This document discusses concepts in vector differentiation including: 1. It defines scalar and vector point functions, with examples such as temperature and fluid velocity. 2. It introduces the vector differential operator del (∇) and defines the gradient of a scalar function. 3. It explains normal and directional derivatives, and gives examples of finding the gradient and directional derivative of functions. 4. It defines the divergence of a vector function and gives examples of applying vector differentiation concepts.

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
𝑟⃗ = 𝑟⃗(𝑡)
If a particle is moving along a curved path then the position vector 𝑟⃗ of the
particle is a function of 𝑡. If the component of 𝑓(𝑡) along 𝑥 − 𝑎𝑥𝑖𝑠,
𝑦 − 𝑎𝑥𝑖𝑠, 𝑧 − 𝑎𝑥𝑖𝑠 are𝑓1(𝑡), 𝑓2(𝑡), 𝑓3(𝑡) respectively.
Then,
𝑓(𝑡)⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 𝑓1(𝑡) 𝑖̂ + 𝑓2(𝑡) 𝑗̂ + 𝑓3(𝑡) 𝑘̂
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 𝑃(𝑥, 𝑦, 𝑧) of a region 𝑅 in space there corresponds a
unique scalar 𝑓(𝑃) , then 𝑓 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 𝑃(𝑥, 𝑦, 𝑧) of a region 𝑅 in space there corresponds a
unique vector 𝑓(𝑃) , then 𝑓 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 𝒊. 𝒆. 𝛁

3. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 3
The vector differential operator Del is denoted by 𝛁. It is defined as
𝛁 = 𝒊̂
𝝏
𝝏𝒙
+ 𝒋̂
𝝏
𝝏𝒚
+ 𝒌̂
𝝏
𝝏𝒛
Note: 𝛁 is read Del or nebla.
3.1 Gradient of a Scalar Function:
If ∅(𝑥, 𝑦, 𝑧) be a scalar function then 𝑖̂
𝜕∅
𝜕𝑥
+ 𝑗̂
𝜕∅
𝜕𝑦
+ 𝑘̂ 𝜕∅
𝜕𝑧
is called the gradient
of the scalar function ∅.
And is denoted by grad ∅.
Thus, grad ∅ = 𝒊̂
𝝏∅
𝝏𝒙
+ 𝒋̂
𝝏∅
𝝏𝒚
+ 𝒌̂ 𝝏∅
𝝏𝒛
grad ∅ = (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) ∅(𝑥, 𝑦, 𝑧)
grad ∅ = 𝛁 ∅
4.1 Normal and Directional Derivative:
1) Normal:
If ∅(𝑥, 𝑦, 𝑧) = 𝑐 represents a family of surfaces for different values of
the constant 𝑐. On differentiating ∅, we get 𝑑∅ = 0
But 𝑑∅ = ∇ ∅ . 𝑑𝑟⃗
So ∇∅. 𝑑𝑟 = 0
The scalar product of two vectors ∇∅ and 𝑑𝑟⃗ being zero, ∇ ∅ and 𝑑𝑟⃗
are perpendicular to each other. 𝑑𝑟⃗ is in the direction of tangent to the
giving surface.
Thus ∇∅ is a vector normal to the surface ∅(𝑥, 𝑦, 𝑧) = 𝑐.
2) Directional derivative:
The component of ∇∅ in the direction of a vector 𝑑⃗ is equal to ∇∅. 𝑑⃗
and is called the directional derivative of ∅ in the direction of 𝑑⃗.
𝜕∅
𝜕𝑟
= lim
𝛿𝑟→0
𝛿∅
𝛿𝑟
Where, 𝛿𝑟 = 𝑃𝑄

4. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 4
𝜕∅
𝜕𝑟
is called the directional derivative of ∅ at 𝑃 in the direction of 𝑃𝑄.
4.2 Examples:
Example 1: If ∅ = 3𝑥2
𝑦 − 𝑦3
𝑧2
; find grad ∅ at the point (1,-2, 1).
