By the end of this chapter, you will be able to:
These skills will help you solve real-world problems with precision and confidence in technical fields that rely on geometry.
Coordinate geometry is fundamental in water engineering for mapping, analyzing, and designing water infrastructure. Precise location and spatial relationships of features such as pipelines, reservoirs, and pump stations are essential for efficient water resource management. This chapter focuses on applying coordinate geometry principles, specifically polar and Cartesian equations, to solve practical problems encountered by water engineers in Kenya.
Polar coordinate systems are widely used in water engineering for problems involving circular or radial symmetry, such as locating wells, designing circular tanks, and analyzing flow around cylindrical structures. The polar coordinate system describes points by their distance from a reference point (origin) and the angle from a reference direction, simplifying calculations involving curves and angles.
A point in the plane can be represented by its polar coordinates \((r, \theta)\), where \(r\) is the distance from the origin and \(\theta\) is the angle measured from the positive x-axis.
$$ x = r \cos \theta, \quad y = r \sin \theta $$
Example 1:
A water sampling station is located 30 m from a pump station at an angle of \(45^\circ\) from the east. Find its Cartesian coordinates.
Given:
\(r = 30\,m\), \(\theta = 45^\circ\)
$$ x = r \cos \theta $$$$ x = 30 \times \cos 45^\circ $$$$ x = 30 \times 0.7071 = 21.213\,m $$
$$ y = r \sin \theta $$$$ y = 30 \times \sin 45^\circ $$$$ y = 30 \times 0.7071 = 21.213\,m $$
Answer: (21.213 m, 21.213 m)
Example 2:
A borehole is drilled at a location 50 m from a reservoir, at an angle of \(120^\circ\) from the positive x-axis. Find the Cartesian coordinates.
Given:
\(r = 50\,m\), \(\theta = 120^\circ\)
$$ x = 50 \times \cos 120^\circ = 50 \times (-0.5) = -25\,m $$$$ y = 50 \times \sin 120^\circ = 50 \times 0.8660 = 43.3\,m $$
Answer: (-25 m, 43.3 m)
Example 3:
Locate a valve at \(r = 40\,m\), \(\theta = 300^\circ\). Calculate its Cartesian coordinates.
Given:
\(r = 40\,m\), \(\theta = 300^\circ\)
$$ x = 40 \times \cos 300^\circ = 40 \times 0.5 = 20\,m $$$$ y = 40 \times \sin 300^\circ = 40 \times (-0.8660) = -34.64\,m $$
Answer: (20 m, -34.64 m)
Example 4:
A circular water tank has radius 25 m. Find the coordinates of a point on the tank’s perimeter at an angle of \(210^\circ\).
Given:
\(r = 25\,m\), \(\theta = 210^\circ\)
$$ x = 25 \times \cos 210^\circ = 25 \times (-0.8660) = -21.65\,m $$$$ y = 25 \times \sin 210^\circ = 25 \times (-0.5) = -12.5\,m $$
Answer: (-21.65 m, -12.5 m)
Example 5:
A pipeline extends from the origin to a point 60 m away at an angle of \(75^\circ\). Find the Cartesian coordinates.
Given:
\(r = 60\,m\), \(\theta = 75^\circ\)
$$ x = 60 \times \cos 75^\circ = 60 \times 0.2588 = 15.53\,m $$$$ y = 60 \times \sin 75^\circ = 60 \times 0.9659 = 57.95\,m $$
Answer: (15.53 m, 57.95 m)
Converting coordinates allows integration of data from different surveying and mapping methods. From Cartesian \((x,y)\), polar coordinates are found by:
$$ r = \sqrt{x^2 + y^2}, \quad \theta = \tan^{-1}(\frac{y}{x}) $$
Example 1:
Find the polar coordinates of a water treatment plant located at (24 m, 7 m).
Given:
\(x=24\,m\), \(y=7\,m\)
$$ r = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\,m $$$$ \theta = \tan^{-1} (\frac{7}{24}) = \tan^{-1}(0.2917) = 16.26^\circ $$
Answer: \((25\,m, 16.26^\circ)\)
Example 2:
Convert Cartesian coordinates (-15 m, 20 m) of a storage tank to polar form.
Given:
\(x = -15\,m\), \(y = 20\,m\)
$$ r = \sqrt{(-15)^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\,m $$$$ \theta = \tan^{-1}(\frac{20}{-15}) = \tan^{-1}(-1.3333) = -53.13^\circ $$
Since \(x < 0\) and \(y > 0\), \(\theta = 180^\circ - 53.13^\circ = 126.87^\circ\)
Answer: \((25\,m, 126.87^\circ)\)
Example 3:
Convert (0 m, -10 m) to polar coordinates.
