Sem4 sscz fizik math note

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1

Lecture 2

Scalar product

Vector product

SSCP 2613- Mathematical Physics

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2.1 The DOT product

• The scalar product of two vectors is written as

– It is also called the dot product

•

– q is the angle between Aand B

• Applied to work, this means

A B cos q A B

A B

cosW F r q F r

BA

BA1cosq

- q measured counterclockwise .

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Algebraic definition

If

Then

kBjBiBB

kAjAiAA

zyx

zyx

ˆˆˆ

ˆˆˆ

zzyyxx BABABABA

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If A, B, C are vectors, and c is a scalar, then(1) A · A = |A|2

(2) A · B = B · A(3) A · (B + C) = (A · B) + (A · C) (4) (cA) · B = c(A · B) = A · (cB)(5) 0 · A = 0

2.2 Properties of the Dot Product

i.i= j.j = k.k = 1

i.j = i.k = j.k = 0

Dot product for unit vector

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If q (between 0 and ) is the angle between A and B, then

A× B = |A||B| sinq

i.e. the magnitude is the area of the parallelogram form by

the vectors A and B.

B

Aq

i x i = j x j = k x k = 0

i x j = k , j x i = - k k x i = j , i x k = - jj x k = i , k x j = - i

i

jk

2.3 The CROSS product

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2.4 The CROSS product – Right Hand Rule

c has DIRECTION to a and b

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Given two vectors A = a1, a2, a3 and B = b1, b2, b3,

the cross product of A and B can be expressed as:

zyx

zyx

bbb

aaa

kji

BA A x B = -B x A

yx

yx

zx

zx

zy

zykji

bb

aa

bb

aa

bb

aa

= (aybz – azby)i - (axbz -azbx)j + (axby – aybx)k

2.5 Vector Products Using Determinants

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2.6 Properties of the Cross product

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2.7 Torque as a Cross Product

The torque is the cross productof a force vector with theposition vector to its point ofapplication

a force F acting on a body atposition r produces a torquewith magnitude:

Right Hand Rule: curl fingersfrom r to F, thumb points alongtorque.

Fr

sinrF F causes a torque τ that rotates the object about an axis perpendicular to BOTH r AND F.

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Lets TRY…

mjirNjiF )ˆ5ˆ4( )ˆ3ˆ2(

BA

Solution:

jiBjiA ˆ2ˆ ˆ3ˆ2

k

jijiBA

ˆ7

)ˆ2ˆ()ˆ3ˆ2(

2. Calculate torque given a force and its location

(Nm) k

jijiFr

ˆ2

)ˆ3ˆ2()ˆ5ˆ4(

Solution:

1. Find: Where:

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THANK YOU…

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3/18/2014

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Lecture 3

Vector Differentiation


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2.1 WHAT IS VECTOR CALCULUS?

It answers questions like how to define and measure the

variation of temperature, fluid velocity, force, magnetic

flux space.

In the real 3D engineering world, one wants to know

things like the stress and strain inside a structure, the

vortices of the air flow over a wing, or the induced

electromagnetic field around an aerial.

Vector calculus provides the necessary mathematical

notation and techniques for dealing with such issues.

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3.2 SCALAR FUNCTION AND VECTOR FUNCTION

A scalar function (of one variable) f(x) or f(t) is a

formula that takes a scalar and returns a scalar.

- to describe the spatial variation of

temperature T(x) along a one-dimensional

bar heated at one end, or the time variation

of the DC current i(t) across a certain

component in an electrical circuit.

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3.2 SCALAR FUNCTION AND VECTOR FUNCTION

A vector function (of one variable) v(x) or v(t)

takes a scalar and returns a vector:

Such functions might be used to describe the

motion of a particle r(t) at time t; or the external

forces F(x). .

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3.3 VECTOR FIELD

A vector field v(x, y, z) is a vector-valued quantity defined

over a region of space. It is defined by a function that

takes a vector (of positions) and returns a vector.

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The vector function represented by A(u) is said to be

continuous at u0 if given any positive number.

