hw1_solutions

hw1_solutions - actual theoretical Figure 2: Error vs. #...

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ME 140A, Fall 2007 Homework #1 Solutions October 16, 2007 1. Problem 18.2 a) Analytical solution: I = b x + 1 2 e - 2 x B 4 0 = 4 + 1 2 ( e - 8 - 1 ) = 3 . 50016 b) Single application of trapezoid rule: I = ( b - a ) f ( a ) + f ( b ) 2 with a = 0 , b = 4 f ( a ) = 0 , f ( b ) = 1 - e - 8 gives I = 4 1 - e - 8 2 = 1 . 999329 1
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10 0 10 1 10 2 10 -4 10 -3 10 -2 10 -1 10 0 Slope = -2 Slope = -1.980 actual theoretical Figure 1: Error vs. # segments (n) for Composite trapezoid method c) see attached m-±le problem1c.m d) Single application of Simpson’s 1 3 rule: I = h 3 [ f (0) + 4 f (2) + f (4)] with f (0) = 0 f (2) = 1 - e - 4 f (4) = 1 - e - 8 gives I = 3 . 2843 e) see posted m-±le problem1e.m f) For part c), plotting the logarithm of the error versus the number of segments should result in a straight line with a slope around -2. An example of this is shown in Fig. 1. The composite Simpson rule, shown in Figure 2 should have a slope around -4. 2
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10 0 10 1 10 2 10 -8 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 Slope = -4 Slope = -3.868
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Unformatted text preview: actual theoretical Figure 2: Error vs. # segments (n) for composite Simpson method 2. Problem 17.10 For MATLAB implementation details, see posted m-le problem2.m . a) Using the composite trapezoid rule with n=6 gives F=996.14 and d=13.14. b) Using the composite Simpson rule with n=6 gives F = 1042.29 and d = 12.68. 3. Problem 17.11 Using composite trapezoid rule with unequal segments gives F = 526200 N and d = 158.62 m. See problem3and4.m for implementation details. 4. Problem 17.15 The average value for the data given is -0.8333. See posted le question3and4.m . 5. Problem 17.17 Using the composite trapezoid rule gives Area = 9.05 m 2 Flow rate = .69625e-01 m 3 /s . See posted problem5.m for details. 3...
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This note was uploaded on 11/05/2009 for the course ME 140A taught by Professor Meiburg during the Fall '08 term at UCSB.

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hw1_solutions - actual theoretical Figure 2: Error vs. #...

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