Calculus

A LITTLE BIT OF CALCULUS

by

S. VADAKKAN

For we are in God�s hands . . .our understanding and technical knowledge.

Wisdom VII : 16

I have held many thingsin my hands, and lost them all.

But whatever I have placedin God�s hands I still possess.

Martin Luther King Jr.(1929 - 1968)

Title : A LITTLE BIT OF CALCULUSAuthor : Stephen VADAKKANabout author: http://www.staloysius.ac.in/campus/alumni/stephenv.phpSpecial International Edition : January 2006

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http://www.staloysius.ac.in/campus/alumni/stephenv.php

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This book is dedicated to theThis book is dedicated to theThis book is dedicated to theThis book is dedicated to theThis book is dedicated to the

Ancient and Noble House of Bou-RisliAncient and Noble House of Bou-RisliAncient and Noble House of Bou-RisliAncient and Noble House of Bou-RisliAncient and Noble House of Bou-Risli

of the Bani Khalid (the Pure)of the Bani Khalid (the Pure)of the Bani Khalid (the Pure)of the Bani Khalid (the Pure)of the Bani Khalid (the Pure)

of Arabiaof Arabiaof Arabiaof Arabiaof Arabia

i

Pr e f aceThis is not a Text Book, Primer or Guide Book on Calculus. It is just an introductionto the fundamental concepts. Particular words and symbols are used to expressthese concepts. These concepts and words take time to sink in. It is hopedthat in studying this booklet the student will become familiar with the concepts,terminology, notation and the kind of calculations that can be done: what to lookfor and what to expect.

Generally speaking, in most of the Sciences there are usually two main branches.Biology is divided into Botany (plant life) and Zoology (animal life). Chemistryis divided into Organic (carbon compounds) and Inorganic (non-carbon compounds).In Light we have the discrete particle photon and the continuous wave. Likewise,Mathematics is divided into two main branches: Algebra and Analysis.

In Algebra we deal with discrete operands, be they numbers like 1,2,3, . . . orsymbols for numbers like x, y, z, . . . . The operands take on discrete values. The7 operations are + and its inverse -- , and its in its inverse , exponentiationy = xn and its two inverses ny = x and logxy = n. Usually the set of differentvalues the operands may take on is finite, or at least the set of values can be countedor enumerated.

In Analysis we deal with continuous expressions or operands or functions. Takefor example a bouncing ball: each position it bounces is discrete and can be counted.The ball may bounce indefinitely. On the other hand a ball rolling in a straight line hasinfinitely many positions in a continuous manner. We cannot even begin to count allthe different positions. So rather than discrete identification of position, we have acontinuous expression or function to describe its position or motion over time.

ii

The fundamental concepts in Calculus are INFINITESIMAL and LIMIT which are usedto develop the concepts of INSTANT, INSTANTANEOUS, CONTINUITY andDIFFERENTIABLITY. Once these concepts are in place we can talk of functions thatare WELL BEHAVED: SINGLE VALUED, CONTINUOUS and DIFFERENTIABLE. We canthen do two very beautiful calculations or operations:

1 . Given a SINGLE VALUED, CONTINUOUS and DIFFERENTIABLE functionthat expresses CHANGE we can DIFFERENTIATE the function and get thenew function that expresses the INSTANTANEOUS RATE OF CHANGE.

2 . Given the function that expresses the INSTANTANEOUS RATE OFCHANGE, we can INTEGRATE the function and get the new function thatexpresses CHANGE.

Working with polynomials is easy since they are CONTINUOUS everywhere. Thefundamental concepts and calculations (Differentiation and Integration) canbe taught with ease and clarity. Getting more information about the function(increasing, decreasing, maximum, minimum, inflexion) from its Higher OrderDerivatives can also be shown.

I have deliberately chosen very simple examples to illustrate the concepts. I haveavoided all the rigorous details under which calculations are done. The theory ofReal Analysis and Calculus deals with this. Rigour does not necessarily mean clarityand ease of understanding.

At the Class XI level this should be the approach. This approach will providethe confidence to deal with more difficult functions and the motivation tostudy the theory.

The notes are intended as thought evoking pointers for students who wish tostudy "A Litti le More Calculus " by the author.

iii

(0, 0)

u

u.cos u.

sin

(0, 0)

u

Range x(t)

He

igh

t y

(t)

x

y

Throughout the book we make use of one main example, the projection of a ball, toillustrate all the concepts and demonstrate all the calculations. Non-science studentsmay not be familiar with the projectile equation. So a brief explanation follows.

An object may be projected into space with an initial velocity u and an angle ofprojection . Let T be the time the object takes to go up and come down.

POSITION (2-Dimension) SPEED (2-Dimension)

The initial speed u has two components: a ver tical component u . s i n and ahorizontal component u.cos . We know that gravity g acts in the downward ornegative direction. So, with the appropriate units of measure approach in mind,we may see that the ver tical speed must be u . s i n -- g t [ meters/sec].

timeTT/ 2 (0,0)

VERTICAL SPEED (1-Dimension)

u . s i n -- g t [ meters/sec]

+ u . s i n

u . s i n

iv

VERTICAL POSITION (1-Dimension)

y(t) = u . s i n . t -- 1--2

g t 2

(0, 0)Time

T t

Hei

ght

y(t

)

y(t1)

h1

t1 T/2

Since DISTANCE = SPEED X TIME, the ver tical position or height must be of the

form A1.(u.sin ).t -- A2 g t 2 [meters]. The coefficients A1 and A2 may be

determined Analytically (as we shall see in integration) or by physical experiments.

It turns out that A1= 1 and A2 = 1/2 . So the function or expression that describesthe ver tical position or height is u . s i n . t -- 1--

2 g t 2 .

Even if the object is projected ver tically up, this is what the graph of theVERTICAL POSITION wil l look l ike over the time inter val [0, T]. Thedif ferent iat ion operation allows us to find the VERTICAL SPEED from theVERTICAL POSITION. And the integrat ion operation allows us to find theCHANGE in the VERTICAL POSITION from the VERTICAL SPEED.

The book is divided into 4 parts. On initial reading the student may skip part 3.

v

C O N T E N T S

1. INTRODUCTION 1

Par t 1 : CONTINUITY

2. NUMBERS AND THE NUMBER LINE 8

3. REALS, COMPLETE, CONTINUOUS 13

4. TENDS TO and LIMIT 14

5. INFINITESIMALS 16

6. x and INSTANT 19

7. SINGLE VALUED FUNCTIONS 21

8. LIMIT of a Function 24

9. Analysis of Limits 34

10. VALUE versus LIMIT 37

11. CONTINUITY of a Function 39

Par t 2 : DIFFERENTIATION

12. INSTANTANEOUS RATE OF CHANGE of y(t) 57

13. INSTANTANEOUS RATE OF CHANGE of f(x) 61

14. INSTANTANEOUS RATE OF CHANGE of xn 65

15. Differentiation from FIRST PRINCIPLES 71

16. Tables and Rules 79

17. Units of Measure 84

18. DIFFERENTIABILITY 86

vi

Par t 3 : ANALYTICAL GEOMETRY

19. FIRST DERIVATIVE = Slope of Tangent 98

20. Angle of Intersection 104

21. Increasing , MAXIMUM, Decreasing 108

22. Decreasing , MINIMUM, Increasing 113

23. MAXIMA and M IN IMA 116

24. Po ints o f INFLEXION 121

Par t 4 : INTEGRATION

25. F(x) = ANTIDERIVATIVE {f(x)} 129

26. F(x) = area under f(x) = f(x)dx 131

27. F(x) = ANTIDERIVATIVE {f(x)} = f(x)dx 136

28. a

b f(x)dx = CHANGE in F(x) 138

29. Direction of Integrationand CHANGE in F(x) 140

30. Area under f(x) and Plotting FC (x) 157

31. Constant of Integration 163

32. INDEFINITE and DEFINITE INTEGRAL 167

33. INTEGRABLE 170

34. Applications of Integration 178

1

11111 . . . . . I N T R O D U C T I O NI N T R O D U C T I O NI N T R O D U C T I O NI N T R O D U C T I O NI N T R O D U C T I O N

Microbiologists work with small things called cells that form the basic unit ofmatter they are deal ing with. Physicists and chemists work with moleculesrepresentative of a compound or atoms representative of an element. Veryoften scientists work with entities smaller than cells or atoms. However, no matterhow small it is, it is always something tangible.

While molecules, atoms and particles are DISCRETE objects, in Calculus we work withCONTINUOUS entities like the functions that describe the motion of an object - itsposition, speed, acceleration and so on. We call these CONTINUOUS operands.

Mathematicians like to study the behavior of a function at a pointpointpointpointpoint or instantinstantinstantinstantinstant andover a very small interval around a pointpointpointpointpoint or instantinstantinstantinstantinstant. The interval is so small that wehave the special word infinfinfinfinfinitesimalinitesimalinitesimalinitesimalinitesimal to describe it. Understanding the behaviorof a function over an infinfinfinfinfinitesimalinitesimalinitesimalinitesimalinitesimal interval will help us to:::::

1. Formulate an expression to describe its behavior at a pointpointpointpointpoint or instantinstantinstantinstantinstant.

2. Use this expression to extract more information about the behavior of thefunction, i.e. is the function increasing or decreasing, at a maximum orminimum, or changing direction.

Let us see an example of a CONTINUOUS operand and the kind of operations wewould like to perform.

2

height

h1h2

h3

h4

y(t1)

y(T/////2) = maximum

(0, 0)time

t4t3 tn = Tt1 t2t0 T/////2

y(t2) y(t3)

y(t4)

M OM OM OM OM O T I VT I VT I VT I VT I V AAAAA T I O NT I O NT I O NT I O NT I O N

Throw a ball up into the air. Suppose we know the height at any instant intime, i.e. we know the function y(t) that describes the height. From this wecan estimate the CHANGE in height with respect to time.

Examp le :E xamp le :E xamp le :E xamp le :E xamp le :CHANGE in height from time t1 to t2 is simply : h2 ---------- h1

which is : y(t2) ---------- y(t1)From the function y(t) can we find out the RATE OF CHANGE in height at anychosen INSTANT ?What is the ver tical speed or RATE OF CHANGE in height at time instant t1 ?What is the eeeeexprxprxprxprxpression ession ession ession ession of the INSTANTANEOUS RATE OF CHANGE in height ?

We know that ver tical speed =

Ver tical speed between t1 and t2 =y(t2) ---------- y(t1)

t2 ------ - - - - t1

CHANGE in height CHANGE in time

3

This is the AVERAGE VERTICAL SPEED over the time interval t1 to t2.

No matter how close we take t2 to t1, i.e. no matter how small the timeinterval is, we still get only an AVERAGE VERTICAL SPEED. We do not get thever tical speed at time instant t1.

WWWWWe ask twe ask twe ask twe ask twe ask two fundamenta l quest ionso fundamenta l quest ionso fundamenta l quest ionso fundamenta l quest ionso fundamenta l quest ions .....

Quest ion 1 :Quest ion 1 :Quest ion 1 :Quest ion 1 :Quest ion 1 :

Quest ion 2 :Quest ion 2 :Quest ion 2 :Quest ion 2 :Quest ion 2 :

With Calculus, for functions that are WELL BEHAVED: SINGLE VALUED,CONTINUOUS and DIFFERENTIABLE, we are able to find the ver tical speed atany chosen instantinstantinstantinstantinstant, i . e . f rom the func t ion tha t expresses the CHANGEwe can f i nd the INSTANTANEOUS RATE OF CHANGE.

How can we go from interinterinterinterinter vvvvvalalalalal [t1, t2], no matterhow small, to reach the instantinstantinstantinstantinstant t1 ?

How can we find the RATE OF CHANGE at instantinstantinstantinstantinstant t1 fromthe AVERAGE RATE OF CHANGE over the interinterinterinterinter vvvvvalalalalal [t1, t2] ?

4

From the Analysis Analysis Analysis Analysis Analysis point of view we would like to introduce some special notation to

express what we are doing. We may say:

∆ t = t2 −−−−− t

1 . So t

2 = t

1 + ∆ t

∆ y = y(t2) −−−−− y(t

1) = y(t

1 + ∆ t) −−−−− y(t

1)

where the Greek letter ∆ denotes a difference that is measurable.

1.1.1.1.1. AAAAAVERAVERAVERAVERAVERAGE steGE steGE steGE steGE step:p:p:p:p: So the AVERAGE VERTICAL SPEED may be denoted by :

2.2.2.2.2. TENDS TENDS TENDS TENDS TENDS TTTTTO steO steO steO steO step:p:p:p:p: We then fix t1 and let t

2 get closer and closer to t

1. The difference

∆ t between t1 and t

2 becomes smaller and smaller. It becomes infinitely small. This

kind of difference we denote using the Greek symbol δ. When t2 gets closer and

closer to t1 we get a better or more accurate AVERAGE VERTICAL SPEED. This we

denote by :

3.3.3.3.3. LIMIT ste LIMIT ste LIMIT ste LIMIT ste LIMIT step:p:p:p:p: Finally, when we let t2 coincide with t

1, we get the INSTANTANEOUS

VERTICAL SPEED at instant instant instant instant instant t1. This we denote by :

To do th is we need to develop the concept of CONTINUOUS from the concepts

that ar e descr ibed us ing the spec ia l v o c a b u l a r y : T E N D S T O ,

L I M I T, I N F I N I T E S I M A L and I N S TA N T. We shall then show that the TIME

AXIS is a CONTINUOUS set of INSTANTS and that each INSTANT corresponds to a

Real number. We shall then define CONTINUOUS functions.

∆ y =

y(t2) −−−−− y(t

1)

= y(t

1 + ∆ t) −−−−− y(t

1)

∆ t t2 −−−−− t

1 (t

1 + ∆ t) −−−−− t

1

δy=

y(t1 + δ t) −−−−− y(t

1)

δt

(t1 + δ t) −−−−− t

1

dy =

LIMIT δy =

LIMIT δy

dt t2→ t

1 δt δt→ 0 δt

y(t1 + δ t) −−−−− y(t

1)

(t

1 + δ t) −−−−− t

1

LIMIT

δt→ 0=

5

PPPPPararararar t 1 :t 1 :t 1 :t 1 :t 1 : CONTINUITY CONTINUITY CONTINUITY CONTINUITY CONTINUITY

6

OvOvOvOvOvererererer vievievievieviewwwww

To show that the TIME AXIS is a contincontincontincontincontinuous uous uous uous uous set of instants instants instants instants instants and that each instantinstantinstantinstantinstantcorresponds to a Real number we shall proceed in three steps.

SteSteSteSteStep 1:p 1:p 1:p 1:p 1: We know from class 8 Geometry that a straight line is a contincontincontincontincontinuous uous uous uous uous set ofpointspointspointspointspoints. We are also familiar with the different kinds of sets of numbers in Algebra:N (natural numbers), Z (integers), Q (rational numbers), Irrational numbers andR (real numbers). We present the AnalAnalAnalAnalAnalysis ysis ysis ysis ysis concept of the COMPLETENESS propertyof the Real numbers.

We thus show the connection between the set R of real numbers in Algebra and astraight line (a contincontincontincontincontinuous uous uous uous uous set of pointspointspointspointspoints) in Geometry. Each point x

iiiii on the number

line corresponds to a Real number and vice versa. We may call this number line thex-axis. And when we say x1 < x2 for x1, x2 ∈ R , the picture from the Geometry pointof view is :

ALGEBRA GEOMETRY

set R ≡ x-axis

x1 < x2

SteSteSteSteStep 2:p 2:p 2:p 2:p 2: In this step we learn the concepts: TENDS TO, LIMIT and INFINITESIMAL.

We shall use these concepts to take an AnalAnalAnalAnalAnalysis ysis ysis ysis ysis view of the x-axis as a contincontincontincontincontinuousuousuousuousuousset of pointspointspointspointspoints. We shall then define an instant instant instant instant instant on the TIME AXIS. From this we shallshow the equivalence between the x-axis as a continuous continuous continuous continuous continuous set of points points points points points and theTIME AXIS as a continuous continuous continuous continuous continuous set of instantsinstantsinstantsinstantsinstants.

GEOMETRY ANALYSIS

x-axis ≡ time axis

Each point point point point point xi on the x-axis corresponds to an instant instant instant instant instant ti on the TIME AXIS.

x1 x2

t1 t

2x

1 x

2

7

SteSteSteSteStep 3:p 3:p 3:p 3:p 3: Here we relate the TIME AXIS to the set of real numbers R. Now when wesay instant t1 we mean some definite real number t1 ∈ R. And when we say t1 < t2

for t1, t2 in R , the picture from the Analysis Analysis Analysis Analysis Analysis point of view is :

ALGEBRA ANALYSIS

set R ≡ time axis

t1 < t2

Since the TIME AXIS is a continuous continuous continuous continuous continuous set of instantsinstantsinstantsinstantsinstants, we may say t2 TENDS TO t

1 in

a contincontincontincontincontinuous uous uous uous uous manner. This is denoted by t2 → t

1 . And each instant instant instant instant instant on the way to

t1 corresponds to a definite real number in R.

We may even let t2 coincide with t

1 . This is expressed in AnalAnalAnalAnalAnalysisysisysisysisysis terminology as:

L I M I Tt2 → t

1

After we develop the concept of the CONTINUOUS INFINITESIMAL δt, the t2 → t

1

may be used in the calculation as :

t2 = t1 + δt

And the concept of t2 coinciding with t1 may be expressed using the word LIMIT withδt in the calculation as:

L I M I T L I M I Tt2 → t1 δt → 0

We then extend these concepts from a parparparparpar ticularticularticularticularticular instant t1 to a gggggenerenerenerenereneralalalalal instant tsimply by dropping the subscript.

t1 t2

{ expression involving t1 and t2 } ≡ { expression involving t1 and δt }

8

2. NUMBERS AND THE NUMBER LINE2. NUMBERS AND THE NUMBER LINE2. NUMBERS AND THE NUMBER LINE2. NUMBERS AND THE NUMBER LINE2. NUMBERS AND THE NUMBER LINE

NaNaNaNaNaturturturturtural nal nal nal nal numberumberumberumberumbers:s:s:s:s: N = { 0, 1, 2, 3, . . . } *

1. We note that between any two natural numbers there are FINITELYmany natural numbers.e.g. between 1 and 5 there are exactly three natural numbers: 2, 3 and 4.

2. We cannot exhaust counting the natural numbers because there are infinitelymany of them. But we can COUNT or ENUMERATE them in an order lymanner without missing any. This property we call COUNTABLE or ENUMERABLE.

A set of numbers with these two properties is called DISCRETE. The set of naturalnumbers N is DISCRETE.We can draw an infinite line and represent the natural numbers on thisline in an orderly manner as individual points spaced evenly apar t.

0 1 2 3

InteInteInteInteIntegggggererererers:s:s:s:s: Z = { . . . −−−−−3, −−−−−2, −−−−−1, 0, 1, 2, 3, . . . } 1. Between any two integers there are FINITELY many integers. 2. We may COUNT the intergers in an orderly manner without missing any as

follows: 0, +1, −−−−−1, +2, −−−−−2, +3, −−−−−3, . . . . The set Z is COUNTABLE.Hence the set Z is DISCRETE. To continue with our representation of numbers on aninfinite line we can represent the integers in an orderly manner as:

−−−−−3 −−−−−2 −−−−−1 0 1 2 3

with the line extending to infinity,∞, in both directions.

* Note: In some contexts zero is not considered a natural number. For our purposes it does not really matter.All we need to know is whether we can count the natural numbers or not. One number more or less does notmake a difference.

9

RRRRRaaaaationals:tionals:tionals:tionals:tionals: Q = { p

q −−−−− p, q∈Z, q /////= 0 }1. We know from class 4 Algebra between any two rational numbers there are

INFINITELY many rational numbers.2. Even though there are infinitely many rationals between any two rationals

we can still COUNT or ENUMERATE the rationals in an orderly mannerwithout missing any. The rationals are COUNTABLE. This is how we may countthe rational numbers.

0/////1 → ±1/////1 ±2/////1 → ±3/////1 ±4/////1 . . .. . .. . .. . .. . .

±1/////2 ±2/////2 ±3/////2 ±4/////2 ±5/////2 . . .. . .. . .. . .. . .

±1/////3 ±2/////3 ±3/////3 ±4/////3 ±5/////3 . . .. . .. . .. . .. . .

±1/////4 ±2/////4 ±3/////4 ±4/////4 ±5/////4 . . .. . .. . .. . .. . .

. . .. . .. . .. . .. . . and so on.

The set of rational numbers Q is more than DISCRETE.We say that the set of rational numbers Q is DENSE.Let us look again at our representation of numbers on an infinite line.

−−−−−32−−−−−

− − − − −12−−−−−

12−−−−−

32 −−−−−

−−−−−3 −−−−−2 −−−−−1 0 1 2 3

If we try to plot the rationals as co-linear points in an orderly manner we would getan “almost continalmost continalmost continalmost continalmost continuousuousuousuousuous”. The rational numbers by themselves do not exhaust,COMPLETELY cover, the number line.

→

→

→

→

→

→

→

→

10

From the Set Theory point of view we see that N ⊂ Z ⊂ Q .

Can we say that all the POINTS on the line between the rational number

1 and the rational number 2 are rational numbers ? There are numbers that are notrational numbers. Let us consider the square root of 2 denoted byS2 .

If S2 is a rational number then S2 = pq−−−−− p, q ∈ Z, q /////=0 .

Further, let p, q be such that they have no factor in common.

2 = p2

q−−−−−2

2q2 = p2

2q2 is even ⇒ p2 is also even ⇒ p is even.

Let p = 2n

Since 2q2 = p2 ⇒ 2q2 = (2n)2

⇒ q2 ⇒ 2n2 .

2n2 is even ⇒ q2 is even ⇒ q is even.

Thus p is even and q is even, implying they have as common factor 2. This contradictsour condition that p and q have no common factor. Therefore our assumption thatS2 is rational is wrong. We say S2 is irrational. The term IRRATIONAL literallymeans “ having no ratiohaving no ratiohaving no ratiohaving no ratiohaving no ratio ”. For practical mensuration the whole range ofthe rationals is more than sufficient. But from the theoretical point of view thisis not enough. For example, we need to define the length of the diagonal of asquare of side of unit length.

In f act , between any two rrrrr aaaaa t ionalt ionalt ionalt ionalt ional number s there are inf in i te ly manyiririririrrrrrraaaaationaltionaltionaltionaltional numbers. So we cannot say that the set of rationals Q is COMPLETE.We cannot say that each and every point on the number line is some rationalnumber. The set of rational numbers Q is DENSE but not COMPLETE.

11

I rI rI rI rI r rrrrr aaaaa t ional nt ional nt ional nt ional nt ional numberumberumberumberumber s:s :s :s :s :

Look at the number line again and consider a por tion of it, say from 0 to 1...... . .. .. .. .. . ..... ..... .....

−−−−−3 −−−−−2 −−−−−1 0 1 2 3Try to visualize it as a collection of infinitely many points, so many that we cannoteven begin to ENUMERATE them. Infinitely many points represent rrrrraaaaationaltionaltionaltionaltionalnumber s and in f in i te ly many points represent iririririr rrrrr aaaaational tional tional tional tional numbers.

1. Between any two irrational numbers there are INFINITELY many irrationalnumbers. The numbers:

S2, S5, 3SS3 + S2, S3S5 + S7

and many other expressions involving rational numbers under the radicalsign SSSSS are irrational. These irrational numbers are said to beexpressed in terms of radicals.

The decimals help us classify the rational and irrational numbers. Rational numbersare represented by TERMINATING DECIMALS,

e.g. 14−−−−− = 0.25

or INFINITE REPEATING DECIMALS, e.g. 13−−−−− = 0.333 . . .

I r r a t i ona l numbe r s a r e r epresented by NON-----TERMINATING NON-----REPEATING DECIMALS.

Examples :Examples :Examples :Examples :Examples :

S2 = 1.41421356237 . . .

e = 2.718281828459045235360 . . .

π = 3.14159265358979323846264338327950 . . .

12

2. Are the irrationals COUNTABLE ?

Without giving a formal proof let us try to get an intuitive feel for how many iririririrrrrrraaaaationaltionaltionaltionaltionalnumbers there could be as campared to the COUNTABLE set of rrrrraaaaationals tionals tionals tionals tionals Q .

How many rrrrraaaaational tional tional tional tional numbers can you create with TERMINATING (finitely manydigits) decimal expansion? Infinitely many.

How many rrrrraaaaational tional tional tional tional numbers can you create with NON-TERMINATING (infinitelymany digits) and REPEATING decimal expansion? Again, infinitely many.

Now, imagine how many iririririrrrrrraaaaational tional tional tional tional numbers you can create with NON-TERMINATINGNON-REPEATING decimal expansion?

Between the rationals rationals rationals rationals rationals 14−−−−− = 0.25 and 1

3−−−−− = 0.333 . . . we can create

innumerable iririririrrrrrraaaaational tional tional tional tional numbers with NON-TERMINATING NON-REPEATING patternsof decimal digits.

Again, between any two such iririririr rrrrraaaaational tional tional tional tional numbers created above we may createinnumerably more iririririrrrrrraaaaational tional tional tional tional numbers in a similar manner. There is a vir tualflood of iririririr rrrrraaaaational tional tional tional tional numbers. This should give us the feeling that the irrationalsare NOT COUNTABLE. Since we cannot even begin to COUNT them, there is nosense in talking about a subscript to enumerate them.

If the rrrrraaaaationals tionals tionals tionals tionals are DENSE then the iririririrrrrrraaaaationals tionals tionals tionals tionals are more than DENSE. Eachiririririrrrrrraaaaational tional tional tional tional number corresponds to a point on the number line. If we try to plot theiririririrrrrrraaaaationals tionals tionals tionals tionals as co-linear points in an orderly manner we would get an “almost“almost“almost“almost“almostcontincontincontincontincontinuous”uous”uous”uous”uous” line. However, the iririririrrrrrraaaaationals tionals tionals tionals tionals by themselves do not exhaust,COMPLETELY cover, the number line.

−−−−−S2 S2 e π

−−−−−3 −−−−−2 −−−−−1 0 1 2 3

13

3.3.3.3.3. REALS REALS REALS REALS REALS,,,,, COMPLETE COMPLETE COMPLETE COMPLETE COMPLETE,,,,, CONTINUOUS CONTINUOUS CONTINUOUS CONTINUOUS CONTINUOUS

From the decimal expansion point of view, any number must have either a FINITE(terminating) number of digits or an INFINITE (non-terminating) number of digits.And, if it has an INFINITE number of digits, then the pattern of digits must beREPEATING (recurring) or NON-REPEATING. Can you think of any other possibility?Also, we can compare any two numbers (rational or irrational) and plot them on thenumber line in an orderly manner.

RRRRReal Numbereal Numbereal Numbereal Numbereal Numbers:s:s:s:s: It should now be at least intuitively clear that any point onthe number line is either a rrrrraaaaational tional tional tional tional or an iririririr rrrrraaaaational tional tional tional tional number. And togetherthese two sets of numbers COMPLETELY cover or exhaust the whole numberline extending in both directions, positive and negative. The union of the setof rrrrraaaaationaltionaltionaltionaltional numbers and the set of iririririrrrrrraaaaational tional tional tional tional numbers is called the set of realnumbers R .

−−−−−S2 S2 e π

−−−−−3 −−−−−2 −−−−−1 0 1 2 3

COMPLETENESS prCOMPLETENESS prCOMPLETENESS prCOMPLETENESS prCOMPLETENESS properoperoperoperoper ty ofty ofty ofty ofty of the R the R the R the R the Real neal neal neal neal numberumberumberumberumbers:s:s:s:s:Corresponding to every point point point point point on the number line we have a unique rrrrreal neal neal neal neal numberumberumberumberumberand vice versa. Between any two real numbers on the number line each and everypoint corresponds to a real number. The set of real numbers R is COMPLETE. Thereal number line is smooth and CONTINUOUS. There are no gaps, breaks or bumps.This Real number line we call the x-axisx-axisx-axisx-axisx-axis.We now see the connection between the set R of real numbers in Algebra and astraight line (a contincontincontincontincontinuous uous uous uous uous set of pointspointspointspointspoints) in Geometry. Each point x

iiiii on the x-axis

corresponds to a Real number and vice versa. And when we say x1 < x2 for x1, x2 ∈ Rthe picture from the Geometry point of view is :

ALGEBRA GEOMETRY set R ≡ x-axisx1 < x2

x1 x2

−−−−−32−−−−−

− − − − −12−−−−−

12−−−−−

32 −−−−−

14

4.4.4.4.4. TENDS TENDS TENDS TENDS TENDS TTTTTOOOOO and and and and and LIMIT L IMIT L IMIT L IMIT L IMIT

In elementary geometry you learned the definition or meaning of a pointpointpointpointpoint.You also know how to name or label a pointpointpointpointpoint.

In Calculus a pointpointpointpointpoint on the number line or x-axis corresponds to a real number.Sometimes we know its exact value, e.g. 2. Sometimes we will know only theapproximate value. However, we may denote it by a special name or labelor symbol, e.g. S2, e, π . Regardless of knowing the exact value or not weknow which pointpointpointpointpoint we are talking about.

In Calculus we prefer to use the word instantinstantinstantinstantinstant rather than pointpointpointpointpoint. We speak ofthe instant instant instant instant instant 2, or the instantinstantinstantinstantinstant S2 , and so on. In Calculus we have anotherway to define an instantinstantinstantinstantinstant. Before we do this we need to know what anin f in i tes imalin f in i tes imalin f in i tes imalin f in i tes imalin f in i tes imal is .In order to give a formal definition of an infinfinfinfinfinitesimalinitesimalinitesimalinitesimalinitesimal we need to use twoconcepts: TENDS TO and LIMIT.TENDS TO :TENDS TO :TENDS TO :TENDS TO :TENDS TO :Let a = 0.5 and x = 0.4, 0.49, 0.499, 0.4999, . . . progressively.It is clear that x TENDS TO 0.5, we write this as: x → a.Will x = a ? No. We will always have 0 <CCCCC x−−−−− a CCCCCL I M I TL I M I TL I M I TL I M I TL I M I T :::::

If we let b = 0.51 we are correct in saying: x TENDS TO 0.51, denoted by x → b.Will x = b ? No. 0 < CCCCCxxxxx−−−−−bbbbbCCCCCWhat is the difference between xxxxx →→→→→ aaaaa and xxxxx →→→→→ bbbbb ?

In x → b : we cannot choose xxxxx as cas cas cas cas close as wlose as wlose as wlose as wlose as we like like like like likeeeee tototototo bbbbb. The differenceCCCCCx−−−−−bCCCCC cannot be made as small as was small as was small as was small as was small as we like like like like likeeeee. We cannot have CCCCCx−bCCCCC< δfor any small quantity δ as small as small as small as small as small as we likeas we likeas we likeas we likeas we like. For example:

if we choose δ = 0.001 we will have 0 < δ < 0.01 < CCCCCx−bCCCCC.....

15

In x → a : we can choose xxxxx as close as we likeas close as we likeas close as we likeas close as we likeas close as we like tototototo aaaaa. The difference CCCCCx−−−−−aCCCCC canbe made as small as we likeas small as we likeas small as we likeas small as we likeas small as we like. If we choose δ > 0, any small positivequantity as small as we likeas small as we likeas small as we likeas small as we likeas small as we like, we can choose x such that CCCCCx−aCCCCC < δ .

In x = 0.4, 0.49, 0.499, 0.4999, ... progressively with a = 0.5, since x < a, wecan be more precise and say x → a from the left.left.left.left.left. We write this as x → a----------.

We could have let x = 0.51, 0.501, 0.5001, 0.50001, ... progressively witha = 0.5 . Here too we have x → a but from the rightrightrightrightright. We write this as x → a+++++.

Sometimes we will be very specific and say:x TENDS TO a from the leftlef tlef tlef tlef t , written as : x → a----------

x TENDS TO a from the rightrightrightrightright, written as : x → a+++++

Which ever be the case, whether x → a+++++ (from the rightrightrightrightright) or x → a---------- (fromthe leftleftleftleftleft) we insist that x ≠ a, i.e. 0 < CCCCCx−aCCCCC. The case when x = a we willspeak of separately.

Since we know that LIMIT (x → a−−−−−) is aaaaa, we write : Limit { x } = a .

Since we know that LIMIT (x → a+++++) is aaaaa, we write : Limit { x } = a .

x → a----------

x → a+

x → a

Since we can choose xxxxx as close as we likeas close as we likeas close as we likeas close as we likeas close as we like tototototo aaaaa,and only to a a a a a but not to any other pointpointpointpointpoint, we say:

LIMIT as x → a is aaaaa, written as Limit x = a

16

55555. . . . . INFINITESIMALSINFINITESIMALSINFINITESIMALSINFINITESIMALSINFINITESIMALS

Let us now focus on the difference ∆ ∆ ∆ ∆ ∆ = CCCCCa−−−−− xCCCCC as x → a = 0.5, be it fromthe left left left left left or from the rightrightrightrightright.

∆ ∆ ∆ ∆ ∆ = CCCCCa−−−−− xCCCCC = 0.01, 0.001, 0.0001, 0.00001, . . .

After a while it becomes infinfinfinfinfinitelinitelinitelinitelinitely smally smally smally smally small and we denote it by δδδδδ. We may say :

δ δ δ δ δ = 1/////10nnnnn as n → ∞

We can see that { 1/////10nnnnn } = 0 .

Let us look at another example. Imagine A and B located one meter apar t on thex-axis . At each second let B jump 1/////2 distance towards A. We may say that B → A.

The difference ∆ ∆ ∆ ∆ ∆ = CCCCCA−−−−− BCCCCC = 1/////2 , 1/////4 ,

1/////8 , . . .

After a while it becomes infinfinfinfinfinitelinitelinitelinitelinitely smally smally smally smally small and we denote it by δδδδδ. We may say :

δ δ δ δ δ = 1/////2nnnnn as n → ∞

We can see that { 1/////2nnnnn } = 0 .

By definition:

Note that the LIMIT is zero, i.e. the quantityNote that the LIMIT is zero, i.e. the quantityNote that the LIMIT is zero, i.e. the quantityNote that the LIMIT is zero, i.e. the quantityNote that the LIMIT is zero, i.e. the quantity δ δ δ δ δ TENDS TO zero butTENDS TO zero butTENDS TO zero butTENDS TO zero butTENDS TO zero butdoes notdoes notdoes notdoes notdoes not become zbecome zbecome zbecome zbecome zerererererooooo..... δ δ δ δ δ does not vdoes not vdoes not vdoes not vdoes not vanish.anish.anish.anish.anish.

An infinitely smallinfinitely smallinfinitely smallinfinitely smallinfinitely small quantity whose LIMIT is zero iscal led an INFINITESIMAL.

L IMITn → ∞

LIMITn → ∞

A

a

B

a + 1 meter

← 1/////8 1/////4

1/////2 x-axis

17

We may now use the infinitesimal δ and say:x → a+++++ ≡ x = a + δx → a---------- ≡ x = a −−−−− δB → A ≡ B = a + δ

To express the concept x coincides with a, we may say :

from the right right right right right : x = { a + δ }

from the left left left left left : x = { a −−−−− δ }

Likewise, to express B coincides with A we may say : B = { a + δ }

As B jumps from 1 to 1/////2 to 1/////4 to 1/////8 . . . to A it skips a lot points points points points points in between.Likewise, as x → a it skips a lot of points points points points points in between. This is because δ is adiscrete infinitesimaldiscrete infinitesimaldiscrete infinitesimaldiscrete infinitesimaldiscrete infinitesimal.

We can think of the number line or x-axis as smooth and CONTINUOUS.Thereare no gaps or breaks. In Calculus we want to study the behavior of a functionat each and every pointpointpointpointpoint on the x-axis or instantinstantinstantinstantinstant on the TIME AXIS.

So we want x to TEND TO a pointpointpointpointpoint in a smooth and CONTINUOUS manner.For the infinitesimal that TENDS TO zero in a smooth and CONTINUOUSmanner we have the special notation δx.

Geometrically, we can think of δx as a small CONTINUOUS line segmentrepresenting an infinitely small CHANGE in x. The length of this small linesegment δx is always > 0 no matter where we are on the x-axis. δx getssmaller and smaller. The length of δx TENDS TO zero but does not becomezero. δx does not vanish. However, the LIMIT of δx is zero.

Limitδ → 0Limitδ → 0

Limitδ → 0

18

We may let :

δx = 1/////10xxxxx as x → ∞

or δx = 1/////2xxxxx as x → ∞

Now we may express x → a in a continuous continuous continuous continuous continuous manner by :x → a+++++ ≡ x = a + δxx → a---------- ≡ x = a −−−−− δxB → A ≡ B = A + δx

To express the concept x coincides with a a a a a , we may say :

from the right right right right right : x = { aaaaa + δx }

from the left left left left left : x = { aaaaa −−−−− δx }

Likewise, to express B coincides with A we may say : B = { A + δx }

Limitδx → 0

Limitδx → 0

Limitδx → 0

19

6.6.6.6.6. δδδδδxxxxx andandandandand INSTINSTINSTINSTINSTANTANTANTANTANTWe may now take an Analysis view of the parparparparpar ticular ticular ticular ticular ticular point x

11111 on the x-axis and define

it as :

And the general general general general general point x on the x-axis is defined by dropping the subscript.

Likewise, we may also define the parparparparparticular ticular ticular ticular ticular instant t11111 on the time axis as :

And the general general general general general instant t on the time axis is defined by dropping the subscript.

Thus we see the TIME AXIS is identical to the x-axis except for a change in name.

GEOMETRY ANALYSIS

x-axis ≡ time axis

Each point point point point point x on the x-axis corresponds to an instant instant instant instant instant t on the TIME AXIS.

The parparparparparticular pointticular pointticular pointticular pointticular point x11111 on the x-axis is where

(x11111−−−−−δx) = x

11111 = (x

11111 + δx) Limit

δx → 0Limit

δx → 0

A generalgeneralgeneralgeneralgeneral pointpointpointpointpoint x on the x-axis is where

(x−−−−−δx) = x = (x + δx) Limitδx → 0

Limitδx → 0

The parparparparparticular instantticular instantticular instantticular instantticular instant t11111 on the time axis is where

(t11111−−−−−δt) = t

11111 = (t

11111 + δt) Limit

δt → 0Limitδt → 0

A generalgeneralgeneralgeneralgeneral instantinstantinstantinstantinstant t on the time axis is where

(t−−−−−δt) = t = (t + δt) Limitδt → 0

Limitδt → 0

x t

20

When we relate this to the set of real numbers R in Algebra we have :

ALGEBRA GEOMETRY and ANALYSIS

x1 < x2 for x1 , x2 ∈ R ≡ x-axis

t1 < t2 for t1 , t2 ∈ R ≡ time axis

Now when we say t2 TENDS TO t1 on the TIME AXIS, denoted by t2 → t1 , we letinstant instant instant instant instant t2 go instant instant instant instant instant , instant instant instant instant instant , instant instant instant instant instant , . . . all the way to instant instant instant instant instant t1 in acontincontincontincontincontinuous uous uous uous uous manner. And each instant instant instant instant instant on the TIME AXIS corresponds to a definitereal number in R.

Likewise, when we say x2 TENDS TO x1 on the x-axis, denoted by x2 → x1 , we letpoint point point point point x2 go pointpointpointpointpoint, pointpointpointpointpoint, pointpointpointpointpoint, . . . all the way to point point point point point x1 in a continuouscontinuouscontinuouscontinuouscontinuousmanner. And each point point point point point on the x-axis corresponds to a definite real number in R.

From the Algebra point of view R is a COMPLETE set of real numbers. From theGeometry point of view R is a CONTINUOUS line of POINTS. And from the Analysispoint of view R is a CONTINUOUS set of INSTANTS.

We can now answer the fundamental Question 1 on page 3.

What happens at an instant instant instant instant instant is instantaneousinstantaneousinstantaneousinstantaneousinstantaneous.In English, instantaneous instantaneous instantaneous instantaneous instantaneous = occuring or completed in an instantinstantinstantinstantinstant.

x1 x

2

t1 t2

In taking the Limit as t2 → t1 we go from theinterinterinterinterinter vvvvval al al al al [ t1 , t2 ] and reach the instant instant instant instant instant t1.

In general, i f t is any instantinstantinstantinstantinstant, by takingthe L imit as δt → 0 we can go from theinterinterinterinterinter vvvvval al al al al [t, t+δt] and reach the instant instant instant instant instant t.

21

77777..... S INGLE S INGLE S INGLE S INGLE S INGLE VVVVVALALALALALUED FUNCT IONSUED FUNCT IONSUED FUNCT IONSUED FUNCT IONSUED FUNCT IONS

In the next few chapters we apply the concepts TENDS TO, LIMIT and CONTINUOUSINFINITESIMALS to look at the first two properties of WELL-BEHAVED functions,i.e. SINGLE VALUED and CONTINUOUS.

To know whether a function is CONTINUOUS or not, we need to know :

1. The LIMIT of a function.2. The VALUE of a function.

With these two concepts in place we define CONTINUOUS functions.

A function is said to be SINGLE VALUED* if at every pointpointpointpointpoint or instantinstantinstantinstantinstant on thereal line where the function is defined it has one and only one value.

Moreover, for an infinitesimal CHANGE δx at the instantinstantinstantinstantinstant a a a a a (where the functionis defined) there is exactly one corresponding CHANGE δy in the value ofthe function.

* Note: In Algebra the concept of a function as developed from a rrrrrelaelaelaelaelation tion tion tion tion (set of ordered pairs) andmamamamamapping pping pping pping pping is single valued by definition. In Analysis the concept is not so strict. For calculation purposes wehave to enforce this by restricting the range of the function.

f(x)f(x)f(x)f(x)f(x)

δδδδδyyyyy

a a+a a+a a+a a+a a+δδδδδxxxxx


xxxxx(0, 0)(0, 0)(0, 0)(0, 0)(0, 0)

δδδδδxxxxx

22

In the figure below at x = a there are two values of f(x). So, for an infinitesimalCHANGE δx at x = a we have two corresponding CHANGES δy.

We may make f(x) SINGLE VALUED by restricting the range as in either one of thediagrams below.


(0, 0)(0, 0)(0, 0)(0, 0)(0, 0)

δδδδδxxxxx

which δδδδδy do we choose ?

xxxxxaaaaa aaaaa + + + + + δδδδδxxxxx


(0, 0)(0, 0)(0, 0)(0, 0)(0, 0)

δδδδδxxxxx


δδδδδ yyyyy


(0, 0)(0, 0)(0, 0)(0, 0)(0, 0)

δδδδδxxxxx


δδδδδ yyyyy

23

The function y = √x for x ≥ 0 is not SINGLE VALUED. It has two values: +√ xand −−−−− √x . We must specify which root we are using.

Since we are dealing with real real real real real valued functions over the real linereal linereal linereal linereal line we avoid caseswhere the function may take on complex values such as √x for x < 0.

Also, the functions we deal with must take on wwwwwell-defell-defell-defell-defell-definedinedinedinedined real values. We avoidundefined entities such as +∞, −∞, division by zero and ∞/∞.

Functions of the form a n xn + an----------1 xn--1 + . . . + a2 x2 + a1 x + a0 arecalled polynomialspolynomialspolynomialspolynomialspolynomials. These functions are single valuedsingle valuedsingle valuedsingle valuedsingle valued.

y = +√ x

y = −−−−−√ x

x(0, 0)

y

a a a a a aaaaa+++++δδδδδxxxxx

δδδδδxxxxx δδδδδy > 0y > 0y > 0y > 0y > 0

δδδδδxxxxx δδδδδy < 0y < 0y < 0y < 0y < 0

24

88888. L IMIT o f a Funct ion. L IMIT o f a Funct ion. L IMIT o f a Funct ion. L IMIT o f a Funct ion. L IMIT o f a Funct ion

To understand the behaviour of a function f(x) very close to some instant a instant a instant a instant a instant a , thatis to say near instant a instant a instant a instant a instant a and to the LEFT and RIGHT of instant ainstant ainstant ainstant ainstant a, we mustanalyse :

{ f(x) } or { f(a − δ x) }

and { f(x) } or { f(a + δ x) }

If it turns out that there is some definitive real number b such that :

LIMIT from the LEFT = b = LIMIT from the RIGHT

{ f(x) } = b = { f(x) }

{ f(a − δ x) } = b = { f(a + δ x) }

then we say : { f(x) } = b .

Properly speaking, finding or calculating the { f(x) } is a 2 step process.This becomes clear when we use the infinitesimal δ x.

1.1.1.1.1. TENDS TENDS TENDS TENDS TENDS TTTTTO steO steO steO steO step :p :p :p :p : as x → a+++++ we may say x = a + δ x. We may substitute(a + δ x) for x in f(x). Then we do whatever expansion, regrouping of terms,cancellation and simplification possible. We may perform these operationsbecause δ x ≠ 0 . It only TENDS TO zero. TTTTTherherherherhere is no die is no die is no die is no die is no division bvision bvision bvision bvision by zy zy zy zy zererererero in this steo in this steo in this steo in this steo in this steppppp.Likewise, when x → a−−−−− we may substitute (a −−−−− δ x) for x in f(x).

