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Transcript
```1
Vector Calculus
1.1
SCALAR
A scalar is a quantity that has only magnitude, such as distance, mass, temperature or
time. The value of a scalar is an ordinary number. Operations with scalars obey the rules
of elementary algebra. The order of compounding is immaterial.
1.2
VECTORS
A vector is a quantity that has both magnitude and direction, such as displacement, force,
velocity or momentum. A vector is represented either by a bold-faced letter such as P or
Æ
by an arrow over the head of a letter such as P . Graphically, a vector
B
is represented by the directed line segment AB as in Fig. 1.1. The
Æ
vector P has a direction from A to B. The point A is called the initial
P
point (tail of the arrow) and the point B is called the terminal point
ææÆ
Æ
of P (head of the arrow). The length | AB | of the line segment is the
magnitude of P. The magnitude of P is denoted either by P or |P|
|P| > 0 for any P π 0
|P| = 0 if and only if P = 0. It is called a Null Vector. Vectors
are compounded geometrically. The order of compounding is
immaterial.
1.3
EQUIVALENCE OF VECTORS
Two vectors P and Q are equal if they have the same magnitude and
direction regardless of the position of their initial points. Thus, in
Fig. 1.2, P = Q.
Commutative property: If A = B, then B = A
Transitive property: If A = B and B = C, then A = C.
A
Fig. 1.1
P
Q
Fig. 1.2
2 Mechanics of Particles, Waves & Oscillations
1.4
1.4.1
MULTIPLICATION OF VECTORS BY SCALARS
Scalar Multiplication
Let A be any vector and m any scalar. Then the vector mA
as in Fig. 1.3 is defined as follows:
(a) The magnitude of mA is |m||A|; that is |mA|
= |m||A|
(b) If m > 0, the direction of mA is that of A.
(c) If m < 0, the direction of mA is opposite to that of A.
Fig. 1.3
(d) m (nA) = (mn) A for any scalars m, n.
Two non-zero vectors A and B are parallel if and only if there exists a scalar m such
that B = mA
Thus, the result of multiplying a given vector by a scalar is a vector parallel to the
given vector.
1.4.2
Result of Scalar Multiplication
(a) The zero vector
if m = 0, we obtain the zero vector 0; i.e. (0) A = 0, where A is any vector. The zero
vector 0 is a vector that has magnitude 0 and has any direction.
(b) The negative of a vector
If m = –1, we obtain the negative of vector A, indicated by –A; that is, (–1) A = –A.
Thus, a negative vector –A is a vector with magnitude of A, but whose direction is
opposite of that of A.
(c) The unit vector
If A π 0 and m = 1/|A| = 1/A, the unit vector, eA =
1
A.
|A|
Thus, the vector eA is a vector whose magnitude is |eA| = 1, with the direction same
as that of A. The vector A may be represented by the product of its magnitude and the
unit vector eA; that is;
A = AeA
1.5
The sum or resultant of two vectors A and B is a vector C
which can be determined geometrically, Fig. 1.4. If the tail
of B is placed at the head of A, the resultant C is the vector
whose tail is at the tail of A and whose head is at the head
of B. This geometric method of vector addition is known as
the triangle law.
C
B
A
Fig. 1.4
Vector Calculus
1.6
3
SUBTRACTION OF VECTORS
If A and B are two vectors, the difference (A – B) is the sum
C, of A and (–B); that is:
C = A – B = A + (–B)
The vector subtraction is shown geometrically in Fig. 1.5.
1.7
Fig. 1.5
A+ B= B+ A
(Commutative law)
A + (B + C) = (A + B) + C
(Associative law)
m(A + B) = mA + mB
(Distributive law)
(m + n) A = mA + nA
(Scalar distributive law)
A+ 0= A
A– A= 0
z
1.8
RECTANGULAR UNIT VECTORS i, j, k
^
Consider a right-handed, rectangular co-ordinate system.
j, k are shown in the directions of x,
The unit vectors i,
y and z axes respectively, Fig. 1.6.
A vector A in three dimensions, can be represented by
its initial point at the origin 0 and rectangular components
(A1, A2, A3) for its terminal point, Fig. 1.7. Since the
resultant of A1 i + A2 j + A3 k , is the vector A,
A = A i + A j + A k ,
1
2
k
^
j
x
Fig. 1.6
3
z
The magnitude of A is
A = |A| =
A
A12 + A22 + A32
Let A = ax i + ay j + az k and B = bx i + by j + bz k be two
vectors where ax, ay, az are the x, y and z components
respectively, and bx, by, bz, are the corresponding components of B. Then the resultant of A and B is given by:
R =
y
0
^
i
2
2
( a x + bx ) + (a y + by ) + ( a z +
^
A3 k
^
A1 i
y
0
^
A2 j
x
Fig. 1.7
The three unit vectors may be written as:
i = (1, 0, 0); j = (0, 1, 0); k = (0, 0, 1).
The Position Vector or Radius Vector r from O to the point (x, y, z) is written as
r = x i + y j + z k
and has magnitude r = |r| =
x 2 + y2 + z 2
4 Mechanics of Particles, Waves & Oscillations
Example 1
Forces A and B expressed by the equation A = A1 i + A2 j + A3 k and B = B1 i + B2 j
+ B3 k, act on an object. Find the magnitude of the resultant of these forces.
Resultant force
R =A+ B
= (A1 + B1)i + (A2 + B2)j + (A3 + B3)k
R =
( A1 + B1 )2 + ( A2 + B2 )2 + ( A3
The result can be extended to any number of forces.
Example 2
Determine the vector having initial points P (x1, y1,
z1) and the terminal points Q(x2, y2, z2) and find its
magnitude.
The position vector of P is r = x i + y j + z k
1
1
1
1
The position vector of Q is r2 = x2 i + y2 j + z2 k
In Fig. 1.8, r1 + R = r2
or
R = r2 – r1 = (x2 i + y2 j + z2 k ) – (x1 i + y1 j + z1 k )
= (x – x ) i + (y – y ) j + (z – z ) k
2
R =
1.9
1
2
1
2
Fig. 1.8
1
( x2 - x1 )2 + ( y2 - y1 )2 + ( z2 -
SCALAR OR DOT PRODUCT
The Scalar or Dot Product, also called the Inner Product, of two vectors A and B is
denoted by A.B and is defined as a scalar given by,
A . B = |A| |B| cos q = AB cos q
(1)
where q is the acute angle between A and B. We may
regard the scalar product of two vectors as the product
of the magnitude of one vector and the component of
A
the other vector in the direction of the first Fig. 1.9.
As the scalar product A.B is represented by a dot
q
between two vectors, it is called the dot product. From
B
A cos q
the definition of the scalar product (1), the angle
between A and B is found from the formula
Fig. 1.9
A.B
(1a)
cos q =
AB
provided A π 0, and B π 0.
If the angle q between two vectors is a right angle then the two vectors are said to be
perpendicular or orthogonal and the condition for the vectors A and B to be orthogonal
(A π 0, and B π 0) is seen from (2) to be
A . B = 0; Condition for orthogonality
(2)
Vector Calculus
1.9.1
5
Properties of the Scalar Product
A . B = B. A
(Commutative law)
A . (B + C) = A . B + A . C
(Distributive law)
(mA) . B = A . (mB) = m(A . B)
A . A = |A|2 = A2 ; for every A.
A . A = 0 ; if and only if A = 0,
where m is an arbitrary scalar.
i . i = j . j = k . k = 1
i . j = j . k = k . i = 0
If
A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k
then
(3)
A . B = A1 B1 + A2 B2 + A3 B3
A . A = A2 = A 21 + A 22 + A 23
B . B = B2 = B 21 + B 22 + B 23
1.10
VECTOR OR CROSS PRODUCT
The vector product of the vector A and B, written
as A × B, is another vector:
C =A× B
z
A ¥B
(4)
y
The magnitude of C is given by
B
q
C = AB sinq
(5)
x
where q is the acute angle between A and B. The
direction of C is that of the advance of a right-hand
screw as A rotates towards B through the angle q
Fig. 1.10.
A
B¥A
Fig. 1.10
1.10.1 Geometric Interpretation
Fig. 1.11 shows the parallelogram completed from
the vectors A and B. Now, the height of the
parallelogram B sinq = h. Thus, by (5), the
magnitude of the vector product C = |A × B|,
represents the area of the parallelogram whose
sides are A and B.
1.10.2
B
q
A
Fig. 1.11
Properties of the Vector Product
(i) A × B = –B × A
h = B sin q
(Anticommutative law)
(ii) A × (B + C) = A × B + A × C
(6)
(Distributive law)
(mA) × B = A × (mB) = m A × B
(iii) A × A = 0
(iv) A × 0 = 0 for any A
UV
W
(7)
6 Mechanics of Particles, Waves & Oscillations
i = j × j = k × k = 0
j = k , j × k = i , k × i = j ,
j × i = – i × j , k × j = – j × k , i × k = – k × i
= B i + B j + B k then,
(vi) If A = A1 i + A2 j + A3 k and B
1
2
3
(v) i ×
i ×
U|
V|
W
i
A × B = A1
B1
1.11
j
A2
B2
(8)
k
A3
B3
SCALAR TRIPLE PRODUCT
The scalar triple product of the three vectors A, B and C is a scalar A . (B × C) or simply,
Scalar Triple Product = A . (B × C)
(9)
Observe that the expression A × (B . C) is meaningless since it implies the vector
product of a scalar and a vector.
If
A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k
C = C i + C j + C k
1
then,
A1
A . (B × C) =
2
3
A2
A3
B1 B2
C1 C2
B3
C3
The scalar triple product A . (B × C) is also written as [ABC].
1.11.1
Geometric Interpretation of A . B × C
Fig. 1.12 shows a parallelepiped whose sides
are A, B and C. By (4), the vector product of
B and C is equal to the area S for the
parallelogram with adjacent sides B and C.
S =
|B × C|
If h is the altitude of the parallelopiped then,
h =
|A| |cos q|
where q is the angle between A and B × C. Therefore,
the volume of the parallelopiped is
Fig. 1.12
V = hS = |A| |B × C| |cos q| = |A . B × C|
1.11.2
Condition for Coplanarity
Vector A, B and C are coplanar if and only if
A . B × C = 0; Condition for coplanarity
(10)
This is understandable since h = 0 in the event of coplanarity and so volume V = 0.
The converse of (10) is also true.
Vector Calculus
7
Example 3
If any two vectors in a scalar triple product are equal, show that the product is zero; that
is,
A . A × C = 0 ; C . B × C = 0; A . B × B = 0
The vector product A × C = P is perpendicular to A. Hence, A . P = 0 by (3), that is,
A . A × C = 0.
Also, since B × C is perpendicular to C, their scalar product is zero, that is, C . B ×
C = 0.
Again, B × B = 0 by (7). Hence A . 0 = 0.
1.11.3
Fundamental identify for the Scalar Triple Product
A . B× C = A× B . C
(11)
This implies that in the scalar triple product the position of the dot and cross is
immaterial.
1.12 VECTOR TRIPLE PRODUCT
The vector triple product of three vectors A, B and C is the vector D given by:
D = A × (B × C)
(12)
Here a parenthesis is essential since A × B × C depends on the result of the product,
whether we form A × B first or B × C first.
Thus, A × (B × C) π (A × B) × C
1.12.1
Fundamental Identities for the Vector Triple Product
A × (B × C) = (A . C) B – (A . B) C
(13)
(A × B) × C = (A . C) B – (B . C) A
(14)
Example 4
Evaluate (a) i . i ; (b) i . k
(a) i . i = | i | | i | cos 0° = (1) (1) (1) = 1
(b) i . k = | i | | k | cos 90° = (1) (1) (0) = 0
Example 5
If A = A1 i + A2 j + A3 k and B = B1 i + B2 j +
B2 + A3 B3.
A . B = (A1 i + A2 j + A3 k ) . (B1 i + B2
= A B i . i + A B i . j + A B
B3 k prove that A . B = A1 B1 + A2
j + B k )
3
i . k + A B j .
1
1
1
2
1
3
2
1
+ A2 B3 j . k + A3 B1 k . i + A3 B2 k . j + A3 B3
= A1 B1 + A2 B2 + A3 B3
where we have used (3).
i + A2 B2 j . j
k . k
8 Mechanics of Particles, Waves & Oscillations
Example 6
A man proceeds from his house 3 Km. due north. He then turns in the north-east direction
and goes further far 4 km. How far is he from his house? What is the direction of the
terminal point?
ax = 0 ; ay = 3 ; bx = 4 sin45° ; by = 4 cos 45°
( a x + bx )2 + ( a y + by )2
r =
2
e0 + 4 2 j + e3 + 4 2 j
=
B
2
= 6.478 km.
a y + by
tan q = ry/rx =
a x + bx
=
3+4
m
4k
by
4 5°
A
C
ay =
3 km
2
= 2.06.
