Independence of Path and Conservative Vector Fields

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Independence of Path and onservative Vector Fields MATH 311, alculus III J. Robert Buchanan Department of Mathematics Summer 2011

Goal We would like to know conditions on a vector field function F(x, y) = M(x, y), N(x, y), so that if A and B are any two points fixed in the plane, then F(x, y) dr is the same regardless of the path taken from A to B. B A

Goal We would like to know conditions on a vector field function F(x, y) = M(x, y), N(x, y), so that if A and B are any two points fixed in the plane, then F(x, y) dr is the same regardless of the path taken from A to B. B A In this case the line integral is independent of path.

onnected Regions Definition A region D R 2 is called connected if every pair of points in D can be connected by a piecewise-smooth curve lying entirely in D. onnected Not connected

Simply onnected Regions Definition A region D R 2 is simply connected if every closed curve in D encloses only points in D Simply connected Not simply connected

Independence of Path (1 of 2) Theorem Suppose that the vector field F(x, y) = M(x, y), N(x, y) is continuous on the open, connected region D R 2. Then the line integral F(x, y) dr is independent of path if and only if F is conservative on D.

Independence of Path (2 of 2) Suppose the vector field is conservative. F(x, y) = f (x, y) = f x, f y = M(x, y), N(x, y) Select any two points A = (x 0, y 0 ) and B = (x 1, y 1 ) in D and let = (x(t), y(t)) for a t b be any path connecting A and B.

Independence of Path (2 of 2) Suppose the vector field is conservative. F(x, y) = f (x, y) = f x, f y = M(x, y), N(x, y) Select any two points A = (x 0, y 0 ) and B = (x 1, y 1 ) in D and let = (x(t), y(t)) for a t b be any path connecting A and B. F(x, y) dr = = b a b M(x, y) dx + N(x, y) dy [ f x (x(t), y(t))x (t) + f ] y (x(t), y(t))y (t) dt d = [f (x(t), y(t))] dt a dt = f (x 1, y 1 ) f (x 0, y 0 )

Fundamental Theorem of alculus for Line Integrals Theorem Suppose that F(x, y) = M(x, y), N(x, y) is continuous in the open, connected region D R 2 and that is any piecewise-smooth curve lying in D, with an initial point (x 0, y 0 ) and terminal point (x 1, y 1 ). Then, if F is conservative on D, with F = f (x, y), we have F(x, y) dr = f (x, y) (x 1,y 1 ) (x 0,y 0 ) = f (x 1, y 1 ) f (x 0, y 0 ).

Example (1 of 4) Suppose F(x, y) = (2xe 2y + 4y 3 )i + (2x 2 e 2y + 12xy 2 )j. 1 Show that the line integral F(x, y) dr is independent of path. 2 Evaluate the line integral along any piecewise-smooth curve connecting (0, 1) to (2, 3).

Example (2 of 4) F(x, y) = (2xe 2y + 4y 3 )i + (2x 2 e 2y + 12xy 2 )j 1 If f (x, y) = x 2 e 2y + 4xy 3, then f (x, y) = F(x, y) which implies F(x, y) is conservative and hence independent of path.

Example (2 of 4) F(x, y) = (2xe 2y + 4y 3 )i + (2x 2 e 2y + 12xy 2 )j 1 If f (x, y) = x 2 e 2y + 4xy 3, then f (x, y) = F(x, y) which implies F(x, y) is conservative and hence independent of path. 2 By the Fundamental Theorem of alculus for Line Integrals F(x, y) dr = f (x, y) (2,3) (0,1) = 216 + 4e 6.

Example (3 of 4) Suppose F(x, y) = x 2 y 3 i + x 3 y 2 j and then find the work done moving from (0, 0) to (2, 1).

Example (4 of 4) F(x, y) = x 2 y 3 i + x 3 y 2 j 1 If f (x, y) = 1 3 x 3 y 3, then f (x, y) = F(x, y) which implies F(x, y) is conservative and hence independent of path.

