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The first (C) is the line segment joining (0, 0) to (1, 1), i.e. x=t,y=t,0stsl. The second (C₂) is the line segment joining (1, 1) to (3, 1), i.e. x=t, y=1, 1sts 3.

The first (C) is the line segment joining (0, 0) to (1, 1), i.e. x=t,y=t,0stsl. The second (C₂) is the line segment joining (1, 1) to (3, 1), i.e. x=t, y=1, 1sts 3.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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How did he get x=t but when I did it I got x=1+2t
Line Integral Along a Set of Discrete Curves Sometimes a path consists of a number of line segments or discrete curves, i.e. C = C₁ + C₂+...+ C. If this is the case, then you should evaluate the line integrals along each discrete section and then add the results to get the final answer. Example. Evaluate f(2x - y)dx + (3x + 2y)dy, where C is the path consisting of the straight line segments from (0, 0) to (1, 1) to (3, 1). x=(1-6) (0,0) + t (1₁₁) t, t Here C consists of two line segments: The first (C) is the line segment joining (0, 0) to (1, 1), i.e. x=t,y=t,0stsl. The second (C₂) is the line segment joining (1, 1) to (3, 1), i.e. x=t, y=1, 1sts 3. for Curve C₂, x = (1-t) (1₁1) + +(3,1) = (1-t) + 3t, (1-t) + t = 1+2t, 1
Transcribed Image Text:Line Integral Along a Set of Discrete Curves Sometimes a path consists of a number of line segments or discrete curves, i.e. C = C₁ + C₂+...+ C. If this is the case, then you should evaluate the line integrals along each discrete section and then add the results to get the final answer. Example. Evaluate f(2x - y)dx + (3x + 2y)dy, where C is the path consisting of the straight line segments from (0, 0) to (1, 1) to (3, 1). x=(1-6) (0,0) + t (1₁₁) t, t Here C consists of two line segments: The first (C) is the line segment joining (0, 0) to (1, 1), i.e. x=t,y=t,0stsl. The second (C₂) is the line segment joining (1, 1) to (3, 1), i.e. x=t, y=1, 1sts 3. for Curve C₂, x = (1-t) (1₁1) + +(3,1) = (1-t) + 3t, (1-t) + t = 1+2t, 1
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