Vector II 2015

Vector Functions and Its Differentiation

1

Sk/EUM114/ Vector 2 Lecture/2015

Vector functions and its differentiation

1.1 Vector function

Def. : A vector function f(t) as a function of a scalar t is defined as:

f(t)=f1(t) i + f2(t) j +f3(t) k

in which f1(t), f2(t) and f3(t) are the scalar functions of t.

Example: The trajectory of a cannon ball fired at 45o to the horizontal ground with

an initial velocity v is given by r(t)= v(cos45o)t i + [v(sin45

o)t-0.5gt

2] k. Find the

velocity and acceleration of the cannon ball.

1.2 Parametric equation of plane curves

For a plane curve in the two dimensional space, we have already known that it

is usually represented by an equation in x and y. For example, a circle is represented

by x2+y

2=r

2 or a parabola is represented by y=ax

2. However, a plane curve is also

possibly represented by the so called parametric equation, which is indeed a vector

function.

Consider the case of a circle with radius r0 in the x-y plane,

x=r0cos and y=r0sin

and we represent the circle locus by the

parametric equation:

f()=r0cos i +r0sin j

r0

x

y

2


Def. : If the vector function f(t)=f1(t) i + f2(t) j + f3(t) k is the parametric equation of a locus in

the three dimensional space, then the coordinates of all the points on the locus satisfy: x=f1(t),

y=f2(t), z=f3(t) for t [a, b], where [a, b] is the range.

Example :

Find the Cartesian equation of the locus represented by the parametric equation: f(t)=t i +et j.

Solution : x=t and y= et

1.3 Differentiation and integration of vector function

Differentiation of vector function

If f(t)=f1(t) i + f2(t) j + f3(t) k is a vector function and f1, f2 and f3 are differentiable, then the

derivative of f(t) is defined as:

d

dt' (t) f ' (t) f ' (t) f ' (t)1 2 3

ff i j k (1)

Example 1: If f()=(cos)i+e2j, find f().

Solution : f()=-sin i+ 2e2 j Example 2: The trajectory of a cannon ball fired at 45

o to the horizontal ground with an initial

velocity v is given by r(t)= v(cos45o)t i + [v(sin45

o)t-0.5gt

2] k. Find the velocity and acceleration

of the cannon ball.

Solution : v(t) = dr(t)/dt = v(cos45o) i + [v(sin45

o)-gt] k

a(t) = dv(t)/dt = -g k

1.3 Integration of vector function

If f(t)=f1(t) i + f2(t) j + f3(t) k is a vector function and f1, f2 and f3 are integrable, then

the indefinite integration of f(t) is defined as:

f i j k C(t)dt f (t)dt] f (t)dt] +[ f (t)dt1 2 3 [ [ ] (2)

where C is a constant vector.

Likewise, the definite integral of f(t) is defined as :

3


f i j k(t)dt [ f (t)dt] [ f (t)dt] [ f (t)dt]a

b

1a

b

a

b

3a

b

2 (3).

In previous example if r(t) is the vector function denoting the position vector of a particle as a

function of the time t. Then, the velocity vector and the acceleration vector is described by the

vector function: r(t) and r(t).

Properties of Vector Differentiation

(i) If f and g are differentiable vector functions,

d

dt

d

dt

d

dtf g

f g

(i) If f is differentiable vector function and is a constant scalar,

d

dt

d

dt f

f

(ii) If f(t) is a vector function, v is a constant vector and vf is differentiable,

d

dt

d

dtv f v

f

(iii) If h(t) is a scalar function, f(t) is a vector function and hf is differentiable,

d

dth h

d

dt

dh

dtf

ff

(iv) If f and g are vector functions and fg are differentiable,

d

dt( )

d

dt

d

dtf g f

gg

f

(v) If f and g are vector functions and fg are differentiable,

d

dt

d

dt

d

dtf g

fg f

g

Example 1: If f(t)=ti+t3j, g(t)=(cost)i+(sint)j and v=2i-3j. Calculate (a) (f+g); (b) (vf) and

(c) (fg). Solution :

(a) f+g= (t+ cos t )i + (t3+sint)j so (f+g)=(1-sin t )i + (3t2+cos t)j (b) vf=2t-3 t3 so (vf)=2-9 t2 ,

you can also check (vf)=v(f)=(2i-3j)( i+3t2j)= 2-9 t2 (c) fg=t cos t + t3 sin t so (fg)=cos t tsin t + 3 t2sin t + t3cos t

4


Try d

dt( )

d

dt

d

dtf g f

gg

f yourself to see if you get the same result.

