1. Vector Fields. f 1 (x, y, z)i + f 2 (x, y, z)j + f 3 (x, y, z)k.

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1 HAPTER 14 Vector alculus 1. Vector Fields Definition. A vector field in the plane is a function F(x, y) from R into V, We write F(x, y) = hf 1 (x, y), f (x, y)i = f 1 (x, y)i + f (x, y)j. A vector field in space is a function F(x, y, z) from R 3 into V 3, We write F(x, y, z) = hf 1 (x, y, z), f (x, y, z), f 3 (x, y, z)i = Example. F(x, y) = h y, xi = yi + xj Some values: F(, 1) = h 1, i = i =) kf(, 1)k = 1. f 1 (x, y, z)i + f (x, y, z)j + f 3 (x, y, z)k. F( 1, 1) = h1, 1i = i j =) kf( 1, 1)k = p. n general, kf(x, y)k = kh y, xik = p x + y. 143

2 VETOR ALULUS The diagram on the left on the previous page is drawn to scale, with F(x, y) placed at (x, y). The diagram on the right is scaled smaller with relative magnitudes to fit in the diagram. This vector field could be called a spin field. Since hx, yi h y, xi = xy + xy =, each F(x, y) is tangent to the circle centered at the origin of radius p x + y, pointing in a counter-clockwise direction with magnitude equalling the radius of the circle.

3 Sample vector fields: 1. VETOR FELDS 145

4 VETOR ALULUS More examples of scaled and unscaled vector fields:

5 1. VETOR FELDS 147 Some vector fields are velocity vector fields, i.e., F(x, y) gives the velocity of a particle at (x, y). Suppose a particle starts to flow at (x, y ) at time t. Then the curve traced out by hx(t), y(t)i, where x(t) and y(t) are solutions of the di erential equations x (t) = f 1 (x(t), y(t) and y (t) = f (x(t), y(t) with initial conditions x(t ) = x and y(t ) = y, is a flow line. We use the chain rule to find a di erential equation for y as a function of x for the velocity vector field: dy dx = dy/dt dx/dt = y (t) x (t) = f (x, y) f 1 (x, y). n our example, dy dx = 4 3 =) y = 4 3 x +. For the flow line through (1, ), = =) = 3. Thus y = 4 3 x + 3 is the equation of the flow line.

6 VETOR ALULUS Example. We consider the vector field F(x, y) = h1, xi and the flow line through (, ). This is a velocity vector field for dy dx = x 1 = x. Then y = 1 x +. For the flow line through (, ), = + =) =. Thus the equation of the flow line is y = 1 x. Definition. For any scalar function f (from R or R 3 to R), the vector field F(x, y) = rf is called the gradient field for the function f. We call f a potential function for F. Whenever F = rf for some scalar function f, we refer to F as a conservative vector field.

7 1. VETOR FELDS 149

8 VETOR ALULUS Problem (Page 996 #36). Determine whether F(x, y) = hy cos x, sin x is conservative, and if so, find its potential function. (We proceed in a manner di erent from the text.) This is the same as determining whether y cos {z x } dx + sin x {z y } dy = M =f? x N =f? y is exact and finding its solution as we did in Math 31. M y = cos x and N x = cos x =) M y = N x =) Thus f(x, y) = the DE is exact =) F is conservative. y cos x dx f(x, y) = (sin x y) dy = y sin x + g(y) = y sin x + h(x) f y (x, y) = sin x + g (y) f x (x, y) = y cos x + h (x) = sin x y =) = y cos x =) g (y) = y =) g(y) = y + h (x) = =) h(x) = y f(x, y) = y sin x + Maple. See vfield(14.1).mw or vfield(14.1).pdf. yi y

9 . LNE NTEGRALS 151. Line ntegrals Oriented curve one from which we have chosen a direction two possible directions. Definition. The line integral of f(x, y, z) with respect to arc length along the oriented curve in three-dimensional space, written f(x, y, z) ds, is defined by f(x, y, z) ds = lim kp k! nx f(x i, yi, zi ) s i, provided the limit exists and is the same for all choices of evaluation points. Note. There is a similar definition for two dimensions. i=1

