By the end of this chapter, you will be able to: - accurately calculate polar equations using coordinate geometry principles - correctly plot graphs of polar equations by applying coordinate geometry techniques - draw normals and tangents with precision, following coordinate geometry standards
Mastering these skills will help you solve real-world engineering problems that require precise geometric analysis and representation.
Coordinate geometry is fundamental in land surveying for accurately locating points and features on the earth’s surface. Surveyors use coordinate systems to express positions in two-dimensional or three-dimensional space, enabling precise mapping and layout of land parcels, infrastructure, and construction projects. This chapter focuses on polar coordinate equations, which relate distances and angles from a fixed point, a method frequently applied in traversing and curve setting in Kenyan surveying practice.
Polar coordinate geometry expresses points in terms of a distance from a reference origin and an angle from a reference direction. This is particularly useful in land surveying when plotting points based on measured distances and bearing angles from a known station. The fundamental polar equation relates the rectangular coordinates \((x, y)\) to the polar coordinates \((r, \theta)\) by the equations:
$$ x = r \cos \theta $$$$ y = r \sin \theta $$
where \(r\) is the distance from the origin to the point, and \(\theta\) is the angle measured from the positive x-axis (usually East in surveying).
Conversion between polar and rectangular coordinates is essential in surveying computations, especially when coordinates obtained in the field as distances and bearings must be converted to Cartesian coordinates for mapping or further calculations.
The governing formulas are:
From polar to rectangular:$$ x = r \cos \theta $$$$ y = r \sin \theta $$
From rectangular to polar:$$ r = \sqrt{x^{2} + y^{2}} $$$$ \theta = \tan^{-1} (\frac{y}{x}) $$
Example 1: A surveyor measures a point 50 m from a reference station at a bearing angle of 30°. Find the rectangular coordinates \(x\) and \(y\) assuming the reference station is at the origin.
Given: \(r = 50\) m, \(\theta = 30^\circ\)
$$ x = 50 \cos 30^\circ $$$$ x = 50 \times 0.866 $$$$ x = 43.3 \text{ m} $$
$$ y = 50 \sin 30^\circ $$$$ y = 50 \times 0.5 $$$$ y = 25.0 \text{ m} $$
Answer: \(x = 43.3\) m, \(y = 25.0\) m
Example 2: A point has rectangular coordinates \(x = 40\) m and \(y = 30\) m. Find the polar coordinates \(r\) and \(\theta\).
Given: \(x = 40\) m, \(y = 30\) m
$$ r = \sqrt{40^{2} + 30^{2}} $$$$ r = \sqrt{1600 + 900} $$$$ r = \sqrt{2500} $$$$ r = 50 \text{ m} $$
$$ \theta = \tan^{-1} (\frac{30}{40}) $$$$ \theta = \tan^{-1} (0.75) $$$$ \theta = 36.87^\circ $$
Answer: \(r = 50\) m, \(\theta = 36.87^\circ\)
Example 3: A survey point is located at a distance of 75 m from the origin with an angle of 135°. Find the coordinates \(x\) and \(y\).
Given: \(r = 75\) m, \(\theta = 135^\circ\)
$$ x = 75 \cos 135^\circ $$$$ x = 75 \times (-0.707) $$$$ x = -53.0 \text{ m} $$
$$ y = 75 \sin 135^\circ $$$$ y = 75 \times 0.707 $$$$ y = 53.0 \text{ m} $$
Answer: \(x = -53.0\) m, \(y = 53.0\) m
Example 4: Given rectangular coordinates \(x = -25\) m and \(y = -25\sqrt{3}\) m, find the polar coordinates \(r\) and \(\theta\).
Given: \(x = -25\) m, \(y = -25\sqrt{3} \approx -43.3\) m
$$ r = \sqrt{(-25)^{2} + (-43.3)^{2}} $$$$ r = \sqrt{625 + 1875} $$$$ r = \sqrt{2500} $$$$ r = 50 \text{ m} $$
$$ \theta = \tan^{-1} (\frac{-43.3}{-25}) $$$$ \theta = \tan^{-1} (1.732) $$$$ \theta = 60^\circ $$
Since both \(x\) and \(y\) are negative, the point is in the third quadrant, so
$$ \theta = 180^\circ + 60^\circ = 240^\circ $$
Answer: \(r = 50\) m, \(\theta = 240^\circ\)
Example 5: A surveyor measures a point at distance 100 m and angle 270°. Calculate the rectangular coordinates.
