math_skills for engineering science and applied mathematics

8/7/2019 Math_Skills for Engineering Science and Applied Mathematics

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,ESSENTIALM A T H E M A T I C A LS K I L L S


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M A T H E M A T I C AS K I L L S

for engineering, scienceand appliedmathematics

Dr Steven Ian Barry+Dr Stephen Alan Davis


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A U NSW P r es s b o okP ub lis he d b y

U n iv ers ity o f N e w S ou th W a le s P re ss L tdU n iv er sit y o f N ew S o ut h Wa l esU NS W S yd ney N SW 2 05 2

AUSTRAL IAwww.unswpress . com.au

S te ve n Ia n B a rry a n d S te ph en A la n D a vis 2 00 2F i rs t p ub li she d 2002

T h is b oo k is c op yrig h t. A p ar t fr om a n y fa ir d ea lin g fo rp u r po ses o f p r iv a te s tu d y , r esear c h, c ri ti ci sm o r

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5 10Prin ter BPA
http://www.unswpress.com.au/http://www.unswpress.com.au/


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CONTENTS

Preface . lX1 Algebra and Geometry

1.1 Elementary Notation1.2 Fractions1.3 Modulus.......1.4 Inequalities ....1.5 Expansion and Factorisation

1.5.1 Binomial Expansion1.5.2 Factorising Polynomials

1.6 Partial Fractions . . .1.7 Polynomial Division .1.8 Surds .

1.8.1 Rationalising Surd Denominators1.9 Quadratic Equation1.10 Summation ...1.11 Factorial Notation.1.12 Permutations1.13 Combinations1.14 Geometry . .

1.14.1 Circles1.15 Example Questions

12334566910101112121 31 3141516

2 Functions and Graphs2.1 The Basic Functions and Curves2.2 Function Properties2.3 Straight Lines2.4 Quadratics.2.5 Polynomials.2.6 Hyperbola..2.7 Exponential and Logarithm Functions2.8 Trigonometric Functions2.9 Circles.......2.1 0 Ellipses . . . . . .2.11 Example Questions

171718212223242526272829


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Vi

3 TranscendentalFunctions3.1 Exponential Function3.2 Index Laws .3.3 Logarithm Rules .3.4 Trigonometric Functions3.5 Trigonometric Identities3.6 Hyperbolic Functions3.7 Example Questions

3131323335363839

4 Differentiation4.1 First Principles4.2 Linearity....4.3 Simple Derivatives4.4 Product Rule4.5 Quotient Rule . . .4.6 Chain Rule ....4.7 Implicit Differentiation4.8 Parametric Differentiation4.9 Second Derivative.4.10 Stationary Points4.11 Example Questions

414142434344454647474850

5 Integration5.1 Antidifferentiation5.2 Simple Integrals ..5.3 The Definite Integral5.4 Areas .5.5 Integration by Substitution5.6 Integration by Parts5.7 Example Questions

5151525355565758

6 Matrices6.1 Addition.6.2 Multiplication6.3 Identity ...6.4 Transpose..6.5 Determinants

6.5.1 Cofactor Expansion.6.6 Inverse............

6.6.1 Two by Two Matrices.6.6.2 Partitioned Matrix6.6.3 Cofactors Matrix

6.7 Matrix Manipulation ...6.8 Systems of Equations . . .6.9 Eigenvalues and Eigenvectors.6.10 Trace .6.11 Symmetric Matrices. . . . . .

59596061626364656666676870737474


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6.12 Diagonal Matrices.6.13 Example Questions

CONTENTS Vll

7475777778798080828386878788888990919192929394959797979899100100102105107107108111114

7 Vectors7.1 Vector Notation .7.2 Addition and Scalar Multiplication .7.3 Length .7.4 Cartesian Unit Vectors7.5 Dot Product . . . . .7.6 Cross Product . . . .7.7 Linear Independence7.8 Example Questions .

8 Asymptotics and Approximations8.1 Limits .8.2 L'H6pital's Rule.8.3 Taylor Series ..8.4 Asymptotics ...8.5 Example Questions

9 Complex Numbers9.1 Definition..........9.2 Addition and Multiplication9.3 Complex Conjugate ..9.4 Euler's Equation ....9.5 De Moivre's Theorem.9.6 Example Questions

10 Differential Equations10.1 First Order Differential Equations

10.1.1 Integrable .10.1.2 Separable .10.1.3 Integrating Factor .

10.2 Second Order Differential Equations10.2.1 Homogeneous .10.2.2 Inhomogeneous.

10.3 Example Questions11 Multivariable Calculus

11.1 Partial Differentiation .11.2 Grad, Div and Curl11.3 Double Integrals. .11.4 Example Questions


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Vlll

12 Numerical Skills12.1 Integration.12.2 Differentiation ..12.3 Newton's Method12.4 Differential Equations.12.5 Fourier Series .....

12.5.1 Even Fourier Series.12.5.2 Odd Fourier Series

12.6 Example Questions

115115116117118119120121122

13 Practice Tests13.1 Test 1: First Year - Semester One .13.2 Test 2: First Year - Semester One .13.3 Test 3: First Year - Semester Two.13.4 Test 4: First Year - Semester Two.13.5 Test 5: Second Year .13.6 Test 6: Second Year .

123124125126127128129

14 Answers 13115 Other Essential Skills 14 3

Index 14 6


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PREFACE lX

PREFACETO THE STUDENT

There are certain mathematical skills that are essential for any of your courses that use mathemat-ics. Your lecturer will assume that you know them perfectly - not just a vague idea, but that youhave completely mastered these skills. Without these necessary skills, you will find present and latersubjects extremely difficult. You may also lose too many marks making 'silly' mistakes in exams.So what skills do you need to have?This book contains the mathematical skills we think are essential for you to not only know but remem-ber. Itis not a textbook and does not attempt to teach you, hence there are no long wordy explanations.This book should act as a reminder to you of material you have already learned. If you are havingtrouble with a section or chapter then we suggest you consult a more thorough textbook. We have lefta number of blank pages at the back of the book for you to add in skills that you or your lecturersthink are important to remember but we did not include.

This book covers the essential mathematics in the first one to two years of a science, engineeringor applied mathematics degree. If you are in a first year undergraduate course you may not havecovered some of the material included in this book.

As a guide, we expect our students at University College to have mastered (by the start of eachsemester) the following:

First Year - Semester One: Chapters 1-3. First Year - Semester Two: Chapters 1-7. Second Year: Chapters 1-10. Third Year: Everything in the book!

There are practice tests in Chapter 13 based on these divisions.Can you do the practice test at the end of these notes?Ifyou can't then perhaps there are some skills you need to do some revision on. Ifyou can then youmay need this book to help you revise those skills later on.

Ifyou want more questions to practice on then see our extensive website:http://www.ma.adfa.edu.aul ...sib/EMS.html

Itcontains extra questions, fully worked solutions, practice tests and also code for the Maple algebraicmanipulation package giving solutions for every example and question.
http://www.ma.adfa.edu.aul/http://www.ma.adfa.edu.aul/


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x

TO THE LECTURER

What do you assume your students know? What material do you expect them to have a vague ideaabout (say the proof of Taylor's Theorem) and what material do you want students to know thoroughly(say the derivative of sin z)? This book is an attempt to define what material students should havecompletely mastered at each year in an applied mathematics, engineering or science degree. Naturallywe would like our students to know more than the bare essentials detailed in this book. However,most students do not get full marks in their previous courses and a few weeks after the exam willonly remember a small fraction of a course. They are also doing many other courses not involvingmathematics and are not constantly using their mathematical skills. This book can then act as guide towhat material should realistically be remembered from previous courses. Naturally both the materialand the year in which the students see this material will vary from university to university. This bookrepresents what we feel is appropriate to our students during their degrees.

We invite you to look at our extensive web site:http://www.ma.adfa.edu.aul ...sib/EMS.html

Itcontains more questions, solutions, practice tests and Maple code. There is a database of questionsin LaTeX and pdf, which you can use to format your own tests and assignments. We are not concernedthat students may access this database; if they can do the questions in the database then they have, ineffect, learned the necessary skills.

Ifyou have any questions or queries please do not hesitate to email us.

Steven Barry and Stephen DavisSchool of Mathematics and StatisticsUniversity College, UNSWCanberra, ACT, 2600email: [email protected]
http://www.ma.adfa.edu.aul/mailto:[email protected]:[email protected]://www.ma.adfa.edu.aul/


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CHAPTER 1ALGEBRA AND GEOMETRY

1.1 E LE ME N TA RY N O TA TIO N

I. {}:A set of objects.2. E: A member of a set. For example 3 E {I, 2, 3}.3. R: The set of real numbers. For example -1,3,3.2,.J2 E R.4. Z: The set of integers. For example -2, 0, 3 E Z.5. : Less than, greater than. For example 5 < 6,7> 5.6. :5,~: Less than or equal to, greater than or equal to.7. ===}: Becomes. For example x - 2 =3 ===} x =5.8 . [a, b ] : Bounds of a variable. For example x E [1,3] means 1 :5 x :5 3.9. (a, b) : Bounds of a variable. For example x E (1,3) means 1 < x < 3.10. -+: Tends to. For example l/x -+ 0 as x -+ 00.11. ~ : Approximately equal to. For example 3.02 ~ 3.

EXAMPLES1. W = {f(x) = a + b : a, bE R} means W is the set of all functions f(x) = a + b where a, b

are real numbers (constants), Hence 1 + 2x E Wand 3 - 1.2x E W.2. S = {x : x 2 : 5, x E R} means that S is the set of all numbers bigger than or equal to 5. This is

also written as x E [5, (0).


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2 ALGEBRA AND GEOMETRY

1.2 FRACTIONS

A fraction is of the formihere a is called the numerator and b is called the denomina-tor.Rules for operating on fractions

a b a+ b1. -+-=--c c c (c ~ 0 )a c ad+bc2. b + d = b e la b ab3. - x - =-c d cd

(b,d ~ 0 )

(c,d~O)a cad ad4. b + d = b x ~ = be (b,c,d ~ 0)

EXAMPLES2 3 6 1I. -x-=-=-9 8 72 121 1 2 1 3 12. - + - = - + - =- =-3 6 6 6 6 2

3 x + 2 _ x - 2 = (x + 2)2 - (x - 2)2 = (x2 + 4x + 4) - (x2 - 4x + 4) =~. x - 2 x + 2 (x - 2)(x + 2) (x2 - 4) x2 - 4

1 1 14. To rearrange the equation - + - = - to find y writex y 101 1 1=Y 10 x1 x -10= = = > =Y l O xlOx= = = > y = x -10'

NOT Y = 10- x


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MODULUS 3

1.3 MODULUS

The absolute value or modulus of x, written lxi , is defined byIx l ={ x, ~fx 2 : : 0-x, lf x < O .

