WSKT 212 PEC

FUNCTIONS, TRIGONOMETRY AND ELEMENTARY
STATISTICS FOR FET TECHNOLOGY
STUDY GUIDE AND MANUAL FOR
WSKT 212 PEC
*WSKT212PEC*
FACULTY OF EDUCATION SCIENCES

Study guide compiled by:
Mr Rudi van de Venter
Page lay-out by Rudi van de Venter, Subject group Mathematics Education of the School of
Education Training: School of Natural Science Subjects
Printing and distribution handled by Distribution Centre
Printed by The Platinum Press (018)294 8879/(016) 981 9401
Copyright © 2011 edition. Revision date 2013.
North-West University, Potchefstroom campus.
Copyright reserved. No part of this book may be reproduced in any form or in any way
without the written permission of the publishers. This includes photocopying of all or part of
this book.

Contents
0 General information ....................................................................................................... 3
0.1 Word of welcome .................................................................................................................. 3
0.2 Contact person ...................................................................................................................... 4
0.3 Why do we study Technical Mathematics? ...................................................................... 4
0.4 Prerequisites ......................................................................................................................... 6
0.5 Study material ....................................................................................................................... 6
0.6 How to use this study guide ................................................................................................ 8
0.7 Monitoring of progress ....................................................................................................... 10
0.8 Assessment ......................................................................................................................... 11
0.9 Warning against plagiarism ............................................................................................... 14
0.10 The module plan ................................................................................................................. 15
0.11 Time schedule and study work program ......................................................................... 16
0.12 Module outcomes ............................................................................................................... 16
0.13 Important arrangements .................................................................................................... 17
1 Mathematical models and functions ........................................................................... 19
1.1 Radians as unit of measurement for angles ................................................................... 21
1.2 Polar co-ordinates .............................................................................................................. 22
1.3 Functions ............................................................................................................................. 23
2 Polynomial functions ................................................................................................... 39
2.1 Linear functions ................................................................................................................... 42
2.2 Quadratic functions ............................................................................................................ 63
2.3 Cubic functions ................................................................................................................... 84
3 Rational functions, exponential functions and logarithmic functions .................. 103
3.1 Rational functions ............................................................................................................. 108
3.2 Exponential functions ....................................................................................................... 122
3.3 Logarithmic functions ....................................................................................................... 132
1

4 Trigonometric functions ............................................................................................ 141
5 Conic sections ............................................................................................................ 161
6 Trigonometry .............................................................................................................. 177
6.1 Application of radian measure ........................................................................................ 178
6.2 Application of trigonometry to vectors ........................................................................... 180
6.3 The sine rule, the cosine rule and the area rule .......................................................... 182
7 Elementary decriptive statistics ................................................................................ 205
7.1 Frequency distributions .................................................................................................... 206
7.2 Mode, median and arithmetic mean .............................................................................. 225
7.3 The administration of computerized mark schedules using Microsoft Excel ........... 237
7.4 Regression using Microsoft Excel .................................................................................. 239
2

0 General information
0.1 Word of welcome
Welcome to this module. This is the second of three Technical Mathematics modules, of
which you will do one per year for the first three years of your study in the B.Ed. (FET
Technology) program. The contents of these three modules focus on real life applications of
mathematics within the technical subject area and in natural science.
As you have already done some mathematics at secondary school level, you will be familiar
with some of the topics we shall study. However, secondary school mathematics has a very
wide focus and you may have learned many things that you will never need again. Those
things were merely background knowledge for possible further study.
On the other hand, a teacher of the technical subject area or of science requires a fairly
advanced knowledge of mathematics in certain topics. Examples of such topics are:
• Units for measure and numbers
• Basic algebraic operations
• The solution of different kinds of equations (formulas)
• Vectors and complex numbers
• Functions and mathematical representations of real life processes (Mathematical
models)
• Graphical and geometrical representations of information
• Trigonometry
• Differential calculus
• Integral calculus
At school not nearly enough time has been devoted to these topics to equip you as a teacher
for the technical subject area. Some of these topics may not even have been treated at
school because it was supposed to be only relevant to those who would work in the technical
and scientific fields.
3

Therefore the North-West University presents you with three modules of Technical
Mathematics. This aims at empowering you, as technical subject teacher, so that you
will be able to handle the mathematics that you will come across in your subject area
with confidence.
0.2 Contact person
Mr. Rudi van de Venter
Office G 07
Educamus-building (B10)
018 2991859
10086013@nwu.ac.za
0.3 Why do we study Technical Mathematics?
There are many different ideas on what exactly mathematics is. Should you ask a few
people what they think mathematics is, the following two answers will often be given:
• Some reason (as the Greek philosopher Plato believed) that mathematics is something
supernatural given by the gods – it already exists “out there” – hidden in nature. Exceptionally
gifted and talented people that have been blessed may discover it and formulate it so that
ordinary people can use it without necessarily understanding it. These people believe that
mathematics is difficult and inaccessible and that it is normal that very few people will
understand it. They believe that a special talent is required in order to do mathematics as it is
always about difficult and abstract concepts, far removed from everyday life and that it does
not have much relevance for our ordinary existence.
• Others think (like the Greek philosopher Aristotle believed) that mathematics is something that
man makes, or designs. They regard mathematics as an attempt of mankind to describe and
understand what they observe and experience. It is a system of knowledge, ideas, rules and
skills that continues to grow and change as man works in his everyday life with mathematical
concepts like numbers, measurements, sketches, equations and relationships.
4

As we live in a world where length, width, height, temperature, time, money and the
interconnections between them influence our lives from day to day, these people think that no
special skill is required to do mathematics – because each of us is doing it subconsciously all
day long.
It is up to you to decide which of these views is the closest to your own ideas. But be warned:
The way you think about mathematics, determines largely what your approach to the study of
mathematics will be. In other words: If you have a negative attitude to mathematics, your
study of mathematics will be negatively influenced. The good news, however, is that any
attitude on earth can be changed, provided that one makes a deliberate decision to
change that attitude.
We suggest that you reflect thoroughly on the above ideas and also on the following:
You started to develop your own ideas about mathematics even when you were still an
infant. Even as a baby you could distinguish between different quantities (hungry, thirsty, hot,
cold, joy, pain, lots and little) although you may not have been able to assign a value to these
quantities. You learned to crawl on a planar surface (the floor); you learned to handle three
dimensional toys and to walk in a space. You became aware of numbers and quantity when
you learned to use your hands and fingers. From infancy you were able to figure out some
connection between cause and effect. As your hand-eye coordination improved, you
participated in sport where you learned to regulate processes by making use of
measurement or estimation.
The questions we leave you with, are the following:
• Are the abstractions encountered in “advanced” mathematics not simply elaborations
of the ideas that every normal person have encountered since childhood?
• If every person who can function well in everyday life and who can live a normal life
does possess a “built-in” basic mathematics ability – should it not then be possible
and even easy for everybody to do mathematics?
• How much of the perceptions that mathematics is difficult and can only be understood
by certain gifted individuals, is simply superstition?
5

What do we use Technical Mathematics for?
In the technical subject area we rely heavily on mathematical ideas and calculations. In the
course of this module you will become aware of the powerful tool mathematics is for solving
many problems in the technical and scientific disciplines.
All the mathematics treated in this module will be illustrated with real, practical examples, to
enable you to experience the mathematics within the context of reality. We shall assist you in
the cultivating of your further knowledge of mathematics by letting you experience the
technical and scientific contexts of the work (just like you have developed your basic
knowledge of mathematics through experience in everyday life).
This module on Technical Mathematics will therefore teach you to:
• Read mathematically
• Write mathematically
• Think mathematically.
Passing this mathematics module is a requirement for completing your degree successfully.
0.4 Prerequisites
Students registering for this module, have to be registered and selected as student teacher in
one of the technical fields. This implies that the student had Mathematics up to grade 12.
For the other prerequisites that may apply, you are referred to the program leader for the B.
Ed. (FET Technology) course.
0.5 Study material
• Prescribed Textbook
WASHINGTON, Allyn J. 2005. 8 th edition. Basic technical mathematics with calculus: SI
version. 8 th edition. Pearson Addison Wesley: Toronto
6

• Additional study material
Any other mathematics textbook, as well as the internet, of course, may be used for further
examples, applications and exercises.
You may perhaps benefit from a textbook for Grade 11 and Grade 12 Mathematics, although
the contents of the prescribed textbook of Washington are at a fairly basic level.
The “Know and Understand” series, with which many of you may be familiar, may also be
useful.
• Scientific calculator
You will require a good scientific calculator for each of the three Technical Mathematics
modules, one with the trigonometric functions, exponential functions and logarithmic
functions. The calculator that you used for Matric mathematics ought to be adequate.
Should you possess over a calculator with graphical facilities (the kind that draws graphs and
that usually is programmable also) you are welcome to use it in contact sessions and for
homework exercises to check your answers. However, note that programmable
calculators are not allowed in tests or in examinations, as we shall be testing specific
calculation skills.
• Access to The Geometer’s Sketchpad and Microsoft Office Excel
The Geometer’s Sketchpad is a graphical computer program which is available on all
computer laboratories of the NWU-PUK. The program may also be purchased for about
R350 via the subject group Mathematics Education. Microsoft Office Excel is part of the
Office package which is available on most Windows-based computers; this program is also
available on all computer laboratories of the NWU-PUK.
• Access to Groupwise-e-pos and eFundi
The lecturer communicates practically daily with you by means of e-mail and eFundi; it is
your responsibility to see to it that you have full access to both these facilities from the first
day of the semester.
7

0.6 How to use this study guide
This is an example of a study guide that was written to be interactive, i.e., it was designed to
go through the work with you and to guide you through the module, almost like another
person would do it. The study guide takes the place of the lecturer when you are working on
your own.
The reason why the study guide was designed in this way may be explained as follows:
This module has a weight of 8 credits, which means that it requires approximately 80 hours
of study to be completed. These 80 hours include:
• Preparation for participation in contact sessions (which includes discussions,
group work, writing of progress tests and tutorial activities);
• attendance of contact sessions during which the lecturer will facilitate the teachinglearning
activities;
• independent working through examples or solving of problems, for instance
work to be submitted;
• preparation for tests and the examination;
• writing of tests and examinations.
Note, however, that the total contact time (lectures or other contact sessions that are part of
this program) comprise of only 33 hours of the total of 80 hours required for the module. This
implies that you have to make up the remaining 50 hours by study activities in your
own time. A primary aim of the study guide is to assist you with these activities.
For your convenience a detailed time schedule and student study program is included in this
study guide. We strongly recommend that you adhere to this schedule.
You must utilise the green textbook by Washington as primary source of study
material. The study guide contains detailed references to those paragraphs in the textbook
pertaining to each study unit or section. You should use these references in order to prepare
for lectures. U must also use these references when you perform exercises or assignments
and when you study for tests or exams. The study guide also contains certain
complementary discussions and examples in order to further facilitate your progress.
8

The study guide was designed to do, inter alia, the following for you:
• To highlight for you the specific outcomes of each study section, so that your academic
goals will be clearly defined;
• to lend structure and order to the study contents;
• to give you the support to keep up with the rate of work of the lecturer;
• to enable you to study largely on your own the contents of the different study sections;
• to work systematically and regularly so as to complete the assignments timely and to
start preparing for tests before it is too late.
Hints
The following points of advice may help you to be successful in your studies:
• Work according to the time schedule suggested in this study guide.
• Always let your study activities by guided by the study outcomes that are set at the
beginning of every study section in the study guide.
• Read the textbook, especially when the study guide refers you to it.
• Do not try to do the exercises directly – first glance over the sections in the textbook
assigned for reading and pay special attention to the examples.
• Make notes (in your textbook or study guide, if you so prefer, or in a special notebook) of
everything new in the work.
• For your own sake, keep a file in which you collect and organise all notes, exercises,
marked assignments, etc.
• Reflect on the meaning of every outcome.
• Form a group (even if you and only one other person will work together) where you will
discuss the work, explain it to each other and work through the exercises together,
especially before tests or examinations.
• Attend the facilitation sessions
• Please feel free to see the lecturer personally immediately when you have difficulties in
achieving any of the outcomes.
9

0.7 Monitoring of progress
Typical questions you should ask yourself in order to plan and monitor your work and
to reflect on the completed assignments are listed below.
• Always start by planning your approach to the problem. What is the problem about?
• What do I know about this?
• What is it related to?
• What is required of me?
• What should I do first?
• Which strategies and techniques should I use?
• Do I know where to get the knowledge or information that I need?
• Which steps are needed to complete the task?
• Is there another way?
• How will I know if I make a mistake?
• When is a solution not relevant?
• Do I fully understand what I’m doing?
• Does it seem to be correct?
• Am I considering all possibilities?
• How can I make adjustments if it appears that I am on a wrong track?
• Where will it take me?
• Have I done it completely and correctly?
• How does mine compare to those of others?
• Have I achieved the learning outcomes?
• What have I learned from this?
• When will I need to do something similar?
• How can I use it in future?
10

0.8 Assessment
Assessment will be performed in the following way:
Self assessment
You should always do the exercises that are given and then mark them yourself
according to the answers that are supplied, even though the work is not formally
submitted and marked by the lecturer or by the assistants.
Formative assessment
Certain assignments have to be completed and submitted for assessment at the due
date arranged by the lecturer. Such assignments are usually given at the end of a
learning section in this study guide. Shortly after submission, a memorandum will be
made available.
We shall also make use of regular progress tests to monitor your progress. Consult the
Time Schedule and Student Work Program for more information about these tests.
Summative assessment
At the end of the semester an examination will be written covering all the work of the
module.
Group work
The amount of work in this module is simply too much for one person to do all of it. We
therefore recommend that you should join up with at least one other person to work through
the exercises. This will save some of your valuable time.
You will be assessed individually – especially in the progress tests, but also in the
examination. It is therefore totally futile to copy the work mindlessly from someone else. On
the occasions that the lecturer will collect the exercises that you have done and will formally
mark them, the mark will make a minor contribution to your participation mark. The progress
test that follows it, however, will make a considerable contribution to it.
11

In this module group work therefore must not be considered to be work that you and
others submit together in order to get a combined mark for the group. On the contrary,
group work in this module simply means that you will be cooperating with at least one other
person in order to achieve the outcomes on your own. The lecturer will hardly ever assign a
joint mark for group work. The outcomes in this module are assessed individually. Nobody
will be awarded marks for the work of somebody else.
Admission to the exam
The General Rules of the University specify the following in this regard:
No student will be admitted to the examination, except if the student can proof by
means of a proof of participation (participation mark) to the satisfaction of the lecturer,
subject group and school director that he adhered to the minimum requirements
regarding the achievement of certain knowledge, skills and attitudes as specified by
the module outcomes.
The following is required for a participation certificate for this module:
Participation mark of at least 40%
The following activities contribute to your participation mark:
1. A number of exercises that may have to be submitted as assignments for marking.
2. A number of progress tests to be written during contact sessions.
The following approximate weights will apply for the calculation of the participation mark:
1. Work sheets and exercises (assignments): 30%
2. Progress tests: 70%
Examination
One two-hour examination paper is written at the end of the semester. This examination
paper will cover all the work of the entire semester. There is a sub minimum of 40% on the
examination mark for passing the nodule, irrespective of your participation mark.
12

Module mark of at least 50%
Your final mark for the module is calculated from the participation mark and the examination
mark in a ratio of 60:40. To pass this module, a module mark of not less than 50% is
required, provided, of course, that you have obtained not les than 40% in the examination
paper.
Calculation of module mark:
Module mark = 0,6 x Participation mark + 0,4 x Examination mark
Should you therefore have a participation mark of 40%, you will have to obtain at least 65%
in the examination in order to pass the module.
Irregularities in the course of assessment
The official rules place the responsibility squarely on the student to see that assignments are
submitted on time.
Late assignments cannot be accepted, because memorandums of the assignments will be
made available directly after the due date of the assignments. Those that are late could
therefore have had access to the worked out solutions.
It is of utmost importance for students to do their own work and to refrain from copying from
any other student and submitting as if it is his / her own work, for assessment purposes. (See
the next section on plagiarism in this regard).
It is possible for students to co-operate in groups of two or more in order to achieve the
outcomes – in fact, we recommend that you do so. However, you may never submit the
same end product as that of any other student. You should therefore avoid looking at the end
product of another student. Your work should reflect your own personal style of writing
and argumentation.
Tests will be announced in advance by means of announcements during lectures as well as
by e-mail/ eFundi. If a student should fail to show up for a test, then he shall receive a zero
mark for that test. So, “valid excuses” will not serve as substitution for zero marks. Tests
are, of course, individual activities.
13

0.9
Warning
againsst
plagiarrism
WORRK
ASSIGGNMENTS
ARE INDIIVIDUAL
TASKS T ANND
NOT GGROUP
ACTIVITIES
A S
(EXCCEPT
WHERE
SPECIFFICALLY
INNDICATED
AS A GROOUP
ACTIVVITY)
Copyying
of texxt
from the work of othher
students
or from oother
sourcees
is unacceptable
–
only brief quotess
may be ussed
and theen
only if cle early acknoowledged.
Existting
text shoould
be refoormulated
in your own n words to explain whhat
you have
read. It iss
not acceptablee
to simplyy
retype ssome
exist ting text/mmaterial/inforrmation
an nd then too
acknnowledge
thhe
source in a foottnote
– Yo ou should be capabble
of pres senting thee
conccept/idea
in your own wwords
withouut
quoting the
original author verbbatim.
The purpose of f assignmennts
is not a simple rep production oof
existing mmaterial
or text, but too
see if you aree
competennt
enough to integra ate existingg
texts, to formulate your ownn
interppretation
annd/or
criticaal
evaluation
and to fin nd a creativve
solution for the pro oblems youu
face. .
Warnning:
Students
submmitting
copied
text wi ill receive a zero for the assign nment andd
disciplinary
stteps
may be taken by the Faculty F or the Univversity
aga ainst suchh
studdents.
It is also unaccceptable
foor
students s to be doing
the worrk
of other rs for themm
or too
allow them
to copy your workk
– do not lend
out yoour
work orr
make it available a too
otheers!
14

0.10 The module plan
Study Unit Topic Hours Assessment
1 Mathematical models and functions 14 Test 1
2 Polynomial functions 14 Test 2
3
Rational functions, exponential functions
and logarithmic functions
4 Trigonometric functions 10
5 Conic sections 8
6 Trigonometry 12
7 Elementary descriptive statistics 10
15
12
Test 3
Test 4
Test 5/
Practical ?
The above topics form an integrated unit: Knowledge, skills and competence acquired in one
study unit are applied in one or more of the other study units. The module can be compared
to an engine, where all the gears of the engine are required to enable the engine to perform
its task.
Throughout your study of the technical and scientific disciplines you will see that
mathematics cannot really be compartmentalised, even though we offer it in modules that are
once again subdivided into study units and study sections. These divisions are only done to
give some structure to the study units and to facilitate your study of mathematics.
In this context mathematics can be compared to a living organism with different limbs – each
limb is attached to the body and none of the limbs can function on its own. The existence and
operation of each limb is to a lesser or greater extent dependent on the existence and
operation of one or more of the other limbs.

0.11 Time schedule and study work program
Study Unit Topic
1 Mathematical models and functions
2 Polynomial functions
3
Rational functions, exponential functions and
logarithmic functions
16
Chapters in
the book of
Washington
4 Trigonometric functions 10
8
21
3
5
21
7
15
3
13
Paragraphs
8.3
21.9 – 21. 10
3.1 – 3.6
5.1 – 5.3
21.1 – 21.2
7.4
15.2 – 15.3
3.4
13.1 – 13.2
10.1 – 10.3
10.5 – 10.6
5 Conic sections 21 21.3 – 21.6
6 Trigonometry
7 Elementary descriptive statistics 22
0.12 Module outcomes
After completing this module the student should be able to do the following:
4
8
4.1 – 4.5
8.1 – 8.4
22.1 – 22.2
22.6 – 22.7
• demonstrate solid knowledge, understanding and insight regarding modelling through
the use of a variety of functions, Cartesian as well as polar coordinate systems,
simple polar curves, Cartesian curves, conic sections in terms of loci and Cartesian
equations, trigonometry as well as elementary descriptive statistics;

• demonstrate skill in the limited modelling of real-world situations and problems, the
application of basic analytical geometry, sketching polar curves using suitable
computer software, Cartesian curves of a variety of functions, graphical solution of
systems of equations, solving problems involving trigonometry and applying basic
descriptive statistics;
• be competent to apply the above-mentioned skills practically and to model real-world
situations from the technical and scientific fields of study and solve associated
problems as well as to apply elementary statistics through the use of suitable
computer technology in order to administrate and interpret data;
• be capable of evaluating the meaning, validity and accuracy of mathematical models
and calculations as applied to real-world situations from the technical and scientific
fields of study.
0.13 Important arrangements
In order to avoid unpleasantness and mutual bitter feelings between lecturer and the student,
the lecturer of this module wishes to make the following arrangements with the student:
1. Communication with the lecturer will take place by e-mail, by land line telephone
(during office hours!) or by a personal discussion during the official consulting hours
of the lecturer – otherwise by appointment made through the school secretary.
2. Special arrangements or apologies regarding tests, assignments or lectures are to be
communicated to the lecturer by e-mail, within a reasonable time after (or before, if
possible) the emergency.
3. Sports-, social- or cultural activities are not classified as emergencies. If such
activities seriously interfere with the academic responsibilities of the student, it is
simply the task of the student to reconsider his or her priorities.
4. The lecturer has no moral or other compulsion to accommodate a student by means
of any special arrangement – should it however happen that a student is assisted, it is
to be considered as a gesture of goodwill by the lecturer.
17

5. In view of point 4 above, we request students to refrain from raising tragic events or
flights of fancy in order to gain the pity of the lecturer. A student who does this is an
embarrassment for himself and for the profession he intends embarking on.
6. A week consists of a full five working days and that includes the entire Friday.
7. We also wish to kindly request students not to waste the time of the lecturer with
unnecessary matters during the examinations and the semester tests, as large
amounts of marking has to be done during these times.
8. The marking and processing of marks during semester tests and (especially)
examinations, last 7 working days. The reason for this is that the scripts have to be
moderated, the marks checked, borderline cases have to be reconsidered and the
marks have to be entered into the system and checked again. These mechanisms
take time, but they are to the advantage of the student
9. Participation marks and examination results will be sent to the students on the
officially scheduled dates. Students should please refrain from enquiring from the
lecturer before these due dates, because such enquiries can only delay the
processing of the marks.
10. No negotiation regarding participation marks will be conducted after the date for
finalising the participation marks. According to the A rules a participation mark is a
mark that the student has compiled during the semester by participating and
acquiring marks in class activities, performance tests and assignments. A lecturer
cannot “give you” a participation mark or a “conditional admission” to the examination.
11. Determining the dates of the second examination opportunity and the arrangements
for it fall outside the responsibilities of the lecturer. No lecturer is empowered to alter
any of these dates or arrangements. Therefore we want to suggest with all due
respect, that overseas trips, visits abroad and the like rather be arranged to
commence after the date of the second examination opportunity.
We can relate a large number of anecdotes to illustrate how misunderstandings regarding
the above matters have led to spoiled relations between people. Please, let us try to
avoid such situations and rather try to cooperate in harmony like civilised people.
18

1 Mathematical models and functions
Estimated time required to master the learning outcomes
14 hours
Previously acquired knowledge which is needed
1. Definitions of the six trigonometric functions (Paragraphs 4.2 to 4.4 in the textbook)
2. The Theorem of Pythagoras
3. Rectangular systems of axes (Paragraph 3.3)
Learning outcomes for this study unit
After completing this study unit the student must be capable of doing the following:
1. Use radian measure with confidence in order to describe angles;
2. Convert angles expressed in degrees to radians and vice versa;
3. Use rectangular co-ordinate systems and polar co-ordinates to represent information
and to interpret information;
4. Convert rectangular co-ordinates to polar co-ordinates and vice versa;
5. Sketch simple polar curves by means of a table and pencil and paper methods, as
well as by means of suitable computer software;
6. Define the concept “function” and to utilise an example in order to explain how
functions are used to describe mathematical models;
7. Correctly use function notation;
8. Correctly use the concepts “domain” and “range” in order to describe the behaviour of
a function;
19

9. Sketch the graph of a given data set (table) on paper;
10. Sketch the graph of a given data set (table) using Microsoft Excel;
11. Use suitable computer software in order to generate graphs of given functions.
20

1.1 Radians as unit of measurement for angles
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Use radian measure with confidence in order to describe angles;
2. Convert angles expressed in degrees to radians and vice versa
Study the following material in the book by Washington
Paragraph Page numbers
8.3 244 - 248
Exercise 1.1 for self-assessment
Exercise in the textbook Page number Problems
8.3 248 5, 7, 9, 11
21
13, 15, 17, 19
21, 25
45, 47 (set your calculator to radian
mode)
63, 67
The final answers of the given problems appear at the back of the text book.

1.2 Polar co-ordinates
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Use rectangular co-ordinate systems and polar co-ordinates to represent information
and to interpret information;
2. Convert rectangular co-ordinates to polar co-ordinates and vice versa;
3. Sketch simple polar curves by means of a table and pencil and paper methods, as
well as by means of suitable computer software
Study the following material in the book by Washington
Paragraph Page numbers
21.9 596 - 598
21.10 600 - 602
Exercise 1.2 for self-assessment
Exercise in the textbook Page number Problems
21.9 599 5, 7, 9, 13
21.10
Do questions 13 – 21 by
means of GSP
602
22
17, 19
21, 23
1
5, 7, 9 ( secθ
= ), 11, 13, 15, 17,
cosθ
21
The final answers of the given problems appear at the back of the text book.

