2.8 3 3.2 4 1.92 2.1 2.2 2.3 Solutions Manual Applied Numerical Met
2.7 2.8
2.9 3
3.1 3.2
3.3 10 20 30 Therefore, the coefficient of the exponent β 5 is −0.0231 and the lead coefficient α 5 is 10 3.2964 = 1978.63, and the fit is t c 0231 . 10 63 . 1978 − = Consequently the concentration at t = 0 is 1978.63 CFU100 ml. b The time at which the concentration will reach 200 CFU100 mL can be computed as t 0231 . 10 63 . 1978 200 − = t 0231 . 63 . 1978 200 log 10 − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ d 08 . 43 0231 . 63 . 1978 200 log 10 = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = t Thus, the results are identical to those obtained with the base- e model. The relationship between β 1 and β 5 can be developed as in t t e 5 1 10 α α − − = Take the natural log of this equation to yield 10 ln 5 1 t t α α − = − or 5 1 302585 . 2 α α = 128 12.12 The power fit can be determined as W kg A m 2 log W log A 70 2.1 1.845098 0.322219 75 2.12 1.875061 0.326336 77 2.15 1.886491 0.332438 80 2.2 1.90309 0.342423 82 2.22 1.913814 0.346353 84 2.23 1.924279 0.348305 87 2.26 1.939519 0.354108 90 2.3 1.954243 0.361728 logA = 0.3799logW - 0.3821 R 2 = 0.97110.31 0.32
0.33 0.34
0.35 0.36
0.37 1.8 1.841.88 1.92
1.96 Therefore, the power is b = 0.3799 and the lead coefficient is a = 10 −0.3821 = 0.4149, and the fit is 3799 . 4149 . W A = Here is a plot of the fit along with the original data:2.05 2.1
2.15 2.2
2.25 2.3
2.35 70 75 80 85 90 The value of the surface area for a 95-kg person can be estimated as 2 3799 . m 34 . 2 95 4149 . = = A 129 12.13 The power fit can be determined as Mass kg Metabolism kCalday log Mass log Met 300 5600 2.477121 3.748188 70 1700 1.845098 3.230449 60 1100 1.778151 3.041393 2 100 0.30103 2 0.3 30 -0.52288 1.477121 logMet = 0.7497logMass + 1.818 R 2 = 0.9935 1 2 3 4 -1 1 2 3 Therefore, the power is b = 0.7497 and the lead coefficient is a = 10 1.818 = 65.768, and the fit is 7497 . Mass 768 . 65 Metabolism = Here is a plot of the fit along with the original data: 2000 4000 6000 8000 100 200 300 400 12.14 Linear regression of the log transformed data yields σ ε log 6363 . 2 log 41 . 5 log + − = B r 2 = 0.9997 130 -3.6 -3.2 -2.8 -2.40.7 0.8
0.9 1
1.1 1.2
Parts
» Solutions Manual Applied Numerical Met
» a The first two steps are 2.3 The first two steps yield a a
» The true value is ln3 = 1.098612
» function root = bisectnewfunc,xl,xu,Ead function root = falseposfunc,xl,xu,es,maxit
» a The graph can be generated with MATLAB
» a The function can be set up for fixed-point iteration by solving it for x in two different
» a a Solutions Manual Applied Numerical Met
» a The formula for Newton-Raphson is
» a The formula for Newton-Raphson is a The formula for Newton-Raphson is
» a Terms can be combined to yield The mass balances can be written as
» The problem can be written in matrix form as The problem can be written in matrix form as a
» a Prob. 8.3: a The equations can be expressed in a format that is compatible with graphing x
» Equation 9.6 is The matrix to be evaluated is The LU decomposition can be computed as
» The system can be written in matrix form as The following solution is generated with MATLAB.
» a The first iteration can be implemented as
» The first iteration can be implemented as The first iteration can be implemented as
» a 2 Solutions Manual Applied Numerical Met
» 2 2 a 7.2 a Solutions Manual Applied Numerical Met
» 2.8 3 3.2 4 1.92 2.1 2.2 2.3 Solutions Manual Applied Numerical Met
» 1.2 1.8 Solutions Manual Applied Numerical Met
» a a a Solutions Manual Applied Numerical Met
» a The simultaneous equations for the clamped spline with zero end slopes can be set up as
» 8.4 8.8 Solutions Manual Applied Numerical Met
» a The analytical solution can be used to compute values at times over the range. For
» Note that students can take two approaches to developing this M-file. The first program
» a Solutions Manual Applied Numerical Met
» a a The explicit Euler can be written for this problem as
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