Vector Calculus CHAPTER 9.5 ~ 9.9. Ch9.5~9.9_2 Contents 9.5 Directional Derivatives 9.5 Directional...

Vector Calculus

CHAPTER 9.5 ~ 9.9

Ch9.5~9.9_2

Contents

9.5 Directional Derivatives9.6 Tangent Planes and Normal Lines9.7 Divergence and Curl9.8 Lines Integrals9.9 Independence of Path

Ch9.5~9.9_3

9.5 Directional Derivative

IntroductionSee Fig 9.26.

Ch9.5~9.9_4

The Gradient of a Function

Define the vector differential operator as

then (1)

(2)

are the gradients of the functions.

,yx

ji

zyx

kji

jiyf

xf

yxF

),(

kjizF

yF

xF

zyxF

),,(

Ch9.5~9.9_5

Example 1

Compute

Solution

235)(),,( yxyyxfyxf

ji )5()5(),( 2323 yxyy

yxyx

yxf

ji2 )25(3),( 32 yxyxyxf

Ch9.5~9.9_6

Example 2

If F(x, y, z) = xy2 + 3x2 – z3, find the gradient at (2, –1, 4).

Solution

kji 22 32)6(),,( zxyxyzyxF

kji 48413)4,1,2( F

Ch9.5~9.9_7

The directional derivative of z = f(x, y) in the direction

of a unit vector u = cos i + sin j is

(4)provided the limit exists.

DEFINITION 9.5Directional Derivatives

hyxfhyhxf

yxfDh

),()sin,cos(lim),(

0

u

Ch9.5~9.9_8

Fig 9.27

Ch9.5~9.9_9

ProofLet x, y and be fixed, then g(t) = f(x + t cos , y + t sin ) is a function of one variable.

If z = f(x, y) is a differentiable function of x and y,

and u = cos i + sin j, then

(5)

THEOREM 9.6Computing a Directional Derivative

uu ．),(),( yxfyxfD

Ch9.5~9.9_10

First

Second by chain rule

hyxfhyhxf

hghg

g

h

h

),()sin,cos(lim

)0()0(lim)0(

0

0

sin)sin,cos(

cos)sin,cos(

)sin()sin,cos(

)cos()sin,cos()(

2

1

2

1

tytxf

tytxf

txdtd

tytxf

txdtd

tytxftg

Ch9.5~9.9_11

Here the subscripts 1 and 2 refer to partial derivatives of f(x + t cos , y + t sin ) w.s.t. (x + t cos ) and (y + t sin ). When t = 0, x + t cos and y + t sin are simply x and y, then (7) becomes

(8)

Comparing (4), (6), (8), we have sin),(cos),()0( yxfyxfg yx

u

j)iji

u

．

．

),(

sin(cos]),(),([

sin),(cos),(),(

yxf

yxfyxf

yxfyxfyxfD

yx

yx

Ch9.5~9.9_12

Example 3

Find the directional derivative of f(x, y) = 2x2y3 + 6xy at (1, 1) in the direction of a unit vector whose angle with the positive x-axis is /6.

Solution

,64 3 yxyxf

xyxyf

66 22

,)66()64(),( 223 ji xyxyxyyxf ji 1210)1,1( f

Ch9.5~9.9_13

Example 3 (2)

Now, = /6, u = cos i + sin j becomes

Then

jiu21

23

63521

23

)1210(

)1,1()1,1(

jiji

uu

．

．ffD

Ch9.5~9.9_14

Example 4

Consider the plane perpendicular to xy-plane and passes through P(2, 1), Q(3, 2). What is the slope of the tangent line to the curve of intersection of this plane and f(x, y) = 4x2 + y2 at (2, 1, 17) in the direction of Q.

SolutionWe want Duf(2, 1) in the direction given by , and form a unit vector

jiu )2/1()2/1(

ji PQ

Ch9.5~9.9_15

Example 4 (2)

Now

then the requested slope is

,28),( ji yxyxf

ji 216)1,2( f

292

12

1)216()1,2(

jijiu ．fD

Ch9.5~9.9_16

Functions of Three Variables

where , , are the direction angles of the vector u measured relative to the positive x, y, z axis. But as before, we can show that

(9)

hzyxFhzhyhxF

zyxfD

h

),,()cos,cos,cos(lim

),,(

0

u

uu ．),,(),,( zyxFzyxFD

Ch9.5~9.9_17

Since u is a unit vector, from (10) in Sec 7.3 that

In addition, (9) shows

),(comp),( yxfyxfD uu

),,(comp),,( zyxFzyxFD uu

zw

zyxFD),,(k

Ch9.5~9.9_18

Example 5

Find the directional derivative of F(x, y, z) = xy2 – 4x2y + z2 at (1, –1, 2) in the direction 6i + 2j + 3k.

