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Differential Vector

Calculus

28.2Introduction

A vector eld or a scalar eld can be differentiated with respect to position in three ways to produceanother vector eld or scalar eld. This Section studies the three derivatives, that is: (i) the gradientof a scalar eld (ii) the divergence of a vector eld and (iii) the curl of a vector eld.

Prerequisites

Before starting this Section you should . . .

be familiar with the concept of a function of two variables be familiar with the concept of partialdifferentiation be familiar with scalar and vector elds

Learning OutcomesOn completion you should be able to . . .

nd the divergence, gradient or curl of avector or scalar eld.

HELM (2005):Section 28.2: Differential Vector Calculus

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1. The gradient of a scalar eldConsider the height above sea level at various points on a hill. Some contours for such a hill areshown in Figure 14.

A

BC D

E

1020 30 40 50 60

Figure 14

We are interested in how changes from one point to another. Starting from A and makinga displacement d the change in height ( ) depends on the direction of the displacement. Themagnitude of each d is the same.

Displacement Change in AB 4030 = 10AC 4030 = 10AD 3030 = 0AE 2030 = 10

The change in clearly depends on the direction of the displacement. For the paths shown

increases most rapidly along AB , does not increase at all along AD (as A and D are both on thesame contour and so are both at the same height) and decreases along AE .The direction in which changes fastest is along the line of greatest slope and orthogonal (i.e.perpendicular) to the contours. Hence, at each point of a scalar eld we can dene a vector eldgiving the magnitude and direction of the greatest rate of change of locally.A vector eld, called the gradient, written grad , can be associated with a scalar eld so thatat every point the direction of the vector eld is orthogonal to the scalar eld contour and is thedirection of the maximum rate of change of .For a second example consider a metal plate heated at one corner and cooled by an ice bag at theopposite corner. All edges and surfaces are insulated. After a while a steady state situation exists inwhich the temperature at any point remains the same. Some temperature contours are shown inFigure 15.

heat source

ice bag510

15202530

35heat source

ice bag510

15202530

35

(a) (b)Figure 15

18 HELM (2005):Workbook 28: Differential Vector Calculus

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The direction of the heat ow is along ow lines which are orthogonal to the contours (see the dashedlines in Figure 15(b)); this heat ow is measured by F = grad .

DenitionThe gradient of the scalar eld = f (x,y,z ) is grad =

=

x

i +

y

j +

z

k

Often, instead of grad , the notation is used. ( is a vector differential operator called del or

nabla dened by x

i + y

j + z

k. As a vector differential operator, it retains the characteristics

of a vector while also carrying out differentiation.)

The vector grad gives the magnitude and direction of the greatest rate of change of at anypoint, and is always orthogonal to the contours of . For example, in Figure 14, grad points inthe direction of AB while the contour line is parallel to AD i.e. perpendicular to AB . Similarly, inFigure 15(b), the various intersections of the contours with the lines representing grad occur atright-angles.

For the hill considered earlier the direction of grad is shown at various points in Figure 16. Notethat the magnitude of grad is greatest when the hill is at its steepest.

1020 30 40 50 60

Figure 16

Key Point 3

is a scalar eld but grad is a vector eld.


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Example 9Find grad for

(a) = x 2

3y (b) = xy 2 z 3

Solution

(a) grad = x

(x 2 3y)i + y

(x 2 3y) j + z

(x 2 3y)k = 2xi + ( 3) j + 0 k = 2xi 3 j

(b) grad = x

(xy 2 z 3 )i + y

(xy 2 z 3 ) j + z

(xy 2 z 3 )k = y2 z 3 i + 2xyz 3 j + 3xy 2 z 2 k

Example 10For f = x 2 + y2 nd grad f at the point A(1, 2). Show that the direction of gradf is orthogonal to the contour at this point.

Solution

grad f = f x i +

f y j +

f z k = 2xi + 2yj + 0k = 2xi + 2yj

and at A(1, 2), this equals 21i + 2 2 j = 2i + 4 j .Since f = x 2 + y2 then the contours are dened by x2 + y2 = constant, so the contours are circlescentered at the origin. The vector grad f at A(1, 2) points directly away from the origin and hencegrad f and the contour are orthogonal; see Figure 17.

grad f

A

1

2

O x

y

Figure 17

To nd the change in a function in a given direction (given in terms of a unit vector a) take thescalar product, (grad ) a .