1.3 Functions
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Define the concept “function” and to utilise an example in order to explain how
functions are used to describe mathematical models;
2. Correctly use function notation;
3. Correctly use the concepts “domain” and “range” in order to describe the behaviour of
a function;
4. Sketch the graph of a given data set (table) on paper;
5. Sketch the graph of a given data set (table) using Microsoft Excel;
6. Use suitable computer software in order to generate graphs of given functions.
Study the following material in the book by Washington
Paragraph Page numbers
3.1 82 - 84
3.2 85 - 88
3.4 92 - 96
3.5 98 - 102
3.6 103 - 105
23

The Geometer’s Sketchpad-program (GSP) operates like a graphical calculator. It is
capable of generating the graph of a given function directly. Microsoft Excel, in
contrast, requires that a table of values be input first; only after that is Excel capable
to produce a graph of the values. Both programs are powerful and useful; we now
supply a list of the actions or facilities offered by GSP. The list was compiled by Mrs.
Heleen Coetzee of the subject group Mathematics Education.
You should experiment with this program whenever you find time.
Check each action off in the column below as soon as you are certain how it works.
ACTION EXPLANATION CHECK
Selection Arrow Tool Select figure.
Point Tool Plot points.
Nothing is selected, if you click on “white”
area.
Construct -segment Draw line segments.
You will always carry on making dots or
whatever you have chosen last, until you
choose this tool. If something does not want
to work, check if you have chosen the correct
objects
Measure-Angle When three points are selected, it measures
the angle of the middle point.
Measure-Calculate Calculates where measurements were chosen
as variables.
Measure-Length Measures the length of the selected line
segment.
Measure -Distance Measures distance between 2 selected points.
Display-Animate point Select a point that has to move randomly.
Straightedge Tool Construct line segments. (Point needs to be
highlighted to be connected to the other line
segment.)
Construct-
Perpendicular Line
The point and the line segment through which
the perpendicular line has to intersect needs
to be selected.
24

Construct-Intersection
Construct-(Triangle)
Interior
Measure-Perimeter
(Area)
Select two geometrical objects to get the
intersection
or
Click with mouse close to intersection
or (not a safe method)
Place point with the Point Tool on the
intersection when both objects are highlighted
Select all the vertices of the polygon. Select
the whole polygon this way.
You can measure the circumference or the
perimeter of a polygon, if the polygon is
selected (fine grid).
File-New Sketch Makes a new document.
Graph-Grid Form-
Square Grid
Graph-Grid Form-
Rectangular Grid
Graph-Grid Form-Polar
Grid
For a square grid.
Drag the origin around to see less or more of
the specific quadrant.
Move the mouse over one of the numerals of
the grid until the mouse becomes a ↔ or a .
When this has occurred, you may change the
axis scale, while staying square
For graph paper that has different scales on
each axis.
Move the mouse over one of the numerals of
the grid until the mouse becomes a ↔. The
scale of the horizontal axis is independent of
the vertical axis.
The vertical axis’ scale can be changed
independently of the horizontal axis, if the
mouse changes into a .
For graph paper which works in polar coordinates.
Your functions are then of the form
r = f ( θ ) .
Graph-Hide(Show) Grid To make the grid invisible or to make it visible
again
25

Graph-Snap Points Points close to the intersections of the grid is
place precisely on the intersection.
Display-Hide Axis(Axes) Select axis and hide it (them).
Display-Hide Points (or
other figure)
Display-Line
Width(dashed, Thin,
thick)
First select the object then hide it.
Select a line segment or function and change
its appearance.
Display-Colour Select an object and change its colour.
Graph-Plot New
Function
sin
cos
tan
sqrt
ln
log
Check that nothing is selected before you use
this tool.
^ for exponents
* for multiplication
π and e is in “Values”
At Equation one can choose between
y = f ( x)
and x = f ( y)
. The latter is
necessary to draw vertical lines as functions,
e.g. x = 3.
Be careful for trigonometric functions. If you
work in degrees you will have to stretch or
shrink the axis
You may not draw a circle like x + y = 25
directly, because it is not a function. First draw
2
= then draw f ( x)
y 25 − x
26
− .
Edit Function Right click on the function to make corrections
of changes without having to retype the
function.
Properties- Plot-Domain Right click on the graph to adjust the domain.
Measure- Coordinates
Measure a selected point’s coordinates.
2
2

Measure-Abscissa Measures only the x − coordinate of the
selected point.
Measure-Ordinate Measures only the y − coordinate of the
selected point.
Edit-Merge Point to
Function Plot
Select a point very close to the graph and the
graph to place the point on the graph.
Merge is unnecessary, if the graph changes
colour when you place the point on it.
The point can now be dragged on the graph.
Text Tool Give the points names, by clicking with the
hand (which becomes black). Change the
name by double clicking on the A inside.
Edit-Select All
Edit-Copy
Edit-Paste of
Edit- Paste Special –
Picture (for better
quality)
Select everything and click on the objects that
you do not want to select. Place them on the
Clipboard.
This can be dragged into WORD or any other
document.
Drag into WORD.
You can stretch or shrink be dragging with a
diagonal arrow at any point.
File-Save As Save as a .gsp file any place you prefer.
File-Document Options-
Add Page-
Blank/Duplicate
File-Document Options-
Add Page-
Blank/Duplicate-Page
name.
File-Print Preview-Fit to
Page
Add different pages to the same document.
This helps to group the sketches that belong
together under one name.
Name the pages appropriately in the above
two white areas.
To print a page that fits on an A4 page. This
helps if you’d like to make transparencies.
Edit-Preferences Set your units with accuracy.
Measure-Slope Select a line segment and measure the
gradient.
Measure-Equation Select a line and determine its equation.
27

Graph - Plot Points –
Plot / Done
Plot a point’s coordinates that will not change
if you were to change the axis’ scale.
Edit - Paste Picture Duplicate something (like a picture or an
Equation Editor equation) from another
programme.
Transform-Mark Centre Indicate the point of rotation or division.
Transform-Mark Mirror Select the axis of symmetry.
Transform-Mark Vector Choose the translation vector.
Transform-Translate Translate the selected object.
Transform-Rotate Rotate the selected object with respect to. the
point that was marked as centre.
Transform-Dilate Divide the segment from the point (mark
centre)
Transform-Reflect Reflect the selected objects with respect to.
the axis of symmetry.
Many others PLEASE experiment and learn from each
other. Remember this best of all....
Edit Undo Reverse a mistake.
28

Exercise 1.3 for self-assessment
Exercise in the textbook Page number Problems
3.1 84 5, 7, 9
3.2
29
13, 15, 17 (what if x = 0?),
19
41
88 7, 8, 13, 15
(in each case, only state at which value
of the independent variable is the
function undefined)
31, 33, 41
3.4 96 Sketch on paper:
5, 7, 9, 13, 15, 17, 21, 25, 31
Check your answers by using a
computer program to sketch the curves
and comparing them with your pencil
and paper sketches.
Sketch using a computer program:
39, 43, 46, 47, 57
3.5 102 Do the following using a computer
program:
37, 43, 44, 50
You then answer the questions by taking
readings off the graph.
3.6 106 With pencil and paper: 1, 3, 5, 7
With Microsoft Excel: 2, 4, 6, 8
The final answers of the given problems appear at the back of the text book.
The rest will be discussed during lectures.

‘n Elliptiese masjienonderdeel word in
die volgende diagram getoon; die
punt O is die middelpunt van die
onderdeel en hierdie punt word as
oorsprong van ‘n XY-assestelsel
gekies:
1.1 Skryf die vergelyking van die ellips in
algemene vorm neer.
1.2 A, B, C en D is gate wat in die
onderdeel geboor is. Bereken die
afstand AD.
1.3 Bereken die grootte van θ , dit is
∠ DAB .
1.4 MPQR is ‘n reghoekige opening wat
in die onderdeel gesny is. Indien M
die middelpunt is van AD, bereken die
oppervlakte van die opening MPQR.
Assignment 1
Vraag 1/ Question 1
30
An elliptical machine part is shown
in the following diagram; the point
O is the centre of of the part and
this point is chosen as the origin of
an XY-system of axes:
Write down the equation of the ellips in
general form.
A, B, C and D are holes which were
drilled through the part. Calculate teh
distance AD.
Calculate the magnitude of θ , that is
∠ DAB .
MPQR is a rectangular hole that was
cut into the part. If M is the midpoint of
AD, calculate the area of the hole
MPQR.
1.5 Bereken die gradiënt (helling) van AB. Calculate the gradient (slope) of AB.

2 Twee vliegtuie, A en B, nader die
lughawe, dit is die oorspong van die
assekruis:
2.1 Skryf die posisies van die twee
vliegtuie in poolkoördinate.
2.2 Bepaal die grootte van die hoek
tussen hulle vlugpaaie in grade.
2.3 Bepaal die grootte van die hoek
tussen hulle vlugpaaie in radiale.
2.4 Skryf die posisies van die twee
vliegtuie in reghoekige (Cartesiese)
koördinate.
2.5 Bepaal die afstand wat die twee
vliegtuie van mekaar af is.
Vraag 2/ Question 2
31
Two aircraft, A and B, are approaching
the airport, that is the origin of the
system of axes:
Write the positions of the two aircraft in
polar co-ordinates.
Determine the magnitude of the angle
between their flight paths in degrees.
Determine the magnitude of the angle
between their flight paths in radians.
Write the positions of the two aircraft in
rectangular (Cartesian) co-ordinates.
Determine the distance between the two
aircraft.

Vraag 3/ Question 3
OA is ‘n minuutwyser wat vanaf punt A OA is a minute hand which moves from
tot by punt B beweeg. Die hoek θ is point A to point B. The angle θ is the
die hoek tussen die wyser se
oorspronklike posisie en sy latere
posisie:
3.1 Bepaal die koördinate van die
wyserpunt A in poolvorm.
3.2 Bepaal die koördinate van die
wyserpunt B in poolvorm.
3.3. Hoeveel sekondes neem die wyser om
van A na B te beweeg?
32
angle between the original posision of
the hand and its later position:
Determine the co-ordinates of the tip of
the hand A in polar vorm.
Determine the co-ordinates of the tip of
the hand B in polar vorm.
How many seconds does the hand take
to move from A to B?
3.4 Bepaal die grootte van θ in grade. Determine the magnitude of θ in
3.5 Skakel θ om na radiale; rond dit af tot
vyf desimale plekke, aangesien u dit
hieronder in ander berekinge moet
gebruik.
degrees.
Convert θ to radians; round it to five
decimal places, since you need to use
it below in other calculations.

3.6 Bepaal die hoeksnelheid van die
wyser, in radiale/sekonde.
33
Determine the angular velocity of the
hand, in radians/second.
3.7 Bereken die lengte van boog AB. Calculate the length of arc AB.
3.8 Bereken die oppervlakte van die
sirkelsektor wat deur boog AB
onderspan word (dit is die
geskakeerde gedeelte).
3.9 Bepaal die afstand AB in cm.
(ons bedoel die lynstuk wat A en B
verbind)
4.1 Skets r = 2cos θ deur die tabel te
voltooi en die punte te plot:
θ in radiale/ radians 0
r in lengte-eenhede /
distance units
4.2 Skets r 2cos θ
4.3 Skets
Vraag 4/ Question 4
π
4
2
= en r = 4sin(
θ )
met behulp van ‘n rekenaarprogram.
y x
2
= 2 − 9 met potlood en
papier. Kontroleer u grafiek met
behulp van ‘n rekenaarprogram.
4.4 Dit is moontlik om die vergelyking
y x
2
= 2 − 9 na poolvorm te
transformeer:
π
2
Calculate the area of the circle sector
subtended by arc AB (that is the
shaded region).
Determine the distance AB in cm.
(we mean the line segment connecting
A and B)
Sketch r = 2cos θ by completing the
given table and plotting the resulting
points:
3π
π
4
5π
4
Sketch r 2cos θ
3π
2
7π
2π
4
2
= and r = 4sin(
θ )
by means of a computer program.
Sketch
y x
2
= 2 − 9 using pencil and
paper. Check your graph using a
computer program.
It is possible to transform the equation
y x
2
= 2 − 9 to polar form:

Skets die kromme van
y = x −
2
2
9
2
( )
rsinθ = 2 rcosθ − 9 uit/ from x = rcos θ en/ and y = rsinθ
2 2
∴rsinθ − 2r cos θ + 9 = 0
2 2
∴− 2r cos θ + rsinθ
+ 9 = 0
∴− ⋅ + ⋅ + =
2 2
2cos θ r sinθ
r 9 0
− b ±
∴ r =
2
b − 4ac
2a
− sinθ ±
=
2
sin θ − 4
2 2
2
−2cos
θ 9
2 ( − cos θ )
− sin ± sin + cos
∴ r =
−4cos
θ
( )( )
θ
2
θ
2
72
2
θ
2 2
− sinθ ± sin θ + 72cos
θ
r =
met
2
−4cos
θ
behulp van GSP en vergelyk dit met
die grafiek wat u in 4.3 hierbo geteken
het.
Beskou die volgende grafiek wat toon
hoe die konsentrasie van ‘n
soutoplossing met verloop van tyd
verander:
Vraag 5/ Question 5
34
Sketch the curve of
2 2
− sinθ ± sin θ + 72cos
θ
r =
using
2
−4cos
θ
GSP and compare it to the graph which
you produced in 4.3 above.
Consider the following graph which
shows how the concentration of a salt
solution changes with the passing of
time:

5.1 Beskryf die grafiek in u eie woorde
deur na die volgende begrippe te
verwys:
• vorm of rigting van die kromme
• waardeversameling
• definisieversameling
5.2 Skryf enige twee afleidings
(gevolgtrekkings) neer wat u uit die
grafiek en u antwoorde op vraag 5.1
kan maak.
5.3 Bereken die konsentrasie na vyf
sekondes.
5.4 Skat die tyd wat die oplossing neem
om ‘n konsentrasie van 0,2 mol/liter te
bereik.
Stefan het sy kar se ligte aan vergeet
en toe hy by sy kar kom, wou dit nie
vat nie. Hy vra toe een van sy vriende
wat ‘n batterylaaier het om dit vir hom
te leen. Die spanningsfunksie van die
karbattery word hieronder grafies
aangetoon:
Vraag 6/ Question 6
35
Describe the graph in your own words
by referring to the following concepts:
• shape or direction of the curve
• range
• domain
Write down any two deductions
(conclusions) which you can make
from the graph and from your answers
to question 5.1.
Calculate the concentration after five
seconds.
Estimate the time the solution takes to
attain a concentration of 0,2 mole/liter.
Steven forgot to switch off his car’s
headlights and when he got to his car,
it refused to start. He then asked one
of his friends who owned a battery
charger to borrow it to him. The
voltage function of the car battery is
represented graphically below:

Beantwoord nou die volgende vrae
deur van die grafiek gebruik te maak:
6.1 Beskryf die grafiek in u eie woorde
deur na die volgende begrippe te
verwys:
• vorm of rigting van die kromme
• waardeversameling
• definisieversameling
36
Answer the following questions by
making use of the graph:
Describe the graph in your own words
by referring to the following concepts:
• shape or direction of the curve
• range
• domain
6.2 Hoe lank is die ligte aan gelaat? How long were the lights left switched
on?
6.3 Hoe lank het dit die vriend geneem
om die laaier te bring?
6.4 Wat was die spanning van die battery
twee en ‘n half uur nadat die laaier
aangeskakel is?
6.5 Op watter interval is die
spanningsfunksie stygend?
How long did it take the friend to bring
the charger?
What was the voltage of the battery two
and a half hours after the charger was
switched on?
On which interval is the voltage
function increasing?

6.6 Skryf die horisontale asimptoot van
die funksie neer en verduidelik wat dit
beteken.
37
Write down the horizontal asymptote of
the function and explain what it means.
The solutions to the problems presented above are discussed during lectures. The
solution of certain problems may be published on eFundi.

2 Polynomial functions
Estimated time required to master the learning outcomes
14 hours
Previously acquired knowledge which is needed
1. The standard forms of linear (first degree) functions, quadratic (second degree)
functions and cubic (third degree) functions as taught in secondary school
mathematics
2. Sketching straight line graphs, parabolas and cubic curves as taught in secondary
school mathematics
3. Study Unit 1
4. WSKT 111 (Study Units 1 – 4)
Learning outcomes for this study unit
After completing this study unit the student must be capable of doing the following:
1. Writing linear equations in the standard form y = mx+ c for a linear (first degree)
function;
2. Represent linear functions graphically (by hand as well as using suitable computer
software);
3. Solve simple real life problems where linear models are involved;
4. Writing quadratic equations in the standard form
(second degree) function;
39
2
y = ax + bx + c for a quadratic
5. Represent quadratic functions graphically (by hand as well as using suitable
computer software);
6. Solve simple real life problems where quadratic models are involved;

7. Writing cubic equations in the standard form
degree) function;
40
3 2
y = ax + bx + cx + d for a cubic (third
8. Represent cubic functions graphically (by hand as well as using suitable computer
software);
9. Solve simple real life problems where cubic models are involved
Introductory remarks
From Study Section 1.3 follows:
A function is a special kind or rule according to which a value (called the dependent
variable) may be calculated by substituting another value (called the independent
variable) into a certain algebraic equation (called the model).
Functions are used to describe processes or situations in the real world. We use function
in order to set up mathematical models.
What is a mathematical model?
Well, it is a mathematical representation of a physical situation. Real life and most
technological applications have to do with complicated problem situations.
Such situations may however be simplified by considering one or two or maybe three of their
measurable aspects at a time. These measurable aspects are called variables. The way in
which one variable depends on another, is expressed in the form of a formula (mathematical
equation).
This formula is called a mathematical model.

There are at least five ways in which a real life situation or problem (or the function which
describes it) may be represented, namely:
• a numeric description (a table of measured or calculated values)
• a graphic description (a curve sketched on a co-ordinate plane with axes)
• a verbal description and/ or a schematic description such as a diagram
• an algebraic description (a formula or equation)
• a practical example which illustrates how the process or problem behaves
In this study unit we study real life problems which may be modelled using one of the
following types of functions:
1. Linear functions (first degree functions)
2. Quadratic functions (second degree functions)
3. Cubic functions (third degree functions)
These types of functions are examples of polynomial functions. A polynomial function is a
n
function of the standard form y = a x
n−1 n−2 n−3
+ a x + a x + a x +
0
+ a x where the symbols
0
1 2 3 ...
a 0 , 1 a , a 2 , ... are numeric co-efficients (constant number values) and where x is the
variable. The natural number n is called the degree of the polynomial function.
In the case of first degree functions the standard form
n
y = a x
0
n−1 n−2 n−3
0
1 0
+ a1x + a2x + a3 x + ... + anx reduces to y = a x + a
0 1x
and it is common
practice to write this as y = mx+ c.
In the case of second degree functions the standard form
n
y = a x
n−1 n−2 + a x + a x
n−3
+ a x +
0
2
+ a x reduces to y = a x
1 0
+ a x + a x and it is common
0
practice to write this as
1 2 3 ...
2
y = ax + bx + c .
Similarly, third degree functions are written in standard form as
n
41
n
0 1 2
3 2
y = ax + bx + cx + d .
In the case of any polynomial function the value of y may be calculated for any realnumbered
value of x ; Therefore we say that polynomial functions are continuous for all
real-numbered values of x .

2.1 Linear functions
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Writing linear equations in the standard form y = mx+ c for a linear (first degree)
function;
2. Represent linear functions graphically (by hand as well as using suitable computer
software);
3. Solve simple real life problems where linear models are involved
Study the following material in the book by Washington
Paragraph Page numbers
5.1 – 5.3 138 – 149
21.1 – 21.2 560 – 567
Linear models and proportionality
When a real life problem or situation may be described by means of a linear equation (or be
represented by means of a straight line graph) then it implies that the quantity on the
vertical axis (the dependent variable) is proportional to the quantity on the horizontal
axis (the independent variable).
In the special case where a straight line graph passes through the origin of the system of
axes, we say that the quantitity on the vertical axis is directly proportional to the quantity on
the horizontal axis. Sometime this relationship is also called direct variation.
42

Also study the following discussion which illustrates how we
proceed in order to develop a linear mathematical model for a real
life situation
To determine the pressure below the surface of a reservoir as function of depth below
the surface
Consider a tank, 20 meters deep and a pressure gauge which is lowered beneath the
surface of the water so that a pressure reading is taken every 4 meters:
Note that the pressure reading depends on how deep below the surface the reading is taken.
The pressure changes as the depth changes. From this observation we may conclude that
the pressure has a specific unique value at each depth and that the pressure therefore
depends on the depth where it is measured. So there are two variables, namely pressure
and depth. The pressure P is called the dependent variable and the depth d is called the
independent variable.
We can now attempt to mathematically express the relationship (the way in which P
depends on d ) between the two variables.
43

For this purpose we employ a formula or equation. Getting hold of this formula usually
requires a bit of scientific or mathematical insight – but by utilising our existing knowledge
regarding graphs (straight lines, parabolas, etc) we are capable of developing a heuristic (a
logical, flexible method or strategy) by which we can obtain the formula which describes a
mathematic relationship.
A good way to investigate the relationship between the dependent and the independent
variable is by setting up a table of measurements:
Depth d
(meters)
Pressure
P (kPa)
0 4 8 12 16 20
101,3 140,5 179,7 218,9 258,1 297,3
The information constitutes six sets (pairs) of values, where each set (ordered pair) consists
of a depth and an associated pressure reading. The next step is to graphically represent the
information.
In order to do so, we need to decide where we are going to place the domain and range and
how we are going to choose our axes.
• Let us agree to always place the domain (domain refers to the values assumed by
the independent variable) on the horizontal axis. So, in this case, we shall put the
depth on the horizontal axis.
• Similarly, let us agree to always place the range (range refers to the values
assumed by the dependent variable) on the vertical axis. So, in this case, we
shall put the pressure P on the vertical axis.
If we are to draw the graph on paper, we next need to consider the scale of the axes.
44

The idea is that the graph must be accurate enough that we may obtain good readings from
it. Therefore, graph paper may come in handy; if not available, we may use a regular sheet
of paper and proceed in a similar way.
Always attempt to use as much of the surface area of the paper as possible. The larger the
graph, the easier it will be to obtain accurate readings from it. See that the highest value of
the independent variable (greatest depth) fits as far as possible to the right on the horizontal
axis. (Similarly, we try to place the highest value of the dependent variable (greatest
pressure) as high as possible on the vertical axis.) Choose a convenient value (“round
number” or integer) close to or equal to the highest value which appears on an axis and mark
it off (see the system of axes below, on the next page).
It is not necessary to make a lot of smaller marks (ticks) on the axes; still, doing so may help
to give us a better “feel” for the scale of the axis. But the highest value on each axis must
lie on a convenient (integer) centimeter or millimetre distance from the origin (point
where the axes intersect).
The scale of each axis may then be determined easily.
By “scale of each axis” we mean that we want to establish what is actually meant in terms of
the units of measurement of that axis by 1 cm or 10 mm (or whatever distance on that axis).
On the graph below, the scale on the horizontal axis is 200 mm = 20 meter which implies
that 10 mm = 1 meter . We may also write it as 1mm = 01 , meter .
Similarly, the scale on the horizontal axis of the graph below is 150 mm = 300 kPa which
implies that 10 mm = 20 kPa.
We may also write it as 1mm = 2kPa.
45

As soon as the scales of the axes have been established, we are ready to plot the
information represented in the table as points on the flat plane defined by the axes. We call
this two-dimensional flat plane defined by the axes of a graph the Cartesian plane.
(See the representation below)
In order to plot the points on the Cartesian plane, we consider each set of values in the
table as a set of co-ordinates. We treat each value of the independent variable as a
vertical dashed line which intersects the horizontal axis perpendicularly; similarly, we treat
each value of the dependent variable as a horizontal dashed line which intersects the
vertical axis perpendicularly. Where the two dashed lines intersect, we plot a point.
For example: The second point which we obtain from the table above is the point ( 4; 140, 5)
.
46

We call the first number in the brackets (it is an element of the domain) the depth coordinate
and we call the second number in the brackets (it is an element of the range)
the pressure co-ordinate. (in school they usually spoke of the x and the y co-ordinates –
the X-axis was always chosen to be horizontally orientated and the Y-axis used to be
orientated vertically.
Everything we are doing here is exactly what they did in school; what is different here, of
course, is that we assign names and symbols which make sense in terms of the specific
problem situation we are investigating to the axes and variables.
(See the representation below)
47

By repeating the process above for each set of values in the table, we obtain the following
points:
The next step is now to decide which kind of curve (straight line, parabola, hyperbola,
third degree (cubic) curve, etc) may be drawn on the graph in such a way that it best
represents the shape in which the points are located; We want to draw a curve over
(through) the points in order that we may use this curve to draw conclusions and make
predictions regarding the real life situation which we are investigating.
At first glance, it is perfectly clear (and we may verify this by placing a straightedge over the
points) that the points on our graph lie in a straight line.
48

Therefore, we are justified in drawing a straight line through the points – and we do it in such
a way that the line passes “as well as possible” through all of the points:
We call this line which may be drawn over (through) the points in such a way that it passes
as well as possible through all of the points a regression line.
The fact that we obtained a straight line from the graphical representation implies that
pressure is proportional to depth.
We may use the regression line in order to obtain conclusions and predictions regarding the
real life situation which we are investigating. – that is the main objective of the entire exercise
we are engaged in.
Example 1:
What would the pressure be at a depth of 10 m below the surface of the water?
49

Solution:
Construct a vertical dashed line perpendicular to the depth axis upwards until it intersects the
regression line; then project to the left, perpendicular to the pressure axis. Where this
dashed line intersects the pressure-axis, you may read off the depth at 10 m:
The pressure would be approximately 200,8 kPa at a depth of 10 meter. (conclusion)
Example 2:
How deep below the surface of the water would the depth gauge register a pressure of
160 kPa ?
50

Solution:
Construct a horizontal dashed line perpendicular to the pressure axis to the right until it
intersects the regression line; then project downwards perpendicular to the depth axis.
Where this dashed line intersects the depth axis, you may read off the depth where the
pressure would be 160 kPa .
It is clear that the depth would be approximately 6, where a pressure of 160 kPa would be
measured. (conclusion)
The two examples above are good examples of interpolation. Interpolation is when we
utilise a regression line in order to conclude what the measurements would have been under
circumstances for which we do not possess actual measurements.
51

This comes down to the fact that you determine readings by means of the regression
regarding points which lie on the regression line.
We may even make predictions regarding the pressure if the reservoir were deeper than 20
m.
Example 3:
What would the pressure have been at a depth of 100m?
Solution:
A depth of 100 m is a value which does not appear on the horizontal axis of our graph (100 is
not an element of the domain). Therefore, we have to extend the regression line until it is
long enough for us to locate the point 100 m on the horizontal axis so that we may project
perpendicularly upwards to the curve and to the left, perpendicularly to the pressure axis in
order to make a pressure reading for a depth of 100 m:
It is clear that the pressure would have been 1081,3 kPa at a depth of 100 m. (prediction)
52

Example 4:
At which depth would the pressure have been 700 kPa?
Solution:
A pressure of 700 kPa is a value which does not appear on the vertical axis (700 is not an
element of the range). Therefore, we have to extend the regression line until it is long
enough for us to locate the point 700 kPa on the vertical axis so that we may project
perpendicularly to the right to the curve and downwards, perpendicularly to the depth axis in
order to make a depth reading for a pressure of 700 kPa:
It is clear that at a depth of approximately 61,1 m a pressure of exactly 700 kPa would be
measured. (prediction)
The two examples above are good examples of extrapolation. Extrapolation is when we
utilise a regression line in order to predict readings under circumstances which fall outside
the circumstances for which we took actual readings (see the table several pages ago). It
comes down to the fact that you use the regression line in order to obtain readings regarding
points which lie outside the regression line.
53

Therefore we have to extend the regression line whenever we perform extrapolation.
The question which next emerges is: Can we calculate the answers to the questions in
the four examples above, instead of reading it off the graph? Does there exist a way in
which we may algebraically calculate the answers to the four questions above?
The answer is: YES – definitely. All we need to do so, is the equation of the straight
line which we constructed through the points (the equation of the regression line, in
other words).
In order to determine this equation, we need our foreknowledge from High School Algebra:
The equation of any straight line is always y = mx + c where m is the gradient (slope) of the
line and c is the intercept on the vertical axis. y and x are the quantities (variables) which
are to be represented on the vertical axis and the horizontal axis, respectively.
Now, by considering the graphs above it is clear that the intercept on the vertical axis is
101, 3 kPa ; therefore c = 101, 3 .
54

Similarly, the gradient (slope) of the regression line may be obtained by constructing any
right triangle upon the regression line and then dividing the length of its opposite (vertical)
side by the length of its adjacent (horizontal) side:
ΔP
vertical difference between two points
m = Altways: gradient =
Δd
horizontal difference between two points
196
=
20
= 98 , kPa/m
Therefore, the equation of the regression line is:
y = mx+ c True for any straight line
∴ P = 9, 8d +
101, 3 Substitute the applicable symbols and values
55

Note that P = 9, 8d + 101, 3 is the mathematical model which describes the real life situation:
It is a function (formula, in other words) with which the pressure P may be calculated for any
given value of d which lies inside the domain of the function (soon, we shall say more about
domain and range).
Let us now use the mathematical model (that is in spirit and essence the equation of the
regression line) in order to calculate the pressure when the depth is 10 m:
P = 98 , d + 1013 ,
∴ P = 9, 8 10 + 101, 3 The notation P means P when d = 10 m
( )
d= 10 m d=
10 m
= 199, 3 kPa
Here we performed interpolation using our mathematical model.
(Note that this value is very close to the value which we obtained on using graphical methods
Let us now use the mathematical model (that is in spirit and essence the equation of the
regression line) in order to calculate the depth where the pressure would have been 700 kPa:
P = 98 , d + 1013 ,
∴ 700 = 9, 8d + 101, 3
∴− 98 , d = − 700+ 1013 , Manipulate the formula; it is first year work
−598,
7
∴ d =
−98
,
= 61, 092 m
Here we performed extrapolation using our mathematical model.
(Note that this value is very close to the value we obtained using graphical methods.)
56

A little more about domain and range
Now that we have the equation of the regression line, we may refer to the function
P = 9, 8d + 101, 3 and ask questions such as:
• What is the domain of the function?
• What is the range of the function?
The domain of the function is the set of actual measured values which lie between and
including the smallest actual depth and the greatest possible depth where actual
measurements have been made. (See the graphical representation)
We write it as Df{ d 0 d 20; d }
= ≤ ≤ ∈ .
This literally means in words: “The domain of the function is the set of all d -values such
that d lies between 0 and 20 m, but d may also be equal to 0 or 20 m; d may be any real
number which lies inside the closed interval from and including 0 m until and including 20 m.”
The range of the function is the set of actual measured values which lie between and
including the smallest actual pressure and the greatest possible pressure where
actual measurements have been made. (See the graphical representation below)
We write it as Wf{ P101, 3 P 297, 3;
P }
= ≤ ≤ ∈ .
This literally means in words: “The range of the function is the set of all P -values such
that P lies between 101,3 and 297,3 kPa, but P may also be equal to 101,3 kPa or 297,3
kPa; P may be any real number which lies inside the closed interval from and including
101, 3 kPa until and including 297,3 kPa.”
57

