Analytical ToolboxAnalytical Toolbox

Introduction to arithmetic & algebraIntroduction to arithmetic & algebraByBy

Dr J. WhittyDr J. Whitty

2

Why study mathematicsWhy study mathematics

3

The Laws of MathematicsThe Laws of Mathematics

AsscociativeAsscociative 3+(4+5)=(3+4)+53+(4+5)=(3+4)+5 a+(b+c)=(a+b)+ca+(b+c)=(a+b)+c a(bc)=(ab)ca(bc)=(ab)c

CommutativeCommutative 4+5=5+44+5=5+4 a+b=b+aa+b=b+a ab+=baab+=ba

DistributiveDistributive 3(4+5)=3x4+4x53(4+5)=3x4+4x5 a(b+c)=ab+aca(b+c)=ab+ac (a+b)/c=a/c+b/c(a+b)/c=a/c+b/c

4

Module Leaning ObjectivesModule Leaning Objectives

To achieve this unit a learner must:

1. Determine the fundamental algebraic laws and apply algebraic manipulation techniques to the solution of problems involving algebraic functions, formulae and graphs

2. Use trigonometric ratios, trigonometric techniques and graphical methods to solve simple problems involving areas, volumes and sinusoidal functions

3. Use statistical methods to gather, manipulate and display scientific and engineering data

4. Use the elementary rules of calculus arithmetic to solve problems that involve differentiation and integration of simple algebraic and trigonometric functions.

5

Assessment MethodsAssessment Methods

Assignment 1: Assignment 1: Introductory mathematical statisticsIntroductory mathematical statistics

Milestone test 1: Milestone test 1: Mock examination mathematics questionsMock examination mathematics questions

Assignment 2: Assignment 2: Introductory mathematical engineering systems modellingIntroductory mathematical engineering systems modelling

Milestone test 2:Milestone test 2: Three mock-exam science questionsThree mock-exam science questions

Milestone test 3: Milestone test 3: Computer modelling methods tutorialComputer modelling methods tutorial

ExaminationExamination (informal) (informal)

6

Recommended ReadingRecommended Reading

Engineering Engineering mathematics KE mathematics KE Stroud & DJ BoothStroud & DJ Booth

Palgrave Palgrave macmillian,macmillian,

Fifth Edition, Fifth Edition, London, 2001London, 2001

7

Session Learning objectivesSession Learning objectives

After this session you should be able to:After this session you should be able to: Compute numeric expressions using BIDMASCompute numeric expressions using BIDMAS Evaluate numeric expressions in standard form Evaluate numeric expressions in standard form Transpose Transpose simplesimple formulae formulae Solve simple linear equations inc those involving Solve simple linear equations inc those involving

fractionsfractions Solve elementary non-linear equations inc. squares Solve elementary non-linear equations inc. squares

and square rootsand square roots

8

BIDMAS BIDMAS Can you remember Can you remember

what this stands for?what this stands for? BracketsBrackets IndicesIndices DivisionDivision MultiplicationMultiplication AdditonAdditon SubtractionSubtraction

1.1. 2a+5b+3c2a+5b+3c 2x3 + 5x(-1) +3x22x3 + 5x(-1) +3x2 =7=7

2.2. 6a-7b6a-7b 6x3 - 7x(-1) =6x3 - 7x(-1) = 18 - -718 - -7 =25=25

3.3. 2a2a22 - 3b - 3b22

2xaxa - 3xbxb2xaxa - 3xbxb 2x3x3 -3x(-1)x(-1)2x3x3 -3x(-1)x(-1) =15=15

Let a=3, b=-1, c=2Let a=3, b=-1, c=2

9

Remove the bracketsRemove the brackets

3(a+b)3(a+b) =3a + 3b=3a + 3b

5(2a-5c)5(2a-5c) =10a-25c=10a-25c

4(2a-3b)-5(a-6b)4(2a-3b)-5(a-6b) =8a-12b-5a=8a-12b-5a++30b30b =3a+18b=3a+18b

6c-(3a+2b-5c)6c-(3a+2b-5c) =6c-3a-3b+5c=6c-3a-3b+5c =11c-3a-3b=11c-3a-3b

4(5a-b)-2(3a-b-c)4(5a-b)-2(3a-b-c) =20a- 4b-6a+2b+2c=20a- 4b-6a+2b+2c =14a-2b+2c=14a-2b+2c