1.4 Geometric Interpretation of f(t)

Suppose f(t) is the parametric equation of a locus. The derivative of f is:

ff f

' (t) lim(t t) (t)

tt 0

and its geometric meaning is indicated in the figure. Notice that as t0, the vector f(t+t)-f(t) will become parallel to the tangent. Therefore, f(t) is indeed has the same direction with the tangent vector to the curve.

1.4.1 Unit tangent vector

If f(t) is a vector function representing the parametric equation of a curve, then the unit tangent

vector T is defined as:

Tf

f(t)

' (t)

' (t) (4)

The unit tangent vector indeed represents the direction of the tangent to the curve.

Example :

Find the unit tangent vector of the following curves:

(i) f=(ln t)i + (1/t)j at t=1 Solution: f=(1/t) i + (-1/t2)j; f=(1/t2+1/t4)1/2 when t=1, we T(t=1)=(1/(2)1/2) i + (-1/(2)1/2)j (ii) f(t)=(cost)i +(sint)j +(sint)k. Solution : f(t)=(-sin t)i +(cos t)j +(cos t)k and f(t)= (sin2t+cos2t + sin2t)1/2 = (1 + sin

2t)

1/2

f(t+t)-f(t)

f(t+t)

f(t)

5


Theorem 2

If f(t)=f1(t)i+f2(t)j+f3(t)k be a vector function and the magnitude

f (t) [f (t)] [f (t)] [f (t)]12

2

2

3

2 is a constant for all value of t, then

f(t)f(t)=0 for all t.

Proof: Suppose ff=|f|2=C=constant,

d

dt( ) = 2(

d

dt

dC

dt0f f f

f )

Therefore, ff=0.

Example: Prove that Theorem 2 is valid for the case f(t)=cost i + sint j.

Solution : f2 (t)= (sin

2t+cos

2t ) = 1 then f(t)=(-sin t)i +(cos t)j and ff=-sint(cos t)+cost(sint)=0

1.4.2 Unit normal vector

As we know that the magnitude of the unit tangent vector T is always unity, by Theorem 2,

TT=0, which means that the derivative T is always perpendicular to the unit tangent vector T.

Def. : The unit normal vector of a curve represented by a parametric equation is defined as:

(t)

(t)'(t)

T'

Tn (5)

N.B. The direction of the unit normal vector is always pointing in the direction of the concave

side of the curve.

Examples: Calculate the unit normal vectors to the following curves:

(i)f(t)=cos t i + sin t j

Solution : f(t)=(-sin t)i +(cos t)j and f=1 since Tf

f(t)

' (t)

' (t) = f(t)

T(t)= f(t)=-cost i +(-sin t)j and T(t)=1, therefore (t)

(t)'(t)

T'

Tn =-cost i +(-sin t)j

We can see explicitly that Tn=sint cost-cos t sin t =0

6


(ii)f(t)=[(t3/3)-t]i+t

2j

Try this yourself and make sure that you get Tn=0

1.5 Vector field

Vector field is a function that assigns a unique vector pF for each point p in a region. Imagine the flow of a stream, at each point, the water has a certain velocity, which can be represented as a

vector. Thus the stream is a vector field with velocity vectors. There are a lot of applications of

vector field used in our daily life. For example, gravitational force and Coulomb force are vector

fields as the force vector is a function of the position vector.

We will use bold face character as vector and the conventional symbols: i is the unit vector in

the x-axis direction, j is the unit vector in the y-axis direction and k is the unit vector in the z-

axis direction.

Example

(a) jiF yx, .

In this vector field, the same vector ji is assigned to every point.

(b) jiF yxyx , .

7


(c) jiF44

,2222

yx

y

yx

xyx .

(c) jiF xyyx sin, .

8


1.5.1 Gradient and directional derivatives

The gradient of a (scalar) function yxff , is defined by

ji yxfyxfyxf yx ,,, .

As we can see, the gradient of a function is a vector field. It is the vector field which will points

in the direction of the greatest rate of change of the function f. The magnitude of the gradient is

the greatest rate of change.