3.4 LIMITS, CONTINUITY, AND DERIVATIVES OF VECTOR FUNCTIONS.

)()(lim 00

uuuu

AA

The derivatives of A(u) is defined as

u

uuu

du

d

u

)()(lim

0

AAA

provides this limit exists.

In case A(u) = A1(u)i + A2(u)j + A3(u)k, then

kA

jA

iAA

du

d

du

d

du

d

du

d 321

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If A(x, y, z) = A1(x, y, z)i + A2(x, y, z)j + A3(x, y, z)k, then

dzdz

ddy

dy

ddx

dx

dd

AAAA is the differential of A.

Derivatives of product obey rules similar to those for

scalar functions. However, when cross products are involved

the order may be important.

AA

Adu

d

du

d

du

d )(

BAB

ABA ..).(dydydy

xBAB

AxAxBdydydy

)(

Ex:a)

b)

c)

(Maintain the order of A and B)

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3.5 DIFFERENTIATION OF VECTORS

Consider the following figure.

If r represents the position vector of an object which is

moving along a curve C, then the position vector will be

dependent upon the time, t. We write r = r(t)

Suppose that the object is at

the point P with position

vector r at time t and at the

point Q with position vector r(t

+ t) at the later time (t + t),

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The average velocity during the time from t to (t + t) is:

dt

d

t

ttt

t

rrrv

)()(lim

0

and acceleration a along the curve: adt

d

dt

d

dt

d

dt

d

vrr2

2

kdt

zdj

dt

ydi

dt

xd

kdt

dvj

dt

dvi

dt

dv

dt

tvdta

kvjvivkdt

dzj

dt

dyi

dt

dx

dt

rdtv

ktzjtyitxtr

zyx

zyx

ˆˆˆ

ˆˆˆ)()(

ˆˆˆˆˆˆ)(

ˆ)(ˆ)(ˆ)()(

2

2

2

2

2

2

:physics In

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Example:

Solution:

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du

d

du

d

du

d BABA )(

du

dA

du

dA

du

d..).( B

BBA

xBAB

AxAxBdu

d

du

d

du

d)(

AA

Adu

d

du

d

du

d )(

BxCA

xCB

AC

BxABxCA ....du

d

du

d

du

d

du

d

xBxCA

xCB

AxC

AxBxAxBxCdu

d

du

d

du

d

du

d

3.6 Differentiation Formula

2)

3)

4)

5)

6)

1)

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Example:

Solution:

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Lets TRY:find j,ikji tttGandttttFIf cossin)(5)( 32

)() F.Gdt

di )() FxG

dt

dii )() F.F

dt

diii

Answer:

53

22323

2

62100

cos11sinsin5sin3coscos3sin)

sin11cos)15()

ttt

tttttttttttttii

tttti

iii)

kji

THANK YOU…

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3/18/2014

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Lecture 4

Partial Derivatives of a vector


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4.1 PARTIAL DERIVATIVES OF VECTOR

If A is a vector depending on more than one scalar variable say x, y, z:

A = A (x, y, z)

The partial derivative of A with respect to x is defined as:

x

zyxzyxxA

x xx

),,(),,(lim

0

AAA

y

zyxzyyxA

y yy

),,(),,(lim

0

AAA

z

zyxzzyxA

z zz

),,(),,(lim

0

AAA

Similarly,

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If A and B are functions of x, y, z then

BAB

ABA ...xxx

1)

2) xBAB

AxAxBxxx

B

ABABABA ..).(.

2

xxyxyxy

3)

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4.2 Higher order derivatives

xxx

AA2

2

yyy

AA2

2

zzz

AA2

2,

,

yxyx

AA2

xyxy

AA2

2

2

2

3

zxzx

AA

,

,

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Example 1:

For the function

i) Find the partial derivatives of f with respect to x and y ii) Compute the rates of change of the function in the x and

y directions at the point (-1,2).