4 2
q = 64.1°, north of east.
q
O
Example 7
bx
Fig. 1.13
Show that (a) i × j = k; (b) j × j = 0
(a)
i j k
i × j = 1 0 0 = k
0 1 0
(b)
i j k
j × j = 0 1 0 = 0
0 1 0
Example 8
= x i + y j + z k and B
= x i + y j + z k , prove that :
If A
1
1
1
2
2
2
i
A × B = x1
x2
j
y1
y2
k
z1
z2
A × B = (x1 i + y1 j + z1 k ) × (x2 i + y2 j + z2 k )
= x1 x2 i × i + x2 y2 i
+ y1 x2 j × i + y1 y2
+ z x k × i + z y
1
2
1
2
× j + x1 z2 i × k
j × j + y z j × k
1 2
k × j + z z k × k
1
2
= x1 y2 k – x1 z2 j – y1 x2 k + y1 z2 i + z1 x2 j – z1 y2 i
= (y1 z2 – z1 y2) i + (z1 x2 – x1 z2) j + (x1 y2 – y1 x2) k
D
Vector Calculus
i
= x1
x2
j
y1
y2
9
k
z1
z2
where we have used (8).
Example 9
Prove that A × (B × C) = (A . C) B – (A . B) C
j
i
A × (B × C) = A × B1 B2
C1 C2
k
B3
C3
= (A1 i + A2 j + A3 k ) ×
+ k (B1 C2 – B2 C1)]
= – k A (B C – B C ) –
– k
+ j
[ i (B2 C3 – B3 C2) – j (B1 C3 – B3 C1)
j A (B C – B C )
1
1
2
2
1
A2 (B2 C3 – B3 C2) + i A2 (B1 C2 – B2 C1)
A (B C – B C ) + i A (B C – B C )
1
1
3
3
2
3
3
1
3
2
3
1
3
3
1
where we have used (8).
\
A × (B × C) = i [B1 (A2 C2 + A3 C3) – C1 (A2 B2 + A3 B3)]
+ j [B2 (A1 C1 + A3 C3) – C2 (A1 B1 + A3 B3)]
+ k [B (A C + A C ) – C (A B + A B )]
3
1
1
2
3
2
2
3
3
3
1
1
2
2
= i [B1 (A1 C1 + A2 C2 + A3 C3) – C1 (A1 B1 + A2 B2 + A3 B3)]
+ j [B2 (A1 C1 + A2 C2 + A3 C3) – C2 (A1 B1 + A2 B2 + A3 B3)]
+ k [B3 (A1 C1 + A2 C2 + A3 C3)– C3 (A1 B1 + A2 B2 + A3 B3)]
= (A C + A C + A C ) ( i B + j B + k B )
1
1
1
2
3
– (A1 B1 + A2 B2 + A3 B3) ( i C1 + j C2 + k C3)
– (A . C) B – (A . B) C
Example 10
Verify the Jacobi identity:
A × (B × C) + B × (C × A) + C × (A × B) = 0
Using the fundamental identity (13),
A × (B × C) = (A . C) B – (A . B) C
B × (C × A) = (B . A) C – (B . C) A
C × (A × B) = (C . B) A – (C . A) B
Add these identities to obtain the Jacobi identity.
Example 11
Prove that (A × B) × (C × D) = [A B D] C – [A B C] D
10 Mechanics of Particles, Waves & Oscillations
Let A × B = F, then by (13),
(A × B) × (C × D) = F × (C × D)
= (F . D) C – (F . C) D
= (A × B . D) C – (A × B . C) D
= [A B D] C – [A B C] D
Example 12
Find the angle between the vectors A = 2 i + 2 j – k and B = 4 i + 2 j +4 k
A =
(2)2 + (2)2 + ( -1)2 = 3 ; B =
(4)2 + (2)2 + (4)2
= 6
A . B = (2 i + 2 j – k) . (4 i + 2 j + 4 k) = (2) (4) + (2) (2) + (–1) (4)
= 8.
cosq =
8
A.B
=
= 0.4444 ; q = 63.6°.
(3) (6)
AB
Example 13
Find the angles which the vector A = 2 i – 2 j + k makes with the co-ordinate axes.
Let a, b, g be the angles which A makes with the positive x, y, z axes respectively.
A . i = (A) (1) cos a =
(2)2 + ( -2)2 + (1)2 cos a = 3 cos a.
A . i = (2 i – 2 j + k ) . i = 2 i . i – 2 j . i + k . i = 2
cos a = 2/3 = 0.6667 and a = 48.2°.
Similarly, cos b = – 0.6667 and b = 131.8°,
and
cos g = 0.3333 and g = 70.5°.
The Cosines of a, b, g are called the direction cosines of A.
Example 14
Show that cos2a + cos2b + cos2g = 1, where cos a, cos b and cos g are the direction cosines
of a vector.
Let
A = A1 i + A2 j + A3 k .
The direction cosines are given by
cos a = A1/A, cos b = A2 /A, cos g = A3 /A.
cos2a + cos2b + cos2g =
=
A 21
A2
+
A 22
A2
+
A 23
A2
A 21 + A 22 + A 23
A2
=
A2
A2
= 1.
1.13 APPLICATION OF VECTOR MULTIPLICATIONS
1.13.1 Scalar Product
In general, when the force and displacement are not parallel, then the quantity of work
Vector Calculus 11
is given by the component of the force parallel to the
displacement, multiplied by the displacement:
F
W = (F cos q) d = F . d.
1.13.2
Fig. 1.14
F
r
0
q
nq
r si
Fig. 1.15
w
rs
in
v
q
P
p
r
q
0
Fig. 1.16
Triple Scalar Product
Consider the vector r and the force F lying in the xy
plane, as in Fig. 1.17. Then the torque of the force F
about z-axis (which is perpendicular to the xy plane) is
given by the vector product r × F. This special case was
actually considered under 1.13.2. However, if the axis L
is inclined to the z axis then the result must be modified.
Consider the axis L which is inclined to the z-axis. Take
a unit vector n along the axis L. Then the torque about
L is given by the Triple Scalar product, n . (r × F),
since it gives the component along the axis L.
1.13.4
d
Vector Product
(a) Torque: In general, the torque (or moment of a
perpendicular to the paper) is defined as the magnitude of the force times its arm which is the perpendicular distance between the axis of rotation and
the line of action of force; that is t = Fr sin q = |r
× F|. Thus r × F represents the torque of F about
an axis through O and perpendicular to the plane of
the paper.
(b) Angular Velocity: Let a particle P move in a circular orbit about an axis through the point O with
angular velocity w, Fig. 1.16. Since the radius of the
circle is r sin q, the magnitude of the linear velocity
w × r|.
v is w (r sin q) = |w
Also, v must be perpendicular to both w and r.
The quantities w, r and v form a right-handed coordinate system. We thus have the vector relation
v = w × r.
(c) Angular Momentum: With reference to Fig. 1.16
Angular momentum (J) is defined by the vector
produced J = r × p, where p the linear momentum
is in the direction of v. The magnitude of J is given
by |J| = rp sin q.
1.13.3
q
Triple Vector Product
L
z
n
y
0
r
x
F
Fig. 1.17
(a) Angular Momentum: Let a particle of mass m be at rest on a rotating rigid body
such as the earth, Fig. 1.18. Then the angular momentum of m about point 0
12 Mechanics of Particles, Waves & Oscillations
is defined as, J = r × (mv) = mr × v. But since
v = w × r, we find
J = mr × (w
w × v)
w
v
(b) Centripetal Acceleration: The centripetal accelw × v). For the
eration of m in Fig. 1.18 is a = w × (w
special case when r is perpendicular to w, the above
w2 r so
result reduces to the familiar formula, a = –w
that the acceleration is towards the centre of the
circle and of magnitude w2r.
m
r
0
1.14 DIFFERENTIATION OF VECTORS
Fig. 1.18
Let R(u) be a vector which is a function of a single
scalar variable u. Then,
DR = R ( u + Du ) – R
where D u denotes an increment in u, as in Fig. 1.19.
The derivation of the vector R (u) with respect to the
scalar u is given by:
O
R
(u
+
Du
)
DR
R (u + D u) - R (u)
=
Du
Du
R
(u )
dR
= Lim
Du Æ 0
du
DR
R (u + D u) - R (u)
Fig. 1.19
= Lim
Du Æ 0
Du
Du
provided the limit exists.
Since dR/du is itself a vector depending on u, if its derivative with respect to u exists,
then it is denoted by d2R/du2. Similarly, higher order derivatives may be defined.
If r = xi + yj + zk, is the position vector of a moving particle P (x, y, z) in space, then,
dr = dx i + dy j + dz k
(15)
and the velocity is
v =
dr
dx i
dy j
dz k
=
+
+
dt
dt
dt
dt
(16)
and the acceleration is
a =
d2r
dt2
=
d 2 x i
+
d 2 y j
+
d 2 z k
dt 2
dt 2
dt 2
Here we have assumed that the unit vectors, i, j and k, remain fixed in space.
Example 15
Using unit vectors show that the acceleration of a particle p moving on a circle or radius
r with constant angular velocity dq/dt is given by a = –w2r.
Referring to Fig. 1.20,
r = r cosq i + r sinq j
therefore, v =
dr
dr dq
.
=
dt
dq dt
Vector Calculus 13
y
dq
= (– r sinq i + r cosq j )
dt
dv
d2 r
a =
=
dt
dt2
and
= (– r cosq i – r sinq j )
therefore,
a = – w2 r
v
P
FG dq IJ
H dt K
r
2
q
x
1.14.1 Differentiation Formulae
Fig. 1.20
d
dA dB
(A + B) =
+
du
du du
(i)
(18)
(ii)
d
dB dA
(A . B) = A .
. B
+
du
du du
(19)
(iii)
d
dB dA
(A × B) = A ×
× B
+
du
du du
(20)
d
dA df
A
(fA) = f
+
du
du du
(iv)
(v)
(vi)
(21)
dC
dB
dA
d
+A .
¥C+
(A . B × C) = A . B ×
. B× C
du
du
du
du
d
{A × (B × C)} = A ×
du
FG B ¥ dC IJ
H du K
+ A×
FG dB ¥ CIJ
H du K
+
dA
× (B × C)
du
(22)
(23)
The order in the above products may be important.
Example 16
A particle moves in a circle of radius r at constant speed v. Obtain the formula of
centripetal accleration using the fact that r2 = r . r = Constant, and v2 = v . v = Constant.
Differentiating the given equation with respect to time,
2 r . r = 0
= 0
2 v . v
or
r . v =0
(i)
or
v . a =0
(ii)
or
r . a = – v2
Differentiating (i),
r . a+ v . v = 0
(iii)
By (i), r is perpendicular to v and by (ii), a is perpendicular to v. It follows that a and
r are either parallel or antiparallel since the motion is in a plane. The angle q between
a and r is either 0° or 180°. Using the definition of scalar product in (iii),
r . a = |r| |a| cosq = –v2
Since cosq is negative, q = 180°. By (iv),
|r| |a| (–1) = –v2
or
a = v2/r.
(iv)
14 Mechanics of Particles, Waves & Oscillations
Example 17
If A and B are differentiable functions of a scalar u, prove that
d
dB dA
(A . B) = A .
. B
+
du
du du
Let A = A1 i + A2 j + A3 k and B = B1 i + B2 j + B3 k.
d
d
(A . B) =
(A1 B1 + A2 B2 + A3 B3)
du
du
Then
=
FG A dB + A dB + A dB IJ
H du
du
du K
F dA B + dA B + dA B IJ
+ G
H du
K
du
du
1
1
1
=A .
2
2
2
1
3
3
2
3
3
dB dA
. B
+
du du
Example 18
If A and B are differentiable functions of a scalar u, prove that
dB dA
d
+
(A × B) = A ×
× B
du
du du
d
d
(A × B) =
du
du
i
j
k
A1
B1
A2
B2
A3
B3
i
=
j
k
A1
A2
A3
dB1
du
dB2
du
dB3
du
=A×
+
i
dA1
du
B1
j
dA2
du
B2
k
dA3
du
B3
dB dA
+
× B
du du
where we have used a theorem of differentiation of determinants
1.14.2
Rules for Partial Differentiation of Vector Functions
If A and B are differentiable vector functions of u, v and w, and f is a differentiable scalar
function of u, v and w, then using u as an example,
∂
∂A ∂B
+
(A + B) =
∂u
∂u ∂u
∂
∂A ∂f
A
(f A ) = f
+
∂u
∂u ∂u
(24)
(25)
Vector Calculus 15
∂
∂B ∂A
+
(A . B) = A .
. B
∂u
∂u ∂u
∂
∂B ∂A
+
(A × B) = A ×
× B
∂u
∂u ∂u
Observe that the order of factor in (27) is important.
(26)
(27)
1.15 UNIT VECTORS IN PLANE POLAR CO-ORDINATES
So far we have expressed vectors in terms of their rectangular components using the unit
vectors i, j and k. In many situations it is convenient to use other co-ordinate systems
such as polar co-ordinates in two dimensions and spherical or cylindrical co-ordinates in
three dimensions. Here we shall be concerned with the plane polar co-ordinates as in Fig.
