Unit-6.PDF Engg Math

7/24/2019 Unit-6.PDF Engg Math

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Vector Differential

CalculusUNIT 6 VECTOR DIFFERENTIAL CALCULUS

Structure

6.1 Introduction

Objectives

6.2

Scalar and Vector Fields6.3

Vector Calculus

6.3.1 Limit and Continuity

6.3.2 Differentiability

6.3.3 Applications of Derivatives

6.4 Directional Derivatives and Gradient Operator

6.5 Divergence of a Vector Field

6.5.1 Physical Interpretation

6.5.2 Formulae on Divergence of Vector Functions

6.6

Curl of a Vector Field6.6.1 Physical Interpretation

6.6.2 Rotation of a Rigid Body

6.6.3 Formulae on Divergence Gradient, and Curl

6.7

Summary

6.8 Answers to SAQs

6.1 INTRODUCTION

In physical problems we often come across quantities such as temperature of a liquid,

distance between two points, density of a gas, velocity and acceleration of a particle or a

body, tangent to a curve and normal to a surface. In Unit 5, you have learnt that physical

quantities can be categorized either as a scalar or as a vector. You might have noticed

that some physical quantities, whether scalars or vectors, are variable. That is, their

values are not constant or static but change with the change in variable. For example, the

density of a gas, which is a scalar quantity, changes from place to place and is different at

different places. Similarly, tangent to a curve may have different directions at different

points of the curve. This variable character of scalars and vectors give rise to scalar

functions and vector functions. Further, you may notice that at different positions the

temperature or velocity of a body does not remain same. The distribution of temperature

or velocity is therefore defined at each point of a given domain in space which leads to

the idea of scalar fields and vector fields. We shall discuss about scalar functions and

scalar fields, vector functions and vector fields in Section 6.2.

In Section 6.3, we shall extend, in a very simple and natural way, the basic concepts of

differential calculus to vector-valued functions. We shall also discuss about physically

and geometrically important concepts related to scalar and vector fields namely,

directional derivatives, in Section 6.4 and give their applications.

We shall introduce the vector operator in Section 6.5 and give the physicalinterpretation of the divergence of a vector field and some basic formulas involving it.

Finally, the concept of curl of a vector field and its invariance is discussed inSection 6.6. Formulas involving curl, divergence, gradient and Laplacian operator, 2,are also developed here.


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Engineering Objectives

After studying this unit you should be able to

define a scalar function, a scalar field, a vector function and a vector field,

state conditions for the convergence of a sequence of vectors,

define limit, continuity and differentiability of vector functions,

differentiate sum, difference and products involving vectors,

describe the notion of directional derivative and compute directionalderivatives,

define and compute gradient of a scalar field and divergence and curl ofvector fields,

interpret physically the gradient, divergence and curl of a vector,

define conditions for solenoidal and irrotational vector fields, and

solve problems on application of del operator and product rules involvingthe del operator.

6.2 SCALAR AND VECTOR FIELDS

Let us first talk about the scalar field.

Scalar Fields

Consider the distance of a point Pfrom a fixed point P0which will be a real

number. As we vary the point Pits distance from a fixed point also changes. It

depends only on the location of point Pin space and may be regarded as function

f(P). If we consider cartersian coordinate system in space and take coordinates of

fixed point P0as (x0,y0,z0) and the variable point as (x,y,z) then the distance

between fixed point and the variable point is given by the well-known formula2

02

02

0 )()()(),,()( zzyyxxzyxfPf ++==

Next, consider a room fitted with an air-conditioner (A.C.). Once A.C. is switched

on for its cooling effect; the temperature of room falls down. Now, if we put off

the A.C., the temperature starts rising up till it reaches the room temperature. The

temperature further rises up if we now switch on the A.C. for its heating effect.

Thus, the temperature of the room depends on the switching system of the air

conditioner.

In this case, the temperature of the room can be considered as a function of

switching system of the A.C.

In both the exmaples taken above, you may note that distance, as well as,

temperature give us only the magnitude and not the direction. Hence both are

scalar quantities. These quantities depend either on the position of Por on the

switching system of the A.C. In both the situations, we get a function. Functions of

this type are called Scalar Functions. It may also be noted here that distance or

temperature functions do not depend on the choice of coordinate system or brand

of A.C., but only on the physical situations such as actual distance or actual

duration of switching on the A.C.

Formally, we give the following definition of a scalar function.

Definition

A Scalar functionis a function which is defined at each point of certainregion (domain) in space and whose values are real numbers depending

only on the points in space but not on particular choice of the coordinate

system.


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Vector Differential

CalculusIn most of the applications, the domain of a scalar function is a curve, a surface, or

a three-dimensional region in space.

In the case of temperature of a room fitted with A.C., the domain of temperature

function is the set of points on the regulator of A.C., which controls the

temperature of the room.

The functionfassociated with each point in domainDis a scalar (a real number)

and we say that a scalar field is obtained. More formally, we have the followingdefinition :

Definition

If be a function which associates a unique scalar with each point in agiven region, then is called ascalar field function,or simply ascalar-

field.

In a plane, for instance, the equation of a curve is given by constant),( =yxf , i.e.these are curves along whichfhas a constant value for all points in the xy-plane.

Similarly, a surface may be given by constant),,( = zyx , i.e. these are surfaces

for which has a constant value for all points in space. In these examples scalarfunctionsfand have plane and space as their respective domains and are scalarfields.

Some more examples of scalar fields are the density of the air of the earths

atmosphere and the pressure within a region through which a compressible fluid is

flowing.

We can represent a scalar field by a formula as well as pictorially. For a pictorial

representation, we may use the curves and surfaces. The pictorial representation of

scalar fields are maps showing physical geography of a region (indicating hills,

lakes, land above or below sea level, etc.).

In the same manner when we assign a vector to each point of a certain region we

obtain a vector field. Let us now talk about vector fields.

Vector Fields

Consider a curve in a plane or in space. At each point of the curve we can draw a

tangent to the curve. These tangents may have different directionsat different

points.

We can assign to each point Pof the curve, a tangent vector )(Pt (Figure 6.1).

Figure 6.1 : Tangent Vectors of a Curve

Similarly at each point of a surface, we can draw a normal (Figure 6.2). These

normals may have different directions at different points of the surface. Thus to

each point Qof the surface, a normal vector )(Qn may be assigned.


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Engineering

Figure 6.2 : Normal Vectors of a Surface

Moreover, a vector may depend on one or more than one independent scalar

variables. The velocity of a particle, for instance, depends on the position of the

particle as well as time. The time is true for position vector and acceleration of a

particle.

We now give the following definition :

Definition

If to each point P of a certain region G in space a vector V(P)is assigned,

then V(P) is called a vector function.

There are many examples of vector functions in physics. For example velocity,

acceleration, and force are all vector functions. The gravitational force exerted by

the sun on a unit mass is also a variable vector, depending on the position of the

mass, and thus represents a vector function.

The collection of all such vector functions V(P) is called a vector field on G.More

precisely, we give the following definition :

Definition

IfF be a function which assigns a vector to each point x in its domain, then

Fis called a vectorfield functionor a vector field.

A simple example of a vector field is the field defined by the vector

)( kjir zyx ++= . A physical example of vector field is given by the particles ofa fluid under flow.

At any instant the velocity vector V(P) of rotating body constitutes a vector field,

called the velocity field of rotation. If we introduce a cartesian coordinate system

having the origin on the axis of rotation, then

)( kjiwrwV ),,( zyxzyx ++== ,

where,x,y,zare the coordinates of any point Pof the body in a planeperpendicular to the axis of rotation and wis the rotation vector or constant angular

velocity of the body (Figure 6.3).

Figure 6.3 : Field of a Body Rotating with Constant Angular Velocity in the Positive

(Counter Clockwise) Direction


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Vector Differential

CalculusNext consider a particleAof massM, which is placed at a fixed point P0and let a

particleBof mass m be free to take up various positions Pin space (Figure 6.4).

Figure 6.4 : Some of the Vectors of Gravitational Field

Then particleAattracts particleB. According to Newtons law of

attraction/gravitation, the corresponding gravitational forcepis directed from Pto

P0and its magnitude is proportional to 21

r, where ris the distance between Pand

P0. Then

|p| ,2r

mMG=

where Gis the gravitational constant. Hencepdefines a vector field in space.

By now you must have clearly understood what we mean by scalar functions,

scalar fields, vectors functions and vector fields. You can test your knowledge by

attempting the following exercise.

SAQ 1

Which of the following are scalar functions, scalar fields, vector functions and

vector fields?

(i)

The gravitational force on a particle of massMat distance rdue to another

particle of mass m.

(ii)

A constant force applied to a particle.

(iii)

The temperature at every point of a mass of heated liquid.

(iv)

Potential of an electric charge placed at the origin.

(v)

The force of a unit charge placed at a point Pdue to an electric charge e

placed at the origin.

You have already learnt about the basic concepts of limit, continuity, differentiabilityand partial differentiationof scalar functions in Block 1. We shall now, in the next

section, introduce these basic concepts of calculus for vector functions in a simple and

natural way.


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Engineering 6.3 VECTOR CALCULUS

We begin by introducing the idea of limitand continuityof a vector function. We know

that the concept of limit is a basic and useful tool for the analysis of functions. It enables

us to study derivatives, improper integrals and other important features of functions. We

shall introduce the concepts of limit and continuity in a most informal manner without

going into the intricacy of delta, epsilon method.

