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Calculus

Prepared by Pr. Ralph W.P. Masenge

African Virtual universityUniversit Virtuelle AfricaineUniversidade Virtual Africana

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African Virtual University

Notice

This document is published under the conditions of the Creative Commons

http://en.wikipedia.org/wiki/Creative_Commons

Attribution

http://creativecommons.org/licenses/by/2.5/

License (abbreviated cc-by), Version 2.5.

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I. Mathematics3,Calculus_____________________________________3

II. PrerequisiteCourseorKnowledge_____________________________3

III. Time____________________________________________________4

IV. Materials_________________________________________________4

V. ModuleRationale __________________________________________5

VI. Content__________________________________________________6

6.1 Overwiew___________________________________________6

6.2 Outline_____________________________________________6

6.3 GraphicOrganizer_____________________________________8

VII. GeneralObjective(s)________________________________________9

VIII. SpecificLearningObjectives__________________________________9

IX. TeachingAndLearningActivities _____________________________10

9.1 Pre-Assessment_____________________________________10

9.2 Pre-AssessmentAnswers______________________________17

9.3 PedagogicalCommentForLearners______________________18

X. KeyConcepts(Glossary)____________________________________19

XI. CompulsoryReadings______________________________________26

XII. CompulsoryResources_____________________________________27

XIII. UsefulLinks _____________________________________________28

XIV. LearningActivities_________________________________________31

XV. SynthesisOfTheModule ___________________________________77

XVI. SummativeEvaluation_____________________________________120

XVII.MainAuthoroftheModule ________________________________131

Table of ConTenTs

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I. Mthmtic 3, Ccuu

Prof. Ralph W.P.Masenge, Open University of Tanzania

Figure 1 : Flamingo family curved out of horns of a Sebu Cow-horns

II. Prrquiit Cur r Kwdg

Unit 1: Elementary differential calculus (35 hours)

Secondary school mathematics is prerequisite. Basic Mathematics 1 is co-re-

quisite.

This is a level 1 course.

Unit 2: Elementary integral calculus (35 hours)

Calculus 1 is prerequisite.


Unit 3: Sequences and Series (20 hours)

Priority A. Calculus 2 is prerequisite.


Unit 4: Calculus of Functions of Several Variables (30 hours)

Priority B. Calculus 3 is prerequisite.


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III. Tim

120 hours

IV. Mtri

The course materials for this module consist of:

Study materials (print, CD, on-line)

(pre-assessment materials contained within the study materials)

Two formative assessment activities per unit (always available but with speci-

ed submission date). (CD, on-line)

References and Readings from open-source sources (CD, on-line)

ICT Activity les

Those which rely on copyright software

Those which rely on open source software

Those which stand alone

Video les

Audio les (with tape version)Open source software installation les

Graphical calculators and licenced software where available

(Note: exact details to be specied when activities completed)

Figure 2 : A typical internet caf in Dar Es Salaam

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V. Mdu Rti

The secondary school mathematics syllabus covers a number of topics, including

differentiation and integration of functions. The module starts by introducing

the concept of limits, often missed at the secondary school level, but crucial in

learning these topics. It then uses limits to dene continuity, differentiation and

integration of a function. Also, the limit concept is used in discussing a class of

special functions called sequences and the related topic of innite series.

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VI. Ctt

6.1 Overview

This is a four unit module. The rst two units cover the basic concepts of the

differential and integral calcualus of functions of a single variable. The third unit

is devoted to sequences of real numbers and innite series of both real numbers

and of some special functions. The fourth unit is on the differential and integralcalculus of functions of several variables.

Starting with the denitions of the basic concepts of limit and continuity of

functions of a single variable the module proceeds to introduce the notions of

differentiation and integration, covering both methods and applications.

Denitions of convergence for sequences and innite series are given. Tests for

convergence of innite series are presented, including the concepts of interval

and radius of convergence of a power series.

Partial derivatives of functions of several variables are introduced and used in

formulating Taylors theorem and nding relative extreme values.

6.2 Outline

Unit 1: Elementary differential calculus (35 hours)

Level 1. Priority A. No prerequisite. Basic Mathematics 1 is co-requisite.

Limits (3)

Continuity of functions. (3)

Differentiation of functions of a single variable. (6)

Parametric and implicit differentiation. (4)Applications of differentiation. (6)

Taylors theorem. (3)

Mean value theorems of differential calculus. (4)

Applications. (6)

Unit 2: Elementary integral calculus (35 hours)

Level 1. Priority A. Calculus 1 is prerequisite.

Anti derivatives and applications to areas. (6)

Methods of integration. (8)Mean value theorems of integral calculus. (5)

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Numerical integration. (7)

Improper integrals and their convergence. (3)

Applications of integration. (6)

Unit 3: Sequences and Series (20 hours)

Level 2. Priority A. Calculus 2 is prerequisite.

Sequences (5)

Series (5)

Power series (3)

Convergence tests (5)

Applications (2)

Unit 4: Calculus of Functions of Several Variables (30 hours)

Level 2. Priority B. Calculus 3 is prerequisite.

Functions of several variables and their applications (4)

Partial differentiation (4)

Center of masses and moments of inertia (4)

Differential and integral calculus of functions of several variables:

Taylors theorem (3)

Minimum and Maximum points (2)

Lagranges Multipliers (2)

Multiple integrals (8)

Vector elds (2)

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6.3 Graphic Organizer

This diagram shows how the different sections of this module relate to each

other.

The central or core concept is in the centre of the diagram. (Shown in red).

Concepts that depend on each other are shown by a line.

For example: Limit is the central concept. Continuity depends on the idea of

Limit. The Differentiability depend on Continuity.

Limit

Sequence

Functionsof severalvariable

Functionsof a singlevariable

Integrability

Differentiability

Continuity

InfiniteSeries

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VII. Gr ojctiv()

You will be equipped with knowledge and understanding of the properties of

elementary functions and their various applications necessary to condently teach

these subjects at the secondary school level.

You will have a secure knowledge of the content of school mathematics to con-

dently teach these subjects at the secondary school level

You will acquire knowledge of and the ability to apply available ICT to improvethe teaching and learning of school mathematics

VIII. spcic lrig ojctiv(Itructi ojctiv)

You should be able to demonstrate an understanding of

The concepts of limits and the necessary skills to nd limits.

The concept of continuity of elementary functions.

and skills in differentiation of elementary functions of both single andseveral variables, and the various applications of differentiation.

and skills in integration of elementary functions and the various appli-cations of integration.

Sequences and series, including convergence properties.

You should secure your knowledge of the following school mathematics:

Graphs of real value functions.

Idea of limits, continuity, gradients and areas under curves using graphs offunctions.

Differentiation and integration a wide of range of functions.

Sequences and series (including A.P., G.P. and notation).

Appropriate notation, symbols and language.

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African Virtual University 0

IX. Tchig ad lrig activiti

9.1 Pre-assessment

Module 3: Calculus

Unit 1: Elementary Differential Calculus

1. Which of the following sets of ordered pairs ( , )x y represents a function?

(a) (1,1), (1,2), (1,2), (1,4)

(b) (1,1), (2,1), (3,1), (4,1)

(c) (1,1), (2,2), (3,1), (3,2)

(d) (1,1), (1,2), (2,1), (2,2)

The set which represents a function is

)(

)(

)(

)(

d

c

b

a

2. The sum of the rst n terms of a GP, whose rst term is a and common

ratio is r , is11

1

n

n

rS a

r

+ =

. For what values of r will the GP converge?

The GP will converge if

=

, however small, there exists a corresponding positive integer

)(N , such that . It can be proved that if asequence is a Cauchy sequence, then it converges.

(iii) Series

DivergenceTest: If the series

=1k

na converges, then 0lim = nn a . If

0lim

nn

a then the series diverges. We note, in passing, that 0lim =

nn

a is only

a necessary condition for a series to converge but it is not a sufcient condition.

This means that there divergent series for which the condition applies. Indeed,

one simple series, the harmonic series

=1

1

k kcan be shown to be divergent al-

though clearly .0

1

lim =

nn

IntegralTest:Let )(xf be continuous and positive for all ),1[ x . Then

the innite series

=1

)(k

nf converges if, and only if, the improper integral

converges.

