Mathematics > SOLUTIONS MANUAL > University of Texas M 427L HW 10- Solutions: huynh (lth436) – HW10 – gilbert – (56690) (Homewo (All)

University of Texas M 427L HW 10- Solutions: huynh (lth436) – HW10 – gilbert – (56690) (Homework 10 Solutions)

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huynh (lth436) – HW10 – gilbert – (56690) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. ... 001 10.0 points If f(x, y) is a potential function for the evaluate 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 + D0(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 . 002 10.0 points Which vector field F has graph We determine F(x, y) by looking at a few points on the graph. Now on the x-axis, F(x, 0) = −x j , This study source was downloaded by 100000806173026 from CourseHero.com on 01-03-2022 06:59:49 GMT -06:00 https://www.coursehero.com/file/9179361/HW10-solutions/huynh (lth436) – HW10 – gilbert – (56690) 2 while on the y-axis, F(0, y) = y i . The only one of the given vector fields satisfying these conditions is 003 10.0 points Which vector field F has graph 6. F(x, y = x i − y j correct Explanation: We determine F(x, y) by looking at a few points on the graph. Now on the x-axis, F(x, 0) = x i , while on the y-axis, The only one of the given vector fields satisfying these conditions is . 004 10.0 points Which vector field F has graph Explanation: We determine F(x, y) by looking at a few points on the graph. Now on the x-axis, F(x, 0) = x i , while on the y-axis, F(0, y) = y j . The only one of the given vector fields satisfying these conditions is F(x, y) = x i + y j . This study source was downloaded by 100000806173026 from CourseHero.com on 01-03-2022 06:59:49 GMT -06:00 https://www.coursehero.com/file/9179361/HW10-solutions/huynh (lth436) – HW10 – gilbert – (56690) 3 005 10.0 points Which vector field F has graph Explanation: We determine F(x, y) by looking at a few points on the graph. Now on the x-axis, F(x, 0) = 0 i+? j , while on the y-axis, F(0, y) = y i + const j . The only one of the given vector fields satisfying these conditions is F(x, y) = y i + cos x j . 006 10.0 points Which vector field F has graph We determine F(x, y) by looking at a few points on the graph. Now on the x-axis, The only one of the given vector fields satisfying these conditions is F(x, y) = 3ey i + (3xey − 2) 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 06:59:49 GMT -06:00 https://www.coursehero.com/file/9179361/HW10-solutions/huynh (lth436) – HW10 – gilbert – (56690) 4 along the curve C given parametrically by Explanation: Since F is a gradient vector field, the Fundamental Theorem for Line Integrals says that I = Z01 F(c(t)) · c0(t) dt = f(c(1)) − f(c(0)) for any function f such that F = ∇f. To write F(x, y) = 3ey i + (3xey − 2) j as a gradient vector field, we have to find a function f(x, y) such that ∂f ∂x = 3ey, ∂f ∂y = 3xey − 2. Now the first equation says that f(x, y) = 3xey + D(y) , for some arbitrary function D(y), where by the second equation ∂f ∂y = 3xey + D0(y) = 3xey − 2 , i.e., D0(y) = −2. Thus D(y) = −2y + K, so f(x, y) = 3xey − 2y + K , for some arbitrary constant K. On the other hand, when c(t) = tet i + (1 + t) j , 0 ≤ t ≤ 1 , then 008 10.0 points A gradient vector field F = ∇f and points P, Q, . . . , U on contour lines of z = f(x, y) are shown in [Show More]

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