MATHEMATICS I (Engineering) · reasoning. In mathematics we are not only interested in the correct answer, but also in the method you used to obtain the answer. Pay attention to the

MATHEMATICS I

(Engineering) STUDY GUIDE 1 for MAT1581

L E Greyling Department of Mathematical Sciences UNIVERSITY OF SOUTH AFRICA PRETORIA

© UNISA 2016 First edition 2004 Second edition 2005 Third edition 2017 All rights reserved Printed and published by the University of South Africa Muckleneuk, Pretoria MAT1581 Layout done by the Department

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CONTENTS PAGE Introduction iii

STUDY GUIDE 1

MODULE 1 Binomial theorem 1

MODULE 2 Systems of equations and determinants

Learning unit 1 Properties of determinants 15 Learning unit 2 The value of a determinant 23 Learning unit 3 Cramer’s rule 30 Post-test 40

MODULE 3 Partial fractions

Learning unit 1 Introduction 46 Learning unit 2 Proper fractions 52 Learning unit 3 Improper fractions 70 Post-test 82

MODULE 4 Complex numbers

Learning unit 1 Imaginary and complex numbers 90 Learning unit 2 Operations with complex

numbers 98

Learning unit 3 Polar and exponential form 108 Learning unit 4 Operations in polar and

exponential form 117

Post-test 132 MODULE 5

Analytic geometry

Learning unit 1 The straight line 143 Learning unit 2 The parabola 157 Learning unit 3 The hyperbola 166 Learning unit 4 The circle 172 Learning unit 5 The ellipse 177 Learning unit 6 The central hyperbola 187 Post-test 192

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INTRODUCTION:

MATHEMATICS I (Engineering) Welcome to this module Mathematics I for Engineering. This study guide material has been compiled to serve the mathematical needs of students engaged in a first course in engineering. Students from other fields who want to broaden their mathematical knowledge will also benefit from this course.

Purpose of this module

This module will be useful to students in developing basic skills which can be applied in the natural and engineering sciences. Students credited with this module will have an understanding of basic ideas of algebra and calculus in handling problems related to Cramer’s rule to solve systems of linear equations, complex number system, binomial theorem, basic differentiation and integration.

The focus is on building strong algebraic skills that will support the development of analytical skills that are crucial in problem solving in more advanced mathematics and related subjects.

This module will support you in your studies in the field of engineering and the physical sciences as part of a diploma.

You must have mastered all mathematical operations with exponential, logarithmic and trigonometric functions before starting this module. You will find revision material on these topics and other basic mathematical concepts on the myUnisa page for MAT1581. The style of the study guide makes it suitable for self-study. To achieve success requires discipline and hard work. You require a framework for effective studying. The next section assists you with this framework.

Where do you start?

Learn the following off by heart:

your student number the module (subject) code: MAT1581 Do the following on the internet: Log on to https://my.unisa.ac.za Register on myUnisa and claim your myLife e-mail address. Log on weekly to check for any new announcements. You must register as

the lecturer posts announcements and you are alerted via e-mail to read them.

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Organise your workspace. You need

a place to write paper and a pen to try examples and do activities and exercises a non-programmable scientific calculator as used for high school

mathematics, for example a Sharp EL-531 LH or a Casio fx-82L ordinary tools for drawing the tutorial letter for MAT1581

Consult your tutorial letter to obtain information on

assignments how to obtain a year mark examinations due dates prescribed and recommended books contact details of your lecturer or tutor

Set up a study plan

Identify your goals. Make a weekly timetable. Indicate work and family responsibilities on your timetable. Remember to allow time for rest and relaxation. Evaluate the time available for studying. Before you can complete your plan, we must explain the format of the study material.

Format of study material

Modules The work is divided into modules. Each module deals with a major subject area and is divided into learning units. The end of each unit provides a natural break. This enables you to plan your time. You will find a complete list of modules and units in the table of contents. Your tutorial letter will give guidance as to the importance of each module.

Learning units Each learning unit starts with OUTCOMES. These outcomes list what you should be able to do after you have mastered the content of the unit. To explain the content you will find examples. An example gives both the question and the answer. To develop your understanding you also will find activities. The activities give a list of questions that you should attempt immediately. The answers to activities are given at the end of each unit. When you have completed the activity, check your answer. An average student should be able to study a unit in an evening. Contact the lecturer/tutor the next day by telephone or e-mail if necessary to clarify any points. Contact details are given in the tutorial letter and on the myUnisa page for your lecturer or tutor.

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Post-test At the end of each module you need to assess your progress. The post-test should be used for self-assessment. These questions include questions from past papers and questions similar to what you can expect in future examination papers. The solutions to all the questions are given and should be used to mark your own work. An average student should be able to complete the test in an evening. Try to do the test without referring to your study notes. Warning: Do not look at the answers before attempting a solution! When answering a question your writing should be clear and legible. That is, the marker or any other person reading your answer must be able to follow your reasoning. In mathematics we are not only interested in the correct answer, but also in the method you used to obtain the answer. Pay attention to the correct use of symbols. You will lose marks if your writing is not mathematically correct. Mathematics is a language in which we use symbols in a specific way to communicate.

Use of computer software We do not mention any specific programs in the notes for two reasons: The first is the rapid development in this field, which results in any reference becoming outdated very quickly. The second is that this is an entry-level course and we would like as many students as possible to have the opportunity to study mathematics at this level.

However, we would encourage its use and hope to introduce such programs at second- or third-year level.

Study success The study of mathematics, as you well know, requires you to understand the various units. In most cases the work is sequential and you need to master a particular unit before you can move on to a more advanced unit. Studying mathematics requires you to engage with the material, make notes and practise many examples. You need to continuously evaluate your work as well as reflect to consider your thought processes and logic. With the required planning, dedication and hard work, you should be successful in your studies. For your exam preparation, refer to the tutorial letters, assignments and myUnisa. This is a blended module which means you have to log on to myUnisa for additional information regarding the module. You must have worked through all the learning units and engaged with all the examples and activities as well as the post-tests. Work through the assignments as well as past examination papers (which are available on myUnisa).

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You need to pace your studies for this module, which is a 12-credit module, and this means you need to spend at least 120 notional hours of actively engaging with the study material. Manage your time as set out in your study plan as much as possible to ensure study success. We wish you all the very best in studying this interesting module.

USEFUL INFORMATION

MATHEMATICAL SYMBOLS + plus minus plus or minus multiply by multiply by divide by = is equal to is identically equal to is approximately equal to is not equal to is greater than is greater than or equal to is less than is less than or equal to n! factorial n = 1 2 3 ….. n k modulus of k, that is the size of k

irrespective of the sign is a member of set set of natural numbers set of integers set of real numbers set of rational numbers

therefore infinity e base of natural logarithms (2,718…) ln natural logarithm log logarithm to base 10 sum of terms limn

limiting value as n

integral dy

dx derivative of y with respect to x

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GREEK ALPHABET Greek letter Greek name Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mu Nu Xi Omicron Pi Rho Sigma Tau Upsilon Phi Chi Psi Omega

Formula sheets The following pages contain the information sheets and table of integrals that will be included with the examination paper.

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INFORMATION SHEET ALGEBRA Laws of indices

n

nn

n

n

n

n

n mn

m

mnmnnm

nm

n

m

nmnm

b

a

b

a

baab

a

aa

aa

aa

aaa

aa

a

aaa

.8

.7

1.6

1and

1.5

.4

.3

.2

.1

0

Logarithms

Definitions If xay then yx alog

If xey then ynx

Laws

log n

1. log log log

2. log log log

3. log log

log4. log

log

5. a

n

ba

b

f f

A B A B

AA B

B

A n A

AA

a

a f e f

Factors

2233

2233

babababa

babababa

Partial fractions

dx

C

cbxax

BAx

dxcbxax

xf

bx

D

ax

C

ax

B

ax

A

bxax

xf

cx

C

bx

B

ax

A

cxbxax

xf

22

323

Quadratic formula

a

acbbx

cbxax

2

4then

0If

2

2

DETERMINANTS

223132211323313321122332332211

3231

222113

3331

232112

3332

232211

333231

232221

131211

aaaaaaaaaaaaaaa

aaaa

aaaaa

aaaaa

aaaaaaaaaa

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SERIES Binomial theorem

11and

...!3

21

!2

111

and

..3

21

2

1

32

33221

x

xnnn

xnn

nxx

ab

..ba!

nnnba

!

nnbnaaba

n

nnnnn

Maclaurin’s theorem

11

32

!1

0

!3

0

!2

0

!1

00 n

n

xn

fx

fx

fx

ffxf

Taylor’s theorem

afn

haf

haf

hafhaf

axn

afax

afax

afax

afafxf

nn

nn

112

11

32

!1!2!1

!1!3!2!1

COMPLEX NUMBERS

212

1

2

1

212121

22

2

:Division.6

:tionMultiplica.5

andthen,If.4

:nSubtractio.3

:Addition.2

tanarg:Argument

:Modulus

1where

,sincos.1

r

r

z

z

rrzz

qnpmjqpjnm

dbjcajdcjba

dbjcajdcjba

a

barcz

bazr

j

rerjrbjaz j

1

1 1

7. De Moivre's theorem

cos sin

8. has distinct roots:

360with 0, 1, 2, , 1

9. cos sin

cos and sin

10. cos sin

11.

n n n

n

n n

j

j j

a jb a

j

r r n r n j n

z n

kz r k n

n

re r j

re r re r

e e b j b

n re n r j

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GEOMETRY

1. Straight line

11 xxmyy

cmxy

Perpendiculars, then 2

1

1

mm

2. Angle between two lines

21

21

1tan

mm

mm

3. Circle

222

222

rkyhx

ryx

4. Parabola cbxaxy 2

axis at a

bx

2

5. Ellipse

12

2

2

2

b

y

a

x

6. Hyperbola

axis- round1

axis- round1

2

2

2

2

2

2

2

2

yb

y

a

x

xb

y

a

x

kxy

MENSURATION 1. Circle: ( in radians)

2

2

2

Area

Circumference 2

Arc length

1 1Sector area

2 21

Segment area sin2

r

r

r

r r

r

2. Ellipse

ba

ab

nceCircumfere

Area

3. Cylinder

2

2

22area Surface

Volume

rrh

hr

4. Pyramid

height base area3

1Volume

5. Cone

r

hr

surfaceCurved3

1Volume 2

6. Sphere

3

2

3

4

4

rV

rA

7. Trapezoidal rule

0 1 11

2 22 n n

b af x f x f x f x

n

8. Simpson’s rule

0 1 2 3

4 2 1

1[ 4 2 4

3

2 2 4 ]n n n

b af x f x f x f x

n

f x f x f x f x

9. Prismoidal rule

1 2 1n nb a

f m f m f m f mn

(x)

HYPERBOLIC FUNCTIONS Definitions

xx

xx

xx

xx

ee

eex

eex

eex

tanh

2cosh

2sinh

Identities

x

x

xxx

xxx

xx

xx

xx

xx

xx

2

2

22

2

2

22

22

22

sinh21

1cosh2

sinhcosh2cosh

coshsinh22sinh

12cosh2

1cosh

12cosh2

1sinh

cosech1coth

sechtanh1

1sinhcosh

TRIGONOMETRY Compound angle addition and subtraction formulae sin(A + B) = sin A cos B + cos A sin B sin(A - B) = sin A cos B - cos A sin B cos(A + B) = cos A cos B - sin A sin B cos(A - B) = cos A cos B + sin A sin B

BA

BABA

BA

BABA

tantan1

tantantan

tantan1

tantantan

Double angles sin 2A = 2 sin A cos A cos 2A = cos2A – sin2A = 2cos2A - 1 = 1 - 2sin2A sin2 A = ½(1 - cos 2A) cos2 A = ½(1 + cos 2A)

A

AA

2tan1

tan22tan

Products of sines and cosines into sums or differences sin A cos B = ½(sin (A + B) + sin (A - B)) cos A sin B = ½(sin (A + B) - sin (A - B)) cos A cos B = ½(cos (A + B) + cos (A - B)) sin A sin B = -½(cos (A + B) - cos (A - B)) Sums or differences of sines and cosines into products

2sin

2sin2coscos

2cos

2cos2coscos

2sin

2cos2sinsin

2cos

2sin2sinsin

yxyxyx

yxyxyx

yxyxyx

yxyxyx

TRIGONOMETRY Identities

cos

sintan

tan- = )(-tan

cos = )(- cos

sin - = )sin(-

cosec = 1 +cot

sec = tan+ 1

1 cos sin

22

22

22

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DIFFERENTIATION

h 0

1

2

1

1. lim

2. 0

3.

4. . . ' . '

. ' . '5.

6. ( ) ( ) . '( )

7. . .

n n

n n

f x h f xdy

dx hd

kdxd

ax anxdxd

f g f g g fdxd f g f f g

dx g g

df x n f x f x

dxdy dy du dv

dx du dv dx

8. Parametric equations

2

2

dydy dt

dxdxdt

d dyd y dt dx

dxdxdt

9. Maximum/minimum For turning points: f '(x) = 0

Let x = a be a solution for the above If f '(a) > 0, then a minimum If f '(a) < 0, then a maximum For points of inflection: f " (x) = 0 Let x = b be a solution for the above

Test for inflection: f (b h) and f(b + h) Change sign or f '"(b) if f '"(b) exists

'( )110. sin ( )21 ( )

'( )111. cos ( )21 ( )

'( )112. tan ( )21 ( )

'( )113. cot ( )21 ( )

'( )114. sec ( )2( ) 1

'( )115. cosec ( )2( ) 1

'( )116. sinh ( )

d f xf x

dxf x

d f xf x

dxf x

d f xf x

dx f x

d f xf x

dx f x

d f xf x

dxf x f x

d f xf x

dxf x f x

d f xf x

dxf

2( ) 1

'( )117. cosh ( )2( ) 1

'( )118. tanh ( )21 ( )

'( )119. coth ( )2( ) 1

'( )120. sech ( )21 ( )

'( )121. cosech ( )2( ) 1

22. Increments: . . .

x

d f xf x

dxf x

d f xf x

dx f x

d f xf x

dx f x

d f xf x

dxf x f x

d f xf x

dxf x f x

z z zz x y w

x y w

23. Rate of change:

. . .z z zdz dx dy dw

dt x dt y dt w dt

INTEGRATION

b

a

b

a

b

a

dxyb-a

dxyb-a

F(aF(b)dxf(x)vduuv-udv

22 1)R.M.S.(.4

1= Mean value.3

).2:partsBy.1

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TABLE OF INTEGRALS

1

1

1 11

2 ' 11

'3

4 '

5 '

6 ' sin cos

7 ' cos sin

8 '

(n )n

nn

f(x) f(x)

f(x)f(x)

a x. ax dx c, n

n

f(x). f(x) .f (x) dx c, n

n

f (x). dx n f(x) c

f(x)

. f (x).e dx e c

a. f (x).a dx c

n a

. f (x). f(x) dx f(x) c

. f (x). f(x) dx f(x) c

. f (x)

2

2

tan sec

9 ' cot sin

10 ' sec sec tan

11 ' cosec osec cot

12 ' sec tan

13 ' cosec cot

14

. f(x) dx n f(x) c

. f (x). f(x) dx n f(x) c

. f (x). f(x) dx n f(x) f(x) c

. f (x). f(x) dx n c f(x) f(x) c

. f (x). f(x) dx f(x) c

. f (x). f(x) dx f(x) c

' sec tan sec

15 ' cosec cot cosec

. f (x). f(x). f(x) dx f(x) c

. f (x). f(x). f(x)dx f(x) c

MAT1581 Mathematics 1 (Engineering)

1

M O D U L E 2

M O D U L E 1

MODULE 1: BINOMIAL THEOREM CONTENTS PAGE

LEARNING UNIT 1 BINOMIAL THEOREM 2

1. INTRODUCTION ................................................................................................... 3 2. THE EXPANSION OF (a + b)n FOR n A POSITIVE INTEGER ........................ 4 3. THE EXPANSION OF (a + b)n FOR n A NEGATIVE INTEGER ...................... 6 4. THE EXPANSION OF (a + b)n FOR n A FRACTION ........................................ 7 5. THE rth TERM OF A BINOMIAL SERIES .......................................................... 8 4. RESPONSES TO ACTIVITIES .............................................................................. 9 4.1 Activity 1 ................................................................................................................. 9 4.2 Activity 2 ............................................................................................................... 10 4.3 Activity 3 ............................................................................................................... 10 4.4 Activity 4 ............................................................................................................... 10

POST-TEST: BINOMIAL THEOREM 12

POST-TEST SOLUTIONS 13


2

BINOMIAL THEOREM

CONTENTS PAGE

1. INTRODUCTION ................................................................................................... 3 2. THE EXPANSION OF (a + b)n FOR n A POSITIVE INTEGER ........................ 4 3. THE EXPANSION OF (a + b)n FOR n A NEGATIVE INTEGER ...................... 6 4. THE EXPANSION OF (a + b)n FOR n A FRACTION ........................................ 7 5. THE rth TERM OF A BINOMIAL SERIES .......................................................... 8 6. RESPONSES TO ACTIVITIES .............................................................................. 9 6.1 Activity 1 ................................................................................................................. 9 6.2 Activity 2 ............................................................................................................... 10 6.3 Activity 3 ............................................................................................................... 10 6.4 Activity 4 ............................................................................................................... 10

MODULE 1

LEARNING UNIT 1

OUTCOMES

At the end of this learning unit, you should be able to write down a binomial expansion using the binomial theorem determine any term in a binomial expansion

Module 1 Learning unit 1 BINOMIAL THEOREM


3

1. INTRODUCTION

A mathematical expression which consists of only two terms, say a + b, is called a binomial (“bi” means two). In this module we are investigating the powers of a binomial, that is (a + b)n. The binomial theorem gives us a quick way to raise an expression comprising two terms to any given power. This theorem is used to work out annuity formulae in financial management and differential formulae in this course. Your pocket calculator uses this theorem in its calculations, for example to extract roots. If n in (a + b)n is small, we can easily use multiplication to expand the series, but if n becomes bigger, say (a + b)20, multiplication becomes tedious. Examine the following expansions:

2 2 2

3 3 2 2 3

4 4 3 2 2 3 4

2

3 3

4 6 4

a b a b a b a ab b

a b a b a b a b a a b ab b

a b a b a b a b a b a a b a b ab b

We can conclude that

5a b would begin with a5 and end with b5

20a b would begin with a20 and end with b20

na b would begin with an and end with bn

Note this:

2

3

4

expands to three terms

expands to four terms

expands to five terms

a b

a b

a b

We can conclude that

5a b would expand to six terms

20a b would expand to twenty-one terms

na b would expand to (n + 1) terms

We notice that

the coefficients read the same backwards as forwards the first coefficient (and the last one) is 1 in all the expansions the powers of a are descending and the powers of b are

ascending the sum of the indices of a and b is n



4

We will now state the binomial theorem without proof. The theorem is on your formula sheets under series. The binomial theorem

1 2 2 3 31 1 2.. and

2 3n n n n nn n n n n

a b a na b a b a b .. b a! !

The special case with a = 1 is also given:

2 31 1 21 1 ... and 1 1

2! 3!n n n n n n

x nx x x x

.

Remember, n! stands for factorial n and per definition n! = ( 1)( 2)( 3)...1n n n n Thus 2! 2.1 2

3! 3.2.1 6

and

2. THE EXPANSION OF (a + b)n FOR n A POSITIVE INTEGER

Example 1

Expand 8a b .

The value of n is 8; thus there will be nine terms in the expansion. The easiest way to write down the expansion is to break the theorem on your information sheet into separate terms and substitute n.

81

1 72

2 2 6 2 6 23

3 3 5 3 5 34

4 4 4 4 4 45

5 5 3 5 3 56

8

1 8.728

2 2.1

1 2 8.7.656

3 3.2.1

1 2 3 8.7.6.570

4 4.2.3.1

1 2 3 ( 4) 8.7.6.5.456

5 5.4.2.3.1

n

n

n

n

n

n

T a a

T na b a b

n nT a b a b a b

!

n n nT a b a b a b

!

n n n nT a b a b a b

!

n n n n nT a b a b a b

!

6 6 2 6 2 67

1 2 3 ( 4)( 5) 8.7.6.5.4.328

6 6.5.4.2.3.1nn n n n n n

T a b a b a b!

7 7 7 78

1 2 3 ( 4)( 5) 6 8.7.6.5.4.3.28

7 7.6.5.4.2.3.1nn n n n n n n

T a b ab ab!



5

8 8 0 8 89

8 8 7 6 2 5 3 4 4 3 5 2 6 7 8

1 2 3 ( 4)( 5) 6 7 8.7.6.5.4.3.2.1

8 8.7.6.5.4.2.3.1

Thus

8 28 56 70 56 28 8

nn n n n n n n nT a b a b b

!

a b a a b a b a b a b a b a b ab b

Note: If we try to continue the series beyond term 9, the next and all succeeding terms will be 0. To save time, you may use some of the observations that were made in the introduction. For example, the coefficients read the same backwards and forwards, so after term 5 you can reverse the coefficients. Also, the powers of a are descending, while the powers of b are ascending.

Example 2

Expand 4 4 and a b a b

The value of n is 4, thus there will be five terms in the expansion.

41

1 32

2 2 2 2 2 23

3 3 3 34

4 4 4 45

4 4 3 2 2 3 4

4

1 4.36

2 2.1

1 2 4.3.24

3 3.2.1

1 2 3 4.3.2.1

4 4.2.3.1

Thus

4 6 4

n

n

n

n

n

T a a

T na b a b

n nT a b a b a b

!

n n nT a b a b a b

!

n n n nT a b b b

!

a b a a b a b ab b

To find 4a b notice that 4 4

( )a b a b . We obtain the answer by replacing b

with – b in the expansion for 4a b .

4 4 2 3 44 3 2

4 3 2 2 3 4

Thus

( ) 4 6 4

4 6 4

a b a b a a b a b a b b

a a b a b ab b

Example 3

Write down the first three terms in the expansion of 53

2

1

3

x

x



6

Substitute 3

21

3 , and 5xx

a b n into the binomial theorem.

53 15

1

43 12 101

2 2 2

3 23 9 52 2

3 2 4

53 15 10 5

2

3 243

1 1 55 5

3 81 81

1 5.4 1 1 1010

2 2.1 3 27 27

Thus

1 5 10...

3 243 81 27

n

n

n

x xT a

x x xT na b

x x

n n x x xT a b

! x x

x x x x

x

ACTIVITY 1

1. Write expansions of 4 42 and 2x x .

2. Write down the first four terms in the expansion of 103x y .

Remember to check the response on page 9.

3. THE EXPANSION OF (a + b)n FOR n A NEGATIVE INTEGER

If n is a negative integer, the expansion is infinite.

Example 4

Expand to three terms 22

1

xx

Rewrite the given expression to expand:

2222

1x

x

xx

Substitute 2, and 2xa x b n into the binomial theorem.

21 2

1 3 42 4

22 2 4 4 6

3 2 6

222 2 4 62

1

2 42 4

1 2. 3. 2 4 123 12

2 2.1

Thus

1 1 4 12...

n

n

n

x

x

T a xx

T na b x xx x

n nT a b x x x

! x x x

xx x xx



7

ACTIVITY 2

Expand to four terms 33x a

.


4. THE EXPANSION OF (a + b)n FOR n A FRACTION

If n is a fraction, the expansion is infinite.

Example 5

Expand to four terms 24

2x

x

Rewrite the given expression

12

2 24 4

2 2x x

x x

Substitute 4

2 12

2, and

xa x b n into the binomial theorem.

4

12

12

21

1 212 2

1 4

5

2

( )

1

n

n

x

T a x x

T na b x

x x

x

4

32

2 23

21 122 2

38 11

2

1

2

2.1

1 1 4 1

2 4 2

n

x

n nT a b

!

x

xx x



8

4

52

12

3 34

331 122 2 2

512

5

12 17

2 24 4 5 11 17

2

1 2

3

3.2.1

1 3 8

6 8

3 1

6 2

Thus

2 2 1 1 1...

