1

Further Engineering Mathematics K.A.Stroud

Vector Analysis (Part 1)

Summary of prerequisites 1. A scalar quantity has magnitude only; a vector quantity has both magnitude and

direction. 2. The axes of reference, OZ,OY,OX, form a right-handed set. The symbols

kji ,, denote unit vectors in the direction OZOYOX ,, respectively.

If kjirOP zyx aaa ++== then 222zyx aaa ++== rOP

3. The direction cosines [ ]nml ,, are the cosines of the angels between the vector

r and the axes OZ,OY,OX, respectively.For any vector kjir zyx aaa ++=

rrr

zyx ana

mal === ;;

and 1222 =++ nml

4. Scalar product (‘dot product’) 0cosAB=⋅ BA where 0 is the angle between A and B

If kjiA zyx aaa ++= and kjiB zyx bbb ++= then

zzyyxx bababa ++=⋅ BA 5. Vector product (‘cross product’) 0sinAB=× BA in a direction perpendicular to A and B

So that )(,, BABA × form a right-handed set Therefore 0sinAB=× BA

Also

zyx

zyx

bbbaaakji

BA =×

2

6 Angle between two vectors 2121210cos nnmmll ++=

Where 111 ,, nml and 222 ,, nml are the direction cosines of vectors 1r and 2r respectively

For perpendicular vectors 0212121 =++ nnmmll For parallel vectors 1212121 =++ nnmmll

One or two examples will no doubt help to recall the main points. Example 1 (Direction cosines)

If kji ,, are unit vectors in the directions OZOYOX ,, respectively, then any position vector ( )rOP = can be represented in the form

kjirOP zyx aaa ++== Then KK=r ________________________________________________________________________

222zyx aaa ++=r

The direction of OP is denoted by stating the direction cosines of the angles made by OP and the three coordinate axes.

ral x===

OPOLαcos

ra

m y===OPOMβcos

ran z===

OPONγcos

3

γβα cos,cos,cos,, =∴ nml

So, if P is the point (3,2,6), then ;KK=r

KKKKKK === nml ;; ________________________________________________________________________

;7=r 857.0;286.0;429.0 === nml

Since ( ) 4936492 =++=r 7=∴ r

4286.073cos === αl

2857.072cos === βm

8571.076cos === γn

Example 2 (angle between two vectors) If the direction cosines of A are 111 ,, nml and those of B are ,,, 222 nml then the angle between the vectors is given by 2121210cos nnmmll ++= If kjiA 432 ++= and kjiB 32 +−= , we can find the direction cosines of each and hence 0 which is………. ________________________________________________________________________

'36660 o= For 291694: 1 =++=rA

294;

293;

292

11 ===∴ nml

For 14941: 2 =++=rB

4

143;

142;

141

222 =−

==∴ nml

Then { } 3970.012622914

10cos =+−×

=

'36660 o=∴ Let us now look at the question of scalar and vector products. ________________________________________________________________________ Example 3 (scalar product) If A and B are two vectors, the scalar product of A and B is defined as 0cosAB=⋅BA Where 0 is the angle between the two vectors. If we consider the scalar products of the unit vectors kji ,, which are mutually perpendicular, then ( )( ) 090cos11 ==⋅ oji 0=⋅=⋅=⋅∴ ikkjji and ( )( ) 10cos11 ==⋅ oii 1=⋅=⋅=⋅∴ kkjjii in general, if kjiA zyx aaa ++= and kjiB zyx bbb ++= then

zzyyxx bababa ++=⋅ BA which is, of course, a scalar quantity so, if kjiA 432 +−= and kjiB 52 ++= , then KK=⋅BA ________________________________________________________________________

162062 =+−=⋅ BA Also, since 0cos. AB=BA ,we can determine the angle 0 between the vectors. In this case .....................0 = ________________________________________________________________________

'9570 o=

kjiA 432 +−= 291694 =++==∴ AA

kjiB 52 ++= 302541 =++==∴ BB We have already found that 16=⋅BA and 0cosAB=⋅BA

5

0cos302916 =∴ 5425.00cos =∴ '9570 o=∴ So , the scalar product of kjiA zyx aaa ++= and kjiB zyx bbb ++= is zzyyxx bababa ++=⋅ BA and 0cosAB=⋅BA where 0 is the angle between the vectors. It can also be shown that ( ) ABBA ⋅=⋅i and ( ) ( ) CABACBA ⋅+⋅=+⋅ii ________________________________________________________________________ Example 4 (vector product) If kjiA zyx aaa ++= and kjiB zyx bbb ++= the vector product BA× is defined as

0sinAB=× BA in the direction perpendicular to A and B such that ,A B and ( )BA× form a right-handed set. We can write this as ( ) ( )30sin nBA AB=× Where n is defined as a unit vector in the positive normal direction to the plane of A and B, i.e forming a right-handed set.

Also ( )4

zyx

zyx

bbbaaakji

BA =×

If we consider the vector products of the unit vectors, i,j,k, then ( )( ) 190sin11 ==× oji 1=×=×=×∴ ikkjji Note that ( ) 1−=×−=× jiij 1−=×=×=×∴ kijkij Also ( )( ) 00sin11 ==× oii 0=×=×=×∴ kkjjii It can also be shown that ( ) ( ) CABACBA ×+×=+×i and ( ) ( ) ( )5ABBA ×−=×ii Make a note of these result ( )3 , ( )4 and ( )5 Then, if kjiA 423 +−= and kjiB 232 −−= KK=× BA ________________________________________________________________________

kjiBA 51416 −+=×

6

We simply evaluate the determinant

232

423−−

−=×kji

BA

( ) ( ) ( ) kjikji 514164986124 −+=+−+−−−+= ________________________________________________________________________ We have seen therefore that The scalar product of two vectors is a scalar But that the vector product of two vectors is a vector. We know also that 0sinAB=× BA Therefore, the angle between the vectors A and B given in Example 4 is KK=0 ________________________________________________________________________

'40790 o= For ;232;423 kjiBkjiA −−=+−= and kjiBA 51416 −+=×

84.2147751416 222 ==++=×∴ BA

385.529423 222 ==++== AA

123.417232 222 ==++== BB ( )( ) 0sin123.4385.584.21 =∴

'407909838.00sin o=∴=∴ So, to recapitulate: If kjiA zyx aaa ++= and kjiB zyx bbb ++= and 0 is the angle between them (i) Scalar product zzyyxx bababa ++=⋅= BA 0cosAB=

(ii)vector product

zyx

zyx

bbbaaakji

BA =×

and 0sinABBA =× ________________________________________________________________________

7

Triple products We now deal with the various products that we form with three vectors. Scalar triple product of three vectors. If A,B,C are three vectors, the scalar formed by the product ( )CBA ×⋅ is called the scalar triple product. If ;kjiA zyx aaa ++= ;kjiB zyx bbb ++= ;kjiC zyx ccc ++=