1.21. The point P has the co-ordinates r and q. The co-ordinate r is the radial distance
of P from the origin O and the angle q is measured by the line OP with x-axis. The unit
vector (that is, a vector of unit length) er is along the line q = Constant, in the direction
of increasing r (r-direction). The other unit vector eq is along the circle r = Constant, in
the direction of increasing q(q – direction) and tangential
to the circle. These two unit vectors are mutually
perpendicular to each other and they continuously
change their orientation, although maintaining constant
magnitude as the point P moves, unlike the rectangular
unit vectors i and j which have fixed orientation in
space.
We can express the given vector in terms of its components in the direction er and eq simply by finding its
projection in these directions. But since their directions
change from point to point, the derivatives of a vector
in polar co-ordinates are obtained by differentiating the
unit vectors as well as the components. This is in contrast with the rectangular unit vectors where we differFig. 1.21
entiate the components only.
Example 19
Let (r, q) be the polar co-ordinates describing the position of a particle. If er is a unit vector
in the direction of the positive vector r and eq is a unit vector perpendicular to r and in
the direction of increasing q, show that:
er = cosq i + sinq j,
(i)
eq = – sinq i + cosq j.
Since ∂r ∂r is a vector tangent to the curve q = Constant, a unit vector in the
direction of r (increasing r) is given by,
er =
∂r
∂r
∂r
∂r
Since r = xi + yj = r cosq i + r sinq j
(ii)
16 Mechanics of Particles, Waves & Oscillations
as in Fig. 1.21,
so that
∂r
= cosq i + sinq j ,
∂r
er = cosq i + sinq j .
∂r
∂r
= 1
(iii)
Further, ∂r ∂q is a vector tangent to the curve r = Constant. A unit vector in this
direction is given by,
eq =
By (ii),
∂r
∂q
∂r
∂q
(iv)
∂r
= –r sinq i + r cosq j ,
∂q
∂r
∂q
= r
so that (iv) yields:
eq = – sinq i + cosq j
(v)
Example 20
.
.
.
.
Show that (a) er = q eq (b) eq = – q er
(a) By (iii) of Ex. 19,
der
∂er dr ∂er dq
.
+
=
er =
dt
∂r dt
∂q dt
.
= (0) (r) + (– sinq i + cosq j) (q )
.
= q eq
.
dr
dq
.
where r and q mean
and
, respectively.
dt
dt
(b) By (v) of Ex. 19,
deq
∂e q dr ∂e q dq
.
=
eq =
+
dt
∂r dt
∂q dt
.
.
.
= (0) (r) + (– cosq i – sinq j ) (q) = – q er
Example 21
.
.
Prove that in polar coordinates (a) the velocity is given by v = r er + r q eq and (b) the
acceleration is given by
..
.2
..
.. .
a = (r - r q )er + (r q + 2 r q) e q
(a)
r = rer
v =
dr dr
de
er + r r
=
dt dt
dt
.
.
.
.
= r er + r er = r er + r q e q
where use has been made of Example (20).
Vector Calculus 17
..
dv d .
(r er + r q e q )
=
dt dt
..
.
. .
..
. .
= r er + r er + r q e q + r q e q + r q e
..
..
. .
. .
= r e r + r (q e q ) + r q e q + r q e q + r
.
..
..
..
= (r - r q2 ) er + (r q + 2 r q) e q
(b)
a =
where we have used the results of Ex. 20 and Ex. 21 (a). The above expressions for
velocity and acceleration will be used in Chapter 3 & 4.
1.16 FIELDS
In general, physical quantities have different values at different points in space. Thus,
for example, the temperature in a room varies from one place to another, being higher
near a fire place and lower near an open window. Similarly, the electric field near a point
charge is larger than at points farther from it. The expression field is used to imply both
the region and the value of the physical quantity in the region (electric field, gravitational
field etc.). If the physical quantity is a scalar (for example temperature) then we are
concerned with a scalar field. If the quantity is a vector (for example electric field,
velocity etc.) then we speak of a vector field.
1.17.1
The Del Operator
The vector differential operation Del, symbol —, is defined by
∂ ∂ ∂ ∂ ∂
∂
i+
j+
k = i
(28)
+j
+ k
∂x
∂y
∂z
∂x
∂y
∂z
Del is also called Nabla. It enjoys the properties of both a vector as well as a differential
operator. The operator del is not a vector in the geometrical sense since it has no scalar
magnitude but it does transform properly so that it may be treated formally as a vector.
It is found useful in defining Gradient, Divergence, Curl and Laplacian.
— =
1.17.1.1 Properties of the Del Operator
If we multiply a scalar quantity u with this vector operator or —, we obtain
FG
H
IJ
K
∂ ∂
∂
∂u ∂u ∂u
+j
+ k
u = i
—u = i
+j
+k
∂x
∂y
∂z
∂x
∂y
∂z
(i) If we form the scalar product of the del operator with a vector v, we obtain, according
to the definition of scalar product, the sum of the products of corresponding
components:
FG
H
∂ ∂
∂
— . v = i
+j
+ k
∂x
∂y
∂z
=
IJ . ev i + v
K
x
∂ v x ∂v y ∂v z
+
+
= div v.
∂x
∂y
∂z
+ v k
z
yj
j
18 Mechanics of Particles, Waves & Oscillations
(ii) If we form the vector product of the del operator with v then we obtain
FG
H
IJ × (v i + v j + v k )
K
∂v I F ∂v
∂v I
+jG
H ∂z - ∂x JK +
∂z JK
∂ ∂
∂
+j
+ k
— × v = i
∂x
∂y
∂z
F
GH
∂v z
= i
∂y
y
x
y
z
z
x
= Curl v.
(iii) If we form the scalar product of the del operator with itself then we obtain,
FG i ∂ + j ∂ + k ∂ IJ ◊ FG i ∂ + j ∂ + k ∂ IJ =
H ∂x ∂y ∂z K H ∂x ∂y ∂z K
∂2
∂x 2
+
∂2
∂y 2
+
∂2
∂z 2
= —2 = Laplacian
(iv) The operations with del operator are distributive with respect to addition. That is,
—(v1 + v2) = —v1 + —v2
(v) The definition of del operator is independent of the co-ordinate system.
1.17.2
Let f (x, y, z) be a differentiable scalar field in a certain region of space x, y, z. Then the
—f =
FG ∂ i + ∂ j + ∂ k IJ f =
H ∂x ∂y ∂z K
∂f ∂f ∂f i+
j+
k
∂x
∂y
∂z
(29)
Observe that —f defines a vector field. The component —f in the direction of a unit
vector n is given by —f . n and is called the direction derivative of f in the direction n.
This is the rate of change of f at (x, y, z) in the direction n.
From the calculus,
∂f
∂f
∂f
(30)
dx +
dy +
dz.
∂x
∂y
∂z
Let r be the position vector to the point P (x, y, z).
r = x i + y j + z k
df =
r+
If we move to the point Q (x + dx, y + dy, z + dz),
as in Fig. 1.22,
(31)
dr = dx i + dy j + dz k
Using (29) and (31), we can take the dot product of
dr and —f to yield df as in (30),
d f = dr . —f
(32)
Q
z
Dr
Dr
P
0
r
y
x
Fig. 1.22
1.17.2.1 Geometrical Interpretation of —f
Consider a surface
f (x, y, z) = c
(33)
Vector Calculus 19
where c is a particular constant. Let us select an infinitesimal displacement dr of r, and
consider only those displacements which are tangential to the surface described by (33).
As long as we move along this surface, f has the constant value and df = 0.
Consequently from (32),
dr . —f = 0
(34)
Now —f is a vector which is completely determined once f has been differentiated,
and since neither dr is zero nor in general —f , according to (34), —f is perpendicular
to dr where dr denotes a change from P to Q with Q remaining on the surface
f = Constant. We therefore conclude that f is normal to all possible tangents to the
surface at P so that —f must necessarily be normal to the surface f (x, y, z) = Constant,
as in Fig. 1.23. In general,
dr . —f = |dr| |—f| cosq
where q is the angle between the unit vector n and
the vector —f. Let |dr| = ds so that
df
= n . —f
ds
is the directional derivative along n.
Since |n| = 1,
(35)
df
= |—f| cosq
ds
Thus, df/ds
(namely |—f|)
if we go in the
that is df/ds =
(36)
Fig. 1.23
is the projection of —f on the direction n. The largest value of df/ds
occurs if we go in the direction of —f (that is q = 0). On the other hand
opposite direction (that is q = 180°) f has the largest rate of decrease,
–|—f|.
Example 22
If f = x2 y – xz3, find —f.
FG
H
∂ ∂
∂
+j
+ k
—f = i
∂x
∂y
∂z
IJ ex y - xz j
K
2
3
= (2xy – z3) i + x2 j – 3xz2 k .
Example 23
If f =
1
, where r =
r
FG
H
x 2 + y 2 + z 2 , show that —f = –
∂ ∂
∂
+j
+ k
—f = i
∂x
∂y
∂z
=
-
IJ ex
K
2
r
r3
.
+ y2 + z 2
1
1
1
. 2 x i - . 2 y j - . 2 z k
r
- ( xi + yj + zk )
2
2
2
=- 3.
=
2
2
2 32
2
2 32
2
(x + y + z )
r
(x + y + z )
20 Mechanics of Particles, Waves & Oscillations
Example 24
Find a unit vector normal to the surface, x2y + xz = 2 at the point (1, –1, 1).
FG
H
IJ
K
∂ ∂
∂
(x2y + xz)
—( x 2 y + xz ) = i
+j
+ k
∂x
∂y
∂z
= (2xy + z) i + x2 j + x k
= – i + j + k
at the point (1, –1, 1).
A unit vector normal to the surface is obtained by dividing the above vector by its
magnitude. Hence the unit vector is
j
- i + j + k
k
- i
+
+
=
3
3
3
( -1)2 + (1)2 + (1)2
Example 25
Find the directional derivative of f = x2yz + 3xz2 at (1, –1, 2) in the direction 2i + j – 2k.
FG
H
IJ e
K
∂ ∂
∂
—f = i
x 2 yz + 3xz
+j
+ k
∂x
∂y
∂z
= (2xyz + 3z2) i + x2z j + (x2 y + 6xz) k
= 8 i + 2 j + 11 k
at (1, –1, 2)
The unit vector in the direction of 2 i + j – 2 k is
n =
2 i + j - 2k
(2)2 + (1)2 + ( -2)2
=
2 1 2 i + j - k.
3
3
3
The required directional derivative is
—f . n = (8 i + 2 j + 11 k ) .
FG 2 i + 1 j - 2 k IJ
H3 3 3 K
4
16 2 22
+ = - .
3
3 3
3
Since this is negative, f decreases in this direction.
=
— (C f) = C —f
— (f + Y) = —f + —Y
— ( fY) = f—Y + Y—f
(37)
(38)
(39)
where f and Y are differentiable scalar functions in some region in space and C is a
constant.
1.17.2.3 An Example of a Gradient
Consider the lines of equal pressure (isobars) marked on a weather map. The direction
of the wind is then given, apart from the earth’s rotation, by the direction of greatest
pressure drop, which is perpendicular to the isobars, and the strength of the wind is
given by the magnitude of the pressure drop.
Vector Calculus 21
Example 26
Prove that
—( fY ) = f—Y + Y—f
—( fY ) =
∂
∂
∂
( fY)i +
( fY)j +
( fY)k
∂y
∂z
∂x
FG
H
= f
= f
IJ FG
K H
∂Y
∂f ∂Y
∂f
i+ f
+Y
+Y
∂x
∂x
∂y
∂y
IJ
K
FG ∂Y i + ∂Y j + ∂Y k IJ + Y FG ∂f i + ∂f j + ∂f k IJ
H ∂x ∂y ∂z K H ∂x ∂y ∂z K
= f—Y + Y—f .
Example 27
Find the angle between the surfaces x2 + y2 + z2 = 4 and z = x2 + y2 – 1 at the point
(1, –1, 1).
The angle between the surfaces at the point is the angle between the normal to the
surfaces at the point.
A normal to x2 + y2 + z2 = 4 at (1, –1, 1) is
—f1 = — (x2 + y2 + z2) = 2xi + 2yj + 2zk = 2i – 2j + 2k
A normal to z = x2 + y2 – 1 or x2 + y2 – z = 1 at (1, –1, 1) is
—f 2 = — (x2 + y2 – z) = 2xi + 2yj – zk = 2i – 2j – k.
b—f g . b—f g =
1
2
—f1 —f2
cosq,
where q is the required angle. Then
(2i – 2j + 2k) . (2i – 2j – k) = |2i – 2j + 2k| |2i – 2j – k| cosq
4 + 4 + –2 =
or
cosq =
(2)2 + ( -2)2 + (2)2
6
12
9
=
1
3
(2)2 + ( -2)2 + ( -1)2 cosq
= 0.5773.
Thus the acute angle is q = 54.7°.
1.17.3
The Divergence
Let V(x, y, z) = V1 i + V2 j + V3 k be a differentiable vector field at each point (x, y, z) in
a certain region of space. Then the divergence of V, symbol — . V or div V is defined by
FG ∂ i + ∂ j + ∂ k IJ . (V i + V
H ∂x ∂y ∂z K
F ∂ V + ∂ V + ∂ V IJ
= G
H ∂x ∂y ∂z K
—.V =
1
1
Observe that — . V π V . —.