Example (4 of 4) F(x, y) = x 2 y 3 i + x 3 y 2 j 1 If f (x, y) = 1 3 x 3 y 3, then f (x, y) = F(x, y) which implies F(x, y) is conservative and hence independent of path. 2 By the Fundamental Theorem of alculus for Line Integrals W = F(x, y) dr = f (x, y) (2,1) (0,0) = 8 3.

losed urves (1 of 4) Theorem Suppose that F(x, y) is continuous in the open, connected region D R 2. Then F is conservative on D if and only if F(x, y) dr = 0 for every piecewise-smooth closed curve lying in D.

losed urves (2 of 4) Suppose F(x, y) dr = 0 for every piecewise-smooth closed curve lying in D. Let P and Q be any two points in D. Let 1 and 2 be any two piecewise-smooth curves connecting P and Q as shown next.

losed urves (3 of 4) 2 1 y 0 1 2 2 1 0 1 2 x = 1 ( 2 ) is a closed curve in D.

losed urves (4 of 4) 0 = F(x, y) dr = F(x, y) dr + F(x, y) dr 1 2 = F(x, y) dr F(x, y) dr 1 2 F(x, y) dr = F(x, y) dr 1 2

Simply-connected Regions Recall: a region is simply-connected if it contains no holes. Theorem Suppose that M(x, y) and N(x, y) have continuous first partial derivatives on a simply-connected region D. Then M(x, y) dx + N(x, y) dy is independent of path in D if and only if M y (x, y) = N x (x, y) for all (x, y) D.

Example Show that the following line integral is independent of path. e 2y dx + (1 + 2xe 2y ) dy

Example Show that the following line integral is independent of path. e 2y dx + (1 + 2xe 2y ) dy Let M(x, y) = e 2y and N(x, y) = 1 + 2xe 2y. Then M y (x, y) = 2e 2y = N x (x, y) and thus by the previous theorem the line integral is independent of path.

onservative Vector Fields Summary: suppose that F(x, y) = M(x, y), N(x, y) where M(x, y) and N(x, y) have continuous first partial derivatives on an open, simply-connected region D R 2. In this case the following statements are equivalent. 1 F(x, y) is conservative in D. 2 F(x, y) is a gradient field in D (i.e. F(x, y) = f (x, y) for some potential function f, for all (x, y) D. 3 F(x, y) dr is independent of path in D. 4 F(x, y) dr = 0 for every piecewise-smooth closed curve lying in D. 5 M y (x, y) = N x (x, y) for all (x, y) D.

Graphical Interpretation (1 of 2) 1 Suppose F(x, y) = y, x. x 2 +y 2 1.0 0.5 y 0.0 0.5 1.0 1.0 0.5 0.0 0.5 1.0 x The vector field is not conservative since along the closed path the vector field is always oriented with the curve (or always against it).

Graphical Interpretation (2 of 2) Suppose F(x, y) = y, x. 1.0 0.5 y 0.0 0.5 1.0 1.0 0.5 0.0 0.5 1.0 x The vector field is conservative since along the closed path as much of the vector field is oriented with the curve as against it.

Three Dimensions (1 of 3) Remark: much of what we have stated for two-dimensional vector fields can be extended to three-dimensional vector fields. Example Evaluate the line integral y 2 dx + (2xy + e 3z ) dy + 3ye 3z dz along any piecewise-smooth path connecting (0, 1, 1/2) to (1, 0, 2).

Three Dimensions (2 of 3) F(x, y, z) = y 2, 2xy + e 3z, 3ye 3z 1 If f (x, y, z) = xy 2 + ye 3z then F(x, y, z) = f (x, y, z) which implies F(x, y, z) is conservative.

Three Dimensions (2 of 3) F(x, y, z) = y 2, 2xy + e 3z, 3ye 3z 1 If f (x, y, z) = xy 2 + ye 3z then F(x, y, z) = f (x, y, z) which implies F(x, y, z) is conservative. 2 By the Fundamental Theorem of alculus for Line Integrals F(x, y, z) dr = xy 2 + ye 3z (1,0,2) (0,1,1/2) = e3/2.

Three Dimensions (3 of 3) Theorem Suppose that the vector field F(x, y, z) is continuous on the open, connected region D R 3. Then, the line integral F(x, y, z) dr is independent of path in D if and only if the vector field F is conservative in D, that is, F(x, y, z) = f (x, y, z), for all (x, y, z) in D, for some scalar function f (a potential function for F). Further, for any piecewise-smooth curve lying in D, with initial point (x 1, y 1, z 1 ) and terminal point (x 2, y 2, z 2 ) we have F(x, y, z) dr = f (x, y, z) (x 2,y 2,z 2 ) (x 1,y 1,z 1 ) = f (x 2, y 2, z 2 ) f (x 1, y 1, z 1 ).

Homework Read Section 14.3. Exercises: 1 51 odd