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Task skGiven = x 2 y2 z 2 , nd

(a) grad

(b) grad at (1, 1, 1) and a unit vector in this direction.(c) the derivative of at (2, 1, 1) in the direction of (i) i (ii) d = 35 i +

45 k.

Your solution

Answer

(a) grad = x

i + y

j + z

k = 2xy 2 z 2 i + 2x 2 yz 2 j + 2x 2 y2 zk

(b) At (1, 1, 1), grad = 2i + 2 j + 2kA unit vector in this direction isgrad

|grad | = 2i + 2 j + 2k

(2)2 + 2 2 + 2 2 = 12 3(2i + 2 j + 2k) =

1

3 i + 1

3 j + 1

3k(c) At (2, 1, 1), grad = 4 i + 8 j 8k

(i) To nd the derivative of in the direction of i take the scalar product(4i + 8 j 8k) i = 4 1 + 0 + 0 = 4 . So the derivative in the direction of d is 4.

(ii) To nd the derivative of in the direction of d = 35

i + 45

k take the scalar product

(4i + 8 j 8k) (35

i + 45

k) = 4 35

+ 0 + ( 8) 45

= 12

5 32

5 = 4.

So the derivative in the direction of d is

4.


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Exercises

1. Find grad for the following scalar elds

(a) = y x . (b) = y x 2 , (c) = x 2 + y2 + z 2 , (d) = x 3 y2 z .2. Find grad for each of the following two-dimensional scalar elds given that r = xi + yj and

r = x 2 + y2 (you should express your answer in terms of r ).(a) = r , (b) = ln r , (c) =

1r

, (d) = r n .

3. If = x 3 y2 z , nd,

(a) (b) a unit vector normal to the contour at the point (1, 1, 1).

(c) the rate of change of at (1, 1, 1) in the direction of i.(d) the rate of change of at (1, 1, 1) in the direction of the unit vector n = 1 3 (i + j + k).

4. Find a unit vector which is normal to the sphere x2 + ( y 1)2 + ( z + 1) 2 = 2 at the point(0, 0, 0).5. Find unit vectors normal to 1 = y x 2 and 2 = x + y 2. Hence nd the angle betweenthe curves y = x 2 and y = 2 x at their point of intersection in the rst quadrant.

Answers

1. (a) x (y x)i +

y (y x) j = i + j

(b) 2xi + j(c) [

x

(x 2 + y2 + z 2 )]i + [ y

(x 2 + y2 + z 2 )] j + [ z

(x 2 + y2 + z 2 )]k = 2xi + 2yj + 2zk ,

(d) 3x 2 y2 zi + 2x 2 yzj + x3 y2 k

2. (a) rr

, (b) rr 2

, (c) rr 3

, (d) nr n 2 r

3. (a) 3x2

y2

zi + 2x3

yzj + x3

y2

k, (b) 1

14(3i + 2 j + k), (c) 3, (d) 2 34. (a) The vector eld where = x 2 + ( y 1)2 + ( z + 1) 2 is 2xi + 2( y 1) j + 2( z + 1) k

(b) The value that this vector eld takes at the point (0, 0, 0) is 2 j + 2 k which is a vectornormal to the sphere.(c) Dividing this vector by its magnitude forms a unit vector:

1 2( j + k)

5. 108 or 72 (intersect at (1, 1))


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2. The divergence of a vector eldConsider the vector eld F = F 1 i + F 2 j + F 3 k.

The divergence of F is dened to be

div F = F 1

x + F

2

y + F

3

z .

Note that F is a vector eld but div F is a scalar.In terms of the differential operator , div F = F since

F = ( i x

+ j y

+ k z

) (F 1 i + F 2 j + F 3 k) = F 1

x +

F 2y

+ F 3

z .

Physical Signicance of the Divergence

The implication of the divergence is most easily understood by considering the behaviour of a uid

and hence is relevant to engineering topics such as thermodynamics. The divergence (of the vectoreld representing velocity) at a point in a uid (liquid or gas) is a measure of the rate per unit volumeat which the uid is owing away from the point. A negative divergence is a convergence indicatinga ow towards the point. Physically divergence means that either the uid is expanding or that uidis being supplied by a source external to the eld. Conversely convergence means a contraction orthe presence of a sink through which uid is removed from the eld. The lines of ow diverge froma source and converge to a sink.