2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step : here we let δ x = 0 so that x coincides with a.

We combine both steps in one expression as: {f(x)} or {f(x)}.

Then we combine both the above into one expression: {f(x)} or {f(x)}.

Note: More precisely, near near near near near ===== as c as c as c as c as close as wlose as wlose as wlose as wlose as we like like like like likeeeee.

Limit x → a+++++

Limit x → a----------

Limitδx → 0

Limitδx → 0

Limit x → a+++++

Limit x → a----------

Limitδx → 0

Limitδx → 0

Limit x → a

Limit x → a+++++

Limit x → a+++++

Limit x → a----------

Limit x → a

Limitδx → 0

25

L I M I Tx → a−−−−−

L I M I Tx → a−−−−−

L I M I Tδx → 0

L I M I Tδx → 0

L I M I Tx → a+

L I M I Tx → a+

L I M I Tδx → 0

L I M I Tδx → 0

L I M I Tx → a−−−−−

L I M I Tx → a+

L I M I Tx → a

VALUEx = a

L I M I Tx → a

VALUEx = a

VALUEx = a

L I M I Tx → a

VALUEx = a

Example 1:Example 1:Example 1:Example 1:Example 1: Consider the function f(x) = C, a constant.

What is the LIMIT of f(x) as x TENDS TO a ?

Limit from the LEFTLimit from the LEFTLimit from the LEFTLimit from the LEFTLimit from the LEFT

{ f(x) } = { C } = C.

{ f(a − δ x) } = { C } = C.

Limit from the RIGHTLimit from the RIGHTLimit from the RIGHTLimit from the RIGHTLimit from the RIGHT

{ f(x) } = { C } = C

{ f(a + δ x) } = { C } = C

Since, { f(x) } = C = { f(x) } , we say { f(x) } = C .

{ f(x) } = f(a) = C .

Here we have : { f(x) } = C = { f(x) } .

f(x) at instant x = a is f(a). This we call the VALUE of f(x) at x = a, written as :

{ f(x) } = f(a)

Sometimes we may have : { f(x) } = b = { f(x) }

But this is not always the case as we shall see from the examples that follow.

26

Example 2:Example 2:Example 2:Example 2:Example 2: What is the LIMIT of f(x) = x + a as x TENDS TO a ?


{ f(x) } = { x + a } = { a + a } = 2a

Let us find {f(x)} in 2 steps using δx.

1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step : near aaaaa and to the left left left left left of aaaaa : x = a −−−−− δx f(x) = { (a − δ x) + a } = 2a − δ x

2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : {2a − δ x } = 2a


{ f(x) } = { x + a } = { a + a } = 2a


1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step : near aaaaa and to the right right right right right of aaaaa : x = a + δx

f(x) = { (a + δx) + a } = 2a + δ x

2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : {2a + δ x} = 2a

Since, { f(x) } = 2a = { f(x) } , we say { f(x) } = 2a .

{ f(x) } = f(a) = 2a .

Here we have : { f(x) } = 2a = { f(x) } .

VALUEx = a

L I M I Tx → a

VALUEx = a

Limitδx → 0

Limit x → a----------

Limit x → a----------

Limit x → a----------

Limit x → a+++++

Limit x → a+++++

Limit x → a+++++

Limitδx → 0

Limit x → a+++++

Limit x → a----------

Limit x → a

Limitx → a----------

Limit x → a+++++

27

Example 3:Example 3:Example 3:Example 3:Example 3: What is the LIMIT of f(x) = x2 as x TENDS TO a ?


{ f(x) } = { x2 } = { a2 } = a2


1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step : near aaaaa and to the left left left left left of aaaaa : x = a −−−−− δx

f(x) = (a − δ x)2 = a2 − 2aδ x + δ x2

2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : {a2 − 2aδ x + δ x2 } = a2


{ f(x) } = { x2 } = { a2 } = a2



f(x) = (a + δ x)2 = a2 + 2aδ x + δ x2

2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : {a2 + 2aδ x + δ x2 } = a2

Since, { f(x) } = a2 = { f(x) } , we say { f(x) } = a2 .

{ f(x) } = f(a) = a2 .

Here we have : { f(x) } = a2 = { f(x) } .

We should begin to get the general feeling that for polpolpolpolpolynomials ynomials ynomials ynomials ynomials of the form :

f(x) = an xn + an----------1 xn--1 + . . . + a 2 x2 + a1 x + a0

we have: { f(x) } = { f(x) } .

VALUEx = a

L I M I Tx → a

VALUEx = a

L I M I Tx → a

VALUEx = a

Limitδx → 0

Limit x → a----------

Limit x → a----------

Limit x → a----------

Limit x → a----------

Limit x → a+++++

Limit x → a+++++

Limit x → a+++++

Limit x → a+++++

Limitδx → 0

Limit x → a----------

Limit x → a+++++

Limit x → a

28

Example 4:Example 4:Example 4:Example 4:Example 4: What is the LIMIT of f(x) = as x TENDS TO a ?


{ f(x) } = { } = { 1 } = 1

Note how we simplify fsimplify fsimplify fsimplify fsimplify fiririririrst and then takst and then takst and then takst and then takst and then take the limite the limite the limite the limite the limit. We can simplify first becausewhen x → a we have (x −−−−− a) ≠ 0. Let us do this in 2 steps using δx.


f(x) = { } = { } = 1

Note how we simplify fsimplify fsimplify fsimplify fsimplify fiririririrst and then takst and then takst and then takst and then takst and then take the limite the limite the limite the limite the limit. We can simplify first becauseδx only TENDS TO zero. δx ≠ 0. So therSo therSo therSo therSo there is no die is no die is no die is no die is no division bvision bvision bvision bvision by zy zy zy zy zererererero in this steo in this steo in this steo in this steo in this steppppp.....

2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step : {1} = 1


{ f(x) } = { } = { 1 } = 1


f(x) = { } = { } = 1

Note how we simplify fsimplify fsimplify fsimplify fsimplify fiririririrst and then takst and then takst and then takst and then takst and then take the limite the limite the limite the limite the limit. We can simplify first becauseδx only TENDS TO zero. δx ≠ 0. So therSo therSo therSo therSo there is no die is no die is no die is no die is no division bvision bvision bvision bvision by zy zy zy zy zererererero in this steo in this steo in this steo in this steo in this steppppp.....

2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step : {1} = 1

Since, { f(x) } = 1 = { f(x) } , we say { f(x) } = 1.

{ f(x) } = { } = 0///// 0 , which is something undefined.

L I M I Tx → a−−−−−

L I M I Tx → a−−−−−

L I M I Tx → a−−−−−

L I M I Tx → a+

L I M I Tx → a

VALUEx = a

L I M I Tx → a−−−−−

L I M I Tx → a+

L I M I Tx → a+

L I M I Tx → a+

x ---------- ax -- a

x ---------- ax -- a

(a ---------- δ x) ---------- a(a ---------- δ x) ---------- a

---------- δ x---------- δ x

x ---------- ax -- a

(a +δ x) ---------- a(a +δ x) ---------- a

+δ x+δ x

x ---------- ax -- a VALUE

x = a

L I M I Tδx → 0

L I M I Tδx → 0

29

Example 5:Example 5:Example 5:Example 5:Example 5: What is the LIMIT of f(x) = as x TENDS TO a ?


{f(x)} = { } = { } = {x+a} = 2a

Note how we simplify fsimplify fsimplify fsimplify fsimplify fiririririrst and then takst and then takst and then takst and then takst and then take the limite the limite the limite the limite the limit. We can simplify first becausewhen x → a we have (x −−−−− a) ≠ 0. Let us do this in 2 steps using δx.


f(x) = { } = { } = 2a ---------- δ x

Note how we simplify fsimplify fsimplify fsimplify fsimplify fiririririrst and then takst and then takst and then takst and then takst and then take the limite the limite the limite the limite the limit. We can simplify first becauseδx only TENDS TO zero. δx ≠ 0. TTTTTherherherherhere is no die is no die is no die is no die is no division bvision bvision bvision bvision by zy zy zy zy zererererero in this steo in this steo in this steo in this steo in this steppppp.....

2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step : { 2a ---------- δ x } = 2aLimit from the RIGHTLimit from the RIGHTLimit from the RIGHTLimit from the RIGHTLimit from the RIGHT

{f(x)} = { } = { } = {x+a} = 2a


f(x) = { } = { } = 2a+ δ x

2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step : { 2a+ δ x } = 2a

Since, { f(x) } = 2a = { f(x) } , we say { f(x) } = 2a.


The reader must have guessed by now that { } = n a n−−−−−1 .

We shall see this very special limit in its proper context when we study the derideriderideriderivvvvvaaaaatititititivvvvveeeeeof f(x) = xn .

x2 ---------- a2

x -- a

x2 ---------- a2

x -- a

Limit x → a----------

x2 ---------- a2

x -- a Limit x → a----------

Limit x → a----------

(x----------a)(x+a)x -- a

Limit x → a----------

(a ---------- δ x)2 ---------- a2

(a ---------- δ x) ---------- a----------2a.δ x+δ x2

---------- δ x

Limit x → a+++++

x2 ---------- a2

x -- a Limit x → a+++++

Limit x → a+++++

(x----------a)(x+a)x -- a

Limit x → a+++++

(a +δ x)2 ---------- a2

(a +δ x) ---------- a2a.δ x+δ x2

δ x

Limit x → a+++++

Limit x → a----------

Limit x → a

Valuex = a

Valuex = a

Limit x → a

xn ---------- an

x -- a

Limitδx → 0

Limitδx → 0

30

f(x) = +√ x

x(0, 0)

Example 6Example 6Example 6Example 6Example 6 : What is the LIMIT of f(x) = + √x as x TENDS TO 0 ?


1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step : near 00000 and to the left left left left left of 00000 : x = 0 −−−−− δx

f(x) = √ (0 −−−−− δx) = √ −−−−− δx

A negative real number under the radical sign is a complex number. We are dealingwith rrrrreal eal eal eal eal numbers and rrrrreal eal eal eal eal valued functions. Here √ −−−−− δx is not rrrrreal eal eal eal eal valued andso it is not defined. f(x) = + √x is not definednot definednot definednot definednot defined for x < 0.

2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step :2. LIMIT step : The LIMIT from the left left left left left does not exist.


1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step : near 00000 and to the right right right right right of 00000 : x = 0 + δx

f(x) = √ (0 +δx) = √δx

2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : {√δx} = 0

Since f(x) ≠ f(x) we say : f(x) does NOT exist.

f(x) = f(0) = √0 = 0

Limitx → 0−−−−−

Limitx → 0+

Valuex = 0

Limit x → 0

Limitδx → 0

31

Example 7:Example 7:Example 7:Example 7:Example 7: What is the LIMIT of f(x) = 1/////x as x TENDS TO 0 ?


1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step : near 0 0 0 0 0 and to the left left left left left of 00000 : x = 0 −−−−− δx

f(x) = f(0 − δ x) = { 1/////(0 − δ x) } = −−−−− 1/////δ x . Note that δx ≠ 0 .

2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : { −−−−− 1/////δ x }= −−−−− 1/////0 = −∞ , something undefined.


1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step :1. TENDS TO step : near 00000 and to the right right right right right of 00000 : x = 0 + δx

f(x) = f(0 + δ x) = { 1/////(0 + δ x) } =+1/////δ x . Note that δx ≠ 0 .

2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : {+1/////δ x }= =+1/////0 = +∞ , something undefined.

The LIMIT from the left left left left left does not exist. And, the LIMIT from the right right right right right does not exist.

{ f(x) } = f(0) = 1/////0 = ∞ , something undefined. VALUEx = 0

Limitδx → 0

Limitδx → 0

−∞ ½ 1 2 . . . +∞

+∞

2

1

½

0

−1

−2

−∞

f(x) = 1/////x

32

Example 8:Example 8:Example 8:Example 8:Example 8: What is the LIMIT of f(x) = as x TENDS TO 0 ?

We shall use the trigonometric identity sin (A ± B) = sin A cos B ± cos A sin B.Also, when x is very small, we may say* : sin (x) = x .


1. TENDS TO step : 1. TENDS TO step : 1. TENDS TO step : 1. TENDS TO step : 1. TENDS TO step : near 00000 and to the left left left left left of 00000 : x = 0 −−−−− δx

f(x) = = = = 1

Note how we simplify fsimplify fsimplify fsimplify fsimplify fiririririrst and then takst and then takst and then takst and then takst and then take the limite the limite the limite the limite the limit. We can simplify first becauseδx only TENDS TO zero. δx ≠ 0. TTTTTherherherherhere is no die is no die is no die is no die is no division bvision bvision bvision bvision by zy zy zy zy zererererero in this steo in this steo in this steo in this steo in this steppppp.

2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : {1} = 1

sin(x)x

* Note: we may infer this from the basic definition of sin (θ) = opposite side / hypoteneuse.When θ is infinitely small, say δ θ , then sin ( δ θ ) = r δ θ/r = δ θ . Another way is to look at the expansionof sin(x). When x becomes infinitely small, say δ x , then only the first term in the expansion matters.

{ sin(0 − δ x)

}(0 − δ x)

{ sin(− δ x)

}− δ x

{ − δ x

}− δ x

Limitδx → 0

−−−−−4π −−−−−3π −−−−−2π − − − − −1π 0 +1π +2π +3π +4π . . .. . .. . .. . .. . . x

f(x) = sin (x)

x

33


1. TENDS TO step : 1. TENDS TO step : 1. TENDS TO step : 1. TENDS TO step : 1. TENDS TO step : near 00000 and to the right right right right right of 00000 : x = 0 + δx

f(x) = = = = 1

Note how we simplify fsimplify fsimplify fsimplify fsimplify fiririririrst and then takst and then takst and then takst and then takst and then take the limite the limite the limite the limite the limit. We can simplify first becauseδx only TENDS TO zero. δx ≠ 0. TTTTTherherherherhere is no die is no die is no die is no die is no division bvision bvision bvision bvision by zy zy zy zy zererererero in this steo in this steo in this steo in this steo in this steppppp.....

2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : 2. LIMIT step : {1} = 1

Since, { f(x) } = 1 = { f(x) } , we say { } = 1 .


Example 9 :Example 9 :Example 9 :Example 9 :Example 9 : We now present an example of a function f(x) where f(x)

exists everywhere (i.e. a can be any real number), but the f(x) does NOT

exist anywhere.

f(x) =

Recall what we said about the rationals and irrationals. Between any two rationalsthere are infinitely many rationals. And between any two rationals there are alsoinfinitely many irrationals. So we can see that as x TENDS TO a, the function f(x) willkeep fluctuating. The function f(x) will be either 0 or 1. It will not take on a singlespecific value. Hence we can say that the f(x) does NOT exist. Since a canbe any point on the real number line, the Limit does NOT exist anywhere.

On first reading it is not necessary to study the Analysis of Limits. The reader mayskip the next chapter without loss of continuity.

{ sin(0 + δ x)

}(0 + δ x)

{ sin(+ δ x)

}+ δ x

{ δ x

}δ x

Limit x → 0+++++

Limit x → 0----------

Limit x → 0

Valuex = 0

Valuex = 0

sin(x)x

sin(x)x

Valuex = a Limit

x → a

{+1 if x is rationalrationalrationalrationalrational 0 if x is iririririrrrrrraaaaationaltionaltionaltionaltional

Limitx → a

Limitδx → 0

34

9. Analysis of Limits9. Analysis of Limits9. Analysis of Limits9. Analysis of Limits9. Analysis of Limits*

In general, we can talk about the LIMIT of a function f(x) as x → a .

Does f(x) TEND TO any par ticular value b as x TENDS TO a ?

Can we choose f(x) as close as we like toas close as we like toas close as we like toas close as we like toas close as we like to b by choosing x sufficientlysufficientlysufficientlysufficientlysufficientlyclosecloseclosecloseclose to a ?

Does f(x) = b = f(x) ?

Does f(a----------δx) = b = f(a+δx) ?

If it does, then we call this b the LIMIT of f(x) as x TENDS TO a and we writethis as:

f(x) = b or f(a+δx) = b

We are not concerned with if 0 < CCCCCf(x)----------bCCCCC or 0 = CCCCC f(x) ---------- b CCCCCWhat we are concerned with is when x → a , for any discrdiscrdiscrdiscrdiscrete infete infete infete infete infinitesimalinitesimalinitesimalinitesimalinitesimalε as small as we like :

can we have CCCCCf(x) ---------- bCCCCC< ε or equivalently CCCCC f(a + δx) ---------- b CCCCC<ε ?

Again, we stress that x /////= a. f(x) at the instant instant instant instant instant x = a as we said earlier, is

f(x) = f(a), if it exists.

We may combine x → a−−−−− and x → a+ into one expression : CCCCCx ---------- aCCCCC< δ .

Formally, we may define the LIMIT of a function f(x) as x TENDS TO a :

*Note : A more detailed and rigorous presentation is in A LITTLE MORE CALCULUS by the author.

Limitx → a

Limitx → a−−−−−

Limitx → a+

Limitδx → 0

Limitδx → 0

Limitδx → 0

Valuex = a

f(x) = b if for any discrete infinitesimaldiscrete infinitesimaldiscrete infinitesimaldiscrete infinitesimaldiscrete infinitesimal εas small as we like, we can find δ such that :::::CCCCCf(x) ---------- bCCCCC< ε when CCCCCx ---------- aCCCCC< δ .

Limitx → a

35

Based on the formal definition of the “limit of“limit of“limit of“limit of“limit of a function” a function” a function” a function” a function” , we may analyse thebehaviour of a function near some instant a, and see if it has a Limit or not. Let uslook at the example 2 of the previous chapter.

ExamplExamplExamplExamplExampleeeee 2 2 2 2 2::::: Consider the function f(x) = x + a

What is the Limit of f(x) as x TENDS TO a ? As x → a we have f(x) → 2aSince 0 < CCCCCx−−−−−aCCCCC we have 0 < CCCCCf(x)−−−−−2aCCCCC, i.e. f(x) /////===== 2a.

We can choose f(x) as cas cas cas cas close as wlose as wlose as wlose as wlose as we like like like like likeeeee tototototo 2a by choosing x sufsufsufsufsuf ffffficientlicientlicientlicientlicientlyyyyyccccc loseloseloseloselose to a . S S S S Sufufufufuf fffff ic ient lic ient lic ient lic ient lic ient lyyyyy c c c c c loseloseloseloselose means as cas cas cas cas close as necessarlose as necessarlose as necessarlose as necessarlose as necessaryyyyy.

If we want 0 < CCCCCf(x)−−−−−2aCCCCC < ε some discrete infintesimal as small as we like,we can choose x sssssufficientlyufficientlyufficientlyufficientlyufficiently close close close close close to a. For example, we may choose :

0 < CCCCCx−−−−−aCCCCC < δ , where δ = ε2−−−−−

Work it out:

0 <CCCCCx−−−−−aCCCCC < δ ⇒ 0 < CCCCCx−−−−−aCCCCC < ε2−−−−− ⇒ a−−−−−

ε2−−−−−< x < a +

ε2−−−−−

f(x) = x + a = ( a +--ε2−−−−−) + a = 2a +--

ε2−−−−−

Now: 0 < CCCCC f(x) --2aCCCCC = CCCCC(2a +--ε2−−−−−) --2aCCCCC= ε

2−−−−− which is < ε

So when CCCCCx -- aCCCCC < ε2−−−−− we will have CCCCC f(x) -- 2aCCCCC < ε

We can say : f(x) = 2a

Instead of saying : f(x) which is (x+a)

we can use the infinitesimal δx and say:

(x+a) = ((a--δ x ) + a ) = 2a (from the lefleflefleflefttttt)

and (x+a) = ((a+δx)+a) = 2a (from the rightrightrightrightright)

Limitx → a

Limitx → a

Limitx → a

Limitδx → 0

Limitδx → 0


Limitx → a+

36

Let us now look at the example 9 of the previous chapter.Example 9 :Example 9 :Example 9 :Example 9 :Example 9 :

f(x) =

Suppose the f(x) = b . Let us try to choose an εεεεε such that CCCCC f(x) --bCCCCC > εεεεεfor some x very close to a, that is to say for 0 < CCCCCx−−−−−aCCCCC < δ .

Recall what we said about the rationals and irrationals. Between any two rationalsthere are infinitely many rationalsrationalsrationalsrationalsrationals. And between any two rationals there are alsoinfinitely many iririririrrrrrraaaaationalstionalstionalstionalstionals.

No matter how close x is to a there will always be infinitely many rationalrationalrationalrationalrationalnumbers and infinitely many iririririrrrrrraaaaational tional tional tional tional numbers. So we can see that as x TENDSTO a, the function f(x) will keep fluctuating : f(x) will be 0 or 1.

There are two possibilities : b = 0 or b ≠ 0.

Case b = 0 : choose Case b = 0 : choose Case b = 0 : choose Case b = 0 : choose Case b = 0 : choose εεεεε = = = = = 11111/////22222No matter how close x is to a , that is to say no matter how smallthe δ we choose, there will always be some rrrrraaaaational tional tional tional tional numberssuch that 0 < CCCCCx−−−−−aCCCCC < δ .

If x is rational then CCCCC f(x) --bCCCCC = CCCCC1 --0CCCCC = 1 > εεεεε = 1/////2 .

So b = 0 cannot be the f(x) .

Case b Case b Case b Case b Case b ≠≠≠≠≠ 0 : choose 0 : choose 0 : choose 0 : choose 0 : choose εεεεε =====

No matter how close x is to a , that is to say no matter how smallthe δ we choose, there will always be some rrrrraaaaational tional tional tional tional numberssuch that 0 < CCCCCx−−−−−aCCCCC < δ .

If x is rational then CCCCC f(x) --bCCCCC = CCCCC1 -- bCCCCC > εεεεε = .

So b ≠ 0 cannot be the f(x) .

Since a can be any point on the real number line, f(x) does not exist anywhere.


Limitx → a

Limitx → aCCCCC1 -- b CCCCC

2

CCCCC1 -- b CCCCC2Limit

x → a

Limitx → a

37

10.10.10.10.10. VVVVVALALALALALUE vUE vUE vUE vUE vererererer sus LIMITsus LIMITsus LIMITsus LIMITsus LIMIT

So far when we took Limits of functions as x → a or δx → 0 we ASSUMEDthat these Limits exist. But this is not so simple.

1. Does the Limit exist ?

2. Is the Limit from the right = Limit from the left ?

i.e. is f(x) = f(x)

x > a x < aright left

3. Is the f(x) = f(x)

In finding the Limit we first simplify and then put x = a or x = 0 or whateverand evaluate. We can simplify because:

δx only TENDS TO 0, δx ≠ 0,

x only TENDS TO a, x ≠ a,

x only TENDS TO 0, x ≠ 0.

Whereas in finding the Value we must directly put x = a or x = 0 or whateverand evaluate.

Limit x → a+++++

Limit x → a----------

Value x = a

Limit x → a

38

(ii) VVVVValuealuealuealuealue and LimitLimitLimitLimitLimit exist but are not equal.

Let f(x) =

f(x) = 1 = f(x) , f(x) = 2

Try drawing the graph of this function.

(iii) VVVVValuealuealuealuealue exists but LimitLimitLimitLimitLimit does not.

Let f(x) =

f(x) = 0 ≠ f(x) = 2 , f(x) = 0

Try drawing the graph of this function.

(iv) VVVVValuealuealuealuealue does not exist but LimitLimitLimitLimitLimit does.

Let f(x) = , f(x) = 2a, f(x) = ?

(v) VVVVValuealuealuealuealue and LimitLimitLimitLimitLimit do not exist.

Let f(x) = 1----------x ,

Value x = a

x2 ---------- a2

x -- a Limit x → a

Limit x → a-----

Limit x → a+

Value x = a

Value x = 0

Limit x → 0

sin x xLet f(x) = , f(x) = 1, f(x) = ?

Limit x → 1----- Limit

x → 1+ Value x = 1

Value x = 0

Limit x → 0

f(x) = ? , f(x) = ?

Let us compare f(x) and f(x)

We have the following possibilities:

(i) VVVVValuealuealuealuealue and LimitLimitLimitLimitLimit both exist and are equal : f(x) = x

Value x = a

Limit x → a

{ 2 for x = a 1 for x ≠ a

{ 1 ---------- x for x ≤ 1 3 ---------- x for 1 < x

39

11111 11111. . . . . CONT INU I TY o fCONT INU I TY o fCONT INU I TY o fCONT INU I TY o fCONT INU I TY o f aaaaa F un c t i onFunc t i onFunc t i onFunc t i onFunc t i onOur common perception of continuous continuous continuous continuous continuous is so taken for granted that we seldompause to give it a precise mathematical definition. In a string of beads, the string iscontincontincontincontincontinuous uous uous uous uous and the beads are discrdiscrdiscrdiscrdiscreteeteeteeteete. Yet, from ancient times philosopher-mathematicians were aware of the concept “continuous”“continuous”“continuous”“continuous”“continuous” and tried to define it.

In his PhysikPhysikPhysikPhysikPhysik, Aristotle (384-322 BC), Greek philosopher and student of Plato andtutor to Alexander the great, explained “continuous”“continuous”“continuous”“continuous”“continuous” as: ‘ I say that something iscontinuous whenever the two extremities of their contiguous parts coincide, and asthe name itself implies, they are kept together. ’

Gottfried Wilhelm von LEIBNIZ (1646-1716), co-inventor of Calculus and librarianand historian under Duke Johann Friedrich of Hanover, defined “contin“contin“contin“contin“continuous”uous”uous”uous”uous” as:‘The whole is said to be continuous, when any two component parts thereof (or moreprecisely any two parts which together make up the whole) have something in common,... at the very least a common boundary.’

It is interesting that Sir Isaac Newton did not have much to say on “continuous”“continuous”“continuous”“continuous”“continuous” .More recently R. DEDEKIND (1872) defined “continuous”“continuous”“continuous”“continuous”“continuous” as: ‘If the points of aline are divided into two classes, in such a way that each point of the first class lies tothe left of every point of the second class, then there exists one and only one pointof division which produces this particular sub-division into two classes, this cutting ofthe line into two parts.’

We saw the connection between the set of Real numbers being complete complete complete complete complete and theReal line being continuouscontinuouscontinuouscontinuouscontinuous. The absence of a single point causes a cut or break ordiscontinuity discontinuity discontinuity discontinuity discontinuity in the Real line.

In Calculus we deal with contincontincontincontincontinuous uous uous uous uous operands or functions. We now carry forwardthe concept of the Real line being contincontincontincontincontinuous uous uous uous uous to functions over the Real line. Weneed to know the behaviour or vvvvvalue alue alue alue alue of a function at each and every point point point point point orinstantinstantinstantinstantinstant. Also, we need to know the behaviour of a function near any chosen instanta a a a a : a little to the left left left left left of aaaaa or (aaaaa −−−−− δx) and a little to the right right right right right of aaaaa or (aaaaa + δx).

40

GeometricallGeometricallGeometricallGeometricallGeometricallyyyyy when we say that a SINGLE VALUED function is CONTINUOUSaround the point or instant ainstant ainstant ainstant ainstant a, we mean that the curve of the function is withoutbreaks or gaps.

f ( x ) i s CONT INUOUS af ( x ) i s CONT INUOUS af ( x ) i s CONT INUOUS af ( x ) i s CONT INUOUS af ( x ) i s CONT INUOUS a ttttt t h e po i n t o r i n s t an t a t he po i n t o r i n s t an t a t he po i n t o r i n s t an t a t he po i n t o r i n s t an t a t he po i n t o r i n s t an t a

f(x) is f (x) is f (x) is f (x) is f (x) is CONTINUOUSCONTINUOUSCONTINUOUSCONTINUOUSCONTINUOUS at a t a t a t a t instant instant instant instant instant a bua bua bua bua buttttt on ly f rom the on ly f rom the on ly f rom the on ly f rom the on ly f rom the le f tle f tle f tle f tle f t

f (x) is f (x) is f (x) is f (x) is f (x) is CONTINUOUSCONTINUOUSCONTINUOUSCONTINUOUSCONTINUOUS at a t a t a t a t instant instant instant instant instant a bua bua bua bua buttttt on ly f rom the on ly f rom the on ly f rom the on ly f rom the on ly f rom the rightrightrightrightright

f(x)

aaaaa----------δδδδδxxxxx aaaaa+++++δδδδδxxxxxaaaaaxxxxx

( ( ( ( (0,00,00,00,00,0)))))


aaaaa----------δδδδδxxxxx aaaaa+++++δδδδδxxxxxaaaaa xxxxx(((((0 ,00 ,00 ,00 ,00 ,0 )))))


aaaaa ----------δδδδδ xxxxx aaaaa+++++δδδδδ xxxxxaaaaa xxxxx (((((0,0,0,0,0,0)0)0)0)0)


41

AnalyticallyAnalyticallyAnalyticallyAnalyticallyAnalytically: Assume f(x) = f(a) exists and is something definite. Then

1. if f(x) exists and is = f(x)

we say that f(x) is CONTINUOUS at the point x = a from the leftleftleftleftleft.

2. if f(x) exists and is = f(x)

we say that f(x) is CONTINUOUS at the point x = a from the rightrightrightrightright.

Using the infinitesimal δx we can write this as:

If f(x) is not CONTINUOUS at a we say f(x) is DISCONTINUOUS at a.We may test the contincontincontincontincontinuity ofuity ofuity ofuity ofuity of a function a function a function a function a function using the decision tree below.

f(x) CONTINUOUS at a ?f(x) CONTINUOUS at a ?f(x) CONTINUOUS at a ?f(x) CONTINUOUS at a ?f(x) CONTINUOUS at a ?

LIMIT LIMIT LIMIT LIMIT LIMIT =====????? VVVVVALALALALALUE UE UE UE UE f(x) = f(a)f(x) = f(a)f(x) = f(a)f(x) = f(a)f(x) = f(a)

xxxxx →→→→→ aaaaa x = ax = ax = ax = ax = a

directly evaluate with x = adirectly evaluate with x = adirectly evaluate with x = adirectly evaluate with x = adirectly evaluate with x = a

LIMITLIMITLIMITLIMITLIMIT =====????? LIMIT LIMIT LIMIT LIMIT LIMIT

xxxxx →→→→→ aaaaa ---------- xxxxx →→→→→ aaaaa+++++

simplify fsimplify fsimplify fsimplify fsimplify f iririririrst then takst then takst then takst then takst then take the limite the limite the limite the limite the limit

Limitx → a+

Limitx → a----------

Value x = a

Value x = a

Value x = a

Limitx → a+

Limitx → a----------

f(x) is said to be CONTINUOUS at the pointpointpointpointpoint or instantinstantinstantinstantinstant x = a

if f(x) = f(a) = f(x)

f(x) is said to be CONTINUOUS at the pointpointpointpointpoint or instantinstantinstantinstantinstant x = a

if f(a -- δx) = f(a) = f(a + δx)Limitδx → 0

Limitδx → 0

42

Are the polpolpolpolpolynomials ynomials ynomials ynomials ynomials CONTINUOUS everywhere ?

We can easily see that f(x) = xn for n = 1, 2, 3, . . .

f(x) = f(x) = an

Since a can be any point on the real number line, these f(x) are continuouscontinuouscontinuouscontinuouscontinuouseverywhere.

Likewise, f(x) = an xn for n = 0, 1, 2, 3, . . .

where the coefficients an are real numbers, are also contincontincontincontincontinuous uous uous uous uous everywhere.

Combining both we get the general polynomial f(x) :

f(x) = an xn + an----------1 xn--1 + . . . + a2 x2 + a1 x + a0

These are singsingsingsingsingle vle vle vle vle valuedaluedaluedaluedalued and contincontincontincontincontinuous uous uous uous uous everywhere.

y( t ) = u . s i n θ . t ---------- 1--2 g t 2 is a polynomial of the for m a0 + a1t + a2t2

with coefficients a0 = 0, a1 = u . s i n θ and a2 = ---------- 1--2 g . So the height

function y(t) is single valuedsingle valuedsingle valuedsingle valuedsingle valued and continuous continuous continuous continuous continuous .

It is important to know that polynomials are to functions what rational numbers areto real numbers. We can approximate any real number to any required degree ofaccuracy by a suitable rational number. Likewise we can approximate almost any realvalue function by a suitable polynomial. For this we have Lagrange’s method ofinterpolation.

Limitx → a

Valuex = a

43


Limitδx → 0

Limitx → a+

Limitδx → 0

Limitδx → 0

Limitδx → 0


Limitx → a+

Valuex = a


Limitx → a+

Valuex = a

Example 1Example 1Example 1Example 1Example 1: is f(x) continuous continuous continuous continuous continuous at x = a ?

Approaching aaaaa from the left left left left left : f(x) = x

Near aaaaa and to the left left left left left of aaaaa : x = a −−−−− δx

f(x) = { a −−−−− δx} = a

Approaching aaaaa from the right right right right right : f(x) = 2a −−−−− x

Near aaaaa and to the right right right right right of aaaaa : x = a + δx

f(x) = { 2a −−−−−(a +δx)} = { 2a −−−−−a −−−−−δx }

= { a −−−−− δx } = a

Hence, f(x) = a = f(x)

f(x) = f(a) = a

Since, f(x) = f(x) = f(x) = a

we say : f(x) is continuous continuous continuous continuous continuous at x = a.

( ( ( ( (0,00,00,00,00,0)))))xxxxx

x for 0 x for 0 x for 0 x for 0 x for 0 ≤≤≤≤≤ xxxxx ≤≤≤≤≤ a a a a a

fffff(((((xxxxx) = ) = ) = ) = ) = { { { { { 2a 2a 2a 2a 2a ------ x for a - x for a - x for a - x for a - x for a ≤≤≤≤≤ xxxxx ≤≤≤≤≤ 22222aaaaa

aaaaa

aaaaa 22222 aaaaa

fffff (((((xxxxx)))))

44

Example 2Example 2Example 2Example 2Example 2 : is f(x) = |x| continuous continuous continuous continuous continuous at x = 0 ?

Approaching 00000 from the left left left left left : f(x) = |x| = −−−−−x

Near 00000 and to the left left left left left of 00000 : x = 0 −−−−− δx

f(x) = { −−−−− (0 −−−−− δx)} = { +δx} = 0

Approaching 00000 from the right right right right right : f(x) = x

Near 00000 and to the right right right right right of 00000 : x = 0 + δx

f(x) = { (0 +δx) = { +δx } = 0

Hence, f(x) = 0 = f(x)

f(x) = f(0) = 0

Since, f(x) = f(x) = f(x) = 0

we say : f(x) is continuous continuous continuous continuous continuous at x = 0.


Limitδx → 0

Limitx → 0+

Limitδx → 0

Limitδx → 0


Limitx → 0+

Valuex = 0


Limitx → 0+

Valuex = 0

Limitδx → 0

(((((0 ,00 ,00 ,00 ,00 ,0 ))))) x x x x x

f(x) = |x|f(x) = |x|f(x) = |x|f(x) = |x|f(x) = |x|


45

ExampleExampleExampleExampleExample 3 3 3 3 3::::: v(t) is the speed of a car weighing one ton that star ts from rest(0 kmph) and accelerates steadily due East along the x-axis for 10 secs andthen levels off at 90 kmph (25 m/sec). Is v(t) continuous continuous continuous continuous continuous at t = 10 secs ?

What acceleration are we going to use ? We still do not know how to difdifdifdifdifffffferererererentiaentiaentiaentiaentiatetetetetethe speed fuction v(t) to find the acceleration (rate of changerate of changerate of changerate of changerate of change of speed) at anyanyanyanyanyinstantinstantinstantinstantinstant. Since the acceleration is steady or constant we may use the averageaverageaverageaverageaverageacceleration between any two instants , say t = 0 secs and t = 10 secs, where thespeeds are known.

a a a a a =

Approaching t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs from the left left left left left : v(t) = aaaaa.t = 2.5 t meter/sec2

Near t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs and to the left left left left left of t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs : t = 10 −−−−− δt

v(t) = {a a a a a (10 −−−−− δt)} = {10aaaaa −−−−−10δt} = 10aaaaa = 25 m/sec Limitt → 10−−−−− Limit

δ t → 0Limitδ t → 0

v(t = 10 sec) −−−−− v(t = 0 sec) =

25 m/sec = 2.5 m/sec22222

10 sec − 0 sec 10 sec

55555 1 01 01 01 01 0 1 51 51 51 51 5 2 02 02 02 02 0 2 52 52 52 52 5 ( ( ( ( (0,00,00,00,00,0))))) ttttt

3 03 03 03 03 0

55555

1 01 01 01 01 0

1 51 51 51 51 5

2 02 02 02 02 0

2 52 52 52 52 525 m/sec25 m/sec25 m/sec25 m/sec25 m/sec

SSSSSPPPPPEEEEEEEEEEDDDDD

[met

er[m

eter

[met

er[m

eter

[met

ers s s s s

//// / se

c]se

c]se

c]se

c]se

c]

T IMETIMETIMETIMETIME [sec] [sec] [sec] [sec] [sec]

v(t) =

a .a .a .a .a . t for t ≤ 10 secs where aaaaa is the constant acceleraton.

25 m/secm/secm/secm/secm/sec for t ≥ 10 secs.

v(t)v(t)v(t)v(t)v(t)

46

ttttt 00000 ttttt22222 ttttt11111 ttttt33333

ttttt

n

y(t)y(t)y(t)y(t)y(t)

hhhhheeeeeiiiiiggggghhhhhttttt

Approaching t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs from the rightrightrightrightright: v(t) = 25 meters/sec, constant.

Near t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs and to the right right right right right of t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs : t = 10 + δt

v(t) = { 25 meter/sec } = 25 m/sec

Hence, v(t) = 25 m/sec = v(t)

v(t) = v(t = 10 secs) = 25 meters / sec

Since, v(t) = v(t) = v(t) = 25 m/sec

we say : v(t) is continuous continuous continuous continuous continuous at t = 10 secs.

Example 4:Example 4:Example 4:Example 4:Example 4: Is the function y(t), that describes the height of a bouncing ball,continuous continuous continuous continuous continuous at instants t0, t1, t2, t3, . . . ?

In general, y(t) = u.sinθ.t −−−−− 1--2 g t22222 where u is the initial velocity and θ is the angleof projection. In this example, over each interval [t0, t1], [t1, t2], [t2, t3] andso on, we have different initial velocities u0, u1, u2, u3, . . . and angles of projectionθ0, θ1, θ2, θ3, . . . respectively. So over each interval [t0, t1], [t1, t2], [t2, t3]and so on, yi(t) = u i.s inθ i. t −−−−− 1--2

g t 22222 for i = 0, 1, 2, 3, . . . respectively.

Valuet = 10

Limitt → 10−−−−−

Limitt → 10+

Valuet = 10

Limitt → 10+

Limitδ t → 0

Limitt → 10+ Limit

t → 10−−−−−

a bouncing balla bouncing balla bouncing balla bouncing balla bouncing ball

46

47

Let us take the general equation y(t) = u.....sinθ.....t −−−−− 1--2 gt22222 that describes the height

of a projectile and adapt it to each interval [t0, t

1], [t

1, t

2], [t

2, t

3].

Over [t0, t1] the height function y(t) is : y0(t) = u0.....sinθ0.....(t −−−−− t0) −−−−− 1--2 g(t −−−−− t0)

22222 .


22222 .


22222

and so on. Let us first look at the behaviour around t = t0.

Approaching t0 from the left left left left left : y(t) is not defined. We do not know the position orheight of the ball. Hence y(t) is not continnot continnot continnot continnot continuousuousuousuousuous from the leftleftleftleftleft. We say that y(t) isdiscontinuousdiscontinuousdiscontinuousdiscontinuousdiscontinuous at t = t0.

We may remove this discontindiscontindiscontindiscontindiscontinuityuityuityuityuity by defining the height of the ball when at reston the ground to be zero. In this case y(t) = 0 for t ≤ t

0. So, approaching t

0 from

the leftleftleftleftleft y(t) = 0.

Now, approaching t0 from the right right right right right : y0(t) = u0.....sinθ0.....(t −−−−− t0) −−−−− 1--2 g(t −−−−− t0)

22222 .

Near t0 and to the right right right right right of t0 : t = t0 + δt . So (t −−−−− t0) = δt .

y0(t) = y0(t0 + δt) = { u0.....sinθ0.....(δt) −−−−− 1--2 g(δt)22222 } = 0

Hence, y0(t) = 0 = y

0(t)

y0(t) = y0(t0) = 0

Since, y0(t) = y0(t) = y0(t) = 0

we say : y0(t) is continuous continuous continuous continuous continuous at t = t

0.

Limitt → t

00000+

Limitδ t → 0

Limitδ t → 0

Limitt → t

00000−−−−−

Limitt → t

00000+

Valuet = t

00000

Limitt → t

00000−−−−−

Limitt → t

00000+

Valuet = t

00000

48

Let us now look at the behaviour around t = t1.

Approaching t1 from the left left left left left y(t) is : y0(t) = u0.....sinθ0.....(t −−−−− t0) −−−−− 1--2 g(t −−−−− t0)

22222.

Near t1 and to the left left left left left of t1 : t = t1 −−−−− δt .

So, y0(t) = u

0.....sinθ0.....(t1

−−−−− δt −−−−− t0) −−−−− 1--2

g(t1 −−−−− δt −−−−− t

0)22222.

y0(t) = y0(t) = u0.....sinθ0.....(t1 −−−−− t0) −−−−− 1--2 g(t1 −−−−− t0)

22222 - - - (1)- - - (1)- - - (1)- - - (1)- - - (1) .

We know from the projectile equation for t0 = 0 we have t

1 = 2u

0.....sinθ0 ///// g.

Substituting t0 = 0 and t1 = 2u0.....sinθ0 ///// g in equation (1) above, we get :

y(t) = y0(t) = y

0(t) = 0 - - - (2)- - - (2)- - - (2)- - - (2)- - - (2) .

Approaching t1 from the right right right right right y(t) is : y1(t) = u1.....sinθ1.....(t −−−−− t1) −−−−− 1--2 g(t −−−−− t1)

22222.

Near t1 and to the right right right right right of t1 : t = t1 + δt .

So, y1(t) = u

1.....sinθ1.....(t1

+ δt −−−−− t1)−−−−− 1--2

g(t1 + δt −−−−− t

1)22222 = u

1.....sinθ1.....( δt ) −−−−− 1--2

g( δt)22222

y(t) = y1(t) = { u1.....sinθ1.....( δt ) −−−−− 1--2 g( δt)22222 } = 0 .

Hence, y(t) = 0 = y(t)

y(t) = 0

Since, y(t) = y(t) = y(t) = 0

we say : y(t) is continuous continuous continuous continuous continuous at t = t1.

Limitt → t

11111−−−−−

Limitt → t

11111+

Valuet = t

11111

Limitt → t

11111−−−−−

Limitt → t

11111+

Valuet = t

11111

Limitt → t

11111−−−−−

Limitδ t → 0

Limitt → t

11111−−−−−

Limitδ t → 0

Limitt → t

11111−−−−−

Limitt → t

11111+

Limit t → t

11111+

Limitδ t → 0

49

y = +√ x

x(0, 0)

y

Example 5Example 5Example 5Example 5Example 5 : is f(x) = + √x continuous continuous continuous continuous continuous at x = 0 ?


f(x) = √ (0 −−−−− δx) = √ −−−−− δx

A negative real number under the radical sign is a complex number. We are dealingwith rrrrreal eal eal eal eal numbers and rrrrreal eal eal eal eal valued functions. Here √ −−−−− δx is not rrrrreal eal eal eal eal valued andso it is not defined. f(x) = + √x is not definednot definednot definednot definednot defined for x < 0.

f(x) = √ (0 +δx) = √+δx = 0

Since, f(x) ≠ f(x) , we say f(x) does not exist.

f(x) = f(0) = √0 = 0

Since f(x) does not exist we say : f(x) is discontinuous discontinuous discontinuous discontinuous discontinuous at x = 0.

We may remove this discontinuity by redefining f(x) to be : f(x) =


Limitδx → 0

Limitx → 0+

Limitδx → 0

Limitδx → 0


Limitx → 0+

Valuex = 0

Limit x → 0

Limitδx → 0

{ 0 for x < 0 √x for x ≥ 0

Limit x → 0

50

Valuex = 0

Example 6:Example 6:Example 6:Example 6:Example 6: Is the function f(x) = 1/////x continuous continuous continuous continuous continuous at x = 0 ?


{ f(x) } = { 1/////x } = −−−−− 1/////0 = −∞ , something undefined.

{ f(x) } = {f(0 − δ x)} = { 1/////(0 − δ x) } = −−−−− 1/////0 = −∞


{ f(x) } = { 1/////x } = +1/////0 = +∞ , something undefined.