2 2

n

x

n n nT a b

!

x

xx

x

x x

x x xx x x x x

ACTIVITY 3

Expand to four terms 34

1 x .


5. THE rth TERM OF A BINOMIAL SERIES

If we write the binomial series as

1 2 2 3 31 1 2...

2 3n n n n nn n n n n

a b a na b a b a b! !

Then the rth term of this expansion is given by

1 11 2 ....( 2)

( 1)n r rn n n n r

a br !

Example 6

Calculate the sixth term in the expansion of 112 2x y

2 2

1 1

6 52 2 12 106

, ( ), 11, 6,

Now 1 5

2 11 6 2 7

1 11 6 1 6

1 2 ....( 2)

( 1)

11.10.9.8.7Thus 462

5.4.3.2.1

n r rr

a x b y n r

r

n r

n r

n n n n rT a b

r !

T x y x y



9

ACTIVITY 4

Find the middle term in the expansion of 10x y .


4. RESPONSES TO ACTIVITIES

4.1 Activity 1

1. The value of n is 4, thus there will be five terms in the expansion. In the binomial theorem substitute a = x, b = 2 and n = 4.

41

1 3 32

22 2 2 2 23

33 34

4 2 8

1 4.32 6 4 24

2 2.1

1 2 4.3.22 4 8 32

3 3.2.1

n

n

n

n

T a x

T na b x x

n nT a b x x x

!

n n nT a b x x x

!

44 45

4 4 3 2

4 4 3 2

1 2 3 4.3.2.12 16

4 4.2.3.1

Thus

2 8 24 32 16

and

2 8 24 32 16

nn n n nT a b

!

x x x x x

x x x x x

2. In the binomial theorem substitute a = x, b = 3y and n = 10

101

1 9 92

22 2 8 8 2 8 23

33 3 7 7 3 7 34

10 10 9 8 2 7 3

10 3 30

1 10.93 45 9 405

2 2.1

1 2 10.9.83 120 27 3240

3 3.2.1

Thus

3 30 405 3240 ...

n

n

n

n

T a x

T na b x y x y

n nT a b x y x y x y

!

n n nT a b x y x y x y

!

x y x x y x y x y



10

4.2 Activity 2

Substitute , 3 and 3a x b a n into the binomial theorem.

31 3

1 4 42 4

22 2 5 2 5 2

3 5

1

93 (3 ) 9

1 3. 4 54(3 ) 6 9

2 2.1

n

n

n

T a xx

aT na b x a ax

x

n n aT a b x a x a

! x

33 3 6 3 6 3

4 6

2 33

3 4 5 6

1 2 3. 4. 5 270(3 ) 10 27

3 3.2.1

Thus

1 9 54 2703 ...

nn n n aT a b x a x a

! x

a a ax a

x x x x

4.3 Activity 3

Substitute 341, and a b x n .

Instead of using the general binomial theorem, we can use the special case. From the information sheet:

2 31 1 21 1 ... and 1 1

2! 3!n n n n n n

x nx x x x

34

1

32 4

23 12 2 24 4

3

3 33 513 3 34 4 4

4

2 334

1

1 1 3 3.

2! 2.1 2 16 32

1 2 1 15 15 5

3! 3.2.1 6 64 384 128

Thus

3 51 1 ...

32 128

T

T nx x

n n xT x x x

n n n x xT x x x

x xx x

4.4 Activity 4

The value of n is 10, thus there will be eleven terms in the expansion. The term in the middle would be the sixth term.



11

1 1

5 5 5 56

, , 10, 6,

Now 1 5

2 10 6 2 6

1 10 6 1 5

1 2 ....( 2)

( 1)

10.9.8.7.6Thus 252

5.4.3.2.1

n r rr

a x b y n r

r

n r

n r

n n n n rT a b

r !

T x y x y



12

POST-TEST: BINOMIAL THEOREM 1. Expand using the binomial theorem:

1132

5

32

2

4

a) 3

b)

c) 2

y

x y

x

x y

2. Expand to four terms:

23

3

11

3

23

a) 3

b) 2

c) 1

3d) 1

x

x a

x

x

y

x

3. Calculate the fourth term of 112 2x y



13

POST-TEST SOLUTIONS

1.

1 3 1 2 41 13 3 3 32 2 2

5 5 4 3 2 2 3 4 5

32 6 4 2 2 33 3 1

2 2 4 8

42

a) 3 243 405 270 90 15

b)

c) 2 8 24 32 16

y

x y x x y x y x y xy y

x x x y x y y

x y x x y xy x y y

2.

23

13

2 33

3 4 5 6

113 5 7

3 3 6 95 4023 9 81

2 2 2 4 63

2 3

1 9 54 270a) 3 ...

1 1 1 1b) 2 ...

2 4 8 16

c) 1 1 ...

3 3 5d) 1 1 1 ...

3

x

a a ax a

x x x x

xx x x x

x x x x

y y y y y

x x x x x

3. Fourth term 16 64 165T x y

You have now completed learning unit 1: Binomial theorem. You should be able to write down a binomial expansion using the binomial theorem determine any term in a binomial expansion

We now move on to module 2: Systems of equations and determinants. Learning unit 1 deals with the properties of determinants.


14

M O D U L E 2

MODULE 2: SYSTEMS OF EQUATIONS AND DETERMINANTS CONTENTS PAGE

LEARNING UNIT 1 PROPERTIES OF DETERMINANTS 15

1. WHAT IS A DETERMINANT? ........................................................................... 16 2. PROPERTIES OF DETERMINANTS ................................................................. 18 3. RESPONSES TO ACTIVITIES ............................................................................ 20 3.1 Activity 1 ............................................................................................................... 20 3.2 Activity 2 ............................................................................................................... 21

LEARNING UNIT 2 THE VALUE OF AN n n DETERMINANT 23

1. MINORS ................................................................................................................ 24 2. COFACTORS ........................................................................................................ 24 3. EVALUATING A DETERMINANT ................................................................... 25 4. RESPONSES TO ACTIVITIES ............................................................................ 28 4.1 Activity 1 ............................................................................................................... 28 4.2 Activity 2 ............................................................................................................... 28

LEARNING UNIT 3 CRAMER’S RULE 30

1. SOLVING A SYSTEM OF TWO LINEAR EQUATIONS ................................ 31 2. CRAMER’S RULE FOR SOLVING A SYSTEM OF LINEAR EQUATIONS .. 33 3. RESPONSES TO ACTIVITIES ............................................................................ 38 3.1 Activity 1 ............................................................................................................... 38 3.2 Activity 2 ............................................................................................................... 39

POST-TEST: SYSTEMS OF EQUATIONS AND DETERMINANTS 40



15

SYSTEMS OF EQUATIONS AND DETERMINANTS Properties of determinants

CONTENTS PAGE

1. WHAT IS A DETERMINANT? ........................................................................... 16 2. PROPERTIES OF DETERMINANTS ................................................................. 18 3. RESPONSES TO ACTIVITIES ............................................................................ 20 3.1 Activity 1 ............................................................................................................... 20 3.2 Activity 2 ............................................................................................................... 21

MODULE 2

LEARNING UNIT 1

OUTCOMES

At the end of this learning unit, you should be able to find the value of a 2 2 and a 3 3 determinant use the properties of determinants to evaluate a determinant identify an element according to its position in a determinant

Module 2 Learning unit 1 SYSTEMS OF EQUATIONS AND DETERMINANTS: Properties of determinants


16

Whenever you have to solve more than two equations in two variables, the substitution and the addition methods you have used till now become very long and difficult. There are other methods that are easier. This unit gives the background material for one of the easier methods.

1. WHAT IS A DETERMINANT?

If a, b, c and d are any four real numbers, then the symbol a b

c d is called a 2 2 (read

2 by 2) determinant or determinant of the second order. Each determinant has a numerical value. Evaluating a 2 2 determinant

If a, b, c and d are any four real numbers, then a b

ad cbc d

As a memory aid you can draw the diagonals of the determinanta b

ad cbc d

Example 1

Evaluate the determinant2 5

3 2

Solution: 2 5

2 2 3 5 4 15 193 2

A 3 3 determinant or determinant of the order 3 can be represented by the symbol

1 2 3

1 2 3

1 2 3

a a a

b b b

c c c

where 1 2 3 1 2 3 1 2 3, , , , , , , and a a a b b b c c c are any real numbers.

The horizontal lines are called rows and the vertical lines columns. The rows are numbered from top to bottom and the columns from left to right. Evaluating a 3 3 determinant Rewrite the first two columns on the right of the determinant as follows:

1 2 3 1 2

1 2 3 1 2

1 2 3 1 2

a a a a a

b b b b b

c c c c c

(1) Form the products of the elements in each of the three diagonals shown which run

down from left to right, and precede each of these three terms with a positive sign. (2) Form the products of the elements in each of the three diagonals shown which run

down from right to left, and precede each of these three terms with a negative sign.



17

(3) The algebraic sum of the six products of (1) and (2) is the required expansion of the determinant.

The main or principal diagonal of a determinant is the diagonal from upper left to lower right.

For example, in

a b c

e f g

h i j

the entries on the main diagonal are a, f and j.

A determinant is in triangular form if all the entries above or below the main diagonal are 0.

For example,

0 0

0 and 0

0 0

a b c a

f g e f

j h i j

Example 2

Evaluate the determinant:

3 2 2

6 1 1

2 3 2

Solution: Rewrite first two columns: 3 2 2 3 2

6 1 1 6 1

2 3 2 2 3

Draw diagonals to help. Construct the sum of the six products. Group the positive and the negative products together:

(3)(1)(2) (-2)(-1)(-2) (2)(6)(-3) (2)(1)(-2) + (3)(-1)(-3) +(-2)(6)(2)

6 4 36 4 9 24

34 19

34 19

15

In general a square arrangement of real numbers can be called a determinant. Element ija

denotes a typical element of a matrix. The first subscript (i) always refers to the row in which the element occurs and the second subscript (j) always refers to the column. Thus a determinant of order n n can be written as

11 1

1

n

n nn

a a

a a

We will discuss a method to determine the value of any determinant in the next unit.



18

ACTIVITY 1 1. Evaluate the following determinants:

3 1 1 1a) b)

2 0.4 1 1

5 6 71 2

c) d) 8 9 03 4

3 4 2

1 3 1

e) 3 4 1

1 6 2

2. Use the determinant

4 3 9

1 2 3

1 2 1

and write down the element in the

indicated position: a) 13a

b) 21a

c) 32a


2. PROPERTIES OF DETERMINANTS

The properties are stated without proof. An example of each property is given. 1. Interchanging corresponding rows and columns of a determinant does not

change the value of the determinant. Thus any theorem proved true for rows holds for columns, and conversely.

Example:

7 9

42 72 308 6

and 7 8

42 72 309 6

2. If each element in a row (or column) is 0, the value of the determinant is 0.

Examples:

1 0

08 0

or

1 3 4

0 0 0

4 1 2

3. Interchanging any two rows (or columns) reverses the sign of the determinant. Examples:



19

If 2 3

151 9

then 1 9

152 3

, because the first and second rows were

swapped. Thus 2 3 1 9

1 9 2 3

If

3 1 5

2 0 2 14

5 1 4

then

5 1 3

2 0 2 14

4 1 5

, because the first and third

columns were swapped. 4. If two rows (or columns) of a determinant are identical, the value of the

determinant is 0. Examples:

2 3 1

9 8 7 0

2 3 1

, because rows 1 and 3 are the same.

6 9 9

1 3 3 0

7 4 4

, because columns 2 and 3 are the same.

5. If each of the elements in a row (or column) of a determinant is multiplied by

the same number p, the value of the determinant is multiplied by p. Example:

If you multiply each entry in the second column of the determinant 2 3

151 9

by

5, you get 2 15

751 45

, and 75 5 15 .

If you combine property 4 and 5, you can see that if any row (or column) is a multiple of any other row (or column), then the value of the determinant is 0. 6. If to each element of a row (or column) of a determinant is added m times the

corresponding element of any other row (or column), the value of the determinant is not changed. Example:

In the determinant 7 3

5 1, if you multiply each number in the first row by 4 and

add these new numbers to the second row, then 7 3 7 3

28 5 12 1 23 13

We see that the value of the determinant is unchanged because7 3

91 69 2223 13



20

7. If a determinant is in triangular form, its value is the product of the main diagonal. Example:

131

3

9 3 1 2

0 4 2 79 4 1 12

0 0 17

0 0 0 1

ACTIVITY 2 Evaluate each determinant using the properties of determinants:

2 5 8

a) 16 4 3

2 5 8

1 2 5

b) 1 2 3

3 6 15

9 18 -1

c) 0 -7 5

0 0 2

1 -3 -1 -1 -3 1

d) If 3 4 1 3 find the value of 1 4 3

1 6 2 2 6 1



3.1 Activity 1

1.

3 1a) 3 0.4 2 1 1.2 2 0.8

2 0.4

1 1b) 1 1 1 1 1 1 0

1 1

1 2c) 1 4 3 2 4 6 2

3 4



21

5 6 7 5 6

d) 8 9 0 8 9

3 4 2 3 4

5 9 2 6 0 3 7 8 4 7 9 3 5 0 4 6 8 2

90 0 224 189 0 96

314 93

221

1 3 1 1 3

e) 3 4 1 3 4

1 6 2 1 6

(1)(4)(2) ( 3)(1)(1) ( 1)(3)(6) ( 1)(4)(1) (1)(1)(6) ( 3)(3)(2)

8 3 18 4 6 18

13 16

13 16

3

2. a) 13a = 9

b) 21a = 1

c) 32a = 2

3.2 Activity 2

2 5 8

a) 16 4 3 0, because rows 1 and 3 are the same.

2 5 8

1 2 5

b) 1 2 3 0, because row 3 is a multiple of row 1 (or column 2 is a multiple of column 1).

3 6 15

9 18 1

c) 0 7 5 (9)( 7)(2) 126

0 0 2

1 3 1

d) 1 4 3 3, becau

2 6 1

se the first and third columns are swapped.



22

You have now completed learning unit 1: Properties of determinants.

You should be able to

find the value of a determinant

use the properties of determinants to evaluate a determinant

identify an element according to its position in a determinant

We now move on to learning unit 2: The value of a determinant.


23

SYSTEMS OF

EQUATIONS AND DETERMINANTS The value of an n n

determinant

CONTENTS PAGE

1. MINORS ................................................................................................................ 24 2. COFACTORS ........................................................................................................ 24 3. EVALUATING A DETERMINANT ................................................................... 25 4. RESPONSES TO ACTIVITIES ............................................................................ 28 4.1 Activity 1 ............................................................................................................... 28 4.2 Activity 2 ............................................................................................................... 28

MODULE 2

LEARNING UNIT 2

OUTCOMES

At the end of this learning unit, you should be able to find the minor and cofactor of an element in a determinant use cofactors to find the value of an n n determinant

Module 2 Learning unit 2 SYSTEMS OF EQUATIONS AND DETERMINANTS: The value of an n x n determinant


24

The methods discussed in learning unit 1 of this module to evaluate determinants are specific to second- and third-order determinants. In this unit we will develop a general method suitable to evaluate any determinant.

1. MINORS

Each element in a determinant has a minor associated with it. The minor of a given element is the determinant that is formed by deleting the row and column in which the element lies.

Example 1

Consider the determinant

2 4 4

5 6 7

1 2 4

a) The minor of the first element, 2, is found by crossing out the first row and first

column

2 4 4

5 6 7

1 2 4

The minor of 2 is 6 7

24 14 102 4

.

b) The minor of 7 is found by crossing out the second row and third column 2 4 4

5 6 7

1 2 4


4 4 01 2

.

c) The minor of 1 is found by crossing out the third row and the first column 2 4 4

5 6 7

1 2 4


28 ( 24) 526 7

.

2. COFACTORS

Each element also has a cofactor. The value of the cofactor is determined by first adding the number of the row and the number of the column where the element is located. If this sum is even, the value of the cofactor is equal to the value of the minor for that element. If the sum is odd, the value of the cofactor is then –1 times the value of the minor for that element.

Example 2 Consider the same determinant as in example 1.



25

a) The cofactor of 2 is 10. Element 2 is in row 1 column 1. Since 1 + 1 is an even number, the value of the cofactor is the same as that of the minor.

b) The cofactor of 7 is 0. c) The cofactor of 1 is 52. Element 1 is in row 3 column 1. Since 1 + 3 is an even

number, the value of the cofactor is the same as that of the minor. d) Find the element, minor and cofactor in the position: row 2 column 1.

Element = 5

Minor = 4 4

16 8 242 4

Cofactor = 24. Since 2 + 1 = 3 is an odd number, the cofactor is –1 times the value of the minor.

ACTIVITY 1

Use the determinant

4 2 1

3 7 4

2 1 1

and for the indicated position find (a) the

element in that position, (b) the minor of that element and (c) the cofactor of that element. 1. Row 1 column 2 2. Row 2 column 1 3. Row 3 column 2 4. Row 2 column 3 Remember to check the response on page 28.

3. EVALUATING A DETERMINANT

Steps to evaluate any determinant: 1. Select any row or column of the determinant. 2. Multiply each element of that row or column by its cofactor. 3. Add the results. To illustrate that this method is valid for all determinants we will start with examples of 2 2 and 3 3 determinants before moving to higher order determinants.

Example 3

a) Find the value of 1 2

3 4

Develop the determinant along the second column. The minor of element 2 is 3 and the cofactor is (-1)(3). The minor of 4 is 1 and the cofactor is 1.

Therefore 1 2

2( 1) 3 4 1 6 4 23 4

This is the same answer as previously obtained in learning unit 1, activity 1(c), page 20.



26

The method in learning unit 1 for second-order determinants is easier to use, so we prefer to use that method.

b) Find the value of

5 6 7

8 9 0

3 4 2

Develop along the second row. Instead of first finding the minors and cofactors of each element in row 2, we will write down the sum immediately and then simplify.

5 6 76 7 5 7

8 9 0 8( 1) ( 9) 04 2 3 2

3 4 2

Note: The third term becomes 0 as that element is 0 and 0 multiplied by any number is 0.

Since we can select any row or column, you should always try to select the row or column with the most zeros.

To develop the 2 2 determinants we will use the method in unit 1.

5 6 76 7 5 7

8 9 0 8( 1) ( 9) 04 2 3 2

3 4 2

8 12 28 9 10 ( 21)

8 40 9 11

320 99

221

This is the same answer as previously obtained in unit 1, activity 1(d) page 138. For third-order determinants choose the method that will be the easiest.

c) Evaluate

2 1 5 2

1 0 4 5

7 2 1 0

5 0 3 2

The second column of this determinant has two zeros, so we will develop along column 2. 2 1 5 2

1 4 5 2 5 21 0 4 5

1( 1) 7 1 0 0 2( 1) 1 4 5 07 2 1 0

5 3 2 5 3 25 0 3 2

Now we need to evaluate each of the 3 3 determinants. Develop the first determinant along column 3.



27

1 4 57 1 1 4

7 1 0 5 0 25 3 7 1

5 3 2

5 21 5 2 1 28

5 26 2 27

130 54

184

Develop the other determinant along row 1.

2 5 24 5 1 5 1 4

1 4 5 2 ( 5)( 1) ( 2)3 2 5 2 5 3

5 3 2

2 8 ( 15) 5 2 25 2 3 20

2 23 5 23 2 23

46 115 46

23

So, the value of the original determinant is 2 1 5 2

1 4 5 2 5 21 0 4 5

1( 1) 7 1 0 0 2( 1) 1 4 57 2 1 0

5 3 2 5 3 25 0 3 2

1( 184) 2( 23)

184 46

230

ACTIVITY 2 Evaluate the following determinants:

12

1 12 2

1 12 2

4 2 1

a) 3 0 4

3 1 1

1 0 0

0 0b)

0 1

0 1 0 1




28


4.1 Activity 1

1. a) 2

b) 3 4

3 8 52 1

c) Since 1 + 2 = 3 is odd, the cofactor is the negative of the minor or –( 5) = 5. 2. a) 3

b) 2 1

2 ( 1) 31 1

c) Since 1 + 2 = 3 is odd, the cofactor is the negative of the minor, or –3. 3. a) 1

b) 4 1

16 ( 3) 133 4

c) Since 3 + 2 is odd, the cofactor is – (–13) = 13. 4. a) – 4

b) 4 2

4 4 82 1

c) 2+3 is odd, so the cofactor is 8.

4.2 Activity 2

4 2 1

a) 3 0 4 Develop along second column.

-3 1 1

3 4 4 12( 1) 0 1( 1)

3 1 3 4

2 3 12 1 16 3

2 15 19

49



29

12

1 12 2

1 12 2

12

1 12 2

12

1 12 2

12

1 112 22

1 14 2

1 14 2

14

1 0 0

0 0b) Develop along third column.

0 1

0 1 0 1

1 0

0 0 1 0 0

0 1 1

1 0

0 Develop along third row.

0 1 1

1 01= 0 1 1 1

0

0 0

You have now completed learning unit 2: The value of an n x n determinant.

You should be able to find the minor and cofactor of an element in a determinant use cofactors to find the value of an n n determinant We now move on to learning unit 3: Cramer’s rule.


30

SYSTEMS OF

EQUATIONS AND DETERMINANTS Cramer’s rule

CONTENTS PAGE

1. SOLVING A SYSTEM OF TWO LINEAR EQUATIONS ................................ 31 2. CRAMER’S RULE FOR SOLVING A SYSTEM OF LINEAR EQUATIONS .. 33 3. RESPONSES TO ACTIVITIES ............................................................................ 38 3.1 Activity 1 ............................................................................................................... 38 3.2 Activity 2 ............................................................................................................... 39

MODULE 2

LEARNING UNIT 3

OUTCOME

At the end of this learning unit, you should be able to solve a system of simultaneous

equations using Cramer’s rule.

Module 2 Learning unit 3 SYSTEMS OF EQUATIONS AND DETERMINANTS: Cramer’s rule


31

Determinants can be applied to solving a system of n linear equations in n unknowns. This method is very useful as it enables us to solve for one unknown without having to solve for the others. Remember, if you are asked to solve the system, you must find the value of all the unknowns.

1. SOLVING A SYSTEM OF TWO LINEAR EQUATIONS

Consider a system of two linear equations with two unknowns: ax by h

cx dy k

If we multiply the first equation by d and the second equation by b, we get adx bdy hd

cbx bdy bk

We can eliminate the y terms if we subtract the second equation from the first equation, thus

adx cbx hd bk

x ad cb hd bk

hd bkx

ad cb

If we use the definition of a second-order determinant on the solution for x, we get h b

k dhd bkx

a bad cb

c d

provided that 0a b

c d

Similarly, we can show that

a h

c kak chy

a bad cb

c d

These last two equations are cumbersome, so we will abbreviate the determinants by a b

c d , where the elements of the determinant are the coefficients of x and y

x

h b

k d , where the column of x-values is replaced by the constants

y

a h

b k , where the column of y-values is replaced by the constants

We can now state Cramer’s rule for solving two linear equations in two variables.



32

Cramer’s rule for solving a 2 2 linear system

The unique solution to the linear system 1 1 1

2 2 2

a x b y k

a x b y k

is

1 1 1 1

2 2 2 2

1 1 1 1

2 2 2 2

and yx

k b a k

k b a kx y

a b a b

a b a b

provided that 1 11 2

2 2

0 and , 0a b

k ka b

If 1 2, 0k k we have a homogeneous system and Cramer’s rule cannot be applied as the

determinants become 0. If 0 , the equations in the system are either inconsistent or dependent.

Example 1

Solve the following system: 2 5 0

3 6

x y

y x

Rewrite the system in the appropriate form: variables in the same order on the left-hand side and constants on the right-hand side:

2 5

3 6

x y

x y

The determinant of the coefficient matrix is 2 1

6 1 51 3

Since 0 , a unique solution exists.

Solving for x, we have

5 1

6 3 ( 15 6) 21

5 5 5x

Solving for y, we have

2 5

1 6 (12 ( 5)) 17

5 5 5y

Thus the solution is 21 17

and 5 5

x y

(Check the answer by substituting the values in the original system of equations.)



