Vector Differential Calculus Theory Unit V

Engineering Mathematics - I Semester – 1 By Dr N V Nagendram

UNIT – V Vector Differential Calculus Gradient, Divergence and Curl

1.01 Introduction

1.02 Vector Differentiation

1.03 Problems Exercise and solutions

1.04 Directional derivative, Gradient of a scalar function and conservative field

1.05 Divergence

1.06 Curl

1.07 Related properties of Gradient, Divergence and curl of sums

1.08 second order differential operator

1.09 Curvilinear coordinates: Cylindrical and spherical coordinates

1.10 Summary


UNIT – V Vector Differential Calculus Gradient, Divergence and Curl Class 1

1.1 Introduction

In this section, vector differential calculus is considered, which extends the

basic concepts of ordinary differential calculus to vector functions, by

introducing derivative of a vector function and the new concepts of gradient,

divergence and curl.

1.2 Vector Differentiation

1.2.1 Definition: Scalar function. Is a function of a scalar variable t is a

function F F(t) which uniquely associate a scalar F(t) for every value of the

scalar t in an interval [a, b.

1.2.2 Definition: Scalar field. Is a region in space such that for every point P

in this region, the scalar function f assoiates a scalar f(P).

1.2.3. Definition: Scalar function. Scalar function of a vector variable u is a

function F F(u ) which uniquely associates a scalar F(u ) for every vector u .

1.2.4 Definition: Vector function. Vector function of a scalar variable t is a

function F F (t) which uniquely associates a vector F for each scalar t.

1.2.5 Definition: Vector field. Is a region in space such that with every point

P in that region, the vector function V associates a vector V (P).

1.2.6 Definition: Derivative. Derivative of a function F (u) w.r.t. a scalar

variable u is denoted by and is defined as du

Fd

u

uFuuFLimu

)() (

0.

Let i , j , k be three mutually orthogonal unit vectors in the direction of the x,

y, z – coordinate axes such that i , j , k from a right handed triad that is

i. I 1, i.j 0, i.k 0, i.j 1….. etc.

1.2.7 Definition: Derivative in the component form.

Let F (u) F1(u) i F2(u) j F3(u) k in the component form with F1(u), F2(u),

F3(u) as components of F in the x,y, and z – ordinate axes.

Then du

Fd

du

Fd 1 i du

Fd 2 j du

Fd 3 k .

Thus the derivative of a vector function F w.r.t. a scalar variable u is the

vector whose components are the derivatives of the components F1, F2, F3 of

F w.r.t. u.

1.2.8 Results : Most of the basic rules of differentiation that are true for a

scalar function of a scalar variable hold good for vector function of a scalar

variable, provided the order of factors in vector products is maintained.

1.2.8.1 du

cd 0 where c is a constant vector

1.2.8.2 )()( uGuFdu

d

du

Fd

du

Gd

1.2.8.3 )()( uFudu

d

du

FduF )(

du

dF

1.2.8.4 )(.)( uGuFdu

d G

du

Fd. F .

du

Gd

1.2.8.5 )()( uGuFdu

d F

du

Gd G

du

Fd

1.2.8.6 )()(.)( uCuBuAdu

d BA .

du

Cd A .

du

BdC CB

du

Ad.

1.2.8.7 CBAdu

d

du

CdBA A

C

du

BdC CB

du

Ad.

1.2.9 Definition: Velocity and Acceleration.

Let r be the position vector of a point P(x(t), y(t), z(t)) in space where t is the

scalar time. Then r In the component form is r r (t) x(t) i y(t) j z(t) k .

Derivative of r denoted by r is defined as

du

rd r

t

trttrLimt

)()(

0

dt

dxi

dt

dyj

dt

dzk r and

..

r denote the velocity

and acceleration of a particle with position vector r .

1.2.10 Definition: Unit tangent vector. Let s be the arc length reckoned or

measured from a fixed point M0 of a space curve c whose equation is r r (s).

Then the unit tangent vector of c is ds

rd

ds

dxi

ds

dyj

ds

dzk such that

ds

rd

222

ds

dz

ds

dy

ds

dx 1.

1.2.11 Definition: Partial derivatives of a vector function F .

Partial derivatives of a vector function F which depends on more than one

scalar variables u, v, w. the partial derivate of F w. r. t. u is

u

F

u

wvuFuuFLimt

),,()(

0.

In the component form, if ),,( wvuF ),,(1 wvuF i ),,(2 wvuF j ),,(3 wvuF k .

Then the partial derivative of F w. r. t. say u is obtained by taking the partial

derivatives of the components 1F ,

2F ,3F of F w. r. t. u i.e.,

u

F

u

F

1 i u

F

2 j u

F

3 k .

