Polking Differential Equations Chapter 3fliacob/An2/2014-2015/Resurse_MCM_2015... · Polking/Boggess/Arnold: Differential Equations 1/E [Second Pages] 1/25/97 10:29 Page 3 126 Chapter3

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Modeling and Applications

The discovery of the calculus occurred at the beginning of the scientific revolutionin the seventeenth century. This discovery was not a side issue in the revolution.Rather, it was the linchpin on which much of what followed was based. For the firsttime, humankind had a systematic way to study how things changed. In many cases,the study of change has led to a differential equation, or to a system of differentialequations through the process known as modeling.

We have explored a few applications, and we have constructed the correspond-ing models in Chapter 2. In this chapter, we will look carefully at the modelingprocess itself. The process will then be used in several applications. Along the way,we will also consider some examples of modeling that are faulty.

The main idea in the modeling process is explained easily. Suppose x is aquantity that varies with respect to the variable t . We want to model how it changes.From the mathematical point of view, the rate of change of x is the derivative

x ′ = dx

dt.

Building a model of the process involves finding an alternate expression for the rateof change of x as a function of t and x , say f (t, x). This leads us to the differentialequation

dx

dt= f (t, x).

This equation is the mathematical model of the process.The problem, of course, is discovering how the rate of change varies, and this

means discovering the function f (t, x). Let’s look at some examples.

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124 Chapter 3 Modeling and Applications

3.1 Modeling Population GrowthThe process of modeling is exemplified by the modeling of the growth of popula-tions. We will study the growth of a population of protozoa, which are single-celledorganisms such as amoebae or paramecia. We will assume that there is no lack ofnutrients or of room to grow. Let P(t) denote the number of cells at time t .

Protozoa multiply by cell division. We may assume that any cell is as likelyto divide as any other. Let b be the probability that a cell will divide in a unit oftime, where time is measured in some convenient unit such as days or hours. Thenbetween times t and t + 1t , there will be approximately

bP(t)1t

divisions, resulting in the same number of new cells. Furthermore, this approxima-tion gets better as 1t gets smaller. We will call b the birth rate.

Like any living creature, single-celled organisms die for a variety of reasons.Let d denote the probability that any individual cell will die in a unit of time. Justas in our discussion of births, the number of deaths between t and t + 1t will beapproximately

d P(t)1t,

and this approximation gets better as 1t gets smaller. We will call d the death rate.Putting together the results on births and deaths, we see that the change in the

population between times t and t + 1t is

P(t + 1t) − P(t) ≈ bP(t)1t − d P(t)1t = (b − d)P(t)1t.

Hence, by the limit quotient definition of the derivative,

P ′(t) = lim1t→0

P(t + 1t) − P(t)

1t= (b − d)P(t).

If we let

r = b − d (1.1)

and call r the reproductive rate, the equation is

P ′ = r P. (1.2)

Equation (1.2) is a first-order differential equation involving the function P(t).It is a mathematical model of the growth of a population. The nature of the predic-tions of the model depend on the nature of the reproductive rate r .

The Malthusian modelIf we assume that the population has plenty of resources like food and space, thenit makes sense that the birth and death rates do not depend on time nor on the sizeof the population. Hence both are constants, and the reproductive rate r is also aconstant. Then equation (1.2) is the exponential equation, and solutions are easilyfound. Separation of variables leads us to the solution

P(t) = Cer t


3.1 Modeling Population Growth 125

for any constant C . Furthermore, if the population at time t = 0 is P0, then we seethat C = P0 and our solution is

P(t) = P0er t . (1.3)

We can use our model (1.2) and the solution (1.3) to predict the population ofour colony of protozoa. There are two cases. If the death rate exceeds the birth rate,then r = b − d < 0, and the population given by (1.3) will decline. The colonywill eventually disappear. On the other hand, if r = b − d > 0, the population willgrow. In fact, it will grow exponentially.

The economist Thomas Malthus is usually given credit for the model of popu-lation growth in (1.2) and (1.3). It is referred to as the Malthusian model.1

As an example, suppose we have a population of bacteria, which has 1000

0 5 100

1

2

3

Popu

latio

n in

thou

sand

s

Time in hours

Figure 1 Exponential growth of apopulation.

individuals at time t = 0, and that the reproductive rate is r = 0.1. Suppose alsothat we are measuring time in hours. Then according to (1.3), the population at latertimes will be given by

P(t) = 1000e0.1t .

The graph of this solution over the first 10 hours is shown in Figure 1. Noticethat the population almost triples in that time. Figure 2 shows the same population,but over the first 100 hours. The difference is quite striking and illustrates howrapidly exponential growth really is. After 100 hours there are about 22,000,000paramecia.

Thus we see that the exact solution is important because it allows us to extractquantitative information about a population obeying the Malthusian law of growth.

Evaluating the parameters in the Malthusian modelIn the Malthusian models there occurs the parameter r , the reproductive rate. Thisconstant must be chosen to reflect reality. Usually this is done on the basis of exper-imentation or observation. Let’s look at an example.