Then

zyx

zyx

cccbbbkji

CB =×

( ) ( )zyx

zyxzyx

cccbbbaaakji

kjiCBA ⋅++=×⋅∴

Multiplying the top row by the external bracket and remembering that 0=⋅=⋅=⋅ ikkjji and 1=⋅=⋅=⋅ kkjjii

We have ( ) ( )6

zyx

zyx

zyx

cccbbbaaa

=×⋅ CBA

Example If ;432 kjiA +−= ;32 kjiB −−= kjiC 22 ++=

Then ( )212321

432−−

−=×⋅ CBA

KK= ________________________________________________________________________

For ( )212321

432−−

−=×⋅ CBA

( ) ( ) ( ) 42414623342 =+++++−=

( ) 42=×⋅ CBA

8

Properties of scalar triple products

(a) ( )zyx

zyx

zyx

zyx

zyx

zyx

bbbcccaaa

aaacccbbb

−==×⋅ ACB

Since interchanging two rows in a determinant reverses the sign. If we now interchange rows 2 and 3 and again change the sign, we have

( ) ( )CBAACB ×⋅==×⋅

zyx

zyx

zyx

cccbbbaaa

( ) ( ) ( )BACACCBA ×⋅=×⋅=×⋅∴ B )7( i.e. the scalar triple product is unchanged by a cyclic change of the vectors involved

(b) ( )zyx

zyx

zyx

zyx

zyx

zyx

cccbbbaaa

cccaaabbb

−==×⋅ CAB

( ) ( )CBACAB ×⋅−=×⋅∴ )8( i.e. a change of vectors not in cyclic order, changes the sign of the scalar triple product.

(c) ( ) 0==×⋅

zyx

zyx

zyx

aaabbbaaa

ABA since two rows are identical

( ) ( ) ( ) 0=×⋅=×⋅=×⋅∴ CACBCBABA )9( Example If ;32 kjiA ++= ;32 kjiB +−= kjiC 23 −+= ( ) KK=×⋅ CBA ( ) KK=×⋅ ABC ________________________________________________________________________

( ) ( ) 52;52 −=×⋅=×⋅ BACCBA

For ( ) ( ) ( ) ( ) 52923342161213

132321

=++−−−−=−

−=×⋅ CBA

9

( )ABC ×⋅ is not a cyclic change from the above. Therefore

( ) ( ) 52−=×⋅−=×⋅ CBAABC Coplanar vectors The scalar triple products provides a test to show whether three given vectors lie in the same plane.

By definition, ( )CB × is a vector product of B and C , of magnitude CB acting in a direction perpendicular to the plane of B and C and forming a right-handed set.

The scalar product of two vectors A and D is DA ⋅ where 0,0cosDADA =⋅ being the angle between A and D

If 02

cos,2

0 ==⋅=ππ DADA

0=⋅∴ DA

10

Therefore, if ( )CBD ×= , combining the two results above, we have this conclusion: If A is third vector Perpendicular to ( )CB × , then

(i) A,B and C are coplanar (ii) ( ) 0=×⋅ CBA

Therefore, three vectors A,B,C are coplanar if ( ) 0=×⋅ CBA Example 1 Show that ;32 kjiA −+= ;22 kjiB +−= and kjiC −+= 3 are coplanar. We just evaluate ( ) KK=×⋅ CBA and apply the test ________________________________________________________________________

( ) 0=×⋅ CBA

For ( ) ( ) ( ) ( ) 0323622211113

212321

=+−−−−−=−

−−

=×⋅ CBA

Therefore A,B,C are coplanar Example 2 If ;32 kjiA +−= ;23 kjiB ++= kjiC 4++= p are coplanar, find the value of p . The method is clear enough. We merely set up and evaluate the determinant and solve the equation ( ) 0=×⋅ CBA KK=p ________________________________________________________________________

3−=p

11

Since ( ) 0=×⋅ CBA 041123312=

−∴

p

( ) ( ) ( ) 0233112182 =−+−+−∴ pp 217 −=∴ p 3−=∴ p One more. Example 3 Determine whether the three vectors ;23 kjiA −+=

;32 kjiB +−= kjiC 22 +−= are coplanar. Work through it on your own. The result shows that….. ________________________________________________________________________

A,B,C are not coplanar

In this case ( ) 13221312123=

−−

−=×⋅ CBA

CBA ,,∴ are not coplanar

________________________________________________________________________ Vector triple products of three vectors If A,B and C are three vectors, then

( )CBA ×× are called the vector triple products )10( And ( ) CBA ×× Consider ( )CBA ×× where ;kjiA zyx aaa ++= ;kiiB zyx bbb ++= and

kjiC zyx ccc ++= Then ( )CB × is a vector perpendicular to the plane of B and C and ( )CBA ×× is a vector perpendicular to the plane containing A and ( )CB × ,i.e coplanar with B and C. Now

( )yx

yx

zx

zx

zy

zy

zyx

zyx ccbb

ccbb

ccbb

cccbbb kjikji

CB +−==×

( ) 0=/×⋅∴ CBA

12

Then ( )

yx

yx

zx

zx

zy

zy

zyx

ccbb

ccbb

ccbb

aaa

−

=××kji

CBA

yx

yx

xz

xz

zy

zy

zyx

ccbb

ccbb

ccbb

aaakji

=

In symbolic form, further expansion of the determinant becomes somewhat tedious. However a numerical example will clarify the methods. Example 1 If ;32 kjiA +−= ;2 kjiB −+= ;33 kjiC ++= Determine the vector triple product ( )CBA ×× We start off with K=×CB ________________________________________________________________________

kjiCB 567 −−=×

For ( ) ( ) ( )

kjikji

kjiCB

567613316

313121

−−=−++−+=

−=×

Then ( ) KK=×× CBA ________________________________________________________________________

( ) kjiCBA 91721 ++=××

13

For ( )567132

−−=××

kjiCBA

( ) ( ) ( )

kjikji

917212112710615

++=+−+−−−+=

That is fundamental enough. There is, however, an even easier way of determining a vector triple product. It can be proved that

( ) ( ) ( )CBABCACBA ⋅−⋅=×× ( )11 And ( ) ( ) ( )ABCBACCBA ⋅−⋅=×× The proof of this is given in the Appendix. For the moment, make a careful note of the expressions: then we will apply the method to the example we have just completed. ________________________________________________________________________

;32 kjiA +−= ;2 kjiB −+= kjiC 33 ++= and we have ( ) ( ) ( )CBABCACBA ⋅−⋅=×× ( )( ) ( )( )kjikji 331622336 ++−−−−++−= ( ) ( ) kjikjikji 9172133526 ++=+++−+= Which is, of course the result we achieved before. Here is another. Example 2 If ;223 kjiA −+= ;34 kjiB +−= kjiC +−= 32 determine ( ) CBA ×× using the relationship. ( ) ( ) ( )ABCBACCBA ⋅−⋅=×× ( ) KK=×× CBA ________________________________________________________________________

kji 222650 +−− For ( ) ( ) ( )ABCBACCBA ⋅−⋅=××

( )( ) ( )( )( ) ( )

kjikjikji

kjikji

22265022314342

22333834266

+−−=−+−+−−=

−+++−+−−−=

Now one more.

