14 a Let a t be the amount of pollutant and the concentration c t is the

14 a let a t be the amount of pollutant and the

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14. a. Let a ( t ) be the amount of pollutant, and the concentration c ( t ) is the concentration of pollutant (in ppb). The change in amount = the amount entering - the amount leaving. The change in amount, a 0 ( t ), has units (mass/day). The amount entering is f 1 Q 1 + f 2 Q 2 = 4000 · 18+2500 · 4 = 82 , 000 ppb · m 3 /day (mass/day), while the amount leaving is ( f 1 + f 2 ) c ( t ) = 6500 c ( t ) ppb · m 3 /day (mass/day). Thus, the differential equation for the change in amount is da ( t ) dt = 82 , 000 - 6500 c ( t ) . The relation between the amount and concentration is c ( t ) = a ( t ) V = a ( t ) 3000000 and c 0 ( t ) = a ( t ) 3000000 , so the concentration differential equation: dc ( t ) dt = 82 , 000 - 6500 c ( t ) 3 , 000 , 000 - 13 6000 c - 164 13 ≈ - 0 . 0021667( c - 12 . 615) .
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With the initial condition c (0) = 0, we make the substitution z ( t ) = c ( t ) - 12 . 615, so z (0) = - 12 . 615 and the differential equation is dz dt = - 0 . 0021667 z, z (0) = - 12 . 615 , which gives z ( t ) = - 12 . 615 e - 0 . 0021667 t = c ( t ) - 12 . 615. Thus, c ( t ) = 12 . 615 1 - e - 0 . 0021667 t . b. We solve c ( t ) = 12 . 615 ( 1 - e - 0 . 0021667 t ) = 4, so e 0 . 0021667 t = 12 . 615 8 . 615 = 1 . 4643. Thus, t = ln(1 . 4643) 0 . 0021667 176 . 01 days. Hence, the concentration reaches 4 ppb at t 176 . 01 days. The limiting concentration is lim t →∞ c ( t ) = 12 . 615 ppb . This easily follows because the exponential tends to zero for large t . 15. a. The DE dP dt = ( b - at ) P is both a linear and a separable equation. From our linear techniques, we write dP dt + ( at - b ) P = 0, which has an integrating factor μ ( t ) = exp Z ( at - b ) dt = e at 2 2 - bt . Thus, d dt e at 2 2 - bt P ( t ) = 0 or P ( t ) = Ce bt - at 2 2 . With the initial condition, P (0) = 6 . 94 = C , so P ( t ) = 6 . 94 e bt - 1 2 at 2 . From the data, P (20) = 7 . 47 and P (40) = 7 . 72. 7 . 47 = 6 . 94 e 20 b - 200 a and 7 . 72 = 6 . 94 e 40 b - 800 a . Taking logarithms, ln 7 . 47 6 . 94 = 20 b - 200 a and ln 7 . 72 6 . 94 = 40 b - 800 a. Taking the 2 nd equation times 0.5 and adding to the first gives 200 a = ln 7 . 47 6 . 94 - 0 . 5 ln 7 . 72 6 . 94 a = 0 . 005 ln 7 . 47 7 . 72 × 6 . 94 0 . 000101685 . It follows that b = 0 . 0046965. b. The year 2000 is t = 50. P (50) = 6 . 94 e 0 . 0046965(50) - 0 . 000050843(50) 2 7 . 7293 million. The actual census data is 8.13 million, so the percent error is 100 (7 . 73 - 8 . 13) 8 . 13 % = - 4 . 93% from the actual census data.
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c. The maximum population occurs when dP dt = ( b - at ) P = 0, so t = b a 46 . 19 yr. This gives P max = 6 . 94 e 0 . 004697(46 . 19) - 0 . 00005084(46 . 19) 2 7 . 735 . Therefore, according to the model, the maximum population occurs in about 1996, and is about 7.735 million. This is clearly not correct based on the census data in 2000. 17. The electric circuit with R = 10Ω, C = 0 . 1F, and E ( t ) = 40V is a linear differential equation: 10 dQ dt + 1 0 . 1 Q = 40 or dQ dt + Q = 4 , which has an integrating factor μ ( t ) = e t . Thus, d dt e t Q = 4 e t , so e t Q ( t ) = 4 e t + C. With the initial condition, Q (0) = 0, the solution becomes Q ( t ) = 4 - 4 e - t . 18. Programs are provided to solve this problem along with solutions expected in the written work. a. The least squares best fit to the data is found with MatLab. The sum of square errors is computed with: 1 function J = sum vonB(p , tdata , ldata ) 2 % von Bertalanffy eqn 3 model = p (1) * (1 - exp ( - p (2) * tdata ) ) ; % Model eqn with parameters 4 error = model - ldata ; % Error between model and data 5 J = error * error ’ ; % Sum of square e r r o r s 6 end The least sum of square errors uses the command line: [p1,J,flag] = fminsearch( @ sum vonB,[1.9,0.2],[],tdfish,ldfish) where tdfish and ldfish are the age and length data for the fish.
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