Higher order partial derivatives can be obtained similarly.



1.3 Problems Vs Solutions

Problem 1.3.01: Prove that dt

AdA. 0 if A is a constant vector.

Solution: For any vector A , A . A A 2

On differentiation w.r.t. t

AAdt

d. A .

dt

Ad

dt

Ad. A 2A

dt

Ad and 2A

dt

Ad 2 A .

dt

Ad

If A is a vector of constant magnitude dt

Ad 0 so that A .

dt

Ad 0 is required

solution.

Problem 1.3.02: If A Cos xy i (3xy – 2x2) j (3x 2y) k then find x

A

,

y

A

,

2

2

x

A

,

2

2

y

A

,

yx

A

2

.

Solution: Given A Cos xy i (3xy – 2x2) j (3x 2y) k

x

A

y Sin xy i (3y – 4x) j 3 k

y

A

x Sin xy i 3x j 2 k

2

2

x

A

y2 Cos xy i - 4 j

2

2

y

A

x2Cos xy i

yx

A

2

( Sin xy xy Cos xy ) i 3 j is required solution.

Problem 1.3.03: If A t2 i -t j (2t 1) k , B ( 2t – 3 ) i j - t k find (a)

t

BA

.

(b)

t

BA

(c)

t

BA

(d)

t

BA

t at t 1.

Solution: (a) To find

t

BA

.

Let BA. t2( 2t – 3 ) t t (2t 1)

dt

BAd . 6 t2 6 t – 1 – 4 t – 1 |at t 1 6

(b) BA

tt

ttt

kji

132

)12(2 ( t2 2t – 1 ) i ( t3 4 t2 – 4t – 3 ) j (3t2 – 3t) k

dt

BAd (2t – 2 ) i ( 3t2 8 t – 4 ) j ( 6t – 3 ) k at t 1,

dt

BAd 7 j 3 k

To find (c)

dt

BAd

Let BA (t2 2t – 3 ) i (1 – t ) j (t 1) k

| BA | 2222 )1()1() 3 -2t (t tt 1121 - 4t (t 34 t

dt

BAd ||

1121 - 4t (t2

12124

34

23

t

tt at t 1 is 1.

(d) To find

t

BA

t at t 1 for that let

dt

Bd 2 i 0 k

t

BA

tt

ttt

kji

132

)12(2 t i ( t2 4 t – 2 ) j 2t k

t

BA

dt

d i ( 2t 4 ) j 2 k at t 1 is i 6 j 2 k is required solution.

Problem 1.3.04: Find the magnitude of the velocity and acceleration of a

particle which moves along the curve x 2 Sin 3t, y 2 Cos 3t, z 8t at any

time t > 0. Find unit tangent vector to the curve.

Solution: the position vector r of the particle is r (t) x(t) i y(t) j z(t) k

r (t) 2 Sin 3t i 2 Cos 3t j 8t k

Velocity V t

r

6 Cos 3t i – 6 Sin 3t j 8 k

Acceleration a V 2

2

t

r

– 18 Sin 3t i – 18 Cos 3t j 0 k

|V | 64336336 22 tSintCos 64 36 10

| a | tCostSin 318318 2222 218 18

Unit tangent vector

dt

rd

dt

rd

kjtSinitCos 83636[10

1 is required solution.



1.3 Exercise Try Urself……

Problem 1.3.05: Prove that

dt

BAd C ( A B ) if

du

Ad C A ;

du

Bd C B

Problem 1.3.06: Find the angle between the directions of the velocity and

acceleration vectors at time t of a particle with position vector r (t2 1) i 2t j

(t2 1) k .

Problem 1.3.07: If a, b, w are constants show that the acceleration of a

particle with displacement vector r a Cos t b Sin t is always directed

towards the origin.

Problem 1.3.08: A particle moves along the curve x 2t2, y t2 – 4t, z 3t – 5

where t is the time. Find the components of its velocity and acceleration at time

t 1 in the direction i 3 j 2 k .

Problem 1.3.09: Determine the magnitude of velocity and acceleration at t 0

of a particle moving along a curve whose parametric equations are x e- t , y

2 Cos 3t, z 2 Sin 3t where t is the time.

Problem 1.3.10: Find the angle between the tangents to the curve r t2 i 2 t

j t3 k at the points t 1.

Problem 1.3.11: Given the curve x t2 2, y 4t 5, z 2t2 – 6t find unit

tangent vector at the point t 2.

Problem 1.3.12: Prove that A dt

Ad 0 if A (t) has constant or fixed direction.