CmparisnTest: Let nn ba

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if

=1k

na diverges then

=1k

nb also converges.

RatiTest: If 0>na for all 1n and if La

a

n

n

n=+

1lim , then the series

converges if 1L . If 1=L the test fails.

thn RtTest:If 0>na for all n and if La nnn

=

1

lim then

the series

=1k

na converges if 1L . If 1=L then

the test fails.

Learnng Actvty

introducton

1. Look carefully at the images ( 1I and 2I ) of objects (sticks) arranged sequen-tially.

Both are meant to represent the rst few terms of an innite sequence of numbers

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{ }na . In each case the arrow points in the direction of increasing values of theindex n of the elements of the sequence.

Qestins

(i) Is the sequence { }na represented by the Image 1I increasing or de-creasing?

(ii) Is the sequence converging or not converging (diverging)?

Pose and attempt to answer the same two questions for the sequence represented

by the image 2I .

2. Consider the sequence of numbers whose rst ve terms are shown here:

1,2

1,

3

1,

4

1,

5

1,

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Key Defntons

A clear understanding by the learner, of terms used in this learning activity, is

very essential. The learner is therefore strongly advised to learn and carefully

note the following denitions.

(i) Seqences

A sequence { }na is said to be mntneif it is increasing, decreasing, non-increasing or non-decreasing.

A sequence { }na is said to be bndedif there exist two numbers Mm,

such that Mam n for all n larger than some integerN.

(ii) Series

Absltecnvergence:A series

=1k

ka is said to converge absolutely if the

series

=1k

ka converges.

Cnditinalcnvergence:If the series

=1k

ka converges but the series

=1k

ka diverges, then

=1k

ka is said to converge conditionally.

Psitivetermseries:If 0>na for all n , then the series

=1k

ka is called

a positive term series.

Alternatingseries:If 0>na for all n then the series

( )

=

+ ++=1

4321

1.....1

k

k

kaaaaa

or

( )

=

++=1

4321 .....1k

k

kaaaaa

is called an alternating series.

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Mathematcal Analyss

The principal reference text for the learner in the topic of Sequences and Series

is S.G.Gill. The material is contained in Chapter 8 (pp 315 354)

Sequences

For a sequence, the mathematical problem is

Either: to determine whether it converges or diverges,

or: to nd the limit if it converges.

After giving a mathematical denition of the concept of convergence of a se -

quence, a set of standard algebraic properties of convergent sequences is given

on page 316. The set includes limits of sums, differences, products and quotients

of convergent sequences. The learner is urged to know those properties and read

carefully the proofs that immediately follow (pp 316 319).

Conditions for convergence of a sequence are very well spelt out in the main

reference sighted above.

DoTHISStudy carefully the set of key denitions given on page 320 of the text. The

denitions are immediately followed by a key theorem which gives criteria for

convergence. In a group work, go over the given proofs to ensure you understand

and accept the premises (conditions stated) and the conclusion of the theorem.

ErrrAlert:Please note an unfortunate printing error that appears on page 323

of the reference, where the thn partial sum =

n

k

kt

1

is referred to as an innite

series. This should be corrected to

=1k

kt .

DoTHIS

On the basis of the denition of convergence given on page 315, the properties

of convergent sequences given on page 316 and the denitions of terms and

convergence criteria stated and proved starting from page 320, slveasmany

prblemsasycanfrom Exercise 8.1 (pp 326 327).

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infnte seres

In general, innite series are derived from sequences. An innite series

=1k

ka can

be viewed as a sum of the terms of a sequence { }k

a and convergence of the series

is dened in terms of convergence of the associated sequence { },...,,, 4321 SSSS

of the series, where ==n

k

kn aS1

is the sum of the rst n terms of the series, also

called the n th partial sum.

Example

Let

+ )1(

1

nn be a sequence of numbers from which we can form the se-

ries

= +1 )1(

1

k kk. To determine whether or not this series converges, we need

to form its n th partial sum = +

=n

k

nkk

S1 )1(

1. Since

1

11

)1(

1

+=

+ kkkk one f inds

1

11

1

11

)1(

1

1 +=

+=

+=

= nkkkkS

n

k

n , and s ince

now 11

11limlim =

+=

nS

nn

n, we conclude that the infinite series

= +1 )1(

1

k kkconverges, and that its sum is unity.

Presentation of the material on innite series is done in frmajrstages.

Sectin8.4 (pp 329 334) is devoted to psitivetermseries

Under positive term series some very important theorems are stated, including:

(i) The comparison test (page 339)

(ii) The ratio test (page 331) and

(iii) The n th root test.

These are stated and quite elegantly proved.

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DoTHIS

Study carefully the assumptions made for each of the tests, read carefully the

proofs given and then check the level of your understanding by solving as pro-

blems as you can in Exercise8.2(pp335341).

Sectin8.5(pp 341 - 346) presents alternatingseries

Alternating series are quite common when evaluating some power series at

a xed point. Important results related to alternating series include the twin

concepts ofabslteand cnditinalconvergence associated with such series.

A classical example of an alternating series which is conditionally convergent

is the series ( )k

k

k

11

1

1

+

=

which can be shown to converge. However, the seriesobtained by taking absolute values of the terms is the harmonic series which we

know diverges. Its being divergent can easily be proved using the integral test.

Sectin8.6(pp 347 354) deals with pwerseries

A power series is a series of the form

k

k

kxa

=1

rk

k

kcxa )(

1

=

, where c is a constant.

The cnvergencefapwerseries can be established by applying the rati

test. Here one considers

n

n

n

n

n xa

xa1

1lim

+

+

=

n

n

n a

ax 1lim +

r

n

n

n a

acx 1lim +

.

If La

a

n

n

n=+

1lim then for convergence one demands that 1

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Lx

1< or

Lcx

1

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XV. SynthesisoftheModule

Completion of Unit Four marks the end of this module. At this juncture, and before

presenting oneself for the summative evaluation, it is desirable and appropriate

that the learner reects back on the mission, objectives, activities and attempts

to form a global picture of what he/she is expected to have achieved as a result

of the collective time and efforts invested in the learning process.

As laid out in the module overview in Section 6, the learner is expected to have

acquired knowledge of the basic concepts of differential and integral calculus of

functions of one and several independent variables. Inherently imbedded in the

key word calcls is the underlying concept oflimitswhich underpins the

entire spectrum of coverage of the module.

The learner is now expected to be able to comfortably dene the limit concept,

evaluate limits of functions, sequences and innite series. One is also expected

to appreciate the use of the limit concept in dening the concepts of continuity,

differentiation (ordinary and partial) and integration (Riemann integral). Thus, the

signposts (salient features) on the roadmap for this module are limits,cntinity,

differentiatinandintegratinas applied to functions of both single and several(specically two) independent variables.

The key areas of application of the knowledge acquired are in determining local

maxima and minima, areas, volumes, lengths of curves, rates of change, moments

of inertia and centers of mass.

The module has been structured with a view to escorting and guiding the learner

through the material with carefully selected examples and references to the core

references. The degree of mastery of the module contents will largely depend on

the learners deliberate and planned efforts to monitor his/her progress by solving

the numerous self-exercise problems that are pointed out or given throughout

the module.

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Mathematics Module 3: Calculus

Unit 4: Calculus of Functions of Several Variables (30 Hrs)

Activity 1: Limits, Continuity and Partial Derivatives (11 Hrs)

Specific Learning Objectives

At the end of this activity the learner should be able to:

Find out whether or not a limit exists;

Evaluate limits of functions of several variables;

Determine continuity of a function at a point and over an interval.

Find partial derivatives of any order.

Expand a function in a Taylor series about a given point

Summary

The rst three units of this module dealt exclusively with the elementary dif-

ferential and integral calculus of functions of a single independent variable,

symbolically expressed by the equation )(xfy = . However, only a very smallminority of real life problems may be adequately modeled mathematically using

such functions. Mathematical formulations of many physical problems involve

two, three or even more variables.

In this unit, and without loss of generality, we shall present the elementary dif-

ferential and integral calculus of functions of two independent variables. Speci-cally, in this learning activity, we shall discuss the concepts of limit, continuity

and partial differentiation of functions of two independent variables.