{ f(x) } = {f(0 + δ x)} = { 1/////(0 + δ x) } =+1/////0 = +∞

The LIMIT from the left left left left left does not exist. And, the LIMIT from the right right right right right does not exist.

Also { f(x) } = f(0) = 1/////0 = ∞ , something undefined.

So on any one of the three counts, be it LIMIT from the left left left left left or LIMIT from the rightrightrightrightrightor the VALUE, f(x) is discontindiscontindiscontindiscontindiscontinuous uous uous uous uous at x = 0. Moreover, we cannot remove thisdiscontinuity.

Limitδx → 0

Limitδx → 0

Limitδx → 0

Limitx → 0+

Limitx → 0+

Limitδx → 0

Limitδx → 0

Limitδx → 0



51

Example 7:Example 7:Example 7:Example 7:Example 7: Is the function f(x) = continuous continuous continuous continuous continuous at x = 0 ?

We shall use the trigonometric identity sin (A ± B) = sin A cos B ± cos A sin B.Also, when x is very small, we may say : sin (x) = x .


{f(x)} = =

= = {1} = 1 .

Note how we simplify fsimplify fsimplify fsimplify fsimplify fiririririrst and then takst and then takst and then takst and then takst and then take the limite the limite the limite the limite the limit. We can simplify first becauseδx only TENDS TO zero. δx ≠ 0.


{f(x)} = =

= = {1} = 1 .

Note how we simplify fsimplify fsimplify fsimplify fsimplify fiririririrst and then takst and then takst and then takst and then takst and then take the limite the limite the limite the limite the limit. We can simplify first becauseδx only TENDS TO zero. δx ≠ 0.

Since, { f(x) } = 1 = { f(x) } , we say { } = 1 .


Since f(x) ≠ f(x) we say f(x) is discontinuous discontinuous discontinuous discontinuous discontinuous at x = 0.

We may remove this discontinuity by redefining f(x) to be : f(x) =

sin(x)x

Limitδx → 0

Limitδx → 0

{ sin(0 − δ x)

}(0 − δ x)

Limitδx → 0

{ sin(− δ x)

}− δ x

Limitδx → 0

{ − δ x

}− δ x

Limitδx → 0

Limitδx → 0

Limitδx → 0 {

sin(0 + δ x) }

(0 + δ x) Limitδx → 0

{ sin(+ δ x)

}+ δ x

Limitδx → 0

Limitδx → 0

{ δ x

}δ x

Limit x → 0+++++

Limit x → 0----------

Limit x → 0

Valuex = 0

Valuex = 0

sin(x)x

sin(x)x

Limit x → 0

Valuex = 0 { for x ≠ 0

1 for x = 0

sin(x)x

52

Example 8 :Example 8 :Example 8 :Example 8 :Example 8 : Is the function f(x) defined below continuous continuous continuous continuous continuous anywhere ?

f(x) =

We saw that the f(x) does NOT exist anywhere.

However, f(x) is well-defined everywhere.

Since f(x) ≠ f(x) we say that f(x) is not continuous not continuous not continuous not continuous not continuous at x = a.

But a can be any point on the Real number line. So f(x) is not continuous not continuous not continuous not continuous not continuous anywhere.


Limitx → a

Valuex = a

Limitx → a

Valuex = a

53

PPPPPararararar t 2 :t 2 :t 2 :t 2 :t 2 : DIFFERENTIA DIFFERENTIA DIFFERENTIA DIFFERENTIA DIFFERENTIATIONTIONTIONTIONTION

54

VERTICAL POSITION (1-Dimension)VERTICAL POSITION (1-Dimension)VERTICAL POSITION (1-Dimension)VERTICAL POSITION (1-Dimension)VERTICAL POSITION (1-Dimension)

y(t) = y(t) = y(t) = y(t) = y(t) = u . s i nu . s i nu . s i nu . s i nu . s i n θθθθθ . t. t. t. t. t ---------- 11111----------22222

g ttttt 22222

(0, 0)Time

T t

He

igh

t y

(t)

y(t1)

h1

t1 T

/////2


Let us continue with our main example.

When an object is projected into the air with a given initial velocity uuuuu and

angle of projection or direction θ , the curve that describes its path is a

parabola. There are two functions that describe the two components of its

position. A VERTICAL function y(t) which describes its HEIGHT at any chosen

INSTANT and a HORIZONTAL function x(t) which describes its RANGE at any

chosen INSTANT (refer Preface iii).

We know from Dynamics that:

RANGE function x(t) = u . cos θ . t

HEIGHT function y(t) = u . s i n θ . t ---------- 1--2

g t2

Let us concentrate on the VERTICAL function y(t). Let t1 the time for the

object to travel from t = 0 to y(t1). Let h

1 be the height or ver tical displacement

described in time t1.

y(t1) = u . s i n θ . t

1---------- 1--

2 g t

1

2

This is a polynomial of the form: f(x) = a2x2 + a

1x + a

0 which is CONTINUOUS

everywhere.

Now that we know the TIME AXIS is a CONTINUOUS set of INSTANTS and y(t) is a

SINGLE VALUED and CONTINUOUS function over [A , B], we may proceed to describe

the DIFFERENTIATION operation and perform the calculation.

(0, 0)

uuuuu

θ

u.cos θ

u.si

n θ

SPEED (2-Dimension)SPEED (2-Dimension)SPEED (2-Dimension)SPEED (2-Dimension)SPEED (2-Dimension)

55

Before we do any calulation (differentiation differentiation differentiation differentiation differentiation ) we will develop the concept and

definition of the FIRST DERIVATIVE or the INSTANTANEOUS RATE OF CHANGE.

Using our main example we will illustrate the concept to find the INSTANTANEOUS

RATE OF CHANGE of the parparparparpar ticular ticular ticular ticular ticular height function y(t) at some parparparparpar ticulaticulaticulaticulaticularrrrr

instant instant instant instant instant t1.

By dropping the subscript of t1 , we extend this concept of the INSTANTANEOUS RATE

OF CHANGE of the parparparparpar ticular functionticular functionticular functionticular functionticular function y(t) at parparparparpar ticular instantticular instantticular instantticular instantticular instant t1 to any

general instantgeneral instantgeneral instantgeneral instantgeneral instant t on the time axis.

With the concept now clear, we define the INSTANTANEOUS RATE OF CHANGE of

some gggggenerenerenerenereneral functional functional functional functional function f(x) at some parparparparpar ticular instantticular instantticular instantticular instantticular instant x1

.

Again, dropping the subscript of x1 , we extend this concept of the INSTANTANEOUS

RATE OF CHANGE of the gggggenerenerenerenereneral functional functional functional functional function f(x) at parparparparpar ticular instantticular instantticular instantticular instantticular instant x1 to any

general instantgeneral instantgeneral instantgeneral instantgeneral instant x on the x-axis.

Here too we have to take the LIMIT both from the left left left left left and from the right right right right right . Even with

SINGLE VALUED and CONTINUOUS functions, it is quite possible when taking the

limits to end up with something undefined such as +∞, −∞, division by zero and

∞/∞ , or the left left left left left LIMIT ≠ right right right right right LIMIT. In this case we say the function f(x) is not

DIFFERENTIABLE at that instant.

f(x + δ x) −−−−− f(x)

(x + δ x) − − − − − x

df =

LIMIT δf

dx δx→ 0 δx

LIMIT

δt→ 0

dy

t1

=

LIMIT δy =

LIMIT δy

dt t2→ t

1 δt δt→ 0 δt( ) =

y(t

1 + δ t) −−−−− y(t

1)

(t

1 + δ t) −−−−− t

1

dy =

LIMIT δy

dt δt→ 0 δt

y(t + δ t) −−−−− y(t)

(t + δ t) −−−−− t LIMIT

δt→ 0=

f(x1 + δ x) −−−−− f(x

1)

(x1 + δ x) −−−−− x

1

df

x1

=

LIMIT δf =

LIMIT δf

dx x2→ x

1 δx δx→ 0 δx( )

LIMIT

δx→ 0=

LIMIT

δx→ 0=

56

With the definition of the FIRST DERIVATIVE in mind we proceed to difdifdifdifdifffffferererererentiaentiaentiaentiaentiate te te te te thegeneral polynomial f(x) = an xn + an----------1 xn--1 + . . . + a2 x2 + a1 x + a0 .

In fact we need to know only how to difdifdifdifdifffffferererererentiaentiaentiaentiaentiate te te te te xn , the general variable termof the general polynomial. We shall do this two ways : the Vedic way and the Westernway. Once we know how to difdifdifdifdifffffferererererentiaentiaentiaentiaentiate te te te te xn , we may differentiate any polynomial.Also, we will be able to difdifdifdifdifffffferererererentiaentiaentiaentiaentiate te te te te WELL-BEHAVED functions. Finding theDERIVATIVE by directly applying the definition is known as difdifdifdifdifffffferererererentiaentiaentiaentiaentiation frtion frtion frtion frtion fromomomomomfffff iririririr st principlesst principlesst principlesst principlesst principles.

It is not necessary to always difdifdifdifdifffffferererererentiaentiaentiaentiaentiate te te te te from FIRST PRINCIPLES. We maydifferentiate each of the various functions just once and build a TTTTTaaaaabbbbble ofle ofle ofle ofle of Deri Deri Deri Deri Derivvvvvaaaaatititititivvvvveseseseses.We then look at more complicated functions made up from elementary functions. Todifferentiate these combinations of other functions we have a Set of Rules Set of Rules Set of Rules Set of Rules Set of Rules .

We also look at the Units ofUnits ofUnits ofUnits ofUnits of Measur Measur Measur Measur Measureeeee in difdifdifdifdifffffferererererentiaentiaentiaentiaentiationtiontiontiontion and inteinteinteinteintegggggrrrrraaaaationtiontiontiontion. Acalculation usually has two parts: an oper oper oper oper operaaaaationtiontiontiontion and a units ofunits ofunits ofunits ofunits of measur measur measur measur measureeeee. Theoperation with a function function function function function or expressionexpressionexpressionexpressionexpression as an input operand produces as outputanother functionfunctionfunctionfunctionfunction or expressionexpressionexpressionexpressionexpression. When we evaluate this output expression we get avaluevaluevaluevaluevalue. The output expressionexpressionexpressionexpressionexpression or value value value value value has a units of measureunits of measureunits of measureunits of measureunits of measure associated withit. For example, we may perform an operation to find the expression πr2 that tellsus the area of a circle. When we evaluate this expression πr2, say with r = 2, we getthe value 4π. The expression πr2 or the value 4π has units of units of units of units of units of measur measur measur measur measureeeee[meter[meter[meter[meter[meter22222 ] ] ] ] ] associated with it.

We assume we are dealing with WELL-BEHAVED functions : SINGLE VALUED,CONTINUOUS and DIFFERENTIABLE. We have seen the first two properties ( SINGLEVALUED and CONTINUOUS ) and why they are necessary. We conclude this part bylooking at the DIFFERENTIABLE property.

57

height

h1h2

y(t1)

(0, 0) timeTt1 t2

y(t2)

1111122222..... INST INST INST INST INSTANTANTANTANTANTANEOUS RAANEOUS RAANEOUS RAANEOUS RAANEOUS RATE OF CHANGE ofTE OF CHANGE ofTE OF CHANGE ofTE OF CHANGE ofTE OF CHANGE of y(t)y(t)y(t)y(t)y(t)

If we measure the height at any chosen instant, say t1, then in terms ofthe height function y(t) = u . s i n θ . t ---------- 1--2

g t 2 the AVERAGE SPEED overtime t = 0 to t1 is :

= u . s i n θ ---------- 1--2 g t

We may ask: what is the actual ver tical speed at the time instantinstantinstantinstantinstant t1 ?We shall do this in three steps.

1.1.1.1.1. AAAAAVERAVERAVERAVERAVERAGE steGE steGE steGE steGE step:p:p:p:p: If we take a par ticular small time interval, say t1 to t2 ,we could be more precise. AVERAGE VERTICAL SPEED from time t1 to t2 is :

y (t ) t

CHANGE in height = CHANGE in time

∆ y =

y(t2) −−−−− y(t1)∆ t t2 −−−−− t1

distance = time

58

2.2.2.2.2. TENDS TENDS TENDS TENDS TENDS TTTTTO steO steO steO steO step:p:p:p:p: the smaller the time interval t2 ---------- t1 the more accurate

the calculation of the AVERAGE VERTICAL SPEED at time instant t1. So we fix t1

and let t2 get closer and closer to t

1. The difference ∆ t between t

1 and t

2 becomes

smaller and smaller. It becomes infinitely small. This kind of difference we denote

using the Greek symbol δ. As t2→ t

1 we may say : t

2 = t

1 + δ t .

So : δ t = t2 −−−−− t

1 = (t

1 + δ t) −−−−− t

1

and δ y = y(t2) −−−−− y(t

1) = y(t

1 + δ t) −−−−− y(t

1)

Hence,

3.3.3.3.3. LIMIT ste LIMIT ste LIMIT ste LIMIT ste LIMIT step:p:p:p:p: Finally, when we let t2 coincide with t

1, we get the INSTANTANEOUS

VERTICAL SPEED at instant instant instant instant instant t1. This we denote by :

This ver tical speed or “r“r“r“r“raaaaate ofte ofte ofte ofte of c c c c changhanghanghanghange”e”e”e”e” in height got by taking the LIMIT

is called the INSTANTANEOUS SPEED or INSTANTANEOUS RATE OF CHANGE

in height at the time instant t1 . We can now answer Question 2 on page 3.

This is called the FIRST DERIVATIVE of y(t) at instant instant instant instant instant t1. There are several

other notations for the FIRST DERIVATIVE :dy---dt

t1

, y ’( t1) , Dy( t

1) , y

.(t

1)

δy=

y(t1 + δ t) −−−−− y(t

1)

δt (t1 + δ t) −−−−− t

1

Limit of AVERAGE RATE OF CHANGE of y(t) over the interval [ t1, t2 ] as t2

TENDS TO t1 = INSTANTANEOUS RATE OF CHANGE of y(t) at the instant t1

( )

dy =

LIMIT δy =

LIMIT δy

dt t2→ t

1 δt δt→ 0 δt

y(t1 + δ t) −−−−− y(t

1)

(t

1 + δ t) −−−−− t

1

LIMIT

δt→ 0=

59

height

h1h2

h3

h4

y(t1)

y(T/////2) = maximum

(0, 0)time

t4t3 tn = Tt1 t2t0 T/////2

y(t2) y(t3)

y(t4)

We know from observation that the INSTANTANEOUS VERTICAL SPEED orINSTANTANEOUS RATE OF CHANGE in height is not the same at differentinstants t1, t2, t3 and t4 in time.

Using this same method of taking the we can find the INSTANTANEOUSRATE OF CHANGE in height at any parparparparpar ticular instantticular instantticular instantticular instantticular instant t2 or t3 or t4, in the timeinterval t0 to tn.

Can we find the INSTANTANEOUS RATE OF CHANGE in height at ananananany instanty instanty instanty instanty instant t in thetime interval t0 to tn, rather than at parparparparpar ticular instants ticular instants ticular instants ticular instants ticular instants t1, t2, t3 or t4 ? Wemay drop the subscript and let t be a general instant general instant general instant general instant general instant in [t0 , tn].

dy---dt

, y ’( t ) , Dy( t ) , y.(t)

So far we have only a NOTATION, some symbols that express the concept ofwhat we are trying to do: find the INSTANTANEOUS RATE OF CHANGE of theparticular function y(t) AT ANY INSTANT.

Limitδt→0

60

When we perform the calculation to find the INSTANTANEOUS RATE OF CHANGE of

y(t) at any instant we get (refer Preface iii) :

INSTANTANEOUS VERTICAL SPEED = dy---------------dt

= u ..... s i n θ ---------- g t

Compare this expression of INSTANTANEOUS VERTICAL SPEED to the

ear lier calculation of AVERAGE VERTICAL SPEED:

AVERAGE VERTICAL SPEED =

When we compare the AVERAGE SPEED with the INSTANTANEOUS SPEED we

make the profound distinction between an INTERVAL (no matter how small)

and an INSTANT. Only with this distinction and the calculation using INSTANT

are we able to get the INSTANTANEOUS SPEED.

δt is an INTERVAL. (a + δt) is the INSTANT a.

δx is an INTERVAL. (x + δx) is any INSTANT x.

From the AnalAnalAnalAnalAnalysis ysis ysis ysis ysis point of view : an interinterinterinterintervvvvval al al al al δt or δx, no matter how small, is

a continuous continuous continuous continuous continuous set of instants instants instants instants instants or pointspointspointspointspoints.

From the AlgAlgAlgAlgAlgeeeeebrbrbrbrbra a a a a point of view : an interinterinterinterintervvvvval al al al al corresponds to a subset of the real

numbers R. This subset has UNCOUNTABLY many real numbers (both rational and

irrational numbers) in a contiguous contiguous contiguous contiguous contiguous manner, ie; with none missing in between.

Since we cannot even begin to COUNT them, there is no sense in talking about a

subscript to enumerate them.

We shall use the words DIFFERENTIATE or FIND THE DERIVATIVE to denote

the process of finding the expression for the INSTANTANEOUS RATE OF

CHANGE of a function. Let us now use this concept to define the INSTANTANEOUS

RATE OF CHANGE of a general function f(x).

yt(t)

= u ..... s i n θ ---------- 1--2 g t

Limitδt → 0

Limitδx → 0

61

f ( x )f ( x )f ( x )f ( x )f ( x )22222

xxxxx 11111 xxxxx 22222 xxxxx


(((((0 ,00 ,00 ,00 ,00 ,0 )))))

f ( x )f ( x )f ( x )f ( x )f ( x ) 11111

1111133333..... INST INST INST INST INSTANTANTANTANTANTANEOUS RAANEOUS RAANEOUS RAANEOUS RAANEOUS RATE OF CHANGE ofTE OF CHANGE ofTE OF CHANGE ofTE OF CHANGE ofTE OF CHANGE of f(x)f(x)f(x)f(x)f(x)

What is the instantaneous rinstantaneous rinstantaneous rinstantaneous rinstantaneous raaaaate ofte ofte ofte ofte of c c c c changhanghanghanghangeeeee of f(x) at particular instant x1 ?We shall define this in three steps.

1.1.1.1.1. AAAAAVERAVERAVERAVERAVERAGE steGE steGE steGE steGE step:p:p:p:p: If we take a par ticular small inter val, say x1 to x2 , wecould be more precise. AVERAGE RATE OF CHANGE from instant x1 to x2 is :

2.2.2.2.2. TENDS TENDS TENDS TENDS TENDS TTTTTO steO steO steO steO step:p:p:p:p: the smaller the interval x2 ---------- x1 the more accurate thecalculation of the AVERAGE RATE OF CHANGE at parparparparpar ticularticularticularticularticular instant instant instant instant instant x1. So we fixx1 and let x2 get closer and closer to x1. The difference ∆ x between x1 and x2

becomes smaller and smaller. It becomes infinitely small. This kind of difference wedenote using the Greek symbol δ. As x2→ x1 we may say : x2 = x1 + δ x .

So : δ x = x2 −−−−− x1 = (x1 + δ x) −−−−− x1

and δ f = f(x2) −−−−− f(x

1) = f(x

1 + δ x) −−−−− f(x

1)

Hence,

CHANGE in function =CHANGE in variable

∆ f =

f(x2) −−−−− f(x

1)

∆ x x2 −−−−− x

1

δf=

f(x1 + δ x) −−−−− f(x1)δx (x1 + δ x) −−−−− x1

62

3.3.3.3.3. LIMIT ste LIMIT ste LIMIT ste LIMIT ste LIMIT step:p:p:p:p: Finally, when we let x2 coincide with x1, we get the INSTANTANEOUSRATE OF CHANGE of f(x) at parparparparparticular instant ticular instant ticular instant ticular instant ticular instant x1. This we denote by :

This “r“r“r“r“raaaaate ofte ofte ofte ofte of c c c c changhanghanghanghange”e”e”e”e” got by taking the LIMIT is called the INSTANTANEOUSSPEED or INSTANTANEOUS RATE OF CHANGE at the particular instant x1 .

This is denoted by f ’(x1).

f ’(x1) =

This is called the FIRST DERIVATIVE of f(x1). There are several other notations

for the FIRST DERIVATIVE :

df---dx

x

1

, f’(x1), Df(x

1), f.(x

1)

Now, r ather than choos ing the parparparparpar t i cu lar ins tant t i cu lar ins tant t i cu lar ins tant t i cu lar ins tant t i cu lar ins tant x1, we want theINSTANTANEOUS RATE OF CHANGE of the funct ion f(x) a t any gggggenerenerenerenereneralalalalaliiiiins tantns tantns tantns tantns tant x .

Limit of AVERAGE RATE OF CHANGE of f(x) over the interval [ x1, x2 ] as x2

TENDS TO x1 = INSTANTANEOUS RATE OF CHANGE of f(x) at the instant x1

df =

LIMIT δf =

LIMIT δfdx x2→ x1 δx δx→ 0 δx

( )

Limit f (x1 + δx) ---------- f (x1) =δx → 0 (x1 + δx) ---------- x1

Limit f (x1 + δx) ---------- f (x

1)

δx → 0 δx

Limit f (x1 + δx) ---------- f (x1)δx → 0 δx

63

f ( x )f ( x )f ( x )f ( x )f ( x )

xxxxx 00000 xxxxx 11111 xxxxx 22222 xxxxx 3 3 3 3 3 . . .. . .. . .. . .. . . xxxxx nnnnnxxxxx


(((((0 ,00 ,00 ,00 ,00 ,0 )))))

Using this same method of taking the we can find the INSTANTANEOUSRATE OF CHANGE at any parparparparpar ticular instantticular instantticular instantticular instantticular instant x2 or x3 or x4, in the time intervalx0 to xn. We may drop the subscript and let x be a gggggenerenerenerenereneral instant al instant al instant al instant al instant in [x0 , xn].

Definition : instantaneous rate of change of f(x)Definition : instantaneous rate of change of f(x)Definition : instantaneous rate of change of f(x)Definition : instantaneous rate of change of f(x)Definition : instantaneous rate of change of f(x)

This is called the FIRST DERIVATIVE of f(x1). There are several other notationsfor the FIRST DERIVATIVE :

df---dx

, f’( x), D f ( x), f.(x)

So far we have only a NOTATION, some symbols that express the idea ofwhat we are trying to do: find an expression for the INSTANTANEOUS RATEOF CHANGE AT ANY INSTANT.

Limitδt→0

Limit f (x + δx) -- f (x)δx → 0 δx

f ’(x) =

Limit f (x + δx) ---------- f (x) =δx → 0 (x + δx) ---------- x

Limit f (x + δx) ---------- f (x)δx → 0 δx

64

We shall use the words DIFFERENTIATE or FIND THE DERIVATIVE to denote

the process of finding the expression for the INSTANTANEOUS RATE OF

CHANGE of a function.

Properly speaking we should say :

LEFT DERIVATIVE of f(x) =

RIGHT DERIVATIVE of f(x) =

If the LEFT DERIVATIVE = the RIGHT DERIVATIVE and is something wwwwwell-defell-defell-defell-defell-definedinedinedinedined,

then and only then we may combine both the LEFT DERIVATIVE and the RIGHT

DERIVATIVE into one expression and say :

Even with SINGLE VALUED and CONTINUOUS functions, it is quite possible when taking

the LIMITS to find the derivatives, we may end up with something undefined such as

+∞, −∞, division by zero and ∞/∞ , or the LEFT DERIVATIVE ≠ RIGHT DERIVATIVE.

In this case we say the function f(x) is not DIFFERENTIABLE at that instant.

Note : In A LITTLE MORE CALCULUS the student will learn the MEAN VALUE THEOREM

which states that : : : : :

=====

AVERAGE RATE OF CHANGE

over the interval [a,b]

INSTANTANEOUS RATE OF CHANGE at some

MEAN point (we do not know which)

denoted by x---------- in the interval [a,b]

f9( x-- )

= f(b) ---------- f(a)

b ---------- a


Limit f (x -- δx) -- f (x)δx → 0 --δx


f ’(x) =DEFINITION :

65

14.14.14.14.14. INST INST INST INST INSTANTANTANTANTANTANEOUS RAANEOUS RAANEOUS RAANEOUS RAANEOUS RATE OF CHANGE ofTE OF CHANGE ofTE OF CHANGE ofTE OF CHANGE ofTE OF CHANGE of x x x x xnnnnn

L e t u s c o n s i d e r h o w t o d e r i v e t h e g e n e r a l e x p r e s s i o n f o rI N S TA N TA N E O U S R AT E O F C H A N G E of the general polynomial :

an xn + an ---------- 1 xn ---------- 1 +. . . + a1x + a0

Obser ve that the terms of this function are of the form anxn

for n = 0, 1, 2, 3, . . .

To DIFFERENTIATE the general polynomial we should be able to f ind theINSTANTANEOUS RATE OF CHANGE of the general variable term xn . We canthink of xn as the function xn. In other words, we should be able toDIFFERENTIATE the function f(x) = xn.

As usual we will first find the INSTANTANEOUS RATE OF CHANGE of the functionf(x) = xn at parparparparpar ticular pointticular pointticular pointticular pointticular point aaaaa . Then we will find the general expression of theINSTANTANEOUS RATE OF CHANGE f(x) = xn at ananananany pointy pointy pointy pointy point x.

We note that :

We can differentiate f(x) = xn at x = a in two ways :

1. The Vedic way - without the use of δx .

2. The Western way - using the infinitesimal δx .

Value f(x) = an x = a

66

VVVVVedic wedic wedic wedic wedic waaaaay :y :y :y :y : fffff ’ (a) = (a) = (a) = (a) = (a) = LimitLimitLimitLimitLimit xxxxx nnnnn----------aaaaa nnnnn xxxxx→→→→→a a a a a xxxxx----------aaaaa

AVERAGE RATE OF CHANGE close to a . =

CHANGE in functionCHANGE in variable

So : f(x) −−−−− f(a) =

xn −−−−− an

x −−−−− a x −−−−− aProperly speaking we have to compute both :

But in the Vedic way the Algebraic notation is lacking. For n = 1 and n = 2 thereader may wish to review example 4 (page 28) and example 5 (page 29).

FFFFFor or or or or nnnnn = 3 = 3 = 3 = 3 = 3 : : : : : x3--a3

= (x--a) (x2+xa+a2) = (x2+xa+a2) x--a (x--a)

Substitute for x = a to get this Limit = 3a2

Let us do this in three steps for general n.

1.1.1.1.1. AAAAAVERAVERAVERAVERAVERAGE steGE steGE steGE steGE step :p :p :p :p : f(x) −−−−− f(a) =

xn −−−−− an

x −−−−− a x −−−−− a

2. TENDS TO step :2. TENDS TO step :2. TENDS TO step :2. TENDS TO step :2. TENDS TO step : xn--an

= (x −−−−− a)(xn--1 + xn--2 a + xn--3 a2 + . . . + xan----------2 + an----------1) x--a x--a

= (xn--1 + xn--2 a + xn--3 a2 + . . . + xan----------2 + an----------1)We may simplify because (x −−−−− a) ≠ 0 .

3. LIMIT step : 3. LIMIT step : 3. LIMIT step : 3. LIMIT step : 3. LIMIT step : (xn--1 + xn--2 a + xn--3 a2 + . . . + xan----------2 + an----------1)

Substitute for x = a to get nan----------1

Limitx → a

Limitx → a

Limitx → a

Limitx → a

f(x) −−−−− f(a) and

f(x) −−−−− f(a) x −−−−− a x −−−−− a

Limit x → a+++++

Limit x → a----------

67

1.1.1.1.1. AAAAAVERAVERAVERAVERAVERAGE steGE steGE steGE steGE step :p :p :p :p : f(x) −−−−− f(a) =

(a+δx)n----------an (from the right)x −−−−− a (a+δx) ----------a

In the numerator ::::: (a+δx)n = an(1+δxa )n

So::::: (a + δx)n ---------- an = an[ (1 + δxa )n ----------1]

In the denominator ::::: (a+δx ) ---------- a = δx

So :::::

2. TENDS TO step : 2. TENDS TO step : 2. TENDS TO step : 2. TENDS TO step : 2. TENDS TO step : expanding by the Binominal Theorem :::::

terms with higher powers of δx ) ---------- 1]The +++++1 and ----------1 will cancel out. We are left with :

terms with higher powers of δx ]

Since δx → 0 but δx /////===== 0 we may simplify the terms with δx to get :

terms with higher powers of δx ]

3. LIMIT step : 3. LIMIT step : 3. LIMIT step : 3. LIMIT step : 3. LIMIT step : now take the LIMIT as δx → 0. δx = 0, and (terms with higher powers of δx) = 0.

We are left with

Limitδx → 0

Limitδx → 0

an n----------a = nan----------1 .

(a+δx )n ---------- an =

an

[ (1+ δxa )

n ----------1]

(a + δx) ---------- a δx

Limit δxδx → 0 a22222 = 0

Limitδx → 0

(a+δx)n----------an = nan----------1

(a+δx)----------a (from the right)

WWWWWesteresteresteresterestern wn wn wn wn waaaaay :y :y :y :y : fffff ’ (a) = (a) = (a) = (a) = (a) = L i m i tL i m i tL i m i tL i m i tL i m i t ( a +( a +( a +( a +( a + δδδδδ xxxxx ))))) nnnnn ---------- aaaaa nnnnn

δδδδδ xxxxx →→→→→ 00000 ( a +( a +( a +( a +( a + δδδδδ x )x )x )x )x ) ---------- aaaaa

= an [ (1+ n δx

a + n( n ----------1)

[ δxa ]

2 2 2 2 2 + . . . δx

2!

= an [ nδx

a + n( n ----------1)

[ δxa ]

2 2 2 2 2 + . . . δx 2!

= annnnn [ n----------a

+ n( n ----------1)

[ δx ] + . . .

2! a22222

68

In calculating f ’(a) when we wrote we meant , i.e. from the rightrightrightrightright.

We could have taken , i.e. from the leftleftleftleftleft.

What is (a ----------δx)n ---------- an? ( from the lefleflefleflefttttt )

i.e. what is (a ----------δx)n ---------- an ? ( from the lefleflefleflefttttt )

IMPORTANT: notice the ----------δx in the denominator when approaching theinstant a from the leftleftleftleftleft :

x → a-- = a ---------- δx as δx → 0 and (a ---------- δx) ---------- a = ----------δx .

Now, r ather than choos ing the par t i cu lar ins tant ins tant ins tant ins tant ins tant aaaaa, we may want tofind the INSTANTANEOUS RATE OF CHANGE of the funct ion f(x) = xxxxxnnnnn atanyanyanyanyany i ns tan ti ns tan ti ns tan ti ns tan ti ns tan t x . In general, if we let x be any instant instant instant instant instant over which thefunction is WELL- BEHAVED (SINGLE VALUED, CONTINUOUS and DIFFERENTIABLE):

f ’(x) =

Instead of using the symbols δx we will use the letter h. So, by ourunderstanding, we can define the DERIVATIVE of a function to be:

f ’(x) =

Now, let us use our definition of the FIRST DERIVATIVE to prove that :

Limit x → a

Limitx → a+

Limit x → a----------

Limitδx → 0

Limitδx → 0

----------δx

( a----------δx ) ---------- a

Limit f (x + h) -- f (x) h → 0 h


dxxxxx nnnnn

dx = n xxxxx n ---------- 1

69

PrPrPrPrProofoofoofoofoof :::::f( x+h ) = ( x+h )n = xn (1 + h--x )

f(x) = xn

1.1.1.1.1. AAAAAVERAVERAVERAVERAVERAGE steGE steGE steGE steGE step :p :p :p :p :

2. TENDS TO step : 2. TENDS TO step : 2. TENDS TO step : 2. TENDS TO step : 2. TENDS TO step : expanding by the Binomial Theorem :

The +1 and --1 will cancel out.

Since h → 0 but h /////= 0 we may cancel h in the numerators and denominator.

3. LIMIT step :3. LIMIT step :3. LIMIT step :3. LIMIT step :3. LIMIT step : now we can take the LIMIT as h → 0 .

, and (terms with higher powers of h) = 0.

We are left with : xn n--x = n xn ---------- 1

= xn [ (1 + h--x )n ---------- 1] ---h

= xn [ (n h--x + n( n ---------- 1 ) ((((( h--x )))))2 + . . . terms with higher powers of h ]

---h 2!

= xn [ ( n--x + n( n ---------- 1) ((((( h---x2 )))))

+ . . . terms with higher powers of h ] 2!

Limit h = 0 h → 0 Limit h---

x2 = 0 h → 0 Limit h → 0

f (x + h) ---------- f (x) = 1 [ xn (1 + h--x )n ---------- xn ] h

--h

= xn [ (1 + n h--x + n( n ---------- 1 ) ( h--x )2

+ . . . terms ---h 2!

with higher powers of h ) ----------1 ]

We denote this INSTANTANEOUS RATE OF CHANGE as :

d---dx

xxxxxnnnnn = nxn--1

This is also called the FIRST DERIVATIVE of xxxxxnnnnn.

70

Most of the confusion in Calculus arises out of the fear of dividing by zero.Limit δx TENDS TO zero is zero. Therefore, it appears in to bedividing by zero.

We must always keep in mind that δx ia attached to some parparparparpar ticular instantticular instantticular instantticular instantticular instant

aaaaa or gggggenerenerenerenereneral instantal instantal instantal instantal instant xxxxx . As δx → 0 we approach that INSTANT computing aswe go along the AVERAGE RATE OF CHANGE .

When we take of this AVERAGE RATE OF CHANGE we get the

INSTANTANEOUS RATE OF CHANGE at the parparparparparticular instant ticular instant ticular instant ticular instant ticular instant aaaaa orgeneral instantgeneral instantgeneral instantgeneral instantgeneral instant xxxxx.

ExExExExExererererercise :cise :cise :cise :cise : Prove the INSTANTANEOUS RATE OF CHANGE of a nx n = na nxn−−−−−1 ?

The same rule applies. The proof is not required at this level.

Limit δyδx → 0 δx

δy δx

Limitδx → 0

δy δx

dy dx

You have now learned to DIFFERENTIATE xn when the exponent nis an integral or whole number. What if n is a RATIONAL number ?

How would you DIFFERENTIATE x1/2 or x------3/2 ?Mechanically, we may say:

But we must be careful and add : for x > 0

dx1/2 = 1----------2

x----------1/2 and dx----------3/2 = ----------3----------2

x----------5/2dx dx

Let f(x) = annnnnxnnnnn + a

nnnnn−−−−−11111xnnnnn−−−−−11111 + . . . + a

22222x22222 + a

11111x11111 + a

00000

be any polynomial of degree n, then : d d d d df(x) dddddx = f’(x) = n .a

nnnnnxnnnnn−−−−−11111 + (n−−−−−1) .a

nnnnn−−−−−11111xnnnnn−−−−−22222 + . . . + 2a

22222x11111 + a

11111

71

Example 1Example 1Example 1Example 1Example 1: What is the instantaneous rate of changeinstantaneous rate of changeinstantaneous rate of changeinstantaneous rate of changeinstantaneous rate of change of f(x) = CCCCC ?

1.1.1.1.1. AAAAAVERAVERAVERAVERAVERAGE steGE steGE steGE steGE step :p :p :p :p :

2. TENDS TO step : 2. TENDS TO step : 2. TENDS TO step : 2. TENDS TO step : 2. TENDS TO step : since x2 only TENDS TO x1 : x2 ≠ x1, we may simplify .

3. LIMIT step : 3. LIMIT step : 3. LIMIT step : 3. LIMIT step : 3. LIMIT step : now we may take the LIMIT by letting x2 coincide with x1 .

f ,(x) =

Since f(x) is CONSTANT its rrrrraaaaate ofte ofte ofte ofte of c c c c changhanghanghanghangeeeee is zero.

15. 15. 15. 15. 15. Differentiation from FIRST PRINCIPLESDifferentiation from FIRST PRINCIPLESDifferentiation from FIRST PRINCIPLESDifferentiation from FIRST PRINCIPLESDifferentiation from FIRST PRINCIPLES

Let us now find the FIRST DERIVATIVE of some simple functions by directly applyingthe definition. This is known as difdifdifdifdifffffferererererentiaentiaentiaentiaentiation frtion frtion frtion frtion from fom fom fom fom fiririririrst principlesst principlesst principlesst principlesst principles. Before weDIFFERENTIATE a function we must ensure that it is WELL-BEHAVED : SINGLE VALUED,CONTINUOUS and DIFFERENTIABLE.

CCCCC PPPPP11111 PPPPP22222

xxxxx11111 xxxxx22222xxxxx

(((((0,00,00,00,00,0)))))

f(x) = f(x) = f(x) = f(x) = f(x) = CCCCC constantconstantconstantconstantconstant


df(x) = dCCCCC = 0 dx dx

P2 ---------- P1 = f(x2) ---------- f(x1) = CCCCC ---------- CCCCC = 0

x2 ---------- x1 x2 ---------- x1 x2 ---------- x1 x2 ---------- x1

L imit 0 = 0x2 → x1

0 = 0 x2 ---------- x1

72

ExampleExampleExampleExampleExample 2 2 2 2 2: What is the instantaneous rate of changeinstantaneous rate of changeinstantaneous rate of changeinstantaneous rate of changeinstantaneous rate of change of y = bx ?

It is not necessary to find the derivative in 3 steps. We may combine all 3 steps inone expression.

2b2b2b2b2bPPPPP 11111

PPPPP 22222

xxxxx(((((0 ,00 ,00 ,00 ,00 ,0 ))))) 11111 22222

bbbbb

y = bx

y = bx

y = bx

y = bx

y = bxyyyyy

Since x2 only TENDS TO x1 : x2 /////= x1, we can cancel out the ( x2 -- x1) /////= 0.We simplify first and then take the LIMIT. Observe how we may factor out thecoefficient b.

The instantaneous rinstantaneous rinstantaneous rinstantaneous rinstantaneous raaaaate ofte ofte ofte ofte of c c c c changhanghanghanghangeeeee of y is CONSTANT.

The general linear function is : y = mx +y = mx +y = mx +y = mx +y = mx + C C C C C where mmmmm is the slope slope slope slope slope .

So the DERIVATIVE of a linear function is its slope slope slope slope slope .

= L imit b(x2 ---------- x1) = L imit b = b . x2 → x1 x2 ---------- x1 x2 → x1

dy = d(bx) = b

dx dx

= L imit P2 ---------- P1 = L imit y(x2) ---------- y(x1) = L imit bx2 ---------- bx1 x2 → x1 x2 ---------- x1 x2 → x1 x2 ---------- x1 x2 → x1 x2 ---------- x1

dy dx

73

Example 3:Example 3:Example 3:Example 3:Example 3: Let us find the general expression for the INSTANTANEOUS RATE

OF CHANGE in height of the ball from the function of its height.

We need to difdifdifdifdifffffferererererentiaentiaentiaentiaentiate te te te te y(t) = u ..... s i n θ . t ---------- 1--2 g t 2

at gggggenerenerenerenereneral instant al instant al instant al instant al instant t .

1.1.1.1.1. AAAAAVERAVERAVERAVERAVERAGE steGE steGE steGE steGE step:p:p:p:p:

2. TENDS TO step:2. TENDS TO step:2. TENDS TO step:2. TENDS TO step:2. TENDS TO step:

Here we may simplify since δt ≠ 0.

3.3.3.3.3. LIMIT ste LIMIT ste LIMIT ste LIMIT ste LIMIT step:p:p:p:p: Finally, we take the LIMIT as δt → 0 to get the INSTANTANEOUS

VERTICAL SPEED at general instant general instant general instant general instant general instant t .

∆ y =

y(t + ∆ t) −−−−− y(t)

∆ t (t + ∆ t) −−−−− t

= {u ..... s i n θ . (t + ∆ t) ---------- 1--

2 g (t + ∆ t)

2 } ---------- {u ..... s i n θ . t ---------- 1--

2 g t2}

∆ t

= {u ..... s i n θ . (t + ∆ t) ---------- 1--

2 g (t

2 + 2 t ∆ t + ∆ t

2)} ---------- {u ..... s i n θ . t ---------- 1--

2 g t2}

∆ t

= u ..... s i n θ . ∆ t ---------- g t ∆ t ---------- 1--

2 g ∆ t

2

∆ t

δy

δt=

u ..... s i n θ . δ t ---------- g t δ t ---------- 1--2

g δ t2

δ t

δy

δt= u ..... s i n θ ---------- g t ---------- 1--

2 g δ t

dy =

LIMIT δy =

LIMIT

dt δt→ 0 δt δt→ 0{u ..... s i n θ ---------- g t ---------- 1--

2 g δ t }

y ’(t) = u ..... s i n θ ---------- g t (refer Preface iii & iv)

74

tTT///// 2 (0,0)

height y(t) = u ..... s i n θ ..... t ---------- 1--2

g t2

y ’(t) = u ..... s i n θ ---------- g t

When we evaluate this expression for the INSTANTANEOUS RATE OF CHANGE at

a par ticular instant in time we get a numerical value. In Calculus we also learn

how to interpret this value. We will be able to say things such as the height is

increasing, decreasing, is at a maximum or a minimum and so on.

If the duration of flight is T, then from the physics of the situation we know the

maximmaximmaximmaximmaximum um um um um height is reached when t = T///// 2. We also know that at the maximum

height the vertical speed should be zero. Substituting t = T///// 2 in the vertical speed

equation u ..... s i n θ ---------- g t = 0, we get T = 2u.u.u.u.u. sinθ/////g .

y (t )

t

distance = time

We may compare this with the AVERAGE SPEED. If we measure the height at any

chosen instant t , then in terms of the height function y(t) = u . s i n θ . t ---------- 1--2

g t2

the AVERAGE SPEED over time t = 0 to t is :

= u . s i n θ ---------- 1--2 g t

y( t ) = u . s i n θ . t ---------- 1--2

g t 2 is a polynomial of the for m a0 + a1 t + a2 t2

with coefficients a 0 = 0, a1 = u . s i n θ and a2 = ----------

1--2 g . We may directly

apply our formula for differentiating polynomials to get :

Vertical speed y ’(t) = dddddy

dddddt =

ddddd

dddddt (u ..... s i n θ . t ---------- 1--

2 g t 2

) = u ..... s i n θ ---------- g t

75


ttttt

n

y(t) = vy(t) = vy(t) = vy(t) = vy(t) = vererererer tical positiontical positiontical positiontical positiontical position

yyyyy’(t) = v(t) = v(t) = v(t) = v(t) = vererererer tical speedtical speedtical speedtical speedtical speed

Example 4:Example 4:Example 4:Example 4:Example 4: What is the instantaneous vinstantaneous vinstantaneous vinstantaneous vinstantaneous vererererer tical speedtical speedtical speedtical speedtical speed of the bouncing ball

(i) over (t0, t

1) ? (ii) over (t

1, t

2) ? (iii) at t

1 ?

Over [t0, t

1] the height y(t) is : y

0(t) = u

0.....sinθ0.....(t −−−−− t

0) −−−−− 1--2

g(t −−−−− t0)22222 .

Over [t1, t2] the height y(t) is : y1(t) = u1.....sinθ1.....(t −−−−− t1) −−−−− 1--2 g(t −−−−− t1)

22222 .

(i) Over (t0, t

1) the instantaneous vinstantaneous vinstantaneous vinstantaneous vinstantaneous vererererer tical speedtical speedtical speedtical speedtical speed y

0’(t) = u

0.....sinθ0 −−−−− g(t −−−−− t

0).

(ii) Over (t1, t2) the instantaneous vinstantaneous vinstantaneous vinstantaneous vinstantaneous vererererer tical speedtical speedtical speedtical speedtical speed y1’(t) = u1.....sinθ1 −−−−− g(t −−−−− t1).

(iii) At t1 we have two instantaneous vinstantaneous vinstantaneous vinstantaneous vinstantaneous vererererer tical speedstical speedstical speedstical speedstical speeds : the speed on impact at t

1

while descending descending descending descending descending and the speed after impact at t1 while risingrisingrisingrisingrising. We know fromthe previous example the duration of flight over [t0, t1] is T0 = 2u0.....sinθ/////g.To find the speed on impact at t

1 while descending descending descending descending descending we substitute t

0 = 0 and t = T

0

in y0’(t) = u0.....sinθ0 −−−−− g(t −−−−− t0) to get y0’(t−−−−−1 ) = −−−−−u0.....sinθ0 .

After impact at t1 we know the speed on rising rising rising rising rising is y

1’(t+

1) = +u

1.....sinθ1. Alternatively,

we may substitute t = t1 in y1’(t) = u1.....sinθ1 −−−−− g(t −−−−− t1) to get the same result.