To conclude
The discussion on the previous pages illustrates in a step by step manner the heuristic by
which we may determine the formula with which aspects of a real life situation or process
may be described.
We summarise the steps of the heuristic:
1. Obtain two sets of values (data) through measurement. (The one set forms the
independent variable; the other set forms the dependent variable).
2. Represent the values (usually measurements) in the form of a table.
3. Use the table in order to obtain an accurate graph.
4. Use the pattern (shape) in which the data points lie on the graph and construct a best line
or curve through the points in such a way that the line or curve follows the shape of the
points.
58

5. Use this “best line” or curve in order to do interpolation and extrapolation by means of
making readings off the graph.
6. In the event that accurate inter- and extrapolation is required, obtain the equation of the
“best line” or curve which you obtained in step 4. This equation is the function
(mathematical model) which describes the problem situation you are investigating.
7. As soon as you obtained the equation of the regression line or regression curve, you may
use it in order to perform interpolation and extrapolation, by means of substitution and
calculation.
8. The behaviour of the function may now be described from the graph and from the
equation (formula) in terms of concepts like:
• Increasing/ decreasing
• Concave up/ concave down (see Fig. 24.31 (a) on p. 709 of the textbook)
• Maximum values/ minimum values as obtained from turning points and edge values
• Vertical asymptotes (we discuss this later – this is where the graph performs “jumps”)
• Horiszontal asymptotes (these are horizontal lines which the graph approach, but never
intersects or crosses)
• Domain
• Range
We shall employ the heuristic above in order to investigate any experiment or suitable
problem situation or physical process from the real life world. We shall utilise this same
heuristic time and again in every discussion in Study Section 2.1 to Study Section 3.3.
As you could see in Step 8, we may obtain an incredibly good understanding for the situation
or process – this enables us to make interpretations and evaluations regarding the
situation or process.
For example:
1. What is the value of atmospheric pressure? (that is the pressure at the surface of the
water in the tank)
2. Try to find out what the actual value of atmospheric pressure is (use Google and search
for “atmospheric pressure”). What do you conclude?
59

3. In text books the formula for hydrostatic pressure is given as P = ρ gh+ Patmospheric
.
Compare this formula with the mathematical model we developed above. If the value of
2
3
g = 98 , m/s , calculate the value of ρ (this is the density of water, measured inkg/m
).
4. What is the actual density of pure fresh water? (Use Google and search for “density of
fresh water”.)
5. Would you say that our mathematical model is accurate? Supply as many reasons as
you can.
Exercise 2.1 for self-assessment
Exercise in the textbook Page number Problems
5.2 144 Check all your answers using GSP:
61
5, 7, 9, 11
21, 23, 25, 27
29, 31, 33
5.3 148 11, 15 (with pencil and paper)
23, 27, 31 (using only GSP)
33, 35
21.1 563 Determine the length of the segments
connecting the points given in nr 5 and
nr 7
Determine the gradient of the segments
connecting the points given in nr 5 and
nr 7
25, 27

21.2 568 45, 51, 53, 55
The final answers of the given problems appear at the back of the text book.
62

2.2 Quadratic functions
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
4. Writing quadratic equations in the standard form
(second degree) function;
63
2
y = ax + bx + c for a quadratic
5. Represent quadratic functions graphically (by hand as well as using suitable
computer software);
6. Solve simple real life problems where quadratic models are involved
Previously acquired knowledge which is needed
1. Paragraphs 7.1, 7.3 (covered in WSKT 111, Study Unit 4)
Study the following material in the book by Washington
Paragraph Page numbers
7.4 228 – 232

Also study the following discussion which illustrates how we
proceed in order to develop a quadratic mathematical model for a
real life situation when the form of the equation (formula) is known
To mathematically model the motion of a rubber ball which is thrown vertically
upwards
Let us consider a rubber ball which is thrown vertically upwards and which is allowed to fall
back to earth (to perhaps bounce upwards again, or to come to rest). For the purposes of
this example, we only consider the up and down motion; we shall not be interested in what
happens after the ball returned back to earth. We want to represent this process
mathematically. The rubber ball has, of course, many measurable properties, for example
volume, mass, density, colour and position above the earth (altitude). The way in which the
ball moves when simply thrown vertically upwards and then allowed to fall back has,
however, very little or nothing to do with the properties mentioned above – any object simply
thrown upwards and then allowed to fall back, moves in more or less the same way.
(except, of course, objects with special shapes, such as parachutes and aircraft or certain
types of seeds – the shape of these objects allow them to float to the earth and in these
cases air plays an important part). All ordinary objects (such as balls, stones, bricks, cannon
shells, etc.) has however, at least one property in common, namely that their altitude varies
at all times during their motion until they come to rest (or cease to exist for practical
purposes, as in the case of a cannon shell).
From this observation we may say that the object (rubber ball) has a particular specific
altitude at every instant in time and that the altitude of the object therefore depends on the
time when it is measured. So, there are two variables involved, namely altitude and time.
The altitude h is called the dependent variable and the time t is called the independent
variable.
We may now try to express the relationship (that is the way in which h depends on t )
between the two variables mathematically.
64

In order to do so, we use a formula or equation. Obtaining this formula normally requires a
bit of scientific or mathematical insight – but in the case of most real life situations relevant to
mathematic literacy you will not need to derive any formula or equation.
The following few paragraphs explain how we may utilise knowledge from the field of
physics in order to set up a formula for hte altitude of the ball in terms of time. It is a
piece of applied mathematics. Students who took Physical Science as subject in
school might recognise the arguments we shall employ. For the sake of everyone
who had not taken Physical Science at school level I shall endeavour to write out the
arguments and reasoning followed in order to set up the formula.
In the case of objects which are simply thrown vertically upwards without air resistance
playing any part at all in their motion (such as rubber balls, stones, bricks, etc but not
parachutes, aircraft, balloons, etc) their are certain equations of motion which are valid, such
as the formula
1
s s v t at
2
2
= 0 + 0 + .
Note that the formula consists of certain symbols, where each symbol has a particular fixed
value for each given situation (almost like a setting). We refer to such symbols as constants.
In the case of an object which is thrown vertically upwards there are several properties of the
motion influencing the way in which the object moves, namely the altitude from which it is
thrown, the velocity at which it is thrown and the acceleration at which it moves, because
these properties determine how high the object can rise before starting to fall downwards, as
well as how soon it reaches a certain altitude. Therefore we call these properties constants.
1 2
In the equation of motion s = s0 + v0t + at the symbol s represent the distance travelled by
2
an object t seconds after the motion started measured from its starting point, s 0 represents
the distance between the object and some point of reference at the instant the motion
started, v 0 represents the initial velocity and a represents the acceleration of the object at all
times.
65

In school physics problems, the initial position of the object usually was at the point of
reference so that s 0 = 0 and then the equation reduced to:
66
1
s vt at
2
used the symbol u instead of v 0 and then the formula looked as follows:
2
= 0 + . Some text books
1
s ut at
2
2
= + .
Now, since we are throwing a rubber ball vertically upwards, the distance of the ball above
the ground is actually its altitude, so instead of s we may write h :
1
h= h0 + v0t + at
2
2
But because the arms of the person throwing the ball most definitely do not reach to the
ground level, the ball is not really moving upwards from ground level, but in fact from an
altitude of about one meter above ground level, except if he is throwing the ball vertically
upwards from shoulder height (about 1,5 m above the ground level). Let us, for the purpose
of this discussion, assume that the ball is at an altitude of one meter at the moment it leaves
the hand of the thrower. Then, the initial altitude h 0 = 1:
1
h= 1 + v0t + at
2
2
Let us further assume that the initial velocity, that is the velocity at which it leaves the
thrower’s hand, is 12 m/s; so v 0 = 12 :
1
h= 1+ 12t
+ at
2
2
Every object close to the surface of the earth is, however, subject to gravity. Gravity causes
constant downward acceleration of approximately -9,8 m/s 2 . The negative sign which is often
used simply means that the accelerating force is acting downwards. (This value is called
gravitational acceleration and it is indicated by a symbol g instead of a ). Nonetheless, for
all objects which moves freely in proximity to the earth, it holds that a =− 9,8 :
1
h= 1+ 12t + −
9,8 t
2
( ) 2

The formula which describes the altitude of the ball above the ground at any instant t is
therefore:
h= 1+ 12t − 4,9t
2
Mathematically, we may say that h is a function of t ; some mathematicians write it
as: h= f ( t)
which simply means that a corresponding h -value may be calculated for
any given t -value by substituting the t -value into the formula for h .
Remark: Note that all positions (distances and altitudes) are expressed in meters, all
velocities in m/s, all accelerations in m/s 2 and all time values in seconds – this is always true
whenever we work with equations of motion – the measuring units must be consistent.
It is interesting that the mathematical model
2
67
2
h= 1+ 12t − 4,9t
may be written as
h=− 4,9t + 12t + 1.
Note that this formula stands in the standard form for the equation
of a parabola, namely
2
y = ax + bx+ c.
In this situation, only, we are not dealing with the
variables y and x , but instead we are working with h and t . Therefore, the formula is of the
form:
2
h= at + bt + c where a =− 4,9 and b = 12 and c = 1
Of course we may now represent the formula (equation) above graphically. This means that
we shall place the dependent variable h on the vertical axis and that we shall place the
independent variable t on the horizontal axis. You have already learned how to sketch
graphs.

We may employ any one of a variety of methods in order to sketch a graph of
2
h=− 4,9t + 12t + 1;
The result of our endeavours will look as follows:
2
The curve of the function h() t =− 4,9t + 12t + 1 is clearly concave down.
Note that this graph DOES NOT show the trajectory of the ball; it shows the altitude of the
ball at any instant from t = 0 until t = 2,530 . In this way, the horizontal co-ordinate of the
turning point of the curve gives the time that it takes the ball to reach its maximum altitude
and the vertical co-ordinate of the turning point give the maximum altitude achieved.
It is interesting that the domain of the function
f
68
2
h= − 4,9t + 12t + 1 may be written as
= { 0≤ ≤2,530; ∈ } and that the range may be written as Wf{ h 0 h 8,347; h }
D t t t
Do you still remember what this means and how we obtain it from the graph?
= ≤ ≤ ∈ .
Now we are able to calculate the altitude of the ball at any instant (time) within the
domain 0 ≤t≤ 2,530 .

Example: How high is the ball after 2 seconds?
Solution:
2
69
2
( ) ( )
( ) ( ) ( ) ( )
Set t = 2 in the formula h=− 4,9t + 12t+ 1: h=−
4,9 2 + 12 2 + 1
∴ h = −4,9 = 5,4 m
⋅ 4 + 12 ⋅ 2 + 1
Note that the answer to any word problem MUST have a correct unit of measurement.
Also note that you could have read this value off the graph:
We are also able to calculate the time that it takes the ball to reach any altitude within
the range 0≤h≤ 8,347.

Example: How long does it take the ball to achieve an altitude of 7 meters?
Solution:
2
Set h= 7 in the formula h=− 4,9t + 12t+ 1and
solve for t:
2
7=− 4,9t + 12t+ 1
⎧This
is a quadratic equation which we may solve by using
⎪
2
⎪
− b± b −4ac
⎪the
formula t =
provided, of course, that
⎨
2a
⎪the
equation is written in the standard form, which implies
⎪
2
⎩⎪
that the equation must be written in the form at + bt + c = 0.
⎧We
have written all the terms on the same side of the
2
∴ 0=− 4,9t + 12t−6 ⎪
⎪equation
in order that one side
of the equation is zero;
⎨
⎪now
the equation is in the standard form and we may apply
⎪
⎩the
quadratic formula entirely without incident.
− b± ∴ t =
2
b −4ac
2a
where a=− 4,9 b= 12 c=−6
− ( 12) ±
∴ t =
2
( 12) −4( −4,9)( −6)
2 −4,9
( )
−12
± 144 −117,6
∴ t =
−9,8
−12 − 26,4
∴ t =
−9,8 or
− 12 + 26,4
t =
−9,8
∴ t = 1,749 seconds or t = 0,7 seconds
⎧Always
be extremely careful regarding the calculation within
⎪
⎪
the square root sign; especially as far as the sign between
⎪
⎨the
two terms is concerned. DO NOT BE CAUGHT ASLEEP.
⎪Just
one single
error and the entire calculation goes down
⎪
⎪⎩ the drain!
Note that the answer to any word problem MUST have a correct unit of measurement.
It is interesting that there are two moments in time when the ball is at an altitude of 7 meters;
one moment (time-value) is when the ball is ascending; the other time value is when the ball
is descending; on its way downwards it also reaches, for a moment, an altitude of 7 meters.
70

Note that you could also have obtained these values by reading them off the graph:
Remark: The ball falls to the ground; the person who threw it upwards, does not catch it from
the air. Can you see from the graph why I say so?
There are many more very interesting questions which may be asked regarding the
behaviour of the ball:
Example: Calculate the time that the ball takes to reach its maximum altitude and also
calculate this maximum altitude. (see the first graph above where these values have already
been indicated after reading them off the graph)
71

Solution:
There are several ways to answer this question. Perhaps some of the students in the class
are familiar with Differential Calculus; for the rest, however, I gladly give the following hint:
Determine by calculation the co-ordinates of the turning point of the parabola.
2
For the parabola h= at + bt + c the turning point is always located on the vertical dashed
b
line, t =− , which we call the axis of symmetry. This formula gives the exact value of the
2a
horizontal co-ordinate of the turning point.
b
t =− with a=− 4,9 and b=
12
2a
12
∴ t =−
2( −4,9)
= 1, 224 seconds
In order to obtain the vertical co-ordinate of the turning point, simply substitute for t = 1,224
into the formula
2
h =− 4,9t + 12t+ 1 and calculate h :
2
( ) ( )
h =− 4,9 1,224
= 8,347 meter
+ 12 1,224 + 1
(You may also use the hateful formula
catastrophic mess of epic proportions)
72
2 ( b 4ac)
− −
h = , which usually produces a
4a
It takes the ball 1,224 seconds in order to achieve an altitude of 8,347 m.
(the values obtained above closely resemble the values indicated on the graph above – the
small differences in value result from the fact that it is impossible to make perfectly accurate
readings off a graph.)

Example: Determine the total flight time of the ball (how long it was in the air)
Solution:
Simply calculate how long the ball will take to reach the ground, in other words how long it
would take for the ball to reach an altitude h = 0.
Set h= 0 in the formula h=− 4,9t + 12t+ 1and
solve for t:
2
0=− 4,9t + 12t+ 1
2
− b± ∴ t =
b −4ac
2a
where a=− 4,9 b= 12 c=
1
−
∴ t =
± − −
( 12) 2
( 12) 4( 4,9)( 1)
2( −4,9)
2
− 12 ± 144 + 19,6
∴ t =
−9,8
⎧Be
extremely careful with the calculation inside the square
⎪
⎪root
sign, especially with regard to the sign between the
⎨
⎪two
terms. DO NOT BE CAUGHT ASLEEP. One single
⎪
⎩error
and your entire calculation is doomed!
−12 − 163,6
∴ t =
−9,8 or
− 12 + 163,6
t =
−9,8
∴ t = 2,530 seconds or t =− 0,081 seconds
Time may not assume negative values, therefore the negative root we just obtained in the
calculation above is an invalid solution (such a value for t is anyway not an element of the
domain of the function). Only the positive root is therefore a valid solution.
So, the ball takes 2,530 seconds to reach the ground from the moment it left the hand of the
thrower; this is the total flight time.
You are more than welcome to check the graphs above to see if our calculated values agree
with the intersection of the parabola with the time axis.
73

Also study the following discussion which illustrates how we
proceed in order to develop a quadratic mathematical model for a
real life situation when a table of data is given
Example: Determine the equation of the parabola which would fit best through a given
set of data
It can be shown by means of advanced mathematics that the shape of a chain or cable which
supports only its own weight is not quite a parabola. If, however, the cable or chain is
subjected to a distributed load such as a beam suspended from it, then the cable does
indeed hang in the shape of a parabola.
This is the case with the main support cables of a suspension bridge, such as the Golden
Gate Bridge in San Francisco. The driving surface (road) is suspended beneath the main
support cables:
The support cables of the main span hang in the shape of a perfect parabola.
74

Now suppose that die co-ordinates of five points on the main support cables of the bridge is
given:
The representation above represents a simple graphical model and a limited numeric
model for the shape of the cable. The Y-axis passes through the left hand tower and the Xaxis
represents the driving surface (road) for the traffic.
Suppose, however, that we wish to determine the equation of the parabolic curve ABCDE.
This equation (formula) will then be an algebraic model for the shape of the cable. What
follows, is a discussion on how we may determine the equation of a parabola by hand, as
well as by means of Microsoft Excel.
How to determine the equation of a parabola if at least three points on the curve is
given (manual method, by means of pen and paper)
In essence, this comes down to finding the values of a , b and c in the formula
2 ( ) = + + . In this specific case y f ( x)
f x ax bx c
75
= represents the altitude of the cable
above the driving surface and x represents a horizontal distance right of the left hand tower.
Since our problem requires that we solve for three unknown variables, namely a , b and c , it
follows that we must set up a system of three linear equations in three unknowns. These
three equations are obtained by substitution of the co-ordinates of any three points on the
curve into the equation
equations for the three unknowns a , b and c .
2
y = ax + bx + c . Then we simply solve the resulting system of
What follows, is a discussion of how these calculations may proceed:

Solution:
Choose any three convenient points in the table (if possible, choose a point of which the
independent co-ordinate is zero – such a point represents the intercept of the curve on the
vertical axis – if you can identify such a point, then you automatically know the c -value in the
equation
2
y = ax + bx + c ). We may, for example, choose from the given data the following
three points (in order to illustrate the power of the technique, I deliberately choose not to use
the point A ( 0; 152)
which would actually have been a logical point to use):
B ( 320; 39, 5)
, ( 640 2)
C ; and E ( 1 280; 152)
Now substitute the x - and y -co-ordinates of B into the formula
2
( ) ( )
39, 5 = a 320 + b 320 + c
∴ 102 400a+ 320b+ c = 39, 5 [ 1]
Also substitute the x - and y -co-ordinates of C into the formula
2
( ) ( )
2 = a 640 + b 640 + c
∴ 409 600a+ 640b+ c = 2 [ 2]
Also substitute the x - and y -co-ordinates of E into the formula
2
( ) ( )
152 = a 1 280 + b 1 280 + c
∴ 1 638 400a+ 1 280b+ c = 152 [] 3
76
2
y = ax + bx + c ; that yields:
2
y = ax + bx + c ; that yields:
2
y = ax + bx + c ; that yields:
The third-order linear system which consists of equations [1], [2] and [3] may now easily be
solved in any of a number of ways; I gladly illustrate the method known as substitution as
well as the method known as the Rule of Cramer:
Method 1: Substitution (Discussed in WSKT 111, in 2007 and 2008)
Strategy:
Use substitution and eliminate one variable from each of the first two equations in order to
obtain a new equation [4] which contains only two unknowns. Again, use substitution and
eliminate the same variable as before from the last two equations in order to obtain a new
equation [5] which contains only two unknowns. Next, solve the system consisting of [4] and
[5] and use the solutions to solve the other unknown from equation [1].
The strategy proceeds as follows when put into action:

Combine [1] and [2] by doing the following:
From [1] we have that c =−102 400a− 320b+ 39, 5
From [2] we have that c =−409600a− 640b+ 2
So: −102 400a− 320b+ 39, 5 = −409 600a− 640b+ 2
∴ 307 200a+ 320b = −37,
5 [ 4]
Combine [2] and [3] by doing the following:
From [2] we have that c =−409600a− 640b+ 2
From [3] we have that c =−1638 400a− 1 280b+ 152
So:
−409 600a− 640b+ 2 = −1638 400a− 1 280b+ 152
∴ 1 228 800a+ 640b = 150 [] 5
Solve for a and b using any method of your choice; I shall again employ substitution:
From [4] we have that
From [5] we have that
So:
−320b −37,
5
a =
307 200
a =
− 640b + 150
1228800
−320b−37, 5 − 640b+ 150
=
307 200 1 228 800
−320b−37, 5 − 640b+ 150
∴ × = ×
307 200 1 1 228 800 1
∴( −320b− 37, 5)( 1 228 800) = ( − 640b+ 150)( 307 200)
∴−393 216 000b− 46 080 000 = − 196 608 000b+ 46 080 000
∴− 196 608 000b = 92 160 000
92 160 000
∴ b =
−196
608 000
∴ b = −0,
468 750
( 307 200)( 1 228 800) ( 307 200)( 1 228 800)
Substitute b =− 0, 468 750 into [4] and solve for a :
From [ 4 ] we have that
−320b − 37, 5
a = with b = −0,
468 750 so that
307 200
−320( −0, 468 750) −37,
5
a =
307 200
So: a =
0, 000 366 211
77

Substitute a = 0, 000 366 211 and b = − 0, 468 750 into [1] and solve for c :
( ) ( )
102 400 0, 000 366 211
∴ c = 151, 999 994
+ 320 − 0, 468 750 + c = 39, 5
From the analysis above it is apparent that a = 0, 000 366 211,
b =− 0, 468 750 and
c = 151, 999 994 .
The algebraic model for the shape of the cable is therefore
y = x − x+
2
0, 000 366 0, 468 750 151, 999 994
(Note that in this case, where several of the values are bitterly small, we choose not to round
our final answers to three decimal places – it is also mathematically preferable to round all
our answers to the same problem to the same number of decimal places, although it is quite
clear that c = 152 exactly (it is the height of the left hand tower).)
It is left to the reader to check the calculations above by choosing one or more of the other
points, for example the points A, B and C.
See if you can discover how simple the entire calculation becomes when one the
points you use is the intercept on the vertical axis (point A in this case).
Method 2: Rule of Cramer (Discussed in WSKT 111, in 2009)
Strategy:
Calculate by hand or by means of Microsoft Excel the determinants Δ , a Δ , b
78
Δ and Δ c .
Then calculate by hand or by means of Excel the values of a , b and c according to the
following formulas:
Δa
a =
Δ
Δb
b =
Δ
Δc
c =
Δ
The strategy proceeds as follows when put into action:

102 400a+ 320b+ c = 39, 5 [ 1]
409 600a+ 640b+ c = 2 [ 2]
1 638 400a+ 1 280b+ c = 152 [] 3
The augmented matrix of the system is then
⎡ 102 400 320 1 39, 5⎤
⎢ ⎥
⎢ 409 600 640 1 2 ⎥
⎢1638 400 1 280 1 152 ⎥
⎣ ⎦
The four determinants may now easily be calculated:
102 400 320 1
Δ= 409 600 640 1 =− 196 608 000
1 638 400 1 280 1
39, 5 320 1
Δ a = 2 640 1 = − 72 000
152 1 280 1
102 400 39, 5 1
Δ b = 409 600 2 1 = 92 160 000
1 638 400 152 1
102 400 320 39, 5
Δ c = 409 600 640 2 = − 29 884 416 000
1 638 400 1 280 152
The values of the determinants may now be substituted into the formulas for a , b and c :
Δa −72
000
a = = = 0, 000 366 211
Δ −196
608 000
b 92 160 000
b 0, 468 750
196 608 000
Δ
= = = −
Δ −
Δc −29
884 416 000
c = = = 152
Δ −196
608 000
From the analysis above it is apparent that a = 0, 000 366 211,
b = − 0, 468 750 and c = 152 .
The algebraic model for the shape of the cable is therefore
y = x − x+
2
0, 000 366 0, 468 750 152, 000 000
79

How to determine the equation of a parabola if at least three points on the curve is
given (regression by means of Microsoft Excel)
Simply type the three points in table form into an Excel spreadsheet and select all six cells.
Then select “Insert” “Scatter plot” and select all the points on the graph. Select “Add
Trendline” and find the type of curve which best fit through all the points (that will be
“polynomial order 2” if all the selected points lie on a parabola). Then check “Display
Equation on chart”. Excel will then display the equation of the curve for you on the graph.
Remark: Excel employs extremely complicated numeric methods (which follows from
advanced mathematics) in order to do regression..
In the last study unit of this module we shall again encounter regression as a powerful
statistical technique by which the equation of the curve which best fit a given set of data
points may be determined.
80

Now that we have to our disposal an algebraic model for the shape of the cable, we are able
to perform interpolation and we calculate the altitude of the cable at any point between the
two towers. We may, of course, also calculate where (how far from the left hand tower) the
cable has a particular altitude.
Example 1: Determine the altitude of the cable 800 m left of the right hand tower.
Solution:
800 left of the right hand tower implies, of course, that x = 480 must be substituted into the
equation
y x x
2
= 0, 000 366 − 0, 468 750 + 152, 000 000 , since x represents in this model the
distance to the right of the left hand tower. (Recall that 1 280 − 800 = 480 m)
2
So: Set x = 480 into y = 0, 000 366x − 0, 468 750x+ 152, 000 000 :
2
( ) ( )
y = 0, 000 366 480
∴ y = 11, 326
− 0, 468 750 480 + 152, 000 000
The cable is therefore 11,326 m above the driving surface at a distance 800 m left of the right
hand tower.
Example 2: Determine how far from the left hand tower the cable has an altitude of 100 m
above the road.
Solution:
2
Set y = 100 into y = 0, 000 366x − 0, 468 750x+ 152, 000 000 :
2
100 = 0, 000 366x − 0, 468 750x+ 152, 000 000
2
∴ 0 = 0, 000 366x − 0, 468 750x+ 52, 000 000
Solve any quadratic equation easily by making use of the quadratic formula:
− b± x =
2
b − 4ac
with a = 0, 000 366, b = − 0, 468 750 and c = 52
2a
−( − 0, 468 750) ±
∴ x =
2
( −0, 468 750) −4(
0, 000 366)( 52)
2( 0, 000 366)
0, 468 750 ± 0, 219 727 −0,
076128
=
0, 000 732
∴ x = 122, 686 or 1158, 052
The cable is therefore has an altitude of 100 m above the road at a distance of 122,686 m to
the right of the left hand tower and also at a distance of 1 158,052 m right of the left hand
tower.
81

For the sake on interest:
In WSKT 311 we shall encounter a method by which will enable us to determine the exact
length of the cable, by applying Calculus (differentiation and integration) on the equation of
the parabolic curve:
0
1280
⎛dy ⎞
length of the curve = 1 + ⎜ dx
dx
⎟
⎝ ⎠
∫
dy
where = 0, 000 732x − 0, 468 750
dx
which is obtained by differentiating
2
y x x
2
= 0, 000 366 − 0, 468 750 + 152, 000 000 .
It is, however, only possible to employ this advanced calculation after the equation of the
curve is known – which follows from ordinary algebra.
82

Exercise 2.2 for self-assessment
Exercise in the textbook Page number Problems
7.4 231 Do with pencil and paper but check your
answers using GSP:
83
45, 47
Using only GSP: 49, 51
Do with pencil and paper but check your
answers using GSP:
53, 55, 59, 61
Using only GSP: 67, 68
The final answers of the given problems appear at the back of the text book.

2.3 Cubic functions
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Writing cubic equations in the standard form
degree) function;
84
3 2
y = ax + bx + cx + d for a cubic (third
2. Represent cubic functions graphically (by hand as well as using suitable computer
software);
3. Solve simple real life problems where cubic models are involved
Previously acquired knowledge which is needed
1. Study section 1.3
Page through the following material in the book by Washington
You only need to acquire a cursory idea about the content of this section.
Paragraph Page numbers
15.1 – 15.3 416 – 431
Study the following discussion which addresses important,
pertinent points from the reading material given above
Certain processes or situations from the technical or scientific fields of study may be
modelled using cubic (third degree) functions.