6.3.1 Limit and Continuity

We say that the limit of the vectorf(t) as tais the vector liff(t) moves into theposition occupied by las ta.In the limit, the length and direction offshould matchthe length and direction of l; more precisely, we define limits of vector valued functions

in terms of the familiar limits of real-valued functions in the following way.

Definition

Let kjif )()()()( 321 tftftft ++= be a vector valued function of t defined insome neighbourhood of a (possibly except at a). The limit of ( )tf as t approaches

the number a is the vector liff the limit of |)( lf t| as t approaches a is zero. Insymbols

0|)(|lim)(lim ==

lffl ttatat

. . . (6.1)

Observethatf(t) having las a limit means that the components offhave the

corresponding components of las limits. In other words, if

kjilkjif and)()()()( 321321 llltftftft ++=++=

then

332211 limandlim,lim)(lim lflflft ====lf . . . (6.2)

This equivalence says that we may calculate limits of vector-valued functionscomponent wise, i.e., one component at a time.

The limit of a vector valued functions can also be defined in the usual , manneras we do for the real-valued functions in the following way :

A vector functionf(t) of a real variabletis said to tend vector las t approaches a,

if to any pre-assigned positive number , however small, there corresponds apositive number such that

||when|)(|


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Vector Differential

CalculusTherefore, kjif 2)(lim ++=t

The definition of continuity of a vector functionf(t) is the same as the definition

for continuity of a real-valued function as can be seen below.

In the definition of a limit of a vector function, we mentioned thatf(t) is defined in

the neighbourhood of a(possibly except at a). If the vector functionf(t) is also

defined at aand its value at ais equal to vector l, the limit off(t) as ta, then we

say that the vector function is continuous at t= a. We now give below the precisedefinition of continuity of a vector function.

Definition

A vector valued functionf(t) is continuous at t = a iffis defined at a and

)()(lim atat

ff =

. . . (6.3)

In view of the equivalence in Eq. (6.2) above, we may say that f(t) is continuous at

t= aif each component offis continuous at t= a. Thus we may test a vector

function for continuity by applying our knowledge of real-valued functions to each

component off.

Alsof(t) is continuous function if it is continuous at every point of its domain.

Just as in the case of real valued functions, the sum, the difference, scalar product

and vector product of two continuous vector functions are also continuous. We

shall not be proving these results here. You can check them yourself.

Let us consider the following example.

Example 6.2

Discuss the continuity of the function

kjif )1(ln)(sin1

)( 2tt

t

t +++=

Solution

The functionf(t) is continuous at every value of 0t> because each component is

continuous for t> 0. However,fis discontinuous for t0 becauset

1, the first

component off, is not defined for t0.

You may now try the following exercises.

SAQ 2

(a) Find )(lim

0

t

t

f

if

(i) jif )(tt etet +=

(ii) kjif )cos()sin()( ttt etetet +=

(iii) kjif cos1

sin)( +

+=

t

t

t

tt

(b) At what values oftare the following vector functionsf(t) continuous.

(i) kjif )(sin)(cos)( ++= ttt

(ii) kjif |1|ln1

1cos)( tt

et t ++

++=


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Engineering In mechanics, if the position vector of a particle is given by r(t) and if we wish to find its

velocity, we will have to differentiater(t) with respect to time. Similarly, if the potential

of an electric charge is given, then to determine force due to an electric charge, we shall

have to take a recourse to differentiation. We now discuss the differentiation of a vector

function.

6.3.2 Differentiability

We define the derivative of a vector-valued functionf(t) at a point t= aby the same type

of limit equation as we use for scalar functions. Thus

h

ahaa

h

)()(lim)(

0

fff

+=

. . . (6.4)

provided the limit on the right exists. We then expect thatfis differentiable at t= aiff

each of its components is differentiable at t= a. In this connection we prove the

following result :

Theorem 1

A vector function kjif )()()()( 321 tftftft ++= is differentiable at t= aiffeach of its component function is differentiable at t= a. If this condition is met,

then

kjif )()()()( 321 afafafa ++= . . . (6.5)

Proof

Consider the difference quotient

jiff )()()()()()( 2211

h

afhaf

h

afhaf

h

aha ++

+=

+

k)()( 33

h

afhaf ++ . . . (6.6)

The left hand side of Eq. (6.6) has a limit as h0 iff each component on the righthand side has a limit as h0. The first component on the right has a limit ifff1isdifferentiable at a.

Finally, if each component is differentiable at a, then taking the limit h0 inEq. (6.6) of each of the quotient, we get Eq. (6.5), thus proving the result.

Note that the differential coefficient )(af is itself a vector and is called the

derivative off(t) at t= a. From the definition it is clear that every derivable vectorfunction is continuous. Consider the following example :

Example 6.3

Obtain the derivative of kjif )3(tan)(ln)(sin)(12 tttt ++=

Solution

The function and its components are defined at every positive value of tand

possess derivatives for all t. Thus

kjif

91

31)cossin2()(2

tt

ttt

+

++=

We now give the geometrical interpretation of the derivative at a vector valued

function.


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Vector Differential

CalculusGeometrical Representation of Derivative

Draw the vectorf(t) for values of the independent variable tin some interval

containing tand t+ tfrom the same initial point 0. Then the locus of head ofarrows representingffor different values of ttraces out a space curve (Figure 6.5).

Figure 6.5 : Derivative of a Vector Function

Let OP=f(t) and OQ=f(t+ t),

then f(t+ t) f(t) = OQ OP=PQ= f, (say).

Hencetdt

d

t

=

ff

0lim

The direction ofdt

dfis the limiting direction of

tf

or of f . But as Qtends to

tends to P,PQtends to the tangent line at P. Hence the direction ofdt

dfis along the

tangent to the space curve traced out by P. Let sdenote the length of the arc of this

curve from a fixed point on it up to P. Then the magnitude ofdt

dfis given by

t

s

t

s

stdt

d

tt

=

=

=

.||

lim||

lim00

fff,

since the ratio 0as1arc

chord||=

tPQ

PQ

s

f.

Thus derivative of a vector function represents a vector whose direction is tangent

to the space curve traced by the vector function and the magnitude isdt

ds, where s

is the arc length from a fixed point on the curve to the variable point representing

the vector function.

You may note here that a vector will change if either its magnitude changes or

direction changes or both direction and magnitude changes. In this regard, the

following results may be remembered :

(a)

The necessary and sufficient condition forf(t) to be constant is 0=dt

df.

(b) The necessary and sufficient condition forf(t) to have constant magnitude

is 0. =dt

dff .

(c) The necessary and sufficient condition forf(t) to have constant/uniform

direction is 0=dt

dff .


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Engineering The familiar rules of differentiation of real functions yield corresponding rules for

differentiating vector functions; for example,

(i) ff = CC)( (Ca constant).

(ii) vuvu = )(

(iii)dtdu

dtduu fff +=)( (uis a scalar function of t).

(iv) vuvuvu += ..).(

(v) vuvuvu += )(

(vi) ],,[],,[],,[],,[ wvuwvuwvuwvu ++=

In (v) above the order of the vectors must be carefully observed, as cross

multiplication of vectors is not commutative.

The chain rule of differentiation is also valid for vector valued functions. That is, if

f(t) is a differentiable function of t, and t= g(s) is a differentiable function of s,

then the composite functionf(g(s)) is a differentiable function of sand

)())(( sgsgds

d= f

f . . . (6.7)

We can write Eq. (6.7) in the form

ds

dt

dt

d

ds

d.

ff= . . . (6.8)

The chain rule given for vector functions by Eq. (6.8) is an immediate

consequence of the chain rule for scalar functions that applies to the componentsf1,f2andf3.

Consider the following example.

Example 6.4

Expressds

dfin terms of sif kjif )1(sin)( 1++++= tett and

1)( 2 == ssgt .

Solution

From the chain rule we have

)2())1((cos 1 setds

dt

dt

d

ds

d tkj

ff +++==

kj 2)(cos22

2 sesss +=

We would obtain the same result if we first substitute 1)( 2 == ssgt in theformula forf(t) and then differentiate w.r.t. s.

kjif )(sin))((22 sessg ++=

kjf 2)(cos2))((22 sessssg

ds

d+=

And now a few exercises for you.


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Vector Differential

CalculusSAQ 3

(a) Find the derivative of the vector functionf(t) in each case and give the

domain of derivative :

(i) jif )(2 tt etet +=

(ii) kjif 3)( +=t

(iii)

++= t

ttt13tan2sin)( 11 kjif

(b)

Find

++=

uuuux

1

1tancos)(if)( 1 kjiff and

122 ++= xxu .

As we have already mentioned, not all vector functions are functions of one variable. The

velocity of a fluid particle in motion is a function of time and position. The position

vector of a fluid particle at any time and at any position is a vector function of four

variablesx,y,zand t. Thus, in any physical problem involving a vector function of two

or more scalar variables, we may be required to find the partial derivatives of this vector

function. In other words, we may be required to find the derivative of the vector function

w.r.t. one scalar variable treating the other scalar variables as constant. Partial derivatives

can be calculated for vector functions by applying the rules we already know for

differentiating vector functions of a single scalar variable.