E X A M P L E 1 . 4 u A biologist starts with 10 cells in a culture. Exactly 24 hours later he counts 25.Assuming a Malthusian model, what is the reproductive rate? What will be thenumber of cells at the end of 10 days?

The model is given by equation (1.2), with a constant reproductive rate r . This

0 50 1000

10

20

30

Time in hours

Popu

latio

n in

mill

ions

Figure 2 Exponential growth overa longer period of time.

is the exponential equation, and the general solution is

P(t) = Cer t . (1.5)

Let’s measure time in days. Then for our initial condition we have

10 = P(0) = C.

1 Thomas Robert Malthus (1766–1834) was an English clergyman and political economist. In his famoustreatise, An Essay on the Principle of Population, he postulated that population tended to increase “ge-ometrically” (by which he meant exponentially, as we see in equation (1.3)) while resources, especiallyfood, increased only linearly. As a result, human and other populations will increase until checked bynatural limitations, principally to do with food supply. In the case of humans he predicted that populationgrowth would be limited by wars, pestilence, and other natural disasters.

Of course the supply of food and other resources (except space) has tended to increase exponentiallysince the time of Malthus. Nevertheless, there are modern Malthusians who think he was on to something.



Then from the second data point we have

25 = P(1) = 10er .

Solving for r , we getr = ln(2.5) ≈ 0.9163.

After 10 days, the population is predicted to be

P(10) = 10er×10 ≈ 95,367. u

In practice, the biologist in Example 1.4 would probably count the cells at leastonce each day. After each count, he would reestimate r using all of his data. Hewill do this using a process called linear regression. It is our experience that manystudents are able to perform the process of linear regression on calculators, eventhough they do not know how the calculator does the job.

The key to the method is found by taking the logarithm of equation (1.5), getting

ln P(t) = ln C + r t.

Thus ln P(t) is a linear function of t with coefficients ln C and r . Suppose we makemeasurements at N different times t0, t1, . . . , tN−1, and the populations at thesetimes are P0, P1, . . . , PN−1. The method of linear regression chooses ln C and r tominimize the function

N−1∑i=0

(ln Pi − ln P(ti ))2 =

N−1∑i=0

(ln Pi − ln C − r ti )2 .

The method of linear regression is a special case of the method of least squares,since it minimizes the sum of the squares of the errors.

If you do not know how to compute a linear regression, you may skip the fol-lowing example.

E X A M P L E 1 . 6 u After each of the first four days, the biologist in Example 1.4 counts 25, 62, 154, and375 cells. What is the reproductive rate r? What does he estimate the populationwill be after 10 days?

We have estimated values for t = 0, 1, 2, 3, 4. We will choose the coefficientsusing linear regression. Using a calculator or a computer, this is an easy process. Indoing so we find that ln C = 2.3091 or C = 10.0658, and r = 0.9067.

After 10 days the population is predicted to be

P(10) = Cer×10 ≈ 87, 195. u

The use of linear regression in the previous example allows the biologist to useall of his data to estimate the parameters. This will usually result in better estimates.



The logistic model of growthWhenever we have a mathematical model, we should examine its implications care-fully to discover to what extent the model correctly predicts what happens in thelaboratory or in the real world. In the case of the Malthusian model the predictionof unlimited exponential growth is clearly impossible. If a colony of protozoa grewexponentially, it would cover the earth in times that are observable, and this justdoes not happen. Nevertheless, laboratory experiments do verify the exponentialgrowth of populations that are relatively small.

If we reexamine our assumptions, we quickly see the problem. We have as-sumed that the colony has no lack of nutrients and no lack of space in which togrow. These assumptions are valid for relatively small populations but become un-true as the population grows. Let’s look again at our derivation of (1.2) and takeinto account limits of growth.

Let’s first take a closer look at the death rate. A lack of food means that thecells will have to compete with each other for what food there is, and some cellswill die of malnutrition as well as of natural causes. Similarly, a lack of spacewill cause increased interaction between cells, which may increase the number ofdeaths. Both of these increase the death rate by increasing the number of interactionsbetween cells. The number of interactions that an individual cell will have will beproportional to the probability of an interaction. Assuming that the population isevenly distributed over the area occupied by the species, the probability will beproportional to the size of the population. Hence a more realistic formulation of thedeath rate would have it increase proportionally to the size of the population. Thisleads us to the formula

d + a P

for the death rate. The constant d is the same as before, the death rate for smallpopulations. The new term a P measures additional deaths that are due to interac-tions between a cell and other cells. The constant a is a measure of the impact ofinteractions on the rate at which deaths occur.

Similarly, the birth rate will decrease proportionally to the size of the popula-tion. The modified birth rate will have the formula

b − cP.

Arguing as we did before, we see that the change of population between t and1t is

P(t + 1t) − P(t) ≈ (b − cP(t))P(t)1t − (d + a P(t))P(t)1t

= (b − d − (a + c)P(t))P(t)1t.