The absolute value is the magnitude of a number and ignores whether it is positive ornegative.

EXAMPLES1 . 1 + 5 1 = 52. 1 - 3 1 = 33. I-xl lyl = Ix l ly l = I xY I

1.4 INEQUALITIES

I. Ifx > y then x + a > y + a for any a.2. Ifx > y then ax > ay if a is positive, but ax < ay if a is negative.3. Ifx > y and u > v, then x + u > y + v.

EXAMPLESI. To find x such that -5x - 2 :5 3 write

-5x - 2:5 3= = = > -5x:5 5= = = > x 2 : : -1.

2. To find values of x such that x + 1 > 2x - 5 we writex+1>2x-5

= = = > x < 6.


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Inequalities with modulusI. The inequality I x - b l < a can be written as -a < x - b < a.2. The inequality I x - b l > a can be written as x - b > a or (x - b ) < -a.

EXAMPLESI. To find x such that 12x - 11 : s 3 write

12x - 11 s 3 ===} -3:S 2x - 1 : s 3===} -2:S 2x:S 4===} -1:S x S; 2.

1 3 x - 1 1. To find x such that -4- ;:::2 write3x-lor --


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EXPANSION AND FACTORISATION 5

2. (x - 3)(x + 5)2(x + 3 ) = (x - 3)(x + 3)(x + 5 )2= (X2 - 9)(x2 + lO x + 25 )" " " "

3. s2-4 (s-2)(s+2)= -'-----:-'-'--_-'-2+s 2+s=s-2

4. (a + 1)3 = (a + l)(a2 + 2a + 1)= a3 + 3a2 + 3a +1

1.5.1 BINOMIAL EXPANSION

n(n -1)(a + b)n =an + nan-1b +" "an-2b2 + ... + nabn-1 + bn2 !(See also Section 1.13). To remember the coefficients of each term use Pascal's trianglewhere each number is the sum of the two numbers above it.

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

Each term in a row represents the coefficients of the corresponding term in the expansion.

EXAMPLES

2. (1+ X)4 = 1+ 4x + 6x2 + 4x3 + X43. The coefficient of x3 in (2 + x )5 is 10 X 22 = 40.


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1.5.2 F A CT O RIS IN G P O LY N O M IA L S

Factorising a polynomial is the opposite of the expansion described above, that is, splittingthe polynomial into its factors:

p(x) = (x - al)(x - < l2 ) . . . (x - an) .

E X A M P L E SI. x2 - 1= (x - 1)(x + 1)2. x2 - 3x + 2 = (x - 2)(x - 1)3. 3x2 - 7x + 2 = (3x - l)(x - 2)4. x3 - 4x2 + 4x = x(x - 2) 25. a3 + 3a2 + 3a + 1 = (a + 1)3

1.6 PARTIAL FRACTIONS

Itis sometimes convenient to writeex+d A B..,---::-;-----,:-:--- + --(x + a)(x + b) x + a x + b

where A and B are constants found by equating the numerators of both sides once th elight hand side is written as one fraction:

ex + d =A(x + b)+ B(x + a).Some similar partial fraction expansions are

1 ABC= --+ +--x+a (x+aF x+bAx+B C

x2 + bx + c + x + a(x + aF(x + b)

1(x2 + bx + c)(x + a)


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PARTIAL FRACTIONS 7

EXAMPLES.. 1 . cAB. .1. Wntmg ( ) ( ) m the rorm -- + -- impliesx+l x-I x+l x-I

A(x - 1 ) + B(x + 1 ) =1.The constants A and B can be found two simple ways. First, setting

x=1 ===}x=-1 ===}

Alternatively the equation could be expanded asAx + Bx - A + B =1

and the coefficients of Xl and xO equated givingA+B=O-A+B=1.

Solving these equations simultaneously gives A = -1/2 and B = 1/2. Thus1 1 ( 1 1 )(x + 1)(x - 1 ) = " 2 x-I - x + 1 .2 'T d 3x + 1 . ial fracti .. 10 expan (x + 7)(x _ 3) using parti ractions wnte

3x+l A B(x + 7) (x - 3) = x + 7 + x - 3giving

A(x - 3) + B(x + 7) =3x + 1.Setting x = 3 implies B = 1 and setting x = -7 implies A = 2. Alternatively, equating thecoefficients of

Ax + Bx - 3A + 7B =3x + 1gives

A+B=3-3A+ 7B = 1.These simultaneous equations are solved for A and B to give A =2 and B =1. Hence

3x+l 2 1--;-(-+----,7:-;-)--;-(-_-3: : - - c - ) = -x-+ -7 + -x---3.


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3 0 The partial fraction for ( ) ! ( ) isx+l x+21 ABC--,------~------,--=--+ +--(x+l)2(x+2) x+l (x+l)2 x+2

giving1= A(x + 1)(x + 2)+ B(x + 2)+ C(x + 1)2

so thatx =-1 ===} I=Bx =-2 ===} I=Corder x2 ===} O=A+C ===} A =-1.

Thus1 1 1 1--,------~------,--= ---+--(x+l)2(x+2) (x+l)2 x+l x+2

4 0 The partial fraction for ( 2 3 ) ( ) isx +x+l x+23 Ax+B C--,----;;,-----------::-;:---;--------:::-;-+ --(x2+x+l)(x+2) x2+x+l x+2

giving3 = (Ax + B)(x + 2)+ C(x2 + X + 1)0

Hencex =-2 ===} C=1x=O ===} 3=2B+C ===} B= 1

order x2 ===} O=A+C ===} A= -1.Thus

3 1 x-I(x2 + X + 1)(x + 2) -- - ----0----x+2 x2 +x+ 10


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POLYNOMIAL DIVISION 9

1.7 P OL YN O M IA L D IV IS IO N

Polynomial division is a type of long division for polynomials best illustrated by thefollowing examples.

EXAMPLES

1. When dividing x2 + 3x + 4 by x +1consider only the leading order terms to begin with. Thus xgoes into x2, x times. Thus x(x + 1) = x2 + z, which is subtracted from x2 + 3x + 4. The firststep is therefore

xx + 1) x2 + 3x + 4

x2 +x2x+4The division is completed by considering that x (the leading order of x + 1) goes into 2x + 4 twotimes. Subtracting 2(x + 1) from 2x + 4 gives

x+1)x2+3x+4x2 +xx+2

2x+42x+22

Thus x2 + 3x+ 4 2---1- = (x+2)+--1'X + " X +2. Dividing 3x3 + 2X2+ X + 1by x-I gives

32 . .3x+2x+x+l 3256 71 = x+ x+ +--1'x- x-

3. 4x3+6x2+4x+l 22 2 1------- = x + x+2x +1


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1 0 A LG EB RA A ND G EO ME TR V

1.8 SURDS

A surd is of the form nJii (= a1/n):

I. Jii x Vb = VQi ,2. ~ = J 3. bJii cVa = (b c)Jii

EXAMPLESI. v '5 x v '2 =v'iO2 . . . . / 2 7 = .J9X3 = 3 V 33 . 3v 'iO - 2v 'W = v 'W4. - V i 4 . 14 = {l4= .,J7v '2 V2

1.8.1 RATIONALISING SURD DENOMINATORS

For an expression of the forma

b + ..jCit may be preferable to have a rational denominator. A surd denominator is rationalisedby multiplying the expression by : = ~ (= 1):

ab + ..jC

a b-..jC= --x--b+v'c b-v'ca(b - v 'c )= 1 ? - c .


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QUADRATIC EOUATION II

EXAMPLES1 . 5 5 I-V5--=--x--1+V5 1+V5 I--V5

5 - 5-V5= (1)2 - (J5)25 - 5-V5= (-4)5V5-5=

2.4

6x 6x 1- 2y'X--= = x-----=-=1 + 2JX 1 + 2JX 1 - 2JX6x -12xy'x

= 1-4x

1.9 QUADRATIC EaUATION

A quadratic equation is of the formy =ax2 + bx +c

where a,b, c are constants. The roots of a quadratic equation (when y = 0) a re

A quadratic is factorised if it is written in the form

EXAMPLES1 . The solutions to x2 + 3x + 1=0 are

x= -3 +-V52 or-3 --V5

22. The quadratic y =x2 + X - 6 is factorised into y = (x + 3)(x - 2).3. The quadratic y = x2 + 2x + 1 is factorised into y = (x + 1)2.


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12 A LG EB RA A ND G EO ME TR V

4. The solutions to 3x2 + 5x + 1 = 0 are-5 y'25 -12

x= 6so that

-5+ vTIx= 6 or -5-V136

1.10 S U M M A T I O N

nL:f(i) = f(l) + f(2) + f(3) +.. .+ f(n - 1 ) + f(n).i=l

The summation sign Lis defined as

4EXAMPLE L:i2= 12+ 22+ 32+ 42 = 30

i=l

1.11 F AC TO R IA L N O T A T IO N

n! = n.(n - 1).(n - 2) ... 3.2.1 where n is an integer.The factorial notation is defined as follows:

EXAMPLESI. 5! = 5 x 4 x 3 x 2 x 1 = 1202. O!= 1 by definition.3. n! =n(n - I)!4. 2.4.6.8 ... 2n = (2.2.2 ... 2)(1.2.3 ... n) = 2nn!


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PERMUTATIONS 13

1.12 P E R M U T A T I O N S

pn= n !r (n - r)!

A permutation is a particular ordering of a set of unique objects. The number of per-mutations of r unique objects, chosen from a group of n, is given by

EXAMPLEThe number of ways a batting lineup of 3 can be chosen from a squad of 8 cricket piayers is given by

8 8! 8!P3 = (8 _ 3)1 = 5! = 8 x 7 x 6 = 336.

1.13 C O M B I N A T I O N S

cn= n !r r!(n-r)!

If order is not important when choosing r things from a group of n then the number ofpossible combinations is given by

EXAMPLES1. The number of possible groups of 4 delegates chosen from a group of 11 is given by

11 11! 11! 11 x 10 x 9 x 8C4 = 4!(11 _ 4)! = 4!7! = 4 x 3 x 2 x 1 = 330.

