vector calculus 1st 2
TRANSCRIPT
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Vector Algebra
Mr. HIMANSHU DIWAKARAssistant professor
ECED
Mr. Himanshu Diwakar
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VECTOR CALCULUS
Introduction
Scalars And Vectors
Gradient Of A Scalar
Divergence Of A Vector
Divergence Theorem
Curl Of A Vector
Stokes’s Theorem
Laplacian Of A Scalar
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Introduction• Electromagnetics(EM):-
Interaction between electric charges at rest and
in motion.
• It entails the analysis, synthesis, physical interpretation, and application of electric and magnetic fields.
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Scalars and vectors
• A scalar is a quantity that has only magnitudeEx:- time, mass, distance, temperature, entropy etc.
• A vector is a quantity that has both magnitude and direction• Velocity, force, displacement and electric field intensity.
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Unit vectors
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x
z
y
az
ay
ax
Unit vectorsaz ,ay ,az
Similarly a A vector in Cartesian co-ordinate
A=Ax.ax+Ay.ay+Az.az
So Unit vector of A
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Position and distance vector
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x
z
y
az
ay
ax
P (3, 4, 5)
O (0, 0, 0)
The distance vector is the displacement from one point to another.
A= 3.ax+ 4.ay+ 5.az
OP distanceOP=
= =7.071
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Vector multiplication
• Scalar (or dot) product = A.B
• Vector (or cross) product = AB
• Scalar triple product :
• Vector triple product :
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Angels
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If A and B are vectors then the angle between these vectors can be find
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Differential Length (Cartesian Coordinates )
• Differential elements in length, area, and volume are useful in vector calculus. They are defined in the Cartesian, cylindrical, and spherical coordinate systems.
1. Differential displacement is given by
2. Differential normal area is given by
3. Differential volume is given by
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Cylindrical Coordinates• Notice from Figure that in cylindrical coordinates, differential
elements can be found as follows:1. Differential displacement is given by
2. Differential normal area is given by
3. Differential volume is given by
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Spherical Coordinates
From Figure, we notice that in spherical coordinates,1. The differential displacement is
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2. The differential normal area is
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3. The differential volume is
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Que:- Consider the object shown in Figure and CalculateThe distance BC, The distance CD, The surface area ABCD,
The surface area ABO, The surface area A OFD, The volume ABDCFO
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C(0, 5, 0)
B(0, 5, 0)
D(5, 0, 10)
A(5, 0, 0)
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Line, Surface And Volume Integrals
• The familiar concept of integration will now be extended to cases when the integrand involves a vector. By a line we mean the path along a curve in space. We shall use terms such as line, curve, and contour interchangeably.• The line integral is the integral of the tangential component of A
along curve L. Or simplyAnd For closed surface
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Example:- Given that , calculate the circulation of F around the (closed) path shown in Figure
Ans:
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Dell operator
• The del operator, written , is the vector differential operator. In Cartesian coordinates,
• This vector differential operator, otherwise known as the gradient operator, is not a vector in itself, but when it operates on a scalar function, for example, a vector ensues. The operator is useful in defining
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1. The gradient of a scalar V, written, as 2. The divergence of a vector A, written as (• A)3. The curl of a vector A, written as ( X A)4. The Laplacian of a scalar V, written as
Each of these will be denned in detail in the subsequent sections.
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So the solution for above equations is
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CLASSIFICATION OF VECTOR FIELDS
• A vector field is uniquely characterized by its divergence and curl. Neither the divergence nor curl of a vector field is sufficient to completely describe the field.
• All vector fields can be classified in terms of their vanishing or non vanishing divergence or curl as follows:
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Typical fields with vanishing and non vanishing divergence or curl.
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•A vector field A is said to be solenoidal (or divergenceless) if .
•A vector field A is said to be irrotational (or potential) if .
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The differential distances are the components of the differential distance vector :
dzdydx ,,
zyx dzdydxd aaaL Ld
However, from differential calculus, the differential temperature:
dzzTdy
yTdx
xTTTdT
12
GRADIENT OF A SCALAR
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But,
z
y
x
ddz
ddyddx
aL
aLaL
So, previous equation can be rewritten as:
Laaa
LaLaLa
dzT
yT
xT
dzTd
yTd
xTdT
zyx
zyx
GRADIENT OF A SCALAR (Cont’d)
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The vector inside square brackets defines the change of temperature corresponding to a vector change in position .This vector is called Gradient of Scalar T.