10

Try some yourselvesTry some yourselves

4);2)(2)(( xxxa

10,98,7

13;2)(

gl

g

lb

5.1,3);23)(12)(( xxxc

1,3;))(( 2 qpqpd

yxyxyxe 3;32)32)(( 2222

11

IndicesIndices

Consider 2Consider 26 6

The 2 is called the base, the 6 is the power or The 2 is called the base, the 6 is the power or the index. The above is said “2 to the power the index. The above is said “2 to the power of 6” and is calculated 2 x 2 x 2 x 2 x 2 x 2 = of 6” and is calculated 2 x 2 x 2 x 2 x 2 x 2 = 6464

On the calculator press 2 x On the calculator press 2 x y y 6 =6 = 226 6 DOES NOT EQUAL 12DOES NOT EQUAL 12

12

Negative IndicesNegative Indices

Consider 2 Consider 2 -- 66

The rule to eliminate the negative power is : The rule to eliminate the negative power is : write 1 divided by the old base, to the positive write 1 divided by the old base, to the positive powerpower

This is equal to This is equal to

13

Fractional IndicesFractional Indices

440.50.5 is the same as the square root of 4 is the same as the square root of 4 4 4 1/8 1/8 is the same as the eighth root of 4 and is is the same as the eighth root of 4 and is

again calculated using the x again calculated using the x y y button on the button on the calculator (= 1.1892)calculator (= 1.1892)

14

Rules of IndicesRules of Indices

If two powers are multiplied, then if the bases If two powers are multiplied, then if the bases are the same, the powers are ADDEDare the same, the powers are ADDED

2255 x 2 x 277 = 2 = 21212

If two powers are divided, then if the bases If two powers are divided, then if the bases are the same, the powers are SUBTRACTEDare the same, the powers are SUBTRACTED

3388 3 355 = 3 = 333

15

Examples on IndicesExamples on Indices

4477 4 488 ==

44-1-1 = 1/4 = 1/4

55-2-2 x 5 x 5- 4- 4 = =

5 5 -6-6 = =

1/156251/15625

33-3-3 3 3-1-1 = =

33-2-2=1/9=1/9

5544 x 5 x 522 5 5-3-3 = =

5599 = 1953125 = 1953125

101000==

11

9900==11 aa00 = = 11

33-7-7 3 3-5-5 x 27= x 27=

33

16

Standard FormStandard Form

Standard Form is used to write very large or Standard Form is used to write very large or very small numbers in a more convenient wayvery small numbers in a more convenient way

A number in standard form is written: A number in standard form is written: a x 10 a x 10 nn

where a is a number between 1 and 10, and where a is a number between 1 and 10, and n is an integer, positive or negativen is an integer, positive or negative

17

Standard & Engineering Form examplesStandard & Engineering Form examples

2,000,000 2,000,000

= 2x10 = 2x10 66

35,800 35,800

= 3.58x10 = 3.58x10 4 4 (35.8 x10(35.8 x1033))

5,927,000,000 5,927,000,000

= 5.927x10= 5.927x1099 (5.9x10(5.9x1099))

43 43

= 4.3 x 10 = 4.3 x 10 11 (43)(43)

0.000 0040.000 004

= 4x10= 4x10-6-6 (4x10(4x10-6-6))

0.000 4580.000 458

= 4.58x10= 4.58x10- 4- 4 (458x10(458x10-6-6))

0.000 000 000 0210.000 000 000 021

= 2.1x10= 2.1x10-11-11 (21x10(21x10-12-12))

0.350.35

= 3.5x10= 3.5x10-1 -1 (350x10(350x10-3-3))

18

Using a calculatorUsing a calculator

To calculate 3.98 x 10 To calculate 3.98 x 10 1212 x 4.2 x 10 x 4.2 x 10 1111

press the following keys:press the following keys:

3.98 exp 12 x 4.2 exp 11 =3.98 exp 12 x 4.2 exp 11 =

Do NOT type Do NOT type x10x10, the exp button does this , the exp button does this automatically. automatically.

The answer should appear as The answer should appear as 1.6716 1.6716 2424 on on screen.screen.