If a vector functionf(u, v) be a differentiable function of two scalar variables u,vgiven

in the component form as

kjif ),(),(),(),( 321 vufvufvufvu ++= ,

then partial derivatives offw.r.t. uand vare denoted by

vu

ff

,

respectively and are defined as

kf

jf

iff 321

uuuu

+

+

=

and kf

jf

iff 321

vvvv

+

+

=

Similarly, kf

jf

iff

2

32

22

2

21

2

2

2

uuuu

+

+

=

kf

jf

iff 3

22

21

22

vuvuvuvu

+

+

=

etc. are the second order partial derivatives.

The derivatives


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Engineering

uvvu

ff 22

and

are called mixed partial derivativesand they are equal ifvu

ff

and are continuous

functions.

Physically,u

fgives the rate of change offw.r.t. to uat a given point (u, v) in space.

Thus, partial derivatives

zyx

fff

,,

of a functionf(x,y,z, t) give us the rate of change offin the directions ofx,y,zaxes at a

given instant andt

fgives the rate of change offwith respect to time at a given point in

space.Consider the following example.

Example 6.5

Find the first order partial derivatives of kjir sincos),( 21121 ttatatt ++= .

Solution

We have jir cossin 11

1

tatat

+=

kr

2

=

t

Note that ),( 21 ttr is a position vector. It represents a cylinder of revolution of

radius a, having thez-axis as axis of rotation.

You may now try the following exercise.

SAQ 4

For each of the vector functionf, find the first partial derivatives w.r.t.x,y,z,

(i) jif zyyx +=

(ii) jif zy ee =

(iii) kjif 222 xzzyyx ++=

You know that curves occur in many considerations in calculus as well as in physics; for

example, as paths of moving particles. Let us consider some basic facts about curves in

space as an important applicationof vector calculusabout which we are going to talk in

our next section.


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Vector Differential

Calculus6.3.3 Applications of Derivatives

The simplest application of vector calculus is provided by curves in space. Given a

cartesian coordinate system, we may represent a curve Cby a vector function

kjir )()()()( tztytxt ++=

Here to each value of the real variable t, there corresponds a point of Chaving position

vectorr(t) (Figure 6.6).

Figure 6.6 : Parametric Representation of a Curve

For example, any straight lineLcan be represented in the form

bar tt +=)(

whereaandbare constant vectors and lineLpasses through the pointAwith position

vectorr=aand has the direction ofb(Figure 6.7).

Figure 6.7 : Parametric Representation of Straight Line

The vector function

jir sincos)( tbtat +=

represents an ellipse in thexy-plane with centre at origin and axes in the directions of

xandyaxes.

Further, if a curve Cis represented by a continuously differentiable vector functionr(t),

where tis any parameter, then the vector

t

ttt

dt

d

t +

=

)()(lim

0

rrr

has the direction of the tangent to the curve at )(tr (Figure 6.8).


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Engineering

Figure 6.8 : Representation of the Tangent to a Curve

Thus, the position vector of a point on the tangent is the sum of the position vector rof a

point Pon the curve and a vector in the direction of the tangent. Hence the parametric

representation of the tangent is

dt

drrq +=)( ,

where bothrand dt

drdepend on Pand the parameter is a real variable.

Let now consider the following example.

Example 6.6

Ifa(t) be a variable unit vector, show that

(i)dt

dais a vector normal toa.

(ii)d

dais a unit normal vector toa, being the angle through whichaturns.

Solution

(i) Sincea(t) is a unit vector,

a2= 1 . . . (6.9)

Differentiating Eq. (6.9) w.r.t. to t, we get

02 =dt

da.a

Thusaanddt

daare at right angles, i.e.,

dt

dais a vector normal toa.

(ii) Let OP=aand OQ=a+ abe two neighbouring values of the given vectormaking an angle with each other (Figure 6.9).

Figure 6.9

Then aaa +== OPOQPQ

= a


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Vector Differential

Calculusand

=

aa

0lim

d

d

Since

ais normal toain the limiting position when 0, therefore

dda

is normal toa.

Also 1.

limlim00

==

=

=

OPOPaa

d

d

Henced

dais a unit vector normal toa.

Let us now look into some of the applications of derivatives to dynamics.

Let the position vector of a point moving on a curve be given by r(t). Its

displacement in time tis

rrr =+= )()( ttt , say.

Since the velocity Vof the moving point is the rate of change of its

displacement w.r.t. to time, therefore

dt

drV=

Again, the acceleration of the point, being the rate of the change of velocity,

is given by

2

2

dt

d

dt

d rVa ==

Consider the following examples.Example 6.7

Show that ifr=asin t+bcos twhereaandbare constants, then

)(and22

2

bar

rrr

==dt

d

dt

d

Solution

We haver=asin t+bcos t

Differentiating w.r.t. to t, we get

ttdt

d= sincos ba

r

and )cossin(cossin 2222

2

ttttdt

d+== baba

r

r2=

Also )sincos()cossin( ttttdt

d+= baba

rr

ba+= )cossin( 22 tt

)0and0( === bbaa

)( ba=


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Engineering Let us take up another example.

Example 6.8

A particle Pmoves on a disk towards the edge, the position vector being

r(t) = tb,

wherebis a unit vector, rotating together with the disk with constant angularvelocity in the counter clockwise direction. Find the accelerationaof P.

Solution

Since the particle is rotating with constant angular velocity , thereforebis of theform

jib sincos)( ttt += . . . (6.10)

The position vector of particlePis

r(t) = tb . . . (6.11)

Differentiating Eq. (6.11) w.r.t. to t, we get

bbrV t+== . . . (6.12)

Obviouslyb is the velocity of P relative to the disk and t b is the additional

velocity due to the rotation (Figure 6.10).

Figure 6.10 : Motion in Example 6.8

Differentiating Eq. (6.12) one more w.r. to t, we obtain

bbVa t+== 2 . . . (6.13)

In the last term of Eq. (6.13), using Eq. (6.10), we have bb 2= . Hence the

acceleration bt is directed towards the centre of the disk and is called the

Centripetal Accelerationdue to the rotation.

The most interesting term in Eq. (6.13) is b2 , which results from the interaction of

the rotation of the disk and the motion of Pon disk. It has the direction ofb, i.e., it

is tangential to the edge of the disk and it points in the direction of rotation. This

form, b2 , is called Coriolis Acceleration.


SAQ 5

(a) A particle moves along the curve tztyex t 3sin2,3cos2, === where tis the time variable. Determine its velocity and acceleration at t= 0.


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Vector Differential

Calculus(b)

A particle moves so that its position vector is given by

jir sincos tt +=

Show that the velocity Vof the particle is perpendicular torandrVis aconstant vector.

(c) Find the Coriolis acceleration when the particle moves on a disk towards the

edge with position vector

btr2)( =t

wherebis a unit vector, rotating together with the disk with the constant

angular speed in the anti-clockwise sense.

If we consider a scalar fieldf(x,y,z) in space, then we know that

z

f

y

f

x

f

,,

are the rates of change offin the directions ofx,yandzcoordinate axes. It seems

unnatural to restrict our attention to these three directions and you may ask the

natural question. How to find the rate of change offin any direction? The answer

to this question leads to the notion of directional derivative which we shall try to

answer in the next section.

6.4 DIRECTIONAL DERIVATIVES AND GRADIENTOPERATOR

Let us consider a scalar field in space given by the scalar function f (P) =f(x,y,z), where

we have chosen the point Pin space. Let us choose direction at P, say given by vectorb.

Let Cbe a ray from Pin the direction ofband let Qbe a point on C, whose distance from

Pis s(Figure 6.11).

Figure 6.11 : Directional Derivative

The limit

,)()(

lim

)(

0 s

PfQf

s

f

PQ

s

=

. . . (6.14)

if it exists, is called the directional derivative of the scalar function f at P in the direction

ofb.


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Engineering It this way, there can be infinitely many directional derivatives of fat P, each

corresponding to a certain direction. An interesting question arises Can we represent

any such directional derivative in terms of some derivative or derivatives offat P? The

answer is yes and it is achieved as follows :

Let a cartesian coordinate system be given. Letabe the position vector of Prelative to

the origin of this system. Then any point on ray C, or ray Citself, can be represented in

form.

)0()()()()( +=++= ssszsysxs bakjir . . . (6.15)

Nows

fis the derivative off[x(s),y(s),z(s)] with respect to arc-lengths of ray C.

Hence assuming thatfhas continuous first partial derivatives and applying the chain rule,

we obtain

ds

dz

z

f

ds

dy

y

f

ds

dx

x

f

s

f

+

+

=

. . . (6.16)

where ds

dz

ds

dy

ds

dxand, are evaluated at s= 0.

Also from Eq. (6.15), we have

bkjir

=++= ds

dz

ds

dy

ds

dx

ds

d

This suggests that we introduce the vector

grad kji z

f

y

f

s

ff

+

+

=

and write Eq. (6.16) in the form of a scalar product

fx

fgrad.b=

The vector gradfis called the gradientof the scalar functionf. More precisely, we give

the following definition :

Definition

The vector function

z

f

y

f

x

f

+

+

kji

is called the gradient of the scalar function f and is written as grad f, viz.,

gradz

f

y

f

x

ff

+

+

= kji

Here we have assumed that scalar function is a continuously differentiable function. Thus

the directional derivative of the scalar point functionfalong the direction of vectorbcan

be written as

fs

f=

.b

In other words, the directional derivative s

f

is the resolved part of grad f in the

direction ofb.