Solution:

i)

Plug in the values x= -1 and y=2 into the equations. fx(-1,2) = 10 and fy(-1,2) = 28.

ii)

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Example 2:

Solution:

tex ty 3cos2 tz 3sin2

A particle moves along a curve whose parametric equation are

where t is the time.a) Determine its velocity and acceleration at any timeb) Find the magnitude of the velocity and acceleration at t = 0

kjikjir ttezyx t 3sin23cos2

kjir

v ttedt

d

dt

d t 3sin23cos2

kji tte t 3cos63sin6

kjir

a ttedt

d t 3sin183cos182

2

a) The position vector r of the particle is

The velocity,

The acceleration,

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ki

kjir

v

6

3cos63sin6

ttedt

d t

ji

kjir

a

18

3sin183cos182

2

ttedt

d t

b) at t = 0,

3761|||| 22 v

325)18(1|||| 22 a

The magnitude i) velocity at t = 0:

ii) acceleration at t = 0:

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zxyzyx 2),,(

)(2

Azx

Example 3:

and A = xzi – xy2 j + yz2k, find

at the point (2, -1, 1)

If

kjiA 222 yzxyxzzxy

kji 3342222 zxyzyxzyx

Solution:

kjiA 3342222)( zxyzyxzyxzz

kji 234222 32 zxyyxzyx

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kjiA234222 32)( zxyyxzyx

xzx

kji 2342 324 zyxyzxy

)(2

Azx

At point (2,-1,1)

= 8i - 4j – 3k

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Example 4:

If r is the position vector of a particle of mass m relative to point O and F is the external forces on the particle, then r x F = M, is the torque or moment of F about O.

Show that M = dH/dt, where H = r x mv and v is the velocity of the particle.

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Solution:

vmdt

d

vxr

vxrvxr mdt

dm

dt

dm

dt

d

vxvvxr mmdt

d

vxr mdt

d

vxrM mdt

d

dt

dH

M = r x F = r x

But

THANK YOU…

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Lecture 5

Curve:

Arc Length

Tangent ands Curvature


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5.1 THE ARC LENGTH

dudu

rd

du

rds

du

rd

du

rd

du

ds

du

rd

du

rd

du

ds

dzdydx

rdrdds

kdzjdyidxurd

kuzjuyiuxur

u

u

2

1

2

222

2

)(

)()()(

)(

ˆˆˆ)(

ˆ)(ˆ)(ˆ)()(

is length arc The

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5.2 TANGENTS

.

)(

t

Cdt

rdtrC

increasing of direction the inpoint that at

totangent vector a is , by described is curve a If

dt

rd

)(tr

The unit tangent vector to the

curve r, denoted by T, as

'( )( )

'( )

tt

t

rT

r

This indicates the direction of the curve.

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The unit normal vector N to the curve r is

'( )( )

'( )

tt

t

TN

T

The binormal unit vector B of a curve is

B = T x N

The three vectors are mutually perpendicular ,

or they are orthogonal.

The direction of B follow the RH rule.

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Find the unit normal and binormal vectors

for the circular helix

r(t) = cost i + sin t j + t k

Example:

Solution:

Find unit tangent vector

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1( ) ( ) ( ) sin cos 1

2cos sin 0

1sin , cos ,1

2

t t t t t

t t

t t

i j k

B T N

Find the unit normal vector

Find the unit binormal vector

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The curvature of a curve , is:

where T is the unit tangent vector.

d

ds

T

This indicates the magnitude of the rate at which

unit tangent vector turns with respect to the arc

length along the curve.

5.3 CURVATURE

However, ds/dt = |r’(t)| , after using chain rule:

'( )( )

'( )

tt

t

T

r

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5.4 TORSION

dtd

dtd

ds

d

/

/

r

BB

This indicates the magnitude of the rate at which

binormal vector turns with respect to the arc length

along the curve.

