Answers and Solutions to Section 13.3 Homework Problems 1-23 (odd) and S. F. Ellermeyer. f dr

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1 Answers and Solutions to Section 13.3 Homework Problems 1-23 (odd) and S. F. Ellermeyer 1. By looking at the picture in the book, we see that f dr For the vector field Fx,y 6x 5yi 5x 4yj, we have P y 5 Q x 5. Since the (implied) domain of F is 2, which is a simply connected set, we conclude that F is conservative. To find f such that f F, we begin with f xx,y 6x 5y. This gives us fx,y 3x 2 5xy hy. Differentiation with respect to y then gives us f yx,y 5x h y. However, we must also have f yx,y 5x 4y. Thus h y 4y which means that hy 2y 2. Since we are only looking for a single potential function, we might as well take. We thus obtain fx,y 3x 2 5xy 2y 2. Let us check that this is correct: fx,y 6x 5yi 5x 4yj Fx,y. 5. For the vector field we have Fx,y xe y i ye x j, 1

2 P y xey Q x yex. Since P/y Q/x, we conclude that the vector field F is not conservative. 7. For the vector field Fx,y 2xcosy ycosxi x 2 siny sinxj, we have P 2xsiny cosx y Q 2xsiny cosx. x Since the (implied) domain of F is 2, which is a simply connected set, we conclude that F is conservative. To find f such that f F, we begin with f xx,y 2xcosy ycosx. This gives us fx,y x 2 cosy ysinx hy. Differentiation with respect to y then gives us f yx,y x 2 siny sinx h y. However, we must also have f yx,y x 2 siny sinx. Thus h y which means that hy. Since we are only looking for a single potential function, we might as well take. We thus obtain fx,y x 2 cosy ysinx. Let us check that this is correct: fx,y 2xcosy ycosxi x 2 siny sinxj Fx,y. 9. For the vector field Fx,y ye x sinyi e x xcosyj, we have P y ex cosy Q x ex cosy. 2

3 Since the (implied) domain of F is 2, which is a simply connected set, we conclude that F is conservative. To find f such that f F, we begin with f xx,y ye x siny. This gives us fx,y ye x xsiny hy. Differentiation with respect to y then gives us f yx,y e x xcosy h y. However, we must also have f yx,y e x xcosy. Thus h y which means that hy. Since we are only looking for a single potential function, we might as well take. We thus obtain fx,y ye x xsiny. Let us check that this is correct: fx,y ye x sinyi e x xcosyj Fx,y. 11. The vector field Fx,y 2xyi x 2 j is conservative, so all line integrals of F are path independent. A potential function for F is fx,y x 2 y so, by the Fundamental Theorem for line integrals, the integral of F over any path beginning at 1,2 and ending at 3,2 is F dr f dr f3,2 f1, First we find a potential function, f, for the vector field Fx,y x 3 y 4 i x 4 y 3 j. The potential function must satisfy f xx,y x 3 y 4 and thus fx,y 1 4 x4 y 4 hy and f yx,y x 4 y 3 h y. 3

4 f yx,y x 4 y 3, we obtain h y and thus hy. Therefore, a potential function for F is fx,y 1 4 x4 y 4. The path rt t i 1 t 3 j t 1 has initial point r,1 and terminal point r1 1,2. Thus, by the Fundamental Theorem for Line integrals, we have F dr f1,2 f, First we find a potential function, f, for the vector field Fx,y,z yzi xzj xy 2zk. The potential function must satisfy f xx,y,z yz and thus fx,y,z xyz hy,z and f yx,y,z xz h y. f yx,y,z xz we see that h/y and hence that hy,z gz. We now have fx,y,z xyz gz This tells us that f zx,y,z xy g z f zx,y,z xy 2z, we obtain g z 2z and thus gz z 2. Therefore, a potential function for F is fx,y,z xyz z 2. By the Fundamental Theorem for Line integrals, we have 4