Compulsory Reading

All of the readings for the module come from Open Source text books. This means

that the authors have made them available for any to use them without charge.

We have provided complete copies of these texts on the CD accompanying this

course.

The Calculus Bible, Prof. G.S. Gill: Brigham Young University, Maths Depart-

ment, Brigham Young University USA.

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Internet and Software Resources

Software

Extend your expertise with mxMaxima. Investigate any functions that you nd.

Check in the wxmaxima manual to nd if wxMaxima can do them. If it can

practice and explore.

Wolfram Mathworld (Visited 07.11.06)http://mathworld.wolfram.com/PartialDerivative.html

Read this entry for Partial Derivatives.

Follow links to explain specic concepts as you need to.

Wikipedia (visited 07.11.06)

http://en.wikipedia.org/wiki/Partial_derivative

Read this entry for Partial Derivatives.

Follow links to explain specic concepts as you need to.

Use Wikipedia and MathWorld to look up any key technical terms that you come

across in this unit.

Key Concepts

Domain and Range

If ),( yxfz= is a function of two independent variables then the set2),( = yxD of points for which the function ),( yxfz= is dened is

called the Domain of z, and the set { }DyxyxfzR == ),(:),( of valuesof the function at points in the domain is called the Range of z.

Graph of a function

The graph of a function ),( yxfz= is the set { }DyxzyxG = ),(:),,(While the graph of a function of a single variable is a curve in they plane, the

graph of a function of two independent variables is usually a surface.

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Level curve

A level curve of a function of two variables ),( yxfz= is a curve with an equa-

tion of the form Cyxf =),( , where Cis some xed value ofz.

Limit at a point

Let ),( ba be a xed point in the domain of the function ),( yxfz= and L be

a real number. L is said to be the limit of ),( yxf as ),( yx approaches ),( ba

written Lyxfbayx

=

),(lim),(),(

, if for every small number 0> one can nd

a corresponding number )(= such that

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Key Theorems and/or Principles

Criteria for Nonexistence of a Limit

If a function ),( yxf has different limits at a point ),( ba as the point ),( yx

approaches point ),( ba along different paths (directions), then the function has

no limit at point ),( ba .

Remark: Taking limits along some specic paths is not a viable strategy for

showing that a limit exists. However, according to this theorem, such an approach

may be used to show that a limit does not exist.

Sufcient Condition for Existence of a Limit

Let D be the domain of a function ),( yxf and assume that there exists a

number L such that ),(),(),( yxyxgLyxf that satisfy the inequa-

lity

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Equality of Mixed Partial Derivatives

If the mixed second partial derivativesxyf

yx

f=

2and

yxf

xy

f=

2are

continuous on an open disc containing point ),( ba , then ),(),( bafbaf yxxy =

Necessary Condition for Local Extreme Values

If ),( yxf has a relative extreme value (local maximum or local minimum)

at a point ),( ba and ifx

ffx

= ,

y

ffy

= exist, then 0),( =bafx and

0),( =bafy .

Second Derivative Test for Local Extreme Values

Let ),( yxf possess continuous rst and partial derivatives at a critical point

),( ba . At point ),( ba we dene the quantity

2),( xyyyxxyyyx

xyxxfff

ff

ffbaD ==

Then,

If 0>D and 0>xxf , then ),( baf is a local minimum.

If 0>D and 0

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Learning Activity

1. The volume of a cube with sides of length zyx ,, is given by the formula

.xyzV =

2. The volume of a right circular cone of height h and base radius r is given

by hrV2

3

1= .

Mathematical Problem

Definitions of Function, Domain and Range

What is common in all two examples given above is that, in each case, a rule is

given for assigning a unique value to the respective dependent variable (Area

or Volume) whenever a set of the corresponding independent variables is spe-

cied.

Typically, if the radius of the circular cone in Example 3 is xed to be 2 and theheight is3 , then the volume of the cone has the value 4=V .

The learner is strongly encouraged to recall the corresponding denitions for

a function of a single independent variable and see the close similarity or even

outright identity in the denitions.

In general, we may define a function ),...,,( 321 nxxxxf of n indepen-dent variables as a rule or a formula which assigns a unique number

=z ),...,,( 321 nxxxxf for every given set{ } nn Dxxxx ,...,, 321 .

Symbolically, such a mapping is represented by writing f : nD .

The domain of a function of several independent variables is the set D of all

elements },...,,{ 321 nxxxx for which the function f has a unique value.

The range of a function of several independent variables is the set ofvalues of

the function f for all points },...,,{ 321 nxxxx taken from the domainD . Fora function of two independent variables, the range is often a connected surface

(roof or cover) spanned above the domain.

Please Note This Carefully

For purposes of maintaining simplicity of presentation and thereby hopefully

gain clarity of understanding, but without loss of generality, we shall limit our

treatment of functions of several independent variables to functions of two

independent variables.

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Example

Consider the relationship =z 4ln),( 22 += yxyxf.For this relationship

to represent a function of the two independent variables ),( yx the logarithmicfunction must be dened (have a unique value) at such points. This is only pos-

sible if 422 + yx 0> .

This requirement restricts the choice of pairs ),( yx , namely to the region or

domain 4),(22 += yxyxD

, implying that ),( yx must lie outside the

open disc 422

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Do This Exercise

1. Give as many examples as you can (from real life, mathematics and science

in general) of functions of more than one independent variable.

2. For each of the following functions

(i) Find and sketch the domain of the function, and(ii) Find the range.

(a)322),( yxyyxyxf +=

(b)22

2

1

2),(

yx

yxyxf

=

(c)

229

),(yx

eyxf

=

(d) )sin(),( xyyxf = .

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Level curves

One does not need to be a geographer to know that the position of any point on

our planet earth is completely determined once its latitude and longitude are

known. Equally true is the fact that the altitude (height above mean sea level) of

any point depends on its position on the globe.

Geographers use contours to obtain a visual impression of the steepness of a

given landscape.

What are contours?

Contours are lines (curves) on a map connecting all points on a given landscape

which have the same height above mean sea level.

The idea of contours has been adopted by mathematicians to gain some insight

of the shape of a function of two independent variables. Instead of the word

contour mathematicians use the more descriptive phrase level curves.

What are Level Curves?

A level curve of a given function ),( yxf is any curve described by an equa-

tion of the form Cyxf =),( , where C is a parameter that may take any valuein the range of f . In other words, a level curve is the locus of all points ),( yx

in the domain D of ),( yxf for which the function has the constant valueC

Examples

1. The level curves of the function22),( yxyxf += are circles given by the

equation kyx =+ 22 in which k is a positive constant. For different choices

of the parameter k the circles are concentric with centre at the origin and haveradius kr= .

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Concentric circles of radii r= 1k , 2k and 3k .

Do this

Sketch some level curves for the function yxyxf += 2),( .

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Limit of a function of two variables

As pointed out in Unit 1 of this same Module in connection with a function

)(xf of a single variable, the limit concept is also fundamental for the analysisof functions of several variables. The denition of a limit is given in Section 1.4

as one of the key concepts of this unit.

As is true elsewhere in mathematics, it is one thing to dene a concept, such as

that of a limit, and quite another to use it. It is particularly difcult for the limit

concept, in that the denition inherently assumes that one already knows the limit!

Here we shall lean heavily on the rst two key theorems on limits listed in Section

1.1: Key Theorems / Principles. We shall discuss three illustrative examples and

let the learner do the same in a group work on some problems.

Example 1 (An application of Theorem 1)

Let22

),(yx

xyyxf

+= . We now pose the question: Does

+ 22)0,0(),(lim

yx

xy

yx

exist? In an attempt to nd an answer to this question we let point ),( yx approach

)0,0( along the liney=mx where the slope m of the line is purposely left as aparameter. In this case

+ 22)0,0(),(lim

yx

xy

yx

=2222

2

0 1lim

m

m

xmx

mx

x +=

+.