33

ACTIVITY 1 Solve the following systems of equations:

a) 2 7

3 2 7

x y

x y

b) 3 3

2 13 3

x y

x y

c) 2,5 3,8 9,3

0,5 0,76 2, 44

x y

x y


2. CRAMER’S RULE FOR SOLVING A SYSTEM OF LINEAR EQUATIONS

The rule can be extended to a system of n equations in n unknowns. Cramer’s rule Let a system of n linear equations in n unknowns be given by

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

n n

n n

n n nn n n

a x a x a x k

a x a x a x k

a x a x a x k

If the determinant of the coefficients 11 1

21

n

nn

a a

a a

is not equal to 0, the system has

a unique solution.

The solution is 1 2

1 2, , nxx xnx x x

where mx , the numerator of mx , is the determinant obtained by replacing the mth

column of by the column of constants 1 nk k .

We will usually work with 3 equations and 3 unknowns.

Example 2

a) Solve the system

2 3

3 4

2 6

x y z

x y z

x y z



34

1 2 1

3 1 1

1 1 2

Rewrite the first two columns:

1 2 1 1 2

3 1 1 3 1

1 1 2 1 1

1 1 2 2 1 1 1 3 1 1 1 1 1 1 1 2 3 2

2 2 3 1 1 12

7 10

3

3 2 1

4 1 1

6 1 2


3 2 1 3 2

4 1 1 4 1

6 1 2 6 1

x

3 1 2 2 1 6 1 4 1 1 1 6 3 1 1 2 4 2

6 12 4 6 3 16

10 13

3

31

3xx

1 3 1

3 4 1

1 6 2


1 3 1 1 3

3 4 1 3 4

1 6 2 1 6

y



35

1 4 2 3 1 1 1 3 6 1 4 1 1 1 6 3 3 2

8 3 18 4 6 18

13 ( 16)

3

31

3yy

1 2 3

3 1 4

1 1 6


1 2 3 1 2

3 1 4 3 1

1 1 6 1 1

z

1 1 6 2 4 1 3 3 1 3 1 1 1 4 1 2 3 6

6 8 9 3 4 36

23 29

6

62

3zz

b) Solve the following system for z only

5 6

2 4

2 6

3 4 2

x y w

x y z

y z w

x w

1 1 0 5

1 2 1 0

0 2 1 1

3 0 0 4

Develop along the third column:

1 1 5 1 1 5

=0+1(-1) 0 2 1 1 1 2 0 0

3 0 4 3 0 4

Develop the first determinant along the second row:



36

1 1 51 5 1 1

0 2 1 0 2 1( 1)3 4 3 0

3 0 4

2 4 15 1 0 3

38 3

35

Develop the second determinant along the second row:

1 1 51 5 1 5

1 2 0 1( 1) 20 4 3 4

3 0 4

1 4 0 2 4 15

4 2( 19)

34

1 1 5 1 1 5

=0+1(-1) 0 2 1 1 1 2 0 0

3 0 4 3 0 4

( 1)( 35) 1( 34)

35 34

1

1 1 6 5

1 2 4 0

0 2 6 1

3 0 2 4

Develop along the first column

2 4 0 1 6 5 1 6 5

=1 2 6 1 1( 1) 2 6 1 0 3( 1) 2 4 0

0 2 -4 0 2 4 2 6 1

z

z

Develop the first determinant along the first column:.

2 4 06 1 4 0

2 6 1 2 2( 1) 02 4 2 4

0 2 -4

2 24 2 2 16 0

2 26 32

20



37

Develop the second determinant along the first column:

1 6 56 1 6 5

2 6 1 1 2( 1) 02 4 2 4

0 2 4

24 2 2 24 10

26 2( 34)

26 68

42

Develop the last determinant along the second row:

1 6 56 5 1 5

2 4 0 2( 1) 46 1 2 1

2 6 1

2 6 30 4 1 10

48 36

12

Now

2 4 0 1 6 5 1 6 5

=1 2 6 1 1( 1) 2 6 1 0 3( 1) 2 4 0

0 2 -4 0 2 4 2 6 1

20 1(42) 3(12)

20 42 36

98

9898

1

z

zz

ACTIVITY 2 Solve the following systems of equations:

a)

3 7

5 7 3

2 2 0

x y z

x y z

x y z

b)

5 3 2 5

3 4 3 13

6 4 8

x y z

x y z

x y z




38


3.1 Activity 1

a) 2 1

( 4 3) 73 2

Since 0 , a unique solution exists. 7 1

( 14 ( 7)) 14 7 77 2

71

7

2 7( 14 21) 35

3 7

355

7

x

x

y

y

x

y

The solution is x = 1 and y = 5.

b) Rewrite the system in the appropriate form. 3 3

2 3 13

x y

x y

3 1( 9 2) 11

2 3

Since 0 , a unique solution exists. 3 1

( 9 13) 2213 3

222

11

3 3(39 6) 33

2 13

333

11

x

x

y

y

x

y

Thus the solution is x = 2 and y = -3.

c) 2,5 3,8

(1,9 1,9) 00,5 0,76

Since 0 , the system is inconsistent or dependent and no solution exists.



39

4.2 Activity 2

a) 3 7

5 7 3

2 2 0

1 3 1

5 7 1 42

2 1 2

7 3 1 1 7 1 1 3 7

3 7 1 126, 5 3 1 84, 5 7 3 84

0 1 2 2 0 2 2 1 0

126 84 843 2 2

42 42 42

x y z

yx z

x y z

x y z

x y z

x y z

b)

5 3 2 5

3 4 3 13

6 4 8

5 3 2

3 4 3 9

1 6 4

5 3 2 5 5 2 5 3 5

13 4 3 18, 3 13 3 9, 3 4 13 9

8 6 4 1 8 4 1 6 8

18 9 92 1 1

9 9 9

x y z

x y z

x y z

x y z

x y z



40

POST-TEST: SYSTEMS OF EQUATIONS AND DETERMINANTS 1. Evaluate the following determinants:

3 2 23 2

a) b) 1 4 51 4

6 1 4

2 1 1c) 2 3 1 d)

2 24 1 2

x y zx x

x x

2. a) For what value of k does

3 1 2

3 2 1 0

3 3

k

k

?

b) Find the values of x for which 2 1 2 1

01 4 2

x x

x x

3. Use Cramer’s rule to do the following: a) Solve for x and y:

4 2 5

3 4 1

x y

x y

b) Solve for x, y and z: 3 2 1

2 3 2

2 2 10

x y z

x y z

x y z

c) Solve for x and y: 3 6 1

6

2 3 1

2

x y

x y

4. If

3 2 4

2 1 5

1 5 2

A

write down the minors of the elements of the second row.


41

POST-TEST SOLUTIONS

3 21. a)

1 4

3 4 1 2 (Cross multiply.)

10

3 2 2

b) 1 4 5

6 1 4

Expand along the first column.

3 4 4 1 5 1 2 4 1 2 6 2 5 4 2

63 6 108

39

c) 2 3 1

4 1 2

Expand along the first row.

6 1 4 4 2 12

5 8 14

2d)

x y z

x y z

x y z

x

2 2

2

1 1

2 2

2 1 2 2 1 Cross multiply.

2 5 2 3 2

8

x

x x

x x x x

x x x x

x x

Module 2 SYSTEMS OF EQUATIONS AND DETERMINANTS: Post-test solutions

MAT1581 Mathematics 1(Engineering)

42

3 1 2

2. a 3 2 1 0

3 3

Evaluate determinant by expanding along the first row.

3 6 3 1 9 2 9 2 0

3 9 9 18 4 0

0 0

The equation is valid for all values of , where .

k

k

k k k

k k k

k k

2 2

2

2

2 1 2 1b 0

1 4 2

2 1 4 2 1 2 1 0

8 2 2 3 1 0

6 3 3 0

2 1 0

2 1 1 0

1of/or 1

2

x x

x x

x x x x

x x x

x x

x x

x x

x x

3. a 4 2 5

3 4 1

4 222

3 4

5 2

1 4 221

22

4 5

3 1 11 1

22 2

x y

x y

x

y



43

b 3 2 1

2 3 2

2 2 10

All the deterninants below are calculated by expanding along the first row.

3 1 2

2 3 1 3 4 1 5 2 7 21

1 2 2

1 1 2

2 3 1

10 2 2 1 4 1 6 2 26 422

21 21

3 1 2

2 2 1

1 10 2 3 6 1 5 2 22

2

x y z

x y z

x y z

x

y

211

1 21

3 1 1

2 3 2

1 2 10 3 26 1 22 1 7 633

21 21z

161 12 2

161 72 6

3 6 1c

6

2 3 1

2

3 621

2 3

6

3 31 1

21 6

6

3

21 1

21 18

18

x y

x y

x

x

y

y


MAT1581 Mathematics 1(Engineering)

44

3 2 4

4. 2 1 5

1 5 2

The minor for the element in :

2 4row 2, column 1 is 4 20 16

5 2

3 4row 2, column 2 is 6 4 10

1 2

3 2row 2, column 3 is 15 ( 2) 17

1 5

A

You have now completed learning unit 3: Cramer’s rule.

You should be able to solve a system of simultaneous equations using Cramer’s rule.

We now move on to module 3: Partial fractions. Learning unit 1 is the introduction to this module.


45

M O D U L E 3 3

MODULE 3: PARTIAL FRACTIONS CONTENTS PAGE

LEARNING UNIT 1 INTRODUCTION 46

1. ALGEBRAIC FRACTIONS ............................................................................ 47 2. PROPER AND IMPROPER FRACTIONS ...................................................... 47 3. SIMPLIFYING FRACTIONS BY CANCELLING COMMON FACTORS .. 48 3.1 Dividing by a monomial ................................................................................ 48 3.2 Factorising numerator and denominator ....................................................... 48 4. SIMPLIFYING FRACTIONS BY FINDING PARTIAL FRACTIONS. ........ 49 5. RESPONSES TO ACTIVITIES ................................................................... 50 5.1 Activity 1 ....................................................................................................... 50 5.2 Activity 2 ....................................................................................................... 50 5.3 Activity 3 ....................................................................................................... 50

LEARNING UNIT 2 PROPER FRACTIONS 52

1. LINEAR FACTORS ......................................................................................... 54 2. REPEATED LINEAR FACTORS ................................................................... 57 3. QUADRATIC FACTORS ................................................................................ 60 4. RESPONSES TO ACTIVITIES ....................................................................... 63 4.1 Activity 1 ....................................................................................................... 63 4.2 Activity 2 ....................................................................................................... 63 4.3 Activity 3 ....................................................................................................... 65 4.4 Activity 4 ....................................................................................................... 67

LEARNING UNIT 3 IMPROPER FRACTIONS 70

1. FIND PROPER FRACTION BY LONG DIVISION ....................................... 71 2. INCLUDE EXTRA TERMS ............................................................................ 73 3. SUMMARY ...................................................................................................... 75 4. RESPONSES TO ACTIVITIES ....................................................................... 76 4.1 Activity 1 ....................................................................................................... 76 4.2 Activity 2 ....................................................................................................... 80

POST-TEST: PARTIAL FRACTIONS ...................................................................... 82


46 MAT1581 Mathematics 1 (Engineering)

PARTIAL FRACTIONS Introduction

CONTENTS PAGE

1. ALGEBRAIC FRACTIONS ................................................................................ 47 2. PROPER AND IMPROPER FRACTIONS .......................................................... 47 3. SIMPLIFYING FRACTIONS BY CANCELLING COMMON FACTORS ...... 48 3.1 Dividing by a monomial ....................................................................................... 48 3.2 Factorising numerator and denominator ............................................................... 48 4 SIMPLIFYING FRACTIONS BY FINDING PARTIAL FRACTIONS. ............ 49 5. RESPONSES TO ACTIVITIES ........................................................................... 50 5.1 Activity 1 .............................................................................................................. 50 5.2 Activity 2 .............................................................................................................. 50 5.3 Activity 3 .............................................................................................................. 50

MODULE 3

LEARNING UNIT 1

OUTCOMES

At the end of this learning unit, you should be able to distinguish between a proper and an improper algebraic fraction use an example to describe the term "partial fractions" use factorisation to simplify fractions

Module 3 Learning unit 1 Partial fractions: Introduction


47

1. ALGEBRAIC FRACTIONS Algebraic fractions are expressions in the form of a ratio or quotient of two polynomials:

polynomial expression numeratoralgebraic fraction =

polynomial expression denominator

For example 2

2 3 2

2 42 19 or

4 4 4

x x x

x x x x

2. PROPER AND IMPROPER FRACTIONS The degree of a polynomial is the highest power to which the variable is raised in the expression. Thus, the degree of 7 24 9 2 1x x x is 7.

Thus, the degree of 5 4 3 2ax bx cx dx ex f is 5. When presented with a fraction, we can note the degree of the numerator with, say, n and that of the denominator with d. A fraction is proper if the degree of the denominator is greater than the degree of the numerator, that is, d > n. A fraction is improper if the numerator is of the same degree as, or higher degree than, the denominator, that is, d n .

ACTIVITY 1 Classify the following fractions as either proper or improper. In each case state the degree of both numerator and denominator.

3 2

2

2

2

2

2

2

2

4

42 19a)

4 43

b)1 2

2c)

4

5d)

1

5 2e)

3 1

3 7f )

1

x

x x xx

x x

x x

x

x x

x

x

x x x

x

x




3. SIMPLIFYING FRACTIONS BY CANCELLING COMMON FACTORS

3.1 Dividing by a monomial Divide each term of the numerator by the denominator.

Example 1

2 215 9 15 9

6 6 65 3

2 2

x x x x

x x xx

Example 2

2 2 2 2 2 2 2 26 4 9 12 6 4 9 12

3 3 3 3 3

42 3 4

3

xy xy x y x y xy xy x y x y

xy xy xy xy xy

yx xy

Be careful! A typical error here is cancelling non-common factors. Do not make either of the following mistakes:

1

41

4

x yy

x

yy

ACTIVITY 2 Simplify the following:

2 3

2

4 2

2

2 2

96a)

16

4b)

7 7c)

a xy

axy

x x

x

a b ab

ab


3.2 Factorising numerator and denominator

Example 1

23 2

2

2

22

4 2 2

provided 22

x xx x

x x x

xx

x



49

Remember that the one operation we cannot perform is division by 0. Consequently, cancelling the common factor (x + 2) cannot be performed if x takes on the value -2 because that makes the common factor equal to 0.

Example 2

2

2

2 68 12

3 10 2 3 5

6 provided 2

3 5

x xx x

x x x x

xx

x


2

2

4 4

2 2

2 8a)

5 4

b)

x x

x x

x a

x a


4. SIMPLIFYING FRACTIONS BY FINDING PARTIAL FRACTIONS

What are partial fractions? In elementary Algebra we learnt how to add and subtract fractions to get a single equivalent fraction.

Thus,2 3 2

2 11 2 42 19 can easily be converted to

41 4 4

x x

xx x x x

It is often easier to deal with a few relatively easy fractions than with one complicated fraction. We go the opposite way and reduce a given fraction into its components. These components are called partial fractions. You might find that you want to return to units 2 and 3 of this module to find partial fractions when dealing with limits, integration and Laplace transforms.

For example

To reduce the fraction

3

1 2

x

x x

we need to find the components 4 5

1 2x x

The denominators of the partial fractions are the factors of the original denominator polynomial. The rules for finding partial fractions depend on the form of the denominator. The denominator of an algebraic fraction can be factorised into a product of linear and quadratic factors. Linear factors are those of the form , for example, 4,ax b x 2 3 and 7 2x x .

Quadratic factors are those of the form 2 , for exampleax bx c 23 7x x . Quadratic factors are those that cannot be factorised into linear factors.



In learning unit 2 we will find partial fractions of proper fractions and in learning unit 3 of improper fractions.


5.1 Activity 1 a) The degree of the numerator n is 1. The degree of the denominator d is 3. The

fraction is proper since d > n. b) Expanding 21 2 produces 3 2.x x x x The fraction is proper because

n = 1 and d = 2. c) Expanding the numerator produces 3 22x x . The fraction is improper because n

= 3, d = 2, so d < n. d) The fraction is improper because n = 2, d = 2, so d = n. e) The fraction is proper because n = 2, d = 3, so d > n. f) The fraction is proper because n = 2, d = 4, so d > n.

5.2 Activity 2

2 3

2

4 2 4 22

2 2 2

2 2 2 2

96a) 6 0, 0, 0

16

4 4b) 4 1 0

7 7 7 7c) 7 7 0, 0

a xyay a x y

axy

x x x xx x

x x x

a b ab a b aba b a b

ab ab ab

5.3 Activity 3

2

2

4 4

2 2

2 2 2 2

2 2

2 2 2 2

2 8)

5 44 2

4 1

2, 4 0

1

)

, 0

x xa

x xx x

x x

xx

x

x ab

x a

x a x a

x a

x a x a



51

You have now completed learning unit 1, which is an introduction to partial fractions.


distinguish between a proper and an improper algebraic fraction use an example to describe the term "partial fractions" use factorisation to simplify fractions We next engage with learning unit 2: Partial fractions – proper fractions.


52

PARTIAL FRACTIONS Proper fractions

CONTENTS PAGE

1. LINEAR FACTORS ............................................................................................. 54 2. REPEATED LINEAR FACTORS ....................................................................... 57 3. QUADRATIC FACTORS .................................................................................... 60 4. RESPONSES TO ACTIVITIES ........................................................................... 63 4.1 Activity 1 .............................................................................................................. 63 4.2 Activity 2 .............................................................................................................. 63 4.3 Activity 3 .............................................................................................................. 65 4.4 Activity 4 .............................................................................................................. 67

MODULE 3

LEARNING UNIT 2

OUTCOMES

At the end of this learning unit, you should be able to factorise the denominator into prime factors recognise linear, repeated linear and quadratic factors in the denominator express an algebraic fraction in partial fractions

Module 3 Learning unit 2 PARTIAL FRACTIONS: Proper fractions


53

In this unit we will deal with proper fractions, that is, the degree of the numerator must be less than the degree of the denominator. The first step towards finding partial fractions is to factorise the denominator into prime factors. The denominator can contain the following types of factors: 1. Linear factors ax b 2. Repeated linear factors n

ax b

3. Quadratic factors 2ax bx c A linear factor that occurs more than once is called a repeated linear factor.

The quadratic factor is a quadratic expression that does not factorise without containing surds or imaginary terms.

Example 1

2

3 8 3 8

( 7)( 5)12 35

x x

x xx x

The denominator contains two linear factors.

Example 2

22

3 2 3 2 3 2

2 1 ( 1)( 1) 1

x x x

x x x x x

The denominator contains a repeated linear factor.

Example 3

3 2

5 5

8 2 2 4x x x x

The denominator contains a linear and a quadratic factor. Hint: See the information sheet to factorise the sum of two third powers. To call attention to the fact that we are dealing with identities, we write three lines instead of two (=), which is the symbol for equality.

An identity requires that the two sides be equal for all values of x.

ACTIVITY 1 Factorise the denominator of the following fractions and describe the form of the factors:

2

3 8a)

2 35

x

x x



2

2

3

2

3 2

2

4

3 19b)

15 8 63

3 1c)

1

5 2d)

3

3 7e)

1

x

x x

x

x

x

x x x

x

x


1. LINEAR FACTORS These are the steps to calculate partial fractions if the denominator is a product of linear factors: 1. Factorise the denominator. 2. Each factor of the denominator produces a partial fraction. A factor of the

form ax b produces a fraction of the form ( )

A

ax b, where A is an unknown

constant. 3. Evaluate the unknown constants:

3.1 Add the partial fractions together to form a single algebraic fraction. The numerator contains the unknown constants and the denominator is identical to that of the original expression.

3.2 Equate the numerator so obtained with the numerator of the original algebraic fraction.

3.3 Substitute appropriate values of x to determine the values of the unknowns.

Example 1

Resolve 2

3 8

2 35

x

x x

into partial fractions.

The first step is to factorise the denominator.

2

3 8 3 8

7 52 35

x x

x xx x

(x + 7) is a linear factor, so produces a partial fraction of the form 7

A

x

(x 5) is a linear factor, so produces a partial fraction of the form 5

B

x



55

3 8Therefore

7 5 7 5

Writing the right-hand side using a common denominator we have

3 8 ( 5) ( 7)

7 5 7 5 7 5

Equating numerators gives

3 8 ( 5) ( 7)

x A B

x x x x

x A B A x B x

x x x x x x

x A x B x

13

12

By appropiate choice of we can eliminate one of the unknowns:

Let 7, then the term that contain vanishes:

3( 7) 8 ( 7 5)

21 8 ( 12)

13 12

x

x B

A

A

A

A

13 2312 12

2

23

12

Let 5, then 3(5) 8 (5 7)

23 12

3 8Hence

7 52 3513 23

12 7 12 5

x B

B

B

x

x xx x

x x

You should test the result. Combining the partial fractions should give the original expression. You can also choose a value for x and show that the two sides are equal.

Example 2

Find partial fractions for 2

3 19

15 8 63

x

x x

Factorise the denominator:

2

3 19 3 19

5 9 3 715 8 63

x x

x xx x

We have two linear factors, so we assume that the expression can be separated into partial fractions of the form:

2

3 19 3 19

5 9 3 7 5 9 3 715 8 63

x x A B

x x x xx x



Writing the right-hand side using a common denominator we have

3 19 (3 7) (5 9)

5 9 3 7 5 9 3 7 5 9 3 7


3 19 (3 7) (5 9)

By appropiate choice of we can eliminate

x A B A x B x

x x x x x x

x A x B x

x

one of the unknowns:

7

3

7 7

3 3

Let , then the term that contain vanishes:

3 19 5 9

35 277 19

3

6212

3

36 18

62 31

9 9 9Let , then 3 19 3 7

5 5 5

27 95 27 35

5 5

122

x A

B

B

B

B

x B

B

61 1831 31

2

(62)

122 61

62 31

3 19Hence

15 8 63 5 9 3 761 18

31 5 9 31 3 7

A

B

x

x x x x

x x

ACTIVITY 2 Find partial fractions for

2

2

11 3a)

2 3

2 9 35b)

1 2 3

x

x x

x x

x x x




57

2. REPEATED LINEAR FACTORS If a factor occurs twice, this will generate two partial fractions; if it occurs three times,

it will generate three partial fractions and so on. The factors 2ax b in a

denominator generate partial fractions of the form 2

A B

ax b ax b

. The rest of the

method is the same as for linear factors. Instead of substituting values for x to solve for the unknowns, we can also use the following principle: If two algebraic functions are identically equal, then the coefficients of like powers of x must be equal.

3 2 2

3 2

If

0

then 0; ; and .

ax bx cx d Kx Mx P

x Kx Mx P

a b K c M d P

Example 1

Resolve

2

4

3 7

1

x

x

in partial fractions:

The factor (x 1) is repeated four times, thus generating four partial fractions:

3 22

1

By appropiate choice of we can eliminate one of the unknowns:

Let , then 4

Thus 3 7 1 1 ( 1) 4

To find the values of , and we have a choice of methods.

Method 1:

Choosing other value

x

x D

x A x B x C x

A B C

s of will lead to a system of simultaneous equations

with three unknowns , and .

We can choose any values of as we have an identity.

Let 0, then 7 4 ( )

Let 2, then 5 4 ( )

Let 1,

x

A B C

x

x A B C a

x A B C b

x

then 4 8 4 2 4 ( )

These can then be solved with any of the methods for solving simultaneous equations.