The rate of change of the function yxf , at 00 , yx in the direction u is define to be

uu 0000 ,, yxfyxfD .

This is called the directional derivative of f at 00 , yx in the direction of u.

Note: u is a unit vector.

Example

9


Find the directional derivative of yyxyxf 3, 2 in the direction of 2,1 at the point 2,1 .

Solution

The gradient of f is jiji 32 2 xxyfff yx . At the point 2,1 ,

jiji 44312122,1 2 f .

The unit vector in the direction ji 2 is

jiji

ji

jiu

5

1

5

2

12

2

2

2

22

The directional derivative is

5

4

5

14

5

24

5

1

5

244

5

1

5

22,12,1

jijijiu ffD .

Example

Find the direction in which the function yyxyxf 3, 2 has the highest rate of change at the

point 2,1 .

Solution

The directional derivative at the point 00 , yx in the direction of u is

cos,cos,,, 00000000 yxfyxfyxfyxfD u uu

since u is a unit vector and is the angle between the two vectors 00 , yxf and u. From the

right side of the equation, the quantity 00 , yxfDu is largest when cos reaches its maximum,

i.e., 1cos . This happens when 0 . Since is the angle between the two vectors

00 , yxf and u, that means u is parallel to 00 , yxf .

Thus yyxyxf 3, 2 has the highest rate of change at the point 2,1 in the direction

jiji 44312122,1 2 f .

10


11


12


1.5.2 Divergence and curl

We have been working on real valued functions: functions where the output is just a number.

Although we had defined what a vector field is, we have not done any operation yet. We will

next define two operations on vector fields in 3-D space.

Del or Nabla operator zr x y z

i j k

Let kjiF zyxhzyxgzyxfzyx ,,,,,,,, be a vector field in the 3-D space. Then the

divergence of F, denote div F, is defined to be

z

h

y

g

x

fdiv

F .

The curl of F, denote curl F, is defined to be

kji

kji

FF

y

f

x

g

x

h

z

f

x

g

y

h

hgf

zyx

curl

.

Divergence is the flux density across a rectangular region:

y

Q

x

P

area Rectangle

boundary rectangle acrossFlux

13


This is the net rate of change of the mass of fluid flowing from a point. If 0div F (for all

points), then the field F is said to be incompressible. If 0),,(div 000 zyxF , the point

),,( 000 zyx is called a point source because the field is gaining fluid at the point. If

0),,(div 000 zyxF , the point ),,( 000 zyx is called a point sink because the field is losing fluid

at the point.

Curl is the circulation density around a rectangular region:

Circulation around rectangle boundary

Rectangle area

Q P

x y

.

This is a measure of the rate of rotation of the field about a point. If (curl F) = 0 at a point P,

then the fluid is free of rotations at P and F is called irrotational at P.

14


Derivation of Curl in the Plane

Lets look at a simple development of curl (circulation density or spin) at a point (x, y). Consider

the circulation around a small rectangle with corners ),,(),,( yxxyx ),,( yyxx and

),( yyx as shown below.

Circulation and curl are oriented concepts. ),( yyx ),( yyxx

Following the right hand rule, we regard

counter-clockwise travel or rotation as being

positive orientation. We imagine the rotation (x,y) ),( yxx

at the point (x, y) in the plane about a vertical axis.

In this manner, we consider the effect of the field

jiF ),(),(),( yxQyxPyx along the positively k

oriented rectangle. We make the following

15


observations:

Across the bottom of the rectangle: xyxPd ),(rF

Across the right side of the rectangle: yyxxQd ),(rF

Across the top of the rectangle: xyyxPd ),(rF

Across the left side of the rectangle: yyxQd ),(rF

The careful reader should note that we have chosen to evaluate F at the point closest to (x,y) in

each case. Therefore the circulation around the rectangle is

yyxQxyyxPyyxxQxyxP ),(),(),(),(

If we factor out yx , we get the circulation about the rectangular area:

yxy

yxPyyxP

x

yxQyxxQ

),(),(),(),(

Taking the limit as y and x go to zero yields

y

P

x

Q omitting the area differential

We regard this quantity as the circulation density or curl at the point (x, y).

Curl in 3 Dimensions

To develop the curl about a point in three dimensions, we also consider the curl components

about the axis in the i direction with respect to y and z and about the axis in the j direction with

respect to x and z.