2
3
2
j + V k )
3
(40)
22 Mechanics of Particles, Waves & Oscillations
1.17.3.1 Physical Significance of Divergence
Consider the flow of a fluid of density r(x, y, z)
with velocity v(x, y, z), through a small
parallelpiped ABCDEFGH (Fig. 1.24) of dimensions dx, dy, dz. We will first calculate the
amount of fluid passing along the x-direction
through the face EFGH per unit time. The y
and z components of the velocity v contribute
nothing to the flow through EFGH. The mass of
fluid entering EFGH per unit time is given by
rvx dy dz. The mass of the fluid leaving the face
ABCD per unit time is
LMrv
N
x
+
Fig. 1.24
OP
Q
∂
(rv x )dx dy dz.
∂x
The net rate of flow out for these two faces is simply the difference between these two
flows, or
∂
(rv x ) dxdydz.
∂x
If we also take into consideration the other two faces we find that the total loss of mass
per unit time is
Net rate of flow out =
LM ∂ (rv
N ∂x
x)
+
∂
∂
( rv y ) +
( rv z )
∂y
∂z
O
Q
dxdydz.
so that the quantity within square brackets represents the loss of mass per unit time per
unit volume and is called the Divergence — . (rv). This is the physical meaning of
divergence.
div (rv) may not be zero either because of the time variation of density or the existence
of sources and sinks.
Let
Y = source density minus sink density
= net mass created per unit time per unit volume.
∂r
= time rate of increase of mass per unit volume.
∂t
Then rate of increase of mass per unit volume
= rate of creation minus rate of outward flow.
In symbols, the balance equation becomes
∂r
= Y – — . (rv)
∂t
Calling
V = rv,
∂r
∂t
In the absence of sources and sinks, Y = 0, and (41) reduces to
—. V = Y-
(41)
Vector Calculus 23
∂r
= 0; Equation of continuity
(42)
∂t
If there is no gain of fluid anywhere then — . V = 0. This is called the continuity
equation for an incompressible fluid. The fluid is said to have no sources or sinks since
it is neither created nor destroyed at any point. A vector such as V whose divergence is
zero is called solenoidal.
—. V +
∂r
= 0, then (41) reduces to
∂t
If
—. V = Y
(43)
The above treatment is equally applicable to electric and magnetic fields where v is
replaced by E or B and the quantity corresponding to outflow of a metal substance is
called flux. In the case of an electric field, the so-called sources and sinks are the electric
charges and the equation analogous to (43) is
div D = Y
(44)
where Y is the charge density and D is the electric displacement. For the magnetic field
the sources are assumed to be magnetic poles. However, free magnetic poles do not exists
so that
div B = 0
(45)
Equation (44) and (45) constitute two of the celebrated equations in Electromagnetism,
originally due to Maxwell.
The continuity equation (42) also has an application in the interpretation of the wave
function in Quantum Mechanics.
Example 28
Calculate — . (ff), where f = u(x, y, z) i + v(x, y, z) j + w(x, y, z) k .
— . (ff) =
∂
∂
∂
(fu) +
(fv) +
(fw)
∂x
∂y
∂z
=f
FG ∂u + ∂v + ∂w IJ + FG u ∂f + v ∂f + w ∂f IJ
H ∂ x ∂ y ∂ z K H ∂x ∂y ∂z K
= f— . f + f . —f
Example 29
Calculate — . f if f = r/r3. (inverse-square force)
— . (r–3 r) = r–3 — . r + r . —r–3
But
FG
H
∂ ∂
∂
+j
+ k
— . r = i
∂x
∂y
∂z
=
∂x ∂y ∂z
= 3
+
+
∂x ∂y ∂z
and by problem (16), —rn = nrn–2 r
IJ . ( i x + j y + k z)
K
24 Mechanics of Particles, Waves & Oscillations
— r–3 = –3 r–5 r
so that
\
— . (r–3 r) = 3r–3 – 3r–5 r . r = 3r–3 – 3r–3 = 0
Thus the divergence of an inverse square force is zero.
1.17.4
The Laplacian
The Laplacian, symbol —2 is the divergence of a gradient.
FG
H
∂ ∂
∂
+j
+ k
— . —f = i
∂x
∂y
∂z
=
—2 =
where
∂2 f
∂x
∂
+
2
∂2 f
∂y
2
+
∂x 2
∂
2
+
∂2 f
∂z
2
+
∂y 2
2
∂2
IJ . FG i ∂f + j ∂f + k ∂f IJ
K H ∂x ∂y ∂z K
= — 2f
= Laplacin.
∂z 2
Example 30
Prove that — . (fA) = (—f). A + f(— . A), where f is a scalar and A is a vector.
— . (fA) = — . fA i + fA j + fA k
e
1
2
3
FG
H
∂ ∂
∂
+j
+ k
= i
∂x
∂y
∂z
=
IJ . efA i + fA j + fA k j
K
1
2
3
∂
∂
∂
(fA1 ) +
(fA2 ) +
(fA3 )
∂x
∂y
∂z
= f
=
j
∂A1
∂A
∂A
∂f
+ f 2 + f 3 + A1
+
∂x
∂y
∂z
∂x
FG ∂f i + ∂f j + ∂f k IJ . e A i + A j + A k j
H ∂x ∂y ∂z K
F ∂ + j ∂ + k ∂ IJ . e A i + A j + A k j
+ f G i
H ∂x ∂y ∂z K
1
2
1
3
2
3
= (—f) . A + f (— . A)
Example 31
f = x2 y – 2xz3, find —2 f.
If
F∂
GH ∂x
2
—2 f =
2
+
∂2
∂y
2
+
∂2
∂z
2
I x y - 2xz
j
JK e
2
3
= 2y – 12xz.
1.17.5
The Curl
If V(x, y, z) is a differentiable vector field then the Curl or rotation of V, symbol — × V,
Curl V or rot V is defined by
Vector Calculus 25
— × V=
FG ∂ i + ∂ j + ∂ k IJ × (V
H ∂x ∂y ∂z K
i
∂
=
∂x
Vx
j
∂
∂y
Vy
i + V2 j + V3 k )
k
∂
∂x
Vz
∂
∂
∂
= ∂y ∂z i - ∂x
Vx
V y Vz
F ∂V
GH ∂y
1
∂
∂
j + ∂x
∂z
Vz
Vx
I FG
JK H
∂
∂y
V
∂V y ∂V x ∂V z i+
j+
∂z
∂z
∂x
Curl V represents the rotation or vorticity in the fluid. If the flow is irrotational,
=
z
-
IJ
K
— × V = 0; Condition for irrotationality.
The curl may be used to describe the motion of a rigid body rotating about an axis with
uniform angular velocity w.
v = w × R, is the linear velocity of any point in the body with radius vector R.
Curl v = — × (w
w × R).
By identity (13), A × (B × C) = B (A . C) – C(A . B).
Therefore, Curl v = w(— . R) – R(— . w).
But, — . R = 3, by Example (29). Also, R (— . w) = (w
w . —)R since w is a constant vector.
RS
T
UV
W
∂
∂
∂
Therefore, (w
w . —)R = w x ∂x + w y ∂y + w z ∂z ( i x + j y + k z)
= i wx + j wy + k wz = w
Therefore, Curl v = 3w – w = 2w.
Thus the curl of the linear velocity of any point of a rigid body is equal to twice the
angular velocity.
1.17.5.1 The Physical Significance of the Curl
The physical significance of the curl is brought about by considering the circulation of
fluid around a differential loop in the xy-plane, Fig 1.25.
Circulation1234 =
z
z
z
v x ( x, y) dx + v y ( x, y) dy - v
1
2
3
Use the Taylor expansion about the point (x0, y0), taking into account the displacement
of line segment 3 from 1 and 2 from 4.
v y ( x0 + dx, y0 ) = v y ( x0, y0 ) +
F ∂v I dx + . . .
GH ∂x JK
y
x0 , y0
26 Mechanics of Particles, Waves & Oscillations
The higher-order terms will drop off in the
limit dx Æ 0. A correction term for the
variation of vy with y is cancelled by the
corresponding term in the fourth integral.
Circulation1234 = vx (x0, y0)dx
y
∂v
L
O
dx P dy
+ Mv ( x , y ) +
∂x
N
Q
L
O
∂v
dy P ( - dx ) + v
+ Mv ( x , y ) +
y
∂
N
Q
F ∂v - ∂v I dxdy
= G
H ∂x ∂y JK
y
x
0
0
y
0
0
x 0 y 0 + dy
(x 0 +
2
4
x 0, y0
y
x
3
(– x 0
1
Fig. 1.25
y
x
Dividing by dxdy, we have
Circulation per unit area = — × v z
The circulation about the differential area in the xy-plane is given by the z-component
of — × v. In fluid dynamics — × v is called the “vorticity”. In principal, the curl — × v at
(x0, y0) can be determined by inserting a (differential) paddle wheel into the moving fluid
at the point (x0, y0). The rotation of the small paddle wheel would be a measure of the
curl, and its axis along the direction of — × v which is perpendicular to the plane of
circulation. Whenever the curl of a vector v vanishes,
—× v = 0
(47)
and v is called irrotational.
1.17.5.2 Examples of the Curl of Vector Field
(i) The curl of a vector field implies circulation or vortex motion (rotation). If the fluid
velocity v has a ‘curl’ at some point then that signifies the existence of vorticity at
that point. Since the line integral of a conservative field A around any closed path
z
is zero, that is A . dr = 0, the conservative fields have zero curl at all points of
space.
(ii) An example of the conservative vector field is the electrostatic field E. Therefore,
curl E = 0.
(iii) For waves in an elastic medium, if the displacement U is irrotational, — × U = 0,
plane waves (or spherical waves at large distances) become longitudinal. If V is
solenoidal, — . U = 0, then the waves becomes transverse. The displacement of a
seismic wave may be resolved into a solenoidal part and an irrotational part. The
irrotational part corresponds to the longitudinal P (primary) earthquake waves. The
solenoidal part yields the slower S (secondary waves) (See 13.1.2)
(iv) If v is the linear velocity then curl v may be used to describe the motion of a rigid
body rotating about an axis with uniform angular velocity w. It can be shown that
curl v = 2w (see 1.17.5). Thus the curl of the linear velocity of any point of a rigid
body is equal to twice the angular velocity.
Vector Calculus 27
Example 32
Prove that Curl of a gradient is zero.
— × (—f) =
= i
i
∂
∂x
j
∂
∂y
k
∂
∂z
∂f
∂x
∂f
∂y
∂f
∂z
F ∂ f - ∂ f I - jF ∂ f - ∂
GH ∂y ∂z ∂z ∂y JK GH ∂x ∂z ∂z
2
2
2
2
= 0
since terms in brackets cancel in pairs.
Example 33
Prove that Curl Curl V = grad div V – —2 V.
By identity (13), A × (B × C) = B (A . C) – (A . B) C.
Putting A = —, B = — , C = V,
— . —) V = grad div V – —2 V.
— × (—
— × V) = — (—
— . V) – (—
Example 34
— × V) = 0
Show that the divergence of a curl is zero, that is — . (—
FG i ∂ + j ∂ + k ∂ IJ
H ∂ x ∂ y ∂z K
.
=
FG i ∂ + j ∂ + k ∂ IJ
H ∂ x ∂ y ∂z K
.
=
∂
∂
∂
∂
∂
∂y ∂z ∂x
∂x V V
∂y V
x
y
z
=
∂
∂
∂
∂x ∂y ∂z
∂
∂
∂
∂x ∂y ∂z
Vx V y Vz
i
∂
∂x
Vx
LM
MMi
N
∂
∂y
Vy
j
∂
∂y
Vy
k
∂
∂z
Vz
∂
∂
∂z - j ∂x
Vx
Vz
∂
∂
∂z +
∂z
Vz
= 0
since two rows of the determinant are identical.
∂
∂
k
+
x
∂
∂z
Vz
Vx
28 Mechanics of Particles, Waves & Oscillations
Example 35
If A and B are irrotational, prove that A × B is solenoidal.
By problem — × A = 0 and — × B = 0
It follows that B . (—
— × A) = 0
A . (—
— × B) = 0
— × B) = 0
Subtracting B . (—
— × A) – A . (—
By problem (28), left hand side of the above equation is equal to — . (A × B).
Therefore, — . (A × B) = 0, so that (A × B) is solenodial.
Example 36
A central field A in space is given by A = r F(r). (a) Show that the field is irrotational.
(b) What should be the function F(r) so that the field is solenoidal?
(a) That Curl A = 0 for the central field (criterion for the conservative forces) is proved
in Chapter 4.
(b) If the field is solenoidal, then,
— . r F(r) = 0
∂
∂
∂
[ xF (r )] +
[ yF (r )] +
[ zF
∂x
∂y
∂z
=0
∂F
∂F
∂F
+F+y
+F+z
=0
∂x
∂y
∂z
∂F x
∂F y
+y
+z
3F (r ) + x
=0
∂r r
∂r r
F+x
FG IJ
H K
3F(r) +
FG IJ
H K
FG ∂F IJ FG x
H ∂r K H
2
+ y2 + z 2
r
I
JK
=0
But x2 + y2 + z2 = r2
∂F
∂r
= -3
F
r
Integrating, ln F = – 3 ln r + In C, where ln C = constant.
