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Consider F and C below. F(x, y) = x2 i + y² j C is the arc of the parabola y = 5x2 from (1, 5) to (2, 20) Exercise (a) Find a function f such that F = Vf. Step 1 If Vf(x, y) = F(x, y) = x²i + y²j, then f(x, y) = and f,(x, y) = y2 y? Step 2 Since f(x, y) = x², find f(x, y). f(x, y) = 3 + g(y) Step 3 Now, we know the following. fy(x, y) = Step 4 Since we must have f,(x, y) = y² = g'(y), then v° g(y) = 3 y3 Substituting this back into f(x, y), we have f(x, y) = 3 3 3 Exercise (b) Use part (a) to evaluate Vf · dr along the given curve C. Step 1 Since we want a specific potential function, we will choose K = 0, and so f(x, y) = | 1/3 . 1/3 V We know that if the curve C is given by r(t) with a sts b, then F. dr = f(r(b)) – f(r(a)). Step 2 Our curve C is the arc of the parabola y = 5x2 that begins at (1, 5) and ends at (2, 20). Therefore, we can conclude the following. F. dr = f(2, 20) – f(1, 5) - ( -2456 Submit Skip (you cannot come back)

Consider F and C below. F(x, y) = x2 i + y² j C is the arc of the parabola y = 5x2 from (1, 5) to (2, 20) Exercise (a) Find a function f such that F = Vf. Step 1 If Vf(x, y) = F(x, y) = x²i + y²j, then f(x, y) = and f,(x, y) = y2 y? Step 2 Since f(x, y) = x², find f(x, y). f(x, y) = 3 + g(y) Step 3 Now, we know the following. fy(x, y) = Step 4 Since we must have f,(x, y) = y² = g'(y), then v° g(y) = 3 y3 Substituting this back into f(x, y), we have f(x, y) = 3 3 3 Exercise (b) Use part (a) to evaluate Vf · dr along the given curve C. Step 1 Since we want a specific potential function, we will choose K = 0, and so f(x, y) = | 1/3 . 1/3 V We know that if the curve C is given by r(t) with a sts b, then F. dr = f(r(b)) – f(r(a)). Step 2 Our curve C is the arc of the parabola y = 5x2 that begins at (1, 5) and ends at (2, 20). Therefore, we can conclude the following. F. dr = f(2, 20) – f(1, 5) - ( -2456 Submit Skip (you cannot come back)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Consider F and C below. F(x, y) = x2 i + y² j C is the arc of the parabola y = 5x2 from (1, 5) to (2, 20) Exercise (a) Find a function f such that F = Vf. Step 1 If Vf(x, y) = F(x, y) = x²i + y²j, then f(x, y) = and f,(x, y) = y2 y? Step 2 Since f(x, y) = x², find f(x, y). f(x, y) = 3 + g(y) Step 3 Now, we know the following. fy(x, y) = Step 4 Since we must have f,(x, y) = y² = g'(y), then v° g(y) = 3 y3 Substituting this back into f(x, y), we have f(x, y) = 3 3 3 Exercise (b) Use part (a) to evaluate Vf · dr along the given curve C. Step 1 Since we want a specific potential function, we will choose K = 0, and so f(x, y) = | 1/3 . 1/3 V We know that if the curve C is given by r(t) with a sts b, then F. dr = f(r(b)) – f(r(a)). Step 2 Our curve C is the arc of the parabola y = 5x2 that begins at (1, 5) and ends at (2, 20). Therefore, we can conclude the following. F. dr = f(2, 20) – f(1, 5) - ( -2456 Submit Skip (you cannot come back)
Transcribed Image Text:Consider F and C below. F(x, y) = x2 i + y² j C is the arc of the parabola y = 5x2 from (1, 5) to (2, 20) Exercise (a) Find a function f such that F = Vf. Step 1 If Vf(x, y) = F(x, y) = x²i + y²j, then f(x, y) = and f,(x, y) = y2 y? Step 2 Since f(x, y) = x², find f(x, y). f(x, y) = 3 + g(y) Step 3 Now, we know the following. fy(x, y) = Step 4 Since we must have f,(x, y) = y² = g'(y), then v° g(y) = 3 y3 Substituting this back into f(x, y), we have f(x, y) = 3 3 3 Exercise (b) Use part (a) to evaluate Vf · dr along the given curve C. Step 1 Since we want a specific potential function, we will choose K = 0, and so f(x, y) = | 1/3 . 1/3 V We know that if the curve C is given by r(t) with a sts b, then F. dr = f(r(b)) – f(r(a)). Step 2 Our curve C is the arc of the parabola y = 5x2 that begins at (1, 5) and ends at (2, 20). Therefore, we can conclude the following. F. dr = f(2, 20) – f(1, 5) - ( -2456 Submit Skip (you cannot come back)
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