Brush Up Your Maths 3

7/24/2019 Brush Up Your Maths 3

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Brush up your maths

An access course for H. E. Engineering.by

Dr. D. Baxterof

Mid Cheshire College

Acknowledgements

A work of this kind is necessarily deriatie and the author acknowledgesthe influence of established writers in the field of engineeringmathematics! such as A "reer and " # $aylor! % & Bird and A % C May!

% C 'ates and ( )utherland! D Howell and A #olf! whose books arerecommended to the reader for further study. Eery effort has been made to obtain permission to reproduce copyright!but if any copyright material has been inadertently included please get intouch with Aimhigher *Cheshire and #arrington + with a iew tocorrecting this.$he deelopment time was funded by Aimhigher of Cheshire and#arrington and proided by Mid Cheshire College! whose studentsrepresented a captie audience,

Introduction

$his E-book is intended to proide a grounding in some of themathematics needed to study Engineering at degree leel. As such it isery focussed upon the basic ideas in each area of mathematics coered.

t is noel in as much as the reader is expected to use modern methods

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of calculation and graphing such as Excel and to access interactieinternet mathematics sites in order to check calculations. n each sectionthere are worked examples ! practice examples and appropriateengineering problems.

CONTENTS

Chapter One$his consists of an introduction to an important topic in basic algebra!rearranging e/uations.

Chapter TwoComprises a graphical approach to the solution of engineering problemsusing Excel.$opics coered are straight line e/uations! simultaneous e/uations!/uadratics and cubics. Algebraic solutions are also coered.

Chapter Three

ntroduces extensions to the standard number system which hae greatpractical application in engineering and includes complex numbers andectors.

Chapter FourConcerns the arithmetic of determinants and matrices and concludes withan application to simultaneous e/uations. $he use of Excel to do matrixmanipulation is explained.

Chapter Five$his is an introduction to the ideas behind calculus and its two branches!differentiation and integration.

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Chapter &neREARRANGING EQUATIONS

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SECTION ONEH&# $& (EA((A0"E E31A$&0)

An algebraic e/uation consists of /uantities on the left hand side and/uantities on the right hand side connected by an e/uals sign ! which

means that both sides hae the same alue.

t is useful to think of an e/uation in terms of an old fashioned pair ofscales. $he left hand side /uantities are in the left hand side scale panand the right hand side /uantities are in the right hand side one. $he twosides are only e/ual if the scales balance exactly.

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magine that you had : kilos of flour on the left hand side and : kilos onthe right hand side. the scales would balance and you could say : ; : ,f you now changed one side * say you added one kilo to the left pan +! youwould hae to carry out exactly the same change to the other side in

order to keep the scales in balance. t would now be true to say that :<= ;:<=! that is if you add = to both sides the scales still balance.

Exactly the same thing applies to e/uations. #hateer you do to one sidemust be done to the other in order to maintain the balance and make thee/uals sign true.

WORKED EA!"#ES

Rearran$in$ %or Tran&po&in$ ' e(uation& where

(uantitie& are connected )* a p+u& or a ,inu& &i$n

$ranspose C ; c < =:> for c.

$his means rearrange the e/uation so that c e/uals something.

!ethod- f we subtract =:> from the right hand side *rhs+! then c

will be by itself! which is what we want. n order to keep thebalance howeer we need to do exactly the same thing to the otherside *lhs+ that is! subtract =:>.

#e get C-=:> ; c < =:> -=:>&r C -=:> ; c&r c ; C -=:> * t does not matter if the c is on theleft or right +

$ranspose ' ? f ; =@ for '

!ethod- $o get ' by itself we need to ADD f to the lhs. $omaintain the balance we need to add f to the rhs also.

#e get '-f < f ; =@ <f&r ' ; =@ < f

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Tran&po&in$ e(uation& where the (uantitie& are

connected a& a product.

$ranspose = ×4 ( for (

!ethod- n this case we want ( by itself but we hae times toomany (s! so what we need to do is diide by . f we diide the rhsby then we must also diide the lhs by in order to maintain thebalance.

Diiding both sides by we get×

=4 (

which becomes =

4 (


connected a& a (uotient.

$ranspose =4

(

for 4

!ethod8 Here we hae 4 diided by so we need to multiply bothsides by to get 4 by itself.

#e get ×× = 4 (

! so × =( 4

$ranspose ρ ×

=5

( A

for A

!ethod8 7irst of all we get rid of the fraction by multiplying bothsides by A

#e get ρ × ×

× =5 A

( AA

so ρ × = ×( A 5

0ow we diide both sides by ( to get A by itself. ρ × ×

=( A 5

( (

7inally we hae ρ ×

=5

A(

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contained in )rac/et&.

$ranspose = − =p

x p for p

!ethod8 Multiply both sides by * p-=+. #e get ( )− ==x p p

0ow multiply out the brackets − =xp x p

"roup the terms containing p on the lhs. xp-p ; x

$herefore p*x-=+ ;x. Diide both sides by *x-=+. #e get =− =

x p

x

EA!"#ES FOR 0OU TO TR0$ranspose the following8

= C ; d for d: ) ;dn for d@ 24 ; c for 4 A ; rl for l ' ; x < = for x

F ' ; ax < b for x D ; B -=.:Gd for dG 5 ; a < *n - =+d for n A ; r*r < h+ for h

=> =a

x y for y

== =E

(

for (

=: =M E

(

for (

=@ =−

:( 4

( r for r

A""#ICATION QUESTIONS

= f p is the pressure in a thin pipe of outside diameter dand thickness t! the greatest tensile stress being f! then

=

+:

pd t

p f . 7ind d when f ; >>>! p ; >> and t ; .

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: f =+E

c ( r

. 7ind r if c ; .>:! E ; =F. and ( ; :.G

SECTION TWO$(A0)2&)0" E31A$&0) C&0$A00"

2&#E() A0D (&&$)

WORKED EA!"#ES

E(uation& containin$ power&-

Consider the following e/uation 4

:

; 1

:

< :A)#e wish to transpose it to find 1.

$he first step is to get the 1: by itself.)ubtract :A) from both sides.

4: ? :A) ; 1:

$he next step is to take the s/uare root of both sides

− = :: :A) u

$his gies us

− =::A) 1

E(uation& containin$ root&-

Consider the following e/uation8 π = :5

$ g

#e wish to transpose this for g .$he first step is to s/uare both sides.

π =

:: C 5

g $

0ow multiply each side by g

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π =: :

C$ g 5

7inally diide both sides by :$

π =:

: 5 g $

EA!"#ES FOR 0OU TO TR0

(earrange the following

= = :4 gh for h

: π = :4 d h for d

@ = :.DE m4 for 4

+

=−

f p D

d f p for f


= f =:.>@

:54

h dg

7ind 4 when h ; .F=! 5 ; =FG!d ; .: and g ; =>

: f = :4e

m

7ind 4 if eIm ;=.Gx=>== and ;>.Fx=>G

)ac/ to content&

Chapter $woSO#1ING "RO2#E!S USING GRA"3S

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$H) $HE0 HA) $& BE )&54ED.

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)&ME E31A$&0) CA0 BE )&54ED 1)0" A5"EB(A! B1$ &7$E0!0 E0"0EE(0"! &$HE( ME$H&D) HA4E $& BE 1)ED A0D"(A2HCA5 $ECH031E) A(E 4E(' 2&#E(715.

"(A2H) HAD $& BE D(A#0 B' HA0D 10$5 (ECE0$5' B1$ 0&##E CA0 1)E )2(EAD)HEE$) $& 2(&D1CE H"H 31A5$' "(A2H)(E5A$4E5' EA)5'.