Hence

P ′(t) = lim1t→0

P(t + 1t) − P(t)

1t= (b − d − (a + c)P(t))P(t).

We set r0 = b − d, but now we call it the natural reproductive rate. To put theequation into a nicer form, we set a + c = r0/K , where K = r0/(a + c) is a newconstant. Then

P ′ = (r0 − r0 P/K )P = r0(1 − P/K )P. (1.7)



Equation (1.7) is called the logistic equation. The model of population growthembodied in the logistic equation is called the logistic model.2 Referring back toequation (1.2), we see that the reproductive rate assumed in the logistic model is

r = r0(1 − P/K ).

This is no longer a constant but rather depends on the population. Notice that thereproductive rate becomes negative for P > K .

The direction field for the logistic equation is shown in Figure 3.

0

0

K

t

P

Figure 3 The direction field for the logistic equation.

Qualitative analysis of the logistic equationThe logistic equation in equation (1.7) is autonomous, since the right-hand side

P

f (P ) = r0(1 − P/K )P

(K , 0)

Figure 4 The graph of theright-hand side of the logisticequation.

f (P) = r0 P(1 − P/K )

does not depend on t . The graph of f is shown in Figure 4. The equilibrium points

0

K

t

P

Figure 5 The equilibriumsolutions for the logisticequation.

are where f (P) = 0, or at P1 = 0, and P2 = K . Consequently,

P1(t) = 0 and P2(t) = K

are equilibrium solutions. The graphs of these solutions are shown in Figure 5.Next we notice that P ′ = f (P) > 0 if 0 < P < K , and P ′ = f (P) < 0 if

P > K or if P < 0. Hence P1 = 0 is an unstable equilibrium point, and P2 = K isstable. In particular, we see that if P(t) is any solution with a positive population,then it must stay positive, and it must tend to K as t → ∞. Consequently, every

2 The logistic model was first constructed by Pierre-Francois Verhulst (1804–1849). His work was mostlyignored until around 1920, when it was published again by R. Pearl and L. J. Reed. Since then there havebeen many experiments that verify the logistic model in carefully controlled circumstances. In particular,we mention the Russian biologist G. F. Gause, who in 1934 published The Struggle for Existence, aclassic text with many examples of logistic growth, especially of paramecia.



population governed by the logistic equation tends to K as time increases. For thisreason K is called the carrying capacity.

Hence without solving the initial value problem we know that the solutioncurves have the appearance shown in Figure 6. We cannot say how fast P(t) → Kas t → ∞. For this reason we have not included any tick marks along the t axis inFigure 6.

0

K

t

P

Figure 6 Several solutions to the logistic equation.

Solution of the logistic equationTo make our notation a bit simpler, let’s drop the subscript in r0 in the logisticequation. Then it becomes

P ′ = r P(1 − P/K ),

and r is now a constant. The logistic equation, being autonomous, is separable. Wecan write

K dP

P(K − P)= r dt.

To solve the equation, we must integrate both sides. On the left, we can use partialfractions to discover that

K

P(K − P)= 1

P+ 1

K − P.

Hence we want to solve (1

P+ 1

K − P

)d P = r dt. (1.8)

Integrating both sides of equation (1.8), we get

ln |P| − ln |K − P| = r t + C

ln

∣∣∣∣ P

K − P

∣∣∣∣ = r t + C,



or ∣∣∣∣ P

K − P

∣∣∣∣ = er t+C = eC er t .

The constant eC is positive. If we replace it by A, and allow it to be negative orpositive we can drop the absolute values and write

P

K − P= Aer t . (1.9)

If we let P0 represent the population at time t = t0, then P(t0) = P0 and

Aer t0 = P0

K − P0. (1.10)

Next we solve equation (1.9) for P to get

P(t) = K Aer t

1 + Aer t. (1.11)

Using (1.10) and (1.11) and a little algebra, we can eliminate A and find that

P(t) = K P0er(t−t0)

(K − P0) + P0er(t−t0), (1.12)

or

P(t) = K P0

P0 + (K − P0)e−r(t−t0). (1.13)

We already know as a result of our qualitative analysis that P(t) → K ast → ∞. This can also be seen from (1.13). As t → ∞, e−r(t−t0) → 0 and

limt→∞ P(t) = lim

t→∞K P0

P0 + (K − P0)e−r(t−t0)= K P0

P0= K .

It is interesting to notice that the qualitative analysis that leads to the notion andinterpretation of K as the carrying capacity is so much easier than finding the exactsolution.

Evaluation of the parameters in the logistic equationHowever, the exact solution in (1.13) is important because it allows us to be quantita-tive about our conclusion in well-controlled circumstances. There are two constantsin the logistic equation, the carrying capacity K and the rate r . To find a specificsolution we need in addition the initial population P0. With just three observationswe ought to be able to compute these constants and use (1.13) to tell us the completehistory.



E X A M P L E 1 . 1 4 u Suppose we start at time t0 = 0 with a sample of 1000 cells. One day later we seethat the population has doubled, and sometime later we notice that the populationhas stabilized at 100,000.

