13.5. Click here for answers. Click here for solutions. CURL AND DIVERGENCE. 17. F x, y, z x i y j x k. 2. F x, y, z x 2z i x y z j x 2y k
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1 SECTION CURL AND DIVERGENCE 1 CURL AND DIVERGENCE A Click here for answers. S Click here for solutions Find (a the curl and (b the divergence of the vector field. 1. F x, y, xy i y j x k. F x, y, x i x y j x y k 3. F x, y, xe y j ye k 4. F x, y, x i y j 1 k 5. F x, y, xe y i ye x j e xy k 6. F x, y, x i y j k 7. F x, y, x y i y j x k 8. F x, y, y i x j xy k 9. F x, y, y i x y k 10. F x, y, xy j xy k 11. F x, y, sin x i cos x j k 1. F x, y, e x i e y j 3xe y k 13. F x, y, x 3y 5 i 3y j 5x 6y k 14. F x, y, xe y i e y j y ln k 15. F x, y, e xy i sin x y j xy k 16 7 Determine whether or not the given vector field is conservative. If it is conservative, find a function f such that F f. 16. F x, y, y i x j k 17. F x, y, x i y j x k 18. F x, y, y i j x k 19. F x, y, i y j x y k 0. F x, y, cos y i sin x j tan k 1. F x, y, x i e y sin j e y cos k. F x, y, y i y x j xy k 3. F x, y, x i xy j y k 4. F x, y, x i y j k 5. F x, y, xy 3 i x y 3 j 3x y k 6. F x, y, e x i e j e y k 7. F x, y, ye x i e x j xye x k
2 SECTION CURL AND DIVERGENCE ANSWERS E Click here for exercises. S Click here for solutions. 1. (a y i j x k (b x + y (a 5 i 0 j 3 k (b 3. (a 3 i 3 j + k (b 14. (a ( e y +ln i xe y k 3. (a e i + e y k (b xe y + ye 4. (a y i x +1 j (b 5. (a x (e xy ye x i + y (xe y e xy j + (ye x xe y k (b e y + e x + e xy 6. (a 0 (b 3 7. (a ( y i +x j + x k (b xy + + x 8. (a 0 (b 0 9. (a x i + ( y +xy j y k (b x y 10. (a x i y j + y k (b x (1 + y 11. (a sin x k (b cos x + 1. (a (3xe y +ye y i +(xe x 3e y j (b e x e y (b e y + e y + y 15. (a x ( i + xye xy + y j +[cos(x y xe xy ] k (b ye xy cos (x y+ xy 16. f (x, y, xy + + K 17. Not conservative 18. Not conservative 19. f (x, y, x + y + K 0. Not conservative 1. f (x, y, 1 x + e y sin + K. f (x, y, xy y3 + K 3. Not conservative 4. f (x, y, 1 x + 1 y K 5. Not conservative 6. Not conservative 7. f (x, y, ye x + K
3 SECTION CURL AND DIVERGENCE 3 SOLUTIONS E Click here for exercises. 1. (a curl F F xy y x [ y (x ] [ (y i x (x ] (xy j [ + x (y ] y (xy k (0 y i ( 0 j +(0 x k y i j x k x (xy+ y (y+ (x y + + x x + y +. (a curl F F x x+ y + x y ( 1 i (1 + j +(1 0 k 3 i 3 j + k x (x + y (x + y + + (x y (a curl F F 0 xe y ye (e 0 i (0 0 j +(e y 0 k e i + e y k x (0 + y (xey + (ye xe y + ye 4. (a curl F F x/ y/ 1/ (0+ y i (0+ x j +(0 0 k y i x j ( x + ( y + ( 1 x y (a curl F F xe y ye x e xy (xe xy xye x i (ye xy xye y j +(ye x xe y k x (e xy ye x i + y (xe y e xy j + (ye x xe y k x (xey + y (yex + (exy e y + e x + e xy 6. (a curl F F x y (0 0 i +(0 0 j +(0 0 k 0 x (x+ y (y+ ( (a curl F F x y y x (0 y i +(0 x j + ( 0 x k ( y i +x j + x k ( x y + ( y + ( x x y xy + + x 8. (a curl F F y x xy (x x i +(y y j +( k 0 x (y+ y (x+ (xy (a curl F F y 0 x y ( x 0 i + ( y +xy j +(0 y k x i + ( y +xy j y k ( y + x y (0 + ( x y 0+0 x y x y