Since the limit depends on the slope of the line, we conclude that the func-

tion22

),(yx

xyyxf+

= has no limit at the origin. This result follows from

Theorem 1

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Example 2 (A second application of Theorem 1)

Consider now the function42

22),(

yx

xyyxf

+= and pose the same question: Does

limit

+ 42

2

)0,0(),(

2lim

yx

xy

yxexist? Adopting the same linear approach with

y = mx we get

+=

+=

)1(

2

1

2),(

22

2

24

2

xm

mx

xm

xmyxf , and hence conclude

that

+ 22

2

)0,0(),(

2lim

yx

xy

yx= 0

1

2lim

22

2

0

=

+x

xm

m

xregardless of the direction one

chooses in approaching the origin.

This result may tempt one to conclude that zero is the limit of the function at the

origin. However, let us be careful. Key Theorem No.1 warns us that we must

be very careful in using this approach, for we have not exhausted all possible

directions of approach. Indeed we soon nd out that if we take the path dened

by the curve2yx = then

+ 42

2

)0,0(),(

2lim

yx

xy

yx= 1

2lim

44

4

0

=

+ yy

y

y

This proves that the function has no limit at the origin because we get two different

limits depending on the path we take to approach the origin.

What reliable method can we then adopt in trying to determine limits of

functions of several (two) variables?

Example 3 (An application of Theorem 2)

An example taken from Robert T. Smith and Roland B. Minton: Multivariable

Calculus, p. 44.

Let22

2

),(yx

yxyxf

+= , and consider ),(lim

)0,0(),(

yxfyx

. The learner is

requested to show that, in this case, any approach to point the )0,0( along any

straight limey = mx will give lead to the limit 0=L . This is equally true

if one takes the paths2xy

=or

2yx=

. We may therefore reasonably suspect

that .0lim22

2

)00(),(=

+ yx

yx

yxWe consider applying Theorem 2 to prove that indeed

the limit of the function is zero as

)0,0(),( yx .

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We begin by noting that since for all :),( yx x2 + y2 x2 , i t fol lows that

f(x, y) L =x2 y

x2 + y2

x2 y

x2= y . The function ),( yxg sighted in

Theorem 2 is in this case yyxg =),( . But 0),(lim)0,0(),(

=

yxgyx

. Therefore,

by Theorem 2, it follows that .0lim 22

2

)00(),(=

+ yxyx

yx

Do This

Using the above three examples and working with a colleague whenever and

wherever possible, determine whether or not the following limits exist. Where

a limit exists, nd it.

1. )4(lim22

)0,0(),(

+

yxyx

2.

+

22)0,0(),(lim

yx

yxyx

3.

+

22

2

)0,1(),( )1(

)ln()1(lim

yx

xx

yxConsider limits as )0,1(),( yx along

the three: along 1=x ; along 0=y , and along 1= xy . Do you suspectanything?

Continuity of a function of two variables

In Unit 1 of this Module we came across the concept of continuity of a function

of one independent variable on two occasions.

The rst time was in Section 1.4 on Key Concepts where the concept of conti-

nuity at a point ax = was rst dened. At that juncture we listed three necessary

and sufcient conditions for a function )(xf to be continuous at a point ax =namely:

(i) The function f must have a value (dened) at the point, meaning that )(af must exist;

(ii) The function f must have a limit as x approaches a from right or left,

meaning that )(lim xfax

must exist (let the limit be L ); and

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(iii) The value of the function at ax = and the limit at ax = must be equal,

meaning )(afL = .

In short, we summed up these conditions with the single equation )()(lim afxfax

=

The second time we came across the concept of continuity was in Section 1.7.2,

where we learnt how to determine whether or not a given function ),( yxf is

continuous at a point or over an entire interval bxa , including what wasreferred to as removable discontinuity.

In the current unit (Unit 4) we are dealing with functions of more than one inde-

pendent variable. Specically, we are considering a function f of two indepen-

dent variables: ),( yxfw =

We want to dene the concept of continuity for a function of two independent

variables. Again we note that we already have dened the concept in Section 1.4

of this Unit. The denition is exactly the same as that for a function of a single

variable, namely, ),( yxf is said to be continuous at a point ),( ba in its domain

if, and only if ),(),(lim),(),(

bafyxfbayx

=

. In short,

(i) If either ),( yxf is not dened at point ),( ba or

(ii) If ),(lim),(),(

yxfbayx

does not exist, or

(iii) If ),(lim),(),(

yxfbayx

exists but is different from ),( baf

then we can conclude that ),( yxf is not continuous (discontinuous) at ),( ba .

In the special case where the function is dened at the point and the limit exists,

but the limit and the function value are different, one speaks of a removable dis-

continuity, for one can under such circumstances dene a new function

=

=

),(),(),(lim

),(),(),(

),(

),(),(bayxifyxf

bayxifyxf

yxF

bayx

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that clearly satises all the three stated conditions for continuity at ),( ba .

Examples

We wish to discuss continuity of each of the following three functions at the

points given. The three examples have purposely been chosen to demonstrate the

three possible situations: continuity, removable discontinuity and non-removable

discontinuity.

1. xyxxyyxf 32),( 22 += at )1,2(P .

We observe that f is a polynomial of degree two in two variables. A polyno-

mial function is continuous everywhere, and therefore it is continuous at )1,2( .Specically we note that:

The function is dened at the point: 0)1,2( =f

The function has a limit at the point: 032lim 22)1,2(),( =+ xyxxyyx

The limit is equal to the value of the function at the point. Therefore, the

function is continuous at (2,1)

2.x

xyyxf

)sin(1),(

2+= at )0,0(P .

We note that this function is not dened at point: )0,0(f does not exist, andtherefore, the function is not continuous at the origin. However, we note that the

function has a limit at the origin, since

( ) 111)sin(lim1lim)sin(1lim0

2

0

2

)0,0(),(==

+=

+ x

xy

x

xy

xyyx

Because the l imit exists but the function is discontinuous, we

have a typical case of a removable discontinuity. The derived function

=

=

)0,0(),(1

)0,0(),(),(

),(

yxif

yxifyxf

yxF

is continuous at )0,0( .

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3.22

),(yx

xyxf

+= at )0,0(P .

Clearly, this function is not dened at the origin and therefore, it is discontinuous

at the point. As to whether the function has a limit at the origin, we note the

following:

If we approach the point (0,0) along the along the y-axis (x = 0) we see that thefunction is identically zero and hence it has zero limit.

If, on the other hand, one approaches the origin along the x-axis (y = 0) we note that

the function takes the formx

yxf1

),( = , which has no limit as 0x . We

therefore conclude that the function has no limit and hence is not continuous at

)0,0( . In this case the discontinuity is not removable.

Now Do This

Discuss the continuity of the following three functions at the points given. If

the function is discontinuous, state giving reasons, whether the discontinuity is

removable or not removable.

1.2

2

1),(

yx

yxyyxf

= ; )1,2(

2.yx

yyxf

=),( ; )0,0(

3.22

1),(

yx

yxyxf

+

+= ; )0,0(

Partial Differentiation

One of the key concepts covered in the second learning activity of Unit one is the

derivative of a function )(xf of a single independent variable. In Section 1.4

of this Unit a key concept of partial derivatives of a function ),...,,( 321 nxxxxf of several independent variables is mentioned. We now make a formal denition

of this important concept specically in connection with functions ),( yxf oftwo independent variables.

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Partial Derivatives

Let ),( yxfw = be a function of the two variables ),( yx in a domain 2D

of the xy plane and ),( ba be a point inD . The function ),()( bxfxg = ,

obtained from ),( yxf by holding y at the constant value by = is a functionofx alone and is dened over an interval on thex axis that contains the value

ax = . One may therefore wish to nd out whetherg is differentiable at ax =

This involves nding

+=

+ h

bafbhaf

h

aghag

hh

),(),(lim

)()(lim

00

If the limit exists it is called the partial derivative of ),( yxf with respect to

the variable x at point ),( ba and is denoted by

),( bax

f

or ),( bafx or

x

w

The partial derivative ),( bay

f

is dened similarly by holding x constant at

the value ax = and differentiate the function ),()( yafyh = at by = .

If we allow the point ),( ba to vary within the domain D then the resultingpartial derivatives will also be functions of the variable point ),( yx . We there-fore have

),(),(),(

lim0

yxx

f

h

yxfyhxf

h

=

+

;

),(),(),(

lim0

yxy

f

k

yxfkyxf

k

=

+

.