14

Example 3 If ;23 kjiA ++= ;52 kjiB −+= kjiC 32 ++=

( )

( ) KK

KK

=××=××

CBACBA

Finish them both. ________________________________________________________________________

( )( ) kjiCBA

kjiCBA313817583511

−+=××−+=××

For ( ) ( ) ( )CBABCACBA ⋅−⋅=××

( )( ) ( )( )( ) ( )

kjikjikji

kjikji

58351132155213

32215252661

−+=++−−+=

++−+−−+++=

And ( ) ( ) ( )ABCBACCBA ⋅−⋅=××

( )( ) ( )( )( ) ( ) kjikjikji

kjikji3138172395213

23310252661−+=++−−+=++−+−−+++=

As these two result clearly show ( ) ( ) CBACBA ××=/×× so beware! Before we proceed, note the following concerning the unit vectors.

( ) ( )

( )( )

( ) ( )( ) 0

00)(=××∴

=×=××

−=××∴−=×=××∴

=×

jiijjii

jjiijkijii

kji

ii

i

And once again, we see that ( ) ( ) jiijii ××=/×× ________________________________________________________________________

15

Finally, by way of revision: Example 4 if ;325 kjiA +−= ;23 kjiB −+= ;43 kjiC +−= determine (a) the scalar triple product ( )CBA ×⋅ (b) the vector triple products

( ) ( )( )( ) .CBA

CBA××

××iii

___________________________________________________________________

( ) ( )( ) ( )

kjiCBAkjiCBA

CBA

227109)(744462

12

−+=××−+=××

−=×⋅ba

Here is the working.

(a) ( )431213

325

−−

−=×⋅ CBA

( ) ( ) ( ) 121932122645 −=−−+++−= (b)(i) ( ) ( ) ( )CBABCACBA ⋅−⋅=××

( )( ) ( )( )( ) ( ) kjikjikji

jikji7444624372323436215231265

−+=+−−−+=+−−−−−+++=

(ii) ( ) ( ) ( )ABCBACCBA ⋅−⋅=×× ( ) ( )( )kjikji 32582323 +−−−−+= kji 227109 −+=

16

Differential of vectors In many practical problems, we often deal with vectors that change with time, e.g. velocity, acceleration, etc. If a vector A depends on a scalar variable t , then A can be represented as ( )tA and A is then said to be a function of t . If kjiA zyx aaa ++= then zyx aaa ,, will also be dependent on the parameter .t i.e. ( ) ( ) ( ) ( )kjiA tatatat zyx ++= Differentiating with respect to t gives….. ________________________________________________________________________

( ){ } ( ){ } ( ){ } ( ){ }tadtdta

dtdta

dtdt

dtd

zyx kjiA ++=

In short .dt

dadt

dadt

dadtdA zyx kji ++=

The independent scalar variable is not, of course, restricted to t . In general, if u is the paramater, then

KK=dudA

_____________________________________________________________

duda

duda

duda

dud zyx kjiA

++=

If a position vector OP moves to OQ when u becomes uu δ+ , then as 0⎯→⎯uδ , the direction of the chord PQ becomes that of the tangent to the curve at P , i.e. the

direction of dudA is along the tangent to the locus of P .

17

Example 1 If ( ) ( ) kjiA 32 45243 uuu +−++= , then

................=dudA

________________________________________________________________________

kjiA 21226 uudud

++=

If we differentiate this again, we get kiA udud 2462

2

+=

When 2=u kjiA 48212 ++=dud and kiA 4862

2

+=dud

Then

............=dudA and .......2

2

=dud A

________________________________________________________________________

;52.49=dudA and 37.482

2

=dud A

For { } { } 52.49245248212 2121222 ==++=dudA

And { } { } 37.482340486 2121222

2

==+=dud A

18

Example 2 If , ( )ttet t 42sin 33 −++= kjiF then when 1=t

;KK=dtdF ...........2

2

=dt

d F

_______________________________________________________________________

kjiF

kjiF

692sin4

32cos2

32

2

3

++−=

−+=

edt

d

edtd

From these, we could if required find the magnitudes of dtdF and 2

2

dtd F

................... 2

2

==dt

ddtd FF

________________________________________________________________________

9.180;27.60 2

2

==dt

ddtd FF

For ( ){ } 2162 192cos2 ++= edtdF

{ } 27.60136316927.0 21 =++=

And ( ){ } 21622

2

36812sin4 ++−= edt

d F

{ } 9.180363267823.13 21 =++= One more example. Example 3 If ( ) ( ) kjiA 32 223 uuu ++−+= determine

(a)dudA (b) 2

2

dud A (c)

dudA (d) 2

2

dud A at 3=u

Work through all sections and then check with the next frame

________________________________________________________________________

19

Here is the working. ( ) ( ) kjiA 32 223 uuu ++−+=

(a) kjiA 262 uudud

+−= at kjiA 546,3 +−==dudu

(b) kjA udud 1222

2

+−= at kjA 362,3 2

2

+−==dudu

(c) { } ( ) 34.5429532916361 2121 ==++=dudA

(d) { } ( ) 06.36130012964 21212

2

==+=dud A

The next example is of a rather different kind, so move on. ________________________________________________________________________ Example 4 A particle moves in space so that at time t its position is stated as

232 2,3,32 ttzttytx +=+=+= . We are required to find the components of its velocity and acceleration in the direction of the vector kji 432 ++ when 1=t First we can write the position as a vector r ( ) ( ) ( )kjir 232 2332 ttttt +++++= Then, at 1=t

........2

2

=dtd r

________________________________________________________________________

;752 kjir++=

dtd

kjr 1022

2

+=dtd

For ( ) ( )kjir tttdtd 43322 2 ++++=

∴ at ,1=t kjir 752 ++=dtd

And ( )kjr 4622

2

++= tdtd

........;=dtdr

20

∴At ,1=t kjr 1022

2

+=dtd

Now, a unit vector parallel to kji 432 ++ is …………………….. ________________________________________________________________________

( )kjikji 432291

1694432

++=++++

Denote this unit vector by I . Then the component of dtdr in the direction of I

________________________________________________________________________

73.8

Since ( ) ( ) ( )28154291432752

291

++=++⋅++ kjikji

73.829

47==

Similarly, the component of 2

2

dtd r in the direction of I is

.................... ________________________________________________________________________

54.8

For ( ) ( )kjikjIrr 4321022910cos 2

2

2

2

++⋅+=⋅=dtd

dtd

54.829

46)406(291

==+=

( ) ( )

......................

