Three major steps in modeling: Construction of the Model Analysis of the Model Comparison with Experiment or Observation

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1 Section 2.3 Modeling : Key Terms: Three major steps in modeling: Construction of the Model Analysis of the Model Comparison with Experiment or Observation Mixing Problems Population Example Continuous Compounding

2 Mixing Mixing problems involve the mixing of substances and have particular significance to chemistry and biology. The idea is to build a model that predicts the amount of a substance (salt, drugs, etc) in a container. The substance flows into the container at some given rate (input rate), is mixed with the ingredients in the container, and then follows out of the container at some given rate (output rate). We assume the container is thoroughly mixed. We will assume that the substance is neither created nor destroyed in the container. Thus variations in the amount of the substance are due solely to the flows in and out of the tank. If we denote the amount of substance in the container at time t by Q(t), the rate of change of Q, dq/dt, is equal to the rate at which the substance flows in minus the rate at which the substance flows out. dq rate in of substance - rate out of substance dt

3 A usual situation is that we are dealing with a fluid. An amount of substance is dissolved in the fluid that comes into the container, is mixed with the fluid that is in the container (which may contain some of the substance), and then the fluid that exits the container also contains some of the substance dissolved in the fluid. Lets consider a general situation: Let Q(t) be the number of pounds of salt in a tank at time t measured in minutes. Then dq/dt has units the number of pounds of salt per minute. Since we are dealing with a fluid we have so many gallons/minute coming in and so many gallons/minute going out. But we must keep in mind that dq rate in of substance - rate out of substance dt has units pounds/minute.

4 dq rate in of substance - rate out of substance dt We can determine the expression for the RATE IN and the RATE OUT of the substance as follows: RATE IN = (Concentration In) times (Fluid Flow Rate In) (lb/min) (lb/gal) (gal/min) RATE OUT = (Concentration Out) times (Fluid Flow Rate Out) (lb/min) (lb/gal) (gal/min)

5 Example: At time t = 0 a tank contains Q 0 Ib of salt dissolved in 100 gal of water. Assume that water containing 1/4 Ib of salt/gal is entering the tank at a rate of r gal/min and that the well-stirred mixture is draining from the tank at the same rate. (a) Construct the IVP that describes Q the amount of salt in the tank at any time t. Find the amount of salt Q(t) in the tank at any time. (b) Find the limiting amount Q L, that is present after a very long time. RATE IN = (Concentration In) times (Fluid Flow Rate In) = ¼ lb/gal r gal/min = ¼ r lb/min RATE OUT = (Concentration Out) times (Fluid Flow Rate Out) amt of salt in tank at time t gal = r number of gallons in tank at time t min IVP = (Q/100) lb/gal r gal/min = (rq/100) lb/min

6 IVP Is the DE is 1 St order linear? Is it separable? Is it 1st order linear autonomous? Rewriting as a first order linear DE we get It is also 1 st order linear autonomous. In standard form we have For this DE we have the form dy ay dt dq r r r r Q so a, b dt Then the solution is r/ 4 Q Ce 25 Ce r/ 100 Applying the initial condition we find C = Q So the solution of the IVP is ( r / 100) t ( r / 100) t b 100 Q 25 Q 25 e ( r / ) t For completeness the next page shows the work using the1st order linear technique. 0

7 Rewriting as a first order linear DE we get Here is the work for 1 st order linear. Integrating factor is The general solution of the DE is obtained from the following steps. Solving for Q we get Applying the initial condition we find C = Q So the solution of the IVP is

8 (b) Find the limiting amount Q L ; that is, amount present after a very long time. As t we have Q 25; that is, Q L = 25. Let s look at the solution of the IVP closely: The larger the value of r the faster Q 25. Rewrite the solution as The amount of salt in the tank due to the flow processes. Portion of the original salt that remains in the tank at time t.

9 Let s modify things a bit: set r = 3 gal/min and Q 0 = 50 lbs. In this case the solution of the IVP is Find the time T at which Q is within 2% of Q L = 25. (recall that Q(t) > 25) 2% of 25 is 0.5 so we want T when Q = 25.5 Solve for t Next let s find the rate r so that the time T when Q(t) is within 2% of 25 lbs does not exceed 45 min. Here we use We set t = 45, Q = 25.5 and solve for r. We have

10 Although this particular example has no special significance, models of this kind are often used in problems involving a pollutant in a lake, or a drug in an organ of the body, for example, rather than a tank of salt water. In such cases the flow rates may not be easy to determine or may vary with time. Similarly, the concentration may be far from uniform in some cases. Finally, the fluid rates of inflow and outflow may be different, which means that the variation of the amount of liquid in the problem must also be taken into account.

11 Example: A 1000 gallon tank contains 400 gallons of pure water. A valve is opened so that fluid containing 2lbs of salt per gallon enters the tank at the rate of 4 gallons per minute and at the same time a drain valve is opened so fluid exits the tank at 2 gallons per minute. Construct an IVP for this situation. dq rate in of substance - rate out of substance dt 4 gal / min (2lbs / gal) dq dt Q(t) 2gal / min t Q(t) = 4 gal / min (2lbs / gal) -2gal / min, Q(0) = t dq dt Q(t) = 8-2, Q(0) = t As time goes on what happens to the tank in this case? Is the DE is 1 St order linear? Is it separable? Is it 1st order linear autonomous? How do you find the time that this occurs? Do you need to solve the IVP?