Problem 1.3.13: Find

yx

BA

2

at (1,0,2) if A x2y z i 2 xz3 j xz2 k ,

B 2 z i y j x2 k .

Problem 1.3.14: Find

yx

Af

2

3

at the point (2, - 1, 1) if f xy2z,

A xz i - xy2 j yz2 k .

Problem 1.3.15: If A (2x2 y – x4) i (exy – y Sin x) j (x2 Cos y) k then find

(i) x

A

(ii)

y

A

(iii)

2

2

x

A

(iv)

2

2

y

A

(v)

yx

A

2

(vi) xy

A

2

Problem 1.3.16: If A Sin u i Cos u j u k , B Cos u i Sin u j 3 k

and C 2 i 3 j k then find )( CBAdu

d at u 0.

Problem 1.3.17: If A 2t i t2 j t3 k , B t i t2 k and C t3 j 2t k

then find CBAdu

d. at t 1.

Problem 1.3.18: Find (i) BAdu

d. (ii) BA

du

d (iii) AA

du

d. .

Problem 1.3.19: If A 5t2 i t j t3 k , B Sin t i Cos t j then find (i)

BAdt

d. (ii) BA

dt

d (iii) AA

dt

d.



1.3 Exercise Vs Answers …….

1.3.05: **** 1.3.06: Cos-1

12

2

2t

t 1.3.07: ***

1.3.08:

7

148;

7

14

1.3.09: V - e- t i - 6 Sin 3t j 6 Cos 3t k ; a e- t i - 18 Cos 3t j 18 Sin 3t k ;

|V | at t 0 is 37 ; | a | at t 0 is 325.

1.3.10: Cos -1(9/17)

1.3.11:

13125

)32(2

2 tt

ktjit;

3

22 kji

1.3.12: ****

1.3.13: 4 i 8 j

1.3.14: 4y2z i - 2 y4 j ; 4 i - 2 j

1.3.15: (i) (4xy – 4x3 ) i + (yexy – y cos x) j + 2x Cos y k

(ii) 2x2 i + (x exy – Sin x) j - x2 Sin y k

(iii) (4y – 12 x2) i +(y2 exy + y Sin x) j + 2 Cos y k (iv) 0 + x2 exy j - x2 Cos y k

(v), (vi) 4x i +(xy exy + exy – Cos x) j - 2 x Sin y k

1.3.16: 7 i + 6 j 6 k

1.3.17: - 12 t5 + 8 t3 – 7 t6, 11

1.3.18: 2 Sin u + 2u cos u – 4 u3; 9u2 i +(u3 Cos u + 3 u2 Sin u + 4u) j + (3 u2

Cos u + 6u Sin u ) k

1.3.19: (i) (5t2 – 1 ) Cos t + 11 t Sin t

(ii) (t3 Sin t –3 t2 Cos t ) i (t3 Cos t + 3t2 Sin t) j +(5 t2 Sin t –11t Cos t – Sin t ) k

(iii) 100 t3 + 2t + 6t3.



1.4 Directional derivative, Gradient of a scalar function and conservative field

1.4.1 Introduction.

In vector differential calculus, it is very convenient to introduce the symbolic

linear vector differential “Hamiltonian” operator del defined and equation

denoted as below = ix

+ j

y

+ k

z

…………………………………………….. (1)

this operator read as del or nabla is not a vector neither has magnitude nor

direction but combines both differential and vectorial properties analogous to

those of ordinary vectors.

1.4.2 Definition: Directional derivative. If f = f(x, y, z) the the partial

derivatives x

f

,

y

f

,

z

f

are the derivatives rates of change of f in the “direction”

of the co-ordinate axes OX, OY, OZ respectively. This concept can be extended

to define a derivative of f in a “given” direction PQ as shown below:

S

P b O G

Figure

Let P be a point in space and b be a unit vector from P in the given direction.

Let s be the arc lengths measured from P to another point Q along the ray C in

the direction of b . Now consider f(s) = f(x, y, z) = f((s), y(s), z(s)).

Then ds

df=

x

f

ds

dx+

y

f

ds

dy+

z

f

ds

dz……………………….………………………………. (2)

The directional derivative of f at the point P in the given direction b isds

df given

by (2). ds

df gives the rate of change of f in the direction of b . Since

ds

dxi +

ds

dyj +

ds

dzk = b = Unit Vector ……………………………………………………. (3)

Using the del operator defined by (1) ds

df given by (2) can be written as below:

ds

df =

z

fk

y

fj

x

fi .

k

ds

dzj

ds

dyi

ds

dx

ds

df =

zk

yj

xi f. b = f. b …………………………………………….……… (4)

Thus the directional derivative of f at P is component or dot product of f in

the direction of unit vector b .