Given: \(r = 100\) m, \(\theta = 270^\circ\)
$$ x = 100 \cos 270^\circ $$$$ x = 100 \times 0 $$$$ x = 0 \text{ m} $$
$$ y = 100 \sin 270^\circ $$$$ y = 100 \times (-1) $$$$ y = -100 \text{ m} $$
Answer: \(x = 0\) m, \(y = -100\) m
In land surveying, traversing involves measuring a series of connected lines with known distances and bearings. Polar equations are used to compute the coordinates of traverse points from a known starting point.
The formulas for coordinates of a traverse point \(i\) relative to the previous point \(i-1\) are:
$$ x_i = x_{i-1} + d_i \cos \theta_i $$$$ y_i = y_{i-1} + d_i \sin \theta_i $$
where \(d_i\) is the length of the traverse line and \(\theta_i\) is its bearing.
Example 1: Starting at point \(P_0\) with coordinates \(x_0 = 100\) m, \(y_0 = 200\) m, a traverse line is measured with length 50 m and bearing 45°. Find coordinates of point \(P_1\).
Given: \(x_0 = 100\) m, \(y_0 = 200\) m, \(d_1 = 50\) m, \(\theta_1 = 45^\circ\)
$$ x_1 = 100 + 50 \cos 45^\circ $$$$ x_1 = 100 + 50 \times 0.707 $$$$ x_1 = 100 + 35.35 $$$$ x_1 = 135.35 \text{ m} $$
$$ y_1 = 200 + 50 \sin 45^\circ $$$$ y_1 = 200 + 50 \times 0.707 $$$$ y_1 = 200 + 35.35 $$$$ y_1 = 235.35 \text{ m} $$
Answer: \(x_1 = 135.35\) m, \(y_1 = 235.35\) m
Example 2: From point \(P_1\) at \(x_1 = 135.35\) m, \(y_1 = 235.35\) m, another traverse line is measured with length 40 m and bearing 120°. Find coordinates of point \(P_2\).
Given: \(x_1 = 135.35\) m, \(y_1 = 235.35\) m, \(d_2 = 40\) m, \(\theta_2 = 120^\circ\)
$$ x_2 = 135.35 + 40 \cos 120^\circ $$$$ x_2 = 135.35 + 40 \times (-0.5) $$$$ x_2 = 135.35 - 20 $$$$ x_2 = 115.35 \text{ m} $$
$$ y_2 = 235.35 + 40 \sin 120^\circ $$$$ y_2 = 235.35 + 40 \times 0.866 $$$$ y_2 = 235.35 + 34.64 $$$$ y_2 = 269.99 \text{ m} $$
Answer: \(x_2 = 115.35\) m, \(y_2 = 269.99\) m
Example 3: Given point \(P_2\) at \(x_2 = 115.35\) m, \(y_2 = 269.99\) m, a traverse line of length 60 m is measured at bearing 210°. Calculate coordinates of point \(P_3\).
Given: \(x_2 = 115.35\) m, \(y_2 = 269.99\) m, \(d_3 = 60\) m, \(\theta_3 = 210^\circ\)
$$ x_3 = 115.35 + 60 \cos 210^\circ $$$$ x_3 = 115.35 + 60 \times (-0.866) $$$$ x_3 = 115.35 - 51.96 $$$$ x_3 = 63.39 \text{ m} $$
$$ y_3 = 269.99 + 60 \sin 210^\circ $$$$ y_3 = 269.99 + 60 \times (-0.5) $$$$ y_3 = 269.99 - 30 $$$$ y_3 = 239.99 \text{ m} $$
Answer: \(x_3 = 63.39\) m, \(y_3 = 239.99\) m
Example 4: From \(P_3\) at \(x_3 = 63.39\) m, \(y_3 = 239.99\) m, a traverse line 70 m long is measured on bearing 300°. Find coordinates of point \(P_4\).