2. The number ways of choosing a team of 5 people from 7 is cl =21.


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14 A LG EB RA A ND G EO ME TR V

1.14 GEOMETRY

The trigonometric ratios can be expressed in terms of the sides of a right-angled triangle:

c a()

b. 0 asin =-,e

b a sinOcosO = -, tan 0 = - b = --0e cosThe longest length, opposite the right angle, is called the hypotenuse.Pythagoras' Theorem states

The sine, cosine and tangent of the common angles can be related to the followingtriangles:

1v '21

Jrj41J2


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GEOMETRV 15

EXAMPLE

The three common triangles are theI. isosceles: any two sides are of equal length.2. equilateral: all three sides of of equal length.3. right angled: one of the angles is i.

All triangles have three angles that sum to 1f.

EXAMPLES1. A right angled triangle has one other angle - i . Hence the third angle is i.2. An equilateral triangle must have three identical angles ofi.1.14.1 C I R C L E S

A circle of radius r has

2, circumference =21fr

EXAMPLESI. The area of a circle with diameter d = 6 is 11"32= 911".2. The circumference of the circle with diameter d = 7 is 711".


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1.15 EXAMPLE QUESTIONS 5. Write the following expressions as partial fractions.(i) 3(Answers are given in Chapter 14) (x - 2)(x - 4)(ii) 4x -11. Simplify the following. (x - 1)(x + 2)

11 5 1 (iii) x2+5x+6i) : 2 - ( ; + 10 3xx 5 (iv) (x - 2)(x + 4)(ii) -----x-3 x+2 1(v) (x + 3)2(X - 2)(iii) x-I 2x-----x+2 x-2 6. Simplify the following.

(iv) 2x+ 1 x-I (i) y'27v'3-----x-4 x+2 v '5(ii) y'45x2 - 2 x+ 3(v) --+-- v'I7 + 5v'I7-I x (iii)x3 _ x2 2v'I7(vi) x(x2 - 1) 2(iv) 3+v'3

2. Find the solution set for the following inequalities. 7. Factorise the following quadratic equations.

2d+2:::;4d-3 (i) Y = x2 + 6x + 5(i) (ii) Y = x2 - 6x + 5(ii) 3d-2>4d+6 (iii) Y = x2 +4x - 5(iii) Ix -101 < 5 (iv) y = x2 - 4x - 5(v) y = 2X2 +x-l(iv) Iz+31 ~ 8 8. Find the zeros of the following quadratics.(v) la+41 > 1 (i) Y = x2 +4x+4(vi) 1~-~1


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CHAPTER 2FUNCTIONS AND GRAPHS

.2.1 THE BASIC FUNCTIONS AND CURVES

The standard functions and shapes areI. Straight Lines: y =mx + c2. Quadratics (parabolas): y = ax2 + bx + c3. Polynomials: y =anxn +.. .+ atX + ao

14. Hyperbola: y = -x5. Exponential: y = eo; = = expx6. Logarithm: y = I n x7. Sine: y = sin x8. Cosine: y =cos x9. Tangent: y =tan x10. Circles: y 2 + x2 = r2

( y ) 2 ( X " ) 211. Ellipses: a : + b =1.


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18 FUNCTIONS AND GFiAPHS

2.2 FUNCTION PROPERTIES

A function is a IUle for mapping one number to another. For example: f(x) = x2 is amapping from x to x2 so that f(3) =32 =9.

EXAMPLESI. If f(x) =3x + 1 then f(2) =7 and f(a) =3a + 1.2. If f(z) =Z2 - 1 then f(l) =0.

The domain of a function is the set of all possible input values for that function.

EXAMPLESI. Y =x2 + 4 has domain of all real numbers.2. Y = 1j(x - 1) has domain x t- 1 . That is, all real numbers except x = 1 can be used in this

function. If x = 1 then the function is undefined because of division by zero.Sometimes the domain is defined as part of the function such as y = x2 for 0 < x < 3 so that thedomain is restricted to be in the interval zero to three.

The range of a function is the set of all possible output values for that function.

EXAMPLESI. Y = x2 has range y ~ 0 since any squared number is positive.2. Y = sin x has range -1 :5 y :5 1 since the sine function is always between positive and negative

one.3. Y =x2, < x < 3 (so the domain is restricted to x E (0,3) has range 0 < y < 9.


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FUNCTION PROPERTIES 1 9

The argument of a function could be the value of another function. For example ifI(x) =x2 and g(x) =x + 1 then

I(g(x)) = (g(X))2 = (x + 1)2.

EXAMPLES1 . If f(x) =3x - 1 then I( x + 1)=3 (x + 1) - 1=3x + 2.2. If f(x) =2x + 1and g(x) = cos(x) then f(g(x)) =2 cos(x) + 1and g(1(x)) =cos(2x + 1).

The inverse of a function is denoted 1-1x ) and has the property that

EXAMPLES1 . 1 x) =x2 and g( x) =.jX are inverses since ..jX'i = (vfx)2 =x.2. IfJ(x) = 3x2 + 1 then the inverse is found by rearrangement:

f(x) = 3x2 +1= = = > ~ = x

= = > f-l(X) = ~ .

The zeros of a function, I(x), are the values of x when I(x) =O .

EXAMPLES31. f(x) =2x+3haszerox =--.. 2

2. f(x) = x2 + 3x + 2 has zeros x = -1, -2.


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y - b = f(x - a).A graph y = f(x) shifted from being centred on (0,0) to being centred on (a , b ) is writtenin the form

EXAMPLESI. A circle with centre (1,2) has form (x - 1)2 + (y - 2)2 = r2.2. A parabola y = x2 with turning point (0,0) if shifted to having turning point (3,4) has equation

(y - 4) = (x - 3)2.y

11

to

8

6

4

2

0 X0 2 3 4 5 6

A function is even if f(-x) = f(x) and odd if f(-x) = -f(x).

EXAMPLESI. Y = f(x) = x3 is odd since f( -x) = (-x)3 = -x3 = - f(x).2. Y = f(x) = X4 is even since f ( -x ) = (_X)4 = X4 = f(x).


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STRAIGHT LINES 21

.2 .3 STRAIGHT LINES

y=mx+aA line has the general form

where a and m are real numbers and m is the slope of the line.

EXAMPLESI. Part of the straight line y = 0.6 - x is drawn in the following diagram:

Y1.0

~--~--~----~--~----~--~x1.0

2. The line y =2x + 1 cuts the x axis when y =0 giving x =- ~as the zero.1 x 13. The line 5y = x-I has slope m = " 5 since it can be rewritten as y = " 5 - " 5 '

4. The equation of a line that passes through the points (0, -1) and (3,0) is y =~-1. The gradient.. 3is found fromY2-Yl 0+11m= =--=-.X2 - Xl 3 - 3


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2.4 Q U A D R A T I C S

A quadratic (parabola) has the general formy = ax2 +bx+ c

and can have either no real zeros, one real zero or two real zeros.If the quadratic has two real zeros, C!, C2 then it can also be written as

y = a(x - ct}(x - C 2 ) .

EXAMPLESections of the three quadratic functions

y = (x - 1)2 + 1, y = (x - 3)2, Y= (x - 5)(x - 6)are drawn in the following diagram:

Y2.5

2.0

1.5

1.0

.5

.0 +---,----,--------"-f"------,,-------\,--------f------, Xo 7


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POLVNOMIALS 23

.2.5 P O L Y N O M I A L S

A polynomial has the general form

where ai,i= O . . . n, are real numbers, and has the following properties.I. The polynomial has degree n if its highest power is xn.2. A polynomial of degree n has n zeros (some of which may be complex).3. The constant term in the above polynomial is n o .4. The leading order term in the above polynomial is anxn since this is the term thatdominates as x -+ 00.

EXAMPLES1. Y= 2x3 + 4x2 + 1 has degree 3, constant term I and leading order term 2x3.2. y =x2 + 5x + 6has two zeros x =-3 and x =-2.3. The third degree polynomial y = (x - l)(x - 2)(x - 3) = x3 - 6x2 + llx - 6 is plotted below

for x E [0,4]:y

6

x

6


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2 4 FUNCTIONS AND GFiAPHS

2.6 HYPERBOLA

A hyperbola centred on the origin is usually written in the formky= -x

although other orientations of hyperbolas can be written as

or

EXAMPLEThe hyperbola y = 0.15 is drawn in the following diagram:x

y1.5

1.0

0.5

0.5 1.0X

1.5

The hyperbola above is not defined for x = o .


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EXPONENTIAL AND LOGARITHM FUNCTIONS 2 5

.2.7 EXPONENTIAL .AND LOGA.RITHM FUNCTIONS

y =e~ = = expxThe exponential function is

with its inverse the logarithm functiony = lnx.

The general properties of the exponential are listed in the next chapter on transcendentalfunctions.

EXAMPLEThe exponential function y = e~ (upper curve) and logarithm function y = Inx (lower curve) aredrawn in the following diagram:

y8

6

-4

The logarithm function is not defined for x :5 o .


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2.8 TRIGONOMETRIC FUNCTIONS

The main trigonometric functions are sinx and cos z, which are cyclic with period 27rthus sin(x + 27r) = sin x. Sine and Cosine can be defined in terms of angles as discussedin sections 1.14 and 3.4.

EXAMPLESI. The function y =sin 2x will have a period of 1f.2. The functions sin x and cos x are plotted below for the first period x E [0, 27r], while

tan x = sin x/cos x is plotted for x E [-7r / 2 ,1 1 : / 2 ] .Y1.0

-1.0 sin x

- r rJ2

y1.0

0.5

O+----4----~--~~--~x 2~

cos zY6

6tan x

,------.------~-=----_,------_. x1 1 1 2


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CIRCLES 27

.2.9 CIRCLES

x(t) =r c ost, y (t) =rsint, t E [0,2111

A circle centred on the origin has the general equationx2 + y2 = r2

where r is the radius. This is often written in parametric form

EXAMPLESI. The circles x2 + y2 = 1 and (x - 2)2 + (y - 1.5)2 = 1 are drawn in the following diagram:

y3

2

~----~----~~----~----~ X-1 2 3

2. The curve x2 + 2x + y2 + 4y = -4 can be written as (x + 1)2 + (y + 2)2 = 1, which is a circlecentred on (-1,- 2) with radius I.

3. The curve represented by x(t) = 2 cos t + 1, y(t) = 2 sin t - 3 , t E [0,211")is the circle radius 2centred on (1, -3).


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2.10 ELLIPSES

A n ellipse centr ed o n the o rigin has the gener al equatio nC1X2 + C2XY + C3y2 = 1.

If th e x an d y ax es are the ax es o f the ellipse then it is usually w ritten in the fo rmx 2 y 2a2+62=1

where 2a is the length o f the ellipse in th e x d ir ec ti o n a nd 2b the length o f the ellipse inth e y d ir ec tio n. A n e llips e is o fte n w ritte n in p ar ame tr ic f o rm

x(t) = asint, y(t) = bcost, t E [ 0 , 2 x ] .