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Given the space curve x = t, y = t2, z = 2/3 t3

Find (a) curvature,

(b) the torsion,

Example:

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kjir 2' 22 tt

dt

dr

221

.

t

dt

d

dt

d

dt

d

rrr

i) The position vector is r = ti + t2j + 2/3t3 k,

Solution:

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2

2

21

22

/

/

t

tt

dtdr

dtdand

kjirT

22

2

)21(

4)42(4

t

ttt

dt

d

kjiTT'

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22

32

2222

)21(

2

)21(

)4()42()4(

/

/

t

t

ttt

dtd

dtd

ds

d

r

TT

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)21(

2)21(22

2

t

ttt

kjiN

dtd

dtd

ds

dii

/

/)

r

BB

Then

2

2

21

22

t

tt

kjiTxNB

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22

2

)21(

4)24(4

t

ttt

dt

d

kjiB

22)21(

2

t

In this case, for this curve.

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5.4 Frenet-Serret formulae

NT

ds

d

TBN

ds

d

NB

ds

d

THANK YOU…

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Lecture 6

GRADIENT, DIVERGENCE AND CURL


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6.1 DEL OPERATOR OR NABLA ,The vector differential operator, is defined by

kjizyx

6.2 THE GRADIENT

),,( zyx

zyx

kji

Let , then the gradient of , written

or grad , is defined by

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3

Example 2:

(1,3,2).Pat grad determine , If 2232 zxyyzx

Solution:

4

.ˆ)23(ˆ)2(ˆ)2(

ˆˆˆ

222232223 kzxyyzxjxyzzxizyxyz

kji

zyx

Therefore,

.ˆ72ˆ32ˆ84 kji

have we (1,3,2),PAt

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5

Exercise 3:

(1,2,3).Ppoint at grad determine

, If 323

zxyyzx

.110111126(1,2,3),PAt

Grad

z

y

x

then,Given

~~~

323

kji

zxyyzx

Solution:

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A(x,y,z) = A1 i + A2 j + A3 k

6.3 THE DIVERGENCE

Then the divergence of A, written A.

or div A, is defined by

kjikjiA 32 AAAzyx

1..

z

A

y

A

x

A

321

(a Scalar Product)

A is the net flux of A per unit volume at the point

considered.

Divergence

(a) Positive divergence, (b) negative divergence, (c) zero divergence.

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Example 4:

(1,2,3).point at determine

If

Adiv

kyzjxyziyxA

,ˆˆˆ 22

.13

)3)(2(2)3)(1()2)(1(2

.22

.

Adiv

yzxzxy

z

a

y

a

x

aAAdiv zyx

3), 2, (1,point At

Solution:

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Then the curl or rotation of V, written xV, curl V is defined by

6.4 THE CURL

V(x,y,z) = V1 i + V2 j + V3 k

VVV 321 kjixkjixV

zyx

321 VVV

zyx

kji

kji

213132 VV

yxVVzx

VVzy

kji

y

V

x

V

x

V

z

V

z

V

y

V 123123

(a vector)

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curl V 0

curl V 0

curl V 0

curl V = 0

curl V = 0

MEANING OF CURL

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Example:

.)2,3,1(at determine

,)()( If

~

~

2

~

22

~

224

~

Acurl

kyzxjyxizxyA

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Solution:

yzxyxzxy

zyx

kji

AAcurl

222224

~~~

~~

~

3

~

2

~

2 )42()22( kyxjzxxyzizx

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Solution:

.10682

))3(4)1(2(

))2()1(2)2)(3)(1(2()2()1(

~~~

~

3

~

2

~

2

~

kji

k

jiAcurl

(1,3,-2),At

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)(.1

6.5 RELATED FORMULA

BABA ..).(.2

xBxABAx )(.3

).().().(.4 AAA

)()()(.5 xAxAAx

).().().(.6 xBAxABAxB

).().().().()(.7 BABAABABAxBx

)()().().().(.8 xBAxxABxBAABBA

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2

2

2

2

2

22).(.9

zyx

2

2

2

2

2

22

zyxwhere

is called the Laplacian

operator

0)(.10 x the curl of the gradient is zero

0).(.11 xA the divergence of the curl of A is zero

AAxAx 2).()(.12

THANK YOU…

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Lecture 7

VECTOR INTEGRATION: LINE INTEGRAL


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The concept of vector integral is the same as the

integral of real-valued functions except that the result of

vector integral is a vector.

.

ˆ)(ˆ)(ˆ)()(

kj i

then

If

b

az

b

ay

b

ax

b

a

zyx

duaduaduaduA

kuajuaiuauA

7.1 INTEGRALS OF VECTOR

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3

1

32

.

,ˆ4ˆ)52(ˆ)43(

dtF

ktjtittF

calculate

If

Example 1:

.ˆ80ˆ2ˆ42

ˆ4ˆ)52(ˆ)43(3

1

33

1

3

1

23

1

kji

kdttjdttidtttdtF

Solution:

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If r(t) = 2 cos t i + sin t j + 2t k, find:

22 2

00

2

( ) [2sin cos ]

24

t dt t t t

r i j k

i j k

2/

0)())()

dttriidttri

Solution:

Example 2:

kj ir dttdttdttdtti 2sincos2)()

Cttt kji 2cossin2

ii)

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7.2 LINE INTEGRALS OF A VECTOR FIELD

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. c

zc

yc

x dzFdyFdxF

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Find the total work done in moving a particle in

a force field given by F = 3xy i - 5z j+10x k

along the curve x = t2 + 1, y = 2t2 , z = t3 from

t =1 to t =2.

Example 3:

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Solution:

))(1053(. kjikjirF dzdydxxzxydCC

Total work:

C

dzxdyzdxxy 1053

For x = t2 +1, y = 2t2 , z = t3 from t =1 to t =2.

= 303

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Example 4:

Solution:

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Example 5:

Solution:

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.y2y

if,2,4curve thealong

(4,2,1)B to(0,0,0)A from .Calculate

~~~

2

~

32

~~

kzjxzixF

tztytx

rdFc

TRY:

30

2326

Answer:

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7

3/18/2014 SSCP 2613 - RZ@JFUTM 13

.344

And

.4432

)()2(2)()4()2()4(

2yGiven

~

2

~~

~~~~

~

5

~

4

~

4

~

32

~

3

~

22

~~~

2

~

kdttjdttidt

kdzjdyidxrd

ktjtit

kttjttitt

kyzjxzixF

Solution:

3/18/2014 SSCP 2613 - RZ@JFUTM 14

.)1216128(

)344)(4432(.

75

~

2

~~~

5

~

4

~~~

dtttt

kdttjdttidtktjtitrdF

4

4

Then

.1

,1,22,44 (4,2,1),Bat and,

.0

,0,02,04 (0,0,0),AAt

32

32

t

ttt

t

ttt

3/18/2014

8

3/18/2014 SSCP 2613 - RZ@JFUTM 15

.30

2326

)1216128(.1

0

754

~~

t

t

B

AdttttrdF

THANK YOU…

3/18/2014 SSCP 2613 - RZ@JFUTM 16

3/18/2014

1

Lecture 8

VECTOR INTEGRATION:

SURFACE INTEGRAL


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8.1 SURFACE INTEGRALS

dS

S

n

SS

dSd nFSF ..

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2

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THEOREM:

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Example 1:

Solution:

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3

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4

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Example 2:

Example 1:

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5

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Solution:

THANK YOU…

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6

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1

Lecture 9

VOLUME INTEGRAL


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8.1 VOLUME INTEGRALS

S

dVA S

dV

Consider a closed surface in space enclosing a volume V.

Then

and

Let F = 2xzi – xj + y2k. Evaluate S

dVF

where V is the region bounded by the surfaces x =0, y = 0,

y = 6, z =x2, z =4.

Example 1:

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2

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Solution:

4

2

6

0

2

0 2

)2(

xzyx

dzdydxyxxz kjiS

dVF

46

0

2

0 2

)2(

xzyx

dzdydxxzi

46

0

2

0 2

)(

xzyx

dzdydxxj

4

2

6

0

2

0 2

)(

xzyx

dzdydxyk

= 128i – 24j + 384k

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Example 2:

Solution:

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3

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TRY:

Solution:

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4

THANK YOU…

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3/18/2014

1

Lecture 10

DIVERGENCE THEOREM

STOKE’s THEOREM

GREENS THEOREM


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9.1 The DIVERGENE THEOREM OF GAUSS.