If there is no gain or loss of uid anywhere then div v = 0 which is the equation of continuityfor an incompressible uid.

The divergence also enters engineering topics such as magnetic elds. A magnetic eld (denoted byB ) has the property B = 0 , that is there are no sources or sinks of magnetic eld.

Key Point 4

F is a vector eld but div F is a scalar eld.


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Task skFind the divergence of F 2 = x 2 yi 2xy 2 j , in general terms and at (1, 0, 3).

Your solution

Answer

2xy , 0,

Task sk

Find the divergence of F 3 = x2

zi 2y3

z 3

j + xyz 2

k, in general terms and at thepoint (1, 0, 3).

Your solution

Answer2xz 6y2 z 3 + 2 xyz , 6

3. The curl of a vector eldThe curl of the vector eld given by F = F 1 i + F 2 j + F 3 k is dened as the vector eld

curl F = F =

i j k

x

y

z

F 1 F 2 F 3

=F 3y

F 2z

i +F 1z

F 3x

j +F 2x

F 1y

k

Physical signicance of curlThe divergence of a vector eld represents the outow rate from a point; however the curl of a vectoreld represents the rotation at a point.

Consider the ow of water down a river (Figure 18). The surface velocity v of the water is revealedby watching a light oating object such as a leaf. You will notice two types of motion. First theleaf oats down the river following the streamlines of v, but it may also rotate. This rotation maybe quite fast near the bank but slow or zero in midstream. Rotation occurs when the velocity, and


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hence the drag, is greater on one side of the leaf than the other.

bank bank

Figure 18

Note that for a two-dimensional vector eld, such as v described here, curl v is perpendicular to themotion, and this is the direction of the axis about which the leaf rotates. The magnitude of curl vis related to the speed of rotation.For motion in three dimensions a particle will tend to rotate about the axis that points in the directionof curl v, with its magnitude measuring the speed of rotation.

If, at any point P, curl v = 0 then there is no rotation at P and v is said to be irrotational at P . If curl v = 0 at all points of the domain of v then the vector eld is an irrotational vector eld .

Key Point 5

Note that F is a vector eld and that curl F is also a vector eld.

Example 13Find curl v for the following two-dimensional vector elds(a) v = xi + 2 j (b) v = yi + xjIf v represents the surface velocity of the ow of water, describe the motion of aoating leaf.

Solution

(a ) v =i j k

x

y

z

x 2 0

= y

(0) z

(2) i + z

(x) x

(0) j + x

(2) y

(x) k = 0

A oating leaf will travel along the streamlines without rotating.


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Solution (contd.)

(b)

v =i j k

x

y

z

y x 0=

y

(0) z

(x) i + z

(y) x

(0) j + x

(x) y

(y) k= 0 i + 0 j + 2k = 2k

A oating leaf will travel along the streamlines (anti-clockwise around the origin ) and will rotateanticlockwise (as seen from above).

Example 14(a) Find the curl of u = x 2 i + y2 j . When is u irrotational?

(b) Given F = ( xy xz )i + 3 x2

j + yzk , nd curl F at the origin (0, 0, 0)and at the point P = (1 , 2, 3).

Solution

(a)

curl u = F =i j k

x

y

z

x 2 y2 0

= y

(0) z

(y2 ) i + z

(x 2 ) x

(0) j + x

(y2 ) y

(x 2 ) k

= 0 i + 0 j + 0k = 0

curl u = 0 so u is irrotational everywhere.


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Solution (contd.)

(b)

curl F = F =i j k

x

y

z

xy xz 3x 2 yz =

y

(yz ) z

(3x 2 ) i + z

(xy xz ) x

(yz ) j

+ x

(3x 2 ) y

(xy xz ) k= zi xj + 5xk

At the point (0, 0, 0), curl F = 0. At the point (1, 2, 3), curl F = 3i j + 5k .

Engineering Example 1

Current associated with a magnetic eld

Introduction

In a magnetic eld B , an associated current is given by:

I = 10

(B )Problem in words

Given the magnetic eld B = B 0 xk nd the associated current I .

x

y

Figure 19

Mathematical statement of problemWe need to evaluate the curl of B .


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Mathematical analysis

B =

i j k

x

y

z

0 0 B0 x

= 0 i B 0 j + 0k= B 0 j

and so I =

B 0

0 j .