$HE 21(2&)E &7 $H) CHA2$E( ) $& )H&# '&1 H&# $& 1)EMC(&)&7$ E9CE5 $& )&54E E0"0EE(0" 2(&B5EM)"(A2HCA55'.

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CONTENTS

2(&D1C0" A CHA($ 7&( A )$(A"H$ 50E

)&540" )M15$A0E&1) 50EA( E31A$&0)

)&540" 31AD(A$C E31A$&0)

25&$$0" 2&5'0&MA5)

(E7E(E0CE )EC$&0

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SECTION ONE2(&D1C0" A CHA($ 7&( A )$(A"H$ 50E

WORKED EA!"#E

A )$(A"H$ 50E ) (E2(E)E0$ED MA$HEMA$CA55' B' $H)E31A$&08

' ; mx < c

m is the gradient or slope of the line

c is the point on the y-axis where the line crosses.

m and c are numbers which uni/uely specify a particular line.

' ; x ? :is an example of a straight line with a gradient of and a crossing point or intercept of -:.

31E)$&0

#hat are the gradients and intercepts of the followinglinesJ

' ; @9-! ' ; -9 < ! ' ; 9I: -=

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$his time we shall choose alues of 9 from > to G.nput these alues! as shown.

0ext we wish to calculate the ' alues . $o do this we need to input aformula into the appropriate cell! in this case! B.

$his is the formula we need8 * t is the E9CE5 ersion of y ; m9 < c+

; NDN= O A < NDN:

5et us look at each part of this formula8

; * A formula always begins with an e/uals sign+

NDN= *$his means A5#A') the number in D=+

OA *multiplied by the number in A to start with +

< NDN: *plus A5#A') the number in D:+

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0ow enter the aboe formula into B ! shown below.

0ext click the on the formula bar to enter the first calculation

$o complete the calculation of the ' alues! click and drag on the little

s/uare on the bottom right hand corner of cell B.

$he result is as shown.

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$he final stage is to use the chart wiPard to plot the graph in exactly theway we want it.

)elect the x and y data alues with a click and drag and click on the chartwiPard icon!as below

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Click on the highlighted options below! then 0extQ to continue8

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1se the Chart options sheet shown below to customise your graph! thatis! add titles! choose labels! scales and colours. Click next.

Choose either to embed your graph into the spreadsheet table or hae itcompletely separate.

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Click finish and you should hae something like the one below

#E55 D&0E,,

$he great adantage of using E9CE5 to plot your graph is not only

)2EED and ACC1(AC'

but alsoRRf you simply change the alues of m and C

$he computer will recompute eerything!including the graph ,

EA!"#ES FOR 0OU TO DO

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2lot the graphs of the following lines on separate charts.

2rint out your results and the data tables that go with the graphs

=. ' ; 9 - :. ' ; - 9 -:@. 9 ; :' < * you will hae to rearrange this first, +


=.

$he olume of a gas 1 at a temperature T degrees C is gien by

1 4 .567T 8 97.:6

2lot a graph of 1 against T for alues of T from => to F> degrees

*1 ertically+. 1se your graph to find the olume at @ degrees.

ContdR

:.A force 7 applied to a crane lifts a load 5$he connection between the force and the load is-

F 4 5.9# 8 ::

2lot a graph of F *ertically+ against # for load alues between @> and=:>> 0ewtons. 7ind the force needed to lift =>>> 0ewtons from yourgraph.

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SECTION TWOSO#1ING SI!U#TANEOUS #INEAR EQUATIONS

WORKED EA!"#E

f! instead of one straight line! we hae two! it follows that they willcross at some point.

$he only exception to this is if the two lines are parallel.

At the point that the* cro&&; the* wi++ )oth have the &a,e and 0

va+ue&.

f we hae the e/uations of the two lines we can work out the 9 and 'alues at the crossing point and we call this 8

SO#1ING T3E TWO EQUATIONS SI!U#TANEOUS#0

T here are two ways of doing this8

a+ 1sing algebra * the method is explained in the reference section +

b+ By plotting their graphs and reading off the 9 and ' alues of the crossing point.

WE S3A## 2E USING ECE# TO CARR0 OUT !ET3OD )'.

Consider the two straight lines

' ; - 9 < *m; -=! c ; +! and' ; 9 ? : *m ; =! c ; -:+

#e shall plot these in E9CE5 and find the crossing point.

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&2E0 E9CE5 &0 '&1( C&M21$E(

nput both sets of alues of m and c! as shown below.

0ow we select a suitable range of alues for 9! in this case > to shoulddo. $his gies us8

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0ext we need to enter the two formulae to calculate the ' alues! asbefore in section =. Do this to obtain the following result.

0ext fill down as before8

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0ow select the data columns and chart the two lines together! producingsomething like8

#e can read off the 9 and ' alues from the chart

)& $HE )&51$&0 7&( $HE $#& )M15$A0E&1)

E31A$&0) ) ' ; =.! 9 ;@.

f we re/uire a more accurate result then we simply input some moreprecise alues of 9 near the crossing point e.g. @.>! @.=! @.:! @.@! @.! @[email protected] and rechart the graphs in this region.

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)ole the following simultaneous e/uations using both an algebraic and agraphical method. Can you think of another way of checking if youranswers are correctJ

a+3

8

3

1+−= X Y Hint 8 7irst conert the fractions to decimals

and

2

3

2

1−= X Y

b+ 52 =+ Y X

and72 =−Y X Hint 8 7irst you must rearrange each e/uation

into the form ' ; m9 < c

A""#ICATION QUESTION

$he forces 7 acting on a beam are such that

:7= < @7: ; =@ and

7= - :7: ; :

7ind the alues of 7= and 7:

SECTION T3REESO#1ING QUADRATIC EQUATIONS

WORKED EA!"#E

#e saw in the preious sections that a straight line can be representedby the e/uation ' ; m9 < c.

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0ow consider the following e/uation8 ' ; @9: < :9 < . Here the highestpower of 9 is :! that is! 9: .

$his type of e/uation always represents a cure and is called a /uadratice/uation.

n fact the cure is a parabola.

$he general e/uation for a /uadratic is8

' ; a9: < b9 < c #here a! b and c are any numbers! commonly called

Coefficients. $hey can be positie or negatie.

)o ! if ' ; 9: < 9 < ! then! for this /uadratic e/uation!

a ; b ; c ;

#e can use E9CE5 to plot any /uadratic cure.

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&2E0 E9CE5 &0 '&1( C&M21$E(

#e wish to plot the following /uadratic cure8

' ; 9: < @9 < : . )o a ; =! b ; @ and c ; :

nput the alues of a!b and c as shown.

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)elect an appropriate range of alues for 9 ! - to <: is suitable.nput these alues next

0ow we must enter the formula to calculate the ' alues as before.Can you see how the formula is constructedJ Consult the referencesection if not.

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7inally! fill down. $he data can be charted in exactly the same way asbefore.

Do this and you should obtain something like the chart below.

0otice that the cure crosses the 9 axis at two points. 9 ; -: and 9 ; -=

T3ESE 1A#UES OF ARE CA##ED T3E SO#UTIONS TO

T3E QUADRATIC EQUATION 5 4 9

8 < 8 9

T3AT IS ; T3E SO#UTIONS ARE T3E 1A#UES OF FOR

W3IC3 0 4 5.

$hese alues can be obtained as accurately as re/uired by selecting fineralues of 9 near the crossing points and recharting.