SolutionSince

we have

,42,8 22 xxyyF

xyyxF

z

zF

2

kji

kji

469)2,1,1(

2)42()8(),,( 22

F

zxxyxyyzyxF

Ch9.5~9.9_19

Example 5 (2)

Since ‖6i + 2j + 3k‖= 7, then u = (6/7)i + (2/7)j + (3/7)k is a unit vector. It follows from (9) that

754

73

72

76

)469(

)2,1,1(

kjikji

u

．

FD

Ch9.5~9.9_20

Maximum Value of the Direction Derivative

From the fact that

where is the angle between and u. Because

then

)1||(||,cos||||cos|||||||| uuu fffD

f

1cos1

|||||||| ffDf u

Ch9.5~9.9_21

In other words, The maximum value of the direction derivative is and it occurs when u has the same direction as (when cos = 1) , (10)and The minimum value of the direction derivative is and it occurs when u has opposite direction as (when cos = −1) (11)

|||| f

|||| f

Ch9.5~9.9_22

Example 6

In Example 5, the maximum value of the directional derivative at (1, −1, 2) is

and the minimum value is .

)2,1,1(,133||)2,1.1(|| FDf u

133

Ch9.5~9.9_23

Gradient points in Direction of Most Rapid Increase of f

Put another way, (10) and (11) stateThe gradient vector points in the direction in which f increase most rapidly, whereas points in the direction of the most rapid decrease of f.

ff

Ch9.5~9.9_24

Example 7

Each year in L.A. there is a bicycle race up to the top of a hill by a road known to be the steepest in the city. To understand why a bicyclist with a modicum of sanity will zigzag up the road, let us suppose the graph ofshown in Fig 9.28(a) is a mathematical model of the hill. The gradient of f is

,3/24),( 22 yxyxf ,40 z

rji222222

3/232

),(yxyx

y

yx

xyxf

Ch9.5~9.9_25

Example 7 (2)

where r = – xi – yj is a vector pointing to the center of circular base. Thus the steepest ascent up the hill is a straight road whose projection in the xy-plane is a radius of the circular base. Since

a bicyclist will zigzag a direction u other than to reduce this component. See Fig 9.28.

ffD uu compf

Ch9.5~9.9_26

Fig 9.28

Ch9.5~9.9_27

Example 8

The temperature in a rectangular box is approximated by

If a mosquito is located at ( ½, 1, 1), in which the direction should it fly up to cool off as rapidly as possible?

30,20,10

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

zyx

zyxxyzzyxT

Ch9.5~9.9_28

Example 8 (2)

SolutionThe gradient of T is

Therefore, To cool off most rapidly, it should fly in the direction −¼k, that is, it should dive for the floor of the box, where the temperature is

T(x, y, 0) = 0

k

ji

)23)(2)(1(

)22)(3)(1()21)(3)(2(

),,(

zyxxy

yzxxzxzyyz

zyxT

k4/1)1,1,2/1( T

Ch9.5~9.9_29

9.6 Tangent Plane and Normal Lines

Geometric Interpretation of the Gradient : Functions of Two VariablesSuppose f(x, y) = c is the level curve of z = f(x, y) passes through P(x0, y0), that is, f(x0, y0) = c.If x = g(t), y = h(t) such that x0 = g(t0), y0 = h(t0), then the derivative of f w.s.t. t is

(1)

When we introduce

0

dtdy

yf

dtdx

xf

,),( jiyf

xf

yxf

jir

dtdy

dtdx

t )(

Ch9.5~9.9_30

then (1) becomes When at t = t0, we have

(2)

Thus, if , is orthogonal toat P(x0, y0). See Fig 9.30.

0)(),( 000 tyxf r．

0)( 0 tr ),( 00 yxf

Ch9.5~9.9_31

Fig 9.30

Ch9.5~9.9_32

Example 1

Find the level curves of f(x, y) = −x2 + y2 passing through (2, 3). Graph the gradient at the point.