432752291

0cos

=

++⋅++=

⋅==

kjikji

Idtdr

dtdr

21

Differentiation of sums and products of vectors If ( )uAA = and ( )uBB = ,then

(a) { }dudcc

dud AA =

(b) { }dud

dud

dud BABA +=+

(c) { } BABABA ⋅+⋅=⋅dud

dud

dud

(d) { } .BABABA ×+×=×dud

dud

dud

These are very much like the normal rules of differentiation. However, if ( ) kjiA zyx aaau ++= ( ).constai

i.e. constant

Also ( ) ( ){ } ( ) ( ){ } ( ) ( ){ }ududuu

duduuu

dud AAAAAA ⋅+⋅=⋅

( ) ( ){ } { } 02 2 ==⋅= AAAdudu

dudu

Assuming that ( ) 0=/uA then since ( ) ( ){ } 0=⋅ ududu AA it follows that ( )uA and

( ){ }udud A are perpendicular vectors because…….

________________________________________________________________________

( ) 22222)( Aaaauu zyx ==++=⋅ AAA

22

( ) ( ){ } ( ) ( ){ } 00cos ==⋅ ududuu

dudu AAAA

00cos =∴ 20 π=∴

Now let us deal with unit tangent vectors.

Unit tangent vectors We have already established in frame 30 of this programme that if OP is a position

vector ( )uA in space, then the direction of the vector denoting ( ){ }udud A is………

________________________________________________________________________

Parallel to the tangent to the curve at P Then the unit tangent vector T at P can be found from

( ){ }

( ){ }udud

udud

A

AT =

In simpler notation, this becomes: If kjir zyx aaa ++= then the unit tangent vector T is given by

duddud

//

rrT =

23

Example 1 Determine the unit tangent vector at the point (2,4,7) for the curve with parametric equations ;2ux = ;32 += uy 52 2 += uz First we see that the point (2,4,7) corresponds to 1=u The vector equation of the curve is ( ) ( )kjikjir 5232 22 ++++=++= uuuaaa zyx

...........=∴dudr

________________________________________________________________________

kjir uudud 422 ++=

And at ,1=u kjir 422 ++=dud

Hence KK=dudr

and KK=T

________________________________________________________________________

;62=dudr { }kjiT 2

61

++=

For { } 62241644 2121 ==++=dudr

{ }kjikjir

r

T 26

162

422++=

++==

duddud

Let us do another. Example 2 Find the unit tangent vector at the point ( )π,0,2 for the curve with parametric equations ;0sin2=x ;0cos3=y .02=z We see that the point ( )π,0,2 corresponds to 20 π= Writing the curve in vector from KK=r ________________________________________________________________________

24

kjir 020cos30sin2 ++=

Then, at ,20 π= .............0=

ddr

KK=0d

dr

KKK=T Finish it off ________________________________________________________________________

( )kjT

rkjr

23131

130

;230

+−=

=+−=dd

dd

And now Example 3 Determine the unit tangent vector for the curve ;3tx = ;2 2ty = ttz += 2 At the point ( )6,8,6 On your own. .................=T ________________________________________________________________________

( )kjiT 583981

++=

The point ( )6,8,6 corresponds to 2=t ( )kjir tttt +++= 2223

( )kjir 1243 +++=∴ ttdtd

At ,2=t kji 686 ++=r and kjir 583 ++=dtd

( ) 9825649 21 =++=∴dtdr

( )kjirrT 583

981

++==∴dtddtd

________________________________________________________________________

25

Partial Differentiation Of Vectors If a vector F is a function of two independent variables u and v , then the rules of differentiation follow the usual pattern. If kjiF zyx ++= then zyx ,, will also be function of u and v

Then kjiFuz

uy

ux

u ∂∂

+∂∂

+∂∂

=∂∂

kjiFvz

vy

vx

v ∂∂

+∂∂

+∂∂

=∂∂

kjiF2

2

2

2

2

2

2

2

uz

uy

ux

u ∂∂

+∂∂

+∂∂

=∂∂

kjiF2

2

2

2

2

2

2

2

vz

vy

vx

v ∂∂

+∂∂

+∂∂

=∂∂

kjiFvu

zvu

yvu

xvu ∂∂

∂+

∂∂∂

+∂∂

∂=

∂∂∂ 2222

And for small finite changes du and dv in u and v , we have

vv

uu

d ∂∂∂

+∂∂∂

=FFF

Example If ( ) ( )kjiF 22 22 vuvuuv ++−+=

KKKKKKKKK =∂∂

∂=

∂∂

=∂∂

=∂∂

vuuvuFFFF 2

2

2

;;;

________________________________________________________________________

;22 kjiF++=

∂∂ uv

u kjiF vuv

222 +−=∂∂

iFjF 2;22

2

2

=∂∂

∂=

∂∂

vuu

This is straightforward enough.

26

Integration of vector functions The process is the reverse of that for differentiation. If a vector kjiF zyx ++= where

zyx ,,,F are expressed as functions of u , then ∫ ∫ ∫ ∫++=

b

a

b

a

b

a

b

azduyduxdudu kjiF

Example 1 if ( ) ( ) ,45243 32 kjiF tttt +−++= then

( ) ( ) dttdttdtttdt ∫∫∫∫ +−++=3

1

33

1

3

1

23

145243 kjiF

KKK= ________________________________________________________________________

kji 80242 +−

For ( ) ( )[ ]31

3

1

4223 52∫ +−++= tttttdt kjiF

( ) ( ) kjikjikji 802424381645 +−=+−−+−= Here is slightly different one. Example 2 If ( )kjiF 23 2 +++= uuu And ( )kjiV 232 −+−= uuu Evaluate ( )du∫ ×

2

0VF

First we must determine VF × in terms of u KKK=×VF ________________________________________________________________________

( ) ( ) ( )kjiVF 23223 92106 uuuuuuu +−−−++=×

For ( )( )232

23 2

−−+=×

uuuuuu

kjiVF

Which gives the results above. Then ( ) .................

2

0=×∫ duVF

27

{ }kji 24131434

−+

( ) kjiVF ⎟⎟⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛++=×∫ 3

42

32

34

32

53

334

uuuuuuudu

( ) ( ) ( )kjiVF 2482038124 3

82

0+−⎟

⎠⎞

⎜⎝⎛ −−++=×∴∫ du

{ }kji 2413143

4 −+= Example 3 If ( )CBAF ××= where ( ) kjiA ttt 4323 2 +−+= ( )kjiB tt −++= 1342 kjiC ttt 232 2 −−=

Determine ∫1

0dtF

First we need to find ( ).CBA ×× The simplest way to do this is to use the relationship ( ) KK=×× CBA ________________________________________________________________________

( ) ( ) ( )CBABCACBA ⋅−⋅−×× So KK=⋅CA And KK=⋅BA ________________________________________________________________________

2222

22233

2121212868966

ttttttttttt

=−+−+=⋅

=−+−=⋅

BACA

Then ( ) ( ){ } { }kjikjiCBAF ttttttt 23221342 222 −−−−++=××=

∫ =∴1

0KKFdt

Finish off the simplification and complete the intergration ________________________________________________________________________

28

{ }kji 7513220601 ++−

For ( ) ( ) ( ) ( )kjiCBAF 324332 36442 tttttt ++++−=××= Integration with respect to t then gives the result stated above. ________________________________________________________________________

Scalar and vector fields

If every points ( )zyx ,,P of a region R of space has associated with it a scalar quantity ( )zyxo ,,/ then ( )zyxo ,,/ is a scalar function and a scalar field is said to exist in the

region R Examples of scalar fields are temperature, potential, etc.

