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
οΆ Introduction:
If vector r is a function of a scalar variable t, then we write
πβ = πβ(π‘)
If a particle is moving along a curved path then the position vector πβ of the
particle is a function of π‘. If the component of π(π‘) along π₯ β ππ₯ππ ,
π¦ β ππ₯ππ , π§ β ππ₯ππ areπ1(π‘), π2(π‘), π3(π‘) respectively.
Then,
π(π‘)βββββββββ = π1(π‘) πΜ + π2(π‘) πΜ + π3(π‘) πΜ
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 π(π₯, π¦, π§) of a region π in space there corresponds a
unique scalar π(π) , then π 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 π(π₯, π¦, π§) of a region π in space there corresponds a
unique vector π(π) , then π 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 π. π. π
3. Unit-4 VECTOR DIFFERENTIATION
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The vector differential operator Del is denoted by π. It is defined as
π = πΜ
π
ππ
+ πΜ
π
ππ
+ πΜ
π
ππ
Note: π is read Del or nebla.
3.1 Gradient of a Scalar Function:
If β (π₯, π¦, π§) be a scalar function then πΜ
πβ
ππ₯
+ πΜ
πβ
ππ¦
+ πΜ πβ
ππ§
is called the gradient
of the scalar function β .
And is denoted by grad β .
Thus, grad β = πΜ
πβ
ππ
+ πΜ
πβ
ππ
+ πΜ πβ
ππ
grad β = (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) β (π₯, π¦, π§)
grad β = π β
4.1 Normal and Directional Derivative:
1) Normal:
If β (π₯, π¦, π§) = π represents a family of surfaces for different values of
the constant π. On differentiating β , we get πβ = 0
But πβ = β β . ππβ
So ββ . ππ = 0
The scalar product of two vectors ββ and ππβ being zero, β β and ππβ
are perpendicular to each other. ππβ is in the direction of tangent to the
giving surface.
Thus ββ is a vector normal to the surface β (π₯, π¦, π§) = π.
2) Directional derivative:
The component of ββ in the direction of a vector πβ is equal to ββ . πβ
and is called the directional derivative of β in the direction of πβ.
πβ
ππ
= lim
πΏπβ0
πΏβ
πΏπ
Where, πΏπ = ππ
4. Unit-4 VECTOR DIFFERENTIATION
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πβ
ππ
is called the directional derivative of β at π in the direction of ππ.
4.2 Examples:
Example 1: If β = 3π₯2
π¦ β π¦3
π§2
; find grad β at the point (1,-2, 1).
Solution:
grad β = ββ
= (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) (3π₯2
π¦ β π¦3
π§2
)
= πΜ
π
ππ₯
(3π₯2
π¦ β π¦3
π§2) + πΜ
π
ππ¦
(3π₯2
π¦ β π¦3
π§2) + πΜ π
ππ§
(3π₯2
π¦ β π¦3
π§2
)
= πΜ(6π₯π¦) + πΜ(3π₯2
β 3π¦2
π§2) + πΜ(β2π¦3
π§)
grad β at (1, β2,1)
= πΜ(6)(1)(β2) + πΜ[(3)(1) β 3(4)(1)] + πΜ(β2)(β8)(β1)
= β12πΜ β 9πΜ β 16πΜ
Example 2: Find the directional derivative of π₯2
π¦2
π§2
at the point (1, β1,1)
in the direction of the tangent to the curve
π₯ = π π‘
, π¦ = sin 2π‘ + 1, π§ = 1 β πππ π‘ ππ‘ π‘ = 0.
Solution: Let β = π₯2
π¦2
π§2
Direction Derivative of β = ββ
= (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) (π₯2
π¦2
π§2)
ββ = 2π₯π¦2
π§2
πΜ + 2π¦π₯2
π§2
πΜ + 2π§π₯2
π¦2
πΜ
Directional Derivative of β at (1,1,-1)
= 2(1)(1)2
(β1)2
πΜ + 2(1)(1)2
(β1)2
πΜ + 2(β1)(1)2(1)2
πΜ
= 2πΜ + 2πΜ β 2πΜ _________ (1)
πβ = π₯πΜ + π¦πΜ + π§πΜ = π π‘
πΜ + (π ππ2π‘ + 1)πΜ + (1 β πππ π‘)πΜ
πββ =
ππβ
ππ‘
= π π‘
πΜ + 2 πππ 2π‘ πΜ + π πππ‘ πΜ
6. Unit-4 VECTOR DIFFERENTIATION
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4.3 Exercise:
1) Find a unit vector normal to the surface π₯2
+ 3π¦2
+ 2π§2
=
6 ππ‘ π(2,0,1).
