By the end of this chapter, you will be able to:
These skills are essential because they help you understand and visualize engineering problems clearly, making your work more precise and effective in the real trade.
Coordinate geometry is fundamental in land surveying for accurately determining positions and distances on the earth’s surface. Surveyors use coordinate systems to translate field measurements into precise plot points, enabling the mapping and layout of parcels, roads, and infrastructure. This chapter focuses on the application of polar equations in coordinate geometry, a method vital for converting between angular and linear measurements commonly encountered in surveying tasks across Kenya.
Polar coordinate geometry expresses points in terms of a distance from a fixed origin and an angle from a reference direction. This system is particularly useful in land surveying when dealing with bearings and distances from a known station. Understanding how to manipulate polar equations allows surveyors to compute coordinates from field observations and vice versa.
Polar coordinates define a point by its distance \( r \) from the origin and the angle \( \theta \) measured from the positive x-axis (or a reference direction such as north in surveying). The relationship between polar and Cartesian coordinates \((x,y)\) is given by:
$$ x = r \cos \theta $$
$$ y = r \sin \theta $$
where \( r \geq 0 \) and \( \theta \) is measured in degrees or radians.
Example 1: A surveyor measures a point 50 m from the origin at a bearing of 30°. Find the Cartesian coordinates.
Given: \( r = 50\,m \), \( \theta = 30^\circ \)
$$ x = r \cos \theta $$
$$ x = 50 \cos 30^\circ $$
$$ x = 50 \times 0.8660 = 43.30\,m $$
$$ y = r \sin \theta $$
$$ y = 50 \sin 30^\circ $$
$$ y = 50 \times 0.5 = 25.00\,m $$
Answer: (x, y) = (43.30\,m, 25.00\,m)
Example 2: Convert the point with Cartesian coordinates (35 m, 35 m) to polar coordinates \((r, \theta)\).
Given: \( x = 35\,m \), \( y = 35\,m \)
$$ r = \sqrt{x^2 + y^2} $$
$$ r = \sqrt{35^2 + 35^2} = \sqrt{1225 + 1225} = \sqrt{2450} $$
$$ r = 49.50\,m $$
$$ \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \frac{35}{35} = \tan^{-1} 1 = 45^\circ $$
Answer: \(r = 49.50\,m, \theta = 45^\circ\)
Example 3: A point lies 70 m from the origin at an angle of 120°. Find its Cartesian coordinates.
Given: \( r = 70\,m \), \( \theta = 120^\circ \)
$$ x = 70 \cos 120^\circ = 70 \times (-0.5) = -35.00\,m $$
$$ y = 70 \sin 120^\circ = 70 \times 0.8660 = 60.62\,m $$
Answer: (x, y) = (-35.00\,m, 60.62\,m)
Surveyors frequently convert between polar and Cartesian forms to plot points on maps or calculate distances and bearings from coordinates. The general formulas are:
From polar to Cartesian:
$$ x = r \cos \theta $$
$$ y = r \sin \theta $$
From Cartesian to polar:
$$ r = \sqrt{x^2 + y^2} $$
$$ \theta = \tan^{-1} \frac{y}{x} $$
where \( \theta \) is adjusted to the correct quadrant based on the signs of \( x \) and \( y \).
Example 1: A point has polar coordinates \( r = 100\,m \), \( \theta = 225^\circ \). Find its Cartesian coordinates.
Given: \( r = 100\,m \), \( \theta = 225^\circ \)
$$ x = 100 \cos 225^\circ = 100 \times (-0.7071) = -70.71\,m $$
$$ y = 100 \sin 225^\circ = 100 \times (-0.7071) = -70.71\,m $$
Answer: (x, y) = (-70.71\,m, -70.71\,m)
Example 2: Convert Cartesian coordinates (-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 \( y > 0 \), the point lies in the second quadrant, so:
$$ \theta = 180^\circ - 36.87^\circ = 143.13^\circ $$
Answer: \( r = 50\,m, \theta = 143.13^\circ \)
Example 3: A surveyor records Cartesian coordinates (0 m, -25 m). Find the polar coordinates.