Similarly, if every point ( )zyx ,,P of a region R has associated with it a vector quantity ( )zyx ,,F , then ( )zyx ,,F is a vector function and a vector field is said to exist in the region R Examples of vector fields are force, velocity, acceleration, etc. ( )zyx ,,F can be defined in terms of its components parallel to the coordinate axes, .,, OZOYOX That is, ( ) kjiF zyx FFFzyx ++=,,

________________________________________________________________________

29

Grad (gradient of a scalar function) If a scalar function ( )zyxo ,,/ is continuosly differentiable with respect to its variables

zyx ,, throughout the region, then the gradient of o/ written grad o/ , is defined as the vector

( )12kjizo

yo

xoograd

∂/∂+

∂/∂+

∂/∂=/

Note that, while o/ is a scalar function, grad o/ is a vector function. For example, if o/ depends upon the position of P and is defined by 322 yzxo =/ , then

kji 22323 624 yzxzxxyzograd ++=// Notation The expression (12) above can be written

ozyx

ograd /⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

+∂∂

=// kji

Where ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

zyxkji is called a vector differential operator and is denoted by the

symbol ∇ (pronounced ‘del’or sometimes ‘nabla’)

i.e. ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∇zyx

kji

Beware! ∇ cannot exist alone:it is an operator and must operate on a stated scalar function ( )zyxo ,,/ If F is a vector function, F∇ has no meaning. So we have:

( )13ozyx

ogrado /⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=/=/∇ kji

zo

yo

xo

∂/∂+

∂/∂+

∂/∂= kji

________________________________________________________________________ Example 1 If 2232 zxyyzxo +=/ , determine grad o/ at the point ( )2,3,1P

By the definiton, grad .kjizo

yo

xooo

∂/∂+

∂/∂+

∂/∂=/∇=/

30

All we have to do then is to find the partial derivatives at 2,3,1 === zyx and insert their values. ________________________________________________________________________

( )kji 188214 ++

Since 2232 zxyyzxo +=/ 2232 zyxyz

xo

+=∂/∂∴

232 2xyzzxyo

+=∂/∂

zxyyzxzo 222 23 +=∂/∂

Then at ( )2,3,1 3648 +=∂/∂xo

84=∂/∂∴xo

248 +=∂/∂yo

32=∂/∂∴yo

3636 +=∂/∂zo

72=∂/∂∴zo

∴grad ( )kjikji 188214723284 ++=++=/∇=/ oo Example 2 if kjiA zyxyzx 22 ++= And kjiB zxxzyz 22 ++= determine an expression for grad ( ).BA ⋅ This we can soon do since we know that BA ⋅ is a scalar function of yx, and z First then, ...................=⋅ BA

222232 zyxyzxyzx ++=⋅ BA Then ( ) KK=⋅∇ BA ________________________________________________________________________

( ) ( ) ( )kji yzzyxyzzzxyzzxyz 2132112 22222 ++++++++ For if ( ) ( )kjikjiBA zxxzyzzyxyzxo 2222 ++⋅++=⋅=/ 222232 zyxyzxyzxo ++=⋅=/ BA

( )yzzxyzzxyxyzxyzxo

++=++=∂/∂ 12222 2223

..........=/∇∴ o

31

( )yzzzxyzxzxzxyo 212 2222232 ++=++=∂/∂

( )yzzyxzyxyxyzxzo 21323 2222222 ++=++=∂/∂

( ) ( ) ( ) ( )kjiBA yzzyxyzzzxyzzxyz 2132112 22222 ++++++++=⋅∇∴

Now let us obtain another useful relationship. If OP is a position vector r where

kjir zyx ++= and rd is a small displacement corresponding to changes dzdydx ,, in zyx ,, respectively, then kjir dzdydxd ++=

If ( )zyxo ,,/ is a scalar function at P , we know that

kjizo

yo

xooograd

∂/∂+

∂/∂+

∂/∂=/∇=/

Then grad KKK=⋅/ rdo ________________________________________________________________________

zzoy

yox

xodograd ∂

∂/∂+∂

∂/∂+∂

∂/∂=⋅/ r

For grad ( )kjikjir dzdydxzo

yo

xodo ++⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛∂/∂+

∂/∂+

∂/∂=⋅/

zzoy

yodx

xo

∂∂/∂+∂

∂/∂+

∂/∂=

= the total different d o/ of o/

( )14ograddod /⋅=/∴ r ________________________________________________________________________

32

Directional derivatives We have just established that ograddod /⋅=/ r

If ds is the small element of arc between ( )rP and ( )rr dQ + then rdds =

rrr

dd

dsd

=

and dsdr is thus a unit vector in the direction of .rd

φgraddsd

dsod

⋅=/∴r

If we denote this unit vector by ,a i.e ,adsd

=r the result becomes

ograddsod

/⋅=/ a

dsod /

is thus the projection of grad o/ on the unit vector a and is called the directional

derivative of o/ in the direction of a . It gives the rate of change of o/ with distance

measured in the direction of a and ograddsod

/⋅=/ a will be a maximum when a and

grad o/ have the same direction of ,since then 0cosˆˆ ogradograd /=/⋅ aa and 0 will be zero Thus the direction of grad o/ gives the direction in which the maximum rate of changes of o/ occurs Example 1 Find the directional derivative of the function 222 2 yzxyzxo ++=/ at the point ( )1,2,1 − in the direction of the vector .432 kjiA −+= We start off with 222 2 yzxyzxo ++=/ ...........=/∇∴ o ________________________________________________________________________

33

( ) ( ) ( )kji yzxzxyyxzo 2422 222 +++++=/∇

Since ;22 2yxzxo

+=∂/∂

;4 2zxyyo

+=∂/∂

yzxzo 22 +=∂/∂

Then at ( )1,2,1 −

( ) ( ) ( ) kjikji 396411882 −+=−++++−=/∇o Next we have to find the unit vector a where kjiA 432 −+=

..................=a) ________________________________________________________________________

( )kjia 432291ˆ −+=

For kjiΑ 432 −+= 291694 =++=∴ A

( )kjiAAa 432

291ˆ −+==

So we have kji 396 −+=/∇o and ( )kjia 432291ˆ −+=

............ˆ =/∇⋅=/∴ odsod a

________________________________________________________________________

47.929

51==/

dsod

Since ( ) ( )kjikjia 396432291ˆ −+⋅−+=/∇⋅=/ o

dsod

( ) 47.929

51122712291

==++=

That is all there is to it (i) from the given scalar function o/ , determine o/∇ (ii)Find the unit vector a in the direction of the given vector A