C
ln F = ln C – ln r3 = ln 3
r
Therefore F = C/r3. The field is A = r/r3 (inverse square law).
therefore
1.17.6
Table of Vector Identities Involving —
Here A and B are differentiable vector functions, and f and Y are differentiable scalar
functions of position (x, y, z) and r is the position vector.
1. —(f + Y) = —f + —Y
2. —(fY) = f—Y + Y—f
3. — . (A + B) = — . A + — . B
Vector Calculus 29
—f) . A + f(—
— . A)
4. — . (fA) = (—
5. — × (A + B) = — × A + — × B
—f) × A + f(—
— × A)
6. — × (fA) = (—
— × A) – A . (—
— × B)
7. — . (A × B) = B . (—
— . A) – (A . —)B + A(—
— . B)
8. — × (A × B) = (B . —)A – B(—
— × A) + A × (—
— × B)
9. — (A . B) = (B . —)A + (A . —) B + B × (—
—f) = div grad f = Laplacian f
10. — . (—
=
∂2 f
2
+
∂2 f
2
∂2 f
+
∂x
∂y
∂z 2
—f) = 0. The curl of gradient of f is zero.
11. — × (—
— × A) = 0 The divergence of curl of A is zero.
12. — . (—
— × A) = —(—
— . A) – —2A
13. — × (—
—f × —Y) = 0
14. — . (—
15. — . r = 3
16. — × r = 0
17. (A . —) r = A
1.18 VECTOR INTEGRATION
1.18.1
Line integral
This is an extension of the line integral described in Scalar calculus. Any integral which
is to be evaluated along a curve is called a line integral.
Examples of Line integral in vector calculus are:
(a)
z
C
f dr
(b)
z
C
A . dr
(c)
z
C
A × dr
where f is a scalar, A is a vector, and r is the position vector r = xi + yj + zk. Each of
these is a line integral along the curve C. The result of integration is a vector for (a), a
scalar for (b) and a vector for (c).
When the space curve C forms a closed path which is assumed to be a simple closed
curve, that is, a curve which does not intersect itself anywhere, the line integral (a) is
z
f dr. We can similarly write for (b) and (c). The movement along the closed
written as
C
curve C is said to be positive or counterclockwise if the enclosed region always lies to the
left, and negative or clockwise if the enclosed region lies always to the right. A line
integral means an integral along a curve or a line, that is a single integral in contrast
to a double integral over a surface or area, or a triple integral over a volume. It must be
emphasised that in a line integral there is only one independent variable because we are
constrained to remain on a curve. In two dimensions, the equation of a curve becomes y
= F(x), where x is the independent variable. In three dimensions, we could take x as the
independent variable and find y and z as functions of x. Alternatively, the parameter t
may be taken as independent variable so that x, y, z are all functions of t. Thus, the line
integral is evaluated by writing it as a single integral using one independent variable.
30 Mechanics of Particles, Waves & Oscillations
1.18.1.1
Calculation of Work Done by a Varying Force on a Body Using the Line
Integral (b)
dr
Work done by a force on an object which undergoes an infinitesimal vector displacement
dr can be written as dW = F . dr. In general, the force F acting on the object varies from
point to point. For example, the force on a charged
z
particle in an electric field would be a function of x, y,
z. However, along a curve x, y, z are related by the
B
equation of the curve. Since along a curve there is only
one independent variable, we can write F and dr = i dx
F
+ j dy + k dz as functions of a single variable. The
A
integral of dW = F . dr along the given curve is then
y
reduced to an ordinary integral of a function of one
variable, and the total work done by F in moving an
x
object say from A to B, can be determined (Fig. 1.26).
Fig. 1.26
1.18.1.2 Examples of Line Integral
(i) The line integral,
z
F . dr represents the work done by the force along the curve C.
(ii) The quantitative relationship between current i and the magnetic field B is given
by Ampere’s law
z
B . dl = m0i
(iii) The potential difference between two points A and B is related to the electric field
by the line integral
B
z
- E . dl = VB – VA
A
B
(iv) If A = grad f, then the integral
z
A . dr depends only on initial and final values of
A
f and is independent of the path. Also
z
A . dr = 0.
z
Conversely, if A . dr = 0, then there must exist a scalar point function f such that
A = grad f. The vector field is said to be conservative. Examples of conservative
fields are gravitational field and electric field (See Example 43).
(v) Magnetostatics also provides a physical example of the third type of line integral of
a loop of wire C carrying a current I placed in a magnetic field B, then the force dF
on a small length dr of the wire is given by dF = I dr × B, so that the total vector
force on the loop is
F = I
z
dr ¥ B.
C
Example 37
Evaluate
z
C
f dr if f = f(x, y, z).
Vector Calculus 31
Using the differential displacement vector dr = dx i + dy j + dz k , we find.
z
f dr =
c
z
z
z
z
f(dx i + dy j + dz k ) = i fdx + j fdy + k fdz
c
c
c
c
where we have taken i , j and k outside the integral as they have constant magnitude
and direction.
Example 38
Evaluate
vector.
z
C
z
C
A . dr if A = A1(x, y, z) i + A2(x, y, z) j + A3(x, y, z) k and r is the radius
A . dr =
=
z
z
C
C
( A1 i + A2 j + A3 k ) . (dxi + dyj +
( A1dx + A2 dy + A3 dz).
If A is the force F on a particle moving along C, this line integral represents the work
done by the force.
Example 39
z
Evaluate C A . dr from the point P (0, 0, 0) to Q (1, 1, 1) along the curve r = i t + j t2
+ k t3 with x = t, y = t2, z = t3, where A = xy i + z j + xyz k .
Since x = t, y = t2, z = t3, y = x2, z = x3, dy = 2x dx, dz = 3x2 dx.
z
C
A . dr =
z
z
1
=
(xy i + z j + xyz k ) . ( i dx + j dy + k dz) =
1
1
z
z
x 3 dx + 2 x 4 dx + 3 x 8 dx =
0
0
0
59
60
z
(xydx + zdy + xyzdz)
y
(1 , 1 )
2
Example 40
Evaluate
z
C
y =x
A . dr around the closed curve C defined by
y= x
y = x2 and y2 = x, with A = (x – y)i + (x + y)j.
z
C
A . dr =
=
z
z
z
[(x – y) i + (x + y) j ] . [ i dx + j dy]
C
1
=
Fig. 1.27
1
(x – x2)dx +
0
z
(x + x2) . 2xdx
(along y = x2)
0
0
+
x
0
(x – y)dx + (x + y)dy
C
2
z
1
0
(y2 – y) . 2ydy +
z
(y2 + y)dy (along y2 = x)
1
= 2/3
If the movement was opposite to the indicated direction then the result would have
2
been - .
32 Mechanics of Particles, Waves & Oscillations
Example 41
Evaluate
z
C
A × dr if A = A1 i + A2 j + A3 k
i
A × dr = A1
dx
j
A2
dy
k
A3
dz
= i(A2dz – A3dy) + j(A3dx – A1dz) + k(A1dy – A2dx).
So the integral is:
z
C
z
A × dr = i ( A2 dz - A3 dy) + j
z
C
( A3 dx - A
Example 42
Solve Kepler’s problem by the Vector method.
The solution to Kepler’s problem can be obtained by
solving a differential equation for the motion of a planet
in a gravitational field. Here we demonstrate the power of
the Vector method in solving the same problem.
Consider a planet of mass m around a heavy Sun of
mass M at S Fig. 1.28. Let k be a fixed vector along an
arbitrary reference line which is taken as the polar axis.
The co-ordinates of the planet at any instant are defined
by r, the radial distance from the Sun, and the polar angle
q, measured with respect to the polar axis.
The force acting on the planet due to the Sun is
F = – (GmM/r3)r
P
m
r
q
M
k
S
Fig. 1.28
By Newton’s second law,
GmM
r
so that
Now
3
r = m
d2 r
dt
2
= m
dv
dt
dv
GM
= - 3 r
dt
r
d
dv dr
+
(r × v) = r ×
× v
dt
dt
dt
(46)
(47)
dv
d
(r × v) = r ×
(48)
dt
dt
since the second term on the right hand side of (47) vanishes, being the cross product of
parallel vectors. Using (46) in (48)
Therefore
d
(r × v) = r ×
dt
FG - GM rIJ
H r K
3
= 0
r × v = h = const . vector
Vector Calculus 33
or
r×
dr
dt
=h
(49)
dr
= twice the area of the sector, we get 2(dA/dt) = |h|, that is equal areas
dt
are swept out in equal intervals of time. This is Kepler’s second law of planetary motion.
Further, using the fact that if two vectors in a triple scalar product are identical, the
product is zero, we find r . [r × dr/dt] = r . h = 0. Thus r remains perpendicular to the
fixed vector h, and the motion is planar.
Since r ¥
Taking the cross product of (46) with h,
GMr
dv
¥ h = - 3 ¥ h = - GMr × (r × v)
dt
r
r3
where we have used (49). Also,
(50)
d
dv
dh
dv
¥h +
¥h
(v × h) =
× v=
dt
dt
dt
dt
(51)
since h is constant. Combining (50) and (51),
d
GMr
(v × h) = - 3 × (r × v)
dt
r
Now, r = r er, where er is the unit vector. Hence
der dr
dr
er
+
v=
= r
dt
dt
dt
(52)
so that (52) is reduced to
FG
H
de r
GM
d
(v × h) = - 3 r ¥ r ¥ r
dt
r
dt
= - GM
LMFG e
NH
r
.
IJ
K
= – GM er ×
FG e
H
r
¥
de r
dt
IJ
K
IJ
K
der
de
er - (er . er )
dt
d
where we have used the identity (13), and the relations er . er = 1 and er .
der
d
(v × h) = GM
dt
dt
der
dr
= 0.
(53)
Integrating (53), we find
v × h = GM er + k
whence
(54)
r . (v × h) = r . (GM er + k)
= GM r . er + r . k
= GMr + rk cosq
where k is an arbitrary constant vector with magnitude k, and q is the angle between
k and er. Using the result of problem 9.
34 Mechanics of Particles, Waves & Oscillations
r . (v × h) = (r × v) . h = h . h = h2
h2
GM + K cosq
This is the polar equation of conic section. For planetary motion these conic sections
are closed curves so that we deduce Kepler’s first law which states that the orbits of the
planets are ellipses with the Sun as one of the foci. Kepler’s third law can be deduced
without further use of vectors as indicated in Chapter 4.
r =
Example 43
(a) If F = —f, where f is single-valued and has continuous partial derivatives, show that
the work done is moving a particle from point A [x1, y1, z1] to B[x2, y2, z2] in this field
is independent of the path joining the two points.
z
z
z FGH
z
z
(b) Conversely, if C F . dr is independent of the path C joining any two points show that
there exists a function f such that F = —f.
B
(a) Work done =
B
z
F . dr =
A
B
=
A
B
=
A
—f . dr
A
∂f ∂f ∂f i+
j+
k
∂x
∂y
∂z
IJ
K
∑ (dx i + dy j + dz k )
∂f
∂f
∂f
dx +
dy +
dz
∂x
∂y
∂z
B
=
A
df = f(B) – f(A) = f(x , y , z ) – f(x , y , z ).
2
2
2
1
1
1
B
II
Thus the integral depends only on the points A
and B, not on the path joining them. It is assumed
that f(x, y, z) is single valued at A and B. In Fig.
I
A
1.29 two paths I and II are shown. The above
statement would then imply that the result of
integration along these two paths would be
Fig. 1.29
identical.
(b) Suppose the line integral is independent of the path C, that is, F is a conservative
field, then
( x , y, z )
f(x, y, z) =
z
( x1 , y1 , z1 )
( x , y, z )
F . dr =
z
( x1 , y1 , z1 )
F .
dr
ds
ds
df
dr
=F .
.
ds
ds
dr
df
dr
= 0.
But
= —f .
; so that (—f - F) .
ds
ds
ds
Since this result must be valid irrespective of (dr/ds), we deduce F = —f.
Differentiating,
Vector Calculus 35
Example 44
(a) If F is a conservative field, prove that Curl F = — × F = 0, that is F is irrotational.
(b) Conversely, if — × F = 0, prove that F is conservative.
(a) If F is a conservative field then by Example 43, F = —f.
