Let F be a field defined on an open region D in space, and suppose that the (work) integral A

Transcription

1 Path Independence and Conservative Fields Definition. Path Independence Let F be a field defined on an open region D in space, and suppose B that the work) integral A F dr is the same for all paths from A to B in D). Then the integral F dr is path independent in D and the field F is conservative on D. It turns out that a field F is conservative if and only if F = f, that is, if and only if F is a gradient vector field for some differentiable function f. Definition. Potential Functions If F is a field defined on D and F = f for some scalar function f on D, then f is called a potential function for F.

2 Some important assumptions: 1. All curves are piecewise smooth. 2. If F = M i+n j+p k then M, N, P have continuous first partials. 3. D is an open, connected region in space. Theorem 1. The Fundamental Theorem of Line Integrals 1. Let F = M i+n j+p k be a vector field with continuous components throughout an open connected region D in space. Then there exists a differentiable function f such that F = f = f f i+ x y j+ f z k B if and only if for all points A and B in D the integral A independent in D. F dr is path 2. In this case 1) ˆ B A F dr = ˆ B A f dr = fb) fa)

3 Example 1. Let F be the force field F = ysinzi+xsinzj+xycosz +sinz)k and let A = 1,1,π/6) andb = 2,3,π/2). Find the work done along the straight line connecting A to B. Notice that fx,y,z) = xysinz cosz is a potential function for F since f = ysinzi+xsinzj+xycosz +sinz)k In other words, the field F is conservative. So by equation 1) ˆ B A F dr = ˆ B A f dr = fb) fa) = f 2,3,π/2) f 1,1,π/6) = 6 1/2 ) 3/2 = 11/2+ 3/2 Remark. We will see how to find f below.

4 Notice that if F is conservative then the line integral around any closed curve is ˆ A A F dr = ˆ A A f dr = fa) fa) = Theorem 2. The following statements are equivalent: 1. F dr = around every closed curve in D. 2. The field F is conservative on D.

5 Definition. Del Notation and Curl We define a new object...the del operator: = i x + j y + k z This is just a convenient notation to help remember some formulas below.) We also define the curl of the vector field F by curlf = F i j k = x y z M N P P = y N ) i z M + z P ) j+ x N x M ) k y Remark. We will discuss curl in more detail in the next section.

6 Example 2. Find the curl of the vector field F = ysinzi+xsinzj+xycosz +sinz) k curlf = F i j k = x y z ysinz xsinz xycosz +sinz) xycosz +sinz) = xsinz) ) i y z ) ysinz) xycosz +sinz) + j z x xsinz) + ysinz) ) k x y = xcosz xcosz)i+ycosz ycosz)j +sinz sinz)k =

7 The result from the last example was no coincidence. It turns out that a field is conservative if the curl is zero. More precisely, suppose that F = M x,y,z) i+n x,y,z) j+p x,y,z) k is a field whose component functions have continuous first partials and D is a simply connected region in space. Then F is conservative on D if and only if curlf = F =.

8 Example 3. Show that F = y 2 z 3 i+2xyz 3 j+3xy 2 z 2 k is conservative. i j k F = x y z y 2 z 3 2xyz 3 3xy 2 z 2 = i 3xy 2 z 2) y 2xyz 3) ) j z 3xy 2 z 2) x y 2 z 3) ) z + k 2xyz 3 ) x y 2 z 3) ) y = i 6xyz 2 6xyz 2) j 3y 2 z 2 3y 2 z 2) + k 2yz 3 2yz 3) =

9 Finding Potential Functions Suppose that F = M i+n j+p k is conservative. How do we find a function f such that f = F? To find the function f, observe that it must satisfy the following partial differential equations PDEs). 2) f x = M, f y = N, f z = P We illustrate below. Example 4. Find the potential function f from Example 1. Recall that F = ysinzi+xsinzj+xycosz +sinz)k. So by 2) we must solve the following partial differential equations, simultaneously. 3) f x = ysinz 4) f y = xsinz 5) f z = xycosz +sinz

10 Antidifferentiating 3) with respect to x yields fx,y,z) = xysinz +C 1 where C 1 does not depend on x. Thus fx,y,z) = xysinz +gy,z) for some differentiable function g. We now have a candidate function to work. Specifically, f must satisfy the remaining partial differential equations, 4) and 5). Now 4) implies xsinz = xysinz +gy,z)) y = xsinz + g y It follows that g does not depend on y. In other words, fx,y,z) = xysinz +hz) for some differentiable function h. Finally, 5) implies xycosz +sinz = xysinz +hz)) z = xycosz +h z) It follows that hz) = cosz +C and hence fx,y,z) = xysinz cosz +C, Here C is an arbitrary constant. Note: As usual, we chose to let C = in Example 1).