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Worked Example

Find the partial derivatives of the function3223 232),( xyyxyxyxf ++=

By considering y as a constant and differentiating f as if it were only a

function of the variable x we get 322 266),( yxyyxyxx

f++=

.

Similarly, by considering x as a constant and differentiating f as if it were

only a function of the variable y one gets 223 662),( xyyxxyxx

f++=

.

Now Do This

Find the partial derivatives of the following functions:

1. ),( yxf =

3

2

lny

x

2. ),( rf =2

)2sin(

r

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Higher Order Partial Derivatives

We have seen that the partial derivativesx

f

and

y

f

of a function ),( yxf

of two independent variables are in turn functions of the same two independent

variables. Therefore, one can consider their partial derivatives in the same way we

did for the original function. One may therefore attempt to nd the derivatives

x

f

x;

y

f

x;

y

f

x;

y

f

y.

If limits of the associated difference quotients exist, then we call them the second

partial derivatives of f and denote them by

x

f

x=

2

2

x

f

or

xxf ;

x

f

y=

xy

f

2or

xyf ;

y

f

x

=yx

f

2or yxf ;

y

f

y=

2

2

y

f

or yyf .

Note Carefully

The learner should note carefully the notation used for the two mixed second

partial derivativesxy

f

2and

yx

f

2. The notation

xy

f

2or

xyf means that

one differentiates rst with respect to x and then with respect to y , while

yx

f

2or yxf means differentiation rst with respect to y and then with

respect to x .

Worked Example

Find all four second partial derivatives of the function ),( yxf = )cos( xyy .

One nds the following results

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x

f

= xf = )sin(

2xyy ;

y

f

= yf = )sin()cos( xyxyxy ;

2

2

x

f

=

xxf = )cos(3 xyy ;

xy

f

2

=xyf = )cos()sin(2 2 xyxyxyy ;

yx

f

2

= yxf = )cos()sin()sin(2 xyxyxyyxyx ;

2

2

y

f

= yyf = )cos()sin(2

2 xyyxxyx

Now Do This

Find the second partial derivatives of the following functions

1. ),( txf = 3223 432 xyyxyx ++ .

2. ),( yxf = )ln( 32yx .

Equality of Second Mixed Partial Derivatives

From the above worked example and from the results the Learner will get by

solving the last two self exercise problems, do you detect any relationship between

the two mixed partial derivatives fxy

andfyx

? State the relationship and look

back at the key theorem stated in item 5 of Section 1.5 on Key Theorems and/orPrinciples.

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What we nd is that in all cases the mixed second partial derivatives are equal.

This result is not by accident but a well established result based on the theorem

referred to and which states that iffxy

andfyx

are both continuous over the domain

2D of ),( yxf then fxy andfyx are equal at all points Dyx ),( .

C l e a r l y , p a r t i a l d e r i v a t i v e s o f o r d e r h i g h e r t h a n t w o c a n

be computed in exactly the same way by di fferen tiat ing second par-

tial derivatives. This will yield eight third partial derivatives, namely

xxxf , xxyf , xyxf , xyyf , yxxf , yxyf , yyxf , yyyf .

Do this

1. Find all eight third partial derivatives of the function ),( yxf = )32cos( xx + note the equality of the following sets of mixed third partial derivatives:

yxxxyxxxy fff ,, ; yyxyxyxyy fff ,, .

2. Show that the function ),( yxf = )cos(ye x satisfies the equation

0=+ yyxx ff .

3. Find the second partial derivatives of the function ),( yxf =

y

x

tdtte

.

Partial Increments, Total Increment and Total Differential

Partial Increments

In deriving the partial derivativesx

f

and

y

f

of a function ),( yxfz= of two

independent variables we kept one of the variables constant while giving the other

variable a small increase. Specically, in the case of the partial derivativexf

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we kept y constant and gave x a small increment x . We call the corres-

ponding change in the value of the function ),(),( yxfyxxf + the partialincrement ofz with respect to x , and denote it by

zx

= ),(),( yxfyxxf +

Similarly, one denes the partial increment of z with respect to y , writtenas

zy

=

The quantities zx

and zy

are called partial increments because, in each case,only one of the variables is allowed to change, the other being kept constant.

Total Increment

If we allow both variables x and y to change by small amounts x and

y respectively, then the change ),(),( yxfyyxxf ++ in the function

value is called the total increment inzwith respect to both its variables andis denoted by

z = ),(),( yxfyyxxf ++

Do this

Find the partial increments zx

and zy

of the function yxz 2= and use

the results to show that, in general, zx

+ zy

z .

),( yyxf + ),( yxf

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Total Differential

Let us attempt to relate the total increment with the partial derivativesx

f

andy

f

at ),( yx . By adding and subtracting the term ),( yyxf + we can

express z in the form

z = ),(),( yxfyyxxf ++

= ),(),(),(),( yxfyyxfyyxfyyxxf +++++

= xx

yyxfyyxxf

+++ ),(),(+ yy

yxfyyxf

+ ),(),(

= xyyxxx

f++

),( + yyyx

y

f+

),(

where the quantities and lie in the interval )1,0( .

If we now assume that the function ),( yxfz= has continuous partial deriva-tives, then

),(),(lim0,0

yxx

fyyxx

x

f

yx

=++

and

),(),(lim0,0

yxy

fyyx

x

f

yx

=+

We can therefore express the total increment in terms of the partial derivatives

in the form

z = xyxxf

+

1),( + yyxyf

+

2),(

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= xyxx

f

),( + yyx

y

f

),( + x

1 + y

2

where the quantities1

and2

tend to zero as x and y approach zero.

The term xyxx

f

),( + yyxy

f

),( in the expression for the above total

increment and which is linear in x and y is called the total differentialofz and is denoted by

dz = xyxx

f

),( + yyx

y

f

),( .

Example

Calculate the total differential and the total increment of the function

yxyxfz 2),( == at point )3,2( given that 2.0=x and 1.0=y .

Solution

The expressions for the total increment and the total differential are:

( ) ( ) yxxyyxxyxxxyz ++++= 222 22

yxxxydx +=2

2

Substituting the values 2=x , 3=y , 2.0=x and 1.0=y into both ex-pressions gives

dz = 8.2 and 004.3=z .

We observe from these results that the total differential gives a very good ap-

proximation of the total increment of a function. In the example we have solved,

the difference between dz and z is less than %8 .

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Now Do This

Using the total differential, approximate the total increment in calculating the

volume of material needed to make a cylindrical can of radius 2=R cm, height

10=H cm and whose walls and bottom are of thickness 1.0== HR cm.

Chain Rule for Differentiation

The variables x and y in the function ),( yxfz= may themselves be func-tions of either a single variable s or of two variables s and t.

The case of a single variable

If )(sxx = and )(syy = then

))(),((),( sysxfyxfz ==

is implicitly a function of the single variable s and therefore one may consider

nding the ordinary derivative of z with respect to s , that is, nding ds

df

.

The expression for the derivative is found using the chain rule for differentiating

a composite function (a function of a function):

ds

dz

=

ds

dx

x

f

+

ds

dy

y

f

.

Note carefully the different notations used in this expression, partial differenti-

ation of ),( yxf and the ordinary derivatives of both x and y .

The case of two variables

If ),( tsxx = and ),( tsyy = then

)),(),,((),( tsytsxfyxfz ==

is implicitly a function of the two variables s and t. One can therefore consider

nding the partial derivatives s

f

and t

f

. Again, these are found by the chain

rule, but this time, partial derivatives are involved throughout.

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s

z

=

s

x

x

f

+

s

y

y

f

;

t

z

=

t

x

x

f

+

t

y

y

f

.

Example

Given the function )ln(),( yxyxfz +== where vux 2= and 2uvy = ,

express the partial derivativesu

z

and

v

z

as functions of u and v .

Solution

Direct application of the second set of equations for the chain rule gives

u

z

=

yx

uv

+

2+

yx

v

+

2

=22

22

uvvu

vuv

+

+=

vu

v

+

+2

v

z

=

yx

u

+

2

+yx

uv

+

2=

22

2 2

uvvu

uvu

+

+=

vu

u

+

+ 2.