Solution Since f(2, 3) = 5, we have −x2 + y2 = 5.Now

See Fig 9.31.

,22),( ji yxyxf ji 64)3,2( f

Ch9.5~9.9_33

Fig 9.31

Ch9.5~9.9_34

Geometric Interpretation of the Gradient : Functions of Three Variables

Similar concepts to two variables, the derivative of F(f(t), g(t), h(t)) = c implies

(3)

In particular, at t = t0, (3) is

(4)See Fig 9.32.

0

dtdz

zF

dtdy

yF

dtdx

xF

0

kjikjidtdz

dtdy

dtdx

zF

yF

xF ．

0)(),,( 0000 tzyxF r．

Ch9.5~9.9_35

Fig 9.32

Ch9.5~9.9_36

Example 2

Find the level surfaces of F(x, y, z) = x2 + y2 + z2 passing through (1, 1, 1). Graph the gradient at the point.

Solution Since F(1, 1, 1) = 3 ， then x2 + y2 + z2 = 3

See Fig 9.33.

kji zyxzyxF 222),,(

Ch9.5~9.9_37

Fig 9.33

Ch9.5~9.9_38

Let P(x0, y0, z0) is a point on the graph of F(x, y, z) = c,

where F is not 0. The tangent plane at P is a plane

through P and is perpendicular to F evaluated at P.

DEFINITION 9.6Tangent Plane

Ch9.5~9.9_39

That is, . See Fig 9.34.

Let P(x0, y0, z0) is a point on the graph of F(x, y, z) = c,

where F is not 0. Then an equation of this tangent

plane at P isFx(x0, y0, z0)(x – x0) + Fy(x0, y0, z0)(y – y0) + Fz(x0, y0, z0)(z – z0) = 0 (5)

THEOREM 2.1Criterion for an Extra Differential

0)(),,( 0000 rr．zyxF

Ch9.5~9.9_40

Fig 9.34

Ch9.5~9.9_41

Example 3

Find the equation of the tangent plane to x2 – 4y2 + z2 = 16 at (2, 1, 4).

SolutionF(2, 1, 4) = 16, the did graph passes (2, 1, 4). Now Fx(x, y, z) = 2x, Fy(x, y, z) = – 8y, Fz(x, y, z)= 2z, then

From (5) we have the equation: 4(x – 2) – 8(y – 1) + 8(z – 4) = 0 or x – 2y + 2z = 8.

,282),,( kji zyxzyxF kji 884)4,1,2( F

Ch9.5~9.9_42

Surfaces Given by z = f(x, y)

When the equation is given by z = f(x, y), then we can set F = z – f(x, y) or F = f(x, y) – z.

Ch9.5~9.9_43

Example 4

Find the equation of the tangent plane to z = ½x2 + ½ y2 + 4 at (1, –1, 5).

SolutionLet F(x, y, z) = ½x2 + ½ y2 – z + 4. This graph did pass (1, –1, 5), since F(1, –1, 5) = 0. Now Fx = x, Fy = y, Fz = –1, then

From (5), the desired equation is(x + 1) – (y – 1) – (z – 5) = 0

or –x + y + z = 7

,),,( kji yxzyxF

kji )5,1,1(F

Ch9.5~9.9_44

Fig 9.35

Ch9.5~9.9_45

Normal Line

Let P(x0, y0, z0) is on the graph of F(x, y, z) = c, where F 0. The line containing P that is parallel to F(x0, y0, z0) is called the normal line to the surface at P.

Ch9.5~9.9_46

Example 5

Find parametric equations for the normal line to the surface in Example 4 at (1, –1, 5).

SolutionA direction vector for the normal line at (1, –1, 5) is

F(1, –1, 5) = i – j – kthen the desired equations are

x = 1 + t, y = –1– t, z = 5 – t

Ch9.5~9.9_47

9.7 Divergence and Curl

Vector Functions

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

F(x, y, z) = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k

Ch9.5~9.9_48

Fig 9.37 (a) ~ (b)

Ch9.5~9.9_49

Fig 9.37 (c) ~ (d)

Ch9.5~9.9_50

Example 1

Graph F(x, y) = – yi + xj

SolutionSince

let

For and k = 2, we have

(i) x2 + y2 = 1 ： at (1, 0), (0, 1), (–1, 0), (0, –1), the corresponding vectors j, –i,– j , i have the same length 1.