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77

Vector Differential

CalculusWe see that the gradient of a scalar fieldfis obtained by operating onfby the vector

operator

zyx

+

+

kji

This operator is denoted by the symbol (read as del or nabla) and it operatesdistributively.

In terms of , we write

gradz

f

y

f

x

ff

+

+

= kji

fzyx

+

+

= kji

fs

f=

.b

The operator is also known asdifferential operatoror gradient operatorand| |f gives the greatest rate of change off.

The interest and usefulness of introducing this symbol lies in the fact that it can beformally assumed to have the character of a vector and as such it facilitates the

manipulations with expressions involving differential operators. Thus formally, fbeingproduct of a vector by a scalarfis a vector.

Before we take up the properties of gradient of a scalar field and operator , we take up afew examples to illustrate how directional derivatives are calculated.

Example 6.9

Find the directional derivative 222 32),,,(for, zyxzyxfs

f ++= at the point

P(2, 1, 3) in the direction of the vector kia 2 = .

Solution

Here grad )32( 222 zyxzyx

f ++

+

+

= kji

kji 264 zyx ++=

At P(1, 2, 3), (gradf )P= 3,1,2)

2

6

4( ===++ zyxzyx kji

kji 668 ++=

Now kia 2 =

521|| 22 =+=a

Unit vector in the direction of ais

kikiaa

a

5

2

5

1)2(

5

1

||

1 ===

Therefore,5

4

5

2

5

1.)668( =

++=

kikjis

f

The minus sign indicates thatfdecreases in the direction under consideration.


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Engineering Let us take another example.

Example 6.10

Find the directional derivative off(x,y,z) =x2y

2z

2at the point (1, 1, 1) in the

direction of the tangent to the curvex= et,y= 2sin t+ 1,z= t cos t, 1 t1.

Solution

Here grad 222 zyxzyx

f

+

+

= kji

kji 222222222 zyxzyxzyx ++=

At P(1, 1, 1), (gradf )p kji 222 +=

Now any point on the given curve has position vector

kjir )cos()1sin2( tttet +++=

Direction of tangenttto the given curve is

kjir

t )sin1(cos2 ttedt

d t +++==

ttette tt sin2cos32)sin1()cos2(|| 22222 +++=+++=t

The point (1, 1, 1) on the curve corresponds to the value t= 0.

6

2

||

1)(

0at

)1,1,1(at

kjit

tt

++=

==

t

Required direction derivative

)1,1,1(at)1,1,1(at )(.)( = tf

+++=

6

2.)222(

kjikji

6

4

6

242=

+=


SAQ 6

(a) Find the directional derivative of zyxyx 422 ++ at (1, 2, 2) in the

direction of kji 22 + .

(b) Find the direction in which the directional derivative of

),(

)(),(

22

yx

yxyxf

= at (1,1) is zero.

(c) Find the directional derivative of 2223 34 zyxzx at (2, 1, 2) along thez-axis.


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79

Vector Differential

CalculusWe shall now discuss some of the important properties of gradient of a scalar fieldfunctions.

Consider a differentiable scalar functionf(x,y,z) in space. For each constant Cthe

equation

f(x,y,z) = C= constant

represents a surface Sis space. Thus, by letting Cassume all values, we obtain a family

of surfaces, which are called level surfacesof the functionf. Since, by the definition of afunction, our functionfhas a unique value at each point in space, it follows that through

each point in space there passes one, and only one, level surface off.

If (x,y,z) denotes the potential, the surface

(x,y,z) = C

is called an equipotential surface. The potential of all points on this surface is equal to theconstant C.

Important geometrical characterization of the gradient of a scalar functionfis in terms of

a vector normal to a level surface or an equipotential surface. This property can also beused in obtaining the normal to a given surface at a given point. We shall now take up

this property.

Property 1 : Gradient as Normal Vector to Surfaces

Let Pbe a point on the level surface

f(x,y,z) = C= constant

Let Qbe a neighbouring point of Pon this surface. With reference to some base

point O, letrandr+ rbe the position vectors of Pand Qrespectively, thenPQ= r(Figure 6.12).

Figure 6.12

Now )(.. zyx

z

f

y

f

x

ff ++

+

+

= kjikjir

zz

fy

y

fx

x

f

+

+

=

f= (by differential calculus) . . . (6.17)

where f is the difference in values offat Qand P.

Hence if Qlies on the same level surface as P, f. r= 0.

This means that f is perpendicular to every rlying in the surface. Thus fisnormal to surface

f(x,y,z) = C

Moreover, let f = | f | n where n is a unit vector normal to the surface.


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80

Engineering Let Qbe a point on a neighbouring level surfacef+ fand let n be theperpendicular distance along PNbetween the two surfaces. Then the rate of change

offnormal to the surface

n

f

n

f

n

=

0lim

,.lim0 n

rf

n =

using (6.17)

n|f

n

=

rn .|lim

0

n

NPQ|f

n

=

!)(cos||lim

0

r

n

n|f

n

=

|lim0

|f= |

Hence the magnitude of fis equal ton

f

. Thus the gradient of a scalar field f is a

vector normal to the surface f = constant and having a magnitude equal to the rate

of change of f along this normal.

Let us take up an example for the better understanding of what we have discussed above.

Example 6.11

Find a unit vector normal to surface 422 == zxyx at the point (2, 2, 3).

Solution

Let f= 422 == zxyx

Then a vector normal to the surface is

grad )2()2()2( 222 zxyxz

fzxyx

yzxyx

xf +

++

++

= kji

kji 2)22( 2 xxzyx +++=

At (2, 2, 3), (gradf )at (2, 2, 3) kji 442 ++=

616164|)grad(| )3,2,2(at =++=f

Hence a unit vector normal to the surface

kjikji 3

23

23

1)442(

6

1++=++=

Some of the vector fields occurring in physics and engineering are given by vector

functions which can be obtained as the gradients of suitable scalar functions. Such a

scalar function is then called apotential functionorpotentialof the corresponding vector

field. The use of potentials simplifies the investigation of those vector fieldsconsiderably. To understand this let us consider the gradient of a potential due to an

electric charge.


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81

Vector Differential

CalculusProperty 2 : Gradient of a Potential due to an Electric Charge

Consider an electric charge eplaced at the origin.

The potential of this charge at a point P(x,y,z) isr

ewhere OP= r(Figure 6.13).

Figure 6.13

The force on a unit charge placed at Pwill be

2r

e

r

e

dr

d=

. . . (6.18)

in the direction OP.

The component of the force in the direction ofxwill be

33222222 )(

2.

2

1

)( r

ex

zyx

xe

zyx

e

xr

e

x=

++=

++

=

Similarly, the component of the force in y and z directions can be obtained as

33and

r

ze

r

ye respectively.

The resultant force may then be written as

kji 333 r

ze

r

ye

r

xe

)(3

kji zyxr

e++=

3

r

er=

This result also follows from Eq. (6.18).

If we denote the potential byr

e, then the force on the particle is

+

grad.e.i,.zyx

kji

The physical definition of potential shows that this will be true for all cases.

Similarly, the gradient of the potential due to a number of charges placed at

various points will give force due to these charges.

In the fluid flow, the gradient of the velocity potential will give the velocity at apoint.

Also the gradient of gravitational potential will give the force due to the gravity.


24/56


25/56

83

Vector Differential

Calculuszyx

+

+ f

kf

jf

i ...

is called the divergence offor divergence of the vector field defined byfand is

denoted by divf.

In terms of operator , we write

zyx ++= fkfjfif ...div

fkji .

+

+

=zyx

f.=

The symbol . is pronounced as del dot.

Let us consider a cartesian coordinate system Oxyz. Letfhave the scalar field

componentsf1,f2,f3along the directions ofx,y,zaxes respectively, so that

321 fff kjif ++=

)(..div 321 fffzyx

kjikjiff ++

+

+

==

z

f

y

f

x

f

+

+

= 321

While writing in the above form, we have the understanding that in the dot product

)(. 1fx

ii

means the partial derivativex

f

1 etc. (It may be understood that kji ,,

are constant vectors and their partial derivatives w. r. to x,y,zare zero.) It is a convenientnotation that is being used.

From the definition and the notation used, i.e..f, it is clear that the divergence of avector field function is itself a scalar field. Thus we can construct a scalar field from a

vector field by taking its divergence.

The meaning of divergence of a vector field is indicated in the name itself. div fis a

measure of how much the vector fieldfdiverges (or spreads out) from a point.

We shall now give the physical interpretation of divergence. But before that we would

like to mention that the function div divfis apoint function. By a point function we

mean that the value of divfis independent of the particular choice of coordinates, i.e., its

value is invariant w. r. t. coordinate transformation. Learner interested in knowing thedetails about the invariance of the divergence may see Appendix-III.


We consider the motion of a compressible fluid in a regionRhaving no sources or sinks

inR, i.e., no points inRat which fluid is produced or disappears.

Let (x,y,z, t) be the density of fluid and v= v(x,y,z, t) be the velocity of fluid particleat a point (x,y,z) at time t.