11 Exact Differential Forms Definition. The expression 6) M dx+n dy +P dz is called a differential form. It is called exact on a region D if there is a real-valued function f defined on D such that 7) f f f dx+ dy + x y z dz }{{} df = M dx+n dy +P dz Now if the differential form 6) is exact on a region D in space and f is a scalar function defined on D satisfying 7), and A, B D then ˆ B A M dx+n dy +P dz = ˆ B A df = fb) fa) as a direct consequence of the Fundamental Theorem of Line Integrals Theorem 1). Notice that equation 7) is equivalent to the statement that the field F = M i+n j+p k is conservative. In other words, the differential form 6) is exact if and only if there a real-valued function f defined on D such that f f f 8) i+ j+ x y z k = M i+n j+p k }{{} f

12 Example 5. Show that y 2 z 3 dx+2xyz 3 dy +3xy 2 z 2 dz is exact and compute the integral ˆ 1,1/2, 3),,) Let fx,y,z) = xy 2 z 3. Then y 2 z 3 dx+2xyz 3 dy +3xy 2 z 2 dz df = y 2 z 3 dx+2xyz 3 dy +3xy 2 z 2 dz It follows that the given form is exact. Thus ˆ 1,1/2, 3),,) y 2 z 3 dx+2xyz 3 dy +3xy 2 z 2 dz = ˆ 1,1/2, 3),,) df = fx,y,z) 1,1/2, 3),,) = f 1,1/2, 3) f,,) = 1)1/2) 2 ) 3) 3 ) = 27/4

13 What do conservative vector field look like? We know that gradient vector fields are conservative. Consider the following example. Example 6. Let fx,y) = x 2 +xy. The gradient field, f = 2x+y)i+xj, is shown in the sketch below. a) Now let C be any smooth simple, closed) curve in R 2. Is it believable that the integral C F dr =?

14 b) As a specific example, let C be a circle of radius 2 centered at the origin. Verify that Theorem 2 holds. C: rt) = 2 costi+2 sintj, t 2π M = 2x+y = 4cost+2sint N = x = 2cost dx = 2sintdt dy = 2costdt A direct calculation of the circulation integral yields ˆ 2π M dx+n dy = 4cost+2sint) 2sint)+2cost) 2cost)) dt C = 4 = 4 ˆ 2π ˆ 2π cos 2 t 2sint cost sin 2 t ) dt cos2t sin2t) dt = 2 sin2t+cos2t) as expected. 2π =

15 c) Verify that Theorem 2 holds for the ellipse defined by C: rt) = 1+cost)i+ 1+ sint ) j, t 2π 2 M = 2x+y = 3+2cost+ sint 2 N = x = 1+cost dx = sintdt dy = cost 2 dt Once again a direct calculation of the circulation integral yields C M dx+n dy = = = ˆ 2π ˆ 2π ˆ 2π 3+2cost+ sint 2 ) sint)+1+cost) 3sint 2sintcost+ cos2 t sin 2 t 2 3sint sin2t+ cos2t + cost 2 2 ) dt + cost ) 2 )) cost 2 dt dt =. = as expected.

16 It s not clear whether the vector field in the next example is conservative or not. Example 7. Evaluate the circulation integral of the region So let F = 3yi+2xj. C 3ydx+2xdy) where C is boundary x π, y sinx So let C 1 : r 2 t) = ti, t π C 2 : r 1 t) = π1 t)i+sinπ1 t)) j, t 1 For C 1 we have M = 3y =, dx = dt N = 2x = 2t, dy =

17 So that ˆ C 1 3ydx+2xdy = A quick inspection of the vector field suggest that the flow integral along C 1 should be, so this result isn t a surprise.

18 Now for C 2 we have M = 3y = 3sinπ1 t), dx = πdt N = 2x = 2π1 t), dy = πcosπ1 t)dt Soˆ C 2 3ydx+2xdy = ˆ 1 3sinπ1 t)) π)+2π1 t)) πcosπ1 t))) dt and after a change of variables we obtain = ˆ π 3sinu+2ucosu) dt = 3cosu+2usinu+2cosu) = 2usinu cosu) = 2 A quick inspection of the vector field suggests that the flow integral along C 2 should be negative. It now follows that the vector ˆ field is not conservative ˆ since 3ydx+2xdy) = 3ydx+2xdy + 3ydx+2xdy C C 1 C 2 = 2 π π Remark. We will revisit many of these examples in the next section.