Solution:
grad ∅ = ∇∅
= (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) (3𝑥2
𝑦 − 𝑦3
𝑧2
)
= 𝑖̂
𝜕
𝜕𝑥
(3𝑥2
𝑦 − 𝑦3
𝑧2) + 𝑗̂
𝜕
𝜕𝑦
(3𝑥2
𝑦 − 𝑦3
𝑧2) + 𝑘̂ 𝜕
𝜕𝑧
(3𝑥2
𝑦 − 𝑦3
𝑧2
)
= 𝑖̂(6𝑥𝑦) + 𝑗̂(3𝑥2
− 3𝑦2
𝑧2) + 𝑘̂(−2𝑦3
𝑧)
grad ∅ at (1, −2,1)
= 𝑖̂(6)(1)(−2) + 𝑗̂[(3)(1) − 3(4)(1)] + 𝑘̂(−2)(−8)(−1)
= −12𝑖̂ − 9𝑗̂ − 16𝑘̂
Example 2: Find the directional derivative of 𝑥2
𝑦2
𝑧2
at the point (1, −1,1)
in the direction of the tangent to the curve
𝑥 = 𝑒 𝑡
, 𝑦 = sin 2𝑡 + 1, 𝑧 = 1 − 𝑐𝑜𝑠𝑡 𝑎𝑡 𝑡 = 0.
Solution: Let ∅ = 𝑥2
𝑦2
𝑧2
Direction Derivative of ∅ = ∇∅
= (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) (𝑥2
𝑦2
𝑧2)
∇∅ = 2𝑥𝑦2
𝑧2
𝑖̂ + 2𝑦𝑥2
𝑧2
𝑗̂ + 2𝑧𝑥2
𝑦2
𝑘̂
Directional Derivative of ∅ at (1,1,-1)
= 2(1)(1)2
(−1)2
𝑖̂ + 2(1)(1)2
(−1)2
𝑗̂ + 2(−1)(1)2(1)2
𝑘̂
= 2𝑖̂ + 2𝑗̂ − 2𝑘̂ _________ (1)
𝑟⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂ = 𝑒 𝑡
𝑖̂ + (𝑠𝑖𝑛2𝑡 + 1)𝑗̂ + (1 − 𝑐𝑜𝑠𝑡)𝑘̂
𝑇⃗⃗ =
𝑑𝑟⃗
𝑑𝑡
= 𝑒 𝑡
𝑖̂ + 2 𝑐𝑜𝑠2𝑡 𝑗̂ + 𝑠𝑖𝑛𝑡 𝑘̂

5. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 5
Tangent vector,
Tangent (𝑎𝑡 𝑡 = 0) = 𝑒0
𝑖̂ + 2 (cos 0)𝑗̂ + (𝑠𝑖𝑛0)𝑘̂
= 𝑖̂ + 2𝑗̂ __________ (2)
Required directional derivative along tangent = (2 𝑖̂ + 2𝑗̂ − 2𝑘̂)
(𝑖̂+2𝑗̂)
√1+4
[𝑓𝑟𝑜𝑚 (1)𝑎𝑛𝑑 (2)]