A B C c

Add equations ( ) and ( ) : 2 2 8

6 2

3

a b B

B

B



2

4 2 3 4

3 22

4 4

3 7

11 1 1 1

Adding the right-hand side using a common denominator we have

1 1 ( 1)3 7

1 1

x A B C D

xx x x x

A x B x C x Dx

x x

3 22


3 7 1 1 ( 1)x A x B x C x D

2

4

Now equation (b) becomes: 5 3 4

that is, 6 ( )

and equation (c) becomes: 4 8 4(3) 2 4

12 8 2

that is, 6 4 ( )

Subtract equations ( ) and ( ) : 0 3

0

Substitute in ( ) : 6

3 7 3Hence

1

A C

A C d

A C

A C

A C e

d e A

A

d C

x

x x

2 3 4

6 4

1 1 1x x

3 2

2 2

3 2 2 2

Method 2:

Expand the right-hand side of equation (*)

1 1 ( 1) 4 (*)

1 2 1 ( 2 1) 4

2 2 2 4

and rearrange terms so that we can compare coefficients

A x B x C x

A x x x B x x Cx C

Ax Ax Ax Ax Ax A Bx Bx B Cx C

3 2 3 2

3

2

2

4 2 3

0 3 0 7 3 3 2 ( 4)

Comparing terms in gives: 0

Comparing terms in gives: 3 3

3 0

3

Comparing constant terms gives: 7 4

3 3

6

3 7 3 6 4Hence

1 1 1 1

x x x Ax x A B x A B C A B C

x A

x A B

B

B

A B C

C

C

x

x x x x

4



59

Example 2

Find partial fractions for 2

3 2

1

x

x x

2

The linear factor generates a partial fraction of the form

The linear factor 1 is repeated twice, thus generating

two partial fractions and .1 1

Ax

xx

B C

x x

2 2

2

2 2

2

2 2

2 2

3 2Thus

11 1

Combine the right-hand side:

1 13 2

1 1

Equating numerators gives:

3 2 1 1

2 1

2

and rearrange terms so that we

x A B C

x xx x x

A x Bx x Cxx

x x x x

x A x Bx x Cx

A x x Bx Bx Cx

Ax Ax A Bx Bx Cx

2 2

2

2

can compare coefficients

0 3 2 ( ) (2 )

Comparing constant terms gives 2


0 2

2


3 2( 2) 2

3 2

5

3 2Hence

1

x x A B x A B C x A

A

x A B

B

B

x A B C

C

C

C

x

x x

2

2 2 5

1 1x x x

ACTIVITY 3 Find partial fractions for

2

2

3 2a)

1

3b)

2 1

x

x

x

x x




3. QUADRATIC FACTORS

Sometimes a denominator is factorised producing quadratic factors that cannot be

factorised any further. For example, for 3 2 23 3 1x x x x x x , the quadratic

expression 2 3 1x x cannot be factorised any further.

In general a quadratic factor of the form 2ax bx c produces a partial factor of the

form 2

Ax B

ax bx c

Example 1

Resolve 2

3

3 1

1

x

x

in partial fractions.

First factorise the denominator:

2 2

3 2

2

2

3 1 3 1[Use information sheet to factorise]

1 1 1

The linear factor ( 1) generates a partial fraction of the form1

The quadratic factor 1 generates a partial factor

of the form

x x

x x x x

Ax

x

x x

Bx C

x x

2 2

3 22

22

2 2

1

3 1 3 1Thus

11 11 1


1 13 1

1 1 1 1

x x A Bx C

xx x xx x x

A x x Bx C xx

x x x x x x

2 2


3 1 1 1 (*)

Let 1, then 2 = A 3

2

3

x A x x Bx C x

x

A



61

2 2 2

2 2

2 22 2 23 3 3

Expanding equation (*) gives

3 1


3 0 1 ( ) ( ) ( )

Now 3 0 1 ( ) ( ) ( )

Comparing constant ter

x Ax Ax A Bx Bx Cx C

x x A B x A B C x A C

x x B x B C x C

23

23

2 23

23

2 7 523 3 3

3 2

2

ms gives 1

1

5

3


3

7

3

3 1Hence

11 12 7 5

3 1 3 1

C

C

C

x B

B

xx

xx x xx

x x x

Example 2

Find partial fractions for

2

2

5 2

3 1

x

x x x

2

2

2

22

The linear factor generates a partial fraction of the form

The quadratic factor 3 1 generates a partial factor

of the form3 1

5 2Thus

3 13 1


5

Ax

x

x x

Bx C

x x

x A Bx C

x x xx x x

22

2 2

3 12

3 1 3 1

A x x Bx C xx

x x x x x x



2 2


5 2 3 1 (*)

Let 0, then 2

x A x x Bx C x

x A

2 2

2 2

2 2

2


5 2 2 3 1

2 6 2


5 0 2 ( 2 ) ( 6 ) 2


7

Comparing term

x x x Bx C x

x x Bx Cx

x x B x C x

x B

B

2

22

s in gives: 0 6

6

5 2 2 7 6Hence

3 13 1

x C

C

x x

x x xx x x

ACTIVITY 4

3

2

2

5a)

8

5 3 2b)

2 1 3

x

x x

x x




63


4.1 Activity 1

2

2

2 2

3 2

2 2

3 2 2

3 8 3 8a) Two linear factors

2 35 7 5

3 19 3 19b) Two linear factors

15 8 63 5 9 3 7

3 1 3 1c) One linear and one quadratic factor

1 1 1

5 2 5 2d). One linear and on

3 3 1

x x

x x x x

x x

x x x x

x x

x x x x

x x

x x x x x x

2

4

e quadratic factor

3 7e) Repeated linear factor

1

x

x

4.2 Activity 2

2

2

11 3a)

2 3

11 3 11 3Factorise the deniminator:

2 3 1 3


The linear factor 3 generates a partial fraction of the form 3

x

x x

x x

x x x x

Ax

xB

xx

2

2

11 3 11 3Thus

2 3 1 3 1 3


3 111 3 11 3

2 3 1 3 1 3


11 3 3 1 )

Let 1, then 8 (4)

2

Let 3, then 20 ( 4)

x x A B

x x x x x x

A x B xx x

x x x x x x

x A x B x

x A

A

x B

B

2

5

11 3 11 3 2 5Hence

2 3 1 3 1 ( 3)

x x

x x x x x x



22 9 35c)

1 2 3


The linear factor 2 generates a partial fraction of the form 2

The linear factor 3 generates a partial fr

x x

x x x

Ax

xB

xx

x

2

2

2

action of the form 3

2 9 35Thus

1 2 3 1 2 3


2 3 1 3 1 22 9 35

1 2 3 1 2 3


2 9 35 2 3 1 3 1 2

C

x

x x A B C

x x x x x x

A x x B x x C x xx x

x x x x x x

x x A x x B x x C x x

2

Let 1, then -24 ( 3) 2

244

6Let 2, then 45 (3) 5

453

15Let 3, then 10 ( 2) 5

101

10

2 9 35 4 3 1Hence

1 2 3 1 2 3

x A

A

x B

B

x C

C

x x

x x x x x x



65

4.3 Activity 3

2

2

2 2

2 2

3 2a)

1



3 2Thus

11 1


13 2

1 1


3

x

x

x

A B

x x

x A B

xx x

A x Bx

x x

2 2

2 1

and compare coefficients

3 2


Comparing terms in constants gives:

2

2 3

5

3 2 3 5Hence

11 1

x A x B

Ax A B

x Ax A B

x A

A B

B

B

x

xx x



2

2

2

3c)

2 1




3Thus

1 2 (2 1

x

x x

Ax

xx

B C

x x

x A B C

x x xx x

2

2

2 2

2

2 2

2 2

2)


2 2 1 ( 1)3

2 1 2 1


3 2 2 1 ( 1)

4 4 ( 2)

4 4 2

and rearrange terms so that we ca

A x B x x C xx

x x x x

x A x B x x C x

A x x B x x Cx C

Ax Ax A Bx Bx B Cx C

2 2

2

n compare coefficients

0 3 0 ( ) (4 ) (4 2 )

Comparing constant terms gives 0 4 2 ( )

Comparing terms in gives: 0 ( )

Comparing terms in gives: 3 4 ( )

Add equations and ( ) : 3

x x A B x A B C x A B C

A B C a

x A B b

x A B C c

a c

2 2

8 ( )

Add equations and ( ) : 3 9

3 1( )

9 31

Substitute in : 03

1

31 1

Substitute and in (c): 3 43 3

3 1

2

3 1 1 2Hence

3( 1) 3( 2) ( 2)2 1

A B e

b e A

A f

A b B

B

A B C

C

C

x

x x xx x



67

4.4 Activity 4

3 3 23

2

2

5 5 5[Use information sheet to factorise]

8 2 2 42



of the form

x x x xx

Ax

x

x x

Bx C

x

3 2 2

2

2 2

2

2

2 4

5 5Thus

28 2 2 4 2 4


2 4 25

2 2 4 2 2 4


5 2 4 2 (*)

Let 2, then 5 = A (-2) 2( 2) 4

5 (12)

x

A Bx C

xx x x x x x

A x x Bx C x

x x x x x x

A x x Bx C x

x

A

A

2 2

2 2

2 25 5 512 12 12

5

12Expanding equation (*) gives

5 2 4 2 2


0 0 5 ( ) ( 2 2 ) (4 2 )

Now 0 0 5 ( ) ( 2 2 ) (4 2 )

Compar

Ax Ax A Bx Cx Bx C

x x A B x A B C x A C

x x B x B C x C

512

53

2 512

5 5512 312

3 2 2

2

ing constant terms gives 5 4 2

2 5

102

310 5

6 3


5

125 5

Hence28 2 2 4 2 4

5 5 20

12 2 12 2 4

C

C

C

C

x B

B

x

xx x x x x x

x

x x x



2

2

2

2

2

2 2

5 3 2b)

2 1 3



of the form2 1

5 3 2Thus

32 1 3 2 1

Combin

x x

x x

Ax

x

x

Bx C

x

x x A Bx C

xx x x

22

2 2

2 2

22

2 2

e the right-hand side:

2 1 35 3 2

2 1 3 2 1 3


5 3 2 2 1 3 (*)

Let 3,

then 5(-3) 3 3 2 2 3 1

38 19

2


5 3 2 2 2

A x Bx C xx x

x x x x

x x A x Bx C x

x

A

A

A

x x x

2 2

2 2

2

2

1 3

4 2 3 3


5 3 2 (4 ) ( 3 ) (2 3 )


1

Comparing terms in gives: 3 3(1)

0

5Hence

Bx C x

x Bx Cx Bx C

x x B x C B x C

x B

B

x C

C

x

2 2

3 2 2

32 1 3 2 1

x x

xx x x



69

You have now completed learning unit 2: Partial fractions – proper fractions.


factorise the denominator into prime factors recognise linear, repeated linear and quadratic factors in the denominator express an algebraic fraction in partial fractions Now that you have completed learning unit 2, we will move on to learning unit 3:

Partial fractions - improper fractions.


70

PARTIAL FRACTIONS Improper fractions

CONTENTS PAGE

1. FIND PROPER FRACTION BY LONG DIVISION ........................................... 71 2. INCLUDE EXTRA TERMS ................................................................................ 73 3. SUMMARY .......................................................................................................... 75 4. RESPONSES TO ACTIVITIES ........................................................................... 76 4.1 Activity 1 .............................................................................................................. 76 4.2 Activity 2 .............................................................................................................. 80

MODULE 3

LEARNING UNIT 3

OUTCOMES

At the end of this learning unit, you should be able to use long division to simplify improper fractions add an extra term to the partial fractions to find partial fractions of

improper fractions

Module 3 Learning unit 3 Partial fractions: Improper fractions


71

When calculating partial fractions of improper fractions we can use long division to first simplify the fraction or we can add appropriate terms to find the correct form of the partial fraction.

1. FIND PROPER FRACTION BY LONG DIVISION

Let’s consider 2

2

3 2 7

2 3

x x

x x

, that is, 2 2(3 2 7) 2 3x x x x .

The result of this division is called the quotient of the two expressions. Write both expressions in decreasing order of powers. If there are any missing terms in the dividend (numerator), write them with a coefficient of 0. We set up the division as for the long division of numbers:

2 23 2 3 2 7x x x x

To make 3x2, x2 must be multiplied by 3, so we insert this as the first term in the quotient.

2 2

3

3 2 3 2 7x x x x

Multiply the divisor ( 2 3 2x x ) by 3 and subtract the answer from the dividend.

2 2

2

3

3 2 3 2 7

3 9 6

7 1

x x x x

x x

x

The degree of the answer is less than that of the divisor. We cannot go any further.

Thus we can write 2

2 2

3 2 7 7 13

2 3 3 2

x x x

x x x x

The fraction is proper and can be resolved in partial fractions using the methods discussed in unit 2.

Example

Resolve 2

2

1

3 2

x

x x

into partial fractions.

The denominator is of the same degree as the numerator. The fraction is improper and therefore we must divide first:

Module 3 Learning unit 3 PARTIAL FRACTIONS: Improper fractions


72

2 2

2

1

3 2 0 1 Because there is no term in we need to insert 0

3 2

3 1 Be careful doing the subtraction.

x x x x x x

x x

x

2

2 2

1 3 1 3 1Therefore 1 1

1 23 2 3 2

The proper fraction can now be resolved into partial fractions.

( 2) 13 1Let

1 2 1 ( 2) 1 2


3 1 ( 2) 1

Let

x x x

x xx x x x

A x B xx A B

x x x x x x

x A x B x

2

2 2

1. Then 2

2

Let 2. Then 5

3 1 2 5Therefore

1 2 1 ( 2)

1 3 1and 1

3 2 3 23 1

11 2

2 51

1 ( 2)

x A

A

x B

x

x x x x

x x

x x x xx

x x

x x

ACTIVITY 1 Resolve into partial fractions:

2

2

2

2

3 2

2

2

3 8a)

4

3 2 5b)

6 5 1

2 4 4c)

2

1d)

1 2

x

x

x x

x x

x x x

x x

x x

x x




73

2. INCLUDE EXTRA TERMS The extra term is a polynomial of degree n - d, where n is the degree of the numerator and d is the degree of the denominator. Recall that a polynomial of degree 0 is a constant, a polynomial of degree 1 has the form Ax B , a polynomial of degree 2 has the form 2Ax Bx C , and so on. For example, if the numerator has degree 3 and the denominator has degree 2, the partial fractions will include a polynomial of degree 1, that is, of the form Ax B .

Example 1

Express as partial fractions 23 2

1

x x

x

The degree of the numerator, n is 2 and the degree of the denominator d is 1. Thus the fraction is improper. The partial fractions will include a polynomial of degree 1 because n - d = 1, that is, Ax B .

2

2

2

2

2

3 2Hence

1 1( 1) ( 1)

1Equating numerators gives:

3 2 ( 1) ( 1)

Let 1, then 3 2

1

Now 3 2 ( 1) ( 1) 1 [because 1]

1

( ) ( 1)

Comparing c

x x CAx B

x xAx x B x C

x

x x Ax x B x C

x C

C

x x Ax x B x C

Ax Ax Bx B

Ax A B x B

2

2

oefficients in gives

3

Comparing the constant term gives

0 1

1

3 2 1Therefore 3 1

1 1

x

A

B

B

x xx

x x

Note: This method saves us the long division step but finding the values of A, B or C may be more difficult.



74

Example 2

Resolve in partial fractions 3 2

2

4 12 13 7

4 4 1

x x x

x x

The fraction is improper, with n - d = 1. The denominator has repeated linear factors of 2 1x .

3 2 3 2

2 2

2

2 2

2

3 2 2 2

3 21 1 12 2 2

4 12 13 7 4 12 13 7Hence

4 4 1 2 1

2 1 2 1

(2 1) (2 1) 2 1

2 1


4 12 13 7 (2 1) (2 1) 2 1

Let , then 4 +12 +1

x x x x x x

x x x

C DAx B

x x

Ax x B x C x D

x

x x x Ax x B x C x D

x

12

3 2 2 2

2 2

3 2 2

3 2

3

3 +7

3

Now 4 12 13 7 (2 1) (2 1) 2 1 3

4 4 1 4 4 1 2 3

4 4 4 4 2 3

4 4 4 ( 4 2 ) ( 3)

Comparing coefficients in gives

4 = 4A

1

Compar

D

D

x x x Ax x B x C x

Ax x x B x x Cx C

Ax Ax Ax Bx Bx B Cx C

Ax A B x A B C x B C

x

A

2

3 2

2 2

ing coefficients in gives

12 4 4

12 4 4

8 4

2

Comparing the constant term gives

7 3

7 2 3

2

4 12 13 7 2 3Therefore 1 2

2 14 4 1 2 1

x

A B

B

B

B

B C

C

C

x x xx

xx x x



75

ACTIVITY 2 Express as partial fractions:

2

3 2

3

9 12 5a)

3 2

6 5 1b)

y y

y

x x x

x x


3. SUMMARY The rules of partial fractions are as follows for proper fractions: 1. The numerator of the given function must be of a lower degree than that of the

denominator. 2. Factorise the denominator into prime factors. This is important as the factors

obtained determine the shape of the partial factors:

2.1 A linear factor ax b gives a partial fraction of the formA

ax b

2.2 Repeated linear factors 2a b give partial fractions

2( )

A B

ax b ax b

and 3ax b give partial fractions

2 3( )

A B C

ax b ax b ax b

2.3 A quadratic factor 2ax bx c gives a partial fraction 2

Ax B

ax bx c

3. Add the partial fractions together to form a single algebraic fraction. The numerator contains the unknown constants and the denominator is identical to that of the original expression.

4. Equate the numerator so obtained with the numerator of the original algebraic fraction.

5. Substitute appropriate values of x to determine the values of the unknowns. 6. If some values remain unknown, return to the equation in step 4. Simplify and

rearrange the numerator of the algebraic fraction. Substitute all values obtained in step 5. Compare coefficients to find the remaining unknowns.

The rules of partial fractions are as follows for improper fractions: Divide out by long division. The resulting remainder is a proper fraction. Now

follow the rules for proper fractions. OR

Add an extra term. The extra term is a polynomial of degree n - d, where n is the degree of the numerator and d is the degree of the denominator. Remember that a polynomial of degree 0 is a constant, a polynomial of degree1 has the form Ax B , a polynomial of degree 2 has the form

2Ax Bx C , and so on.



76


4.1 Activity 1

2

2

2 2

2

2

2 2

2

3 8a)

4This is an improper fraction, so we need to divide first.

34 3 8

3 12

20

3 8 20 Therefore 3

4 420

32 2

20Now,

4 2 2

thus 20 2 2 [multiply by ( 2)( 2)]

If 2,

x

x

x x

x

x

x x

x x

A B

x x x

A x B x x x

x

2

2

then 20 4

5

If 2, then 20 4

5

3 8 5 5Thus 3

4 2 2

B

B

x A

A

x

x x x



77

2

2

12

2 2

2 5 12 2

12 2

2 112

2

112

3 2 5b)

6 5 1This is an improper fraction, so we need to divide first

6 5 1 3 2 5

3

5

3 2 5 1Therefore

6 5 1 2 1 6 1

Now 1 6 1 1 6 1

6 1 1

1 6 1

Equa

x

x

x

x x

x x

x x x x

x x

x x

x x x x

A B

x x x x

A x B x

x x

16

112

2

2

ting numerators gives:

116 1 1

2Let 1, then 6 =7

6

71 1 1

Let , then - 11 12 6 6

1 1 66 7

2 6 6

65 14

65

14

6 65Therefore,

1 6 1 7 1 14 6 1

3 2 5and

6 5

x

xA x B x

x A

A

x B

B

B

B

x x x x

x x

x

1 6 65

1 2 7 1 14 6 1x x x



78

3 2

2

2 3 2

3 2

2

2

3 2

2 2

2

2 4 4)

2This is an improper fraction, so we need to divide first.

3

2 2 4 4

2

3 2 4

3 3 6

10

2 4 4 10Thus 3

2 210 10

Now2 2 1

10and

2 1

x x xc

x x

x

x x x x x

x x x

x x

x x

x

x x x xx

x x x xx x

x x x x

x

x x

3 2

2

2 1

( 1) ( 2)

2 ( 1)


10 ( 1) ( 2)

Let 1, then -9 = 3

3

Let 2, then 12 3

4

10 4 3Therefore

2 1 2 1

2 4 4 4 3and 3

2 2 1

A B

x x

A x B x

x x

x A x B x

x B

B

x A

A

x

x x x x

x x xx

x x x x



79

2 2

2

2 2

2

2

1 1)

1 2 3 2

This is an improper fraction, so we need to divide first.

1

3 2 1

3 2

2 1

1 2 1Therefore 1

1 2 1 2

2 12 1Now

1 2 1 2 2 1

Equating nume

x x x xd

x x x x

x x x x

x x

x

x x x

x x x x

A x B xx A B

x x x x x x

2

rators gives:

2 1 ( 2) ( 1)

Let 1, then 1

Let 2, then 3

3

2 1 1 3Therefore

1 2 1 2

1 1 3and 1

1 2 1 2

x A x B x

x A

x B

B

x

x x x x

x x

x x x x



80

4.2 Activity 2 a) The fraction is improper, n - d = 1. The denominator has only one linear factor.

2

2

22 2 23 3 3

2

2

2

9 12 5Hence

3 2 3 2

3 2) (3 2

3 2


9 12 5 3 2) (3 2

Let , then 9 +12 +5

1

Now 9 12 5 3 2) (3 2 1

3 2 3 2 1

3 (2

y y CAy B

y y

Ay y B y C

y

y y Ay y B y C

x C

C

y y Ay y B y

Ay Ay By B

Ay A

2

2

3 ) (2 1)


9 3

3

Comparing the coefficients gives

12 2(3) 3

12 6 3

3 6

2

9 12 5 1Therefore 3 2

3 2 3 2

B y B

y

A

A

y

B

B

B

B

y yy

y y



81

b) The fraction is improper, n - d = 0. The partial fraction will include a constant.

The denominator is factorised to 2 1x x . The linear factor x produces a

partial fraction of the form B

x, and the quadratic factor x2 +1 produces a

partial fraction of the form 2 1

Cx D

x

3 2 3 2

3 2

2

2 2

2

3 2 2 2

3 2 2 2

3

6 5 1 6 5 1Hence

1

1

1 ( 1)

1


6 5 1 1 ( 1)

Let 0, then -1

Now 6 5 1 1 1( 1)

x x x x x x

x x x x

B Cx DA

x x

Ax x B x x Cx D

x x

x x x Ax x B x x Cx D

x B

x x x Ax x x x Cx D

Ax

2 2

3 2

3

2

3 2

3

1

(1 ) ( ) 1


6 = A


1 (1 )

1 1

2

Comparing the coefficients gives

5

5 6

1

6 5 1Therefore

Ax x Cx Dx

Ax C x A D x

x

x

C

C

C

x

A D

D

D

x x x

x x

3 2

2

2

6 5 1

1

1 2 16

1

x x x

x x

x

x x


82

POST-TEST: PARTIAL FRACTIONS Find the partial fractions for/Resolve into partial fractions/Express as partial fractions:

2

2

2

2

2

2

2

3

2

2

3 2

2

1a)

25

b) 25

3 1c)

3 2

2 3d)

3 2

3 1e)

1

4 3f)

1

14 31 5g)

1 2 3

2 3 54 50h)

2 24

x

x

x

x

x x

x

x x

x x

x

x

x

x x

x x

x x x

x x


83

POST-TEST SOLUTIONS

2

2

a) Factorise the denominator

5 51 1

25 5 5 5 5 5 5


1 5 5

Let 5, then 1 10

1

10Let 5, then 1 10

1

101 1 1

25 10 5 10 5

A x B xA B

x x x x x x x

A x B x

x A

A

x B

B

x x x

2

2

5 5b)

25 5 5 5 5 5 5


5 5

Let 5, then 5 10

1

2Let 5, then 5 10

1

21 1

Therefore 25 2 5 2 5

A x B xx x A B

x x x x x x x

x A x B x

x A

A

x B

B

x

x x x

2

2

2 2

2

3 1c)

3 2 This is an improper fraction. We must divide first

3

3 2 3 1

3 6 9

5 9

x

x x

x x x

x x

x

Module 3 Partial fractions: Post-test solutions


84

2

2 2

3 1 9 5Hence 3

3 2 3 2Now find partial fractions for the proper fraction

1 29 5

2 1 2 1 2 1


9 5 1 2

Let 1, then 4

4

Let 2, then 13

Theref

x x

x x x x

A x B xx A B

x x x x x x

x A x B x

x B

B

x A

2

2

3 1 13 4ore 3

2 13 2

x

x xx x

2

2

1 3 22 3 2 3d)

3 2 3 2 1 3 2 1 3 2 1


2 3 1 3 2

Let 1, then 5 5

1

2 5 5Let , then

3 3 31

2 3 1 1Therefore

3 2 1 3 2

A x B xx x A B

x x x x x x x x

x A x B x

x B

B

x A

A

x

x x x x

2 2

2 2

2 2

2

2

2 2

3 1 3 1e)

2 11

This is an improper fraction. We must divide first

(or add a constant term to the partial fractions).