The directional derivative in the direction

of any non-unit vector a is ds

df=f.

a

a.............................................................................. (5)

Equation (4) introducs the vector quantity, the gradient of a scalar function

f(x, y, z) or gradient f denoted by f and defined as below

f =

zk

yj

xi f =

z

fk

y

fj

x

fi = grad f = Vector

f =

z

f

y

f

x

f,, .

1.4.3 Note: Properties of Gradient.

1.4.3.1 Projection of f in any direction is equal to the derivative of f in that

direction.

1.4.3.2 The gradient of f is in the direction of the normal to the level surface

f(x, y, z) = C = constant. So the angle between any two surfaces f(x, y, z) = C1

and g(x, y, z) = C2 is the angle between their corresponding normals given by f

and g respectively.

1.4.3.3 The gradient at P is in the direction of maximum increase of f at P.

1.4.3.4 The modulus of the gradient is equal to the largest directional

derivative at a given point P.

i.e., max Patds

df = | f | Pat = Pat

z

f

y

f

x

f222

.

These properties thus state that the vector gradient f indicates the direction

and magnitude of maximum change of scalar function f at a given point.

1.4.3.5 Definition: Normal derivative. dn

df= f. n where n is the unit normal

to the surface f = constant.

1.4.3.6 Definition: Conservative. A vector function A is said to be a

conservative vector field if A = f i.e., A is the gradient of a scalar function f.

In this case f is known as the potential function of A .

Instead of applying or operating on a scalar function f, if del is applied to a

vector function A , we get divergence and curl.



1.4.5 Problems Vs Solutions on Directional derivative, Gradient of a scalar

function and conservative field Try Urself…..

1.4.5.01 Problem: Find f(r) such that f = 5r

r and f(1) = 0

1.4.5.02 Problem: If f = (y2 – 2xyz3 ) i +(3+2xy – x2z3) j +(6z3-3x2yz2) k

1.4.5.03 Problem: Find the constants a and b so that the surface

ax2 – b yz = (a + 2) x will be orthogonal to the surface 4x2y + z3 = 4 at the point

(1, -1, 2).

1.4.5.04 Problem: Find the values of the constants a, b, c so that the

directional derivative of f = axy2 + byz + cz2x3 at (1, 2, -1) has a maximum of

magnitude 64 in a direction parallel to the z-axis.

1.4.5.05 Problem: Find the directional derivative of f(x, y, z) = 4 e2x-y+z at the

point (1, 1, -1) in the direction towards the point (-3, 5, 6).

1.4.5.06 Problem: Evaluate (i) nr (ii) 3

r (iii) (3 r 2 - 4r +6r-1/3) (iv) r

(v) Log r) (vi) 1r

1.4.5.07 Problem: If A = 2x2 i - 3 yz j +xz2 k and f = 2z – x3y find A. f and

(ii) A f at the point (1, -1, 1).



1.4.6 Exercise on Directional derivative, Gradient of a scalar function and

Conservative field Try yourself…..

1.4.6.01 Problem: Prove that

g

f =

2g

gffg

[Hint: Use quotient formula for derivative of

g

f.

1.4.6.02 Problem: (i) Determine f when f = (zx i + xz j -xy k )/z2;

(ii) If f = xy i + 2xy j then find f.

1.4.6.03 Problem: Find f given f = 2x i + 4y j + 8z k

1.4.6.04 Problem: If f = 2xyz3 i + x2z3 j + 3x2yz2 k then find f(x, y, z) if

f(1, -2, 2) = 4.

1.4.6.05 Problem: Calculate the angle between the normals to the surface

2x2 + 3y2 = 5z at the points (2, -2, 4) and (-1, -1, 1).

1.4.6.06 Problem: Determine the angle between the normals to the surface

xy = z2 at the points (4, 1, 2) and (3, 3, -3).

1.4.6.07 Problem: Find the angle of intersection of the spheres x2 + y2 + z2 =

29 and x2 + y2 + z2 + 4x – 6y – 8z = 47 at (4, -3, 2).

1.4.6.08 Problem: Find the angle between the surfaces x2 + y2 + z2 = 9 and z =

x2 + y2 – 3 at the point (2, 1, -2).

1.4.6.09 Problem: In what direction from (3, 1, -2) is the directional derivative

of f = x2 y2 z4 maximum. Find also the magnitude of the maximum.