=
2+4+0
√5
=
6
√5
_______ Ans.
Example 3: Verify ∇̅ × [𝑓(𝑟)𝑟⃗] = 0
Solution:
∇̅ × [𝑓(𝑟)𝑟⃗] = [𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
] × [𝑓(𝑟)𝑟⃗]
= [𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
] × 𝑓(𝑟)(𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂)
= [𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
] × [𝑓(𝑟)𝑥𝑖̂ + 𝑓(𝑟)𝑦𝑗̂ + 𝑓(𝑟)𝑧𝑘̂]
= |
𝑖̂ 𝑗̂ 𝑘̂
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑓(𝑟)𝑥 𝑓(𝑟)𝑦 𝑓(𝑟)𝑧
|
= [𝑧
𝜕𝑓(𝑟)
𝜕𝑦
− 𝑦
𝜕
𝜕𝑧
𝑓(𝑟)] 𝑖̂ − [𝑧
𝜕
𝜕𝑥
𝑓(𝑟) − 𝑥
𝜕
𝜕𝑧
𝑓(𝑟)] 𝑗̂ + [𝑦
𝜕
𝜕𝑥
𝑓(𝑟) − 𝑥
𝜕
𝜕𝑦
𝑓(𝑟)] 𝑘̂
= [𝑧
𝑑
𝑑𝑟
𝑓(𝑟)
𝜕𝑟
𝜕𝑦
− 𝑦
𝑑
𝑑𝑟
𝑓(𝑟)
𝜕𝑟
𝜕𝑧
] 𝑖̂ − [𝑧
𝑑
𝑑𝑟
𝑓(𝑟)
𝜕𝑟
𝜕𝑥
− 𝑥
𝑑
𝑑𝑟
𝑓(𝑟)
𝜕𝑟
𝜕𝑧
] 𝑗̂ + [𝑦
𝑑
𝑑𝑟
𝑓(𝑟)
𝜕𝑟
𝜕𝑥
−
𝑥
𝑑
𝑑𝑟
𝑓(𝑟)
𝜕𝑟
𝜕𝑦
] 𝑘̂
=
𝑓′(𝑟)
𝑟
[(𝑦𝑧 − 𝑦𝑧)𝑖̂ − (𝑥𝑧 − 𝑥𝑧)𝑗̂ + (𝑥𝑦 − 𝑥𝑦)𝑘̂] = 0 ____ Proved.

6. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 6
4.3 Exercise:
1) Find a unit vector normal to the surface 𝑥2
+ 3𝑦2
+ 2𝑧2
=
6 𝑎𝑡 𝑃(2,0,1).
2) Find the direction derivative of
1
𝑟
in the direction 𝑟⃗ where
𝑟⃗⃗⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂.
3) If 𝑟⃗⃗⃗ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂, show that
a) grad 𝑟 =
𝑟⃗⃗⃗
𝑟
b) grad (
1
𝑟
) = −
𝑟⃗⃗⃗
𝑟3
4) Find the rate of change of ∅ = 𝑥𝑦𝑧 in the direction normal to the
surface 𝑥2
𝑦 + 𝑦2
𝑥 + 𝑦𝑧2
= 3 at the point (1, 1, 1).
5.1 DIVERGENCE OF A VECTOR FUNCTION:
The divergence of a vector point function 𝐹⃗ is denoted by div 𝐹 and is
defined as below.
Let 𝐹⃗ = 𝐹1 𝑖̂ + 𝐹2 𝑗̂ + 𝐹3 𝑘̂
div 𝐹⃗ = ∇⃗⃗⃗. 𝐹⃗ = (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) (𝑖̂ 𝐹1 + 𝑗̂ 𝐹2 + 𝑘̂ 𝐹3)
=
𝜕𝐹1
𝜕𝑥
+
𝜕𝐹2
𝜕𝑦
+
𝜕𝐹3
𝜕𝑧
It is evident that div 𝐹 is scalar function.
Note: If the fluid is compressible, there can be no gain or loss in the volume
element. Hence
div 𝑽̅ = 𝟎
Here, V is called a Solenoidal vector function. And this equation is called
equation of continuity or conservation of mass.
Example 1: If 𝑢 = 𝑥2
+ 𝑦2
+ 𝑧2
, 𝑎𝑛𝑑 𝑟̅ = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂, then find div (𝑢𝑟̅)
in terms of u.

7. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 7
Solution: div (𝑢𝑟̅) = (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) [(𝑥2
+ 𝑦2
+ 𝑧2)(𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧𝑘̂)]
= (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) . [(𝑥2
+ 𝑦2
+ 𝑧2)𝑥𝑖̂ + (𝑥2
+ 𝑦2
+ 𝑧2)𝑦𝑗̂ +
(𝑥2
+ 𝑦2
+ 𝑧2)𝑧𝑘̂]
=
𝜕
𝜕𝑥
(𝑥3
+ 𝑥𝑦2
+ 𝑥𝑧2) +
𝜕
𝜕𝑦
(𝑥2
𝑦 + 𝑦3
+ 𝑦𝑧2) +
𝜕
𝜕𝑧
(𝑥2
𝑧 + 𝑦2
𝑧 + 𝑧3)
= (3𝑥2
+ 𝑦2
+ 𝑧2) + (𝑥2
+ 3𝑦2
+ 𝑧2) + (𝑥2
+ 𝑦2
+ 3𝑧2)
= 5(𝑥2
+ 𝑦2
+ 𝑧2)
= 5𝑢 ___________ Ans.