Cubic functions give rise to graphs known as cubic curves or third degree curves. Such a
curve may possess a maximum of three intercepts on the horizontal axis – in that case the
curve will also exhibit two turning points. It is also possible that such a curve possess only
two intercepts on the horizontal axis (and therefore only one turning point), or even only one
intercept on the horizontal axis.
How to solve a cubic equation
If we want to solve the equation
3 2
ax bx cx d
+ + + = 0 we may attempt to do so algebraically
by means of factorising and long division (see, for the sake of interest, Paragraph 15.10). In
most cases this is a considerably tedious process which requires theory which we do
not study in the course of the Technical Mathematics modules. The only case we can
indeed handle given our existing knowledge is the case where d = 0 in the equation
3 2
ax bx cx d
+ + + = 0 – in this case we may employ factorising and also the quadratic formula
2
− b± b −4ac
x = .
2a
Example of a cubic model which may easily be handled algebraically:
Suppose that a certain particle is free to move from side to side on a flat plane. Also
suppose that the distance d (in meter) at which the particle is located from its starting point
is given by the formula
d t , t , t
3 2
= 2 − 15 4 + 25 44 where t is expressed in seconds. If 0
85
d > ,
then the particle is to the right of its starting position. Determine at which points in time (that
means, for which values of t ) the particle will be located at its starting point if the motion of
the particle continues for 65 , seconds .
Solution:
Note that we must calculate the value(s) of t when the particle is located at its starting point,
which means when d = 0 . The question therefore means to ask of us:
Solve for t if
− , + , = .
3 2
2t 15 4t 25 44t 0
Since t is a common factor, we may rewrite the left hand side of the equation as follows:
2 ( , , )
t 2t − 15 4t + 25 44 = 0
Note that this implies that t = 0 or that
− , + , = .
2
2t 15 4t 25 44 0

But
− , + , = is a quadratic equation, which we may solve in the usual manner:
2
2t 15 4t 25 44 0
2
− b± b −4ac
t = with a = 2 , b =− 15, 4 and c = 25, 44 . Therefore it follows that:
2a
( 15, 4) (
2
15, 4) 2( 2)
4( 2)( 25, 44)
− − ± − −
t =
15, 4 ±
∴ t =
237, 16 −203,
52
4
15, 4 − 33, 64
∴ t =
4
or
15, 4 + 33, 64
t =
4
∴ t = 24 , or t = 53 ,
Note that there are three values for t , namely t = 0 , t = 24 , and t = 53 , .
That means the particle was initially (when t = 0 seconds) located at the starting point of its
motion, as well as at 24 , seconds after it started moving and then again at 53 , seconds after
the motion started.
This implies that the particle had to turn around and move backwards twice – this must have
happened sometime between t = 0 and t = 24 , and again sometime between t = 24 , and
t = 53 , .
Remark: Further questions may be posed, such as:
1. What is the farthest that the particle travels from its starting point?
2. On which side (left or right) of the starting point does that happen?
3. What is the domain and range of this function?
The first two of these questions may be answered algebraically by means of differentiation
(differentiation is only introduced in the next Technical Mathematics module). The third
question we may already answer in part, since the domain of the function
3 2
= 2 − 15, 4 + 25, 44 is clearly in this case Df{ t 0 t 6,; 5 t }
d t t t
86
= ≤ ≤ ∈ .

In order to determine the range, however, we need to devise a plan to obtain the highest and
lowest values of d . That may be done algebraically by means of differentiation application.
We do not, however, possess these techniques as yet. Therefore, in this module we shall
rather employ a graphical approach. That boils down to simply sketching
d t , t , t
3 2
= 2 − 15 4 + 25 44 accurately and then using the graph to make accurate readings.
How to solve a cubic equation graphically
When we represent the cubic function
the following information first:
• the intercept on the vertical axis
3 2
y = ax + bx + cx + d graphically, we need to establish
• the intercepts on the horizontal axis (these are called the roots of the cubic equation
3 2
ax bx cx d
+ + + = 0 )
• the position of the turning point(s)
We may also, of course, employ a table of values in order to sketch an accurate graph.
In order to obtain the most accurate graph possible, we may use the intercepts, turning
points as well as table of values.
Example of a cubic model which is handled graphically:
Suppose that a certain particle is free to move from side to side on a flat plane. Also
suppose that the distance d (in meter) at which the particle is located from its starting point
is given by the formula
d t , t , t
3 2
= 2 − 15 4 + 25 44 where t is expressed in seconds. If 0
87
d > ,
then the particle is to the right of its starting position. Determine at which points in time (that
means, for which values of t ) the particle will be located at its starting point if the motion of
the particle continues for 65 , seconds .

Solution:
Note that we have to read off the value of t when the particle is located at its starting point,
that is where d = 0 . The question, therefore, actually states:
Read off from the graph of
d t , t , t
3 2
= 2 − 15 4 + 25 44 where the graph intersects the t -axis.
In order to obtain an accurate graph of the function, we may use GSP or Excel.
If however, you opt to use a pencil and paper, you would need to set up a table such as the
following (bear in mind: the more points you include in the table, the more accurate the
resulting plot will be):
If these points are consequently plotted on paper, the result would look as follows:
If you utilise a table, the graph shown above may also be produced using Excel.
(method: First generate the table (You may type the formula into the cell where the first d -
value should appear and then simply copy the formula into the rest of the cells)
Then you high light the entire data set (both rows of numbers) and select “insert”
“chart” “scatter plot”.
88

You should experiment with the other available options and features in Excel in
order to label the axes, draw a smooth curve through the points, adjust the grid,
etc.)
However, when we have the equation of a graph available, it is easier to employ GSP to
sketch the curve. Then we may construct a point on the curve, measure the co-ordinates of
the point and then move the point to whichever location we are interested in, anywhere on
the curve, in order to make an accurate reading from our graph. In this way, we obtain:
From the graphical analyses above, it is clear that the particle had been at its starting
position initially (when t = 0 seconds), as well as after 24 , seconds and after that again
when t = 53 , seconds .
In other words, the particle turned around twice and moved backwards – sometime between
t = 0 and t = 24 , it turned around and again at some time between t = 24 , and t = 53 , .
The facts we obtained above from the graphs, agree with what we obtained from the
algebraic approach followed earlier in the discussion.
89

From the discussion presented above it is apparent that the solutions of the equation
− , + , = is, in fact, the same as the roots of the equation. The roots may be
3 2
2t 15 4t 25 44t 0
obtained graphically by reading off the intercepts on the horizontal axis.
In general: Whenever we wish to solve an equation
we simply sketch the function
90
3 2
ax bx cx d
+ + + = 0 graphically,
3 2
y = ax + bx + cx + d and then read off where the curve
intersects the horizontal axis; these intercepts yield the solutions of the equation
3 2
ax bx cx d
+ + + = 0 .
Furthermore, note that it is clear from the graphs that the range of the function is given by
f
{ 16, 73 64, 3;
}
W = d − ≤ d ≤ d ∈ .
The farthest that the particle moves to the left is therefore 16,73 m and the farthest that it
moves tot the right is 64,3 m.
Exercise 2.3 for self-assessment
Exercise in the textbook Page number Problems
15.3 432 Do with the aid of GSP:
29, 31, 35
The final answers of the given problems appear at the back of the text book.

‘n Battery het ‘n emk (maksimum
moontlike vermoë om stroom te
lewer) van 16 Volt en lewer ‘n
konstante stroom van
I = 10 Ampere . Namate die battery
arbeid verrig, neem die interne
weerstand r (gemeet in Ω ) toe (die
die battery word pap).
Die klemspanning V (dit is die
werklike “sterkte” van die battery),
gemeet in Volt, word gegee deur die
formule V = E − Ir.
1.1 Gebruik die gegewe inligting en toon
aan dat die formule V = 16 − 10r
vir
hierdie battery geld.
1.2 Gebruik die volgende skaal en skets
die kromme met vergelyking
V = 16 − 10r:
1cm = 1Volt
1cm = 01 , Ω
1.3 By watter interne weerstandswaarde
sal die battery heeltemal pap wees?
Assignment 2
Vraag 1/ Question 1
91
A battery has an emf (maximum
possible ability to supply current) of
16 Volt and supplies a constant current
of I = 10 Ampere . While the battery
does work, its internal resistance r
(measured in Ω ) increases (the battery
runs down or becomes “flat”).
The terminal voltage V (that is the
actual “strength” of the battery),
measured in Volt, is given by the
formula V = E − Ir.
Use the given information and show
that for this battery the formula
V = 16 − 10r
holds.
Use the following scale and sketch the
curve of the equation V = 16 − 10r:
1cm = 1Volt
1cm = 01 , Ω
At which internal resistance will the
battery be completely flat?

1.4 Wat is die klemspanning wanneer die
battery splinternuut is (wanneer die
interne weerstand nul is)?
Die temperatuur in grade Celcius op
‘n hoogte y meter bo die
aardoppervlak word soos volg
grafies voorgestel:
2.1 Bepaal die vergelyking van die
grafiek in standaardvorm.
2.2 Bereken hoe hoog bo die
aardoppervlakte sal die
temperatuur 13° C wees.
Vraag 2/ Question 2
92
What is the terminal voltage when the
battery is brand new (when the internal
resistance is zero)?
The temperature in degrees
Centigrade at an altitude y meters
above the surface of the earth is
represented graphically as follows:
Determine the equation of the
graph in standard form.
Calculate how high above the
surface of the earth the
temperature would be 13° C .

‘n Argitek ontwerp ‘n venster sodat dit
die vorm het van ‘n reghoek met ‘n
halfsirkel bo-op:
Die hoogte van die reghoekige
gedeelte is 10 cm meer as die wydte
van die venster.
3.1 Toon aan dat die omtrek P van die
venster as funksie van die radius r
geskryf kan word as:
( ) ( π )
P r = + 6 r+
20
3.2 Teken ‘n grafiek van P teenoor r vir
0≤r≤ 50.
Vraag 3/ Question 3
93
An architect designs a window such
that it has the shape of a rectangle
with a semicircle on top:
The height of the rectangular part of
the window is 10 cm more that its
width.
Show that the circumference
(perimeter) P of the window may
expressed as a function of the radius
r as:
( ) ( π )
P r = + 6 r+
20
Sketch a graph of P against r for
0≤r≤ 50.

3.3 Skryf die waardeversameling van die
funksie ( ) ( π )
P r = + 6 r+
20 neer.
3.4 Bereken die omtrek van die venster
indien die radius 35 cm is en toon
duidelik op die grafiek aan waar u
hierdie waarde sou aflees.
Die vorm van ‘n betonboog in die
struktuur van ‘n gebou word beskryf
deur die vergelyking
1
y x
4
2
=− + 7 .
Die vloer van die gebou (grondvlak)
word as die X-as gekies; alle
afmetings is in meter.
4.1 Skets die boog en toon sy hoogte en
die breedte van die basis duidelik
aan.
Vraag 4/ Question 4
94
Write down the range of the function
( ) ( π )
P r = + 6 r+
20.
Calculate the circumference of the
window when the radius is 35 cm and
clearly indicate on the graph where
you would read off this value.
The shape of a concrete arch in the
structure of a building is described by
1 2
the equation y =− x + 7 .
4
The floor of the building (ground
level) is taken as the X-axis; all
dimensions are given in meters.
Sketch the arch and clearly indicate its
height and the width of its base.

Die vorm van die draagkabels van ‘n
hangbrug word deur middel van ‘n
funksie h= f ( x)
voorgestel, waar
( )
2
f x ax bx c
= + + :
5.1 Bepaal die waardes van a , b en c in
2
die funksie f ( x) = ax + bx+ c.
Vraag 5/ Question 5
95
The shape of the supporting cables of
a suspension bridge is represented by
a function h= f ( x)
where
( )
2
f x ax bx c
= + + :
Determine the values of a , b and c
2
in the function f ( x) = ax + bx+ c.

Die hoogte van ‘n projektiel word deur
middel van ‘n funksie h = f ( t)
2
voorgestel, waar f () t = at + bt+ c:
6.1 Bepaal die waardes van a , b en c in
2
die funksie f () t = at + bt+ c.
'n Silinder wat
3
5m vloeistof kan hou
word vervaardig deur ‘n vel
plaatmetaal op te rol in ‘n buis en ‘n
sirkelvormige skyf vir die bodem te
sny en dit onderaan die buis vas te
sweis:
Vraag 6/ Question 6
Vraag 7/ Question 7
96
The altitude of a projectile is
represented by a function h = f ( t)
2
where f ( t) = at + bt+ c:
Determine the values of a , b and c
2
in the function f ( t) = at + bt+ c.
A cylinder which must be able to hold
3
5m of liquid is manufactured by
rolling a sheet of sheet metal in a tube
and cutting a circular disk for the
bottom and welding it onto the tube:

7.1 Van watter twee veranderlikes hang
die grootte van die buite-oppervlakte
af?
7.2 Bewys dat die buite-oppervlakte van
die silinder in die vorm van ‘n funksie
uitgedruk kan word as
2 10
( ) π
A r = r + .
r
7.3 Voltooi die tabel hieronder en gebruik
dit om ‘n netjiese, akkurate grafiek
van die funksie ( ) 2 10
A r = π r +
r
te
skets. Gebruik ‘n skaal van
3 cm = 1eenheid op die x -as en ‘n
skaal van 1cm = 5 eenhede op die
vertikale as.
97
On which two variables does the
magnitude
depend?
of the surface area
Show that the surface area of the
cylinder may be expressed in the
form of a function as ( ) 2 10
.
A r = π r +
Complete the table below and use it
to sketch a neat, accurate graph of
the function ( ) 2 10
A r = π r +
r
. Use a
scale of 3 cm = 1unit on the x -axis
and a scale of 1cm = 5 eenhede on
the vertical axis.
r

U hoef nie die tabel in u skrif oor te
teken nie.
7.4 Gebruik u grafiek en skat die waarde
van r waarvoor die buite-oppervlakte
‘n minimum sal wees en dui dit
duidelik op u grafiek aan.
7.5 Gebruik u antwoord op vraag 2.4 en
bereken die minimum moontlike
buite-oppervlakte van die silinder.
Toon hierdie waarde duidelik op u
grafiek aan.
7.6 Wat is die grootste aantal draaipunte
wat ‘n tweedegraadse grafiek kan hê?
7.8 Noem enige vier maniere maniere
waarop ‘n werkliksgetroue situasie
wiskundig voorgestel kan word.
98
You do not need to copy the table into
your answering script.
Use your graph to estimate the value
of r so that the surface area is a
minimum and clearly indicate it on
your graph.
Use your answer to question 2.4 and
calculate the minimum possible surface
area of the cylinder. Clearly indicate
this value on your graph.
What is the highest number of turning
points that a quadratic curve can have?
List any four ways in which a real life
situation my be represented
mathematically.

'n Kartonhouer word vervaardig deur
'n reghoekige vel karton, 45 cm by 30
cm te neem en uit elke hoek 'n
vierkantige stuk karton te sny. Die
kante word dan boontoe gevou om die
wande van die houer te vorm:
8.1 Laat x die sylengte wees van elke
vierkant wat uit die karton gesny word.
Toon aan dat die volume van die
houer gegee word deur die formule
3 2
V 4x 150x 1350x
= − + .
Vraag 8/ Question 8
99
A cardboard container is manufactured
from a rectangular sheet of cardboard,
45 cm by 30 cm. A square piece of
cardboard is cut from each of the four
corners and the sides are then folded
upwards to form the sides of the
container:
Let x be the side length of each of the
squares which are cut out of the sheet
of cardboard. Show that the volume of
the container is given by the formula
3 2
V 4x 150x 1350x
= − + .

8.2 Dit is moontlik om die formule hierbo
soos volg te skryf:
( ) 3 2
V x = 4x − 150x + 1350x,
Wat presies beteken hierdie
skryfwyse?
Wenke:
• Wat is die onafhanklike
veranderlike?
• Op watter as van ‘n grafiek sal ons
die onafhanklike veranderlike
aandui?
• Wat is die afhanklike veranderlike
en hoe sien u dit?
• Op watter as van ‘n grafiek sal ons
die afhanklike veranderlike
aandui?
• Wat moet ons weet voordat ons
die volume-waarde kan bereken?
8.3 Wat word bedoel met die
definisieversameling van die
funksie?
(Wees spesifiek in u verduideliking)
100
It is possible to write the formula above
as follows:
( ) 3 2
V x = 4x − 150x + 1350x,
What is the precise meaning of this
notation?
Hints:
• What is the independent variable?
• On which axis of the graph shall we
indicate the independent variable?
• What is the dependent variable and
how do you see that?
• On which axis of the graph shall we
indicate the dependent variable?
• What must we know before we can
calculate the volume-value?
What do we mean by the domain of
the function?
(Be explicit in your explanation)

8.4 Skryf die definisieversameling van
hierdie funksie neer deur middel van
die korrekte notasie en simbole.
8.5 Voltooi die volgende tabel van
waardes en stip (“plot”) die grafiek van
volume teen x :
8.6 Gebruik die grafiek en skat watter
waarde van x lewer die grootste
volume.
Wat is die maksimum moontlike
volume van die houer?
8.7 Skryf die waardeversameling van
hierdie funksie neer deur van die
korrekte notasie en simbole gebruik te
maak.
8.8 Gebruik Microsoft Excel en doen nr
7.5 hierbo op rekenaar.
8.9 Gebruik GSP en teken die grafiek van
die funksie
( ) 3 2
V x = 4x − 150x + 1350x.
101
Write down the domain of this function
by utilising the correct natation and
symbols.
Complete the following table and plot
the graph of volume against x :
Use the graph and estimate which
value of x yields the highest volume.
What is the maximum possible volume
of the container?
Write down the range of this function by
utilising the correct notation and
symbols.
Use Microsoft Excel and do nr 7.8
above on the computer.
Use GSP and sketch the graph of the
function ( ) 3 2
V x = 4x − 150x + 1350x.
The solutions to the problems presented above are discussed during lectures. The
solution of certain problems may be published on eFundi.

102

3 Rational functions, exponential functions and
logarithmic functions
Estimated time required to master the learning outcomes
12 hours
Previously acquired knowledge which is needed
1. Study Unit 1
2. How to solve an equation which contains fractions by multiplying each of the terms by
the LCM of the denominators (WSKT 111, Study Section 3.5)
3. How to solve exponential and logarithmic equations (WSKT 111, Study Unit 5)
Learning outcomes for this study unit
After completing this study unit the student must be capable of doing the following:
1. Write simple rational equations in the standard form for a rational function
103
k
y = ;
x
2. Represent rational functions graphically (manually (pencil and paper) as well as using
suitable computer software);
3. Solve simple real life problems where rational models are involved;
4. Write exponential equations in the standard form for an exponential function, namely
kx
y = Ae or
kx
y Ae −
kx
= y A( 1 e ) −
= − ;
5. Represent exponential functions graphically (manually (pencil and paper) as well as
using suitable computer software);

6. Solve simple real life problems where exponential models are involved;
7. Write logarithmic equations in the standard form for a logarithmic function, namely
y = klogb ax;
8. Represent logarithmic functions graphically (manually (pencil and paper) as well as
using suitable computer software);;
9. Solve simple real life problems where logarithmic models are involved.
Revision of previous study units
From Study Section 1.3 follows:
A function is a special kind or rule according to which a value (called the dependent
variable) may be calculated by substituting another value (called the independent
variable) into a certain algebraic equation (called the model).
Functions are used to describe processes or situations in the real world. We use function
in order to set up mathematical models.
What is a mathematical model?
Well, it is a mathematical representation of a physical situation. Real life and most
technological applications have to do with complicated problem situations.
Such situations may however be simplified by considering one or two or maybe three of their
measurable aspects at a time. These measurable aspects are called variables. The way in
which one variable depends on another, is expressed in the form of a formula (mathematical
equation).
This formula is called a mathematical model.
104

There are at least five ways in which a real life situation or problem (or the function which
describes it) may be represented, namely:
• a numeric description (a table of measured or calculated values)
• a graphic description (a curve sketched on a co-ordinate plane with axes)
• a verbal description and/ or a schematic description such as a diagram
• an algebraic description (a formula or equation)
• a practical example which illustrates how the process or problem behaves
In this study unit we study real life problems which may be modelled using one of the
following types of functions:
1. Rational functions (power functions where the exponent is negative, typically
1
something like y k x −
k
= ⋅ , which effectively means the same as y = where k is any
x
real-valued constant)
2. Exponential functions (
kx
y = Ae or
growth functions and decay functions)
3. Logarithmic functions ( y = klogb ax )
kx
y Ae −
105
kx
= or y A( 1 e ) −
= − , which represent
These types of equations are not polynomial functions. Thus, they cannot be written in
n
the form y = a x
n−1 n−2 n−3
+ a x + a x + a x +
0
+ a x .
The functions
0
k
y = ,
x
1 2 3 ...
kx
y = Ae ,
kx
y Ae −
kx
= or y A( 1 e ) −
n
= − and y = klogb ax possess certain
specific algebraic and graphical properties which make them useful for representing certain
types of problems for which polynomial functions are not well suited.
Examples of such problems that polynomial functions are not well suited for are the
following:
• Inverse or indirect proportionality (also called: inverse variation)

• Situations which are described by discontinuous functions (functions which are such
that there exist certain values of the independent variable for which we are unable to
calculate a corresponding value for the dependent variable – functions of which the
graphs exhibits a “jump” or an “opening”)
• Situations where steady growth or decay are involved
• Situations where restricted or limited growth are involved
The functions mentioned above are discussed in the following study sections. In each case
we consider a typical situation or problem from the technical or scientific fields of study where
that specific type of function is used in order to model the situation or problem.
Note that the heuristic which we encountered in Study Section 2.1, is still used in exactly the
same way as before whenever we model any real life situation or problem.
For the sake of revision, we repeat here the steps of the heuristic:
The steps or the heuristic by which we may analyse any real life problem situation
1. Obtain two sets of values (data) through measurement. (The one set forms the
independent variable; the other set forms the dependent variable).
2. Represent the values (usually measurements) in the form of a table.
3. Use the table in order to obtain an accurate graph.
4. Use the pattern (shape) in which the data points lie on the graph and construct a best line
or curve through the points in such a way that the line or curve follows the shape of the
points.
5. Use this “best line” or curve in order to do interpolation and extrapolation by means of
making readings off the graph.
6. In the event that accurate inter- and extrapolation is required, obtain the equation of the
“best line” or curve which you obtained in step 4. This equation is the function
(mathematical model) which describes the problem situation you are investigating.
7. As soon as you obtained the equation of the regression line or regression curve, you may
use it in order to perform interpolation and extrapolation, by means of substitution and
calculation.
8. The behaviour of the function may now be described from the graph and from the
equation (formula) in terms of concepts like:
• Increasing/ decreasing
106

9. Concave up/ concave down (see Fig. 24.31 (a) on p. 709 of the textbook)
• Maximum values/ minimum values as obtained from turning points and edge values
• Vertical asymptotes (we discuss this later – this is where the graph performs “jumps”)
• Horizontal asymptotes (these are horizontal lines which the graph approach, but never
intersects or crosses)
• Domain
• Range
Note how we use this heuristic in one form or another as we work through the sections which
follow.
107

3.1 Rational functions
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Write simple rational equations in the standard form for a rational function
108
k
y = ;
x
2. Represent rational functions graphically (manually (pencil and paper) as well as using
suitable computer software);
3. Solve simple real life problems where rational models are involved
Study the following material in the book by Washington
Paragraph Page numbers
3.4 92 – 96
Rational function models and indirect proportionality (also called
inverse proportionality or inverse variation)
Whenever a real life problem or situation may be described by a power function with an
exponent of − 1 (in other words, an equation such as
1
y k x −
= ⋅ ) or be represented by a
rectangular hyperbola then it is an indication of the fact that the quantity represented on
the vertical axis (the dependent variable) is inversely proportional to the quantity
represented on the horizontal axis (the independent variable).
k
The equation of such a rectangular hyperbola was in school written as . y = .
x

An example of inverse proportionality from the theory of electrical circuits
A simple experiment is performed in order to investigate the relationship between the current
strength I and the resistance R in an electrical circuit while the voltage V is kept constant.
The experimental setup consists of a battery (container of cells connected in series), a light
bulb with an adjustable brightness, an ammeter (to measure current) and a volt meter (to
measure voltage).
Initially, at maximum brightness, the light bulb has a resistance value of 10 Ω which causes
the ammeter to register a current of 0,6 A :
The brightness of the light bulb is now steadily decreased by increasing the resistance of the
light bulb in increments (steps) of 10 Ω until the resistance is 70 Ω .
The ammeter reading is recorded during each increment and tabulated together with the
resistance values.
The following results are obtained:
109

In order to obtain a better impression regarding the relationship between the current strength
(brightness of the light) and the resistance of the light bulb, we may represent the information
graphically..
Since the resistance values at which we measured the current were independent of any
factor except our own choice, the resistance R is the independent variable and therefore it
will be represented on the horizontal axis; according to a similar argument the current I
(of which the value depended on the resistance value at which it was measured) must be
the dependent variable and therefore it will be represented on the vertical axis.
Consequently, we may consider current as a function of resistance and it may be written as
follows:
I = f ( R)
It is already clear from the table of data that the domain of the current function is:
f
{ 10 70; }
D = R ≤ R≤ R∈
It is equally clear from the table of data that the range of the current function is:
f
{ 0,086 0,6; }
W = I ≤I ≤ I∈
110

The resulting graph contains seven data points and we may perform interpolation by joining
the points by means of a smooth curve; the shape of this curve leads us to suspect that it is
a rectangular hyperbola:
In order to confirm that a given curve is indeed a rectangular hyperbola we may apply
the following simple test:
Establish whether the product of the dependent and independent co-ordinates of each point
on the curve yields a constant value k ; if so, then we are without doubt dealing with a
rectangular hyperbola.
It is clear that the product of I and R does indeed in the case of each of the seven data
points in the table yield the same value, namely 6. Therefore, we conclude that the curve is
indeed a rectangular hyperbola with constant k = 6 .
Furthermore, the shape of the curve clearly suggests that if the resistance increases in a
certain ratio, then the current decreases in the same ratio. If, for example, the resistance
increases from 20 Ω to 60 Ω (triples in magnitude), then the current decreases from 0,3 A
to 0,1 A (becomes three times smaller in magnitude). This kind of relationship between
dependent and independent variable is called inverse proportionality.
111

The representations above are, however, only schematic, numeric, graphical and verbal
models; in order to perform accurate and proper interpolation and extrapolation, we require
an algebraic model for the situation – in other words, we must determine the equation of the
curve.
It is known from school mathematics that the equation of a rectangular hyperbola is
in terms of our situation the equation is
6
I = .
R
So it is very easy to determine the equation of a simple rational algebraic model.
112
k
y = ;
x
We may, of course, also make use of the built-in regression feature of Microsoft Excel in
order to obtain the equation of the curve which best fit the data points on the graph. As
before (in Study Section 2.2) we may proceed as follows:
Simply type the data points in table form into an Excel spreadsheet and select all the data
cells. Then select “Insert” “Scatter plot” and select all the points on the graph. Select “Add
Trendline” and find the type of curve which best fit through all the points (that will be “power”
if all the selected points lie on a rectangular hyperbola). Then check “Display Equation on
chart”. Excel will then display the equation of the curve for you on the graph.

Remark: You will recall from your knowledge of exponent laws that the equation
which is displayed by Excel may be written as
6
y = .
x
113
y 6x
−
=
Also note that Excel is unfortunately unable to display the equation in terms of our symbols I
and R ; it uses the symbol x for the independent variable and the symbol y for the
dependent variable. So in terms of our symbols, Excel agrees that
6
I = .
R
6
Now that we have the algebraic model I = to our disposal, we may perform interpolation
R
(calculate, for example, the current value when the resistance is 24 Ω , or calculate the
resistance whereby the current has a value of 0,5 A ) as well as extrapolation (calculate, for
example, what the value of the current would be if the resistance could assume the value 1Ω
, or determine the resistance which would correspond to a current of 0,0001 A ).
If, however, we sketch the function
6
I = by means of GSP (of course, we shall have to type
R
6
it in as y = ) and we sketch the graph on the interval −10 ≤ x ≤ 100 then we obtain the
x
following interesting curve:
1

It is clear that the function
6
y = exhibits extraordinary behaviour in the vicinity of the point
x
x = 0 ; apparently the curve “breaks” or “jumps” where x = 0 and this means that there exists
no y -value where x = 0 . Such a point (in this case the point x = 0 ) where a curve “breaks”
or “jumps” is called a discontinuity.
6
The question now arises: What information regarding our mathematical model I = ,
R
which states that the current in the circuit is inversely proportional to the resistance,
may we deduce from the occurrence of a discontinuity?
There are at least three important observations which may be made:
• In the first place, we must note that the resistance can never physically assume a
zero value; if that were to happen, the current value would tend to infinity. Anyone who
possesses some scientific knowledge knows that the heating effect of an electrical
current strongly depends on the magnitude of the current; a very large current typically
causes the conducting wires and components to fuse, resulting in the destruction of the
circuit. Mathematically we write I →∞ if R → 0 . We may also write
means the same.
114
lim I = ∞ ; that
k
• In the second place we must note that the graph of a rational function I = never
R
intersects the vertical line R = 0 (in this case it is the same as the vertical axis).
Therefore we call the line R = 0 a vertical asymptote for the function
R→0
k
I = .
R
• If R = 0 , we are unable to calculate a corresponding I -value, since the equation of the
6
function I = would then require us to divide by zero:
R
6
I R=
0 0
which means an infinity large, undefined value
=
=∞
In essence then, a discontinuity at a certain point indicates that division by zero
would occur at that value in the equation of the graph.

