By the end of this chapter, you will be able to:
These skills will help you confidently analyze and visualize engineering problems, making your work more precise and effective in the trade.
Graphs of polar equations are essential in refrigeration and air conditioning engineering, particularly when dealing with components and systems that exhibit rotational symmetry or cyclical behavior. Understanding how to plot these graphs allows professionals to visualize pressure variations, temperature distributions, and flow patterns in circular ducts or rotating machinery such as compressors and fans. This chapter focuses on applying coordinate geometry to draw graphs of given polar equations, enabling precise analysis and design in practical refrigeration and air conditioning contexts.
The polar coordinate system represents points in a plane by a distance from a fixed origin and an angle from a fixed direction, differing from Cartesian coordinates which use horizontal and vertical distances. The fundamental relationship between polar coordinates \((r, \theta)\) and Cartesian coordinates \((x, y)\) is given by the formulas:
$$ x = r \cos \theta $$
$$ y = r \sin \theta $$
where \(r\) is the radius (distance from origin) and \(\theta\) is the angle measured in radians or degrees from the positive x-axis.
Example 1: Convert the polar point \(r=5\) m, \(\theta=60^\circ\) to Cartesian coordinates.
Given: \(r=5\) m, \(\theta=60^\circ\)
$$ x = 5 \cos 60^\circ $$
$$ x = 5 \times 0.5 = 2.5 \text{ m} $$
$$ y = 5 \sin 60^\circ $$
$$ y = 5 \times 0.866 = 4.33 \text{ m} $$
Answer: \( (x, y) = (2.5 \text{ m}, 4.33 \text{ m}) \)
Example 2: Find Cartesian coordinates of a point with \(r=8\) m and \(\theta=135^\circ\).
Given: \(r=8\) m, \(\theta=135^\circ\)
$$ x = 8 \cos 135^\circ $$
$$ x = 8 \times (-0.707) = -5.656 \text{ m} $$
$$ y = 8 \sin 135^\circ $$
$$ y = 8 \times 0.707 = 5.656 \text{ m} $$
Answer: \( (x, y) = (-5.656 \text{ m}, 5.656 \text{ m}) \)
Example 3: Convert the polar point \(r=10\) m, \(\theta=210^\circ\) to Cartesian coordinates.
Given: \(r=10\) m, \(\theta=210^\circ\)
$$ x = 10 \cos 210^\circ $$
$$ x = 10 \times (-0.866) = -8.66 \text{ m} $$
$$ y = 10 \sin 210^\circ $$
$$ y = 10 \times (-0.5) = -5 \text{ m} $$
Answer: \( (x, y) = (-8.66 \text{ m}, -5 \text{ m}) \)
Polar equations often describe simple shapes such as circles and lines important in the design of circular ducts and piping in refrigeration systems. The general forms include:
Plotting these helps visualize components like circular coils and flow paths.
Example 1: Plot the circle \(r = 4\) m by calculating points at \(\theta = 0^\circ, 90^\circ, 180^\circ, 270^\circ\).
Given: \(r=4\) m constant
Calculate Cartesian points:
At \(\theta=0^\circ\):
$$ x = 4 \cos 0^\circ = 4 $$
$$ y = 4 \sin 0^\circ = 0 $$
At \(\theta=90^\circ\):
$$ x = 4 \cos 90^\circ = 0 $$
$$ y = 4 \sin 90^\circ = 4 $$
At \(\theta=180^\circ\):
$$ x = 4 \cos 180^\circ = -4 $$
$$ y = 4 \sin 180^\circ = 0 $$
At \(\theta=270^\circ\):
$$ x = 4 \cos 270^\circ = 0 $$
$$ y = 4 \sin 270^\circ = -4 $$
Answer: The circle passes through points (4,0), (0,4), (-4,0), (0,-4).
Example 2: Plot the line \(\theta = 45^\circ\) for \(r = 0\) to \(5\) m by calculating Cartesian points at \(r = 1, 3, 5\) m.
Given: \(\theta=45^\circ\)
At \(r=1\) m:
$$ x = 1 \cos 45^\circ = 0.707 $$
$$ y = 1 \sin 45^\circ = 0.707 $$
At \(r=3\) m:
$$ x = 3 \cos 45^\circ = 2.121 $$
$$ y = 3 \sin 45^\circ = 2.121 $$
At \(r=5\) m:
$$ x = 5 \cos 45^\circ = 3.536 $$
$$ y = 5 \sin 45^\circ = 3.536 $$
Answer: The line passes through (0.707, 0.707), (2.121, 2.121), and (3.536, 3.536).
Example 3: Plot the circle \(r = 6 \cos \theta\) by calculating points for \(\theta = 0^\circ, 60^\circ, 90^\circ, 120^\circ, 180^\circ\).