76

ExampleExampleExampleExampleExample 5 5 5 5 5: Let us differentiate differentiate differentiate differentiate differentiate f(x) = sin (x)

1.1.1.1.1. AAAAAVERAVERAVERAVERAVERAGE steGE steGE steGE steGE step:p:p:p:p:

2. TENDS TO step:2. TENDS TO step:2. TENDS TO step:2. TENDS TO step:2. TENDS TO step:

When x is infinitely small, say δx , then as we said earlier, we may let :

sin ( δx) = δx and cos ( δx) = 1

Here we may simplify since δx ≠ 0 :

3. LIMIT step:3. LIMIT step:3. LIMIT step:3. LIMIT step:3. LIMIT step: Finally we take the LIMIT as δx → 0 .

f’(x) = cos(x)

∆ f(x) =

f(x + ∆ x) −−−−− f(x) =

f(x + ∆ x) −−−−− f(x) ∆ x (x + ∆ x) −−−−− x ∆ x

=

sin(x + ∆ x) −−−−− sin(x) ∆ x

=

{sin(x) . cos(∆ x) + cos (x) . sin (∆ x)} −−−−− sin(x)∆ x

δ f(x) =

{sin(x) . cos(δ x) + cos (x) . sin (δ x)} −−−−− sin(x) δ x δ x

δ f(x) =

{sin(x) . 1+ cos (x) . δ x} −−−−− sin(x) δ x δ x

= cos (x) . δ x

δ x

δ f(x) = cos(x)

δ x

{ cos (x)}df(x) =

LIMIT δf =

LIMIT dx δx→ 0 δx δx→ 0

77

ExampleExampleExampleExampleExample 6 6 6 6 6: Let us now difdifdifdifdifffffferererererentiaentiaentiaentiaentiate te te te te the function y = sin (θ) the Vedic way.

Let us now magnify the two diagrams and superimpose them.

With the concept of arc length = r θ , we can see from the diagram that :

x2 (θ2 −−−−− θ1) ≤ y2 −−−−− y1 ≤ x1 (θ2 −−−−− θ1)

Since (θ2 −−−−− θ1) ≠ 0 we may divide throughout by (θ2 −−−−− θ1) .

AVERAGE RATE OF CHANGE of y with respect to θ : x2 ≤ y2 −−−−− y1

≤ x1 θ2 −−−−− θ1

y

−−−−−1 +1 x

+1

−−−−−1

x 22222

θ 2 2 2 2 2 −−−−− θ

11111

x 11111

)y11111 = = = = = sin θ11111

y22222 = = = = = sin θ22222

} y2 −−−−− y1 = sin θ2 −−−−− sin θ1

θ22222

y

−−−−−1 +1 x

+1

−−−−−1

x 22222

y22222 = = = = = sin θ22222

θ11111

y

−−−−−1 +1 x

+1

−−−−−1

x 11111

y11111 = = = = = sin θ11111

78

Now we may take the LIMIT as θ2 → θ1 to get the INSTANTANEOUS RATE OFCHANGE of y with respect to θ :

As θ2 → θ1 we can see that x2 → x1 :

x1 ≤ dy---dθ

at θ1 ≤ x1

So : dy---dθ

at θ1 = d---dθ

(sin θ) at θ1 = x1 = cos θ at θ1.

We may drop the subscript of particular angle θ1 to generalise and get :

d---dθ

(sin θ) = cos θ

There is no need to DIFFERENTIATE a function from FIRST PRINCIPLES each andevery time. We may differentiate the functions that we encounter most frequentlyand build a table. Then all we have to do is look up the TABLE OF DERIVATIVES.

L imit x2 ≤ L imit y2 −−−−− y1

≤ L imit x1θ2 → θ1 θ2 → θ1 θ2 −−−−− θ1 θ2 → θ1

79

16.16.16.16.16. TTTTTaaaaabbbbbles and Rles and Rles and Rles and Rles and Rulesulesulesulesules

A function f(x) expresses CHANGE. The CHANGE could be in position, velocity,acceleration, temperature, length, area, volume, pressure, charge on acapacitor, cur rent, voltage, resistance, energy, cost, population, time,probability, or any other quantity that CHANGES. The calculation to find thederideriderideriderivvvvvaaaaatititititivvvvveeeee f’(x) , the expression of the INSTANTANEOUS RATE OF CHANGE,from the function f(x), the expression of CHANGE, is almost mechanical.Below is a par tial TTTTTaaaaabbbbble ofle ofle ofle ofle of Deri Deri Deri Deri Derivvvvvaaaaatititititivvvvveseseseses of the more frequently encounteredfunctions in standard form.

function function function function function f(x)f(x)f(x)f(x)f(x) derivative derivative derivative derivative derivative fffff’(x)(x)(x)(x)(x)

xn (n is a rational number) n xn−−−−−1

ex ex

log (x) 1///// x

sin (x) cos (x)

cos (x) −sin (x)

tan (x) sec2 (x)

cosec (x) −−−−−cosec (x)..... cot (x)

sec (x) sec (x) ..... tan (x)

cot (x) −−−−−cosec2 (x)

1sin −1(x) √1 −−−−− x 2

1cos −1(x) √1 −−−−− x 2

1tan −1(x) √1+x 2

−−−−−

80

We may difdifdifdifdifffffferererererentiaentiaentiaentiaentiate te te te te various combinations of functions using the following rules.Let u(x) and v(x) be functions of x, the independent variable.

functionfunctionfunctionfunctionfunction derivativederivativederivativederivativederivative

C (constant) 0

C ..... u C du---------------dx

u + vdu---------------dx

+ dv---------------dx

u − vdu---------------dx

−−−−− dv---------------dx

u.v udv---------------dx

+ vdu---------------dx

u/////v udv---------------dx

−−−−− vdu---------------dx

v2

Chain rChain rChain rChain rChain r uleuleuleuleule

Let u(v) be a function of v and v(x) be a function of x, the independent variable.

Higher order derivativesHigher order derivativesHigher order derivativesHigher order derivativesHigher order derivatives

If y(x) is a function of x , then y’(x) = dy---------------dx

is also a function of x. If y’(x) is a

well-behaved well-behaved well-behaved well-behaved well-behaved function, then the derivative derivative derivative derivative derivative of y’(x) is :

y” = dy’---------------dx

= d---------------dx

(dy---------------dx) =

d2y--------------------dx2 .

y” is the second derivativesecond derivativesecond derivativesecond derivativesecond derivative of y(x) with respect to x. This is known as successivesuccessivesuccessivesuccessivesuccessivedifferentiationdifferentiationdifferentiationdifferentiationdifferentiation. In general :

dddddu(v) d d d d du(v) d d d d dv(x) dddddx = d d d d dv

. . . . . dddddx

dny--------------------dxn = d---------------

dx ( )dn−−−−−1y

dxn−−−−−1

81

Example : Example : Example : Example : Example : y(t) is the function that expresses CHANGE in height.y’(t) is the INSTANTANEOUS RATE OF CHANGE in height or ver tical speed.

We can think of y’(t) as the function that expresses CHANGE in ver ticalspeed and y”(t) the INSTANTANEOUS RATE OF CHANGE in ver tical speed,i.e. ver tical acceleration, which in our case is g for gravity.

Vertical acceleration : y ”(t) = (u ..... s i n θ . t ---------- 1--2 g t 2 ) = ---------- g

As with any calculation there are two par ts: the operoperoperoperoperaaaaationtiontiontiontion par t and theunits ofunits ofunits ofunits ofunits of measur measur measur measur measureeeee par t. Here the height is measured in [meters] and time tttttis measured in [secs]:

yyyyy = vertical position or height [meter[meter[meter[meter[meters]s]s]s]s]

yyyyy ’ = speed [meter[meter[meter[meter[metersssss ///// sec sec sec sec sec]]]]]

yyyyy” = acceleration [meter[meter[meter[meter[metersssss ///// se se se se seccccc 22222 ]]]]]

ddddd22222y ddddd22222

dddddt = dddddt2

tTT///// 2 (0,0)

height y(t) = u ..... s i n θ ..... t ---------- 1--2 g t 2

y ’(t) = u ..... s i n θ ---------- g t

---------- g = gravity y ”(t) = ---------- g(the minus sign indicates the downward direction of gravity)

82

DeriDeriDeriDeriDerivvvvvaaaaatititititivvvvve ofe ofe ofe ofe of the in the in the in the in the invvvvverererererse functionse functionse functionse functionse functionLet y = x2 . Then the inininininvvvvverererererse functionse functionse functionse functionse function is : x = y½. We may now differentiate theinininininvvvvverererererse functionse functionse functionse functionse function x with respect to y to get dx/////dy = 1/////2y½. Alternatively, we knowthat dy/////dx = 2x. Hence:

Now we may substitute x = y½ to get the derivative of the inininininvvvvverererererse functionse functionse functionse functionse function x withrespect to y : dx/////dy = 1/////2y½ . So the derivative of the inininininvvvvverererererse functionse functionse functionse functionse function x is :

DifDifDifDifDif ffffferererererentiaentiaentiaentiaentiation in partion in partion in partion in partion in parametric fametric fametric fametric fametric fororororormmmmmWe have now learned to find the INSTANTANEOUS RATE OF CHANGE of afunction. The reader should ask the question: what about the INSTANTANEOUSRATE OF CHANGE of one function with respect to another function of the sameparameter ?

A plane flying horizondally at a speed of 600 kms / hour descends at the rate of1200 feet / minute. What is the glide ratioglide ratioglide ratioglide ratioglide ratio ?

horizontal speed speed speed speed speed u’(t) = 600 kms / hour ≅ 500 feet / sec

downward speed speed speed speed speed v’(t) = 1200 feet / minute = 20 feet / sec

That is to say, for every 25 feet forward the plane looses height (descends) by 1 foot.Notice the absence of the time time time time time units of measure. If we know the present altitude wecan tell how far the plane will glide before it touches the surface.

Here the speed speed speed speed speed functions u’(t) and v’(t) are given. How will we find the glide ratioglide ratioglide ratioglide ratioglide ratiogiven only the horizontal positionpositionpositionpositionposition function u(t) and vertical positionpositionpositionpositionposition function v(t) ?

1 dx 1dy/////dx = dy = 2x

dx 1dy

= dy/////dx

glide ratioglide ratioglide ratioglide ratioglide ratio = horizontal speed speed speed speed speed u’(t)

= 500 feet / sec

= 25 downward speed speed speed speed speed v’(t) 20 feet / sec

83

The glide ratioglide ratioglide ratioglide ratioglide ratio does have to be constant. Using our example of flight path ofthe ball one could ask: what is the INSTANTANEOUS RATE OF CHANGE ofRANGE with respect to HEIGHT, both functions of the same parameter t.

HEIGHT = y(t) = u.....sin θ ..... t ---------- 1--2 g t2

RANGE = x(t) = u.....cos θ ..... t

We have Le HOSPITAL’S RULE :

ExExExExExerererererceise 1ceise 1ceise 1ceise 1ceise 1: Differentiate term by term the following:

a) sin (x) = x ---------- x3---3!

+ x5

---5! ---------- x

7---7!

+ . . .. . .. . .. . .. . .

b) cos (x) = 1 ---------- x2

---2! + x

4---4!

---------- x6

---6! + . . .. . .. . .. . .. . .

c) ex = 1 + x---1! + x

2---2!

+ x3

---3! + . . .. . .. . .. . .. . .

ExExExExExererererercise 2cise 2cise 2cise 2cise 2: Differentiate term by term sin (ax), cos (ax), and eax .ExExExExExererererercise 3cise 3cise 3cise 3cise 3: Differentiate sin (ax), cos (ax), and eax using the ccccchain rhain rhain rhain rhain ruleuleuleuleule.

It is fundamental for the student to understand how to find the INSTANTANEOUS RATEOF CHANGE. He should feel at ease finding the INSTANTANEOUS RATE OF CHANGE ofelementary functions where the wwwwwell-behaell-behaell-behaell-behaell-behavvvvvededededed proper ties SINGLE VALUED,CONTINUOUS and DIFFERENTIABLE are easily satisfied.

du(x) =

u ,(x)

dv(x) v ,(x)

A f unc t i on f ( x ) i s an exp ress ion o f A f unc t i on f ( x ) i s an exp ress ion o f A f unc t i on f ( x ) i s an exp ress ion o f A f unc t i on f ( x ) i s an exp ress ion o f A f unc t i on f ( x ) i s an exp ress ion o f CHANGECHANGECHANGECHANGECHANGE .....G i ven f ( x ) t he exp ress ion o f G i ven f ( x ) t he exp ress ion o f G i ven f ( x ) t he exp ress ion o f G i ven f ( x ) t he exp ress ion o f G i ven f ( x ) t he exp ress ion o f CHANGECHANGECHANGECHANGECHANGE ,,,,,

wwwww e can D IFFERENT IAe can D IFFERENT IAe can D IFFERENT IAe can D IFFERENT IAe can D IFFERENT IATE f ( x ) and fTE f ( x ) and fTE f ( x ) and fTE f ( x ) and fTE f ( x ) and f i nd i nd i nd i nd i nd fffff ’( x ) t he( x ) t he( x ) t he( x ) t he( x ) t heeeeeexp rxp rxp rxp rxp r ess ion o fess ion o fess ion o fess ion o fess ion o f INST INST INST INST INSTANTANTANTANTANTANEOUS RAANEOUS RAANEOUS RAANEOUS RAANEOUS RATE OF CHANGE .TE OF CHANGE .TE OF CHANGE .TE OF CHANGE .TE OF CHANGE .

84

17. 17. 17. 17. 17. Units of MeasureUnits of MeasureUnits of MeasureUnits of MeasureUnits of MeasureA function may be dependent on more than one variable. For example f(x, y, z) maybe a function of three independent variables x, y and z along the three orthogonalx-axis, y-axis and z-axis respectively. The function f(x, y, z) may describe the positionof an object in 3-dimensional space.

With any calculation there are two par ts: the operoperoperoperoperaaaaationtiontiontiontion par t and the unitsunitsunitsunitsunitsofofofofof measur measur measur measur measureeeee par t. The difdifdifdifdifffffferererererentiaentiaentiaentiaentiation tion tion tion tion operation gives the EXPRESSION of theINSTANTANEOUS RATE OF CHANGE. The inteinteinteinteintegggggrrrrraaaaation tion tion tion tion operation gives us theEXPRESSION of CHANGE.

From the calculation calculation calculation calculation calculation point of view the symbol dxdxdxdxdx in the denominator of dfdfdfdfdf/////dxdxdxdxdxtells us with respect to which independent variable we differentiated the function*. Ifthis independent variable happens to have a units of measureunits of measureunits of measureunits of measureunits of measure associated with it,then dxdxdxdxdx has the same units of measure units of measure units of measure units of measure units of measure . For example, dtdtdtdtdt has the standard unitsof measure [second][second][second][second][second] along the time axis. Likewise, dddddttttt 22222 has [se[se[se[se[seccccc 22222]]]]] as units ofunits ofunits ofunits ofunits ofmeasure measure measure measure measure .

From the Analysis Analysis Analysis Analysis Analysis point of view the symbol dxdxdxdxdx denotes the calculation at an instantinstantinstantinstantinstantrather than over an interinterinterinterintervvvvval al al al al . Later in inteinteinteinteintegggggrrrrraaaaationtiontiontiontion we shall see the symbol dxdxdxdxdx ina similar role.

Let us look at some examples to understand the concept. Let sssss = positionor distance, vvvvv = speed, aaaaa = acceleration, with distance measured in [meters]and time ttttt measured in [secs].

* Note: In A LITTLE MORE CALCULUS we shall see how to differentiate a function ofmore than one independent variable. This is known as parparparparpar tial diftial diftial diftial diftial difffffferererererentiaentiaentiaentiaentiation tion tion tion tion .

SPEED = d d d d d POSITION

[meter [meter [meter [meter [meter ///// sec] sec] sec] sec] sec] dtdtdtdtdt

ACCELERATION = d d d d d SPEED

[meter [meter [meter [meter [meter ///// sec sec sec sec sec22222]]]]]dtdtdtdtdt

85

Area = ∫ length ..... ddddd breadth [meters]2 = ∫[meters] ..... ddddd[meters]

∫ FORCE . . . . . ddddds = ENERGY

= POWER

∫ POWER . . . . . ddddd t = ENERGY

This format of units ofunits ofunits ofunits ofunits of measur measur measur measur measureeeee calculation should serve as a template inany application.

units of measureunits of measureunits of measureunits of measureunits of measure

[meters / sec] = ddddd[meters]ddddd[sec]

[meters / sec2] = ddddd[meters / sec] = ddddd22222[meters] d d d d d[sec] ddddd[sec]22222

[meters / sec] = ∫ [meters / sec2] ..... d d d d d[sec]

[meters] = ∫ [meters / sec] ..... d d d d d[sec]

ddddd ENERGY ddddd t

operationoperationoperationoperationoperation

v = dddddsdddddt

a = dddddvdddddt = ddddd

2sdddddt2

v = ∫a ..... ddddd t

s = ∫v ..... dddddt

86

11111 88888. . . . . D I FFERENT IAB IL I TYD IFFERENT IAB IL I TYD IFFERENT IAB IL I TYD IFFERENT IAB IL I TYD IFFERENT IAB IL I TY

If you take a c loser look at how we took the LIMIT in the FIRST DERIVA-TIVE of xn, you wi l l notice that when we took

We were actually taking the limit from the right.

In other words, we were taking i.e. approaching the instant afrom the right.

We should ask: what would happen if we took i.e. approaching theinstant a from the left ?

this is

because x → a---------- in terms of δx is :

x → a---------- = a ---------- δx as δx → 0 and (a -- δx ) ---------- a = ----------δx .

Will we get the same INSTANTANEOUS RATE OF CHANGE ?

In general, for any function f(x) we expect:

INSTANTANEOUS RATE OF CHANGE =

INSTANTANEOUS RATE OF CHANGEapproaching a from the left approaching a from the right

If they are equal then we say that f(x) is DIFFERENTIABLE at x = a.


Limit f (x + h) -- f (x) or h → 0 h

Limit x → a+++++

Limit x → a----------

Limit f (x -- δx) -- f (x)δx → 0 --δx

Limit f (x -- h) -- f (x) or h → 0 --h

Limit f (x) ---------- f (a) = x → a---------- x ---------- a

Limit f (x) ---------- f (a) x → a+++++ x ---------- a

Limit f (a ---------- δx) ---------- f (a) =δx → 0 ----------δx

Limit f (a + δx) ---------- f (a)δx → 0 δx

87

Example 1Example 1Example 1Example 1Example 1: is f(x) differentiable differentiable differentiable differentiable differentiable at x = a ?

Approaching aaaaa from the left left left left left : f(x) = x

Near aaaaa and to the left left left left left of aaaaa : x = a −−−−− δx

Approaching aaaaa from the right right right right right : f(x) = 2a −−−−− x

Near aaaaa and to the right right right right right of aaaaa : x = a + δx

left derivative left derivative left derivative left derivative left derivative ≠ right derivativeright derivativeright derivativeright derivativeright derivative

we say : f(x) is not differentiable differentiable differentiable differentiable differentiable at x = a.

Limit f(x) −−−−− f(a) =

Limit f (a −−−−− δx) −−−−− f(a)

x → a−−−−− x − a δx → 0 (a − δx) − a

=

Limit (a −−−−− δx) −−−−− a

= Limit −−

−−− δx =

+1 δx → 0 − δx δx → 0 − δx

Limit f(x) −−−−− f(a) =

Limit f (a + δx) −−−−− f(a)

x → a+ x − a δx → 0 (a + δx) − a

= Limit { 2a −−−−−(a +δx)} −−−−− { 2a −−−−−a } =

Limit −−−−− δx

= −−−−−1 δx → 0 +δx δx → 0 + δx

Limit f(x) −−−−− f(a)x → a−−−−− x − a

Limit f(x) −−−−− f(a)x → a+ x − a

xxxxx

x for 0 x for 0 x for 0 x for 0 x for 0 ≤≤≤≤≤ xxxxx ≤≤≤≤≤ a a a a a

fffff(((((xxxxx) = ) = ) = ) = ) = { { { { { 2a 2a 2a 2a 2a ------ x for a - x for a - x for a - x for a - x for a ≤≤≤≤≤ xxxxx ≤≤≤≤≤ 22222aaaaa

aaaaa

aaaaa 22222 aaaaa

fffff (((((xxxxx)))))

(((((0,00,00,00,00,0)))))

88

Example 2Example 2Example 2Example 2Example 2 : is f(x) = |x| differentiable differentiable differentiable differentiable differentiable at x = 0 ?

Approaching 00000 from the left left left left left : f(x) = |x| = −−−−−x


Approaching 00000 from the right right right right right : f(x) = x

Near 00000 and to the right right right right right of 00000 : x = 0 + δx


we say : f(x) is not differentiable differentiable differentiable differentiable differentiable at x = 0.

Limit f(x) −−−−− f(0) =

Limit f(0 −−−−− δx) −−−−− f(0)

x → 0−−−−− x − 0 δx → 0 (0 − δx) − 0

=

Limit −−−−−(0 −−−−− δx) −−−−− 0

= Limit

+ δx =

Limit (−−−−−1) =

−−−−−1 δx → 0 − δx δx → 0 − δx δx → 0

Limit f(x) −−−−− f(0) =

Limit f (0 + δx) −−−−− f(0)x → 0+ x − 0

δx → 0 (0 + δx) − 0

Limit f(x) −−−−− f(0)x → 0−−−−− x − 0

Limit f(x) −−−−− f(0)x → 0+ x − 0

= Limit (0 +δx) −−−−− 0 =

Limit + δx =

Limit (+1) =

+1 δx → 0 +δx δx → 0 + δx δx → 0

(((((0 ,00 ,00 ,00 ,00 ,0 ))))) x x x x x

f(x) = |x|f(x) = |x|f(x) = |x|f(x) = |x|f(x) = |x|


89

Example 3:Example 3:Example 3:Example 3:Example 3: v(t) is the speed of a car weighing one ton that star ts from rest(0 kmph) and accelerates steadily due East (along the x-axis) for 10 secs andthen levels off at 90 kmph (25 m/sec). Is v(t) differentiable differentiable differentiable differentiable differentiable at t = 10 secs?

Approaching t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs from the left left left left left : v(t) = aaaaa.t = 2.5 t meter/sec2

Near t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs and to the left left left left left of t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs : t = 10 −−−−− δt

= 2.5 m/sec2

So the acceleration acceleration acceleration acceleration acceleration from the left or left derivativeleft derivativeleft derivativeleft derivativeleft derivative = 2.5 m/sec2 .

Limit v(t) −−−−− v(10) =

Limit {2.5 (10 −−−−− δt)} −−−−− 25 m/sec

t → 10−−−−− t − 10 δt → 0 (10 − δt) − 10

=

Limit (25 −−−−− 2.5δt) −−−−− 25 =

Limit −−−−− 2.5δt

= Limit

(2.5) δt → 0 − δt δt → 0 − δt δt → 0

55555 1 01 01 01 01 0 1 51 51 51 51 5 2 02 02 02 02 0 2 52 52 52 52 5(((((0,00,00,00,00,0))))) ttttt

3 03 03 03 03 0

55555

1 01 01 01 01 0

1 51 51 51 51 5

2 02 02 02 02 0

2 52 52 52 52 525 m/sec25 m/sec25 m/sec25 m/sec25 m/sec

SSSSSPPPPPEEEEEEEEEEDDDDD

[met

er[m

eter

[met

er[m

eter

[met

ers s s s s

//// / se

c]se

c]se

c]se

c]se

c]

T IMETIMETIMETIMETIME [sec] [sec] [sec] [sec] [sec]

v(t) =

a .a .a .a .a . t for t ≤ 10 secs where aaaaa is the constant acceleraton.

25 m/secm/secm/secm/secm/sec for t ≥ 10 secs.

v(t)v(t)v(t)v(t)v(t)

90

Approaching t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs from the rightrightrightrightright: v(t) = 25 meters/sec, constant.

Near t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs and to the right right right right right of t = 10 secst = 10 secst = 10 secst = 10 secst = 10 secs : t = 10 + δt

So the acceleration acceleration acceleration acceleration acceleration from the right or right derivativeright derivativeright derivativeright derivativeright derivative = 0 m/sec2, which is whatwe expect since the speed or v(t) = 25 m/sec for t ≥ 10 seconds is constant.


we say : v(t) is not differentiable differentiable differentiable differentiable differentiable at t = 10 secs.

Even though v(t) is not differentiable differentiable differentiable differentiable differentiable at t = 10 secs, we can still compute the leftleftleftleftleftderideriderideriderivvvvvaaaaatititititivvvvve e e e e and right deriright deriright deriright deriright derivvvvvaaaaatititititivvvvveeeee and interpret them according to the situation.

ExExExExExererererercise1:cise1:cise1:cise1:cise1: Given f ’(x) = AVERAGE RATE OF CHANGE of f(x), what is f(x) ?

a) exxxxx

b) |x|

c) constant

d) x

e) ax + b

Limit v(t) −−−−− v(10) =

Limit 25 −−−−− 25 m/sec

= Limit 0

= 0 m/sec2t → 10+++++ t − 10 δt → 0 (10 + δt) − 10 δt → 0 δt

Limit v(t) −−−−− v(10)t → 10−−−−− t − 10

Limit v(t) −−−−− v(10)t → 10+ t − 10

91

n

y(t)y(t)y(t)y(t)y(t)

hhhhh

eeeee

iiiii

ggggg

hhhhh

ttttt

ttttt00000 ttttt

11111 ttttt

22222 ttttt

33333

ttttt

SLOPE of tangent at t11111 from the left left left left left is negativenegativenegativenegativenegative

SLOPE of tangent at t11111 from the right right right right right is postivepostivepostivepostivepostive

Example 4:Example 4:Example 4:Example 4:Example 4: Is the function y(t), that describes the height of a bouncing ballbouncing ballbouncing ballbouncing ballbouncing ball,

differentiable differentiable differentiable differentiable differentiable at instants t0, t

1, t

2, t

3, . . . ?

At t1 we have two instantaneous vinstantaneous vinstantaneous vinstantaneous vinstantaneous vererererer tical speedstical speedstical speedstical speedstical speeds : the speed on impact at t

1 while

descendingdescendingdescendingdescendingdescending and the speed after impact at t1 while risingrisingrisingrisingrising. We know the duration of

flight T = 2u0.....sinθ/////g (see page 75). To find the speed on impact at t

1 while

descendingdescendingdescendingdescendingdescending we substitute T = 0 and t = T0 in y

0’(t) = u

0.....sinθ

0 −−−−− g(t −−−−− t

0)

to get y0’(t−−−−−

11111) = −−−−−u

0.....sinθ

0 .

After impact at t1 we know this speed on rising rising rising rising rising is y

1’(t+

1) = +u

1.....sinθ

1. Alternatively,

we may substitute t = t1 in y

1’(t) = u

1.....sinθ

1 −−−−− g(t −−−−− t

1) to get the same result.

GeometricallGeometricallGeometricallGeometricallGeometricallyyyyy, we can see that we have two tangents at t11111.

left derivative left derivative left derivative left derivative left derivative y0’(t−−−−−

11111) = −−−−−u

0.....sinθ

0 ≠ y

1’(t+

1) = +u

1.....sinθ

1 right derivativeright derivativeright derivativeright derivativeright derivative

we say : y(t) is not differentiable differentiable differentiable differentiable differentiable at t = t11111 secs.

Even though y(t) is not differentiable differentiable differentiable differentiable differentiable at t = t0, t

1, t

2, t

3, . . . , we can still

compute the left derileft derileft derileft derileft derivvvvvaaaaatititititivvvvve e e e e and right deriright deriright deriright deriright derivvvvvaaaaatititititivvvvveeeee and interpret them according to

the situation.

92

f(x) = +√ x

x(0, 0)

f(x)Example 5Example 5Example 5Example 5Example 5 : is f(x) = + √x differentiable differentiable differentiable differentiable differentiable at x = 0 ?

We know that f(x) = + √x is not defined for x < 0 and hence f(x) is not contincontincontincontincontinuousuousuousuousuousfor x < 0. So we may straight away say that f(x) is not differentiable differentiable differentiable differentiable differentiable at x = 0.Another way to think about this is : since f(x) is not defined for x < 0 then certainlyf’(x) is not defined for x ≤ 0.

Let us make f(x) continuous continuous continuous continuous continuous at x = 0 by defining f(x) = 0 for x < 0. Now :

We may now apply the method of finding the derivative from fffffiririririrst principlesst principlesst principlesst principlesst principles as wehave been doing so far. Alternatively, we may mechanically apply the formula to findthe derivative of xn when n is rational. Either way we get :

At x = 0 : the right deriright deriright deriright deriright derivvvvvaaaaatititititivvvvveeeee f’(x) = 1///// 2√x is not defined. So at x = 0 :


We say : f(x) is not dif dif dif dif difffffferererererentiaentiaentiaentiaentiabbbbble le le le le at x = 0.

When finding the derivatives of functions with rational exponentsrational exponentsrational exponentsrational exponentsrational exponents, we must becautious and check to see if f(x) and f’(x) are defined at x = 0 and x < 0.

f(x) = { + √x for x ≥ 0 0, a constant for x < 0

f’(x) = { 1///// 2√x for x ≥ 0 is the derideriderideriderivvvvvaaaaatititititivvvvve e e e e from the rightrightrightrightright 0 for x < 0 is the derivative derivative derivative derivative derivative from the leftleftleftleftleft

93

ExExExExExerererererciscisciscisciseeeee ::::: A car weighing one ton and s tar t ing f rom r est (0 kmph)accelerates steadily due EAST (along the x-axis) for 10 secs and levelsoff at 90 kmph (25 m/sec). Its position function x(t) and the graph is shownbelow.

What is the speed x’(t) over speed over [0, 10] ?

What is the speed x’(t) over speed over t ≥ 10 ?

Is x’(t) DIFFERENTIABLE at t = 10 secs ?

Draw the speed graph x’(t) over [0, 25].

What is the acceleration x”(t) over [0, 10] ?

What is the acceleration x”(t) over t ≥ 10 ?

Is x”(t) DIFFERENTIABLE at t = 10 secs ?

Draw the acceleration graph x”(t) over [0, 25].

( Hint : see example 3 page 89 and example 2 page 181) .

0 10 20 [secs] t

x(t) = 2.5t2/////2 over [0, 10]

25t − 125 for t ≥ 10{

375

250

125

[met

ers]

94

Before finding the INSTANTANEOUS RATE OF CHANGE of a function thewwwww e l l - behae l l - behae l l - behae l l - behae l l - beha vvvvv ededededed p r ope r t i e s S INGLE VALUED, CONT INUOUS andDIFFERENTIABLE must be satisfied. The first par t of the study of Calculusdeals with this.

VEDIC mathematicians understood the need for finding the INSTANTANEOUSRATE OF CHANGE known in Sanskrit as Chal-na KChal-na KChal-na KChal-na KChal-na Kal-naal-naal-naal-naal-na. This is illustrated in theproof to find the INSTANTANEOUS RATE OF CHANGE OF xn at x = a.

The genius of this approach is in its simplicity. First DIVIDE to find the rrrrraaaaate ofte ofte ofte ofte ofccccchanghanghanghanghangeeeee then SUBSTITUTE for the instantinstantinstantinstantinstant. This method works almost always,especially if we note that almost all functions can be approximated by finitepolynomials or infinite polynomial series.

PPPPPolololololynomials arynomials arynomials arynomials arynomials are to functions we to functions we to functions we to functions we to functions whahahahahat rt rt rt rt raaaaationals artionals artionals artionals artionals are to re to re to re to re to real neal neal neal neal numberumberumberumberumbersssss.....

Without elaborating by analysis the concept, they developed formulas orsutrassutrassutrassutrassutras: Gunaka-samuccaya-sutrasGunaka-samuccaya-sutrasGunaka-samuccaya-sutrasGunaka-samuccaya-sutrasGunaka-samuccaya-sutras for differentiating various functions.

The methods were encoded in mnemonics (mental one-liners) known asslokasslokasslokasslokasslokas or stotrasstotrasstotrasstotrasstotras (aphorisms). The more difficult functions could bedecomposed using the PPPPParararararaaaaavvvvvararararar tytytytytya Sutra Sutra Sutra Sutra Sutraaaaa.

It was up to the genius of the western mind, the great Newton and Leibniz, todevelop the concepts of INFINITESIMAL and LIMIT. These concepts are illustratedin the proof to find the INSTANTANEOUS RATE OF CHANGE of xn.

It is only with these concepts of INFINITESIMAL and LIMIT that we candevelop the concepts of CONTINUOUS and DIFFERENTIABLE. While theconcept of CONTINUOUS can be intuitively understood the concept ofDIFFERENTIABLE is not so obvious. The reader should take hear t thateven the great Cauchy, master of rigorous analysis, failed to grasp thedifference. Put another way :

f(x)f(x)f(x)f(x)f(x) is DIFFERENTIABLE i f is DIFFERENTIABLE i f is DIFFERENTIABLE i f is DIFFERENTIABLE i f is DIFFERENTIABLE i f and only if fand only if fand only if fand only if fand only if f’(x)(x)(x)(x)(x) is CONTINUOUS. is CONTINUOUS. is CONTINUOUS. is CONTINUOUS. is CONTINUOUS.

95

PPPPPararararar t 3 :t 3 :t 3 :t 3 :t 3 : ANANANANANALALALALALYTICAL GEOMETRYTICAL GEOMETRYTICAL GEOMETRYTICAL GEOMETRYTICAL GEOMETRYYYYY

96


We are familiar with the concept of the tangtangtangtangtangent ent ent ent ent to a curve: a straight line that

touches the curve at a point. As a rule, for WELL-BEHAVED functions, to each point on

the graph or curve there is exactly one tangtangtangtangtangent ent ent ent ent . However, there are exceptions to

this rule. We are also familiar with the sign of the trigonometric functions in the

4 quadrants with the angle θ measured in the counter-clockwise (positive) direction.

sin allsin allsin allsin allsin all

tan costan costan costan costan cos

Let y(x) be a function and let z1(x) be the tangtangtangtangtangent ent ent ent ent to the curve of y(x) at the

point x1. Let θ1 be the angle that the tangent tangent tangent tangent tangent z1(x) makes with the x-axis. From

the sign of tan tan tan tan tan θθθθθ11111 we can gain more information about y(x) at x1. We can tell if y(x)

is increasing, decreasing, at an extremum extremum extremum extremum extremum (maximum or minimum), or changing

direction (flexing) at the point x1.

From the AnalAnalAnalAnalAnalysisysisysisysisysis point of view, since the tangtangtangtangtangent ent ent ent ent z1(x) is a straight line, its

equation must be of the form z1(x) = m1x + c1 , where m1 = tan θ1 is the

SLOPE of the tangent tangent tangent tangent tangent .

So, if we know θ1, then we may say that: z1(x) = tan tan tan tan tan θθθθθ11111 . x + c1.

With a little AnalAnalAnalAnalAnalysisysisysisysisysis we can show that FIRST DERIVATIVE y’(x1) = tan tan tan tan tan θθθθθ11111.

So instead of geometrically geometrically geometrically geometrically geometrically :

1. drawing the graph of y(x) .

2. drawing the tangtangtangtangtangent ent ent ent ent z1(x) to y(x) at some particular point y(x1) .

3. measuring the angle θ1 of the tangent tangent tangent tangent tangent z1(x) at y(x1) .

4. computing tan tan tan tan tan θθθθθ11111

to tell the behaviour of y(x) at x1 , we may straight away use the FIRST DERIVATIVE

y’(x1) and analyse the behaviour of y(x) at x1.

97

We may apply this method of AnalAnalAnalAnalAnalysisysisysisysisysis to two intersecting curves and analytically

compute the angle between two intersecting curves.

The most useful result of this part is to be able to use the FIRST DERIVATIVE to find

the points at which y(x) is a maximum or minimum.

Another imporAnother imporAnother imporAnother imporAnother impor tant insight is:tant insight is:tant insight is:tant insight is:tant insight is:

if we know the tangtangtangtangtangentsentsentsentsents z1(x), z2(x), z3(x), . . . , zn (x), (first derivatives of y(x))

at sufficiently many points x1, x2, x3, . . . , xn , we may reconstruct the curve y(x).

From the AnalAnalAnalAnalAnalysis ysis ysis ysis ysis point of view, this insight tells us that we should be able to

calculate y(x) from y’(x). This inininininvvvvverererererse operse operse operse operse operaaaaationtiontiontiontion to find y(x) from its derideriderideriderivvvvvaaaaatititititivvvvveeeee

y’(x) is known as integrationintegrationintegrationintegrationintegration.

The reader may skip this part on first reading and go directly to Part 4 without loss of

continuity.

98

Limit δyδx → 0

--------------------δx

1111199999..... FIRST DERIV FIRST DERIV FIRST DERIV FIRST DERIV FIRST DERIVAAAAATIVE TIVE TIVE TIVE TIVE ===== Slope Slope Slope Slope Slope of of of of of TTTTTangangangangangententententent

We obser ve that the line z(x) thru P and Q makes an angle θ with the x-axis.

We also observe that : tan θ = δy--------------------δx

What is the LimitLimitLimitLimitLimit of the line z(x) as Q → P ?

The tangtangtangtangtangententententent to the curve of the function y(x) at the point P or at x = x1.

Let us call it : z1(x) = m1x + c1 , where m1 is the SLOPE .

Note that at P :

This tangtangtangtangtangententententent to the curve of the function y(x) at the point P intersects the x-axisat an angle. Let us call this angle θ

1 .

So, when we take the LimitLimitLimitLimitLimit as Q → P the line z(x) becomes the tangenttangenttangenttangenttangent to thecurve of y(x) at the point P. So : z1(x) = tan θ1 . x + c1 .

This is equivalent to saying: at the point P = tan θ1

= SLOPE of TANGENT z1(x) to the curve of y(x) at x = x1.

z1(x1) = y(x1)

xxxxx11111

δδδδδxxxxx

δδδδδyyyyy

xxxxx11111 + + + + + δδδδδxxxxx(((((00000,,,,, 00000)))))

yyyyy

θθθθθ

PPPPP

xxxxx θθθθθ11111

PPPPP

yyyyy

xxxxxz 1

(x) = y’(x 1

) ..... x + c 1tangent

(((((0,00,00,00,00,0))))) xxxxx11111

QQQQQ z(x)

z(x)

z(x)

z(x)

z(x)

99

From the Analysis Analysis Analysis Analysis Analysis point of view :δy--------------------δx is the AVERAGE RATE OF CHANGE of the function y(x) over the

small inter val δx.

y ,(x) = is the INSTANTANEOUS RATE OF CHANGE.

y ,(x1) = FIRST DERIVATIVE of the function y(x) at x = x1

= GRADIENT or SLOPE of the TANGENT z1(x)to the curve of y(x) at x = x1.

Hence :

On closer observation of the diagrams, as δx → 0 it would appear that δy alsoTENDS TO zero. Hence:

should = 0///// 0 .

What we can depict in a diagram has its limitations. So let us look at what is happeningfrom an AnalAnalAnalAnalAnalysis ysis ysis ysis ysis point of view.

Let y(x) = x3

What is δy ? (x + δx)3 −−−−− x3

= { x3 + 3x2 δx + 3x δx2 + δx3 } −−−−− x3

= 3x2 δx + 3x δx2 + δx3

So :

= 3x2 + 3x δx + δx2

We may simplify because δx ≠ 0 . It only TENDS TO zero as P → Q . Now we takethe LIMIT as P → Q .

y ,(x1) = tan θ

1

Limit δyδx → 0

--------------------δx

Limit δyδx → 0

--------------------δx

δy =

3x2 δx + 3x δx2 + δx3

δx δx

Limit δy = = { 3x2 + 3x δx + δx2 } = 3x2

P → Q --------------------δx Limit δyδx → 0

--------------------δx Limitδx → 0

Limit δyP → Q --------------------δx

=

100

I t i s imporI t i s imporI t i s imporI t i s imporI t i s impor tant to note thatant to note thatant to note thatant to note thatant to note tha t in no wt in no wt in no wt in no wt in no w aaaaay wy wy wy wy we are are are are ar e sae sae sae sae say ing the F IRSTy ing the F IRSTy ing the F IRSTy ing the F IRSTy ing the F IRSTDERIVDERIVDERIVDERIVDERIVAAAAATIVE is alTIVE is alTIVE is alTIVE is alTIVE is alwwwwwaaaaays a strys a strys a strys a strys a straight l ine function ofaight l ine function ofaight l ine function ofaight l ine function ofaight l ine function of the f the f the f the f the forororororm m m m m mmmmmx + x + x + x + x + ccccc.....

Example:Example:Example:Example:Example: Let y(x) = x3. Then y ,(x) = 3x2 is not a straight line function.

Draw the curve of the function y(x) = x3. Let P be a point on the curve of thefunction x3. Let xxxxx 11111

be the x co-ordinate of the point P. The value of the FIRSTDERIVATIVE at x = xxxxx 11111

of the function x3 is:

y ,(xxxxx 11111

) = 3xxxxx 111112.

This numerical value 3xxxxx 111112 = tan θ

11111, where θ

11111 is the angle that the tangenttangenttangenttangenttangent

at the point P on the curve of the function x3 makes with the x-axis.

Draw the tangtangtangtangtangententententent z11111(x) to the curve at the point y(x

11111 = 1) = 13 = 1.

Measure the angle θ11111 that the tangenttangenttangenttangenttangent z

11111(x) makes with the x-axis.

It should be just about 72 . And tan θ11111 = tan 72 = 3.078.

At the point on the curve x3 = 1 the x co-ordinate is x11111 = 1.

Evaluate the FIRST DERIVATIVE of x3 at x11111 = 1.

tan 72tan 72tan 72tan 72tan 72 = y = y = y = y = y ,(x(x(x(x(x11111) = 3 = SL) = 3 = SL) = 3 = SL) = 3 = SL) = 3 = SLOPE ofOPE ofOPE ofOPE ofOPE of TTTTTANGENT to the curANGENT to the curANGENT to the curANGENT to the curANGENT to the curvvvvve ae ae ae ae at xt xt xt xt x11111 = 1. = 1. = 1. = 1. = 1.

Try this again at the point on the curve y(x22222 = 2) = 23 = 8.

Draw the tangtangtangtangtangententententent z22222(x) to the curve at this point.

Measure the angle θ22222 that the tangenttangenttangenttangenttangent z

22222(x) makes with the x-axis.

It should be a little more than 85 . And tan θ22222 = tan 85 = 12.

At the point on the curve x3 = 8 the x co-ordinate is x22222 = 2.

Evaluate the FIRST DERIVATIVE of x3 at x22222 = 2.

tan 85tan 85tan 85tan 85tan 85 = y = y = y = y = y ,(x(x(x(x(x

22222))))) ===== 12 12 12 12 12 ===== SLSLSLSLSLOPE ofOPE ofOPE ofOPE ofOPE of TTTTTANGENT to the curANGENT to the curANGENT to the curANGENT to the curANGENT to the curvvvvve ae ae ae ae at t t t t xxxxx22222 = 2. = 2. = 2. = 2. = 2.

101

At x = 1 the equation of tangenttangenttangenttangenttangent to y = x3 is : z1(x) = 3x −−−−− 2




Excercise 1Excercise 1Excercise 1Excercise 1Excercise 1 : Draw the graph of y = y = y = y = y = xxxxx33333 and complete the following table.

ziiiii (xiiiii) = y (xiiiii)

At x = xiiiii slope of tangentslope of tangentslope of tangentslope of tangentslope of tangent ziiiii (x) = y ’(xiiiii)

ziiiii (x) = y ’(xiiiii) ..... x + c iiiii is the tangenttangenttangenttangenttangent to y (x) at x = xiiiii

The constant c iiiii = ziiiii (xiiiii) −−−−− y ’(xiiiii) ..... xiiiii

In pointpointpointpointpoint-----slopeslopeslopeslopeslope form the linear equation of the tangtangtangtangtangententententent is :

ziiiii (x) −−−−− ziiiii (xiiiii) = y ’(xiiiii) ..... (x −−−−− xiiiii)

xi approx. θi tanθi y’(xi) y(xi) ziiiii (x) = y ’(xiiiii) ..... x + c iiiii

1 72ooooo 3.078 3 3 3x − 2

2 85ooooo 12 8 12x − 16

3 87ooooo 27 27 27x − 54

4 88ooooo 48 64 48x − 128

102

tTT///// 2 (0,0)

height y(t) = u ..... s i n θ ..... t ---------- 1--2 g t 2

y ’(t) = u ..... s i n θ ---------- g t

t3

( θ3

z3(t3) = y(t3)

y’(t3) = tan(θ3)

Exercise 2:Exercise 2:Exercise 2:Exercise 2:Exercise 2: Repeat exercise 1 with y(x) = 1/////2 x2.

What if tan θ = 0 ?, that is to say the tangent to the curve is parallel to thex-axis. We shall deal with this in the next few chapters.

Exercise 3:Exercise 3:Exercise 3:Exercise 3:Exercise 3: In our main example of the projectile depicted below with u = 2√2and θ = π/////4 draw the graph and complete the table as in exercise 1 forinstants t1 = T///// 4 , t2 = T/////2 , and t3 = 3T/////4 .