(iii) Then odsod

/∇⋅=/ a

34

Example 2 Find the directional derivative of xzzyyxo 222 ++=/ at the point ( )2,1,1 − in the direction of the vector kji 524 −+=A . Same as before.Work through it and check the result with the next frame. ________________________________________________________________________

43.353

23−=

−=/

dsod

For xzzyyxo 222 ++=/

( ) ( ) ( )kji zxyyzxzxyo 222 222 +++++=/∇∴ ∴At ( ),2,1,1 − kji 532 +−=/∇o

534525416 ==++=∴ A

( )kjia 52453

1ˆ −+=∴

( ) ( )kjikjia 53252453

1ˆ +−⋅−+=/∇⋅=/∴ odsod

( ) 43.353

23256853

1−=

−=−−=

Example 3. Find the direction from the point 0,1,1 which gives the greatest rate of increase rate of increase of the function ( ) ( )22 23 zyyxo −++=/ This appears to be different, but it rests on the fact that the greatest rate of increase of o/ with respect to distance is in….. ________________________________________________________________________

The direction of o/∇ All we need then is to find the vector o/∇ , which is ………. ________________________________________________________________________

( )kji −+=/∇ 824o For ( ) ( )22 23 zyyxo −++=/

( );32 yxxo

+=∂/∂∴ ( ) ( );2436 zyyx

yo

−++=∂/∂

( )zyzo

−−=∂/∂ 22

∴At ( ),0,1,1 ;8=∂/∂xo ;32=

∂/∂yo 4−=

∂/∂zo

( )kjikji −+=−+=/∇∴ 8244328o ∴greatest rate of increase occurs in direction kji −+ 82

kjiA 524 −+=

35

________________________________________________________________________ Unit normal vectors If ( ) =/ zyxo ,, constant, this relationship represents a surface in space, depending on the value ascribed to the constant.

If rd is a placement in this surface, then 0=/od since o/ is constant over the surface. Therefore our previous relationship odogradd /=/⋅r becomes 0=⋅ φgraddr For all such displacements rd in the surface. But 00cos ==⋅ φφ graddgradd rr

∴=∴2

0 π grad o/ is perpendicular to rd , i.e. grad o/ is a vector perpendicular to the

surface at P , in the direction of that maximum rate of change of o/ . The magnitude of that maximum rate of change is given by φgrad The unit vector N in the direction of grad o/ is called the unit normal vector at P .

∴unit normal vector

( )15oo/∇/∇

=N

Example 1 Find the unit normal vector to the surface 024 223 =+++ zxyxzyx at the point ( )1,3,1 −

Vector normal ...........=/∇= o ________________________________________________________________________

( ) ( ) ( )kji 23222 8243 xyxzxyzxzyzyxo ++++++=/∇

36

Then, at ( )1,3,1 − kji +−=/∇ 54o And the unit normal at ( )1,3,1 − is ________________________________________________________________________

( )kji +− 54421

Since 4212516 =++=/∇o

And ( )kjiN +−=/∇/∇

= 54421

oo

One more Example 2 Determine the unit normal to the surface 0552 =−−+ yzyxxyz at the point ( )2,1,3 All very straightforward. Complete it. ________________________________________________________________________

Unit normal ( )kjiN 258931

−+==

For 552 −−+=/ yzyxxyzo

( ) ( ) ( )kji yxyzxxzxyyzo 552 2 −+−+++=/∇∴ At ( ) 9342564;258,2,1,3 =++=/∇−+=/∇ oo kji

∴unit normal ( )kjiN 258931

−+=/∇/∇

==oo

Collecting our results so far, we have, for ( )zyxo ,,/ a scalar function

(a) grad kjizo

yo

xooo

∂/∂+

∂/∂+

∂/∂=/∇=/

(b) ograddod /⋅=/ r where zzoy

yox

xood ∂

∂/∂+∂

∂/∂+∂

∂/∂=/

(c) directional derivative ograddsod

/⋅=/ a

(d)unit normal vector oo/∇/∇

=N

copy out this brief summary for future reference. It will help. ________________________________________________________________________

37

Grad of sums and products of scalars

(a) ( ) ( ) ( ) ( )⎭⎬⎫

⎩⎨⎧ ++∂∂

+⎭⎬⎫

⎩⎨⎧

+∂∂

+⎭⎬⎫

⎩⎨⎧ +∂∂

=+∇ BAz

BAy

BAx

BA kji

⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

+∂∂

+⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

+∂∂

= kjikjizB

yB

xB

zA

yA

xA

( ) BABA ∇+∇=+∇∴

(b) ( ) ( ) ( ) ( )⎭⎬⎫

⎩⎨⎧∂∂

+⎭⎬⎫

⎩⎨⎧∂∂

+⎭⎬⎫

⎩⎨⎧∂∂

=∇ ABz

ABy

ABx

AB kji

⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

+∂∂

+⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

+∂∂

=

⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

+∂∂

+⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

+∂∂

=

⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

+⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

+⎭⎬⎫

⎩⎨⎧

∂∂

+∂∂

=

kjikji

kjikji

kji

zA

yA

xAB

zB

yB

xBA

zAB

yAB

xAB

zBA

yBA

xBA

zAB

zBA

yAB

yBA

xAB

xBA

( ) ( ) ( )ABBAAB ∇+∇=∇∴ Remember that in these results A and B are scalars. The operator ∇ acting on a vector ________________________________________________________________________

Has no meaning Example 1 If 22 xzyzxA += and 32 zzxyB −= evaluate ( )AB∇ at the point ( )3,1,2 We know that ( ) ( ) ( )ABBAAB ∇+∇=∇ At ( ),3,1,2 ....;............;.......... =∇=∇ AB ________________________________________________________________________

kjikji 161221;25123 ++=∇−+=∇ AB

( )kjikji 222 32 zxyxyzzyzB

yB

xBB −++=

∂∂

+∂∂

+∂∂

=∇

kji 25123 −+= at ( )3,1,2 ( ) ( )kjikji xzyxzxzxyz

zA

yA

xAA 22 222 −+++=

∂∂

+∂∂

+∂∂

=∇

( )3,1,2161221 atkji ++=

38

Now ( ) ( ) ( ) .............=∇+∇=∇ ABBAAB Finish it ________________________________________________________________________

( ) ( )kji 362361173 −+−=∇ AB

( ) ( ) ( )ABBAAB ∇+∇=∇ ( ),3,1,2at∴ 301812 =+=A

32 zzxyB −= ( ),3,1,2at∴ 21276 −=−=B ( ) ( ) ( )kjikji 161221212512330 ++−−+=∇∴ AB

( )kjikji 3623611731086108351 −+−=−+−= So add these to the list of results.

