Applied_Numerical_Methods_with_Matlab_for_engineers_Solutions_Manual

# Applied_Numerical_Methods_with_Matlab_for_engineers_Solutions_Manual

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Solutions Manual to accompany Applied Numerical Methods With MATLAB for Engineers and Scientists Steven C. Chapra Tufts University

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1 CHAPTER 1 1.1 You are given the following differential equation with the initial condition, v ( t = 0) = 0, 2 v m c g dt dv d = Multiply both sides by m / c d 2 v g c m dt dv c m d d = Define d c mg a / = 2 2 v a dt dv c m d = Integrate by separation of variables, dt m c v a dv d = 2 2 A table of integrals can be consulted to find that a x a x a dx 1 2 2 tanh 1 = Therefore, the integration yields C t m c a v a d + = 1 tanh 1 If v = 0 at t = 0, then because tanh –1 (0) = 0, the constant of integration C = 0 and the solution is t m c a v a d = 1 tanh 1 This result can then be rearranged to yield = t m gc c gm v d d tanh 1.2 This is a transient computation. For the period from ending June 1:
2 Balance = Previous Balance + Deposits – Withdrawals Balance = 1512.33 + 220.13 – 327.26 = 1405.20 The balances for the remainder of the periods can be computed in a similar fashion as tabulated below: Date Deposit Withdrawal Balance 1-May \$ 1512.33 \$ 220.13 \$ 327.26 1-Jun \$ 1405.20 \$ 216.80 \$ 378.61 1-Jul \$ 1243.39 \$ 350.25 \$ 106.80 1-Aug \$ 1586.84 \$ 127.31 \$ 450.61 1-Sep \$ 1363.54 1.3 At t = 12 s, the analytical solution is 50.6175 (Example 1.1). The numerical results are: step v(12) absolute relative error 2 51.6008 1.94% 1 51.2008 1.15% 0.5 50.9259 0.61% where the relative error is calculated with % 100 analytical numerical analytical error relative absolute × = The error versus step size can be plotted as 0.0% 1.0% 2.0% 00 . 511 . 522 . 5 relative error Thus, halving the step size approximately halves the error. 1.4 (a) The force balance is

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3 v m c g dt dv ' = Applying Laplace transforms, V m c s g v sV ' ) 0 ( = Solve for m c s v m c s s g V / ' ) 0 ( ) / ' ( + + + = ( 1 ) The first term to the right of the equal sign can be evaluated by a partial fraction expansion, m c s B s A m c s s g / ' ) / ' ( + + = + ( 2 ) ) / ' ( ) / ' ( ) / ' ( m c s s Bs m c s A m c s s g + + + = + Equating like terms in the numerators yields A m c g B A ' 0 = = + Therefore, ' ' c mg B c mg A = = These results can be substituted into Eq. (2), and the result can be substituted back into Eq. (1) to give m c s v m c s c mg s c mg V / ' ) 0 ( / ' ' / ' / + + + = Applying inverse Laplace transforms yields t m c t m c e v e c mg c mg v ) / ' ( ) / ' ( ) 0 ( ' ' + = or
4 () t m c t m c e c mg e v v ) / ' ( ) / ' ( 1 ' ) 0 ( + = where the first term to the right of the equal sign is the general solution and the second is the particular solution. For our case, v (0) = 0, so the final solution is t m c e c mg v ) / ' ( 1 ' = (b) The numerical solution can be implemented as 62 . 19 2 ) 0 ( 1 . 68 5 . 12 81 . 9 0 ) 2 ( = + = v 2087 . 6 2 ) 62 . 19 ( 1 . 68 5 . 12 81 . 9 62 . 19 ) 4 ( = + = v The computation can be continued and the results summarized and plotted as: t v dv / dt 0 0 9.81 2 19.6200 6.2087 4 32.0374 3.9294 6 39.8962 2.4869 8 44.8700 1.5739 10 48.0179 0.9961 12 50.0102 0.6304 0 20 40 60 048 1 2 Note that the analytical solution is included on the plot for comparison.

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5 1.5 (a) The first two steps are Bq/L 8 . 9 1 . 0 ) 10 ( 2 . 0 10 ) 1 . 0 ( = = c Bq/L 604 . 9 1 . 0 ) 8 . 9 ( 2 . 0 8 . 9 ) 2 . 0 ( = = c The process can be continued to yield t c dc / dt 0 10.0000 -2.0000 0.1 9.8000 -1.9600 0.2 9.6040 -1.9208 0.3 9.4119 -1.8824 0.4 9.2237 -1.8447 0.5 9.0392 -1.8078 0.6 8.8584 -1.7717 0.7 8.6813 -1.7363 0.8 8.5076 -1.7015 0.9 8.3375 -1.6675 1 8.1707 -1.6341 (b) The results when plotted on a semi-log plot yields a straight line 2 2.1 2.2 2.3 2.4 0 0.2 0.4 0.6 0.8 1 The slope of this line can be estimated as 20203 . 0
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## This note was uploaded on 11/17/2011 for the course ENGINEEIRI 210 taught by Professor Idunno during the Spring '11 term at University of Toronto- Toronto.

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Applied_Numerical_Methods_with_Matlab_for_engineers_Solutions_Manual

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