It does not make sense to talk about a tang tang tang tang tangent ent ent ent ent to a WELL-BEHAVED linear function.However, its SLOPE = FIRST DERIVATE = tan θ, when θ is the angangangangangle ofle ofle ofle ofle of inter inter inter inter intersectionsectionsectionsectionsectionthe linear function makes with the x-axis.

With the concept of the tangent tangent tangent tangent tangent in mind, the reader may wish to review the bouncingbouncingbouncingbouncingbouncingball ball ball ball ball example on page 91 and gain an AnalAnalAnalAnalAnalytical gytical gytical gytical gytical geometreometreometreometreometr yyyyy view ofdifdifdifdifdif ffffferererererentiaentiaentiaentiaentiabil itybil itybil itybil itybil ity.....

When a SINGLE-VALUED and CONTINUOUS function is NOT difdifdifdifdifffffferererererentiaentiaentiaentiaentiabile bile bile bile bile at aparticular point or instant, it may have more than one tangent at that instant. Inexample 1 on page 87, f(x) is SINGLE-VALUED and CONTINUOUS. But f(x) is NOTdifferentiabledifferentiabledifferentiabledifferentiabledifferentiable at x = a. At the co-ordinates ( a, f(a) ) we may draw infinitely manytangents.

103

tangent z 1(x) = y’(x 1

) ..... x + c 1

y(xy(xy(xy(xy(x11111)))))

yyyyy

(0, 0) (0, 0) (0, 0) (0, 0) (0, 0) xxxxx11111 x x x x x22222 x x x x x

y(xy(xy(xy(xy(x22222)))))

tang

ent

z 2(x)

= y’

(x 2) .....

x + c 2

We may relate the AnalAnalAnalAnalAnalysis ysis ysis ysis ysis view of difdifdifdifdifffffferererererentiaentiaentiaentiaentiabbbbble le le le le and the AnalAnalAnalAnalAnalytical gytical gytical gytical gytical geometreometreometreometreometryyyyyproperty of the tangtangtangtangtangent ent ent ent ent and say :

Suppose we know the tangents z1(x), z2(x), z3(x), . . . at x1, x2, x3, . . . respectively,can we find the curve y(x)?

We may consider all the tangents z i (x) to a par ticular cur ve y(x) as belongingto a single family. We say that the curve y(x) ENVELOPES this family z i (x) ofstraight lines. The curve y(x) is touched by all the straight lines of this family.

From the AnalAnalAnalAnalAnalytical gytical gytical gytical gytical geometreometreometreometreometryyyyy point of view we ask: if we know the tangtangtangtangtangentsentsentsentsentsat each point to a curve y(x) can we determine the cur ve ?

If we know the tangents z i (x) = y ’(x i) ..... x + ci for i = 1, 2, 3, . . ., then we can findy (xi) = z i (xi) for i = 1, 2, 3, . . . .

If we know the tangtangtangtangtangents ents ents ents ents z1(x), z2(x), z3(x), . . . , zn (x), (first derivatives of y(x)) atsufficiently many points x1, x2, x3, . . . , xn , we may reconstruct the curve y(x). Fromthe AnalAnalAnalAnalAnalysis ysis ysis ysis ysis point of view, this insight tells us that we should be able to calculatey(x) from y’(x). This inininininvvvvverererererse operse operse operse operse operaaaaationtiontiontiontion to find y(x) from its derideriderideriderivvvvvaaaaatititititivvvvve e e e e y’(x) isknown as integrationintegrationintegrationintegrationintegration.

f(x) is DIFFERENTIABLE at a ⇔ f(x) has a UNIQUE TANGENT at f(a).

104

2020202020..... AngAngAngAngAngle ofle ofle ofle ofle of Inter Inter Inter Inter IntersectionsectionsectionsectionsectionLet y1(x) = m1x + c1 and y2(x) = m2x + c2 be two linearfunctions.

The angangangangangle ofle ofle ofle ofle of inter inter inter inter intersectionsectionsectionsectionsection between y1(x) and y2(x) is φ . Usuallythe smaller angle is taken. Hence :

φ = minimumminimumminimumminimumminimum (|θ11111−−−−−θ22222|, 180o −−−−− |θ11111−−−−−θ22222|).

With our understanding that SLSLSLSLSLOPE = tanOPE = tanOPE = tanOPE = tanOPE = tanθθθθθ , we may say that :

m1 = tan θ11111 and m2 = tan θ22222 .

Now if y1(x) and y2(x) are ororororor thothothothothogggggonalonalonalonalonal (perperperperperpendicularpendicularpendicularpendicularpendicular to each other)then:

xxxxx 11111

yyyyy

θ22222xxxxx

yyyyy11111(x)(x)(x)(x)(x)

) φ

)) θ11111

yyyyy22222(x)(x)(x)(x)(x)

m1 ..... m2 = −−−−−1

105

Proof :Proof :Proof :Proof :Proof : tan θ11111 = sin θ11111 ///// cos θ11111

If y1 and y2 are ororororor thothothothothogggggonalonalonalonalonal then : θ22222 = 90o + θ11111

So : tan θ22222 =

=

Therefore m1 ..... m2 = tan θ11111 ..... tan θ22222 =

Hence :

From the AnalysisAnalysisAnalysisAnalysisAnalysis point of view :

y1’(x1) = SLOPE m1 = tan θ11111

y2’(x1) = SLOPE m2 = tan θ22222

Hence, without drawing any graphs, we may directly analyseanalyseanalyseanalyseanalyse and say :

if y1’(x1) ..... y2’(x1) = −−−−−1

then y1(x) and y2(x) must be ororororor thothothothothogggggonalonalonalonalonal (perperperperperpendicularpendicularpendicularpendicularpendicular to each other)at x1.

Also

sin (90o + θ11111) =

sin 90o ..... cosθ11111 + cos 90o ..... sinθ11111

cos (90o + θ11111) cos 90o ..... cosθ11111 −−−−− sin 90o ..... sinθ11111

cosθ11111

−−−−−sin θ11111 sinθ11111 . . . . .

cosθ11111 = −−−−−1

cosθ1 1 1 1 1 −−−−−sinθ11111

m2 = −−−−−1///// m1

y2’(x1) = −−−−−1///// y1’(x1)

106

xxxxx 11111

xxxxx

zzzzz11111(x)(x)(x)(x)(x)

) θ11111

yyyyy nnnnn11111(x)(x)(x)(x)(x) y(x)y(x)y(x)y(x)y(x)

EquaEquaEquaEquaEquations oftions oftions oftions oftions of the tang the tang the tang the tang the tangent and norent and norent and norent and norent and normalmalmalmalmal

Let y(x) be a general function with tangtangtangtangtangententententent z1(x) and nornornornornormalmalmalmalmal n1(x) at (x1, y(x1))

as depicted below.

From the AnalAnalAnalAnalAnalysisysisysisysisysis point of view the equations of the tangtangtangtangtangententententent and nornornornornormalmalmalmalmal are :

tangenttangenttangenttangenttangent z1(x) −−−−− y(x1) = y’(x1) (x −−−−− x1)

nornornornornormalmalmalmalmal n1(x) −−−−− y(x1) = −−−−−1///// y’(x1) (x −−−−− x1)

There are two special cases for the tangent tangent tangent tangent tangent .

1. If y’(x1) = 0 then the tangtangtangtangtangententententent at (x1, y(x1)) is PARALLEL to the x-axis.

So z1(x) = y(x1). Try to imagine what the picture of y(x) will look like

around x = x1 . Can we say that y(x1) is either a maximummaximummaximummaximummaximum or a mimimummimimummimimummimimummimimum ?

2. If y’(x1) = ∞ then the tangtangtangtangtangententententent at (x1, y(x1)) is PARALLEL to the y-axis.

So z1(x) = x1 .

Likewise, there are two special cases for the nornornornornormal mal mal mal mal .

1. If y’(x1) = 0 then the nornornornornormalmalmalmalmal at (x1, y(x1)) is PARALLEL to the y-axis.

So n1(x) = x1 .

2. If y’(x1) = ∞ then the nornornornornormalmalmalmalmal at (x1, y(x1)) is PARALLEL to the x-axis.

So n1(x) = y(x1) .

107

aaaaaxxxxx) θ

zzzzz fffff (x)(x)(x)(x)(x)zzzzzggggg(x)(x)(x)(x)(x) f(x)f(x)f(x)f(x)f(x)

g(x)g(x)g(x)g(x)g(x)

) 9090909090ooooo + + + + + θ

We may extend this thinking to find the angle of intersection between any two functionsf(x) and g(x).

We may draw the tangtangtangtangtangents ents ents ents ents to f(x) and g(x) at the point of intersection aaaaa . Let :

zzzzz fffff ( x ) = m(x) = m(x) = m(x) = m(x) = mfffff ..... x + cx + cx + cx + cx + c fffff and and and and and zzzzz ggggg(x) = m(x) = m(x) = m(x) = m(x) = m g g g g g ..... x + cx + cx + cx + cx + c ggggg

be the equations of these tangentstangentstangentstangentstangents . Then :

mmmmmfffff = tan = tan = tan = tan = tan θθθθθ fffff and m and m and m and m and mggggg = tan = tan = tan = tan = tan θθθθθ ggggg

So : φφφφφ fffff,,,,, g g g g g = minimminimminimminimminimumumumumum (|θ fffff −−−−− θggggg|, 180o −−−−− |θ fffff −−−−− θggggg|).

And, if f(x) and g(x) intersect at right angles at aaaaa, then : mmmmmfffff . . . . . m m m m mggggg = = = = = − − − − −11111

This is the AnalAnalAnalAnalAnalytical gytical gytical gytical gytical geometreometreometreometreometryyyyy point of view. From the AnalAnalAnalAnalAnalysisysisysisysisysis point of view, wenote that at the point of intersection a :

fffff’(a) = (a) = (a) = (a) = (a) = mmmmm fffff = tan = tan = tan = tan = tan θθθθθ fffff

ggggg’(a) = m(a) = m(a) = m(a) = m(a) = mggggg = tan = tan = tan = tan = tan θθθθθ ggggg

So, to know if the two curves f(x) and g(x) intersect at right angles at a , all we haveto do is verify if :

fffff’(a) . g(a) . g(a) . g(a) . g(a) . g’(a)(a)(a)(a)(a) = = = = = − − − − −11111

108

2121212121..... Incr Incr Incr Incr Increasingeasingeasingeasingeasing,,,,, MAXIMUM, MAXIMUM, MAXIMUM, MAXIMUM, MAXIMUM, Decr Decr Decr Decr Decreasingeasingeasingeasingeasing

M OM OM OM OM OT I VT I VT I VT I VT I VAAAAAT I O NT I O NT I O NT I O NT I O NYou are standing on the ground with a height measuring instrument. Yourfriend is projected into the air in a module equipped with only a ver ticalspeedometer. Using your height measuring instrument you can measurethe height of the module at any instant: so many meters above the ground.Your friend, by looking at the ver tical speedometer, can know his ver ticalspeed at any instant: climbing at so many meters per second or descendingat so many meters per second.

Without communicating with your friend and using only the information youhave - the height at any instant, how can you know what your friend knows -the ver tical speed of the module at any instant ? Can you tell when thever tical speed is zero ?

Vice versa, how can your friend, with only the information he has - his ver ticalspeed at any instant, find out what you know - his height at any instant ? Canyour friend tell when he has reached a maximum height and exactly what isthe maximum height ?

y(t) = u . s i n θ . t -- 1--2

g t 2

1-Dimension1-Dimension1-Dimension1-Dimension1-Dimension

(0, 0) T t

Hei

ght

y(t

)

y(t1)

h1

t1 T/////2

VERTICAL POSITIONVERTICAL POSITIONVERTICAL POSITIONVERTICAL POSITIONVERTICAL POSITION

θ

(0, 0)

u

Range x(t)

He

igh

t y

(t)

y 2-Dimension2-Dimension2-Dimension2-Dimension2-Dimension

x

POSITIONPOSITIONPOSITIONPOSITIONPOSITION

109

When an object is projected into the air with a given initial velocity u andangle of projection θ, there are two functions that describe its position.A VERTICAL function y(t) which describes its HEIGHT at any INSTANT and aHORIZONTAL function x(t) which describes its RANGE at any INSTANT.

We know from Dynamics that :

Range function x(t) = u. cos θ . t

Height function y(t) = u. s in θ . t ---------- 1----------2 g t2

GeometricallGeometricallGeometricallGeometricallGeometricallyyyyy, by looking at the picture that describes the flight path, wecan see that the height increases, reaches a maximum and then decreases.

AnalAnalAnalAnalAnalyticallyticallyticallyticallyticallyyyyy, by looking at the expression of the HEIGHT function y(t), howcan we tell when the height is increasing, decreasing and at a maximum ?

With a little thinking you can infer that when the height is maximum thever tical speed must be zero.

Conversely, your friend, by closely watching the ver tical speedometer, caninfer that when the ver tical speed is zero (while he is still in the air) he hasreached the maximum height.

AnalAnalAnalAnalAnalyticallyticallyticallyticallyticallyyyyy, can you and your friend tell at which instant in time themaximum height is reached ?

AnalAnalAnalAnalAnalyticallyticallyticallyticallyticallyyyyy, can your friend find out the maximum height ?

110

Height

h1h2

h3

h4

p1

p2p3

p4

= maximum

(0, 0)Time

t4t3 tnt1 t2t0

φ1 φ4

The expression for the INSTANTANEOUS RATE OF CHANGE in height is

d y--------------------dt

(t) = y’(t) = u . sin θ ---------- g t

We know from observation that the INSTANTANEOUS RATE OF CHANGE inheight or vvvvvererererer tical speed tical speed tical speed tical speed tical speed is not the same at different INSTANTS t1, t2, t3and t4 in time and this is reflected in the expression.

When we evaluate this expression for the INSTANTANEOUS RATE OF CHANGEin height at any par ticular instant tiiiii we get a numerical value = tan tan tan tan tan φi i i i i . Weshall now learn how to interpret the SIGN of this value.

We shall make use of the SIGN of the tantantantantan φiiiii in the 4 quadrants with the angle φiiiiimeasured in the counter-clockwise (positive) direction.

sin allsin allsin allsin allsin all

and the fact thattan costan costan costan costan cos

y ,(tiiiii ) = tan φiiiii

111

From the AnalAnalAnalAnalAnalytical gytical gytical gytical gytical geometreometreometreometreometryyyyy point of view, if we draw the tangtangtangtangtangent ent ent ent ent to thecurve at the point P1 it will intersect the x-----axis at an acute angle. Let uscall this angle φ1.

The sign of tan φ1 is > 0, positive.positive.positive.positive.positive.

The fffff iririririrst derist derist derist derist derivvvvvaaaaatititititivvvvveeeee of y(t) evaluated at time instant t1y’(t1) = tan φ1 = u. sin θ ---------- g t1 must be posititveposititveposititveposititveposititve.

Over the interval [t0, t3), from t0 upto but not including t3, we note that theheight y(t) is incrincrincrincrincreasingeasingeasingeasingeasing. We also note that the SIGN of the tangtangtangtangtangent ent ent ent ent orequivalently the value of the fffff iririririrst derist derist derist derist derivvvvvaaaaatititititivvvvveeeee y’(t) evaluated at any instantover [t0, t3) is positive.positive.positive.positive.positive.

At P3, where the object reaches its maximummaximummaximummaximummaximum height, the tangent tangent tangent tangent tangent to thecurve is PARALLEL to the x-----axis . Hence φ3 = 0 .

The fffff iririririrst derist derist derist derist derivvvvvaaaaatititititivvvvveeeee of y(t) evaluated at time instant t3 is

y’(t3) = tan φ3 = u. sin θ ---------- g t3 = 0

From the AnalAnalAnalAnalAnalytical gytical gytical gytical gytical geometreometreometreometreometryyyyy point of view, if we draw the tangtangtangtangtangent ent ent ent ent to thecurve at the point P4 it will intersect the x-----axis at an obtuse angle φ4 .

The sign of tan φ4 is < 0, negative.negative.negative.negative.negative.

The fffff iririririrst derist derist derist derist derivvvvvaaaaatititititivvvvveeeee of y(t) evaluated at time instant t4

y’(t4) = tan φ4 = u. sin θ ---------- g t4 must be negativenegativenegativenegativenegative.

Over the interval ( t3, tn ], from but not including t3 to tn, we note that theheight y(t) is decrdecrdecrdecrdecreasingeasingeasingeasingeasing. We also note that the SIGN of the tangtangtangtangtangent ent ent ent ent orequivalently the value of the fffff iririririrst derist derist derist derist derivvvvvaaaaatititititivvvvveeeee y’(t) evaluated at any instantover ( t3, tn ] is negativenegativenegativenegativenegative.

We can say : if y(t) is incrincrincrincrincreasingeasingeasingeasingeasing then y’(t) is positipositipositipositipositivvvvveeeee (>0).Conversely : if y’(t) is positipositipositipositipositivvvvveeeee then y(t) must be incrincrincrincrincreasingeasingeasingeasingeasing.....

112

In general, let f(x) be a function and aaaaa the point where f(x) is maximummaximummaximummaximummaximum.

At a maximummaximummaximummaximummaximum point aaaaa the value of the function is greatergreatergreatergreatergreater than the valuesaround it. Let δx be a small change in x.

Then f (aaaaa ---------- δx) < f(aaaaa) and f (aaaaa + δx) < f(aaaaa)

What can you say about f”(aaaaa) ?

ExercisExercisExercisExercisExercise e e e e 1:1:1:1:1: Given that you know the initial velocity uuuuu and the angle ofprojection θ, how long will it take to reach a maximummaximummaximummaximummaximum height ?

Exercise 2:Exercise 2:Exercise 2:Exercise 2:Exercise 2: What is the duration of flight ?

Exercise 3:Exercise 3:Exercise 3:Exercise 3:Exercise 3: What is the range ?

We can say : if y(t) is decrdecrdecrdecrdecreasingeasingeasingeasingeasing then y ’(t) is nenenenenegggggaaaaatititititivvvvveeeee (< 0).Conversely : if y’(t) is nenenenenegggggaaaaatititititivvvvveeeee then y(t) must be decrdecrdecrdecrdecreasingeasingeasingeasingeasing.....

At a maximummaximummaximummaximummaximum point f ’(x) isposi t iposi t iposi t iposi t iposi t i vvvvve befe befe befe befe beforororororeeeee,,,,, z z z z zerererererooooo,,,,, ne ne ne ne negggggaaaaat it it it it i vvvvve aftere aftere aftere aftere after.....

113

Consider the function y = x2

d y-------------------- dx = y’(x) = 2x

The tangtangtangtangtangententententent to the cur ve of y = x2 at P1 wil l inter sect the x-----axisat obtuse angle φ1.

The SIGN of tan φ1 is < 0, negative.negative.negative.negative.negative.

The fffff iririririrst derist derist derist derist derivvvvvaaaaatititititivvvvveeeee of y = x2 evaluated at instant x1

y’(x1) = tan φ1 = 2x1 must be negativenegativenegativenegativenegative.....

When x < 0 we note that the function y = x2 is decrdecrdecrdecrdecreasingeasingeasingeasingeasing..... We also notethat the SIGN of the tangtangtangtangtangententententent or equivalently the value of the fffff iririririrst derist derist derist derist derivvvvvaaaaatititititivvvvveeeeey’(x) evaluated at any instant x < 0 is negative.negative.negative.negative.negative.

x1 x2 = 0 x3

x

y

φ1 φ3

P3P1

P2

2222222222..... Dec r Dec r Dec r Dec r Dec r eas ingeas ingeas ingeas ingeas ing ,,,,, M IN IMUM, M IN IMUM, M IN IMUM, M IN IMUM, M IN IMUM, I nc r I nc r I nc r I nc r I nc r eas ingeas ingeas ingeas ingeas ing

114

At P2, where y = x2 is minimminimminimminimminimumumumumum, the tangtangtangtangtangent ent ent ent ent to the curve is PARALLEL tothe x-axis. Hence φ2 = 0 . So the fffffiririririr st derist derist derist derist derivvvvvaaaaatititititivvvvveeeee of y = x2 evaluated atinstant x2 is :

y’(x2) = tan φ2 = 2x2 = 0

The tangtangtangtangtangent ent ent ent ent to the cur ve of y = x2 at P3 wil l inter sect the x-axisat acute angle φ3.

The SIGN of tan φ3 is > O, positipositipositipositipositivvvvveeeee.

The fffffiririririrst derist derist derist derist derivvvvvaaaaatititititivvvvveeeee of y = x2 evaluated at instant x3

y’(x3) = tan φ3 = 2x3 must be positivepositivepositivepositivepositive.

When x > 0 we note that the function y = x2 is incrincrincrincrincreasingeasingeasingeasingeasing. We also notethat the SIGN of the tangent or equivalently the value of the fffffiririririr st derist derist derist derist derivvvvvaaaaatititititivvvvveeeeey’(x) evaluated at any instant x > 0 is positivepositivepositivepositivepositive.

Let f(x) be a function and bbbbb the point where f(x) is minimumminimumminimumminimumminimum.

At a m in imummin imummin imummin imummin imum po in t bbbbb the va lue o f the func t ion i s l esserlesserlesserlesserlesser thanthe values around it. Let δx be a small change in x.

Then f(bbbbb ---------- δx) > f(bbbbb) and f(bbbbb + δx) > f(bbbbb)

What can you say about f”(bbbbb) ?

We can say: if y(x) is decrdecrdecrdecrdecreasingeasingeasingeasingeasing then y’(x) is nenenenenegggggaaaaatititititivvvvveeeee (<0).Conversely : if y’(x) is nenenenenegggggaaaaatititititivvvvveeeee then y(x) must be decrdecrdecrdecrdecreasingeasingeasingeasingeasing.

We can say : if y(x) is incrincrincrincrincreasingeasingeasingeasingeasing then y’(x) is positipositipositipositipositivvvvveeeee (>0).Conversely : if y’(x) is positipositipositipositipositivvvvveeeee then y(x) is incrincrincrincrincreasingeasingeasingeasingeasing.

At a minimumminimumminimumminimumminimum point f ’(x) isnenenenenegggggaaaaat it it it it i vvvvve befe befe befe befe beforororororeeeee,,,,, z z z z zerererererooooo,,,,, posit i posit i posit i posit i posit i vvvvve aftere aftere aftere aftere after.

115

Exercise 1:Exercise 1:Exercise 1:Exercise 1:Exercise 1: Find an instant where f(x) = x2 + x has an extreme value(maximum or minimum).

Exercise 2:Exercise 2:Exercise 2:Exercise 2:Exercise 2: Find an instant where f(x) = ----------x2 + x has an extreme value(maximum or minimum).

ExExExExExererererercise 3:cise 3:cise 3:cise 3:cise 3: Draw the curves of the functions below and answer the following :

a) f(x) = x2 for x < 0 . The shape of the curve is concave upwardsand falling.

b) f(x) = ----------x2 for x > 0 . The shape of the curve is concave downwardsand falling.

c) f(x) = x2 for x > 0 . The shape of the curve is concave upwardsand rising.

d) f(x) = ----------x2 for x < 0 . The shape of the curve concave downwardsand rising.

i) What kind of angle does the tangent to the curve make ?(acute, obtuse) . What is the sign of the tangent ?

ii) What is the sign of the first derivative of the function ?(positive, negative)

iii) Is the function increasing or decreasing ?

Exercise Exercise Exercise Exercise Exercise 44444::::: Repeat exercise 3 for the functions below and describe the shape ofthe curve.

a) f(x) = x3 for x < 0 .

b) f(x) = x3 for x > 0 .

c) f(x) = ---------- x3 for x < 0 .

d) f(x) = ---------- x3 for x > 0 .

116

2222233333. MAXIMA and MINIMA. MAXIMA and MINIMA. MAXIMA and MINIMA. MAXIMA and MINIMA. MAXIMA and MINIMA

Given a function f(x) how can we find the instants where f(x) has maximummaximummaximummaximummaximumor minimum minimum minimum minimum minimum values ?

If the point aaaaa is an eeeeextrxtrxtrxtrxtremeemeemeemeeme then the first derivative must equal zero.

Solve the equation f’(aaaaa) = 0 to find a a a a a .

This information is necessary but not sufficient to know whether the point aaaaais a maximummaximummaximummaximummaximum point or a minimumminimumminimumminimumminimum point. There are three methods.

Method Method Method Method Method 11111 : : : : : We need to fur ther check around a a a a a if :

f (aaaaa ---------- δx) < f(aaaaa) and f (aaaaa+δx) < f(aaaaa)implying that aaaaa is a maximummaximummaximummaximummaximum point

orf (aaaaa ---------- δx) > f(aaaaa) and f (aaaaa+δx) > f(aaaaa)implying that aaaaa is a minimumminimumminimumminimumminimum point.

These calculations may be tedious.

If we know the graph of the function then we can determine whether aaaaa ismaximum maximum maximum maximum maximum or minimum minimum minimum minimum minimum. This GGGGGeometric eometric eometric eometric eometric method may not be practical.So let us try to find an AnalAnalAnalAnalAnalytical ytical ytical ytical ytical method.

We shall use the results :

if f(x) is increasingincreasingincreasingincreasingincreasing then f’(x) is positivepositivepositivepositivepositive.

if f’(x) is positivepositivepositivepositivepositive then f(x) is increasingincreasingincreasingincreasingincreasing.

if f(x) is decreasingdecreasingdecreasingdecreasingdecreasing then f’(x) is negativenegativenegativenegativenegative.

if f’(x) is negativenegativenegativenegativenegative then f(x) is decreasingdecreasingdecreasingdecreasingdecreasing.....

117

Method Method Method Method Method 22222 ::::: At the maximummaximummaximummaximummaximum point aaaaa: f(x) is increas ing beforeincreas ing beforeincreas ing beforeincreas ing beforeincreas ing before anddecrdecrdecrdecrdecr eas ing af tereas ing af tereas ing af tereas ing af tereas ing af ter.....

Correspondingly, f’(x) is positipositipositipositipositivvvvve befe befe befe befe beforororororeeeee and nenenenenegggggaaaaatititititivvvvve aftere aftere aftere aftere after.....

We need to check around a a a a a if :

f’(aaaaa ---------- δx) > 0 and f’(aaaaa + δx) < 0.

At the m in imummin imummin imummin imummin imum po in t aaaaa: f (x) i s dec reas ing be fo re decreas ing be fo re decreas ing be fo re decreas ing be fo re decreas ing be fo re andincrincrincrincrincr eas ing a f tereas ing a f tereas ing a f tereas ing a f tereas ing a f ter.....

Correspondingly, f’(x) is ne ne ne ne negggggaaaaatititititivvvvve befe befe befe befe beforororororeeeee and positipositipositipositipositivvvvve aftere aftere aftere aftere after.....

We need to check around a a a a a if :

f’(aaaaa ---------- δx) < 0 and f’(aaaaa + δx) > 0.

This again may be tedious.

Method 3:Method 3:Method 3:Method 3:Method 3: We know that at the maximmaximmaximmaximmaximumumumumum point aaaaa: f’(x) is positipositipositipositipositivvvvve befe befe befe befe beforororororeeeeeand nenenenenegggggaaaaatit it it it i vvvvve aftere aftere aftere aftere after.....

Going from positivepositivepositivepositivepositive to negativenegativenegativenegativenegative means f’(x) must be decreasing.decreasing.decreasing.decreasing.decreasing.

We also know that if a function is decrdecrdecrdecrdecreasingeasingeasingeasingeasing its first derivativemust be nenenenenegggggaaaaatititititivvvvveeeee..... We take the first derivative of f’(x) which is f”(x),the second derivative of f(x), and compute f”(aaaaa).

f”(aaaaa) < 0 implies aaaaa is maximum.maximum.maximum.maximum.maximum.

Similar ly, we know that at the minimumminimumminimumminimumminimum point aaaaa: f’(x) is negativenegativenegativenegativenegativebefbefbefbefbefororororore e e e e and posit iposit iposit iposit iposit i vvvvve aftere aftere aftere aftere after.....

Going from negativenegativenegativenegativenegative to positivepositivepositivepositivepositive means f’(x) must be increasing.increasing.increasing.increasing.increasing.

We also know that if a function is incrincrincrincrincreasingeasingeasingeasingeasing its first derivative mustbe positipositipositipositipositivvvvveeeee..... We take the first derivative of f’(x) which is f”(x), thesecond derivative of f(x), and compute f”(aaaaa).

f”(aaaaa) > 0 implies aaaaa is minimum.minimum.minimum.minimum.minimum.This is usually the easiest and most useful method. But there are exceptions.

118

ExerciseExerciseExerciseExerciseExercise 1 1 1 1 1::::: For the function f(x) = x3 f’(x) = 3x2.

At x = 0 : f’(0) = 0 implies zero is an extreme point.

Is x = 0 a maximum point or minimum point ?

Use all three methods to determine this.

Exercise Exercise Exercise Exercise Exercise 22222::::: For the following functions find the extreme points and

determine if they are maximum or minimum.

a) x + 2

b) x2

c) ---------- x2

d) x3

e) ---------- x3

f) x4

g) ---------- x4

From the number of times a function intersects the x-----axis (the number of

real roots) can you estimate the number of extreme points ?

ExExExExExererererercise cise cise cise cise 33333::::: Divide 10 into two par ts such that the product is a maximum.


Draw the graph and verify.


Draw the graph and verify.

119

f(t) = sin (+3ω0t)

T

a

m

p

l

i

t

u

d

e

Absolute Maximum and Absolute MinimumAbsolute Maximum and Absolute MinimumAbsolute Maximum and Absolute MinimumAbsolute Maximum and Absolute MinimumAbsolute Maximum and Absolute Minimum

It is possible to have more than one maximmaximmaximmaximmaximum um um um um and one minimminimminimminimminimum um um um um as in f(t) below.

In this case we will find the overall or absolute maximumabsolute maximumabsolute maximumabsolute maximumabsolute maximum and the overall or

absolute minimumabsolute minimumabsolute minimumabsolute minimumabsolute minimum . By solving f(t) = 0 we find all the maximum maximum maximum maximum maximum and minimumminimumminimumminimumminimum

points. Each maximum maximum maximum maximum maximum point is called a local maximum maximum maximum maximum maximum . And each minimumminimumminimumminimumminimum

point is called a local minimum minimum minimum minimum minimum .

absolute maximum absolute maximum absolute maximum absolute maximum absolute maximum = maximum maximum maximum maximum maximum (all locallocallocallocallocal maximumaximumaximumaximumaximummmmm )

absolute minimumabsolute minimumabsolute minimumabsolute minimumabsolute minimum = minimumminimumminimumminimumminimum (all local minimulocal minimulocal minimulocal minimulocal minimummmmm )

Exercise :Exercise :Exercise :Exercise :Exercise : Find the maximum and minimum points of f(t) = sin (+3ω0t) .

Acknowledgement : the diagrams on this page are taken from “A PrA PrA PrA PrA Preeeeevievievievieview ofw ofw ofw ofw of Con Con Con Con Convvvvvolutions and olutions and olutions and olutions and olutions and TTTTTrrrrransfansfansfansfansfororororormsmsmsmsms”

by permission of the author.

0 t0 t

1 t

2 t

3 t

4 t

5 t

6 t

7 t

8 t

f(t)

120

At a maximmaximmaximmaximmaximumumumumum, minimminimminimminimminimumumumumum or infinfinfinfinflelelelelexion xion xion xion xion point, the tangtangtangtangtangententententent to the curve is

PARALLEL to the x-axis and hence tan θ = 0 . Conversely, if the tangtangtangtangtangententententent to the

curve at a point is zero (tan θ = 0) then that point must be a maximmaximmaximmaximmaximumumumumum,

minimum minimum minimum minimum minimum or inf lexion inf lexion inf lexion inf lexion inf lexion point.

We mentioned ear lier that polpolpolpolpolynomials arynomials arynomials arynomials arynomials are to functions we to functions we to functions we to functions we to functions whahahahahat rt rt rt rt raaaaationalstionalstionalstionalstionals

ararararare to re to re to re to re to real neal neal neal neal numberumberumberumberumbersssss. We can do all our calculations of reals to any degree

of accuracy using the rationals. Likewise we may approximate any single-valued

function by a suitable polynomial. In fact, thinking more deeply on what we

just covered in the last few chapters, we may be more specific. We may

piece-wise approximate any single-valued function by a suitable choice of

polynomials of maximum degree 3 or cubic.

We just saw how a function is either increasing, decreasing or changing

direction. Polynomials of degree 0,1,2, and 3 are sufficient to cover all

these variations. We only need to identify the maximmaximmaximmaximmaximumumumumum, minimminimminimminimminimumumumumum, and

infinfinfinfinf lelelelelexionxionxionxionxion points of the function thru which the approximating polynomials

must interpolate.

Note :Note :Note :Note :Note : Later the student will learn ROLLE’S THEOREM, which is a special case of

the MEAN VALUE THEOREM. In the special case when f(a) = f(b) then there exists

some MEAN point _x in the interval [a, b] such that f ’(

_x ) = 0. Is it possible to find

the point _x now ? (see note page 64)

121

2222244444..... P P P P Po ints o fo ints o fo ints o fo ints o fo ints o f INFLEXION INFLEXION INFLEXION INFLEXION INFLEXION

A point at which a curve changes its shape in known as a point of infinfinfinfinflelelelelexion.xion.xion.xion.xion.

infinfinfinfinf lelelelelexionxionxionxionxion = a bending in the curve, a change in curvature or shape, achange in direction, modulation of sound. (flexible, inflexible)

We have seen in the previous sections that at a maximmaximmaximmaximmaximumumumumum or minimminimminimminimminimumumumumumpoint a curve changes its shape.

maximummaximummaximummaximummaximum : concave down and going left to right from rising to falling.minimumminimumminimumminimumminimum : concave up and going left to right from falling to rising.

There is another type of change in shape possible: concave up to concavedown or vice versa concave down to concave up.

Consider the function f(x) = x3 + 1. Please see the graph. f ’(x) = 3x2 and f ” (x) = 6x

At x = 0 : f(0) = +1 and f ’(0) = 0, implying x=0 may be a maximummaximummaximummaximummaximumor a minimminimminimminimminimumumumumum point. From the graph we can see it is neither. f(x) iscontinuously increasing. However, the curve is changing shape from concaconcaconcaconcaconcavvvvveeeeedododododownwnwnwnwn to concaconcaconcaconcaconcavvvvve upe upe upe upe up.

At x=0 : f ”(0) = 0 and f ”(x) is nenenenenegggggaaaaatititititivvvvve befe befe befe befe beforororororeeeee and positipositipositipositipositivvvvve aftere aftere aftere aftere after.....

Consider the function f(x) = ----------x3 + 1. Please draw the graph. f ’(x) = ----------3x2 and f ”(x) = ----------6x

At x = 0 : f(0) = +1 and f ’(0) = 0, implying x = 0 may be a maximummaximummaximummaximummaximumor a minimminimminimminimminimumumumumum point. From the graph we can see it is neither. f(x) iscontinuously decreasing. However, the curve is changing shape from concaconcaconcaconcaconcavvvvveeeeeupupupupup to concave downconcave downconcave downconcave downconcave down.

At x = 0: f” (0) = 0 and f” (x) is positipositipositipositipositivvvvve befe befe befe befe beforororororeeeee and nenenenenegggggaaaaatititititivvvvve aftere aftere aftere aftere after.

122

4

3

2

1

----------1

----------2

----------3

----------4

x

f(x) = x3 + 1f '(x) = 3x2

f "(x) = 6x

----------3 ----------2 ----------1 0 1 2 3

y

122

1

123

In both cases, f(x) = x3 + 1 and f(x) = −−−−−x3 + 1, we have f’(0) = 0.

But x = 0 is neither a maximmaximmaximmaximmaximumumumumum nor a minimminimminimminimminimumumumumum. How do we determinewhat kind of point x is ?

f’ (x) = 0 and f” (x) negative ⇒ maximummaximummaximummaximummaximum.

f’ (x) = 0 and f” (x) positive ⇒ minimumminimumminimumminimumminimum.

f’ (x) = 0 and f” (x) = 0 and changing sign ⇒ inflexioninflexioninflexioninflexioninflexion.

Let us now summarize in tabular form the results of these last few chapters. Forany well-behavedwell-behavedwell-behavedwell-behavedwell-behaved function y = f(x), with dy/////dx = f’ (x) and d2y/////dx2 = f” (x),we have:

Type of pointdydx = tan θ = slope of tangentslope of tangentslope of tangentslope of tangentslope of tangent d2y

dx2

Maximum 0 + − − or 0

Minimum 0 − + + or 0

Inflexion (turning) 0 ± ± 0 and

before after

changing sign

124

Reviewing the example of the bouncing ball at the instants t1, t2, t3, ............... wecan analyse and say:

1. The function (height) is decreasingdecreasingdecreasingdecreasingdecreasing before and increasingincreasingincreasingincreasingincreasing after .

2. The first derivatives are nenenenenegggggaaaaatititititivvvvveeeee before and positipositipositipositipositivvvvveeeee after i.e. changing sign.

3. f ’(x ) = 0 . Wha t can you say abou t f ”(x) ? I s the cur ve changing shape ?

It makes sense to say that some object is going up and coming down atshor ter and shor ter time inter vals.

If we now note that the maximum height in succeeding time intervals (t0, t

1),

(t1, t

2), (t

2, t

3), ... is decreasing, we may reasonably interpret this to mean that

the object is bouncing in some kind of force (gravitational) field which is causingit to slow down.

If on the other hand the time intervals were of the same duration (t0, t1) = (t1, t2)= (t2, t3) = ... and the maximum height in succeeding intervals kept decreasing,how would you interpret this ? A pendulum oscillating ?

Is it not amazing that Galileo knew so much about falling objects, pendulums,projectiles and planetary motion, yet missed discovering gravity !!!!!

125

PPPPPararararar t 4 :t 4 :t 4 :t 4 :t 4 : INTEGRA INTEGRA INTEGRA INTEGRA INTEGRATIONTIONTIONTIONTION

126

VEDIC mathematicians treated Integration in the Ekadikha Sutra as the naturalinverse of Differentiation: the ANTIDERIVATIVE. The application was esotericin nature and more of aesthetic value ( Bijaganita Bijaganita Bijaganita Bijaganita Bijaganita ===== Algebra Algebra Algebra Algebra Algebra )rather thangeometric ( PPPPPaaaaatigtigtigtigtiganitaanitaanitaanitaanita ): finding the area under the curve.

Whi le var ious approx imat ions were employed in d i f ferent anc ientcivi l izations to f ind the areas enc losed by cer tain special cur ves, thecredit is usually given to Archimedes (c .287-212B.C.), of unsinkable fame,for the embryonic idea of the general method of INTEGRATION for findingthe area under a cur ve.

This concept remained dormant ti l l the Renaissance mathematiciansCavalieri (1591-1647) and Torricelli (1608-1647), both students ofGalileo (1564-1642), made it germinate.

Issac Barrow (1630-1677), a teacher of the genius Sir Isaac Newton (1642-1727), was the fir st to make the connection between the slope of theTANGENT (ddddderierierierierivvvvvaaaaatititititivvvvveeeee) and “arararararea under the curea under the curea under the curea under the curea under the cur vvvvveeeee ” in 1659. This ideawas abstracted and developed both by Newton and Leibniz to establ ishf i r mly the r e la t ionsh ip between D i f fer ent ia l and In tegr a l Ca lcu lus.

Gottfried Wilhelm Leibniz (1646-1716) in his manuscript dated 29 October1673, was the first to use the integral sign ∫, the elongated or stretched Sas we know and use it today, to denote contincontincontincontincontinuous uous uous uous uous summation. Leibniz calledthe expression ∫ f(x) dx the INTEGRAL from the Latin integralisintegralisintegralisintegralisintegralis or whole.

However, both the concepts of def in ing the re lat ionship betweenDif ferential and Integral Calculus: DERIVATIVE & ANTIDERIVATIVE andTANGENT and “arararararea under the curea under the curea under the curea under the curea under the cur vvvvveeeee ”, are useful but l imited. Thecorrect concept that cor responds to the relationship is INSTANTANEOUSRATE OF CHANGE and CHANGE.

127


We may define a WELL-BEHAVED function F(x) and on difdifdifdifdifffffferererererentiaentiaentiaentiaentiation tion tion tion tion let :

Hence f(x) is the DERIVATIVE of F(x). If we write f(x)dx = dF(x) then we call f(x)the DIFFERENTIAL COEFFICIENT.

We may now define the inininininvvvvverererererse operse operse operse operse operaaaaation tion tion tion tion of difdifdifdifdifffffferererererentiaentiaentiaentiaentiation tion tion tion tion as finding theANTIDERIVATIVE . So :

F(x) = ANTIDERIVATIVE { f(x)}

Just like with difdifdifdifdifffffferererererentiaentiaentiaentiaentiation tion tion tion tion we may construct a table of ANTIDERIVATIVES. Thecalculation becomes almost mechanical. This is purely from the calculation calculation calculation calculation calculation pointof view.

From the Analysis Analysis Analysis Analysis Analysis point of view we may infer that : if f(x) is the expression of theINSTANTANEOUS RATE OF CHANGE of F(x), then F(x) must be the expression ofCHANGE.

From the Geometric Geometric Geometric Geometric Geometric point of view we have an even more interesting relationship. Itseems almost magical.

area under f(x) =

CHANGE in F(x) = F(b) −−−−− F(a)over interval [a, b] over interval [a, b]

Integral Calculus developed out of the need for a general method to findareas, volumes and centres of gravity. Computing the area under f(x) involves acontinuous summationcontinuous summationcontinuous summationcontinuous summationcontinuous summation . It is from this process of continuous summation continuous summation continuous summation continuous summation continuous summation thatwe have the operation expressed as integration integration integration integration integration derived from the Latin integralisintegralisintegralisintegralisintegralisand the Mathematical notation or symbol ∫ to define it.

f(x) =

d F(x) dx

128

In AlgebraAlgebraAlgebraAlgebraAlgebra we are familiar with the concept of discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summation of a finitenumber of terms. For example :

We are also familiar with the discrdiscrdiscrdiscrdiscrete summaete summaete summaete summaete summationtiontiontiontion of an infinite number of terms.For example, the geometric series :

We could have defined a more general case like :

Notice that even though the summation involves infinitely many terms, they areCOUNTABLE. Hence we may identify each term using a discrdiscrdiscrdiscrdiscrete ete ete ete ete subscript. Recallthe difference between the set of natural numbers N that is COUNTABLE, and thecontiguous contiguous contiguous contiguous contiguous set of real numbers R that is UNCOUNTABLE. In the contincontincontincontincontinuousuousuousuousuoussummasummasummasummasummation tion tion tion tion over an interval [a, b] we must include each and every real number orpoint or instant in [a, b]. Since there are UNCOUNTABLY many of them it does NOTmake sense to use a subscript. We know only the end points aaaaa and bbbbb of the interval[a, b]. So we evolve the notation for the continuous summation continuous summation continuous summation continuous summation continuous summation in 3 steps :

For the continuous summationcontinuous summationcontinuous summationcontinuous summationcontinuous summation to be of practical value it must converge to somedefinite finite quantity. This fundamental property, in an intuitive way, is known asinteinteinteinteintegggggrrrrraaaaabbbbblelelelele. We give an informal definition of this.

1/////2 iiiii

nnnnn

Σiiiii ===== 0 0 0 0 0

1/////2 iiiii

∞

Σiiiii ===== 0 0 0 0 0

f(x i)∞

Σiiiii ===== 0 0 0 0 0

n n n n n →→→→→∞∞∞∞∞

Σiiiii ===== 0 0 0 0 0

nnnnn

Σiiiii ===== 0 0 0 0 0

bbbbb

∫aaaaa

⇒ ⇒

discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summationof FINITELY many terms

discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summationof COUNTABLY many terms

continuous summationcontinuous summationcontinuous summationcontinuous summationcontinuous summationof UNCOUNTABLY many terms

where n is finite.

129

2525252525..... F(F(F(F(F(xxxxx))))) = = = = = ANTIDERIVANTIDERIVANTIDERIVANTIDERIVANTIDERIVAAAAATIVETIVETIVETIVETIVE { { { { {f(x)f(x)f(x)f(x)f(x)}}}}}Suppose we know the function that expresses the INSTANTANEOUS RATE OFCHANGE, are we able to determine the function that expresses the CHANGE ?Suppose we know that the function y’(t) that describes the ver tical speed i.e.the INSTANTANEOUS RATE OF CHANGE in height is :

y’(t) = dy

dt(t)

= u.....si n θ ---------- g t

Are we able to determine the function y(t) = u.....si n θ ..... t ---------- 1 -- 2 g t 2

that describes the CHANGE in height ?