From this data, we immediately see that the initial population is P0 =1000, andthe carrying capacity is K =100,000. To find r we use the solution and the fact thatP(1) = 2000. Although we could use (1.13), it is easier to combine equations (1.9)and (1.10), giving

P(t)

K − P(t)= P0

K − P0er t .

With P0 = 1000, K = 100,000, and P(1) = 2000 this becomes

2000

100,000 − 2000= 1000

100,000 − 1000er(1).

Thuser = 198/98,

and

r = ln

(198

98

)≈ 0.7033.

Consequently, the population model is captured by substituting the appropriate pa-rameters into equation (1.13)

P(t) = (100,000)(1000)

1000 + (100,000 − 1000)e−0.7033t,

or, better yet,

0 5 10 15 200

2

4

6

8

10

Time (days)

Popu

latio

n

104×

Figure 7 The population P(t )from Example 1.14

P(t) = 100,000

1 + 99e−0.7033t. (1.15)

The plot of P(t) is shown in Figure 7. Note that the population eventually nears itscarrying capacity of 100,000 in about 15 days.

Of course, now that we have an equation modeling the population, there are anumber of interesting quantitative questions that can be posed and answered. Forexample, the population after 5 days is projected to be

P(5) = 100,000

1 + 99e−0.7033(5)≈ 25,377 cells.

In addition, if you wish to know how long it takes the population to reach a level of50,000 cells, then the equation

50,000 = 100,000

1 + 99e−0.7033t

and a little algebra will produce the solution t = ln 99/0.7033 ≈ 6.5 days. u



It is somewhat more difficult to estimate the parameters for a logistic modelwhen we are not far enough along in the growth of the population to know thecarrying capacity K . There are three constants in the solution given in equation(1.13), r , the reproductive rate for small populations, K , the carrying capacity, andP0, the initial population. Estimates of these three can be computed from any threemeasurements of the population. However, the calculations are somewhat difficult.

The situation is somewhat easier if the times of the measurements are equallyspaced, say at 0, h, and 2h. Suppose the populations at these times are P0, P1, andP2. If we solve equation (1.13) for K , we get

K = P0 P(t)(1 − e−r(t−t0))

P0 − P(t)e−r(t−t0).

Let’s apply this equation twice, first with t0 = 0 and t = h, and then with t0 = hand t = 2h. We get two equations for K :

K = P0 P1(1 − e−rh)

P0 − P1e−rhand

K = P1 P2(1 − e−rh)

P1 − P2e−rh.

(1.16)

Setting these equal and solving for r , we get

r = 1

hln

(P2(P1 − P0)

P0(P2 − P1)

). (1.17)

Then we can substitute this into one of the expressions in (1.16) to find K .Let’s see how well this works.

E X A M P L E 1 . 1 8 u Early in the twentieth century, the German biologist T. Carlson performed exper-iments in which he measured the growth of a yeast culture.3 He dropped a fewyeast cells into an appropriate nutritive sugar solution, and over the next 18 days hemeasured carefully the size of the population of yeast cells. His data are shown inTable 1. Find a logistic curve that fits this data.

This problem is not well posed. We have not defined what it means to fit thedata. We will look at a number of possible answers. First we take the first three datapoints and use the method indicated in equations (1.16) and (1.17). Undoubtedly,the experimenter would do this computation after the second day to get some ideaof what will happen to the yeast culture. However, when we compute the carryingcapacity, we get K = 48.5. One look at the rest of data shows that this figure iscompletely unrealistic. The data show that the carrying capacity should be largerthan 660.

What went wrong? The answer is twofold. First, we are dealing with experi-mental data, and there is always some error associated with real data. Second, and

3 Uber Geschwindigkeit und Grosse der Hefevermehrung in Wurze, Biochem. Ztschr., Bd. 57, pp. 313–334, 1913. The data in Table 1 labeled “Quantity of yeast” are not actually the number of yeast cells.Instead Table 1 contains a quantity that Carlson measured that is proportional to the number of cells.



Table 1 The growth of a population of yeast cells

Day Quantity of yeast Day Quantity of yeast Day Quantity of yeast

0 9.6 7 257.3 13 629.41 18.3 8 350.7 14 640.82 29.0 9 441.0 15 651.13 47.2 10 513.3 16 655.94 71.1 11 559.7 17 659.65 119.1 12 594.8 18 661.86 174.6

perhaps more important, is the fact that a logistic curve is fairly complicated. Lookat Figure 8. We are trying to get at the entire S-shaped logistic curve by lookingonly at its three starting points in the lower part of the S-shaped curve. To expecthigh accuracy is unrealistic.

Next let’s look at what happens if we choose three data points that are fartherapart, so that they represent what is happening in more of the logistic curve. Let’suse the data after 5 and 10 days in addition to the initial data. Then using (1.16) and(1.17) we find that r = 0.5396 and K = 674.4. The results are shown in the blackcurve in Figure 8. The black dots are the experimental data in Table 1. Clearly thematch of the data and the computed curve is quite good.