3

2 1 3 1

3 6 3

7 2

3 1 7 2Therefore 3 3

11 1

x x x x

x xx

x x x x

x x

x

x x x A B

xx x

21x



85

2 2 2

Resolve the proper fraction in partial fractions

17 2

1 ( 1)1 1


7 2 1

7 2

Compare coefficients of terms in :

7

Compare constant terms

2

A x Bx A B

x xx x

x A x B

x Ax A B

x

A

A B

B

2

2 2

5

3 1 7 5 Therefore 3

11 1

x x

xx x

3 2

3 2 3 3

2

2

2 2

1 1 14 3)

11 1 1 1


4 3 1 1

2

0 4 3 ( 2 ) (

Comparing coefficients gives:

0

2 4

4

3

1

Therefore

A x B x C xx A B Cf

xx x x x

x A x B x C

Ax Ax A Bx B C

x x Ax A B x A B C

A

A B

B

A B C

C

3 2 3

4 3 4 1

1 1 1

x

x x x

g)

2

2

2

2

14 31 521 2 31 2 3 2 3

2 3 1 2 3 1

1 2 3

A B Cx xx xx x x

A x B x x C x

x x



86

Compare coefficients:


2214 31 5 2 3 2 3 ( 1) ( 1)

2 24 12 9 2 3

2(2 4 ) (12 ) (9 3 )

14 4 2 (1)

31 12 (2)

5 9 3 (3)

(2) (3): 36 21

x x A x B x x C x

Ax Ax A Bx Bx B Cx C

B A x A B C x A B C

A B

A B C

A B C

2 (4)

(4) (1): 50 25

2

Substitute in (1):

2 14 -8

6

3

Substitute in (2):

31 24 3 C

C 31-27

A B

A

A

B

B

4

2 2 3 414 31 52 21 2 31 2 3 2 3

x xx xx x x



87

h)

3 2 3 2

2

3 2

2 3 54 50 2 3 54 50

6 42 24

This is an improper fraction.

1

Add an expression of degree 1 to the partial fractions.

2 3 54 50

6 4 6 4

6 4 6 4 ( 4) ( 6)

x x x x x x

x xx x

n d

x x x C DA Bx

x x x x

A x x Bx x x C x D x

x

3 2

3 2

3 2

6 4


2 3 54 50 6 4 6 4 ( 4) ( 6) (*)

Let 6, then 2( 6) 3( 6) 54( 6) 50 ( 6 4)

50 10

5

Let 4, then 2(4) 3(4) 54(4) 50 (4 6)

10 10

1

Thus * become

x

x x x A x x Bx x x C x D x

x C

C

C

x D

D

D

3 2

2 2

2 3 2

3 2

3

s:

2 3 54 50 6 4 6 4 ( 4) ( 6)

2 24 2 24 4 6

2 24 2 24 4 6

( 2 ) (2 24 ) ( 24 4 6 )

Compare coefficients in : 2

Compare coef

x x x A x x Bx x x C x D x

A x x Bx x x Cx C Dx D

Ax Ax A Bx Bx Bx Cx C Dx D

Bx A B x A B C D x A C D

x B

2

3 2

2

ficients in : 3 2

3 4

1

Therefore

2 3 54 50 5 11 2

6 42 24

x A B

A

A

x x xx

x xx x

You have now completed learning unit 3: Partial fractions – improper fractions. You

should be able to use long division to simplify improper fractions add an extra term to the partial fractions to find partial fractions of improper fractions Next is module 4: Complex numbers. Learning unit 1 deals with imaginary and complex numbers.

MAT1581 Mathematics 1 (Engineering) 88

M O D U L E 4

COMPLEX NUMBERS CONTENTS PAGE

LEARNING UNIT 1 IMAGINARY AND COMPLEX NUMBERS 90

1. IMAGINARY NUMBERS ................................................................................... 91 2. COMPLEX NUMBERS ........................................................................................ 93 2.1 Equality of complex numbers ................................................................................ 93 2.2 Conjugates of complex numbers ........................................................................... 93 3. ARGAND DIAGRAM .......................................................................................... 94 4. RESPONSES TO ACTIVITIES ............................................................................ 95 4.1 Activity 1 ............................................................................................................... 95 4.2 Activity 2 ............................................................................................................... 95 4.3 Activity 3 ............................................................................................................... 95 4.4 Activity 4 ............................................................................................................... 96

LEARNING UNIT 2 OPERATIONS WITH COMPLEX NUMBERS 98

1. ADDITION AND SUBTRACTION ..................................................................... 99 2. MULTIPLICATION ............................................................................................. 99 3. DIVISION ........................................................................................................... 100 4. COMPLEX EQUATIONS ................................................................................. 102 5. RESPONSES TO ACTIVITIES .......................................................................... 102 5.1 Activity 1 ............................................................................................................. 102 5.2 Activity 2 ............................................................................................................. 103 5.3 Activity 3 ............................................................................................................. 104 5.4 Activity 4 ............................................................................................................. 106

LEARNING UNIT 3 POLAR AND EXPONENTIAL FORM 108

1. POLAR FORM OF A COMPLEX NUMBER ................................................... 109 2. EXPONENTIAL FORM .................................................................................... 112 3. RESPONSES TO ACTIVITIES .......................................................................... 113 3.1 Activity 1 ............................................................................................................. 113 3.2 Activity 2 ............................................................................................................. 115

Module 4 Learning unit 1 COMPLEX NUMBERS: Imaginary and complex numbers


89

LEARNING UNIT 4 OPERATIONS IN POLAR AND EXPONENTIAL FORM

117

1. ADDITION AND SUBTRACTION IN POLAR AND EXPONENTIAL FORM ... ............................................................................................................................ 118 2. MULTIPLYING AND DIVIDING IN EXPONENTIAL FORM ...................... 118 3. MULTIPLYING AND DIVIDING IN POLAR FORM ..................................... 120 3.1 Multiplication ...................................................................................................... 120 3.2 Division ............................................................................................................... 121 4. POWERS, ROOTS AND DEMOIVRE’S FORMULA ................................... 122 5. RESPONSES TO ACTIVITIES .......................................................................... 127 5.1 Activity 1 ............................................................................................................. 127 5.2 Activity 2 ............................................................................................................. 127 5.3 Activity 3 ............................................................................................................. 128 5.4 Activity 4 ............................................................................................................. 128 5.5 Activity 5 ............................................................................................................. 129

POST-TEST 132


MAT1581 Mathematics 1 (Engineering) 90

COMPLEX NUMBERS Imaginary and complex

numbers

OUTCOMES

At the end of this learning unit, you should be able to express an imaginary number in standard form find the complex conjugate of a complex number represent a complex number on an Argand diagram solve any quadratic equation

CONTENTS PAGE

1. IMAGINARY NUMBERS ................................................................................... 91 2. COMPLEX NUMBERS ........................................................................................ 93 2.1 Equality of complex numbers ................................................................................ 93 2.2 Conjugates of complex numbers ........................................................................... 93 3. ARGAND DIAGRAM .......................................................................................... 94 4. RESPONSES TO ACTIVITIES ............................................................................ 95 4.1 Activity 1 ............................................................................................................... 95 4.2 Activity 2 ............................................................................................................... 95 4.3 Activity 3 ............................................................................................................... 95 4.4 Activity 4 ............................................................................................................... 96

MODULE 4

LEARNING UNIT 1



91

1. IMAGINARY NUMBERS

If the quadratic equation 2 2 5 0x x is solved using the quadratic formula, then 2

2

4

2

(2) (2) 4(1)(5)

2(1)

2 16

2

2 4 1

2

1 2 1

b b acx

a

It is not possible to evaluate 1 in real terms. To solve this problem, a new number was invented to correspond to the square root of –1. The name for this number is the imaginary unit, and it is represented by the symbol i or j. Thus we have

1i j

Another popular name for the imaginary unit is the j-operator. In pure mathematics the symbol i is used but in science and technology j is preferred, because i represents electrical current. Definition Pure imaginary number

If b is a real number, 0, then b b is a pure imaginary number and we have

( 1) 1 where 1b b b j b j

We call or j b b j the standard form for a pure imaginary number.

ACTIVITY 1 Simplify and express each of the following radicals (roots) in the standard

form:

a) 9

b) 0.25

c) 3

d) 18




92

Since 1j , we have some interesting relationships:

22

3 2

4 2 2

1 1

. 1.

. 1. 1 1

j

j j j j j

j j j

Any larger power of j can be reduced to one of these basic four. Thus 5 4 1 4

15 4 4 4 3 4 4 4 3

. 1.

1.1.1.( )

j j j j j j

j j j j j j j j

2

3 2

4 2 2

5 4

1Also Important

11 1

11

1 1 1 1( )

1 1 11

( 1)( 1).

1 1 1

...

i ii

i i i

i

i ii ii i i

i i i

iii i i

We need to be careful when working with imaginary numbers:

Consider the problem 9 4 . We know that 9 3 and 4 2j j , so

29 4 3 2 6 6j j j .

But we got used to the property a b ab . If we use it in this problem, we get

9 4 36 6 . If our new set of numbers is going to be successful, we cannot get two different answers when simplifying. The rule is that whenever you work with square roots of negative numbers, express each number as a complex number in i or j form before you proceed with any calculations.

Therefore the correct answer to the above problem is 29 4 3 2 6 6j j j .


2a) 4

b) 3 12

c) 2 8




93

2. COMPLEX NUMBERS

When an imaginary number and a real number are added, we get a complex number. A complex number is of the form a jb , where a and b are real numbers. When a = 0 and

0b , we have a pure imaginary number. When b = 0, we get a number of the form a, which is a real number. Definition Rectangular or Cartesian form of a complex number The form a bj is known as the rectangular form of a complex number, where a is the real part and b is the imaginary part.

2.1 Equality of complex numbers

Two complex numbers are equal if both the real parts are equal and the imaginary parts are equal. Thus

If a bj = c dj then a = c and b = d

2.2 Conjugates of complex numbers

Every complex number has a conjugate. This is particularly useful when you are dividing by a complex number, as you will see in unit 2. The conjugate of a complex number a bj is the complex number a bj .

To form the conjugate of a complex number you need to change only the sign of the imaginary part of the complex number. For example, the conjugate of 2 + 3j is 2 – 3j and the conjugate of 2 – 3j is 2 + 3j. Thus each number is the conjugate of the other.

ACTIVITY 3 1. Write down the conjugates of the following complex numbers: a) 3 4

b) 5 2

c) 7j

d) 12

j

j

2. Solve the following quadratic equations: 2

2

2

a) 4 0

b) 2 3 5 0

c) 2 10 0

x

x x

x x




94

3. ARGAND DIAGRAM

A complex number may be represented pictorially on rectangular or Cartesian axes. The horizontal (or x-axis) is used to represent the real axis and the vertical (or y-axis) is used to represent the imaginary axis. This diagram is called an Argand diagram. In figure 1 the following complex numbers are represented:

A 3 2 ; B 2 3 ; C 4 3 and D 4 5j j j j

FIGURE 1

ACTIVITY 4 Represent the following complex numbers on an Argand diagram:

A 2 3 ; B 4 2 and C 1 6j j j




95


4.1 Activity 1

a) 9 9 1 3

b) 0.25 0.25 1 0.5

c) 3 3 1 3 or 3

d) 18 9 2 1 3 2 or 3 2

j

j

j j

j j

Many people prefer to write the symbol j in front of the radical sign to reduce the danger of thinking it is under the radical sign.

4.2 Activity 2

2 2

2

22

a) 4 2

4

4

b) 3 12 3 12

3 2 3

2 3

2( 1)(3)

6

c) 2 8 2 8

16

4

j

j

j j

j j

j

j

j

j

4.3 Activity 3

1. a) 3 4

b) 5 2

c) 7j

d) 12

j

j

Since 12 = 12 0 j and the conjugate is 12 0 j 2. 2

2

a) 4 0

4

4

2

x

x

x

j



96

2

2

b) 2 3 5 0

3 (3) 4(2)(5)

2(2)

3 31

4

3 31

4 40.750 1.392

x x

x

j

j

2c) 2 10 0

2 4 4(1)(10)

2

2 36

22 6

2 21 3

x x

x

j

j

4.4 Activity 4



97

You have now completed learning unit 1 on imaginary and complex numbers and you

should be able to express an imaginary number in standard form find the complex conjugate of a complex number represent a complex number on an Argand diagram solve any quadratic equation We now move on to learning unit 2 on operations with complex numbers.


98

COMPLEX NUMBERS Operations with complex

numbers

CONTENTS PAGE

1. ADDITION AND SUBTRACTION ..................................................................... 99 2. MULTIPLICATION ............................................................................................. 99 3. DIVISION ........................................................................................................... 100 4. COMPLEX EQUATIONS ................................................................................. 102 5. RESPONSES TO ACTIVITIES .......................................................................... 102 5.1 Activity 1 ............................................................................................................. 102 5.2 Activity 2 ............................................................................................................. 103 5.3 Activity 3 ............................................................................................................. 104 5.4 Activity 4 ............................................................................................................. 106

MODULE 4

LEARNING UNIT 2

OUTCOMES

At the end of this learning unit, you should be able to add, subtract, multiply and divide complex numbers solve complex equations

Module 4 Learning unit 2 COMPLEX NUMBERS: Operations with complex numbers


99

1. ADDITION AND SUBTRACTION

Two complex numbers are added/subtracted by adding/subtracting separately the two real parts and the two imaginary parts.

Example 1

Simplify and write in the form a bj :

a) 2 3 3 4j j

b) 2 3 3 4j j

Solution a) 2 3 3 4 2 3 3 4 5j j j j j

b) 2 3 3 4 2 3 3 4 1 7j j j j j .

ACTIVITY 1 1. Find the following sums:

a) 9 2 8 6

b) 6 3 5 7

c) 4 3 1 4

j j

j j

j

2. Find the following differences:

a) 3 4 2

b) 5 7 3 10

c) 8 4 3 10

j j

j j

j j


2. MULTIPLICATION

Multiplication of complex numbers is achieved by assuming that all quantities involved

are real and then using 2 1j to simplify.

Example 2

Simplify 3 2 4 5j j .

Solution

23 2 4 5 12 15 8 10

12 7 10( 1)

12 10 7

22 7

j j j j j

j

j

j



100

ACTIVITY 2 Multiply and write the answer in the form a bj :

2

a) 2 5 3 4

b) 5 3

c) 7 3 7 3

d)

j j

j

j j

a bj a bj


3. DIVISION

Division of complex numbers is achieved by multiplying both numerator and denominator by the complex conjugate of the denominator. This results in a real denominator.

Example 3

Divide and express each answer in the form a bj : 10 4

a)1

8 6b)

2

3 2c)

4 2

0.5 3d)

2.4

j

j

j

j

j

j

j

j

Solution

2

10 4 10 4 1a) .

1 1 1

10 10 4 4

1 16 14

23 7

j j j

j j j

j j j

j

j

2

8 6 8 6 2b) .

2 2 2

16 8 12 6

4 110 20

52 4

j j j

j j j

j j j

j

j



101

2

3 2 3 2 4 2c) .

4 2 4 2 4 2

12 6 8 4

16 48 14

208 14

20 200, 4 0,7

j j j

j j j

j j j

j

j

j

0,5 3 0,5 3 2.4d) .

2.4 2.4 2.4

0,5( 2.4 ) 3 2.4

5.76

1.2 2.4 3

5.76

2.4 3 1.2

5.76 5.76

3 1

2.4 4.80.72 0.21

j j j

j j j

j j j

j

j

j

ACTIVITY 3 Simplify and write the answers in the form a ib :

a) 4 3

7 2

i

i

b) 1 3 7

1 2 1i i i

c) 3 386

17 1i i i

i

d) 2

1 2 1

1 2

i i

i




102

4. COMPLEX EQUATIONS

To solve equations we use the equality of complex numbers as stated in unit 1. If two complex numbers are equal, then their real parts and their imaginary parts are equal:

If then and .a bj c dj a c b d

Example 4

Solve 4 3 7 2j j x yj for x and y. Solution The best way is to rearrange the terms so that the known values are on one side of the equation and the unknowns x and y are on the other.

4 3 7 2

4 3 7 2

2 4

j j x yj

j j x yj

j x yj

Thus x = 2 and y = 4, since the real parts must be equal and the imaginary parts must be equal.

ACTIVITY 4 1. Solve the following equations for x and y:

a) 2 6 3

b) 2 3 2 3

x jy j

x j y y j x j

2. Solve the following equations for a and b:

a) 2 3

b) 1 2 2 3

j a jb

j j a jb



5.1 Activity 1

1.

a) 9 2 8 6

9 8 2 6

17 8

b) 6 3 5 7

6 5 3 7

11 4

j j

j

j

j j

j

j



103

c) 4 3 1 4

4 3 1 4

4 3 1 2

4 1 3 2

5

j

j j

j j

j

j

2.

a) 3 4 2

3 2 4 1

1 3

b) 5 7 3 10

5 3 7 ( 10)

2 17

c) 8 4 3 10

8 3 4 10

11 6

j j

j

j

j j

j

j

j j

j

j

5.2 Activity 2

2

a) 2 5 3 4

6 8 15 20

6 7 20( 1)

26 7

j j

j j j

j

j

2

2

b) 5 3

5 3 5 3

25 15 15 9

25 9 30

16 30

j

j j

j j j

j

j

2

c) 7 3 7 3

49 21 21 9

49 9

58

j j

j j j



104

2 2

2

2 2

d)

( 1)

a bj a bj

a abj abj bj

a b

a b

Note in answers c) and d) that the product of a complex number and its conjugate is a real number. This is always true.

5.3 Activity 3

a) 4 3 7 24 3

7 2 7 2 7 2

i ii

i i i

2

2

28 8 21 6

49 14 14 428 8 21 6

49 422 29

5322 29

53 53

i i i

i i ii i

i

i

b) Simplify each, then combine the answers to obtain a final answer:

Term 1:

1 1

1 1 1

1

1 11 1

2 2

i

i i i

i

i

Term 2:

3 23

2 2 2

6 3

4 16 3

5 5

i

i i i

i

i

Term 3:

7 17

1 1 1

7 7

1 17 7

2 2

i

i i i

i

i



105

Thus 1 3 7 1 1 6 3 7 7

1 2 1 2 2 5 5 2 2i i i

i i i

1 1 6 3 7 7

2 2 5 5 2 21 6 7 1 3 7

2 5 2 2 5 2

4 22 2

5 514 12

5 5

i i i

i

i

i

c) Term 1: 8 4 4

1 1

1

i i i

Term 2: 6

1 1

11

i

Term 3: 3 3 3

3

1i i

i

i

Term 4:

3 2 37 1 7 1 3 3

7 1 3 3

14 14

i i i i

i i

i

Thus 3 386

17 1 1 1 14 14

16 15

i i i i ii

i

d)

2

2 2

1 2 1 1 2 2

1 4 41 2

i i i i i

i ii



106

2

2 2

1 2 2

1 4 43

3 43 3 4

3 4 3 4

9 12 3 4

3 49 12 3 4

9 1613 9

25 25

i i

ii

ii i

i i

i i i

i i

i

5.4 Activity 4

1. a) 2 6 3

2 2 6 3

2 6 and 2 3

33

2

x jy j

x j y j

x y

x y

b) 2 3 2 3

2 3 2 3

2 and (1)

2 3 3 (2)

We now have two simultaneous equations to solve:

Eq. (1) 2 : 2 2 4 (3)

(2) (3) 7

7

Substitute in (1) to obtain

7 2

9

x j y y j x j

x y j y x j

x y

y x

x y

x

x

y

y

2.

2

2

a) 2 3

2 3

2 3 2 3

4 12 9

5 12

5 and b 12

j a jb

j a jb

j j a jb

j j a jb

j a jb

a



107

2

b) 1 2 2 3

2 4 3 6

4 7

4 and 7

j j a jb

j j j a jb

j a jb

a b

Now that you have completed learning unit 2, you should be able to add, subtract, multiply and divide complex numbers solve complex equations Now move on to learning unit 3: Complex numbers: polar and exponential form.


108

COMPLEX NUMBERS Polar and exponential

form

CONTENTS PAGE

1. POLAR FORM OF A COMPLEX NUMBER ................................................... 109 2. EXPONENTIAL FORM .................................................................................... 112 3. RESPONSES TO ACTIVITIES .......................................................................... 113 3.1 Activity 1 ............................................................................................................. 113 3.2 Activity 2 ............................................................................................................. 115

MODULE 4

LEARNING UNIT 3

OUTCOMES

At the end of this learning unit, you should be able to write a complex number in polar form write a complex number in exponential form determine the modulus and argument of a complex number

Module 4 Learning unit 3 COMPLEX NUMBERS: Polar and exponential form


109

1. POLAR FORM OF A COMPLEX NUMBER

Sometimes it is convenient to express a complex number z = a + jb in a different form.

Let OP

be the complex number z = a + jb on the Argand diagram in figure 1:

FIGURE 1

If is the angle which OP

makes with the x-axis and r is its length, then

2 2 2 2 2 and tan which implies that tan .

Also cos and sin .

b br a b r a b arc

a aa r b r

Since z = a + jb this can be written as cos sin cos sinz r jr r j .

This is called the polar form of the complex number a + jb. Polar form

cos sinz r j where 2 2 and tan .b

r a b arca

The expression cos sinr j is often abbreviated as cis or rr . In the

abbreviation cis r , the c represents cosine, the s represents sine and the i represents the mathematical operator i or j. The symbol r is read “ r at angle ”. We have special names for the values of r and r is called the modulus (or magnitude) of the complex number z and is often

abbreviated to “mod z” or indicated by z

is the argument (or amplitude) of the complex number z and can be abbreviated to “arg z”

Example 1 Determine the modulus and the argument of the complex number z = 4 + j3 and express z in polar form. Start with a diagram as in figure 2.



110

FIGURE 2

2 2

34

Modulus 4 3

25

5

Argument arg arctan 36,87 36 52 '

Thus cos sin

5 cos36 52 ' sin 36 52 '

5 cis 36 52 '

5 36 52 '

z r

z

z r j

j

Example 2 Find the polar form of the complex number z = 2 + j3 Start with an Argand diagram as drawn in figure 3.

FIGURE 3

2 22 3

13

3.606

r

r

3tan

21,5

56.31 56 18'



111

Thus 3.606 cos56 18' sin56 18'

3.606 cis 56 18'

3.606 56 18'

z j

ACTIVITY 1 1. Determine the argument of the complex number z = 3 –j4 2. Express the following complex numbers in polar form: a) 3 + j4 b) 3 + j4 c) 3 –j4 d) 3 –j4 Remember to check the response on page 113.

We have already used the shorthand version r to denote the polar form of a complex

number, where cos sinr r j . However, we sometimes find complex numbers

of the form cos sinr j and then we denote this as r .

It is easy to remember:

resembles the first quadrant and indicates measuring angles in the positive direction while

resembles the fourth quadrant and indicates measuring angles in the negative direction.

Example 3

Thus 3 cos 45 sin 45 3 45 3 45

and 2 cos 150° sin150 2150 2 210

j

j

Convert from polar form to rectangular form

A complex number written in polar form as or cis or cos sinr r r j has

the rectangular coordinates a + bj, where cos and sina r b r To convert is thus a simple matter of substitution.

Example 4

Convert 5(cos35 sin35 )j to rectangular form. cos 5cos35 4.096

sin 5sin 35 2.868

Thus 5(cos35 sin 35 ) 4.096 2.868

a r

b r

j j



112

2. EXPONENTIAL FORM

There is yet another way of expressing a complex number. It is called the exponential form because it involves exponents of the number e. Exponential form

The exponential form of a complex number uses Euler’s formula, cos sinje j , and states that

jz re where is in radians.