Then curl F = — × —f = 0 where we have used the fact that the curl of a gradient
of f is zero.
i
∂
(b) If — × F = 0, then
∂x
F1
so that
j
∂
∂y
F2
k
∂
= 0,
∂z
F3
∂F3
∂F2
=
∂y
∂z
(55)
∂F1
∂F3
=
∂z
∂x
(56)
∂F1
∂F2
=
∂y
∂x
(57)
Required to prove that F = —f
Work done to move the particle along a path C joining (x1, y1, z) and (x, y, z) in the
force field F is
z
C
F1 (x, y, z)dx + F2 (x, y, z)dy + F3 (x, y, z)dz
With the choice of a path consisting of straight line segments from (x1, y1, z1) to (x, y1,
z1) to (x, y, z1) to (x, y, z) and calling f(x, y, z) the work done along this particular path,
we have
y
x
f(x, y, z) =
z
x1
F1 ( x, y1 , z1 )dx +
z
F2 ( x, y, z1 )
y1
∂f
= F3 (x, y, z), since differentiation of the first two terms on the right hand
∂z
side give zero.
whence
∂f
= F2(x, y, z1) +
∂y
z
z
z
z1
z
= F2(x, y, z1) +
z1
∂F3 ( x, y, z ) dz
∂y
∂F2 ( x, y, z ) dz
∂z
z
= F2(x, y, z1) + F2(x, y, z) |z1
36 Mechanics of Particles, Waves & Oscillations
= F2(x, y, z1) + F2(x, y, z) – F2(x, y, z1)
= F2(x, y, z)
where we have used (55).
∂f
= F1(x, y1, z1) +
∂x
y
z
z
y1
y
= F1(x, y1, z1) +
y1
∂F2 ( x, y, z1 )dy
+
∂x
∂F1 ( x, y, z1 ) dy
+
∂y
z
z
z1
∂F3 ( x, y, z ) dz
∂x
z
z
z1
∂F1 ( x, y, z ) dz
∂z
y
z
= F1(x, y1, z1) + F1(x, y, z1) |y1 + F1(x, y, z) |z1
= F1(x, y1, z1) + F1(x, y, z1) – F1(x, y1, z1) + F1(x, y, z) – F1(x, y, z1)
= F1(x, y, z)
where we have used (56) and (57).
∂f ∂f ∂f i+
j+
k = —f.
If follows that F = F1 i + F2 j + F3 k =
∂x
∂y
∂z
We conclude that a necessary and sufficient condition that a field F be conservative
is that Curl F = — × F = 0.
Example 45
(a) Show that F = (3x2y + z3) i + x3 j + 3xz2 k is a conservative force field,
(b) Find the scalar potential,
(c) Find the Work done in moving an object in this field from (1, –1, 2) to (2, 1, 1).
(a) It is sufficient to show that Curl F = 0.
—× F =
i
∂
∂x
3x 2 y + z 3
j
∂
∂y
x3
k
∂
∂z
3xz 2
= 0
Thus F is a conservative force field.
(b) df = F. dr = (3x2y + z3)dx + x3dy + 3xz2dz
= (3x2ydx + x3dy) + (z3dx + 3xz2dz)
= d(x3y) + d(z3x) = d(x3y + z3x)
f = x3y + z3x + constant.
(c) Work done = f(2, 1, 1) – f(1, –1, 2) = 3.
1.18.2 Surface Integrals
Let a surface S be divided into infinitesimal elements each of which may be considered
as a vector dS. Integrals that involve the differential element dS of the surface area are
called Surface Integrals. There are three types of surface integrals.
Vector Calculus 37
(a)
zz
fdS
zz
(b)
S
A . dS
(c)
S
zz
A ¥ dS
S
The result of (a) is a vector, that of (b) is a scalar and that of (c) is a vector.
If we are dealing with a closed surface, the surface integrals are written as
zz
zz
or
f dS
S
A . dS
or
S
zz
A ¥ dS
S
For closed surfaces, it is usual to assume that the positive direction of the normal
extends outward from the surface. (A closed surface is one that has no boundary and
completely encloses a bounded region in space). The surface integral
zz
V . dS is called
S
the flux of V through the surface, for if V is the product of density and velocity of fluid,
the integral represents the amount of fluid flowing through a surface in unit time.
Alternatively, the vector V may refer to electric, magnetic, or gravitational force.
1.18.2.1 Examples of Surface integral
(1) The total electric charge on the surface of a shell is given by
z
S
r(r) ds.
(2) The electric flux fE is given by the surface integral,
fE =
z
E . ds
Also gauss’ law is given by
e0
z
E . ds = q
where E is the electric field, e0 the permittivity and q the electric charge.
(3) The electromagnetic flux of energy out of a given volume V bounded by a surface S
is
z
S
( E ¥ H ) . ds .
1.18.3 Volume Integrals
Let dV = dx dy dz be the volume element. Since this is a scalar we can form only two
volume integrals.
(a)
zzz
V
(b)
f dV
zzz
V
A dV
(a) being a scalar and (b) a vector.
1.18.3.1 Examples of Volume Integral
(1) Total mass of a fluid contained in a volume V is give by
z
z
V
r(r ) dV .
(2) Total linear momentum of a fluid is given by
r( r ) v(r) dV, where v(r) is the
V
velocity field in the volume element dV.
(3) Consider
a small volume element dV situated at position r; its linear momentum rdV
.
r , where r = r(r) is the density distribution and its angular momentum J is
.
(r ¥ r) r dV
J =
z
V
38 Mechanics of Particles, Waves & Oscillations
.
putting r = w × r yields
J =
z
V
[r ¥ (w ¥ r )] rdV -
z
V
( r . w )r r
y
1.19 GREEN’S THEOREM IN THE PLANE
d
zz
S
d
∂v ( x, y)
dxdy =
∂x
b
∂v ( x, y)
dxdy =
x
∂
a
zz
c
z
S
d
z
C
c
a
Fig. 1.30
v (b, y) - v (a, y) dy
b
x
Let u(x, y) and v(x, y) be functions with continuous
first partial derivatives. Consider the double integral of
∂
v(x, y) over the rectangle S as in Fig. 1.30. We shall
∂x
show that the double integral is equal to a line integral
around the boundary of the rectangle. We first perform
the x integration to find.
(58)
c
We now evaluate
v(x, y) dy in the counterclockwise direction so that S is always
towards left as we move around the closed curve. We note that along the horizontal sides
of S, integrals are zero since dy = 0. Along the right side, x = b, and y ranges from c to
d. Along the left side, x = a, and y ranges from d to c. Hence,
d
z
v( x, y) dy =
c
z
c
z
v (b, y) dy + v (a, y) dy =
c
d
Combining (58) and (59),
zz
S
∂v
dx dy =
∂x
d
z
v (b, y) - v (a, y) dy
(59)
c
z
v dy.
(60)
z
u dx.
(60a)
c
Similarly,
-
zz
S
∂u
dx dy =
∂y
c
z
c
(u dx + v dy) =
zz FGH
S
IJ
K
∂v ∂ u
dx dy
∂x ∂y
Formula (62) is called Green’s theorem in the plane.
This result which was proved for a rectangle can be
generalized to any arbitrary figure in the xy-plane bounded
by a closed curve as in Fig. 1.31. Let the figure S be
divided into a set of small rectangles. For each rectangle
the result of (62) is valid. The double integral over the
whole area S is equal to the sum of all the double integrals
in (62). However, it is seen that for two typical small
rectangles P and Q, the line integrals are in the opposite
(61)
Fig. 1.31
Vector Calculus 39
sense over the common boundary LM and therefore get cancelled. What is left over is
simply the line integral around C. Hence the result (62) is equally valid for an arbitrary
figure in a plane. Thus, with the use of Green’s theorem, the line integral around a closed
path can be evaluated or a double integral over the area enclosed can be found out.
Example 46
z
Verify Green’s theorem in the plane for ( x - y) dx + ( x + y) dy where C is the closed curve
C
of the region bounded by y = x and y = x2.
Let
u =x– y
∂u
= –1
∂y
;
v =x+ y
∂v
=1
∂x
The curves y = x and y = x2 intersect at (0, 0) and (1, 1)
z z FGH
;
IJ
K
∂v ∂u
dx dy =
∂x ∂y
zz
1 - ( -1) dxdy
S
1
x
z z
= 2
x=0y=x
1
= 2
ze
0
dxdy = 2
2
x - x 2 dx =
j
LM OP
MN dyPQ dx
1 O
L2
2M x - x P
3 Q
N3
1
x
0
x2
zz
32
3
1
0
=
2
3
This is in agreement with the result obtained for the line integral in Ex. 40.
1.20 STROKES’ THEOREM
This theorem states that if S is an open, two-sided surface bounded by a simple closed
curve C, then, if A has continuous derivatives,
z
C
A . dr =
zz
S
( — ¥ A ) . n dS =
zz
( — ¥ A ) . dS
S
where C is traversed in the positive direction.
In words, “The line integral of the tangential component of a vector A taken around
a simple closed curve C is equal to the surface integral of the normal component of the
curl of A taken over any surface S having C as its boundary”. This theorem relates an
integral over an open surface to the line integral around the curve bounding the surface.
Proof
Let S be a surface bounded by closed contour C. Let C¢ be its projection on the xy-plane
as in Fig. 1.32. Let a vector at each point on the surface be defined as A = A1i + A2j + A3 k.
We associate a point P(x, y) on the plane with every point on the surface. Since on
the surface, z is a function of x and y, that is z = f(x, y), a function A1(x, y, z) becomes
A1(x, y, z) = f(x, y), since the value of f must be equal to the value of A1 on C. We may
similarly consider projections on yz and xz-planes.
40 Mechanics of Particles, Waves & Oscillations
We are required to prove that
zz
(— ¥ A ) . n dS =
zz
z
z
b
dS
S
g
— ¥ A1i + A2 j + A3 k . n dS
S
S
=
C
A . dr
y
0
R
x
dx dy
— ¥ ( A1 i) . n dS is transformed.
S
Now,
i
∂
— × (A1 i) =
∂x
A1
P ¢( x, y, z )
C
where C is boundary of S and n is a unit vector
perpendicular to the surface at any point. We can find
out how a typical term
zz
n
j
∂
∂y
0
FG ∂A
H ∂z
1
P ( x, y )
Fig. 1.32
k
∂A
∂A1
∂
= j 1 -k
∂z
∂y
∂z
0
= [— × (A1i) . n dS] =
C¢
n. j-
IJ
K
∂A1
n . k dS
∂y
(62)
Since z = f(x, y) is taken as the equation of S, the position vector to any point of S is
r = xi + yj + zk = xi + yj + f(x, y)k.
∂z
∂f
∂r
k = j+
k
= j+
∂y
∂y
∂y
But
∂r
is a vector tangent to S and thus perpendicular to n, so that
∂y
∂r
∂z
n.
n.k = 0
= n . j +
∂y
∂y
or
n . j = -
∂z
n.k
∂y
(63)
Substitute (63) in (62)
FG ∂A
H ∂z
1
n.j -
FG
H
IJ
K
[— × (A1 i] . ndS = -
or
Now, on S,
IJ
K
∂A1 ∂z
∂A1
∂A1
n . k dS = - ∂z ∂y n . k - ∂y n . k dS
∂y
FG ∂A
H ∂y
1
+
∂A1 ∂z
∂ z ∂y
IJ
K
n . k dS
(64)
A1(x, y, z) = A1[x, y, f(x, y)] = F(x, y)
Therefore,
Using (65) in (64)
∂F
∂A1 ∂A1 ∂z
=
+
∂y
∂y
∂z ∂y
[— × (A1 i)] . n dS = -
∂F
∂F
n . k dS = dx dy
∂y
∂y
(65)
Vector Calculus 41
Therefore,
zz
z
S
[— × (A1 i] . n dS =
zz
R
∂F
dx dy
∂y
–
where R is the projection of S on the xy-plane. By Green’s theorem for the plane, the last
integral equals
C¢
F dx where C' is the boundary of R. Since at each point (x, y) of C'
the value of F is the same as the value of A1 at each point (x, y, z) of C, and since dx
is the same for both C and C', we must have
z
C¢
zz
or
S
F dx =
[— ¥ ( A1 i)] . n dS =
z
z
C¢
C
A1dx
A1dx
(66)
Similar equations are obtained by considering the projections of S on the xz and yzplanes.
zz
zz
S
S
[— ¥ ( A2 j)] . n dS =
[— ¥ ( A3 k )] . n dS =
z
z
C
C
A2 dy
(67)
A3 dz
(68)
Adding (66), (67) and (68), we get the desired result
zz
(— ¥ A ) . n dS =
z
C
A1dx + A2 dy + A3 dz =
z
C
A . dr
(69)
Green’s theorem in the plane is a special case of Stoke’s theorem.
1.21 DIVERGENCE THEOREM
A method of reducing triple integrals to double integrals is provided by the Divergence
theorem of Gauss.
z
To prove:
zzz
V
— . A dV =
zz
zzz
dS 1
S
If A is the flux density of an incompressible fluid
then we have shown that — . AdV gives the total
amount of fluid flowing out of the volume dV per
second. The total flow from a large volume is
— . AdV which must be equal to the rate of flow
zz
A . dS
S1
q1
n^ 1
A . n dS
q2
dS 2
S2
n^ 2
0
R
across all the surfaces of the given volume
x
This is the proof of Divergence (Gauss) theorem. An
Fig. 1.33
analytical proof follows.
Let S be a closed surface which is such that any line parallel to the co-ordinate axes
cuts S in utmost two points. Let the equation of the upper portion S1 be z = f1(x, y) and
that of the lower portion S2, z = f2(x, y). Let the vector field be given by A = A1 i + A2 j
+ A3 k. Consider the projection of the surface on the xy-plane, called R.