2) Find the direction derivative of
1
π
in the direction πβ where
πβββ = π₯πΜ + π¦πΜ + π§πΜ.
3) If πβββ = π₯πΜ + π¦πΜ + π§πΜ, show that
a) grad π =
πβββ
π
b) grad (
1
π
) = β
πβββ
π3
4) Find the rate of change of β = π₯π¦π§ in the direction normal to the
surface π₯2
π¦ + π¦2
π₯ + π¦π§2
= 3 at the point (1, 1, 1).
5.1 DIVERGENCE OF A VECTOR FUNCTION:
The divergence of a vector point function πΉβ is denoted by div πΉ and is
defined as below.
Let πΉβ = πΉ1 πΜ + πΉ2 πΜ + πΉ3 πΜ
div πΉβ = ββββ. πΉβ = (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) (πΜ πΉ1 + πΜ πΉ2 + πΜ πΉ3)
=
ππΉ1
ππ₯
+
ππΉ2
ππ¦
+
ππΉ3
ππ§
It is evident that div πΉ is scalar function.
Note: If the fluid is compressible, there can be no gain or loss in the volume
element. Hence
div π½Μ = π
Here, V is called a Solenoidal vector function. And this equation is called
equation of continuity or conservation of mass.
Example 1: If π’ = π₯2
+ π¦2
+ π§2
, πππ πΜ = π₯πΜ + π¦πΜ + π§πΜ, then find div (π’πΜ )
in terms of u.
7. Unit-4 VECTOR DIFFERENTIATION
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Solution: div (π’πΜ ) = (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) [(π₯2
+ π¦2
+ π§2)(π₯πΜ + π¦πΜ + π§πΜ)]
= (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) . [(π₯2
+ π¦2
+ π§2)π₯πΜ + (π₯2
+ π¦2
+ π§2)π¦πΜ +
(π₯2
+ π¦2
+ π§2)π§πΜ]
=
π
ππ₯
(π₯3
+ π₯π¦2
+ π₯π§2) +
π
ππ¦
(π₯2
π¦ + π¦3
+ π¦π§2) +
π
ππ§
(π₯2
π§ + π¦2
π§ + π§3)
= (3π₯2
+ π¦2
+ π§2) + (π₯2
+ 3π¦2
+ π§2) + (π₯2
+ π¦2
+ 3π§2)
= 5(π₯2
+ π¦2
+ π§2)
= 5π’ ___________ Ans.
Example 2: Find the directional derivative of the divergence of πΜ ( π₯, π¦, π§) =
π₯π¦πΜ + π₯π¦2
πΜ + π§2
πΜ at the point (2, 1, 2) in the direction of the outer normal to
the sphere π₯2
+ π¦2
+ π§2
= 9.
Solution: πΜ ( π₯, π¦, π§) = π₯π¦πΜ + π₯π¦2
πΜ + π§2
πΜ
Divergence of πΜ ( π₯, π¦, π§) = β.Μ πΜ = (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) . (π₯π¦πΜ + π₯π¦2
πΜ + π§2
πΜ)
= π¦ + 2π₯π¦ + 2π§
Directional derivative of divergence of (π¦ + 2π₯π¦ + 2π§)
= (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) (π¦ + 2π₯π¦ + 2π§)
= 2π¦πΜ + (1 + 2π₯)πΜ + 2πΜ __________(i)
Directional derivative at the point (2,1,2) = 2πΜ + 5πΜ + 2πΜ
Normal to the sphere = (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) (π₯2
+ π¦2
+ π§2
β 9)
= 2π₯πΜ + 2π¦πΜ + 2π§πΜ
Normal at the point (2,1,2) = 4πΜ + 2πΜ + 4πΜ __________ (ii)
Directional derivative along normal at (2,1,2) = (2πΜ + 5πΜ + 2πΜ).