Let V= vthen Vis a vector having the same direction as vand a magnitude| V| = | v|. It is known as flux. Its direction gives the direction of the fluid flow andits magnitude gives the mass of the fluid crossing per unit time a unit area placed

perpendicular to the flow.

If the unit area is placed with its normal at an angle to the flow (Figure 6.14), then themass of the fluid crossing it per unit time is

(1 . cos ) v= Vcos


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84

Engineering = Resolved part of V in the direction of the normal to the area.

Figure 6.14 : Flux

Consider a small fixed rectangular parallelopiped of sides x, y, zparallel to thecoordinate axes as shown in Figure 6.15 enclosing a point P(x,y,z).

Let kVjViVV zyx ++=

Figure 6.15

The velocity component parallel toy-axis at any point of the faceABCD.

+= zyyx ,

21,yV

,2

1),,(

y

Vyzyx

y

= yV

omitting powers of yhigher than one.

Thus the mass of fluid that moves out of the face ABCD in time t

tzxy

VyzyxV

yy

+=

2

1),,(

Similarly, the mass of the fluid that enters through the face DCBA in time t.

tzxy

VyzyxV

yy

=

2

1),,(

Thus the net mass of the fluid that moves out through the faces ABCD and DCBA perpendicular toy-axis.

tzxy

VyzyxVtzx

y

VyzyxV

yy

yy

=

2

1),,(

2

1),,(

tzyxy

Vy

=

Similarly, considering the other two pairs of faces, we see that the total mass of fluid

flowing out of the parallelopiped in time t


27/56

85

Vector Differential

Calculustzyxz

V

y

V

x

V zyx

+

+

=

The volume of the parallelopiped is xyz. On taking the limit, when x, y, z, talltend to zero, the amount of fluid per unit time that passes through a point P(x,y,z).

Vdiv=

+

+

=

z

V

y

V

x

V zyx

Thus div V= div (v) gives the net rate of fluid outflow per unit volume per unit time ata point of the fluid.

The outflow will cause a decrease in the density of the fluid inside the parallelopiped, say

in time t. Thus loss of mass per unit time per unit volume at a point

t

= .

Equating the loss of mass to the outflow, we get

Vdiv= t

0)(div =

+

tv . . . (6.19)

This important relation is called the condition for conservation of mass or the continuity

equation of a compressible fluid flow.

Similarly, we can discuss the flow of electricity or the flow of heat or flow of particles

from a radioactive source or water flowing into a drain. Thus, in general, ifFis any

vector field defined at all points in a given region, then the divergence ofFat any point

represents the flux per unit volume out of the volume dVenclosing the point, as dVismade smaller and smaller, i.e., dV0.

You may note that the net rate of fluid outflow at a point Pis positive i.e., div V> 0

when the fluid has the tendency to diverge away from P, but if the fluid flows towards

the point, then div V< 0. Thus a point of positive divergence means that there is a net

outflow from that point. Similarly a point of negative divergence implies a net inward

flow.

If we consider the steady fluid motion of an incompressible fluid, so that 0=

t

and

is constant, then Eq. (6.19) becomes

div v= 0

i.e. the rates of outflow and inflow are equal for any given volume at any times, i.e. the

amount of the material in a volume remains constant.

We know that for a magnetic field, the lines of force are closed they neither flow out of

a point nor into a point. Thus for a magnetic field B, we have

divB= 0

Thus there exists vector fields where divergence is zero. Such vector fields are called

divergence free or solenoidal. We give below the formal definition of solenoidal vector

field.

Definition

A vector fieldFis called divergence free or solenoidal in a given region if for all

points in that region


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86

Engineering .F= 0

Thus magnetic field or velocity of a steady flow of compressible fluid are examples

of solenoidal vector fields.

We now apply the concepts discussed in this section to some examples.

Example 6.12

Find the divergence of the vector jkiA 222 zxzyyx += .

Solution

From definition,

div )2()2()( 2 zyz

zyy

yxx

+

+

=A

yyx 202 ++=

)1(2 += xy

Example 6.13

Show that 2)1()grad(div += nn rnnr .

Solution

Let rbe the distance of a point P(x,y,z) from a fixed pointA(x0,y0,z0)

202

02

0 )()()( zzyyxxr ++=

2202

02

0 ])()()[(

n

n zzyyxxr ++=

Now grad 2202

02

0 ])()()[()(

n

n zzyyxxx

r ++

= i

)(2.})()(){(2

0

220

20

20 xxzzyyxx

nn

++= i

div (grad rn) )(})()(){([ 01

220

20

20 xxzzyyxxn

x

n

++

=

20

1122

02

02

0 )(2.})()(){(12

xxzzyyxxnn

n

++

=

+++

1.})()(){(

122

02

02

0

n

zzyyxx

+=

12

22

0

22

2

)()2(

nn

rnxxrnn

220

20

20

4 3])()()[()2( +++= nn rnzzyyxxrnn

]3)2([2 nnnrn +=

]32[ 22 nnnrn +=


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87

Vector Differential

Calculus ]32[ 22 nnnrn +=

2)1( += nrnn

Hence the result.

Example 6.14

Show that the vector kjiA )2()3()3( zxzyyx +++= is solenoidal.

Solution

We know that A is solenoidal if divA= 0.

Now div )2()3()3( zxz

zyy

yxx

+

++

=A

= 1 + 1 2 = 0.

Hence A is solenoidal vector.

Example 6.15

A rigid body is rotating about a fixed axis with a constant angular speed . Thevelocity vector field Vof the rigid body at any pointris given by V= r. Showthat Vis a divergence free vector.

Solution

If Vis a divergence free vector, then

div V= 0

Letz-axis be the axis of rotation for the rigid body.

k=

Ifris the position vector of any particle Pof the rigid body, then

kjir zyx ++=

V= velocity ofP= r

)( kjik zyx ++=

ij yx =

Now, by definition,

div )0()()( zxyyx

+

+

=V

= 0 + 0 + 0 = 0

Hence velocity vector Vis a divergence free vector.

How about trying a few exercises now.

SAQ 8

(a) If kjir zyx ++= show that

(i) divr= 3

(ii) div 03

=

r

r

(iii) rr += 3)(div . grad , where is a scalar function ofx,y,z.


30/56

88

Engineering(iv) dir

222

2)(

zyx ++=r

(b) The gravitational forcepof attraction of two particles is the gradient of the

scalar functionr

czyxf =),,,( . Show that for r> 0,pis a solenoidal

vector.

(c) Determine the electric field and charge distribution corresponding to

potential = r2and

+=

r

aa

32 2 .

(Hint :Electric field,E= ) and charge distribution = .E.)

We now give some formulas on divergence of vector functions.

6.5.2 Formulae on Divergence of Vector Functions

(i) div (Kf ) = Kdivf, where Kis a constant andfis a vector function.

(ii) div (f ) = divf+f. grad , where is a scalar function andfis a vectorfunction.

(iii) =

+

+

= 2

2

2

2

2

2

2

)grad(div

zyx

, where is a scalar field.

Learner interested in knowing the proofs of these formulas may see Appendix-IV.

As mentioned earlier, we now discuss the derivative of vector field involving the rate ofchange of components of a vector field in directions other than their own, i.e. we discuss

the curl of a vector field.

6.6 CURL OF A VECTOR FIELD

Curl of a vector field helps us in constructing a vector point function from a vector field.The formal definition of curl of a vector field is as follows :

Definition

Iff(x,y,z) be given continuously differentiable vector function, then the function

zyx

+

+

f

kf

jf

i

is called the curl offor curl of the vector field defined byf and is denoted by

curlf.

In terms of operator , we write

zyx

+

+

=

fk

fj

fif curl

fkji

+

+

=zyx


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89

Vector Differential

Calculus = f

The symbol is pronounced as del cross. Curlfcan also be obtained in the form ofa determinant as follows :

Letx,y,zbe right-handed Cartesian coordinates in space and let

kjif ),,(),,(),,(),,( zyxfzyxfzyxfzyx zyx ++=

be a differentiable vector function, then the function curlfis

zyx fff

zyx

==

kji

ff

Curl

+

+

=y

f

x

f

x

f

z

f

z

f

y

f xyzxzz kji

In the case of a left handed Cartesian coordinate system, the determinant for curlfis

preceded by a minus sign.

What does curlfrepresents physically? We shall answer this question in the next

subsection.


LetFbe a continuously differentiable vector field.

Then

+

+

==y

F

x

F

x

F

z

F

z

F

y

F xyzxyz kjiFF Curl

Let us consider thez-component of f, i.e.

y

F

x

Fxyz

= )( F

Now z)( F will be always positive if

(i)x

Fy

increases and

y

Fx

decreases or

(ii)y

F

x

Fxy

>

, when both

y

F

x

Fxy

and are positive.

The projection ofFonxy-plane will be jiOA yx FF += . At pointA(x,y, 0), the

y-component ofFincreases by the factor dxx

Fy

and thex-component ofF decreases

by the factor dyy

Fx

due to displacement (dx, dy, 0) inA. Hence at the point

B(x+ dx,y+ dy, 0),

jiOB

++

= dxx

FFdy

y

FF

yy

xx

and the resultant displacement isAB, which has turned left. If we give a further

displacement, Fywould increase and Fxwould decrease, giving a further resultantBC.Thus in going from pointAto C, the field vector has rotated anti-clockwise.