42 Mechanics of Particles, Waves & Oscillations
zzz
zzz FGH
— . A dV =
IJ
K
∂A1 ∂A2 ∂A3
+
+
dV
∂x
∂y
∂z
Now,
zzz
V
∂A3
dV =
∂z
zzz
zz
zz
V
=
∂A3
dzdxdy =
∂z
LM
MN
f1 ( x , y )
R
f2
OP
PQ
∂A3
dz dx dy
z
∂
( x, y)
zz z
f1
A3 ( x, y, z ) | dx dy
=
z = f2
[ A3 ( x, y, f1 ) - A3 ( x, y, f2 )]dy dx
R
For the upper portion S1, dy dx = cosq1 dS1 = k . n1 dS1 since the normal n1 to S1
makes an acute angle q1 with k.
For the lower portion S2, dy dx = –cosq2 dS2 = –k . n2 dS2 since the normal n2 to S2
makes an obtuse angle q2 with k.
Then,
zz
zz
A3 ( x, y, f1 )dy dx =
R
zz
zz
A3 k . n1 dS1
S1
A3 ( x, y, f2 )dy dx = -
A3 k . n 2 dS2
S2
R
Therefore,
zz
zz
zz
A3 ( x, y, f1 )dy dx –
=
A3 k . n 1 dS1 +
S1
=
zz
A3 ( x, y, f2 )dy dx
R
R
zz
A3 k . n 2 dS2
S2
A3 k . n dS
S
So that
zzz
V
∂A3
dV =
∂z
zz
A3 k . n dS
(70)
S
Similarly from considerations of projections of S on the yz and xz-planes we have
zzz
zzz
V
V
∂A1
dV =
∂x
∂A2
dV =
∂y
zz
zz
dS
A1 i . n
(71)
A2 j . n dS
(72)
S
S
zzz FGH
V
IJ
K
∂A1 ∂A2 ∂A3
dV =
+
+
∂x
∂y
∂z
zz
S
( A1 i + A2 j + A3 k ) . n dS
Vector Calculus 43
zzz
or
— . A dV =
V
zz
dS
A.n
(73)
S
The Divergence theorem gives a passage from volume integral to surface integral. It
is a generalization of Green’s theorem in the plane and is called Green’s theorem in
space.
1.22 GREEN’S THEOREMS
1.22.1
Green’s First Identity or Theorem
zzz
[ f— 2 Y + (—f) . (—Y ] dV =
zz
(74)
( f—Y ) . dS
S
V
This is easily obtained by substituting A = f—Y in (73).
1.22.2
Green’s Second Identity or Theorem
zzz
( f — 2 Y - Y— 2 f) dV =
V
zz
(75)
( f—Y - Y—f). dS
S
This is obtained by interchanging f and Y in (74) and subtracting the result from
(74). Green’s identities are most frequently encountered as transformation formulae in
mathematical physics.
Example 47
Verify Stokes’ theorem for A = (x – y) i + yz2 j – y2z2 k where S is the upper half surface
of the sphere x2 + y2 + z2 = 1 and C its boundary.
The boundary C of S is a circle in xy-plane of radius unity and centre at the origin.
Let x = cos q, y = sin q, z = 0, 0 £ q £ 2 p be the parametric equations of C. Then,
z
C
A . dr =
z
C
( x - y) dx + yz 2 dy - y 2 z 2 dz
2p
=
z
(cos q - sin q) ( - sin q dq) = p
n
0
Also,
—× A =
S
i
j
∂
∂
∂x
∂y
x - y yz 2
k
∂
∂z
- y2 z 2
R
C
Fig. 1.34
= i(–2yz2 – 2yz2) + j(0) + k(1) = k
since z = 0.
Then
zz
S
(— ¥ A ) . n dS =
zz
S
k . n dS =
zz
dx dy
R
since n . k dS = dx dy and R is the projection of S on the xy-plane. But the area of the
circle of unit radius is p. Thus Stokes’ theorem is verified.
44 Mechanics of Particles, Waves & Oscillations
Example 48
Given A = 3y i + x j + 2z k , find
z ≥ 0.
z
(— × A) . n dS over the hemisphere x2 + y2 + z2 = a2,
First method
i
∂
∂x
3y
—× A =
j
∂
∂y
x
k
∂
∂z
2z
= – 2 k .
Since this plane area is in the xy-plane, we have
n = k ; (— × A) . n = –2 k . k = –2
so the integral is
–2
z
ds = – 2 . pa2 = – 2pa2
Second method
Using Stoke’s theorem we evaluate
z
C
A . dr =
=
z
z
z
C
A . dr around the circle x2 + y2 = a2 in the xy-plane.
(3 y i + x j + 2 z k ) . (dx i + dy j + dz k )
3 y dx + x dy + 2 z dz.
Put x = a cosq, dx = – a sinq dq ; y = a sinq, dy = a cosq dq;
z = 0, with 0 £ q £ 2p.
z
A . dr = – 3a2
2p
2p
0
0
z
sin 2 q dq + a 2
z
cos2 q dq = – 2pa2.
Example 49
Use the divergence theorem to show that
radius R and A = i x3 + j y3 + k z3.
zz
But
A . dS =
— . A =
zzz
z
S
A . dS =
12
pR5 where S is the sphere of
5
∂ 3
∂ 3
∂ 3
x +
y +
z
∂x
∂y
∂z
= 3x2 + 3y2 + 3z2 = 3(x2 + y2 + z2) = 3R2,
since
x2 + y2 + z2 = R2.
z
A . dS =
zzz
3R 2 dV =
z
= 12 p R 4 dR =
z
(3R 2 ) (4 p R 2 dR)
12
pR5 .
5
Example 50
Evaluate
z
A . n dS over the closed surface of an open cylinder, of height h and radius R.
Vector Calculus 45
z
By the divergence theorem, this is equal to
—. A =
z
A . n dS =
Surface of
cylinder
— . AdV over the volume of the cylinder.
∂x ∂y ∂z
+
+
= 3.
∂x ∂y ∂z
z
z
— . A dV =
3 dV = 3V
Volume of
cylinder
= 3 times volume of cylinder = 3pR2h.
Example 51
Show that the radius vector r = xi + yj + zk is irrotational.
Solution
To show
—× r =0
—× r =
i
∂
∂x
x
j
∂
∂y
y
k
∂
∂z
z
FG ∂z - ∂y IJ
H ∂y ∂ z K
F ∂z - ∂x IJ + k FG ∂y - ∂x IJ
– j G
H ∂x ∂z K H ∂x ∂y K
= i
=0 + 0 + 0 = 0
Example 52
If f = 4x3 + 3yz2 – z3, find —2f at (1, –1, –1).
Solution
— 2f =
∂2 f
+
∂2 f
+
∂2 f
∂x 2 ∂y 2 ∂z 2
= 24x + 0 + (6y – 6z)
= [(24) (1) + (6) (–1) – (6) (–1)]
= 24
Example 53
Show that A . (B ¥ C) = ( A ¥ B) . C
Solution
A . (B ¥ C) = (A1 i + A2 j + A3 k) ∑
i
B1
C1
j
B2
C2
k
B3
C3
46 Mechanics of Particles, Waves & Oscillations
A1
=
=
A2
A3
B1
B2
B3
C1
C2
C3
C1
C2
C3
A1
A2
A3
B1
B2
B3
=–
C1
C2
C3
B1
B2
B3
A1
A2
A3
= C . ( A ¥ B) = ( A ¥ B ) . C
Example 54
Find —
FG 1 IJ , where r is position vector.
Hr K
2
Solution
—
FG 1 IJ = FG i ∂ + j ∂ + k ∂ IJ (x
H r K H ∂x ∂y ∂z K
2
2
+ y2 + z 2
=
- 2 ( xi + yj + zk)
- (2 xi + 2 yj + 2 zk)
=
2
2
2 2
r4
(x + y + z )
=
-2r
r4
Example 55
Prove that — . r = 3 (r is a position vector). [OU March’97] [OU March’99]
Solution
—. r =
=
FG i ∂ + j ∂ + k ∂ IJ . (xi + yj + zk
H ∂x ∂ y ∂z K
∂x ∂y ∂z
+
+
= 1 + 1 + 1 = 3
∂x ∂y ∂z
[Refer Example 29]
Example 56
For scalar function f(x, y, z) and a vector A = Ax i + Ay j + Az k show that — × (fA)
= f (— × A) + —f × A.
Solution
— × (f A) = — × (fA1 i + fA2 j + fA3 k)
=
i
∂
∂x
fA1
j
∂
∂y
fA2
k
∂
∂z
fA3
Vector Calculus 47
LM ∂ (fA ) - ∂ (fA )OP i + L ∂ (fA
∂z
N ∂y
Q MN ∂z
L ∂A + ∂f A - f ∂A - ∂f A OP
= Mf
∂z
∂z
N ∂y ∂y
Q
L ∂A + ∂f A - f ∂A - ∂fA OP j + LMf ∂A
+ Mf
∂x
∂x Q
N ∂z ∂z
N ∂x
LF ∂A - ∂A IJ i + FG ∂A - ∂A I
= f MG
NH ∂y ∂z K H ∂z ∂x K
LF ∂f A - ∂f A IJ i + FG ∂f A - ∂f
+ MG
NH ∂y ∂z K H ∂z ∂x
=
3
3
2
3
1
2
3
3
1
3
2
= f(— × A) +
i
∂f
∂x
A1
1
j
∂f
∂y
A2
k
∂f
∂z
A3
= f(— × A) + (—f) × A
Example 57
If f is a scalar quantity show that curl (grad f) = 0.
Solution
i
∂
∂x
=
∂f
∂x
FG ∂f i + ∂f j + ∂f kIJ .
H ∂x ∂y ∂ z K
j
∂
∂y
∂f
∂y
k
∂
∂z
∂f
∂z
LM ∂ FG ∂f IJ - ∂ FG ∂f IJ OP i - L ∂ FG ∂f
N ∂y H ∂z K ∂z H ∂y K Q MN ∂x H ∂z
L ∂ F ∂f I ∂ FG ∂f IJ OP k
+ M G JN ∂x H ∂y K ∂y H ∂x K Q
L ∂ f - ∂ f OP i - LM ∂ f - ∂ f OP
= M
N ∂y∂z ∂z∂y Q N ∂x∂z ∂z∂x Q
=
2
=0
2
2
LM ∂ (fA ) - ∂ (fA )OP k
∂y
N ∂x
Q
2
1
2
3
1
3
— × (—f) = — ×
+
2
2
2
+
OP
Q
∂fA2
∂A
∂f
-f 1 A1 k
∂x
∂y
∂y
48 Mechanics of Particles, Waves & Oscillations
Example 58
Find —f where f = log|r|
Solution
|r| = xi + yj + zk. Then |r| =
—f =
=
and f = log |r| =
1
— log (x2 + y2 – z2)
2
RS
T
1 R
=
Si
2 T x
=
x 2 + y2 + z 2
1
∂
∂
i
l
log ( x 2 + y 2 + z 2 ) + j
2 ∂x
∂y
2
2x
2y
+
+j 2
+ y2 + z 2
x + y2 + z 2
r
x i + y j + zk
= 2
2
r
r
Example 59
Prove that — . ( A + B) = — . A + — . B .
Solution
Let A = A1i + A2 j + A3 k, B = B1i + B2 j + B3 k.
Then,
— . ( A + B) =
FG ∂ i + ∂ j + ∂ k IJ .
H ∂x ∂y ∂z K
( A1 + B1 ) i + ( A2 + B2 ) j + ( A3 +
=
∂
∂
∂
( A1 + B1 ) +
( A2 + B2 ) +
∂x
∂y
∂z
=
∂A1 ∂B1 ∂A2 ∂B2 ∂A3
+
+
+
+
+
∂x
∂x
∂y
∂y
∂z
=
FG ∂ i + ∂ j + ∂ kIJ . (A i + A j + A k)
H ∂x ∂y ∂z K
F ∂ ∂ j + ∂ kIJ . (B i + B j + B k)
+ G i+
H ∂x ∂y ∂z K
1
2
1
= —. A + —.B
Example 60
Show that div ( fA) = f div A + A . grad f .
3
2
3
1
log (x2 + y2 +z2)
2
Vector Calculus 49
Solution
div ( fA ) = — ( fA ) = — . ( fA1i + fA2 j + fA3 k )
=
∂
∂
∂
(fA1 ) +
( fA2 ) +
( fA3 )
∂x
∂y
∂z
=
∂A
∂A
∂f
∂f
A1 + f 1 +
A2 + f 2 +
∂x
∂x
∂y
∂y
=
FG ∂f i + ∂f j + ∂f k IJ . (A i + A j + A k )
H ∂x ∂y ∂z K
F ∂ ∂ j + ∂ k IJ . (A i + A j + A
+ f G i +
H ∂x ∂y ∂z K
1
2
1
3
2
3
k )
= (—f) . A + f (— . A )
Example 61
If A = iy + j(x2 + y2) + k(yz + zx), then find the value of curl A at (1, –1, 1)
Solution
Curl A = — ¥ A
=
i
∂
∂x
y
j
∂
∂y
x 2 + y2
k
∂
∂
∂ 2
∂
∂y
( yz + zx ) ( x + y 2 ) – j
( yz + zx ) = i
∂z
∂y
∂z
∂x
∂z
yz + zx
LM
N
+ k
OP
Q
LM ∂ (x
N ∂x
2
+ y2 ) -
∂y
∂y
LM
N
OP
Q
OP
Q
= (z – 0) i – z j + (2x – 1) k
= z i – z j + (2x – 1) k
= (1) i – (1) j + [(2)(1) – 1] k
= i – j + k
Example 62
Find the value of the constant ‘a’ for which the vector A = i(x + 3y) + j(y – 2z)
+ k(x + az) is a solenoidal vector.