It is, however, also clear that the function
value of x becomes very large:
6
y = exhibits extraordinary behaviour where the
x
Apparently the value of y tends to zero, but never achieves zero (if so, then we would
observe the curve intersecting the X-axis). It seems as if the value of y simply shrinks away
as x increases. We may use a computer to check what becomes of y as x increases to a
very large value:
115

The computer-generated table above supplies interesting insight regarding our algebraic
6
model I = .
R
The following two remarks are worth considering:
• In the first place, the fact that the curve always lies above the current-axis implies that
I → 0 if R →∞. Physically, this implies that the current becomes smaller and smaller as
the resistance becomes larger and larger, but that the current never actually reaches a
value of zero. We may express this mathematically as lim I = 0 .
k
• In the second place we should note that the graph of a rational function I = never
R
intersects the horizontal line I = 0 (in this case it is the same as the horizontal
axis). Therefore we call the horizontal line I = 0 a horizontal asymptote for the function
k
I = .
R
So we see that our investigation brought several extremely interesting aspects of indirect
proportionality and rational functions to light.
116
R→∞

Also study the following discussion which illustrates how more
complicated rational functions may be used for mathematical
models
Suppose a patient is injected intravenously with a medicine and the concentration C of the
medicine in his blood stream is measured every half an hour for the consecutive ten hours.
The measurements are represented numerically and graphically:
The asymmetrical shape of the curve betrays the fact that the algebraic model for this
situation is cannot be a linear function, or a quadratic function, or a cubic polynomial function,
or a simple rational function. Complicated mathematical techniques (which we do not study
in WSKT 221) must be employed in order to determine the equation of this curve. In this way
25t
it may be established that the curve of the function C() t = fits the data points on
2
t + 2t + 1
the curve almost perfectly.
So, rational functions may possess more complicated equations than
example
y
f ( x)
= where both ( )
g ( x)
117
k
y = , such as for
x
f x and g ( x ) are polynomial functions. In the case of our
25t
example the equation of the rational function model is C = 2
t + 2t + 1
.

Such complicated rational functions (as well as simple rational functions, such as for
6
example I = ) exhibit the interesting property that there exist values of the independent
R
variable for which it is simply impossible to calculate a corresponding value of the dependent
variable.
It is true of all rational functions that they are defined in terms of algebraic fractions; the
25t
function C() t = , for example, is defined in terms of a numerator (namely 25t ) and
2
t + 2t + 1
a denominator (namely 2
t + 2t+ 1).
Because the division operation may only be performed as
25t
long as division by zero does not occur, it is clear that the function C() t = 2
t + 2t + 1
only
exists as long as 2
t + 2t+ 1≠ 0;
whenever 2
t + 2t+ 1= 0,
then division by zero takes place and
then it is impossible to calculate the value of C .
Since the value t = − 1 does not appear in the table above, the function seems to be
continuous (and it is indeed continuous for t ≥ 0 ). Also, recall that time may actually not
25t
assume negative values. Therefore, the model C() t =
is valid for all practical
2
t + 2t + 1
purposes.
Check, however, what happens if you attempt to calculate C when t =− 1 in the function
() = 2
C t
25t
t + 2t + 1
.
25t
We say that the rational function C() t = 2
t + 2t + 1
118
is undefined for t =− 1.
That means that
the function is discontinuous for t = − 1 and by that we mean that the graph of the function
() = 2
C t
25t
t + 2t + 1
executes a jump or an opening where t = − 1.
25t
The question now arises: What does the graph of C() t = 2
t + 2t + 1
look like where t = − 1?
25t
On the next page a graph of C() t = for 3 8
2
t + 2t + 1
t − ≤ ≤ is represented; the graph was
created using GSP:

It is clear that the graph exhibits extraordinary behaviour in the vicinity of the point t = − 1 on
the time axis.
This type of behaviour is, as before, referred to as a discontinuity.
• Graphically a discontinuity is characterised by a place where the curve is broken (or
where the curve “jumps”)
• Algebraically it is characterised by an illegal mathematical operation with the
equation of the curve (in this case, division by zero occurs at t =− 1;
if this value is
25t
substituted into the equation C() t = we obtain (as we have already seen):
2
t + 2t + 1
C
( 1)
25( −1)
(
2
) ( )
− =
− 1 + 2 − 1 + 1
−25
=
1− 2+ 1
25
=−
0
Some computers/ calculators give the answer of the calculation above as C ( − 1)
= −∞; it
is apparent from the graph that this answer means that the value of C tends to an
infinitely large negative value as the value of t approaches − 1.
119

It is also interesting that the time axis is the horizontal asymptote of the function
() = 2
C t
25t
t + 2t + 1
:
From this fact we may draw the following conclusions:
• Graphically it means that the curve approaches the time-axis as the time value
increases, but that the curve never touches or intersects the time-axis.
• Physically it means that the concentration of the medicine in the blood of the patient
continually decreases at an ever decreasing rate as time proceeds but that the
concentration of the medicine (mathematically speaking) never reaches a value of zero.
120

Exercise 3.1 for self-assessment
Exercise in the textbook Page number Problems
3.4 96 Sketch manually by means of pencil and
paper and check your results using GSP:
121
29, 30, 31
53 (GSP only)
3.6 106 In the following problems you must
sketch the graph in Microsoft Excel.
Label the axis and let the computer draw
a smooth curve through the data points:
3, 5, 7
The final answers of the given problems appear at the back of the textbook.

3.2 Exponential functions
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Write exponential equations in the standard form for an exponential function, namely
kx
y = Ae or
kx
y Ae −
kx
= or y A( 1 e ) −
= − ;
2. Represent exponential functions graphically (manually (pencil and paper) as well as
using suitable computer software);
3. Solve simple real life problems where exponential models are involved
Study the following material in the book by Washington
Paragraph Page numbers
13.1 371 – 372
13.2 373 – 377
Exponential models as a means to mathematically model growth
and decay
Most people are familiar with the concept of compound interest – when a client invests a
certain amount of money in a special account so that the money earns interest, in other
words: the amount of money “grows” at a rate dependent on the interest rate. The final
value of the investment depends on the magnitude of the initial deposit, the interest rate and
the duration of the investment period.
The most interesting feature of investment at compound interest is the fact that the money
earns “interest” upon “interest” – so, the longer the money remains in the account, the faster
it grows.
At any point in time the growth rate of the investment is proportional to the magnitude
(value) of the investment at that moment.
122

A typical graph of an investment at compound interest rate is depicted below:
Note that the curve depicted above seems to resemble part of a parabola or hyperbola; it is,
however, most certainly neither a parabola nor a hyperbola, because:
• This real life situation deals with steady growth and not with a situation where a turning
point (extreme value) is involved.
• Also, if we analyse the behaviour of the function for all real values of time, we notice that
the curve does not show any turning points; neither does it execute any jumps or breaks
which means that the function is, in fact, continuous for all real values of the independent
variable.
123

While it is true that the model depicted above only has physical meaning for positive and
zero values of time, the graph is convincing: there are no turning points or breaks or jumps
(discontinuities) in the curve – so this is neither a quadratic function nor a rational function.
Similarly, most people are aware of the fact that the value of certain assets, such as motor
vehicles, loses value (depreciates) at a certain rate as it ages. Very interesting is the fact
that the vehicle loses value quickly when it is new, while the older it gets, the better it keeps
its value.
A typical graph of the depreciation of a car is depicted below:
Note that the second curve depicted above again seems to resemble a rectangular
hyperbola; it is, however, most certainly not a rectangular hyperbola, because:
• This context deals with steady decay and not inverse proportionality.
• Also, if we analyse the behaviour of the function for all real values of time, we notice that
the curve executes no jumps or breaks which means that the function is, in fact,
continuous for all real values of the independent variable:
124

Once again: while it is true that the model depicted above only have physical meaning for
positive and zero values of time, the graph is convincing: there are no turning points or
breaks or jumps (discontinuities) in the curve – so this is neither a quadratic function nor a
rational function.
So we see that the functions depicted above differ radically from any of the other functions
studied so far.
These particular types of behaviour are not restricted to economical contexts; we only used
the examples above to introduce the concept of exponential growth and exponential decay
because most people already have an intuitive feeling for these situations.
Next, we shall proceed to consider contexts from the technological and scientific fields of
study where exponential growth and exponential decay occurs; in doing so we shall discover
the relevant characteristics of exponential functions which make these functions so useful.
125

Investment at compound interest rate
If you invest an amount P in a bank account for t years at an interest rate of r % interest
per year, calculated yearly, then the bank uses the following formula to calculate the final
amount A which you will receive from the bank after the investment period is complete:
⎛ r ⎞
A= P⎜1+ ⎟
⎝ 100 ⎠
t
(This is a mathematical model – a formula which describes a real life
process)
As you can see, the final amount A depends on various variables, namely the magnitude of
the initial deposit P , the interest rate r and the duration t of the investment. But in real life
P is a fixed amount (for example R3000) which you deposit at the bank, which in turn agrees
to give you an interest rate r (for example 10%). So the time t is the only variable which
influences the growth of the investment (except of course if the bank changes the interest
rate sometime during the investment period or if you draw some money or deposit an
additional amount – but for the sake of this discussion we will keep matters simple).
So, with the amount of R3000 and an interest rate of 10%, we may write the mathematical
⎛ r ⎞
model (formula) A= P⎜1+ ⎟
⎝ 100 ⎠
t
as follows:
3000 11 t
Of course, this simplifies to: A = ( , )
⎛ 10 ⎞
A = 3000⎜1+ ⎟
⎝ 100 ⎠ .
Note that the final value A of the investment is a variable which depends only on the value of
the variable t (the number of years the investment takes); so we may say that A is a
function of t :
() = ( )
3000 11 t
A t ,
Note that the right hand side of the equation consists of a co-efficient (namely 3000), as well
as a base (namely 1,1) and a symbolic exponent or independent variable , namely t .
Therefore, the right hand side of the equation is an exponential expression. That is why
we call the equation A() t ( , )
3000 11 t
= an exponential function.
126
t

Let us now consider a graph of this function for the first 10 years of the investment:
Suppose now that we consider the following depreciation problem.
Depreciation on the value of a vehicle, calculated at compound rate
If you purchase a vehicle for a price P and you own the vehicle for t years, then
depreciation occurs at a rate of r % per year. The value of r here depends on economic
factors such as inflation, but also on the durability of the vehicle and on how often you use it.
Vehicle salesmen normally use the following formula to calculate the value A of the vehicle
after t years:
⎛ r ⎞
A= P⎜1− ⎟
⎝ 100 ⎠
t
(This is a mathematical model – a formula which describes a real life
process)
As you can see, the final value A of the vehicle depends on several matters, namely the
magnitude of the purchasing price P , the depreciation rate r and the duration t for which
you owned the vehicle.
127

Now assume that the purchasing price P of a vehicle was R110 000 and that the
depreciation rate r was 10% per year. Then the time t is the only variable which influences
the process by which the vehicle loses value (except if during the time you owned the vehicle
you made modifications which increased its value, had an accident which damaged the car
and decreased its value, or something like that – for the sake of this discussion, however, we
shall not focus on such complications).
So, with a purchasing price of R110 000 and a depreciation rate of 10%, we may write the
⎛ r ⎞
mathematical model (formula) A= P⎜1− ⎟
⎝ 100 ⎠
110 000 0 9 t
Of course, this simplifies to: A = ( , )
t
128
as follows:
⎛ 10 ⎞
A = 110 000⎜1− ⎟
⎝ 100 ⎠ .
Note that the final value A of the vehicle therefore only depends on the time t (the number
of years the vehicle remained in your possession); therefore, we may say that A is a function
of time t :
() = ( )
110 000 0 9 t
A t ,
Note that the right hand side of the equation consists of a co-efficient (namely 110 000), as
well as a base (namely 0,9) and a symbolic exponent or independent variable , namely t .
Therefore, the right hand side of the equation is an exponential expression. That is why
we call the equation A() t ( , )
110 000 0 9 t
= an exponential function.
t

Let us now consider a graph of this function for the first 10 years that the vehicle remained in
your possession:
By comparing the examples above, it is possible to make several interesting observations;
among other matters, it is clear that an exponential equation with a base greater than 1 is
an increasing function (think: “investment at compound interest”), while an exponential
function with a base smaller than 1 is a decreasing function (think: “depreciation on a
vehicle”).
Just bear in mind that the value of t must always be greater or equal to 0 – time may not
assume negative values.
We may apply these observations to any exponential function of the form
is a positive number and kx is a positive number.
• If b < 1,
then
• If b > 1,
then
kx
y = a⋅ b decreases as x increases
kx
y = a⋅ b increases as x increases
129
kx
y = a⋅ b where a

Next we shall consider situations from technical and scientific contexts where exponential
growth and exponential decay are involved; as we do so, we shall discover the extremely
useful features of exponential functions which we exploit in order to model certain real life
situations.
Exponential functions with the natural base e
You will notice that we prefer to work with the exponential functions
kx ( 1 )
130
kx
y = Ae or
kx
y Ae −
= or
y A e −
= − whenever we model real life processes from technical or scientific contexts.
In these formulas the base of the exponent is always the so-called natural base e .
This number (the natural base e ) is a fundamental constant in nature. The exact value of it,
accurate to nine decimal places, is accepted as e = 2, 718 281828 . This value is derived in
advanced mathematics courses – we are not interested now in the exact theory behind the
value and derivation of the number e . The exact value of e may in any case be easily
obtained from your calculator.
For our purposes e is a convenient base to use whenever we model gradual growth or
decay in nature. The main reason for this is that the equations
kx ( 1 )
kx
y = Ae and
kx
y Ae −
= and
y A e −
= − lead to simple computational procedures in advanced mathematics; also, the
curves of the equations
kx
y = Ae or
kx
y Ae −
kx
= or y A( 1 e ) −
= − fit the data points obtained
from experiments where gradual growth or decay is involved extremely well.
Since e > 1 the graph of the function
characteristic of exponential growth. Similarly, the graph of
always be decreasing; this is characteristic of exponential decay.
kx
y = Ae with k > 0 will always be increasing; this is
kx
y Ae −
= with k > 0 will
So, we agree that the constant k in these formulas will always be a positive value; the
closer the value of this constant is to zero, the slower the growth or decay modelled by the
particular function. The bigger the value of k , the faster the growth or decay modelled by
the function. That is why mathematicians often refer to k as the growth constant.

Exponential models in technical and scientific contexts
An example of exponential growth in biology or natural science:
Bacteria which grows in a fertile medium
Note that this discussion will be held in class.
131

3.3 Logarithmic functions
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Write logarithmic equations in the standard form y = klogb ax of a logarithmic
function;
2. Graphically represent logarithmic functions (by hand as well as by utilising suitable
computer software);
3. Solve simple real life problems where logarithmic models are involved.
Study the following material in the book by Washington
Paragraph Page numbers
13.1 370 – 372
13.2 373 – 377
Background
Logarithmic functions occur in many contexts from the technical and scientific fields of study,
such as:
• Operational amplifiers (electronics)
• pH-values (The chemistry of acids and bases)
• the intensity of earthquakes
One of the best uses of logarithmic functions is the way it enables us to express very large or
very small numbers in a manageable format.
132

Consider, for example, the following situation:
Let a assume the value
12
3, 456 × 10 and let b assume the value
133
11
1, 713 × 10 . Now
represent a and b according to scale on the same number line. Let 1 cm represent one
unit.
Solution:
Goodness! If 1 cm represents one unit then a will be located 34 560 000 km right of the
zero point and b will be located 1713 000 km right of the zero point.
It is clearly impossible to represent these numbers according to scale on the same number
line.
Logarithms enable us to solve this problem as follows: Let us represent the numbers loga
and logb graphically on the same number line. Let 1 cm represent one unit.
12
( )
log a = log 3,456 × 10
= 12,539 Using your calculator!
11
( )
logb = log 1,713 × 10
= 11,234 Using your calculator!
Representation:
You can see how easily be may plot the positions of the logarithms of these numbers. Note,
however, that the scale of the number line is now 1 cm = 10, 2 cm = 100, 3 cm = 1000 and
so on.
When dealing with operational amplifiers, pH-values and the intensity of earthquakes we also
work with values or measurements which are very large or very small; therefore, in the case
of these types of problems we use logarithms in order to “convert or transform” very large or
very small numbers to more convenient values.

Example of a logarithmic function in electronics: Operational
amplifiers
The gain A (measured in decibel, with the symbol dB) of an operational amplifier is given by
⎛P⎞ out
the formula A = 10log⎜ ⎟ where P in is the input signal (measured in Watt) of the amplifier
⎝ Pin
⎠
and P out represents the output signal.
By regulating the strength of the input signal P in it is possible to control the gain A for a
certain output signal strength which we desire. So the gain A is the dependent variable and
P in is the independent variable.
Suppose that for a certain operational amplifier we wish to obtain an output signal of 3 Watt .
The algebraic model for the behaviour of the amplifier is therefore:
134
⎛ 3 ⎞
A = 10log⎜
⎟
⎝Pin ⎠
Let us now generate a numerical model for the behaviour of the amplifier by completing the
following table (the values for P in in the table were chosen randomly, because it is in any
case the independent variable and therefore under our control):
(The values of A were calculated using a regular scientific calculator.)
Let us now generate a graphical model for the behaviour of the amplifier by plotting the data
from the table on graph paper; note that we once again, as before, place the dependent
variable on the vertical axis and the independent variable on the horizontal axis:

(It is not necessary to connect the points on the graph by a smooth curve; it is, however, not
wrong either to do so.)
Let us now compose a verbal model for the behaviour of the amplifier.
= ≤ ≤ ∈ (Note: P in may not be zero, because in
• Domain” Df { Pin 0,5 Pin 6; Pin
}
⎛ 3 ⎞
that case division by zero would occur in the formula A = 10log⎜
⎟.
Consult the
⎝Pin ⎠
table from where we obtained the data points)
• Range Wf= { A − 3,01 < A≤7,782; A∈
}
• The curve is everywhere concave up
• The curve is decreasing which means that the function values A decrease as the
value of P in increases, but the decrease diminishes gradually
• The curve has a horizontal intercept at the point P in = 3 so the function has a root or
Conclusion:
zero point at P in = 3
• For input signals smaller than 3 Watt the amplifier causes an amplification gain
(positive gain)
135

• For input signals exceeding 3 Watt the amplifier causes an amplification loss
(negative gain)
• If the input signal and the output signal are equal (in this case 3 Watt each) then
there is no amplification gain and also no amplification loss ( A = 0 when
Pin = Pout
= 3Watt)
Let us now do some calculations with the model which we developed:
1. Calculate the amplification gain if an input signal of 5 Watt is used. Explain the
meaning of your answer.
Solution:
Set P in = 5 in
⎛ 3 ⎞
A = 10log⎜
⎟ and compute A :
⎝Pin ⎠
⎛3⎞ A = 10log ⎜
5
⎟
⎝ ⎠
= 10log( 0,6)
= 10 ⋅( −0,221849)
=−2,218
dB and this agrees with what may be seen on the graph
For this value of the input signal we obtain amplification loss (“weakening of the
signal) since the algebraic sign of the answer is negative.
2. Compute the input signal which would cause a gain of 4 dB .
Solution:
Set A = 4 in
∴ 0, 4 = log 10 ⎜ ⎟
Pin
⎛ 3 ⎞
A = 10log⎜
⎟
⎝Pin ⎠
and compute P in :
⎛ 3 ⎞
4 = 10log⎜
⎟
⎝Pin ⎠
4 ⎛ 3 ⎞
∴ = log⎜ ⎟
10 ⎝Pin ⎠
x
according to the information sheet: p = a means x = loga
p
Note that log actually means log 10 when there is "nothing" written beside the g:
⎛ 3 ⎞
⎝ ⎠
136

Now we simply read the relationship on the formula sheet from right to left in order
to write the following:
3
∴
Pin
0,4
= 10
0,4
the operation 10 is sometimes called the "antilog of 0,4"
3
∴
Pin
= 2,511886 the "to the power of" calculation on your calculator!
∴ 3 = 2,511886 ⋅Pin
3
∴ Pin
=
2,511886
multiplied both sides of the equation by the denominator
= 1,194 Watt and this agrees well with what may be inferred from the graph
Note that a little algebra is involved when we want to calculate the value of the independent
variable for a given value of the dependent variable.
Here is another example of the same type of calculation:
3. Calculate the input signal which will produce a gain of − 1dB
Solution:
Set A =− 1 in
⎛ 3 ⎞
A = 10log⎜
⎟
⎝Pin ⎠
and calculate P in :
⎛ 3 ⎞
− 1= 10log⎜
⎟
⎝Pin ⎠
−1
⎛ 3 ⎞
∴ = log⎜ ⎟
10 ⎝Pin ⎠
⎛ 3 ⎞
∴− 0,1 = log 10 ⎜ ⎟
⎝Pin ⎠
x
according to the information sheet: p = a means x = loga
p
3
∴
Pin
−0,1 = 10
−0,1
the operation 10 is also called the "antilog of -0,1"
3
∴
Pin
= 0,794 328
∴ 3 = 0,794
328 ⋅ Pin
3
∴ Pin
=
0,794 328
=
3,777 Watt and this agrees well with what may be inferred from the graph
137

Example of a logarithmic function in Chemistry: pH-values
(Read this carefully – but this is a difficult example which I shall not assess)
The pH-value (measured as a unit less value ranging from 0 to 14) of an aquatic solution is
expressed as the relationship pH log(
H ) +
=− .
The symbol H + refers to the concentration of hydrogen ions (measured in moles per litre or
moles per cubic decimetre). This concentration is a very small number – it refers to the
number of hydrogen ions present in one litre of solution. The more hydrogen ions present in
a solution, the more acidic is that solution. So the pH-value is the dependent variable and
the hydrogen concentration H + is the independent variable.
There exists a natural law which states that the hydrogen ion concentration may never
exceed1 mole per liter ; Also, it may never drop below
138
−14
10 moles per liter .
Using the given information we may compose a numeric model (table):
Note that 1.E-14 means:
14
1 10 −
× .
If we generate a graph of the data from the table, we obtain:

This graph clearly shows that the pH-value of a solution is not linearly dependent on
the concentration of hydrogen ions present. Because the numbers on the horizontal axis
are so bitterly small, the graph is practically unreadable for most of the domain. The values
on the horizontal axis are also approximated rather drastically in order to fit in on the graph.
Still, we may describe the relationship between pH and hydrogen concentration surprisingly
well in verbal form by consulting the graph:
+ − 14 + +
• Domain Df= { H 10 ≤ H ≤1; H ∈ }
• Range Wf= { pH 0< pH ≤14; pH ∈ }
• The curve is everywhere concave up
• The curve is decreasing which means that the function values pH decrease as the
value of the hydrogen concentration increases, but the decrease diminishes
gradually
• The curve has a horizontal intercept at H 1
+ = so the function has a root or zero
Conclusion:
point at H 1
+ = (that is theoretical – such a situation is physically impossible)
The pH-value is low when the concentration of the hydrogen ions is large – then the
solution is acidic; similarly, when the pH-value is high (close to 14) then the solution
is basic (alkaline).
Let us now do some calculations using our model:
1. Calculate the pH of a solution if the concentration of hydrogen ions is
Is the solution acidic or basic?
Solution:
Set H 0,003
+ = in pH log(
H ) +
=− and calculate pH :
pH =−
=− −
= 2,523
log( 0,003)
( 2,522 879)
Since the pH has a small value (not far from zero) the solution is acidic.
139
3
0,003 mol/dm .

2. Compute the H + -concentration of a solution if its pH is 7. (By the way: A pH-value
of 7 indicates that a solution is neither acidic nor basic, but rather neutral)
Solution:
Set pH = 7 in pH log(
H ) +
=− and compute H + :
7 =−log
∴− 7 = log
10
+ −7
+ ( H )
+ ( H )
∴ H = 10 according to the information sheet x = log a p implies p = a
+
∴ H
3
= 0,000 000 1 moles/dm
This is a very small concentration of hydrogen ions.
140
x

4 Trigonometric functions
Learning outcomes for this study unit
After completing this study unit the student must be capable of doing the following:
1. Describe a rotational motion (turning motion of a pulley or motor or generator or
similar machine) as a trigonometric function;
2. Describe a periodic motion (simple harmonic motion, that is a regularly repeated
motion such as a pendulum swinging from side to side or a weight which is bobbing
up and down suspended from a spring) as a trigonometric function;
3. Describe a wave as a trigonometric function;
4. Explain the meaning of each of the following concepts as they are encountered in the
theory of rotational motion, periodic motion and waves: amplitude, angular
frequency, period, frequency and vertical translation (displacement);
5. Determine the values of the amplitude A , the angular frequency ω , the period T , the
frequency f and the vertical translation (displacement) d from a given graphical
representation of a rotational motion, periodic motion or a wave;
6. Express the period, the frequency and the angular frequency in the correct units of
measurement;
7. Graphically represent the functions y = Asin( ω t) + d and cos(
)
of a table or other means;
8. Apply the equations y = Asin( ω t) + d and cos(
)
the technical field of study.
141
y = A ω t + d by means
y = A ω t + d to real life problems from
Study the following material in the book by Washington
Paragraph Page numbers
10.5 306 - 309
10.6 309

Study the PowerPoint “Betekenisvolle Trigonometriese Grafieke”
which may be found on eFundi.
Also work through the following examples:
Trigonometric model for the human heartbeat
Suppose the human heart is considered as a spherical structure with radius r which
oscillates between the values r = 4 cm and r = 6 cm as the cardiac muscle contracts and
relaxes. If the human heart beats at 58 beats per minute, compose a mathematical model
which describes the radius of the heart as a function of time. Assume the initial state of the
heart (when t = 0 minutes ) as the average of the minimum value and the maximum value of
the radius (therefore, a neutral or equilibrium state). Also sketch your model as a graphical
representation of r against t .
Solution:
From experience we know that each cycle of the heart represents a contraction and a
relaxation, so 58 beats per minute actually implies 58 contractions and relaxations per
minute, each contraction followed by a relaxation.
This process may readily be represented by a sine function or cosine function of time t .
Let us now construct a rough auxilliary sketch to help us visualise the amplitude, period,
phase shift (if applicable) and vertical translation of the motion:
142

Is it clear that the model () sin(
ω φ)
rt = a t+ + d will be suitable?
The radius has initial value of 5 cm (when t = 0 the radius is the average of 4 cm and 5 cm);
if, however, the radius started with a maximum or minimum value then () cos(
ω φ)
would have been a better choice.
143
rt = a t+ + d
Let us now think and decide about the values which we should assign to a , ω , φ and d :
Amplitude a : Maximum deviation from the neutral position is 1 cm, since the radius
Vertical
max=6
gemiddelde= 6+4
2 =5
min=4
Radius r in cm
7
6
5
4
3
2
1
-0.2 0.2 0.4 0.6 0.8 1
-1
oscillates between 4 cm and 6 cm and we agreed to define as neutral
position the average of the minimum 4 cm and the maximum 6 cm.
So a = 1cm .
aanvangsradius as t=0
58 siklusse
1 minuut
Tyd t in minute
displacement d : We agreed to define as neutral position the average of the minimum,
which is 4 cm and the maximum, which is 6 cm. Therefore, the neutral
position is 5cm above the horizontal axis. In the case of the basic sine
function y = sin x and the basic cosine function y = cos x we have that
the neutral position coincides with the horizontal axis. In the case of
the given function the neutral position has been displaced 5 cm
upwards.

