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# [Solution Manual] - Applied Numerical Methods with MATLAB for Engineers and Scientists - Chapra.pdf

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Solutions Manual to accompany Applied Numerical Methods With MATLAB for Engineers and Scientists Steven C. Chapra Tufts University CHAPTER 1 1.1 You are given the following differential equation with the initial condition, v(t = 0) = 0, dv c = g − d v2 dt m Multiply both sides by m/cd m dv m = g − v2 c d dt c d Define a = mg / c d m dv = a2 − v2 c
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## Calculus

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Solutions Manual to accompany Applied Numerical Methods With MATLAB for Engineers and Scientists Steven C. Chapra Tufts University   1 CHAPTER 1 1.1 You are given the following differential equation with the initial condition, v ( t   = 0) = 0,   2 vmcgdt dv  d  −=  Multiply both sides by m / c d    2 vgcmdt dvcm d d  −=  Define d  cmga / =   22 vadt dvcm d  −=  Integrate by separation of variables, dt mcvadv  d  ∫∫  =− 22  A table of integrals can be consulted to find that a xa xadx 122 tanh1  − =− ∫  Therefore, the integration yields C t mcava d  += − 1 tanh1 If v  = 0 at t   = 0, then because tanh  –1 (0) = 0, the constant of integration C   = 0 and the solution is t mcava d  = − 1 tanh1 This result can then be rearranged to yield ⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ =  t mgccgmv  d d  tanh 1.2  This is a transient computation. For the period from ending June 1:   2Balance = 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 WithdrawalBalance 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 %100analyticalnumericalanalytical error relativeabsolute  ×−=  The error versus step size can be plotted as 0.0%1.0%2.0%0 0.5 1 1.5 2 2.5relative error   Thus, halving the step size approximately halves the error. 1.4   (a)  The force balance is   3 vmcgdt dv ' −=  Applying Laplace transforms, V mcsgvsV  ')0(  −=−  Solve for mcsvmcss gV  /')0()/'(  +++=  (1) The first term to the right of the equal sign can be evaluated by a partial fraction expansion, mcs Bs Amcss g /')/'(  ++=+  (2) )/'( )/'( )/'(  mcss  Bsmcs A mcss g +++=+  Equating like terms in the numerators yields  Amcg B A '0 ==+  Therefore, ' '  cmg Bcmg A  −==  These results can be substituted into Eq. (2), and the result can be substituted back into Eq. (1) to give mcsvmcscmgscmgV  /')0(/''/'/ +++−=  Applying inverse Laplace transforms yields t mct mc evecmgcmgv )/'()/'( )0('' −− +−=  or
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