Calculus Multivariable McCallum, Hughes-Hallett, Gleason 4th Ed ch18 solutions

Calculus: Multivariable

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18.1 SOLUTIONS 1257 CHAPTER EIGHTEEN Solutions for Section 18.1 Exercises 1. Positive, because the vectors are longer on the portion of the path that goes in the same direction as the vector field. 2. Negative because the vector field points in the opposite direction to the path. 3. Positive, because the vector field points in the same direction as the path. 4. Zero, because, by symmetry, the positive integral along the left half of the path cancels the negative integral along the right half. 5. At every point, the vector field is parallel to segments Δ ~ r = Δ x ~ i of the curve. Thus, Z C ~ F · d~ r = Z 6 2 x ~ i · dx ~ i = Z 6 2 xdx = x 2 2 6 2 = 16 . 6. Since ~ F is perpendicular to the curve at every point along it, Z C ~ F · d~ r = 0 . 7. At every point along the curve, ~ F = 2 ~ j and is parallel to the curve. Thus, Z C ~ F · d~ r = 2 · Length of curve = 2 · 5 = 10 . 8. Since ~ F is a constant vector field and the curve is a line, R C ~ F · d~ r = ~ F · Δ ~ r , where Δ ~ r = 7 ~ j . Therefore, Z C ~ F · d~ r = (3 ~ i + 4 ~ j ) · 7 ~ j = 28 9. Since ~ F is perpendicular to the line, the line integral is 0. 10. The ~ j -component of ~ F does not contribute to the line integral. Since Δ ~ r = Δ x ~ i , we have Z C ~ F · d~ r = Z 6 2 ( x ~ i + y ~ j ) · dx ~ i = Z 6 2 xdx = x 2 2 6 2 = 16 . 11. Only the ~ i -component contributes to the integral, so Z C ~ F · d~ r = 6 · Length of path = 6 · (7 - 3) = 24 . 12. At every point on the path, ~ F is parallel to Δ ~ r . Suppose r is the distance from the point ( x, y ) to the origin, so k ~ r k = r. Then ~ F · Δ ~ r = k ~ F kk Δ ~ r k = r Δ r. At the start of the path, r = 2 2 + 2 2 = 2 2 and at the end r = 6 2 . Thus, Z C ~ F · d~ r = Z 6 2 2 2 rdr = r 2 2 6 2 2 2 = 32 . 13. The path is along the y -axis, so only the ~ j -component contributes to the line integral. Since C is oriented in the - ~ j direction, we have Z C ( x ~ i + 6 ~ j - ~ k ) · d~ r = - 6 · Length of path = - 6 · 8 = - 48 .
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1258 Chapter Eighteen /SOLUTIONS Problems 14. (a) See Table 18.1. Table 18.1 ( x, y ) ~ F ( x, y ) (0 , - 1) - ~ i (1 , - 1) - ~ i + ~ j (2 , - 1) - ~ i + 4 ~ j (3 , - 1) - ~ i + 9 ~ j (4 , - 1) - ~ i + 16 ~ j (4 , 0) 16 ~ j (4 , 1) ~ i + 16 ~ j (4 , 2) 2 ~ i + 16 ~ j (4 , 3) 3 ~ i + 16 ~ j 1 2 3 x - 1 1 2 3 y I K O O 6 Figure 18.1 (b) See Figure 18.1. (c) From the point (0 , - 1) to the point (4 , - 1) , the x -component of the force field is always - 1 , i.e., it is pushing the object backward with a constant force of 1 . Thus, the work done on that part of the path is - 1 · 4 = - 4 , because only the horizontal component of the force field contributes to work. From the point (4 , - 1) to the point (4 , 3) , the y -component of the force field is always 16 , so it is pushing the object forward with force of 16 . Thus, the work done on that part of the path is 16 · 4 = 64 , because only the vertical component of the force field contributes to work. So the total work done is - 4 + 64 = 60 . 15. Since it appears that C 1 is everywhere perpendicular to the vector field, all of the dot products in the line integral are zero, hence R C 1 ~ F · d~ r 0 . Along the path C 2 the dot products of ~ F with Δ ~ r i are all positive, so their sum is positive and we have R C 1 ~ F · d~ r < R C 2 ~ F · d~ r . For C 3 the vectors Δ ~ r i are in the opposite direction to the vectors of ~ F , so the dot products ~ F · Δ ~ r i are all negative; so, R C 3 ~ F · d~ r < 0 . Thus, we have Z C 3 ~ F · d~ r < Z C 1 ~ F
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