10 VETOR ALULUS Theorem (Evaluation Theorem). Suppose that f(x, y, z) is continuous in a region D containing the curve and that is described parametrically by x(t), y(t), z(t) for a apple t apple b where x(t), y(t), and z(t) all have continuous first derivatives. Then b f(x, y, z) ds = f x(t), y(t), z(t) p [x (t)] + [y (t)] + [z (t)] dt. n the two-dimensional case, f(x, y) ds = a b a f x(t), y(t) p [x (t)] + [y (t)] dt. Definition. A space curve is smooth if it can be described parametrically by x = x(t), y = y(t), and z = z(t) for a apple t apple b, where x(t), y(t), and z(t) all have continuous first derivatives and [x (t)] + [y (t)] + [z (t)] 6= on [a, b]. (Similarly for plane curves.) Example. Find 3y ds where is the quarter-circle x + y = 4 from (, ) to (, ). 1 / x = cos t =) x (t) = sin t and y = sin t =) y (t) = cos t. 3y ds = 3(4 sin t) p 4 sin t + 4 cos t dt = 4 sin t dt = / h (1 cos t) dt = 1 t / 1 i h sin t = 1 / i = 6.

11 Problem (Page 11 # 4). Find from (, 1, ) to (,, ). x =, and for apple t apple 1, Thus xz ds =. LNE NTEGRALS 153 xz ds, where is the line segment y = 1 t, z = t =) x (t) =, y (t) = 1, z (t) =. 1 ds = p + ( (t) p 5 dt = 4 p 5 1 1) + dt = p 5 dt t dt = 4 p h t 5 i 1 = 4p 5 1 = p 5. Theorem. Suppose f(x, y, z) is a continuous function in some region D containing the oriented curve. Then, if is piecewise smooth, with = 1 [ [ [ n, where 1,,..., n are all smooth and where the terminal point of i is the same as the initial point of i+1, for i = 1,,..., n 1, we have f(x, y, z) ds = f(x, y, z) ds and f(x, y, z) ds = f(x, y, z) ds + 1 f(x, y, z) ds + f(x, y, z) ds. n Note. There is a similar statement for two dimensions.

12 VETOR ALULUS Example. Find (x + y) ds over = 1 [ where 1 is the quarter circle from (1, ) to (, 1) and is the line segment from (, 1) to ( 1, ). 1 : x(t) = cos t, y(t) = sin t, apple t apple =) x (t) = sin t, y (t) = cos t. : x(t) = t, y(t) = 1 t, apple t apple 1, x (t) = 1, y (t) = 1. Thus (x + y) ds = (x + y) ds + 1 (x + y) ds = / (cos t+sin t) p ( h sin t / cos t sin t) + (cos t) dt+ (cos t + sin t) dt + p i / 1 1 t+(1 t) p ( 1) + ( 1) dt = (1 t) dt = + p h i 1 t t = p ( ) =. Theorem. For any piecewise smooth curve (in two or three dimensions), 1 ds gives the arc length of the curve.

13 Line integrals with respect to x nx f(x, y, z) dx = lim f(x i, yi, zi ) kpk! f(x, y, z) dx = f(x, y, z) dx = f(x, y, z) dx = b a i=1. LNE NTEGRALS 155 x i f x(t), y(t), z(t) x (t) dt f(x, y, z) dx f(x, y, z) dx + 1 f(x, y, z) dx + f(x, y, z) dx n Note. We have similar results for y and z and two dimensions. Notation. We write f(x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz = f(x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz Example. Find 3y dy where is the line segment from (, ) to (1, 3). Parameterize the line by x = t, y = 3t, apple t apple 1. Then dy = 3 dt and 1 1 3y dy = 3(3t) (3dt) = 81 t dt = 7t 3 1 = 7.

14 VETOR ALULUS Example. Find (, ). 3y dy where is the portion of y = x from (, 4) to 1) Parameterize by x = t, y = t from apple t apple. Then dy = t dt and 3y dy = 3t 4 (t) dt = 6 t 5 dt = t 6 = ( )6 = 64 ) ould also parameterize by x = t, y = ( t) from apple t apple. Then dy = ( t) dt and 3y dy = 3( t) 4 ( )( t) dt = Line ntegrals of Vector Fields 6 ( t) 5 dt = ( t) 6 = 64 = 64. Let F(x, y, z) = F 1 (x, y, z), F (x, y, z), F 3 (x, y, z) be a vector field along the curve defined by x = x(t), y = y(t), and z = z(t), a apple t apple b. Let Define the line integral F(x, y, z) dr = Work r = hx, y, zi =) dr = hdx, dy, dzi. F 1 (x, y, z) dx + F (x, y, z) dy + F 3 (x, y, z) dz = F 1 (x, y, z) dx + F (x, y, z) dy + F 3 (x, y, z) dz f F(x, y, z) is a force field, the work done by F in moving a particle along the curve can be written as W = F(x, y, z) dr.