0ot all /uadratics hae two solutions . )ome hae only one and some haenone,

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2lot the graphs of the following /uadratic e/uations for the alues of 9

shown.

n each case find the alues of 9! if any! for which ' ; >.*$hat is ! find the solutions. +

=. ' ; 9: ? 9 < @. * 9 from >.> to <. in >. steps +

:. ' ; 9: < @9 < :.: * 9 from -:. to <=. in >. steps +

@. ' ; :9: < 9 < = * 9 from-=. to < :.> in >. steps +

1se an algebraic method to sole /uestion one and erify your answerfrom the graph. Consult the reference section to remind you how to dothis.


$he power in watts deeloped in an electrical circuit is gien by 8

2 ; :> ? =F: where is the current in Amperes.

7ind the current necessary to produce a power of @ watts in the circuit.

Hint82ut 2 ; @ into the e/uation! then rearrange in the form ' ; a: < b < c

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SECTION FOUR25&$$0" 2&5'0&MA5)

WORKED EA!"#E

A polynomial is an e/uation which looks like this8

' ; a < b9 < c9: < d9@ < e9 < RR

Here a! b! c! d and e are any numbers! also known as coefficients.

f c! d! e! etc are >! then the e/uation becomes8' ; a < b9 that is! a straight line e/uation , Compare it with' ; m9 < c

f d!e! etc are >! then the e/uation becomes8' ; a< b9 < c9: that is! a /uadratic e/uation ,

7 e etc are >! then the e/uation becomes8' ; ; a < b9 < c9: < d9@ that is! a cubic e/uation,

$o plot a straight line we had to input two coefficients *we called them mand c+

$o plot a /uadratic we had to input three coefficients *we called them a!

b and c+

$o plot a cubic * where the highest power of 9 is 9 cubed+ we will hae toinput four coefficients a! b! c and d.

#e could easily continue to higher power polynomials.

n this case! howeer! we shall limit ourseles to a cubic.

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#e wish to plot a graph of the cubic8

' ; - - G9 < 9: < 9@

Here a ; -! b ; -G! c; = and d ; =

&2E0 E9CE5 &0 '&1( C&M21$E(

nput a! b! c and d alues as shown.

0ow we must select a range of 9 alues8 * from - to <@ will do +

nput these alues into the A column! starting with - in AF

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Enter the formula into BF as before to calculate the ' alues.#e enter8

; NDN= < NDN:OAF < NDN@OAFS: < NDNOAFS@

Here8

NDN= is KaL

< $his is always positie. f the coefficient is negatie! this will be

automatically taken into account.

NDN:OAF is Kb9L

NDN@OAFS: is Kc9:

NDNOAFS@ is K d9@ L

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0ow fill down as before and chart the data table produced.

t should look like this8

0otice that this cure crosses the 9 axis @ times.

$here can be up to @ solutions to a cubic e/uation.

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EA!"#ES FOR 0OU TO TR0

2lot the graphs and find the solutions! if any! for the following cubice/uations. *written using the S key for powers eg 9S: is 9:+

=. ' ; = < 9 < 9S: < 9S@

:. ' ; = < >.9 ? 9S: < >.9S@


Applications of the cubic are beyond the scope of this manual! but here isan unusual /uestion for you to try ,

1sing a graphical method sole the following two e/uations

&i,u+taneou&+*.

' ; F ? 9 ? :9S: ? 9S@ and

' ; 9 < =

*1se alues of 9 from -@ to < in >. steps+

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SECTION FI1E(E7E(E0CE

2A($ &0E8 )&540" )M15$A0E&1) E31A$&0)1)0" A0 A5"EB(AC ME$H&D.

WORKED EA!"#E

$here are seeral different algebraic methods for soling simultaneouse/uations! but we shall use only one! called the method of substitution.

7or example! we wish to sole the following e/uations8

9 < :' ; *=+And :9 < ' ; *:+

$he first step is to rearrange e/uation *=+ so that 9 is by itself on theleft hand side.

7irst 9 ; ? :' *#e hae taken :' to the (H).+

0ow we can )1B)$$1$E 7&( 9 in e/uation *:+! that is instead of9 we can write ? :'.

)o e/uation *:+ becomes8

:*-:'+ < ' ; $his e/uation only contains '! so we can rearrange it and find '.

7irst => ? ' < ' ; ! so

=> ? ; @'! andFI@ ; '

$hat is ' ; :

#e know that9 ; ? :'

)o putting ' ; : into this gies us 99 ; ? ; =

$he e/uations are now soled. 9 ; =! ' ; :

- 35 -



- 36 -



2A($ $#& 8 )&540" 31AD(A$C E31A$&0) 1)0"

A5"EB(A

WORKED EA!"#E

Again! there are seeral algebraic methods for soling /uadratice/uations! but here we will use only one! the formula method.5et us sole the following e/uation using the formula method8

' ; @9S: < =:9 - =Here a ; @! b ; =:! c ; -=

$he /uadratic formula is

2a

4ac- b^2 b- ±= X

0ow#e substitute our alues of a!b and c into the formula! so8

3*2

1*3*42^1212 −−±−= X

6

1214412 −−±−= X

6

15612±−= X

6

5.1212±−= X

0ow! the < option gies us one solution for 9 and the ? option the other.

Hence! 9 ; >.IF or 9 ; -:.IF

$herefore 9 ; .>G@ or 9 ; - .>G@

A popular mistake when using this method is not to take into account thenegatie signs! so take care when substituting negatie coefficients intothe e/uation ,

- 37 -



2A($ $H(EE8 HE52 #$H E9CE5

AC60&#5ED"EME0$8 $he author is indebted to the publishers Hodder

Headline 25C for their kind permission to reproduce the following section

on E9CE5 HE52.

- 38 -



- 39 -



- 40 -



- 41 -



Back to contents

Chapter ThreeC&M25E9 01MBE() A0D 4EC$&()

t is easy to forget that number systems are not naturally ocurring! butwere inented by mathematicians. Complex numbers ectors can bethought of as extensions to our more familiar number system.

AmaPingly! een though these things were dreamt up by mathematicians!they actually turn out to be 4E(' 1)E715 in engineering ,

Also! Tust as we had to learn how to do arithmetic *add! subtract! diideand multiply+with ordinary numbers! we also need to be able to do these things withcomplex numbers and ectors.

CONTENTS

C&M25E9 01MBE()

- 42 -



4EC$&()

EACH )EC$&0 0C51DE)8

#&(6ED E9AM25E)

E9AM25E) 7&( '&1 $& D&

A225CA$&0 2(&B5EM) 7&( '&1 $& )&54E #$H '&1(0E# )655)

SECTION ONEC&M25E9 01MBE()

Introduction

Complex numbers were inented to gie a meaning to the s/uare root of anegatie number. n the ordinary number system this is not possible.

7or example8*<:+: ; <

And *- :+: ; <

$herefore! whether we s/uare a plus : or a minus :! we always end up

with a plus .

$here is no number we could s/uare and end up with a negatie , $hat howeer! is what would hae to happen if we wanted the s/uare rootof a negatie number.

$he chosen solution to this problem is partly to sidestep it ,

$he ery cunning plan is to let 1− ; =

- 43 -



)o for example we could write 4− as 4 x 1− ; : x T

0ow we can write any negatie root as the positie root times T

T has some odd properties.

T: ; 1− x 1− ; - =

T@ ; 1− x 1− x 1− ; T: x T; -= x T; -T

)ee if you can continue. 7ind T! T! TF! T! TG etc.

#hat is the patternJ

Carte&ian >or,

#e are now in a position to write down a complex number.

t consists of a real part a and an imaginary part b.

f P is a complex number then P ; a UTb ! for example P ; @ < T:or P ; ? T. $here are seeral ways of writing a complex number and thisone is called the CA($E)A0 or (EC$A0"15A( form. $his name is gienbecause we can plot complex numbers on a special graph with a on the xaxis *the real axis + and b on the y axis *the imaginary axis+. $he graph iscalled the complex plane or an Argand diagram.

