm. Converting between Cartesian and polar forms is a core skill in industrial control system design.

4.1.1 Understanding Polar Coordinates and Equations

Polar coordinates define a point’s location by its distance from a fixed origin (\(r\)) and the angle (\(\theta\)) measured from a reference direction (usually the positive \(x\)-axis). The polar equation relates \(r\) and \(\theta\) directly.

The conversion formulas are:$$ x = r \cos\theta $$$$ y = r \sin\theta $$$$ r = \sqrt{x^2 + y^2} $$$$ \theta = \arctan(\frac{y}{x}) $$

Worked Examples

Example 1: Convert the polar point \( (r = 5\, \text{m},\ \theta = 60^\circ) \) to Cartesian coordinates.

Given: \( r = 5\, \text{m} \), \( \theta = 60^\circ \)

$$ x = r \cos\theta $$$$ x = 5 \times \cos(60^\circ) $$$$ x = 5 \times 0.5 $$$$ x = 2.5\, \text{m} $$

$$ y = r \sin\theta $$$$ y = 5 \times \sin(60^\circ) $$$$ y = 5 \times 0.866 $$$$ y = 4.33\, \text{m} $$

Answer: \( (x, y) = (2.5\, \text{m},\ 4.33\, \text{m}) \)

Example 2: A sensor is mounted at \( (r = 8\, \text{cm},\ \theta = 120^\circ) \). Find its Cartesian coordinates.

Given: \( r = 8\, \text{cm} \), \( \theta = 120^\circ \)

$$ x = 8 \times \cos(120^\circ) $$$$ x = 8 \times (-0.5) $$$$ x = -4\, \text{cm} $$

$$ y = 8 \times \sin(120^\circ) $$$$ y = 8 \times 0.866 $$$$ y = 6.93\, \text{cm} $$

Answer: \( (x, y) = (-4\, \text{cm},\ 6.93\, \text{cm}) \)

Example 3: A point on a control dial is at \( (r = 10\, \text{mm},\ \theta = 210^\circ) \). Find the \(x\) and \(y\) coordinates.

Given: \( r = 10\, \text{mm} \), \( \theta = 210^\circ \)

$$ x = 10 \times \cos(210^\circ) $$$$ x = 10 \times (-0.866) $$ x = -8.66\, \text{mm}$$

$$ y = 10 \times \sin(210^\circ) $$$$ y = 10 \times (-0.5) $$$$ y = -5\, \text{mm} $$

Answer: \( (x, y) = (-8.66\, \text{mm},\ -5\, \text{mm}) \)

Example 4: Convert Cartesian coordinates \( (x = 3\, \text{m},\ y = 4\, \text{m}) \) to polar form.

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

$$ r = \sqrt{x^2 + y^2} $$$$ r = \sqrt{3^2 + 4^2} $$$$ r = \sqrt{9 + 16} $$$$ r = \sqrt{25} $$$$ r = 5\, \text{m} $$

$$ \theta = \arctan(\frac{y}{x}) $$$$ \theta = \arctan(\frac{4}{3}) $$ \theta = 53.13^\circ$$

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

Example 5: A robotic arm tip is at \( (x = -6\, \text{cm},\ y = 8\, \text{cm}) \). Find its polar coordinates.

Given: \( x = -6\, \text{cm} \), \( y = 8\, \text{cm} \)

$$ r = \sqrt{(-6)^2 + 8^2} $$$$ r = \sqrt{36 + 64} $$$$ r = \sqrt{100} $$$$ r = 10\, \text{cm} $$

$$ \theta = \arctan(\frac{8}{-6}) $$ \theta = \arctan(-1.333)$$ \theta = -53.13^\circ $$

Since the point is in the second quadrant, add \(180^\circ\):

$$ \theta = 180^\circ - 53.13^\circ $$$$ \theta = 126.87^\circ $$

Answer: \( (r, \theta) = (10\, \text{cm},\ 126.87^\circ) \)

4.1.2 Plotting Polar Equations and Curves

Polar equations are commonly used to describe circles, spirals, and other shapes relevant in industrial automation. The most typical forms are:

• Circle: \( r = a \)
• Spiral: \( r = a\theta \)
• Line through origin: \( \theta = \text{constant} \)

Worked Examples

Example 1: Plot the circle \( r = 4\, \text{m} \).

Given: \( r = 4\, \text{m} \) for all \( \theta \)

• For any value of \( \theta \), \( r \) remains 4 m.
• The plot is a circle centered at the origin, radius 4 m.

Answer: The locus is a circle of radius 4 m centered at the origin.

Example 2: Plot the spiral \( r = 2\theta \) for \( \theta \) from \( 0 \) to \( 2\pi \).

Given: \( r = 2\theta \)

• At \( \theta = 0 \), \( r = 0 \)
• At \( \theta = \frac{\pi}{2} \), \( r = 2 \times \frac{\pi}{2} = \pi \approx 3.14 \)
• At \( \theta = \pi \), \( r = 2\pi \approx 6.28 \)
• At \( \theta = 2\pi \), \( r = 4\pi \approx 12.57 \)

Answer: The curve is a spiral starting at the origin and increasing in radius as \( \theta \) increases.

Example 3: Plot the line \( \theta = 45^\circ \).

Given: \( \theta = 45^\circ \), \( r \) can be any value.

• For any \( r \), the point lies on a straight line at \( 45^\circ \) from the \( x \)-axis.

Answer: The locus is a straight line through the origin at \( 45^\circ \) to the positive \( x \)-axis.