The correct answer is then The correct answer is then 1.6716 x 10 1.6716 x 10 2424

19

Evaluate, in standard formEvaluate, in standard form

3.2 x 10 3.2 x 10 66 x 4 x 10 x 4 x 10 44

= 1.28x10 = 1.28x10 1111

4.5 x 10 4.5 x 10 1616 + 4 x 10 + 4 x 10 1515

=4.9x10 =4.9x10 1616

2.8 x10 2.8 x10 2727 x 3.5x 10x 3.5x 10-25-25

= 980= 980

= 9.8x10 = 9.8x10 22

4.4 x10 4.4 x10 -10-10 x 5.2x10 x 5.2x10 -4-4

= 2.288x10 = 2.288x10 -13-13

5 x 10 5 x 10 -8 -8 2 x 10 2 x 10 -7-7

= 0.25= 0.25

2.5x10 2.5x10 -1-1

20

Rules of IndicesRules of Indices

mnmn xxx

mnmnm

n

xxxx

x /

nmmn xx

21

CorollariesCorollaries

11 0 xxx

x nnn

n

mnm n xx /

10 x

mn

m xxx nm n 1

nn

xx

1

nnnn

xx

xxx

100

111

1 :NB xxx

xxxx nn

n

n

22

Let’s use them:Let’s use them:

4

32

3

333)(

a )2)(6)(( 0xb

21

)36)(( c 43

)16)(( d

4

3223

3

32)(

e

23

They still work algebraically!They still work algebraically!

yx

yxa

2

23

4

12)(

2

2

4

423

)(c

a

bca

cbab

202323)( acabcbc

24

Recap: Make x the subjectRecap: Make x the subject

Equation:Equation:

3x+2 = 233x+2 = 23

3x+23x+2 - 2= 23= 23 -2

3x = 23 - 23x = 23 - 2

x = (23-2)/3x = (23-2)/3

x=7x=7

Formula:Formula:

gx + h = kgx + h = k

gx + hgx + h - h = k= k - h

gx = k- hgx = k- h

x = (k - h)/gx = (k - h)/g

Last lecture we examined the differences between equations and formulae and their subsequent solution protocols:

25

Transposition Of FormulaeTransposition Of Formulae

The rules are exactly the same as for The rules are exactly the same as for algebra, except the final result is an algebraic algebra, except the final result is an algebraic expression instead of a numerical answer.expression instead of a numerical answer.

26

Simple TranspositionSimple Transposition

In the Science units you In the Science units you will come across very will come across very simple formulae for simple formulae for instance instance DensityDensity Newton’s second law Newton’s second law

(mechanics)(mechanics) Electrical chargeElectrical charge Ohms LawOhms Law

maF

It

Q

IRV

V

m

27

Simple TranspositionSimple Transposition

Here the same Here the same rules apply as the rules apply as the letters in the letters in the formulae are just formulae are just numbers in numbers in disguise disguise

maF

am

F m

a

F

Fm a

28

ActivityActivity

In groups use the systematic transposition (or In groups use the systematic transposition (or otherwise) approach to develop calculation otherwise) approach to develop calculation transposition triangles for the formulae, describing:transposition triangles for the formulae, describing: DensityDensity Electrical ChargeElectrical Charge VoltageVoltageIn each case define each of the variables you have In each case define each of the variables you have

usedused

29

Solution of Linear EquationsSolution of Linear Equations

MathematicalMathematical 6x+1=2x+96x+1=2x+9

6x+16x+1-1-1=2x+9=2x+9-1-1 6x=2x+86x=2x+8 6x6x-2x-2x =2x =2x-2x-2x+8+8 4x=84x=8 4x4x/4/4=8=8/4 /4 =2=2

SystematicSystematic 6x+1=2x+96x+1=2x+9 6x6x-2x-2x= +9= +9-1-1 4x=84x=8 x=8x=8/4/4=2=2

Note: The approaches are exactly the same however in the systematic approach we MOVE numbers & variables

30

What about brackets?What about brackets?

The rule with The rule with brackets is simply brackets is simply MULTIPLY (that’s all MULTIPLY (that’s all they mean).they mean).

MULTIPLY MULTIPLY everything inside the everything inside the bracket by everything bracket by everything outside the bracket.outside the bracket.