- 44 -



Here the complex numbers -: < T ! @ < T ! - -:T and @ - :T are plottedon an Argand diagram.

Co,p+e? arith,etic in Carte&ian >or,

WORKED EA!"#ES

#e need to be able to add! subtract! multiply and diide complexnumbers.

Addition

f we wish to add two complex numbers together we add the real partsand add the imaginary parts * 0ot too tricky is it,+

Example8 f P= ; : < T and P: ; @ < :T! thenP= < P: ; <

FT

Su)traction

f we wish to subtract two complex numbers then we subtract the realparts and subtract the imaginary parts.

- 45 -

Ar$and dia$ra,

3, 4

-2, 5

3, -2-4, -2

-3

-2

-1

0

1

2

3

4

5

6

-5 -4 -3 -2 -1 0 1 2 3 4

rea+ a?i&

i , a

$ i n a

r *

a ? i &



Example8 f P= ; @ < T and P: ; : - @T! then P= - P: ; *@-:+ < *- -@+T; = < T

* $ake care with the two minuses, +

!u+tip+ication

$o multiply two complex numbers we hae to be able to multiply out twobrackets

Example8 P= ; @ < T and P: ; : < @T! then P= x P: ; *@ < T+* :< @T+

&ne way of doing this is to multiply the first bracket by the :! then thefirst bracket by the @T8

P= x P: ; *@ < T+* :< @T+; @x: < Tx: < @x@T <

Tx@T; F < GT < T <

=:T:

*0B T: ; - = $herefore =:T: ; =: x -= ; -=:+

P= x P: ; F - =: < =T; -F < =T

Divi&ion

#e want to diide two complex numbers. E?a,p+e- P= ; < FT and P: ; @ ? T

)o we hae P= V P: ;+−

F

@

T

T

$he first step is to multiply top and bottom of this fraction by @ < T. #eget this by changing the minus sign on the bottom line into a plus sign . @ <T is the complex conTugate of @ ? T. #e get the complex conTugate ofan* complex number by changing the sign in the middle.

)o P= V P: ;*< FT+*@ <T+

*@-T+*@< T+

0ow we multiply out top and bottom brackets Tust as we did formultiplication.

- 46 -



)o P= V P: ;=:<=G <=F -:

-=: <=: <=F

T T

T T ;− +=: @

:

T ;

−+

=: @: :

T

0otice that the bottom line becomes a real number . $his always happenswhen you multiply a complex number by its complex conTugate.


: Add the following complex numbers

a+ @ <T! <@Tb+ :-T! @<GTc+ -:T! --FT

9 )ubtract the following complex numbers

a+ :<T minus :<GTb+ @-T minus -FTc+ -GT minus @<FT

< Multiply the following

a+ *: < @T+*@<T+b+ *:-FT+*@-T+c+ *:<T+*:-T+

6 Diide the followinga+ -:-@T by -:T

b+ :<T by :-T

0ou can chec/ *our an&wer& )* +o$$in$ on to the >o++owin$ h*per+in/ed

&ite

http-@@,athin&ite.),th.ac.u/@app+et@co,p+e?@co,p+e?.ht,+

"o+ar >or,

Another way of writing complex numbers is in polar form8

Consider the complexnumber P ; x < Ty! as shown.

- 47 -

I,a$inar* a?i&

Rea+ a?i&

r

θ

x

Ty

&rigin! &

complex number! P

http://mathinsite.bmth.ac.uk/applet/complex/complex.html




t can also be described by the distance of P from the origin! r and the angle W that r makes with the horiPontal! as shown.

$he polar form of P is written8 P ; r ∠ W.#e say this Kr angle thetaL . #e call r the M&D151) and W theA("1ME0$.By conention! W is measured from the horiPontal anticlockwise up to <=G>X! then from the horiPontal clockwise up to ? =G>X. $his idea is shownbelow8

Co,p+e? arith,etic in "o+ar >or,

WORKED EA!"#ES

#e do not normally add or subtract complex numbers in polar form! it istoo inoled.

!u+tip+ication

Consider P= ; ∠ @>X and P: ; @∠ X#e want to find P= x P:.

$o do this we M15$25' the moduli and ADD the arguments

)o P= x P: ; x@∠ *@>X < X+; =:∠ X

Divi&ion

- 48 -

>X >X

>X

=G>X

>X=G>XX =G>XX

=G>X

< FX <=F>X -=>X -X

P ; rFX P ; r=F>X P ; r-=>X P ; r-X



Consider P= ; :∠ =X and P: ; ∠ ::X.

#e want to find P=IP:

$o do this we D4DE the moduli and )1B$(AC$ the arguments.

)o P=IP: ; :I ∠ *= X- :: X+; >. ∠ - X


=. Multiply the following complex numbers8a+ @ ∠ :@ X x ∠ F Xb+ G ∠ ==> X x ∠ - Xc+ : ∠ -:= X x F ∠ -G X

:. Diide the following complex numbers8

a+ ∠

∠ :>@ >

b+

∠ −

∠

=> >

:>

c+ ∠ −

∠ −@ =>F

Conver&ion )etween >or,&

t is possible to conert from polar to cartesian and ice ersa. )ome

calculators hae this facility pre-programmed into them. Consult youroperating manual to find out how to do it on your calculator. #e shall do itby calculation.

a+ Cartesian to 2olar

#e know x and y and we want to find r and W#e use these two e/uations8

= +: :y r x and W ; −=tan * +y x

- 49 -



b+ 2olar to Cartesian

#e know r and W and we want to find x and y

x ; rcos W and y ; rsin W

WORKED EA!"#ES

Carte&ian to po+ar

Conert P ; @ < T to polar form

Here x ; @ and y ; #e want r and W

1sing = +: :y r x

= +: :@ r ;

1sing W ; −=tan * +y x

; tan-= *I@+ ; @.=X

$he polar form is therefore P ; ∠ @.=X

"o+ar to carte&ian

Conert @∠ F>X into cartesian form

Here r ; @ and W ; F>X

#e want x and y

1sing x ; rcos W and y ; rsin W

9 ; @x . ; =.

' ; @x .GFF ; :.F

- 50 -



)o P ; =. < :.FT


Conert the following into polar forma+ : < @Tb+ -Fc+ - < Td+ -T

Conert the following into cartesian forma+ ∠ @>Xb+ @∠ >Xc+ F∠ -=:>X

'ou can check your answers using the same hyperlink as before.


=. 7ind the impedance! P of a circuit if the oltage of > < FT oltsproduces a current of > ? :>T amps.

:. "ien that the oltage applied to a circuit is = ; :T olts and the

current in the circuit is @ < T amps! find the power dissipated inthe circuit.* Hint8 for 3=! impedance is e/ual to oltage diided by current. n3:! power e/uals oltage times current.+

@. $hree forces of magnitude F! G and =>k0 are inclined to thepositie x-axis

at angles of @>! F> and =:> degrees respectiely! measuredclockwise.

Calculate the magnitude and direction of the resultant force byusing complex numbers.* Hint8 Express each force as a complex number in polar form! rbeing the magnitude and W being the angle. Conert into cartesian!add! then conert your answer back to polar to obtain the finalsum.+

1se the hyperlink to check your answers. http-@@,athin&ite.),th.ac.u/@app+et@co,p+e?@co,p+e?.ht,+

)ac/ to content&

- 51 -





SECTION TWO4EC$&()

Introduction

A scalar /uantity is one which is completely specified by a magnitude* siPe + only.Examples are mass! length! temperature! olume etc.)calars are simply added together! for example the total olume of twoobTects

of =m

@

< :m

@

; @m

@

.