15 . LNE NTEGRALS 157 Example. Find the work done by F(x, y, x) = hz,, 3x i along, the quarter ellipse given by x = cos t, y = 3 sin t, z = 1, from (,, 1) to (, 3, 1). We have F dr = apple t apple, dx = sin t dt, dy = 3 cos t dt, dz =. / z dx + dy + F 3 (x, y, z) dz = h i (1)( sin t) + (3 cos t) + 3(4 cos t)() dt = cos t / = =. Recall F dr = kfk kdrk cos, where is the angle between F and dr, apple apple. Thus the line integral of a vector field measures the extent to which is going with the vector field (+) or against it ( ).

16 VETOR ALULUS Using the diagram below, arrange R i F dr, i = , and the number in order from left to right (smallest to largest). F dr < F dr < < F dr < F dr and have the opposite direction of the vector field with the vectors on having the greater magnitude. 3 and 4 have the same direction of the vector field with the vectors having a similar magnitude, but 4 is longer. Maple. See lineintegral(14.).mw or lineintegral(14.).pdf.

17 3. NDEPENDENE OF PATH AND ONSERVATVE VETOR FELDS ndependence of Path and onservative Vector Fields Definition. (1) A region D R n (for n ) is called connected if every pair of points in D can be connected by a piecewise-smooth curve lying entirely in D. connected not connected () A path is a piecewise-smooth curve traced out by the endpoint of the vector-valued function r(t) for a apple t apple b. (3) The line integral F dr is independent of path in the domain D if the integral is the same for every path contained in D that has the same beginning and end points. Theorem. Suppose the vector field F(x, y) = M(x, y), N(x, y) is continuous on the open, connected region D R. Then the line integral F(x, y) dr is independent of path if and only if F is conservative on D. Note. A similar result is valid in any number of dimensions.

18 VETOR ALULUS Theorem (Fundamental Theorem for Line ntegrals). Suppose F(x, y) = M(x, y), N(x, y) is continuous on the open, connected region D R and is any piecewise-smooth curve lying in D, with initial point (x 1, y 1 ) and terminal point (x, y ). Then, if F is conservative on D, with F(x, y) = rf(x, y), F(x, y) dr = f(x, y) (x,y ) = f(x, y ) f(x 1, y 1 ). (x 1,y 1 ) Example. ompute hye xy, xe xy i dr for the curve shown below. We need to find f(x, y) such that rf = hf x, f y i = hye xy, xe xy i. Assume f x = ye xy. Then f(x, y) = ye xy dx = e xy + g(y) =) f y (x, y) = xe xy + g (y). Then rf = F if g(y) = for some constant. hoose = =) f(x, y) = e xy. Then F is conservative on R, so hye xy, xe xy (3,1) xy i dr = e = ( 1, 1) e3 e.

19 3. NDEPENDENE OF PATH AND ONSERVATVE VETOR FELDS 161 Definition. A curve is closed if its two endpoints are the same. For defined by x = g(t), y = h(t), a apple t apple b, this means g(a), h(a) = g(b), h(b). Theorem. Suppose F is continuous on the open, connected region D R. Then F is conservative on D if and only if F(x, y) dr = for every piecewise-smooth closed curve lying in D. Definition. A region D is simply-connected if every closed curve in D encloses only points in D. simply-connected not simply-connected Theorem. Suppose M(x, y) and N(x, y) havecontinuous 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) in D. Example. F(x, y) = M(x, y), N(x, y) = hx y, x i. M y = 1 and N x = 1. Thus F(x, y) dr = (x y) dx + (x ) dy is not independent of path.

20 VETOR ALULUS Theorem (onservative Vector Fields). Suppose F(x, y) = M(x, y), N(x, y) and M(x, y) and N(x, y) have continuous first partial derivatives on an open, simply-connected region D R. Then the following are equivalent: (1) F(x, y) is conservative in D. () F(x, y) is a gradient field in D, i.e., F(x, y) = rf(x, y) for some potential function f for all (x, y) in D. (3) F(x, y) dr is independent of path in D. (4) F(x, y) dr = for every piecewise-smooth closed curve lying in D. (5) M y (x, y) = N x (x, y) for all (x, y) in D. Example. Are the following vector fields conservative? F is constant =) M y = N x = =) conservative.