Example 4: Plot the curve \( r = 5 \sin\theta \) for \( \theta = 0^\circ, 30^\circ, 60^\circ, 90^\circ \).

Given: \( r = 5 \sin\theta \)

• \( \theta = 0^\circ \): \( r = 5 \times 0 = 0 \)
• \( \theta = 30^\circ \): \( r = 5 \times 0.5 = 2.5 \)
• \( \theta = 60^\circ \): \( r = 5 \times 0.866 = 4.33 \)
• \( \theta = 90^\circ \): \( r = 5 \times 1 = 5 \)

Answer: The curve starts at the origin, increases to a maximum at \( \theta = 90^\circ \).

Example 5: Plot the curve \( r = 3 + 2\cos\theta \) for \( \theta = 0^\circ, 90^\circ, 180^\circ \).

Given: \( r = 3 + 2\cos\theta \)

• \( \theta = 0^\circ \): \( r = 3 + 2 \times 1 = 5 \)
• \( \theta = 90^\circ \): \( r = 3 + 2 \times 0 = 3 \)
• \( \theta = 180^\circ \): \( r = 3 + 2 \times (-1) = 1 \)

Answer: The curve varies in radius with \( \theta \), forming a limaçon shape.

4.1.3 Applications of Polar Equations in Industrial Control and Installation

Polar equations are practical for describing the movement and placement of equipment that rotates or pivots, such as sensors, arms, and actuators. They are also used for calculating cable lengths, sweep areas, and coverage zones.

Worked Examples

Example 1: A rotating sensor sweeps a circle with \( r = 7\, \text{m} \). What is the area covered?

Given: \( r = 7\, \text{m} \), Area of a circle \( = \pi r^2 \)

$$ \text{Area} = \pi r^2 $$$$ \text{Area} = \pi \times (7)^2 $$$$ \text{Area} = \pi \times 49 $$$$ \text{Area} = 153.94\, \text{m}^2 $$

Answer: Area covered is \( 153.94\, \text{m}^2 \)

Example 2: A robotic arm extends as \( r = 1.5\theta \) (with \( \theta \) in radians, up to \( \theta = 2 \) rad). What is the maximum reach?

Given: \( r = 1.5\theta \), \( \theta_{\text{max}} = 2 \)

$$ r_{\text{max}} = 1.5 \times 2 $$$$ r_{\text{max}} = 3\, \text{m} $$

Answer: Maximum reach is \( 3\, \text{m} \)

Example 3: A control panel knob rotates from \( \theta = 0^\circ \) to \( \theta = 120^\circ \) at a fixed radius \( r = 6\, \text{cm} \). What arc length does the knob travel?

Given: \( r = 6\, \text{cm} \), \( \theta = 120^\circ = \frac{2\pi}{3} \) radians

Arc length \( s = r\theta \)

$$ \theta = 120^\circ = \frac{120}{180}\pi = \frac{2}{3}\pi $$$$ s = 6 \times \frac{2}{3}\pi $$$$ s = 4\pi $$$$ s = 12.57\, \text{cm} $$

Answer: Arc length is \( 12.57\, \text{cm} \)

Example 4: Find the distance between two points in polar coordinates: \( (r_1 = 4\, \text{m},\ \theta_1 = 30^\circ) \) and \( (r_2 = 4\, \text{m},\ \theta_2 = 150^\circ) \).

Distance formula:

$$ d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2 \cos(\theta_2 - \theta_1)} $$

$$ d = \sqrt{4^2 + 4^2 - 2 \times 4 \times 4 \cos(150^\circ - 30^\circ)} $$$$ d = \sqrt{16 + 16 - 32 \cos(120^\circ)} $$$$ \cos(120^\circ) = -0.5 $$$$ d = \sqrt{32 - 32 \times (-0.5)} $$$$ d = \sqrt{32 + 16} $$$$ d = \sqrt{48} $$$$ d = 6.93\, \text{m} $$

Answer: The distance is \( 6.93\, \text{m} \)

Example 5: A technician needs to install a camera to cover a sector of radius \( 10\, \text{m} \) and angle \( 60^\circ \). What is the area covered?

Area of sector: \( A = \frac{1}{2} r^2 \theta \) (with \( \theta \) in radians)

Given: \( r = 10\, \text{m} \), \( \theta = 60^\circ = \frac{\pi}{3} \)

$$ A = \frac{1}{2} \times (10)^2 \times \frac{\pi}{3} $$$$ A = \frac{1}{2} \times 100 \times \frac{\pi}{3} $$$$ A = 50 \times \frac{\pi}{3} $$$$ A = \frac{50\pi}{3} $$$$ A = 52.36\, \text{m}^2 $$

Answer: Area covered is \( 52.36\, \text{m}^2 \)

Chapter Summary

This chapter provided a comprehensive introduction to coordinate geometry, beginning with an exploration of polar equations and their representation of points using distance and angle from a fixed origin. It then covered Cartesian equations, emphasizing their use in describing geometric objects through coordinates on the \(x\) and \(y\) axes. The chapter progressed to plotting and interpreting graphs of polar equations, illustrating the unique shapes and curves that result from varying parameters. Techniques for finding normals and tangents to curves were discussed, with step-by-step calculations relevant to engineering applications. The definition of a point was clarified in both polar and Cartesian systems, forming the foundation for more complex geometric analysis. The concept of the locus of a point, especially in relation to a circle, was examined through calculation and graphical methods. Finally, loci of points generated by specific mechanisms were analyzed, demonstrating how movement and constraints in engineering systems can be described mathematically using coordinate geometry.