A ector /uantity is one which is completely specified by both

a magnitude and a direction.

Examples are force! acceleration and ac oltages and currents *inpractice+.

f two forces are acting upon an obTect then the oerall effect dependsnot only upon the siPe of the forces but also the direction they are actingin.

Here the same two forces are acting on the obTect! but the effect isdifferent due to the changes in direction.

4ector /uantities can be represented on a diagram by arrows whoselength represents the magnitude and whose direction corresponds to thatof the ector.

- 52 -

=>0

F0

=>0

F0



1ector addition

$o add two ectors we must take into account the possibility of themhaing two different directions.

2arallelogram (ule

$he 0E$ effect of 2 and 3 is the ector ( which is the diagonal of a

parallelogram formed by 2 and 3 . t is called the (E)15$A0$.

2 < 3 = (

mportant special case

2 and 3 at right angles

$he Magnitude of ( = +r ::

32

$he direction of ( is gien by tan W ; 2I3

WORKED EA!"#E

#hat is the resultant of the followingJ

$he Magnitude ; +: :=> = ; =.:0

$he Direction ; tan-=*=>I=+ ; @.X anticlockwise from thehoriPontal.

- 53 -

2

3

( *Add and )

=>0

=0

2

3

(

θ



0otice that this is a ery similar process to conerting from cartesian topolar form in complex numbers.


= An obTect is acted upon by two forces of magnitude F0 and G0.

How would these forces be applied to gie a resultant force ofa+ :0 b+ =0 c+=>0 J

: 7ind the resultant of the two forces shown below8

Re&o+ution o> vector&

$he reerse of adding two ectors together is called resoling or taking

components.

f a force 7 acts at an angle W to the horiPontal as shown below! itseffect can be represented by two components at right angles to eachother! one horiPontal and one ertical.

7H ; 7cosW

74 ; 7sinW

$his process is called (E)&540"

)imple rule8 (E)&540" $H(&1"H $HE A0"5E8 use cosW(E)&540" A#A' 7(&M $HE A0"5E8 use sinW

- 54 -

0

=:0

θ

74

7H

7



0otice that this is a ery similar process to conerting from polar tocartesian form in complex numbers.


Calculate the horiPontal and ertical components of the force shownbelow on an obTect A.

7ind the ertical components of8

7ind the horiPontal components of8

- 55 -

=>0

@>X

A

=>0

X

=>0

@>X

0

=>0

F>X

0



1se the hyperlink to check your answers.http8IIwww.frontiernet.netIYimagingIectorZcalculator.htm


f the frictional force between the obTect below and the ground is =>>0!what is the least force 7 which is re/uired to moe it horiPontallyJ

$he obTect & below is not moing8

= n terms of 7 write an expression for the total horiPontalforce on &.

: 0ow calculate force 7 since the total horiPontal forcemust e/ual Pero.

@ (epeat the aboe for the total ertical force andcalculate #.

back to contents

Chapter Four

Matrices and Determinants

- 56 -

7

F>XA

7

F>X

A

X

#

&

http://www.frontiernet.net/~imaging/vector_calculator.html

http://www.frontiernet.net/~imaging/vector_calculator.html



see how to find this.

Determinants can only be s/uare and are written like this8

1 27 4

#e need to be able to add! subtract! multiply and diide matrices and alsobe able to ealuate determinants.

:' Eva+uation o> deter,inant&

#orking out a :9: determinant is ery easy.

Any :x: determinant can be written8a b

c d

ts alue is gien by *axd ? )xc+ .$ake care with double minus signs here!for example if b or c are negatie.

WORKED EA!"#ES

2 3

3 6 ; *:xF ? @x@+ ; *=: ? + ; @

3 2

3 4

−− ; *@x - - -:x@+ ; -=: < F ; -F * 0ote the double minus here +


Ealuate a+3 2

4 5 b+

1 2

3 4

−− c+

3 2

4 2

−−

#orking out larger determinants *@x@! x etc+ is a little trickier.

7irst of all! each po&ition in a determinant has a sign *positie or

negatie+!

- 58 -



starting with 8 in the top left hand corner.

)o we hae! for a @x@ determinant!

+ − +− + −+ − +

$hese signs are used solely when working out the alue of adeterminant and are in addition to any signs which the indiidual elementsmay hae.

$here are seeral different ways of doing the ealuation! but we shallstick to only one! expanding along the top row.As an example! consider the following determinant8

3 4

2 6

2 5

2

3

1

−−

− and remember

− +− ++

−+ − +

)tarting with the blue : in the top left hand corner! this is a 8 position.

$he blue : is at the intersection of the first row and first column asshown below8

6

2

3

3

2

2

5

4

1

−−

− $his leaes a :x: determinant

6 3

2 5− ! shown in red.

The >ir&t part o> wor/in$ out the va+ue o> the <?< deter,inant i&

89 ?6 3

2 5−

0ext we use the blue -@ in the top center! this is a position

2 4

2 6

5

3

3

1 2

−−

− remember

+ +− +

−−

+ − +

- 59 -



$he blue -@ is at the intersection of the first row and the second columnas shown below8

2 3 4

62

52

3

1

−−


2 3

1 5

−! shown in red.

The &econd part o> wor/in$ out the va+ue o> the <?< deter,inant i&

< x2 3

1 5

−

7inally we use the blue four in the top right hand corner! this is a 8position.

2 3

2 6

2

4

3

1 5

−−

− remember

+ −− +

+−

+ − +

$he blue four is at the intersection of the first row and the third column!as shown below8

2 6

2 3 4

3

51 2

−−


2 6

1 2

−− ! shown in red.

The third part o> wor/in$ out the va+ue o> the <?< deter,inant i&

8 6 x 2 61 2− −

$herefore! to work out the alue of the @x@ determinant by

expanding along the top row! we get8

89 ?6 3

2 5− < x2 3

1 5

−8 6 x

2 6

1 2

−−

; :*Fx -@x-:+ < @*-:x ? @x=+ < *-:x-: ? Fx=+

- 60 -



; :*@> < F+ < @* -=> -@ + < * -F+

; : -@ -G

; :

n general! to ealuate any @x@ determinant! we can expand along the toprow as follows8

a b c

d e f

g h i

; < a xe f

h i

- b xd f

g i

< c xd e

g h


Ealuate the following @x@ determinants

a+

3 5 7

11 9 1315 17 19

b+

13 2 23

30 7 5339 9 70

c+

6 10 14

2 1 19 15 12

−− − −

'ou should get =! @ and :@.

9' !atri? Arith,etic

#e need to be able to add! subtract! multiply and diide matrices in orderto use them to sole simultaneous e/uations.

ADD$&0 and )1B$(AC$&0

t is only possible to add or subtract same siPe matrices!since themethod is simply to add or subtract each indiidual element in thematrices.

WORKED EA!"#ES

Add the following matrices

- 61 -



1 4

6 3

÷

<

2 3

5 8

÷

;

1 2 4 3

6 5 3 8

+ + ÷ + + ;

3 7

11 11

÷

2

3

6

÷÷

÷

<3

4

6

÷÷

÷

;

2 3

3 4

6 6

+ ÷ + ÷ ÷+

;

5

7

12

÷÷

÷

1

4

÷

< ( )3 6 ; 0ot possible! since the matrices are different.