Given: \( x = 0\,m \), \( y = -25\,m \)
$$ r = \sqrt{0^2 + (-25)^2} = 25\,m $$
Since \( x = 0 \) and \( y < 0 \), the angle is:
$$ \theta = 270^\circ $$
Answer: \( r = 25\,m, \theta = 270^\circ \)
Traverse surveying involves measuring a series of connected lines with known lengths and bearings. Polar equations allow surveyors to compute the coordinates of traverse points starting from a known control point. This is essential for plotting boundaries, roads, and infrastructure layouts.
Given a starting point \((x_0, y_0)\), bearing \(\theta\), and distance \(r\), the coordinates of the next point \((x_1, y_1)\) are:
$$ x_1 = x_0 + r \cos \theta $$
$$ y_1 = y_0 + r \sin \theta $$
Example 1: Starting at point (100 m, 100 m), a traverse line is measured at a bearing of 60° for 40 m. Find the coordinates of the next point.
Given: \( x_0 = 100\,m \), \( y_0 = 100\,m \), \( r = 40\,m \), \( \theta = 60^\circ \)
$$ x_1 = 100 + 40 \cos 60^\circ = 100 + 40 \times 0.5 = 120\,m $$
$$ y_1 = 100 + 40 \sin 60^\circ = 100 + 40 \times 0.8660 = 134.64\,m $$
Answer: (x_1, y_1) = (120\,m, 134.64\,m)
Example 2: From point (150 m, 200 m), a traverse line is measured with distance 75 m and bearing 135°. Find the next point coordinates.
Given: \( x_0 = 150\,m \), \( y_0 = 200\,m \), \( r = 75\,m \), \( \theta = 135^\circ \)
$$ x_1 = 150 + 75 \cos 135^\circ = 150 + 75 \times (-0.7071) = 150 - 53.03 = 96.97\,m $$
$$ y_1 = 200 + 75 \sin 135^\circ = 200 + 75 \times 0.7071 = 200 + 53.03 = 253.03\,m $$
Answer: (x_1, y_1) = (96.97\,m, 253.03\,m)
Example 3: A traverse starts at (0, 0). The first line is 30 m at 270°, the second line is 40 m at 0°. Find the coordinates after the second line.
Step 1: First point
Given: \( x_0 = 0 \), \( y_0 = 0 \), \( r = 30\,m \), \( \theta = 270^\circ \)
$$ x_1 = 0 + 30 \cos 270^\circ = 0 + 30 \times 0 = 0\,m $$
$$ y_1 = 0 + 30 \sin 270^\circ = 0 + 30 \times (-1) = -30\,m $$
Step 2: Second point
Given: \( x_1 = 0 \), \( y_1 = -30 \), \( r = 40\,m \), \( \theta = 0^\circ \)
$$ x_2 = 0 + 40 \cos 0^\circ = 0 + 40 \times 1 = 40\,m $$
$$ y_2 = -30 + 40 \sin 0^\circ = -30 + 40 \times 0 = -30\,m $$
Answer: (x_2, y_2) = (40\,m, -30\,m)
Once the coordinates of two points are known, the distance and bearing between them are found using polar equations. This is crucial for verifying field measurements and for map plotting.
Distance \( d \) between points \((x_1, y_1)\) and \((x_2, y_2)\):
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
Bearing \( \theta \) from point 1 to point 2 is:
$$ \theta = \tan^{-1} \frac{y_2 - y_1}{x_2 - x_1} $$
Adjust \( \theta \) to the correct quadrant based on the signs of \( (x_2 - x_1) \) and \( (y_2 - y_1) \).
Example 1: Find the distance and bearing from (100 m, 100 m) to (130 m, 160 m).