Example 2: Find the directional derivative of the divergence of 𝑓̅( 𝑥, 𝑦, 𝑧) =
𝑥𝑦𝑖̂ + 𝑥𝑦2
𝑗̂ + 𝑧2
𝑘̂ at the point (2, 1, 2) in the direction of the outer normal to
the sphere 𝑥2
+ 𝑦2
+ 𝑧2
= 9.
Solution: 𝑓̅( 𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑖̂ + 𝑥𝑦2
𝑗̂ + 𝑧2
𝑘̂
Divergence of 𝑓̅( 𝑥, 𝑦, 𝑧) = ∇.̅ 𝑓̅ = (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) . (𝑥𝑦𝑖̂ + 𝑥𝑦2
𝑗̂ + 𝑧2
𝑘̂)
= 𝑦 + 2𝑥𝑦 + 2𝑧
Directional derivative of divergence of (𝑦 + 2𝑥𝑦 + 2𝑧)
= (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) (𝑦 + 2𝑥𝑦 + 2𝑧)
= 2𝑦𝑖̂ + (1 + 2𝑥)𝑗̂ + 2𝑘̂ __________(i)
Directional derivative at the point (2,1,2) = 2𝑖̂ + 5𝑗̂ + 2𝑘̂
Normal to the sphere = (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) (𝑥2
+ 𝑦2
+ 𝑧2
− 9)
= 2𝑥𝑖̂ + 2𝑦𝑗̂ + 2𝑧𝑘̂
Normal at the point (2,1,2) = 4𝑖̂ + 2𝑗̂ + 4𝑘̂ __________ (ii)
Directional derivative along normal at (2,1,2) = (2𝑖̂ + 5𝑗̂ + 2𝑘̂).
(4𝑖̂+2𝑗̂+4𝑘̂ )
√16+4+16

8. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 8
[𝐹𝑟𝑜𝑚 (𝑖), (𝑖𝑖)]
=
1
6
(8 + 10 + 8)
=
13
3
____________ Ans.
5.2 Exercise:
1) If 𝑣̅ =
𝑥𝑖̂+ 𝑦𝑗̂+ 𝑧𝑘̂
√𝑥2+𝑦2+𝑧2
, find the value of div 𝑣̅.
2) Find the directional derivative of div (𝑢⃗⃗) at the point (1, 2, 2) in the
direction of the outer normal of the sphere 𝑥2
+ 𝑦2
+ 𝑧2
= 9 for 𝑢⃗⃗ =
𝑥4
𝑖̂ + 𝑦4
𝑗̂ + 𝑧4
𝑘̂.
6.1 CURL:
The curl of a vector point function 𝐹 is defined as below
Curl 𝐹̅ = ∇̅ × 𝐹̅ (𝐹̅ = 𝐹1 𝑖̂ + 𝐹2 𝑗̂ +
𝐹3 𝑘̂)
= (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) × 𝐹1 𝑖̂ + 𝐹2 𝑗̂ + 𝐹3 𝑘̂
= |
𝑖̂ 𝑗̂ 𝑘̂
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝐹1 𝐹2 𝐹3
|
= 𝑖̂ (
𝜕𝐹3
𝜕𝑦
−
𝜕𝐹2
𝜕𝑧
) − 𝑗̂ (
𝜕𝐹3
𝜕𝑥
−
𝜕𝐹1
𝜕𝑧
) + 𝑘̂ (
𝜕𝐹2
𝜕𝑥
−
𝜕𝐹1
𝜕𝑦
)
Curl 𝐹̅ is a vector quantity.
Note: Curl 𝑭̅ = 𝟎, the field 𝐹 is termed as irrotational.
