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Evaluate ſ y?dx + x dy along the following paths. EXAMPLE 4 (a) C = C, is the line segment from (-13, -7) to (o, 6) (b) C = C2 is the arc of the parabola x = 36 - y from (-13, -7) to (0, 6). SOLUTION (a) A parametric representation for the line segment is Video Example ) x = 13t - 13 y = -13t –7 Ostsi Then dx = 13dt, dy = 13 dt and the formulas for line integrals give Sy?dx + xdy = )?(13dt) + (13t - 13)(13dt) = 13 - 169t + 36)dt = 13 (b) Since the parabola is given as a function of y, let's take y as the parameter and write C, as x = 36 - y2 y = y -7 sys6 Then dx = -2y dy and so we have Sy'dx + xdy = )dy + (36 - y?)dy - y? + 36)dy

Evaluate ſ y?dx + x dy along the following paths. EXAMPLE 4 (a) C = C, is the line segment from (-13, -7) to (o, 6) (b) C = C2 is the arc of the parabola x = 36 - y from (-13, -7) to (0, 6). SOLUTION (a) A parametric representation for the line segment is Video Example ) x = 13t - 13 y = -13t –7 Ostsi Then dx = 13dt, dy = 13 dt and the formulas for line integrals give Sy?dx + xdy = )?(13dt) + (13t - 13)(13dt) = 13 - 169t + 36)dt = 13 (b) Since the parabola is given as a function of y, let's take y as the parameter and write C, as x = 36 - y2 y = y -7 sys6 Then dx = -2y dy and so we have Sy'dx + xdy = )dy + (36 - y?)dy - y? + 36)dy

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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Evaluate ſ y?dx + x dy along the following paths. EXAMPLE 4 (a) C = C, is the line segment from (-13, -7) to (o, 6) (b) C = C2 is the arc of the parabola x = 36 - y from (-13, -7) to (0, 6). SOLUTION (a) A parametric representation for the line segment is Video Example ) x = 13t - 13 y = -13t –7 Ostsi Then dx = 13dt, dy = 13 dt and the formulas for line integrals give Sy?dx + xdy = )?(13dt) + (13t - 13)(13dt) = 13 - 169t + 36)dt = 13 (b) Since the parabola is given as a function of y, let's take y as the parameter and write C, as x = 36 - y2 y = y -7 sys6 Then dx = -2y dy and so we have Sy'dx + xdy = )dy + (36 - y?)dy - y? + 36)dy
Transcribed Image Text:Evaluate ſ y?dx + x dy along the following paths. EXAMPLE 4 (a) C = C, is the line segment from (-13, -7) to (o, 6) (b) C = C2 is the arc of the parabola x = 36 - y from (-13, -7) to (0, 6). SOLUTION (a) A parametric representation for the line segment is Video Example ) x = 13t - 13 y = -13t –7 Ostsi Then dx = 13dt, dy = 13 dt and the formulas for line integrals give Sy?dx + xdy = )?(13dt) + (13t - 13)(13dt) = 13 - 169t + 36)dt = 13 (b) Since the parabola is given as a function of y, let's take y as the parameter and write C, as x = 36 - y2 y = y -7 sys6 Then dx = -2y dy and so we have Sy'dx + xdy = )dy + (36 - y?)dy - y? + 36)dy
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