32/56

90

Engineering We can relate this rotation ofFto thez-component of CurlF. From the right-hand rule,

the direction of z)( F will be along thez-axis. The magnitude of z)( F tells ushow the magnitude of the field vectorFchanges as it rotates.

We can extend this argument to thexandycomponents of F and can say that thexandycomponents of F represent the rotation aboutxandy-axes, respectively.

Thus curl of a vector field gives us an idea of its rotation about an axis. It is termed asVORTEX field. The direction of F (i.e.,curlF) is along the axis about which thevector fieldFrotates (or curls) most rapidly and | F | is a measure of speed of thisrotation.

The sense of rotation (clockwise or anti-clockwise) is determined by the right-hand rule.

The curl of a vector functions plays an important role in many applications. Its

significance will be explained in more detail in Unit 7. At present we confine ourselves to

some simple examples.

6.6.2 Rotation of a Rigid Body

A rotation of a rigid bodyBin space can be simply and uniquely described by a vector .The direction of is that of the axis of rotation and is such that the rotation appears

clockwise if we look from the initial point of to its terminal point. The magnitude of

is equal to the angular speed (> 0) of the rotation, i.e. the linear (or tangential) speed ofa point ofBdivided by its distance from the axis of rotation (Figure 6.16).

Figure 6.16 : Rotation of a Rigid Body

Let Pbe any point of the bodyBand let dbe the distance of Pfrom the axis of rotation.

Letrbe the position vector of Preferred to some origin Oon the axis of rotation. Then

d= |r| sin , where is the angle between andr.

From above and the definition of vector product, the velocity vector Vof Pis given by

V= r

Let the axis of rotation be along the z-axis and we choose right-handed Cartesian

coordinates such that k= .

Then )()( kjikrV zyx ++==

ji xy +=

Now

0

Curl

xy

zyx

=

kji

V


33/56


34/56

92

Engineering

where

1

2

(4 )

Q

QK=

and is the dielectric constant. Ispan irrotational field?

Solution

pis an irrotation field, if curlp= 0.

)()( 2/32223

kjirp zyxzyx

K

r

K++

++==

Then, by definition,

2/32222/32222/3222 )()()(

curl

zyx

Kz

zyx

Ky

zyx

Kx

zyx

++++++

=

kji

p

++

++

=2/32222/3222 )()(

zyx

Ky

xzyx

Kz

yi

++

++

+2/32222/3222 )()(

zyx

Kz

xzyx

Kx

zj

++

++

+2/32222/3222 )()(

zyx

Kx

yzyx

Ky

xk

++

++

=

2/52222/5222 )(

2.2

3

)(

2.2

3

zyx

zky

zyx

ykz

i

++

++

+

2/52222/5222 )(

2.2

3

)(

2.2

3

zyx

xkz

zyx

zkx

j

++

++

+ 2/52222/5222 )(

2.

2

3

)(

2.

2

3

zyx

ykx

zyx

xky

k

= 0 + 0 + 0 = 0. Hencepis an irrotational field.

Let us take up another example.

Example 6.17

Show that the vector field defined by

kjiF 3222323 zyxzxzyx ++=

is irrotational. Find a scalar potential u such that F= grad u.Solution

By def.,


35/56

93

Vector Differential

Calculus

22323 32

curl

zyxzxzyx

zyx

=

kji

F

)22()66()33( 33222222 zxzxzyxzyxzxzx ++= kji

= 0 + 0 + 0 = 0.

HenceFis irrotational and henceFcan be expressed as grad u.

Let,z

u

y

u

x

u

+

+

= kjiF

Also kjiF 32 22323 zyxzxzyx ++=

Comparing the two expressions forF, we get

22323 3,,2 zyxz

uzx

y

uzyx

x

u=

=

=

Now dzz

udy

y

udx

x

udu

+

+

=

dzzyxdyzxdxzyx 22323 32 ++=

)()( 323223 zdyxdyzxxdzy ++=

)( 32 zyxd=

constant32 += zyxu

You may now try the following questions and see whether you have understood theconcepts given in this section.

SAQ 9

(a) Find CurlF, where )3( 333 zyxzyx ++=F .

(b) A fluid motion is given by kjiq )()()( yxxzzy +++++= . Is thismotion irrotational? If so, find the velocity potential.

(c)

If rV = show that V= 0.

You know that the operator is a vector operator. You can use this operator to provesome formulae on gradient, divergence and curl. We shall state these formulae in the next

section.6.6.3 Formulae on Gradient, Divergence and Curl

You already know from Sections 6.4, 6.5 and 6.6 that


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94

Engineering (i) The operator, when operated on a scalar fieldfgives rise to a vector fieldf(gradient off).

(ii)

Scalar product of with a vector fieldFgives a scalar field, divergence ofFor (.F).

(iii)

Vector product of with a vector fieldFgives a vector field, curlFor

(F).We now give the following formulas :

(i)

grad () = grad + grad .

(ii)

grad (A.B) =AcurlB+BcurlA+ (A.)B+ (B.)A.

(iii) grad (divA) = Curl CurlA+ 2A.

(iv) div (AB) =B. curlAA. curlB

(v)

div (curlA) = 0

(vi) div (fgrad ) =f2+ f. g

(vii)

curl (A) = (grad ) A+ curlA

(viii)

curl (AB) =AdivBBdivA+ (B. )A (A. )B.

For the proofs of formulas (i)-(viii) see Appendix-V.

Using the above formulas, you may now try the following exercises.

SAQ 10

(a) Ifais a constant vector andrdenotes the position vector of any point in

space and iff= (ar) rn, show that

divf= 0 and curlf= (n+ 2) rna n r

n 2

(a.r)r.

(b) Show that

(i) div [(ra) b] = 2 (a.b)

(ii) grad [r,a,b] =ab

(iii) curl (ra) = 2a

(iv) div (ra) = 0

(v) grad (a.r) =a,

whereaandbare a constant vector andris the radius vector.

6.7 SUMMARY

We will now summarise the result of this unit.

A function which is defined at each point of a certain region in space andwhose values are real numbers, depending on the points in space but not on

particular choice of coordinate system, is called ascalar function.

If be a function which associates a unique scalar with each point in a givenregion, then is called ascalar field.


37/56

95

Vector Differential

Calculus If to each point Pof a certain region in space, or if to each set Pof variables

of a certain sets of variables, a vector V(P) is assigned, then V(P) is called

a vector function.

IfFbe a function which associates a unique vector with every point in agiven region, thenFis called a vector field.

A vector functionf(t) of a real variables tis said to tend to its limit las t

approaches aiff(t) moves into the position occupied by las ta.

t alim ( )t

= =f l

The vector functionf(t), of scalar variables t, is said to be differentiable at apoint tif the limit

t

ttt

t +

=

)()(lim

0

ff

exists and it is then denoted by )(or tdt

df

f .

The derivative of a vector function represents a vector in the direction of thetangent to the curve traced by the vector function.

If a vector function is expressed in terms of its components, then the limit,continuity and differentiability of the function exists provided limit,

continuity and differentiability respectively of each component exists.

The partial derivativeu

fgives the rate of change off(u, v) w.r.t. uat a

given point (u, v) in space.

The vector

fz

f

y

f

x

f

f =

+

+

= kji

grad

is called the gradient of scalarf, wherefis a continuously differentiable

function. Here the symbol is read as deland represents vector operator

+

+

zyxkji .

The directional derivative of a scalar functionfalong the vectorbcan be

written as

f

s

fgrad.b=

The gradient of a scalar fieldfis a vector normal to the level surfacef= constant and has a magnitude equal to the rate of change of falong this

normal.

IfF(x,y,z) be any given continuously differentiable vector function, thenthe function

zyx

+

+ F

kF

jF

i ...

is called the divergenceofFand is written as divFor .F.

The divergence of a vector field represents the net outward flux per unittime at any point of the vector field.

A vector fieldFis called divergence free or solenoidal field in a givenregion if for all points in that region div F= 0.


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96

Engineering IfF(x,y,z) be any continuously differentiable vector function, then thefunction

zyx

+

+

Fk

Fj

Fi

is called the curl ofFand is denoted by curlFor F.

The curl of a vector field representsa rotation about an axis.

A field that has a vanishing curl everywhere is called irrotational (orconservative) field, i.e., if curlA= 0, thenAis called irrotational.

6.8 ANSWERS TO SAQs

SAQ 1

(i)

Vector function

(ii)

Vector function

(iii)

Scalar field

(iv)

Scalar function

(v) Vector function

SAQ 2

(a) (i) The limits of the components off(t) as t0 are

0lim,1lim00

==

t

t

t

tete

if )(lim0

=

tt

(ii) The limits of the components off(t) as t0 are

1)(lim,11.1)cos(lim,00.1)sin(lim 00 ===== t

t

t

t

t

etete

kjf )(lim0

=

tt

(iii) The limits of the components off(t) as t0 are

11lim,01

sin0lim

cos1lim,1

sinlim

0000==

+=

=

tttt

t

t

t

t

t

kif )(lim0

+=

tt

(b) (i) The function

kjif )(sin)(cos)( ++= ttt is continuous at every value of t, because each component is

continuous for all t.