22|||| yx F

kyx 22

2,1 kk

Ch9.5~9.9_51

Example 1 (2)

(ii) x2 + y2 = 2 ： at (1, 1), (–1, 1), (–1, –1), (1, –1), the corresponding vectors – i + j, – i – j, i – j, i + j have the same length .

(iii) x2 + y2 = 4 ： at (2, 0), (0, 2), (–2, 0), (0, –2), the corresponding vectors 2j, –2i, –2j, 2i have the same length 2.

See Fig 9.38.

2

Ch9.5~9.9_52

Example 1 (3)

Ch9.5~9.9_53

In practice, we usually use this form:

(1)

The curl of a vector field F = Pi + Qj + Rk is the vector field

DEFINITION 9.7Curl

kjiF

yP

xQ

xR

zP

zQ

yR

RQPzyx

kji

FFcurl

Ch9.5~9.9_54

Observe that we can also use this form:

(4)

The divergence of a vector field F = Pi + Qj + Rk is the scalar function

DEFINITION 9.8Divergence

zR

yQ

xP

Fdiv

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

div

zyxRz

zyxQy

zyxPx

FF ．

Ch9.5~9.9_55

Example 2

If F = (x2y3 – z4)i + 4x5y2zj – y4z6 k, find curl F and div F 。Solution

kji

kji

FF

)320(4)44(

4

curl

222432563

6425432

yxzyxzyxzy

zyzyxzyxzyx

Ch9.5~9.9_56

Example 2 (2)

5453

6425432

682

)()4()(div

zyyzxxy

zyz

zyxy

zyxx

FF ．

Ch9.5~9.9_57

Please prove

If f is a scalar function with continuous second partial derivatives, then

(5)

If F is a vector field with continuous second partial derivatives, then

(6)

0 ff )grad(curl

0)()curl(div FF ．

Ch9.5~9.9_58

Physical Interpretations

See Fig 9.41. The curl F is a measure of tendency of the fluid to turn the device about its vertical axis w. If curl F = 0, then the flow of the fluid is said to be irrotational. Also see Fig 9.42.

Ch9.5~9.9_59

Fig 9.41

Ch9.5~9.9_60

Fig 9.42

Ch9.5~9.9_61

From definition 9.8 we saw that div F near a point is the flux per unit volume. (i) If div F(P) > 0: source for F.(ii) If div F(P) < 0: sink for F.(iii) If div F(P) = 0: no sources or sinks near P.

Besides, If F = 0: incompressible or solenoidal.

See Fig 9.43.

Ch9.5~9.9_62

Fig 9.43

Ch9.5~9.9_63

9.8 Line Integrals

TerminologyIn Fig 9.46, we show four new terminologies.

Fig 9.46.

Ch9.5~9.9_64

Line Integral in the Plane

If C is a smooth curve defined by x = f(t), y = g(t), a t b. Since dx = f’(t) dt, dy = g’(t) dt, and

(which is called the differential of arc length), then we have

(1)

(2)

(3)

tdtgtfsd 22 )]([)]([

b

adttftgtfG )())(),((C dxyxG ),(

CdyyxG ),(

b

adttgtgtfG )())(),((

b

aCdttgtftgtfGdsyxG 22 )]([)]([))(),((),(

Ch9.5~9.9_65

Example 1

Evaluate (a) (b) (c)on the ¼ circle C defined by Fig 9.47:

Cdxxy2 C

dyxy2 Cdsxy2

2/0,sin4,cos4 ttytx

Ch9.5~9.9_66

Example 1 (2)

Solution(a)

64sin41

256

cossin256

)cos4()sin16()cos4(

2/

0

4

2/

0

3

2/

0

2

22

t

dttt

dx

dtt

y

t

x

tdxxyC

Ch9.5~9.9_67

Example 1 (3)

(b)

2/

0

2

2/

0

22

2/

0

2

22

2sin41

256

cossin256

)cos4()sin16()cos4(

dtt

dttt

dy

dtt

y

t

x

tdyxyC

164sin41

32

)4cos1(21

64

2/

0

2/

0

tt

dtt

Ch9.5~9.9_68

Example 1 (4)

(c)