4 4 SECTION CURL AND DIVERGENCE 10. (a curl F F 0 xy xy x i y j + y k x (0 + y (xy+ (xy 0+x + xy x (1 + y 11. (a curl F F sin x cos x ( sin x +0k sin x k x (sin x+ y (cos x+ ( cosx + 1. (a F e x e y 3xe y (3xe y +ye y i +(xe x 3e y j x (ex + y ( ey + (3xey e x e y 13. (a F x +3y 5 3y 5x +6y (6 1 i +( 5 5 j +(0 3 k 5i 0 j 3 k x (x +3y 5+ ( 3y y (a curl F xe y e y y ln ( e y +ln i xe y k + (5x +6y (b div F x (xey + ( e y + (y ln y e y + e y + y 15. (a F e xy sin (x y xy/ x ( i + xye xy + y j +(cos(x y xe xy k x (exy + y sin (x y+ ( ye xy cos (x y+ xy xy 16. curl F 0 and F is defined on all y x 1 of R 3 with component functions which have continuous partial derivatives, so by (4, F is conservative. Thus there exists f such that F f. Thenf x (x, y, y implies f (x, y, xy + g (y, and f y (x, y, x + g y (y,. But f y (x, y, x, sog (y, h ( and f (x, y, xy + h (. Thus f (x, y, h ( but f (x, y, 1so h ( + K. Hence a potential for F is f (x, y, xy + + K. 17. curl F j 0 x y x 18. curl F y x i +(y x j k Since curl F (y y i +(1 1 j +(0 0 k 0, F is defined on all of R 3, and the partial derivatives of the f x (x, y, implies f (x, y, x + g (y, and f y (x, y, g y (y,. But f y (x, y, y, so g (y, y + h ( and f (x, y, x + y + h (. Then f (x, y, x + y + h (, but f (x, y, x + y so h ( K and f (x, y, x + y + K. 0. curl F (cosx sin y k 0. cos y sin x tan
5 1. Since curl F x e y sin e y cos (e y cos e y cos i +(0 0 j +(0 0 k 0 F is defined on R 3, and since the partial derivatives of the f x (x, y, x implies f (x, y, 1 x + g (y, and f y (x, y, g y (y,. But f y (x, y, e y sin, so g (y, e y sin + h ( and f (x, y, 1 x + e y sin + h (. Thus f (x, y, e y cos + h (. Butf (x, y, e y cos implies h ( K and f (x, y, 1 x + e y sin + K is a potential for F.. Since curl F y y + x xy (x x i +(y y j +( k 0 F is defined on R 3, and since the partial derivatives of the f x (x, y, y implies f (x, y, xy + g (y, and f y (x, y, x + g y (y,. Butf y (x, y, x + y so g (y, 1 3 y3 + h ( and f (x, y, xy y3 + h (. Then f (x, y, xy + h (. Butf (x, y, xy so h ( K. Hence f (x, y, xy y3 + K is a potential for F. 3. curl F i + x j + y k 0, so x xy y F isn t conservative. 4. curl F 0 by Problem 6(a, F is defined on all of R 3,and the partial derivatives of the component functions are continuous, so F is conservative. Thus there exists a function f such that f F. Thenf x (x, y, x implies f (x, y, 1 x + g (y, and f y (x, y, g y (y,. But f y (x, y, y, sog (y, 1 y + h ( and f (x, y, 1 x + 1 y + h (. Thusf (x, y, h ( but f (x, y, so h ( 1 + K and f (x, y, 1 x + 1 y K. SECTION CURL AND DIVERGENCE 5 5. curl F F xy 3 x y 3 3x y ( 6x y 6x y i ( 6xy 3xy j + ( 4xy 3 xy 3 k 0 so F isn t conservative. 6. curl F F e x e e y (e y e i (0 0 j +(0 0 k 0 so F isn t conservative. 7. curl F F ye x e x xye x (xe x xe x i [(xye x + ye x (xye x + ye x ] j +(e x e x k 0 F is defined on all of R 3, and the partial derivatives of the component functions are continuous, so F is conservative. Thus there exists a function f such that f F. Then f x (x, y, ye x implies f (x, y, ye x + g (y, and f y (x, y, e x + g y (y,. Butf y (x, y, e x,so g (y, h ( and f (x, y, ye x + h (. Thus f (x, y, xye x + h ( but f (x, y, xye x so h ( K and f (x, y, ye x + K.
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