)ubtract the following matrices

2 1

7 4

− ÷ − -

3 0

7 4

− ÷ − ;

5 1

14 8

− ÷ −

3 1 4

4 3 1

1 4 3

− ÷÷

÷

-

2 7 5

2 1 0

6 3 4

− ÷ − ÷

÷

;

1 6 1

6 2 1

5 1 7

− ÷÷

÷− −

M15$25CA$&0

Matrix multiplication is a strange process, )ome matrices cannot bemultiplied together. 'ou can only multiply two matrices together if[

the siPe of the rows in the first one is the same as the siPe of thecolumns in the second one.

$he examples below show alid matrix combinations

b e h

a d g

c f i

÷÷

÷

x

u

v

w

÷÷

÷

*$he first matrix has @ elements in a row! the second matrix has @elements in a column+

- 62 -



n t

x y

÷

x

l s

t m

÷

*$he first matrix has : elements in a row and the second has : elements

in a column.+

f the aboe conditions are satisfied then the multiplication method is 8multiplyeach row element by the corresponding column element and add theresults.

WORKED EA!"#ES

2 4

3 5

÷

x

6

7

÷

;

3 6

6

5

42 7

7

+ ÷ × + ×

×

× ;

40

53

÷

1 3

2 4

÷

x

2 7

5 8

÷

;

1 7 3 8

2 2 4 5 2 7 4

1 2 3 5

8

× + × ÷ × + × × + ×

+ ×

× ;

10 31

24 46

÷

2 3 4

1 5 73 6 8

÷ ÷ ÷

x

2

13

÷ ÷ ÷

;

2 2 3 1 4 3

1 2 5 1 7 33 2 6 1 8 3

× + × + ×

÷ × + × + × ÷ ÷× + × + × ;

19

2836

÷ ÷ ÷

1 3 4

2 1 5

3 2 6

÷÷

÷

x

2 5

3 6

4 7

÷÷

÷

;

1 2 3 3 4 4 1 5 3 6 4 7

2 2 1 3 5 4 2 5 1 6 5 7

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

× + × + × × + × + × ÷ × + × + × × + × + × ÷

÷× + × + × × + × + ×

;

27 51

27 51

36 69

÷÷

÷

Multiplying a matrix by a single number is comparatiely straightforward.#e simply multiply eery element in the matrix by the number,

7inally a further strange property of matrix multiplication is 8

*MA$(9 A x MA$(9 B+ i& not e(ua+ to *MA$(9 B x MA$(9 A+

n other words! the order is important.


- 63 -



5et A ;1 4

2 3

÷

[ B ;

2 4

6 5

− ÷

[ C ;

4

5

÷

[

D ;

2

4

6

÷÷

÷

[ E ;

3 2 4

1 3 5

4 1 6

− ÷ − ÷ ÷

[ 7 ;

1 2

7 4

9 6

− ÷ − ÷ ÷

7ind8=. A < B:. A ? B@. :A

. BxC. AxBF. Ex7. ExD

5ater we will look at using Excel to perform matrix arithmetic and youcould use this to check your answers.

D4)&0

)trangely! matrix diision inoles multiplying ,

$he idea is as follows8

−= × = × ==AA A B

B B

$hat is! A diided by B is the same thing as A times the inerse of B.

* = oer anything is called the inerse+

7or example8

= × = ×@ =

@ @ .:

Here the diision by four has been changed into a multiplication by theinerse of four !that is .:.

- 64 -



)o! to diide by a matrix! the method is8

7ind the 04E()E of the matrix and M15$25'.

$he only problem now is! how do we find the inerse of a matrix J

70D0" $HE 04E()E &7 A MA$(9

$his is rather like producing a witches brew! without the eye of newt. $hefollowing steps are inoled8

= 7ind the cofactors *whatJ- see later+ of the matrix

: $ranspose *again! later+ the matrix of cofactors! producing theAD%&0$ matrix.

@ 7ind the determinant of the original matrix Diide the AD%&0$ matrix by the determinant

$hese four steps produce the re/uired inerse.

WORKED EA!"#E

5et us find the inerse of matrix A! where

A ;

÷÷

÷

= : =

: > :

= = >(emember place alues

+ − + ÷ − + −÷

÷+ − +

)tep one,

Each element in the matrix has a cofactor. A cofactor is the :x:determinant left after remoing the row and column that the element isin. $he :x: determinant carries the sign associated with the position ofthe element! Tust like in the ealuation of a determinant.

Hence the cofactors of the first row are[

- 65 -



< > :

= > ; -: ! -

: :

= > ; <: ! <

: >

= = ; <:

)imilarly the cofactors of the second row are[

- : =

= > ; <= ! <

= =

= > ; -= ! -

= :

= = ; <=

$he cofactors of the third row are[

< : =

> : ; < ! -

= =

: : ; > ! <

= :

: >; -

$he matrix formed from these cofactors is therefore[

− ÷÷

÷−

−

>

= = =

: : :

)tep two,$his inoes transposing the new matrix. #e swap rows for columns. 'oucan see below that the first row becomes the first column! the secondrow becomes the second column! etc.

− ÷÷

− ÷−

>

=:

=:

=:

$his is called the AdToint matrix.

)tep three, is to work out the determinant of the original matrix.

A ;

÷÷

÷

= : =

: > :

= = > becomes

= : =

: > :

= = >

Expanding along the top row we hae[

- 66 -



<= x> :

= > - : x

: :

= > < = x

: >

= =

;

)tep four,

n this last step we diide the adToint matrix by the determinant alue.

$o do this we diide eery element of the matrix by ! or

=I x − ÷÷

− ÷−

>

=:

=: =: ;

− ÷

÷ − ÷÷÷÷ − ÷

: =

: = > : =

7inally we hae got it ! this is the inerse matrix of the one we startedwith,

t seems ery complicated! but all you hae to do is follow the four stepsexplained aboe.


)ee if you can find the inerse of the following matrices. $he answers tothe first two are gien. 5ater you will be able to use Excel to work out orcheck your answers.

− − ÷ − ÷ ÷

= : @: =

@ = Ans =I x

− − ÷ − ÷ ÷−

= = => G :

== => @

÷÷

÷

: = =

= : :

= @ : Ans - =I@ x

− ÷ − ÷

÷−

: = >

> @ @

= @

- 67 -



÷÷

÷−

= : >

=: F

@ @

(emember! the purpose of finding the inerse of a matrix is to do matrixdiision.

−= × = × ==AA A B

B B

$hat is! A diided by B is the same thing as A times the inerse of B.

'ou now hae the skills needed to do diision.


f A ;

÷÷

÷

@@

: and B ;

÷ − ÷

÷− −

= = =

: @

@ : : 7ind

A

B

f A ;

÷ − ÷

÷−

=@

:

= and B ;

− ÷ − ÷

÷−

: @

@ = :

= : =

U&in$ E?ce+ to carr* out ,atri? arith,etic.

magine we hae the following two matrices! which we wish to diide.

A ;

÷ − ÷

÷

=

=

@ and B ;

− ÷ − ÷

÷−

: = =

: : @

= = :

#e need to find the inerse of matrix B and then multiply it by A.

- 68 -



&pen Excel. nput the matrix B as shown into! say! cells A= up to [email protected] select a region which is the same siPe as the matrix.* n this case Ato C.+ 0ow input the matrix formula as shown onto the formula bar.

7inally! to actiate the formula and find the inerse of the gien matrixin Excel! press Control and )hift and (eturn.

- 69 -



$he inerse matrix magically appears,

7inally!we wish to multiply this inerse matrix by the other matrix.$o do this we insert the second matrix as shown into cells A to A==! then

select suitable cells for the answer *A=@ to A=+ and type in the formula shown in the formula bar.

