B.tech ii unit-4 material vector differentiation

B.Tech -II sem E.M-II

1. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 1

Unit-IV: VECTOR DIFFERENTIATION

Sr. No. Name of the Topic Page No.
1 Scalar and Vector Point Function 2
2 Vector Differential Operator Del 3
3 Gradient of a Scalar Function 3
4 Normal and Directional Derivative 3
5 Divergence of a vector function 6
6 Curl 8
7 Reference Book 12

2. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 2
VECTOR DIFFERENTIATION

 Introduction:
If vector r is a function of a scalar variable t, then we write
�⃗= �⃗(�)
If a particle is moving along a curved path then the position vector �⃗ of the
particle is a function of �. If the component of �(�) along �−��??????�,
�−��??????�, �−��??????� are�
1(�), �
2(�), �
3(�) respectively.
Then,
�(�)⃗⃗⃗⃗⃗⃗⃗⃗⃗=�
1(�) �̂+�
2(�) �̂+�
3(�) �̂

1.1 Scalar and Vector Point Function :

Point function: A variable quantity whose value at any point in a region of
space depends upon the position of the point, is called a point function.

There are two types of point functions.
1) Scalar point function:
If to each point �(�,�,�) of a region � in space there corresponds a
unique scalar �(�) , then � is called a scalar point function.
For example:
The temperature distribution in a heated body, density of a body and
potential due to gravity are the examples of a scalar point function.
2) Vector point function:
If to each point �(�,�,�) of a region � in space there corresponds a
unique vector �(�) , then � is called a vector point function.
For example:
The velocities of a moving fluid, gravitational force are the examples
of vector point function.


2.1 Vector Differential Operator Del ??????.??????. ??????

3. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 3
The vector differential operator Del is denoted by ??????. It is defined as
??????=�̂
??????
??????�
+�̂
??????
??????�
+??????̂
??????
??????�

Note: ?????? is read Del or nebla.

3.1 Gradient of a Scalar Function:
If ∅(�,�,�) be a scalar function then �̂
??????∅
??????�
+�̂
??????∅
??????�
+�̂
??????∅
??????�
is called the gradient
of the scalar function ∅.
And is denoted by grad ∅.
Thus, grad ∅=�̂
??????∅
??????�
+�̂
??????∅
??????�
+??????̂
??????∅
??????�

grad ∅=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
) ∅(�,�,�)
grad ∅=?????? ∅

4.1 Normal and Directional Derivative:
1) Normal:
If ∅(�,�,�)=� represents a family of surfaces for different values of
the constant �. On differentiating ∅, we get �∅=0
But �∅=∇ ∅ .��⃗
So ∇∅.��=0
The scalar product of two vectors ∇∅ and ��⃗ being zero, ∇ ∅ and ��⃗
are perpendicular to each other. ��⃗ is in the direction of tangent to the
giving surface.
Thus ∇∅ is a vector normal to the surface ∅(�,�,�)=�.

2) Directional derivative:
The component of ∇∅ in the direction of a vector �⃗ is equal to ∇∅.�⃗
and is called the directional derivative of ∅ in the direction of �⃗.

??????∅
????????????
=lim
????????????→0
??????∅
????????????
Where, ??????�=��

4. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 4

??????∅
????????????
is called the directional derivative of ∅ at � in the direction of ��.

4.2 Examples:

Example 1: If ∅=3�
2
�−�
3
�
2
; find grad ∅ at the point (1,-2, 1).
Solution:
grad ∅=∇∅
=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
)(3�
2
�−�
3
�
2
)
=�̂
??????
??????�
(3�
2
�−�
3
�
2
)+�̂
??????
??????�
(3�
2
�−�
3
�
2
)+�̂
??????
??????�
(3�
2
�−�
3
�
2
)
=�̂(6��)+�̂(3�
2
−3�
2
�
2
)+�̂(−2�
3
�)
grad ∅ at (1,−2,1)
=�̂(6)(1)(−2)+�̂[(3)(1)−3(4)(1)]+�̂(−2)(−8)(−1)
=−12�̂−9�̂−16�̂
Example 2: Find the directional derivative of �
2
�
2
�
2
at the point (1,−1,1)
in the direction of the tangent to the curve
�=�
??????
,�=sin2�+1,�=1−���� �� �=0.
Solution: Let ∅=�
2
�
2
�
2