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Now Solve

1. If ),( yxfz= = 3223 432 xyyxyx ++ and cos=x ,ndd

dz.

2. If ),( yxf =22 yx + , where and )cos(rx = nd the partial derivatives

rz

and

z .

Exercise Questions (From Gill: The Calculus Bible)

Do This

Solve all the problems given in

Exercise 2.1 (pp 60-61)

Exercise 2.2 (pp 71-72)

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Activity 2: Applications of partial derivatives (19 Hrs)

Specific Learning Objectives

At the end of this activity the Learner should be able to:

State Taylors theorem for a function of two variables

WritedowntherstthreetermsofaTaylorsseriesexpansion

Findthecoordinatesofthecenterofgravity(mass)ofaplanelamina

Deneandclassifycriticalpointsforafunctionoftwovariables

Determinelocal(relative)maxima,minimaandsaddlepoints

UseLagrangemultipliersinconstrainedoptimization

Determine coordinates of center of gravity (mass)

Summary

Thislearningactivityisdevotedtosomeaspectsofapplicationsofpreceding

calculus(limits,continuityandpartialdifferentiation)ofafunctionoftwoin-

dependentvariables.Facedwithalonglistofpossibleareasofapplicationwehavebeenforcestoselectonlyafew.Wehavelimitedourpresentationtofour

areasofapplication,namely

Center of gravity

Taylors theorem

Localmaxima,minimaandsaddlepoints

Lagrangemultipliersforsolvingconstrainedoptimizationproblems.

Presentationoftheseareasofapplicationfollowsourfamiliarpatternofgivinga

shorttheoreticalnote,illustrationofsuchtheoryonanexampleortwofollowed

byatleasttwoselfexercisequestionstobesolvedbythelearner,aloneorwor-

kingwithcolleagues.

Compulsory Reading

AllofthereadingsforthemodulecomefromOpenSourcetextbooks.Thismeans

thattheauthorshavemadethemavailableforanytousethemwithoutcharge.

WehaveprovidedcompletecopiesofthesetextsontheCDaccompanyingthis

course.

Multivariable Calculus, George Cain and James Herod

Chapter7and8

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Internet and Software Resources

Wolfram Mathworld (Visited 07.11.06)

http://mathworld.wolfram.com/

Searchfor:Centerofgravity,Taylorstheorem,Localmaxima,Localminima,

Saddlepoints,Lagrangemultipliers[Ifyouareusingthisdocumentonacomputer,

then the links can be clicked directly]

Key Concepts

Center of gravity

Thecenterofgravityisalsoreferredtoasthecenterofmass.Thisisapointat

whichverticalandhorizontalmomentsofagivensystembalance.

Taylors formula

ThisseekstoextendtheTaylorseriesexpansionofafunction )( xf of a single

variableatapoint ax=

to a function ),( yxf oftwovariablesatapoint

).,( ba

Relative Extrema

Relativeextremaisacollectiveterminologyfortherelativemaximumandmini -

mumvaluesofafunction.Thesingularformofthewordisrelativeextremum,

whichmaybearelativemaximumorarelativeminimumvalue.

Lagrange Multipliers

Lagrangemultipliersarethenumbers(orparameters)associatedwithamethod

knownastheLagrangemultipliersmethodforsolvingproblemsofoptimization

(extrema)subjecttoagivensetofconstraints.

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Key Theorems and/or Principles

Necessary Condition for Local Extreme Values

If ),( yxf hasarelativeextremevalue(localmaximumorlocalminimum)at

apoint ),( ba and ifx

ff

x

= ,

y

ffy

= exist,then 0),( =bafx

and 0),( =bafy

.


Let ),( yxf possesscontinuousrstandpartialderivativesatacriticalpoint

),( ba .Atpoint ),( ba wedenethequantity

2),( xyyyxx

yyyx

xyxxfff

ff

ffbaD == .

Then,

If 0>D and 0>xxf , then ),( baf is a local minimum.

If 0>D and 0

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Thepoint ),( yxC is called the center of gravity or the center of mass of the

lamina.

Calculation of the coordinates x and y of the center of mass involves calcu-

latingthreerelatedquantities,namely:

The total mass M of the region D ;

The vertical moment xM of the lamina about the x axis. xM is a measureof the tendency of the lamina to rotate about the x axis.

Thehorizontalmomenty

M of the lamina about the y axis.y

M is a

measure of the tendency of the lamina to rotate about the y axis.

Aftercalculatingthesethreequantities,thecoordinatesofthecenterofgravity

are then obtained using the formulas:

x =

M

My

; y =

M

Mx.

Calculation of the Total Mass M

The total mass M of a region D in the xy planeandhavingsurfaceden-

sityfunction(massperunitarea) ),( yx is given by the value of the doubleintegral

M = D

dxdyyx ),( .

Example

Ifthemassofacircularplateofradius r isproportionaltoitsdistancefromthe

centerofthecircle,ndthetotalmassoftheplate.

Solution

Thesurfacedensityoftheplateisindirectlygivenas22),( yxkyx += where

k isaconstantofproportionality.Thetotalmassisthencalculatedfrom

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M = dxdyyxkD

+22

= r

0

drdr

2

0

2=

3

3

2rk .

Calculation of the Vertical Momentx

M

The vertical moment of the lamina is given byx

M = D

ydxdyyx ),(

Calculation of the Horizontal Momenty

M

The vertical moment of the lamina is given byy

M = D

xdxdyyx ),(

Example

Determinethecoordinatesofthecenterofgravityofthesectionoftheellipse

12

2

2

2

=+b

y

a

xwhichliesintherstquadrant,assumingthatthesurfacedensity

1),( =yx atallpoints.

Solution

The total mass

M = D

dxdy = dxdya

ba xa

0 0

22

Using the substitution cosax = the above iterated integral gives the value

of the total mass as abM4

1=

Similarly,usingthesamesubstitution,thevaluesofthehorizontalandverticalmoments

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yM =

D

xdxdy = xdxdya

ba xa

0 0

22

and

xM =

D

ydxdy = ydxdya

ba xa

0 0

22

are found to bey

M =3

2ba

andx

M =3

2ab

These results lead to the coordinates x =3

4aand y =

3

4bfor the center

ofgravityoflamina.

Self Exercise

1. Assumingaconstantsurfacedensity kyx =),( ndthecoordinatesof thecenterofmassoftheupperhalfofacircularlaminaofradius r .

2. Findthecenterofgravityofathinhomogeneousplatecoveringtheregion

D enclosed between the x axisandthecurve 22 yyx = .

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Taylors Formula

Onpage354ofourbasicreference:The Calculus Bibleby Brigham Young we

encountered Taylors formula for a function )( xfy = of a single variable about

apoint ax = .Therelevanttheorem(Theorem8.7.1)statesthat,if f and all its

derivatives1(''''''

,...,,+n

ffff are continuous for all x in a small neighborhood

ofapoint ax = ,then

)( xf = ( ) )(!

1 )(

0

afaxk

kkn

k

=

+ nR

where nR = ( ) )()!1(

1 )1(1 ++

+

nn faxn

,with ),( xa .

Wenowwishtoextendthistheoremtofunctionsoftwovariables.Specically,wewishtoexpandafunction ),( yxfz = of two variables in a Taylor series

aboutapoint ),( ba in the domain2D ofthefunction.

Let ),( yxfz = beafunctionoftwovariableswhichiscontinuous,togetherwith

allitspartialderivativesuptothestn )1( + orderinclusive,insomeneighborhood

ofthepoint ).,( ba Then,likeinthecaseofafunctionofasinglevariable,the

function ),( yxf canberepresentedintheformofa sumofann-thdegree

polynomialinpowersof )( ax , )( by ,andaremainderterm nR .

Underthisassumptionthefunction ),( yxf canbeexpressedinaTaylorformula

aboutthepoint ),( ba as follows:

),( yxf =

kn

k yby

xax

k

+

=

)()(!

1

0

),( baf + nR ,

where nR = ),()()(

)!1(

1)1(

f

y

by

x

ax

n

n+

+

+

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forsomepoint ),( lying in the disc centered at ),( ba and containing the

point ),( yx .

