15 Copyright © Cengage Learning. All rights reserved. Vector Analysis.

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Presentation transcript:

15 Copyright © Cengage Learning. All rights reserved. Vector Analysis

Conservative Vector Fields and Independence of Path Copyright © Cengage Learning. All rights reserved. 15.3

3 Understand and use the Fundamental Theorem of Line Integrals. Understand the concept of independence of path. Understand the concept of conservation of energy. Objectives

4 Fundamental Theorem of Line Integrals

5 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.

6 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 a. C 1 : y = xb. C 2 : x = y 2 c. C 3 : y = x 3 Example 1 – Line Integral of a Conservative Vector Field Figure 15.19

7 Example 1(a) – Solution C 1 : y = x => Let r(t) = ti + tj for 0 ≤ t ≤ 1, so that Then, the work done is

8 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

9 C 2 : x = y 2 Let r(t) = ti + for 0 ≤ t ≤ 1, so that Then, the work done is Example 1(b) – Solution cont’d

10

11 Let r(t) for 0 ≤ t ≤ 2, so that Then, the work done is So, the work done by a conservative vector field is the same for all paths. Example 1(c) – Solution cont’d C 3 : y = x 3

12 t = linspace(0,2,100); x1 = t/2; y1 = t.^3/8;

13 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 Fundamental Theorem of Line Integrals

14 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.

15 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 Fundamental Theorem of Line Integrals

16 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 + (x 2 − y)j as shown in Figure Figure Do my example instead

17 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 Example 2 – Solution

18 My example 2D

19

20 My example 3D

21

22

23 Independence of Path

24 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. Independence of Path

25 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 In open regions that are connected, the path independence of is equivalent to the condition that F is conservative. Figure Independence of Path

26 Independence of Path

27 1 st integral depends on x1 – fixed, but not on x.

28 Example 4 – Finding Work in a Conservative Force Field For the force field given by F(x, y, z) = e x cos yi − e x 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 = e x cos y, N = –e x sin y, and P = 2,

29 and it follows that So, F is conservative. If f is a potential function of F, then Example 4 – Solution cont’d M = e x cos y, N = –e x sin y, and P = 2,

30 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) = e x cos y + 2z + K. Example 4 – Solution cont’d

31 Example 4 – Solution Therefore, the work done by F along any curve C from (0,  /2, 1) to (1, , 3) is cont’d

32

33 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. Independence of Path

34

35 Example 5 – Evaluating a Line Integral Evaluate, where F(x, y) = (y 3 + 1)i + (3xy 2 + 1)j and C 1 is the semicircular path from (0, 0) to (2, 0), as shown in Figure Figure 15.24

36 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.

37

38 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) = xy 3 + x + y + K, and, by the Fundamental Theorem, cont’d Example 5 – Solution

39

40 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 C 2 from (0, 0) to (2, 0). Then, r(t) = ti, where 0 ≤ t ≤ 2. Example 5 – Solution cont’d

41 So, dr = i dt and F(x, y) = (y 3 + 1)i + (3xy 2 + 1)j = i + j, so that Of the three options, obviously the third one is the easiest. cont’d Example 5 – Solution

42

43 Conservation of Energy

44 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.

45 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. Conservation of Energy

46 Consequently, the work done by F along a smooth curve C from A to B is as shown in Figure Figure Conservation of Energy

47 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. Conservation of Energy

48 So, by Newton’s Second Law of Motion, F = ma(t) = m(v′(t)), and the work done by F is Conservation of Energy

49 Conservation of Energy

50 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. Conservation of Energy