(ii) The function

kjif |1|log1

1cos)( t

tet t ++

++=

is continuous at all texcept t= 1, because the functiont+1

1cos is

not defined at t= 1and at all other values of teach component iscontinuous.

SAQ 3


39/56

97

Vector Differential

Calculus(a) (i) The derivative of jif )( 2

tt etet += is

jif )(2)( 2 ttt eteet += . The function )(tf and its derivative

)(tf are defined at every value of t.

(ii) Here ,)( 0= tf becausef(t) is a constant vector. Domain is all t.

(iii) Here kjif

1

91

3

41

2

)( 222 tttt +=

Domain : For | 2t| < 1 and t0.

(b) From the chain-rule, we have

dx

duu

du

dx

dx

d.)()( ff =

)22(

)2

1.

)1(

1

)1(

1

2

1.)(sin

22 +

++

++= x

uuuuu kji

222

2

)12(1

)1(2

122

)1(2.)12(sin

+++

++

++

+++=

xx

x

xx

xxx ji

222 )121(

1.

122

)1(2

+++++

++

xxxx

xk

24 )2(

1

)1(1

)1(2)1(sin

++

++

+++=

xx

xx kji

This result is valid for ( 2) 0x+ .

SAQ 4

(i) jf

jif

if ,, y

zzx

yy

x=

+=

=

(ii) jf

iff ,,0 zy e

ze

yx

=

=

=

(iii) kjf

jif

kif 2,2,2 222 xzy

zyzx

yzxy

x+=

+=

+=

SAQ 5(a) The position vector of the particle at any time tis

kjir 3sin23cos2)( ttet t ++=

Velocity kjirV 3cos63sin6)()( ttetdt

dt t +==

Hence kiV 6)0( +=

Acceleration kjiVa 3sin183cos18)()( ttetdt

dt t ==

jia 18)0( =

(b) Here jir sincos tt +=


40/56

98

Engineering ji

rV cossin tt

dt

d+==

0cossincossin. =+= ttttVr

Vis perpendicular tor

Also

0cossin

0sincos

tt

tt

=kji

Vr

kk )sincos( 22 =+= tt

which is independent of t.

HencerVis a constant vector.

(c) Here br 2)( tt =

bbrV 22Velocity tt +===

Here 2 tbis the velocity of any point Prelative to the disk and b2t is the

additional velocity due to rotation.

Also bbbVa 242onAccelerati tt ++===

bbb 242 tt ++=

Also because of rotation,bis of the form

jib sincos tt +=

jib cossin tt +=

and bjib 222 sincos == tt

Hence the Coriolis acceleration bt4= , which is due to the interaction ofrotation of the disk and the motion of Pon the disk. It is in the direction of

b , i.e. tangential to the edge of the disk and it points in the direction of

rotation.

SAQ 6

(a) Here xyzyxf 422 ++=

grad kjii 4)42(42 xyxzyyzxf ++++=

kji )2(1.4}2.1.4)2(.2{}2.)2(.41.2{)(grad )2,2,1( ++++=f

kji 8414 +=

Here kjia 22 +=

3144|| =++=a

kjia 3

13

23

2 +=

Required directional drivative

s

f

=


41/56


42/56

100

Engineering(a) Here 222 zyx|| ++=r , where (x,y,z) is any point in space

2222 )(m

m zyxr ++=

Hence grad kji

z

r

y

r

x

rr

mmmm

+

+

=

]222[)(2

12222 kji zyxzyx

mm

++++=

][1

22

kji zyxrm

m

++=

r2= mrm )( kjir zyx ++=

(b) Here parametric representation of the curve is

kjir )( 32 tttt ++=

kjir 32 2tt ++=

If t is the unit tangent to the curve, then

42

2

941

32

tt

tt

++

++=

kjit

At (1, 1, 1) for the given curve t= 1

)32(14

1

941

32)()( )1()1,1,1( kji

kjitt ++=

++

++== =t

At ( 1, 1, 1) for the given curve t= 1

)32(14

1

941

32)()( )1()1,1,1( kji

kjitt +=

++

+== = t

Now 222 zxyzxyf ++=

kji )2()2()2()(grad 222 yzxxyzxzyf +++++=

)(3)(grad )1,1,1( kji ++=f

and kji 3)(grad )1,1,1( =f

Directional derivative at (1, 1, 1) along the tangent

)32(14

1.)(3 kjikji ++++=

14

18)321(

14

3=++=

Also directional derivative at ( 1, 1, 1) along the tangent

)32(14

1.)3( kjikji +=

14

2)323(14

1 =+=

(c) Let 3333 ++= xyzyxf


43/56

101

Vector Differential

CalculusA vector normal to the surface is

kji 3)33()33()grad( 22 xyxzyyzxf ++++=

At (1, 2, 1)

kji )2.1.3())1(.1.32.3()1(.2.31.3()(grad 22)1,2,1( ++++=f

kji 693 ++=

Also 14312636819|)grad(| )1,2,1( ==++=f

Hence a unit vector normal to the surface

)23(14

1

143

693kji

kji++=

++=

(d) Here force of attraction

=

3r

rwhere is a constant of

proportionality.

If is the gravitational potential for this force, then

=

+

+

=3

gradrzyx

rkji

kji

333 r

z

r

y

r

x

=

2/3222 )(

(

zyx

zyx

++

++=

kji

ji

)(

)(222222

++

+

++

=

zyxyzyxx

k)( 222

++

+zyxz

rzyx

=

++

=

222

SAQ 8

(a) Here kjir zyx ++= and div FkjiFF ..

+

+

==zyx

(i) div 3111 =++=

+

+

=z

z

y

y

x

xr

(ii) div3 2 2 2 3/ 2

.

( )

x y z

r x y z x y z

+ + = + + + +

r i j ki j k

++

+

++

=2/32222/3222 )()( zyx

y

yzyx

x

x

++

+ 2/3222 )( zyx

z

z

2/52222/3222 )(.2.2

3.1.)( ++

+++= zyxxxzyx


44/56

102

Engineering2/52222/3222 )(.2.

2

3.1.)( ++

++++ zyxyyzyx

2/52222/3222 )(.2.2

3.1.)( ++

++++ zyxzzzyx

)()(3)(3

2222/52222/3222

zyxzyxzyx ++++++=

= 0

(iii) div )(.)( kjir ++= zyx

)()()(

+

+

= zz

yy

xx

zz

yy

xx

++

++

+=

z

z

y

y

x

x

+

+

+= 3

+

+

+++=zyx

zyx kjikji .)(3

+= grad.3 r

(iv) Here 222| zyx| ++=r

222

zyx

zyx

++

++=

kjir

++

+

++

=)()(

div222222 zyx

y

yzyx

x

xr

++

+)( 222 zyx

z

z

++

++

+++= ...)()2(

2

1.)( 2/32222/1222 zyxxxzyx

)()()(3 2222/32222/1222 zyxzyxzyx ++++++=

222

2/1222 2)(2zyx

zyx

++=++=

(b) Here

+

+

=

=r

c

zr

c

yr

c

xr

ckjip grad

zr

cy

r

cx

r

c2

2

12

2

12

2

1333

kji =

)(

2222

zyxr ++=

=

+

=333

divr

cz

zr

cy

yr

cx

xp


45/56

103

Vector Differential

Calculus535353

2.

2

32.

2

32.

2

3

r

zcz

r

c

r

ycy

r

c

r

xcx

r

c+++=

+++=

5

222

33

3

r

zyxc

r

c

033

33

=+=r

c

r

c

pp = ,0div is a solenoidal vector.

(c) When = r2

Electric field )()()( 222 rz

ry

rx

E

== kji

)22]2[( kji zyx ++=

)(2 kji zyx ++=

r= 2

Charge distribution

+

+

= )2()2()2(. zz

yy

xx

E

= )222(

= 6

when

+=

r

aa

32 2

Electric field

+++

+

+

==222

32 2

zyx

aa

zyxE kji

kji 2.2

1.

22.2

1.

22.2

1.

23

3

3

3

3

3

zr

ay

r

ax

r

a++=

rkji

3

3

3

3 2)(

2

r

azyx

r

a=++=

Charge distribution

+

+

=3

3

3

3

3

3 222.

r

za

zr

ya

yr

xa

xE

++= z

r

zay

r

yax

r

xa

r

a

r

a

r

a2

2

322.

2

3.

22.

2

3.

22225

3

5

3

3

3

3

3

3

3

3

3

0)(66 222

5

3

3

3

=

++= zyx

r

a

r

a

SAQ 9

(a) Here )3( 333 xyzzyx ++=F


46/56

104

Engineering curl )]3([curl 333 xyzzyx ++=F

= 0 [curl (grad (any function)) = 0]

(b) Here kjiq )()()( yxxzzy +++++=

curl

yxxzzy

zyx

+++

=

kji

q

)11()11()11( += kji

= 0

Hence the fluid motion is irrotational.

Let be the velocity potential

zyx

== kjiq

Comparing the two expression for q , we get

yxz

xzy

zyx

+

+

+=

,,

Now dzz

dyy

dxx

d

+

+

=

][ ydzxdzxdyzdyzdxydx +++++=

)]()()[( ydzzdyxdzzdxxdyydx +++++=

)( zxyzxyd ++=

)( zxyzxy ++= , which is required.

