Mathematics > SOLUTIONS MANUAL > University of Texas EXAM 03 EXAM03-Solutions. Version 039 – EXAM03 – gilbert – (56690). (All)

Version 039 – EXAM03 – gilbert – (56690) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. ... 001 10.0 points Which of the following is a parametrization of the surface 1. Φ = (u, u cos v, u sin v) correct 2. Φ = (u, v3, v) 3. Φ = (u, u + v, v) 4. Φ = (u, cos v, sin v) 5. Φ = (cos u sin v, 3 cos u sin v, cos v) Explanation: Cross-sections of the surface by the plane x = cperpendicular to the x-axis are circles of varying radius centered on the x-axis, i.e., the graph of y2 + z2 = r2 where the radius r varies with c. The only parametrization with this property is Φ(u, v) = (u, u cos v, u sin v) . 002 10.0 points Find the divergence of the vector field F(x, y, z) = xexy i + 3yexy j + 2xyzexy k . 1. div F = 4(1 + xy)exy 2. div F = (4 − 6xy)exy 3. div F = 4xyexy 4. div F = 6(1 − xy)exy 5. div F = 6xyexy 6. div F = (4 + 6xy)exy correct Explanation: The div of F is given symbolically by div F = ∇ · F = ∂ ∂x(xexy) + 3∂y ∂ (yexy) + 2∂z ∂ (xyzexy) . Now ∂ ∂x(xexy) = exy + xyexy = (1 + xy)exy , while ∂ ∂y(yexy) = exy + xyexy = (1 + xy)exy , and ∂ ∂x(xyzexy) = xyexy . Thus div F = (1 + 3)exy + (1 + 3 + 2)xyexy . Consequently, div F = (4 + 6xy)exy . 003 10.0 points Determine the area element dS for the surface given parametrically by Φ(u, v) = (−6u + 3v, u − 3, −2u + 2v) . 1. dS = 4 dudv This study source was downloaded by 100000806173026 from CourseHero.com on 01-03-2022 07:18:54 GMT -06:00 https://www.coursehero.com/file/9179349/EXAM03-solutions/Version 039 – EXAM03 – gilbert – (56690) 2 2. dS = 7 dudv correct 3. dS = 8 dudv 4. dS = 3 dudv 5. dS = 5 dudv 6. dS = 6 dudv Explanation: The area element for a surface parametrized by Φ(u, v) is dS = kTu × Tvk dudv . But when Φ(u, v) = (−6u + 3v, u − 3, −2u + 2v) , we see that Tu = ∂Φ ∂u = (−6, 1, −2) , while Tv = ∂Φ ∂v = (3, 0, 2) , In this case Tu × Tv = i j k −6 1 −2 3 0 2 = 2 i + 6 j + 3 k , and so, kTu × Tvk = q(2)2 + (6)2 + (3)2 . Consequently, dS = √49 dudv = 7 dudv . 004 10.0 points Evaluate the integral I = ZC y ds when C is parametrized by c(t) = t2 i + t j, 0 ≤ t ≤ √2 . 1. I = 5 6 2. I = 3 2 3. I = 13 6 correct 4. I = 11 6 5. I = 7 6 Explanation: Since kc′(t)k = k2t i + jk = p4t2 + 1 and y = t on C, we see that I = Z √2 0 t kc′(t)k dt = Z √2 0 tp4t2 + 1 dt . This last integral can be evaluated by substitution: set u2 = 4t2 + 1. Then 2u du = 8t dt , 1 4 u du = t dt , so I = 1 4 Z13 u2 du = h 12 1 u3 i3 1 . Consequently, I = 13 6 . 005 10.0 points Use the fact that F = (2xy + 4 cos y) i + (x2 − 4x sin y) j , is a gradient vector field to evaluate the line integral I = ZC F · ds This study source was downloaded by 100000806173026 from CourseHero.com on 01-03-2022 07:18:54 GMT -06:00 https://www.coursehero.com/file/9179349/EXAM03-solutions/Version 039 – EXAM03 – gilbert – (56690) 3 along a smooth curve C from (1, π) to 2, π2 . 1. I = π + 4 correct 2. I = 2π + 4 3. I = π − 4 4. I = 2π − 4 5. I = π + 2 6. I = 2π − 2 Explanation: If f(x, y) is a potential function for the gradient vector field F = (2xy + 4 cos y) i + (x2 − 4x sin y) j , then ∂f ∂x = 2xy + 4 cos y , ∂f ∂y = x2 − 4x sin y . Now by the first equation, f(x, y) = x2y + 4x cos y + D(y) for an arbitrary function D(y), which by the second equation satisfies x2 − 4x sin y + D′(y) = x2 − 4x sin y , i.e., D(y) = K for an arbitrary constant K. Thus f(x, y) = x2y + 4x cos y + K . But then f(1, π) = π − 4 + K , while f 2, π2 = 2π + K . Consequently, f 2, π2 − f(1, π) = π + 4 . 006 10.0 points Evaluate the integral I = ZC (3y dx − x dy) when C is the graph of y = cos x on [0, π/2]. Explanation: When C is the graph of y = cos x, then 3y dx − x dy = (3 cos x + x sin x) dx . Thus I = Z0π/2 (3 cos x + x sin x) dx = I1 + I2. Now I1 = Z0π/2 3 cos x dx = 3h sin x iπ/ 0 2 = 3 . On the other hand, to evaluate I2 we integrate by parts: Consequently, I = I1 + I2 = 4 . This study source was downloaded by 100000806173026 from CourseHero.com on 01-03-2022 07:18:54 GMT -06:00 https://www.coursehero.com/file/9179349/EXAM03-solutions/Version 039 – EXAM03 – gilbert – (56690) 4 007 10.0 points Which vector field F has graph Explanation: We determine F(x, y) by looking at a few The only one of the given vector fields satisfying these conditions is F(x, y) = (x − y) i + (x + y) j . 008 10.0 points Find the work done by the force field to move a particle along the curve given parametrically by 1. Work Done = 11 4 units correct 8 units [Show More]

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