3256

sin31

256

cossin256

)sin(cos16)sin16()cos4(

2/

0

3

2/

0

2

2/

0

22

2

22

t

dttt

ds

dttt

y

t

x

tdsxyC

Ch9.5~9.9_69

Method of Evaluation

If the curve C is defined by y = f(x), a x b, then we have dy = f ’(x) dx andThus

(4)

(5)

(6)Note: If C is composed of two smooth curves C1 and

C2, then

b

aCdxxfxGdxyxG ))(,(),(

dxxfds 2)]([1

b

aCdxxfxfxGdyyxG )())(,(),(

b

aCdxxfxfxGdsyxG 2)]([1))(,(),(

21

),(),(),(CCC

dsyxGdsyxGdsyxG

Ch9.5~9.9_70

Notation: In many applications, we have

we usually write as

or(7)

A line integral along a close curve:

CC

dyyxQdxyxP ),(),(

C

dyyxQdxyxP ),(),( C

dyQdxP

C

dyQdxP

Ch9.5~9.9_71

Example 2

Evaluate where C:

Solution See Fig 9.48. Using dy = 3x2 dx,

C

dyxdxxy 2 21,3 xxy

5132

54

4

)3()(

2

1

5

2

1

4

2

1

2232

x

dxx

dy

dxxxdx

y

xxdyxdxxyC

Ch9.5~9.9_72

Fig 9.48

Ch9.5~9.9_73

Example 3

Evaluate , where C:

Solution C dxx 20,sin,cos xtytx

0]11[21

cos21

)sin(cos

2

0

2

2

0

t

dtttdxxC

Ch9.5~9.9_74

Example 4

Evaluate , where C is shown in Fig 9.49(a).

Solution Since C is piecewise smooth, we express the integral as

See Fig 9.49(b).

C

dyxdxy 22

321 CCCC

Ch9.5~9.9_75

Fig 9.49

Ch9.5~9.9_76

Example 4 (2)

(i) On C1, we use x as a parameter. Since y = 0, dy = 0,

(ii) On C2, we use y as a parameter. Since x = 2, dx = 0,

0)0(02

0

222

1 xdxdyxdxy

C

1644

4)0(

4

0

4

0

4

0

222

2

ydy

dyydyxdxyC

Ch9.5~9.9_77

(iii) On C3, we use x as a parameter. Since y = x2, dy = 2x dx,

Hence,

Example 4 (3)

5

8

2

1

5

1

)2(

)2(

0

2

45

2

0

34

0

2

2422

3

xx

dxxx

dxxxdxxdyxdxyC

5

72

5

816022 C dyxdxy

Ch9.5~9.9_78

Note: See Fig 9.50, where −C denote the curve having opposite orientation, then

Equivalently,

(8)

For example, in (a) of Example 1,

CC

dyQdxPdyQdxP

0 CCdyQdxPdyQdxP

Cdxxy 642

Ch9.5~9.9_79

Fig 9.50

Ch9.5~9.9_80

Lines Integrals in Space

n

kkkkk

PCzzyxGdzzyxG

1

***

0||||),,(lim),,(

Ch9.5~9.9_81

Method to Evaluate Line Integral in Space

If C is defined by

then we have

Similar method can be used for

btathztgytfx ),(),(),(

b

aCdtththtgtfGdzzyxG )())(),(),((),,(

CCdyzyxGdxzyxG ),,(,),,(

Ch9.5~9.9_82

and

We usually use the following form

b

a

C

dtthtgtfthtgtfG

dszyxG

222 )]([)]([)]([))(),(),((

),,(

C

dzzyxRdyzyxQdxzyxP ),,(),,(),,(

Ch9.5~9.9_83

Example 5

Evaluate , where C is

Solution Since we have

we get

C

dzzdyxxdy

20,,sin2,cos2 ttztytx

20,,cos2,sin2 ttddzdttdydttdx

22

2sin2

)2cos4(

)cos4sin4(

2

0

2

2

0

2

0formula angle-double

22

tt

dttt

dttdtttdzzdyxdxyC

Ch9.5~9.9_84

Another Notation

Let r(t) = f(t)i + g(t)j, then dr(t)/dt = f’(t)i + g’(t)j = (dx/dt)i + (dy/dt)j

Now if F(x, y) = P(x, y)i + Q(x, y)jthus

(10)When on a space

(11)

