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Course: M 442, Fall 2009
School: Texas A&M
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Word Count: 359

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Assignment M442, 2, due Friday Sept. 14 1. [5 pts] For a pendulum under the inuence of linearly modeled air resistance, the angle y(t) solves the equation g y (t) = sin y by , l where b is a coecient of air resistance. Taking l = 1, g = 9.81, and b = .1, solve this equation in MATLAB with initial conditions y(0) = /4 and y (0) = 0 for a time interval [0, 10]. Turn in a plot of y(t). 2. [5 pts] For the equation...

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Assignment M442, 2, due Friday Sept. 14 1. [5 pts] For a pendulum under the inuence of linearly modeled air resistance, the angle y(t) solves the equation g y (t) = sin y by , l where b is a coecient of air resistance. Taking l = 1, g = 9.81, and b = .1, solve this equation in MATLAB with initial conditions y(0) = /4 and y (0) = 0 for a time interval [0, 10]. Turn in a plot of y(t). 2. [5 pts] For the equation from Problem 1, use event location to determine a sequence of times at which y(t) is 0. Counting the second zero as the rst (since we are not starting at zero), plot the time between these zeros as a function of the number of zeros. Does the plot make sense? 3. [5 pts] Use MATLAB to solve the boundary value problem y (x) 2y (x) + y(x) = x2 y(0) = 2; y(1) = 1. Turn in both your MATLAB code and a plot of y(x). 4a. [5 pts] The logistic equation p dp = rp(1 ); dt K is solved by pexact (t) = p(0) = p0 , p0 K . p0 + (K p0 )ert For parameter values r = .0215, K = 446.1825, and for value initial p0 = 7.7498 (consistent with the U. S. population in millions), solve the logistic equation numerically for t [0, 200], and compute the error 200 E= t=1 (p(t) pexact (t))2 . 4b. Repeat Problem 4a with RelTol = 1010 . 5. [10 pts] Write a MATLAB function M-le that takes as input a partition P = [x0 , x1 , . . . , xn ] (i.e., a vector) and employs the Taylor method of order 2 to solve the dierential equation dy = sin(xy); dx y(0) = , on the interval [0, 1]. Use your M-le to solve the equation with P =linspace(0,1,10) and with P =linspace(0,1,25). Turn in both your M-le and a lis...

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Texas A&M - M - 442
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31 .001309 .043208 .004177 .040258 .001487 .041342 .004515 .044896 .010865 .069105 .001759 .053671 .002594 .046613 .004950 .045492 .007115 .053634 .003186 .046923 .002093 .047043 .005202 .042955 .004489 .039754 .003642 .050224 .003960
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boy = 0.250000000000000000 Number of quadrature points = 21 gamma = -0.693147180559945286 eta= 0.100000000000000006 nu = 0.299999999999999989 alpha = 0.490907905324351579 beta= 0.509092094675648421 kappa0= -1 M2 = 18 NJ= 3
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = -2.30258509299404590 eta= 0.299999999999999989 nu = 0.299999999999999989 alpha = 0.472782819701179080 beta= 0.527217180298820920 kappa0= -1 M2 = 18 NJ= 30
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = 0.693147180559945286 eta= 0.299999999999999989 nu = 0.299999999999999989 alpha = 0.472782819701179080 beta= 0.527217180298820920 kappa0= -1 M2 = 18 NJ= 30
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = 2.30258509299404546 eta= 0.100000000000000006 nu = 0.299999999999999989 alpha = 0.490907905324351579 beta= 0.509092094675648421 kappa0= -1 M2 = 18 NJ= 30
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = -2.30258509299404590 eta= 0.000000000000000000E+00 nu = 0.299999999999999989 alpha = 0.500000000000000000 beta= 0.500000000000000000 kappa0= -1 M2 = 18 NJ
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = 2.30258509299404546 eta= 0.000000000000000000E+00 nu = 0.299999999999999989 alpha = 0.500000000000000000 beta= 0.500000000000000000 kappa0= -1 M2 = 18 NJ=
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Michigan State University - PHY - 252
APPENDIX A: DEALING WITH UNCERTAINTY1. OVERVIEW An uncertainty is always a positive number x &gt; 0. If the uncertainty of x is 5%, then x = .05x. If the uncertainty in x is x, then the fractional uncertainty in x is x/x. If you measure x with a de
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