(4πΜ+2πΜ+4πΜ )
β16+4+16
8. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 8
[πΉπππ (π), (ππ)]
=
1
6
(8 + 10 + 8)
=
13
3
____________ Ans.
5.2 Exercise:
1) If π£Μ =
π₯πΜ+ π¦πΜ+ π§πΜ
βπ₯2+π¦2+π§2
, find the value of div π£Μ .
2) Find the directional derivative of div (π’ββ) at the point (1, 2, 2) in the
direction of the outer normal of the sphere π₯2
+ π¦2
+ π§2
= 9 for π’ββ =
π₯4
πΜ + π¦4
πΜ + π§4
πΜ.
6.1 CURL:
The curl of a vector point function πΉ is defined as below
Curl πΉΜ = βΜ Γ πΉΜ (πΉΜ = πΉ1 πΜ + πΉ2 πΜ +
πΉ3 πΜ)
= (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) Γ πΉ1 πΜ + πΉ2 πΜ + πΉ3 πΜ
= |
πΜ πΜ πΜ
π
ππ₯
π
ππ¦
π
ππ§
πΉ1 πΉ2 πΉ3
|
= πΜ (
ππΉ3
ππ¦
β
ππΉ2
ππ§
) β πΜ (
ππΉ3
ππ₯
β
ππΉ1
ππ§
) + πΜ (
ππΉ2
ππ₯
β
ππΉ1
ππ¦
)
Curl πΉΜ is a vector quantity.
Note: Curl πΜ = π, the field πΉ is termed as irrotational.
Example 1: Find the divergence and curl of
π£Μ = (π₯π¦π§)πΜ + (3π₯2
π¦)πΜ + (π₯π§2
β π¦2
π§)πΜ ππ‘ (2, β1,1)
11. Unit-4 VECTOR DIFFERENTIATION
RAI UNIVERSITY, AHMEDABAD 11
= πΜ(0) β πΜ(0) + πΜ(2π¦ β 2π¦)
= 0
Hence, π΄β is irrotational.
To find the scalar potential function π.
π΄β = β π
ππ =
ππ
ππ₯
ππ₯ +
ππ
ππ¦
ππ¦ +
ππ
ππ§
ππ§
= (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) . (πΜ π π₯ + πΜ π π¦ + πΜ ππ§)
= (πΜ
π
ππ₯
+ πΜ
π
ππ¦
+ πΜ π
ππ§
) π. ππβ
= βπ. ππβ
= π΄β. ππβ (π΄ = β π)
= (π¦2
πΜ + 2π₯π¦πΜ β π§2
πΜ). (πΜ π π₯ + πΜ π π¦ + πΜ ππ§)
= π¦2
ππ₯ + 2π₯π¦ ππ¦ β π§2
ππ§
= π(π₯π¦2) β π§2
ππ§
π = β« π(π₯π¦2) β β« π§2
ππ§
β΄ π = π₯π¦2
β
π§3
3
+ π ________ Ans.
6.2 Exercise:
1) If a vector field is given by πΉβ = (π₯2
β π¦2
+ π₯)πΜ β (2π₯π¦ + π¦)πΜ. Is this
field irrotational? If so, find its scalar potential.
12. Unit-4 VECTOR DIFFERENTIATION
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2) Determine the constants a and b such that the curl of vector π΄Μ =
(2π₯π¦ + 3π¦π§)πΜ + (π₯2
+ ππ₯π§ β 4π§2)πΜ β (3π₯π¦ + ππ¦π§)πΜ is zero.
3) A fluid motion is given byπ£Μ = (π¦ + π§)πΜ + (π§ + π₯)πΜ + (π₯ + π¦)πΜ. Show
that the motion is irrotational and hence find the velocity potential.
4) Given that vector field πΜ = (π₯2
β π¦2
+ 2π₯π§)πΜ + (π₯π§ β π₯π¦ + π¦π§)πΜ +
(π§2
+ π₯2
)πΜ find curl π. Show that the vectors given by curl π at
π0(1,2, β3) πππ π1(2,3,12) are orthogonal.
5) Suppose that πβββ, πββ and π are continuously differentiable fields then
Prove that, div(πβββ Γ πββ) = πββ. ππ’ππ πβββ β πβββ. ππ’ππ πββ.
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