1.4.6.10 Problem: Find the direction in which temperature changes most

rapidly with distance from the points (1, 1, 1) and determine the maximum rate

of change if the temperature at any point is given by f(x, y, z) = xy + yz + zx.

1.4.6.11 Problem: (i) In what direction from the point (2, 1, -1) is the

directional derivative of f = x2yz3 a maximum? (ii) what is the magnitude of this

maximum?

1.4.6.12 Problem: Find the maximal directional derivative of x3y2z at (1, -2, 3).

1.4.6.13 Problem: Determine the directional derivative of f = xy2 + yz3 at the

point (2, -1, 1) in the direction of the vector i + 2 j + 2 k .

1.4.6.14 Problem: Determine the directional derivative of f = xy + yz + zx at the

point (1, 2, 0) in the direction of the vector i + 2 j + 2 k .

1.4.6.15 Problem: Determine the directional derivative of f = x2 yz + 4xz2 at the

point (1, -2, -1) in the direction of the vector 2 i j 2 k .

1.4.6.16 Problem: What is the the directional derivative of f = xy2 + yz3 at the

point (2, -1, 1) in the direction of the normal to the surface x Log z – y2 = 4 at

point (-1, 2, 1).

1.4.6.17 Problem: Determine a unit vector normal to the surface xy3z2 = 4 at

the point (-1, -1, 2).

1.4.6.18 Problem: Find the unit outward drawn normal to the surface (x – 1)2

+ y2 + (z + 2)2 = 9 at the point (3, 1, -4).

1.4.6.19 Problem: Find a unit normal to the surface x2y + 2xz = 4 at the point

(2, -2, 3).

1.4.6.20 Problem: If U = 3x2 y, V = xz2 – 2y evaluate [U.V]

1.4.6.21 Problem: Find f when f = (x2 + y2 + z2) e222 zyx

1.4.6.22 Problem: If f = 2xz4 – x2y, find f and | f | at (2, -2, -1).

1.4.6.23 Problem: If f(x, y, z) = 3x2y – y3z2, find f and | f | at (1, -2, -1).

1.4.6.24 Problem: Find f if f = Log (x2 + y2 + z2).



1.05.1 Divergence

1.05.1.1 Definition: Divergence. Divergence of a vector function A (x, y, z) is

written as divergence of A or div of A and denoted by . A is defined as

. A =

zk

yj

xi ( A ) .

If A =A1 i + A2 j + A3 k , . A =

zk

yj

xi (A1 i + A2 j + A3 k )=

z

A

y

A

x

A 321

. A = is a scalar quantity.

1.05.1.2 Note: . A A . because L.H.S. . A is a scalar quantity whereas

the R.H.S. A . =

zA

yA

xA 321 is a scalar differential operator.

1.05.1.3 Note: The divergence of A at a point P constitutes the volume density

of the flux of A at P i.e., divergence measures outflow minus inflow.

1.05.1.4 Definition: Source or sink. A point P in a vector field A is said to be

source (or sink) if divergence A i.e. A > (or <) 0.

1.05.1.5 Definition: Solenoidal function. A is said to be solenoidal if

divergence A = 0 at all points of function.



1.05.2 Problems Vs Solutions on Divergence

1.05.2.1 Problem: Find (. A ) where A = r

r.

1.05.2.2 Problem: if f and g are solutions of the Laplace equation show that

. )( fggf = 0.

1.05.2.3 Problem: Find the directional derivative of . U at the point (4, 4, 2)

in the direction of the corresponding outer normal of the sphere x2 + y2 + z2 =

36 where U = x4 i + y4 j + z4 k .

1.05.2.4 Problem: .( rr 3 )

1.05.2.5 Problem: Determine the constant b such that A = (x + 3y) i + (y – 2z)

j + (x + bz) k is solenoidal.

1.05.2.6 Problem: Evaluate divergence of (2 x2z i x y2z j +3y z2 k ) at the

point (1, 1, 1).



1.05.3 Exercise Problems on Divergence ….Try your self

1.05.3.01 Problem: Prove that . r = 3


UNIT – V Vector Differential Calculus - Curl Class 10

1.06 Curl:

1.06.01 Definition: Curl. Curl of A , denoted by A also known as rotation

V or not of V is defined as Curl of A = A

=

zk

yj

xi (A1 i + A2 j + A3 k )

=

321 AAA

zyx

kji

=

z

A

y

Ai 23 +

x

A

z

Aj 31 +

y

A

x

Ak 12

A = is a vector quantity.

1.06.01 Definition: Irrotational field. A vector point function A is said to be

Irrotational, if curl of A is zero at every point where A is defined. Otherwise it

is said to be rotational. The curl of any vector point function. In general, gives

the measure of the angular velocity at any point of the vector field.


