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This is a review sheet for test 2 in math 222, a vector calculus course. It covers topics such as real-valued and vector-valued functions, paths and vector fields, curvature and torsion, gradient and curl, line and surface integrals, and change of variables. It includes hints and explanations for various concepts and problems.
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(1) Real-valued functions look like f : Rn^ → R, while vector-valued functions look like f : Rn^ → Rm, where m > 1. The two main types are paths (c : R → Rm) and vector fields (F : Rm^ → Rm). Look in your notes for how to graph them. (2) By Newton’s second law there is no force acting on it. (3) It is a unit speed path if c′(t) = 1 for all t. In this case we have the tangent vector
T (t) = c′(t), the principal normal vector N (t) =
T ′(t) ||T ′(t)||
, and the binormal vector B(t) = T (t) × N (t). (4) It is not possible to have a nonzero binormal vector if the principal normal vector is 0. Mathematically, this is by definition of the binormal as T × N : if N = 0 then B = 0 as well. Intuitively this is because N (t) tells you how the direction of the tangent vector is changing and B(t) tells you how the path is bending out of the plane of T and N. If N = 0 , then the direction is not changing and so obviously the path isn’t bending at all, especially not out of the plane of T and N. (5) This is the trick that starts with the fact that ||T (t)|| = 1 implies ||T (t)||^2 = 1. (6) The curvature function is given by k(t) = ||T ′(t)|| = ||c′′(t)||. It is a measure of how much the curve is bending in the plane given by T and N (the osculating plane). The torsion function is given by the equation B′(t) = τ (t)N (t), and is a measure of how much the curve is bending out of the osculating plane. (7) The curvature being zero means that the path is a line. The torsion being zero means that the curve lies in a plane. (8) It means there is an f : R^3 → R such that F =
∂f ∂x
∂f ∂y
∂f ∂z
. It is an interesting point that not every vector field is a gradient vector field. Find the example in your notes or just make one up. (9) We must have c : [a, b] ⊆ R → R^3 , and then we have that c′(t) = F(c(t)). The tangent to the path always belongs to the vector field. (10) (a) False. The zero vector field, F = 0 , is a trivial example. (b) False. Any vector field of the form F(x, y) = (F 1 (y), F 2 (x)) has gradient 0 but will usually have nonzero curl. (c) False. Consider ∇f where f (x, y, z) = x^2 + y^2 + z^2.
(d) True. If F = ∇f for some f : R^3 → R, then F =
∂f ∂x
∂f ∂y
∂f ∂z
. Set up the matrix to compute the curl and then use that the mixed partials must be equal. (11) In both methods, you are slicing the region into small pieces that estimate the volume, then refining the slices to get better and better approximations, which in the limit result in the exact volume. But the pieces are very different in the two cases. In Cavalieri’s principle one slices the solid as if it were a loaf of bread, while Riemann sums slice the solid as if it were french fries. (12) Double and triple integrals are defined in terms of Riemann sums, but Fubini’s theorem tells us that they can be computed as iterated integrals, which means we can use one-variable integration techniques to compute them. Moreover, Fubini’s theorem says that we can switch the order of integration if we want to.
(13) You can write
∫ (^) b
a
∫ (^) d
c
∂y
∂f ∂x
dydx =
∫ (^) b
a
∂f ∂x
(x, y)
d
c
dx =
∫ (^) b
a
∂f ∂x
(x, d) −
∂f ∂x
(x, c)dx
= f (x, d) − f (x, c)
∣∣b a =^ f^ (b, d)^ −^ f^ (a, d)^ −^ f^ (b, c) +^ f^ (a, c).
2
(14) We can write the integral as the sum of two integrals, one over the right half of the disk and one over the left half of the disk. The function f (x, y) = x is symmetric and we have f (−x, y) = −x = −f (x, y) for x > 0, so the integral over the left half is equal to minus the integral over the right half. (15) The area of D is given by
D
dA and the volume of W is given by
W
dV. (16) The mean value inequality gives the roughest possible estimates for your integral: you underestimate by taking the minimum m of your function and multiplying it by the area of D; you overestimate by taking the maximum M of your function and multiplying it by the area of D. The pictures are in your notes. (17) A function T : Rn^ → Rn^ is one-to-one if whenever T (w) = T (v) we have that w = v. We need T to be one-to-one except possibly on the boundary for the change of variables theorem. (18) For polar coordinates, T (r, θ) = (r cos θ, r sin θ). Compute the Jacobian determinant your- self: your answer should come out to be r. (19) For cylindrical coordinates, T (r, θ, z) = (r cos θ, r sin θ, z). Compute the Jacobian determi- nant yourself: your answer should come out to be r again. (20) For spherical coordinates, T (ρ, θ, φ) = (ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ). Compute the Ja- cobian for yourself: you should get ρ sin φ. (21) The change of variables formula converts from integrating over a region D to integrating over a region D∗, where T (D) = D∗. Those regions can have wildly different areas. The Jacobian determinant keeps track of the change in area. If it wasn’t there, the area element would be inappropriate for the new coordinates and the value of the integral would be wrong. (22) In this scenario
c
f ds represents the total mass of the wire.
(23) In this scenario
c
F · ds represents the total work done on the bumblebee by F. (24) It’s like the usual, one-variable idea of the integral as area, except the path c might be curved instead of straight. The arclength element ds keeps track of how the curving affects the area. See pages 424 and 425 for pictures. (25) First note that T (t) =
c′(t) ||c′(t)||
. Then ∫
c
F · ds =
∫ (^) b
a
F(c(t)) · c′(t) dt =
∫ (^) b
a
F(c(t)) ·
c′(t) ||c′(t)||
||c′(t)|| dt =
c
F · T ds
(26) The key to this proof is to note that ∇f (c(t)) · c′(t) =
d dt
(f (c(t))). Then ∫ (^) b
a
∇f (c(t)) · c′(t)dt =
d dt
(f (c(t))) dt = f (c(t))
∣∣b a =^ f^ (c(b))^ −^ f^ (c(a)). (27) Just use the last problem because c(b) = c(a) if the curve is closed. (28) Note that p′(t) = −c′(−t). Then ∫
p
F · ds =
∫ (^) −a
−b
F(p(t)) · p′(t) dt = −
∫ (^) −a
−b
F(c(−t)) · c′(−t) dt
Now do a u-substitution: u = −t, du = −dt to get
=
∫ (^) a
b
F(c(u)) · c′(u) du = −
∫ (^) b
a
F(c(t)) · c′(t) dt
Try the same argument with the path integral: you’ll see that the negative sign drops away. It’s not a typo!