Solution Manual –Applied Numerical Methods With Matlab For Engineers And Scientists Complete Revised Test VERIFIED 100%

Solution Manual –Applied Numerical Methods With Matlab For Engineers And Scientists Complete Revised Test VERIFIED 100% lOMoAR cPSD| gm cd gcd  m t  CHAPTER 1 1.1 You are given the following differential equation with the initial condition, v(t = 0) = 0, dv  g  cd v 2 dt m Multiply both sides by m/cd m dv  cd dt m g  v 2 cd Define a  m dv  a 2  v 2 cd dt Integrate by separation of variables,  dv   cd dt a 2  v 2 m A table of integrals can be consulted to find that dx  1 tanh1 x  a 2  x 2 a a Therefore, the integration yields 1 tanh 1 v  cd t  C a a m If v = 0 at t = 0, then because tanh–1 (0) = 0, the constant of integration C = 0 and the solution is 1 tanh 1 v  cd t a a m This result can then be rearranged to yield   v  tanh   1.2 This is a transient computation. For the period from ending June 1: 1 mg / cd  lOMoAR cPSD| analytical numerical analytical 2.0% 1.0% 0.0% 0 0.5 1 1.5 2 2.5 relative error 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 absolute relative error  100% The error versus step size can be plotted as Thus, halving the step size approximately halves the error. 1.4 (a) The force balance is lOMoAR cPSD| dv  g  c' v dt m Applying Laplace transforms, sV  v(0)  g  c' V s m Solve for V  g  v(0) (1) s(s  c'/ m) s  c'/ m The first term to the right of the equal sign can be evaluated by a partial fraction expansion, g s(s  c'/ m)  A  s B s  c'/ m (2) g s(s  c'/ m)  A(s  c' / m)  Bs s(s  c'/ m) Equating like terms in the numerators yields A  B  0 g  c' A m Therefore, A  mg B   mg c' c' These results can be substituted into Eq. (2), and the result can be substituted back into Eq. (1) to give V  mg / c'  s mg / c'  s  c'/ m v(0) s  c'/ m Applying inverse Laplace transforms yields v  mg  mg e (c'/ m)t  v(0)e (c'/ m)t c' c' or 3 lOMoAR cPSD| v  mg 1  e (c'/ m)t  c' 60 40 20 0 0 4 8 12 v  v(0)e (c'/ m)t  mg 1  e (c'/ m)t  c' 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 (b) The numerical solution can be implemented as v(2)  0   9.81 12.5 (0) 2 19.62   68.1   v(4) 19.62   9.81 12.5 (19.62) 2  6.2087   68.1   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 Note that the analytical solution is included on the plot for comparison. lOMoAR cPSD|  2.4 2.3 2.2 2.1 2 0 0.2 0.4 0.6 0.8 1 1.5 (a) The first two steps are c(0.1) 10  0.2(10)0.1 9.8 Bq/L c(0.2)  9.8  0.2(9.8)0.1 9.604 Bq/L 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 The slope of this line can be estimated as ln(8.1707)  ln(10)  0.20203 1 Thus, the slope is approximately equal to the negative of the decay rate. 1.6 The first two steps yield y(0.5)  0   400 sin 2 (0)  400  0.5  0  0  0.333330.5  0.16667  3   y(1)  0.16667  sin 2 (0.5)  0.3333330.5  0.21841 5 lOMoAR cPSD|       0.8 0.4 0 0 1 2 3 4 5 -0.4 y m The process can be continued to give t y 0 0 0.5 -0.16667 1 -0.21841 1.5 -0.03104 2 0.299793 2.5 0.546537 3 0.558955 3.5 0.402245 4 0.297103 4.5 0.416811 5 0.727927 gm   c  t 1.7 v(t)  (1 e   ) c 9.8(68.1)   12.5  68.1 10 jumper #1: v(t)  12.5 (1 e   )  44.87 m/s 9.8(75)   14  75 t jumper #2: 44.87  (1 e   ) 14 44.87  52.5  52.5e 0.18666 t 0.14533  e 0.18666 t ln 0.14533  ln e 0.18666 t t = 10.33 sec 1.8 Qin = Qout Q1 = Q2 + Q3 lOMoAR cPSD| 30 = 20 + vA3 10 = 5 A3 A3 = 2 m2 1.9  Min   M out  0 1000 1200  MP  50 400  200 1400  200  350  0 Metabolic production = 300 grams 1.10 % body weight  60 4.5  4.5 12  4.5 1.5  IW  60 % Intracellular water body weight = 33 % 4.5  4.5 12  4.5 1.5  IW  60 % body water  100 7.5  7.5  20  7.5  55  TW  100 % Transcellular water of body water = 2.5 %