exam2_sol - XZ + R 2 = 0 Z, horizontal: cos(60°)F XZ + F...

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CE 335 Exam 2 Solutions 1] (a) Setting f(x) = sin(x) - x 2 , we find that f(0.88) = -0.004 < 0 while f(0.872) = +0.005 > 0. So [0.872 0.88] is a bracketing interval, and since f(x) is continuous, there is a root in that interval. (b) Iteration 1: c = (0.88 + 0.872) / 2 = 0.876, f(c) = +0.0008 > 0 new bracket: [0.876 0.88] Iteration 2: c = (0.876 + 0.88) / 2 = 0.878, f(c) = -0.001 < 0 new bracket: [0.876 0.878] Iteration 3: c = (0.876 + 0.878) / 2 = 0.877, f(c) = -0.0003 < 0 new bracket: [0.876 0.877] So the root is at 0.8765 ± 0.0005 2] The minimum is near 1 (f(1) = -3 while f(0) = 0, f(2) = 50). The Newton's method iteration is x ← x - f'(x)/f''(x) = x - (6x 5 - 6x - 1)/(30x 4 - 6) Starting with x = 1, Iteration 1: x = 1.0417 Iteration 2: x = 1.0380 Iteration 3: x = 1.0379 So x = 1.0379 would be an estimate of the minimum. 3] (a) X, vertical: sin(60°)F XY + sin(60°)F XZ = -500 N X, horizontal: cos(60°)F XY - cos(60°)F XZ = 0 Y, vertical: sin(60°)F XY + R 1 = 0 Y, horizontal: cos(60°)F XY + F YZ = 0 Z, vertical: sin(60°)F
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Unformatted text preview: XZ + R 2 = 0 Z, horizontal: cos(60°)F XZ + F YZ + R 3 = 0 Letting the vector of unknowns be [F XY F XZ F YZ R 1 R 2 R 3 ] T , the coefficient matrix is sin(60°) sin(60°) cos(60°)-cos(60°) sin(60°) 1 cos(60°) 1 sin(60°) 1 cos(60°) 1 1 with right-hand side [-500 0 0 0 0 0] T . (b) Because each equation only involves two or three unknown forces, we can solve for them one by one: From the second equation, F XY = F XZ so, from the first equation, F XY = F XZ = -500 / √3 N. From the third equation, R 1 = 250 N From the fifth equation, R 2 = 250 N From the fourth equation, F YZ = 250 / √3 N Finally, from the sixth equation, R 3 = 0 N. 4] The row-sum norm of A is the maximum sum of each row's absolute values, which is 23. Similarly, the row-sum norm of A-1 is 6.125. The row-sum-norm condition number of A (or A-1 ) is the product of these two norms, or 140.875....
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This note was uploaded on 08/30/2011 for the course CGN 3350 taught by Professor Lybas during the Spring '11 term at University of Florida.

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exam2_sol - XZ + R 2 = 0 Z, horizontal: cos(60°)F XZ + F...

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