7/24/2019 Brush Up Your Maths 3 http://slidepdf.com/reader/full/brush-up-your-maths-3 1/87 Brush up your maths An access course for H. E. Engineering. by Dr. D. Baxter of Mid Cheshire College Acknowledgements A work of this kind is necessarily deriatie and the author acknowledges the influence of established writers in the field of engineering mathematics! such as A "reer and " # $aylor! % & Bird and A % C May! % C 'ates and ( )utherland! D Howell and A #olf! whose books are recommended 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 in touch with Aimhigher *Cheshire and #arrington + with a iew to correcting this. $he deelopment time was funded by Aimhigher of Cheshire and #arrington and proided by Mid Cheshire College! whose students represented a captie audience, Introduction $his E-book is intended to proide a grounding in some of the mathematics needed to study Engineering at degree leel. As such it is ery 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 - 1 -
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
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
$his E-book is intended to proide a grounding in some of themathematics needed to study Engineering at degree leel. As such it isery focussed upon the basic ideas in each area of mathematics coered.
t is noel in as much as the reader is expected to use modern methods
of calculation and graphing such as Excel and to access interactieinternet mathematics sites in order to check calculations. n each sectionthere are worked examples ! practice examples and appropriateengineering problems.
CONTENTS
Chapter One$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.
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. $he twosides are only e/ual if the scales balance exactly.
magine that you had : kilos of flour on the left hand side and : kilos onthe right hand side. the scales would balance and you could say : ; : ,f you now changed one side * say you added one kilo to the left pan +! youwould hae to carry out exactly the same change to the other side in
order to keep the scales in balance. t would now be true to say that :<= ;:<=! that is if you add = to both sides the scales still balance.
Exactly the same thing applies to e/uations. #hateer you do to one sidemust be done to the other in order to maintain the balance and make thee/uals sign true.
WORKED EA!"#ES
Rearran$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. $omaintain the balance we need to add f to the rhs also.
!ethod- n this case we want ( by itself but we hae times toomany (s! so what we need to do is diide by . f we diide the rhsby then we must also diide the lhs by in order to maintain thebalance.
Diiding both sides by we get×
=4 (
which becomes =
4 (
Tran&po&in$ e(uation& where the (uantitie& are
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 × ×
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.
)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.
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 ,
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.
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 ,
)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.
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.
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
$o do this we D4DE the moduli and )1B$(AC$ the arguments.
)o P=IP: ; :I ∠ *= X- :: X+; >. ∠ - X
EA!"#ES FOR 0OU TO DO
=. 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
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.
A""#ICATION QUESTIONS
=. 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,+
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.
0otice that this is a ery similar process to conerting from cartesian topolar form in complex numbers.
EA!"#ES FOR 0OU TO DO
= 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
$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
*$he first matrix has : elements in a row and the second has : elements
in a column.+
f the aboe conditions are satisfied then the multiplication method is 8multiplyeach row element by the corresponding column element and add theresults.
WORKED EA!"#ES
2 4
3 5
÷
x
6
7
÷
;
3 6
6
5
42 7
7
+ ÷ × + ×
×
× ;
40
53
÷
1 3
2 4
÷
x
2 7
5 8
÷
;
1 7 3 8
2 2 4 5 2 7 4
1 2 3 5
8
× + × ÷ × + × × + ×
+ ×
× ;
10 31
24 46
÷
2 3 4
1 5 73 6 8
÷ ÷ ÷
x
2
13
÷ ÷ ÷
;
2 2 3 1 4 3
1 2 5 1 7 33 2 6 1 8 3
× + × + ×
÷ × + × + × ÷ ÷× + × + × ;
19
2836
÷ ÷ ÷
1 3 4
2 1 5
3 2 6
÷÷
÷
x
2 5
3 6
4 7
÷÷
÷
;
1 2 3 3 4 4 1 5 3 6 4 7
2 2 1 3 5 4 2 5 1 6 5 7
3 2 2 3 6 4 3 5 2 6 6 7
× + × + × × + × + × ÷ × + × + × × + × + × ÷
÷× + × + × × + × + ×
;
27 51
27 51
36 69
÷÷
÷
Multiplying a matrix by a single number is comparatiely straightforward.#e simply multiply eery element in the matrix by the number,
7inally a further strange property of matrix multiplication is 8
*MA$(9 A x MA$(9 B+ i& not e(ua+ to *MA$(9 B x MA$(9 A+
$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.
$he matrix formed from these cofactors is therefore[
− ÷÷
÷−
−
>
= = =
: : :
)tep two,$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.
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.
EA!"#ES FOR 0OU TO DO
)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.
&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.
*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.
$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.
EA!"#ES FOR 0OU TO DO
5ook at the following graphs and put into words what the gradientrepresents8
)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.
$here are some special cases .
A horiPontal line has a gradient of Pero. n this case any change in xproduces no change in y! so the gradient is Pero.
f y ; @ is the e/uation of the line then the gradient is Pero,f y ; :@= is the e/uation of the line then the gradient is Pero,f y ; K a constant L is the e/uation of the line then the gradient is Pero,
A ertical line has an infinite gradient! because no change in x producesan infinite change in y.
7inally we can now mention ca+cu+u&, A third way to find the gradient of
an e/uationis to differentiate the e/uation.$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.
$he formula used aboe to getdy
dx ; @ was!
f y ; axn is the e/uation ! where a is any number and n is any number!
0ote an*thin$ 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 >.
EA!"#ES FOR 0OU TO DO
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
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. $his is time consuming and of limitedprecision.
f we hae the e/uation of the cure we can obtain an e/uation for
working out its gradient at any point by differentiating the e/uation asbefore.
magine we hae been asked to find the gradient of the cure y ; x: attwo points!x ; : and x ; .&ne way of doing this is to plot the graph of y ; x: ! draw in the tangentsat x ; : and x ; ! then work out the gradients at these points.
Alternatiely! we can differentiate y ; x:
and obtain an e/uation forworking out the gradient at any point on the cure,
(emember the formula8
f y ; axn is the e/uation ! where a is any number and n is any number!
thendy
dx ; naxn-=
Comparing y ; x: with y ; axn we can see that a ; and n ; : in this case.
)o! the gradient at x ; : is G times : ; =FAnd the gradient at x ; is G times ; >$hese are e?act answers and we hae obtained them /uickly withoutdrawing,
EA!"#ES FOR 0OU TO DO
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.
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
EA!"#ES FOR 0OU TO DO
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.
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.
EA!"#ES FOR 0OU TO DO
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.
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
$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.