where F(x, y, z) = Pi + Qj + Rk dr = dx i + dy j + dz k,

CC

ddyyxQdxyxP rF．),(),(

C

C

d

dzzyxRdyzyxQdxzyxP

rF．

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

Ch9.5~9.9_85

Work

If A and B are the points (f(a), g(a)) and (f(b), g(b)). Suppose C is divided into n subarcs of lengths ∆sk. On each subarc F(x*

k, y*k) is a constant force. See Fig

5.91(a).If, as shown in Figure 9.51(b), the length of the

vector is an approximation to the length of the kth subarc, then the approximate work done by F over the subarc is

jijir kkkkkkk yxyyxx )()( 11

kkkkkk

kkkkkk

yyxQxyxP

yxyx

),(),(

),(||||cos||),((||****

**** rFrF ．

Ch9.5~9.9_86

The work done by F along C is as the line integral

or (12)

Since

we let dr = Tds, where T = dr / ds is a unit tangent to C.(13)

The work done by a force F along a curve C is due entirely to the tangential component of F.

C

dyyxQdxyxPW ),(),( C

dW rF．

dtds

dsd

dtd rr

sFTFrF T CCC

ddsdW comp．．

Ch9.5~9.9_87

Fig 9.51

Ch9.5~9.9_88

Example 6

Find the work done by (a) F(x, y) = xi + yj(b) F = (¾ i + ½ j) along the curve C traced by r(t) = cos ti + sin tj, from t = 0 to t = .

Solution (a) dr(t) = (− sin ti + cos tj) dt, then

0)cossinsincos(

)cossin(sin(cos

0

0

dttttt

dttttt

dyxdWCC

jij)i

rjirF

．

．．

Ch9.5~9.9_89

Fig 9.52

Ch9.5~9.9_90

Example 6 (2)

(b) See Fig 9.53.

23

sin21

cos43

cos21

sin43

0

0

tt

dttt

CCddW rjirF ．．

2

1

4

3

0)cossin(

21

43

dttt jiji ．

Ch9.5~9.9_91

Fig 9.53

Ch9.5~9.9_92

Note:

Let dr = T ds, where T = dr / ds, then

Circulation of F around C

dt

ds

ds

d

dt

d rr

sFTFrF T CCC

ddsdW comp．．

Circulation

CC

dsd TFrF

Ch9.5~9.9_93

Fig 9.54

Ch9.5~9.9_94

9.9 Independence of Path

Differential For two variables:

For three variables:

dyy

dxx

d

dyyxQdxyxPd ),(),(

dzz

dyy

dxx

d

dzzyxRdyzyxQdxzyxPd ),,(),,(),,(

Ch9.5~9.9_95

Path Independence

A line integral whose value is the same for every curve or path connecting A and B.

Ch9.5~9.9_96

Example 1

has the same value on each path between (0, 0) and (1, 1) shown in Fig 9.65. Recall that

dyxdxyC

1Cdyxdxy

Ch9.5~9.9_97

Fig 9.65

Ch9.5~9.9_98

Suppose there exists a (x, y) such that d = Pdx + Qdy,

that is, Pdx + Qdy is an exact differential. Then

depends on only the endpoints A and

B, and

THEOREM 9.8

Fundamental Theorem for Line Integrals

C

dyQdxP

)()( ABdyQdxPC

Ch9.5~9.9_99

THEOREM 9.8 Proof

Let C be a smooth curve: The endpoints are (f(a), g(a)) and (f(b), g(b)), then

)()(

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

))(),((

AB

agafbgbf

tgtfdtdtd

dtdtdy

ydtdx

xdyQdxP

b

a

b

a

b

aC

btatgytfx ),(),(

Ch9.5~9.9_100

Two facts

(i) This is also valid for piecewise smooth curves.(ii) The converse of this theorem is also true.

is independent of path iff P dx + Q dy is an exact differential. (1)

Notation for a line integral independent of path:

C

dyQdxP

B

AdyQdxP

Ch9.5~9.9_101

Example 2

Since d(xy) = y dx + x dy, y dx + x dy is an exact differential. Hence is independent of path. Especially, if the endpoints are (0, 0) and (1, 1), we have

C

dyxdxy

1)( )1,1()0,0(

)1,1(

)0,0(

)1,1(

)0,0( xyxyddyxdxy

Ch9.5~9.9_102

Simply Connected Region in the Plane

Refer to Fig 9.66. Besides, a simply connected region is open if it contains no boundary points.