Example 1: Find the divergence and curl of
𝑣̅ = (𝑥𝑦𝑧)𝑖̂ + (3𝑥2
𝑦)𝑗̂ + (𝑥𝑧2
− 𝑦2
𝑧)𝑘̂ 𝑎𝑡 (2, −1,1)

9. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 9
Solution: Here, we have
𝑣̅ = (𝑥𝑦𝑧)𝑖̂ + (3𝑥2
𝑦)𝑗̂ + (𝑥𝑧2
− 𝑦2
𝑧)𝑘̂
Div 𝑣̅ = ∇∅
Div 𝑣̅ =
𝜕
𝜕𝑥
(𝑥𝑦𝑧) +
𝜕
𝜕𝑦
(3𝑥2
𝑦) +
𝜕
𝜕𝑧
(𝑥𝑧2
− 𝑦2
𝑧)
= 𝑦𝑧 + 3𝑥2
+ 2𝑥𝑧 − 𝑦2
𝑣̅(2,−1,1) = −1 + 12 + 4 − 1 = 14
Curl 𝑣̅ = |
𝑖̂ 𝑗̂ 𝑘̂
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑥𝑦𝑧 3𝑥2
𝑦 𝑥𝑧2
− 𝑦2
𝑧
|
= −2𝑦𝑧𝑖̂ − (𝑧2
− 𝑥𝑦)𝑗̂ + (6𝑥𝑦 − 𝑥𝑧)𝑘̂
= −2𝑦𝑧𝑖̂ + (𝑥𝑦 − 𝑧2)𝑗̂ + (6𝑥𝑦 − 𝑥𝑧)𝑘̂
Curl at (2, -1, 1)
= 2(−1)(1)𝑖̂ + {2(−1) − 1}𝑗̂ + {6(2)(−1) − 2(1)}𝑘̂
= 2𝑖̂ − 3𝑗̂ − 14𝑘̂ __________ Ans.
Example 2: Prove that
(𝑦2
− 𝑧2
+ 3𝑦𝑧 − 2𝑥)𝑖̂ + (3𝑥𝑦 + 2𝑥𝑦)𝑗̂ + (3𝑥𝑦 − 2𝑥𝑧 + 2𝑧)𝑘̂ is both
Solenoidal and irrotational.
Solution:
Let 𝐹̅ = (𝑦2
− 𝑧2
+ 3𝑦𝑧 − 2𝑥)𝑖̂ + (3𝑥𝑦 + 2𝑥𝑦)𝑗̂ + (3𝑥𝑦 − 2𝑥𝑧 + 2𝑧)𝑘̂
For Solenoidal, we have to prove ∇.̅ 𝐹̅ = 0.
Now,

10. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 10
∇.̅ 𝐹̅ = [𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
] . [(𝑦2
− 𝑧2
+ 3𝑦𝑧 − 2𝑥)𝑖̂ + (3𝑥𝑦 + 2𝑥𝑦)𝑗̂ +
(3𝑥𝑦 − 2𝑥𝑧 + 2𝑧)𝑘̂]
= −2 + 2𝑥 − 2𝑥 + 2
= 0
Thus, 𝐹̅ is Solenoidal.
For irrotational, we have to prove Curl 𝐹̅ = 0
Now, Curl 𝐹̅ =
|
𝑖̂ 𝑗̂ 𝑘̂
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
(𝑦2
− 𝑧2
+ 3𝑦𝑧 − 2𝑥) (3𝑥𝑦 + 2𝑥𝑦) (3𝑥𝑦 − 2𝑥𝑧 + 2𝑧)
|
= (3𝑧 + 2𝑦 − 2𝑦 + 3𝑧)𝑖̂ − (−2𝑧 + 3𝑦 − 3𝑦 + 2𝑧)𝑗̂ + (3𝑧 + 2𝑦 − 2𝑦 − 3𝑧)𝑘̂
= 0𝑖̂ + 0𝑗̂ + 0𝑘̂
= 0
Thus, 𝐹̅ is irrotational.
Hence, 𝐹̅ is both Solenoidal and irrotational. __________ Proved.
Example 3: Find the scalar potential (velocity potential) function 𝑓 for 𝐴⃗ =
𝑦2
𝑖̂ + 2𝑥𝑦𝑗̂ − 𝑧2
𝑘̂.
