Solutions to Homework 10

15.8

#2 ∂(x, y)/∂(u, v) = 2u, so

∫ ∫_{T} |\frac{∂(x, y)}{∂(u, v)}| du dv = ∫_{0}^{2} ∫_{0}^{2} 2u du dv = \frac{8}{3}.

Also R = \{(x, y) : 0 ≤ x ≤ 4, \sqrt{x} ≤ y ≤ 2\}, so that

∫ ∫_{R} dx dy = ∫_{0}^{4} ∫_{√\overline{x}}^{2} dx dy = \frac{8}{3}.

#4

\frac{∂(x, y)}{∂(u, v)} = 3 · 2 − 5 · (−4) = 26.

And solving the equations for u and v we get u = (x + 2y)/13 and v = (3y − 5x)/26, so that:

\frac{∂(u, v)}{∂(x, y)} = (3/26) · (1/13) − (−5/26) · (2/13) = 1/26,

establishing the claim. Note that this relationship between the Jacobian and the Jacobian of the reverse transformation always holds, although we will neither prove this nor use it.

#5 ∂(x, y)/∂(u, v) = −3, so that we get that:

∫ ∫_{R} x + y dA = ∫ ∫_{T} 3u| − 3| du dv = ∫_{0}^{3} ∫_{0}^{1} 9u du dv = 27/2,

since the region is a rectangle in uv-coordinates: Change the corner points into uvcoordinates: (0, 0) → (0, 0), (3, −3) → (0, 3), (2, 1) → (1, 0) and (5, −2) → (1, 3).

#7 ∂(x, y)/∂(u, v) = 1/v and thus:

∫ ∫_{R} xy^{2} dA = ∫_{1}^{4} ∫_{1}^{4} v|1/v| du dv = 9.

15.R

#9

∫_{0}^{10} ∫_{0}^{0.1} xe^{xy} dy dx = ∫_{0}^{10} e^{0.1x} − e^{0} dx = 10(e − 2).

#10

∫ 0^{1} ∫_{0}^{y}(sin^{3} x)(cos x)(cos y) dx dy = ∫_{0}^{1}(cos y)\frac{sin^{4} x}{4}|_{0}^{y} dy = \frac{sin^{5} 1}{20}.

#16

∫_{0}^{2π} ∫_{0}^{π/2} ∫_{0}^{3}(ρ sin ϕ cos θ)^{2}ρ^{2} sin ϕ dρ dϕ dθ = 162π/5.

At some stage in this calculation you will probably have to refer to the integral tables on pg.467.

#20

∫_{0}^{5} ∫_{−2y/5}^{2y/5+6}(10 − 2y) dx dy = 550/3 ≈ 183.

16.1

#5 Let f(t) = t^{3} − t. Then the particle moves clockwise exactly when f is decreasing. So since f′(t) = 3t^{2} − 1 the particle will move clockwise when 3t^{2} − 1 > 0, that is −1/\sqrt{3} < t < 1/\sqrt{3}, it is at rest when t = −1/\sqrt{3} or t = 1/\sqrt{3} and moves counterclockwise otherwise.

#8 In all three cases y = x^{2}, so that the motion takes place on this parabola. The first and third motion come from x = −∞ and go to x = ∞, whereas the second motion comes from x = ∞ and goes back there. Also notice that in general the third particle is faster then the second which in turn is faster then the first.

16.2

#10 x = 2 + t and y = 4 + 3t, so that we can eliminate t and get that y − 3x = −2 or y = 3x − 2. Eliminating t in the second motion results in the same line. So both particles move on the same line with slope m = 3 and y-intercept b = −2.

#13 In general k is the radius of the circle and (a, b) its center:

a) k = 5, (a, b) = (0, 0).

b) k = 5, (a, b) = (0, 5).

c) k = 10\sqrt{2}, (a, b) = (10, −10).

#28 r⃗ = ⟨2, 3, −1⟩ + t⟨3, −1, 1⟩, or x = 2 + 3t, y = 3 − t and z = −1 + t.

#29 r⃗ = ⟨1, 2, 3⟩ + t⟨3, −3, 1⟩, or x = 1 + 3t, y = 2 − 3t and z = 3 + t.

#33 If the lines intersect then there are times t_{1}, t_{2} for which the two parametrizations yield the same point, namely the intersection point. But it can be easily seen that x = 2 + 3t_{1} = 1 + 3t_{2}, y = 3 − t_{1} = 2 − 3t_{2} and z = −1 + t_{1} = 3 + t_{2} doesn’t have a simultaneous solution.

16.3

#16 v⃗ = dr⃗/dt = − sin ti⃗ + cos tj⃗ + 2k⃗, so

a) The particle always has a positive velocity in k⃗- direction, hence it is never moving downwards.

b) The k⃗-component of r⃗ is 10 at t = 5.

c) v⃗(t = 5) = − sin 5i⃗ + cos 5j⃗ + 2k⃗.

d) r⃗ = cos 5i⃗ + sin 5j⃗ + 10k⃗ + (t − 5)(− sin 5i⃗ + cos 5j⃗ + 2k⃗).

#22

a) Consider a small time Δt passing, that is a small angle Δα being swept out. The speed will be highest where most coast line will be swept out during that time interval; this is at C, and similarly the speed will be lowest at E.

b) If on the other hand in a small time we sweep out the same area ΔA near any point, then the longest part of coast line will be swept out near E and the shortest near C.

c) Yes, when the light beam is tangent to the shore line, which would happen if you would place the light house on the shore.

d) The speed is not defined at the corners, since the velocity isn’t defined there (direction jumps from pointing say up/down to left/right). Otherwise dA/dt=constant implies that (since the height k of the triangle being swept out is constant also) v = dx/dt is constant via A = kx/2:

c = dA/dt = k/2dx/dt.