EXAMPLESI. T he ellipse (~ ) 2 + x2 = 1 is dr aw n in the fo llo w ing diagr am :

v

-2 2x- ,

2. The curv e (x - 2)2 + 16y2 = 1 is an ellipse centred o n (2,0) w ith m ajo r ax is o f length 2 in the xdirectio n and m ino r ax is o f length ~ .


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EXAMPLE QUESTIONS 29

2.11 EXAMPLE QUESTIONS(Answers are given in Chapter 14)1. If I(x) = x3 + 1 what is 1(2)?2. If I(x) = x3 + 1 what is I (g )?3. If I(x) = x3 + 1 and g(x) = (x - 1) what is I(g(x))?4. If I(x) = x3 + 1 and g(x) = (x - 1) what is I(g(b))?5. If I(x) = x2 and g(z) = sin z find I (g (a)) and g(l(x)). 27.6. If I(x) = x2 + 1 find 1(I(x)).7. If I(x) = (x - 1)2 and g(x) = x2 - 1 find I(g(x))

and g(l(x)).1 ._8. If I(x) = - + 1 find the mverse I lex).x1 . _9. If I(x) = -- find the mverse I lex).x+ 11 ._10. If I(x) = 2" + 1 find the inverse I lex).x

Lines11. Draw the line y = -2 x + 1 for x E [0,1].12. Where is the zero of the line y = x-I?13. Where does the line 2y + x- I = 0 cross the y axis?

What is the slope of the line?14. Draw 3 y - x+ 3 =Ofor x E [0,4].

Quadratics15. Draw the quadratic y = x2 - 2x + 1 for x E [0,2].16. Where are the zeros of the curve

y = (x - 3 )(x - 4)?(For more questions on manipulation of quadratics seeChapter 1.)

Sines and cosines17. Draw the curve y = 2sin3x from x = 0 to x = n,18. Draw the curve y = cos :z ; from x = 0 to

2x = 47r.19. Draw the curve y = cos 2x + 1 from x = 0 to x = 27r.20. What is the period of y = sin(x + I)?21. What is the period of y = cos 3 x?22. What is the period of y = sin(3x + I)?

Circles and ellipses23. Draw the circle y2 + (x - 2)2 =4.24. Draw the ellipse y2 + 2x 2 = 1.25. Draw the ellipse 4 y2 + (x - 1)2 = 1.26. Where does the ellipse (x - 1)2 + 2y 2 = 1 cut the x

axis?What is the equation for an ellipse centred on (0,0) withx axis twice as long as the y axis?

28. What is the equation for a circle centred on (1, 2) withradius 2?

29. What is the equation for a circle centred on (a,2) withradius 3?

General30. What type of curve has equationy2 + (x - 1)2 - 2 =O?31. What type of curve has equation2y 2 + (x - 1)2 - 2 =O?32. What type of curve has equation2y+(x-1 )2-2=0?33. What type of curve has equation

2y + (x - 1) =O?34. What type of curve has equation2-- + (x -1) =O?y -135. What is the equation of the quadratic below:

432

0+----,---,-------,-----, xo 23436. What is the equation of the shape below:

y

x


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30



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CHAPTER 3TRANSCENDENTAL FUNCTIONS

3.1 EXPONENTIAL FUNCTION

I(x) =a;l:, so that x =log, I, a> 0,An exponential function is defined by

where a is the base and x is the index.

EXAMPLESI. If 8 = x3 then x = 81/3 = 2.2. If3 = log2 y then y = 23 = 8.3. If 2 = 10glO Y then y = 102 = 100.4. If y = log2 16 then since 16 = 24, Y = 4.

The most useful exponential function is I(x) = e" = = exp e where e = 2.71828 ...


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32 TRANSCENDENTAL FUNCTIONS

3.2 INDEX LAWS

I. a/ = 0,.0, ... a, for ian integer .. . _ , _ _ . .itimes2. aman = am+n

am3. - = am-nan} Equal Bases Rule

4. ambm = (ab)m5. :: = G ) m } Equal Indices Rule

16. a-n=-an7. (am)n = amn8. 0,0 =1

Power of a Power Rule

EXAMPLES

3. To simplify y = 3293 write 9 = 32 so that

4- = 641/3 = 4ysoy = 1 .


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LOGARITHM RULES 33

3.3 L OG A .R IT HM R U LE S

LQ g of a Product. lo~(xy) = log; x + log, y2. log, (;) = logax -logaY3. log, x P = p log; x4. lo~(a;l:) =x

Log of a QuotientLQ g of a Power

5. a1og"z=x6. log, 1 = 0 and log; a = 1

EXAMPLES

3. Ifloga = 4, log b = 9 then log(a2b3) = 210ga + 3 log b = 8 + 27 = 35.4. log2 (~:) = x - Y

The natural logarithm of z, the inverse of the exponential function eX, is loge X = = In X(also denoted log x):

lnz = c means eC = z.Note that:I. Inez = x2. e1nz = x3.1ne=14. In! = 0


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EXAMPLES

1. exp(3In 2) =exp(In 23) =exp(In 8) =82. e",+ln2 =e"'eln2 =2e'"

In",3 eln",-2Iny =_e_ =~. e21ny y2

4. Ifln X =2 and In y =5 then to find In(x3y2) we writeIn(x3y2) =Inx3 + Iny2

=3Inx + 2Iny=3(2) + 2(5)=16.

5. If In y =3In 2x + c then to find y writeIny =In (2x)3 + c

===} y =exp(In (2x)3 + c)=kexp(In (2x)3), where eC =k=k(2x)3.

6. If x =In 3 and y =In 4 then to find exp( x + 2y ) writee",+2y =e"'(ey)2

=3 x 42=48.

8. Ify = a '" thenIn y =x In a ===} y =e'" In a .


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TRIGONOMETRIC FUNCTIONS 3 5

3.4 TRIGONOMETRIC FUNCTIONS

The unit circle can be used as an aid for finding the sin and cos of common angles. Forexample, cos 7 r /6 =../3/2. By symmetry all the other major angles can be found.

sinO

-1

EXAMPLES1. From the diagram we see that

11 " v acosij =2 7r 1cos " 4 = .J2 ' 11 " 1cos " 3 = 2' CQS7r = -1.? .51r=.~_!~. sm 6 - sm 6 - 2

3 7 r 11 "3. tan4= -tan " 4 =-14. cos(n7r) = (-1)'\ n =0, 1, 2, ...5. sin(n7r) = 0, n = 0, 1, 2, .... (2n+ 1 ) 7 r n6.SlD 2 =-(-I),n=0,I,2, ...


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sin(-a:) = -sin(a:), cos(-a:) = cos eSine is an odd function while cosine is an even function.

1cosec a:= -.-,sm z1see a:= --,cos a:

1cot e = --.tan a:

The Reciprocallligonometric Functions are

3.5 TRIGONOMETRIC IDENTITIES

A fundamental trigonometric identity is

EXAMPLESI. To prove the identity tan a:+ cot a: = sec a:cosec a:consider the left hand side:

sin a: co s ztan x + cot a:=-- + -.-cose sineSi82 x + cos2 x

co s x sin x1---- =see x cosec x.cosxsinx

2 . It is easy to prove1+ tan2 x = sedl xcot2 x + 1 = cosec2 x

by simply dividing sin2 a:+ cos2 a:= 1 by either sin2 a: or cos" a:.


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TRIGONOMETRIC IDENTITIES 37

sin(a: + y) = sin e cosp + co s e sinycos(a: + y) = CQ Sa: cos y - s in a : si n y

sin2a: 2 sin a: co s a:cos2x co s2 x - si n2 a:si n2a: 1-cos2a:= 2CQ S2a: 1+ cos2x= 2

EXAMPLES1 . sin(x - y) =si n x co sy - co s x siny

2. co s(a: - y) = c o s X C Q sy + s in a : s in y3 . sin ( x + ~) = si n X co s ~ + co sx sin ~ = co s z4. co s(x + 1 1 ) =co s x C OS 1r- sin x si n 1 1 =- co s x5 . To find sin ~ co nsider

6 . A l te r na ti v el y th e fo llo w ing m etho d can be used:

1 1 : J l - CQs1rj6sm-= .12 2- J l - V3j2- . 2 .


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3.6 HYPERBOLIC FUNCTIONS

EXAMPLESI. It is easy to show that

sinh 0 =0cosh 0 =1

and thatCOSh2X - sinh2 x =1

since

2. The plots of sinh x and cosh x are illustrated below on the interval x E [- 2, 2 1 .

-2.5sinh x

-5.0


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EXAMPLE QUESTIONS 39

3.7 EXAMPLE QUESTIONS(Answers are given in Chapter 14)1. Simplify as much as possible

1(i) 6 x 3y-2 x _ x - 5y4242(ii) 8-"3

(iii) 210g105 + 10glO 8 - 10glO 2(iv) 3-1og3 p(v) Inx2 + Iny -Inx _lny2(vi) e21nx

2. Solve for t using natural logarithms:(i) 5t = 7(ii) 2 = (1.02)t(iii) 3t7 = 2t5(iv) Q = Qoant(v) y=3-2Int(vi) 3y = 1 + 2e4t

3. If Ins = 2 and Int = 3 calculate(i) In(st)(ii) In(st2)(iii) In(Vsi)(iv) In~ts(v) In- t3

4. If x = In3 and y = In5 then find(i) eXeY(ii) eX+Y(iii) e2x(iv) eX + eY

5. Evaluate(i) tan(n)(ii) . ( 6 n )m 8(iii) cos C ~ n )(iv) sec ( 4; )

6. Simplify(i) _1 __ tan2 (Jcos2 (J(ii) (sin x + COSX)2 + (sin x - COSX)2

tan(J(iii) VI + tan2 (J

7. Solve the following for values of (J between 0 and 2n(i) cos2 (J + 3sin2 (J = 2(ii) 2 c os2 (J = 3 sin (J

8. Prove the following identities:1 + sin(J(i) --. -( J = (sec (J + tan ( J ) 21-sm

(ii) 3sin2 (J - 2 = 1- 3cos2 (J(iii) sinh x - cosh x = _e-x(iv) sinh x + cosh x = e"

9. Use the trigonometric addition of angle formulae to showcos'lr_ = ~(v'6 +h).12 4

10. For the following angles find cos (J , sin (J , tan (J , and sec (J :

(i) ( J = 7 r _ 4(ii) (J = 137r_6(iii) (J = 2n3(iv) (J = _ 5n3(v) (J = 5n4

11. Use the multiple angle formulae to find cos 'lr_.1212. In an experiment you have to calculate the time to melt a

block of ice using the formulat= _ l ( , - - A _ : _ P _ - _ c _ T , - - - , o , -, p _ , _ )t c r ;

wherel =0.1, A = 3 X 105,c = 2 X 103, To = -20,Ta = 20, h = 10,P = 1 X 103.

