1 Construct the di ff erential equation for Q t 2 Solve the di ff erential

# 1 construct the di ff erential equation for q t 2

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(1) Construct the di ff erential equation for Q ( t ). (2) Solve the di ff erential equation. (3) Predict the long term quantity of the salt in the reservoir. Solution: (1) Rate of change of total quantity of salt = Rate In - Rate Out; Rate of change of total quantity of salt = dQ dt ; Rate In = salt IN per day = salt water volume IN per day × concentration of salt water IN; Rate Out = salt OUT per day = salt water volume OUT per day × concentration of salt water OUT. We have dQ dt = 0 . 1 × 50 - 1 × Q 100 , dQ dt = 500 - Q 100 , Q (0) = 0 . (2) Q ( t ) = 500(1 - e - 0 . 01 t ) . (3) lim t →∞ Q ( t ) = 500 . 23
7.4 Exponential Growth and Decay If the population growth dP dt is proportional to the population size P ( t ), then dP dt = kP. The solution is P ( t ) = P (0) e kt . k > 0 growth, k < 0 decay. Absolute growth rate = rate of change of population = dP/dt , Relative growth rate = percent change per unit time = ( dP/dt ) /P, Doubling time ( D ) of exponential growth = the time required for it to double: P ( t ) = P (0)2 t/D . Half life ( H ) of exponential growth = the time required for it to be half: P ( t ) = P (0) 1 2 t/H . Newtons Law of Cooling states that the rate of change of the temperature of a cooling body is proportional to the di ff erence between its temperature T and the temperature of its surrounding medium T s . The model is: dT dt = - k ( T - T s ) , where k is a constant. Example. A bacteria culture growth at a rate proportional to its size. After 2 hours there are 40 bacteria and after 4 hours the count is 120. Find an expression for the population after t hours. Solution: We measure the time t in hours. Let P ( t ) be the population at t hours, then we have dP dt = kP. The solution of the equation is P ( t ) = P (0) e kt . 24
Note that P (2) = 40 and P (4) = 120, we obtain 40 = P (0) e 2 k , 120 = P (0) e 4 k . These imply that P (0) = 40 3 and e 2 k = 3 , or k = ln 3 / 2 . We thus have P ( t ) = 40 3 3 t/ 2 = 40 3 3 t = 40 3 e (ln 3 / 2) t . Example. The half-life of Sodium-24 is 15 hours. Suppose you have 100 grams of Sodium- 24. How many grams remaining after 27 minutes (keep three decimals)? Solution: Assume m ( t ) be the amount after t hours. Then m ( t ) = m (0)( 1 2 ) t/H , where m (0) = 100, H = 15 hours. Note that 27 minutes = 27/60=0.45 hours. Thus m (0 . 45) = 100( 1 2 ) 0 . 45 / 15 = 100( 1 2 ) 0 . 03 = 97 . 942 g. Example. Use Newton’s Law of Cooling to determine the time of death of a healthy man. He died in his room some time before noon; At noon, his body temperature was found to be 70 degrees; His body cooled another 5 degrees in 1 hour after noon; The room temperature was a constant 60 degrees; Normal temperature of people’s body is 98.6 degree. Solution: Let T ( t ) be the temperature of the body at time t . Taking noon as t = 0, we have T (0) = 70. Note that T s = 60, we have dT dt = - k ( T - 60) , this equation is separable. We obtain dT T - 60 = - kdt, T ( t ) = 60 + Ce - kt . 25
From T (0) = 70 we get 70 = 60 + C , C = 10 and T ( t ) = 60 + 10 e - kt . At 1:00pm, his body temperature is 70 - 5 = 65. Hence T (1) = 65. 65 = 60 + 10 e - k (1) , 5 = 10 e - k , k = ln 2 , T ( t ) = 60 + 10 e - t ln 2 . Thus 98 . 6 = 60 + 10 e - t ln 2 , t = - ln 3 . 86 / ln 2 = - 1 . 95 , which means 1 hour and 0.95(60) minutes before noon, or 1 hour and 57 minutes before noon. Or the time of death is 10:03AM.
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