Interpretation

The current is perpendicular to the eld and to the direction of variation of the eld.

Task skFind the curl of the following two-dimensional vector eld (a) in general terms and(b) at the point (1, 2).

F 2 = y2 i + xyj

Your solution

Answer

(a) F 2 =

i j k

x

y

z

y2 xy 0

= 0 i + 0 j + ( y 2y)k = yk

(b) 2k


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Exercises

1. Find the curl of each of the following two-dimensional vector elds. Give each in general termsand also at the point (1, 2).

(a) F 1 = 2xi + 2yj(b) F 3 = x 2 y3 i x 3 y2 j

2. Find the curl of each of the following three-dimensional vector elds. Give each in generalterms and also at the point (2, 1, 3).

(a) F 1 = y2 z 3 i + 2xyz 3 j + 3 xy 2 z 2 k

(b) F 2 = ( xy + z 2 )i + x2 j + ( xz 2)k3. The surface water velocity on a straight uniform river 20 metres wide is modelled by the vector

v = 150 x(20

x) j where x is the distance from the west bank (see diagram).

i

j

x

20 m

(a) Find the velocity v at each bank and at midstream.

(b) Find v at each bank and at midstream.4. The velocity eld on the surface of an emptying bathroom sink can be modelled by two

functions, the rst describing the swirling vortex of radius a near the plughole and the seconddescribing the more gently rotating uid outside the vortex region. These functions are

u(x, y ) = w(yi + xj ),

x 2 + y2 a

v(x, y ) =wa 2 (yi + xj )

x 2 + y2 x 2 + y2 aFind (a) curl u and (b)curl v.

Answers

1. (a) 0; (b) 6x 2 y2 k , 24k2. (a) 0; (b) zj + xk , 3 j + 2k

3. (a) 0; (b) 2 j , 0.4k, 04. (a) 2wk ; (b) 0


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4. The LaplacianThe Laplacian of a function is written as

2 and is dened as: Laplacian = div grad , that is

2 =

= x i + y j + z k=

2 x 2

+ 2 y 2

+ 2 z 2

The equation 2 = 0, that is 2 x 2

+ 2 y 2

+ 2 z 2

= 0 is known as Laplaces equation and has

applications in many branches of engineering including Heat Flow, Electrical and Magnetic Fieldsand Fluid Mechanics.

Example 15Find the Laplacian of u = x 2 y2 z + 2 xz .

Solution

2 u =

2 ux 2

+ 2 uy 2

+ 2 uz 2

= 2y2 z + 2x 2 z + 0 = 2( x 2 + y2 )z


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5. Examples involving grad, div, curl and the LaplacianThe vector differential operators can be combined in several ways as the following examples show.

Example 16If A = 2yzi x 2 yj + xz 2 k, B = x 2 i + yz j xyk and = 2x 2 yz 3 , nd(a) (A ) (b) A (c) B (d) 2

Solution

(a)

(A ) = (2yzi x2

yj + xz 2

k) ( x i +

y j +

z k)

= 2yz x x

2 y y

+ xz 2 z

2x 2 yz 3

= 2 yz x

(2x 2 yz 3 ) x2 y

y

(2x 2 yz 3 ) + xz 2 z

(2x 2 yz 3 )

= 2 yz (4xyz 3 ) x2 y(2x 2 y3 ) + xz 2 (6x 2 yz 2 )

= 8 xy 2 z 4 2x4 y4 + 6 x 3 yz 4

(b)

=

x

(2x 2 yz 3 )i + y

(2x 2 yz 3 ) j + z

(2x 2 yz 3 )k

= 4 xyz 3 i + 2x 2 z 3 j + 6x 2 yz 2 k

So A = 2yzi x2 yj + xz 2 k (4xyz

3 i + 2x 2 z 3 j + 6x 2 yz 2 k)= 8 xy 2 z 4 2x

4 yz 3 + 6 x 3 yz 4

(c) = 4xyz 3 i + 2x 2 z 3 j + 6x 2 yz 2 k so

B =i j k

x 2 yz xy4xyz 3 2x 2 z 3 6x 2 yz 2

= i(6x 2 y2 z 3 + 2 x 3 yz 3 ) + j (4x2 y2 z 3 6x

4 yz 2 ) + k(2x 4 z 3 4xy2 z 4 )

(d) 2 =

2

x 2(2x 2 yz 3 ) +

2

y 2(2x 2 yz 3 ) +

2

z 2(2x 2 yz 3 ) = 4yz 3 + 0 + 12 x 2 yz


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Example 17For each of the expressions below determine whether the quantity can be formedand, if so, whether it is a scalar or a vector.