At \(\theta=0^\circ\):
$$ r = 6 \cos 0^\circ = 6 $$
$$ x = 6 \cos 0^\circ = 6 $$
$$ y = 6 \sin 0^\circ = 0 $$
At \(\theta=60^\circ\):
$$ r = 6 \cos 60^\circ = 3 $$
$$ x = 3 \cos 60^\circ = 1.5 $$
$$ y = 3 \sin 60^\circ = 2.598 $$
At \(\theta=90^\circ\):
$$ r = 6 \cos 90^\circ = 0 $$
$$ x = 0 $$
$$ y = 0 $$
At \(\theta=120^\circ\):
$$ r = 6 \cos 120^\circ = -3 $$
Negative radius means point in opposite direction at \(300^\circ\):
$$ x = 3 \cos 300^\circ = 1.5 $$
$$ y = 3 \sin 300^\circ = -2.598 $$
At \(\theta=180^\circ\):
$$ r = 6 \cos 180^\circ = -6 $$
Opposite direction at \(0^\circ\):
$$ x = 6 \cos 0^\circ = 6 $$
$$ y = 0 $$
Answer: Points are (6, 0), (1.5, 2.598), (0,0), (1.5, -2.598), and (6, 0).
More advanced polar equations such as rose curves \(r = a \cos n\theta\) or spirals \(r = a \theta\) model cyclical and spiral flow patterns in refrigeration fans or coil layouts. Understanding these helps in analyzing airflow and heat exchanger designs.
The general rose curve formula is:
$$ r = a \cos n\theta $$
where \(a\) controls the size and \(n\) controls the number of petals.
Spiral equation:
$$ r = a \theta $$
where \(a\) controls the tightness of the spiral.
Example 1: Plot \(r = 3 \cos 2\theta\) for \(\theta = 0^\circ, 45^\circ, 90^\circ, 135^\circ, 180^\circ\).
At \(\theta=0^\circ\):
$$ r = 3 \cos 0^\circ = 3 $$
At \(\theta=45^\circ\):
$$ r = 3 \cos 90^\circ = 0 $$
At \(\theta=90^\circ\):
$$ r = 3 \cos 180^\circ = -3 $$
At \(\theta=135^\circ\):
$$ r = 3 \cos 270^\circ = 0 $$
At \(\theta=180^\circ\):
$$ r = 3 \cos 360^\circ = 3 $$
Answer: The rose curve has petals at \(r = \pm 3\) and crosses origin at intermediate angles.
Example 2: Plot the spiral \(r = 2 \theta\) for \(\theta = 0^\circ, 45^\circ, 90^\circ, 135^\circ, 180^\circ\) (convert degrees to radians).
Convert degrees to radians:
Calculate \(r\):
At \(0\):
$$ r = 2 \times 0 = 0 $$
At \(0.785\):
$$ r = 2 \times 0.785 = 1.57 $$
At \(1.571\):
$$ r = 2 \times 1.571 = 3.142 $$
At \(2.356\):
$$ r = 2 \times 2.356 = 4.712 $$
At \(3.142\):
$$ r = 2 \times 3.142 = 6.284 $$
Calculate Cartesian points for \(\theta=45^\circ\) (\(0.785\) rad):
$$ x = 1.57 \cos 45^\circ = 1.11 $$
$$ y = 1.57 \sin 45^\circ = 1.11 $$
Answer: Spiral points increase linearly with \(\theta\), demonstrating expanding radius.
Example 3: Plot \(r = 4 \sin 3\theta\) for \(\theta = 0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ\).
At \(\theta=0^\circ\):
$$ r = 4 \sin 0^\circ = 0 $$
At \(\theta=30^\circ\):
$$ r = 4 \sin 90^\circ = 4 $$
At \(\theta=60^\circ\):
$$ r = 4 \sin 180^\circ = 0 $$
At \(\theta=90^\circ\):
$$ r = 4 \sin 270^\circ = -4 $$
At \(\theta=120^\circ\):
$$ r = 4 \sin 360^\circ = 0 $$
Answer: The rose curve with 3 petals appears, with peak radius 4 and zero at multiples of 60°.
Plotting polar graphs systematically ensures accuracy in interpreting refrigeration system designs involving circular components.
Identify the polar equation: Write down the given equation \(r = f(\theta)\).
Determine the domain of \(\theta\): Usually from \(0^\circ\) to \(360^\circ\) or \(0\) to \(2\pi\) radians.
Select values of \(\theta\): Choose increments such as \(15^\circ\), \(30^\circ\), or smaller for smooth curves.
Calculate corresponding \(r\) values: Substitute each \(\theta\) into the equation to find \(r\).
Convert polar to Cartesian coordinates: Use \(x = r \cos \theta\), \(y = r \sin \theta\) for plotting on Cartesian graph paper.
Plot points and connect: Mark each \((x, y)\) point and connect smoothly to reveal the curve.