Thus cos sin can now be written as .jz r j z re The exponential form can be obtained from the polar form easily since the r value and the angle are the same for both. It is important to note that in the exponential form, the angle must be in radians.

Example 5

Change the polar form 5 cos60 sin60j into exponential form.

We need to change the angle to radians first: 60 60180

3

Thus 35 cos 60 sin 60 5j

j e

.

ACTIVITY 2

1. Express 6 cos180 sin180j in exponential form.

2. Write 8 225 in exponential form. 3. Express –4 + 3j in exponential form.

4. Express 5.74.6 je in polar and rectangular form. Remember to check the response on page 115.



113


3.1 Activity 1

1. Start with a diagram as shown in figure 4.

Figure 4

is measured from OX to OP. We first find the acute reference angle E from the diagram.

4tan

31.333

53.13 53 7 '

Because is in the third quadrant: 180

233 7'

Thus arg 233 7 '

E

E

E

z

2. We will use figure 5 to answer this question.

a) 2 23 4 25 5

4tan 1.333

353.13 53 7 '

Thus 3 4 5 53 7 '

r

j



114

FIGURE 5

b) 2 2( 3) 4 25 5

4tan 1.333

353.13 53 7 '

Because is in the second quadrant =180° 53 7 ' 126 53'

Thus 3 4 5126 53'

r

E

E

j

c) Refer to question 1.

By convention we use the principal value for the argument that is numerically the least value, such that .

2 2( 3) ( 4) 25 5

4tan 1.333

353.13 53 7 '

Because is in the third quadrant = 233°7' = 126 53'

Thus 3 4 5 126 53'

r

E

E

j



115

d) 2 2(3) ( 4) 25 5

4tan 1.333

353.13 53 7 '

Because is in the fourth quadrant = 53 7

Thus 3 4 5 53 7 '

r

E

E

j

By convention we use the principal value that is numerically the least value, such that .

3.2 Activity 2

1. r = 6 = 180° = radians

Thus 6 cos180 sin180 6 jj e

2. r = 8

54225 3,927 rad

Thus 5

3,92748 225 8 8j

je e

Both of these answers are correct. Because you will probably be using a calculator to convert from degrees to radians, the last version is the one that you will most likely use. But remember it is rounded off and is therefore the least accurate.

3. We must first determine r and .

2( 4) 3 25 5

3tan

4

r

Use your calculator in radian mode to find the value of in radians.

2.498

R ef = 0.644

is in the second quadrant

0.644 2.498

Thus 4 3 5 jj e

4. r = 4.6 = 5.7

Polar form: 5.74.6 je = 4.6(cos5.7 + j sin5.7) Rectangular form:

With the calculator in radian mode we get 4.6cos5.7 3.840

4.6sin5.7 2.533

a

b

and

5.74.6 je = 3.840 j 2.533



116

You have reached the end of this learning unit, so you should be able to write a complex number in polar form write a complex number in exponential form determine the modulus and argument of a complex number We next examine learning unit 4: Complex numbers: operations in polar and exponential form.


117

COMPLEX NUMBERS Operations in polar and exponential form

CONTENTS PAGE

1. ADDITION AND SUBTRACTION IN POLAR AND EXPONENTIAL FORM ... ............................................................................................................................ 118 2. MULTIPLYING AND DIVIDING IN EXPONENTIAL FORM ...................... 118 3. MULTIPLYING AND DIVIDING IN POLAR FORM ..................................... 120 3.1 Multiplication ...................................................................................................... 120 3.2 Division ............................................................................................................... 121 4. POWERS, ROOTS AND DEMOIVRE’S FORMULA ................................... 122 5. RESPONSES TO ACTIVITIES .......................................................................... 127 5.1 Activity 1 ............................................................................................................. 127 5.2 Activity 2 ............................................................................................................. 127 5.3 Activity 3 ............................................................................................................. 128 5.4 Activity 4 ............................................................................................................. 128 5.5 Activity 5 ............................................................................................................. 129

MODULE 4

LEARNING UNIT 4

OUTCOMES

At the end of this learning unit, you should be able to add, subtract, multiply and divide complex numbers in exponential and polar form raise a complex number to a power use DeMoivre’s theorem to find the nth roots of a complex number

Module 4 Learning unit 4 COMPLEX NUMBERS: Operations in polar and exponential form


118

1. ADDITION AND SUBTRACTION IN POLAR AND EXPONENTIAL FORM

Addition and subtraction in polar or exponential form is not possible directly. Each complex number must be converted into rectangular form first.

Example 1

Evaluate in polar form 2 30 5 45 4120 . Convert all numbers to rectangular form:

2 30 2cos30 2sin 30 1.732

5 45 5cos( 45 ) 5sin 45 3.536 3.536

4 120 4cos120 4sin120 2 3.464

j j

j j

j j

Arrange numbers so that they can be added easily: 1.732

3.536 3.536

2 3.464

7.268 6

Thus 2 30 5 45 4 120 7.268 6

j

j

j

j

j

Convert the answer in polar form:

22(7.268) 6

88.824

9.425

6tan

7268

Ref = 39.54°

is in the fourth quadrant

39.4°

Thus 2 30 5 45 4 120 7,.68 6 9.425 39.54°

r

j

2. MULTIPLYING AND DIVIDING IN EXPONENTIAL FORM

One advantage of using the exponential form for complex numbers is that complex numbers written in exponential form obey the laws of exponents. We will use the basic rules for multiplication, division and powers:



119

m n m n

mm n

n

nm mn

m m m

a a a

aa

a

a a

ab a b

If we have two complex numbers 11 1

jz r e and 22 2

jz r e , we can multiply, divide or

take powers of them using these rules:

1 2 1 2

1

1 2

2

1 1

( )1 2 1 2 1 2

( )1 1 1

2 22

1 1 1

j j j

jj

j

nn j jnn

z z r e r e r r e

z r e re

z rr e

z r e r e

Example 2

a) Multiply 4.27 je by 2.52 je .

b) Divide 3.29 je by 4.32 je .

c) Calculate 52.34 je .

Solution

a) 4. 2.5 (4.2 2.5) 6.77 2 (7)(2) 14j j j je e e e

b) 3.2

(3.2 4.3) 1.1924.3

94.5

2

jj j

j

ee e

e

c) 52.3 5 5(2.3) 11.54 4 1024j j je e e

ACTIVITY 1 Perform the indicated operations:

a) 3 22 6j je e

b) 38 2j je e

c) 3.4 5.38.5 2j je e

d) 423 je

e) 1

6 416 je




120

3. MULTIPLYING AND DIVIDING IN POLAR FORM

Remember the relationship between the exponential and the polar forms:

cos sinjre r j

3.1 Multiplication

Let 11 1

jz r e and 22 2

jz r e be two complex numbers; then we know that

1 2( )1 2 1 2

jz z r r e .

In polar form this would be 1 1 1 2 2 2cos sin . cos sinr j r j .

Using trigonometric identities as in module 6 we can show that this is equal to

1 2 1 2 1 2cos sinr r j .

Using the alternative way of writing, we have

1 2 1 1 2 2 1 2 1 2z z r r r r

Notice that the angles can be either in radians or degrees when working in polar form. Remember to multiply two complex numbers in polar form: Multiply the moduli (r’s) Add the arguments (’s)

Example 3

a) 2 cos30 sin 30 3 cos 40 sin 40

2 3 cos 30 40 sin 30 40

6 cos70 sin 70

b) 4 cos 20 sin 20 3 cos70 sin 70

12 cos90 sin 90

12 sin 90

12

j j

j

j

j j

j

j

j

ACTIVITY 2 Determine:

2

a) cos sin cos sin

b) 3 cos30 sin 30

a j b j

j




121

3.2 Division

Let 11 1

jz r e and 22 2

jz r e be two complex numbers; then we know that

1 2( )1 1

2 2

jz re

z r .

In polar form this would be

1 1 11

2 2 2 2

cos sin

cos sin

r jz

z r j

.

Using trigonometric identities we can show that this is equal to

11 2 1 2

2

cos sinr

jr

.

Using the alternative way of writing, we have

1 11 11 2

2 2 2 2

rz r

z r r

Notice that the angles can be either in radians or degrees when working in polar form. Remember to divide two complex numbers in polar form: Divide the moduli (r’s) Subtract the arguments (’s)

Example 4

6 cos 78 sin 78

3 cos30 sin 30

2 cos 78 30 sin 78 30

2 cos 48 sin 48

2 48

j

j

j

j

ACTIVITY 3 Calculate:

15135°a)

3 75

3 cos 20 sin 20 5 cos10 sin10b)

2 cos30 sin 30

j j

j




122

4. POWERS, ROOTS AND DEMOIVRE’S FORMULA

Finding powers of complex numbers in polar form also uses the same process we developed for powers in exponential form. DeMoivre’s formula

For any complex number cis jz r r re :

or cos sin cos sin

nn j n jn

n n

z re r e

r j r n j n

This theorem is true for all positive, negative and fractional values of n, and is thus useful for determining powers and roots of complex numbers. DeMoivre’s theorem says that to raise a complex number in polar form to any power n raise the modulus r to the power n multiply the argument by n

Example 5 Simplify

a) 44 cos 30 sin 30i

b) 1

24 cos 60 sin 60i

c)

4

5

cos3 sin 3

cos sin

j

j

d) 93 j

Solution

a)

4

4

4 cos30 sin 30

4 cos120 sin120

256cis120

i

i

b)

12

124 cos 60 sin 60

4 cos30 sin 30

2 cis 30

i

i



123

c)

4

5

cos3 sin 3

cos sin

cos 4 3 sin 4 3

cos 5 sin 5

cos12 sin12

cos5 sin 5

cos 7 sin 7

j

j

j

j

j

j

j

d) Write 3 j in polar form and then apply DeMoivre’s theorem.

9 9

9

9

9

9

3 1 2

1tan

3

Ref 30

150 ( is in the second quadrant)

3 2 150

3 2 150

2 1 350

2 cos1 350 sin1 350

2 0 1

2

512

r

j

j

j

j

j

j

3



124

ACTIVITY 4

Express

3 8

4

1 3 1

2 2

j j

j

in the form a + jb.


Finding the roots of a complex number

DeMoivre’s formula can be used to help find all of the roots of a complex number. For

example, the equation 3 1z has three roots. One of the roots is –1. What are the other two? DeMoivre’s formula can be used to find the other two roots.

First write –1 in polar form: 1 1 cos180 sin180j

Using DeMoivre’s formula with 13n , we get

1133

180 1801 1 cos sin

3 3

1 cos60 sin 60

1 3

2 2

0,5 0,866

j

j

j

j

We can see that this is different from –1. You can check that 1

30,5 0,866 1j . So

this is a valid answer. Why did we get a different answer? If you divide any number between 0° and 1080° (0 and 6) by 3, we find an answer between 0° and 360° (0 and 2). Now 180°, 540° and 900° can all be represented by the same terminal side on an Argand diagram. So we could have written –1 as 1 cos540 sin 540j or as 1 cos900 sin 900j . Let’s find the cube root of these numbers:

1133

540 5401 1 cos sin

3 3

1 cos180 sin180

1

j

j

and

1133

900 9001 1 cos sin

3 3

1 cos300 sin 300

0,5 0,866

j

j

j



125

Now we have found three different cube roots of –1. The first and last are conjugates of each other. We have also given an example of a process we can use to find all of the nth roots of a number, where n is a positive integer. Roots of a complex number

If cos sinz r j , then the nth roots of z are given by the formula

360 . 360 .cos sin

where 0,1, 2,..., 1.

nk

k kw r j

n n n n

k n

If is in radians, substitute 2 for 360°.

Example 6 Find the cube root of z = 8(cos120° + jsin120°). The three roots will be called 0 1 2, and w w w .

Using the formula above we find

30

120 360 .0 120 360 .08 cos sin

3 3 3 3

2 cos40 sin 40

1.532 1.286

w j

j

j

31

120 360 .1 120 360 .18 cos sin

3 3 3 3

2 cos160 sin160

1.879 0.684

w j

j

j

32

120 360 .2 120 360 .28 cos sin

3 3 3 3

2 cos280 sin 280

0.347 1.970

w j

j

j

The three roots are shown in figure 1. Notice that the roots are equally spaced around the diagram. Whenever you graph the nth roots of a number, they should be equally spaced around the diagram. If they are not, you have made a calculation error.



126

FIGURE 1

Although there are different roots of a complex number, we are sometimes asked to find the principal root. Principal root This is the root whose vector is the nearest to the positive x-axis. In some cases it may be the first root and in other cases the last root. In example 6 it is the first root w0.

Example 7 Solve for x if x3 + 1 = 0. Give the answer in polar form with the angles in radians. Solution

3

13 3

1

1 1

x

x

We need to find the cube roots of –1 to solve for x. Write –1 as a complex number in polar form.

1 1 cos sinj

Thus 30

2 .0 2 .01 cos sin

3 3 3 3

1 cos sin3 3

13

x j

j



127

31

32

2 .1 2 .11 cos sin

3 3 3 3

1 cos sin

1

2 .2 2 .21 cos sin

3 3 3 3

5 51 cos sin

3 3

51

3

5Thus the values of are 1 , 1 and 1 .

3 3

x j

j

x j

j

x

ACTIVITY 5 1. Find the cube roots of z = 5(cos225° + jsin225°). 2. Solve for x if x4 + 1 = 0. Give the answer in the form a + jb. Remember to check the response on page 129.


5.1 Activity 1

a) 3 2 3 2 52 .6 (2.6) 12j j j j je e e e

b) 3 3 2828 2 4j j j j je e e e

c) 3.4 5.3 3.4 5.3 1.98.528,5 2 4.25j j j j je e e e

d) 42 4 2.4 83 3 81j j je e e

e) 1 11

66 1,54 4416 16 2jj je e e

5.2 Activity 2

a) cos sin cos sin

cos sin

a j b j

ab j

2b) 3 cos30 sin 30

3 cos30 sin 30 3 cos30 sin 30

9 cos 60 sin 60

j

j j

j



128

5.3 Activity 3

15135° 15a) 135 75 5 60

3 75 3

3 cos 20 sin 20 5 cos10 sin10b)

2 cos30 sin 30

15 cos30 sin 30

2 cos30 sin 30

15

2

7,5

j j

j

j

j

5.4 Activity 4

Write all the complex numbers in polar form.

31 3 : 3 1 and tan

1

2 Ref. 60

300

Thus 1 3 2 300

j r

j

11 : 1 1 and tan

1

2 45

Thus 1 2 45

j r

j

2 2 22 2 : ( 2) (2) and tan

2

8 ( 1)

2 2 Ref = 45°

135

Thus 2 2 2 2 135

j r

j

We are now ready to simplify the given expression:



129

12

32

3 88 3

4 4

1 3 1 2 300 2 45

2 2 2 135

8 900 16 360

64 540

2 720

2 0

2cis 0

2

j j

j

5.5 Activity 5

1.

30

225 2255 cos sin

3 3

1,71 cos75 sin 75

1,71 75

0.443 1.652

w j

j

j

31

225 360 225 3605 cos sin

3 3 3 3

1,71 cos195 sin195

1,71 195

1.652 0.443

w j

j

j

32

225 360 .2 225 360 .25 cos sins

3 3 3 3

1.71 cos315 sin 315

1.71 315

1.209 1.209

w j

j

j

Represented on a Argand diagram as seen in figure 2.



130

Figure 2

2.

4

40

1 Let 1 1 cos sin

2 .0 2 .01 cos sin

4 4 4 4

1 cos sin4 4

1 1 11

2 2 2

x j

x j

j

j j

41

2 .1 2 .11 cos sin

4 4 4 4

3 31 cos sin

4 4

1 1 11

2 2 2

x j

j

j j

42

2 .2 2 .21 cos sin

4 4 4 4

5 51 cos sin

4 4

1 1 11

2 2 2

x j

j

j j



131

43

2 .3 2 .31 cos sin

4 4 4 4

7 71 cos sin

4 4

1 1 11

2 2 2

1Thus the values of are 1

2

x j

j

j j

x j


132

POST-TEST: COMPLEX NUMBERS 1. Write down the conjugates of the following numbers:

a) 2 5

b) 3 2

c) 5

j

j

j

2. Write the following numbers in polar form:

a) 5 2

b) 4 3

j

j

3. Represent the following numbers on an Argand diagram: A 4 2 ; B 3 3 and C 2 5j j j

4. Express in the form a ib :

8 1a)

24 7

b)5 61 2

c)13 5 2 6

d)2 4

4e)

1 3 2

i

i

ii

ij j

j

j

j j

5. Add the following numbers: 2 cos30 sin 30 ; 5 cos50 sin 50 and 2 cos100 sin100j j j

6. Find the solutions to the quadratic equation 3x2 4x + 5 = 0. 7. Simplify the following:

a) 2 3 4 5 6

b) 9 18 6 2

j j

c) 1 2 4 3i i

8. If x and y are real, solve the equations:

a) 2 3 3 4j j x jy

b) 3 4

1 3

jx x j

jy x y

Module 4 TRIGONOMETRY: Post-test


133

9. If 22 5 2 3a b j a b j j j , find the values of a and b.

10. Express –5 – 8j in exponential form. 11. Calculate the following:

5.4 6.1

4.3 2.8

12.1 3

a) 3.6 2.5

b) 17 4

c) 1.728

j j

j j

j

e e

e e

e

12. Find the roots of the equation 5 1 0x . Which is the principal root? Give the

answer in polar form with the angles in radians.

13. Prove that

1cos sin

cos sini

i

.

14. Solve for x and y if 2

1 ix iy

i

.

15. Find all fourth roots of 8 8 3i . Write your answer in the form a + jb.

16. Simplify

43

9

1 1 3

1

j j

j

and write in the form a + jb.


134

COMPLEX NUMBERS: POST-TEST SOLUTIONS 1. a) 2 5

b) 3 2

c) 5

j

j

j

2. 225 4a) tan5

29 180 21.8

158.229 cis158.2

r

z

b)

316 9 tan4

5180 36.9

216.95 cis 143.1

r

z

3.

4.a) 8 1 1 1

2 2 21 1i

Module 4 COMPLEX NUMBERS: Post-test solutions


135

b) 4 7 5 6 20 42 11

5 6 5 6 25 3662 11

61

i i i

i ii

c) 1 2 1 3

1 1 2

i i i

i i

d) 3 5 2 6 24 28 2 4

2 4 2 4 2 4

48 112 56 96

4 16160 40

208 2

j j j j

j j j

j j

j

j

e) 4 5 19 9

5 5 25 1

19 9

26

j j j

j j

j

5. Write the numbers in rectangular form.

2 cis 30 1.73

5 cis 50 3.2 3.83

2 cis100 0.35 1.97

Sum 4.28 6.8

i

i

i

i

6. 4 16 60 4 44

6 6 62 1

113 3

2 11

3 3

x

i

The solutions are thus 2 11 2 11

3 3 3 3 and i i .

7.

a) 2 3 4 5 6

2 3 5 4 6

5 2 3 2

j j

j

j

Note that the real part of this complex number is 5 2 3 and the imaginary part is 2 j .


MAT1581

Mathematics 1 (Engineering)

136

b) 9 18 6 2

9 3 2 6 2

9 6 3 2 2

3 2 2

j j

j

j

c) 21 2 4 3 4 3 8 6

4 11 6

2 11

i i i i i

i

i

8. a)

2

2 3 3 4 ,

6 8 9 12

6 ( 1)12

18

18

1

j j x jy

j j j x jy

j x jy

j x jy

x

y

b) 3 4

1 3

jx x j

jy x y

2 2

2

2

cross multiply:

3 1 3 4

3 3 4 3 4

3 4 4 3

Thus 0 3 4 @ and

3 4 3

4

2

Substitute in @: 4 3

6

jx x y jy x j

j x xy x j j xy j y

x y j xy

x y

x xy xy

x

x

y x

y

9. 22 5 2 3a b j a b j j j

2 24 20 25 2 3

4 22 22

18 22

Thus

18

22

2 and 20

j j j j

j

j

a b

a b

a b



137

10. 2 22 2 5 8r a b

85

4.154

25 64

89

9.434

tan 1.6

R ef 1.012

(Remember your answer must be in radians)

is in the third quadrant:

1.012 4.154

Thus 5 8 9.434 jj e

11.

5.4 6.1 11.5

4.3 2.8 1.5

12.1 0.73

a) 3.6 2.5 9

b) 17 4 4.25

c) 1.728 1.2

j j j

j j j

j j

e e e

e e e

e e

12. 5

5

1

1 Let 1 cos0 sin 0

x

x j

50

51

0 2 .0 0 2 .01 cos sin

5 5 5 5

1 cos0 sin 0

1

0 2 .1 0 2 .11 cos sin

5 5 5 5

2 21 cos sin

5 5

x j

j

x j

j

52

0 2 .2 0 2 .21 cos sin

5 5 5 5

4 41 cos sin

5 5

x j

j


MAT1581


138

53

54

0 2 .3 0 2 .31 cos sin

5 5 5 5

6 61 cos sin

5 5

0 2 .4 0 2 .41 cos sin

5 5 5 5

8 81 cos sin

5 5

x j

j

x j

j

xo = 1 is the principal root.

13.

2 2

cos sin1Left-hand side=

cos sin cos sin

cos sin

cos sin

cos sin

Right-hand side

i

i i

i

i

14. 2

2

1

1 2

1

2

0 2

ix iy

i

i i

i

x y

15. Write 8 8 3z i in polar form.

o o

8 364 64 3 16 tan 3

8

180 60

240

8 8 3 16 cis 240

r

z i



139

40

240 360 .0 240 360 .016 cos sin

4 4 4 4

2 cos60 sin 60

1 3

w j

j

j

41

42

43

240 360 .1 240 360 .116 cos sin

4 4 4 4

2 cos150 sin150

3

240 360 .2 240 360 .216 cos sin

4 4 4 4

2 cos240 sin 240

1 3

240 360 .3 24016 cos sin

4 4

w j

j

j

w j

j

j

w j

360 .3

4 4

2 cos330 sin 330

3

j

i

16.

4 33 4

9 9

3 4

9

1 1 3 2 cis 45 2 cis 300

1 2 cis 315

2 cis 3(45 ) 2 cis 4 300

2 cis 9 315

j j

j

2 2 cis 135 16 cis 1200

16. 2 cis 2835

2 cis 1500

2 cis 60

2cos 60 2 sin 60

1 3

i

i


MAT1581


140

At the end of this learning unit 4 on complex numbers - operations in polar and

exponential form, you should be able to

add, subtract, multiply and divide complex numbers in exponential and polar form raise a complex number to a power use DeMoivre’s theorem to find the nth roots of a complex number

We will next engage with module 5. The first learning unit is on the straight line.