Given:
\(x=0\,m\), \(y=-10\,m\)
$$ r = \sqrt{0^2 + (-10)^2} = 10\,m $$
Since \(x=0\), \(\theta = 270^\circ\) (pointing straight down)
Answer: \((10\,m, 270^\circ)\)
Example 4:
Find polar coordinates of a valve at (35 m, -35 m).
Given:
\(x=35\,m\), \(y=-35\,m\)
$$ r = \sqrt{35^2 + (-35)^2} = \sqrt{1225 + 1225} = \sqrt{2450} = 49.5\,m $$$$ \theta = \tan^{-1}(\frac{-35}{35}) = \tan^{-1}(-1) = -45^\circ $$
Because \(x>0\) and \(y<0\), \(\theta = 360^\circ - 45^\circ = 315^\circ\)
Answer: \((49.5\,m, 315^\circ)\)
Example 5:
Convert (-40 m, -30 m) to polar coordinates.
Given:
\(x = -40\,m\), \(y = -30\,m\)
$$ r = \sqrt{(-40)^2 + (-30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\,m $$$$ \theta = \tan^{-1}(\frac{-30}{-40}) = \tan^{-1}(0.75) = 36.87^\circ $$
Since \(x < 0\) and \(y < 0\), \(\theta = 180^\circ + 36.87^\circ = 216.87^\circ\)
Answer: \((50\,m, 216.87^\circ)\)
In water engineering, circular tanks and pipelines often require understanding polar equations of circles and lines.
$$ r = a $$
$$ \theta = \alpha $$
$$ r = \frac{r_0}{\cos(\theta - \alpha)} $$
where \(r_0\) is the perpendicular distance from origin to the line and \(\alpha\) the angle it makes with the x-axis.
Example 1:
Find the Cartesian coordinates of points on a circular water tank of radius 15 m at \(\theta = 30^\circ, 90^\circ, 150^\circ\).
Given:
\(r = 15\,m\), \(\theta = 30^\circ, 90^\circ, 150^\circ\)
For \(\theta=30^\circ\):$$ x = 15 \cos 30^\circ = 15 \times 0.8660 = 12.99\,m $$$$ y = 15 \sin 30^\circ = 15 \times 0.5 = 7.5\,m $$
For \(\theta=90^\circ\):$$ x = 15 \cos 90^\circ = 15 \times 0 = 0\,m $$$$ y = 15 \sin 90^\circ = 15 \times 1 = 15\,m $$
For \(\theta=150^\circ\):$$ x = 15 \cos 150^\circ = 15 \times (-0.8660) = -12.99\,m $$$$ y = 15 \sin 150^\circ = 15 \times 0.5 = 7.5\,m $$
Answer: (12.99 m, 7.5 m), (0 m, 15 m), (-12.99 m, 7.5 m)
Example 2:
Find the polar equation of a pipeline line that passes through the origin at \(60^\circ\).
Given:
\(\alpha = 60^\circ\)
$$ \theta = 60^\circ $$
Answer: \(\theta = 60^\circ\)
Example 3:
Find the distance from origin to a pipeline line given by \(r = \frac{10}{\cos(\theta - 45^\circ)}\) at \(\theta = 45^\circ\).
Given:
\(r_0 = 10\,m\), \(\alpha = 45^\circ\), \(\theta = 45^\circ\)
$$ r = \frac{10}{\cos(45^\circ - 45^\circ)} = \frac{10}{\cos 0^\circ} = \frac{10}{1} = 10\,m $$
Answer: 10 m
Example 4:
Find the Cartesian coordinates of a point on the line \(r = \frac{20}{\cos(\theta - 30^\circ)}\) at \(\theta = 60^\circ\).
Given:
\(r_0=20\,m\), \(\alpha=30^\circ\), \(\theta=60^\circ\)
Calculate \(r\):
$$ r = \frac{20}{\cos(60^\circ - 30^\circ)} = \frac{20}{\cos 30^\circ} = \frac{20}{0.8660} = 23.09\,m $$
Calculate Cartesian coordinates:
$$ x = r \cos \theta = 23.09 \times \cos 60^\circ = 23.09 \times 0.5 = 11.545\,m $$$$ y = r \sin \theta = 23.09 \times \sin 60^\circ = 23.09 \times 0.8660 = 20.0\,m $$
Answer: (11.545 m, 20.0 m)
Example 5:
Determine the polar equation of a line 12 m from the origin at an angle of \(135^\circ\).
Given:
\(r_0=12\,m\), \(\alpha=135^\circ\)
$$ r = \frac{12}{\cos(\theta - 135^\circ)} $$
Answer: \(r = \frac{12}{\cos(\theta - 135^\circ)}\)
Calculating distances between points or lengths of pipeline segments in polar coordinates is essential for cost estimation and design.
