Applications of 1st-Order Equations

Transcription

1 3 Applications of 1st-Order Equations 3.2 Compartmental Analysis For a first example we shall analyze a one-compartment system, and then later consider a two-compartment system. Example 3.1. A brine solution (a mixture of salt and water) flows at a constant rate of 4 L/min into a tank that initially contains 100 L of pure water. The solution inside the tank is kept well-stirred and flows out of the tank at a rate of. If the concentration of salt in the brine entering the tank is 0.2 kg/l, determine the mass of the salt in the tank after t minutes. Find when will the concentration of salt in the tank be 0.1 kg/l. Solution. Let x(t) be the mass of salt, in kilograms, that is in the tank at time t. Since the tank is initially filled with fresh water we know that x(0) = 0. See Figure 1. In order to determine x(t) we will use what we know about the rate at which x(t) changes over time, which is x (t). An expression for x (t) will be given as the rate salt enters the tank minus the rate salt leaves the tank (in kilograms per minute). To do this we first need to know the volume of the solution in the tank at time t. The volume is given to be 100 L initially, and since 4 L of liquid enters the tank and 3 L leaves with each minute, we can see that at time t the volume of solution in the tank must be t. Now, the rate salt enters the tank is easily reckoned: 4 liters of brine is entering per minute, there s 0.2 kg of salt per liter, and so a total of 0.8 kg of salt is entering per minute. As for the rate salt leaves, at time t there s x(t) kg of salt in the tank, and we assume that it is uniformly dissolved throughout the t liters of solution to give a concentration of x(t)/(100 + t) kg of salt per liter. Since 3 liters of solution is leaving per minute, we conclude that 3x(t)/(100 + t) 4 L/min 0.2 kg/l x(t) kg salt x(0) = 0 Figure 1

2 2 kg of salt is leaving per minute. The derivation of x (t) is as follows: x (t) = (rate salt enters Tank 1) (rate salt leaves Tank 1) ( )( ) ( )( ) 0.2 kg 4 L x(t) kg 3 L = 1 L 1 min t L 1 min = 0.8 3x(t) t. Thus we have a linear first-order ODE: x 3 + t x = 0.8. To solve this equation, we multiply by the integrating factor ( ) 3 µ(t) = exp t dt = e 3 ln t+100 +c = (t + 100) 3 to obtain which becomes and thus (t + 100) 3 x + 3(t + 100) 2 x = 0.8(t + 100) 3, (t + 100) 3 x = [ (t + 100) 3 x ] = 0.8(t + 100) 3 0.8(t + 100) 3 dt = (t + 100)4 + c. From this we get a general explicit solution to the ODE, x(t) = 1 c (t + 100) + 5 (t + 100). 3 To determine c we use the initial condition x(0) = 0: 0 = 1 5 ( ) + c ( ) 3 c = 20 c = Thus, the mass of salt in the tank at time t is given by x(t) = (t + 100) 5 (t + 100). 3 The concentration, C(t), of salt at time t is given by the mass x(t) at time t divided by the volume t at time t. That is, C(t) = x(t)/(t + 100), so that C(t) = (t + 100) 4. Now, to find when the concentration of salt is 0.1 kg/l we solve C(t) = 0.1, giving the equation 0.1 = (t + 100) 4. From this we find that t = minutes.

3 The following example illustrates a physical system consisting of two compartments. The general approach is to analyze the first compartment, then use the information garnered to analyze the second compartment. Example 3.2. Beginning at time t = 0, fresh water is pumped at a rate of into a well-stirred tank that is initially filled with 60 L of brine. The increasingly less concentrated salt solution flows at a rate of out a drain that feeds into a second tank initially filled with 60 L of pure water. The resultant mixture of water and salt in the second tank, which is also well-stirred, is pumped into the ocean at a rate of. Find the time when the concentration of salt in the second tank is greatest, and compare the maximal concentration in the second tank to the initial concentration in the first tank. Solution. It is not actually necessary to know how much salt is initially in Tank 1 to determine when the concentration of salt in Tank 2 is greatest. Let x(t) be the number of kilograms of salt in Tank 1 at time t, and let y(t) be the number of kilograms of salt in Tank 2 at time t. We have x(0) = x 0 for some constant x 0, and also y(0) = 0 since the water in Tank 2 is initially pure. See Figure 2. Now, noting that the volume of solution in Tank 1 is a constant 60 L, we have x (t) = (rate salt enters Tank 1) (rate salt leaves Tank 1) ( )( ) x(t) kg 3 L = 0 = 3x(t) 60 L 1 min 60, which yields the equation x = 1 x, also written as dx/dt = x/20. This equation is separable, 20 giving 20 x dx = dt, and hence ln ( x 20) = t + c 0 for arbitrary constant c 0. Exponentiating both sides and letting c 1 = e c 0 be an arbitrary positive constant, we obtain x 20 = e t+c 0 = c 1 e t, and then x(t) = c 1 e t/20. Using the initial condition x(0) = x 0, we substitute t = 0 and x = x 0 into the equation to get x 0 = c 1 e 0 = c 1, and thus x(t) = x 0 e t/20. (1) Now we turn our attention to Tank 2. Since the volume of solution in Tank 2 is always 60 L, we have y (t) = (rate salt enters Tank 2) (rate salt leaves Tank 2) ( )( ) ( )( ) x(t) kg 3 L y(t) kg 3 L = 60 L 1 min 60 L 1 min = x(t) 20 y(t) 20 = x 0e t/20 y(t) 20 3