A scientist who collected data such as that in Table 1, would undoubtedly wantto use all of the data when looking for a matching logistic curve. It would be niceto find the best-fitting curve. Although it takes us beyond the subject matter of thisbook, we will explain briefly how this would be accomplished. First, let Pj be themeasured population in day j . These are the numbers in Table 1. We are lookingfor the function P(t) of the form in equation (1.13) which best matches the data.

0 3 6 9 12 15 180

100

200

300

400

500

600

700

Time in days

Qua

ntity

of

yeas

t

Figure 8 The growth of a yeast culture.



We could look for the function that minimizes the sum of the squares of the errors,

18∑0

(P(t j ) − Pj

)2. (1.19)

(Notice that in our case t j = j .)However, since our populations vary over two orders of magnitude, from 9.6

to 665, it makes more sense to minimize the error relative to size of the individualnumbers. This is done by minimizing

18∑0

(ln P(t j ) − ln Pj

)2 =18∑0

(ln(P(t j )/Pj )

)2. (1.20)

The function P in equation (1.13) has three parameters, P0, r , and K . The taskis to find the values of these parameters that minimize (1.19) or (1.20). This is nota task to be done using a pencil and paper. It is much too difficult. However, itcan be done quite easily on a computer if the required programs are available. Ineither case, the result is called a least squares approximation. Unlike the situationin Example 2, this is a non-linear least squares problem, which is usually an orderof magnitude more difficult that the linear least squares problem found there.

In the case at hand, when we minimized (1.20), our calculations revealed thatthe minimum occurred with P0 = 9.5998, r = 0.5391, and K = 665.0000. Thecolored curve in Figure 8 is the logistic curve with these parameters. u

The fit shown by the colored curve in Figure 8 is quite good. Many experimentssimilar to the one described in Example 1 have shown that the logistic model does avery good job of describing population growth of populations under very carefullycontrolled conditions, such as those found in a well-run laboratory.

................EXERCISES

1. A biologist starts with 100 cells in a culture. After 24 hours, he counts 300.Assuming a Malthusian model, what is the reproduction rate? What will be thenumber of cells at the end of 5 days?

2. A biologist prepares a culture. After 1 day of growth, the biologist counts 1000cells. After 2 days of growth, he counts 3000. Assuming a Malthusian model,what is the reproduction rate and how many cells were present initially?

3. A population of bacteria is growing according to the Malthusian model. If thepopulation triples in 10 hours, what is the reproduction rate? How often doesthe population double itself?

4. A population of bacteria, growing according to the Malthusian model, doublesitself in 10 days. If there are 1000 bacteria present initially, how long will ittake the population to reach 10,000?

5. A certain bacteria, given plenty of nutrient and room, is known to grow accord-ing to the Malthusian model with reproductive rate r . Suppose that the biologistworking with the culture harvests the bacteria at a constant rate of h bacteria perhour. Use qualitative analysis to discuss the fate of the culture.



6. A certain bacteria is known to grow according to the Malthusian model, dou-bling itself every 8 hours. If a biologist starts with a culture containing 20,000bacteria, then harvests the culture at a constant rate of 2000 bacteria per hour,how long until the culture is depleted? What would happen in the same timespan if the initial culture contained 25,000 bacteria?

7. A certain bacteria is known to grow according to the Malthusian model, dou-bling itself every 4 hours. If a biologist starts with a culture of 10,000 bacteria,at what minimal rate does he need to harvest the culture so that it won’t over-whelm the container with bacteria?

8. A biologist grows a culture of fruit flies in a very large enclosure with substan-tial nutrients available. Table 2 contains the data on the numbers for each of thefirst ten days.

Table 2 The growth of a population of fruit flies

Day Number of flies Day Number of flies Day Number of flies

0 10 4 28 8 851 14 5 38 9 1232 19 6 55 10 1363 24 7 72

(a) As was done in Example 1.4, use the first two data points to estimate thereproduction rate. Plot the results of the Malthusian model against the datapoints.

(b) Now use the number of flies after 5 days to estimate the reproduction rate.Plot the results of the Malthusian model against the data points.

(c) If you have a calculator or a computer with the right program, estimate thereproduction rate using linear regression. Plot the results of the Malthusianmodel against the data points.

(d) Explain the differences in the reproduction rates found in yourcomputations.

(e) In your estimation is the Malthusian model a good one for this experiment?

9. Suppose a population is growing according to the logistic equation,

d P

dt= r P

(1 − P

K

).

Prove that the rate at which the population is increasing is at its greatest whenthe population is at one-half of its carrying capacity. Hint: Consider the secondderivative of P .

10. Consider anew the logistic equation,

d P

dt= r P

(1 − P

K

), P(t0) = P0. (1.21)

Typically, mathematical ecologists will introduce dimensionless variables to re-duce the number of parameters in the logistic equation before proceeding withtheir analysis.