(a) grad(div A)

(b) grad(grad )

(c) curl(div F )

(d) div [ curl (Agrad ) ]

Solution

(a) A is a vector and divA can be calculated and is a scalar. Hence, grad(div A) can beformed and is a vector.

(b) is a scalar so grad can be formed and is a vector. As grad is a vector, it is notpossible to take grad(grad ).

(c) F is a vector and hence div F is a scalar. It is not possible to take the curl of a scalarso curl(div F ) does not exist.

(d) is a scalar so grad exists and is a vector. Agrad exists and is also a vector as iscurl Agrad . The divergence can be taken of this last vector to givediv [ curl (Agrad ) ] which is a scalar.

6. Identities involving grad, div and curlThere are numerous identities involving the vector derivatives; a selection are given in Table 1.

Table 11 div(A) = grad A + div A or (A) = () A + ( A)2 curl(A) = grad

A + curl A or

(A) = (

)

A + (

A)

3 div (A B ) = B curl A A curl B or (A B ) = B (A) A (B )4 curl (A B ) = ( B grad ) A (A grad ) B or (A B ) = ( B )A (A )B+ A div B B div A + A B B A5 grad (A B ) = ( B grad ) A + ( A grad ) B or (A B ) = ( B )A + ( A )B+ A curl B + B curl A + A (B ) + B (A)6 curl grad = 0 or () = 07 div curl A = 0 or (A) = 0


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Example 18Show for any vector eld A = A 1 i + A 2 j + A 3 k, that div curl A = 0.

Solution

div curl A = div

i j j

x

y

z

A1 A2 A3

= div A 3y A

2

z i + A 1

z A3

x j + A 2

x A1

yk

= x

A 3y

A 2z

+ y

A 1z

A 3x

+ z

A 2x

A 1y

= 2 A3

xy 2 A2zx

+ 2 A1yz

2 A3yx

+ 2 A2zx

2 A1zy

= 0

N.B. This assumes 2 A3xy

= 2 A3yx

etc.

Example 19Verify identity 1 for the vector A = 2xyi 3zk and the function = xy 2 .

Solution

A = 2x 2 y3 i

3xy 2 zk so

A = 2x2 y3 i 3xy

2 zk = x

(2x 2 y3 ) + z

(3xy2 z ) = 4xy 3 3xy

2

So LHS = 4xy 3 3xy 2 .

=

x

(xy 2 )i + y

(xy 2 ) j + z

(xy 2 )k = y2 i + 2xyj so

() A = ( y2 i + 2xyj ) (2xyi 3zk ) = 2xy

3

A = (2xyi 3zk ) = 2y 3 so A = 2xy 3 3xy 2 giving() A + ( A) = 2xy 3 + (2 xy 3 3xy 2 ) = 4xy 3 3xy 2So RHS = 4xy 3 3xy 2 = LHS.So (A) = () A + ( A) in this case.


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Task skIf F = x 2 yi 2xzj + 2yzk , nd

(a) F (b) F (c) ( F )(d) (F )(e) (F )

Your solution

Answer

(a) 2xy + 2 y,

(b) (2x + 2z )i (x 2 + 2 z )k,(c) 2yi + (2 + 2 x) j (using answer to (a)),

(d) 0 (using answer to (b)),

(e) (2 + 2x) j (using answer to (b))


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Task skIf = 2xz y2 z , nd

(a) (b)

2

= ()(c) ()

Your solution

Answer(a) 2zi 2yzj + (2 x y2 )k, (b) 2z , (c) 0 where (b) and (c) use the answer to (a).

Exercise

Which of the following combinations of grad, div and curl can be formed? If a quantity can beformed, state whether it is a scalar or a vector.

(a) div (grad )

(b) div (div A)

(c) curl (curl F )

(d) div (curl F )

(e) curl (grad )

(f) curl (div A)

(g) div (A B )(h) grad (1 2 )(i) curl (div (A grad ))

Answers

(a), (d) are scalars;

(c), (e), (h) are vectors;

(b), (f), (g) and (i) are not dened.

36 HELM (2005):

28 2 diff vec calc

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