$o carry out the matrix multiplication we now press Control and )hift and(eturn.

$his gies the solution as shown below.

- 70 -



0B in Excel rounding errors can produce a ery! ery small number as ananswer instead of > , n this case =.F@FE-= can be regarded as >.

0ow you can use this method to check your preious answers.

A""#ICATION QUESTION $he tensions in a simple framework! $ =! $ : and $ @ are gien by thee/uations8F$ = <F$ : < F$ @ ; G. $ =< :$ : < $ @ ; :. $ =< :$ : < >$ @ ; .> 7ind $ =! $ : and $ @.

Hint8 $he three e/uations can be rewritten as a matrix e/uation[

- 71 -



÷ ÷ × =÷ ÷

÷ ÷ ÷

=

:

@

F F F G.

= : :.

: > .>

$

$

$

*Check what you get if you multiply out the left hand side+

)o!if we diide the matrix on the right hand side by the @x@ matrix on

the left hand side! we should end up with a matrix e/ual to

÷÷

÷

=

:

@

$

$

$

,,

)ac/ to content&

Chapter FiveMA$H) &0 $HE M&4E 8 CA5C151)

Introduction

n the times of ancient Egypt and "reece! mathematics was mainly aboutthe measurement and description of static obTects! such as the perimeterof a field or the total wages bill for an army.$here were greatdeelopments in geometry and trigonometry during this period.#hen scientists began to study moing obTects! howeer! it becameneccessary to deelop ways of describing speed! acceleration and changesin things using mathematics! so that useful calculations could be done.saac 0ewton played an important part in deeloping calculus! a branch ofmathematics which describes changing situations.

Engineering is full of changing situations[$he position of a piston$he siPe of an AC oltage$he temperature of a furnace$he pressure in a tyre$he charge on a capacitor

t follows that calculus is a ery important part of mathematics forEngineers.

Calculus is split into two main branches! which complement each other!

Differential and ntegral calculus.#e shall look at each in turn.

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SECTION 5NEDifferential calculus

$he straight line graph in blue aboe shows the connection between two/uantities! x and y.

$he gradient of the line is a measure of how much y changes when we

change x.

$he gradient is worked out using the e/uation8

gradient ;∆∆

y

x Delta y *\y in symbols + means K a change in yL

Delta x *\x in symbols + means K a change in x L

$herefore =∆∆

$he change in y$he change in x

y

x

'ou could also say the gradient is the rate of change of y with x.

Consider the following graph of temperature against time[

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y

xΔx * delta x +

Δ y * delta y+

$emperature XC

time I sΔt * delta t +

ΔT * delta $+



"radient ;\$I\t

$he gradient measures how much the temperature changes in a certaintime eg Kchange in tempI second L. #e could also say that the gradientrepresentsKthe rate of change of temperature with timeL.


5ook at the following graphs and put into words what the gradientrepresents8

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4oltage I

time I sΔt

ΔV

2ressureI2a

DepthI mΔD

ΔP

Height Im

AgeI yearsΔA

ΔH



)1MMA('

)o! if we want to know the rate of change of one /uantity with another!one way to find out is to plot the graph and find the gradient.

$here are two problems with this approach. t is time consuming and onlyas accurate as the graph drawing. s there another wayJ

'es! the other way inoles knowing the algebraic e/uation of the line. fwe know this we can immediately obtain the exact gradient of the linewithout haing to draw the graph,

n general any straight line which passes through Pero can be

represented by the e/uation 8' ; mx

n this formula m represents the gradient of the line *see also thesection on straight lines in chapter one+.

)o y ; x is an e/uation which describes a straight line passing throughPero! with a gradient of .

Consider the e/uation 4 ; @t where 4 is the oltage in a particular circuitand t is thetime. $he gradient is @ and this means that ! for this circuit! the rate ofchange of oltage with time is @ 4Is. #e could also say that the change inoltage is @ when the the time changes by one second.

$here are some special cases .

A horiPontal line has a gradient of Pero. n this case any change in xproduces no change in y! so the gradient is Pero.

f y ; @ is the e/uation of the line then the gradient is Pero,f y ; :@= is the e/uation of the line then the gradient is Pero,f y ; K a constant L is the e/uation of the line then the gradient is Pero,

A ertical line has an infinite gradient! because no change in x producesan infinite change in y.

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7inally we can now mention ca+cu+u&, A third way to find the gradient of

an e/uationis to differentiate the e/uation.$his is the purpose of differential calculus! that is! gien an e/uation! to

find an e/uation representing the gradient. $he process is representedbelow8

$o differentiate is to find the gradient * sometimes called thedifferential coefficient or the deriatie+.

(emember! the symbols for the gradient were∆∆

y

x

$he symbol for differentiation isdy

dx

* $his is said as K dee y by dee x L

+

f we start with the e/uation y ; @x! the process is8

' ; @x differentiate dy

dx ; @ *we already know the gradient is @+

n the hundreds of years since calculus was first inented!

mathematicians hae worked out how to differentiate lots of differentkinds of e/uations and hae produced tables of formulae to do it. $heseare called tables of standard differentials.

$he formula used aboe to getdy

dx ; @ was!

f y ; axn is the e/uation ! where a is any number and n is any number!

thendy

dx

; naxn-=

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E/uationDifferentiate

"radient



)o! if y ; @x * ; @x= + dy

dx ; = times @ times x= ? = ;

@x> ; @

Compare with y ; axn * a ; @ and n ; = +dy

dx

; n times a times xn-=

0ote an*thin$ to the power o is =! that is! x> ; =

Also! if the e/uation represents a horiPontal line! we know that thegradient is Pero.

Eg for y ; ! dy

dx ; >

1sing the formula y ; axn

! we can think of y ; as y ; x>

*because x>

; =+

$herefore a ; and n ; > . )incedy

dx ; naxn-= ! then

dy

dx ; > times times

x>-= ; >

)o! if we differentiate a number which is by itself we get >.


Differentiate the following e/uations * finddy

dx +

' ; :x' ; x' ; =:.Fx' ;

' ; =:' ;=:@:.

$his complicated process seems fairly pointless for a straight linee/uation! because we dont need calculus to work out the gradient! butwhat if the e/uation is for a more complicated shape than a straight lineJ

Consider the following graph8

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A cure does not hae a single gradient like a straight line! but has a

different gradient at eery point on it. $he gradient at x= aboe isdifferent from the gradient at x: as shown by the red lines*calledtangents+ . $o find the gradient at a point on a cure ! we would hae todraw the cure! draw in the tangent at the point of interest! and find thegradient of the tangent. $o find the gradient at another point! we wouldhae to repeat the process. $his is time consuming and of limitedprecision.

f we hae the e/uation of the cure we can obtain an e/uation for

working out its gradient at any point by differentiating the e/uation asbefore.

magine we hae been asked to find the gradient of the cure y ; x: attwo points!x ; : and x ; .&ne way of doing this is to plot the graph of y ; x: ! draw in the tangentsat x ; : and x ; ! then work out the gradients at these points.

Alternatiely! we can differentiate y ; x:

and obtain an e/uation forworking out the gradient at any point on the cure,

(emember the formula8

f y ; axn is the e/uation ! where a is any number and n is any number!

thendy

dx ; naxn-=

Comparing y ; x: with y ; axn we can see that a ; and n ; : in this case.