(a) Show that the substitutions ω = αP and s = βt transform equation (1.21)into

dω

ds= r

βω − r

αβKω2. (1.22)

(b) Find values of α and β that transform equation (1.22) into

dω

ds= ω − ω2. (1.23)

(c) Note that equation (1.23) is a variant of Bernoulli’s equation. Use the tech-nique of Exercise 22 in Section 2.4 to show that equation (1.23) has solution

ω = 1

1 − Ce−s. (1.24)

(d) Finally, use the change of variables ω = αP and s = βt , your parametersα and β found in part (b), and the initial condition P(t0) = P0 to show thatequation (1.24) is equivalent to the solution given in equation (1.13).

11. A population, obeying the logistic equation, begins with 1000 bacteria, thendoubles itself in 10 hours. The population is observed to eventually stabilizeat 20,000 bacteria. Find the number of bacteria present after 25 hours and thetime it takes the population to reach one-half of its carrying capacity.

12. A population is observed to obey the logistic equation with eventual population20,000. The initial population is 1000, and 8 hours later, the observed popula-tion is 1200. Find the reproductive rate and the time required for the populationto reach 75% of its carrying capacity.

13. In The Biology of Population Growth, published in 1925, the biologist RaymondPearl reported the data shown in Table 3 for the growth of a population of fruitflies.

Table 3 The growth of another population of fruit flies

Day Number of flies Day Number of flies Day Number of flies

0 6 12 67 24 2653 10 15 104 28 2827 21 18 163 32 3199 52 21 226

(a) Notice that data were not collected systematically. However, data werecollected on days 9 and 18. Use the method of Example 1.18 to estimatethe natural reproductive rate and the carrying capacity for a logistic model.

(b) If you have a computer program with the a least squares program, use itwith all of the data in Table 3 to estimate the natural reproductive rate andthe carrying capacity for a logistic model.

(c) Is the logistic model a good one for this data?



14. G. F. Gause, in his Struggle for Existence, simply estimated the carrying capac-ity of a population from the graph of his data. Plot the data shown in Table 4,and then use your plot to obtain an estimate of the carrying capacity.

Table 4 Estimating carrying capacityfrom data

Day Quantity Day Quantity

0 100 80 858720 476 100 967940 1986 120 993360 5510 140 9986

You now know the carrying capacity and the initial population. You canuse any other point in the table to determine the reproduction rate r . Do so, andthen superimpose the resulting logistic curve on your data plot for comparison.Can you see any problems that could occur with this method?

15. Consider a lake that is stocked with walleye pike and that the population of pikeis governed by the logistic equation

P ′ = 0.1P(1 − P/10),

where time is measured in days and P in thousands of fish. Suppose that fishingis started in this lake and that 100 fish are removed each day.

(a) Modify the logistic model to account for the fishing.

(b) Find and classify the equilibrium points for your model.

(c) Use qualitative analysis to completely discuss the fate of the fish populationwith this model. In particular, if the initial fish population is 1,000, whathappens to the fish as time passes? What will happen to an initial populationhaving 2000 fish?

16. A biologist develops a culture that obeys the modified logistic equation

P ′ = 0.38P

(1 − P

1000

)− h(t), (1.25)

where the “harvesting” is defined by the piecewise function

h(t) ={

200, if t < 3,0, otherwise.

(1.26)

(a) Use your numerical solver to plot solution trajectories for initial bacterialpopulations ranging between 0 and 1000. You’ll note that in some cases,the population “recovers,” but in others, the bacterial count goes to zero.Determine experimentally the critical initial population that separates thesetwo behaviors.

(b) Use an analytic method to determine the exact value of the “critical” initialpopulation found in part (a). Justify your answer.



17. A population, left alone, obeys the logistic law with an initial population of1000 doubling itself in about 2.3 hours. It is known that the environment cansustain approximately 10,000 individuals. Harvesting is introduced into thisenvironment, with 1500 individuals removed per hour, but only during the last4 hours of a 24-hour day. Suppose that the population numbers 6000 at thebeginning of the day. Use a numeric solver to sketch a graph of the populationover the course of the next three days. Approximate the size of the populationat the end of each day.

18. Consider the same lake as in the Exercise 15, but suppose that the fishing isdone for a fixed amount of time every day, with the result that 1% of the fishare caught each day.

(a) Modify the logistic model to account for the fishing.

(b) Find and classify the equilibrium points for your model.

(c) If the initial fish population is 1000, what happens to the fish as time passes?

19. Adjust the “standard” logistic equation

d P

dt= r P

(1 − P

K

)

to reflect the fact that a fixed percentage γ of the population is harvested perunit time. Use qualitative analysis to discuss the fate of the population. In youranalysis, discuss two particular cases: (1) γ < r and (2) γ > r .

20. In Exercise 19, examine the units of the term γ P and explain why the functionY (γ ) = γ P is called the yield. Suppose that the harvesting strategy defined inExercise 19 is kept in place long enough for the population to adjust to its newequilibrium level. What value of γ will maximize the yield at this level? Whatwill be the yield and the new equilibrium level of the population for this valueof γ ?