Given: \(x_3 = 63.39\) m, \(y_3 = 239.99\) m, \(d_4 = 70\) m, \(\theta_4 = 300^\circ\)
$$ x_4 = 63.39 + 70 \cos 300^\circ $$$$ x_4 = 63.39 + 70 \times 0.5 $$$$ x_4 = 63.39 + 35 $$$$ x_4 = 98.39 \text{ m} $$
$$ y_4 = 239.99 + 70 \sin 300^\circ $$$$ y_4 = 239.99 + 70 \times (-0.866) $$$$ y_4 = 239.99 - 60.62 $$$$ y_4 = 179.37 \text{ m} $$
Answer: \(x_4 = 98.39\) m, \(y_4 = 179.37\) m
Example 5: Calculate the coordinates of \(P_5\) from \(P_4\) at \(x_4 = 98.39\) m, \(y_4 = 179.37\) m, with traverse line length 80 m and bearing 15°.
Given: \(x_4 = 98.39\) m, \(y_4 = 179.37\) m, \(d_5 = 80\) m, \(\theta_5 = 15^\circ\)
$$ x_5 = 98.39 + 80 \cos 15^\circ $$$$ x_5 = 98.39 + 80 \times 0.966 $$$$ x_5 = 98.39 + 77.28 $$$$ x_5 = 175.67 \text{ m} $$
$$ y_5 = 179.37 + 80 \sin 15^\circ $$$$ y_5 = 179.37 + 80 \times 0.259 $$$$ y_5 = 179.37 + 20.72 $$$$ y_5 = 200.09 \text{ m} $$
Answer: \(x_5 = 175.67\) m, \(y_5 = 200.09\) m
Surveyors often need to determine the distance and bearing between two known points with given rectangular coordinates. The formulas for distance \(d\) and bearing \(\theta\) between points \(P_1(x_1, y_1)\) and \(P_2(x_2, y_2)\) are:
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$$$ \theta = \tan^{-1} (\frac{y_2 - y_1}{x_2 - x_1}) $$
The bearing \(\theta\) must be adjusted according to the quadrant of the vector from \(P_1\) to \(P_2\).
Example 1: Calculate distance and bearing from \(P_1(100, 200)\) m to \(P_2(150, 250)\) m.
Given: \(x_1 = 100\) m, \(y_1 = 200\) m, \(x_2 = 150\) m, \(y_2 = 250\) m
$$ d = \sqrt{(150 - 100)^2 + (250 - 200)^2} $$$$ d = \sqrt{50^2 + 50^2} $$$$ d = \sqrt{2500 + 2500} $$$$ d = \sqrt{5000} $$$$ d = 70.71 \text{ m} $$
$$ \theta = \tan^{-1} (\frac{250 - 200}{150 - 100}) $$$$ \theta = \tan^{-1} (\frac{50}{50}) $$$$ \theta = \tan^{-1} (1) $$$$ \theta = 45^\circ $$
Since the change in \(x\) and \(y\) is positive, the bearing is \(45^\circ\).
Answer: Distance = 70.71 m, Bearing = 45°
Example 2: Find distance and bearing from \(P_1(200, 300)\) m to \(P_2(150, 250)\) m.
Given: \(x_1 = 200\) m, \(y_1 = 300\) m, \(x_2 = 150\) m, \(y_2 = 250\) m
$$ d = \sqrt{(150 - 200)^2 + (250 - 300)^2} $$$$ d = \sqrt{(-50)^2 + (-50)^2} $$$$ d = \sqrt{2500 + 2500} $$$$ d = \sqrt{5000} $$$$ d = 70.71 \text{ m} $$
$$ \theta = \tan^{-1} (\frac{250 - 300}{150 - 200}) $$$$ \theta = \tan^{-1} (\frac{-50}{-50}) $$$$ \theta = \tan^{-1} (1) $$$$ \theta = 45^\circ $$
Both \(\Delta x\) and \(\Delta y\) are negative, so the bearing is in the third quadrant:
$$ \theta = 180^\circ + 45^\circ = 225^\circ $$
Answer: Distance = 70.71 m, Bearing = 225°
Example 3: Calculate distance and bearing from \(P_1(300, 400)\) m to \(P_2(350, 350)\) m.