Find t.13. Is I(x) = x cos X an odd or even function?


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40



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CHAPTER 4DIFFERENTIATION

4.1 FIRST PRINCIPLES

The definition of a derivative of a function f(x) is:f(x) = df = lim f( x + h) - f(x).dx h-+O h

This is the slope of the tangent to the function f (x ) at the point x . The following diagramiIIustrates:

f ( x + h }y= f ( x }

f ( x )

t a n g e n t

x x+h


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42 DIFFERENTIATION

EXAMPLESI. If lex) = x2 then

2. If f(x) = sin x then

since

f'(x) = lim (x + h)2 - x2. h-->O h

. x2 + 2h x + h 2 - x2= hm------~-----h-->O h= lim 2hx+ h2

h-->O h= lim 2x+ hh-->O= 2x .

f'(x) = lim sin(a; + h ) - sin x. h-->O h

li sin x CQSh + cos x sin h - sin x= m - - - - - - - - - - ~ - - - - - - - - -h-->O h= lim sinx(cos h -1) + cos x sin hh-->O h

= CQSX

lim CQsh-1 =0h-+O h '

(see the Asymptotics chapter for how to evaluate these limits).

4.. 2 LINEARITY

where c is a constant,

!f (x) + g(x))~(c!(x))dx


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SIMPLE DERIVATIVES 4 3

EXAMPLESd. d .1 . dx (3smx) = 3 dx sm z = 3 cosx.

2. If f (x) = sin x + e~ then ! , (x) = ~SinX + ~ e~ = cosx + eX.

4.3 S IM P LE D E RIV A TIV E S

f(x) -+ f'(x)c -+ 0 where c is a constant.xn -+ nx n-1

sin x -+ cos xcos x -+ -sinx

e 1 : -+ e 1 :ln z 1-+ -xsinh x -+ eosh z

cosh x -+ sinh x

The following derivatives of elementary functions are standard:

EXAMPLESd 11 . -d. (41nx) = 4-x . x

2. If J(x) = 5x 2 + sinh x then J'(x) = lO x + cosh x.

4.4 P R O D U C T R U L E

d df dgdx[f(x)g(x)] = dx g (x) + f(x) dx


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44 DIFFERENTIATION

EXAMPLESI. If !(x) = X2 sin x then f(x) = 2x sinx + x2 coax.2. If f(x) = lnx cos x then P(x) = . ! : . CQSX-lnx sin z.x

4.5 QUOTIENT RULE

. ! ! _ [f(X)] = jI(x)g(x) - f(X)gl(X)dx g(x) (g(x))2

EXAMPLESx 2I. If f(x) = -. -. thensmx

2' 2f'(x) = x smx. - x CQSx.sm2x

sin x2 . If f(x) = -- thencos xI. cos X CQSX - sin x( - sin x) 1f (x ) = cos2 X = cos2x '

Thus

3 .j_ (Sinx) = CQs(x)x2 - 2xsinxd x " x2 X4

4.! ! _ ( x 2 - x ) = l i[x2 - x] (x 2 + 2) - (x 2 - x ) ~ [ X 2 + 2]dxx2 + 2, (x2 + 2)2

(2x - 1 )(x2 + 2) - (x 2 - x) 2x= ~--~~~~~--~-(x 2 + 2) 2x2 +4x- 2(x2 + 2)2


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CHAIN RULE 4 5

4.6 CHAIN RULE

d d f dgd x [f(g (x )) ] = dg dx = /,(g(x))g'(x)Differentiate the outer function first then multiply by the derivative of the inner function.

EXAMPLES1. Since ~ sin x = cos x then

d ddx [sin(x2)] = cos(x2) dx [ x 2]=cos(x2)2x.

2. Since!n x =~thend 2' 1 d, 2dx [In(x + x )] = x + x2 dx [x + x ]

1+2x

3. Since ~ x3 =3x2 and ~ sin x = cos x then![sin3x ] = 3sin2x cos z.

4. Since!os x =- sin x and!5 =5x4 then~ [CQSX2+ 3X)5)] = - sinx2 + 3X)5) ~ [(x2 + 3X)5]dx ' dx

= - sinx2 + 3X)5)5(x2 + 3X)4 d~ [(x 2 + 3x)]= - sinx2 + 3x)5)5(x2 + 3X)4(2x + 3)= -5(2x + 3)(x2 + 3X)4 sinx2 + 3X)5).


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46 DIFFERENTIATION

4.7 IMPLICIT DIFFERENTIATION

To find y'(x) where y(x) is given implicitly, differentiate normally but treat each y as anunknown function of x. For example, if given

f (y) = g(x)then differentiating gives

f '(y): =g'(x) ===} dydxg'(x)f '(y)

where the chain rule has been used to obtain the left hand side.

EXAMPLESI. Differentiating

siny =x2with respect to x where y = y(x) gives

dycosy dx =2xor dy 2x

dx CQsy2x

Vl- X4since CQsy = y'l- sin'.:!y = -v ' f="X4.

2. Differentiatingx cosy+y = x3

with respect to x where y =y(x) gives

( dY) dy 2cosy - xsm(y) dx . + dx = 3 x

which can be rearranged to givedy 3 x2 - CQsy=---..:...dx 1-xsiny


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PARAMETRIC DIFFERENTIATION 47

4.8 P A .R A M E T R IC D IF FE R E N T IA T IO N

Given y = f(t) and x = g(t), dyjdx may be calculated asdy _ dy/dt _ /,(t)dx - dxjdt - !I(t)'

EXAMPLES1 . Ify(t) =t2 and x(t) =sin t then

dy = dyjdt = 2tdx dxjdt cos t

2. Ify(t) = sin r and x(t) = cost thendy = dyjdt = cost = _ cottdx dxjdt - sint .

4.9 S EC ON D D ER IV AT IV E

f'(x) = ~; = ~ (!).The second (or double) derivative is the derivative of the derivative:

Higher derivatives are found by repeated differentiation.

EXAMPLES1. If J(x) = X4 then /,(x) = 4x3and r(x) = 12x2.2. If 5(t) = e2t is the position of a particle with time t, then 51(t ) = 2e2t is the velocity and

5"(t) =4 e2t is the acceleration.


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48 DIFFERENTIATION

4.10 S TA TIO N A RY P OIN T S

A stationary point is a point (x , y) where f'(x) =O. At this point the tangent to thegraph is flat.

EXAMPLESI. The function y = x2 + 2x + 2 has a stationary point when

d y =2x+2=Odx x =-1.

2. The function y =2x3 - 9x2 + 12x has stationary points whendy = 6x2 _ 18x + 12= 0dx x= 1,2.

3. The function y =xe-z has a maximum when

x = 1.

A local maximum is when the function at the stationary point is higher than the sur-rounding points. A local minimum is lower than the surrounding points. An inflectionpoint is where the graph is flat but neither a maximum nor minimum.

Y maximum: r>.:"J ~nminimum x+----;---.----;---.-- ......--,o 2 3 5 6


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STATIONARV POINTS 49

At a stationary point x =a the second derivative indicates the type of stationary pointI. if f'(a) > 0 then x = a is a local minimum.2. if f'(a) < 0 then x =a is a local maximum.3. if f'(a) =0 then x =a is an inflection point.Note that x = a is a stationary point so f(a) = O .

EXAMPLES1. The function y =x2 + 2x + 2 has a stationary point at x = -1.The double derivative is

so x =-1s a minimum.2. The function y = 2x3 - 9x2 +12x has stationary points at x = 1 and x = 2. The double derivative

is~yd~ =12x-18x

which is positive at x = 2 (a minimum) and negative at x =1 (a maximum).y

1086420 X

0 2 3

3. The function y = (x - 1)3 + 3 has derivativesdy =3(x _1 )2dx '

~y- = 6(x -1)~2 "which are both zero at x = 1, which is therefore an inflection point.


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50 DIFFERENTIATION

4.11 EXAMPLE QUESTIONS(Answers are given in Chapter 14)1. Use linearity to find dy/ dx:

(i) y = 3sinx - 5cosx(ii) Y = 3e'" - x2(iii) Y = 3lnx(iv) y = 2sinhx - 3coshx

2. Use the chain rule to find dy/ dx:(i) y = sin(2x)(ii) y = sin(x + x3)(iii) Y = (x + 4)3(iv) Y = (x + sinx)5(v) y = sin(lnx2)(vi) y = exp(cos2 x)(vii) y = cosh(2x2)

3. Use the product rule or the quotient rule to find dy/ dx:(i) y= xe'"(ii) cos xy=-- x2(iii) Y = e"'sinx(iv) lnxy=- x4(v) Y = sinx cos x(vi) lnxy=- e'"(vii) y = x2sinx(viii) y = cosh x sinh x

ell'"(ix) y=- x4. Find dy/ dx for these more difficult problems:

(i) y = exp(x cos x2)(ii) Y = e'" cos((2x + 1)2)

5. Use implicit differentiation to find dy/dx:(i) y2 = sin(x - 1)(ii) cos(2y) = (1 _ X2)1/2(iii) In(y) = xe'"(iv) eY = e3 ", + 5(v) y + y3 = x2(vi) y2 + siny = sinx(vii) y(x + 1) - y2 = x

6. Use parametric differentiation to find dy/ dx:(i) yet) = cos t, x(t) = sin(t2)(ii) yet) = et, x(t) = t2(iii) yet) = t2, x(t) = sin t

7. Find the derivative dT/ dt:T=texp(~)

where a is a constant.8. For the following functions find the stationary points and

classify them.(i) Y = (x - 2)2(ii) Y = x3 - 6x2 + 9x + 1(iii) Y = 3x4 - 8x3 + 6x2(iv) y = xe-'"(v) y = x2In(x)(vi) y=sinx+(1-x)cosx,forxE[-1,2](vii) y = (x - 1)2e'"

9. The function y = f(x) is drawn below. Roughly sketchthe function fl(x).

y1(iii) Y = ,j2+x2 X

sinx 5(iv) y= (x+1)2 -1Y = sin(x2 + exp(x3 + x)) -2(v)

expx2 -3(vi) y=-- -4x2


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CHAPTER 5INTEGRATION

5.1 A N T I D I F F E R E N T I A T I O N

!(x) dx = F(x) + cThe indefinite integral (antiderivative) of f with respect to x iswhere F1(x) = f(x) and c is known as the constant of integration.