3.2 Models and the Real WorldMathematical models are meant to explain what is happening in the real world.It is not enough to derive models from theoretical considerations. It is necessaryto check the predictions of our models with what is happening in reality. We didthis in the previous section when we looked at the implications of the Malthusianmodel of population growth and realized that its prediction of unlimited exponentialgrowth was unrealistic. The prediction is only good if the assumption of unlimitedresources is satisfied. While this might be true for small populations, it is certainlynot true in the long run. We then went back to the drawing board and came up withthe logistic model. Our analysis of the experimental data in Example 1.18 showedthat in that case, under controlled circumstances, the logistic model worked verywell.

We should do the same thing for the logistic model. If we want to apply thelogistic model in new circumstances, it is important to know if these circumstancesfit the assumptions behind the logistic model. Let’s recall what those assumptionsare. First we assumed that the population changed due to births and deaths. We thenallowed the birth and death rates to vary with the population in a way that reflected


3.2 Models and the Real World 139

the competition between individuals for limited resources. We did not allow anychange of the reproduction rate with respect to time. Now let’s look at some othersituations.

A Malthusian model of early U.S. populationA number of attempts have been made to model the population of the UnitedStates. Pierre-Francois Verhulst, a Belgian mathematician, argued that the Malthu-sian model could be used to model rapidly growing populations in environmentscontaining seemingly unlimited resources.

The United States (in the late eighteenth and early nineteenth centuries)offers just such an example of a rapidly growing population that is expand-ing as if it had unlimited resources. [Verhulst, Pierre-François. 1845.Recherches mathematiques sur la loi d'accroissement de la population.Noveaux Mémoires de l'Académie Royale des Sciences et Belles-Lettersde Bruxelles 18: 1–38]

Verhulst used United States census data for the years 1790–1840, provided inTable 1. He used the arithmetic mean to estimate the population in intercensalyears, then proceeded to show that the population grew in the geometric progressionpredicted by the Malthusian model.

Table 1 Early population ofthe United States

Year Time Population

1790 0 3,929,8271800 10 5,305,9251810 20 7,239,8141820 30 9,638,1511830 40 12,866,0201840 50 17,062,566

If we use the first two data points in Table 1 and the technique developed inExample 1.4, we get the equation

P = 3,929,827e0.03(t−1790). (2.1)

The plot of equation (2.1) is superimposed on Verhulst’s data in Figure 1. Notethat the graph of equation (2.1) seems a particularly good fit for the early years, butbegins to pull away from the data in the census years 1830 and 1840.

Now we come to the main question of this section. Is Verhulst’s assertion thatthis period of U.S. population growth is a good example of “a rapidly growing pop-ulation that is expanding as if it had unlimited resources” valid? It might be arguedthat, in addition to births and deaths, there was a third factor affecting the growthof the population, namely immigration. However, in the period between 1790 and1820, immigration to the United States was rather small in comparison to the pop-ulation increases that were occurring. Consequently, we can say that Verhulst’s useof the Malthusian model was a valid approximation. Indeed, the agreement of (2.1)with the data in Table 1 as shown in Figure 1 is impressive.



1790 1800 1810 1820 1830 18400

5

10

15

20

Popu

latio

n in

mill

ions

Year

Figure 1 Fitting a Malthusian model to early U.S. population.

Logistic models of U.S. population growthWe have just seen that the early growth of the U.S. population was Malthusian innature. What about the growth since then?

In 1920, Pearl and Reed used the logistic equation to model the United Statespopulation. Their census data, taken from their report to the National Academy ofSciences,4 came from the Bureau of Census figures. It included the data in Table2, up to the year 1910. Table 2 comes from the Statistical Abstract of the UnitedStates, 1998. It shows the population of the United States in thousands. The figurefor 2000 is a projection.5 This data is plotted in Figure 2.

Table 2 Population of the United States (in thousands)

Year Population Year Population Year Population

1790 3,929 1870 39,818 1940 131,6991800 5,308 1880 50,156 1950 151,3261810 7,240 1890 62,948 1960 179,3231820 9,638 1900 75,995 1970 203,3021830 12,866 1910 91,972 1980 226,5421840 17,069 1920 105,711 1990 248,7181850 23,192 1930 122,755 2000 273,4821860 31,443

4 Pearl and Reed, On the Rate of Growth of the Population of the United States Since 1790 and ItsMathematical Representation, Proceedings of the National Academy of Sciences, Volume 6, June 15,1920, Number 6.5 The observant reader will notice that the data in Table 2 differs somewhat from that in Table 1. Thisis simply because the data is from different sources. It frequently happens that different sources reportconflicting data. This is sometimes very disconcerting.


3.2 Models and the Real World 141

1790 1820 1850 1880 19100

20

40

60

80

100

Popu

latio

n in

mill

ions

Year

Figure 2 Fitting the logistic model to U.S. population.