Example 1: Plot \(r = 3\) for \(\theta = 0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ\).
Step 1: Equation given.
Step 2: \(\theta\) from \(0^\circ\) to \(360^\circ\).
Step 3: Select \(\theta = 0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ\).
Step 4: \(r=3\) constant.
Step 5: Calculate Cartesian coordinates:
At \(0^\circ\): \(x=3\), \(y=0\)
At \(90^\circ\): \(x=0\), \(y=3\)
At \(180^\circ\): \(x=-3\), \(y=0\)
At \(270^\circ\): \(x=0\), \(y=-3\)
At \(360^\circ\): \(x=3\), \(y=0\)
Step 6: Plot points and connect to form a circle.
Answer: Circle of radius 3 centered at origin.
Example 2: Plot \(r = 2 + 2 \cos \theta\) for \(\theta = 0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ, 360^\circ\).
Step 1: Equation given.
Step 2: Domain \(0^\circ\) to \(360^\circ\).
Step 3: Select \(\theta\) values.
Step 4: Calculate \(r\):
At \(0^\circ\): \(r=2+2 \times 1=4\)
At \(60^\circ\): \(r=2+2 \times 0.5=3\)
At \(120^\circ\): \(r=2+2 \times (-0.5)=1\)
At \(180^\circ\): \(r=2+2 \times (-1)=0\)
At \(240^\circ\): \(r=2+2 \times (-0.5)=1\)
At \(300^\circ\): \(r=2+2 \times 0.5=3\)
At \(360^\circ\): \(r=2+2 \times 1=4\)
Step 5: Calculate Cartesian coordinates for \(\theta=60^\circ\):
$$ x=3 \cos 60^\circ = 1.5 $$
$$ y=3 \sin 60^\circ = 2.598 $$
Repeat for others.
Step 6: Plot points and connect smoothly forming a cardioid shape.
Answer: Cardioid curve plotted.
Example 3: Plot \(r = 5 \sin 2\theta\) for \(\theta = 0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ, 180^\circ\).
Step 1: Equation given.
Step 2: Domain \(0^\circ\) to \(180^\circ\).
Step 3: Select \(\theta\) values.
Step 4: Calculate \(r\):
At \(0^\circ\): \(r=0\)
At \(30^\circ\): \(r=5 \sin 60^\circ = 5 \times 0.866 = 4.33\)
At \(60^\circ\): \(r=5 \sin 120^\circ = 5 \times 0.866 = 4.33\)
At \(90^\circ\): \(r=5 \sin 180^\circ = 0\)
At \(120^\circ\): \(r=5 \sin 240^\circ = 5 \times (-0.866) = -4.33\)
At \(150^\circ\): \(r=5 \sin 300^\circ = -4.33\)
At \(180^\circ\): \(r=5 \sin 360^\circ = 0\)
Step 5: Convert to Cartesian for \(\theta=30^\circ\):
$$ x = 4.33 \cos 30^\circ = 3.75 $$
$$ y = 4.33 \sin 30^\circ = 2.165 $$
Step 6: Plot points and connect forming four-petal rose.
Answer: Four-petal rose curve plotted.
Convert the polar point \(r = 7\) m, \(\theta = 120^\circ\) to Cartesian coordinates. (4 marks)
Plot points and sketch the circle defined by \(r = 5\) m at \(\theta = 0^\circ, 90^\circ, 180^\circ, 270^\circ\). (6 marks)
Calculate and plot points for the rose curve \(r = 4 \cos 3\theta\) at \(\theta = 0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ\). (8 marks)
Using the spiral \(r = \theta\), plot points for \(\theta = 0, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \pi\) and convert to Cartesian coordinates. (8 marks)
Plot the cardioid \(r = 2 + 2 \cos \theta\) by calculating \(r\) at \(\theta = 0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ, 360^\circ\). (10 marks)
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Create a free accountThis chapter focused on applying coordinate geometry to solve practical problems involving polar equations and curves. It began with methods to draw graphs of given polar equations, illustrating how points are plotted using the radius and angle from the pole. The chapter then explained how to calculate polar equations by converting between Cartesian and polar coordinate systems, enabling accurate representation and analysis of curves. The concepts of normal and tangent lines to curves were introduced, with detailed procedures to determine their equations at specific points using coordinate geometry techniques. Emphasis was placed on understanding the geometric relationship between curves and their tangents or normals. Throughout the chapter, worked examples demonstrated step-by-step calculations to reinforce these concepts. This comprehensive approach ensured mastery of graphing, calculating, and analyzing polar curves and their associated lines using coordinate geometry.
Calculate the radius \(r\) when the polar equation \(r = 4 \cos \theta\) is evaluated at \(\theta = 60^\circ\). (2 marks)
For the polar curve \(r = 3 + 2 \sin \theta\), find the value of \(r\) at \(\theta = 90^\circ\). (2 marks)
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