10 a The change in amount of phosphate P t is found by adding the amount

# 10 a the change in amount of phosphate p t is found

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10. a. The change in amount of phosphate, P ( t ), is found by adding the amount entering and subtracting the amount leaving. dP dt = 200 · 10 - 200 · c ( t ) , where c ( t ) is the concentration in the lake with c ( t ) = P ( t ) / 10 , 000. By dividing the equation by the volume, the concentration equation is given by dc dt = 0 . 2 - 0 . 02 c = - 0 . 02( c - 10) , c (0) = 0 . With the substitution z ( t ) = c ( t ) - 10, the equation above reduces to the problem dz dt = - 0 . 02 z, z (0) = - 10 , which has the solution z ( t ) = - 10 e - 0 . 02 t . Thus, the concentration is given by c ( t ) = 10 - 10 e - 0 . 02 t . b. The differential equation describing the growth of the algae is given by dA dt = 0 . 5(1 - e - 0 . 02 t ) A 2 / 3 . By separating variables, we see Z A - 2 / 3 dA = 0 . 5 Z (1 - e - 0 . 02 t ) dt 3 A 1 / 3 ( t ) = 0 . 5( t + 50 e - 0 . 02 t ) + C A ( t ) = 0 . 5( t + 50 e - 0 . 02 t ) + C 3 ! 3 From the initial condition A (0) = 1000, we have 1000 = 25+ C 3 3 . It follows that C = 5, so A ( t ) = t + 50 e - 0 . 02 t + 10 6 ! 3 . 11. a. Write the differential equation dw dt = - 0 . 2( w - 80), then z ( t ) = w ( t ) - 80. It follows that dz dt = - 0 . 2 z, z (0) = - 80 , with the solution z ( t ) = - 80 e - 0 . 2 t = w ( t ) - 80. Thus, w ( t ) = 80 1 - e - 0 . 2 t .

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For a 40 kg alligator, w ( t ) = 40 = 80 ( 1 - e - 0 . 2 t ) or 40 = 80 e - 0 . 2 t , so e 0 . 2 t = 2 or 0 . 2 t = ln(2). Thus, t = 5 ln(2) 3 . 47 years. b. The pesticide accumulation is given by dP dt = 600 80 1 - e - 0 . 2 t , P (0) = 0 . The solution is given by P ( t ) = 48 , 000 Z 1 - e - 0 . 2 t dt = 48 , 000 t + 5 e - 0 . 2 t + C. The initial condition gives P (0) = 0 = 240 , 000 + C , so C = - 240 , 000. Hence, P ( t ) = 48 , 000 t + 5 e - 0 . 2 t - 240 , 000 . The amount of pesticide in the alligator at age 5 is P (5) = 48 , 000 ( 5 + 5 e - 1 ) - 240 , 000 = 240 , 000 e - 1 88291 μ g. c. The pesticide concentration for a 5 year old alligator is c (5) = P (5) 1000 w (5) = 88 , 291 80 , 000 (1 - e - 1 ) 1 . 75 ppm . 12. a. The differential equation can be written: dc dt = - 0 . 004( c - 15) , so we make the substitution z ( t ) = c ( t ) - 15. Since c (0) = 0, it follows that z (0) = - 15. The solution of the substituted equation is given by: z ( t ) = - 15 e - 0 . 004 t = c ( t ) - 15 c ( t ) = 15 - 15 e - 0 . 004 t . The limiting concentration satisfies: lim t →∞ c ( t ) = 15 mg / m 3 . b. We begin by separating variables, which gives: Z dc c - 15 = - 0 . 001 Z (4 - cos(0 . 0172 t )) dt ln( c ( t ) - 15) = - 0 . 001 4 t - sin(0 . 0172 t ) 0 . 0172 + C c ( t ) = 15 + Ae - 0 . 001 ( 4 t - sin(0 . 0172 t ) 0 . 0172 ) It is easy to see that the initial condition c (0) = 0 implies that A = - 15. Thus, the solution to this problem is given by: c ( t ) = 15 - 15 e - 0 . 001(4 t - 58 . 14 sin(0 . 0172 t ))
13. a. We separate variables, so Z M - 3 / 4 dM = - k Z dt or 4 M 1 / 4 = - kt + 4 C M ( t ) = C - k 4 t 4 From the initial condition, M (0) = 16 = C 4 , it follows that C = 2. From the information that M (10) = 1 = (2 - 10 k/ 4) 4 , we have k = 0 . 4, so M ( t ) = (2 - 0 . 1 t ) 4 . The fruit vanishes in 20 days. b. We separate variables again to find: Z M - 3 / 4 dM = - 0 . 8 Z e - 0 . 02 t dt or 4 M 1 / 4 = 0 . 8 0 . 02 e - 0 . 02 t + 4 C M ( t ) = 10 e - 0 . 02 t + C 4 . From the initial condition, M (0) = 16 = (10 + C ) 4 , it follows that C = - 8, so M ( t ) = 10 e - 0 . 02 t - 8 4 . Solving 10 e - 0 . 02 t = 8, which is when the fruit vanishes, we find t = 50 ln(5 / 4). Thus, the fruit vanishes in 11.157 days. 14. a. The general solution to the Malthusian growth problem with the initial condition P (0) = 60 is P ( t ) = 60 e rt . We are given that 2 weeks later P (2) = 80 = 60 e 2 r , so it follows that r = 1 2 ln 4 3 = 0 . 14384. This gives the solution: P ( t ) = 60 e 0 . 14384 t .

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