UNIT – V Properties Class 11

1.7 Related Properties of Gradient, Divergence and Curl of sums:


1.7.01 Definition: The gradient, divergence and curl are distributive w.r.t. the

sum and difference of functions.

01. (f g) = f g

02. ( A B ) = (. A ) ( . B )

03. ( A B ) = ( A ) ( B )

The above results follow, since derivative of sum or difference of scalars or

vectors is sum or difference of the derivatives of scalars or vectors.

1.7.02 Example:.( A B ) = .( )11 BA ) i .( )22 BA j .( )33 BA k

= )( 11 BAx

+ )( 22 BA

y

+ )( 33 BA

z

=

z

A

y

A

x

A 321

z

B

y

B

x

B 321

= . A . B

1.7.03 Gradient, Divergence and Curl of products.

1.7.03.1 Note: (f g) = fg gf

1.7.03.2 Note: .(f A ) = (f). A (f). A

1.7.03.3 Note: (f A ) = (f) A (f) A

1.7.03.4 Note: ( A . B ) = ( B .) A ( A .) B + B ( A ) A ( B )

1.7.03.5 Note: .( A B ) = B .( A ) A .( B )

1.7.03.6 Note: ( A B ) = ( B . ) A B (. A )( A .) B + A (. B )

1.7.03.7 Note: (f A ) =(f ( kAjAiA 321 ))=(f kAfjAfiA 321 )

=

321 fAfAfA

zyx

kji

=

z

fA

y

fAi 23

z

fA

x

fAj 13 +

y

fA

x

fAk 12

Expanding the product of the derivatives and rearranging the terms, we get

=f

y

fA

x

fAk

z

fA

x

fAj

z

fA

y

fAi 121323 +

z

fA

y

fAi 23

z

fA

x

fAj 13 +

y

fA

x

fAk 12

=(f A ) = (f) A (f) A

1.7.03.08 Problem: Prove that .( A B ) = B .( A ) A .( B )

1.7.03.09 Problem: P.T ( A B ) = ( B . ) A B (. A )( A .) B + A (. B )

1.7.03.10 Problem: P.T ( A . B )= A ( B )+ B ( A ) +( A .) B +( B . ) A


UNIT – V Second Order Differential Operator Class 12


If it is a two-fold application of the operator to function.

1.8.1 Definition: Laplacian operator 2

Div grad f = (f ) =

z

fk

y

fj

x

fi =

x

x

f+

y

y

f+

z

z

f

=

2

2

x

f+

2

2

y

f+

2

2

z

f

=

2

2

2

2

2

2

zyxf

= 2f = f

Div grad f = (f ) =2f = f.

Thus the scalar differential operator read as “nabla squared” or “delta”.

2 = =

2

2

2

2

2

2

zyx is known as Laplacian operator.

1.8.2 Note: f = div grad f = 2f = f =

2

2

2

2

2

2

z

f

y

f

x

f

1.8.3 Note: f = curl grad f = 0

1.8.4 Note: ( A ) = div curl A = 0

1.8.5 Note: ( A ) =curl curl A = ( A ) - 2 A

1.8.6 Note: ( A ) =grad div A = ( A ) + 2 A

1.8.6 Note: Table for the possible combinations of second order differential

operators are tabulated below:

Operator Scalar field Vector Field

Grad Div curl

Grad *** Grad div A ***

Div Div grad f = f *** Div curl A

Curl Curl grad f = 0 *** Curl curl A =

Grad div A = A

1.8.7 Example: Prove that f = 0 for any scalar function f.

1.8.8 Example: Prove that (f g g f ) = 0

1.8.9 Example: P.T (f g) = f g = (g f ) and deduce that (f f).

1.8.10 Example: Show that ( A ) = ( A ) - 2 A .

1.8.11 Example: Prove that ( A ) = 0 for any vector function A .

1.8.12 Example: Prove that (f A ) = f ( A ) for any vector function A .


UNIT – V Second Order Differential Operator Class 13

1.8.13 Problems on 2nd order differential operator – Laplacian 2.

1.8.13.1 Problem: Show that f is both solenoidal and Irrotational if 2f = 0

1.8.13.2 Problem: If F = ra r prove that 2 F = a (a + 3) rn-2 F

1.8.13.3 Problem: Find 2f when f = 4x2 + 9y2 + z2

1.8.13.4 Problem: Find 2f at the point (2, 3, 1) when f = xy/z

1.8.13.5 Problem: Show that .(f) =

2

2

2

2

2

2

z

f

y

f

x

f

1.8.13.6 Problem: Evaluate 2

2.

r

r

1.8.13.7 Problem: Find the directional derivative of (f) at the point (1, -2, 1)

in the direction of the normal to the surface xy2z = 3x + z2 where f = 2x3y2z4.

