Solutions to Homework 11

17.1

#4 The field looks almost like the one from Problem 8, except that the vectors don’t point in the direction of i⃗, but rather of 2i⃗ + j⃗.

#7 Looks somewhat like Figure 20.29, except that the asymptotic line along which the field seems to come in is not y = −x, but y = (−1 − \sqrt{2})x and leaves along y = (−1 + \sqrt{2})x rather then y = x. So the picture is basically just rotated a little.

17.2

#2 Arrows of length 2 pointing up everywhere. The flows are just vertical lines.

#3 Here the field has the vector 3i⃗ − 2j⃗ everywhere and the flows are all lines with slope -2/3.

#7 Here the field is just Figure 20.29 rotated by π/4 clockwise. The flows are hyperbolas, ie of the form y = k/x. This can also be seen from the given flows: by solving for e^{t} and substituting in we get that y = ab/x. The system of DE’s is x′ = x and y′ = y and it can easily be checked that the given parametric equations satisfy the DE’s.

18.1

#4 Positive, since the field points roughly in the same direction as the path.

#5 Zero, since we get a positive contribution along the left half of the circle, which is then cancelled out again by the negative contribution of the right half of the circle.

#9 The field looks like the field 17.1.#13, except that here arrows on the outside are shorter rather then longer, the magnitude of the arrows is 1/∥r∥. So along C_{2} and C_{4} the line integral will be 0, since the field is perpendicular to the path. Along C_{3} the integral will be positive and it will be exactly π/2, since distance travelled is .25(2πr) and the magnitude of the field is 1/r in the direction of C_{3}, so that we only have to multiply the two quantities. Similarly the integral along C_{1} will be −π/2, so that it will cancel with the contribution of C_{3} and the total integral will be 0.

#10 The field looks like 17.1.#8 except that all vectors are pointing to the right, so that all positive and negative contributions along the path will sum to 0 again.

#11 Many correct answers, but see for example Figure 18.18.

18.2

#3 Parametrize the ellipse by x = 2 sin t, y = cos t, 0 ≤ t ≤ π/2. So v⃗ = ⟨2 cos t, − sin t⟩ and thus

∫_{C} F⃗ · dr⃗ = ∫_{0}^{π/2}(e^{2 sin t}i⃗ + e^{cos t}j⃗) · (2 cos ti⃗ − sin tj⃗) dt

= ∫_{0}^{π/2} 2(cos t)e^{2 sin t} − (sin t)e^{cos t} dt = e^{2} − e.

#4 The path can be broken into three line segments C_{1} from (1,0) to (-1,0), C_{2} from (-1,0) to (0,1) and C_{3} from (0,1) to (1,0). Along C_{1} the line integral will be zero, since there y = 0, so that the i⃗ coordinate of F⃗ will be zero, and thus the field is perpendicular to C_{1}. Next parametrize C_{2} by ⟨−1 + t, t⟩ for 0 ≤ t ≤ 1 and thus

∫_{C_{2}} F⃗ · dr⃗ = ∫_{0}^{1}[t(−1 + t)i⃗ + (−1)j⃗] · [i⃗ + j⃗] dt = ∫_{0}^{1} t^{2} − t − 1 dt = −7/6.

And finally parametrize C_{3} as ⟨t, 1 − t⟩, 0 ≤ t ≤ 1 and the corresponding line integral will be

∫_{0}^{1}[t(1 − t)i⃗ + (2t − 1)j⃗] · [i⃗ − j⃗] dt = ∫_{0}^{1} −t^{2} − t + 1 dt = 1/6.

Adding all the contributions together we get that ∫_{C} F⃗ · dr⃗ = −1.

#9 a) ∫_{C} F⃗ · dr⃗ = 11/16.

b) ∫_{C} F⃗ · dr⃗ = 7/6.

#10

a)

∥F⃗ ∥ = ((−y)^{2} + x^{2})^{1/2} = (x^{2} + y^{2})^{1/2} = 1

along the circle x^{2} + y^{2} = 1.

b) If xi⃗+yj⃗ is the position vector of a point on the circle, then xi⃗+yj⃗ is perpendicular to the circle, since it is in the direction of the radius at that point. But now (xi⃗+yj⃗)·F⃗ = x(−y)+yx = 0, so that the field is also perpendicular to the radius and thus in the direction of the circle.

c) The force at every point is exactly one in the direction of C, so that the work done is exactly the distance travelled, which is the length of C. we can also check that the parametrization x = cos t, y = sin t and 0 ≤ t ≤ 2π traces out C, and that ∫_{C} F⃗ · dr⃗ = 2π (which is the length of the unit circle) by doing the algebra.