The divergence theorem establishes equality between triple integral (volume integral) of a function over a region of 3D space and the double integral of a function over the surface that bounds that region.

SV S

dSdSdV ... AnAA

Where n is the positive normal to S.

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2

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The surface integral of the normal component of a vector A taken over a closed surface C is equal to the integral of the divergence of A taken over the volume enclosed by the surface.

IF F = xy i + y2z j + z3 k, evaluate S (F n)dS, where S is the unit cube defined by 0 x 1, 0 y 1, 0 z 1.

Example 1:

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SolutionWe see div F = F = y + 2yz + 3z2. Then

1

0

1

0

1

0

2

2

)32(

)32()(

dxdydzzyzy

dVzyzydSD

SnF

2

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3

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9.2 STOKE’S THEOREM

Stoke’s theorem establishes the equality of the double integral of a vector fields over a portion of a surface and the line integral of a field over a simple closed curve bounding the surface portion.

C S

dSd nxArA ).(.

The line integral of the tangential component of a vector A taken around a closed curve C is equal to the surface integral of the normal component of the curl A taken over any surface S having C as itsboundary.

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Let S be the part of the cylinder z = 1 – x2 for 0 x 1, −2 y 2. Verify Stokes’ theorem if F = xyi + yzj + xzk.

Example 2:

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4

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Surface Integral: From F = xy i + yz j + xz k,

Solution:

kji

kji

F curl xzy

xzyzxy

zyx

14

2

01

2

x

x

g

g

xzzyx

kin is normal the

cylinder, the defines If

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nF (curl Therefore

2

)2(

)2(

14

2)

1

0

2

2

2

dydxxxy

dAxxy

dSx

xxydS

R

SS

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5

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9.3 GREEN’S THEOREM IN THE PLANE

If R is a closed region of the xy plane bounded by a simple closed curve C and if M and N are continuous functions of x and y having continuous derivatives in R, then

dxdydy

M

dx

NNdyMdx

RC

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9.4 RELATED INTEGRAL THEOREMS

V S

ddV S..2

This called Green’s First Theorem

V S

ddV S.22

1.

2.

This is called Green’s second symmetrical theorem

3. S SV

ddSdV xASnxAxA

S SC

ddSd Sxnxr )(4.

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6

THANK YOU…

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Lecture 11


Complex Number

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11.1 Complex Number

Complex Number Representation in Rectangular Form and Trigonometry Form

A complex number Z with magnitude r and direction can be written as:

z = x + jy = r (cos + j sin ) Imaginary

real

r

z

0 x

jy

Rectangular form

Trigonometry form

1j

j is the imaginary unit

The x-axis - real axis The y-axis - imaginary axis

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11.2 Powers of j

Express in terms of j: i) √(-16) ii) √(-2)√(-18)

Example 1:

ii)

i)

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r

jyxjrz )sin(cos

The term r is called the modulus of z. The term θ is

called argument of phase of z, and is denoted by arg z.

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Example 2:

Solution :

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jyxz 22 yxr

)(tan 1

x

y

r < atau rej

x = r cos

z = rej

atau r < y = r sin

CONVERTING from Rectangular to Polar and Polar to Rectangular

Rectangular to Polar

Polar to Rectangular

x + jy

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Example 3:

Solution :

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Suppose we have 2 complex numbers, z1 and z2 given by :

2

1

2222

1111

j

j

erjyxz

erjyxz

2121

221121

yyjxx

jyxjyxzz

Easier with normal form than polar form

Addition and Subtraction of Complex Numbers

2121

221121

yyjxx

jyxjyxzz

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,

The complex conjugate of a complex number, z = x + jy,

denoted by z* , is given by

z* = x – jy.

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)(

21

2121

21

21

j

jj

err

ererzz

magnitudes multiply! phases add!

Easier with polar form than normal form

Multiplying

For a complex number z2 ≠ 0,

)(j

j

j

er

r

er

er

z

z21

2

1

2

1

2

1

2

1

magnitudes divide!

phases subtract!

