Transcript t - leonkag

15
Vector Analysis
Copyright © Cengage Learning. All rights reserved.
15.3
Conservative Vector Fields
and Independence of Path
Copyright © Cengage Learning. All rights reserved.
Objectives
 Understand and use the Fundamental
Theorem of Line Integrals.
 Understand the concept of independence of
path.
 Understand the concept of conservation of
energy.
3
Fundamental Theorem of Line
Integrals
4
Fundamental Theorem of Line Integrals
In a gravitational field, the work done by gravity on an object
moving between two points in the field is independent of the
path taken by the object.
In this section, you will study an important generalization of
this result—it is called the Fundamental Theorem of Line
Integrals.
To begin, an example is presented in which the line integral
of a conservative vector field is evaluated over three
different paths.
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Example 1 – Line Integral of a Conservative Vector Field
Find the work done by the force field
on a particle that moves from (0, 0) to (1, 1) along each path,
as shown in Figure 15.19.
a. C1: y = x
b. C2: x = y2
c. C3: y = x3
Figure 15.19
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Example 1(a) – Solution
C1: y = x => Let r(t) = ti + tj for 0 ≤ t ≤ 1,
so that
Then, the work done is

F
 
r r (t )
xy x 2

2 4
x  x ( t ) t
y  y ( t ) t
t2 t2

2 4
 
 t2 t2 
t2 t2
3t 2
F  dr 
 1,1  dt    dt 
dt
2 4
2
4
4


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xm = 0; xM = 1; dx = xM/8;
[X,Y] = meshgrid(xm:dx:xM);
FX = X.*Y/2; FY = X.^2/4 ;
t = linspace(0,1,100);
x1 = t;
y1 = t;
plot(x1,y1);
hold on
quiver(X,Y,FX,FY)
colormap hsv
hold off
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Example 1(b) – Solution

F
y2
C2: x =
Let r(t) = ti +
so that
for 0 ≤ t ≤ 1,
 
r r (t )
xy x 2

2 4
 
t 3/ 2 t 2
F  dr 
2 4
cont’d
x  x ( t ) t
y  y (t ) t
t 3/ 2 t 2

2 4
 t 3/ 2 t 3/ 2 
5t 3 / 2
dt 
1
dt  

dt
2
8
8
2 t


1
Then, the work done is
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10
Example 1(c) – Solution
C3: y = x3
Let r(t)
so that
for 0 ≤ t ≤ 2,

F

r r (t )
cont’d

2
xy x
2 4
 
t4 t2
F  dr 
32 16
x  x ( t ) t / 2
y  y ( t ) t 3 / 8

4
2
t t
32 16
 t 4 3t 4 
1 3t 2
5t 4


dt   
dt  128 dt
2 8
 64 128 
Then, the work done is
So, the work done by a conservative vector field is the
same for all paths.
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t = linspace(0,2,100);
x1 = t/2;
y1 = t.^3/8;
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Fundamental Theorem of Line Integrals
In Example 1, note that the vector field
is conservative because
where
.
,
In such cases, the following theorem states
that the value of
is given by

xy x 2
x2 y
F
 f ( x, y ) 
2 4
4
2 xy xy
fx 
 ,
4
2
x2
fy 
4
121
1
f ( x(1), y (1))  f ( x(0), y (0))   0 
4
4
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The Fundamental Theorem of Line Integrals states that if the vector field F is conservative, then
the line integral between any two points is simply the difference in the values of the potential
function f at these points – independent of the path taken.
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Fundamental Theorem of Line Integrals
In space, the Fundamental Theorem of Line Integrals takes
the following form.
Let C be a piecewise smooth curve lying in an open region
Q and given by
r(t) = x(t)i + y(t)j + z(t)k,
a ≤ t ≤ b.
If F(x, y, z) = Mi + Nj + Pk is conservative and M, N, and P
are continuous, then
where
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Do my example instead
Example 2 – Using the Fundamental Theorem of Line Integrals
Evaluate
where C is a piecewise smooth curve
from (−1, 4) to (1, 2) and
F(x, y) = 2xyi + (x2 − y)j
as shown in Figure 15.20.
Figure 15.20
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Example 2 – Solution
You know that F is the gradient of f
where
Consequently, F is conservative,
and by the Fundamental
Theorem of Line Integrals,
it follows that
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My example 2D
 
Find  Fdr , where C is piecewise smooth curve
C
from (1,-2) to (3,4)
F  3  2 xy,x2  3 y 2  M , N 
M  3  2 xy  M y  2 x 

  conservati ve
2
2
 N  x  3 y  N x  2 x 
F  M , N  f  f x , f y 
 f x  M  3  2 xy  f  f x dx 3 x  xy2  K ( y ) 