Therefore, d = 5 cm .
Phase shift φ : The radius begins, according to the given data, on exactly 5 cm when
t = 0 minutes. Then the heat beats (so, it contracts once until the
6
5
4
O
radius shrinks to 4 cm and then relaxes (expands) so that the radius
expands to 6 cm):
Note the interesting difference between the behaviour of the function
and the regular, familiar sine function: This function rises where the
regular sine function decreases and vice versa, so we say that this
function is perfectly out of phase with the regular sine function. The
concept “perfectly out of phase” implies a phase shift or phase angle of
180° which simply implies that the sign of the amplitude should be − 1
:
So a =− 1
Angular velocity ω : Since there are 58 beats per minute, there are 58 cycles per minute so
the frequency f of the motion is f = 58 cycles per minute . From this
expression it follows that the period T of the motion (that is, the time
taken for one cycle consisting of a contraction and an expansion) is
1
expressed as T = minute.
58
144

2π
By definition ω = 2π f or ω = (number of revolutions divided by
T
duration of one revolution) and therefore
( )
ω = 2π f = 2π 58 = 116π radians/minute
or
2π 2π
ω = = = 116π
radians/minute .
T 0,017
The entire model is therefore: rt ( πt)
( ) = − 1sin 116 + 5
Let us now graphically represent this algebraic form of the mathematical model and see if the
graph agrees with the verbal description of the process (the information which we used to
compose the formula – so we intend to check our formula for validity:
145

How to sketch the graph on paper using a pencil:
Draw a dashed line where the neutral position (equilibrium state) of the motion is. In this
case the vertical displacement is 5 cm, so the function is translated upwards by 5 cm. Now
treat this dashed line 5 cm above the horizontal axis as a translated horizontal axis
running halfway between the minimum and maximum values of the graph. The
amplitude of the graph is measured up or down from this translated asis (neutral position).
Since the magnitude of the amplitude is 1 cm, the range of the function will be as follows:
{ r 4≤r ≤6; r∈ R}
1 minuut
The period (duration of each full cycle) is T = = 0,017 minuut . Since the regular
58
sine function has one minimum, one maximum and one zero point halfway between its roots,
one cycle of the sine function is divided into four equal parts, each 0,017
= 0,004 minutes
4
long.
So, draw one “upside down” wave with an amplitude of 1 cm on the line r = 5 cm as if this
line is the horizontal axis and take the period of the wave (“wavelenght expressed in
minutes”) as 0,017 minutes.
6
5
4
O
radius r
in cm
146
r(t)=-1sin(116πt)+5
0,004 0,009 0,013
0,017
tyd t in minute

If we use GSP to generate the graph, we obtain the following:
(Simply let the program sketch the function y ( π x)
angular unit set to radians.)
6
5
4
3
2
1
-1
-2
-3
= − sin 116 + 5 on a rectangular grid with the
0. 002 0. 004 0. 006 0. 008 0. 01 0. 012 0. 014 0. 016 0. 018
It is easy to see that our algebraic model produces a graph which fully agrees with the
information originally used to construct the formula.
We may now use GSP to investigate the behaviour of the heart for any time interval, say for
example, for the first 10 seconds of a minute . Just remember that our horizontal axis (timeaxis)
is calibrated in minutes which means that 10 seconds represent 10
= 0,167 minute:
60
147
f x ( ) = -sin 116⋅π⋅x ( )+5

6
5
4
3
2
1
-1
-2
-3
In one full minute there will clearly be just about 60 full cycles, which agrees with what we
already know: this particular heart has a rate of 58 beats per minute.
From the behaviour of the graph we can easily picture the contraction and relaxation of the
heart muscle.
It is interesting how the verbal, algebraic and graphical representations serve to supply us
with a complete, informative “picture” of the entire process which we studied.
Beautiful, don’t you agree?
Tel gerus: net minder
as 10 volledige
siklusse in een sesde
van 'n minuut
0. 02 0. 04 0. 06 0. 08 0. 1 0. 12 0. 14 0. 16 0. 18
You may now compare the steps we followed in our analysis above with the scheme
discussed just before the end of study section 2.1.
148
f x ( ) = -sin 116⋅π⋅x ( )+5

More examples
Example 1:
Analysis of the situation
The average minimum day temperature in Potchefstroom for the previous 10 years
was 11°C and the average maximum day temperature for the same period was 31°C.
The lowest average day temperature occurred between the sixth and the seventh
month and the highest average day temperature occurred between the twelfth month
and the first month of the following year (in the table, t = 0 means the beginning of
January, which is also the end of the previous December. t = 1 means the beginning
of February, which is the end of the previous month, etc):
t (tme in months) 0 1 2 3 4 5 6 7 8 9 10 11 12
T (average day
temperature in °C)
31 29 26 19 16 12 11 13 15 20 25 29 31
Further analysis of the temperatures reveals that the temperatures decreased the
fastest between the second and the fourth month when the average day temperature
was 19°C and that temperatures increased the fastest between the eigth and the
tenth month when the average day temperature was 20°C.
Problem statement:
Represent the information graphically and develop a trigonometric model to describe
the situation.
Solution:
Note that we are dealing with a cyclical (periodic), regular, repeating process: The
seasons run as follows: Winter, Spring, Summer, Autumn, Winter, Spring, ...
Note that the measured temperature values depend on the time of the year that it was
measured (low in winter, high in summer, etc)
The two paragraphs above imply that we need a function whose behaviour reflects
the traits of the situation for modelling this situation mathematically.
149

(If you recall the shape and behaviour of the sinus and cosinus curves, you will see
where we are going.)
Let us now analize the situation graphically by tranferring the numeric information
from the table to a graph defined by a time/ temperature system of axes; let us agree
to place the beginning of January at the point where t = 0 and to place the end of
December at the point where t = 12:
45
40
35
30
25
20
15
10
5
-5
Temperatuur (°C)
Stylste daling
2 4 6 8 10 12
If we nou apply the model () cos(
ω )
Tt = A t + dto
the situation (from the shape of the
pattern of the dots on the graph, it is apparent why we select a cosine function):
Questions we may ask ourselves:
• What is the amplitude A (largest variation between the lowest minumum
temperature value and the highest maximum value)?
• What is the period T of the cycle (how long does the pattern suggested by
the points take to repeat, a period in months?) so that we may calculate the
value of ω from its definition?
150
Steilste styging
tyd (maande)

• What is the vertical displacement d of the function (how high above or below
the horizontal axis of the graph is the average of the maximum and minumum
located?)
Let us now first address the terminology in the questions above and attempt to
answer the questions:
• Regarding the amplitude: From the definition of the amplitude and the
information reflected on the graph it is clear that A = 10 °C . A has a positive
value since the curve begins at a maximum value and then decreases towards
the horizontal axis, the same way a simple cosine function does (PLEASE
NOTE that if the shape of the curve were reversed (beginning at a minimum
and then increasing towards the horizontal axis) then the amplitude would
have been A =− 10° C and we would have said that the cosine function is
180° out of phase or π radians out of phase, or displaced by π radians.)
• Regarding the period: From our experience and from the graphical
representation) it is clear that the seasons repeat every 12 months. The
period T of the function is therefore 12 months.
We also agreed to connect the period and the frequency of a periodically
repeating process to the concept of angular velocity or angular frequency
ω by introducing the relationship ω = 2π f . From our previous knowledge
1
from Grade 10 (Waves, from Physical Science) we recall that f = and we
T
can use this to define the angular frequency in a more useful way as
1
2π
ω = 2π
⋅ which is of course equivalent to ω = .
T
T
2π
π
So ω = = 0,524 which we may also write as ω = . The unit of ω is in
12
6
this case radians/month.
• From the graph it is clear that the average of the minimum temperature and
the minimum temperature, which is 21°C, is located 21 units above the
horizontal axis; so d = 21 ° C .
So we conclude that our mathematical model for the situation is
⎛π⎞ Tt ( ) = 10cos( 0,524t) + 21 (or Tt () = 10cos⎜ t + 21
6
⎟ if you prefer to express it that
⎝ ⎠
way).
151

If we now overlay the function Tt ( t)
the graph, we obtain:
45
40
35
30
25
20
15
10
5
-5
Temperatuur (°C)
( ) = 10cos 0,524 + 21 on top of our data points on
2 4 6 8 10 12
tyd (maande)
It should be clear that our mathematical model agrees rather well with the oberved data. You
may now re-read the discussion above and try to identify all the elements of mathematical
modelling in our reasoning.
152

Example 2:
Analysis of the situation
An alternating voltage signal generator is connected to an oscilloscope. The display of the
oscilloscope shows the following picture:
The scale on the vertical axis is 3V per division and the scale on the horizontal axis is 5
milliseconds per divion (which may be read off the settings on the controls of the
oscilloscope).
Problem statement:
Determine the following information:
1. The peak value (amplitude of the signal) (in Volt)
2. The period of the signal generator (in millisekondes)
3. The frequency of the signal generator (in Hertz, where 1 Hz=1 siklus/second)
Write a trigonometric equation which would describe the situation (derive a trigonometric
model).
Solution:
1. It is immediately clear from the shape of the curve that we are dealing with a regular
positive sine function. The amplitude is simple to determine from the graph: The
maximum deviation from the equilibrium level (horizontal axis) is 4 units on the
153

vertival axis and so 4 × 3 = 12 V (recall that the scale on the vertical axis is 3 V per
division).
The concept “equilibrium level” refers as before to the average of the maximum and
the minimum values, in this case co-inciding with the horizontal axis)
The amplitude A is therefore 12 V. (PLEASE NOTE that if the curve began at the
origin and then decreased towards its minimum value then the amplitude would have
been A =− 12 V and we would have said that the sine function is 180° or π radians
out of phase.)
2. From the graphical representation it is clear that one full cycle (one full wave) of the
signal occupies two and a half horizontal axis divisions. Each division (scale unit)
represents 5 milliseconds, so two and a half divisions mean 2,5 × 5 = 12,5
milliseconds.
The period T of the signal is therefore 12,5 milliseconds.
3. The frequency f of the signal may be calculated entirely without incident by
employing the relationship
per millisecond.
1
1
f = and therefore it follows that f = = 0,08 cycles
T
12,5
1
But we were asked to express the frequency in Hertz, so f = = 80 Hz.
−3
12,5 × 10
We could also have easily obtained the same result by noting that 0,08 cycles per
millisecond actually means that 0,08 × 1000 = 80 cycles takes place per second,
since by definition, one millisecond is one thousandth of a second.
We are now ready to write the trigonometric function Vt () Asin( t)
154
= ω :
A = 12 and by defition ω= 2π f so that ω = 2×π× 0,08 = 0,503.
Then the function is Vt ( ) 12sin( 0,503t) of Vt ( ) 12sin( 0,16 t)
= = π .
NOTE THAT if we wanted to express die voltage in Volts and the time in seconds, then the
function would have taken the form: Vt = ( t) Vt = ( πt)
( ) 12 sin 502,655 or ( ) 12sin 160 .

Example 3
A generator revolves at 60 revolutions per second and generates a peak voltage of 220V
alternating voltage. The alternating voltage generated may mathematically be described as
follows:
( ) = sin(
ω )
V t V t
Questions:
maksimum
1. Determine the amplitude of the generated voltage.
2. Determine the period of one cycle of the generator (one full wave).
3. Determine the angular frequency or angular velocity ω in radians per second.
4. Write the equation V ( t) V sin(
ωt)
= for the generated voltage.
maksimum
5. Represent two cycles of the alternating voltage graphically. Use a vertical axis scale of
110 mm = 220 V and a horizontal axis scale of 50 mm = 0,017 s.
6. Calculate the value of V at the instant 0,005 s after the beginning of a cycle.
7. Calculate how long after the beginning of a cycle the generated voltage attains 180 V.
Solution:
1. 220 V (peak value. The generated voltage alternates between -220 V and 220 V)
2. 60 revolutions per second = 60 cycles per second and that it is the frequency (60 Hz).
i
1
1
From the relationship T = we can calculate the period: T = = 0,016 seconds . This
f
60
is how long one full revolution of the generator armature takes and this is the duration of
one voltage cycle.
3. We have a formula for angular frequency or angular velocity:
ω = 2π
f
= 2× π × 60
= 120π radians/second or 376,991radians/second
4. V ( t) = 220 sin( 376,991t) or V ( t) = 220 sin( 120
πt)
155

5.
6.
() = 220 sin( 120π
)
( 0,005) 220 sin( 120π 0,005)
V t t
∴ V =
= 209,232 V
×
7.
V (Volt)
220
-220
( ) = 220 sin( 120π
)
( πt
)
( πt
)
πt
−1
( )
V t
∴ 180 = 220 sin 120
t
∴ sin 120 = 0,81818
∴ 120 = sin 0,81818
= 0,958
0,958
∴ t = =
120π
∴ t = 0,00254 seconds
0,017
156
T (sekonde)
0,033

Example 4
A crankshaft in a big machine moves up and down according to the following model:
ht () = A⋅ f( ω t) + d where h is the altitude of the end of the shaft above the floor, A is the
vertical distance through which the end of the shaft moves upwards or downwards, ω is the
angular velocity of the motion expressed in radians/second, t is the time in tyd in seconds
and d is the initial distance from the floor.
f ( ω t)
is a trigonometric function (sine or cos) which fits the situation.
A computer simultation of several cycles of the motion looks as follows:
1,8
1,6
1,4
1,2
1,0
0,8
0,6
0,4
0,2
O
h (m)
0,2
Analize the behaviour of the crankshaft by answering the following questions (show
all calculations which you perform):
1. What is the amplitude of the motion?
2. What is the vertical displacement of the motion?
Hint: This is the distance that the equilibrium position or equilibrium level of the curve
of the function is displaced above or below the horizontal axis.
3. Is the function f ( ω t)
a sine or a cosine function? Explain your answer.
0,4
157
0,6
0,8
1,0
t (s)

4. What is the period of the motion (how long does one up and down motion take)?
5. What is the frequency of the motion?
6. What is the angular frequency or angular velocity of the motion?
7. Now write the model ht () = A⋅ f( ω t) + d in terms of the information you developed in
questions 1 to 6 above.
8. Determine how high the end of the crankshaft will be above the floor at the instant 0,7 s
after the beginning of the motion.
Solutions:
1.
maximum vertical displacement − minimum vertical displacement
A =
2
1, 8 − 0, 2
∴ A =
2
= 0,8 m
You could also have derived it directly from the graph.
But the amplitude also carries a sign; if the curve is exactly out of phase with the regular
cosine function, we must assign a negative sign. Although this curve is characteristic of
a cosine graph (note that the curve does not start on the equilibrium level - if that had been
the case, then we would be dealing with a situation best described by a sine function). we
note how the curve begins at a minimum value; so we conclude that we are
dealing with an inverted cosine function which means that the amplitude is negative:
A =−0,8
m
2.
maximum vertical displacement + minimum vertical displacement
equilibrium level =
2
1, 8 + 0, 2
∴ equilibrium level =
2
= 1m
You could also have derived it directly from the graph.
3.
158

Cosine function which is exactly out of phase with the regular fundamental cosine function h= cost.
See the discussion in the answer to Question 1.
4.
T = 0,4 seconds . This is the time interval between any two successive points which are in phase on
the curve.
5.
1
f =
T
1
=
0, 4
∴ f = 2,5 cycles per second, which may also be written as 2,5 Hz.
6.
7.
ω= 2πf
= 2×π× 2,5
= 5πradians/second or 15,708 radians/second
() ( ) () ( )
h t =−0,8⋅cos 5π t + 1 or h t =−0,8⋅ cos 15,708t + 1
8.
( ) ( )
( )
=− ⋅ ( ) +
h 0,7 =−0,8 ⋅cos 5π× 0,7 + 1
=−0,8⋅ cos 10,996 + 1
0,8
=1m
0 1 CALCULATOR IN RADIANS!!!
Does the answer agree with the reading you could make from the graph to answer this
question?
Yes! (Thank goodness)
159

160

5 Conic sections
Learning outcomes for this study unit
After completing this study unit the student must be capable of doing the following:
1. Die definisies van die vier basiese kegelsnitte as lokusse kan gee;
2. Die Cartesiese vergelykings van die vier basiese kegelsnitte in algemene vorm kan
identifiseer;
3. Die Cartesiese vergelykings van die vier basiese kegelsnitte vanuit algemene vorm
na standaardvorm oor te skakel (dit wil sê, u moet y die onderwerp van die
vergelyking kan maak);
4. Die vier basiese kegelsnitte met behulp van ‘n tabel of geskikte rekenaarprogram te
skets;
5. ‘n Getransleerde sirkel in terme van sy Cartesiese vergelyking in middelpuntvorm
2 2 2
kan beskryf, dit is die vorm ( x − a) + ( y− b) = r ;
6. Die Cartesiese vergelyking van ‘n getransleerde sirkel na die algemene vorm om te
skakel, dit is die vorm
2 2
x y px qy c
+ + + + = 0 ;
7. Eenvoudige probleme waar kegelsnitte betrokke is, op te los.
Background
Blaai na p. 595 van die boek van Washington en beskou die paragraaf bo-aan p. 595 tesame
met figuur 21.92. Dit verklaar waarom ons die sirkel, ellips, sentrale hiperbool en parabool
dikwels as kegelsnitte (“conic sections”) beskou.
Bespreking van kegelsnedes (Kegelsnitte)
Sirkels, ellipse, sentrale hiperbole en selfs parabole kan beskou word as figure wat verkry
word deur 'n reghoekige kegel (Engels: "cone") op verskillende maniere met 'n platvlak te
sny. Hierdie bespreking vervang paragrawe 21.3 tot 21.8 in die boek van Washington.
161

Ons sal vervolgens elkeen van die kegelsnedes vlugtig bespreek.
1. Sirkels
162
Let op dat die snyvlak parallel aan die
basisvlakke loop; die snykromme is ‘n
sirkel. Die snyvlak sny die simmetrie-as
van die kegel loodreg.
(Die aansig hiernaas is natuurlik isometries;
om die sirkelvorige snykromme te sien,
moet ‘n mens eintlik loodreg van bo op die
snyvlak afkyk)
'n Sirkel kan gedefinieer word as die lokus van 'n punt wat sodanig beweeg dat die
afstand tussen die punt en ‘n ander, vaste punt (die middelpunt genoem) konstant bly,
Hierdie afstand word die radius van die sirkel genoem (sien hieronder).
Die vergelyking van die sirkel in algemene vorm is
2 2
x y px qy c
+ + + + = 0 .
Hierdie vergelyking kan deur middel van kwadraatsvoltooiing so gemanipuleer word dat
dit in die sogenaamde middelpuntsvorm geskryf kan word:
( x − a) + ( y− b) = r
2 2 2
Hierdie vorm van die vergelyking is nuttig aangesien die koördinate van die middelpunt van
die sirkel dan gegee word deur ( aben ; ) die radius se lengte deur r .
In die spesiale geval waar die middelpunt van die sirkel op die oorsprong van die assestelsel
( 0; 0 ) val, reduseer die vergelykings hierbo na
2 2 2
x + y = r .
Dit is belangrik om enige sirkel te kan skets indien sy vergelyking bekend is. Net so, is dit
net so belangrik om enige sirkel te beskryf in terme van sy vergelyking.
Om 'n bietjie agtergrond te gee oor hierdie twee prosesse, kan u deur die volgende
bespreking werk:

Probleem: Skets die grafiek van die volgende vergelyking:
Opmerkings:
163
2 2
x y x y
+ −6 −8 − 24= 0
Omdat dit gewoon nie moontlik is om hierdie vergelyking maklik in die standaardvorm
y = f( x)
te skryf nie, kan ons nie ‘n tabel van waardes gebruik nie. (Probeer gerus om y
die onderwerp van die vergelyking te maak – u sal dit nie weer doen nie!)
Ook kan mens nie gewoon die afsnitte op die asse bereken deur eers x nul te maak en vir y
op te los en dan vir y nul te maak en vir x op te los nie. (Kyk gerus wat gebeur as ‘n mens
dit probeer doen.)
Die enigste manier om hierdie vergelyking se kromme te skets, is om dit te herken as die
vergelyking van 'n sirkel in die algemene vorm
+ + + + = 0 .
2 2
x y px qy c
Dan sal ‘n mens dit moet manipuleer totdat ons die middelpunt en radius uit die vergelyking
kan verkry.
Laat ons nou ondersoek instel na hoe dit gedoen kan word (Vir interessantheid):
Let eers op hoe die vergelyking
+ + + + = 0 ontstaan het:
2 2
x y px qy c
Gewoonweg as ons ‘n sirkel se vergelyking in die algemene vorm het, lyk dit so:
2 2 2
x + y = r
(1)
Grafies voorgestel:
Maar dit is ‘n sirkel waarvan die middelpunt op die oorsprong (0;0) lê. Onthou nou dat ons
enige afstand tussen twee punte kan beskou as ‘n verskil tussen koördinate.

As daar dus ‘n punt P( x; y ) iewers op ons sirkel lê, sal die horisontale afstand tussen hierdie
punt en die middelpunt ( x − 0 ) eenhede wees en die vertikale afstand tussen hierdie punt en
die middelpunt sal ( y − 0 ) eenhede wees.
Dus sou ons die sirkel se vergelyking kon skryf as
( x 0) ( y 0) r
2 2 2
− + − = (2)
Gestel egter die sirkel se middelpunt skuif nou a eenhede na regs en b eenhede op, sodat
die middelpunt (noem dit punt C) se koördinate nou C( ab ; ) is:
Nou is die horisontale afstand tussen C en P gelyk aan ( x − a)
eenhede en die vertikale
afstand tussen C en P is gelyk aan ( y− b)
eenhede.
Gaan ons nou terug na die vergelyking (2) hierbo en herskryf dit vir ons nuwe middelpunt C(
ab ; ) , kry ons:
( x a) ( y b) r
2 2 2
− + − = (3)
Neem nou hierdie vergelyking en verwyder die hakies; dit lewer
2 2 2 2 2
x − 2ax + a + y − 2by
+ b = r
(4)
Maar dit is ‘n redelik deurmekaar vergelyking en dit val sleg op die oog – so,
gerieflikheidshalwe kies ons om (soos by alle kwadratiese vergelykings die gebruik is) dit in
orde van dalende magte te skryf en om die regterkant nul te maak:
164

2 2 2 2 2
x + y −2ax− 2by+ a + b − r = 0
(5)
Onthou egter dat a ,b en r almal konstantes is en dus opgetel kan word; sodoende kry ons
‘n getal c wat ons in hulle plek kan vervang as ons c so kies dat
Dan word ons vergelyking (5) nou die volgende:
+ −2 − 2 + = 0
(6)
2 2
x y ax by c
Dit lyk al heelwat beter as vergelyking (4) hierbo.
165
2 2 2
c = a + b − r .
Verder, aangesien a en ook b konstantes is, kan ons getalle p en q invoer sodat p = − 2a
en q =− 2b
want dan vereenvoudig ons vergelyking (6) verder na
+ + + + = 0
(7)
2 2
x y px qy c
en dit is presies die algemene vorm van ‘n sirkel.
Bogenoemde vergelyking lyk presies soos die vorm van die vergelyking wat ons wil skets.
Wat ons nou net gedoen het, was eintlik om al die belangrike inligting omtrent die
sirkel (soos hoe lank sy radius is en waar sy middelpunt geleë is) in 'n enkele
kwadratiese vergelyking gekodeer het. (sien vergelyking 7 hierbo).
Indien ons egter 'n sirkel wil skets wanneer 'n vergelyking soos
gegee is, moet ons dus die proses in punt 4 hierbo terugwerk om by
+ + + + = 0
2 2
x y px qy c
( x − a) + ( y− b) = r
2 2 2
(vergelyking (3) in die paragraaf hierbo) uit te kom – want dan het ons die koördinate van die
middelpunt, wat natuurlik tipies ( ab ; ) sal wees en die radius r .
Dit is al die inligting wat nodig is om die sirkel te skets.
Ongelukkig verg dit wat ons in die vorige paragraaf sê, redelik drastiese algebra en meer
spesifiek, die toepassing van ‘n proses genaamd kwadraatsvoltooiing.
Ons gaan nou poog om hierdie proses so deursigtig as moontlik toe te pas op ons probleem.

Ons gaan dus die vergelyking
( x a) ( y b) r
+ −6 −8 − 24 = 0 terugneem na die vorm
2 2
x y x y
2 2 2
− + − = , waaruit ons volkome sonder insident die middelpunt en radius
kan verkry:
Solution:
2 2
x + y −6x−8y− 24= 0
2 2
∴ x − 6x+ y − 8y= 24
−6 −8 −6 −8
∴ x − 6 x+ ( ) + y − 8 y+
( ) = 24 + ( ) + ( )
2 2 2 2
2 2 2 2 2 2
(ons het beide kante van die = -teken die kwadraat van die helfte van die getal voor die x en
voor die y bygetel)
2 2
∴ x − x+ + y − y+
= + +
6 9 8 16 24 9 16
Faktoriseer nou gewoon beide drieterme en vereenvoudig regterkant:
Dit lewer die Cartesiese vergelyking van die sirkel
2 2
− 3 + − 4 = 49
middelpuntvorm: ( x ) ( y )
Vergelyk nou hierdie vergelyking met vergelyking (3) in punt 6 hierbo.
166
+ −6 −8 − 24= 0 in
2 2
x y x y
Kan u wel sien dat die middelpunt (3;4) en die radius 7 eenhede moet wees?
Indien wel, kan die sirkel nou volkome sonder insident m.b.v. ‘n passer en liniaal geskets
word.
U moet bloot die vergelyking van 'n gegewe sirkel in middelpuntvorm kan saamstel en
verwerk na die algemene vorm (geen hakies). Tweedens moet u in staat wees om 'n
sirkel te skets indien sy vergelyking in die middelpuntvorm gegee is; daar sal nie in
toetse of eksamens van u verwag word om deur middel van kwadraatsvoltooiing ‘n
vergelyking vanaf algemene vorm na middelpuntsvorm te herlei nie.

2. Ellipse
167
Let op dat die snyvlak ‘n hoek α van minder
as 90º met die simmetrie-as van die kegel
maak. Hierdie hoek is egter groter as die
tophoek θ van die kegel.
'n Ellips kan gedefinieer word as die lokus van 'n punt wat sodanig beweeg dat die
som van die afstande tussen die punt en twee ander vaste punte (die brandpunte
genoem), 'n konstante waarde aanneem (sien fig. 21.55 op p. 579 van die boek van
Washington).
Die planete en hul mane, asook mensgemaakte satelliete en ruimtetuie, beweeg in elliptiese
bane. Dit is onder meer as gevolg van die elliptiese baan van die aarde dat ons seisoene
beleef.
Vir die doel van WSKT 221 beskou ons slegs ellipse waarvan die middelpunt op die
oorsprong van die Cartesiese Assestelsel geleë is. Sulke ellipse word beskryf deur die
vergelyking
2 2
x y
+ = 1.
2 2
a b
So 'n ellips sny die X-as by –a en +a en die Y-as by –b en +b.
'n Ellips kan met behulp van bogenoemde inligting maklik geskets word indien sy vergelyking
in die algemene vorm
2 2
x y
+ = 1 gegee is.
2 2
a b

Indien 'n meer akkurate skets verlang word, kan die ellips met behulp van 'n tabel geplot
word.
In daardie geval moet die vergelyking
2
1 2
x
y =± b⋅−
omgeskakel word.
a
Dit verloop soos volg:
2 2
x y
+ = 1 eers na die algemene vorm, nl.
2 2
a b
2 2
x y
+ = 1
2 2
a b
2 2 2 2 2 2 2 2
x ab y ab ab
∴ × + × = 1× 2 2
a 1 b 1 1
Vermenigvuldig regdeur met KGV van noemers
2 2 2 2
∴ bx + ay
2 2
= ab
2 2
∴ a y
2 2 2 2
= a b −b
x
2 2 2 2
2 ab − bx
∴ y = 2
a
∴ y =±
2 2 2 2
ab − bx
2
a
Natuurlik kan ons hiermee volstaan, aangesien ons y die onderwerp gemaak het van die
vergelyking
2 2
x y
+ = 1 en dit is wat ons wou doen. Daarom is ons antwoord voldoende.
2 2
a b
Die laaste resultaat hierbo kan natuurlik tog verder vereenvoudig word tot die vorm
2
1 2
x
y =± b⋅−
. Dit is nie vir assesseringdoeleindes nie, maar vir intereressantheid.
a
Dit verloop soos volg:
y =±
2 2 2 2
ab − bx
2
a
=±
ab
2
a
bx
− 2
a
=±
2 2
2 bx
b −
2
a
2 2 2 2
168

2
2 ⎛ x ⎞
⎜12⎟ ∴ y = ± b −
⎝ a ⎠
2
2 ⎛ x ⎞
=± b ⋅ ⎜1− 2 ⎟
⎝ a ⎠
2 ⎛ x ⎞
⎜12⎟ ∴ y = ± b⋅
−
⎝ a ⎠
Ons kan nou bewys dat, wanneer ons die grafiek met ‘n tabel teken, ons slegs x -waardes
tussen − a en a in ons tabel hoef in te sluit. Hierdie bewys is nie vir assessering nie,
maar vir interessantheid:
Let daarop dat die getalle binne vierkantswortels nie negatief mag wees nie. Dus moet dit
2
x
geld dat 1− ≥ 0.
Hieruit volg dan:
2
a
2
x
1− ≥ 0 2
a
2 2
∴ a −x ≥ 0
∴( a− x)( a+ x)
≥ 0
∴ a−x ≥ 0 en a+ x ≥ 0 of a−x ≤ 0 en a+ x ≤ 0
∴ x ≤ a en x ≥ −a of x ≥ a en x ≤ −a
∴ −a ≤ x ≤ a
of geen oplossing
Dus kan u enige x-waardes tussen –a en +a kies en bybehorende y-waardes bereken. Hoe
meer punte u plot, hoe akkurater die skets.
Indien die vergelyking van die ellips in die standaardvorm
169
2 2 2 2
ab − bx
y =± of
2
a
2
1 2
x
y =± b⋅−
bekend is, kan die ellips ook met behulp van 'n rekenaarprogram geteken
a
word.
U moet die vergelyking van 'n gegewe ellips uit die skets kan aflei en tweedens moet u
in staat wees om 'n ellips te skets indien sy vergelyking gegee is.

