This document provides an overview of vector differentiation, including gradient, divergence, curl, and related concepts. It begins with definitions of scalar and vector point functions. It then defines the vector differential operator Del and explores using it to calculate the gradient of a scalar function, directional derivatives, and normal derivatives. The document also covers divergence and curl, providing their definitions and formulas. Examples are given for calculating gradient, divergence, curl, and directional derivatives. The document concludes with exercises and references for further reading.

1. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 1
Course: B.Tech- II
Subject: Engineering Mathematics II
Unit-4
RAI UNIVERSITY, AHMEDABAD

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

3. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 3
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 ( ), ( ), ( ) respectively.
Then,
( )⃗ = ( ) ̂ + ( ) ̂ + ( )
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 . .

4. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 4
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 →
∅
Where, =

5. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 5
∅
is called the directional derivative of ∅ at in the direction of .
4.2 Examples:
Example 1: If ∅ = 3 − ; find grad ∅ at the point (1,-2, 1).
Solution:
grad∅ = ∇∅
= ̂ + ̂ + (3 − )
= ̂ (3 − ) + ̂ (3 − ) + (3 − )
= ̂(6 ) + ̂(3 − 3 ) + (−2 )
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 at the point (1, −1,1)
in the direction of the tangent to the curve
= , = sin 2 + 1, = 1 − = 0.
Solution: Let ∅ =
Direction Derivative of ∅ = ∇∅
= ̂ + ̂ + ( )
∇∅ = 2 ̂ + 2 ̂ + 2
Directional Derivative of ∅ at (1,1,-1)
= 2(1)(1) (−1) ̂ + 2(1)(1) (−1) ̂ + 2(−1)(1) (1)
= 2 ̂ + 2 ̂ − 2 _________ (1)
⃗ = ̂ + ̂ + = ̂ + ( 2 + 1) ̂ + (1 − )

6. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 6
⃗ =
⃗
= ̂ + 2 2 ̂ +
Tangent vector,
Tangent ( = 0) = ̂ + 2 (cos 0) ̂ + ( 0)
= ̂ + 2 ̂ __________ (2)
Required directional derivative along tangent = (2 ̂ + 2 ̂ − 2 )
( ̂ ̂)
√
[ (1) (2)]
=
√
=
√
_______ Ans.
Example 3: Verify ∇ × [ ( )⃗] = 0
Solution:
∇ × [ ( )⃗] = ̂ + ̂ + × [ ( )⃗]
= ̂ + ̂ + × ( )( ̂ + ̂ + )
= ̂ + ̂ + × ( ) ̂ + ( ) ̂ + ( )
=
̂ ̂
( ) ( ) ( )

7. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 7
=
( )
− ( ) ̂ − ( ) − ( ) ̂
+ ( ) − ( )
= ( ) − ( ) ̂ − ( ) − ( ) ̂
+ ( ) − ( )
=
′( )
( − ) ̂ − ( − ) ̂ + ( − ) = 0 ____ Proved.
4.3 Exercise:
1) Find a unit vector normal to the surface + 3 + 2 =
6 (2,0,1).
2) Find the direction derivative of in the direction ⃗ where
⃗ = ̂ + ̂ + .
3) If ⃗ = ̂ + ̂ + , show that
a) grad =
⃗
b) grad = −
⃗
4) Find the rate of change of ∅ = in the direction normal to the
surface + + = 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 ⃗ = ̂ + ̂ +
div ⃗ = ∇⃗. ⃗ = ̂ + ̂ + ̂ + ̂ +
= + +

8. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 8
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 = + + , ̅ = ̂ + ̂ + , then find div ( ̅)
in terms of u.
Solution:div ( ̅) = ̂ + ̂ + ( + + ) ̂ + ̂ + =
̂ + ̂ + . ( + + ) ̂ + ( + + ) ̂ +
( + + ) ]
= ( + + ) + ( + + ) + ( + + )
= (3 + + ) + ( + 3 + ) + ( + + 3 )
= 5( + + )
= 5 ___________Ans.
Example 2: Find the directional derivative of the divergence of ̅( , , ) =
̂ + ̂ + at the point (2, 1, 2) in the direction of the outer normal to
the sphere + + = 9.
Solution: ̅( , , ) = ̂ + ̂ +
Divergence of ̅( , , ) = ∇. ̅ = ̂ + ̂ + . ̂ + ̂ +
= + 2 + 2
Directional derivative of divergence of ( + 2 + 2 )
= ̂ + ̂ + ( + 2 + 2 )

9. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 9
= 2 ̂ + (1 + 2 ) ̂ + 2 __________(i)
Directional derivative at the point (2,1,2) = 2 ̂ + 5 ̂ + 2
Normal to the sphere = ̂ + ̂ + ( + + − 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 .
̂ ̂
√
[ ( ), ( )]
= (8 + 10 + 8)
= ____________ Ans.
5.2 Exercise:
1) If ̅ =
̂ ̂
, 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 + + = 9 for ⃗ =
̂ + ̂ + .
6.1 CURL:
The curl of a vector point function is defined as below
Curl = ∇ × = ̂ + ̂ +
= ̂ + ̂ + × ̂ + ̂ +
=
̂ ̂
= ̂ − − ̂ − + −

10. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 10
Curl is a vector quantity.
Note: Curl = , the field is termed as irrotational.
Example 1: Find the divergence and curl of
̅ = ( ) ̂ + (3 ) ̂ + ( − ) (2, −1,1)
Solution: Here, we have
̅ = ( ) ̂ + (3 ) ̂ + ( − )
Div ̅ = ∇∅
Div ̅ = ( ) + (3 ) + ( − )
= + 3 + 2 −
̅( , , ) = −1 + 12 + 4 − 1 = 14
Curl ̅ =
̂ ̂
3 −
= −2 ̂ − ( − ) ̂ + (6 − )
= −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

11. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 11
( − + 3 − 2 ) ̂ + (3 + 2 ) ̂ + (3 − 2 + 2 ) isboth
Solenoidal and irrotational.
Solution:
Let = ( − + 3 − 2 ) ̂ + (3 + 2 ) ̂ + (3 − 2 + 2 )
For Solenoidal, we have to prove∇. = 0.
Now,
∇. = ̂ + ̂ + . ( − + 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 =
̂ ̂
( − + 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 ̂ − .

12. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 12
Solution: We have, ⃗ = ̂ + 2 ̂ − .
Curl ⃗ = ∇ × ⃗
= ̂ + ̂ + × ̂ + 2 ̂ − .
=
̂ ̂
2 −
= ̂(0) − ̂(0) + (2 − 2 )
= 0
Hence, ⃗ is irrotational.
To find the scalar potential function .
⃗ = ∇
= + +
= ̂ + ̂ + . ( ̂ + ̂ + )
= ̂ + ̂ + . ⃗
= ∇ . ⃗
= ⃗. ⃗ ( = ∇ )
= ̂ + 2 ̂ − . ( ̂ + ̂ + )
= + 2 −
= ( ) −
= ( ) −

13. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 13
∴ = − + ________ Ans.
6.2 Exercise:
1) If a vector field is given by ⃗ = ( − + ) ̂ − (2 + ) ̂. Is this
field irrotational? If so, find its scalar potential.
2) Determine the constants a and b such that the curl of vector ̅ =
(2 + 3 ) ̂ + ( + − 4 ) ̂ − (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 ) ̂ + ( − + ) ̂ +
( + ) find curl . Show that the vectors given by curl at
(1,2, −3) (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

14. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 14