We could mechanically say: dydt(t)

is a polpolpolpolpolynomialynomialynomialynomialynomial of the form ax + b

So y(t) must be a polpolpolpolpolynomialynomialynomialynomialynomial of the form 1 -- 2 ax2 + bx + C

For : dydt(t)

= ---------- g t + u.si n θ

y(t) = ----------1--- 2 g t 2 + u.si n θ . t + may be some constant C.

if at time t = 0 the ball was thrown from the ground level ( = 0) : C = 0

if at time t = 0 the ball was thrown from a height of 5 meters: C = + 5m

This mechanical method of calculation wor ks when the functions areINTEGRABLE. It is known as finding the ANTIDERIVATIVE. There are tableswhich you can look up to find the matching ANTIDERIVATIVE of the givenDERIVATIVE.

AnalAnalAnalAnalAnalyticallyticallyticallyticallyticallyyyyy, if f(x) is the DERIVATIVE of the function F(x), then the functionF(x) is called the ANTIDERIVATIVE of the function f(x). We may write:

If we write f(x)dx = dF(x) then we call f(x) the DIFFERENTIAL COEFFICIENT.

f(x) =

d F(x) dx

130

Below is a partial TTTTTaaaaabbbbble ofle ofle ofle ofle of Inte Inte Inte Inte Integggggrrrrralsalsalsalsals of the more frequently encountered functionsin standard form. The derideriderideriderivvvvvaaaaatititititivvvvve = intee = intee = intee = intee = integggggrrrrrandandandandand and antideriantideriantideriantideriantiderivvvvvaaaaatititititivvvvve = intee = intee = intee = intee = integggggrrrrraaaaalllll .

DERIVDERIVDERIVDERIVDERIVAAAAATIVE TIVE TIVE TIVE TIVE f(x)f(x)f(x)f(x)f(x) ANTIDERIVANTIDERIVANTIDERIVANTIDERIVANTIDERIVAAAAATIVE FTIVE FTIVE FTIVE FTIVE F(x)(x)(x)(x)(x)

xn xn + 1

n + 1 n ≠ −−−−−1

1/////x log |x|ex ex

ax ax

log a a > 0, a ≠ 1

sin (x) − cos xcos (x) sin (x)sec2 (x) tan (x)cosec2 (x) − cot (x)tan (x) ..... sec (x) sec (x)cot (x) ..... cosec (x) − cosec (x)tan (x) log |sec (x)|cot (x) log |sin (x)|sec (x) log|sec (x) + tan (x)|= log|tan (π/////4 + x/////2)|cosec (x) log|cosec (x) − cot (x)|or log|tan x/////2| 1a2 + x2

1/////a tan −−−−−1 ( x/////a) a ≠ 0

1 1--------------- a + xa2 −−−−− x2 2a

log |a − x

|

1 1--------------- x −−−−− ax2 −−−−− a2 2a

log |x + a

|

1√a2 −−−−− x2 sin −−−−−1(

x|a|) x2 < a2

131

2626262626. . . . . F(x) = area under f(x) = F(x) = area under f(x) = F(x) = area under f(x) = F(x) = area under f(x) = F(x) = area under f(x) = ∫∫∫∫∫ f(x)dxf(x)dxf(x)dxf(x)dxf(x)dx

Let us look at the original concept of the integral integral integral integral integral - finding the area under aarea under aarea under aarea under aarea under acurcurcurcurcurvvvvve e e e e . For the time being we denote the area function as F(x).

F(x) = arF(x) = arF(x) = arF(x) = arF(x) = area under the curea under the curea under the curea under the curea under the curvvvvve ofe ofe ofe ofe of f(x) f(x) f(x) f(x) f(x)

In the next chapter we shall prove that :

F(F(F(F(F(xxxxx))))) = = = = = ANTIDERIVANTIDERIVANTIDERIVANTIDERIVANTIDERIVAAAAATIVETIVETIVETIVETIVE { { { { {f(x)f(x)f(x)f(x)f(x)} } } } } = = = = = ∫∫∫∫∫ f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx

We now present 2 methods to find the “““““arararararea under the curea under the curea under the curea under the curea under the cur vvvvveeeee””””” f(x) over theinter val [a,b].

Rectangular methodRectangular methodRectangular methodRectangular methodRectangular method

Let f(x) be the cur ve between x = a and x = b with f(a) = A and f(b) = B.

Divide the interval [a,b] into n sub-intervals ∆x i and construct rectangles instep-like fashion as shown in the figure. We shall compute F(x) in three steps :

123412341234123412341234

1234123412341234123412341234

(((((00000 ,,,,, 00000 )))))


xxxxxxxxxx 00000 ===== aaaaa xxxxx 11111 xxxxx 22222 xxxxx 33333 b =b =b =b =b = xxxxx nnnnnxxxxx 4 4 4 4 4 . . .. . .. . .. . .. . .

123456123456123456123456123456123456123456123456

f(x)f(x)f(x)f(x)f(x)AAAAA

BBBBB

123123123123123123

n n n n n →→→→→∞∞∞∞∞

Σiiiii ===== 0 0 0 0 0

nnnnn −−−−− 1 1 1 1 1

Σiiiii ===== 0 0 0 0 0

bbbbb

∫aaaaa

⇒ ⇒

discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summationof FINITELY many terms

discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summationof COUNTABLY many terms

continuous summationcontinuous summationcontinuous summationcontinuous summationcontinuous summationof UNCOUNTABLY many terms

f(x i).....δx i L i m i tL i m i tL i m i tL i m i tL i m i t f(x i).....∆x i

n n n n n →→→→→∞∞∞∞∞

Σiiiii ===== 0 0 0 0 0

f(x i).....δx i = f(x ).....dx

132

1.1.1.1.1. APPRAPPRAPPRAPPRAPPROOOOOXIMAXIMAXIMAXIMAXIMATE TE TE TE TE AREA steAREA steAREA steAREA steAREA step :p :p :p :p : discrdiscrdiscrdiscrdiscrete summaete summaete summaete summaete summationtiontiontiontion of FINITELY many terms.

An approximation of the actual area function F(x) is Fn(x) where:

Fn(x) = f(x0) (x1 ---------- a) + f (x1) (x2 ---------- x1) + ... + f(xn ---------- 1) (b ---------- xn ---------- 1

) =

2.2.2.2.2. TENDS TENDS TENDS TENDS TENDS TTTTTO steO steO steO steO step :p :p :p :p : discrdiscrdiscrdiscrdiscrete summaete summaete summaete summaete summation tion tion tion tion as n → ∞ of COUNTABLY many terms.

We let n be very, very large by letting n → ∞ . The sub-intervals ∆x i = x1 ---------- a,

x2 ---------- x1, ..., b ---------- xn -- 1 will be very, very small. We can denote ∆x i by δx i

for i = 0, 1,2, . . . , n → ∞ .

δx0 = x1 ---------- a, δx1 = x2 ---------- x1, . . . , δxn -- 1 = b ------ - - - - xn -- 1

So: Fn(x) = f(x0) δx0 + f(x1) δx1 + . . . + f(xn -- 1) δxn -- 1

As n gets larger and larger we can see that the sum Fn(x) is a better andbetter approximation of the area function F(x).

3.3.3.3.3. LIMIT ste LIMIT ste LIMIT ste LIMIT ste LIMIT step :p :p :p :p : contincontincontincontincontinuous summauous summauous summauous summauous summation tion tion tion tion of UNCOUNTABLY many terms.Geometrically Geometrically Geometrically Geometrically Geometrically we can think of δxi as a small continuouscontinuouscontinuouscontinuouscontinuous line segment.As n → ∞ the δx i gets smaller and smaller. The LIMIT of δx i as n → ∞ isa point or instant.Now we take the LIMIT as n → ∞. We can denote δx

iiiii by dx .

Fn(x)= Limit Limit Limit Limit Limit Σ f(xi) .....dx for i = 0, 1, 2, . . . , n → ∞ .

We use the special symbol ∫ which is the initial letter of the word summasummasummasummasumma orsum, to distinguish the cont incont incont incont incont inuousuousuousuousuous summasummasummasummasummationtiontiontiontion of UNCOUNTABLY manyterms from the discrdiscrdiscrdiscrdiscreteeteeteeteete summasummasummasummasummationtiontiontiontion of a finite number of terms using + orCOUNTABLY many terms using Σ .The dxdxdxdxdx in dy/////dx of difdifdifdifdifffffferererererentiaentiaentiaentiaentiation tion tion tion tion refers to the calculation at a point point point point point or instantinstantinstantinstantinstant.The ∫ f(x) ..... dx of inteinteinteinteintegggggrrrrraaaaation tion tion tion tion refers to the contincontincontincontincontinuousuousuousuousuous summasummasummasummasummationtiontiontiontion over aninterinterinterinterintervvvvvalalalalal of contiguouscontiguouscontiguouscontiguouscontiguous instants dxdxdxdxdx .

n n n n n →→→→→∞∞∞∞∞

= Σ f(x i).....δx i

iiiii ===== 0 0 0 0 0

nnnnn −−−−− 1 1 1 1 1

Σiiiii ===== 0 0 0 0 0

f(x i).....∆x i

F(x) = Limit Fn(x) = ∫ f(x)..... dx n → ∞

n n n n n →→→→→∞∞∞∞∞

= Limit Limit Limit Limit Limit Σ f(xi) .....dx

iiiii ===== 0 0 0 0 0

Limitn → ∞

133

BBBBB

AAAAA

f ( x )f ( x )f ( x )f ( x )f ( x )

xxxxx 00000 ===== aaaaa xxxxx 11111 xxxxx 22222 xxxxx 3 3 3 3 3 . . .. . .. . .. . .. . .b =b =b =b =b = xxxxx nnnnnxxxxx

xxxxx---------- 22222xxxxx---------- 11111xxxxx---------- 00000


(((((0 ,00 ,00 ,00 ,00 ,0 )))))

TTTTTrrrrraaaaapepepepepezzzzzoidal moidal moidal moidal moidal methodethodethodethodethod

Instead of approximating the arararararea under the curea under the curea under the curea under the curea under the cur vvvvveeeee f(x) by rectangularstrips we could use trapezoidal strips.

Area of a trapezoid is: (1--2 sum of the parallel sides) * (perpendicular distance)


It is not difficult to see that in sub-interval ∆x0 = [a, x1] there is some MEAN

point x----------0 (see note) such that:

Similar ly, in sub-inter val ∆x 1 = [x1, x2] there some MEAN point x----------1 such that:

and so on. So we can write:Fn(x) = f(x----------0) (x1 ---------- a) + f(x----------1) (x2 ---------- x1) + ............... + f( x----------n ---------- 1

) (b ---------- xn ---------- 1) =

Fn( x) = f(x1) + f(a ) ..... ( x

1 ---------- a ) +

f( x2) + f( x1) ..... ( x2 ---------- x

1) + . . .. . .. . .. . .. . .

2 2

+ f(b ) + f( x n ---------- 1

) ..... (b ---------- x n ---------- 1)

2

f( x----------0)

= f( x1) + f(a )

2

f( x----------1)

= f( x2) + f( x1)

2

nnnnn −−−−− 1 1 1 1 1

Σiiiii ===== 0 0 0 0 0

f(x i).....∆x i

134

2.2.2.2.2. TENDS TENDS TENDS TENDS TENDS TTTTTO steO steO steO steO step :p :p :p :p : discrdiscrdiscrdiscrdiscrete summaete summaete summaete summaete summation tion tion tion tion as n → ∞ of COUNTABLY many terms.All we have to do now is let n get larger and larger by letting n → ∞ . Thesub-intervals ∆x i

will get smaller and smaller. We can denote the sub-intervals∆x i

by δxiiiii for i = 0, 1, 2, . . . , n → ∞ :

δx0 = (x

1 ---------- a), δx

1 = ( x

2 ---------- x1

), . . .. . .. . .. . .. . . , δxn ---------- 1 = (b ---------- xn ---------- 1

)

Fn(x) = f( x----------0) δx0 + f( x----------1) δx1 + . . .. . .. . .. . .. . . + f( x----------n ---------- 1)(δxn ---------- 1

)

As n gets larger and larger we can see that the sum Fn(x) is a better andbetter approximation of the area function F(x).

3.3.3.3.3. LIMIT ste LIMIT ste LIMIT ste LIMIT ste LIMIT step :p :p :p :p : contincontincontincontincontinuous summauous summauous summauous summauous summation tion tion tion tion of UNCOUNTABLY many terms.

Now we take the LIMIT as n → ∞ . We can denote δxiiiii by dx .

Fn(x) = LimitLimitLimitLimitLimit Σ f(x--

i) ..... dx for i = 0, 1, 2, . . . , n → ∞ .

Fn(x) = contincontincontincontincontinuous uous uous uous uous summasummasummasummasummation tion tion tion tion of f(x

i) dx over interval [a, b].

Again: F(x) =

N o t e :N o t e :N o t e :N o t e :N o t e :

We should be careful when we say “ that in the interval [ a, x11111] there is some MEAN point x----------

00000 . . . ”.This may NOT always be the case. Let the set S = (3, 4, 5, 7 ).For the elements 3 and 5 there is the element 4 in s, such that 4 is the MEAN of 3 and 5.

. However, there is no element 6 in s as the MEAN of 5 and 7.

6 is not an element of s.

Since we are dealing only with CONTINUOUS functions we can say “ that in the inter val [a, x11111]

there is some MEAN point x----------00000 . . . ”.

For example, for f(x) = x 22222 : f(2) = 4, and f(4) = 16 .

The mean of f(2) and f(4) is

There is some MEAN point x---------- = √10 in the interval [2, 4 ] such that f( x----------) = 10 .

i.e. 4 = 3 + 5 2

= 5 + 7 2

n n n n n →→→→→∞∞∞∞∞

= Σ f( x-- i) ..... δx i

iiiii ===== 0 0 0 0 0


n n n n n →→→→→∞∞∞∞∞

= Limit Limit Limit Limit Limit Σ f( x-- i) .....dx iiiii ===== 0 0 0 0 0

Limit Fn(x) = ∫ f(x) dx over interval [a, b].n → ∞

Limitn → ∞

= f(2) + f(4) = 4 + 16 = 10 . 2 2

135

It should now be clear that whichever method you choose (rectangular,trapezoidal, . . . ) the is the same and is the area functionF(x) of the “ar“ar“ar“ar“area ea ea ea ea under the curunder the curunder the curunder the curunder the cur vvvvveeeee””””” f(x) over the inter val [a,b].

The area function F(x) = ∫ f(x)dx.

Since we know the inter val over which we are finding the area we could bemore DEFINITE and use the (Fourier refinement of Leibniz) notation:

[F(x)]ba = ∫

ba f(x)dx

With the understanding that F(x) = the area under f(x) = ∫ f(x)dx the reader maywish to review the chapter on Units of Measure Units of Measure Units of Measure Units of Measure Units of Measure .

So far we have only a notation for the area function : F(x) = ∫ f(x)dx

We do NOT know what F(x) is, much less even begin to evaluate it. What isthe area function F(x) ?

Is there a relationship between f(x) and F(x) ? We claim that f(x) = F’(x).

F(x) = ANTIDERIVATIVE of {f(x)}.

And once we prove this, we call f(x) the integrand integrand integrand integrand integrand and F(x) the integral integral integral integral integral .

integrand integrand integrand integrand integrand = the operand of the integration operation.

integralintegralintegralintegralintegral = ∫integrandintegrandintegrandintegrandintegrand .

Thus our Table of DERIVATIVES and ANTIDERIVATIVES becomes a TTTTTaaaaabbbbble ofle ofle ofle ofle of Inte Inte Inte Inte Integggggrrrrralsalsalsalsals.....

Limit Fn(x)n → ∞

136

0 1 2 3 4 x

4

3

2

1

2727272727..... F(F(F(F(F(xxxxx))))) = = = = = ANTIDERIVANTIDERIVANTIDERIVANTIDERIVANTIDERIVAAAAATIVETIVETIVETIVETIVE{{{{{f(x)f(x)f(x)f(x)f(x)} } } } } = = = = = ∫∫∫∫∫ f(x)dxf(x)dxf(x)dxf(x)dxf(x)dxLet us look at two examples to see the relation between :

F(x) = arF(x) = arF(x) = arF(x) = arF(x) = area under the curea under the curea under the curea under the curea under the curvvvvve ofe ofe ofe ofe of f(x) f(x) f(x) f(x) f(x) = = = = = ∫∫∫∫∫ f(x)dx f(x)dx f(x)dx f(x)dx f(x)dxand F(F(F(F(F(xxxxx))))) = = = = = ANTIDERIVANTIDERIVANTIDERIVANTIDERIVANTIDERIVAAAAATIVETIVETIVETIVETIVE { { { { {f(x)f(x)f(x)f(x)f(x)} .} .} .} .} .

Example 1: Example 1: Example 1: Example 1: Example 1: Let f(x) = 1. For bbbbb = 1, 2, 3, . . . we may compute the area function

F(x) = 00000∫∫∫∫∫bbbbb f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx and plot the graph of F(x).

From the diagrams above it is clear that F(x) = x and

Example 2: Example 2: Example 2: Example 2: Example 2: Let f(x) = x. For x = 1, 2, 3, . . . we may compute the area function

F(x) = 00000∫∫∫∫∫bbbbb f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx and plot the graph of F(x).

From the diagrams above it is clear that F(x) = 1/////2 x2 and

F(x) =F(x) =F(x) =F(x) =F(x) = 00000∫∫∫∫∫bbbbb f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx

123456789123456789123456789123456789123456789123456789123456789123456789123456789

123456789123456789123456789123456789123456789123456789123456789123456789123456789

1

0 1 2 x

f(x) = 1

f(f(f(f(f(xxxxx) = ) = ) = ) = ) = xxxxx

1 2 x

2

1

0

123456789123456789123456789123456789123456789123456789123456789123456789123456789

12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890

f(x) =

d F(x) = 1. dx

f(x) =

d F(x) = x . dx

1 2 x

2

1

0

F(x) =F(x) =F(x) =F(x) =F(x) = 00000∫∫∫∫∫bbbbb f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx

137

BBBBB

AAAAA

f ( x )f ( x )f ( x )f ( x )f ( x )

xxxxx 00000 ===== aaaaa b =b =b =b =b = xxxxx nnnnnxxxxx

(((((0 ,00 ,00 ,00 ,00 ,0 ))))) ∆∆∆∆∆ xxxxx iiiii

1234123412341234123412341234123412341234

From these two particular examples it is reasonable to infer that :arararararea function F(x) = ea function F(x) = ea function F(x) = ea function F(x) = ea function F(x) = ANTIDERIVANTIDERIVANTIDERIVANTIDERIVANTIDERIVAAAAATIVE {f(x)}.TIVE {f(x)}.TIVE {f(x)}.TIVE {f(x)}.TIVE {f(x)}.

PrPrPrPrProofoofoofoofoof : : : : : Let f(x) be the cur ve between x = a and x = b with f(a) = Aand f(b) = B. Divide the interval [a,b] into n sub-intervals ∆x i and constructrectangles in step-like fashion as shown in the figure.

Let F(x) = the ar ar ar ar area under the curea under the curea under the curea under the curea under the curvvvvveeeee of f(x).


An approximation of the actual area function F(x) is Fn(x) where:

Fn(x) = f(x0) (x1 ---------- a) + f (x1) (x2 ---------- x1) + ... + f(xn ---------- 1) (b ---------- xn ---------- 1

) =

So we may say : ∆Fiiiii (x) = f(x iiiii).....∆x

iiiii an element element element element element of area .


So: Fn(x) = f(x0) δx0 + f(x1) δx1 + . . . + f(xn -- 1) δxn -- 1

So we may say : δFiiiii (x) = f(x

iiiii).....δx

iiiii an infinitesimal element infinitesimal element infinitesimal element infinitesimal element infinitesimal element of area .


Hence we may say : dF(x) = f(x ).....dx an instantaneous element instantaneous element instantaneous element instantaneous element instantaneous element of area .

So . TTTTThe arhe arhe arhe arhe area function F(x) = ea function F(x) = ea function F(x) = ea function F(x) = ea function F(x) = ANTIDERIVANTIDERIVANTIDERIVANTIDERIVANTIDERIVAAAAATIVE {f(x)}.TIVE {f(x)}.TIVE {f(x)}.TIVE {f(x)}.TIVE {f(x)}.

n n n n n →→→→→∞∞∞∞∞

= Σ f(x i).....δxiiiii

iiiii ===== 0 0 0 0 0

nnnnn −−−−− 1 1 1 1 1

Σiiiii ===== 0 0 0 0 0

f(xiiiii).....∆x

iiiii


n n n n n →→→→→∞∞∞∞∞

= Limit Limit Limit Limit Limit Σ f(xi) .....dx iiiii ===== 0 0 0 0 0

f(x) =

d F(x) dx

138

1 2 x

2

1

0

F(x) =F(x) =F(x) =F(x) =F(x) = xxxxx

123456789123456789123456789123456789123456789123456789123456789123456789123456789

1

0 1 2 x

f(x) = 1


1 2 x

2

1

0

12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890

0 1 2 3 4 x

4

3

2

1

F(x) =F(x) =F(x) =F(x) =F(x) = 1/////2 x2

2828282828. . . . . ∫∫∫∫∫aaaaa

bbbbb f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx = CHANGE = CHANGE = CHANGE = CHANGE = CHANGE in in in in in F(F(F(F(F(xxxxx)))))

From the CALCULATION point of view : F(F(F(F(F(xxxxx))))) = = = = = ANTIDERIVANTIDERIVANTIDERIVANTIDERIVANTIDERIVAAAAATIVETIVETIVETIVETIVE { { { { {f(x)f(x)f(x)f(x)f(x)} .} .} .} .} .

From the GEOMETRIC point of view : F(x) =F(x) =F(x) =F(x) =F(x) = aaaaa∫∫∫∫∫bbbbb f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx = ar = ar = ar = ar = area under the curea under the curea under the curea under the curea under the curvvvvveeeee

of f(x) of f(x) of f(x) of f(x) of f(x) over [a, b].....Now let us see inteinteinteinteintegggggrrrrraaaaationtiontiontiontion from an ANALYSIS point of view. Let us look at two

examples to see : ∫∫∫∫∫aaaaa

bbbbb f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx = CHANGE = CHANGE = CHANGE = CHANGE = CHANGE in in in in in F(F(F(F(F(xxxxx) = F() = F() = F() = F() = F(b) b) b) b) b) −−−−− F(a) . F(a) . F(a) . F(a) . F(a) .

Example 1:Example 1:Example 1:Example 1:Example 1:

11111∫∫∫∫∫22222 f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx = = = = = [[[[[xxxxx]]]]]

11111

22222 = F( F( F( F( F(2) 2) 2) 2) 2) −−−−− F(1) = 2 F(1) = 2 F(1) = 2 F(1) = 2 F(1) = 2 −−−−− 1 = 1 . 1 = 1 . 1 = 1 . 1 = 1 . 1 = 1 .

Example 2:Example 2:Example 2:Example 2:Example 2:

11111∫∫∫∫∫22222 f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx = = = = = [[[[[11111/////22222 xxxxx 22222 ]]]]]

11111

22222 = F( F( F( F( F(2) 2) 2) 2) 2) −−−−− F(1) = F(1) = F(1) = F(1) = F(1) = 11111/////2 2 2 2 2 (2(2(2(2(222222 ))))) −−−−− 11111/////2 2 2 2 2 (1(1(1(1(122222))))) = = = = = 33333/////22222 .....

139

In general:

CHANGE CHANGE CHANGE CHANGE CHANGE in in in in in F(F(F(F(F(xxxxx))))) ===== ∫∫∫∫∫aaaaa

bbbbb f(x)dx = f(x)dx = f(x)dx = f(x)dx = f(x)dx = F(F(F(F(F(b) b) b) b) b) −−−−− F(a) F(a) F(a) F(a) F(a)

What is ∫∫∫∫∫ bbbbb

aaaaa f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx ? ? ? ? ? It must be F(F(F(F(F(a) a) a) a) a) −−−−− F(b) . F(b) . F(b) . F(b) . F(b) .

Apply this to the two examples we just saw.

22222∫∫∫∫∫11111 f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx = = = = = [[[[[xxxxx]]]]]

11111

22222 = F( F( F( F( F(1) 1) 1) 1) 1) −−−−− F(2) = 1 F(2) = 1 F(2) = 1 F(2) = 1 F(2) = 1 −−−−− 2 = 2 = 2 = 2 = 2 = −−−−−1 .1 .1 .1 .1 .

∫∫∫∫∫11111

22222 f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx = = = = = [[[[[11111/////22222 xxxxx22222 ]]]]]

11111

22222 = F( F( F( F( F(1) 1) 1) 1) 1) −−−−− F(2) = F(2) = F(2) = F(2) = F(2) = 11111/////2 2 2 2 2 (1(1(1(1(122222 ))))) −−−−− 11111/////2 2 2 2 2 (2(2(2(2(222222))))) = = = = = −−−−−33333/////22222 .....

Implying that the ar ar ar ar area under the curea under the curea under the curea under the curea under the curvvvvve e e e e of f(x) is NEGATIVE. We know frommiddle school Geometry that there is no such thing as a nenenenenegggggaaaaatititititivvvvve are are are are areaeaeaeaea. So howdo we explain this result ?

In inteinteinteinteintegggggrrrrraaaaation tion tion tion tion the DIRECTION OF THE INTEGRATION has a role to play as weshall see in the next chapter. Moreover, this aspect gives the correct physicalinterpretation.

140

29.29.29.29.29. Dir Dir Dir Dir Direction ofection ofection ofection ofection of InteInteInteInteIntegggggrrrrraaaaationtiontiontiontion and CHANGE in F(x)and CHANGE in F(x)and CHANGE in F(x)and CHANGE in F(x)and CHANGE in F(x)

Let F(x) = ∫ f(x)dx

When we inteinteinteinteintegggggrrrrr aaaaatetetetete the function f(x) over the inter val [a,b] we get avalue or NUMBER. From the Calculus point of view this NUMBER representsa CHANGE in the value of the integral F(x) from a to b.

∫bbbbb

aaaaa f(x)dx = [F(x)]bbbbbaaaaa = F(b) ---------- F(a)

The NUMBER may be positive (> O), meaning an INCREASE in the value of F(x)from a to b. The NUMBER may be negative (< O), meaning a DECREASE in thevalue of F(x) from a to b. The NUMBER may be zero, meaning there is NOCHANGE in the value of F(x) from a to b .

Geometrically Geometrically Geometrically Geometrically Geometrically speaking there is no such thing as a negative area or negative length.We may use the “ar“ar“ar“ar“area under the curea under the curea under the curea under the curea under the curvvvvve”e”e”e”e” f(x) over the interval [ a, b] calculationto find the CHANGE in F(x). We follow the simple rules in the table below to determinethe SIGN of the “ar“ar“ar“ar“area under the curea under the curea under the curea under the curea under the curvvvvve”e”e”e”e” f(x) in our calculation.

This is the usual law of signs under multiplication. So the SIGN of the area underarea underarea underarea underarea underthe curthe curthe curthe curthe curvvvvve e e e e , which is also the SIGN of the CHANGE, depends on both SIGN of thefunction and the direction of integration direction of integration direction of integration direction of integration direction of integration .

SIGN of

integrating f(x) with a < bintegrating f(x) with a < bintegrating f(x) with a < bintegrating f(x) with a < bintegrating f(x) with a < b

“ar“ar“ar“ar“area under the curea under the curea under the curea under the curea under the curvvvvve”e”e”e”e” left left left left left to rightrightrightrightright rightrightrightrightright to leftleftleftleftleft + direction −−−−− direction

curve f(x) is + above the x axis

curve f(x) is −−−−− below the x axis

+++++

+++++−−−−−

−−−−−

∫bbbbb

aaaaa ∫aaaaa

bbbbb

141

InteInteInteInteIntegggggrrrrraaaaation tion tion tion tion has five parts :

1. The integration integration integration integration integration operation to find the integralintegralintegralintegralintegral F(x)

F(x) = the expression of CHANGE = ANTIDERIVATIVE {f(x)} .

2. The interinterinterinterintervvvvvalalalalal, say [a, b] with a < b, over which the operand or inteinteinteinteintegggggrrrrrand and and and and f(x) is

to be integrated.

3. The dirdirdirdirdirection ection ection ection ection ofofofofof inte inte inte inte integggggrrrrraaaaation tion tion tion tion over the interval [a, b] with a < b, where the

integration is being done :

∫bbbbb

aaaaa = positive direction = positive direction = positive direction = positive direction = positive direction = a to b with a < b

∫aaaaa

bbbbb = negative direction = negative direction = negative direction = negative direction = negative direction = b to a with a < b

4. Evaluating the inte inte inte inte integggggrrrrral al al al al F(x) over the interval [a, b] in the given direction to get

something definite definite definite definite definite = CHANGE .

∫∫∫∫∫aaaaa

bbbbb

f(x)dx = f(x)dx = f(x)dx = f(x)dx = f(x)dx = [[[[[F(x)F(x)F(x)F(x)F(x)]]]]]bbbbb

aaaaa = F(F(F(F(F(b) b) b) b) b) −−−−− F(a) F(a) F(a) F(a) F(a)

∫∫∫∫∫bbbbb

aaaaa

f(x)dx = f(x)dx = f(x)dx = f(x)dx = f(x)dx = [[[[[F(x)F(x)F(x)F(x)F(x)]]]]]aaaaa

bbbbb = F(F(F(F(F(a) a) a) a) a) −−−−− F(b) F(b) F(b) F(b) F(b)

5. Finding the CONSTANT OF INTEGRATION.

To get a clear picture and firm grip of the relationship between :

1. dirdirdirdirdirection ofection ofection ofection ofection of inte inte inte inte integggggrrrrraaaaation tion tion tion tion and SIGN of the arararararea under the curea under the curea under the curea under the curea under the curvvvvve e e e e ,

2. SIGN of the arararararea under the curea under the curea under the curea under the curea under the curvvvvve e e e e and the SIGN of the CHANGE ,

3. the correct physical interpretation ,

let us look at all 4 cases in the table using our main example.

142

tT

−−−−− area area area area area under y’(t) over [t1, t2]

= DECREASE in height from t2 to t1

vertical speed y’(t) = u ..... s i n θ ---------- g t (0,0)

y(t11111) y(t

22222)

h11111 h

22222

t11111 t

22222

1234123412341234123412341234

→

tT

+ area+ area+ area+ area+ area under y’(t) over [t1, t2]

= INCREASE in height from t1 to t

2


y(t11111) y(t

22222)

h11111 h

22222

t11111 t

22222

1234123412341234123412341234

→

1. (positive f(x)) 1. (positive f(x)) 1. (positive f(x)) 1. (positive f(x)) 1. (positive f(x)) ..... (positive direction) = positive CHANGE (positive direction) = positive CHANGE (positive direction) = positive CHANGE (positive direction) = positive CHANGE (positive direction) = positive CHANGE

t1

∫t2

y’(t)dt = [y(t)t1

]t2

= y(t2) ---------- y(t1) = INCREASE in height

2. (positive f(x)) 2. (positive f(x)) 2. (positive f(x)) 2. (positive f(x)) 2. (positive f(x)) ..... (negative direction) = negative CHANGE (negative direction) = negative CHANGE (negative direction) = negative CHANGE (negative direction) = negative CHANGE (negative direction) = negative CHANGE

t2

∫t1

y’(t)dt = [y(t)t2

]t1

= y(t1) ---------- y(t2) = DECREASE in height

143

tT

−−−−− area area area area area under y’(t) over [t3, t4]

= DECREASE in height from t3 to t4

(0,0)

h33333 h

44444

t33333 t

44444

12341234123412341234123412341234

→

y(t33333)

y(t44444)

vertical speed y’(t) = u ..... s i n θ ---------- g t

tT

+ area+ area+ area+ area+ area under y’(t) over [t3, t4]

= INCREASE in height from t4 to t

3

(0,0)

h33333 h

44444

t33333 t

44444

1234123412341234123412341234

→

y(t33333)

y(t44444)


3. (negative f(x)) 3. (negative f(x)) 3. (negative f(x)) 3. (negative f(x)) 3. (negative f(x)) ..... (positive direction) = negative CHANGE (positive direction) = negative CHANGE (positive direction) = negative CHANGE (positive direction) = negative CHANGE (positive direction) = negative CHANGE

t3

∫t4

y’(t)dt = [y(t)t3

]t4

= y(t4) ---------- y(t3) = DECREASE in height

4. (negative f(x)) 4. (negative f(x)) 4. (negative f(x)) 4. (negative f(x)) 4. (negative f(x)) ..... (negative direction) = positive CHANGE (negative direction) = positive CHANGE (negative direction) = positive CHANGE (negative direction) = positive CHANGE (negative direction) = positive CHANGE

t4

∫t3 y’(t)dt = [y(t)

t4

]t3

= y(t3) ---------- y(t4) = INCREASE in height

144

tT

zero areazero areazero areazero areazero area under y’(t) over [t1, t4]

= NO CHANGE in height


y(t11111) y(t

44444)

t11111 t

44444

123456123456123456123456123456123456123456123456

123456123456123456123456123456123456

T/////2

y’(t11111)

y’(t44444)

WWWWWhahahahahat does zt does zt does zt does zt does zererererero aro aro aro aro area under the curea under the curea under the curea under the curea under the curvvvvve ye ye ye ye y’(t) impl(t) impl(t) impl(t) impl(t) imply ? NO CHANGE in height.y ? NO CHANGE in height.y ? NO CHANGE in height.y ? NO CHANGE in height.y ? NO CHANGE in height.

In the diagram below let the intervals [ t1, T/////2 ] and [ T/////2 , t4 ] be equal in length.Also let ( t1, y’(t1)) and ( t4, y’(t4)) be equal in length. So from the mensurationmensurationmensurationmensurationmensurationpoint of view the shaded triangles are equal in area. Let us call this area A.

From the CalculationCalculationCalculationCalculationCalculation point of view :

Notice that y’(t) changes SIGN at T/////2 . So we may break up the integral into 2parts.

t1

∫t4 y’(t)dt =

t1

∫T/////2

y’(t)dt + T///// 2

∫t4 y’(t)dt

= [y(t)t1

]T/////2

+ [ y(t)T///// 2

]t4

= { y(T/////2) ---------- y(t1) } + { y(t4) ---------- y(T/////2)}

= y(t4) ---------- y(t1)

145

From the Geometric Geometric Geometric Geometric Geometric point of view :

t1

∫t4 y’(t)dt =

t1

∫T/////2

y’(t)dt + T/////2

∫t4 y’(t)dt

t1

∫T/////2

y’(t)dt = (positive function) ..... (positive direction) = positive area : +++++ A

T/////2∫t4 y’(t)dt = (negative function) ..... (positive direction) = negative area : −−−−− A

So : t1

∫t4 y’(t)dt = +++++ A −−−−− A = 0

From the Analytical Analytical Analytical Analytical Analytical point of view :

t1

∫t4

y’(t)dt = y(t4) ---------- y(t1) = 0

There is NO CHANGE in the height y(t) over interval [ t1, t 4 ] .

NoNoNoNoNow let us REVERSE the dirw let us REVERSE the dirw let us REVERSE the dirw let us REVERSE the dirw let us REVERSE the direction ofection ofection ofection ofection of inte inte inte inte integggggrrrrraaaaation otion otion otion otion ovvvvver interer interer interer interer intervvvvval [al [al [al [al [ t t t t t11111,,,,, ttttt 4 4 4 4 4 ].].].].].

From the CalculationCalculationCalculationCalculationCalculation point of view :

Notice that y’(t) changes SIGN at T/////2 . So we may break up the integral into 2parts.

t4

∫t1

y’(t)dt = t4

∫T/////2

y’(t)dt + T/////2

∫t1

y’(t)dt

= [y(t)t4

]T/////2

+ [ y(t)T///// 2

]t1

= { y(T/////2) ---------- y(t4) } + { y(t

1) ---------- y(T/////2)}

= y(t1) ---------- y(t4)

146

From the Geometric Geometric Geometric Geometric Geometric point of view :

t4

∫t1 y’(t)dt =

t4

∫T/////2

y’(t)dt + T/////2

∫t1 y’(t)dt

t4

∫T/////2

y’(t)dt = (negative function) ..... (negative direction) = positive area : +++++ A

T/////2∫t1 y’(t)dt = (positive function) ..... (negative direction) = negative area : −−−−− A

So : t4

∫t1

y’(t)dt = +++++ A −−−−− A = 0

From the Analytical Analytical Analytical Analytical Analytical point of view :

t4

∫t1 y’(t)dt = y(t1) ---------- y(t4) = 0

There is NO CHANGE in the height y(t) over interval [ t1, t 4 ] .

In general, let f(x) be a contincontincontincontincontinuous uous uous uous uous function over the interval [a, b] andF(x) = ANTIDERIVATIVE {f(x)} . Then :

We note that, even if f(x) changes sign several times over the interval [a, b] , wemay still mechanically look up our TTTTTaaaaabbbbble ofle ofle ofle ofle of Inte Inte Inte Inte Integggggrrrrralalalalalsssss to find F(x) and directlycompute the CHANGE. It is NOT necessary to break up the integration of f(x) intopositive and negative areas as we did above. Let us see a few more examples.

a r ea unde r f ( x ) = 0 a rea unde r f ( x ) = 0 a rea unde r f ( x ) = 0 a rea unde r f ( x ) = 0 a rea unde r f ( x ) = 0 ⇔⇔⇔⇔⇔ CHANGE i n F ( x ) = 0CHANGE i n F ( x ) = 0CHANGE i n F ( x ) = 0CHANGE i n F ( x ) = 0CHANGE i n F ( x ) = 0ooooovvvvver the interer the interer the interer the interer the inter vvvvval [ a,al [ a,al [ a,al [ a,al [ a, b] b] b] b] b] o o o o ovvvvver the interer the interer the interer the interer the intervvvvval [ a,al [ a,al [ a,al [ a,al [ a, b] b] b] b] b]

147

Consider Consider Consider Consider Consider FFFFF(x) = x(x) = x(x) = x(x) = x(x) = x22222..... Then f(x) Then f(x) Then f(x) Then f(x) Then f(x) ===== F'F'F'F'F'(x) (x) (x) (x) (x) ===== 2 x. See the graph. 2 x. See the graph. 2 x. See the graph. 2 x. See the graph. 2 x. See the graph.

Example1:Example1:Example1:Example1:Example1: Integrate contin contin contin contin continuousuousuousuousuous f(x) over the inter val [1, 2] goingLEFT to RIGHT in the positive direction positive direction positive direction positive direction positive direction .

Using the trtrtrtrtr aaaaapepepepepezzzzzoidal methodoidal methodoidal methodoidal methodoidal method a nar row strip or infinfinfinfinfinitesimal elementinitesimal elementinitesimal elementinitesimal elementinitesimal elementof area is :

f( x-- i) ..... δx i

= where δxi = x

i ------ - - - - xi −1

and x--i is the MEAN POINT in [x

i ---------- 1, x i ] such that f( x--

i) = .

This is the area of a narrow trapezium where the parallel sides are f( x i) and

f( x i ---------- 1) , and the perpendicular distance is δxi . We do not need to know x-- i .

n n n n n →→→→→∞∞∞∞∞

F(x) = ∫ f (x)dx = LIMIT Σ f( x-- i) ..... δx i

iiiii ===== 11111

In the LIMIT as n n n n n →→→→→∞∞∞∞∞ each trapezoidal infinitesimal element infinitesimal element infinitesimal element infinitesimal element infinitesimal element o f areaf( x-- i ) ..... δx i becomes an instantaneous element instantaneous element instantaneous element instantaneous element instantaneous element o f area f(x)dx .

F(x) = ∫ f (x)dx = ∫ 2x dx = x2

The AREA under f(x) over the inter va l [1, 2] going LEFT to RIGHT inthe positive directiopositive directiopositive directiopositive directiopositive directionnnnn i s :

∫ 2

1 f(x)dx = ∫

2

1 2x dx = [x2]

2

1

[x2]2

1 = (2)2 ---------- (1)2 = +3

The AREA is 3 in magnitude.The + SIGN+ SIGN+ SIGN+ SIGN+ SIGN is due to (positive f(x)) ..... (positive directionpositive directionpositive directionpositive directionpositive direction ) .

F(2) ---------- F(1) = (2)2 ---------- (1)2 = +3

The CHANGE in F(x) is 3 in magnitude.The + SIGN+ SIGN+ SIGN+ SIGN+ SIGN reflects an INCREASE in F(x) over [1, 2] in the positive directionpositive directionpositive directionpositive directionpositive direction.

[f( x i) + f(x i ---------- 1)] ..... δx i

2f(x

i) + f(x

i ---------- 1)

2

148

We know from mensuration that the area of a trapezoid is = 1/////2 (sum of theparallel sides) . (perpendicular distance). This is the trapezoidal method ofNUMERICAL INTEGRATION. We do not need to know x-- i . However, we must becareful to ensure that the parallel sides f( x i -- 1) and f( x i ) of the trapezium are ofthe same SIGN. So we must watch out for the points where f(x) CHANGES SIGN.In this example f(x) CHANGES SIGN at x = 0.

Note: f(x) δx is an infinitesimalelement of area.

f(x)dx is an instantaneouselement of area. If we try todepict it, it will appear as a line.

123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123

123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123

----------3 ----------2 ----------1 0 1 2 3

4

3

2

1

----------1

----------2

----------3

x

δδδδδx

δδδδδx

f(x) = 2x

F(x) = x2

149

ExampleExampleExampleExampleExample 2:2:2:2:2: Integrate contincontincontincontincontinuousuousuousuousuous f(x) over the inter val [1, 2] goingRIGHT to LEFT in the negative directionegative directionegative directionegative directionegative directionnnnn .

The AREA under f(x) over the inter val [1, 2] going RIGHT to LEFT inthe negat ive direct ion negat ive direct ion negat ive direct ion negat ive direct ion negat ive direct ion is :

∫ 1

2 f(x)dx = ∫

1

2 2x dx = [x2]

1

2

[x2]1

2 = (1)2 ---------- (2)2 = ---------- 3

The AREA is 3 in magnitude.The ---------- SIGN SIGN SIGN SIGN SIGN is due to (positive f(x)) ..... (negative negative negative negative negative directiondirectiondirectiondirectiondirection ) .

F(1) ---------- F(2) = (1)2 ---------- (2)2 = ---------- 3

The CHANGE in F(x) is 3 in magnitude.The ---------- SIGN SIGN SIGN SIGN SIGN reflects a DECREASE in F(x) over [1, 2] in the negative negative negative negative negative directiondirectiondirectiondirectiondirection.

ExampleExampleExampleExampleExample 3 3 3 3 3::::: Integrate contincontincontincontincontinuousuousuousuousuous f(x) over the inter val [----------2, ----------1] goingLEFT to RIGHT in the positive direction positive direction positive direction positive direction positive direction .

The AREA under f(x) over the inter val [----------2, ----------1] going LEFT to RIGHT in theposit ive direct ion posit ive direct ion posit ive direct ion posit ive direct ion posit ive direct ion is :

∫----------1

----------2 f(x)dx = ∫

----------1

----------2 2x dx = [x2]

----------1

----------2

[x2] ----------1

----------2 = (----------1)2 ---------- (----------2)2 = ----------3

The AREA is 3 in magnitude.The ---------- SIGN SIGN SIGN SIGN SIGN is due to (negative f(x)) ..... (positive positive positive positive positive directiondirectiondirectiondirectiondirection ) .

F(----------1) ---------- F(----------2) = (----------1)2 ---------- (----------2)2 = ----------3

The CHANGE in F(x) is 3 in magnitude.The ---------- SIGN SIGN SIGN SIGN SIGN reflects a DECREASE in F(x) over [----------2, ----------1] in the positive positive positive positive positive directiondirectiondirectiondirectiondirection.

150

ExampleExampleExampleExampleExample 4 4 4 4 4::::: Integrate contincontincontincontincontinuous uous uous uous uous f(x) over the inter val [----------2, ----------1] goingRIGHT to LEFT in the negative direction negative direction negative direction negative direction negative direction .

The AREA under f(x) over the interval [----------2, ----------1] going RIGHT to LEFT in the inthe negative direction negative direction negative direction negative direction negative direction is :

∫----------2

----------1 f(x)dx = ∫

----------2

----------1 2x dx = [x2]

----------2

----------1

[x2]----------2

----------1 = (----------2)2 ---------- (----------1)2 = +3

The AREA is 3 in magnitude.The +++++ SIGN SIGN SIGN SIGN SIGN is due to (negative f(x)) ..... (negative direction negative direction negative direction negative direction negative direction ) .

F(----------2) ---------- F(----------1) = (----------2)2 ---------- (----------1)2 = +3

The CHANGE in F(x) is 3 in magnitude.The +++++ SIGN SIGN SIGN SIGN SIGN reflects an INCREASE in F(x) over [----------2, ----------1] in the negative directionnegative directionnegative directionnegative directionnegative direction.

Le t us now cons ide r F (x ) =Le t us now cons ide r F (x ) =Le t us now cons ide r F (x ) =Le t us now cons ide r F (x ) =Le t us now cons ide r F (x ) = ---------- xxxxx22222. . . . . Then f ( x ) = Then f ( x ) = Then f ( x ) = Then f ( x ) = Then f ( x ) = FFFFF ’( x ) = (x ) = (x ) = (x ) = (x ) = ---------- 2x2x2x2x2xand ver i fy our calculat ions with the graph.and ver i fy our calculat ions with the graph.and ver i fy our calculat ions with the graph.and ver i fy our calculat ions with the graph.and ver i fy our calculat ions with the graph.