Distance between two points \((r_1, \theta_1)\) and \((r_2, \theta_2)\) is given by:
$$ d = \sqrt{r_1^2 + r_2^2 - 2 r_1 r_2 \cos(\theta_2 - \theta_1)} $$
Example 1:
Find the distance between two water tanks at \((30\,m, 45^\circ)\) and \((40\,m, 60^\circ)\).
Given:
\(r_1=30\,m\), \(\theta_1=45^\circ\), \(r_2=40\,m\), \(\theta_2=60^\circ\)
$$ d = \sqrt{30^2 + 40^2 - 2 \times 30 \times 40 \times \cos(60^\circ - 45^\circ)} $$$$ d = \sqrt{900 + 1600 - 2400 \times \cos 15^\circ} $$$$ d = \sqrt{2500 - 2400 \times 0.9659} = \sqrt{2500 - 2318.16} = \sqrt{181.84} = 13.49\,m $$
Answer: 13.49 m
Example 2:
Calculate the distance between two inspection points at \((50\,m, 120^\circ)\) and \((60\,m, 150^\circ)\).
Given:
\(r_1=50\,m\), \(\theta_1=120^\circ\), \(r_2=60\,m\), \(\theta_2=150^\circ\)
$$ d = \sqrt{50^2 + 60^2 - 2 \times 50 \times 60 \times \cos(150^\circ - 120^\circ)} $$$$ d = \sqrt{2500 + 3600 - 6000 \times \cos 30^\circ} $$$$ d = \sqrt{6100 - 6000 \times 0.8660} = \sqrt{6100 - 5196} = \sqrt{904} = 30.07\,m $$
Answer: 30.07 m
Example 3:
Find the distance between points \((25\,m, 0^\circ)\) and \((25\,m, 180^\circ)\).
Given:
\(r_1=25\,m\), \(\theta_1=0^\circ\), \(r_2=25\,m\), \(\theta_2=180^\circ\)
$$ d = \sqrt{25^2 + 25^2 - 2 \times 25 \times 25 \times \cos(180^\circ - 0^\circ)} $$$$ d = \sqrt{625 + 625 - 1250 \times \cos 180^\circ} $$$$ d = \sqrt{1250 - 1250 \times (-1)} = \sqrt{1250 + 1250} = \sqrt{2500} = 50\,m $$
Answer: 50 m
Example 4:
Calculate distance between points \((10\,m, 270^\circ)\) and \((20\,m, 90^\circ)\).
Given:
\(r_1=10\,m\), \(\theta_1=270^\circ\), \(r_2=20\,m\), \(\theta_2=90^\circ\)
$$ d = \sqrt{10^2 + 20^2 - 2 \times 10 \times 20 \times \cos(90^\circ - 270^\circ)} $$$$ d = \sqrt{100 + 400 - 400 \times \cos(-180^\circ)} = \sqrt{500 - 400 \times \cos 180^\circ} $$$$ d = \sqrt{500 - 400 \times (-1)} = \sqrt{500 + 400} = \sqrt{900} = 30\,m $$
Answer: 30 m
Example 5:
Find the distance between two points \((35\,m, 10^\circ)\) and \((45\,m, 80^\circ)\).
Given:
\(r_1=35\,m\), \(\theta_1=10^\circ\), \(r_2=45\,m\), \(\theta_2=80^\circ\)
$$ d = \sqrt{35^2 + 45^2 - 2 \times 35 \times 45 \times \cos(80^\circ - 10^\circ)} $$$$ d = \sqrt{1225 + 2025 - 3150 \times \cos 70^\circ} $$$$ d = \sqrt{3250 - 3150 \times 0.3420} = \sqrt{3250 - 1077.3} = \sqrt{2172.7} = 46.61\,m $$
Answer: 46.61 m
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Create a free accountThis chapter explored the application of coordinate geometry principles starting with polar equations, where points are represented by their distance from the origin and angle from a fixed direction. It then examined Cartesian equations, focusing on expressing geometric figures using the familiar \(x\) and \(y\) coordinates. The chapter further developed understanding by analyzing graphs of polar equations, illustrating how these can be plotted and interpreted. Moving into calculus-related geometry, the concepts of normals and tangents were introduced, explaining how to find lines perpendicular or touching a curve at a single point. The definition of a point was revisited to reinforce its fundamental role in geometry. The locus of a point was studied with exercises relating specifically to circles, highlighting the set of points satisfying given conditions. Finally, the chapter applied these ideas to real-world mechanisms by investigating loci of points generated by moving parts, deepening the connection between geometry and engineering contexts.
Find the polar coordinates \((r, \theta)\) of a point located 100 m east and 100 m north from a water pipeline junction. (2 marks)
A water storage tank is positioned at the origin. A leak is detected at a point whose Cartesian coordinates are \((60, 80)\) meters. Calculate the polar coordinates of the leak location. (2 marks)
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