Observation

Thelearnershouldpayspecialattentiontotheoperator

yby

xaxD

+

= )()(

whichappears in theaboveTaylor expansion

formula.Powersofthisoperatorareformedinaccordancewiththebinomial

expansionofthetermnba )( + .Typically,

2

22

2

2

22

2

2)())((2)()()(

yby

yxbyax

xax

yby

xaxD

+

+

=

+

=

Do This

WritedownallthetermsofthedifferentialoperatorD3.

Maximum, Minimum and Saddle Points Functions of Two Variables

Let ),( yxfz = beafunctionoftwovariablesdenedoveraregion2D

We begin by making three observations:1. Themaximumandminimumvaluesofa functioninagivendomainare

collectivelyreferredtoasextremevaluesorsimplyasextremaofthefunction.

Withoutspecifyingthenatureoftheextremevalue,thesingularformoftheword

extremaisextremum.

2. Anextremumofafunction ),( yxfz = maybeabsoluteorrelative.

3. Anextremumofafunction ),( yxfz = can occur only at

i)aboundarypointofthedomainof f ,

ii)atinteriorpointswhere xf = yf = 0 or

iii)atpointswherex

f andy

f failtoexist.

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Denitions

i) Apoint DyxP ),( 00 issaidtobeapointof absolute maximum if

)()( QfPf forallpoints DQ .

ii)Apoint DyxP ),( 00 issaidtobeapointofabsolute minimum if

)()( QfPf forallpoints DQ .

iii)ApointDyxP )

,( 00

issaidtobeapointof absolute maximum if

)()( QfPf forallpoints DQ which are in a small neighborhoodofpoint

iv)Apoint DyxP ),( 00 issaidtobeapointofabsolute minimum if

)()( QfPf forallpoints DQ which are in a small neighborhoodofpoint

Ascanbenotedfromtheabovedenitions,thedifferencebetweenabsoluteand

relativemaximumorminimumliesinthesetofpointsQ onecomparesthexed

pointP with.IfQ isrestrictedtoasmallneighborhoodofpointP then wespeakofrelativeextrema(minimumormaximum).

Basic Question: Howcanonelocatethepoints P at which the function

),( yxf hasextremevalues?

Assuming that ),( yxf hascontinuousrstpartialderivatives,theanswertothe

abovequestionissimple:

All maximum and minimum points which lie in the interior of the domain of

denition of ),( yxf must be among the critical points of the function.

Thecriticalpointsof ),( yxf arethosepointswhichsimultaneouslysatisfythetwoequations

x

f

= 0 ;

y

f

= 0

Oncewehaveobtainedallthecriticalpointsweshallthenhavetoclassifywhich

amongthemaremaximumpointsandamongthemareminimumpoints.Weshall

alsohavetosaysomethingaboutanycriticalpointswhichrepresentneithera

maximumnoraminimumpointofthefunction.Aneasymethodofclassifying

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criticalpointsinvolvesthesecondpartialderivativesof ).,( yxf The method

isgiveninSection2.4aboveandforeaseofreference,wehavereproduceit

here.


Let ),( yxf possesscontinuousrstandsecondpartialderivativesatacritical

point ),( 00 yxP .At P wedenethequantity

2),( xyyyxx

yyyx

xyxxfff

ff

ffbaD == .

If 0>D and 0>xxf ,then ),( baf is a local minimum.

If 0>D and 0

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Solvingthesimultaneousnonlinearequationsx

f

= 0 ;

y

f

= 0 one gets

thefollowingcriticalpoints: )0,0( and )1,1( .Applyingthesecondderivative

testoneachpointwend:

For point )0,0(P : 090330)0,0( =

=D

Since 0)1,1( >xxf we conclude that )1,1(P isapointofrelativeminimum.

Self Exercise

1. Testthesurface444),( yxxyyxfz == formaxima,minimaandsaddle

points.Calculatethevaluesofthefunctionatthecriticalpoints.

2. D e t e r m i n e t h e r e l a t i v e e x t r e ma f o r t h e f u n c t i o n

2933),( 233 +++= yxyyxyxf

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Lagrange Multipliers

Constrained Extrema For Functions Of Two Variables

InSection2.5.3wediscussedtheproblemofdeterminingtheextrema(maximum

and minimum values) of a function ),( yxf .Wedidsowithoutimposingany

condition on the set D ofpossiblevaluesof x and y .

InthisSectionwerevisittheproblemofndinganextremum(singularformof

extrema)ofafunction ),( yxf butwithadifference.Hereweimposeacondi-

tion (or a set of conditions) on the variables x and y .Typicallywemaywish

tondtheextremaof ),( yxf onlyforpointslyingonaparticularcurveinthe

domain D ofthefunction,suchas

0),( =yxg

The problemofndingextremevaluesof a function ),( yxf subjecttoa

constraint of the general form 0),( =yxg isatypicalexampleofaconstrainedoptimizationproblem.

Example

Findthemaximumvolumeofarectangularboxhavingasquarebaseandopen

atthetop,ifthesidelengthofthebaseisx cm,theheightis y cm and the total

area of the material to be used is 108 2

cm

Reduction Method

Apossiblemethod ofsolving thegeneral constrainedminimizationproblem

posedhereisthefollowing:

(i) Solvetheequation 0),( =yxg foroneofthevariables,whicheverismostconvenienttosolvefor.

(ii) Substitute the result into the function ),( yxf beingoptimized.Thesubstitu-

tionwilleliminateoneofthevariablesfromtheexpressionfor ),( yxf reducing

theproblemtoasinglevariableextremumproblem.

Toillustratethisweapplythemethodontheexamplegivenabove.

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Volumetobemaximized: =),( yxV yx2

Areaofthebox: xyxyxA 4),( 2 +=

Constraint: 01084),( 2 =+= xyxyxg

We solve 0),( =yxg for y and get

x

xy

4

1082

= .Wesubstitutethis

expressionfor y into ),( yxV to eliminate y and get

=

x

xxxV

4

108)(

2

2=

3

4

127 xx

Thecriticalpointsof )( xV are found to be 6=x .Sincex representsalength,

wetakethepositivevalue 6=x .Thecorrespondingvalueof y is 3=y .Themaximumvolumeoftheboxsubjecttothegivenconstraintistherefore

108=V 3cm .

Method of Lagrange Multipliers

Thereductionmethodwehaveillustratedhereisapplicableonlytothoseproblems

whereitispossibletosolvetheequation 0),( =yxg foroneofthevariables.Wenowpresentanothermethodwhichisapplicablemoregenerally.

ThemethodofLagrangemultipliersinvolvestheintroductionofanadditional

independentvariable called the Lagrange Multiplier.

Theessentialstagesinitsapplicationarethefollowing:

1. Usingthefunctions ),( yxf , ),( yxg and wedeneanewfunction

),,( yxF = ),( yxf - ),( yxg

2. Wethenndthecriticalpointsof ),,( yxF by solving the three simulta-neousequations

0),,( =

yx

x

F; 0),,( =

yx

y

F; 0),,( =

yx

F.

3. Assumingthatwehaveobtainedallthecriticalpoints( )kkk yx ,, we then

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listthepoints ( )kk

yx , and note that they automatically satisfy the constraint

0),( =yxg .

4. Wenallycomparethevalues ),(kk

yxf of ),( yxf atthepoints ( )kk

yx ,

Thelargest(smallest)valueistheabsolutemaximum(minimum)onthecurve

0),( =yxg .

Example

WeshalldemonstratetheLagrangeMultipliersmethodon theexampleabove

which we had solved using the method of elimination of one of the variables in

),( yxf usingtheconstraintequation 0),( =yxg .

With ),( yxf = ),( yxV = yx2 and ),( yxg = 10842 + xyx =0,the

expressionforthefunction ),,( yxF is:

),,( yxF = yx2 - )1084( 2 + xyx .

Thecriticalpointsof ),,( yxF are obtained by solving the three simultaneousequations

x

F

= yxxy 422 = 0

y

F

= xx 42 = 0

F= 1084

2 + xyx = 0

The solution of this system is 6=x , 3=y and2

3= ,aresultwhichyields

thesamemaximumvalue108 for the volume V oftherectangulartin.

Self Exercise

1. Maximize thefunction xyyxf =),( along the line 022 =+ yx by

applyingboththemethodofeliminatingoneofthevariablesin f as well as

applyingtheLagrangemultipliersmethod.