5 F dr f4,6,3 f1,, First we find a potential function, f, for the vector field Fx,y,z y 2 coszi 2xycoszj xy 2 sinzk. The potential function must satisfy f xx,y,z y 2 cosz and thus fx,y,z xy 2 cosz hy,z and f yx,y,z 2xycosz h y. f yx,y,z 2xycosz we see that h/y and hence that hy,z gz. We now have fx,y,z xy 2 cosz gz This tells us that f zx,y,z xy 2 sinz g z f zx,y,z xy 2 sinz, we obtain g z and thus gz. Therefore, a potential function for F is fx,y,z xy 2 cosz. The path rt t 2 i sintj tk t has initial point r,, and terminal point r 2,,. By the Fundamental Theorem for Line integrals, we have F dr f 2,, f,, 2 2 cos 2 cos. 19. For the vector field Fx,y 2xsinyi x 2 cosy 3y 2 j, 5

6 we have P y 2xcosy Q x 2xcosy. Since the (implied) domain of F is 2, which is a simply connected set, we conclude that F is conservative (and hence that all line integrals of F are path independent). To find f such that f F, we begin with f xx,y 2xsiny. This gives us fx,y x 2 siny hy. Differentiation with respect to y then gives us f yx,y x 2 cosy h y. However, we must also have f yx,y x 2 cosy 3y 2. Thus h y 3y 2 which means that hy y 3. Since we are only looking for a single potential function, we might as well take. We thus obtain fx,y x 2 siny y 3. Now, by the Fundamental Theorem for Line Integrals, for a path with initial point 1, and terminal point 5,1, we have 2xsinydx x 2 cosy 3y 2 dy F dr 21. The force field Fx,y x 2 y 3 i x 3 y 2 j is conservative and has potential function fx,y 1 3 x3 y 3. f5,1 f1, 25sin1 1. Thus, the work done by this force field in moving an object from the point, to the point 2,1 is 6

7 F dr f2,1 f, The vector field, F, shown in the picture (page 944) is not conservative. If it were, then it would have to satisfy P Q x,y y x x,y at all points x,y. However, the picture shows that this is not true. For example, look at the vertical column of vectors that is the second to the left from the y axis. (Hence x x is constant for all vectors in this column.) If we move from bottom to top (in the direction of increasing y) along this column, we see that the vectors make a transition from pointing to the left to pointing to the right. There is some point x,y in the third quadrant at which Fx,y. Also, since Px,y transitions from negative to positive as y increases, we see that P y x,y. However, if we now look at the horizontal row of vectors with y y, we see that these vectors transition from pointing up to pointing down as x increases. This means that Q x x,y. Therefore it is not true that P y x,y Q x x,y. 29. The set x,y x and y is open, connected, and simply connected. 3. The set x,y x is open, but not connected or simply connected. 31. The set x,y 1 x 2 y 2 4 is open and connected, but not simply connected. 32. The set x,y x 2 y 2 1 or 4 x 2 y 2 9 is neither open, nor connected, nor simply connected. 33. onsider the vector field Fx,y y x 2 y i 2 x x 2 y j 2 with domain D x,y x,y,. The vector field F satisfies 7

8 P y x2 y 2 1 y2y x 2 y 2 2 Q x x2 y 2 1 x2x x 2 y 2 2 y 2 x 2 x 2 y 2 2 y 2 x 2 x 2 y 2 2 and thus P y Q x. However, since the domain D is not a simply connected set, we are not guaranteed that the vector field F is conservative. In fact, F is not conservative as we will show by computing line integrals of F over two different paths joining that begin at the point 1, and end at the point 1,. First, we use the path rt costi sintj t. (This path traces out the top half of the unit circle counterclockwise.) For this path, we have F dr Frt r tdt sinti costj sinti costjdt 1dt. Next, we use the path rt costi sintj t. (This path traces out the bottom half of the unit circle clockwise.) For this path, we have F dr Frt r tdt sinti costj sinti costjdt 1dt. 8