4 4 0 kg/l 60 L solution x(t) kg salt x(0) = x 0 x(t) kg/l L solution y(t) kg salt y(0) = 0 y(t) kg/l 1 60 Figure 2 where the last equality follows from (1). Hence we have the equation y y = x 0 20 e t/20, (2) which is a first-order linear ODE and so can be solved by finding an appropriate integrating factor µ(t). We have ( ) 1 µ(t) = exp 20 dt = e t/20 and so, multiplying (2) by e t/20, we obtain which can be written and therefore y e t/ yet/20 = x 0 20, ( ) ye t/20 x 0 = 20, x0 ye t/20 = 20 dt = x 0 20 t + c. Using the initial condition y(0) = 0, we substitute t = 0 and y = 0 into this equation to find that c = 0, and at last we have an expression for y(t): y(t) = x 0 20 te t/20. The concentration of salt in Tank 2 at time t, C(t), is given by C(t) = y(t)/60; that is, C(t) = x te t/20. To determine when the concentration is greatest, we must find t > 0 for which C(t) attains a global maximum value on (0, ). 1 This entails finding t for which C (t) = 0; that is, we must solve x e t/20 x 0 24, 000 te t/20 = 0. But this equation immediately implies that 20 t = 0, and hence t = 20 minutes. 1 Note that, since the volume of solution in Tank 2 is a constant 60 L, we could just as well determine when y(t) attains a maximum.

5 5 Finally, at time t = 20 minutes we find that the mass of salt in Tank 2 is y(20) = x 0 20 (20)e 20/20 = 1 e x 0. That is, Tank 2 is at most 1/e times as salty as Tank 1 was initially. Observe that x(t) 0 and y(t) 0 as t, as is to be expected since fresh water is ultimately displacing brine in the system. Problems concerning the decay of one or more radioactive isotopes also lend themselves to compartmental analysis. In the simplest case there is a single radioactive substance that decays into some other element, so that a compartment is a sample of the substance. No quantity of the substance is entering the compartment, but gradually over time atoms of the substance are leaving (i.e. fissioning into other material). Empirical data make one thing clear: the rate of decay of a radioactive substance is directly proportional to the amount of the substance that is present. Thus, if x(t) is the amount of the substance (in grams, say), then x (t) is the rate of change of the amount of the substance, with x (t) = kx(t) for some constant of proportionality k < 0 that depends on the isotope under consideration. The half-life of a radioactive isotope, which is the time it takes for half of it to decay, can vary from nanoseconds to gigayears. Example 3.3. Cobra Commander has 260 grams of kaboomium-320 (chemical symbol 320 Ka) in the basement of his secret hideout. Upon returning from a carefree five-hour drive with Destro in the countryside in his spiffy new Nissan Cube, he finds that 192 grams remain. After how many hours will only 10 grams remain? Solution. Let x(t) be the amount of 320 Ka in grams at time t in hours, where x(0) = 260 and x(5) = 192. The rate at which 320 Ka decays is proportional to the amount present, which is to say that x (t) = kx(t) for some constant k. This is a separable ODE which becomes 1 x dx = k dt, and thus ln x = kt + c. Of course, x(t) is never negative, so we may write ln(x) = kt + c, whence we obtain x(t) = e kt+c = x 0 e kt (letting x 0 = e c ). From the initial condition x(0) = 260 comes 260 = x 0 e 0, so that x 0 = 260 and hence x(t) = 260e kt. (3) It remains to find k. Fortunately we have another bit of information available: x(5) = 192. Putting this into (3) gives 192 = 260e 5k, whence and we fully determine x(t) to be k = 0.2 ln(192/260) x(t) = 260e t.

6 With this model in hand we can ascertain how many hours it will be before only 10 g of 320 Ka remains for Cobra Commander to play with: set x(t) = 10 to get 260e t = 10, and thus ( ) 1 1 t = ln That is, after about 53.8 hours only 10 g of kaboomium-320 will remain. We assume that Cobra Commander knows his business. And knowing is half the battle! 6