LddT
GRADIENT OF A SCALAR (Cont’d)
For Cartesian coordinate:
zyx zV
yV
xVV aaa
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GRADIENT OF A SCALAR (Cont’d)
For Circular cylindrical coordinate:
zzVVVV aaa
1
For Spherical coordinate:
aaa
V
rV
rrVV r sin
11
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EXAMPLE 1
Find gradient of these scalars:
yxeV z cosh2sin
2cos2zU
cossin10 2rW
(a)
(b)
(c)
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SOLUTION TO EXAMPLE 1
(a) Use gradient for Cartesian coordinate:
zz
yz
xz
zyx
yxe
yxeyxe
zV
yV
xVV
a
aa
aaa
cosh2sin
sinh2sincosh2cos2
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SOLUTION TO EXAMPLE 1 (Cont’d)
(b) Use gradient for Circular cylindrical coordinate:
z
z
zzzUUUU
a
aa
aaa
2cos
2sin22cos2
1
2
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SOLUTION TO EXAMPLE 1 (Cont’d)
(c) Use gradient for Spherical coordinate:
a aa
aaa
sinsin10cos2sin10cossin10
sin11
2
r
rW
rW
rrWW
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DIVERGENCE OF A VECTOR
Illustration of the divergence of a vector field at point P:
Positive Divergence
Negative Divergence
Zero Divergence
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DIVERGENCE OF A VECTOR (Cont’d)
The divergence of A at a given point P is the outward flux per unit volume:
v
dSdiv s
v
AA A lim
0
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DIVERGENCE OF A VECTOR (Cont’d)
What is ?? s
dSA Vector field A at closed surface S
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Where,dSdS
bottomtoprightleftbackfronts
AA
And, v is volume enclosed by surface S
DIVERGENCE OF A VECTOR (Cont’d)
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For Cartesian coordinate:
zA
yA
xA zyx
A
For Circular cylindrical coordinate:
z
AAA z
11A
DIVERGENCE OF A VECTOR (Cont’d)
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For Spherical coordinate:
A
rA
rAr
rr r sin1sin
sin11 2
2A
DIVERGENCE OF A VECTOR (Cont’d)
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EXAMPLE 11
Find divergence of these vectors:
zx xzyzxP aa 2
zzzQ aaa cossin 2
aaa coscossincos12 r
rW r
(a)
(b)
(c)
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(a) Use divergence for Cartesian coordinate:
SOLUTION TO EXAMPLE 11
xxyz
xzzy
yzxx
zP
yP
xP zyx
2
02
P
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(b) Use divergence for Circular cylindrical coordinate:
cossin2
cos1sin1
11
22
Q
zz
z
zQQ
Q z
SOLUTION TO EXAMPLE 11 (Cont’d)
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SOLUTION TO EXAMPLE 11 (Cont’d)
(c) Use divergence for Spherical coordinate:
coscos2
cossin1
cossinsin1cos1
sin1sin
sin11
22
22
W
r
rrrr
Wr
Wr
Wrrr r
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It states that the total outward flux of a vector field A at the closed surface S is the same as volume integral of divergence of A.
VV
dVdS AA
DIVERGENCE THEOREM
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EXAMPLE 12
A vector field exists in the region between two concentric cylindrical surfaces defined by ρ = 1 and ρ = 2, with both cylinders extending between z = 0 and z = 5. Verify the divergence theorem by evaluating:
aD 3
S
dsD
V
DdV
(a)
(b)
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SOLUTION TO EXAMPLE 12
(a) For two concentric cylinder, the left side:
topbottomouterinnerS
d DDDDSD
Where,
10)(
)(
2
0
5
01
4
2
0
5
0 13
z
zinner
dzd
dzdD
aa
aa
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160)(
)(
2
0
5
0 24
2
0
5
02
3
z
zouter
dzd
dzdD
aa
aa
2
1
2
05
3
2
1
2
00
3
0)(
0)(
zztop
zzbottom
ddD
ddD
aa
aa
SOLUTION TO EXAMPLE 12 Cont’d)
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Therefore
150
0016010
SDS
d
SOLUTION TO EXAMPLE 12 Cont’d)
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SOLUTION TO EXAMPLE 12 Cont’d)
(b) For the right side of Divergence Theorem, evaluate divergence of D
23 41
D
So,
150
4
5
0
2
0
2
14
5
0
2
0
2
1
2
zr
zdzdddVD
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CURL OF A VECTOR
The curl of vector A is an axial (rotational) vector whose magnitude is the maximum circulation of A per unit area tends to zero and whose direction is the normal direction of the area when the area is oriented so as to make the circulation maximum.
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maxlim0
aA
A A ns
s s
dlCurl
Where,
CURL OF A VECTOR (Cont’d)
dldldacdbcabs
AA
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CURL OF A VECTOR (Cont’d)
The curl of the vector field is concerned with rotation of the vector field. Rotation can be used to measure the uniformity of the field, the more non uniform the field, the larger value of curl.