Example 5:Example 5:Example 5:Example 5:Example 5: Integrate contincontincontincontincontinuousuousuousuousuous f(x) over the inter val [1, 2] goingLEFT to RIGHT in the positive direction positive direction positive direction positive direction positive direction .

The AREA under f(x) over the inter val [1, 2] going LEFT to RIGHT inthe pos i t i ve d i rect ion pos i t i ve d i rect ion pos i t i ve d i rect ion pos i t i ve d i rect ion pos i t i ve d i rect ion is :

∫2

1 f(x)dx = ∫

2

1 ---------- 2x dx = [----------x2]

2

1

[----------x2]2

1 = ----------(2)2 ---------- ( ---------- (1)2 ) = ----------4 ---------- (----------1) = ---------- 3

The AREA is 3 in magnitude.The ---------- SIGN SIGN SIGN SIGN SIGN is due to (negative f(x)) ..... (positive direction positive direction positive direction positive direction positive direction ) .

151

123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123

123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123123

----------3 ----------2 ----------1 0 1 2 3

4

3

2

1

----------1

----------2

----------3

----------4

x

F(x) = ---------- x2

f(x) = ---------- 2x

δδδδδx

δδδδδx

151

Note: f(x) δx is an infinitesimalelement of area.

f(x)dx is an instantaneouselement of area. If we try todepict it, it will appear as a line.

152

∫ a

b f (x)dx = ---------- ∫

b

a f (x)dx

RIGHT to LEFT = ---------- LEFT to RIGHTover [a,b] over [a,b]

F(2) ---------- F(1) = ----------(2)2 ---------- (---------- (1)2 ) = ---------- 3

The CHANGE in F(x) is 3 in magnitude.The ---------- SIGN SIGN SIGN SIGN SIGN reflects a DECREASE in F(x) over [1, 2] in the positive direction positive direction positive direction positive direction positive direction .

Example Example Example Example Example 66666::::: Integrate contincontincontincontincontinuousuousuousuousuous f(x) over the inter val [1, 2] goingRIGHT to LEFT in the negative negative negative negative negative direction direction direction direction direction .

The AREA under f(x) over the inter val [1, 2] going RIGHT to LEFT inthe negativenegativenegativenegativenegative d i rec t ion d i rec t ion d i rec t ion d i rec t ion d i rec t ion i s :

∫ 1

2 f(x)dx = ∫ 1

2 ---------- 2x dx = [---------- x2]1

2

[---------- x2]1

2 = ---------- (1)2 ---------- ( ---------- (2)2 ) = ---------- 1 + 4 = + 3

The AREA is 3 in magnitude.

The +++++ SIGN SIGN SIGN SIGN SIGN is due to (negative f(x)) ..... (negative direction negative direction negative direction negative direction negative direction ) .

F(1) ---------- F(2) = ----------(1)2 ---------- (---------- (2)2 ) = + + + + + 3

The CHANGE in F(x) is 3 in magnitude.

The +++++ SIGN SIGN SIGN SIGN SIGN reflects an INCREASE in F(x) over [1, 2] in the negative directionnegative directionnegative directionnegative directionnegative direction.

153

In all the previous examples f(x) did not CHANGE SIGN over the interval of integration.Now let us see two examples where f(x) CHANGES SIGN over the interval of integration.

Example Example Example Example Example 77777::::: Integrate contincontincontincontincontinuousuousuousuousuous f(x) = 2x over the inter val [----------1, 2]going LEFT to RIGHT in the positive positive positive positive positive direction direction direction direction direction . See diagram page 148.

The AREA under f(x) = 2x over the inter val [ ----------1, 2] going LEFT to RIGHTin the positive positive positive positive positive d i rec t ion d i rec t ion d i rec t ion d i rec t ion d i rec t ion is :

∫ 2

----------12x dx = ∫

0

----------12x dx + ∫

2

0 2x dx

We must break this up into 2 parts because f(x) CHANGES SIGN at x = 0 .

∫ 0

----------12x dx = [x2]

0

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

The AREA is 1 in magnitude.The ---------- SIGN SIGN SIGN SIGN SIGN is due to (negative f(x)) ..... ( positive direction positive direction positive direction positive direction positive direction ) .

∫2

0 2x dx = [x2]

2

0 = (2)2 ---------- (0)2 = 4 ---------- 0 = +++++4

The AREA is 4 in magnitude.The +++++ SIGN SIGN SIGN SIGN SIGN is due to (positive f(x)) ..... (positive direction positive direction positive direction positive direction positive direction ) .

So: ∫ 2

----------1 2x dx = ----------1 +++++ 4 = + 3

F(2) ---------- F(----------1) = ∫ 2

----------1 2x dx = [x2] 2

----------1

= (2)2 ---------- (----------1)2 = 4 ---------- 1 = +3


The +++++ SIGN SIGN SIGN SIGN SIGN reflects an INCREASE in F(x) over [---------- 1, 2] in the positive directionpositive directionpositive directionpositive directionpositive direction.

154

Example Example Example Example Example 88888::::: Integrate contincontincontincontincontinuousuousuousuousuous f(x) = 2x over the inter val [----------1, 2]going RIGHT to LEFT in the negative negative negative negative negative direction direction direction direction direction . See diagram page 148.

The AREA under f(x) = 2x over the inter val [ ----------1, 2] going RIGHT to LEFTin the negative negative negative negative negative d i rec t ion d i rec t ion d i rec t ion d i rec t ion d i rec t ion is :

∫2

----------12x dx = ∫

2

02x dx + ∫

0

----------12x dx

We must break this up into 2 parts because f(x) CHANGES SIGN at x = 0 .

∫2

02x dx = [x2]

0

----------1 = (0)2 ---------- (2)2 = 0 ---------- 4 = ---------- 4

The AREA is 4 in magnitude.The ---------- SIGN SIGN SIGN SIGN SIGN is due to (positive f(x)) ..... (negative direction negative direction negative direction negative direction negative direction ) .

∫0

----------12x dx = [x2]

2

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

The AREA is 1 in magnitude.The +++++ SIGN SIGN SIGN SIGN SIGN is due to (negative f(x)) ..... (negative direction negative direction negative direction negative direction negative direction ) .

So: ∫2

----------12x dx = ---------- 4 +++++ 1 = ---------- 3

F(----------1) ---------- F(2) = ∫2

----------12x dx = [x2]

2

----------1

= (----------1)2 ---------- (2)2 = 1 ---------- 4 = ----------3


The ---------- SIGN SIGN SIGN SIGN SIGN reflects a DECREASE in F(x) over [----------1, 2] in the negative directionnegative directionnegative directionnegative directionnegative direction.

155

A note on A note on A note on A note on A note on AREA and INTEGRAAREA and INTEGRAAREA and INTEGRAAREA and INTEGRAAREA and INTEGRATIONTIONTIONTIONTION

Several attempts have been made to equate AREA under f(x) over [a, b] from amensuration mensuration mensuration mensuration mensuration point of view with the CHANGE in the INTEGRAL {F(b) −−−−− F(a)} for allpossisble f(x). These attempts have not been successful because they failed to takeinto account the influence of the direction of integrationdirection of integrationdirection of integrationdirection of integrationdirection of integration over [a, b].

WWWWWe mae mae mae mae may use the INTEGRAL F(x) to fy use the INTEGRAL F(x) to fy use the INTEGRAL F(x) to fy use the INTEGRAL F(x) to fy use the INTEGRAL F(x) to find the ind the ind the ind the ind the AREA under continAREA under continAREA under continAREA under continAREA under continuousuousuousuousuousf(x) over [a, b] when :f(x) over [a, b] when :f(x) over [a, b] when :f(x) over [a, b] when :f(x) over [a, b] when :

1. f(x) is positive positive positive positive positive (above the x-axis) and the direction of integration direction of integration direction of integration direction of integration direction of integration is positivepositivepositivepositivepositive.In this case :

AREA under f(x) =

CHANGE in the INTEGRAL F(x) = F(b) −−−−− F(a) over [a, b] over [a, b]

In both SIGN and MAGNITUDE as we saw in the examples 1 and 2.

2. If f(x) is negative negative negative negative negative (below the x-axis) then :

AREA under f(x) =

[ F(b) −−−−− F(a) ] when the direction of integrationdirection of integrationdirection of integrationdirection of integrationdirection of integration over [a, b] over [a, b] is positivepositivepositivepositivepositive


AREA under f(x) =

−−−−− [ F(b) −−−−− F(a) ] when the direction of integrationdirection of integrationdirection of integrationdirection of integrationdirection of integration over [a, b] over [a, b] is negativenegativenegativenegativenegative


156

WWWWWe mae mae mae mae may use the y use the y use the y use the y use the AREA under continAREA under continAREA under continAREA under continAREA under continuous f(x) ouous f(x) ouous f(x) ouous f(x) ouous f(x) ovvvvver [a,er [a,er [a,er [a,er [a, b] to f b] to f b] to f b] to f b] to find theind theind theind theind the

CHANGE in the CHANGE in the CHANGE in the CHANGE in the CHANGE in the INTEGRALINTEGRALINTEGRALINTEGRALINTEGRAL F(x) over [a, b] in the positive direction F(x) over [a, b] in the positive direction F(x) over [a, b] in the positive direction F(x) over [a, b] in the positive direction F(x) over [a, b] in the positive direction

as follows.as follows.as follows.as follows.as follows.

1. We must break up the AREA under f(x) over [a, b] into segments of positipositipositipositipositivvvvveeeee

areasareasareasareasareas and negative areasnegative areasnegative areasnegative areasnegative areas as we did in example 7.

positive areapositive areapositive areapositive areapositive area = area above the x-axis

negative areanegative areanegative areanegative areanegative area = area below the x-axis

Here we implicitly assumed that the direction of integrationdirection of integrationdirection of integrationdirection of integrationdirection of integration is positive positive positive positive positive .

2. We SUM UP all the nenenenenegggggaaaaatititititivvvvve e e e e and positipositipositipositipositivvvvve e e e e areas to get a net result, say A . Then:

CHANGE in F(x) over [a, b] in the positive directionpositive directionpositive directionpositive directionpositive direction = F(b) −−−−− F(a) = A

E X E R C I S E SE X E R C I S E SE X E R C I S E SE X E R C I S E SE X E R C I S E S

The exercises are to show the relationship between the area area area area area under theunder theunder theunder theunder the

curcurcurcurcur vvvvveeeee f(x) = F’(x) and the CHANGE IN VALUE of the Integral F(x).

Exercise Exercise Exercise Exercise Exercise 11111::::: Repeat examples 7 and 8 with f(x) = ----------2x

ExExExExExererererercise cise cise cise cise 22222::::: Integrate f(x) = ----------2x over the inter val [----------2, ----------1] from LEFT to

RIGHT. Verify it against the CHANGE in F(x) = ----------x2 from ----------2 to ----------1. What

is the area under the curve ?

ExExExExExererererercise cise cise cise cise 33333::::: Integrate f(x) = ----------2x over the interval [----------2, ----------1] from RIGHT

to LEFT. Verify it against the CHANGE in F(x) = ----------x2 from ----------1 to ----------2. What

is the area under the curve ?

Exercise Exercise Exercise Exercise Exercise 44444::::: What is the CHANGE in the value of the function sin (x) over

the inter vals [0, π/2], [ π/2, 3π/2], [π , 2π], [0, 2π ]. Use the fact that

the INTEGRAL sin (x) is the ANTIDERIVATIVE of cos (x). Draw the curve of

cos (x) over [0, 2π ]. For each interval verify that the CHANGE IN AREA is

equal to the CHANGE IN VALUE of sin (x) over the interval going LEFT to

RIGHT. Repeat the exercise going RIGHT to LEFT.

157

30. 30. 30. 30. 30. Area under f(x) Area under f(x) Area under f(x) Area under f(x) Area under f(x) and Plottingand Plottingand Plottingand Plottingand Plotting FFFFFCCCCC (((((x)x)x)x)x)In the previous chapter we saw the relation between the arararararea under the curea under the curea under the curea under the curea under the curvvvvveeeeeof contincontincontincontincontinuous uous uous uous uous f(x) over the interval [a, b] and the CHANGE in F(x) over [a, b].

Let: F−−−−−11111(x) = x2 −−−−−1 ⇒ f−−−−−11111

(x) = F’−−−−−11111(x) = 2x

F0(x) = x2 ⇒ f0(x) = F’0(x) = 2x

and F+1+1+1+1+1(x) = x2 +1 ⇒ f

+1+1+1+1+1(x) = F’+1+1+1+1+1(x) = 2x

In all 3 cases : f(x) = F’(x) = 2x.

Also, in all 3 cases: F(x) = ANTIDERIVATIVE {f(x)} = ∫ f(x) dx = x2 .

Again, in all 3 cases the CHANGE over the inter val [a, b] is the same:

F−−−−−11111(b) −−−−− F−−−−−11111

(a) = F00000(b) −−−−− F00000(a) = F+++++11111(b) −−−−− F+++++11111

(a) = b2 −−−−− a2 .

F−−−−−11111(a) −−−−− F−−−−−11111(b) = F00000(a) −−−−− F00000(b) = F+++++11111(a) −−−−− F+++++11111(b) = a2 −−−−− b2 .

So given only f(x) how can we compute FSo given only f(x) how can we compute FSo given only f(x) how can we compute FSo given only f(x) how can we compute FSo given only f(x) how can we compute FCCCCC(x) ?(x) ?(x) ?(x) ?(x) ?

Given contincontincontincontincontinuous uous uous uous uous f(x) we can find F00000(x) and plot the graph of F00000(x) directly. Wesaw the plot of F00000(x) in the 2 examples in chapter 27. But we cannot plot thegraphs of F−−−−−11111(x) and F+++++11111(x) directly. This is because we do not know the constantterm. The constant −−−−− 1 in F−−−−−11111(x) and + 1 in F+++++11111(x) is known as the CONSTANTOF INTEGRATION and is denoted by C C C C C . In the case of F00000(x) the constant CCCCC = 0.

For each value of C C C C C we have a curve similar to the curve of F00000(x). In fact,FC C C C C (x) = F(x) + C C C C C is a family of cur ves. This is the Geometric Geometric Geometric Geometric Geometric view of therole of C C C C C , the CONSTANT OF INTEGRATION.

158

----------3 ----------2 ----------1 0 1 2 3

x

F−−−−−11111(x) = x2 −−−−− 1

F0(x) = x2

F+1+1+1+1+1

(x) = x2 +14

3

2

1

----- 1

----- 2

----- 3

----- 4

f(x) =

2x

159

To know FC C C C C (x) we need to know CCCCC . To calculate C C C C C we need to know FC C C C C (x) at somepoint, say x

0 . Then from F

C C C C C (x) = F(x) + C C C C C we have : C C C C C = FC C C C C (x0

) −−−−− F(x0) .

In F0 0 0 0 0 (x) = x2 the constant C C C C C = 0 since F

0 0 0 0 0 (x) = F(x) everywhere. If weknow F(x) and C C C C C then it is easy to plot F

C C C C C (x) .

Suppose we do NOT know F(x) and C . We only know f(x) , x0 and F

C C C C C (x0).

How can we plot How can we plot How can we plot How can we plot How can we plot FFFFFC C C C C (x) without finding the expression (x) without finding the expression (x) without finding the expression (x) without finding the expression (x) without finding the expression FFFFFC C C C C (x) ?(x) ?(x) ?(x) ?(x) ?

If F(a)F(a)F(a)F(a)F(a) is known, then from aaaaa going LEFT to RIGHT in the positive directiopositive directiopositive directiopositive directiopositive directionnnnn to bbbbb :

∫∫∫∫∫bbbbb

aaaaaf(x)dx f(x)dx f(x)dx f(x)dx f(x)dx ===== F(b) F(b) F(b) F(b) F(b) ---------- F(a)F(a)F(a)F(a)F(a)

F(F(F(F(F(bbbbb) ) ) ) ) ===== ∫∫∫∫∫bbbbb

aaaaaf(x)dxf(x)dxf(x)dxf(x)dxf(x)dx + + + + + F(F(F(F(F(aaaaa)))))

If F(F(F(F(F(bbbbb))))) is known, then from bbbbb going RIGHT to LEFT in the negative directionegative directionegative directionegative directionegative directionnnnn to aaaaa :

∫∫∫∫∫aaaaa

bbbbb f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx ===== F(F(F(F(F(aaaaa) ) ) ) ) ---------- F(F(F(F(F(bbbbb)))))

F(a) F(a) F(a) F(a) F(a) ===== ∫∫∫∫∫aaaaa

bbbbb f(x)dx f(x)dx f(x)dx f(x)dx f(x)dx + F(b)F(b)F(b)F(b)F(b)

Using the relations above we may compute FC C C C C (x) . We must choose a

small inter val ∆x . Then {f(x) ..... ∆x} is an element of arararararea under theea under theea under theea under theea under thecurcurcurcurcur vvvvve e e e e of f(x) over the inter val ∆x . The smaller the ∆x the more accuratethe approximation of the arararararea under the curea under the curea under the curea under the curea under the cur vvvvveeeee of f(x) over the interval∆x . Another reason why we have to choose ∆x very small is that we do notskip or miss the EXTREMUM points (maximum and minimum) in plottingF

C C C C C (x). These are the points where f(x) CHANGES SIGN.

next FC C C C C (x) = CHANGE in F

C C C C C (x) over next ∆x + current value of FC C C C C (x)

CHANGE in FC C C C C (x) over next ∆x = arararararea under the curea under the curea under the curea under the curea under the curvvvvveeeee of f(x) over next ∆x

160

Going Going Going Going Going RIGHT RIGHT RIGHT RIGHT RIGHT ( in the( in the( in the( in the( in the post ive d irect ion post ive d irect ion post ive d irect ion post ive d irect ion post ive d irect ion) f rom x) f rom x) f rom x) f rom x) f rom x0 0 0 0 0 w i th i = 1 :wi th i = 1 :wi th i = 1 :wi th i = 1 :wi th i = 1 :

FFFFFC C C C C (x(x(x(x(xiiiii ))))) = = = = = f(x) dx + f(x) dx + f(x) dx + f(x) dx + f(x) dx + FFFFFC C C C C (x(x(x(x(x

iiiii −−−−− 11111 )))))

FFFFFC C C C C ( x(x(x(x(x iiiii ))))) ≅≅≅≅≅ fffff ( x(x(x(x(x iiiii ))))) .....( +( +( +( +( +∆∆∆∆∆xxxxx) ) ) ) ) + + + + + FFFFFC C C C C ( x(x(x(x(xiiiii −−−−− 11111 ))))) for i = for i = for i = for i = for i = 1, 2, 3, . . .1, 2, 3, . . .1, 2, 3, . . .1, 2, 3, . . .1, 2, 3, . . .

The + + + + + SIGN in ( +∆x) is due to the postive directionpostive directionpostive directionpostive directionpostive direction in the integration.

Going Going Going Going Going LEFT LEFT LEFT LEFT LEFT ( in the( in the( in the( in the( in the negat ive d irect ion negat ive d irect ion negat ive d irect ion negat ive d irect ion negat ive d irect ion) f rom x) f rom x) f rom x) f rom x) f rom x0 0 0 0 0 w i th i = wi th i = wi th i = wi th i = wi th i = −−−−−1:1:1:1:1:

FFFFFC C C C C (x(x(x(x(x iiiii ))))) = = = = = f(x) dx + f(x) dx + f(x) dx + f(x) dx + f(x) dx + FFFFFC C C C C (x(x(x(x(xiiiii +++++11111 )))))

FFFFFC C C C C (x(x(x(x(xiiiii))))) ≅≅≅≅≅ f f f f f(x(x(x(x(xiiiii))))).....(((((−−−−− ∆∆∆∆∆xxxxx) ) ) ) ) + + + + + FFFFFC C C C C (x(x(x(x(xiiiii+++++11111))))) for i = for i = for i = for i = for i = −−−−−1, 1, 1, 1, 1, −−−−−2, 2, 2, 2, 2, −−−−−3, . . .3, . . .3, . . .3, . . .3, . . .

The − SIGN in ( −−−−− ∆x) is due to the negative directionnegative directionnegative directionnegative directionnegative direction in the integration.

This is the rrrrrectangular method ectangular method ectangular method ectangular method ectangular method of NUMERICAL INTEGRATION. In the nextchapter we shall see how to compute and interpret the meaning of theCONSTANT OF INTEGRATION from the nature of the problem.

Wha t i s the d i f f e rence be tween F(x ) and Wha t i s the d i f f e rence be tween F(x ) and Wha t i s the d i f f e rence be tween F(x ) and Wha t i s the d i f f e rence be tween F(x ) and Wha t i s the d i f f e rence be tween F(x ) and FFFFF C C C C C ( x(x(x(x(x) from a) from a) from a) from a) from aprprprprpractical or Phactical or Phactical or Phactical or Phactical or Physics point ofysics point ofysics point ofysics point ofysics point of vie vie vie vie view ? w ? w ? w ? w ? Let us look at our main example.

If F(x) is the expression of CHANGE in positionpositionpositionpositionposition, then the evaluation of F(x) overthe interinterinterinterintervvvvval al al al al [a, b] is the CHANGE in position position position position position from a to b = F(b) −−−−− F(a).We cannot use F(x) to find the position position position position position at any chosen instant instant instant instant instant x i .

Whereas, FC (x) = F(x) + CCCCC evaluated at any chosen instant instant instant instant instant x i gives theposition position position position position FC (x i ) at the instant instant instant instant instant x i .

∫

xxxxxiiiii −−−−− 11111

+++++ ∆ ∆ ∆ ∆ ∆xxxxx

xxxxxiiiii −−−−− 11111

∫

xxxxxiiiii +++++11111

−−−−− ∆ ∆ ∆ ∆ ∆xxxxx

xxxxxiiiii +++++ 11111

161

123456789123456789123456789123456789123456789123456789123456789123456789123456789

123456789123456789123456789123456789123456789123456789123456789123456789123456789

1

0 1 2 x

f(x) = 1

1 2 x

2

1

0

FFFFF00000(x) =(x) =(x) =(x) =(x) = xxxxx

E X E R C I S E SE X E R C I S E SE X E R C I S E SE X E R C I S E SE X E R C I S E S

The exercises are to show how to plot FC C C C C (x) when only f(x), x 0 and F

C C C C C (x 0)

are known and the constant of integration C is not known. Also, we do notneed to know the expression of F(x). We make use of the re la t ionshipbetween the “arararararea ea ea ea ea under the curunder the curunder the curunder the curunder the cur vvvvveeeee” f(x) = F’(x) and the CHANGEI N VA LU E o f t h e I n t e g r a l F(x) . O n a c o m p u t e r t h i s c a n b edemonstrated ver y e legant ly.

Dr aw the cur ve of the F ir st Der i vat i ve f(x). Shade the area you areintegrat ing and s imultaneously p lot i ts value. You wi l l get the cur veof the Integra l F

C C C C C (x ) . Do th is LEFT to RIGHT(in the postive directionpostive directionpostive directionpostive directionpostive direction)

from x 0 and then RIGHT to LEFT (in the negative directionnegative directionnegative directionnegative directionnegative direction) from x 0 .Choose a suitable ∆x .

Exercise Exercise Exercise Exercise Exercise 11111::::: Given f(x) = 1 use the rectangular methodrectangular methodrectangular methodrectangular methodrectangular method to plot F00000

(x)over [ −−−−−2, 0].

162


1 2 x

2

1

0

123456789123456789123456789123456789123456789123456789123456789123456789123456789

12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890

0 1 2 3 4 x

4

3

2

1

FFFFF00000(x) =(x) =(x) =(x) =(x) = 11111/////22222 xxxxx 22222

Exercise Exercise Exercise Exercise Exercise 22222::::: Given f(x) = x use the trapezoidal methodtrapezoidal methodtrapezoidal methodtrapezoidal methodtrapezoidal method to plot F00000

(x)over [ −−−−−2, 0].

ExExExExExererererer c ise c ise c ise c ise c ise 33333::::: On a graph paper draw the curve of the function f(x) = 2xover the inter val [−−−−−4, 4]. Let x0 = 1 and F

C C C C C (x0) = 2 . Plot FC C C C C (x) over

inter val [−−−−−4, 4].

ExExExExExererererer c ise c ise c ise c ise c ise 44444::::: Repeat exerc ise 3 over the inter va l [−−−−−2, 2] us ing asmaller ∆x.

Exerc ise Exerc ise Exerc ise Exerc ise Exerc ise 55555::::: Repeat exercises 3 and 4 this time using f(x) = x2. Comparethe graph you get with the graph of F(x)= x3///// 3 + 1 .

In any kind of NUMERICAL INTEGRATION to find the CHANGE in F(x) or tofind F

C C C C C (x) we must take into consideration :

1. The instants or points where f(x) CHANGES SIGN.

2. The direct ion of integrat iondirect ion of integrat iondirect ion of integrat iondirect ion of integrat iondirect ion of integrat ion.

163

31.31.31.31.31. Constant of Constant of Constant of Constant of Constant of Inte Inte Inte Inte IntegggggrrrrraaaaationtiontiontiontionAny continuous function f(x) has an infinity of ANTIDERIVATIVES. If F(x) is oneof them, then any other one may be given by the expression F(x) + C, whereC is constant. From a Geometric Geometric Geometric Geometric Geometric point of view we have a family of curves.

∫ RATE OF CHANGE f(x) = CHANGE F(x) + C ( a constant )

The constant C is called the CONSTANT OF INTEGRATION. C could be anarbitrary constant. But in general it depends on the situation.

In general: ∫ f n(x)dx = f n ---------- 1(x) + C n ---------- 1

Let us see how we determine the constant of integration C n ---------- 1 by looking at threeexamples.EEEEExample1:xample1:xample1:xample1:xample1: We know y”(t), the function that expresses the INSTANTANEOUSRATE OF CHANGE in ver tical speed (i.e. acceleration). We can integrate y”(t)and get y’(t) the function that expresses the CHANGE in ver tical speed.

We may think of y’(t), the function that expresses the CHANGE in ver ticalspeed, as the function that expresses the INSTANTANEOUS RATE OF CHANGEin height. We can integrate y’(t) and get y(t) the function that expresses theCHANGE in height.

We know : y ’’(t) = ---------- g [ meters / sec 2 ]

y ’(t) = ∫ y ’’( t) dddddt = ∫ ( ---------- g ) dddddt = ---------- g t + C 1 [ meters/ sec ].

The constant C1 = u.....sinθ is the initial ver tical velocity.

So y ’(t) = u.....sinθ ---------- gt [ meters/ sec ].

Now: y (t) = ∫ y ’(t) dddddt = ∫ ( u.....sinθ ---------- gt ) dddddt = u.....si n θ ..... t ---------- 1 -- 2 g t 2

y C 0

(t) = u.....si n θ ..... t ---------- 1 -- 2 g t 2 + C 0 [ meters ].

164

How do you determine the constant of integration C 0 ?

tTT///// 2 + 5

height y5(t) = u ..... s i n θ ..... t ---------- 1--2 g t 2 + 5

y ’(t) = u ..... s i n θ ---------- g t

Case 2: 5 meters above ground level

In both cases y ’(t) is the same. So : y (t) = ∫ y ’( t) dddddt = ∫ ( u.....sinθ ---------- gt ) dddddt

So: y (t) = u.....si n θ ..... t ---------- 1 -- 2 g t 2 is also the same.We need to know something more about the initial conditions. For example :

Case 1: C 0 = 0 at ground levelCase 2: C 0 = + 5 meters above ground level.

tTT///// 2 (0,0)

height y0(t) = u ..... s i n θ ..... t ---------- 1--2 g t 2

y ’(t) = u ..... s i n θ ---------- g t

Case 1: ground level

+5

met

ers

165

y(ti ) = ∫ tiiiii

0 y’(t).....dt gives us only the CHANGE in height. It does NOT tell us the heightat instant ti . We must know the initial height at time t = 0, which in this case is theconstant of integration C 0 . Then :

HEIGHT at instant ti = CHANGE in height + INITIAL height

= y(ti ) + C 0

Example 2Example 2Example 2Example 2Example 2: We know the relationship between temperature t t t t t measuredin oF and oC .

oF (t) = 9/////5 t oC + 32

∫ oF ’(t) dt = 9/////5 t oC + C 0

How do we determine the constant of integration C0 ? We must know oF (t) at t = 0.We may conduct an experiment. We immerse the oF thermometer and the oCthermometer in a beaker of ice which we know is 0 oC . Then we read the oFthermometer to get C 0 = 32. So:

oF (t) = 9/////5 t oC + 32

ExExExExExampleampleampleampleample 33333::::: A pump delivers with a rate of flow f ’(ttttt) = 60 litres/minute.

What is the volume of water in the tank after 5 minutes ?

f(t) = ∫

f ’(ttttt) . dt = 60 ttttt + C0

t=

∫t=0

5f ’(ttttt) ..... dt =

t=

∫t=0

560 ..... dt =[60 ttttt ]

5

0 litres = 300 litres

oF ’(t) = doF(t)

= 9/////5 dt

166

However, this is NOT the volume of water in the tank. This is only the CHANGE involume. We must know the INITIAL volume, say C0 = 500 litres. Then the volume ofwater in the tank at t t t t t = = = = = 5 minutes is :

VOLUME (at t t t t t = = = = = 5 minutes) = CHANGE in volume + INITIAL volume

= 300 litres + 500 litres

= 800 litres.

This example may be easily adapted to compute the charge on a capacitor and alsothe time taken to charge or discharge the capacitor given f ’(ttttt) = the rate of flow ofcharge = current iiiii = dq/////dt. Let us look at this very interesting in example in a moregeneral way .

∫

( RATE of FLOW). dt = CHANGE in VOLUME

There are 4 entities involved: f ’(ttttt) = RATE of FLOW, CHANGE in VOLUME, START TIMEand STOP TIME. Usually we know f ’(ttttt) . Now we have 3 possibilities.

1. If we know the START TIME and STOP TIME we may compute the CHANGE inVOLUME for the given RATE of FLOW f ’(ttttt) .

2. If we know the START TIME and CHANGE in VOLUME we may determine theSTOP TIME. Then ( STOP TIME −−−−− START TIME) will tell us how much time it tookto pump a certain volume of water, or charge capacitor from the given STARTTIME.

3. If we know the STOP TIME and CHANGE in VOLUME, then we may determinethe START TIME for the given RATE of FLOW f ’(ttttt) .

START TIME

STOP TIME

167

32. INDEFINITE and DEFINITE INTEGRAL32. INDEFINITE and DEFINITE INTEGRAL32. INDEFINITE and DEFINITE INTEGRAL32. INDEFINITE and DEFINITE INTEGRAL32. INDEFINITE and DEFINITE INTEGRAL

Let f(x) = x2. When we difdifdifdifdifffffferererererentiaentiaentiaentiaentiate te te te te f(x) to get the derideriderideriderivvvvvaaaaatititititivvvvve e e e e f’(x) = 2x , we donot use the term “indef“indef“indef“indef“indefinite”inite”inite”inite”inite” to say : f’(x) is the “indef“indef“indef“indef“indefinite deriinite deriinite deriinite deriinite derivvvvvaaaaatititititivvvvve”e”e”e”e” of f(x).Also, when we evaluate f’(x) = 2x at some instant aaaaa to get 2a 2a 2a 2a 2a , we do not use theterm “def“def“def“def“definite”inite”inite”inite”inite” to say : 2a2a2a2a2a is the “def“def“def“def“definite deriinite deriinite deriinite deriinite derivvvvvaaaaatititititivvvvve”e”e”e”e” of f(x) at x = a.When it comes to inteinteinteinteintegggggrrrrraaaaationtiontiontiontion we have the terms indefindefindefindefindefinite inite inite inite inite and defdefdefdefdefiniteiniteiniteiniteinite.

F(x) = ANTIDERIVATIVE {f(x)} = ∫ f (x)dx is called the INDEFINITE INTEGRAL ofthe integrand integrand integrand integrand integrand f(x). The INDEFINITE INTEGRAL F(x) of a given function f(x) isthe most general form of its ANTIDERIVATIVE.

FC (x) = F(x) + CCCCC = ∫ f (x)dx+ C C C C C is called the DEFINITE INTEGRAL of theinteinteinteinteintegggggrrrrrand and and and and f(x). C C C C C is called the CONSTANT OF INTEGRATION.

We may evaluate the INDEFINITE INTEGRAL F(x) over and interval [a, b].

∫ b

a f(x)dx = [F(x)]ba = F(b) −−−−− F(a) which is the CHANGE in F(x) from a to b.

We may evaluate the DEFINITE INTEGRAL FC (x) = F(x) + CCCCC over and interval [a, b].In the process the CONSTANT OF INTEGRATION CCCCC disappears.

∫ b

a f(x)dx = [FC (x)]ba = { F(b)+ CCCCC } −−−−− { F(a) + CCCCC } = F(b) −−−−− F(a) which is

the CHANGE in FC (x) from a to b.

If F(x) is the expression of CHANGE in positionpositionpositionpositionposition, then evaluation of the INDEFINITEINTEGRAL F(x) over the interinterinterinterinter vvvvval al al al al [a, b] is the CHANGE in position position position position position froma to b = F(b) −−−−− F(a). We cannot use the INDEFINITE INTEGRAL F(x) to find theposition position position position position at any chosen instant instant instant instant instant x i .

Whereas, the DEFINITE INTEGRAL FC (x) = F(x) + CCCCC evaluated at any choseninstant instant instant instant instant x i gives the position position position position position FC (x i ) at the instant instant instant instant instant x i .

168

tT

vertical position y(t) = u ..... s i n θ ..... t ---------- 1--2 g t 2


y(t11111) y(t

22222)

h11111 h

22222

t11111 t

22222

123123123123123123123

INDEFINITE INTEGRAL y(t) = ∫ y’(t)dt = ∫ { u ..... s i n θ ---------- g t }dt

= u ..... s i n θ ..... t ---------- 1--2 g t 2

= ANTI-DERIVATIVE of y’(t)= area under y’(t)= expression of CHANGE in y(t) .

t1

∫t2

y’(t)dt = y(t2) ---------- y(t1) = shaded area under vertical speed curve y’(t)

= CHANGE in y(t) from t1 to t2

= h2 −−−−− h

1 = CHANGE in height from t

1 to t

2 .

The concept or view of integration as “ar“ar“ar“ar“area under the curea under the curea under the curea under the curea under the curvvvvve”e”e”e”e” is useful ingetting star ted. The most general and correct concept is : the indefindefindefindefindefinite inteinite inteinite inteinite inteinite integggggrrrrralalalalalF(x) is always the expression of CHANGE. When we evaluate the indefinite integralindefinite integralindefinite integralindefinite integralindefinite integralF(x) over a given interval and in a given direction, the value or number we get iscalled the change change change change change .

DEFINITE INTEGRAL yCCCCC(t) = y(t) + C C C C C evaluated at instant instant instant instant instant t i gives the positionpositionpositionpositionposition

yCCCCC(t i ) at instant instant instant instant instant t i .

169

We have three points of view :

CalculaCalculaCalculaCalculaCalculationtiontiontiontion : ANTIDERIVATIVE = ∫ DERIVATIVE

y(t) = ANTI-DERIVATIVE of y’(t) = ∫ y’(t).....dt

F(x) = ∫ f (x)dx where f(x) = F’(x)

FCCCCC(x) = F(x) + CCCCC

Geometric Geometric Geometric Geometric Geometric : arararararea under the curea under the curea under the curea under the curea under the curvvvvveeeee of the INSTANTANEOUS RATE OF CHANGE

y(t) = arararararea under the curea under the curea under the curea under the curea under the curvvvvveeeee y’(t) = ∫ y’(t).....dt

F(x) = arararararea under the curea under the curea under the curea under the curea under the curvvvvveeeee f(x) = ∫ f(x)dx

AnalAnalAnalAnalAnalyticalyticalyticalyticalytical : EXPRESSION OF CHANGE = ∫ INSTANTANEOUS RATE OF CHANGE

CHANGE in y(t) over [t1, t

2] = y(t

2) −−−−− y(t

1) = ∫

t22222

t11111 y’(t).....dt

F(x) = ∫ f (x)dx the indefinite integralindefinite integralindefinite integralindefinite integralindefinite integral

∫ b

a f(x)dx = F(b) −−−−− F(a) = CHANGE in F(x) from a to b

FCCCCC(x) = F(x) + CCCCC the definite integral definite integral definite integral definite integral definite integral .

A func t ion A func t ion A func t ion A func t ion A func t ion FFFFF(x) i s an e(x ) i s an e(x ) i s an e(x ) i s an e(x ) i s an e xp rxp rxp rxp rxp r ess ion o fess ion o fess ion o fess ion o fess ion o f CHANGE .CHANGE .CHANGE .CHANGE .CHANGE . I n w In w In w In w In w o ro ro ro ro r k i ngk ingk ingk ingk ingw i t h INF IN ITES IMALS we have a re l a t i onsh ip be tweenw i th INF IN ITES IMALS we have a re l a t i onsh ip be tweenw i th INF IN ITES IMALS we have a re l a t i onsh ip be tweenw i th INF IN ITES IMALS we have a re l a t i onsh ip be tweenw i th INF IN ITES IMALS we have a re l a t i onsh ip be tween

CHANGE andCHANGE andCHANGE andCHANGE andCHANGE and INSTINSTINSTINSTINSTANTANTANTANTANTANEOUS RAANEOUS RAANEOUS RAANEOUS RAANEOUS RATE OF CHANGE.TE OF CHANGE.TE OF CHANGE.TE OF CHANGE.TE OF CHANGE. If I f I f I f I f w w w w we knoe knoe knoe knoe knowwwwwfffff (x) the e(x) the e(x) the e(x) the e(x) the exprxprxprxprxpression ofession ofession ofession ofession of the INST the INST the INST the INST the INSTANTANTANTANTANTANEOUS RAANEOUS RAANEOUS RAANEOUS RAANEOUS RATE OF CHANGE wTE OF CHANGE wTE OF CHANGE wTE OF CHANGE wTE OF CHANGE weeeee

can INTEGRAcan INTEGRAcan INTEGRAcan INTEGRAcan INTEGRATE TE TE TE TE fffff (x) and f(x) and f(x) and f(x) and f(x) and f ind ind ind ind ind FFFFF(x) the e(x) the e(x) the e(x) the e(x) the exprxprxprxprxpression ofession ofession ofession ofession of CHANGE.CHANGE.CHANGE.CHANGE.CHANGE.

170

33. 33. 33. 33. 33. I N T E G R A B L EI N T E G R A B L EI N T E G R A B L EI N T E G R A B L EI N T E G R A B L E

We have used the word INTEGRABLE without giving a formal definition. Here we givean informal definition with a gggggeometricaleometricaleometricaleometricaleometrical explanation. We also present an intepretationfrom a PhysicsPhysicsPhysicsPhysicsPhysics point of view with the focus on positionpositionpositionpositionposition and energyenergyenergyenergyenergy .

Let us start with a strict definition. Let .

Then f(x) is INTEGRABLE ififififif and onl and onl and onl and onl and only ify ify ify ify if F(x) is WELL-BEHAVED - singsingsingsingsingle vle vle vle vle valued,alued,alued,alued,alued,continuouscontinuouscontinuouscontinuouscontinuous and differentiabledifferentiabledifferentiabledifferentiabledifferentiable. The emphasis here is on the differentiabledifferentiabledifferentiabledifferentiabledifferentiableproperty.

We may be less strict by letting F(x) be just singsingsingsingsingle vle vle vle vle valuedaluedaluedaluedalued and continontinontinontinontinuousuousuousuousuous..... In thiscase it is possible that f(x) will be discontinuousdiscontinuousdiscontinuousdiscontinuousdiscontinuous at some instants as in the examplebelow. At such instants where f(x) is discontindiscontindiscontindiscontindiscontinuousuousuousuousuous we require f(x) to have twodefinite values.

We see that f(x) is defined over the sub-intervals [a, 0 −−−−−) and (0+, b] . Notice that f(x)is discontinuous discontinuous discontinuous discontinuous discontinuous at 0 but it has two definite values.

f(0 −−−−−) = − − − − −1 and f(0+) = +1

With this type of discontinuity we may still integrate f(x) over [a, b] .

∫ b

a f(x)dx = ∫

a 0 −−−−−

f(x) dx + ∫b

0+++++f (x)dx

f(x) =

d F(x) dx

1234567890123456789012345678901234567890123456789012345678901234567

12345671234567123456712345671234567

a0 b x

−−−−−1

+1f(x)

= d F(x) dx

0 x

F(x) = |x|

With the understanding that ∫ is a contincontincontincontincontinuous summauous summauous summauous summauous summationtiontiontiontion , we want the sum tobe FINITE or CONVERGE to some definite value. In this case we can see that both

∫a 0 −−−−−

f(x) dx = S1 = −−−−− a is something finite, and ∫b

0+++++f (x) dx = S2 = +b is

something finite. Also their sum does add up to a definite value .

{ ∫a 0 −−−−−

f(x) dx} + { ∫b

0+++++f (x)dx} = S1 + S2 = −−−−− a + b

We state 4 basic conditions for f(x) to be inteinteinteinteintegggggrrrrraaaaabbbbble le le le le .1. f(x) is defined over [a, b] (that is to say SINGLE VALUED) except at the points

of discontinuity. And f(x) has only FINITELY many n discontinuities over [a, b].2. An each such point of discontinuity discontinuity discontinuity discontinuity discontinuity f(x) has 2 definite values.3. Over each of the (n + 1) sub-intervals the contincontincontincontincontinuous summauous summauous summauous summauous summationtiontiontiontion must be

something finite, say : S0 , S1 , S2 , . . . , Sn .4. The combined discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summation S = {S0 + S1 + S2 + . . . + Sn } must

also be something finite.We may relax the first condition be letting f(x) be discontinuous at COUNTABLY manypoints. But at each such point of discontinuity discontinuity discontinuity discontinuity discontinuity f(x) must have 2 definite values.With this in mind let us review the example of the bouncing ball bouncing ball bouncing ball bouncing ball bouncing ball .


ttttt

n

y(t) = vy(t) = vy(t) = vy(t) = vy(t) = vererererer tical positiontical positiontical positiontical positiontical position

yyyyy’(t) = v(t) = v(t) = v(t) = v(t) = vererererer tical speedtical speedtical speedtical speedtical speed

171

172

At each of the instants t0, t1, t2, . . . , y’(ti) is discontinuous discontinuous discontinuous discontinuous discontinuous and has 2 definitevalues y’(t−−−−−

i ) and y’(t+i ). The ball bounces indefinitely. So there are infinitely many

such instants of discontinuity. However, these instants are COUNTABLE. So we mayintegrate y’(t) over each of the sub-intervals (t0, t1), (t1, t2), (t2, t3), ... to get somethingfinite. (In this example the integration of y’(t)over each sub-intervals is zero, becauseas we can see the CHANGE in height y(t) over each sub-intervals is zero. But thismay not generally be the case). The continuous summation continuous summation continuous summation continuous summation continuous summation will CONVERGE.This means that y(t) will be something definite. From a practical or PhysicsPhysicsPhysicsPhysicsPhysics pointof view we know the vvvvvererererer tical tical tical tical tical positionpositionpositionpositionposition of the ball at any instant.

THEORETICAL CONVERGENCE from the PhysicsPhysicsPhysicsPhysicsPhysics point of view means that the ball willcome to rest. And at each bounce the ball loses a little energyenergyenergyenergyenergy . There are twoways in which the ball may come to rest.

1. The ball may bounce FINITELY many times and then come to rest in a FINITEtime interval [0, T]. Correspondingly, y’(t) will have only FINITELY many pointsof discontinuity.

2. The ball may bounce indefinitely and “eventuallyeventuallyeventuallyeventuallyeventually” or “in the LIMITin the LIMITin the LIMITin the LIMITin the LIMIT” the ballwill come to rest. Here too there are two possibilities.

(i) The ball may bounce indefinitely over a FINITE time interval.Correspondingly, y’(t) will have COUNTABLY many points of discontinuityover a FINITE time interval [0, T].

(ii) The ball may bounce indefinitely over an INFINITE time interval.Correspondingly, y’(t) will have COUNTABLY many points of discontinuityover the INFINITE time interval [0, +∞). So eventually the ball willrun out of enerenerenerenerenergggggyyyyy and come to rest. From a practical point of viewwe may say that the ball disappears into TIME.

We know that enerenerenerenerenergggggy y y y y E is a function of displacement. In simple termsE = m.a.sE = m.a.sE = m.a.sE = m.a.sE = m.a.s, or more precisely E = m.a.dsE = m.a.dsE = m.a.dsE = m.a.dsE = m.a.ds, where m =m =m =m =m = mass, a = a = a = a = a = acceleration,and ds = ds = ds = ds = ds = an element of displacement. In the case of the bouncing ballds = dds = dds = dds = dds = dy(t) = = = = = y’(t)dddddt .