For the spin with respect to y and z about ),( zzy ),( zzyy

the axis in the i direction, we simply replace

x and y in our previous development with y and

z, respectively. Therefore, we obtain the (y, z) ),( zyy

following relationships.

i

Across the bottom of the rectangle: yzyQd ),(rF

Across the right side of the rectangle: zzyyRd ),(rF

Across the top of the rectangle: yzzyQd ),(rF

Across the left side of the rectangle: zzyRd ),(rF

Taking the limit as z and y go to zero gives

16


z

Q

y

R

Due to the right-hand rule, jik , the remaining component of curl is

x

R

z

P

Thus, the 3D version of the curl of F at a given point (x, y, z) is

kjiF

y

P

x

Q

x

R

z

P

z

Q

y

R curl

Furthermore, we have the following notation using the differential operator :

RQP

zyx

kji

FF curl Note: curl F is a vector

and

z

R

y

Q

x

P

FF div Note: div F is a scalar

Rules: f,g scalar functions of position; A,B vector functions of positions

Gradient fg f g g f

A B A B B A A B B A Divergence fA f A A f A B B A A B Curl fA f A A f A B B A A B A B B A

2 2 2

2

2 2 2x y z

Laplacian operator

0curl grad

0A div curl A

17


Note that div F is a scalar function whereas curl F is a vector field.

Example

Find the divergence and curl of the following vector fields

kjiF 2,, yxyxzxyzyx .

Solution

yy

z

yx

y

yxz

x

xy

z

h

y

g

x

fdiv

11

2F .

kji

kji

kjiF

xzz

y

xy

x

yxz

x

yx

z

xy

x

yxz

y

yx

y

f

x

g

x

h

z

f

x

g

y

h curl

1

22

The divergence of a vector field measures how the vector field is expanding (positive value) or

contracting (negative value) at a certain point. For example, the vector field

jiF yxyx 322, has divergence 3Fdiv

18


and the divergence for the vector field in the 3-D space kjiF zyxzyx ,, is 3Fdiv

The curl of a vector field captures how a vector field rotates. It gives the axis of rotation and the

speed of the rotation. The curl of the vector field kjiF xyzyx ,, is kF 2,, zyxcurl .

19


and for a vector field such as kjiF zyxzyx ,, , the curl is zero which says that there is no

rotational action.

A three dimensional vector field is defined as a function mapping from a three

dimensional vector, e.g. position vector r(x,y,z), to a three dimensional vector. i.e.

kzyxRjzyxQizyxPzyxF ),,(),,(),,(),,(

, where P, Q and R are three variable functions.

1.6.1 Line integral Consider a segment of the curve C beginning at points A and B, with the coordinates given by

)(ar

and )(br

respectively. Now we try to partition the curve segment into N partitions so as to

20


obtain the partition points X1, X2,... and XN-1. We can then construct a vector ild

from point Xi-1

to point Xi. The force field vector at this partition can also be obtained as ))(( ii trFF

, where

( )ir t is the function of the line in parametric form. The line integral of the vector field F

over

the curve C from point A to B is given by

i

iidl

B

AldtrFldF

i

))((lim

0

.

Remark :

If the integrating path of the line integral C is made up of several sections of curve C1, C2 and C3

as shown in the figure, then the line integral over the path C from point X1 to X2 is:

321 CCCC

ldFldFldFldF

(2)

Line Integrals in 2-Dimensional Space R2 or 3-Dimensional Space R3

In General

2-Dimensional Space R2

X1 X2

X3

X4

C1

C2

C3

21


22


A vector field is given by:

( , ) ( , ) ( , )F x y P x y i Q x y j

and a curve C is given by the parametric equation :

C : ( ) ( ) ( )r t xi yj f t i g t j , atb.

The line integral of the vector field F

along the curve C is defined as:

, ,

( ), ( ) ( ), ( )

b b

C a a

b b

a a

F dr P x y dx Q x y dy

df dgP f t g t dt Q f t g t dt

dt dt

Or 3-Dimensional Space R3

A vector field is given by:

kzyxRjzyxQizyxPzyxF ),,(),,(),,(),,(

and a curve C is given by the parametric equation :

C : ( ) ( ) ( ) ( )r t xi yj zk f t i g t j h t k , atb.