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For Cartesian coordinate:
CURL OF A VECTOR (Cont’d)
zyx
zyx
AAAzyx
aaa
A
zxy
yxz
xyz
yA
xA
zA
xA
zA
yA aaaA
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z
z
AAAz
aaa
A 1
z
zz
AA
zAA
zAA
a
aaA
1
1
For Circular cylindrical coordinate:
CURL OF A VECTOR (Cont’d)
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CURL OF A VECTOR (Cont’d)
For Spherical coordinate:
ArrAArrr
r
sinsin1
2
aaa
A
a
aaA
r
rr
Ar
rAr
rrAA
rAA
r
)(1
sin11sin
sin1
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EXAMPLE 13
zx xzyzxP aa 2
zzzQ aaa cossin 2
aaa coscossincos12 r
rW r
(a)
(b)
(c)
Find curl of these vectors:
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SOLUTION TO EXAMPLE 13
(a) Use curl for Cartesian coordinate:
zy
zyx
zxy
yxz
xyz
zxzyx
zxzyx
yP
xP
zP
xP
zP
yP
aa
aaa
aaaP
22
22 000
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(b) Use curl for Circular cylindrical coordinate
z
z
zzz
zz
z
z
yQ
xQQ
zQ
zQQ
aa
a
aa
aaaQ
cos3sin1
cos31
00sin
11
3
2
2
SOLUTION TO EXAMPLE 13 (Cont’d)
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(c) Use curl for Spherical coordinate:
a
aa
a
aaW
22
2
cos)cossin(1
coscos
sin11cossinsincos
sin1
)(1
sin11sin
sin1
rr
rr
rrr
rr
r
Wr
rWr
rrWW
rWW
r
r
r
rr
SOLUTION TO EXAMPLE 13 (Cont’d)
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SOLUTION TO EXAMPLE 13 (Cont’d)
a
aa
a
aa
sin1cos2
cossinsin
2cos
sincossin21
cos01sinsin2cossin1
3
2
r
rr
rr
r
rr
r
r
r
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STOKE’S THEOREM
The circulation of a vector field A around a closed path L is equal to the surface integral of the curl of A over the open surface S bounded by L that A and curl of A are continuous on S.
SL
dSdl AA
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STOKE’S THEOREM (Cont’d)
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EXAMPLE 14
By using Stoke’s Theorem, evaluate for
dlA
aaA sincos
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EXAMPLE 14 (Cont’d)
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SOLUTION TO EXAMPLE 14
Stoke’s Theorem,
SL
dSdl AA
where, and
zddd aS Evaluate right side to get left side,
zaA
sin11
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SOLUTION TO EXAMPLE 14 (Cont’d)
941.4
sin110
0
60
30
5
2
aA zS
dddS
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EXAMPLE 15
Verify Stoke’s theorem for the vector field for given figure by evaluating: aaB sincos
(a) over the semicircular contour.
LB d
(b) over the surface of semicircular contour.
SB d
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SOLUTION TO EXAMPLE 15
(a) To find LB d
321 LLL
dddd LBLBLBLB
Where,
dd
dzddd z
sincos
sincos
aaaaaLB
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So
2021
sincos
2
0
2
0
0
00,0
2
01
LBzzL
ddd
4cos20
sincos
0
0,200
2
22
LBzzL
ddd
SOLUTION TO EXAMPLE 15 (Cont’d)
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2021
sincos
0
2
2
00,0
0
23
r
zzLddd
LB
SOLUTION TO EXAMPLE 15 (Cont’d)
Therefore the closed integral,
8242 LB d
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SOLUTION TO EXAMPLE 15 (Cont’d)
(b) To find SB d
z
z
z
zz
a
aaa
a
aa
aaB
11sin
sinsin100
cossin1
0cossin01
sincos
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SOLUTION TO EXAMPLE 15 (Cont’d)
Therefore
821cos
1sin
11sin
0
2
0
2
0
2
0
0
2
0
aaSB
dd
ddd zz
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LAPLACIAN OF A SCALAR
The Laplacian of a scalar field, V written as: V2
Where, Laplacian V is:
zyxzyx zV
yV
xV
zyx
VV
aaaaaa
2
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For Cartesian coordinate:
2
2
2
2
2
22
zV
yV
xVV
For Circular cylindrical coordinate:
2
22
22 11
zVVVV
LAPLACIAN OF A SCALAR (Cont’d)
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LAPLACIAN OF A SCALAR (Cont’d)
For Spherical coordinate:
2
2
22
22
22
sin1
sinsin11
Vr
Vrr
Vrrr
V
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EXAMPLE 16
Find Laplacian of these scalars:
yxeV z cosh2sin 2cos2zU
cossin10 2rW
(a)
(b)
(c)
Try yourself !!
Mr. Himanshu Diwakar
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SOLUTION TO EXAMPLE 16
yxeV z cosh2sin22
02 U
2cos21cos102 r
W
(a)
(b)
(c)
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THANK YOU
Mr. Himanshu Diwakar