1.8.13.8 Problem: Prove that 2(fg) = f 2g + 2g.f + g2f

1.8.13.9 Problem: Prove that (i) 2rn = n(n+1)rn-2 where n is a constant (ii) 2r2

= 6 (iii) 2

r

1 = 0 (iv) 2( Log r) =

2

1

r (v) 2(gh) = g2(h) + h2(g)

1.8.13.10 Problem: Prove that (i) 2 f(r) =

dr

df

rdr

fd 22

2

(ii) Find f(r) such that

2 f(r) = 0.

1.8.13.11 Problem: Calculate 2f when f = 3x2z – y2z3 + 4x3y + 2x – 3y -5 at

the point (1, 1, 0).


UNIT – V Curvilinear co-ordinates Class 14

1.9 Curvilinear coordinates: Cylindrical and spherical coordinates

1.9.1 Definition: Curvilinear co-ordinates.

It is more convenient in many problems to define the position of a point P in

space by three numbers (q1, q2, q3 ) instead of the three Cartesian co-ordinates

(x, y, z). Then q1, q2, q3 are known as curvilinear co-ordinates of the point P.

1.9.2 Definition: co-ordinate surfaces. The three surfaces q1 = c1, q2 = c2,

q3 = c3 where c1, c2, c3 are constants are known as co-ordinate surfaces of the

system of curvilinear co-ordinates. On these co-ordinates surfaces, say q1 = c1

one of the co-ordinates, here q1, remains constant.

1.9.3 Note: The co-ordinate curves or lines or axis are the curves or lines of

intersection of any two co-ordinate surfaces. Thus on the co-ordinate curve say

which is the intersection of q2 = c2 and q3 = c3 , only q1 varies, while q2 and q3

remain constant.

1.9.4 Note: suppose the rectangular co-ordinates (x, y, z) of any point P in

space be expressed as functions of (q1, q2, q3) so that

x = x(q1, q2, q3), y = y(q1, q2, q3) and z = z(q1, q2, q3) …………………………… (1)

on solving (1) for q1, q2, q3 in terms of x, y, z we get

q1 = q1(x, y, z) , q2 = q2 (x, y, z) and q3 = q3(x, y, z)…………………………….. (2)

the set of equations (1) and (2) are known as “transformation of co-ordinates”.

3

Z q3 Curve

q1 q2=c2 q1 = c1

Curve P

2

1 q3 = c3 q2

Curve

O Y

X

1.9.4 Definition: Orthogonal curvilinear system of co-ordinates. If the co-

ordinate surfaces intersect at right angles and therefore lines are at right

angles, then the curvilinear co-ordinate system is known as “orthogonal

curvilinear system of co-ordinates”.

1.9.5 Note: let 1e ,

2e ,3e be unit vectors directed along the tangents to the co-

ordinates axes q1, q2, q3 at the point P in the direction of increasing q1, q2, q3

respectively, such that 1e ,

2e ,3e form a right – handed trihedral (triad).

1.9.6 Example: Express the velocity v and acceleration a of a particle in

spherical co-ordinates.

1.9.7 Example: Represent the vector A = 2y i - z j + 3x k in spherical co-

ordinates and determine rA , A , A .

1.9.8 Example: (i) Find the unit vectors pe ,

0e , e in spherical co-ordinate

system in terms of i , j , k . (ii) Solve for i , j , k in terms of pe ,

0e , e .

1.9.9 Example: Find the Laplacian of f(, , z) = 2 + z23 + - .

1.9.10 Example: Find all solutions of the Laplace’s equation 2f = 0 that

depend solely on the distance .

1.9.11 Example: Find the potential of A = ree

1+

ee

Sinr

log+

ee

log

given in spherical co-ordinates.

1.9.12 Example: Show that the vector field A in spherical co-ordinates

A = rer

Cos3

2 +

e

r

Sin3

is solenoidal.

1.9.13 Example: Compute the curl of A specified in cylindrical co-ordinates

where A = eSin + zezeCos

.

1.9.14 Example: Compute the gradient of the scalar field f = + z Cos

specified in cylindrical co-ordinates (, , z).

1.9.15 Example: A vector field is given in cylindrical co-ordinates as

A (P) = + e . Find the vector lines of the field.

1.9.16 Example: Find the square of the element of arc length in cylindrical co-

ordinates and determine the corresponding scale factors.