( )( ) ( )ABBAAB

BABA∇+∇=∇

∇+∇=+∇

Where A and B are scalars ________________________________________________________________________

Div (divergence of a vector function) The operator ⋅∇ (notice the ‘dot’; it makes all the difference) can be applied to a vector function ( )zyx ,,A to give the divergence of A, written in short as div A If kjiA zyx aaa ++=

div ( )kjikjiAA zyx aaazyx

++⋅⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=⋅∇=

za

ya

xadiv zyx

∂∂

+∂∂

+∂∂

=⋅∇=∴ AA

Note that (a) The grad operator ∇ acts on a scalar and gives a vector (b) The div operation ⋅∇ acts on a vector and gives a scalar Example 1 If kjiA 22 yzxyzyx +−= then .............=⋅∇= AAdiv ________________________________________________________________________

22 xzyzxA +=

39

yzxzxydiv 22 +−=⋅∇= AA We simply take the appropriate partial derivatives of the coefficients of i,j and k. It could hardly be easier. Example 2 If ( ) kjiA 22322 322 zyzyxyyx ++−= determine A⋅∇ i.e. div A. Complete it. ...............=⋅∇ A ________________________________________________________________________

0=⋅∇ A For ( ) kjiA 22322 322 zyzyxyyx ++−=

( )06644

6322422

22

=+−−=

++−=

∂∂

+∂∂

+∂∂

=⋅∇

zyzyxyxyzyzyxyxy

za

ya

xa zyxA

Such a vector A for which 0=⋅∇ A at all points, i.e for all values of zyx ,, is called a solenoid vector. It is rather a special case.

Curl (curl of a vector function) The curl operator denoted by ×∇ ,acts on a vector and gives another vector as a result. If kjiA zyx aaa ++= then curl AA ×∇=

i.e. curl ( )kjikjiAA zyx aaazyx

++×⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=×∇=

zyx az

ay

ax ∂

∂∂∂

∂∂=

kji

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

−∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

=×∇∴ya

xa

xa

za

za

ya xyzxyz kjiA

Curl A is thus a vector function. It is best remembered in its determinant form, so make a note of it. ________________________________________________________________________

40

Example 1 If ( ) ( ) ,222224 kjiA yzxyxzxy −++−= determine curl A at then point ( ).2,3,1 −

Curl

yzxyxzxyzyx222224 −+−∂∂

∂∂

∂∂

=×∇=

kji

AA

Now we expand the determinant

( ) ( ) ( ) ( )

( ) ( )⎭⎬⎫

⎩⎨⎧

−∂∂

−+∂∂

+

⎭⎬⎫

⎩⎨⎧ −

∂∂

−−∂∂

−⎭⎬⎫

⎩⎨⎧

+∂∂

−−∂∂

=×∇

22422

2242222

zxyy

yxx

zxyz

yzxx

yxz

yzxy

k

jiA

All that now remains is to obtain the partial derivatives and substitute the values of

zyx ,, ..................=×∇∴ A ________________________________________________________________________

kji 10682 −− { } { } { }322 4222 yxzxxyzzx −++−−−=×∇ kjiA ∴At ( ),2,3,1 − ( ) { } ( )10824122 −+−−=×∇ kjiA kji 10682 −−= Example 2 Determine curl F at the point ( )3,0,2 given that

( ) .2cos22 kjiF yxyxzze xy +++= In determinant form, curl .............=×∇= FF ________________________________________________________________________

yxyxzzezyx

xy 2cos22 +∂∂

∂∂

∂∂

kji

Now expand the determinant and substitute the values of x,y, and z, finally obtaining curl

.......................=F ________________________________________________________________________

41

Curl ( )kiFF 32 +−=×∇=

{ } { } { }xyxy xzeyzeyx 22 2cos21cos22 −+−−−=×∇ kjiF ∴at ( )3,0,2 ( ) ( ) ( )1261142 −+−−−=×∇ kjiF ( )kiki 3262 +−=−−= Every one is done in the same way.

Summary of grad, div and curl (a) Grad operator ∇ acts on a scalar field to give a vector field (b) Div operator ⋅∇ acts on a vector field to give a scalar field (c) Curl operator ×∇ acts on a vector fields to give a vector field. (d) With a scalar function ( )zyxo ,,/

grad kjizo

dyo

xooo

∂/∂+/∂+

∂/∂=/∇=/

(e) With a vector function kjiA zyx aaa ++=

(i) div za

ya

xa zyx

∂∂

+∂∂

+∂∂

=⋅∇= AA

(ii) curl

zyx aaazyx ∂∂

∂∂

∂∂

=×∇=

kji

AA

Check through that list, just to make sure. We shall need them all ________________________________________________________________________ By way of revision, here is one further example. Example 3 If 2322 yzyzxyxo −+=/ And kjiF xyzyzxy +−= 22 Determine for the point ( )2,1,1 −P (a) ,o/∇ (b)unit normal, (c) F⋅∇ (d) .F×∇ ________________________________________________________________________

42

Here is the working full. 2322 yzyzxyxo −+=/

(a) kjizo

yo

xoo

∂/∂+

∂/∂+

∂/∂=/∇

( ) ( ) ( )kji yzyxzzxyxyzxxy 2232 323222 −+−+++= ∴at ( )2,1,1 − kji 344 +−−=/∇o

(b) φφ

∇∇

=N 4191616 =++=/∇o

( )kjiN 344411

−+−

=∴

(c) kjiF xyzyzxy +−= 22 z

aya

xa zyx

∂∂

+∂∂

+∂∂

=⋅∇ F

xyzy +−=⋅∇∴ 22F

( )2,1,1 − 4141 −=−−=⋅∇ F 4−=⋅∇∴ F

(d)

xyzyzxyzyx

22 −∂∂

∂∂

∂∂

=×∇

kji

F

( ) ( ) ( )xyyzyxz 2002 −+−−+=×∇∴ kjiF ( ) kji xyyzyxz 22 −−+=

kjF 22 +=×∇ ( )kjF +=×∇∴ 2

Now let us combine some of these operations. ________________________________________________________________________ Multiple operations We can combine the operators grad, div and curl in multiple operations, as in the examples that follow. Example 1 If kjiA 332 zxyzyx −+=

Then div ( )kjikjiAA 332 zxyzyxzyx

−+⋅⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=⋅∇=

oxzxy /=++= 332 say

at∴

( ),2,1,1 −∴at

43

Then grad (div A) ( ) kjiAzo

yo

xo

∂/∂+

∂/∂+

∂/∂=⋅∇∇=

( ) ( ) ( )kji 22 3232 zxxy +++= i.e. grad div A ( ) ( ) kjiA 22 3232 zxxy +++=⋅∇∇= Move on for the next example ________________________________________________________________________ Example 2 If 2222 zxzyxyzo +−=/ determine div grad o/ at the point ( )1,4,2 First find grad o/ and then the div of the result. At ( )1,4,2 div grad ( ) ...................=/∇⋅∇=/ oo ________________________________________________________________________

div grad 6=/o We have 2222 zxzyxyzo +−−/

grad kjizo

yo

xooo

∂/∂+

∂/∂+

∂/∂=/∇=/

( ) ( ) ( )kji zxyxyyzxzxzyz 222 2242 +−+−++= ∴div grad ( ) 22 242 xzzoo +−=/∇⋅∇=/ ∴at ( ),1,4,2 div grad ( ) 6842 =+−=/∇⋅∇=/ oo Example 3 If kjiF zyxyzyzx 222 ++= determine curl curl F at the point ( ).1,1,2 Determine an expression for curl F in the usual way, which will be a vector, and then the curl of the result. Finally substitute values Curl curl .....................=F ________________________________________________________________________