The line integral of the vector field F

along the curve C is defined as:

23


, , , , , ,

( ), ( ), ( ) ( ), ( ), ( ) ( ), ( ), ( )

b b b

C a a a

b b b

a a a

F dr P x y z dx Q x y z dy R x y z dz

df dg dhP f t g t h t dt Q f t g t h t dt R f t g t h t dt

dt dt dt

Steps for Line Integrals

Parametrize and Integrate

a) C is a line joining A and B: Use ( ) ( )r t A t B A or ( ) (1 ) ( )r t t A t B

b) C is a circle: Use ( ) ( cos ) ( cos ) ( )r t a t i a t j c k

c) C is an Ellipse: Use ( ) ( cos ) ( cos ) ( )r t a t i b t j c k

d) C is a Hyperbola: Use ( ) ( cosh ) ( cosh ) ( )r t a t i b t j c k

e) C is a Parabola: Use 2( ) ( ) ( )r t t i at bt c j

f) C is a Helix: Use ( ) ( cos ) ( cos ) ( )r t a t i a t j ct k

g) C is given by y=f(x). Use ( ) ( ) ( )r t t i f t j

24


25


26


27


Examples:

(i) A particle is moving along the parabola y=x2 subject to a force given by F( r ) = 2xy i + (x

2+ y

2) j. Calculate the work done when the particle moves from (1,1) to (3,9).

Solution : First we need to express the curve in terms of parametric form, i.e.

( ) ( ) ( )r t f t i g t j , where f(t)=x, g(t)=y. We can choose x=t for simplicity then y= t2

because y=x2. Therefore 2 ( )r t ti t j and the range of integration is 1 t 3. We can also

express the force in terms of parameter t, i.e. 2 2 3 2 4 ( ) 2 ( ) ( ) ( ( ) ( ) ) 2 ( )F t x t y t i x t y t j t i t t j

3 3

1 1

3 3

3 2 4

1 1

( ), ( ) ( ), ( )

2(2 )(1) ( )(2 ) 322

3

C

df dgW F dl P f t g t dt Q f t g t dt

dt dt

t dt t t t dt

(ii) Calculate C

ldF

if jyeixyyxF x ),(

and C is the close rectangle path joining

the points (0, 0), (2, 0), (2, 1) and (0, 1).

Solution: The line can be divided into 4 paths shown in figure below

28


Here C1: r(t)=2ti for 0 t 1 and r(t)=2i , C2 : r(t)=2i + tj for 0 t 1 and r(t)=j, C3 : r(t)=

(2-2t)i + j for 0 t 1 and r(t)=-2i , C4 : r(t)= (1- t)j for 0 t 1 and r(t)=-j

1 2 3 4C C C C CF dl F dl F dl F dl F dl

1

1On C , x=2 and =0 so that =0 and =0, hence

0

x

C

t y xy ye

F dl

2

2

2

12 2

0

On C , =2 and = so that (r)=2 i+ j and r'( )=j, hence

/ 2C

x y t F t te t

F dl te dt e

You can show that

3 4

12 and

2C CF dl F dl

(iii) Find the line integral if kzjyixzyxF ),,(

over the line 2 3 ( )r t ti t j t k joining the points (0, 0, 0) and (1, 1, 1).

Solution : Here P(t)=x(t)=f(t)=t, Q(t)=y(t)=g(t)=t2, R(t)=z(t)=h(t)=t

3 :

1 1 1

2 3 2

0 0 0

( ), ( ), ( ) ( ), ( ), ( ) ( ), ( ), ( )

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

2

b b b

C a a a

df dg dhF dl P f t g t h t dt Q f t g t h t dt R f t g t h t dt

dt dt dt

t dt t t dt t t dt

29


1.6.2 Exact vector field

Consider there are two points A and B in the space and we can construct infinite number

of curves which pass through these two points. If C1 and C2 are two curves which pass through

A and B, and )(rF

is a vector field, then in general the two line integrals 1C

ldF

and 2C

ldF

may not be equal.

Definitions : A vector field )(rF

is said to be an exact field if and only if the line integral of

)(rF

joining the two end points is independent of the path.

30


Example: Let jxiyyxF ),(

. Calculate the line integral if C is as follows:

(i) The straight line from (0, 0) to (1, 1)

(ii) The parabola y=x2 from (0, 0) to (1, 1)

(iii) The curve jtittr )( 52/3

from (0, 0) to (1, 1).