21 3. NDEPENDENE OF PATH AND ONSERVATVE VETOR FELDS 163 For a counter-clockwise circle centered at the origin, conservative. F dr > =) not No rotation =) M y = N x =) conservative. Theorem (Three Dimensions). Suppose the vector field F(x, y, z) is continuous on the open, connected region D R 3. Then F(x, y, z) dr is independent of path in D if and only if F is conservative in D, i.e., F(x, y, z) = rf(x, y, z) for some scalar function f (a potential function for F) for all (x, y, z) in D. Further, for any piecewise-smooth curve lying in D with initial point (x 1, y 1, z 1 ) and terminal point (x, y, z ), F(x, y, z) dr = f(x, y, z) (x,y,z ) = f(x, y, z ) f(x 1, y 1, z 1 ). (x 1,y 1,z 1 )

22 VETOR ALULUS Example. ompute hyz, xz, xyzi dr for the curve shown below. We need to find f(x, y, z) such that Assume Then Thus rf = hf x, f y, f z i = hyz, xz, xyzi. f x = yz =) f(x, y, z) = yz dx = xyz + g(y, z). f y = xz + g y (y, z) =) g y (y, z) = =) g(y, z) = h(z). f(x, y, z) = xyz + h(z) =) f z = xyz + h (z) =) h (z) = =) h(z) =. Take =. Then f(x, y, z) = xyz and rf = hyz, xz, xyzi = F, so F is conservative on R 3, and thus hyz, xz, xyzi dr = xyz (,1,) = =. (,,)

23 Definition. 4. GREEN S THEOREM Green s Theorem (1) A curve is simple if it does not intersect itself, except at the endpoints. simple not simple ()A simple closed curve has positive orientation if the region R enclosed by stays to the left of as the curve is traversed; a simple closed curve has negative orientation if the region R enclosed by stays to the right of as the curve is traversed. positive negative

24 VETOR ALULUS Notation. F (x, y) dr denotes a line integral along a simple closed curve oriented in the positive direction. Theorem (Green s Theorem). Let be a piecewise-smooth simple closed curve in the plane with positive orientation and let R be the region enclosed by. Suppose M(x, y) and N(x, y) are continuous and have continuous first partial derivatives in some open region D, with R D. Then Example. Find M(x, y) dx + N(x, y) dy (x 4 + y) dx + (5x + sin da. With made up of 4 separate continuous curves (lines), direct calculation and parametrizing is cumbersome, so use Green s Theorem. (x 4 + y) dx + (5x + sin y) dy = (5 ) da = R 3 da = 3(area of R) = 3( p ) = 3 = 6 by geometry. R

25 Example. Find 4. GREEN S THEOREM 167 (x y) dx + (x 3 + xy ) dy. Again, for simplicty, use Green s Theorem. (x y) dx + (x 3 + xy ) dy = (3x + y ) x da = R (x + y ) da = (x + y ) da = 1 R r r dr d = r 4 4 R 1 d = d = 4 d = 15 = 15. Notation. We to refer to the boundary of the region R, oriented in the positive direction. Note. We can replace in Green Theorem.

26 VETOR ALULUS Example. Suppose is a piecewise smooth, simple closed curve enclosing the region R. Then (1) x dy = (1 ) da = da. R R () ( y) dx = ( 1) da = da. Therefore, R Area of R = Extending Green s Theorem onsider R as below: R da = 1 R x dy y dx. Green s Theorem doesn t apply since R is not simply connected. But we make two horizontal slits in R, dividing R into two simply-connected regions R 1 and R. Apply Green s Theorem to R 1 and da = M(x, y) dx + N(x, y) dy @R M(x, y) dx + N(x, y) dy =! da =

27 M(x, y) dx + N(x, y) dy + M(x, y) dx + N(x, y) dy = 1 {z } Since the line integrals over the slits cancel M(x, y) dx + N(x, y) dy where = 1 [. 4. GREEN S THEOREM 169 Note. This procedure can be extended to any finite number of holes. D y Example. Suppose F(x, y) = x + y, x E. Show that x + y F(x, y) dr = for every simple closed curve enclosing the origin. Solution. Green s Theorem doesn t apply since F(, ) is not defined. Let be any simple closed curve containing the origin and let 1 be the circle of radius a > centered at the origin, positively oriented, where a is su ciently small so that and 1 do not meet. Let R be the region between and including the curves.

28 VETOR ALULUS Applying the extended Green s F(x, y) dr F(x, y) dr = F(x, y) dr = " # (1)(x + y ) x(x) ( 1)(x + y ) + y(y) da = R (x + y ) (x + y ) F(x, y) dr = F(x, y) dr. 1 Parameterize 1 by x = a cos t, y = a sin t, apple t da = da = =) Then x + y = a, and D y F(x, y) dr = F(x, y) dr = 1 1 a, x E dr = a 1 ( y) dx + x dy = 1 ( sin t)( a sin t) + (a cos t)(a cos t) dt = a 1 a sin t + cos t dt = dt = t =.