Ch9.5~9.9_103

Fig 9.66

Ch9.5~9.9_104

Let P and Q have continuous first partial derivatives in

an open simply connected region. Then is independent of the path C if and only if

for all (x, y) in the region.

THEOREM 9.9Test for Path Independence

C

dyQdxP

x

Q

y

P

Ch9.5~9.9_105

Example 3

Show that is not independent of path C.

SolutionWe have P = x2 – 2y3 and Q = x + 5y, then

　and

Since , we complete the proof.

C

dyyxdxyx )5()2( 32

26yyP

1

xQ

xQyP //

Ch9.5~9.9_106

Example 4

Show that is independent of any path between (−1, 0) and (3, 4). Evaluate.

SolutionWe have P = y2 – 6xy + 6 and Q = 2xy – 3x2, then

and

This is an exact differential.

C

dyxxydxxyy )32()66( 22

xyyP

62

xyxQ

62

Ch9.5~9.9_107

Example 4 (2)

Suppose there exists a such that

Integrating the first, we have

then

we have , g(y) = C.

,66/ 2 xyyx 232/ xxyy

)(63 22 ygxyxxy

22 32)(32 xyxygxyxy

0)( yg

Ch9.5~9.9_108

Example 4 (3)

Since d(y2x – 3x2y + 6x + C) = d(y2x – 3x2y + 6x), we simply usethen

xyxxy 63 22

36)6()1810848(

)63()63(

)32()66(

)4,3(

)0,1(22)4,3(

)0,1(

22

)4,3(

)0,1(

22

xyxxyxyxxyd

dyxxydxxyy

Ch9.5~9.9_109

Another Approach

We know y = x + 1 is one of the paths connecting (−1, 0) and (3, 4). Then

36)726(

]3)1(2[]6)1(6)1[(

)32()66(

3

1

2

3

1

22

22

dxxx

dxxxxdxxxx

dyxxydxxyyC

Ch9.5~9.9_110

Conservative Vector Fields

If a vector field is independent of path, we have

where F = Pi + Qj is a vector field and

In other words, F is the gradient of . Since F = , F is said to be a gradient field and the is called the potential function of F. Besides, we all call this kind of vector field to be conservative.

rFjiji ddydxQP

dyQdxPdyy

dxx

d

．．

)()(

yQxP /,/

Ch9.5~9.9_111

Example 5

Show that F = (y2 + 5)i + (2xy – 8)j is a gradient field. Find a potential function for F.

SolutionSince

then

we have

yxQ

yP

2

,52

yx

82

xyy

xyxy 582

Ch9.5~9.9_112

Let P, Q, and R have continuous first partial derivatives

in an open simply connected region of space. Then

is independent of the path C if and only if

THEOREM 9.10Test for Path Independence

C

dyQdxP

yR

zQ

xR

zP

xQ

yP

,,

Ch9.5~9.9_113

Example 6

Show that is independent of path between (1, 1, 1) and (2, 1, 4).

SolutionSince

it is independent of path.

C

dzxyyzdyxzzxxdyzy )19()3()( 23

yR

xzzQ

xR

yzP

xQ

zyP

29,,1

Ch9.5~9.9_114

Example 6 (2)

Suppose there exists a function , such that

Integrating the first w.s.t. x, then

It is the fact

thus

Rz

Qy

Px

,,

),( zygxyzxy

Qxzzxyg

xzxy

33

,3 3zyg

)(3 3 zhyzg

Ch9.5~9.9_115

Example 6 (3)

Now

and

we have and h(z) = – z + C.

Disregarding C, we get

(2)

)(3 3 zhyzxyzxy

19)(9 22

xyyzzhyzxyz

1)( zh

zyzxyzxy 33

Ch9.5~9.9_116

Example 6 (4)

Finally,

1944198

)3(

)3(

)19()3()(

)4,1,2(

)1,1,1(

3

)4,1,2(

)1,1,1(

3

)4,1,2(

)1,1,1(

23

zyzxyzxy

zyzxyzxyd

dzxyyzdyxzzxdxyzy

Ch9.5~9.9_117

From example 6, we know that F is a conservative vector field, and can be written as F = . Remember in Sec 9.7, we have = 0, thus

F is a conservative vector field iff F = 0