Upload
dominick-stanley
View
241
Download
4
Embed Size (px)
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