 
2
2
2
3
 f y  N  x  3 y  f   f y dy xy  y  L( x)
f  3 x  xy2  y 3
 
( 3, 4 )
F
d
r

f
 (3  3  9  4  64)  (3  2  8)  28

(1, 2 )
C
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19
My example 3D
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Independence of Path
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Independence of Path
From the Fundamental Theorem of Line Integrals, it is clear
that if F is continuous and conservative in an open region
R, the value of
is the same for every piecewise
smooth curve C from one fixed point in R to another fixed
point in R and is equal to f(end) - f(start).
This result is described by saying that the line integral
is independent of path in the region R.
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Independence of Path
A region in the plane (or in space) is connected if any two
points in the region can be joined by a piecewise smooth
curve lying entirely within the region, as shown in
Figure 15.22.
In open regions that are connected,
the path independence of
is equivalent to the condition
that F is conservative.
Figure 15.22
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Independence of Path
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1st integral depends on x1 – fixed,
but not on x.
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Example 4 – Finding Work in a Conservative Force Field
For the force field given by
F(x, y, z) = ex cos yi − ex sin yj + 2k
show that
is independent of path, and calculate the
work done by F on an object moving along a curve C
from (0, /2, 1) to (1, , 3).
Solution:
Writing the force field in the form F(x, y, z) = Mi + Nj + Pk,
you have
M = ex cos y, N = –ex sin y, and P = 2,
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Example 4 – Solution
cont’d
M = ex cos y, N = –ex sin y, and P = 2,
and it follows that
So, F is conservative.
If f is a potential function of F, then
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Example 4 – Solution
cont’d
By integrating with respect to x, y, and z separately, you
obtain
By comparing these three versions of f(x, y, z), you can
conclude that
f(x, y, z) = ex cos y + 2z + K.
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Example 4 – Solution
cont’d
Therefore, the work done by F
along any curve C from
(0,  /2, 1) to (1, , 3) is
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Independence of Path
A curve C given by r(t) for a ≤ t ≤ b is closed if r(a) = r(b).
By the Fundamental Theorem of Line Integrals, you can
conclude that if F is continuous and conservative on
an open region R, then the line integral over every closed
curve C is 0.
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Example 5 – Evaluating a Line Integral
Evaluate
, where
F(x, y) = (y3 + 1)i + (3xy2 + 1)j
and C1 is the semicircular path from (0, 0) to (2, 0), as
shown in Figure 15.24.
Figure 15.24
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Example 5 – Solution
You have the following three options.
a. You can use the method presented in the preceding
section to evaluate the line integral along the given curve.
To do this, you can use the parametrization
r(t) = (1 − cos t)i + sin tj, where 0 ≤ t ≤ .
For this parametrization, it follows that
dr = r′(t) dt = (sin ti + cos tj) dt,
and
This integral should dampen your enthusiasm for this
option.
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37
Example 5 – Solution
cont’d
b. You can try to find a potential
function and evaluate the line
integral by the Fundamental
Theorem of Line Integrals.
Using the technique demonstrated
in Example 4, you can find the
potential function to be
f(x, y) = xy3 + x + y + K,
and, by the Fundamental Theorem,
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39
Example 5 – Solution
cont’d
c. Knowing that F is conservative, you have a third option.
Because the value of the line integral is independent of
path, you can replace the semicircular path with a simpler
path.
Suppose you choose the straight-line path C2 from
(0, 0) to (2, 0).
Then, r(t) = ti, where 0 ≤ t ≤ 2.
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Example 5 – Solution
cont’d
So, dr = i dt and F(x, y) = (y3 + 1)i + (3xy2 + 1)j = i + j,
so that
Of the three options, obviously the
third one is the easiest.
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42
Conservation of Energy
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Conservation of Energy
The English physicist Michael Faraday wrote, “Nowhere is
there a pure creation or production of power without a
corresponding exhaustion of something to supply it.”
This statement represents one of the first formulation of
one of the most important laws of physics—the Law of
Conservation of Energy.
In modern terminology, the law is stated as follows: In a
conservative force field, the sum of the potential and kinetic
energies of an object remains constant from point to point.
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Conservation of Energy
From physics, the kinetic energy of a particle of mass m
and speed v is
.
The potential energy p of a particle at point (x, y, z) in a
conservative vector field F is defined as
p(x, y, z) = −f(x, y, z),
where f is the potential function for F.
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Conservation of Energy
Consequently, the work done by F along a smooth curve C
from A to B is
as shown in Figure 15.25.
Figure 15.25
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Conservation of Energy
Now, suppose that r(t) is the position vector for a particle
moving along C from A = r(a) to B = r(b).
At any time t, the particle’s velocity, acceleration, and
speed are v(t) = r′(t), a(t) = r′′(t), and v(t) = ||v(t)||,
respectively.
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Conservation of Energy
So, by Newton’s Second Law of Motion,
F = ma(t) = m(v′(t)),
and the work done by F is
48
Conservation of Energy
49
Conservation of Energy
Equating these two results for W produces
p(A) − p(B) = k(B) − k(A)
p(A) + k(A) = p(B) + k(B)
which implies that the sum of the potential and kinetic
energies remains constant from point to point.
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