MAT1581 Mathematics I (ENGINEERING 141

M O D U L E 5

ANALYTIC GEOMETRY CONTENTS PAGE

LEARNING UNIT 1 THE STRAIGHT LINE 143

1. INTRODUCTION ................................................................................................. 144 2. THE EQUATION OF A STRAIGHT LINE ......................................................... 144 2.1 Summary of possible cases ................................................................................ 145 2.2 Finding the equation of a straight line given the slope and the coordinates of one

point ....................................................................................................................... 147 2.3 Finding the equation of a line passing through two given points ...................... 147 2.4 Finding the angle between two straight lines whose equations are given ......... 148 2.4.1 Parallel lines ...................................................................................................... 149 2.4.2 Perpendicular lines ............................................................................................ 150 3. THE SKETCH GRAPH OF A STRAIGHT LINE ................................................ 151 3.1 Table method ..................................................................................................... 151 3.2 Intercept method ................................................................................................ 152 3.2 Intercept-gradient method ................................................................................. 153 4. RESPONSES TO ACTIVITIES ............................................................................ 154 4.1 Activity 1 ........................................................................................................... 154 4.2 Activity 2 ........................................................................................................... 155 4.3 Activity 3 ........................................................................................................... 155

LEARNING UNIT 2 THE PARABOLA 157

1. QUADRATIC FUNCTIONS ................................................................................. 158 2. THE PARABOLA ................................................................................................. 158 3. RESPONSES TO ACTIVITIES ............................................................................ 163 3.1 Activity 1 ............................................................................................................. 163 3.2 Activity 2 ............................................................................................................. 164 3.3 Activity 3 ............................................................................................................. 164

LEARNING UNIT 3 THE RECTANGULAR HYPERBOLA 166

1. THE GRAPH OF THE HYPERBOLA xy = k ...................................................... 167 1.1 If k > 0, that is k is positive ................................................................................. 167 1.2 If k < 0, that is k is negative ................................................................................ 168 1.3 Determination of the equation of a hyperbola ..................................................... 169 2. RESPONSES TO ACTIVITY ............................................................................... 170

MAT1581 MATHEMATICS I ( ENGINEERING)

142

M O D U L E 5

LEARNING UNIT 4 THE CIRCLE 172

1. THE EQUATION OF A CIRCLE WITH CENTRE AT THE ORIGIN AND

RADIUS EQUAL TO r .......................................................................................... 173 2. FINDING THE EQUATION OF A CIRCLE WITH CENTRE NOT AT THE

ORIGIN AND RADIUS EQUAL TO r ................................................................. 174 3. RESPONSES TO ACTIVITY ............................................................................... 176

LEARNING UNIT 6 THE ELLIPSE 177

1. THE ELLIPSE 2 2

2 21

x y

a b .................................................................................. 178

2. THE EQUATION OF THE ELLIPSE WITH FOCI ON THE X- or Y-AXIS ...... 178 3. THE EQUATION OF THE ELLIPSE OF WHICH THE FOCI ARE NOT ON THE

X- or Y-AXIS ......................................................................................................... 182 4. RESPONSES TO ACTIVITY ............................................................................... 184

LEARNING UNIT 7 THE CENTRAL HYPERBOLA 187

1. THE GRAPH OF THE CENTRAL HYPERBOLA .............................................. 188 2. RESPONSES TO ACTIVITY ............................................................................... 191

POST-TEST 192


MAT1581 Mathematics I (ENGINEERING)

143

ANALYTIC GEOMETRY The straight line

CONTENTS PAGE

1. INTRODUCTION ................................................................................................. 144 2. THE EQUATION OF A STRAIGHT LINE ......................................................... 144 2.1 Summary of possible cases ................................................................................ 145 2.2 Finding the equation of a straight line given the slope and the coordinates of one

point ....................................................................................................................... 147 2.2 Finding the equation of a line passing through two given points ...................... 147 2.4 Finding the angle between two straight lines whose equations are given ......... 148 2.4.1 Parallel lines ...................................................................................................... 149 2.4.2 Perpendicular lines ............................................................................................ 150 3. THE SKETCH GRAPH OF A STRAIGHT LINE ................................................ 151 3.1 Table method ..................................................................................................... 152 3.2 Intercept method ................................................................................................ 152 3.2 Intercept-gradient method ................................................................................. 153 4. RESPONSES TO ACTIVITIES ............................................................................ 154 4.1 Activity 1 ........................................................................................................... 154 4.2 Activity 2 ........................................................................................................... 155 4.3 Activity 3 ........................................................................................................... 155

MODULE 5

LEARNING UNIT 1

OUTCOMES

At the end of this learning unit, you should be able to find the equation of a straight line interpret different values of the gradient and y-intercept of a straight line find the angle between two straight lines sketch a straight line

Module 5 Learning unit 1 ANALYTIC GEOMETRY: The straight line


144

1. INTRODUCTION

If any equation of the first degree in x and y is graphed, we obtain a straight line. An equation of the first degree in x and y is one that contains only the first power of x and y. The most general equation of the first degree is 0 where , and Ax By C A B C are constants. We will introduce other forms of the equation in this learning unit.

2. THE EQUATION OF A STRAIGHT LINE

The equation to a straight line is an algebraic statement of the relation that exists between the x and y coordinates of every point on the line. It is important to note that no point outside the line can have its coordinates related in the same way as points on the line. The relation is usually written in the form y = mx + c, where m and c are constants, that is numbers independent of the values of x and y. These constants have a very definite significance:

“m” is called the slope or gradient of the line. “c” is the distance from the origin at which the line crosses the y-axis or the y-intercept.

If an equation to a straight line is not given in the form y = mx + c, it can be changed to this form.

Example 1

Convert the general form of an equation of the first degree 0Ax By C to the form y = mx + c Solution

0Ax By C

By Ax C

Ax Cy

B

A Cx

B B

which is of the form y = mx + c, where and .A C

m x cB B

Example 2

Convert the equation 3 4 2x y to the form y = mx + c. Solution


MAT1581 MATHEMATICS 1 (ENGINEERING)

145

3 4 2

4 2 3

2 3

4

2 3

4 4

3 1

4 2

x y

y x

xy

xy

xy

ACTIVITY 1 Convert the following equations to the form y = mx + c: a) 3 4 2x y

b) 1 1

7 9x y


2.1 Summary of possible cases

By arranging equations of the first degree in the form y = mx + c, we can see at a glance how the graph of the line will lie. In figure 1 all the possible cases are depicted. Lines sloping upward to the right have a positive m. Lines sloping downward to the right have a negativem. A positive value of c of a line indicates that the line cuts the y-axis above the origin, while a negative c indicates that this cutting point is below the origin. A line with c = 0 will pass through the origin. The special cases shown in the figure are AB and CD parallel to the axis of y and another pair EF and GR parallel to the axis of x.

AB and CD are at 90o to X1OX, and tan 90o is (infinity). The m is infinity and therefore we cannot use the m in our calculations. The coordinates of any point in these lines are completely defined if we state the following: I. The equation to the line AB is x = k OR x k = 0. II. The equation to the line EF is y = V OR y V = 0. III. The equation to the line GH is y = S OR y + S = 0.



146

FIGURE 1



147

2.2 Finding the equation of a line given the slope and the coordinates of one point

The equation of a straight line can be written in a more useful form called the point-slope form:

1 1y y m x x

Example 3

Find the equation of the line with slope 0.3 drawn through the point (7, 1.5). Solution

1 1

1 1Let , 7 ,1.5 and we know that 0.3

1.5 0.3 7

1.5 0.3 7

1.5 0.3 2.1

0.3 2.1 1.5

0.3 3.6

y y m x x

x y m

y x

y x

y x

y x

y x

2.3 Finding the equation of a line passing through two given points

To find the equation of a straight line passing through two given points, the following formula for the slope is used:

2 1

2 1

y ym

x x

We can combine this formula with the point-slope formula to obtain

2 11 1

2 1

y yy y x x

x x

Further note that this formula for the slope corresponds with the definition of the tangent function, thus

2 1

2 1

tan , where is the smallest angle between the line and the -axisy y

m xx x

.



148

Example 4

Find the equation of the line passing through the points (2,3) and (1,1.7). Solution

1 1 2 2Let 2,3 = , and 1, 1.7 = ,x y x y .

2 11 1

2 1

1.7 33 2

1 2

3 1.567 2

1.567 3.134 3

1.567 0.134

y yy y x x

x x

y x

y x

y x

y x

ACTIVITY 2 Find the equation a) to the line with slope – 0,5 drawn through the point (–8, – 2.5) b) of the straight line PQ through the two points P (2,1) and Q (8,9) Remember to check the response on page 155.

2.4 Finding the angle between two straight lines whose equations are given

FIGURE 2

Refer to figure 2. Let the equation of AB be 1 1y m x c and the equation of CD

2 2y m x c . Then 1 1 2 2tan and tanm m .

It is well known by most students that the sum of the three angles of a triangle = 180°.

A



149

1 2

2 1

2 1

2 1

2 1

2 1

2 1

Thus 180 180

tan tan

tan tan

1 tan .tan

tan1 .

m m

m m

Two very important results follow from this equation.

2.4.1 Parallel lines

If two straight lines are parallel, then the angle between them must be 0. But tan 0 can only be 0 if 2 1m m .

Thus if two lines are parallel, the slopes are equal. Consider the following:

Lines AB and PQ are parallel to each other. If we use ABm to indicate the slope of AB and

PQm to indicate the slope of PQ, it follows that AB PQm m .

Example 5 Find the equation of the straight line through (–5, –3) drawn parallel to the line 2 3 4 0y x . Solution We must first find the slope of the line 2 3 4 0y x .



150

If 2 3 4 0 then 2 3 4

3 4

2 2

1.5 2

y x y x

y x

y x

1m 1.5

Then we use the formula for the equation of a line, remembering that the slope of the given line is equal to the slope of the new line since the two lines are parallel:

1 1

3 1.5 5

3 1.5 5

1.5 7.5 3

1.5 10.5

y y m x x

y x

y x

y x

y x

2.4.2 Perpendicular lines

If two lines are perpendicular, then the angle between them must be a right angle.

2 12 1 2 1

2 1

Thus 90 and tan tan 90

Now if , then must 0

Therefore implies that 1 . 0 and 11 .

ab

b

m mm m m m

m m

Thus if two lines are perpendicular, the product of their slopes is equal to 1. Consider the following:

Lines AB and PQ are perpendicular to each other. If we use ABm to indicate the slope of

AB and PQm to indicate the slope of PQ, it follows that 1AB PQm m .



151

Example 6

Show that the lines 4x 3y 1 = 0 and 6x + 8y + 1 = 0 are perpendicular to each other. Solution

1 2

1 2

4 3 1 0 6 8 1 0

: !

3 4 1 8 6 1

4 1 6 1

3 3 8 8

4 1 3 1

3 3 4 8

4 3

3 4

4 31

3 4

x y x y

y x y x

xy y x

y x y x

m m

m m

REMEMBER First write each equation in standard form

Thus the two lines are perpendicular.

ACTIVITY 3 a) Find the equation to a line drawn through (3, –7) perpendicular to the line

12 3y x

b) If two straight lines 4 3 1 0 and 8 1 0x y Kx y are perpendicular to each other, find the value of K.

c) Find the equation of the straight line through (–5, –3) drawn parallel to the line 2 3 4 0y x .


3. THE SKETCH GRAPH OF A STRAIGHT LINE

An accurate drawing, generally on graph paper, is called a plot. A less accurate drawing, usually on ordinary paper, is called a sketch. When you are required to submit a sketch, you should not use graph paper. The requirements are that salient points should be included, and the derivation of the salient points shown in the calculations. There are three methods to draw a straight line. You should find the one you are most comfortable with.

3.1 Table method

Choose three x-values. Set up a table with the x- and y-values. Plot the points. Connect the points to form a straight line.



152

Example 7 Sketch the graph of y = 3x. Choose three x-values and set up a table: x -1 0 1 y -3 0 3 Use the three points to sketch the graph:

3.2 Intercept method

Determine the x- and y-intercepts by putting x and y equal to 0 alternatively. Plot the values on a graph and connect to form a straight line.

Example 8 Sketch the graph of y + x + 2 = 0. To determine the y-intercept, put x = 0:

0 2 0

2

y

y

To determine the x-intercept, put y = 0: 0 2 0

2

x

x

Use these two values to draw the graph:



153

3.2 Intercept-gradient method

Write the equation in the form y = mx + c.

2 1

2 1

The gradient

change in -values

change in -values

vertical change

horizontal change

and the intercept

y ym

x x

y

x

c y

To sketch the graph we start at c and then count out the appropriate vertical and horizontal change. This is best seen in an example.

Example 9 Sketch the graph of y + 2x – 3 = 0. Rewrite equation in the standard form: 2 3y x .

Then 2 change in vertical change

m 2 and c 31 change in horizontal change

y

x

Start at the y-intercept (A). As the change in y is negative, count down 2 units and as the change in x is positive, count 1 unit to the right, ending at point B. Connect points A and B to form a straight line.



154


4.1 Activity 1

a) 3 4 2x y

4 2 3

2 3

4

2 3

4 4

3 1

4 2

y x

xy

xy

xy

b) 1 1

7 9

1 1

7 9

1 17 7

7 9

77

9

x y

y x

y x

y x



155

4.2 Activity 2

a)

1 1

2.5 0.5 8

2.5 0.5 8

2.5 0.5 4

0.5 6.5

y y m x x

y x

y x

y x

y x

b) 1 1 2 2Let , 2,1 and , 8,9x y x y

2 11 1

2 1

9 11 2

8 2

81 2

6

1 1.33 2

1.33 2.66 1

1.33 1.66

y yy y x x

x x

y x

y x

y x

y x

y x

4.3 Activity 3

a) Let the equation of the required line be y = mx + c. For two lines to be perpendicular 1 2

12

1

1

2

m m

m

m

Since (3, –7) is a point on the line, the required line is

7 2(3)

1

y mx c

c

c

The equation is y = 2x 1.

b) 4 1

If 4 3 1 0 then3 3

1If 8 1 0 then

8 8

x y y x

KKx y y x



156

For perpendicular lines

1 2 1

41

3 8

4. 3 8

24

6

m m

K

K

K

c) 3

If 2 3 4 0 then 22

y x y x .

For parallel lines 1 23

2m m .

The equation of the required line is 1 1y y m x x

Since (-5; -3) is a point on the line

33 5

2

3thus 3 5

2

3 21

2 2

or 2 3 21 0

y x

y x

y x

y x

This is the end of this learning unit and you should be able to find the equation of a straight line interpret different values of the gradient and y-intercept of a straight line find the angle between two straight lines sketch a straight line

We can now move to the next learning unit on parabolas.

MAT181QE / WIM131UE UNISA ENGINEERING

157

ANALYTIC GEOMETRY The parabola

CONTENTS PAGE

1. QUADRATIC FUNCTIONS ................................................................................. 158 2. THE PARABOLA ................................................................................................. 158 3. RESPONSES TO ACTIVITIES ............................................................................ 163 3.1 Activity 1 ............................................................................................................. 163 3.2 Activity 2 ............................................................................................................. 164 3.3 Activity 3 ............................................................................................................. 164

MODULE 5

LEARNING UNIT 2

OUTCOMES

At the end of this learning unit, you should be able to sketch the graph of a parabola find the coordinates of the vertex of the parabola

Module 5 Learning unit 2 ANALYTIC GEOMETRY: The parabola

MAT1581 MATHEMATICS I (ENGINEERING)

158

1. QUADRATIC FUNCTIONS

Definition A function is a quadratic function if, and only if, it can be written in the form

2y ax bx c , where a, b and c are constants and 0a .

For example, 2 23 2 and 3y x x s t are quadratic functions.

However, 2

1

1g x

x

is not a quadratic function, since it cannot be written in the form

2g x ax bx c .

ACTIVITY 1 State whether or not the function is quadratic:

2

2

2 22

a) 26 3 b) 7

c) 4 d) 6 4 1

1e) f) 2 3 4

2 4

4g) h) 1

2

f x x g x x

g x x h s s

h q f t t t tq

sf s g t t


2. THE PARABOLA

The graph of the quadratic function is called a parabola and has a shape such as the curves in figure 1.

(a) (b)

FIGURE 1 If a > 0, the parabola extends upward indefinitely, and we say that the parabola opens upward or is concave up (figure 1(a)). If a < 0, the parabola opens downward or is concave down (figure 1(b)).


MAT1581 MATHEMATICS (ENGINEERING)

159

Figure 1 shows points labelled vertex. If a > 0, the vertex is the "lowest" point on the parabola. This means that at this point y has a minimum value. By performing algebraic manipulations on ax2 + bx + c we can determine not only this minimum value but also where it occurs:

2 2 2

2 2

2

2

2

2 2

24 4 4

244

2

2 4

2 42 4

Adding and subtracting gives b b ba a a

b b ba aa

b ba a

b ac ba a

y ax bx c ax bx c

y ax bx c

a x x c

y a x c

a x

2

2

2

2

2 2 4

2 4

Since 0 and 0, it follows that has a minimum value when

0, that is, when . The minimum value is .

Thus the vertex is the point ; .

ba

b b ba a a

b ba a

x a y

x x c

c

2

4

This is also the vertex of a parabola that opens downward 0 , but in this case

is the maximum value of see figure 1 b .ba

a

c y

Instead of calculating 2

4 bac to find the minimum or maximum value it is easier to

substitute the value of the x coordinate in the given equation. In summary

2The graph of the quadratic function is a parabola.

1. If > 0, the parabola opens upward. If < 0, it opens downward.

2. The vertex occurs at , .2 2

y f x ax bx c

a a

b bf

a a

We can quickly sketch the graph of a quadratic function by first locating the vertex and a few other points on the graph. Frequently it is convenient to choose these other points to be those where the parabola intersects the x- and y-axes. These are called x- and y-intercepts, respectively.

A y-intercept (0,y) is obtained by setting x = 0 in 2 y ax bx c and solving for y.

The x-intercepts (x,0) are obtained by setting y = 0 in 2 y ax bx c and solving for x. Once the intercepts and vertex are found, it is then relatively easy to pass the appropriate parabola through these points. In the event that the x-intercepts are very close to the vertex, or that no x-intercepts exist, we find a point on each side of the vertex so that we can give a reasonable sketch of the parabola.



160

ACTIVITY 2 Answer the questions without sketching the parabola.

1. For the parabola 2 4 8 7f x x x

a) Find the vertex b) Does the vertex correspond to the highest or the lowest point on the

graph?

2. For the parabola 2 2 8y f x x x , find

a) the y-intercept b) the x-intercepts c) the vertex Remember to check your response on page 164.

Example 1 Sketch the following quadratic functions:

a) 212 4y f x x x

Thus a = 1, b = 4 and c = 12. Since a < 0 the parabola opens downward.

If the vertex is (x, y), then

42

2 2 1

bx

a

.

To find y we substitute the value of x in the given equation:

22 12 4 2 2 16y f

Thus the vertex (highest point) is (2,16). Now find the intercepts with the axes:

2If 0, then 12 4 0 0 12. Hence the -intercept is 0,12 .x y y

2If 0, then 0 12 4

0 6 2

6 0 or 2 0

y x x

x x

x x

Thus x = 6 of x = 2, and the x-intercepts are (6,0) and (2,0). Now we plot the vertex and intercepts (see figure 2(a)). Through these points we draw a parabola opening downward. See figure 2(b).



161

(a) (b)

FIGURE 2

b) 22p q

Since p = 2q2 + 0q + 0, p is a quadratic function of q where a = 2, b = 0 and c = 0. The parabola opens upward, since a > 0.

If the vertex is (q, p), then 20

0, and 2 0 02 2 2

bq p

a .

A parabola opening upward with vertex at (0,0) cannot have any other intercepts.

Hence, to draw a reasonable graph we plot a point on each side of the vertex and pass a parabola through the three points. See figure 3.

FIGURE 3

c) 6 7g x x x

Since 2 6 7g x x x , g is a quadratic function

where a = 1, b = 6, and c = 7. The parabola opens upward, since a > 0.

If the vertex is (x, g(x)),



162

then 26

3, and 3 3 6 3 7 2.2 2 1

bx g

a

The vertex is (3; 2). y-intercept = c = 7 To find the x-intercepts we will use the quadratic formula:

2

2

4

2

6 6 4 1 7

2 1

6 8

2

6 4.2

2

b b acx

a

6 2 2

2

6 2 2

2 2

3 2

Thus the -intercepts are 3 2 ,0 and 3 2 ,0 .x

After plotting the vertex and intercepts we draw a parabola opening upward. See figure 4.

FIGURE 4

d) Graph 22 2 3y f x x x and find the range of f.

This function is quadratic with a = 2, b = 2, and c = 3. Since a > 0, the graph is a parabola opening upward.



163

If the vertex is (x,y), then

22 1 1 1 5

and 2 2 32 2 2 2 2 2 2

bx y

a

1 5

Thus the vertex is ,2 2

.

If x = 0, then y = 3 and the y-intercept is (0;3). A parabola opening upward with its vertex above the x-axis has no x-intercepts. In

figure 5 we plotted the y-intercept, the vertex and an additional point to the left of the vertex. Passing a parabola through these points gives the required graph.

From figure 5 we see the range of f is 5

.2

y

FIGURE 5

ACTIVITY 3 Graph the function. Give the vertex and intercepts and state the range:

2

2

a) 6 5

b) 3

y f x x x

y f x x



3.1 Activity 1

2 2

2

2 22 2

a) 26 3 No b) 7 14 49 Yes

c) 4 Yes d) 6 4 1 No

1e) No f) 2 3 4 No

2 4

4 1g) 2 Yes h) 1 No

2 2

f x x g x x x x

g x x h s s

h q f t t t tq

sf s s g t t



164

3.2 Activity 2

1 a) Given: 4, 8 and 7a b c .

If the vertex is the point (x,y), then 8

12 2(4)

bx

a

.

Substitute in the given equation to find y:

21 4( 1) 8( 1) 7 4 8 7 3y f

The vertex is the point (1;3). b) a > 0 , thus the vertex corresponds to the lowest point on the graph. 2 a) If x = 0 then y = 8, thus the y-intercept is –8. b) If y = 0, then

2 2 8 0

4 2 0

4 or 2

x x

x x

x x

Thus the x-intercepts are – 4 and 2.

c) If the vertex is the point (x,y), then 2

12 2(1)

bx

a

.

Substitute in the given equation to find y:

21 ( 1) 2( 1) 8 1 2 8 9y f

The vertex is the point (1,9).

3.3 Activity 3

a) y-intercept : 5 x-intercept:

2 6 5 0

5 1 0

5 or 1

x x

x x

x x

Vertex:

2

( 6)3

2(1)

(3) (3) 6(3) 5 9 18 5 4

x

y f

Range: 4y



165

b) x = 0 and y = 0 Vertex (0,0) Range: 0y

You have completed this learning unit and you should be able to sketch the graph of a parabola find the coordinates of the vertex of the parabola We will now move to the next learning unit on rectangular hyperbolas.


166

ANALYTIC GEOMETRY The rectangular

hyperbola

CONTENTS PAGE

1. THE GRAPH OF THE HYPERBOLA xy = k ...................................................... 167 1.1 If k > 0, that is k is positive ................................................................................. 167 1.2 If k < 0, that is k is negative ................................................................................ 168 1.3 Determination of the equation of a hyperbola ..................................................... 169 2. RESPONSE TO ACTIVITY ................................................................................. 170

MODULE 5

LEARNING UNIT 3

OUTCOMES

At the end of this learning unit, you should be able to sketch a rectangular hyperbola find the equation of a rectangular hyperbola given a point on the hyperbola

Module 5 Learning unit 3 ANALYTIC GEOMETRY: The rectangular hyperbola


167

1. THE GRAPH OF THE HYPERBOLA xy = k

The general equation of a hyperbola is , or k k

xy k y xx y

where k may be positive or

negative.

1.1 If k > 0, that is k is positive

Example 1 Draw the graph of the function defined by xy = k where k = 20.

Thus 20

20 or xy yx



168

Note the following properties where k is positive: The graph consists of two separate branches, one entirely in the first quadrant and the

other entirely in the third quadrant.

If x has a very great positive value, then y has a very small positive value, and conversely.

If x has a very great negative value, then y has a very small negative value, and conversely.

If x increases from , then y decreases, and the graph does not cut the y-axis.

If x > 0 and increases, then y decreases from +, and the graph does not cut the x-axis.

The domain of the function extends from to , 0 excluded.k

y xx

(If a graph approaches the axis until it almost touches the axis at infinity, the axes are called the asymptotes of the graph and since the asymptotes are perpendicular to each other, the graph of xy = k is called a rectangular hyperbola.) The branches are identical and symmetrical about y = x. Each branch is symmetrical with respect to y = x, i.e. if (a; b) is on the graph, then (b;

a) will also be on it. In the graph of xy = 20 we see that both (10; 2) and (2; 10) also both (1; 20) and

(20; 1), etc. are on the curve. Since (x; y) and (x; y) are on the curve, the graph is symmetrical with respect to the origin (0; 0).

1.2 If k < 0, that is k is negative

By drawing the graph of the function defined by xy = 20, the following corresponding properties may be deduced. See figure 2.



169

FIGURE 2

The two branches lie in the second and fourth quadrants. If x has a very great positive value, then y has a very small negative value, and

conversely. If x has a very great negative value, then y has a very small positive value, and

conversely. The branches are symmetrical about y = x. Each branch is symmetrical about y = x and the curve is symmetrical with

respect to the origin.