EXAMPLES

1. If!2 = 2x then !2X dx = x2 + c.2. If ~ In x = _ !_ then !~,x = In I x l + c.dx x ...3. If d~ sin x =cos z then !os x dx =sin x + c.

d. I. If dx sinh x = cosh x then ,cosh x dx = sinh x + c.


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52 INTEGRATION

5.2 SIMPLE INTEGRALS

The following integrals of elementary functions are standard:/ xn d x 1= __ xn+l +c where n f : . -1n+l

/ cosxdx = sin x + c/ sinxdx - cosx + c

/ e Xdx = e" +c

/ 1 I n [z ] + c; dx/ sinhxdx = cosh z + c/ coshxdx = sinh x + c

Integration is l inear so that/U(x) +g(x))dx

/ c f(x) dx= / f(x) dx + / g(x) dx,

c / f(x) dx, where cis a constant,

EXAMPLESI.!sin x + eX) dx = - cos x + e " + c2.!cos x d a : =5 sin x + c3. Simple application of the Chain Rule in differentiation gives

/1coskx dx = I i sin kx + c.


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THE DEFINITE INTEGFIAL 53

4. f 6x dx = 3f 2x dx = 3x2 + C6. f 3sin(2x)dx = -~COS(2X) +C

5.3 THE DEFINITE INTEGRAL

l b f (x) d a : = [F(x)] != F(b) - F (a)

The definite integral with respect to x over the interval [a , b ] is written as:

where F' (x ) =f (z). This is the Fundamental Theorem of Calculus.

EXAMPLES

11 [ 1 ] 1 1 2. sin 7 rxdx = -- co s 7rX = --(COS7r - co s O) = -o 7r 0 7r 7r[3 [ X 4 ] 3 81 13. 11 x3 dx = '4 1= '4 - 4 : = 20

4. r sinh x dx = coshj l)-cosh( -1) = 0 since cosh(x) is even.1 -1


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54 INTEGRATION

l a f(x)dx=Ol b f(x) dx =-ia f(x) dxl b f(x) dx +lf(x) dx = l C f(x) dx(assuming f(x) can be integrated over the required intervals).

EXAMPLESI. r [z] dx = r Ixl dx + r [z] dx}_1 1-1 10

= r -xdx + (2 z d}_1 10= - [ ~ 2 [ 1+ [ ~ 2 J :=- [0 - ~] + [2 - 0] =~

2. If f(x) = {!: :~~'hen

3. If f(x) is odd thenfa {o (ai-a f(x) dx =}_a f(x) dx + 10 f(x) dx =O.


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AREAS 5 5

5.4 A . R EA S

If f is an integrable function thenf a l : > f(x) dx = (area above the z-axis) - (area below the z-axis)

in the region a :5 x :5 b .

EXAMPLES1 . Consider the curve given by f(x) = x3 - 9x 2 + 26x - 24 = (x - 2)(x - 3 )(x - 4). The area

between the curve and the z-axis between x =2 and x =4 is given by:y

0.5

1411 (X)1x =13f(x) dx - f s 4 /(x) dx= 0.25+0.25=0.5.

2. If /(x) is an even function (so f( -x) = f(x)) thenf c f(x) dx = 2 r l(x) dx-c 1 0since the area for x E [ -c ,O) is the same as for x E [0, c ).


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56 INTEGRATION

5.5 INTEGRATION BY SUBSTITUTION

Integrals that can be written in the formf f (g(x))g'(x) dxare solved by the substitution u = g(x) , upon which the integral becomesf f(u) du = F(u) + c = F(g(x)) + c,where F'(x) = f (x). For definite integrals the limits of the integration a re also trans-formed.

EXAMPLESI. To evaluate f 2x sin( x2 + 1)dx let u =x2 + 1 then ~: =2x so thatf 2x sin(x2 + 1) dx = f sin ud u

= -cosu+c= - cos(x2 + 1 ) + c.

f 1 du2. To evaluate r::-71dx let u = x + 1 then -d = 1 so thatvx+ 1 xf 1 f--== dx = u-! du"';x + 1= 2u! +c= 2";x + 1+c.

1,/23 . To find 0 sin" x cos x d$ let 1 = sin x so that the integral becomesliU(.,.-/2) [ 1 5 . ] 1 1u4du= - =-,.sin(O) 5 0 54. To find! hx let x = sin 1 since ";1 - x 2 = cos u and dx = cos 1 d u so thatl-x2 f 1 f cos 1 --===dx = -- d 1 = U = arcsmx+c.~ cos 1


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INTEGRATION BV PARTS 57

5.6 IN T E G R A T IO N B Y P A .R T S

I d V I d U- dx = uv - v- dxdx dxIntegration of a product of two functions can sometimes be solved by integration by parts:

or in short hand, 1u dv = uv - 1v duoEXAMPLES

/dv du .1. To evaluate x cos x dx let u =x and d$ =cos x then dx =1and v =sm z, so that

/ xcosxdx = xsinx - / sinxdx= xsinx - (- cosx + cd=z sin z + cosx + C2.

2. The integral t' xe2:t: dx is performed by setting u =x, ~~ =e2:t: so that ~~ =1and v = !e2:t:.J0 UOl' UOl' 2Then we have

3. Integrating by parts twice we can evaluate/ e2:t:sin xdx = _e2o; cos x + 2 / e2:t:cosxdx

= _e2o; cos x + 2 ( e 2:t:sin x - 2 1 e 2:t:sin x dx)! e2:t:so by rearranging e2:t:sinxdx = 52sinx - cosx) + C.


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58 INTEGRATION

5.7 EXAMPLE QUESTIONS(Answers are given in Chapter 14)1. Find 4. Evaluate the following integrals using a substitut ion.

(i) 12.dx (i) I xe-",2 dxx3(ii) I(~X5) dx (ii) I x+2 dx2 +4x + 5 xl-:+ e-14",) dx I 1iii) (iii) --dyylny(iv) I(x9 + x10 +Xll + X12) dx (iv) I2x3y'7x4 - 1dx(v) Isinh(2x) dx (v) I z cos (_z2) dz(vi) I4cosh(x) - e'" dx lev'svi) -dsV B2. Evaluate (vii) f o 7 r xsin(x2 ) dx(i) 1 0 5 e= dz i4 e-..;y(viii) --dy1 3 7 r / 2 9 2..ffj(ii) sinxdx 5. Evaluate the following integrals using integration by parts.7 r / 2

1 1 / 2 16(iii) -dx f o 7 r x cos xdx1/4 x (i)(iv) 1 0 7 r sin(3x) dx 1 2 1(ii) dxo x2 - 2x - 3

1 5 7 r / 4 X I e2",sin x dx (integrate twice).v) -sin (--) dx (iii)7 r / 4 3(vi) 1 0 3 te= dx (iv) I(x + 1) sinxdx3. Find (v) I x2e'" dx

(i) 1 0 2 I(x) dx where (vi) I In x dx by using u = In x and dv = 1.I(x) ={ 1, x < 1, 6. Find the following integrals using any method.x, x2:1. I sinz dz(i)

j_11 I(x) dx where cosz(ii) I x2 sinxdxii)I(x) ={ _x2, X < 0, l="3 x 2 : o . (iii)(iii) 1 0 00 I(x) dx where (iv) 1v'U(U+1)dUI(x) ={ x~, 1, x < 1, (v) I~ dx (let x = sinhu).x2:1. 1+ x2

(iv) j_11 I(x) dx where I(x) = Ix3 1 .


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CHAPTER 6MATRICES

6.1 A.DDITION

IfA and Bare m x nmatrices such that

[

auauA= .amI

[

buh 2 1B= .bml

and

[ :~~~then A+B = .amI + bmi

ain + bIn 12n + h 2 n. .amn+ bmn

au + buU 22 + h22

Addition of matrices of different sizes is not defined.

EXAMPLES [~~~+ [-~ -~ -~ 1 = [-~ ~ ~13 7 9 3 -14 1 6 -7 101.[~ ~] + [ ~] cannot be done..


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60 MATRICES

6.2 MULTIPLICATION

AB is defined if A is sizem x rand B size r x n. If

[a.ll al2 .. . a .1m 1


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IDENTITY 61

while

[ -2 2 -4] [ 4 2 9]BA = 0 0 1 -3 -1 13 -4 -1 2 1 2[

-8 - 6 - 8 -4 - 2 - 4 -18 + 2 - 8]= 0+0+2 0+0+1 0+0+212+ 12- 2 6 + 4- 1 27 - 4 - 2[

-22 -10 -242]= 2 122 9 21

f: AB.4.

6.3 IDENTITY

The identity matrix, defined only for square matrices (n x n), is

and is defined such that, for all n x nmatrices A,IA=AI= A.

EXAMPLEThe 2 x 2 and 3 x 3 identity matrices are


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62 MATRICES

6.4 TRANSPOSE

The transpose of a matrix is formed by writing its columns as rows. The transpose of anm x n matrix A is an n x m matrix denoted by At, that is, if

[ an al2 a ," ] [ an au amlU2l a22 U2n au a22 a7 , ]= then At = .aml am2 amn aln a2n amn

EXAMPLES

I. IfA ~ [~ _~ 1 men A' ~ [~ ~ _:].

If A and Bare matrices and c is a scalar, thenI. (At)t = A2. (A + B)t = At + Bt3. (cA)t = cAt4. (AB)t = Bt At

EXAMPLE


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DETERMINANTS 63

6.5 DETERMINANTS

The determinant of a 2 x 2 matrix A = [~ :] isdet(A) = IAI =ad - be.

[all

The determinant of a 3 x 3 matrix A= ana:n

(expanding by the first row).

EXAMPLES1. 1 2 1= 4- 6 = -2, 1 2 1= -1- 4 = -5.3 4 2 - 12. -3 2 1

= -31 5 ~ 1- 21 ; ~ 1+ 1 1 ; . : I5 62 -3 1 -3= -3(5 + 18) - 2(4 -12) + (-12 -10)= -75

3. 1 7 5 _ 1 1 2 ~ I + o + o2 60 0 3 - 0=lx2x3=6

4. I ~ 7 : 1 is not possible.


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64 MATRICES

6.5.1 COFACTOR EXPANSION

The determinant of an n x n matrix may be found by choosing a row (or column) andsumming the products of the entries of the chosen row (or column) and their cofactors:

det(A) = o,ljC1j + a2jC2j +.. .+ tlnjCnj,(cofactor expansion along the p h column)

det(A) =


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INVERSE 65

in tum remembering to multiply by (-l)i+j:

Cll = + 11 ~ ~ 1= -2C12=


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66 MATRICES

6.6.1 TWO BY TWO MATRICES

For 2 x 2 matrices, ifA = [~ ~] then

A -1_ 1 [ d - a b ]ad _ be -e providing ad - be f - O.If det(A)= ad - be = 0 then A -1does not exist.