Dividing

)( 21

2

1 r

r

12SSCP 2313- Basic electronics 2012/13-1

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Example 4:

i) ii)

Answer :

i) ii)

Solve:

iii)

iii)

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Example 5: Solve:

i) ii)

Answer :

i) ii)

iii)

iii)

THANK YOU…

3/22/2014 SSCP 2613 - RZ@JFUTM 15

Lecture 12


Conjugate and Absolute Conjugate

De Moivre’s formula

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,

The complex conjugate of a complex number, z = x + jy,

denoted by z* , is given by z* = x – jy.

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zz

2121 zzzz

2121 zzzz

2121 zzzz

2

1

2

1

z

z

z

z

)0( 2 z

12.2 Further results on

ii)

iii)

iv)

v)

i)

LAWS OF CONJUGATES

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iziz 23 ,85 21

Example:

2

12121 ,

z

zandzzzzFind

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4

105

Example:

Solution :

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12.3 De Moivre’s formula

ninin

sincossincos

For n = 2, we have

sincos22sin

sincos2cos

2sin2cossincos2sincos

2sin2cossincos

22

22

2

ii

ii

3/22/2014 SSCP 2613 - RZ@JFUTM 8

5.26sin5.26cos52 ii

Then apply De Moivre’s Formula

Example 1:

Solution:

7.265sin7.265cos)5()2( 1010 ii

)07.0(3125 j

****check

THANK YOU…

3/22/2014 SSCP 2613 - RZ@JFUTM 9

Lecture 13


An Application of Complex Numbers: AC Circuits (part 1)

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,

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,

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,

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, Inductive reactance is : XL = ωL

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Z = R + j(XL - XC)

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,

Solution :

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,

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capacitive reactance is :

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THANK YOU…

3/22/2014 SSCP 2613 - RZ@JFUTM 13

Lecture 14


An Application of Complex Numbers: AC Circuits (part 2)

R L C

VS

The total impedance for the RLC circuit is given by

L CR jX jX Z

In polar form, this is written

22 1tan tot

L C

XR X X

R

Z

14.1 Series RLC circuits

)( CL XX

Z

R

“ Impedance Triangle”

2

2RZ1013@UTM

What is the total impedance for the circuit below?

R L C

VS

330 mH

f = 400 kHz

470 W 2000 pF

786 53.3 W Z

The circuit is inductive.3

3RZ1013@UTM

3/22/2014 SSCP 2613 - RZ@JFUTM 4

,

14.2 Parallel RLC circuits

R L CVS

)11

(1

1111

LC

LC

XXj

R

jXjXRZ

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, ZT =

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,

14.3 Series-parallel RLC circuits

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,

THANK YOU…

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3/22/2014 SSCP 2613 - RZ@JFUTM 10

Lecture 15


Functions of a Complex Variable

The Cauchy-Reinmann equations,

3/22/2014 SSCP 2613 - RZ@JFUTM 2

,

15.1 Functions of a Complex Variable

3RZ1013@UTM

Example 1

4RZ1013@UTM

15.1 Limit of a function

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Example 2

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Example 3

Chain Rule

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Solution

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Example 4:

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Example 5

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Example 6

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Example 7:

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Example 8:

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THANK YOU…

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4/18/2014

1

Lecture 16


Harmonic functions

Complex exponential function

4/18/2014 SSCP 2613 - RZ@JFUTM 2

,

16.1 Harmonic Function

3RZ1013@UTM 4RZ1013@UTM

16.2 Conjugate Harmonic Function

Example 1

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2

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Solution

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16.3 Complex Exponential Function

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Example 2

Solution

2

2

232

)01(

)3sin()3cos(

e

ie

iee i

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ixyyxiyxz 2)( 2222

)2sin()2cos()(

)2(

22

222

xyixye

ee

yx

xyiyxz

)2sin()2cos( )()( 2222

xyiexye yxyx

Example 3

Solution

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3

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THANK YOU…

4/18/2014 SSCP 2613 - RZ@JFUTM 10

Education

Sem4 sscz fizik math note