1.9.17 Example: Prove that curl of gradient of f = 0 in any orthogonal

curvilinear co-ordinate system.

1.9.18 Example: Prove that the surface area of the given region R of the

surface r = r (u, v) is dvduFEGR

2 . Use this to determine the surface area

of a sphere.

1.9.19 Example: Find 1q

r

,

2q

r

,

3q

r

,q1, q2, q3 in cylindrical co-ordinates.

1.9.20 Example: Find the Jacobian J

321 ,,

,,

qqq

zyx for (i) cylindrical (ii) spherical

co-ordinates.

1.9.21 Example: Find the Jacobian of x, y, z w.r.t. the orthogonal curvilinear

co-ordinates q1, q2, q3.

1.9.22 Example: Prove that a spherical co-ordinate system is orthogonal.

1.9.23 Example: Express (i) f, (ii) . A (iii) A and (iv) 2f in spherical

curvilinear co-ordinates.

1.9.24 Example: Express (i) f, (ii) . A (iii) A and (iv) 2f in cylindrical co-

ordinates (r, , z).

1.9.25 Example: Express 2f in orthogonal curvilinear co-ordinates, Deduce

for Cartesian system.

1.9.26 Example: Derive an expression for A in orthogonal curvilinear co-

ordinates. Deduce A for Cartesian co-ordinates.

1.9.27 Example: Derive an expression for . A in orthogonal curvilinear co-

ordinates. Deduce . A in rectangular co-ordinates.

1.9.28 Example: Show that 1e = h2h3q2 q3.

1.9.29 Example: Derive an expression for f in orthogonal curvilinear co-

ordinates. Deduce in rectangular Cartesian co-ordinates.

1.9.30 Example: Spherical co-ordinates q1 = r, q2 = , q3 = we get h1 = Hr = 1,

h2 = H = r, h3 = H = r Sin .

1.9.31 Example: In rectangular co-ordinate system (q1, q2, q3) is replaced by

(x, y, z). Here h1 = Hr = h2 = h3 =1, 1e = i ,

2e = j ,3e = k .

1.9.32 Example: Spherical co-ordinates

q1 = r 0 r ; q2 = 0 ; q3 = 0 2

the co-ordinate surfaces are

r = c1, spheres centered at 0; = c2, circular half-angle cones with z-axis with

vertex at origin ; = c3 half planes adjoining th z-axis through z-axis.

Co-ordinate lines are

r : rays emanating from origin 0

: meridians on a sphere

: parallel on a sphere

Z line r

=c2

Line

r = c1 0

Y

= c2

line

X Figure

Cartesian co-ordinates are related to spherical co-ordinates as x = r cos sin ;

y = r sin cos ; z = r cos .

O


UNIT – V Vector Differential Calculus Class 15

1.10 Summary.

1.10.1 Note: Vector differential calculus extends the differential calculus of

scalar functions to the vector functions, with all the basic rules and formulas

quite similar.

1.10.2 Note: Derivative of a vector function in component form is obtained by

taking derivatives of the components i.e., kFjFiFF 321 .

1.10.3 Note: Then kdu

Fdj

du

Fdi

du

Fd

du

Fd 321 .

1.10.4 Note: A similar definition for partial derivative is obtained by taking the

partial derivatives of the components w.r.t the corresponding independent

variable ku

Fj

u

Fi

u

F

u

F

321 .

1.10.5 Note: The del or nabla operator is = ix

+ j

y

+ k

z

1.10.6 Note: Gradient of a scalar function f is a vector grad f = f =

ix

f

+ j

y

f

+ k

z

f

= vector gives the direction of maximum increase of f.

1.10.7 Note: f is the normal to the surface f(x, y, z) = ‘c’ directional derivative

of f at point p in the direction of a vector a is a

af

ds

df. .

1.10.8 Note: Divergence of a vector function A is a scalar div A = . A =

z

A

y

A

x

A

321 = scalar

1.10.9 Note: when . A =0 the vector field A is solenoidal. Curl of a vector

function A is a vector curl A = A =

321 AAA

zyx

kji

= vector

1.10.10 Note: related properties of sums, difference and products of gradient,

divergence and curl are quite similar to the differential calculus properties,

keeping the order intact.

1.10.11 Note: In the second order differential operators 2f = .f =

2

2

2

2

2

2

z

f

y

f

x

f

and curl (f) = 0 ; div(curl A ) = 0

1.10.12 Note: Expressions for grad, div, curl and Laplacian in curvilinear,

cylindrical and spherical co-ordinates can be presented as per our convenience.

Vector Differential Calculus Theory Unit V

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