EXAMPLESI.IfA=[! ~ ] then A -1= _ _ ! _ [ 4 -2 ]-2 -3 1 .2. IfA = [ ~ ;] then A-1= ~ [ 3 -2 ]3 0 1 .6.6.2 PARTITIONED MATRIX

Inverses can also be found by considering the partitioned matrix

then performing row operations until the final partitioned matrix is of the form

EXAMPLEThe inverse of

[1 2 1 10101 1 0

can be calculated using row reductions where R3 --t R3 - Rl means that Row 3 becomes the old


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INVERSE 67

Row 3 minus Row I.

[ ~ 2 1 1 0 n0 0 11 0 0 01 R3-t R3-Rl[ ~ 2 1 1 0 n0 0 1-1 -1 -1 01 R3-t R3+R2[ ~

2 1 1 0 n0 0 10 -1 -1 11 Rl-t Rl- 2R2R3-t -R3[ ~ 0 1 1 -2 J ]0 0 10 1 1 -11 Rl-t Rl- R3[ ~ 0 0 0 -1 j ]0 0 10 1 1 -1hence [ ~ 2 ~r [ ~ -1 ~ ]= 11 -1 -16.6.3 COFACTORS MATRIX

A-I = _1_CtIA I

The inverse of an x nmatrix A can be found by considering the transpose of the cofactorsmatrix divided by the determinant:

where Gil is the determinant of A with row iand column j deleted., multiplied by(-1 )i+l. The matrix C is called the cofactors matrix.


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68 MATRICES

EXAMPLESl.If

then

Cll = I ~ ~ 1 = 0C21=(-1)1~ ~1=1

and so on. Since IA I =-1e getA-1= _ _ ! _ [ ~ - ~-1 -1 0

_ ~ ] T =-1 [ ~ -~ ~ ] .1-1-12. The matrix

has cofactors matrix

[ - 2 2 0 ]C = 3 -3 14 -2 0hence the inverse

6.7 MATRIX MANIPULATION

Matrices do not behave as real numbers. When manipulating matrix expressions adistinction is made between multiplying from the left (pre-multiplication) and multiplyingfrom the right (post-multiplication).


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MATRIX MANIPULATION 69

EXAMPLES

1. Given that ABC = I find B?

ABC(A-1A)BCIB(CC-1)B

IA-1IA-lIC-1A-1C-1(CA)-l.

pre-multiply both sides by A -1post-multiply both sides by C-1simplifying

2. IfA = PDP-1, then A3 is

A3 (PDP-1) (PDP-1) (PDP-1)PD (P-1 P) DP-1 PDP-1PD2p-1 PDP-1 since pp-1 = IPD3p-1 again since pp-1 = I.

3. IfAv = A V then

A3~= AA A~ = AA(A~)= AAA~=A2Av=A3V.

4. IfAv = A V then if A-1 exists then

A-1 Av = A-1 AV = AA-lVv =AA-lV1 -1:x-~=A ~.

See Section 6.9 on eigenvalues since this example shows that if A has eigenvalue A, with eigen-vector~, then A -1has eigenvalue 1/ A for the same eigenvector.


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70 MATRICES

6.8 SYSTEMS OF EQUATIONS

Systems of m linear equations invol ving n unknowns may be written as a matrix equation.For example,

x+y+2z=12x+4y - 3z =53x+6y - 5z =2

is written as

orAx eb.

Systems of equations are typically solved by Gaussian elimination.IfA is invertible then x = A -1h.

Gaussian Elimination allows a multiple of one row to be added to another row. a row to be multiplied by a (non-zero) number.

Hence R3 -+ R3 - 2Rl means each element in Row 3 becomes the old Row 3 elementminus two times the corresponding Row 2 element.

EXAMPLESI. The augmented matrix is an easy way of writing systems of equations. For the following system

x+y+2z 12x + 4y - 3z 53x+6y - 5z = 2


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SYSTEMS OF EQUATIONS 71

the augmented matrix is

[l 1 2 n-36 -51 R2 --t R2 - 2R1R3 --t R3 - 3 R1[! 1 2 j ]-73 -111 R2 --t R2/2[!

1 2 - :-7/23 -111 R3 --t R3 - 3 R2[! 1 2 1]-7/2 3/20 -1/2 -11/21 R3 --t -2R3[ " 2 3 /~ ]1 -7/2o 0 1 11This gives the straightforward solution by back substitution of x = -61, y =40, z =11.

2. Consider the system2x - 5y -2-x + 3y 4

written as Ax =b such that

The matrix A has inverse A -1= [~ ~] so


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72 M A T R I C E S

After performing Gaussian reduction by row operations the three cases (no solution, infi-nite solutions, one solution) are typically represented by the following:

I. Ifyou perform row operations to obtain

(where a, .. , f are non-zero real numbers) then you get one unique solution.2. Ifyou perform row operations to obtain

[a b C kl 1O de k2o 0 0 k3

then if k3 - you get no solution.3. Ifyou perform row operations to obtain

[a b C kl 1O de k2o 0 0 0

then you get an infinite number of solutions that represent a line where you letz =t, t is some parameter, and then express z , y in terms of t.

EXAMPLETo solve the system:

perform row reductions to obtain

[1 -2 -1011000

and setting z = t gives y = 3 - t and x - 2(3 - t) - t = -1o(e, y,z) = (5- t,3- t, t) = (5,3,0) + t( -1, -1,1)

or a line in three dimensional space.


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EIGENVALUES AND EIGENVECTORS 73

6.9 EIGENVALUES AND EIGENVECTORS

IA - All =o .

IfA is an n x n matrix then a scalar A is called an eigenvalue of A, if associated with itthere is a non-zero vector~, called an eigenvector, such that

Av = AV .'" '"To find the eigenvalues solve the characteristic equation

To find the eigenvectors solve

EXAMPLETo find the eigenvalues and eigenvectors for

set up the characteristic equation

which gives X } . - 1 =0 so Al =1 and A2 = -1 are the eigenvalues. To find the eigenvectors solve[-A 1 ]1 _ A ~ = O .

For Al = 1 let

[ X .l. ]I =' " Y l thenBoth equations give Xl - Y l = 0 so Y l is a free variable, Hence the eigenvector corresponding toAl =1 is t(l, 1), where t is any number, t ERe.For ).2 =-Ilet

~ = [ ;~] then [~ ~] [ ;~ ] = [ ~] .


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74 MATRICES

Both equations give X2 + Y2 = 0 so Y2 is a free variable, Hence the eigenvector corresponding to).2 = -1 is p(l, -1). The length of the eigenvector is unimportant hence it is convenient to write

Vi = (1,1), V2 = (1, -1).'" '"

6.10 TRACE

The trace of a matrix is the sum of its diagonal elements.(Note that the trace is also equal to the sum of the eigenvalues.)

EXAMPLE [ 1 2 3 ]The trace of 4 5 6 = 1+ 5 + 9 = 15.7896.11 S YM M E T RIC M .A TR IC ES

The matrix A is symmetric ifA =At.

EXAMPLE [ 1 2 3 ]The matrix 2 5 6 is symmetric.3696.12 DIAGONAL MATRICES

A diagonal matrix is one with only terms along the main diagonal.

EXAMPLE A 3 x 3 diagonal matrix has the form [~ g ~ ] .


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EXAMPLE QUESTIONS 7 5

6.13 EXAMPLE QUESTIONS(Answers are given in Chapter 14)1. Find A +B, AB, BA and the trace(A):

(i)

A = [ ~B= [ -~

(ii)

A = [ !B = [ 6-7

(iii)

5. Find the determinants of the following matrices.

o3o [ ~(i) 62

[ ~~ii) 649o1

-7-62 1 ]

~ 1-1

o21211ooooo

-1-21-1]-27

~ ]-S1

(iii) [ ! ~ ]6. Solve the system of equations

x+2y-z 2Sx+3y-7z 4

4y - 12z -S.o3ooSo

~ ]J ] 4x -y 04y -z 0-4x + 17y - 4z o .

7. First showing that a non-tr ivial solution does indeed exist,solve

2. FindAB:

(i) A = [ ! ~ ]B = [ ~ 8. For what values of a and c do you get11 (i) one solution,( ii) no solution,(iii) infinite solutions,1 1j ] for the systemx+5y+z 0x+6y-z 22x + ay + z c ?iii) A = [ _~ ] B = [1 -1 1

3. Find At, At A, AAt:(i) A= [ ~

(ii) A= [~ -:]

-13

9. Find the eigenvalues and eigenvectors of A:3-5 (i)

4. Find the inverse, if it exists, of the matrices(i) [ ~

[ ~

1(ii)

!]213

(ii)

~ ] A = [ ~12o ~ ]


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76



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CHAPTER 7VECTORS

7.1 V E CT O R N O TA TIO N

A vector in R2 is represented by an ordered pair ~ = (a, b ) , or geometrically, by adirected line segment in the plane.

y

vb

a~------------------~x

The vector has both length II~II and direction.

EXAMPLES1. The vector (1,0) points in the x direction and has length 1.2. The vector (1, 1) points in a direction with angle 7r / 4 to the x axis.


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78 VECTORS

A vector in R!' is represented by an ordered n-tuple

EXAMPLESI. ~ = (1,3,4) is a three dimensional vector so:e E R3 .2. ~ = (3,5,7,1,2) is a five dimensional vector so:e E R5.

7.2 ADDITION AND SCALAR MULTIPLICATION

V+W = (Vi+Wl>tl2+W2, ... ,Vn+wn)cv (CIJ t,CiJ 2, ... ,CiJ n).

If~= (Vl, V2, ... , vn), ~ = (Wi , W2, ... , wn) and c is a scalar constant then

EXAMPLESI. If~=(1, -1,4) and ~ = (1,-3,3) then,

~+~ = (2, -4,7),and

4~= (4,-4,16).2. If i= (1,0) and j= (0,1) then 'V =(2,3) =2i + 3j ..-v .-v .-v .-v ~3. If~=(l,x,2,x) and ~ = (2,I,x,0) then ~ + ~=(3,1 + x,2 + x,x).4. (1,2) + (4,5,6) is not defined.


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LENGTH 79

7.3 LENGTH

II~II= J v ~ + v i + ... + v ~ .The length of a vector in B" is given by

EXAMPLESI. 11(1,2,1)11= Vip + 22 + 12 = .J62. 11(1,1,2,1,3)11 =V1 + 1 +4+ 1 +9 =..ji6 = 4

The triangle inequality states that

That is the length of the sum of vectors must be less than the length of the two individualvectors added.