Given: \(x_1 = 300\) m, \(y_1 = 400\) m, \(x_2 = 350\) m, \(y_2 = 350\) m
$$ d = \sqrt{(350 - 300)^2 + (350 - 400)^2} $$$$ d = \sqrt{50^2 + (-50)^2} $$$$ d = \sqrt{2500 + 2500} $$$$ d = \sqrt{5000} $$$$ d = 70.71 \text{ m} $$
$$ \theta = \tan^{-1} (\frac{350 - 400}{350 - 300}) $$$$ \theta = \tan^{-1} (\frac{-50}{50}) $$$$ \theta = \tan^{-1} (-1) $$$$ \theta = -45^\circ $$
Since \(\Delta x > 0\) and \(\Delta y < 0\), the bearing is in the fourth quadrant:
$$ \theta = 360^\circ - 45^\circ = 315^\circ $$
Answer: Distance = 70.71 m, Bearing = 315°
Example 4: Find distance and bearing from \(P_1(400, 500)\) m to \(P_2(350, 550)\) m.
Given: \(x_1 = 400\) m, \(y_1 = 500\) m, \(x_2 = 350\) m, \(y_2 = 550\) m
$$ d = \sqrt{(350 - 400)^2 + (550 - 500)^2} $$$$ d = \sqrt{(-50)^2 + 50^2} $$$$ d = \sqrt{2500 + 2500} $$$$ d = \sqrt{5000} $$$$ d = 70.71 \text{ m} $$
$$ \theta = \tan^{-1} (\frac{550 - 500}{350 - 400}) $$$$ \theta = \tan^{-1} (\frac{50}{-50}) $$$$ \theta = \tan^{-1} (-1) $$$$ \theta = -45^\circ $$
Since \(\Delta x < 0\) and \(\Delta y > 0\), the bearing is in the second quadrant:
$$ \theta = 180^\circ - 45^\circ = 135^\circ $$
Answer: Distance = 70.71 m, Bearing = 135°
Example 5: Calculate distance and bearing from \(P_1(0, 0)\) to \(P_2(-30, -40)\).
Given: \(x_1 = 0\), \(y_1 = 0\), \(x_2 = -30\), \(y_2 = -40\)
$$ d = \sqrt{(-30 - 0)^2 + (-40 - 0)^2} $$$$ d = \sqrt{900 + 1600} $$$$ d = \sqrt{2500} $$$$ d = 50 \text{ m} $$
$$ \theta = \tan^{-1} (\frac{-40 - 0}{-30 - 0}) $$$$ \theta = \tan^{-1} (\frac{-40}{-30}) $$$$ \theta = \tan^{-1} (1.333) $$$$ \theta = 53.13^\circ $$
Both \(\Delta x\) and \(\Delta y\) are negative, so bearing is in the third quadrant:
$$ \theta = 180^\circ + 53.13^\circ = 233.13^\circ $$
Answer: Distance = 50 m, Bearing = 233.13°
A point is located 60 m at a bearing of 75°. Find the rectangular coordinates \(x\) and \(y\). (5 marks)
Calculate the coordinates of a traverse point starting at (120 m, 250 m) with a traverse line length of 45 m at bearing 135°. (5 marks)
Given points \(P_1(200, 300)\) m and \(P_2(260, 360)\) m, compute the distance and bearing from \(P_1\) to \(P_2\). (5 marks)
Starting at \(P_0(0,0)\), a traverse has lines: 50 m at 0°, 40 m at 90°, 30 m at 180°, 40 m at 270°. Calculate the coordinates of the final point. (10 marks)
From point \(P_3(80, 100)\) m, a line is measured 70 m long at a bearing of 225°. Find the coordinates of the next traverse point. (5 marks)
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Create a free accountThis chapter introduced the concept of polar equations, explaining how points in the plane can be represented using a distance from the origin and an angle from the positive x-axis. It progressed to graphing various polar equations, demonstrating how to plot curves by converting polar coordinates into Cartesian coordinates and interpreting their shapes. The chapter then explored the methods for finding normals and tangents to curves, emphasizing their importance in understanding the geometry of curves. Techniques for calculating the slope of tangents and equations of normal lines in both Cartesian and polar forms were covered. The relationships between the angle of the tangent and the polar equation were analyzed to provide a comprehensive approach to curve analysis. Through worked examples, the chapter developed skills in applying these concepts to solve real-world engineering problems involving curves and their properties. Overall, the chapter built a strong foundation in coordinate geometry using polar coordinates and the geometric interpretation of curves.