Given: \( (x_1, y_1) = (100, 100) \), \( (x_2, y_2) = (130, 160) \)
$$ d = \sqrt{(130 - 100)^2 + (160 - 100)^2} = \sqrt{30^2 + 60^2} = \sqrt{900 + 3600} = \sqrt{4500} $$
$$ d = 67.08\,m $$
$$ \theta = \tan^{-1} \frac{160 - 100}{130 - 100} = \tan^{-1} \frac{60}{30} = \tan^{-1} 2 = 63.43^\circ $$
Answer: Distance = 67.08 m, Bearing = 63.43°
Example 2: Calculate distance and bearing from (50 m, 75 m) to (20 m, 100 m).
Given: \( (x_1, y_1) = (50, 75) \), \( (x_2, y_2) = (20, 100) \)
$$ d = \sqrt{(20 - 50)^2 + (100 - 75)^2} = \sqrt{(-30)^2 + 25^2} = \sqrt{900 + 625} = \sqrt{1525} $$
$$ d = 39.05\,m $$
$$ \theta = \tan^{-1} \frac{100 - 75}{20 - 50} = \tan^{-1} \frac{25}{-30} = \tan^{-1} (-0.8333) = -39.80^\circ $$
Since \( x_2 - x_1 < 0 \) and \( y_2 - y_1 > 0 \), the point lies in the second quadrant:
$$ \theta = 180^\circ - 39.80^\circ = 140.20^\circ $$
Answer: Distance = 39.05 m, Bearing = 140.20°
Example 3: Determine the distance and bearing from (0 m, 0 m) to (-40 m, -40 m).
Given: \( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (-40, -40) \)
$$ d = \sqrt{(-40 - 0)^2 + (-40 - 0)^2} = \sqrt{1600 + 1600} = \sqrt{3200} $$
$$ d = 56.57\,m $$
$$ \theta = \tan^{-1} \frac{-40}{-40} = \tan^{-1} 1 = 45^\circ $$
Since both differences are negative, the point lies in the third quadrant:
$$ \theta = 180^\circ + 45^\circ = 225^\circ $$
Answer: Distance = 56.57 m, Bearing = 225°
A surveyor measures a point at a distance of 80 m and a bearing of 150°. Calculate its Cartesian coordinates. (5 marks)
Convert the Cartesian coordinates (60 m, -45 m) to polar coordinates. (5 marks)
From a known point (200 m, 150 m), a traverse line is laid at bearing 300° for 90 m. Find the coordinates of the new point. (6 marks)
Calculate the distance and bearing from point (120 m, 100 m) to (80 m, 140 m). (6 marks)
A traverse starts at (0,0). The first line is 50 m at 45°, the second line is 70 m at 135°. Find the coordinates after the second line. (8 marks)
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter explored the fundamentals of polar equations, introducing how points in a plane can be represented using the distance from the origin and the angle from the polar axis. It explained the conversion between polar and Cartesian coordinates, enabling a deeper understanding of spatial relationships. The chapter then focused on the graphs of polar equations, showing how to plot various curves such as circles, spirals, and roses by interpreting their polar forms. Techniques for sketching these graphs were demonstrated, highlighting the unique features of polar plots compared to Cartesian graphs. Finally, the chapter addressed the determination of normals and tangents to curves expressed in polar coordinates, providing methods to calculate their slopes and equations. This included step-by-step procedures for finding the tangent lines and normal lines at specific points on polar graphs. Overall, the chapter provided essential tools for analyzing and graphing curves using polar coordinates and understanding their geometric properties through tangents and normals.
A point in polar coordinates is given by \(r = 5\) m and \(\theta = 30^\circ\). Calculate its Cartesian coordinates \((x, y)\). (2 marks)
Convert the Cartesian coordinates \((x, y) = (4, 4\sqrt{3})\) m to polar coordinates \((r, \theta)\). (3 marks)
At the start of this chapter we promised you would be able to:
Tick each one you can genuinely do.
So, are you there yet?
You're competent when you can confidently do 50% or more of what this chapter promised.
Sign in to record how you're doing.