2 2 a 7.2 a Solutions Manual Applied Numerical Met
0.5 1
1.5 2
Therefore, β 4 = −2.4733 and α 4 = e 2.2682 = 9.661786, and the fit is x xe y 4733 . 2 661786 . 9 − = This equation can be plotted together with the data: 1 20.5 1
1.5 2
12.9 The data can be transformed, plotted and fit with a straight line v, ms F, N ln v ln F 10 25 2.302585 3.218876 20 70 2.995732 4.248495 30 380 3.401197 5.940171 40 550 3.688879 6.309918 50 610 3.912023 6.413459 60 1220 4.094345 7.106606 70 830 4.248495 6.721426 80 1450 4.382027 7.279319 y = 1.9842x - 1.2941 R 2 = 0.9481 2 3 4 5 6 7 8 2 2.5 3 3.5 4 4.5 125 The least-squares fit is 2941 . 1 ln 9842 . 1 ln − = x y The exponent is 1.9842 and the leading coefficient is e −1.2941 = 0.274137. Therefore, the result is the same as when we used common or base-10 logarithms: 9842 . 1 274137 . x y =12.10 a
The data can be plotted 400 800 1200 1600 2000 10 20 30 The plot indicates that the data is somewhat curvilinear. An exponential model i.e., a semi- log plot is the best choice to linearize the data. This conclusion is based on • A power model does not result in a linear plot • Bacterial decay is known to follow an exponential model • The exponential model by definition will not produce negative values. The exponential fit can be determined as t hrs c CFU100 mL ln c 4 1590 7.371489 8 1320 7.185387 12 1000 6.907755 16 900 6.802395 20 650 6.476972 24 560 6.327937 126 y = -0.0532x + 7.5902 R 2 = 0.9887 6 6.46.8 7.2
7.6 10 20 30 Therefore, the coefficient of the exponent β 1 is −0.0532 and the lead coefficient α 1 is e 7.5902 = 1978.63, and the fit is t e c 0532 . 63 . 1978 − = Consequently the concentration at t = 0 is 1978.63 CFU100 ml. Here is a plot of the fit along with the original data: 400 800 1200 1600 2000 2400 10 20 30 b The time at which the concentration will reach 200 CFU100 mL can be computed as t e 0532 . 63 . 1978 200 − = t 0532 . 63 . 1978 200 ln − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ d 08 . 43 0532 . 63 . 1978 200 ln = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = t12.11 a
The exponential fit can be determined with the base-10 logarithm as t hrs c CFU100 mL log c 4 1590 3.201397 8 1320 3.120574 12 1000 3 16 900 2.954243 127 20 650 2.812913Parts
» 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
Show more