A surveyor measures a point in polar coordinates as \(r = 50\,m\) and \(\theta = 60^\circ\). Calculate the Cartesian coordinates \((x, y)\) of the point. (2 marks)
Convert the Cartesian point \((x, y) = (30\,m, 40\,m)\) into polar coordinates \((r, \theta)\). (2 marks)
Type: Individual
| Tools & Equipment | Materials |
|---|---|
| Scientific calculator | Polar graph paper (A3 size) |
| Ruler (scale rule) | Pencil and rubber |
| Protractor | |
| Drawing board |
| S/N | Item | Quantity |
|---|---|---|
| 1 | Polar graph paper (A3 size) | 1 sheet per Candidate |
| 2 | Scientific calculator | 1 per Candidate |
| 3 | Pencil and rubber | 1 set per Candidate |
| 4 | Ruler (scale rule) | 1 per Candidate |
| 5 | Protractor | 1 per Candidate |
| 6 | Drawing board | 1 per Candidate |
| Items to be Evaluated | Marks Available | Marks Obtained | Comments |
|---|---|---|---|
| TASK 1: Preparation and PPE | |||
| Identified and used PPE appropriately (dust coat/apron, closed shoes) (Award 1 mark for correct PPE usage or zero) | 1 | ||
| Selected suitable drawing site and arranged tools (Award 1 mark for correctly selecting a clean, flat drawing surface and arranging tools) | 1 | ||
| Checked and prepared polar graph paper for plotting (Award 1 mark for preparing graph paper properly) | 1 | ||
| Set the scale correctly as 1cm = 1 unit (Award 1 mark for correct scale setting) | 1 | ||
| Identified and understood the two given polar equations (Award 2 marks for correct interpretation of both equations) | 2 | ||
| Sub-Total | 6 | ||
| TASK 2: Plotting the Polar Equation r = 4(1 + cosθ) | |||
| Computed r values correctly for θ = 0°, 30°, 60°, ..., 360° (Award 1 mark for each correct set of computed values, max 3 marks) | 3 | ||
| Plotted points accurately on the polar graph paper (Award 3 marks for accurate plotting within ±1 mm) | 3 | ||
| Connected plotted points smoothly to reveal the cardioid shape (Award 2 marks for neat and continuous curve) | 2 | ||
| Labelled the curve clearly with its equation (Award 1 mark for clear and correct labelling) | 1 | ||
| Sub-Total | 9 | ||
| TASK 3: Plotting the Polar Equation r = 3 sin(2θ) | |||
| Computed r values correctly for θ = 0°, 15°, 30°, ..., 360° (Award 1 mark for each correct set of computed values, max 3 marks) | 3 | ||
| Plotted points accurately on the polar graph paper (Award 3 marks for accurate plotting within ±1 mm) | 3 | ||
| Connected plotted points smoothly to reveal the four-petalled rose curve (Award 2 marks for neat and continuous curve) | 2 | ||
| Labelled the curve clearly with its equation (Award 1 mark for clear and correct labelling) | 1 | ||
| Sub-Total | 9 | ||
| TASK 4: Final Presentation and Clean-up | |||
| Used pencil and rubber neatly to correct any errors (Award 1 mark for neat corrections) | 1 | ||
| Maintained neatness and completeness of the plotted graphs (Award 2 marks for overall neatness and completeness) | 2 | ||
| Organized and cleaned the working area after completion (Award 1 mark for tidying up tools and materials) | 1 | ||
| Sub-Total | 4 | ||
| PRODUCT CHECKLIST | |||
| Polar graph paper correctly plotted with r=4(1+cosθ) curve matching a cardioid shape with labelled points and curve (Award 5 marks for correct shape, scale, and labelling) | 5 | ||
| Polar graph paper correctly plotted with r=3 sin(2θ) curve matching a four-petalled rose shape with labelled points and curve (Award 5 marks for correct shape, scale, and labelling) | 5 | ||