Curl curl ( ) kjiFF 62 ++=×∇×∇=

For curl

zyxyzyzxzyx222∂∂

∂∂

∂∂

=

kji

F

( ) ( )kji zxyzyxxyzyz 22222 −++−=

44

Then curl curl

zxyzyxxyzyzzyx

22222 −−∂∂

∂∂

∂∂

=

kji

F

( ) ( )kji xzzxyxyyxzz 2222222 +−++−−−=

curl curl ( ) kjiFF 62 ++=×∇×∇=

________________________________________________________________________ Remember that grad, div and curl are operators and that they must act on a scalar or vector as appropriate. They cannot exist alone and must be followed by a function. One or two interesting general results appear. (a) curl grad o/ where o/ is a scalar

grad kjizo

yo

xoo

∂/∂+

∂/∂+

∂/∂=/

∴curl grad

zo

yo

xo

zyxo

∂/∂

∂/∂

∂/∂

∂∂

∂∂

∂∂=/

kji

0222222

=⎭⎬⎫

⎩⎨⎧

∂∂/∂

−∂∂/∂

+⎭⎬⎫

⎩⎨⎧

∂∂/∂

−∂∂/∂

−⎭⎬⎫

⎩⎨⎧

∂∂/∂

−∂∂/∂

=xy

oyx

ozx

oxz

oyz

ozy

o kji

( ) 0=/∇×∇=/∴ oogradcurl (b)Div curl A where A is a vector. kjiA zyx aaa ++=

zyx

zyx

aaacurl ∂

∂∂∂

∂∂=×∇=

kjiAA

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

−∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∂∂

=ya

xa

za

xa

za

ya xyxzyz kji

Then div curl ( ) ( )AkjiAA ×∇⋅⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=×∇⋅∇=zyx

0222222

=∂∂

∂−

∂∂∂

+∂∂

∂+

∂∂∂

−∂∂

∂−

∂∂∂

=zy

axz

azy

ayx

axz

ayx

a xyxzyz

( ),1,1,2at∴

45

div curl ( ) 0=×∇⋅∇= AA (c) Div grad o/ where o/ is a scalar

grad kjizo

yo

xoo

∂/∂+

∂/∂+

∂/∂=/

Then div grad ( )oo /∇⋅∇=/

⎟⎟⎠

⎞⎜⎜⎝

⎛∂/∂+

∂/∂+

∂/∂⋅⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

= kjikjizo

yo

xo

zyx

2

2

2

2

2

2

zo

yo

xo

∂/∂

+∂/∂

+∂/∂

=

∴div grad ( ) 2

2

2

2

2

2

zo

yo

xooo

∂/∂

+∂/∂

+∂/∂

=/∇⋅∇=/

________________________________________________________________________ This result is sometimes denoted by o/∇2 So these general results are (a) curl grad ( ) 0=/∇×∇=/ oo (b) div curl ( ) 0=×∇⋅∇= AA

(c )div grad ( ) 2

2

2

2

2

2

zo

yo

xooo

∂/∂

+∂/∂

+∂/∂

=/∇⋅∇=/

That brings us to the end of this particular programme. We have covered quite a lot of new material, so check carefully through the Revision Summary that follows; then you can deal with the test exercise.

∴

46

REVISION SUMMARY

If kjiCkjiBkjiA zyxzyxzyx cccbbbaaa ++=++=++= ;; then we have the following relationships. 1. Scalar product (dot product) θcosAB=⋅ BA ABBA ⋅=⋅ and ( ) CABΑCBA ⋅+⋅=+⋅ 2. Vector product (cross product) ( )NBA θsinAB×

=N unit normal vector where NBA ,, form a right-handed set

zyx

zyx

bbbaaakji

BA =×

( )ABBA ×−=× and ( ) CABACBA ×+×=+×

3. Unit vectors (a) 1=⋅=⋅=⋅ kkjjii 0=⋅=⋅=⋅ ikkjji (b) 0=×=×=× kkjjii 1=×=×=× ikkjji 4. Scalar triple product ( )CBA ×⋅

( )zyx

zyx

zyx

cccbbbaaa

=×⋅ CBA

( ) ( ) ( )BACACBCBA ×⋅=×⋅=×⋅

Unchanged by cyclic change of vectors. Sign reversed by non cyclic change of vectors. 5. Coplanar vectors ( ) 0=×⋅ CBA 6. Vector triple product ( )CBA ×× and ( ) CBA ×× ( ) ( ) ( )CBABCABA ⋅−⋅=×× C

47

And ( ) ( ) ( )ABCBACCBA ⋅−⋅=×× 7. Differentiation of vectors If zyx aaa ,,,A are functions of ,u

kjiduda

duda

duda

dUdA zyx ++=

8. Unit tangent vector T

duddud

A

A

T =

9. Integration of vectors duaduaduadu

b

a z

b

a y

b

a x

b

a ∫∫∫∫ ++= kjiA

10. Grad (gradient of a scalar function φ )

grad kjizyx ∂

∂+

∂∂

+∂∂

=∇=φφφφφ

‘del’= operator ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∇zyx

kji

(a) Directional derivative φφφ∇⋅=⋅= aa ˆˆ grad

dsd where a is a unit vector in a stated

direction. Grad φ gives the direction for maximum rate of changes of φ (b) Unit normal vector N to surface ( ) =zyx ,,φ constant

φφ

∇∇

=N

11. Div (divergence of a vector function A)

za

ya

xadiv zyx

∂∂

+∂∂

+∂∂

=⋅∇= AA

If za

ya

xa zyx

∂∂

+∂∂

+∂∂

=⋅∇= AA

48

12. Curl (curl of a vector function A )

zyx

zyx

aaacurl ∂

∂∂∂

∂∂=×∇=

kjiAA

13. Operators grad ( )∇ acts on a scalar and gives a vector

div ( )⋅∇ acts on a vector and gives a scalar curl ( )×∇ acts on a vector and gives a vector

14. Multiple operations (a) curl grad ( ) 0=∇×∇= φφ (b) div curl ( ) 0=×∇⋅∇= AA

(c) div grad ( ) .2

2

2

2

2

2

zyx ∂∂

+∂∂

+∂∂

=∇⋅∇=φφφφφ