Solution :

(i)For a straight line ( )r t t i t j , P(t)=y(t)=f(t)=t, Q(t)=x(t)=g(t)=t,

1 1

0 0

1 1

0 0

( ), ( ) ( ), ( )

( )(1) ( )(1) 1

C


dt dt

t dt t dt

(ii)For the parabola 2 ( )r t ti t j ,P(t)=y(t)=f(t)=t2 , Q(t)=x(t)=g(t)=t,

1 1

0 0

1 1

2

0 0

( ), ( ) ( ), ( )

( )(2 ) ( )(1) 1

C


dt dt

t t dt t dt

(iii)For the curve jtittr )( 52/3

P(t)=y(t)=f(t)=t5 , Q(t)=x(t)=g(t)=t

3/2 ,

1 1

0 0

1 1

5 4 3/2 1/2

0 0

( ), ( ) ( ), ( )

3( )(5 ) ( )( ) 1

2

C


dt dt

t t dt t t dt

31


Theorem

Let kzyxRjzyxQizyxPzyxF ),,(),,(),,(),,(

be a three dimensional vector. The

following three statements are necessary and sufficient conditions of each other.

(1) ),,( zyxF

is an exact vector field.

(2) There exists a scalar function ( , , )x y z such that ),,(),,( zyxFzyx

.

(3) y

R

z

Q

z

P

x

R

x

Q

y

P

,,

Actually these relations can be proved easily from (2).

Remarks: (i) If the vector space is a two dimensional vector space, then statement (3) becomes:

x

Q

y

P

(1) If F

is an exact vector field and F

, then the line integral of F

over any path C

joining two fixed points ),,( 1,111 zyxr

and ),,( 2,222 zyxr

will be identical and equal to:

32


2

11 2( ) ( )

r

rC

F dl d r r

(2) If F

is an exact vector field, then the line integral of F

over any closed path will be equal

to zero, which is obvious from above expression.

33


In fact exact vector field exist in nature, e.g. gravitational field, static electric field etc. On the

other hand frictional force is not exact vector field.

Examples

(1)Test the following functions to see if they are exact and then find the function such that

F

(a) F i j k , (b)

2 2 ( 1)F xyi x j and (c)

2 2 cos sinF x yi x yj

Solution

By using (3) of Theorem 1, the first two functions are exact and the last one is not. (a)

x y z C

and (b)

2x y y C

(2)In last example we have shown that the work done by a force vector field, jxiyyxF ),(

,

is the same for three different paths. Lets see if this force vector field is exact. We can see that

P=y and Q=x, 1 and 1P Q P Q

y x y x

. Therefore F

is an exact vector field that is

why the line integral of last example is independent of path.

(3) Let jy

yxix

yxyxF 1

31

4),( 2433

. Prove that F

is an exact field and then

calculate the line integral joining the points (1, 1) and (2, 3).

Solution : For this vector field

34


3 3 4 2

3 2 3 2

1 1( , ) 4 and ( , ) 3

12 and 12

P x y x y Q x y x yx y

P Q P Qx y x y

y x y x

Since this is an exact field so we can choose any path to obtain the line integral, the simplest path

for this integration should be C= C1 + C2, where C1 is y=1 (dy=0) and C2 is x=3 (dx=0). The line

integral is given by

1 2

1

2

23 3

1

34 2 4 3

1

( ) ( ) 431 ln 2 ln 3

1( ) 4 (1) 15 ln 2

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

C C C

C

C

W F dl Pdx Qdy Pdx Qdy

Pdx Qdy x dxx

Pdx Qdy y dyy

We can choose another path, for example C= C1 + C2, where C1 is x=1 (dx=0) and C2 is y=3

(dy=0). The line integral becomes

1

2

1 2

34 2 3

1

23 3 4 3 3

1

1( ) 3(1) 3 ln 3 1

1( ) 4 (3) 2 3 ln 2 3

( ) 431 ln 2 ln 3

C

C

C C

Pdx Qdy y dyy

Pdx Qdy x dxx

Pdx Qdy

HW: Repeat the above line integral by using ( ) ( ) ( ) ( 1) (2 1)r t x t i y t j t i t j . You should

obtain the same result.