| Overall neatness, scale accuracy, and completeness of the plotted curves (Award 3 marks for neat, accurate, and complete plotting) | 3 | ||
| Sub-Total | 13 | ||
| GRAND TOTAL | 41 | ||
Type: Individual
| Tools & Equipment | Materials |
|---|---|
| 50 m steel tape | Drawing paper (A3 size) |
| Mallet | Pegs |
| Scientific Calculator | Apron/dust coat |
| Pencil and rubber | Boots/Closed shoes |
| Scale rule | |
| Straight edge |
| S/N | Item | Quantity |
|---|---|---|
| 1 | 50 m steel tape | 2 per Candidate |
| 2 | Mallet | 1 shared per 5 Candidates |
| 3 | Drawing paper (A3 size) | 1 per Candidate |
| 4 | Pegs | 20 per Candidate |
| 5 | Scientific Calculator | 1 per Candidate |
| 6 | Pencil and rubber | 1 per Candidate |
| 7 | Scale rule | 1 per Candidate |
| 8 | Straight edge | 1 per Candidate |
| 9 | Apron/dust coat | 1 per Candidate |
| 10 | Boots/Closed shoes | 1 pair per Candidate |
| Items to be Evaluated | Marks Available | Marks Obtained | Comments |
|---|---|---|---|
| TASK 1: Use of PPE and Preparation | |||
| Identified and wore apron/dust coat appropriately (Award 1 mark for correct use) | 1 | ||
| Wore boots/closed shoes appropriately (Award 1 mark for correct use) | 1 | ||
| Sub-Total | 2 | ||
| TASK 2: Selection of Site and Tools | |||
| Selected a suitable flat site for polar coordinate layout (Award 1 mark for correct site selection) | 1 | ||
| Selected appropriate tools from provided equipment (Award 1 mark for correct tool selection) | 1 | ||
| Sub-Total | 2 | ||
| TASK 3: Interpretation of Drawing | |||
| Correctly interpreted the polar coordinate axes layout drawing (Award 1 mark for correct interpretation) | 1 | ||
| Correctly marked polar axis (0° line) and reference points on paper (Award 1 mark for correct marking) | 1 | ||
| Identified and labelled points at given polar coordinates (Award 1 mark for correct identification) | 1 | ||
| Sub-Total | 3 | ||
| TASK 4: Setting Out Polar Coordinate Axes | |||
| Set out polar coordinate axes on ground with 5 m radius using steel tape and pegs (Award 2 marks for correct setting out) | 2 | ||
| Ensured axes intersect at origin point accurately (Award 1 mark for correct intersection) | 1 | ||
| Sub-Total | 3 | ||
| TASK 5: Locating Points Using Polar Coordinates | |||
| Computed Cartesian coordinates from given polar coordinates correctly (Award 2 marks for correct computation) | 2 | ||
| Measured and marked points on ground at correct radius and angle using tape and scale (Award 3 marks for correct location of points) | 3 | ||
| Sub-Total | 5 | ||
| TASK 6: Plotting Points on Drawing | |||
| Accurately plotted points on A3 drawing using scale rule and straight edge (Award 3 marks for accuracy and neatness) | 3 | ||
| Labelled points clearly with polar coordinates (Award 1 mark for clear labelling) | 1 | ||
| Sub-Total | 4 | ||
| PRODUCT CHECKLIST | |||
| Polar coordinate axes set out with 5 m radius ± 0.05 m accuracy (Award 2 marks for accuracy) | 2 | ||
| Points located on ground within ± 0.05 m radius and ± 2° angle tolerance (Award 3 marks for correct point location) | 3 | ||
| Points plotted on drawing at correct scaled positions within ± 2 mm tolerance (Award 2 marks for drawing accuracy) | 2 | ||
| Sub-Total | 7 | ||
| GRAND TOTAL | 26 | ||
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