Solution
A vector A is solenoidal if its divergence is zero.
div A = — . A =
=
FG i ∂ + j ∂ + k ∂ IJ .
H ∂x ∂y ∂z K
i ( x + y) + j ( y - 2 z ) + k ( x + az )
∂
∂
∂
( x + y) +
( y - 2z) +
(x +
∂x
∂y
∂z
50 Mechanics of Particles, Waves & Oscillations
=1 + 1 + a = 0
a = –2
Example 63
Evaluate div F where F = 2x3z i – xy2z j + 3y2 x .
Solution
FG
H
∂ ∂
∂
+j
+ k
div F = — . F = i
∂x
∂y
∂z
=
IJ
K
. (2x3z i – xy2z j + 3y2x k )
∂
∂
∂
(2 x 3 z ) +
( - xy 2 z ) +
(3 y 2
∂x
∂y
∂z
= 6x2z – 2xyz + 0
= 6x2z – 2xyz
Example 64
If f = xy2 i + 2x2yz j – 3yz2 k find curl f at (1, –1, 1)
Solution
i
∂
Curl f = — × f =
∂x
xy 2
=
j
k
∂
∂
∂y
∂z
2 x 2 yz - 3 yz 2
LM ∂ (- 3yz ) - ∂ (2x yz)OP i – L ∂ (- 3yz ) - ∂ (xy )O j
PQ
∂z
∂z
N ∂y
Q MN ∂x
L∂
O
∂
( xy )P k
+ M (2 x yz ) x
y
∂
∂
N
Q
2
2
2
2
2
= [– 3z2 – 2x2y] i – [0 + 0] j + [4xyz – 2xy] k
= (– 3z2 – 2x2y) i + (0) j + (4xyz – 2xy) k
= – i – 2 k
where we have put x = 1, y = –1 and z = 1.
Example 65
Show that the force F = (y2 – x2) i + 2xy j is conservative.
Solution
To show — ¥ F = 0
—¥F =
i
∂
∂x
y2 - x 2
j
∂
∂y
2 xy
k
∂
∂z
0
LM
N
OP LM
Q N
∂
∂
(2 xy) - j 0 ( y2 = i 0 ∂z
∂z
2
Vector Calculus 51
= + k
LM ∂ (2xy) - ∂ ( y
N ∂x ∂y
2
- x2 )
OP
Q
= 0 + 0 + k (2y – 2y)
=0
Example 66
zz
Find
F . n ds , where F = 4xzi – y2j + yzk and S is the surface of a cube bounded by x
S
= 0, x = 1, y = 0, y = 1, z = 0 and z = 1
Solution
By the divergence theorem, the required integral is equal to
zzz
— . FdV =
zzz LMN
zzz
z z
V
V
=
∂
∂
∂
(4 xz ) +
(- y2 ) +
(y
∂x
∂y
∂z
V
1
=
1
(4z - y) dV =
1
1
1
z z z
(4 z - y) dz dy dx
x=0 y=0 z=0
2 z 2 - yz
x=0 y=0
z =1
z =0
1
dy dx =
1
z z
(2 - y) dy dx =
x=0 y=0
3
2
QUESTIONS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
If A . B = A . C, must we conclude that B = C?
If |A . B| = |A| |B|, what can you say about A and B?
If |A + B|2 = |A|2 + |B|2, what can you say about A and B?
If A . B = |A × B|, what can you say about the angle between A and B?
If A × B = A × C, must we conclude that B = C?
Give the geometric interpretation of the identity
(A – B) × (A + B) = 2A × B.
Give the geomertric interpretation of the identity
(A + B) . (A – B) = A2 – B2
The Work-Energy theorem states that the work done on a particle in moving it from
A to B is equal to the change in Kinetic energies at these points rather than change
in potential energies of these points. Why?
In the vector notation the inverse square law of gravitation is written as F =
GMmr
. How do you accept the use of r3 term in the denominator rather than r2
r3
to describe the inverse square law?
Is the order of multiplication important in forming Scalar products? Vector products?
Give a physical example of (a) Triple scalar product, (b) Triple vector product.
Would you regard Del as a vector, an operator or both?
52 Mechanics of Particles, Waves & Oscillations
13. What is the criterion for a force to be conservative?
14. We can form three types of line integrals and three types of surface integrals but
only two types of volume integrals. Why?
15. Does Stoke’s theorem apply to the Fig. 1.35 shown?
16. For what kind of surface is Green’s theorem applicable?
17. For what kind of surface is Stokes’ theorem applicable?
18. For what kind of surface is the Divergence theorem of
Fig. 1.35
Gauss applicable?
19. Stokes’ theorem gives a passage from
integral to
integral.
20. The Divergence theorem gives a passage from
integral to
integral.
21. (a) Derive an expression for the divergence of a vector field in Cartesian coordinates.
(b) Give the physical significance of divergence of vector field.
22. Explain the terms (a) divergence (b) curl of a vector field. Derive expressions in
terms of Cartesian components for div A and curl A. Give examples from Physics
where these ideas are used.
23. What is meant by curl of a vector function? Derive an expression for it in Cartesian
coordinates.
24. (a) Explain the terms (i) Div (ii) Curl (iii) Gradient, with examples.
(b) Prove the following:
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
(i) — (u A ) = A . —u + u— . A
What are line, surface and volume integrals? Derive Gauss’ divergence theorem.
Explain the terms gradient, divergence and curl bringing out their use in physics.
Explain curl of a vector with two examples.
State and prove Stokes’ theorem.
Define curl and div. of a vector.
State and prove Green’s theorem.
Prove Curl V = 2 w.
Explain what is meant by a scalar field and a vector field.
Define curl of a vector and explain its physical significance.
Explain the physical meaning of divergence of vector and define the curl of a vector.
Prove that ‘the gradient of a scalar field is vector’ and explain the physical significance
Derive an expression for the divergence of a vector field.
Explain how a del operator obeys all the rules of differential calculus.
What are the properties of vector product? What are their physical applications?
Explain the physical significance of divergence with two examples.
Explain the physical significance of line integral.
Explain the physical significance of gradient of a scalar field with two examples.
Prove that if A, B, C are non-zero vectors and (A × B) . C = 0 then A, B, C are
coplanar.
What is a scalar triple product and a vector triple product?
Vector Calculus 53
PROBLEMS
1. Find the work done in moving an object along a vector, r = 2i + 3j – k if the applied
force is F = i – 2j – 6k.
(Hint: Work done = F . r)
[Ans. 2]
2. Prove that |A × B|2 + |A . B|2 = |A|2 |B|2.
3. Evaluate A × B, where A = –3 i – 4 j + 2 k and B = 2 i – j – k .
[Ans. 6 i + j + 11 k ]
4. Prove that A × B = –B × A.
5. If A × B = 0 and if A and B are not zero then show that A is parallel to B.
6. Find the angle between A = i – 2j + 2k and B = 2i – 3j + 6k.
[Ans. q = 17°45¢]
7. Prove that (A × B) × C = (A . C)B – (B . C)A.
8. Prove that (A × B) × C = A × (B × C) if and only if (C × A) × B = 0.
9. Prove that A . (B × C) = B . (C × A) = C . (A × B).
(Hint: Write the triple products in the form of determinants and use the theorem
of determinant which states that the interchange of two rows of a determinant
changes its sign).
10. Prove that (A × B) × (C × D) = [CDA] B – [CDB] A.
11. Prove that if A, B and C are noncoplanar and A × (B × C) = (A × B) × C = 0, then
A, B and C are mutually perpendicular.
12. If a and b are unit vectors and q is the angle between them, show that
1
1
|a - b| = sin q .
2
2
13. Prove that (A – B) × (A + B) = 2A × B.
14. If r = sin t i + cos t j – t k, show that |d2r/dt2| = 1.
15. Show that
16.
17.
18.
19.
20.
21.
22.
d
dC
dB
dA
(A . B × C) = A . B ×
+ A .
× C+
. B × C where A, B, C are
du
du
du
du
differentiable functions of a scalar u.
Show that —r n = nrn – 2r.
A particle moves such that its position vector at any time t is given by r = cos wt i
+ sin wt j, where w is a constant. Show that (a) the velocity v of the particle is
perpendicular to r. (b) r × v = a constant vector w k.
If f(x, y, z) = 2x2y – y2z3, find grad f at the point (2, 1, –1). [Ans. 8i + 10j – 3k]
Prove that —(f + Y) = —f + —Y.
If f = ln|r|, prove that —f = er/r2.
Find the directional derivative of f = x2yz + xz2 at (1, –1, –1) in the direction
i – 2j – 2k.
[Ans. 11/3]
2
Find a unit normal to the surface x y + 2xz = 4 at the point (2, –2, –3).
LM Ans. 1 i - 2 j - 2 kOP
N 3 3 3 Q
54 Mechanics of Particles, Waves & Oscillations
23. Prove that
— × (A × B) = (B . —)A – B(—
— . A) – (A . —)B + A(—
— . B).
24. Prove that
—(A . B) = (B . —
— × A) + A × (—
— × B).
—)A + (A . —)B + B × (—
25. Show that
A = (y2 cos x + z3)i + (2y sin x – 4)j + (3xz2 + 2)k
is a conservative field.
26. If r is the position vector, show that curl r = 0.
27. If f = 3x3 – y2z, find —2f at (1, 1, 8).
—f) . A + f(—
— . A).
28. Prove that — . (fA) = (—
—
—
— × B).
29. Prove that
. (A × B) = B . (— × A) – A . (—
30. Determine the constant C so that the vector
V = (x + 2y)i + (y – z)j + (x – Cz)k is solenoidal.
3
2
31. If f = 1/r, prove that —f = – r/r , where r =
32. If f = 1/r, prove that —2f = 0.
33. Prove that — ¥
LM r OP
Nr Q
2
2
[Ans. 2]
[Ans. 2]
2
x +y +z .
= 0.
34. Prove that Curl (fgrad f) = 0.
35. If a = a xi + b yj + g zk, show that —(a . r) = 2a, where r is the position vector.
36. If a is a constant vector and r is a position vector, prove that —(a . r) = a.
1
— v2 – v × (—
— × v).
37. Prove that (v . —)
—)v =
2
(Hint: In the table of vector identities, entry No, 9, set A = B = V).
38. If f = 2xyz2, F = xyi – zj + x2k and C is the curve x = t2, y = 2t, z = t3 from t = 0
to t = 1, evaluate the line integrals
(a)
z
C
f dr
z
(b)
F × dr
[Ans. (a)
39. If F = 4xyi – y2j, evaluate
z
C
8
4
i+ j+ k
11
5
(b)
2
7
-9
i- j+ k ]
10
3
5
F . dr where C is the curve in the xy-plane, y = 2x2,
from (0, 0) to (1, 2).
40. If A = (2x2 + 4y) i – 7yz j + 10xz2 k , evaluate
the path C with x = t, y = t2, z = t3.
z
C
[Ans. - 2 3 ]
A . dr from (0, 0, 0) to (1, 1, 1) along
[Ans. 3]
z
41. Verify Green’s theorem in the plane C (3xy + 2y2)dx + x2dy where C is the closed
curve of the region bounded by y = x and y = x2.
42. Verify Stokes’ theorem for A = (x – y)i – yz2j – y2zk, where S is the upper half
surface of the sphere x2 + y2 + z2 = 1 and C is its boundary.
[Ans. p]
43. Evaluate
zzz
— ¥ A dV , where A = yi – xj and R is any three-dimensional region
R
with volume V.
[Ans. –2Vk]
Vector Calculus 55
44. Find
zz
r . dS over the surface S of the sphere x2 + y2 + z2 = a2.
[Ans. 4pa3]
S
45. Show that curl (A × B) = A div B – B div A.
46. Determine the angle between the vectors. A = i + 2 j + 3 k and B = 3 i – 2 j + 4 k .
[Ans. 54°54¢]
47. Show that the triangle with sides A = 3 i + 2 j + k, B = 3i + j - 5k and C =
2i + j - 4k is right angled.
[Hint. Show that A . C = 0]
48. If A = 2xi + 2yj + 3zk, find curl A.
[Ans. 0]
2
3 2
49. If f (x, y, z) = 3x y – y z , find the value of grad f at the point (1, –2, –1).
[Ans. –8i – 12j – 16k]
50. If A = 3i – 4j + 5k and B = 2i + 3j – 4k find the magnitudes of ( A + B) and ( A - B).
[Ans. 3 3 and
131 ]
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