(You can also try to find such that W= (2,3)-(1,1). Of course in this example is very easy to be obtained and it may not be true in general. )

(4) Let jyxixyxF )(),(

. Prove that F

is not an exact field and calculate the line integral

joining the points (0, 0) and (1, 1) through the two different paths:

(a) 21 : xyC

(b) 32 : xyC

( , ) and ( , ) , 0 and 1

Therefore this is not an exact vector field.

P Q P QP x y x Q x y x y

y x y x

(a)

1

1 1

0 0

1 1

2

0 0

( ), ( ) ( ), ( )

2( )(1) ( )(2 )

3

C


dt dt

t dt t t t dt

35


(b)

2

1 1

0 0

1 1

3 2

0 0

( ), ( ) ( ), ( )

3( )(1) ( )(3 )

4

C


dt dt

t dt t t t dt

We can see that the line integral is path dependent if the vector field is not exact.

(5) Let kxyjxziyzzyxF ),,(

be a vector field. Prove that F

is an exact field and find the

line integral joining the points (1, 1, 1) and (2, -1, 3).

Solution:

( , , ) , ( , , ) , ( , , )

, , , , ,

, ,

Therefore this is an exact vector field.

P x y z yz Q x y z xz R x y z xy

P P Q Q R Rz y z x y x

y z x z x y

P Q R P Q R

y x x z z y

We choose C= C1 + C2 +C3, where C1 is (1,1,1) to (2,1,1)(it means y=1, z=1, dy=0, dz=0) , C2 is

(2,1,1) to (2,-1,1) (it means x=2, z=1, dx=0, dz=0) and C1 is (2,-1,1) to (2,-1,3) (it means x=2,

y=-1, dx=0, dy=0).

1 2 3

2 1 3

1 1 1

2 1 3

1 1 1

( , 1, 1) (2, ,1) (2, 1, )

(1)(1) (2)(1) (2)( 1) 1 (2)( 2) ( 2)(2) 7

C C C CF dl F dl F dl F dl P x y z dx Q y dy R z dz

dx dy dz

HW: Repeat the above line integral with (1,1,1) to (1,-1,1) to (1,-1,3) to (2,-1,3) to see if you can

the same result.

(Again you can find first and use it to find the line integral)

1.6.3 Flux

36


37


1.6.4 Green's theorem in two dimensional plane

is an area region in the two dimensional plane and the area region is enclosed by the closed

loop . jyxQiyxPyxF ),(),(),(

is a vector field. Green's theorem states that :

dAy

P

x

QldF

Example :

(1) Find the line integral dyyxxydx )( if is the rectangle {(x, y): 0x1, 1y3}. Solution: Using the Greens theorem

1 3

0 1

( , ) , ( , ) , 1, so

(1 ) 1

Q PP x y xy Q x y x y x

x y

Q PF dl dA x dydx

x y

38


You can convince yourself by doing the line integral directly (Be careful about the integration

direction, which must follow (0,1) to (1,1) to (1,3) to (0,3) to (0,1)). The result of the line integral

should be , -2, -3/2 and 4 for these four paths respectively.

(2) Find the line integral C dyxydxyx )2()(3333

where C is a unit circle centered at the

origin.

Solution:

2 2

2 2 2

Since the integration range is a unit circle for the line integral so it

corresponds to a disk in area integration. Next we have

3 3 so

3 3 ( )

Q Px y

x y

Q PF dl dA x y dA r r

x y

2 1

0 0

3

2

drd

You can also carry the line integral and you will find that it is more difficult. You need to use the

following parametric equation : x=cos t and y=sin t

39


40


An Alternative Application of Greens Theorem

In the previous two examples, we have used Green's Theorem to find the line integral by

calculating its corresponding double integral. On the other way around, we can also find the area

of a region by calculating the corresponding line integral.

dA of Area

Construct a vector field jxiyyxF ),(

and apply Green's Theorem:

xdydxyldFdA )(2

xdydxydA )(21

(5)

In fact the function is not unique, either F(x,y)=-yi and F(x,y)=xj can be used. The above

expression can be rewritten as

1

( )2

dA y dx xdy y dx x dy

You can choose any one of these three integrals to get the area.

Example: Use Green's theorem to find the area of the ellipse: 12

2

2

2

b

y

a

x .