12 Now solve the IVP. dq dt Q(t) = 8-2, Q(0) = t dq dt Put the DE into standard form for 1 st order linear: Then p(t) 2 Q(t) t 2 Q(t) dt ln 400+2t so the integrating factor is μ(t) e 400+2t = e = t Q(t) 2 = t Applying theory for 1 st order linear DEs we multiply both sides of the standard form by the integrating factor. d Q μ(t) = t dt Integrate both sides: 2 Q μ(t) = t dt = t + C 2 Solve for Q: t Q = + C t = t + C t t Apply initial condition to get C: C = Solution of IVP: Q = t t

13 SALT t (400+2 t) TIME How do you find the time that the tank contains 800 pounds of salt? Q = t t -1

14 A Population Model The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of other factors, the population doubles each week. There are 200,000 mosquitoes in the area initially, and predators (birds, bats, and so forth) eat 20,000 mosquitoes/day. Determine the population of mosquitoes in the area at any time. Note that we have two time frames in the information given; weeks and days. Since eventually we want the population after predation lets choose time in days. The population doubles every seven days (ignoring predation) so if P(t) is the population function we have IVP dp rp, P( 0 ) 200, 000 and P( 7 ) 2 P( 0 ) dt Note this is equivalent to 1 week. Solving the IVP for P we get P= Ce rt. Applying P(0) = 200,000 we have C = 200,000. So we have P = 200,000 e rt Next using P(7) = 2P(0) so we can 400, , 000 e 7r solve for r. ln( 2) Using logs we get r 7 Now we are ready to include the predation.

15 We were told that predators (birds, bats, and so forth) eat 20,000 mosquitoes/day. Determine the population of mosquitoes in the area at any time. The IVP here is rate of change = input of mosquitoes output of mosquitoes dp ln( 2) rp 20, 000 P 20, 000, P( 0) 200, 000 dt 7 The DE is first order linear autonomous with a = ln(2)/7 and b = 20,000 so the solution is b at 20, 000 P(t) Ce Ce (ln( 2)/ 7)t a ln( 2) / 7 Applying the initial condition P(0) = 200,000 we find that 140, 000 C 200, ln( 2) ln( 2) t P(t) e 7 201, The next slide is the details for the same computation usin first order linear and integrating factors.

16 We were told that predators (birds, bats, and so forth) eat 20,000 mosquitoes/day. Determine the population of mosquitoes in the area at any time. The IVP here is rate of change = input of mosquitoes output of mosquitoes dp ln( 2) rp 20, 000 P 20, 000, P( 0) 200, 000 dt 7 2 This is a first order linear DE. dp ln( ) P 20, 000 dt 7 ln( 2) ln( 2) Integrating factor: dt t (t) e 7 e 7 But it is also 1 st order linear autonomous. Applying the initial condition P(0) = 200,000 we find that So we have ln( 2) t P(t) e 7 201, , 000 C 200, ln( 2)

17 Continuous Compounding Suppose that a sum of money is deposited in a bank or money fund that pays interest at an annual rate r. The value S(t) of the investment at any time t depends on the frequency with which interest is compounded as well as on the interest rate. Financial institutions have various policies concerning compounding: some compound monthly, some weekly, some even daily. In general, if interest is compounded m times per year, then for an initial deposit S(0) = S 0 the amount in the account at time t is given by S(t) S0 1 m r mt where m is the number of times the interest is compounded in a year. If we assume that compounding takes place continuously, then we can set up a simple initial value problem that describes the growth of the investment.

18 COMPARISON: Suppose $1000 is deposited into an account yielding 5% interest compounded at the following frequencies. How much money is in the account after 5 years? S( t) S 0 1 r m m5 Annually.05 5 S(5) ( 1.05) 1 5 $ Semiannually S(5) (1.025) 2 10 $ Quarterly S(5) (1.0125) 4 $ Monthly Daily compounding gives $ S(5) ( ) 12 $

19 If we assume that compounding takes place continuously, then we can set up a simple initial value problem that describes the growth of the investment. Here we use a bit of calculus: mt r lim S(t) lim S rt 01 S0e m m m This formula for continuous compounding of interest is the same as the solution of the IVP ds = rs, S(0) = S0 dt Returning to our Comparison, where $1000 is invested in an account with an interest rate of 5% compounded continuously. How much money will there be in the account after 5 years? (assume no withdrawals) S(t)= $1000 e.05(5) = $ Recall: compounding monthly gave $

20 In the case of continuous compounding, let us suppose that there may be deposits or withdrawals in addition to the accrual of interest. If we assume that the deposits or withdrawals take place at a constant rate (amount) k, then we have that IVP ds rs, S(0) S0 dt is replaced by ds rs k, S(0) S0 dt Also 1 st order linear or, in standard form for 1 st autonomous. order linear, ds rs k, S(0) S0 dt where k is positive for deposits and k negative for withdrawals. This DE is first order linear and its general solution is rt k S(t) Ce, C an arbitrary constant r Appling the initial condition we get the solution (after a rearrangement) of the IVP to be rt k S(t) S rt 0e e 1 r Return on initial deposit Effect of additional deposit or withdrawal.

21 The advantage of stating the problem in this general way without specific values for S 0, r, or k lies in the generality of the resulting formula S(t). rt k S(t) S rt 0e e 1 r With this formula we can readily compare the results of different investment programs or different rates of return. For instance, suppose that one opens an individual retirement account (IRA) at age 25 and makes annual investments of K = $2000 thereafter in a continuous manner. Assuming a rate of return of 8%, what will be the balance in the IRA at age 65? We have S 0 = 0, r = 0.08, and k = $2000. Then S(40) =(25,000)(e 3.2 1) = $588,313 It is interesting to note that the total amount invested is $80,000, so the remaining amount of $508,313 results from the accumulated return on the investment. The balance after 40 years is also fairly sensitive to the assumed rate. For instance, S(40) = $508,948 if r = 0.075, S(40) = $98,36494 if r Alternatively you can have an IRA that is tied to the stock market behavior, rather than a fixed rate of interest. In such a case this formula is no applicable.

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