Consider:Consider: 3(x-2)=93(x-2)=9

3x-6=93x-6=9 Solve as beforeSolve as before

4(2r-3)-2(r-4)=3(r-3)-14(2r-3)-2(r-4)=3(r-3)-1 8r-12-2r8r-12-2r++8=3r-9-18=3r-9-1 Solve as beforeSolve as before

ExerciseExercise

31

Class Discussion/ExerciseClass Discussion/Exercise

2x+5=72x+5=7 2c/3-1=32c/3-1=3 7-4p=2p-37-4p=2p-3 8-3t=28-3t=2 2x-1=5x+112x-1=5x+11 2a+6-5a=02a+6-5a=0 3x-2-5x=2x-43x-2-5x=2x-4 20d-3+3d=11d+5-820d-3+3d=11d+5-8

2(x-1)=42(x-1)=4 16=4(t+2)16=4(t+2) 5(f-2)=3(2f+5)-155(f-2)=3(2f+5)-15 2x=4(x-3)2x=4(x-3) 6(2-3y)-42=-2(y-1)6(2-3y)-42=-2(y-1) 2(g-5)-5=02(g-5)-5=0 4(3x+1)=7(x+4)-2(x+5)4(3x+1)=7(x+4)-2(x+5)

32

Linear Equations with FractionsLinear Equations with Fractions

The systematic method The systematic method is especially useful with is especially useful with fractional coefficients:fractional coefficients:

e.g.e.g.

65

2x

65

2x

562 x

152

56

x

5.262

56

2

5

5

2

xMathematical

Approach

33

Linear Equations with FractionsLinear Equations with Fractions

There are several There are several methods to attempt methods to attempt such problems but by such problems but by far the far the bestbest is to is to attempt to clear the attempt to clear the fractions (some how) fractions (some how) in order to reduce the in order to reduce the equation to something equation to something simpler which can be simpler which can be solvedsolved

Consider: Consider: (math)(math)

2

3

20

15

4

3

5

2 yy

2

320

20

120520

4

320

5

220

yy

yy 301100158

34

Linear Equations with FractionsLinear Equations with Fractions

This is the so called This is the so called mathematical mathematical approachapproach

There is the analogous There is the analogous systematic approach, systematic approach, which most engineers which most engineers find a little easier to find a little easier to applyapply

Thus:Thus:

2

3201520

4

320

5

220 yy

2

3

20

15

4

3

5

2 yy

100151308 yy

11438 y

3y

35

Class ExerciseClass Exercise

1.1.

2.2.

3.3.

4.4.

5.5.

6.6.

6

5

2

31

4

32 yy

2

1514

3

1x

2

1

53xx

2

1312

4

1x

2

3

5

6

4

xxx

343/ d

36

Squares and their rootsSquares and their roots

Proceed as before but at the end to get rid of Proceed as before but at the end to get rid of a square root simply square everythinga square root simply square everything

2x 22 x

44 22 xx55 33 xx

Likewise:82 xWhat About:

37

Squares and their rootsSquares and their roots

A similar approach can be applied to A similar approach can be applied to squares and powers, thussquares and powers, thus

252 x3

2

4

152

tWhat about

38

ExamplesExamples

52 y

61

3

x

x

1

2510

x

916

2t

3

26 a

a

2

85

2

11

x

39

SummarySummary

Have we met our learning objectives, Have we met our learning objectives, specifically, are you specifically, are you able to:able to: Compute numeric expressions using BIDMASCompute numeric expressions using BIDMAS Evaluate numeric expressions in standard formEvaluate numeric expressions in standard form Derive the rules of indices from first principlesDerive the rules of indices from first principles Evaluate and simplify mathematical expressions Evaluate and simplify mathematical expressions

using the rules of indices.using the rules of indices.

40

HomeworkHomework

Evaluate Evaluate or or simplify the followingsimplify the following

3

42

7

77)(

a

2

2

6

66)(

b 2

1

5.0)( c1

7

3)(

d 02876)(e 9

3

6

30)(x

xf

26 12/36)( xxg 26 34)( xxh 26 34)( xxi

063)( xk 7

7

10

5)(x

xl 255)( xn

41

More homeworkMore homework

1.1. 2.2.

3.3.

4.4.

5.5.

6.6.

5

4

4

31

5

33 yy

4

1213

3

1x

4

1

35xx

4

1615

2

1x

6

6

4

3

5

xxx

235/ d

42

Further StudyFurther Study Foundation topics Foundation topics

F1 ArithmeticF1 ArithmeticF2 Introduction to F2 Introduction to

algebraalgebra