Direction Derivative of ∅ =∇∅
=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
)(�
2
�
2
�
2
)
∇∅=2��
2
�
2
�̂+2��
2
�
2
�̂+2��
2
�
2
�̂
Directional Derivative of ∅ at (1,1,-1)
=2(1)(1)
2
(−1)
2
�̂+2(1)(1)
2
(−1)
2
�̂+2(−1)(1)
2
(1)
2
�̂
=2�̂+2�̂−2�̂ _________ (1)
�⃗=��̂+��̂+��̂=�
??????
�̂+(�??????�2�+1)�̂+(1−����)�̂
�⃗⃗=
????????????⃗
????????????
=�
??????
�̂+2 ���2� �̂+�??????�� �̂

5. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 5
Tangent vector,
Tangent (�� �=0)= �
0
�̂+2 (cos0)�̂+(�??????�0)�̂
=�̂+ 2�̂ __________ (2)
Required directional derivative along tangent =(2 �̂+2�̂−2�̂)
(�̂+2�̂)
√1+4

[���� (1)��� (2)]
=
2+4+0
√5

=
6
√5
_______ Ans.
Example 3: Verify ∇̅×[�(�)�⃗]=0
Solution:
∇̅×[�(�)�⃗]= [�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
]×[�(�)�⃗]
=[�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
]×�(�)(��̂+��̂+��̂)
=[�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
]×[�(�)��̂+�(�)��̂+�(�)��̂]
=|
�̂ �̂ �̂
??????
??????�
??????
??????�
??????
??????�
�(�)��(�)��(�)�
|
=[�
????????????(??????)
??????�
−�
??????
??????�
�(�)]�̂−[�
??????
??????�
�(�)−�
??????
??????�
�(�)]�̂+[�
??????
??????�
�(�)−�
??????
??????�
�(�)]�̂
=[�
??????
????????????
�(�)
????????????
??????�
−�
??????
????????????
�(�)
????????????
??????�
]�̂−[�
??????
????????????
�(�)
????????????
??????�
−�
??????
????????????
�(�)
????????????
??????�
]�̂+ [�
??????
????????????
�(�)
????????????
??????�
−
�
??????
????????????
�(�)
????????????
??????�
]�̂
=
??????′(??????)
??????
[(��−��)�̂−(��−��)�̂+(��−��)�̂]=0 ____ Proved.

6. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 6
4.3 Exercise:
1) Find a unit vector normal to the surface �
2
+3�
2
+2�
2
=
6 �� �(2,0,1).
2) Find the direction derivative of
1
??????
in the direction �⃗ where
�⃗⃗⃗=��̂+��̂+��̂.
3) If �⃗⃗⃗=��̂+��̂+��̂, show that
a) grad �=
??????⃗⃗⃗
??????

b) grad (
1
??????
)=−
??????⃗⃗⃗
??????
3

4) Find the rate of change of ∅=��� in the direction normal to the
surface �
2
�+�
2
�+��
2
=3 at the point (1, 1, 1).

5.1 DIVERGENCE OF A VECTOR FUNCTION :
The divergence of a vector point function ??????⃗ is denoted by div ?????? and is
defined as below.
Let ??????⃗=??????
1�̂+??????
2�̂+??????
3�̂
div ??????⃗=∇⃗⃗⃗.??????⃗=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
)(�̂??????
1+�̂??????
2+�̂??????
3)
=
????????????
1
??????�
+
????????????
2
??????�
+
????????????
3
??????�

It is evident that div ?????? is scalar function.
Note: If the fluid is compressible, there can be no gain or loss in the volume
element. Hence
div ??????̅=??????
Here, V is called a Solenoidal vector function. And this equation is called
equation of continuity or conservation of mass.
Example 1: If �=�
2
+�
2
+�
2
,��� �̅=��̂+��̂+��̂, then find div (��̅)
in terms of u.

7. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 7
Solution: div (��̅)=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
)[(�
2
+�
2
+�
2
)(��̂+��̂+��̂)]
=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
).[(�
2
+�
2
+�
2
)��̂+(�
2
+�
2
+�
2
)��̂+
(�
2
+�
2
+�
2
)��̂]
=
??????
??????�
(�
3
+��
2
+��
2
)+
??????
??????�
(�
2
�+�
3
+��
2
)+
??????
??????�
(�
2
�+�
2
�+�
3
)
=(3�
2
+�
2
+�
2
)+(�
2
+3�
2
+�
2
)+(�
2
+�
2
+3�
2
)
=5(�
2
+�
2
+�
2
)
=5� ___________ Ans.
Example 2: Find the directional derivative of the divergence of �̅(�,�,�)=
���̂+��
2
�̂+�
2
�̂ at the point (2, 1, 2) in the direction of the outer normal to
the sphere �
2
+�
2
+�
2
=9.
Solution: �̅(�,�,�)=���̂+��
2
�̂+�
2
�̂
Divergence of �̅(�,�,�)=∇.̅�̅=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
).(���̂+��
2
�̂+�
2
�̂)
=�+2��+2�
Directional derivative of divergence of (�+2��+2�)
=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
)(�+2��+2�)
=2��̂+(1+2�)�̂+2�̂ __________(i)
Directional derivative at the point (2,1,2)=2�̂+5�̂+2�̂
Normal to the sphere =(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
)(�
2
+�
2
+�
2
−9)
=2��̂+2��̂+2��̂
Normal at the point (2,1,2)=4�̂+2�̂+4�̂ __________ (ii)
Directional derivative along normal at (2,1,2)=(2�̂+5�̂+2�̂).
(4�̂+2�̂+4??????̂)
√16+4+16

8. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 8
[??????��� (??????),(????????????)]
=
1
6
(8+10+8)
=
13
3
____________ Ans.
5.2 Exercise:
1) If �̅=
��̂+��̂+�??????̂
√�
2
+�
2
+�
2
, find the value of div �̅.
2) Find the directional derivative of div (�⃗⃗) at the point (1, 2, 2) in the
direction of the outer normal of the sphere �
2
+�
2
+�
2
=9 for �⃗⃗=
�
4
�̂+�
4
�̂+�
4
�̂.

6.1 CURL:
The curl of a vector point function ?????? is defined as below
Curl ??????̅=∇̅×??????̅ (??????̅=??????
1�̂+??????
2�̂+
??????
3�̂)
=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
)×??????
1�̂+??????
2�̂+??????
3�̂
=|
�̂�̂�̂
??????
??????�
??????
??????�
??????
??????�
??????
1??????
2??????
3
|
=�̂(
????????????
3
??????�
−
????????????
2
??????�
)−�̂(
????????????
3
??????�
−
????????????
1
??????�
)+�̂(
????????????
2
??????�
−
????????????
1
??????�
)
Curl ??????̅ is a vector quantity.
Note: Curl ??????̅=??????, the field ?????? is termed as irrotational.


Example 1: Find the divergence and curl of
�̅=(���)�̂+(3�
2
�)�̂+(��
2
−�
2
�)�̂ �� (2,−1,1)

9. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 9
Solution: Here, we have
�̅=(���)�̂+(3�
2
�)�̂+(��
2
−�
2
�)�̂
Div �̅=∇∅
Div �̅=
??????
??????�
(���)+
??????
??????�
(3�
2
�)+
??????
??????�
(��
2
−�
2
�)
=��+3�
2
+2��−�
2

�̅
(2,−1,1)=−1+12+4−1=14
Curl �̅=|
�̂ �̂ �̂
??????
??????�
??????
??????�
??????
??????�
���3�
2
���
2
−�
2
�
|
=−2���̂−(�
2
−��)�̂+(6��−��)�̂
=−2���̂+(��−�
2
)�̂+(6��−��)�̂
Curl at (2, -1, 1)
=2(−1)(1)�̂+{2(−1)−1}�̂+{6(2)(−1)−2(1)}�̂
=2�̂−3�̂−14�̂ __________ Ans.
Example 2: Prove that
(�
2
−�
2
+3��−2�)�̂+(3��+2��)�̂+(3��− 2��+2�)�̂ is both
Solenoidal and irrotational.
Solution:
Let ??????̅=(�
2
−�
2
+3��−2�)�̂+(3��+2��)�̂+(3��−2��+2�)�̂
For Solenoidal, we have to prove ∇.̅??????̅=0.
Now,

10. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 10
∇.̅??????̅=[�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
].[(�
2
−�
2
+3��−2�)�̂+(3��+2��)�̂+
(3��−2��+2�)�̂]
=−2+2�−2�+2
=0
Thus, ??????̅ is Solenoidal.
For irrotational, we have to prove Curl ??????̅=0
Now, Curl ??????̅=
|
�̂ �̂ �̂
??????
??????�
??????
??????�
??????
??????�
(�
2
−�
2
+3��−2�)(3��+2��)(3��−2��+2�)
|
=(3�+2�−2�+3�)�̂−(−2�+3�−3�+2�)�̂+(3�+2�−2�−3�)�̂
=0�̂+0�̂+0�̂
=0
Thus, ??????̅ is irrotational.
Hence, ??????̅ is both Solenoidal and irrotational. __________ Proved.
Example 3: Find the scalar potential (velocity potential) function � for ??????⃗=
�
2
�̂+2���̂−�
2
�̂.
Solution: We have, ??????⃗=�
2
�̂+2���̂−�
2
�̂.
Curl ??????⃗=∇×??????⃗
=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
)×(�
2
�̂+2���̂−�
2
�̂).
=|
�̂�̂�̂
??????
??????�
??????
??????�
??????
??????�
�
2
2��−�
2
|

11. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 11
=�̂(0)−�̂(0)+�̂(2�−2�)
=0
Hence, ??????⃗ is irrotational.
To find the scalar potential function �.
??????⃗=∇ �
��=
????????????
??????�
��+
????????????
??????�
��+
????????????
??????�
��
=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
).(�̂��+�̂��+�̂��)
=(�̂
??????
??????�
+�̂
??????
??????�
+�̂
??????
??????�
)�.��⃗
=∇�.��⃗
=??????⃗.��⃗ (??????= ∇ �)
=(�
2
�̂+2���̂−�
2
�̂).(�̂��+�̂��+�̂��)
=�
2
��+2�� ��−�
2
��
=�(��
2
)−�
2
��
�=∫�(��
2
)−∫�
2
��
∴ �=��
2
−
�
3
3
+� ________ Ans.



6.2 Exercise:
1) If a vector field is given by ??????⃗=(�
2
−�
2
+�)�̂−(2��+�)�̂. Is this
field irrotational? If so, find its scalar potential.

12. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 12
2) Determine the constants a and b such that the curl of vector ??????̅=
(2��+3��)�̂+(�
2
+���−4�
2
)�̂−(3��+���)�̂ is zero.
3) A fluid motion is given by�̅=(�+�)�̂+(�+�)�̂+(�+�)�̂. Show
that the motion is irrotational and hence find the velocity potential.
4) Given that vector field �̅=(�
2
−�
2
+2��)�̂+(��−��+��)�̂+
(�
2
+�
2
)�̂ find curl �. Show that the vectors given by curl � at
�
0(1,2,−3) ��� �
1(2,3,12) are orthogonal.
5) Suppose that �⃗⃗⃗,�⃗⃗ and � are continuously differentiable fields then
Prove that, div(�⃗⃗⃗×�⃗⃗)=�⃗⃗.���� �⃗⃗⃗−�⃗⃗⃗.���� �⃗⃗.


7.1 REFERECE BOOKS :
1) Introduction to Engineering Mathematics
By H. K. DASS. & Dr. RAMA VERMA
S. CHAND
2) Higher Engineering Mathematics
By B.V. RAMANA
Mc Graw Hill Education
3) Higher Engineering Mathematics
By Dr. B.S. GREWAL
KHANNA PUBLISHERS
4) http://elearning.vtu.ac.in/P7/enotes/MAT1/Vector.pdf