173

Depending on how the ball comes to rest, in the integral ∫ b

a to compute the energy

E, b = some finite instant T or b = +∞. Now comes the question:

Does the integral ∫ b

a CONVERGE ?

If it does CONVERGE then the energy function E is INTEGRABLE.

When there is such a loss of energy energy energy energy energy (damping) on each bounce we can see that:

E = E = E = E = E = ∫∫∫∫∫ m.a.ds m.a.ds m.a.ds m.a.ds m.a.ds must CONVERGE.From the PhysicsPhysicsPhysicsPhysicsPhysics point of view this CONVERGENCE means that the energyenergyenergyenergyenergy Erequired for the ball to bounce indefinitely is finite.

What if the ball bounces without loss or gain of energy energy energy energy energy ? The energy will changeform between kinetic and potential. But the total energy will be CONSTANT. Fromthe engineering point of view we will have a perpetual motion machine.

We have a similar situation in the hydrogen atom. The electron revolves around theproton in an elliptical orbit known as the path or curve of CONSTANT energy.Now let us look at the situation where the ball gains energy on each bounce. Eventuallythe ball will disappear from sight. From a practical point of view we may say that theball disappears into SPACE. This is similar to saying a swing or pendulum increases inamplitude with each oscillation. The displacement of such a pendulum is described inthe diagram below.

0 T 2T 3T t

amplitude

174

In this case it is easy to see that the energy energy energy energy energy E, as a function of displacement, willbe infinite. And hence:

E = E = E = E = E = ∫ m.a.ds m.a.ds m.a.ds m.a.ds m.a.ds will not CONVERGE.

Here we implicitly assumed that the length of the pendulum was infinite. What if thependulum was of finite length ? Does it mean that the pendulum will swing all the wayround in a circular orbit ?

What if the ball bounces indefinitely over infinite time interval [0, +∞) with gain inenergy on each bounce ? Will the ball disappear both in TIME and SPACE ?

It is of fundamental importance to see the relationship between the abstract calculationon the one hand and the physical meaning ( the reality - what is happening in nature)on the other hand.

With all this mind we now give informal definition of INTEGRABLE.

1. f(x) is defined over [a, b] (that is to say SINGLE VALUED) except at the points ofdiscontinuity. And f(x) has at most COUNTABLY many discontinuities over [a, b].

So between each pair of consecutive such points of discontinuity we have a

sub-interval.

2. An each such point of discontinuity discontinuity discontinuity discontinuity discontinuity f(x) has 2 definite values.

3. Over each of the sub-intervals in [a, b] the contincontincontincontincontinuous summauous summauous summauous summauous summation tion tion tion tion ∫f(x)dx

must be something FINITE, say : S0 , S

1 , S

2 , . . . , S

n , . . . .

4. The combined discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summation S = {S0 + S1 + S2 + . . . + Sn + . . . }

over all the COUNTABLY many sub-intervals must also be something FINITE

so that ∫ b

a f(x)dx CONVERGES.

Then we say f(x) is INTEGRABLE.

175

A word on A word on A word on A word on A word on CONVERGESCONVERGESCONVERGESCONVERGESCONVERGES

The fundamental requirement in integration integration integration integration integration is that the continuous summationcontinuous summationcontinuous summationcontinuous summationcontinuous summationCONVERGES. Let us compare two examples of discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summation and continuouscontinuouscontinuouscontinuouscontinuoussummationsummationsummationsummationsummation.

The discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summation does NOT CONVERGE.

The integral ∫0 +1

{1/////x} dx is a contincontincontincontincontinuous summauous summauous summauous summauous summation tion tion tion tion over the finite interval [0, +1].

But the summation does not CONVERGE. So it is not INTEGRABLE.

We can see from the diagram that the area under the curve f(x) = 1/////x over the finite

interval [0, +1] is NOT FINITE. From a Geometric Geometric Geometric Geometric Geometric point of view we may define

INTEGRABLE as :

f(x) is INTEGRABLE ⇔ the area under f(x) is FINITE

1/////n

∞

Σn n n n n ===== 0 0 0 0 0

−∞ ½ 1 2 . . . +∞

+∞

2

1

½

0

−1

−2

−∞

f(x) = 1/////x

176

+1

0

f(x) = e−−−−− x

x

The discrdiscrdiscrdiscrdiscrete summaete summaete summaete summaete summation tion tion tion tion or CONVERGES. The terms are

finite in value. However, the subscript i i i i i varies over the infinite interval [0, +∞).

Likewise the discrete summationdiscrete summationdiscrete summationdiscrete summationdiscrete summation also CONVERGES.

It is possible to have an integral or a continuous summaitoncontinuous summaitoncontinuous summaitoncontinuous summaitoncontinuous summaiton over an infiniteinterval.

It is now easy to see that the continuous summationcontinuous summationcontinuous summationcontinuous summationcontinuous summation ∫0

+∞{e−x} dx must CONVERGE

even though the interval [0, +∞) is infinite. Hence f(x) = e−−−−− x is INTEGRABLE.

∫0

+∞{e−x} dx = [−e−x ]

0

+∞ = (−e−∞) − (−e−0) = 0 − (−1) = +1

1/////2 iiiii

∞

Σiiiii ===== 0 0 0 0 0

2− iiiii∞

Σiiiii ===== 0 0 0 0 0

e− iiiii∞

Σiiiii ===== 0 0 0 0 0

177

x

+1

0

e−−−−− xex

−−−−− x

We may also integrate over (−∞, +∞) as in the example below. Let

−∞∫ +∞

f(x) dx = −∞∫0f(x) dx + ∫

0

+∞f(x) dx = +1 +1 = +2

We are familiar with the associaassociaassociaassociaassociatititititivity vity vity vity vity property in addition : (a+b)+c = a+(b+c).This property holds good when the number of terms are finite. It also holds good fora discrdiscrdiscrdiscrdiscrete summaete summaete summaete summaete summationtiontiontiontion of COUNTABLY many terms if all the terms are of samesign. But it breaks down when there are COUNTABLY many terms of both positive andnegative sign. The alteralteralteralteralternananananating seriesting seriesting seriesting seriesting series +1 −−−−−1 +1 −−−−−1 + . . . does NOT CONVERGE.Depending on how we do the summation we have 3 possible answers : −−−−−1, 0, and+1. So one can imagine the difficulties involved in a continuous summationcontinuous summationcontinuous summationcontinuous summationcontinuous summation withUNCOUNTABLY many terms of both signs. Proving CONVERGENCE becomes very,very difficult.

A more formal and rigorous treatment of INTEGRABLE dealing with the problemsof CONVERGENCE - discontinuitiesdiscontinuitiesdiscontinuitiesdiscontinuitiesdiscontinuities, finiteness finiteness finiteness finiteness finiteness and associativity associativity associativity associativity associativity - is coveredin A LITTILE MORE CALCULUS.

e− x for x ≥ 0f(x) = {

ex for x ≤ 0

178

n n n n n →→→→→∞∞∞∞∞

= Σ f(x i).....δxiiiii

iiiii ===== 0 0 0 0 0

nnnnn −−−−− 1 1 1 1 1

Σiiiii ===== 0 0 0 0 0

f(xiiiii).....∆x

iiiii

11111111111

BBBBB

AAAAA

f ( x )f ( x )f ( x )f ( x )f ( x )


00000 δδδδδxxxxx iiiii

BBBBB

AAAAA

f ( x )f ( x )f ( x )f ( x )f ( x )


00000 ∆∆∆∆∆ xxxxx iiiii

123451234512345123451234512345123451234512345

34.34.34.34.34. AAAAApplicapplicapplicapplicapplications oftions oftions oftions oftions of Inte Inte Inte Inte Integggggrrrrraaaaationtiontiontiontion

There are numerous applications of integration. In this chapter we cover a few examplesto illustrate some of the techniques in integration. Let us review the 3 steps inintegration.


Fn(x) = . So : ∆F

iiiii (x) = f(x iiiii).....∆x

iiiii is an element element element element element of area .


So: Fn(x) . So : δFiiiii (x) = f(x

iiiii).....δx

iiiii is an infinitesimalinfinitesimalinfinitesimalinfinitesimalinfinitesimal

element element element element element of area .


Hence : dF(x) = f(x ).....dx is an instantaneous element instantaneous element instantaneous element instantaneous element instantaneous element of area .


n n n n n →→→→→∞∞∞∞∞

= Limit Limit Limit Limit Limit Σ f(xi) .....dx

iiiii ===== 0 0 0 0 0

179

dθ

(0,0)R

x

yR

θ

d arc = Rdθ}}}}}12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890

In the diagrams in the examples that follow we cannot depict an instantaneousinstantaneousinstantaneousinstantaneousinstantaneouselement element element element element f(x)dx . However, in keeping with the standard notation in text bookswe have used the label dx dx dx dx dx of the instantaneous element instantaneous element instantaneous element instantaneous element instantaneous element rather than δδδδδx x x x x ofthe infinitesimal element infinitesimal element infinitesimal element infinitesimal element infinitesimal element . This should NOT give the wrong impression orwrong notion that dxdxdxdxdx is an infinfinfinfinfinitesimal interinitesimal interinitesimal interinitesimal interinitesimal intervvvvvalalalalal . The calculation with respectto dxdxdxdxdx be it differentiation differentiation differentiation differentiation differentiation or integration integration integration integration integration is always instantaneous instantaneous instantaneous instantaneous instantaneous .

From the diagrams the student should be able to construct an infinfinfinfinfinitesimalinitesimalinitesimalinitesimalinitesimalelementelementelementelementelement , say f(x) f(x) f(x) f(x) f(x) δδδδδx x x x x , and then see that indeed f(xf(xf(xf(xf(x))))) dxdxdxdxdx = f(x) f(x) f(x) f(x) f(x) δδδδδxxxxx .

∆θ = 2π///// n

δθ = 2π///// n as n → ∞

dθ = δθ = 2π///// n

In the diagram above an infinitesimal element infinitesimal element infinitesimal element infinitesimal element infinitesimal element of area δδδδδA is a triangle(wedge shape) of magnitude

δδδδδA = 1/////2(R .....δθ).R = 1/////2R 22222δθ

dddddA = { 1/////2(R .....δθ).R } = { 1/////2R 22222δθ } = 1/////2R 22222dθ

is an instantaneuos elementinstantaneuos elementinstantaneuos elementinstantaneuos elementinstantaneuos element of area.

L im i tn → ∞

L im i tn → ∞

L im i tn → ∞

L im i tn → ∞

L im i tn → ∞

180

Examp lE xamp lE xamp lE xamp lE xamp l eeeee 1 1 1 1 1 ::::: Find the arararararea under the curea under the curea under the curea under the curea under the cur vvvvveeeee f ( x ) = 2x ove r t heinter va l [ 0, 1 ].

Us ing Integra l Ca lcu lus: F ’(x) = f(x) = 2x

The INTEGRAL F(x) = ANTIDERIVATIVE of f(x)

What is the ANTIDERIVATIVE of f(x) = 2x ? x2 + C

F(x) = ∫ f(x) dx = ∫ 2xdx = x2 + C

In th is case the constant C = 0. Can you th ink why ?

The ararararar ea under the curea under the curea under the curea under the curea under the cur vvvvveeeee f(x) = 2x f rom x = 0 to x = 1 is :

∫1

0 f(x)dx = ∫

1

0 2xdx = [ x2]1

0 = 12 ---------- 02 = 1.

Evaluation of the INTEGRAL F(x) = x2 in the positive direction positive direction positive direction positive direction positive direction over theinter val [0, 1] is the CHANGE

[F(x)]1

0 = [ x2]1

0 = 12 ---------- 02 = 1.

11111 22222 33333 44444

33333

11111

22222

55555

44444

(((((0 ,00 ,00 ,00 ,00 ,0 )))))


xxxxx

fffff (x) = 2x(x) = 2x(x) = 2x(x) = 2x(x) = 2x

123456123456123456123456123456123456123456123456123456123456123456

181

In Physics the most impor tant entity we are interested in is enerenerenerenerenergggggyyyyy. Any wwwwworororororkkkkkdonedonedonedonedone requires enerenerenerenerenergggggyyyyy and causes CHANGE and any CHANGE involves enerenerenerenerenergggggyyyyy.The sense of the word CHANGE in Calculus is not synonymous with the word CHANGE( = wwwwwororororork donek donek donek donek done = enerenerenerenerenergggggy y y y y ) in Physics.Example 2:Example 2:Example 2:Example 2:Example 2: A car weighing one ton and star ting from rest (0 kmph) acceleratessteadily due EAST (along the x-axis) for 10 secs and levels off at 90 kmph(25 m/sec). What is the distance travelleddistance travelleddistance travelleddistance travelleddistance travelled from t = 10 secs to t = 20 secsand also from t = 0 secs to t = 10 secs ?

55555 1 01 01 01 01 0 1 51 51 51 51 5 2 02 02 02 02 0 2 52 52 52 52 5(((((0 ,00 ,00 ,00 ,00 ,0 ))))) ttttt

25 m/sec25 m/sec25 m/sec25 m/sec25 m/sec3 03 03 03 03 0

55555

1 01 01 01 01 0

1 51 51 51 51 5

2 02 02 02 02 0

2 52 52 52 52 5 DDDDD

AAAAA BBBBB CCCCC

EEEEESSSSSPPPPPEEEEEEEEEEDDDDD

TIMETIMETIMETIMETIME (sec) (sec) (sec) (sec) (sec)

over the interval [ t = 10 secs, t = 20 secs] the SPEED = 25 m/sec ( constant ).

DDDDDistance travelledistance travelledistance travelledistance travelledistance travelled from t = 10 secs to t = 20 secs is 25m/sec ∗ 10 sec = 250mwhich is the arararararea under the speed curea under the speed curea under the speed curea under the speed curea under the speed curvvvvveeeee = area of rectangle BCDE.

In the case of dddddistance istance istance istance istance travelltravelltravelltravelltravellededededed between t = 0 secs and t = 10 secs, theSPEED is NOT constant. What SPEED are you going to use?

DISTANCE = SPEED * TIME [meters = meters * seconds] seconds

182

By looking at the graph we know the SPEED at each and every INSTANTbetween t = 0 secs and t = 10 secs. So we have to take the product ofSPEED and TIME at each and every INSTANT and sum it up. This in effectwill give the arararararea under the speed curea under the speed curea under the speed curea under the speed curea under the speed cur vvvvveeeee from t = 0 secs to t = 10 secswhich is the area of triangle ABE.

This is 1--2 base * height = 1--2 * 10 secs * 25 m / sec = 125 meters .

Conceptually we note that from the speedspeedspeedspeedspeed cur cur cur cur cur vvvvveeeee ( INSTANTANEOUS RATEOF CHANGE of position graph) we got the CHANGE in positionCHANGE in positionCHANGE in positionCHANGE in positionCHANGE in position. CHANGE in positionhas both sign and magnitude. In this case we have denoted going East (right) aspositipositipositipositipositivvvvveeeee and going West (left) as nenenenenegggggaaaaatititititivvvvve e e e e on the x-axis..... We may denote the SPEEDof the car by x’(t) and hence its POSITION on the x-axis will be x(t).

Because the acceleration is steady or constant over [0, 10) we may use the averageacceleration over [0, 10).

acceleration over [0, 10) : x”(t) = = 2.5m/////sec2

acceleration over t ≥ 10 : x”(t) = 0

speed over [0, 10] :

x’(t) = ∫∫∫∫∫x”(t)dt = 2.5t m/////secspeed over t ≥ 10 :

x’(t) = 25 m/////secposition over [0, 10] :

x(t) = ∫∫∫∫∫ 2.5t dt = 2.5t2/////2

position over t ≥ 10 :

x(t) = ∫∫∫∫∫ 25 dt = 25t − 125

From the nature of the problem at t = 10 seconds we may determine that theCONSTANT OF INTEGRATION is −125 meters .

x’(10) −−−−− x’(0)10 secs

0 10 20 [secs] t

x(t) = 2.5t2/////2 over [0, 10]

25t − 125 for t ≥ 10{

375

250

125

[met

ers]

183

We need to be more clear about the terms CHANGE in position, displacementdisplacementdisplacementdisplacementdisplacement,and distance travelled distance travelled distance travelled distance travelled distance travelled in A)A)A)A)A) 1-Dimension in fixed directionfixed directionfixed directionfixed directionfixed direction, B) B) B) B) B) fixed axis fixed axis fixed axis fixed axis fixed axisor 1-Dimension with CHANGE in direction and C) C) C) C) C) 2-Dimensions. These termsneed to be explained from both the CalculusCalculusCalculusCalculusCalculus point of view and the PhysicsPhysicsPhysicsPhysicsPhysics(what is happening in nature) point of view and how they are related.

A)A)A)A)A) 1-Dimension in fixed direction fixed direction fixed direction fixed direction fixed direction : in the example the car travelled East in afixed direction, that is to say moving along a straight line without reversing orchanging direction. So in this very speical case from the PhPhPhPhPhysics ysics ysics ysics ysics point of view:

CHANGE in position = displacement displacement displacement displacement displacement = distance travelleddistance travelleddistance travelleddistance travelleddistance travelled .

Also, from the Calculus Calculus Calculus Calculus Calculus point of view, over a given interval, say [t1, t

2] :

CHANGE in position = t1

∫t2 x’(t)dt = arararararea under the speed curea under the speed curea under the speed curea under the speed curea under the speed curvvvvveeeee

This is because the SPEED function did NOT CHANGE SIGN. In 1-Dimensional 1-Dimensional 1-Dimensional 1-Dimensional 1-Dimensional motionthe SPEED function will CHANGE SIGN only if there is a CHANGE in direction.

B)B)B)B)B) fixed axis fixed axis fixed axis fixed axis fixed axis or 1-Dimension with CHANGE in direction : Let us review an exampleof the projected ball.

tT


(0,0)

y(t11111)

t11111 t

22222

123456123456123456123456123456123456123456

y(t22222)

y(T/////2)

T///// 2


y’(t11111)

y’(t22222)

1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890

184

(0, 0) Range x(t)

He

igh

t y

(t)


P1

P2

CHANGE in position over [t1, t

2] =

t1

∫t2 y’(t)dt = y(t

2) ---------- y(t

1)

dispalcement dispalcement dispalcement dispalcement dispalcement = y(t2) ---------- y(t

1)

distance travelleddistance travelleddistance travelleddistance travelleddistance travelled =|t1

∫T///// 2

y’(t)dt| + |T///// 2

∫t2 y’(t)dt|

= |y(T///// 2 )| ---------- y(t

1) | + | y(t

2) ---------- y(T///// 2)|

The distance travelleddistance travelleddistance travelleddistance travelleddistance travelled (odometer reading or length of path) in the usual (non-mathematical) sense is used to convey the notion “CHANGE in position”“CHANGE in position”“CHANGE in position”“CHANGE in position”“CHANGE in position”. Butin Calculus these two are different. “CHANGE in position”“CHANGE in position”“CHANGE in position”“CHANGE in position”“CHANGE in position” is always with referenceto a co-ordinate system. If an object moves in a circle of radius r thru one completecircle, then the CHANGE in positionCHANGE in positionCHANGE in positionCHANGE in positionCHANGE in position = 0. But the distance travelleddistance travelleddistance travelleddistance travelleddistance travelled = 2πr .

C)C)C)C)C) 2-Dimensions: 2-Dimensions: 2-Dimensions: 2-Dimensions: 2-Dimensions: What if the object is moving in 2-Dimension2-Dimension2-Dimension2-Dimension2-Dimension as in the case ofa projected ball ?

This type of problem may be split into two 1-Dimension 1-Dimension 1-Dimension 1-Dimension 1-Dimension problems.

x’(t) = u ..... cos θ 1-Dimension 1-Dimension 1-Dimension 1-Dimension 1-Dimension in fixed directionfixed directionfixed directionfixed directionfixed direction.

y’(t) = u ..... s i n θ ---------- g t 1-Dimension 1-Dimension 1-Dimension 1-Dimension 1-Dimension with CHANGE in direction.

CHANGE in position is from P1 = (x(t1), y(t1)) to P2 = (x(t2) , y(t2)) = P2 −−−−− P1

displacement displacement displacement displacement displacement = {[x(t2) −−−−− x(t

1)]2 + [y(t

2) −−−−− y(t

1)]2 } ½

185

Example 3Example 3Example 3Example 3Example 3: What is the distance travelled distance travelled distance travelled distance travelled distance travelled (length of path) in 2-Dimension2-Dimension2-Dimension2-Dimension2-Dimension ?

dx = infinitesimal CHANGE in range.dy = infinitesimal CHANGE in height.ds = infinitesimal piece of the path.

dx---------------dt

= x’(t) = u.....cos θ . So dx = x’(t)dt = u.....cos θ..... dt

dy---------------dt

= y’(t) = u ..... sin θ ---------- g t . So dy = y’(t)dt = (u ..... sin θ ---------- g t) ..... dt

By Pythagoras theorem : ds = [(dx)2 + (dy)2 ]½ = [(x’(t))2 + (y’(t))2]½..... dt

s = ∫ ds = ∫ [ (x’(t))2 + (y’(t))2]½..... dt

We should now look up the TTTTTaaaaabbbbble ofle ofle ofle ofle of Inte Inte Inte Inte Integggggrrrrralsalsalsalsals to find the appropriate integralwhere the integrand from the given x(t) and y(t) is in the above form.

In the case of the vertical dimension of the projected ball the range x(t) is absent. So

the above formula reduces to : s = ∫ ds = ∫ y’(t)..... dt

Adapting this technique to find the distance travelleddistance travelleddistance travelleddistance travelleddistance travelled ( length of flight path) of aplane gliding forward at the rate of 500 feet///// second and downward at the rate of 20feet///// second (see page 82) we note that :

dx---------------dt

= x’(t) = 500 feet///// second. So dx = 500 ..... dt

dy---------------dt

= y’(t) = −−−−−20 feet///// second. So dy = −−−−−20 ..... dt

s = ∫ ds = ∫ [(x’(t))2 + (y’(t))2 ]½ ..... dt = ∫ [ (500)2 + (−−−−−20)2 ]½ ..... dt

y(t)

= u

.... .sin

θ .... . t

-------- -- 1 -- 2 g

t2

(0, 0)

ds

dx dy

x(t) = u.....cos θ. t

186

(0, 0)

ds

dx dy

x1

x2

x

y(x)

Let us now apply this method to find the length oflength oflength oflength oflength of a cur a cur a cur a cur a cur vvvvveeeee where y is afunction of x.Example 4Example 4Example 4Example 4Example 4: What is the length of the curve y(x) over the interval [x1, x2] ?

dx = infinitesimal CHANGE along the x-axis.dy = infinitesimal CHANGE along the y-axis.ds = infinitesimal piece of the curve.

dy----dx = y’(x) . So dy = y’(x).....dx

By Pythagoras theorem: ds = [(dx)2 + (dy)2 ]½ = [(dx)2 + ( y’(x).....dx)2 ]½

= [ 1 + (y’(x))2 ]½

..... dx = [ 1 + (dy----dx)

2 ]½

..... dx

We should now look up the TTTTTaaaaabbbbble ofle ofle ofle ofle of Inte Inte Inte Inte Integggggrrrrralsalsalsalsals to find the appropriate integralwhere the integrand from the given WELL-BEHAVED y(x) is in the above form.

Apply this method to find the length of the curve where f(x) = 2x over the interval[0, 1] as in example 1 (see page 180). Here f ’(x) = 2. The result should tallywith the result √5 that we get using the Pythagoras Theorem.

s = ∫ ds = ∫x2

x1

[ 1 + (dy----dx)

2 ]½

..... dx

187

t

y

123456789012123456789012123456789012123456789012123456789012123456789012123456789012123456789012123456789012123456789012123456789012123456789012123456789012

(0,0)

y

y ’

1234567890112345678901123456789011234567890112345678901123456789011234567890112345678901123456789011234567890112345678901123456789011234567890112345678901

T

Example Example Example Example Example 55555::::: What is the CHANGE in height y(t) of the ball over [ 0, T ] ?

Height y(t) = u . s i n θ . t ---------- 1--2 g t 2 .

Vertical speed y ’(t) = u . s i n θ ---------- g t .

What is the shaded area under y ’(t) over [ 0, T ] ?

The area of the two shaded triangles is zero. y( T ) ---------- y ( 0 ) = 0.

The CHANGE in height is zero. This does not necessarily mean that the presentheight above the ground level ( = zero ) at t = T seconds is zero. If initially att = 0 seconds the ball was at height 5 meters above the ground level ( = zero ),then the difference or CHANGE in height from t = 0 secs to t = T secs is zero. So theball is back at its initial height of 5 meters above the ground level ( = zero).

∫ y’(t) ..... dt = y(t) + C

In this case the constant of integration C = + 5 meters.

188

θ

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+1

--1

(0,0 ) Π

y ’

y

3Π2

2Π 3Π 4Π Π2

Examp le Examp le Examp le Examp le Examp le 66666::::: periodic functionsperiodic functionsperiodic functionsperiodic functionsperiodic functions

y = sin θ is shown by the solid line. y ’ = cos θ is shown by the lighter line.

What is the area under y ’ over [ 0, 2π ] ? Zero.

What is the CHANGE in y over [ 0, 2π ] ? Zero.

We observe how y = sinθ repeats itself over every inter val of length 2π. yis sa id to be PERIODIC wi th per iod 2π . S imi lar ly, y ’ is PERIODICwith period 2π.

Let y and y ’ be PERIODIC with period T.Let m and n be whole numbers. nT

∫mT

y ’. dt = [ y nT]mT

= CHANGE in y from mT to nT = O

Geometrically speaking, the area under y’ frommT to nT is zero.

189

ExExExExExampleampleampleampleample 77777::::: A pump delivers at the rate of 60 ttttt litres/minute.

a) How many litres will it pump from ttttt = 5 minutes to ttttt = 10 minutes ?

b) Starting at ttttt = 0 minutes, how long will it take to pump 3000 litres ?

a) f’(t) = 60 t t t t t litres/minute

t=

∫t=5

10f ’(ttttt) ..... dt =

t=

∫t=5

1060 ttttt ..... dt =[ 6

20t2

]10

5 litres

= [ 6000 2−−−−− 1500 ] litres = 2250 litres

b) t=

∫t=0

Tf ’(ttttt) ..... dt =

t=

∫t=0

T60 ttttt ..... dt = [ 6

20t2

]T

0

= 620T2

= 3000

T2 = 100 minutes

T = 10 minutes

This example may be easily adapted to compute the charge on a capacitor and alsothe time taken to charge or discharge the capacitor given the rate of flow of charge= current iiiii = dq/////dt.

190

Examp le Examp le Examp le Examp le Examp le 88888::::: area of a circlearea of a circlearea of a circlearea of a circlearea of a circle

It is not always necessary to work with rrrrrectangular co-orectangular co-orectangular co-orectangular co-orectangular co-ordinadinadinadinadinatestestestestes. Let us findthe area of a circle using polar co-ordinatespolar co-ordinatespolar co-ordinatespolar co-ordinatespolar co-ordinates (around a point). Here we takeadvantage of the symmetry of the elements ofelements ofelements ofelements ofelements of ar ar ar ar areaeaeaeaea around the origin. Thecontiguous elements ofelements ofelements ofelements ofelements of ar ar ar ar areaeaeaeaea when summed up continuously or integrated formthe whole area.

element of area dddddA = 2πrdr

area of circle A = ∫dddddA = R

∫0 2πrdr = [ πr2 ]

R

0 = πR2

element of area dddddA = 1/////2(Rdθ).R = 1/////2R22222dθ

area of ciricle A = ∫dddddA = 2π

∫0 1/////2R22222dθ = [ R

2

2θ ]

2π

0 = πR22222

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r

dr

(0,0)R

x

yR

dθ

(0,0)R

x

yR

θ

d arc = Rdθ}}}}}1234567890112345678901123456789011234567890112345678901123456789011234567890112345678901123456789011234567890112345678901

The elements of elements of elements of elements of elements of areaareaareaareaarea are rings similarin shape but varying in size.

The elements of elements of elements of elements of elements of areaareaareaareaarea are triangularwedges similar in shape and equal in size.

191

ExExExExExampleampleampleampleample 99999::::: volume of a cylindervolume of a cylindervolume of a cylindervolume of a cylindervolume of a cylinder

The “ar“ar“ar“ar“area under the curea under the curea under the curea under the curea under the curvvvvve”e”e”e”e” f(x) = r over the interval [0, h] is :

∫h

0 f (x) ..... d x = ∫

h

0 r d x = [ r x]

h

0 = r ..... h

This is the area of a cross-section. If we rotate this rectangular cross-section arear ..... h thru an infinitesimal angle dθ around the x-axis, we will get a wedge shapedblock of volume dV.

volume of wedge dV = 1/////2 (cross-section area ..... r dθ) = 1/////2 r2 ..... h ..... dθ

We may now rotate this wedge dV from θ = 0 to θ = 2π radians to get the solidcylinder of height h and radius r.

volume of cylinder V = ∫ dddddV = 2π

∫0 1/////2r2 2 2 2 2 h..... dθ = [ r

2

2 h θ ]

2π

0 = π r2 2 2 2 2 h

xxxxx(((((0 ,00 ,00 ,00 ,00 ,0 )))))


height = hheight = hheight = hheight = hheight = h

f(x) = rf(x) = rf(x) = rf(x) = rf(x) = rradius = rradius = rradius = rradius = rradius = r

123456123456123456123456123456123456

1234567890123456789012345678901212345123456789012345678901234567890121234512345678901234567890123456789012123451234567890123456789012345678901212345r dr dr dr dr d θθθθθ

dxdxdxdxdx

192

ExExExExExampleampleampleampleample 1010101010::::: volume of a conevolume of a conevolume of a conevolume of a conevolume of a cone

The cylinder is known as a solid of rotationsolid of rotationsolid of rotationsolid of rotationsolid of rotation. The technique used in example 8to find the volume of a cylinder cannot be extended to other solids of rotationsolids of rotationsolids of rotationsolids of rotationsolids of rotation.Because we cannot get an accurate measure of the element ofelement ofelement ofelement ofelement of v v v v volumeolumeolumeolumeolume dV.We need an accurate measure of dV.

In the case of a cylinder we may take an element of volumeelement of volumeelement of volumeelement of volumeelement of volume dV to be a solid discof radius f(x) and thickness dx. Then :

dddddV = π {f(x)}22222.....dx = πr22222.....dx

volume of a cylinder = ∫dddddV = ∫h

0 π {f(x)}22222.....dx = ∫

h

0 πr22222.....dx = [πr22222x]

h

0 = πr22222h

This technique may now be applied to other solids of rotationsolids of rotationsolids of rotationsolids of rotationsolids of rotation.

volume of cone = ∫dddddV = ∫h

0 π {f(x)}22222.....dx = ∫

h

0 π(rx///// h)22222.....dx

= [π(r22222x33333///// 3h22222)]h

0 = 1///// 3 πr22222h

xxxxx(((((0 ,00 ,00 ,00 ,00 ,0 )))))


height = hheight = hheight = hheight = hheight = h

f(x) = f(x) = f(x) = f(x) = f(x) = rrrrr

xxxxx///// hhhhh

radi

us =

rra

dius

= r

radi

us =

rra

dius

= r

radi

us =

r

d xdxdxdxdx

12345123451234512345123451234512345123451234512345

dxdxdxdxdx

193

The physical entities like positionpositionpositionpositionposition, distance distance distance distance distance, speed speed speed speed speed, acceleration acceleration acceleration acceleration acceleration, area area area area area andvolumevolumevolumevolumevolume are easy to visualize. Let us now look at an application of integration wherethe physical entity being measured is not so obvious.

MMMMMoment ofoment ofoment ofoment ofoment of IIIIInernernernerner t iat iat iat iat ia

In linear motionlinear motionlinear motionlinear motionlinear motion the distribution of the mass (density) of the rigid bodydoes not matter. We may take the center ofcenter ofcenter ofcenter ofcenter of mass mass mass mass mass of the rigid body and treatit as concentrated mass or point masspoint masspoint masspoint masspoint mass. So we have only one position vector forthe rigid body. Hence the orientaorientaorientaorientaorientationtiontiontiontion of the rigid body does not affect theequation of energy of linear motion. In rotational motionrotational motionrotational motionrotational motionrotational motion the orientationorientationorientationorientationorientation ofthe body does affect the equation of energy of rotational motion.

orientation orientation orientation orientation orientation = distance of the individual elements of masselements of masselements of masselements of masselements of mass of the rigid bodyfrom a particular point or axis.

elements of masselements of masselements of masselements of masselements of mass = similar segments or cross-sections of the mass, more thanjust a particle.

Contiguous elements ofelements ofelements ofelements ofelements of mass mass mass mass mass when summed up continuously or integrated formthe entire mass of the rigid body . To keep our calculations simple at the highschool level, we assume the mass of the rigid body to be of unifunifunifunifuniforororororm densitym densitym densitym densitym density.Hence they are of equal mass. Each such element of masselement of masselement of masselement of masselement of mass may be viewed as apoint masspoint masspoint masspoint masspoint mass with its individual position vector. Also, at the high school, we work witheasy examples where a cross-section of the solid is uniform about an axis or at leastsymmetric about an axis.

The moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia I is defined as the measure of the tendenctendenctendenctendenctendency toy toy toy toy torrrrrotaotaotaotaotatetetetete of a rigid body.

About an axis A : IA = the axialaxialaxialaxialaxial moment of iner tia.

About a point P : IP = the polarpolarpolarpolarpolar moment of iner tia.

194

In the diagrams above the same brick of uniform density, depending on itsorientaorientaorientaorientaorientationtiontiontiontion, has a different tendenctendenctendenctendenctendency to ry to ry to ry to ry to rotaotaotaotaotatetetetete or moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatiaabout the x-axis. It is impor tant to note that the brick is not in motion. Thebrick is at rest or in a state of inerinerinerineriner tiatiatiatiatia. Therefore the moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatiaof a rigid body depends only on the orientationorientationorientationorientationorientation or distance of the individualelementselementselementselementselements of ma of ma of ma of ma of massssssssss from the axis or point of rotation.

Measures may be formulated in different ways. Whatever be the measure, itshould give a sense and propor tion of the entity being measured. Thus theformula for the moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia about an axis is defined as:

IA = (mass) ..... (square of perpendicular distance to the axis of rotation)

This has the advantage of taking into account the orientationorientationorientationorientationorientation without beingaffected by which side of the axis, negative or positive, the rigid body ispositioned or located. The quantity I can be used directly in the equationof the kinetic energy of rotation. The moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia is always a positivequantity. The measure of the tendenctendenctendenctendenctendency to ry to ry to ry to ry to rotaotaotaotaotatetetetete does not specify or includethe direction of rotation as in positive for counter-clockwise and negative forclockwise.

By applying the PythagorasTheorem with P as the origin:

IP = IX + IY .

1234512345123451234512345123451234512345123451234512345123451234512345

(0,0)

y

x

greater tendency to rotatetendency to rotatetendency to rotatetendency to rotatetendency to rotate about x-axis

1234567890123412345678901234123456789012341234567890123412345678901234

(0,0)

y

x

lesser tendency to rotatetendency to rotatetendency to rotatetendency to rotatetendency to rotate about x-axis

P

y

x

IP

IY

IX

195

Example 1Example 1Example 1Example 1Example 111111::::: Consider a thin rod AB of mass M of uniform density and length Lalong the x-axis. The rod is free to rotate about the y-axis passing through the centerof mass and perpendicular to its length.

Since the mass M is of uniform density we have: mass per unit length = M///// L .

So an elementelementelementelementelement of the rod has mass dddddM = M///// L ..... dddddx

The moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia of this elementelementelementelementelement of the rod about the y-axis is dependenton the distance x from the y-axis.

Therefore: dddddIY = (M///// Ldddddx) ..... x2 = M///// Lx2dddddx .

The moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia of the entire rod about the y-axis is got by continuouslysumming up or integrating over the range of x from −−−−−L/////2

to +L/////2 .

IY = ∫ dddddIY =

+L/////2

∫−−−−−L/////2

M///// Lx2dddddx = M///// L [x3/////3

+L/////2]

−−−−−L/////2

= M///// 3L[(+L/////2)3 −−−−− (−−−−−L/////2

)3]

IY = ML2///// 12

y

x

1234123412341234

A B

+L/////2−−−−−L///// 2 dxx

196

If we now shift the rod so that the axis of rotation (y-axis) passes thru one end of therod, what is the moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia IY ?

In both cases IY = ML2///// 3

A position vector has both magnitude (distance) and direction. The direction may bepositive or negative depending on which side of the axis the rigid body is located. It isimpor tant to note that the measurmeasurmeasurmeasurmeasureeeee ofofofofof the tendenc the tendenc the tendenc the tendenc the tendency to ry to ry to ry to ry to rotaotaotaotaotate te te te te is definedindependent of the side of the axis the mass is located. The orientationorientationorientationorientationorientation isthe distance of the mass from the axis or point of rotation.

We observe that the greater the distance of the mass from the axis or point ofrotation, the greater the moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia. Hence the easier it is to rotateor turn the rigid body and so lesser energy is required.

If the rigid body happens to be in rotational motion about an axis or point thenthe elementselementselementselementselements of mass of the rigid body, regardless of the distance from the axis orpoint of rotation, have the same angular velocity. If the angular velocity is ω then :

kinetic energy of rotation = 1/////2Iω2 .

This is analogous to the kinetic energy of linear motion = 1/////2mv2 .

y

x

12341234123412341234

A B

+L0dx

y

−−−−−x

12341234123412341234

A B

−−−−−L dx0

197

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R

x

central axis y

ddddd arc = Rdddddθ}}}}}

Example 12Example 12Example 12Example 12Example 12: Consider a ring of mass M of uniform density and radius R. Let usfind the moment of inertia about the y-axis passing thru its center and perpendicularto its plane.

The circumference of the ring is 2πR.

Since the mass is of uniform density we have: mass per unit length = M/////2πR .

The mass of an elementelementelementelementelement (= ddddd arc) of the ring is: dddddM = M/////2πR ..... (Rdddddθ ) .

The moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia of this elementelementelementelementelement about the central y-axis is:

dddddIY = M/////2πR ..... (Rdddddθ ) ..... R2 = MR2/////2π ..... dddddθ

The moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia of the entire ring about the central y-axis is :

IY = ∫ dddddIY = 2π∫0

M/////2πR2dddddθ = MR2/////2π ..... [θ

2π]0

IY = MR2

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Example 13Example 13Example 13Example 13Example 13: Let us now find the moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia of a disc of mass M ofuniform density and radius R.

The area of the disc is πR2.

Since the mass is of uniform density we have: mass per unit area = M/////πR2 .

We may think the disc to be made up of elements elements elements elements elements of concentric rings of radiusr with 0 ≤ r ≤ R and thickness dddddr. The mass of such an element element element element element of the disc is:

dddddM = ( M/////πR2) ..... 2πrdddddr = 2M/////R2 rdddddr

The moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia of such an element element element element element of the disc, we know fromexample 2, is:

dddddIY = dddddM ..... r2 = (2M/////R2 rdddddr)r2 = 2M/////R2 r3dddddr

Therefore the moment ofmoment ofmoment ofmoment ofmoment of iner iner iner iner iner tiatiatiatiatia of the entire disc is:

IY = ∫ dddddIY = 2M/////R2 R

∫0

r3dddddr = 2M/////R2 ..... [r4/////4

R

]0

IY = MR2///// 2 .

To a casual observer a disc rotating at high speed about its central axis may seemstationary. Yet there is enerenerenerenerenergggggy y y y y involved and work being done. The kinetic enerkinetic enerkinetic enerkinetic enerkinetic energggggyyyyyofofofofof r r r r rotaotaotaotaotation tion tion tion tion in this case may be viewed as a form of stored energy or potentialpotentialpotentialpotentialpotentialenerenerenerenerenergggggyyyyy. The rotating disc when used to drive something is known as an impeller.

In comparing ring IY = MR2 with the disc IY = MR2///// 2 about the central y-axis, wesee the reason for using rims (rings) with spokes instead of solid discs as wheels forcycles and bikes.

199

Two satellites A and B, each weighing one tonne, orbit the earth with the sameangular velocity ω. The radii of orbit are RA = 100 kms and RB = 1000 kms.Which of the satellites uses more energy ?

Satellites are usually not of uniform density. Since the dimensions of a satellite aresmall (just a couple of meters across) in comparison to the radius of orbit we mayview the mass of the satellite to be a sphere of uniform density and hence treat it likea point mass.

The satellite in lower orbit will require more energy. To this we must add some moreenergy to keep the satellite in stable orbit to counter the effect of gravity.

In the hydrogen atom the lighter electron revolves around the much heavier proton.What would be the difference if the heavier proton were to orbit the lighter electron ?

In the satellite question the masses of the satellites where the same, but the radii oforbit were different. Here the masses are different:

mass of electron = 9.1096 X 10 −−−−−31 kgmass of proton = 1.6726 X 10 −−−−−27 kg

If the proton is made to orbit the electron, then it must orbit at a much higherangular velocity to prevent it from collapsing inwards. Hence the total energy of thissystem will be greater than the electron orbiting the proton system.

We see that nananananaturturturturture ale ale ale ale alwwwwwaaaaays tends to fys tends to fys tends to fys tends to fys tends to f ind its loind its loind its loind its loind its lowwwwwest staest staest staest staest state ofte ofte ofte ofte of ener ener ener ener energggggyyyyy.If so, why is the universe expanding ? We learn from the Second LaSecond LaSecond LaSecond LaSecond Law ofw ofw ofw ofw ofTTTTT herherherherher modmodmodmodmodynamicsynamicsynamicsynamicsynamics tha t the entrthe entrthe entrthe entrthe entropopopopopyyyyy (r andomness) ofofofofof the un i the un i the un i the un i the un i vvvvvererererer seseseseseincrincrincrincrincr eases w i th eeases w i th eeases w i th eeases w i th eeases w i th e vvvvvererererer y spontaneous cy spontaneous cy spontaneous cy spontaneous cy spontaneous changhanghanghanghangeeeee .

200

We conclude this chapter with a few exercises that may require extra thought.

ExExExExExererererercise1:cise1:cise1:cise1:cise1: Given f(x) = 1/////a f ’(x) and f ’(x) = a f(x) where a ≠ 0 is constant,what is f(x) ?

Exercise2: Exercise2: Exercise2: Exercise2: Exercise2: What is ∫dx ?

a) an interval on the x - axis

b) the independent variable x

c) 1

d) the expression of CHANGE F(x) = x

ExExExExExererererercise cise cise cise cise 33333::::: A pump delivers at the rate of 60T litres/minute. What is thedelivery rate in litres/second ? (T is time).

a) T b) T/120 c) T/60 d) T / 60 2

ExExExExExererererercise cise cise cise cise 44444::::: A car moves at a speed of 60T kms/hour. What is its speed inmeters/second ? (T is time)

a) 50T /3 b) T/(2 x 63) c) T/(60 x 63) d) T/63

ExExExExExererererercise cise cise cise cise 55555::::: 4----------3 πr 3 is the volume of a sol id sphere of radius r. Its

derivative is 4πr 2 which is the surface area of the sphere of radius r. Whatis 8πr, the derivative of the surface area of the sphere of radius r ? (Hint : circles.)ExExExExExererererercise cise cise cise cise 66666::::: A pump delivers at the rate of T

3/////2 litres/minute. What is thedelivery rate in litres/second ? (T is time).

ExExExExExererererercise cise cise cise cise 77777::::: With the bouncing ball example in mind, is it correct to say :

arararararea under curea under curea under curea under curea under curvvvvveeeee of f(x) = 0 ⇔ CHANGE in F(x) = 0 ?

“ GENIUS is ask ing the r ight quest ion. INTELLIGENCE is finding the right answer.”

Stephen Vadakkan

After completing his Bachelor of Science at KuwaitUniversity in 1977 with distinction of rank holder, he wenton to do two masters degree in Computer Science andMathematics at the University of Arizona in Tucson,U.S.A. Later for his Ph.D. he studied Cryptographyand Communications Security under the world renownedProf. F.C. Piper of the University of London. He returnedto Kuwait and worked as a Research Associate at the Kuwait Institute forScientific Research (KISR) and for Kuwait Air Force / Air Defense, and alsoas a consultant for various commercial ventures.

For a brief period he was an Honorary Visiting Faculty member and Reader(Associate Professor) of Computer Engineering at the Manipal Institute ofTechnology, Manipal.

He is a writer and philosopher.He is the author of “A Little Bitof Calculus” - an introductorybook on the fundamentals ofCalculus. He has discoveredand proved the exact formula forthe perimeter of the ellipse .

1 4 perimeter of ellipse = (a −−−−− b) 2 + ( π 2 ) 2 ab/ /

b

a

Prof. Stephen VADAKKAN

Calculus

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