1.3 Determination of the equation of a hyperbola

The equation of a hyperbola can be found if one point on the curve is given.

Example 2

If point (1 , 2) lies on the hyperbola xy = k, find the equation of the hyperbola. Substitute in the general form to find the value of k:



170

1 2

2

The equation is 2

xy k

k

k

xy

ACTIVITY 1 1. Sketch the graphs of a) 2xy b) 2 3 0xy

2. Find the equation of the hyperbola if 2 , 3) is a point on the hyperbola. Remember to check the response below.

2. RESPONSES TO ACTIVITY

1a)



171

1b)

2.

2 3

6

The equation is 6

xy k

k

k

xy

This is the end of this learning unit, and you should be able to sketch a rectangular hyperbola find the equation of a rectangular hyperbola given a point on the hyperbola We can now continue to the next learning unit on the circle.


172

ANALYTIC GEOMETRY

The circle

CONTENTS PAGE

1. THE EQUATION OF A CIRCLE WITH CENTRE AT THE ORIGIN AND RADIUS EQUAL TO R ........................................................................................ 173

2. FINDING THE EQUATION OF A CIRCLE WITH CENTRE NOT AT THE ORIGIN AND RADIUS EQUAL TO r ................................................................. 174

3. RESPONSES TO ACTIVITY ............................................................................... 176

MODULE 5

LEARNING UNIT 4

OUTCOMES

At the end of this learning unit, you should be able to sketch the graph of a circle find the equation of a circle

Module 5 Learning unit 4 ANALYTIC GEOMETRY: The circle


173

1. THE EQUATION OF A CIRCLE WITH CENTRE AT THE ORIGIN AND RADIUS EQUAL TO r

To describe a circle a point must move such that it stays the same distance from a fixed point. This point is the centre. In the figure, P is the moving point with coordinates (x, y). The centre is at the origin with coordinates (0 , 0) and the radius is equal to r. From any position of the point P draw a perpendicular line on the x-axis. The triangle OPQ formed is a right-angled triangle of which the hypotenuse is the side OP. To find the equation we must express the condition that P should stay the same distance from 0, algebraically.

By the theorem of Pythagoras 2 2 2PQ OQ OP .

Hence the equation of the circle is 2 2 2x y r . Thus:

The equation 2 2 2x y a represents a circle with centre at the origin and radius a.

For example, 2 2 25x y is a circle with centre at the origin and radius 5. This equation may sometimes be written in a different form:

2 2 2

2

2

25 is the same circle as 25

25 represents a semi-circle, the lower half of the above circle.

Likewise 25 represents the upper half of the circle.

y x x y

y x

y x



174

2. FINDING THE EQUATION OF A CIRCLE WITH CENTRE NOT AT THE ORIGIN AND RADIUS EQUAL TO r

To find the equation of the circle in this case, we must express the condition that the moving point P should stay at a constant distance from the fixed point (centre) algebraically.

FIGURE 2

In the figure the coordinates of the centre are (h , k). Consider the moving point P in any position. Draw through M lines parallel to the x-axis and y-axis. Draw from P a line PN perpendicular to the x-axis to form the right-angled triangle PMN. In the figure PK = x ; PB = y ; ML = h and MA = k. Hence it follows that MN = x h and that PN = y k. Since the triangle PMN is a right-angled triangle, it follows that MN2 + PN 2 = MP 2 (Pythagoras's theorem). Hence the equation of the circle is

2 2 2x h y k r

This is a very convenient form of the equation of the circle and we may memorise it as follows:

(x x-value of centre)2 + (y y-value of centre)

2 = radius

2

Example 1

Find the equation of the circle with centre at (3,2) and radius 5 units of length.



175

The equation is of the form

2 2 2

22 2

2 2

3 2 5

3 2 5

x h y k r

x y

x y

If the brackets are removed and the terms rearranged, the equation becomes 2 2 6 4 8 0x y x y

Note: If we consider the equation we find that (1) the coefficients of x2 and y2 are equal (2) there are no terms in xy We will find our observations useful to recognise a circle equation.

ACTIVITY 1

1. Find the equation of a circle with centre (-2;3) and radius 3 units.

2. Sketch the graph of 2 2 6 4 8 0x y x y Remember to check the response on page 176.



176


1.

2 2 2

22 2

2 2

2 2

2 3 3

2 3 3

4 6 10 0

x h y k r

x y

x y

x y x y

2. We recognise the equation as that of a circle with the origin not at the centre.

To find the standard form, we rearrange the equation and complete the squares in x and y.

2 2

2 2

2 2

2 2

6 4 8 0

Standard form:

6 4 8

6 9 4 4 8 9 4

3 2 5

Circle with origin ( 3,2) and radius 5 2.23

x y x y

x x y y

x x y y

x y

[To complete a square, divide the coefficient of the middle term by 2 and add the square of your answer: For example, to complete 2 6x x : divide 6 by 2, your answer is 3. Now take 3 and square, the answer is 9. Add 9 to 2 6x x to obtain 2 26 9 ( 3)x x x ]

This ends the learning unit on circles and you should be able to sketch the graph of a circle find the equation of a circle We now move to the next learning unit on ellipses.


177

ANALYTIC GEOMETRY The ellipse

CONTENTS PAGE

1. THE ELLIPSE 2 2

2 21

x y

a b .................................................................................. 178

2. THE EQUATION OF THE ELLIPSE WITH FOCI ON THE X- or Y-AXIS ...... 178 3. THE EQUATION OF THE ELLIPSE OF WHICH THE FOCI ARE NOT ON THE X-

or Y-AXIS .............................................................................................................. 182 4. RESPONSES TO ACTIVITY ............................................................................... 184

MODULE 5

LEARNING UNIT 5

OUTCOMES

At the end of this learning unit, you should be able to sketch the graph of an ellipse find the equation of an ellipse find the vertices, foci and the length of the major and minor axes of the ellipse

Module 5 Learning unit 5 ANALYTIC GEOMETRY: The ellipse


178

1. THE ELLIPSE 2 2

2 21

x y

a b

Definition The ellipse is the set of all points in a plane such that the sum of their distances from two fixed points remains constant. The two fixed points are called the focal points (foci) and the line drawn through them is called the major axis of the ellipse. The points of intersection of the ellipse with the major axis are called vertices.

In figure 1: VV ' is the major axis.

The vertices are V and V. F' and F are the foci.

FIGURE 1

2. THE EQUATION OF THE ELLIPSE WITH FOCI ON THE X- or Y-AXIS

Suppose the two focal points lie on the x-axis at –c and c. FIGURE 2 If the length of the major axis (in figure 2 the x-axis) is given by 2a, and the length of the major axis is given by 2b, we may write the equation of the ellipse as

2 2

2 21

x y

a b

The relation between a, b and c is given by 2 2c a b



179

using the theorem of Pythagoras. See figure 3.

FIGURE 3 If the major axis lies along the y-axis, we can write the equation as

2 2

2 21

x y

b a

See figure 4. [Note: a gives the length of the major axis and b the length of the minor axis. When the major axis is along the y-axis, the x-coordinate of V and F is 0.]

FIGURE 4

Example 1 Find the lengths of the major and minor axes of the following ellipses and sketch them:

2 2

2 2

a) 116 4

b) 25 9 25

x y

x y

Solution

2 2 2 2a) Here we have 16, 4 and 16 4

2 3 3.46

a b c a b

The centre of the ellipse is at the (0,0) and the major axis lies along the x-axis. The points of intersection with the major axis are at x = 4 and x = 4.



180

The points of intersection of the minor axis of the ellipse, that is the values of b, are at y = 2 or 2.

The foci on the major axis lie at 2 3 and 2 3x . [You can use the answer in decimal form, that is foci at 3.46 and 3.46x x .] Length of the major axis = 2a = 8 units. Length of the minor axis = 2b = 4 units.

FIGURE 5

2 2

2 2

259

2 2 2 2

b) 25 9 25

. 11

25 25 25 9Therefore 1 and and 1

9 9 9

16

94

3

x y

x y

b a c a b

FIGURE 6



181

Example 2 Calculate the equation of the ellipse if the following are given:

a) Intercepts (± 8,0) and minor axis = 6

b) One vertex at (0,13), one focal point at (0, 12) and centre at (0,0) Solution a) Draw a sketch.

FIGURE 7

Here we have 2a = 16 and therefore a = 8,

2b = 6 and therefore b = 3

2 2

Equation: 164 9

x y

b) Draw a sketch.

FIGURE 8

The major axis lies along the y-axis, a = 13, c = 12.

Therefore 2 2 2 25b a c .



182

Thus the equation of the ellipse is 2 2

2 2

2 2

1

125 169

x y

b a

x y

3. THE EQUATION OF THE ELLIPSE OF WHICH THE FOCI ARE NOT ON THE X- or Y-AXIS

A further assumption is that the major and minor axes are parallel to the ANALYTIC axis. If the centre of the ellipse is at (h,k) and the length of the major axis is a and that of the minor axis b, we can write the equation of the ellipse as

2 2

2 21

x h y k

a b

FIGURE 9

Example 3

a) Sketch the ellipse represented by the equation 2 29 4 18 23 0x y x y .

b) Write down the equation of the ellipse with vertices (8,3) and (4,3) and one focal point at (6,3).

Solution

Our equation is not yet in the standard form 2 2

2 21

x h y k

a b

.

We alter the given equation as follows: 2 2( 4 ) 9( 2 ) 23x x y y To complete the square, the coefficient of the first term must be 1. Don’t forget to add the same numbers on both sides.

2 2

2 2

( 4 4) 9( 2 1) 23 4 9(1)

2 9 1 36

x x y y

x y

In the standard form the right-hand side must be equal to 1. Divide both sides by 36.



183

2 2

2 2

2 11

36 4

Therefore 36 and 4

and 32 4 2 5.66

x y

a b

c

The centre of the ellipse is at (2,1). The major axis is parallel to the x-axis, that is along the line y = 1. The minor axis is parallel to the y-axis, that is along the line x = 2. The length of the major axis = 2a = 2(6) = 12. The length of the minor axis = 2b = 2(2) = 4. To sketch, start by finding the centre. Draw the axes through the centre. Count the required length along the axes. Mark the foci. Complete the ellipse. This is sufficient for our purposes. You can go further and find the intercepts with the axes.

FIGURE 10

c) The centre of the ellipse is at the midpoint of VV’ that is at (2, 3).

2 2 2

2 2

Therefore 6

and 4

and consequently

36 16

20

Therefore the equation of the ellipse is

2 31

36 20

a

c

b a c

x y



184

ACTIVITY 1 1. For each of the following ellipses, find the length of the major and minor axes and the focal points:

2 2

2 2

2 2

2 2

a) 4 9 36

b) 25 16 400

c) 4 6 32 69 0

d) 16 9 32 36 92 0

x y

x y

x y x y

x y x y

2. Find the equations of the following ellipses:

a) 13,0 and 12,0 b) Centre 0,0 ; 5; 0, 4

c) 7,3 ; ' 3,3 ; 6,3 d) Centre 0,0 ; 0, 5 ; 0, 3

V F a F

V V F V F

Remember to check the response below.


2 2

2 2

2 2

1. a) 19 4

3; 2; focal points 9 4 5 2.24

Length of major axis = 6 and length of minor axis = 4.

b) 116 25

5; 4; focal points 25 16 3

Length of major axis = 10 and length of minor a

x y

a b x

x y

a b y a b

xis = 8. Major axis along -axis.y

2 2

2 2

2 2

c) ( 6 9) 4( 8 16) 69 9 4(16)

3 4 4 4

3 41

4 1

x x y y

x y

x y

2; 1; focal points: 4 1 3

Centre is at (3; 4)

focal points: 3 3; 4

Length of major axis = 4 and length of minor axis = 2.

a b c



185

2 2

2 2

2 2

d) 16( 2 1) 9( 4 4) 92 16 1 9 4

16 1 9 2 144

1 21

9 16

Centre at 1,2 and foci at 1;2 7

Length of the major axis = 8 and length of the minor axis = 6.

x x y y

x y

x y

2 2 2 2 2 2

2

2 2

2. a) Centre at 0,0

Given ( 13,0) thus 13 and the major axis is along the -axis;

Given 12,0 thus 12

thus 12 13

25

1169 25

b) Centre at 0,0 ; 5

Given 0,4 thus 4 and the major axis

V a x

F c

c a b b

b

x y

a

F c

2 2 2 2 2 2

2

2 2

is along the -axis.

4 5

9

19 25

y

c a b b

b

x y



186

c) Given 7,3 ; ' 3,3 and 6,3 thus the major axis is along 3.V V F y

1 12 2

2

The centre is at the midpoint of ' that is at (2,3)

half the length of ' 7 3 10 5

length from centre to 6,3 = 6 2 4

Note that the value of is not necessary to determine the equation.

VV

a VV

c F

c

b a

2 2

2

2 2

25 16

3

2 31

25 9

c

b

b

x y

2 2

2 2

2 2

d) Centre (0,0), 0, 5 and 0, 3

Thus 5, 3 and the major axis is along the

5 3 16 4

116 25

e) Centre (0,0), 0, 5 and 0, 3

Thus 5, 3 and the major axis is along the

5 3 16 4

V F

a c y axis

b

x y

V F

a c y axis

b

x

2 2

116 25

y

Now that you have come to the end of this learning unit, you should be able to sketch the graph of an ellipse find the equation of an ellipse find the vertices, foci and the length of the major and minor axes of the ellipse You can now continue to the next learning unit on the central hyperbola.


187

ANALYTIC GEOMETRY The central hyperbola

CONTENTS PAGE

1. THE GRAPH OF THE CENTRAL HYPERBOLA .............................................. 188 2. RESPONSES TO ACTIVITY ............................................................................... 191

MODULE 5

LEARNING UNIT 6

OUTCOMES

At the end of this learning unit, you should be able to recognise the equation of a central hyperbola sketch the graph of the central hyperbola find the vertices, foci, endpoints of the conjugate axis and the asymptotes of the

central hyperbola

Module 5 Learning unit 6 ANALYTIC GEOMETRY: The Central Hyperbola


188

1. THE GRAPH OF THE CENTRAL HYPERBOLA

Definition A hyperbola is the set of points in a plane for which the difference between the distances of the points from two fixed points is a constant. The fixed points are called the focal points. The transverse axis is the line segment through the two foci with its endpoints on the hyperbola. The endpoints of the transverse axis are called the vertices. The centre C of the hyperbola is the midpoint of the foci. The line segment through the centre that is perpendicular to the transverse axis is called the conjugate axis.

We will discuss both 2 2

2 21

x y

a b and the more general form

2 2

2 21

x h y k

a b

.

The only difference between these two hyperbolas is that the centre of the first is at the origin (0,0), whereas the centre of the second one is at the point (h,k). Standard equation: Hyperbola with a horizontal major axis

The equation 2 2

2 21

x y

a b describes a central hyperbola with the centre at (0,0) and the

major axis on the x-axis:

Please note: The vertices are V(a,0) and V’(a,0). The endpoints of the conjugate axis are W(0,b) and W’(0,-b).

The foci are at F(-c,0) and F’(c,0) where 2 2 2c a b . The lines and b b

a ay x y x are the asymptotes of this hyperbola.

Endpoint of the conjugate axis

Endpoint of the conjugate axis

Asymptote b

y xa

Module 5 Learning unit 6 ANALYTIC GEOMETRY: The central hyperbola


189

[The intercepts b, b, a and a determine the position of the dotted line rectangle which is shown in the figure. This rectangle serves as an aid to draw the asymptotes and finally the hyperbola.] Standard equation: Hyperbola with a vertical major axis

The equation is 2 2

2 21

y x

b a describes the hyperbola with centre at (0,0) and the major

axis on the y-axis. See figure 2.

Please note: The vertices are (0,b) and (0,-b). The endpoints of the conjugate axis are (a,0) and (-a,0).

The foci are at F(0,-c) and F’(0,c) where 2 2 2c a b . The lines - and b b

a ay x y x are the asymptotes of this hyperbola.

Example 1

Sketch the graph of 2 2

2 2

2 11

4 3

x y

The centre of this hyperbola is at (2, 1). Thus we have to draw a rectangle with the point (2, 1) as the centre. The length of the sides parallel to the x-axis is equal to 2a; count 4 units to the left and 4 units to the right of this point. This gives us the vertices V’(2,-1) and V(6,1).

Module 5 Learning unit 6 ANALYTIC GEOMETRY: The Central Hyperbola


190

The lengths of the sides parallel to the y-axis equal to 2b; count 3 units to the left and 3 units to the right of this point. This gives us the endpoints of the conjugate axis W’(2,4) and W(2,2).

The foci: 2 2 16 9 25 5c a b . As the centre is at (2, 1) the foci are at F’( -3, -1) and F(7,-1). To complete the sketch, we can find the intercepts with the x-axis: Put 0y

2

2

2 1 101

16 9 9160

29

2 4.22

6,22 or 2.22

x

x

x

x

FIGURE 3

ACTIVITY 1 For the following hyperbolas find (i) the coordinates of the centre (ii) the foci (iii) the vertices (iv) the ends of the conjugate axis

2 2

2 2

2 2

2 2

1 2a) 1

3 4

3b) 1

12 5

x y

x y


Module 5 Learning unit 6 ANALYTIC GEOMETRY: The central hyperbola


191


a) i Centre 1 , 2

ii 9 16 5

Foci : 4,2 and 6,2

(iii) Vertices: 2, 2 and 4, 2

iv Endpoints of the conjugate axis: (1, 2) and (1,6)

b) i Centre 3,0

ii 144 25 13

Foci : -16,0 and 10,0

iii Vertices: 9 , 0 and 15 , 0

iv End

c

c

points of the conjugate axis: ( 3, 5) and ( 3 , 5)

Now that you have come to the end of this learning unit, you should be able to recognise the equation of a central hyperbola sketch the graph of the central hyperbola find the vertices, foci, endpoints of the conjugate axis and the asymptotes of the

central hyperbola It may be handy to know the term “conic section” in connection with the topics covered in module 5. Collectively we can call the straight line, parabola, rectangular hyperbola, circle, ellipse and central hyperbola conic sections. All of these can be obtained by cutting a cone in different sections. You should now test your knowledge of the straight line, parabola, rectangular hyperbola, circle, ellipse and central hyperbola by doing the post-test for module 5.


192

POST-TEST: ANALYTIC GEOMETRY 1. Show that the two lines 4x 3y 1 = 0 and 6x + 8y + 1 = 0 are perpendicular to

each other. 2. A straight line AB passes through the two points (3, 6) and (2, 3). Find its

equation, and also find the equation of the straight line which bisects the distance between the two points and is perpendicular to it.

3. a) Calculate the distance between the points A (5 , 6) and B (7 , 12). b) Derive the equation of the line passing through these points. 4. a) Find the equation of a line which cuts the x- and y-axis at 3 and 5, respectively. b) Find the equation of the line joining the point (3,2) to the point of intersection of

the lines x y + 4 = 0 and y 2x 5 = 0. 5. Determine the equations of the following straight lines: a) With slope 3 and y-intercept 2 b) Passing through the point (5; 4) and parallel to the line 2x + 3y 12 = 0 6. Determine the angle between the lines 3y = 4x + 2 and y = x. 7. Identify the type of curve of the following equations. Sketch the graph of the

curve. Show the coordinates of all relevant points. a) 4 3 5 0y x

b) 2( ) 2 6y g x x x

c) 2( ) 1y f x x

d) xy = 4

e) 2 24 6 32 69 0x y x y

f) 2 216 1 9 2 144x y

g) 2 29 4 36x y

h) 2 29 4 36x y

i) 2 23 1.21x y


193

ANALYTIC GEOMETRY: POST-TEST SOLUTIONS

1. 1 1

2 2

1 2

4 1If 4 3 1 0 then

3 3

6 1If 6 8 1 0 then

8 8

4 61

3 8

x y y x m x c

x y y x m x c

m m

The two lines are perpendicular.

2. Let the equation of AB be y = mx + c

6 3

3 2

9 3From which we find and

5 5

m c

m c

m c

The equation of AB is 9 3

5 5y x .

To find the coordinates (x1, y

1) of the mid-point of AB we have

2 3 2 31 1

1 1

and 2 2

3 2 6 3

2 2

1 3

2 2

x x y yx y

x y

The slope m of the required perpendicular line

1 2

1

1

1

91

5

5

9

m m

m

m

The midpoint 312 2; is a point on the line, thus 1y m x c

3 5 1

2 9 2

3 5

2 18

27 5 32

18 9

16

9

c

c

c

Module 5 ANALYTIC GEOMETRY: Post-test solutions


194

The equation of the perpendicular is 5 16

9 9

or 9 5 16 0

y x

y x

.

3.

2 22 1 2 1

2 2

a)

7 5 12 6

144 36

180

13.42

ABd x x y y

2 11 1

2 1

b)

12 66 5

7 5

65

12

15

2

2 12 5

2 17

1 17

2 2

y yy y x x

x x

y x

x

x

y x

y x

y x

4. a) Given: 5 and (3;0) is a point on the line

Thus

0 (3) 5

3 5

5

3

c

y mx c

m

m

m

The equation of the line is 5

5 or 5 3 153

y x x y

b) To find the intersection of the lines we must solve the two equations simultaneously.

4 (1)

2 5 (2)

1 (1) (2)

1

x y

x y

x

x

Put in equation (1) to find 3y

Thus (1 , 3) is a point on the line.



195

Therefore 2 11 1

2 1

y yy y x x

x x

2 33 1

3 1

13 1

4

4 12 1

4 11

y x

y x

y x

y x

5. a) 3 2y x

b) Find the slope of the required line: 2 2

If 3 2 12 then 4 and 3 3

y x y x m

Now (5;4) is a point on the line, thus

1 1

24 5

3

3 12 2 10

3 2 22

y y m x x

y x

y x

y x

6. 14 2 4

If 3 4 2 then and 3 3 3

y x y x m

2If then 1y x m .

Therefore

2 1

2 1

43

43

7313

tan1

1

1 1

7 3

3 1

7

81.87

m m

m m

7.



196

a) Straight line: 3 5

If 4 3 5 0, then 4 4

y x y x .

y-intercept = 5

4 = 1.25 x-intercept =

5

3 = 1.67

b) Parabola: y-intercept : (0;0)

x-intercept:

2

2

2 6 0

3 0

3 0

0 or 3 0

3

x x

x x

x x

x x

x

Vertex:

2

( 6) 3

2( 2) 2

3 3 9 18 92 6

2 2 2 2 2

9Range :

2

x

y

y

c) Parabola: y-intercept : (0,1)



197

x-intercepts:

2 1 0

1 1 0

1 or 1

x

x x

x x

Vertex: (0)

02(1)

1

Range : 1

x

y

y

d) 4xy Rectangular hyperbola x -2 -1 1 2 y -2 -4 4 2



198

e)

2 2

2 2

2 2

2 2

2 2

4 6 32 69 0

Standard form:

6 4 8 69

( 6 9) 4( 8 16) 69 9 64

3 4 4 4

3 41

4 1

x y x y

Ellipse

x x y y

x x y y

x y

x y

Centre (h,k) = (3,-4) and focus c = 3

f)

2 2

2 2

2 2

2

2

16( 1) 9( 2) 144

Standard form: Central hyperbola

1 21

9 16intercepts:

1 0 21

9 16

1 41

9 1620

( 1)16

201

42.12 or 0.12

x y

x y

x

x

x

x

x

x x



199

2 2

2 2

g) 9 4 36 Divide by 36 to get standard form

14 9

Central hyperbola The vertices are 2, 0 and ' 2, 0

The foci are at 13, 0 and ' 13, 0

3 3The asymptotes are and

2 2

x y

x y

Y Y

F F

y x y x

2 2

2 2

h) 9 4 36 Divide by 36 to get standard form

14 9

Ellipse The vertices are X 2, 0 and X ' 2, 0

Y(0, 3) and Y'(0,-3)

Focus points 5 and 5

x y

x y



200

15. Circle: Centre(h,k) = (-3, 0) Radius = 1.21 1.1

You have reached the end of study guide 1 and should now be ready to move to study guide 2 on calculus. The first learning unit will refresh your knowledge of functional notation.

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