3. Sentrale Hiperbole (moenie verwar met die Reghoekige Hiperbool nie)
170
Let op dat die snyvlak ‘n hoek α met die
simmetrie-as maak wat kleiner is as die
tophoek θ van die kegel. Die hoek α kan
selfs nul wees (dan is die snyvlak parallel
met die simmetrie-as).
'n Sentrale Hiperbool kan gedefinieer word as die lokus van 'n punt wat sodanig
beweeg dat die verskil van die afstande tussen die punt en twee ander vaste punte
(die brandpunte genoem),'n konstante waarde aanneem (sien fig. 21.67 op p. 584 van
die boek van Washington).
Wanneer ‘n positiefgelaaide deeltjie (soos byvoorbeeld ‘n proton) na ‘n swaar atoomkern
(baie groot massa en baie sterk positiewe lading in vergelyking met ‘n proton s’n) geskiet
word, veroorsaak die elektrostatiese afstotingskrag dat die proton in die vorm van ‘n sentrale
hiperbool gedeflekteer (weggestoot) word.
Vir die doel van WSKT 221 beskou ons slegs sentrale hiperbole waarvan die middelpunt op
die oorsprong van die Cartesiese Assestelsel geleë is en waarvan die wortels op die X-as
voorkom. Sulke hiperbole word beskryf deur die vergelyking
So 'n sentrale hiperbool sny die X-as by –a en +a.
2 2
x y
− = 1.
2 2
a b

Daar is geen snypunte met die Y-as nie. Die sentrale hiperbool het egter asimptote met die
b
b
vergelykings y =− x en y =+ x.
a
a
'n Sentrale Hiperbool kan met behulp van bogenoemde inligting maklik geskets word indien
sy vergelyking in die algemene vorm
2 2
x y
− = 1 gegee is.
2 2
a b
Indien 'n meer akkurate skets verlang word, kan die sentrale hiperbool met behulp van 'n
tabel geplot word. In hierdie geval moet die vergelyking
vorm, nl.
171
2 2
x y
− = 1 eers na die algemene
2 2
a b
2
2 1
x
y =± b⋅
− omgeskakel word (maak seker dat u dit kan doen); daarna kan u enige
a
x-waardes kleiner as of gelyk aan –a en groter as of gelyk aan +a kies en bybehorende y-
waardes bereken. Hoe meer punte u plot, hoe akkurater die skets.
2
Indien die vergelyking van die sentrale hiperbool in die standaardvorm
2 1
x
y =± b⋅
−
a
bekend is, kan die hiperbool ook met behulp van 'n rekenaarprogram geteken word.
U moet die vergelyking van 'n gegewe sentrale hiperbool uit die skets kan aflei en
tweedens moet u in staat wees om 'n sentrale hiperbool te skets indien sy vergelyking
gegee is.

4. Parabole
172
Let op dat die snyvlak ‘n hoek α met die
simmetrie-as maak wat gelyk is aan die
tophoek θ van die kegel. Dit beteken
dieselfde as om te sê dat die snyvlak parallel
is aan die sykant van die kegel.
Soos u kan sien, kan 'n parabool ook as 'n tipe kegelsnede beskou word.
‘n Parabool is die lokus van ‘n punt P wat so beweeg dat dit ewe ver vanaf ‘n vaste
punt (die brandpunt genoem) en ‘n vaste lyn (die riglyn genoem) bly (sien fig. 21.40 op
p.575 van die boek van Washington).
Ons werk egter meestal nie op hierdie manier met parabole in die Tegniese Vakrigtigting
nie; ons hanteer hulle eerder as die krommes van kwadratiese funksies van die vorm
2
y = ax + bx+ c.
Dit is reeds bespreek en ons volstaan daarby.
U mag dit egter interessant vind om op p. 574 tot 579 in die boek van Washington te gaan
kyk hoe ons ’n parabool se vergelyking aflei deur die kromme te beskou as die lokus van ’n
punt wat op ’n sekere manier (onder sekere voorwaardes) beweeg. Hierdie afleiding sal
egter nie geassesseer word nie en is vir blote interessantheid.

Assignment 5
1 Skets die volgende kegelsnit in u
antwoordboek:
2 2
x y
+ = 1
25 49
Dui alle snypunte met die asse aan
en skryf die naam van die kegelsnit
neer.
2 Skryf neer die vergelyking van die
sirkelvormige dam naby die hoek van
Juliusstraat en Smithstraat.
Vereenvoudig die vergelyking totdat
dit geen hakies meer bevat nie:
Julius Str.
O
N
173
A
Sketch the following cone section in
your answering book:
2 2
x y
+ = 1
25 49
Indicate all intercepts with the axes
and write down the name of this cone
section:
Write down the equation of the circular
pond near the corner of Julius Street
and Smith Street. Simplify the
equation so that it contains no
brackets:
C (7; 6)
B (9; 9)
Smith Str.
E

3. Die vorm van ‘n betonsloot in ‘n
besproeiingskema word gegee deur
die vergelyking
2 2
x y
+ = 1 met y ≤ 0 .
144 , 064 ,
Die oppervlak van die grond
(grondvlak) word as die X-as gekies;
alle afmetings is in meter.
Skets die sloot en toon sy diepte en
breedte duidelik aan.
4. Gee die naam en vergelyking van die
volgende kegelsnede:
Wenk: Wat is die vergelyking van die
asimptote? Onthou die asimptote het
b
die vorm y =± x en die vergelyking
a
vir hierdie kegelsnit is
2 2
x y
− = 1.
2 2
a b
174
The shape of a concrete ditch in an
irrigation scheme is described by the
equation
2 2
x y
+ = 1 with y ≤ 0 .
144 , 064 ,
The surface of the ground (ground
level) is taken as the X-axis; all
dimensions are given in meters.
Sketch the ditch and clearly indicate its
depth and width.
Supply the name and equation of the
following conic section:
Hint: What is the equation of the
asymptotes? Recall that the
b
asymptotes have the form y =± x and
a
that the equation for this conic section
is
2 2
x y
− = 1.
2 2
a b

5. Die deursnit van ‘n besproeiingsvoor
word getoon. Die vorm van die
deursnit van die voor word gegee
deur die vergelyking
2
x
y = 6⋅ 1+ −8 met −17 , ≤ x ≤ 17 ,
3
5.1. Bereken die breedte van die
watervlak.
175
The cross-section of an irrigation ditch
is shown. The shape of the crosssection
of the ditch is given by the
equation
2
x
y = 6⋅ 1+ −8 with −17 , ≤ x ≤ 17 , .
3
Determine the width of the water level.
5.2 Bereken die diepte van die voor. Calculate the depth of the ditch.
6. Skryf die definisie van ‘n parabool
neer in terme van die lokus van ‘n
punt wat op ‘n sekere manier
beweeg.
7. Skryf in standaardvorm, met ander
woorde maak vir y die onderwerp
7.1
van die vergelyking en stel ‘n tabel
van minstens 8 waardes op om die
deel van die grafiek bo die X-as te
teken deur punte te stip:
2 2
x y
+ = 1
16 25
Write down the definition of a parabola
in terms of a locus which moves
according to a certain law.
Write in standard form, in other words,
make y the subject of the equation
and set up a table of at least 8 values
in order to sketch the part of the graph
above the X-axis by plotting points:
2 2
x y
+ =
1
16 25

7.2
2 2
x y
− = 1
9 36
7.3 Ons weet dat die basiese ellips en
sentrale hiperbool simmetries om die
X-as is. Gebruik hierdie feit en teken
die volledige grafieke (bo en onder
die X-as) van die kegelsnitte in 7.1 en
7.2.
176
2 2
x y
− = 1
9 36
We know that the basic ellipse and
central hyperbola are symmetrical with
respect to the X-axis. Use this fact and
draw complete graphs (above as well
as below the X-axis) of the conic
sections in 7.1 and 7.2

6 Trigonometry
Estimated time required to achieve the outcomes in this study unit:
12 hours
Essential knowledge required before you attempt this study unit:
1. Knowledge about and skill in the application of the Pythagorean Theorem as
discussed at school level as well as in previous modules;
2. Knowledge and basic skill in the application of the trigonometric functions (covered in
WSKT 121)
3. Study unit 1 of WSKT 212
Learning outcomes for this study unit
After completing this study unit the student must be capable of doing the following:
1. Apply radian measure in order to calculate the length of the arc of a circle as well as
the area of a circle sector;
2. Calculate the angular velocity as well as the linear velocity of a body moving in a
circular trajectory;
3. Apply trigonometry in problems where two-dimensional vectors are involved;
4. Apply the sine rule to problems where scaline triangles are involved;
5. Apply the cosine rule to problems where scaline triangles are involved;
6. Apply the area rule to problems where scaline triangles are involved;
7. Solve simple trigonometric equations.
177

6.1 Application of radian measure
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Apply radian measure in order to calculate the length of the arc of a circle as well as
the area of a circle sector;
2. Calculate the angular velocity as well as the linear velocity of a body moving in a
circular trajectory.
Study the following material in the book by Washington
Paragraph Page numbers
8.4 249 – 251
Important examples in the book of Washington
Example nr Page numbers
1 249
3 250
4 251
5 251
Note that the applicable formulas appear in the information sheet
for WSKT.
178

Excercise 6.1 for self-assessment
Excercise in the text book Page number Problems
8.4 252 5, 7, 9, 11, 13, 15
179
17, 21, 23, 27, 37, 39, 45
The final answers to the given problems appear at the back of the text book.

6.2 Application of trigonometry to vectors
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Apply trigonometry in problems where two-dimensional vectors are involved
Study the following material in the book by Washington
Paragraph Page numbers
9.2 263 – 265
9.3 267 – 270
9.4 272 – 273
Important examples in the book of Washington
Example nr Page numbers
1 264
2 264
3 265
3 (VERY IMPORTANT EXAMPLE) 269
4 (VERY IMPORTANT EXAMPLE) 273
The theory of vectors as well as several applications were
addressed in WSKT 121.
180

Excercise 6.2 for self-assessment
Excercise in the text book Page number Problems
9.2 265 3, 5, 7
181
17, 19, 23
9.3 271 7, 9, 25
9.4 275 20
The final answers to the given problems appear at the back of the text book.

6.3 The sine rule, the cosine rule and the area rule
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Apply the sine rule in real life problems where scaline triangles are involved;
2. Apply the cosine rule in real life problems where scaline triangles are involved;
3. Apply the area rule in real life problems where scaline triangles are involved;
4. Solve simple trigonometric equations
6.3.1 The sine rule
Study the following material in the book by Washington
Paragraph Page numbers
9.5 277 – 282
Important examples in the book of Washington
Example nr Page numbers
1 278
2 279
182

Additional example
Problem 32 on p. 283 of the book by Washington
Point P on the mechanism is driven from side to side.
If the minimum value of angle is θ 32° is, what is the distance between the two furthest
positions of point P? What is the maximum value of angle θ ?
Solution:
Case 1 ( θ = 32°,
what is the distance x ?):
Note that P may be in any one of the two positions P 1 or P 2 with AB and BP still 36 cm and
24,5 cm respectively. Our calculation for the length of AP should therefore yield two different
answers. The reason for this is that the angle BPA may have any of two different values.
Let us calculate these two values:
sinBPA sinθ
=
AB BP
sinBPA sin 32°
∴ =
36 24,5 or in the alternative notation:
36 ⋅ sin32°
∴ sinBPA =
24,5
∴ sinBPA = 0778657 ,
183
sinP sinθ
=
p a
sinP sin32°
∴ =
36 24,5
36 ⋅ sin32°
∴ sinP =
24,5
∴ sinP =
0, 778 657

Let us investigate the two possible equations which we obtained in the previous step:
Clearly there are two possible values for the angle
BPA :
BPA = 51, 137 764° or BPA = 180°− 51, 137 764°
∴ BPA = 51, 137 764° or BPA = 128, 862 236°
The two possible values for AP may now be computed, that is the distance from point A to
P 1 or 2 P , from the fact that there must also be two possible values for ABP :
ABP = 180°−32 −BPA
∴ ABP = 180°−32 − 51,137 764° or ABP = 180°−32 − 128,862 236°
∴ ABP = 96, 862 236° or ABP = 19, 137 764°
So:
AP BP
=
sinABP sinθ
AP 24,5
∴ =
sin96,862 236° sin32°
24, 5 ⋅ sin96,862 236°
∴ AP =
sin32°
∴ AP = 45, 902 256 cm
or
AP BP
=
sinABP sinθ
AP 24,5
∴ =
sin19,137 764° sin32°
24, 5 ⋅ sin19,137 764°
∴ AP =
sin32°
∴ AP =
15, 157 207 cm
184

The distance x which we wish to determine is actually the difference between these two
distances, as may be seen from the diagrams above:
x = 45, 902 256 −15,
157 207
∴ x = 30, 745 cm
Now we may calculate the maximum value which θ may have (Case 2, that is the second
part of the question):
Clearly this happens when BPA = 90°;
then :
sinθ
=
a
p
24, 5
∴ sinθ
=
36
−1
∴ θ = sin 0, 680 556
= 42, 887°
Note that, because of the practical setup of the situation and the fact that we are dealing with
rigid members in the framework, no other values for θ are possible.
This interesting example demonstrated the power of the sine rule as applied to a technical
context.
Note that the applicable formula for the sine rule is given in the
information sheet for WSKT 212.
185

Excercise 6.3.1 for self-assessment
Excercise in the text book Page number Problems
9.5 282 Make a rough sketch of each of the
triangles and compute all the unknown
angles and sides:
186
3, 5, 7, 9
Also do:
29, 31
The final answers to the given problems appear at the back of the text book.

6.3.2 The cosine rule
Study the following material in the book by Washington
Paragraph Page numbers
9.6 284 – 287
Important examples in the book of Washington
Example nr Page numbers
1 285
2 285
3 286
5 286
Additional problem
Example 1
A helicopter uses a laser range finder to measure the angle between its line of sight to the
main pylons of the Humber Bridge in England. The angle between the two laser beams is
exactly 15º:
Calculate BC, that is the main span of the Humber Bridge.
187

Solution:
Using standard notation it follows for ΔABC that
2
a
2 2
= b + c −2bccos
A
2 2 2
( )( )
∴ a = 5 427, 091 + 5 365, 057 − 2 5 427, 091 5 365, 057 cos15°
= 58 237 153, 336 −56
249 053, 364
∴ a = 1 988 099, 972
= 1410
So, the length of BC, the main span of the bridge, is 1 410 m.
Example 2
Compute the angle between the legs of the step ladder:
Solution:
Label all angles and sides in the given triangle according to standard notation:
188

Since the unknown interior angle is located between or included between two known sides
we may strongly suspect that we are dealing with an application of the cosine rule – the
presence of an unknown or known angle included between two given sides is a feature which
often characterizes application involving the cosine rule (as well as the area rule – see later).
So, we may attempt to formulate the cosine rule for the given situation. Since we wish to
determine A , we should formulate the cosine rule in such a way that a appears on the left
hand side of the equation. Then we may make cos A the subject of the equation and solve
for A :
2 2 2
a = b + c −2bccos
A
2 2 2
∴a −b − c = −2bccos
A
2 2 2
∴ 2bc
cos A = c + b −a
2 2 2
c + b −a
∴ cos A =
2bc
Consequently, we may simply substitue the given values for a , b and c :
195 , + 18 , −1
cos A =
218 ( , )( 195 , )
∴ cos A = 0, 860 755
2 2 2
(We could also have proceeded in the following way in order to obtain cos A = 0, 860 755 :
189

2 2 2
a = b + c −2bccos
A
2 2 2
∴1 −18 , − 195 , = −2(
18 , )( 195 , ) cosA
∴− 6, 042 500 = −7,
02cosA
∴ 7, 02cos A = 6, 042 500
6042500 ,
∴ cos A =
702 ,
∴ cos A = 0860755 ,
Still, it is up to you to decide how you wish to proceed – whether by first substituting the
given values before re-arranging the equation or by first manipulating the formula until the
desired symbol is isolated on the left hand side of the equality sign before doing substitution.)
However, next we must examine the expression cos A = 0, 860 755 in order to find valid
values for the angle A :
cos A = 0, 860 755
So (mathematically speaking) there are two possible values for A :
A = 30, 598 541° or A = 360°− 30, 598 541°
∴ A = 30, 599° or A = 329, 401°
190

But because the sum of the interior angles of a triangle may not exceed 180° , the second
solution (in the fourth quadrant) is clearly invalid; the magnitude between the legs of the
ladder is therefore 30, 599°
.
Note that the applicable formula for the cosine rule is given in the
information sheet for WSKT 212.
Excercise 6.3.2 for self-assessment
Excercise in the text book Page number Problems
9.6 282 Make a rough sketch of each of the
triangles and compute all the unknown
angles and sides:
191
3, 5, 7, 9
Also do:
27, 31
The final answers to the given problems appear at the back of the text book.

6.3.3 The area rule
It often happens that we wish to find the area of a triangle. The classic formule
1
A triangle = × base × perpendicular altitude
2
or
1
Atriangle = b⋅⊥ h
2
should be familiar for you. But where will you be if the perpendicular altitude of a triangle is
unknown?
Example:
Calculate the area of ABC :
Please note that all information regarding triangle ABC is known – and yet: if you attempt to
1
apply the formula Atriangle = b⋅⊥ h then it simply does not work.
2
The problem, of course, that we have no information with regard to any of the perpendicular
distances from any vertex to the opposing side (altitudes) of the triangle. If we knew the
altitude, we would have been able to substitute it together with the applicable base into
1
Atriangle = b⋅⊥ h.
2
192

Let us now draw in the altitudes and examine the result to see if we can perhaps calculate
the length of an altitude – if so, we could use it together with any suitable base to substitute
1
the required information into the formula Atriangle = b⋅⊥h and be done:
2
Now consider any altitude, say AE. Is there a way to calculate the length of AE?
Well, in ΔACE it holds that
It follows that AE = 7, 191⋅ sin 56, 251°
AE
sin ACE AE
= , so = sin 56, 251°
.
AC 7,191
The lenght of AE then comes out as 5, 979 168 cm.
We could just as well have determined AE using the other triangle (triangle ΔABE ):
In ΔABE it holds that
AE
sin ABE AE
= , so = sin 44, 874°
.
AB 8,475
193

So it follows that AE = 8, 475 ⋅ sin 44, 874°
The lenght of AE then comes out as 5, 979 537 cm.
Note: The small differences in the obtained values of AE is an artifact of how accurately the
side lengths and interior angles of the triangle were measured. It is not due to any
embarrassing calculation error or approximation that the values of AE are not exactly the
same. It is therefore unnecessary for you to upset yourself about it.
No matter: we succeeded in calculating the value of AE so we now know the altitude of
ΔABC . We are now ready to calculate the area of ΔABC in the usual way:
1
Atriangle = b⋅⊥ h
2
1 ⎧I
used AE=5,979 cm since both obtained values for AE
= ( 10) ⋅( 5, 979)
⎨
2
⎩ agrees to the third decimal position
2
= 29, 895 cm
Conclusion:
• When we wish to calculate the area of a triangle, it is not strictly necessary that we
must be given the altitude of the triangle.
• We may calculate the altitude of the triangle using trigonometry (as above).
• Still, it is interesting and informative to note the pattern:
AE = 7, 191⋅ sin 56, 251°
and AE = 8, 475 ⋅ sin 44, 874°
Note that the side length 7,191 cm and the side length 8,475 cm are side lengths or
bases of ΔABC , namely side AC and side AB.
Also note that the angle 56,251° and the angle 44,874° are base angles on side BC
1
of the triangle for which we used Atriangle = b⋅⊥h. 2
As before, we refer to this type of angle which is located between two known
sides as an included angle. So the angle 56,251° is the included angle of AC and
CB.
194

• The altitude which we ended up using in the formula
number
ACsin ACE or alternatively the number
195
1
Atriangle = b⋅⊥h was actully the
2
AB sin ABE
• Symbolically speaking, we calculated the area of ΔABC as follows:
1
Atriangle = b⋅⊥ h
2
1
= ( 10) ⋅(
5, 979)
2
1
= ⋅BC ⋅AC
sin ACE
2
1
Atriangle = b⋅⊥ h
2
1
10 5 979
2
1
= ⋅BC ⋅AB
sin ABE
2
or = ( ) ⋅(
, )
Let us now label and express our calculations in terms of standard notation:
Then we may write the formulas above as follows:
A =
triangle
1
ab sinC or
2
A =
triangle
1
ac sinB .
2
The results above are kwown as the area rule.
1
A triangle = given side given side ⋅sin
included angle of side and side
2
( )( ) ( )
1 2 1 2
Note: The approach above by which we arrived at the area rule, does not constitute a formal
proof of the formula. Still, it does provide insight into why the area rule does in fact work.
The formal proof may be found in may mathematics text books or on respected internet sites
– we do not need to re-invent the wheel.

Important examples
Example 1
Calculate the area of the given machine part:
Solution:
First label the triangle, its vertexes and its sides according to standard notation; label the
vertexes something like P, Q and R and label the corresponding opposite sides:
Note that the only given angle is located between two known sides. So we see that Q is the
included angle of side p and side r.
We may therefore write the area rule for this particular situation as follows:
1
AΔPQR =
pr sinQ
2
196

The calculation proceeds as follows:
1
AΔPQR = pr sinQ
2
1
= ( 175)( 120) ⋅ sin 17, 5°
2
2
= 3157, 411mm
Example 2
2
The area of a triangular region of land (say ΔLMN) is 3000m
. If the boundaries MN and
ML is 70 m and 90 m respectively, determine the angle α which is included by MN and
ML . Also find the length of the boundary opposite to the angle α .
Solution:
First sketch the triangle according to the given information and label it according to suitable
standard notation (note that it is not a scale drawing):
Clearly we are dealing with the area rule here, since area is given as well as certain
boundary sides and the question involves an included angle. So we write:
1
AΔLMN = ⋅l⋅n⋅sinα 2
1
∴ 3 000 = ( 70)( 90)
sinα
2
∴ 3 000 = 3150 sinα
3000
∴ sinα =
I simply switched the sides of the equation
3150
∴ sin α = 0952381 ,
197

Let us investigate the result of the calculation:
So there are two possible values for α :
α = 72, 247 210° or α = 180°− 72, 247 210°
∴ α = 72, 247° or α = 107, 753°
1
Check: If any of these angles are substituded into AΔLMN = l⋅ nsinα with l = 70 m
2
2
and n = 90 m then you obtain an area of 3000m
. The triangle can therefore
have more than one shape but still the same area.
2
Next we may find the length of the other side which bounds 3000m
; for this purpose we
use the cosine rule (because we are dealing with two sides and an included angle:
198

2
m
2 2
= l + n −2⋅l⋅n⋅cosα 2
∴ m
2 2
= 70 + 90 − 2 70 90 cos 72, 247 210° or
2 2 2
m = 70 + 90 − 2 70 90 cos 107, 752 790°
( )( ) ( )( )
2
∴ m = 13 000 − 12 600 cos 72, 247 210° or
2
m = 13 000 − 12 600cos 107, 752 790°
2 2
∴ m = 13 000 − 3 841, 874 509 or m = 13 000 + 3 841, 874 509
∴ m = 9 158, 125 491
or m = 16 841, 874 509
∴ m = 95, 698 m or m = 129, 776 m
Extremely interesting: Triangles with equal areas do not necessarily have the same shape.
Note that the applicable formula for the area rule is given in the
information sheet for WSKT 212.
Excercise 6.3.3 for self-assessment
1. Calculate the volume of the given prism (note that volume is product of base area and
altitude, provided that the sides forming the altitude are perpendicular to the base area)
Answer: Volume =
45 466, 334 mm
3
199

2. The diagram shows a section of a pavement:
Find the magnitude of the angle θ so that
200
1 690 370
3
, m of concrete will be needed to
make the section of pavement. (hint: θ is an obtuse angle, in other words
90°< θ < 180°).
Answer: θ = 105°
Assignment 6
Vraag 1/ Question 1
1.1 ‘n Helikopter meet met ‘n laserrigtingvinder
die afstande na die
hoofmaste van die Humber Brug in
Engeland. Die hoek tussen die twee
laserstrale word gemeet as presies 15º:
A helicopter measures the distances to
the main pylons of the Humber Bridge
in England using a laser range finder.
The angle between the laser beams is
measured as exactly 15º:

Bereken die grootte van θ , dit is die
hoek wat die een laserstraal by die
oostelike mas maak.
1.2 ‘n Krag van 450 kN werk teen ‘n hoek
van 34º ten opsigte van die vertikale
as.
Maak ‘n diagram van die krag en
ontbind dit in onderling loodregte
komponente.
1.3 Vir ‘n kar wat teen ‘n snelheid v om ‘n
draai met ‘n radius r beweeg, geld dit
dat die ryvlak AB van die pad teen ‘n
hoek θ gekantel moet wees sodat
2
v
tanθ
= :
gr
1.3.1 Bereken die grootte van θ indien
2
v = 33 m/s , g = 9,8 m/s en
r = 150 m .
201
Calculate the magnitude of θ , that is
the angle formed by one of the laser
beams at the eastern pylon.
A force of 450 kN acts at an angle of
34º with respect to the vertical axis.
Construct a diagram of the force and
resolve it in mutually perpendicular
components.
For a car travelling at a velocity v
around a curve of radius r , it holds that
the driving surface AB should be
inclined at an angle θ so that
2
v
tanθ
= :
gr
Calcultate the value of θ if
2
v = 33 m/s , g = 9,8 m/s and
r = 150 m .

1.3.2 Gestel AC, die buitenste rand van
die ryvlak, is 4,762 m hoog.
Bereken AB, die breedte van die
ryvlak.
Vraag 2/ Question 2
2.1 Die grootte van die emk E in Volt wat
in ‘n generator opgewek word deur ‘n
draadlus wat teen ‘n hoeksnelheid ω
(in radiale per sekonde) in ‘n
magneetveld B roteer, word gegee
deur die formule E = NABω⋅ sinθ
waar
N die aantal windings in die lus, A die
oppervlakte van die lus en θ die hoek
in grade wat die normaalvektor met die
magneetveld maak:
Bereken die grootte van θ in
E = NABω⋅ sinθ
indien N = 3,
2
A = 0,02 m , B = 0,5 Tesla ,
ω = 157,08 rad/s en E = 4 V .
Gee twee oplossings.
202
Suppose AC, the outer edge of the
driving surface, is 4,762 m high.
Calculate AB, the width of the
driving surface.
The magnitude of the induced emf E
in Volt which is generated by a
generator when a loop of wire rotates
at an angular velocity ω (in radians
per second) inside a magnetic field B
, is given by the formula
E = NABω⋅ sinθ
where N is the
number of windings in the loop, A is
the area of the loop and θ is the
angle in degrees between the normal
vector and the magnetic field:
Calculate the magnitude of θ in
E = NABω⋅ sinθ
if N = 3,
2
A = 0,02 m ,
B = 0,5 Tesla , ω = 157,08 rad/s and
E = 4 V .
Give two solutions.

2.2 Die diagram hieronder toon ‘n sekere
dakraam:
2.2.1 Bereken die lengte van balk BD, korrek
tot twee desimale plekke, deur slegs
gebruik te maak van die inligting in
driehoek ABD.
2.2.2 Gebruik slegs die inligting in driehoek
BCD en voer ‘n berekening uit om u
antwoord op vraag 2.2.1 te kontroleer.
(Benader u antwoord weer tot twee
desimale plekke)
Vraag 3/ Question 3
3.1 ‘n Krag van 350 kN werk in ‘n plat
horisontale vlak in ‘n rigting 37° wes
van suid.
3.1.1 Bereken die westelike komponent van
die krag.
3.1.2 Bereken die suidelike komponent van
die krag.
203
The diagram below shows a certain
roof frame:
Calculate the length of beam BD,
correct to two decimal places, using
only the information in triangle ABD.
Use only the information in triangle
BCD and perform a calculation to
check your answer to question 2.2.1.
(Once again, approximate your
answer to two decimal places)
A Force of 350 kN acts in a flat
horizontal plane in a direction 37° west
of south.
Calculate the westward component of
the force.
Calculate the southward component of
the force.