EXAMPLE(0,3) + (4,0) = (4,3) and 11(0,3)11= 3, 11(4,0)11= 4,11(4,3)11 = V32 + 42 = 5 < 3 + 4.

y4

(4,3)(0,3)

3

2

(4,0)

2 3 4


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80 VECTORS

7.4 CARTESIAN UNIT VECTORS

The Cartesian unit vectors for R3 are~=(1,O,0), t=(O,1,0), ~=(O,0,1).

Vectors in If 3 are often written as the sum of the components in the direction of theCartesian unit vectors:v = (VI,V2,V3) =VIi+ V2j + V3k.. . . . . . . . ~ ~ . . . . . . . .

EXAMPLESI. (1,2,3) = i + 2j + 3~2. (0,2,0) = 2j

7.5 DOT PRODUCT

If ~ and'!!..are vectors in R!' then the dot product is defined by

This is also called an inner product on H", The result of a dot product is a scalar.

EXAMPLESI. (1,2,3) (1,1,1) =1+ 2+ 3 =62. (1,2,3) (1,2,3) = 12+ 22+ 32= 143 . 1 I ~ 1 I 2 = ' ! . : . ' ! . :


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DOT PRODUCT 8 1

The angle 0 between two vectors is given by

EXAMPLES

1. The angle 0 between (1, 2, 3) and (1, 1,1) is such that6

J42 '2. (1,2,3) and (1,1, -1) are at right angles since (1,2,3) . (1,1, -1) =0 hence cosO = O.

UV =0.Two vectors, ~ and ~ are orthogonal if they are perpendicular to each other and

EXAMPLES

1. ~ = (1,2,1) and ~ = (2,1, -4) are perpendicular since ~ . ' ! ! . . = 2+ 2 - 4 =O.2. To find a vector, (a, b), perpendicular to (1,2) write

(a, b ) . (1,2) = a + 2b =0hence the simplest choice is (a, b ) = (-2,1) although any multiple of this will be perpendicularto (1,2). For example (2, -1) and (-4,2) are still perpendicular to (1,2).


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82 VECTORS

7.6 CROSS PRODUCT

If~and!: are two vectors in R3, then the cross product ~ x ~ is defined in determinantnotation by

~j k11 12 13t X 11 =Vl V2 V3

= i1 'U.2 'U3 1_ j1 11 13 1+ k 1 'Ul 'U.2"" V2 V3 "" Vl V3 "" VI V2= (12V3 - 13V2, 13Vl - 1IV3, 1IV2 - 12Vt) .

EXAMPLESI. (1,0,0) x (0,1,0) =(0,0,1). That is ix j = k ..-v ~ .-v2. If~= (2, -3, 1) and ~ =(12,4, -6), then

i j k"" ""w=uxv= 2 -3 1"" "" "" 12 4 -6=~1 -~ -! 1- ~ 11~ -! 1+~11~ -~ 1= i(18 - 4) - j(-12 -12) + k(8 + 3 6)~ ~ ~=14i + 24j + 44k = (14,24,44).

"" "" ""3. The cross product (1,2,3) x (1,0,1) is:

(1,2,3) x (1,0,1) =i j k""1 2 3 = (2,2,-2).. .1 0 1

4. The cross product (1, 1,2) x (1, -1,0) is:

(1,1,2) x (1,-1,0) =i j k""1 1 2 = (2,2,-2).1 -1


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UNEAAINDEPENDENCE 8 3

Note that the result of taking the cross product of two vectors is another vector where thedirection of ~ x ~ is perpendicular to both ~ and~.

EXAMPLES1 . In a previous example (2, -3, 1) x (12,4, -6) =(14,24,44) and

(2,3, -1) . (14,24,44) =28 - 72+ 44 =O .Similarly (12,4, -6) . (14,24,44) = 0.

2. In a previous example (1,1,2) x (1, -1,0) = (2,2, -2). Note that (2,2, -2) . (1,1,2) =0 and(2,2,-2) (1,-1,0) =0.

7.7 LlNEA.R INDEPENDENCE

11= ClUl + C2U2 + ... + CnUnt " 'o . . J t " 'o . . J t " 'o . . J t " 'o . . J

A vector 11is a linear combination of the vectorsUl, U2, ... , Un if it can be written as~

where Cl, ... ,Cn are constants,

EXAMPLES1 . (2,7,3) is a linear combination of (1,1,0), (0,2,1), (0, 1,0) since

(2,7,3) =2(1,1,0) + 3(0,2,1) - (0,1,0).2. (1,2,1) is not a linear combination of (1,1,0), (2, 1,0), (1,0,0) since we can never combine thethree vectors toget the third component of (1, 2,1).3. Any vector (a, b, c) in R3 canbe found from a linearcombination of {(I, 0,0), (0,1, 0), (0,0, I)}.


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84 VECTORS

ClUJ. + C2U2 + ... + CnUn = 0. . . . . . . . . . . .

A set of vectors U1, U2, ... , Un are linearly independent if the only constants Cl, . .. , Cn. . . .that satisfy

are Cl = C2= ... = Cn = O.

EXAMPLESI. (1,1,0), (0,2,1), (0, 1,0) are independent since

Cl(1,1,0) + C2(0,2,1) + Cs(0,1,0) = (0,0,0)implies

Cl = Cl+C2+CS=0C2=

which gives Cl = C2= Cs= O.2. (1,1,0), (2, 1,0), (1,0,0) are dependent (not linearly independent) since

Cl+ 2C2 + Cs=0Cl+C2=0

0=0has an infinite number of solutions, one being Cl = 1, C2= -1, Cs= 1.

3. The vectors (1,2), (2,1), (1,0) are dependent since

impliesCl+ 2C2 + Cs=0

2Cl+ 2C2 = O .Since we have two equations in three unknowns we can always find a non-zero Cl,C2,Cs to satisfythese equations, for example Cl = 1, C2 = -1, Cs= 1. More than two vectors in R2 can neverbe independent.

4. The vectors i,i.k are independent since for any vector v = (a, b,c) it is possible to write~~~ ~(a, b , c) = Cl i + C2j + csk = ai + bj + ck~ .-v ~ ~ .-v

hence if ~ = ~ then Cl = C2= Cs = O.


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UNEAAINDEPENDENCE 85

A set of vectors is linearly independent if the determinant of the matrix with vectors ascolumns is not zero.

EXAMPLESI. For (1, 1,0), (0,2,1), (0, 1,0) the determinant

1001 2 1 = -1 f : - 0010

hence the vectors are independent.2. For (1,1,0), (2,1,0), (1,0,0) the determinant

1 2 11 1 0 = 0000

hence the vectors are dependent. We can show that(2,1,0) = (1,1,0) + (1,0,0)

so they are not independent of each other.


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86 VECTORS

7.8 EXAMPLE QUESTIONS(Answers are given in Chapter 14)1. Evaluate the sum ~ +~, 3~ and II~II:

(i) ~ = (-2, -1), ~ = (1,1)(ii) ~ = (3,4), ~ = (4,3)(iii) ~=(-2,1),~=(-1,-1)(iv) ~ = (3,4, 2), ~ = (1,1,1)(v) ~ = (3,1,1, 0), ~ = (1,0,1,1)(vi) ~ = 2i_+ 3j +~, ~= i_ - j - ~(vii) u = i+ j, v = i-3j

rV rV rV r V rV

2. For the above vectors verify the triangle inequality thatII~+ ~II:::;I~II+ II~II3. In the diagram below write down the two vectors ~ and

~ in algebraic form then find and draw the vector ~ +~.y

654321

1 3 4 5 64. Evaluate the sum ~ + ~and II~+ ~II:

(i) ~ = (3,2, -1), ~ = (-1, -2, 1)(ii) ~ = (1,0, 9), ~ = (-2, -2, -2)(iii) ~ = (4, -4, -3), ~ = (8,7,1)

5. Find ~ . ~, ~ X ~ and cos e where e is the angle betweenthe u and v:

(i) ~=(1,2,1),~=(-1,3,1)(ii) ~ = (-3,2, -1), ~ = (6,1,1)(iii) ~ = (2,3, 0), ~ = (4, 1,-2)(iv) ~ = (0,0, 0), ~ = (1,4,3)(v) ~ = (3,3, 3),~ = (-1, -1, -1)(vi) ~ = (1,2, 4), ~ = (2,4, -2)

6. For the previous question verify that ~ =~ X ~ isorthogonal (at right angles) to both ~ and ~.

7. Determine whether the following vectors are l inearlyindependent

(i) {(4, 1), (1, 2)}(ii) {(2, 1), (4, 2)}(iii) {(I, 1), (1,2), (3, I)}(~) {(1,1,1),(0,2,0),(1,3,2)}(v) {(I, 1,1), (0,2,0),(1,3, I)}(vi) {(I, 2,0,1), (1, 1,0,1), (2, 1,3,1), (0,2, -3, I)}

8. Find a number c so that (1,2,c) is orthogonal to (2,1,2).9. Find the vector which goes from the point (1,3,1) to thepoint (2, 5, 3). What is the length of this vector?10. Show that the line through the points (1, 1,1) and (2, 3,4)

is perpendicular to the line through the points (1,0,0)and (3, -1,0).

11. Show that a . (b xc) can be written asx

Ial a2 a3

a . (b xc) = bl b2 b3C1 C2 C3

=Ulb2C3 - alb3c2 - a2blc3+ a2b3C1 + a3blc2 - a3b2C1.12. Verify the above equation using the vectors

a = (1,1,2), b = (1,0,1), c= (0,1,1).


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CHAPTER 8ASYMPTOTICS AND APPROXIMATIONS

8.1 LIMITS

I. xn < xm if1< m < n, 0 < x < 1As x -+ 0 then

2. lim !(x) + g(x) = lim !(x) + lim g(x);l)-tO" " z-tO ;l)-tO3. lim !(x)g(x) = lim !(x) lim g(x);l)-tO z-tO ;l)-tOassuming lim !(x) and lim g(x) exist.;l)-tO" z-tO

EXAMPLES" )3 . ( )21 . (0.1 < 0.1

3. lim sin x cos x = lim sinx lim cos z =0 x 1 =0;l)-tO z-tO z-tOx(x - 1) 14. lim = -;l)-tO x(x - 2) 2


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88 ASVMPTOTICS AND APPROXIMATIONS

8.2 L 'H O P IT A L'S R U L E

lim f(X)) = lim f'(x)f lim g'(x),z-tZe 9 .x z-tz< z-tz

math_skills for engineering science and applied mathematics

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