Solution:

41


2

0

2 22

0 0

First we need to find the line equation of the ellipse, which has

the parametric form as

( ) cos sin and 0 2

( cos )( cos )

(cos ) (1 cos 2 )

r t a t i b t j t

d b tA dA xdy a t dt

dt

ab t dt ab t dt ab

42


43


44


Hints on Line Itegrals

Case 1. Parametrize and Integrate

h) C is a line joining A and B: Use ( ) ( )r t A t B A

i) C is a circle: Use ( ) ( cos ) ( cos ) ( )r t a t i a t j c k

j) C is an Ellipse: Use ( ) ( cos ) ( cos ) ( )r t a t i b t j c k

k) C is a Hyperbola: Use ( ) ( cosh ) ( cosh ) ( )r t a t i b t j c k

l) C is a Parabola: Use 2( ) ( ) ( )r t t i at bt c j

m) C is a Helix: Use ( ) ( cos ) ( cos ) ( )r t a t i a t j ct k

n) C is given by y=f(x). Use ( ) ( ) ( )r t t i f t j

Case 2. Stokes Theorem a) In R2, Stokes Theorem is called Greens Theorem. You might as well do every problem

referring to Greens Theorem by using the general Stokes theorem formula.

b) In the xy plane , if F=, then X F=

c) In the xy plane , dS = , so ( )y xR R

W F dS Q P dxdy

d) In the xz plane, ( ) ( ) ( )dS dx i dz k dxdz j

e) In the yz plane, ( ) ( ) ( )dS dy j dz k dydz i

Case 3. Easy. Find the potential f and evaluate at the end points.

45


Summary of Line Integrals

Type

of Int.

Notation (Look closely)

Definition (& Application and/or

Geometric Interpretation) Standard/Direct Way to Compute

Major Theorems that

Indicate How to Compute

Indirectly

lin

e in

tegra

l of

a r

eal-

va

lued

fu

nct

ion

( , )C

f x y ds * *

1

lim [ ( ), ( )]n

i i in

i

f x t y t s

(mass/moments of a wire in the plane/ surface area of a curtain)

2 2

[ ( ), ( )]

b

a

dx dyf x t y t dt

dt dt

,

(i.e., ( ( )) ( )

b

a

f t t dt r r )

( , )C

f x y dx (similar for dy, this is

NOT ds)

* *

1

lim [ ( ), ( )]n

i in

i

f x t y t x

(for convenience in representing

and evaluating line integrals of

vector fields)

( ( )) ( )

b

a

f t x t dt r

( , , )C

f x y z ds * * *

1

lim [ ( ), ( ), ( )]n

i i i in

i

f x t y t z t s

(mass/moments of a wire in space)

2 2 2

[ ( ), ( ), ( )] ,

b

a

dx dy dzf x t y t z t dt

dt dt dt

(i.e., ( ( )) ( )

b

a

f t t dt r r )

( , , )C

f x y z dx (similar for dy and dz.

This is NOT ds)

* * *

1

lim [ ( ), ( ), ( )]n

i i in

i

f x t y t z t x

(for convenience in representing

and evaluating line integrals of

vector fields)

( ( )) ( )

b

a

f t x t dt r

lin

e in

tegra

l

of

a 2

-dim

.

vec

tor

fiel

d

C drF , OR

C dsTF , OR

n

i

iiiin

sttytx1

*** )()(),(lim TF

(work)

Path Independence Thm:

Evaluate the standard way over

a more convenient path

Fund. Thm. of Line Int.:

))(())(( afbfdC

rrrF

Use only when F

is conservative

46


CP dx Q dy

b

a

dttt )('))(( rrF Greens Theorem:

dAy

P

x

QQdyPdx

C D

where C is the boundary of D

lin

e in

tegra

l of

a 3

-

dim

. vec

tor

fiel

d

C drF , OR

C dsTF , OR

CP dx Qdy R dz

n

i

iiiiin

sttztytx1

**** )()(),(),(lim TF

(work)

b

a

dttt )('))(( rrF

Path Independence Theorem:

Evaluate the standard way over

a more convenient path

Fund. Thm. of Line Int.:

))(())(( afbfdC

rrrF Stokes Theorem:

SC dd SFrF curl where C is the boundary of S

Not helpful when F is

conservative (since we

know from Theorem

that the value of the

integral is zero)

Use only when F

is conservative

Not helpful when F is

conservative (since we

know from Theorem

that the value of the

integral is zero)

Vector II 2015

Documents

Transcript of Vector II 2015