Solution: We have, 𝐴⃗ = 𝑦2
𝑖̂ + 2𝑥𝑦𝑗̂ − 𝑧2
𝑘̂.
Curl 𝐴⃗ = ∇ × 𝐴⃗
= (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) × (𝑦2
𝑖̂ + 2𝑥𝑦𝑗̂ − 𝑧2
𝑘̂).
= |
𝑖̂ 𝑗̂ 𝑘̂
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑦2
2𝑥𝑦 −𝑧2
|

11. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 11
= 𝑖̂(0) − 𝑗̂(0) + 𝑘̂(2𝑦 − 2𝑦)
= 0
Hence, 𝐴⃗ is irrotational.
To find the scalar potential function 𝑓.
𝐴⃗ = ∇ 𝑓
𝑑𝑓 =
𝜕𝑓
𝜕𝑥
𝑑𝑥 +
𝜕𝑓
𝜕𝑦
𝑑𝑦 +
𝜕𝑓
𝜕𝑧
𝑑𝑧
= (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) . (𝑖̂ 𝑑 𝑥 + 𝑗̂ 𝑑 𝑦 + 𝑘̂ 𝑑𝑧)
= (𝑖̂
𝜕
𝜕𝑥
+ 𝑗̂
𝜕
𝜕𝑦
+ 𝑘̂ 𝜕
𝜕𝑧
) 𝑓. 𝑑𝑟⃗
= ∇𝑓. 𝑑𝑟⃗
= 𝐴⃗. 𝑑𝑟⃗ (𝐴 = ∇ 𝑓)
= (𝑦2
𝑖̂ + 2𝑥𝑦𝑗̂ − 𝑧2
𝑘̂). (𝑖̂ 𝑑 𝑥 + 𝑗̂ 𝑑 𝑦 + 𝑘̂ 𝑑𝑧)
= 𝑦2
𝑑𝑥 + 2𝑥𝑦 𝑑𝑦 − 𝑧2
𝑑𝑧
= 𝑑(𝑥𝑦2) − 𝑧2
𝑑𝑧
𝑓 = ∫ 𝑑(𝑥𝑦2) − ∫ 𝑧2
𝑑𝑧
∴ 𝑓 = 𝑥𝑦2
−
𝑧3
3
+ 𝑐 ________ Ans.
6.2 Exercise:
1) If a vector field is given by 𝐹⃗ = (𝑥2
− 𝑦2
+ 𝑥)𝑖̂ − (2𝑥𝑦 + 𝑦)𝑗̂. 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 𝐴̅ =
(2𝑥𝑦 + 3𝑦𝑧)𝑖̂ + (𝑥2
+ 𝑎𝑥𝑧 − 4𝑧2)𝑗̂ − (3𝑥𝑦 + 𝑏𝑦𝑧)𝑘̂ is zero.
3) A fluid motion is given by𝑣̅ = (𝑦 + 𝑧)𝑖̂ + (𝑧 + 𝑥)𝑗̂ + (𝑥 + 𝑦)𝑘̂. Show
that the motion is irrotational and hence find the velocity potential.
4) Given that vector field 𝑉̅ = (𝑥2
− 𝑦2
+ 2𝑥𝑧)𝑖̂ + (𝑥𝑧 − 𝑥𝑦 + 𝑦𝑧)𝑗̂ +
(𝑧2
+ 𝑥2
)𝑘̂ find curl 𝑉. Show that the vectors given by curl 𝑉 at
𝑃0(1,2, −3) 𝑎𝑛𝑑 𝑃1(2,3,12) are orthogonal.
5) Suppose that 𝑈⃗⃗⃗, 𝑉⃗⃗ and 𝑓 are continuously differentiable fields then
Prove that, div(𝑈⃗⃗⃗ × 𝑉⃗⃗) = 𝑉⃗⃗. 𝑐𝑢𝑟𝑙 𝑈⃗⃗⃗ − 𝑈⃗⃗⃗. 𝑐𝑢𝑟𝑙 𝑉⃗⃗.
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