3.2 Beskou die volgende dakkap: Consider the following roof truss:
3.2.1 Bereken die lengte van balk PQ,
korrek tot twee desimale plekke.
3.2.2 Gebruik die cosinus-reël en bepaal
die lengte van PR.
204
Calculate the length of beam PQ,
correct to two decimal places.
Use the cosine rule and determine the
lenght of PR.

7 Elementary decriptive statistics
Estimated time required to achieve the outcomes in this study
section
10 hours
Essential knowledge required before you attempt this study unit
1. Training in basic use and functionaltity of Microsoft Excel, as addressed in CMPF 111
or a similar module.
Learning outcomes for this study unit
After completing this study unit the student must be capable of doing the following:
1. Generate a frequency distribution table using Microsoft Excel;
2. Draw a histogram using Microsoft Excel;
3. Determine the mode of a given data set using pen and paper methods as well as
Microsoft Excel;
4. Determine the mean of a certain data set using pen and paper methods as well as
Microsoft Excel;
5. Determine the arithmetic mean of a certain data set using pen and paper methods as
well as Microsoft Excel;
6. Calculate the weighted mean of a set of data where each data point does not
necessary carry the same importance or weight as the others using pen and paper
methods as well as Microsoft Excel;
7. Find the standard deviation of a set of data; using Microsoft Excel
8. Utilize the knowledge and skills above in order to administrate a computerized mark
book like for example in Microsoft Excel;
9. Use the built-in regression function in Microsoft Excel to find the equation of the best
curve fitting a given set of data points.
205

7.1 Frequency distributions
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Generate a frequency distribution table using Microsoft Excel;
2. Draw a histogram using Microsoft Excel;
Study the following material in the book by Washington
Paragraph Page numbers
22.1 609 – 613
Definitions
Term or concept Description
Raw data Information (number values) which have not yet been organized or
ordered or processed, for example a sheet of paper on which the
names, surnames and test marks of 30 children have been recorded.
Ordered data Raw data which has been ordered in some way, for example a sheet
of paper on which the names, surnames and test marks of 30 children
have been ordered in ascending order according to their marks.
Grouped data Whenever data is organized in groups (also named classes)..
An example would be a class of 30 children whose marks have been
classified in the following groups:
Class 1: all marks below 40%
Class 2: all marks from 40% up but less than 50%
Class 3: all marks from 50% up but less than 75%
Class 4: all marks from 75% and up
206

Frequency The number of data values which occur in a specific group or
ordered data.
Frequency
distribution table
Example: Suppose out of a class group of 30 children five of them
achieved less than 40%, 11 children achieved 40% or more but less
than 50% , 10 children achieved 50% or more but less than 75% and
4 children achieved 75% or more.
Then the frequency of the first class is 5, the frequency of the second
class is 11, the frequency of the third class is 10 and the frequency of
the fourth class is 4.
A table which represents the classes in which a grouped data set has
been divided, as well as the frequency for each class. For example:
Class 1 2 3 4
Class interval [0; 40) [40; 50) [50; 75) [75;100]
Frequency 5 11 10 4
Note how we write the class intervals:
For example: In class 2 the class interval [40; 50) means that all
values larger than 40 and including 40 (equal to 40) to all data values
just less than 50 belongs in that class; 50 is not included. A mark of
50 would belong in the third class.
Histogram A histogram is a graphical representation of grouped data in
terms of classes and frequencies. It is effectively simply a
graphical representation of the information contained by a frequency
distribution table:
Frequency
12
10
8
6
4
2
0
5
207
11
[0; 40) [40; 50) [50; 75) [75; 100]
10
Class intervals
4

Histogram Sometimes histograms employ the so-called class mark instead of
indicating class intervals on the horizontal axis:
Frequency
12
10
8
6
4
2
0
5
The class mark is the arithmetic mean (average) of the lower limit
and upper limit of a class interval (add smallest possible and
largest possible numbers in the class interval and divide by 2).
Frequency polygon A frequency polygon is a graphical representation of grouped data in
terms of classes and frequencies, where the frequency values form
the vertexes of a polygon opposite the class marks, for example:
208
11
10
20 45 62.5 87.5
Percentage
4

How to generate a frequency distribution table and histogram in
Microsoft Excel
Suppose you have a data set consisting of 10 children’s names (“Naam”), surnames (“Van”)
and marks (“Punte uit”) for a certain assessment or evaluation, for example a Test (“Toets
1”):
This is a typical raw data set. It may be ordered according to several different criteria, for
example by ordering the entries alphabetically with respect to the surnames (“Van”):
Select (high-light or shade the cells by clicking on the left mouse button and dragging the
mouse marker over the appropriate cells while holding in the button) all the data (not the
headings).
Then click on “Data” in the top horizontal task bar and select “Sort”. When the sorting menu
opens, select the column where the surnames are displayed (it is column B):
209

If you now click on “OK”, then Excel will sort the data set alphabetically according to the
entries in column B; Note that all the data associated with surname (namely the name
(“Naam”) and mark (“Punt uit”) is carried with the surname – so the data does not become
disassociated or jumbled:
210

In order to process and organize our ordered data set further, we may, for example, compute
the percentage (“Persentasie”) achieved by each child. We can do this in Excel by entering
a formula into cells D6 to D15 – this formula must simply divide the mark in die C-column by
the total of the tests in cell C4 and multiply the result by 100 in order to obtain a percentage.
We may enter this formula into cell D6 and then copy it to the other cells by placing the
mouse marker in the right-hand bottom corner of cell D6 until the marker changes into a plus;
then you left-click and drag (holding in the left-mouse button) the formula in D6 down to all
the cells from D6 to D15.
However, before we can do that we must note that each of the calculations which take place
in cells D6 to D 15, is divided by the same value (the value in C4). Because C4 is used in
each of the rows from 6 to 15, we must “anchor” the cell C4 in the formula which we enter
into D6 by employing the dollar-sign. Simply insert a dollar sign before the column letter and
before the row number of the cell you which to anchor:
$C$4
So the formula you have to enter while your mouse marker stands on D6, is:
=C6/$C$4*100
211

Press “Enter” and copy the formula from cell D6 to D 15.
The following will be obtained:
You may format cells D6 to D15 (or any rectangular section of the spreadsheet) to
approximate numbers correct to a certain number of decimal places. If, for example, you
wish to approximate the percentages to one decimal place then you must high-light the cells
D6 to D15 and right-click on the shaded area and select “Format Cells” from the drop-down
menu. Then select “Number” in the next drop-down menu and set the number or decimal
places to 1:
212

If you now click “OK”, the following will happen:
213

The next step is to change the ordered data set into a grouped data set. Note that the term
“grouped data” indicates that we subdivide the data into groups (classes) and then determine
the frequency for each class.
We wish to obtain a statistical picture or view of the percentages achieved by the class of
children.
The question is therefore now: How do we group the data?
In the first place, note that a percentage is a number which belongs in the closed interval
[ 0; 100 ] . Saying this, we mean that a percentage may be zero, but not less than zero – it
may also be 100, but not more than 100 – and all values between 0 and 100 are also
possible percentage values. Now think of the interval [ 0; 100 ] as a horizontal number line.
We may divide this number line in several parts; one way to do this is to divide it into five
parts of equal width:
In this way we obtain five classes of equal class width (namely 20%). In order to determine
the frequencies associated with each class we must simply sort the percentages achieved
into the classes and count how many numbers end up in each class:
We should now be able to construct the following frequency distribution table by hand:
Class 1 2 3 4 5
Class interval [0; 20) [20; 40) [40; 60) [60;80) [80; 100]
Frequency 0 1 4 2 3
214

The entire idea, however, is to do the process described above (the grouping and counting)
by means of Microsoft Excel. In order to achieve this, we proceed as follows:
Create for yourself three additional columns to the right of the percentage column. Label
these columns “class interval”, “Bin” and “Frequency”:
Cells E6 to E 10 you may format as “Text”; cells F6 to G10 you may format as “Number”,
correct to two decimal places.
Now we must fill in the “Bin” column. In order to do so, we must think of the cells E6 to E10
as bins (containers). In the “Bin” column we enter the upper value of the class interval.
Because the upper limit of each class is open (except of the last class) (note the round
brackets), the bin value of each class should be just smaller than the number to the left of the
round bracket in the class interval notation. So we fill in the “Bin” column values as follows:
215

The “Bin” column must contain values which are just smaller that the “next” value which
should be sorted into the class to the right of the current class. If the “Bin” value of the first
class, for example, were 20 it would mean that a percentage of exactly 20% would be sorted
into that class – that is not what we want to happen; 20% should be sorted into the second
class. That is the meaning of the notation [0; 20) – all values smaller that 20% should be
sorted into that class.
We are now ready to employ the built-in “Frequency”-function in Excel in order to compute
and display for us the frequencies of the classes in cells G6 to G10.
To achieve this, we proceed as follows:
• High-light the cells G6 to G10 (where you want to see the frequencies)
• Type into the formula window at the left top of the spreadsheet: =Frequency(
• You will see that Excel now wants you to high-light two columns, namely the column
where the data appear (the percentages, that is cells D6 to D15) as well as the “Bin”
column (cells F6 to F10). When you are done high-lighting columns D6 to D15, type
the comma and then high-light columns F6 to F10. Close the bracket in the top
formula bar:
• In order to automatically display the frequencies (cells G6 to G10) first press
“CTRL” and “SHIFT” simultaneously and then, while you hold both keys in, hit the
“Enter” key. (if you do not follow the instructions above, it will not work.)
The following will happen:
216

This is the same result which we obtained by hand a few pages ago.
So Excel automatically generated a frequency distribution table for us.
Warning: You must never attempt to type anything into the cells where Excel displays
frequency values. If you attempt to enter values by hand into cells G6 to G10, the program
may close and you will lose all your data. If you want to empty the cells G6 to G10, it is also
terribly dangerous to use “Delete”. Rather high-light the region G6 to G10, right-click on the
region and select “Clear Contents”. This is much safer than “Delete”.
The following step is now to generate a histogram from the frequency distribution table.
In order to achieve this, proceed as follows:
• Select the frequencies (cells G6 to G10)
• Select “Insert” in the top horizontal task bar and select “Column”
217

• You may select any of the types of column graphs Excel offers; there is,
incidentally, nothing wrong with the first choice (you are welcome to experiment
with the rest). Select the first possibility in the list. The following should happen:
218

Right-click on “Series 1” and “Delete” it. In order to write the contents of cells E6 to E10
below the horizontal axis of your graph, proceed as follows:
• Right-click on any of the digits or numbers below the horizontal axis.
• Select “Select Data”
• Left-click “Edit” at the top in the right-hand white little window
• When the other small “Axis Labels”-window opens, simply high-light the cells E6
to E10 and click on the tiny red arrow above “OK” on the “Axis Labels”-window:
219

• If you click twice on “OK”, the graph will appear as follows:
• In order to make the vertical bars touch (as they should in a histogram), right-click on
any bar and select “Format Data Series”. Set the “Gap width” as zero.
• In order to vary the colour of the bars, select “Fill” on the left hand side and select
“Vary colours by point”:
220

• To display the frequencies on each column, right-click on any column and select “Add
Data Labels”.
• Select “Labels” in the top horizontal task bar and experiment with the “Labels” and
other features in order to label the axes of the graph and to insert a title for the graph.
You can also obtain interesting effects using the “grid lines” options:
221

If you right-click on any white area inside the frame of the graph and you select “Move
Chart”, then you obtain an option to remove the graph out of the spreadsheet and to attach it
as a large picture to the bottom of the display.
It would now be relatively easy to generate a frequency polygon also of the data – such a
representation does not, however, provide any new insight in the situation and will not be
assessed in this module.
Important:
Practise regularly with Excel and electronic mark books. It saves hours and enables you to
present your records in an accurate, neat and professional way.
Always remember:
Save your data every day (after every work session!) in at least two physically separate
locations (for example: on the hard disk of the computer or laptop AS WELL AS on a
removable data storage device which you store separate from the computer or laptop). The
disadvantage of any electronic data base is that it is extremely vulnerable to negligence. If
your data is lost, it is truly gone and you will never ever see it again.
(People have lost jobs because of that kind of mishap – what will you do if your laptop
gets stolen or if the hard disk breaks?)
Keep daily back-up copies of your work in at least two physically separate locations – or at
some point, you will be truly sorry about your negligence – we guarantee that, out of our own
bitter experience.
Excercise 7.1 for self-assessment
Excercise in the text book Page number Problems
22.1 612 Do with Excel: nr 21
222
Only draw a histogram and see that it is
technically well presented.
Do with Excel: nr 27
Make a frequency distribution table with
six classe of equal width and draw the
histogram.

223
Hint: The first class interval is
[25; 26,83), the class width is
1,83 and the sixth class interval is
[34,15; 36]

Solutions:
21 Note the way we write each class interval.
27. Note that the “Bin”-column contains values which are just smaller than the right endpoint
values of each interval, since the right end point of all intervals (except the last one) is
always chosen as open.
224

7.2 Mode, median and arithmetic mean
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Determine the mode of a given data set using pen and paper methods as well as
Microsoft Excel;
2. Determine the mean of a certain data set using pen and paper methods as well as
Microsoft Excel;
3. Determine the arithmetic mean of a certain data set using pen and paper methods as
well as Microsoft Excel;
4. Calculate the weighted mean of a set of data where each data point does not
necessary carry the same importance or weight as the others using pen and paper
methods as well as Microsoft Excel;
5. Find the standard deviation of a set of data; using Microsoft Excel
Study the following material in the book by Washington
Paragraph Page numbers
22.2 613 – 616
22.3 617 – 618
Definitions
Term or concept Description
Median Suppose a number of data points have been ordered in ascending
order. If the number of data points are odd, then the median is the
data value in the centre of the ordered sequence. If the number of
data points are even, then the median is the arithmetic mean of the
two centre data values.
Example: Given: 23; 44; 10; 7; 0; 15; 80; 70; 15
225

Order the data: 0; 7; 10; 15; 15; 23; 44; 70; 80
Median: 15 (value in the centre)
Example: Given: 23; 44; 10; 7; 0; 15; 80; 70
Order the data: 0; 7; 10; 15; 23; 44; 70; 80
Median: 19 (average of 15 and 23)
Mode Suppose a number of data points are given. The value which occurs
the highest number of times, is the mode. If every data point occurs
only once, there is no mode. There may be more than one mode.
Arithmetic mean or
average
Weighted arithmetic mean
The number of times that the mode occurs, is called its frequency.
Example: Given: 23; 44; 10; 7; 0; 15; 80; 70; 15
Mode: 15 (and its frequency is 2)
Suppose a number of data points are given. The arithmetic mean is
the sum of the data points divided by the number of data points.
Example: Given : 23; 44; 10; 7; 0; 15; 80; 70; 15
sum of the data points
Arithmetic mean =
number of data points
∑
xn
∴ x =
This is the formula in terms of symbols
n
23 + 44 + 10 + 7 + 0 + 15 + 80 + 70 + 15
=
9
264
=
9
= 29,333
Sometimes we wish to calculate the arithmetic mean or average of a sequence of values, but
where each of the values does not have the same or equal importance or weight. If is, for
example, possible that one of the values occurs more frequently than the others. In such a
x1⋅ f1+ x2 ⋅ f2 + ... + xn ⋅ fn
case we may modify our technique by using the formula x =
. The
f + f + ... + f
frequency f of each value is then its weight.
226
1 2
n

Example:
Suppose the following data set is given: 23; 44; 15; 10; 7; 0; 15; 80; 70; 15; 44; 15
Note that each of the data points 23; 10; 7; 0; 80; 70 occurs only once; so their frequencies
are 1. The data point 15 however, occurs four times, so its frequency is 4. The data point 44
occurs twice, so its frequency is 2.
The weight of the data point 15 is therefore 4 and the weight of the data point 44 is
therefore 2.
The weighted arithmetic mean is therefore:
x1⋅ f1+ x2 ⋅ f2 + ... + x8 ⋅ f8
x =
f1+ f2 + ... + f8
231 ⋅ + 101 ⋅ + 71 ⋅ + 01 ⋅ + 801 ⋅ + 701 ⋅ + 154 ⋅ + 442 ⋅
=
1+ 1+ 1+ 1+ 1+ 1+ 4+ 2
338
=
12
= 28,167
Weighted arithmetic means and the calculation of marks
Teachers and lecturers must be capable of calculating term marks, year marks or
participation marks. Often the evaluations or assessments which makes out part of the final
marks, count out of different totals – so the marks does not carry equal weight. The problem
is then to calculate the term mark, year mark or participation mark as a weighted arithmetic
mean – a fair mark which takes into account the relative importance of each mark involved.
Example 1:
A student has a participation mark of 43% and an exam mark of 58%. Calculate the module
mark if the participation mark and the exam counts in the ratio 2:1 respectively.
Solution:
The ratio 2:1 means that the module mark consists of thirds (2+1=3) and that the
participation mark counts two thirds (⅔) while the exam mark counts one third (⅓). If both the
participation mark and the exam mark are expressed as percentages, then the calculation of
the module mark may proceed as follows:
227

Module mark = w1⋅ Participation mark + w2
⋅Exam
mark
2 1
= × 43 + × 58
3 3
= 48 %
Example 2
A student has a participation mark of 58% and an exam mark of 45%. Calculate the module
mark if the participation mark and the exam mark counts in the ratio 60:40.
Solution:
Since all the given information is in terms of percentages (note that 60 + 40 = 100) we may
simply proceed as follows:
Module mark = w1⋅ Participation mark + w2
⋅Exam
mark
60 40
= × 58 + × 45
100 100
= 52,8 %
Example 3
The participation mark of a student consists of three evaluations which count as follows:
35% of the participation mark, 40% of the participation mark and 25% of the participation
mark. The three evaluation marks, expressed as percentages, are respectively 81%, 62%
and 95%. Calculate his participation mark.
Solution:
Participation mark = w ⋅ Evaluation + w ⋅ Evaluation + w ⋅Evaluation
35 40 25
= × 81+ × 62 + × 95
100 100 100
= 76,9 %
Example 4
1 1 2 2 3 3
The participation mark of a student consists of four tests which counts out of the following
totals:
45; 55; 50; 40
His marks attained were as follows:
39 out of 45; 40 out of 55; 19 out of 50; 29 out of 40
Compute the student’s participation mark by utilizing a weighted mean.
228

Solution:
• First convert the four test marks to percentages; this will serve to simplify the
calculations which follow:
39 out of 45 means 39 × 100% which yields 86,667% Similarly, 40 out of 55 means
45
40
× 100% which yield 72,727%; similarly, the other two marks are respectively 38%
55
and 72,5% if we express them as percentages.
• Now we must calculate the weight of each test. For this we use the grand total of the
four tests (sum of totals). The sum of their totals are 45+55+50+40 which is 190.
Note that the weight of each test may be considered the fraction which its total makes
out of the grand total:
45
190
w 1 = , 2
55 50
40
w = , w 3 = en w 4 =
190 190 190
• The formula for the participation mark is now simply:
Participation mark = w1⋅ Test1 + w2 ⋅ Test2 + w3 ⋅ Test3 + w4
⋅Test4
45 55 50 40
= × 86,667 + × 72,727 + × 38 + × 72,5
190 190 190 190
= 66,842 %
Note that since the test marks have first been converted to percentages, the final answer of
our calculation is also automatically a percentage. That is highly desirable.
•
Term or concept Description
Standard deviation The standard deviation of a set of data is a measure of how far the
data points lie from the arithmetic mean. Data points with values
close to each other will have a small standard deviation. Similarly,
data points which differ much from one another, will have a large
standard deviation.
The standard deviation s may be calculated using a calculator or a
suitable computer program (such as Microsoft Excel).
229

How to determine the median, mode, arithmetic mean and weighted
arithmetic mean by using Microsoft Excel
Microsoft Excel offers the functions “Median”, “Mode” and “Average” with which the
abovementioned properties of a data set may be calculated. Even the standard deviation of a
data set may be calculated but that is outside the scope of this module. You will not be
assessed about the standard deviation.
Suppose you have a data set (all the number cells were formatted as “Number” and set to
display numbers correct to one decimal place and the “alignment” was set to centre
numbers):
To display the median of the data points in cells B4 to B9 in cell B11, left-click on B11 and
type into the formula window: =median(
Now simply mark cells B4 to B9 by high-lighting them (left-click with the mouse on B4, hold in
the left mouse button and drag the mouse marker over the cells B4 to B9). Close the bracket
in the formula window:
230

Now hit “Enter” and the median of the data set will appear in cell B11.
Proceed in exactly similar manner and let Excel display the mode, arithmetic mean
(“average”) and standard deviation in the cells B12 to B14:
231

The final result should look as follows:
Note how the computer indicates that there is no mode present in this situation (cell B12).
How to calculate the final mark of a sequence of assessments
which count different weights by using Microsoft Excel
Suppose you have the following data set:
(Note that the white cells are formatted as “Number” and set to display numbers correct to
one decimal place and the “alignment” was set to centre numbers). The columns D to H
contain the information for “Toets 1” (Test 1), “Opdrag 1” (Assignment 1), “Toets 2” (Test 2),
“Opdrag 2” (Assignment 2) and “Toets 3” (Test 3).
We now want to calculate the Final Percentage of the student (“Finale persentasie”) in cell I5.
232

Proceed as follows:
• Convert all marks to percentages. “Insert” a column to the right of each mark so that
you can use it to calculate that mark as a percentage of the assessment total:
To calculate the corresponding percentages for all marks in the cells E5, G5, I5, K5
and M5, you simply need to type the formula “=Punt/Totaal*100” in the appropriate
cells. To calculate the percentage P1 of the first assessment in E5, simply left-click
on E5 and type into the formula window: =D5/D3*100
Hit “Enter” and repeat the process to display the percentages P2 to P5 in the desired
places:
• Next, calculate the grand total (“Groottotaal”) of all the assessment totals in cell B7:
• Next, calculate the weights of the assessments in cells B8 to B12; to calculate the
first weight, simply program into cell B8 the formula “total divided by grand total”:
233

• Repeat the process to calculate all the weights in their appropriate places:
• Next, to calculate the weighted mean of all the assessments in cell, you simply need
to left-click on cell N5 and to type the following formula into the formula window:
=B8*E5+B9*G5+B10*I5+B11*K5+B12*M5
The window above shows the result after you hit “Enter”.
Important:
The process above is exactly the same when you have a mark sheet containing the data for
hundreds of students. You may copy formulas which hold for more than one cell, to other
cells. Just remember to use the $-symbol whenever you copy a formula where the formula
uses a value from a single, fixed cell in more than one cell – for example, you should include
$-symbols in front of the column and in front of the row reference when you copy the
weighted arithmetic mean formula to other rows underneath the first row.
234

Practice Question:
The term mark of a schoolboy consists of four tests which counted out of the following totals:
55; 45; 50; 35
His marks were as follows:
32 out of 55; 38 out of 45; 34 out of 50; 21 out of 35
Calculate his term mark as a percentage by utilizing the idea of a weighted mean. Show all
steps of your calculation. Can you do this calculation on Microsoft Excel? Check your results.
(Answer: 67,568 %)
235

Excercise 7.2 for self-assessment
Excercise in the text book Page number Problems
22.2 616 Do by hand as well as by Excel:
236
33, 34, 37, 38
(“mean” is the arithmetic mean)
The final answers to the given problems appear at the back of the text book.
Practice Question:
The term mark of a schoolboy consists of four tests which counted out of the following totals:
55; 45; 50; 35
His marks were as follows:
32 out of 55; 38 out of 45; 34 out of 50; 21 out of 35
Calculate his term mark as a percentage by utilizing the idea of a weighted mean. Show all
steps of your calculation. Can you do this calculation on Microsoft Excel? Check your results.
(Answer: 67,568 %)

7.3 The administration of computerized mark schedules using
Microsoft Excel
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Utilize the knowledge and skills above in order to administrate a computerized mark
book like for example in Microsoft Excel.
Let us now consider an application of all the knowledge and skills which we acquired so far in
this study unit.
Excercise 7.3 for self-assessment
Suppose you have the following mark schedule:
Use the skills you have developed in this study section and calculate all the omitted values in
the shaded cells.
Also determine the frequency distribution table of the data using the intervals: [0; 50), [50;
80) and [80; 100].
Draw a histogram of the frequency distribution.
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Solution:
Your results should look similar to this (your formatting and the physical placement of your
cells may of course be slightly different):
Remark: Sort your data alphabetically according to name or surname.
What is the meaning of the histogram; what information can you immediately glean from it?.
238

7.4 Regression using Microsoft Excel
Learning outcomes for this study section
After completing this study section the student must be capable of doing the
following:
1. Use the built-in regression function in Microsoft Excel to find the equation of the best
curve fitting a given set of data points
Study the following material in the book by Washington
Paragraph Page numbers
22.6 632 – 640
Background and revision
(Class discussion)
Ons het reeds in vroeëre leereenhede met die volgende probleem kennis gemaak:
Gestel ons het ‘n grafiek waarop datapunte grafies voorgestel is. Watter tipe kromme sou
die beste deur die punte pas en wat is die vergelyking van hierdie kromme?
Hierdie probleem word regressie genoem.
Regressie kan algebraïes gedoen word deur van redelik ingewikkelde analitiese metodes
gebruik te maak. Moderne tegnologie maak dit egter vir ons moontlik om regressie deur
middel van geskikte rekenaarprogrammatuur te doen; Microsoft Excel het ‘n redelik kragtige
regressie-funksie waarmee ons alreeds in vorige leereenhede te doen gekry het.
Ons hersien vervolgens net vlugtig die tegniek aan die hand van ‘n geskikte voorbeeld.
239

Example
Gegee:
Bepaal die vergelyking van die kromme wat die beste deur die datapunte pas.
Solution:
• Voer die data in Excel in
• Selekteer al die waardes en kies “Insert” “Scatter”:
• Kies die eerste keuse in die aftrek-keuselys en Excel sal die volgende grafiek
vertoon:
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• Regskliek op “Series1” en kies “Delete”.
• Regskliek nou op enige datapunt in die grafiek en kies “Add Trendline”
• ‘n Aftrek-keuselys sal oopmaak:
• Soek nou deur die lys van regressietipes deur met u muis op die verligte
moontlikhede te kliek.
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• Indien ‘n ongewenste keuselys opkom (dit gebeur soms as u die “Logarithmic” tipe
kies), kliek gewoon op “Close” en kies “Undo” in die boonste horisontale taakbalk.
Regskliek dan weer op enige datapunt en kies “Add Trendline” sodat u die “Format
Trendline”-aftrek-keuselys terug kry.
• Indien u op die “Polynomial” tipe gekliek het, het u die keuse om die graad (“Order”)
van die polinoomkromme te stel. Eksperimenteer met ‘n reguit lyn (polinoomkromme
van die eerste graad), ‘n parabool (polinoomkromme van die tweede graad) en ‘n
derdegraadse kromme (polinoomkromme van die derde graad) ens. totdat u ‘n
kromme kry wat na u bevrediging deur die datapunte pas.
• Indien u nie geskikte polinoom-tipe vind nie, kan u in die lysie afbeweeg. (“Moving
Average” is nie ‘n tipe funksie waarmee ons vir ons doeleindes sal werk nie)
• In die geval van hierdie voorbeeld, behoort u te vind dat ‘n derdegraadse
polinoomkromme die beste deur die punte pas:
• Merk nou die blokkies “Display Equation on chart” en “Display R-squared value on
chart”. Die volgende sal gebeur:
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• Indien u op die teks langs die kromme regskliek, kan u die lettertipe (“font”),
grootte ensovoorts verander. Indien u op die teks regskliek en “Format Trendline
Label” kies, kan u die wyse waarop die getalle in die teks vertoon word stel
(byvoorbeeld die aantal desimale plekke).
• Die R-kwadraat-waarde wat Excel bepaal, is ‘n statistiese maatstaf van hoe
akkuraat die kromme wel deur die datapunte gaan. Hoe nader aan 1 hierdie
waarde, hoe betroubaarder is die vergelyking wat Excel vir u bepaal het.
• Die vergelyking van die kromme wat die beste deur ons datapunte pas, is dus
(korrek tot drie desimale plekke)
x t t t
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3 2
= 1,990 − 15,034 + 33,506 − 18,6 en dit is
volgens die R-kwadraat-waarde ‘n baie betroubare model vir die data.

Excercise 7.4 for self-assessment
Excercise in the text book Page number Problems
22.6 636 Do using Microsoft Excel like in the
example above. Plot the points, search
for the best fitting curve and measure its
equation and its R-squared value:
244
5, 7, 9
22.7 640 Do using Microsoft Excel like in the
example above. Plot the points, search
for the best fitting curve and measure its
equation and its R-squared value:
5, 7, 9
The final answers to the given problems appear at the back of the text book.