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x=

x:

y

x



t follows thatdy

dx ; : times times x:-= ; Gx

)o! the gradient at x ; : is G times : ; =FAnd the gradient at x ; is G times ; >$hese are e?act answers and we hae obtained them /uickly withoutdrawing,


Differentiate the following expressions8

' ; :

' ; :x

' ; :x:

' ; @ x:

' ; @x@

' ; x

0ote 8 #e hae been using x and y to represent two /uantities! but wecan use any letters we want eg u and ! s and t or whateer. n these

cases we would write the differentials asdu

dv and

ds

dt

7irst!a short note on e/uations8

E/uations of the form y ; mx are like a kind of computer programme.

f you insert alues of x into the right hand side! the programme worksout the alue of the left hand side.$he e/uation gies us the mathematical connection between x and y. f you know x then the programme shows you how to find y.

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n mathematics we call such a programme a KfunctionL. #e say Ky is afunction of xL

#hich is Tust another way of saying K y depends upon xL

n symbols we would write 8

y ; f*x+ $his means Ky is a function of xL)ome examples are8 y ; @xor y ; x: < @x -or y ; @sin* x+

Mathematicians hae worked out formulae to differentiate most of thecommon functions used in Engineering and these are known as &tandard

di>>erentia+&.

A table of some of them is shown below8

$he ] a ] in the table can be any

number! positie or negatie.

WORKED EA!"#ES

#e hae already practised using the firstone in the table! here are some further examples.

n each case we wish to find

dy

dx y ; sin *@x+ f we differentiate sin *@x+ we get @cos *@x+

using the second row in the table aboe.#e hae howeer! got times this! so the

answer is8

dy

dx ; x @cos *@x+ ; =:cos *@x+

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function di>>erentia+axn na?n:

sin *ax+ aco&%a?'

cos *ax+ a&in %a?'

eax aea?

n *x+ :@?



y ; @ex f we differentiate ex we get ex! using the

fourth row in the table.#e hae! howeer! got @ times this! so the

answer is8

dy dx

; @ x ex ; = ex


Differentiate the following8

y ; :cos*x+ y ; e@x

y ; Fsin *-:x+ y ; :e-x

Co,)ination& o> >unction&

$he functions in the standard table can be combined in arious ways! such

as added! subtracted! multiplied and diided.

#e shall only consider the first two combinations here! that is functionsadded or subtracted. $he other possibilities are more complex to dealwith and are outside the scope of this text.

f we need to differentiate a combination of seeral functions! which areadded and Ior subtracted! then we simply differentiate each separatelyand add or subtract the results as appropriate.

WORKED EA!"#ES

f y ; x@< :x:< @x <

7inddy

dx

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dy

dx ; =:x: < x < @ < > Here we hae differentiated each function

separately and added the results.

f y ; @x < :sin*:x+ ? e@x

7inddy

dx

dy

dx

; =:x@ < cos*:x+ -@e@x Here we hae differentiated each

function separately and added orsubtracted as appropriate.


Differentiate the following8

y ; :x < @x -x <

y ; :sin *Fx+ ? cos *x+ < :ex

; @x- eFx <x ? @x <@

SECTION TWOntegral calculus

$he intention in this section is to proide a brief introduction to thisimportant branch of calculus! but the reader is referred to the K7urther)ources of nformationL for an extended discussion and furtherexamples.

ntegral calculus is about the mathematical process of ntegration . tcan be thought of in two ways[ as the reerse of differentiation! or as an

Kadding upL process on a function which enables us to calculatecomplicated areas and olumes.

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#e shall only consider the first iewpoint here! as this is sufficient toac/uaint ourseles with the basic process.

magine we start with the function y ; @x:< :. $hen we differentiate it!

gettingdy

dx ; Fx .

#hat if we had started withdy

dx ; Fx and wanted to go back to the

original function ! @x:< : J

$his reerse process is called integration and is shown diagrammaticallybelow8

0ote that there is no way of knowing that the : was originally there!because any number when differentiated

is >. $o take account of this! a constant! c is always included whenintegrating. $his is known as the constant of integration.

$he process of differentiation has a symbol!dy

dx which means to

differentiate.)imilarly the process of integration has a special symbolwhich means to integrate.

#e would write K#ork out the integral of FxL as =

∫ Fy xdx ! which is as

we know! @x: < c .

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* 4 <?98 9differentiate

7?

7?integrate

* 4 <?98 c



How do we go from Fx to @x: J #e use a table of standard integrals inexactly the same way that we used a table of standard differentials todifferentiate.

* Except for n ; -=+

$he a is any number as before.

$he standard integral we were using is the first one.

#e wish to find the integral of Fx! that is! ∫ Fxdx . Comparing this with

∫ n a dx x we see that a ; F and n ; =

* x by itself is x=+

1sing the aboe table the integral is+

+

=

=

n a x n

< c . $his gies us Fx=<=I=<= ;

Fx:I: ; @x:< c ,

WORKED EA!"#ES

= ntegrate x that is! find ∫ Cdx x

$he appropriate formula is+

+

=

=

n a x n

< c! where in

this case ! a ; = and n ;

$herefore ∫ Cdx x ; =x < =I < =

; xI < c

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function inte$ra+^axndx +

+

=

=

n a x n

8 c

^sin*ax+dx %:@a'co&%a?' 8 c

^cos*ax+dx %:@a'&in%a?' 8 c

^eaxdx %:@a'ea? 8 c

^*=Ix+dx In %?' 8 c



: ntegrate @ that is! find ∫ @dx . $his is a slightly

awkward one until you get used to it.x> ; = ! so we can write @ as @x> *that is @x=+

$he integral now becomes ∫ >@ dx x . n this case a ;

@ and n ; > in+

+

=

=

n a x n

$herefore ∫ @dx ; @x> <=I ><= ; @xI= ;@x < c

n general! therefore ^=dx ; x! ^:dx ; :x! ^@dx ;@x etc

@ ntegrate :sin*@x+ that is! find ∫ :sin*@ +x dx .

1sing the table we see that ∫ sin* +ax dx ; *-=Ia+

cos *ax+

n this case a ; @ . Also it should be realised that

=∫ ∫ :sin*@ + : sin*@ +x dx x dx

)o we hae ∫ :sin*@ +x dx ; :*-=I@+cos*@x+ < c


#ork out the following integrals

: ^@x@

dx9 ^xdx

< ^sin*xI@+

6 ^cos*@x+

B ^e@xdx

7 ^e-:xdx

0ote 8 $he integral of seeral functions added or subracted isobtained by integrating them separately then adding or subtracting

the results. $ry these8 ^*@x< :x -=+dx

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^*:ex- :sin*@x+ < +dx


$he definition of the elocity of an obTect is its rate of change ofdistance with time.f s stands for the distance traelled in t seconds! then the elocity is

gien byds

dt . f we hae the e/uation connecting s with t then we can

obtain an expression for the elocity by differentiation.

2roblem =

f the distance traelled! s ; @t@ ? t: < :t -:7ind the elocity when t ; : seconds.

f we hae already been gien the elocity and we wish to work out thedistance traelled then we must integrate the e/uation for the elocity.n order to find the alue of the constant c we will need to be gien someextra information.

2roblem :

f the elocity of an obTect is gien byds

dt ; @t < and s ; : when

t ; >!

7ind an expression for the distance traelled! s. * 1se the extrainformation to find c+

7ind the distance traelled when t ; :

back to contents

FURT3ER SOURCES OF INFOR!ATION

B(D!%& and MA'!A%C *=G+ $ECH0CA0 MA$HEMA$C).5ongman "roup 5td.

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'A$E)!%C *=@+ 0A$&0A5 E0"0EE(0"MA$HEMA$C).$he Macmillan 2ress 5td.

"(EE(!A and $A'5&(! ".#. *=G@+ MA$HEMA$C) 7&($ECH0CA0). )tanley $hornes 5td.

Brush Up Your Maths 3

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