Math 21a Practice Final Exam AnsersPART A1. y xNote: No points are marked for e.2. a) xyzdz dx dyzx yyy ( )( )///1101 22 2 1 22 1 2− −−∫∫∫. b) r z dz d rdrr20104012 1 2cos sin( )//θ θ θ−π∫∫∫. c) ρ φ φ θ θ θ φ φ ρ ρ3 20402301cos sin cos sin sin//d d dππ∫∫∫( ).3. a) (2, 2, 1) is one such vector. The vector (2, 7, 0) is another. b) -7 x + 2y + 10 z = 0. c) L(x, y, z) = 6 + 19(-7 x + 2 y + 10 z). d) 59 3.4. 196.5. The square of the norm of the gradient is 164x2 + 14y2 + 4 z2. This has its maximum on the boundary at (0, 0, ±1).6. One such equation is t → (1 + 3t, -2t, t).7. 215 y ad a xc bc b8. a) A; if ∇∇∇∇u = 0, then u is constant and uxx and uyy are both zero. b) S; for example u = y2/2 solves the equation as does u = x2/2. c) A; At a global maximum, uxx + uyy ≤ 0. d) N; the origin would then be a local maximum so uxx + uyy ≤ 0 there. e) S; this is the case for 14(x2 + y2), but not the case for 14(x2 + y2) – 3000 x.PART B9. a) The flux through B is bigger. Here is why: First, ∫∫A F•n dS + ∫∫B F•n dS = ∫∫∫V div(F) dV where V is the region between A and B with n being the outward pointingnormal from V. Thus, n•(0, 0, 1) > 0 on B and n•(0, 0, 1) < 0 on A. Meanwhile div(F) = 2+ 2 z > 0 on V. Therefore, the A flux in question (computed with –n) must be less than theB flux computed with n. b) The flux of curl(F) is the same through both since they both have the same boundary (useStokes’ theorem).c) Since the disk where z = 0 and x2 + y2 ≤ 1 has the same boundary as the surface underconsideration, one can just as well compute the flux through this disk. The normal to thisdisk is (0, 0, 1) and curl(F) where z = 0 is (y, x, 0) and so the flux of curl(F) is zero.10. a) Parametrize the surface by (u, v) where 0 ≤ u ≤ 1 and 0 ≤ v ≤ 2π according to the rule thatsends (u, v) → (u, u/2 cos(v), u/2 sin(v)). Then, the surface area is 540102udu dv∫∫π. b) −∫∫π1420102u du dv11. a) v = (z, -z, 0) b) Not possible since a unit radius disk, S, in this plane has normal (1, 0, 0) whose dotproduct with (1, 1, 0) is 1. Thus, the flux of the hypothetical vector field through S wouldequal π and so, by Stokes’ theorem, its line integral over the boundary circle would not bezero. c) v = (2x, 0, 0). d) Not possible by the divergence theorem: The flux through any closed surface would equaltwice the volume of the enclosed region.12. a) By Green’s theorem, v = (0, 13x3). b) 13402cos tdtπ∫ = 14π.PART C9. a) µ = 56 and σ = 16.8. b) 806280807 3mmm m=−∑(. ) (. ). c) 116 856 33 62262π− −∞∫.exp( ( ) / . )x dx10. a) 2627. b) 154. c) 19.11. a) (12.5)k 1k!e-12.5. b) Since µ = 25 and σ = 25, this is 3/5 standard deviations from the mean. c) The probability of 40 deaths in one 4-year span is (25)40 140!e−25. The probability of 40deaths in two consecutive four year spans is the square of this last number.12. a) 10.3 percent. b) Sensitivity = ..013103 ~ 0.13. PV+ = ..0131 = 0.13. c) 0.1.Part D9. a) 16125. b) 61125. c) 5. This is the mean for the probability density on the positive integers that assigns to the integer n the probability, 45115( )−n, of Bernoulli going to jail on the n’th day d) 80369.10. Since G(x) = G(x)1, the assertion is true for n = 1. Now, suppose that the assertion holds forall integers between 1 and n – 1. To show that it then holds for integer n, use the fact that G(x+ y) = G(x)G(y) to write G(nx) = G(x + (n-1)x) = G(x)G((n-1)x). Next, invoke theinduction hypothesis that G((n-1)x) = G(x)n-1 to equate G(nx) with G(x)G(x)n-1 which is thedesired G(x)n.11. a) The program must write each two element set once and only once.b) Here is an algorithm: For each positive integer in turn starting with 2, list that integer witheach of the positive integers that are strictly less than it. Also, number the whole listconsecutively from 1.For those interested in a more ‘computer language’ algorithm, consider thefollowing translation:Step 1: Let m = 2.Step 2: Let k = 1.Step 3: Let n = 1.Step 4: Print ‘k → (n, m)’.Step 5: Let k = k + 1.Step 6: If n = m – 1, go to Step 9.Step 7: Let n = n + 1.Step 8: Go to Step 4.Step 9: Let m = m + 1.Step 10: Go to Step 3.Here are the first six outputs:1 → {1, 2}, 2 → {1, 3}, 3 → {2, 3}, 4 → {1, 4}, 5 → {2, 4}, 6 → {3, 4}. c) For each integer N, let AN denote the collection of n-element sets of distinct, positiveintegers with largest integer N. Note that AN is a set with ( )!( )!( )!Nn N n−− −11 elements. Thisunderstood, the collection = {AN}1≤N<∞ is the collection of all n-element sets of distinctpositive integers, here exhibited as a countable union of finite sets. Thus, assuming theclaim, the collection is itself countable.12. a) Since the integral of f over 2 must equal 1, c = 118. b) 1270. c)
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