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2. ApplytheLagrangeMultipliersmethodtominimizethefunction22

yx +

subjecttotheconstraint 02543),( =+= yxyxg .

Exercise Questions (From Cain and Herod: Multivariable Calculus)

Exercise

Solvealltheexerciseproblemsgivenintheunit.(Chapters7and8)

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XVI. SummativeEvaluation

Summative Assessment Questions

Unit 1: Elementary Differential Calculus

1. Giventhefunction 32)( 2 += xxxf , nd:

(a)

+

h

fhf

h

)2()2(lim

0

(b)

+

h

xfhxf

h

)()(lim

0

2. Without applying LHopitals Rule, nd the following limits if they exist.

(a)

1

13

1

limx

x

x

. (b)

1

13

1

limx

x

x

.

(c)

2

0

1lim

xx.

3. Evaluate the following limits

(a)

+

+

103

13

2

lim x

x

x

(b)

+

323

23

3

lim xx

x

x

.

(c)

x

x

x

sinlim . (d)

x

x

x

coslim

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4. Giventhat

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8. Let dxxIn

n)(sin

0=

.

(a) Find 0I .

(b) By expressing the integrand in the form )(sin)sin()(sin 1 xxx nn =

and integrating by parts, show that2

1

=

nnI

n

nI

(c)Usetheaboveresultstoevaluate. dxx)(sin0

2

9 (a) Explain briey why the integrals dxex

0,

1

0 1 x

dx

are called improperintegrals.(b) In each case, determine whether or not the improper integral converges.

Where there is convergence, nd the value of the integral.

10. Approximate the following integrals using both the trapezium rule and the

Simpson rule. In each case, evaluate the integrand correct to four decimal

places and use two interval lengths: 25.0=h and 1.0=h

(a) +

1

0 1 x

dx (b)

2

1)ln( dxx (c) +

2

1 21 x

dx

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Unit 3: Sequences and Series

11. (a) Dene a Cauchy sequence.

(b) Prove that the sequence of partial sums of the harmonic series

=1

1

k kis

not a Cauchy sequence. What conclusion follows from this result with regard

toconvergenceoftheseries?

12. (a) Write down theth

n partialsumoftheseries =1k

ka .

(b) How is the convergence of a series related to its thn partialsum?

(c) Find the thn partialsumoftheseries

= ++1 )2)(1(

1

k kkand hence

nd the sum of the series, if it converges.

13. (a)Whatismeantbysayingthattheseries

=1k

ka converges

(i)Absolutely,

(ii) Conditionally.

(b)Provethatifaseriesconvergesabsolutely,thenitconverges.

14. By applying the integral test determine the values of the parameter p for

which the series

=1

1k

pk

converges.

15. Expand )cos(x in a Taylor series about the origin, giving the rst threenonzero terms.

Using the sigma notation write down the series for the function. Determine both the

interval of convergence and the radius of convergence of the resulting series.

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Unit 4: Calculus of Functions of Several Variables

16.(a)Givethe ( ) , - denitions of the limit and continuity concepts for a

function ),( yxf ataninteriorpoint ( )ba, in the domain D of f.

(b) Find

+ 22)0,0(),( 2

2lim

yx

xy

yxif it exists or show that the limit does not

exist.

17. Afunction ),( yxT satises the equation 02

2

2

2

=

+

y

T

x

T. Find the

corresponding equation in polar coordinates cosrx = , sinry = .

18. Find the centre of mass (center of gravity) of a laminar with density function

2621),( xyyxf ++= bounded by the region R enclosed between the two

curves2

xy = , 4=y

19. Find all the critical points of the function22

),(yx

xyeyxf= and apply the

second derivative test to classify them.

20.(a)If D is the differential operatory

kx

h

+

, nd

),(2

yxfD , and ),(3 yxfD .

(b) Give the rst three terms of the Taylor series expansion of the function

xy

eyxf =),( aboutthepoint(2,3).

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Summative Assessment Answers

1. (a) 2.

(b) 22 x

2. (a) 3

(b) 0

(c) Limit does not exist

3. (a) Limit does not exist )(

(b)3

1

(c) 0

(d) 0

4. 4=a and 2=b .

5. 3.

6. (a) Cx +)(ln2

1 2

(b)xeC 12

(c) [ ])cos(ln xC

7. (a) -1

(b) 1)2ln(2

(c) 12

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8. (a) =0I

(c)2

)(sin0

2 = dxx .

9. (a) dxex

0 isanimproperintegralbecausetheupperlimitofintegrationis not nite.

1

0 1 x

dxis an improper integral because the integrand is not dened at

.1=x

(b) dxex

0 is divergent.

1

0 1 x

dx

convergesto 2 .

10.

Integral Trapezium Rule Simpsons Rule

h=0.25 h=0.1 h=0.25 h=0.1

+

1

0 1 x

dx

0.8301 0.8287 0.8284 0.8285

2

1

)ln( dxx 0.3837 0.3857 0.3862 0.3863

+

2

1

21 x

dx 0.3235 0.3220 0.3218 0.3217

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(ii) An infinite series

=1k

ka thatconvergesbutisnot

absolutely

Convergent is said to be conditionally convergent.

(b) Sketch of proof that an absolutely convergent series is convergent:

Assumethattheseriesconverges.Notethatforallk ,

kkk

aaa 20 + , and by the comparative test it follows that the

series ( )

=

+1k

kkaa converges.Since

=1k

ka alsoconverges,thenthe

series ( ){ } =

=

=k

k

k

k

kkk aaaa11

isalsoconvergent.

14. Applicationoftheintegraltestontheseries

=1

1

k

px

.Oneanalysesthe

convergencepropertiesoftheimproperintegral

1

px

dx.

1

px

dx

= ( ) .1,11

lim1

1

1limlim

1

1

1

1

=

=

pbppx

x

dxpb

bp

b

b

pb

The limit exists only if 1>p .Therefore,the p -series

=1

1

k

px

converges

for .1>p

15.(a)42

24

1

2

11)cos( xxx += .

(b) ( )

=

=0

2

!2

1)cos(

k

kk

k

xx .Theintervalofconvergenceis

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16(a)Denition of limit of ),( yxf at point ),( ba

),( yxf is said to have limit L as ( )yx, tends to (or approaches) ),( ba iffor

every 0> there exists a )( = such that

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Classication:

Saddle point: )0,0( Localmaxima:

2

1,

2

1

Localminima:

2

1,

2

1

20.(a)2

22

2

2

222 2),(

y

fk

yx

fhk

x

fhyxfD

+

+

=

3

33

2

32

2

32

3

333 33),(

y

fk

yx

fhk

yx

fkh

x

fhyxfD

+

+

+

=

(b)

( ) ( )( ) ( )

+++++= 226 3232

7

12

2

9)3(2)2(31 yyxxyxeexy

.

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XVII. MainAuthoroftheModule

Biography Of The Author

Dr. Ralph W.P. MasengeisProfessorofMathematicsintheFacultyofScience,

Technology and Environmental Studies at the Open University of Tanzania. A

graduate of Mathematics, Physics and Astronomy from the Bavarian University

of Wuerzburg in the Federal Republic of Germany in 1964, Professor Masenge

obtained a Masters degree in Mathematics from Oxford University in the United

Kingdom in 1972 and a Ph. D in Mathematics in 1986 from the University of

Dar Es Salaam in Tanzania through a sandwich research program carried out at

the Catholic University of Nijmegen in the Netherlands.

For almost 33 years (May 1968 October 2000) Professor Masenge was an aca-

demic member of Staff in the Department of Mathematics at the University of Dar

Es Salaam during which time he climbed up the academic ladder from Tutorial

Fellow in 1968 to full Professor in 1990. From 1976 to 1982, Professor Masenge

headed the Department of Mathematics at the University of Dar Es Salaam.Born in 1940 at Maharo Village, situated at the foot of Africas tallest mountain,

The Killimanjaro, Professor Masenge retired in 2000 and left the services of the

University of Dar Es Salaam to join the Open University of Tanzania where he

now heads the Directorate of Research and Postgraduate Studies chairs the Se

44853127-calculus

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