(c) Here

kjir

rV 222222222 zyx

z

zyx

y

zyx

x

r +++

+++

++===

Curl

r

z

r

y

r

x

zyx

=

kji

V

++

+= x

r

z

r

zxz

r

y

r

yz2.

2

12.

2

12.2

12.

2

13333

ji

++

33

2.

2

12.

2

1

r

yx

r

xyk

= 0 + 0 + 0 = 0

SAQ 10

(a) Here nr)( raf =


47/56

105

Vector Differential

Calculus

nrra=

We know that div (AB) =B. curlAA. curlB

div )(div nrraf =

)(curl.curl. nn rr raar =

ais a constant vector, curla= 0

Now kjir zyx ++=

)()( 2222 kjir zyxzyxrn

n ++++=

curl

222222222222 )()()(

)(

nnn

n

zyxzzyxyzyxx

zyxr

++++++

=

kji

r

++++=

zzyxy

nyzyxz

nnn

2.)(2

2.)(2

1

22221

2222i

= 0 + 0 + 0 = 0

div 00 .. arf = nr

= 0 0 = 0

Now curl })({

})({

nn

rxr rairaf

==

Now

+

=

xr

xrr

x

nnn raranra )(})({

1

iaran )(2 += nn rxr

)()(})({ 2 iairainrai +=

nnn rrxrx

]).().[(.. 2 raiarin = xrn

]

).

()

.

[( iaiaiir

n

+ rairnarn

nn).(... 222 xx =

iairar nn ).(+

)(])[( 2222 zyxrrx

nn ++=

anrai

aaarrn nnn rrr + 3).(2

rrana nn ).()2( 2+= rnr

Hence the result.(b) (i) div ])[( bar

]).().[(div rbaabr =


48/56

106

Engineering If kjibkjiakjir ,, 321321 bbbaaazyx ++=++=++=

1 1 2 3[( ) ] [ ( )]a xb yb zbx

= + +

r a b

)]([ 3212 bzbybxay

++

+

+

+

++

+z

z

y

y

x

xbzbybxa

z).()]([ 3213 ba

= (a.b) 3 (a.b)

= 2 (a.b)

(ii) grad [r,a,b]

])(.[( bar =

])()()([componentcomponentcomponent ++= kji bababa zyx

componentcomponentcomponent)()()( ++= kji bakbajbai

= (ab)

(iii) )()(curl ariar

=x

= ar

ix

)( aii =

aiiiai ).().(( =

=a 3a= 2a

(iv) arraar curl.curl.)(div =

Now curl

zyx

zyx

=

kji

r

)00()00()00( ++= kii

= 0

and curla= 0 (ais a constant vector) 0)(div =ar

(v) )().(grad 321 zayaxa ++=ra

if kjia 321 aaa ++=

and kjir zyx ++=

)().(grad 321 zayaxax

++

= ira

1 ai=

=a


49/56


50/56

108

Engineering APPENDIX-II : INVARIANCE OF LENGTH AND DIRECTION

OF GRADf

The gradfis given by

z

f

y

f

x

ff

+

+

= kji grad . . . (1)

where kji ,, are unit vectors in the direction ofx,yandz-axes of a Cartesian coordinate

system, say 0xyzand (x,y,z) are the coordinates of any point Preferred to 0xyz(see Figure

6.12).

Since Eq. (1) involves partial derivatives, which depend on the choice of the coordinates,

hence the result is not obvious.

Now, by the definition of a scalar function, the value offat a point Pdepends on the

location of Pbut is independent of the coordinates. Also the arc length sof ray Cis

independent of the choice of coordinates.

Hences

PfQfs

f

PQ

s

)()(lim0

==

contains only the quantities which are independent of the choice of coordinates.

Also

==

cos|grad|||grad. ffs

fbb

= cos|grad| f ,

where is the angle between gradfand b

. From above, s

f

is maximum when

cos = 1, i.e. = 0 and then |grad| fs

f=

. This shows that the length and the direction

of gradfis independent of the choice of coordinates.


51/56

109

Vector Differential

CalculusAPPENDIX - III : INVARIANCE OF THE DIVERGENCE

Let kji ,, be unit vectors along a cartesian orthogonal coordinate system Oxyz. Let

kji ,, be any other set of mutually orthogonal unit vectors along system zyx O .

Let ),,()( zyxfrf = be a vector point function at any point ),,( zyx .

Let kfjfiff ),,(),,(),,(),,( 321 zyxzyxzyxzyx ++= and suppose that),,( zyxf possesses continuous first order partial derivatives; then scalar functions

f1,f2andf3also possess continuous first order partial derivatives.

Let l, m, nbe the direction cosines of any ray through ),,( zyxP and let

),,( zzyyxxQ +++ be a neighbouring point of Pon this ray. Then we have

PQ

PfQf

PQ

PfQf

PQ

PfQf

PQ

PQ )()()()()()()()( 332211 +

+

=

kjiff

Let QP, then using results of Section 11.4, we get

+

+

+

+

+

=

z

fn

y

fm

x

fl

z

fn

y

fm

x

fl

PQ

PQ

PQ

222111 )()(lim jiff

+

+

+z

fn

y

fm

x

fl 333k

+

+

+

+

+

=y

f

y

f

y

fm

x

f

x

f

x

fl 321321 kjikji

+

+

+

z

f

z

f

z

fn 321 kji

][][ 321321 kjikji fffy

mfffx

l ++

+++

=

][ 321 kji fffz

n ++

+

ffff

+

+

=

+

+

=z

ny

mx

lz

ny

mx

l

)(where,).( kjiafa nml ++==

which gives the directional derivative offalong the direction of a .

Thus the directional derivatives off along the directions of vectors kji ,, are

,).(,).(,).( fkfjfi

i.e. ,).().().(

+

+

zyx

fki

fji

fii

,).().().(

+

+

zyx

fkj

fjj

fij

+

+

zyx

fkk

fjk

fik ).().().(


52/56

110

Engineering Thus the expression for divfin terms of kji ,,

,.x

= fi

+

+

=

zyx

fki

fji

fiii ).().().(.

(sincexf

is directional derivative along x axis, i.e. along i direction)

+

+

=zyx

fkkk

fjjj

fiii .).(.).(.).(

,.x

fi

=

= expression of divfin term of kji ,,

Thus divf is independent of the choice of coordinates i.e. divfis essentially a point

function.


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111

Vector Differential

CalculusAPPENDIX - IV : PROOF OF FORMULAE ON DIVERGENCE

(I) )(.)(div ff kk =

fi kx

.

=

)(.)(. fifix

kkx ==

)(. fix

k

=

fi .

=x

k

f.= k

fdivk=

(II) )(.)(div ff =

)(.. fi

=x

)(. fi

=x

+

=

xx

ffi .

+

=

xx

fifi ..

+

=

xx

fifi ..

fif ..)grad(

+=x

ff .grad. +=

+= grad.div ff

(III) Here is twice differentiable scalar function. By definition,

zyx

+

+

= kji grad

By definition,

+

+

=zzyyxx

)grad(div

2

2

2

2

2

2

zyx

+

+

=

The expression on the right hand side is called the Laplacian of scalar function and is denoted as 2.

Thus = 2)grad(div .


54/56


55/56

113

Vector Differential

Calculus(IV) )()(div BAiBA

=x

+

=xx

BAB

Ai .

xx

+

=

BAiB

Ai ..

AB

iBA

i ..xx

=

(applying property of scalar triple product)

AB

iBA

i ..

=xx

ABBA .)curl(.)curl( =

(V)

+

+

= zyxxA

k

A

j

A

iiA

.

)curl(div

+

+

=

zxyxx

Ak

Aj

Aii

22

2

2.

+

+

= )()(.)(.

22

2

2

kiA

jiA

iiA

zxyxx

=

zxyx

Aj

Ak

22

..

= 0

becausezxzyyxyx

+

+

=

= A

jA

iA

kA

k2222

.... and similar expression

forzx

Aj

2

. .

(VI)

+

+

=zyx

ff kji div)grad(div

+

+

=zyx

fx

kjii .

+

+

=zyxx

fkjii .

+

+

+zyxx

f kjii .

+

+

=zyxx

fkjii . ).1.and.0.(.

2

2

===

+ iikijif

x

2

2

.

xfzyxx

f

+

+

+

= kjii

+= 2)(.).( ff


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Engineering (VII) )()(curl AA =

)( Ai

=x

+

=

xx

AAi

+

=

xx

AiAi

AA curl)grad( +=

(VIII) )()(curl BABA =

Assume BA += , where we consider onlyAas variable while operating A

and onlyBis to be considered as variable when operating by B . Since this takes

care of the product as far as differentiation is concerned and the A and B can

be treated as vectors and rules of multiplication of vectors can be applied. )()(curl BABA =

)()( BABA += BA

BAABBAAB ).().().().( BBAA +=

BABAABAB ).().().().( BBAA +=

BABAABAB ).().().().( +=

BAABABBA ).().(divdiv +=

Note that you should exercise special care to avoid inadvertent mistakes. For thisreason, in the first two steps, the variableA,Bare both written to the right of del

operator. In the next step, we have brought all those vectors which can be treated

as constant to the left of the operator. Since

0.... +=+= AAAA ABA

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