By selecting three data points at equally spaced time intervals and using thetechnique in equations (1.16) and (1.17) of Section 1, Pearl and Reed were able tofit the logistic model to the population data in Table 2 up to 1910. Using the sametechniques as before and the populations for 1790, 1850, and 1910, we fit the datain Table 2 with the logistic model. The estimates of the parameters are6

r = 0.0313 and K = 197,274. (2.2)

These data, together with P0 = 3,939,000, are used in equation (1.13), which is

P(t) = K P0

P0 + (K − P0)e−r(t−t0). (2.3)

The plot of this function is superimposed on the plot of the U.S. population data inFigure 2. Note the excellent fit.

Pearl and Reed then used their model to extrapolate the eventual behavior of theU.S. population. Let’s check how well their model predicted the population of theUnited States. In Figure 3, we extend the plot of equation (2.3) with the parametersin (2.2) to the present, and we show the actual data from Table 2.

The model of Pearl and Reed predicted a carrying capacity of 197,274,000 peo-ple. However, in 1990 the U.S. population was 248,718,301 people, far beyond thatpredicted by Pearl and Reed. Thus, we see that using the logistic model to predictthe population of the United States failed spectacularly after about 1950.

We might ask what would happen if we used all of the data in Table 2 andrecalculated the parameters by finding a least squares approximation. When we do,we get

r = 0.0289 and K = 279,000,000.

6 These computations are a little sensitive, especially to the value of r used to compute K . It is importantto keep the actual computed value of r when computing K . Even then, a calculator may not get the valuefound here due to lack of precision.



1800 1850 1900 1950 20000

50

100

150

200

250

300

Popu

latio

n in

mill

ions

Year

Figure 3 Logistic model projection of U.S. population.

The Bureau of the Census projects that the carrying capacity of 279,000,000 will besurpassed by the year 2010, and that the population will continue to grow after that.It appears that the logistic model is not very good at predicting the population of theUnited States.

Why is this true? If we think about how the population has grown in the UnitedStates, we come to the conclusion that the assumptions of the logistic model aresimply not satisfied. For one thing, since about 1830 immigration has been a veryimportant factor. For example, in the first decade of the twentieth century therewere 8,795,386 immigrants, while the population increased by 15,977,691. Clearlyimmigration cannot be ignored. In addition, great improvements in public hygieneand in health care have had a big impact on the birth and death rates. Agriculturaltechnology has increased the production of food by an incredible amount. Birthrates have decreased over the past century because of individual choice—a factorthat is unrelated to the availability of resources or to the size of the population.

For these reasons, it should not be a surprise that the logistic model gives sucha poor fit with the U.S. population curve. Indeed, it would be highly surprising ifthe logistic model were a good model for any segment of human population over avery long period of time.

For the total human population of the earth, such a model might be accurate forthe period up to about 10,000 years ago, while humans were all hunter/gatherers.However, with the introduction of agriculture, the carrying capacity of the earth in-creased dramatically. The period from then until the start of the industrial revolutionmight be another period when a logistic model would be accurate. However, sincethe start of the industrial revolution changes have occurred at a very high rate, all ofwhich have served to allow the earth to support more people. In addition, there havebeen changes in the attitudes of people that affect the reproduction rate. It is notlikely that a logistic model would be useful for the human population of the earthsince that time.


3.3 Personal Finance 143................EXERCISES

1. Use the method of Example 1.4 on the first two data points in Table 1 to deriveequation (2.1).

2. Use equations (1.16) and (1.17) together with the population data for 1790,1850, and 1910 in Table 2 to verify the estimates of the parameters in (2.2).

3. Find in your library or on the internet the historical census data for one of theUnited States, or for some other country. Attempt to model the data usingthe Malthusian model and the logistic model. Critique the effectiveness of themodels.

3.3 Personal FinanceThere are a number of problems involving personal finance that can be modeled us-ing differential equations. We will start by considering what happens to the balanceof a savings account, or of money invested in a portfolio of stocks and bonds.

Let P(t) be the balance at time t , and suppose the account pays interest at a rateof r percent per year, compounded continuously. This means that the increase in thebalance between times t and t + 1t is approximately

P(t + 1t) − P(t) ≈ r P1t.

Hence

P ′(t) = lim1t→0

P(t + 1t) − P(t)

1t= r P. (3.1)

This is once more the exponential equation, which we have already seen in avariety of applications. We know that the solutions have the form

P(t) = Cer t .

If the initial balance is P(0) = P0, then we see that C = P0 and

P(t) = P0er t .

E X A M P L E 3 . 2 u Suppose you put $1000 in a savings account with a continuously compounded in-terest rate of 5% per year. What will be the balance after 40 years?

In this case, P0 = 1000, and r = 0.05. Hence

P(40) = 1000e0.05×40 = 1000e2 ≈ 7389. u

Under the assumptions we have made, the balance in our savings account willgrow exponentially. The basic assumption is that the interest rate r is a constant.Of course, in practice this is not true. Interest rates change constantly in reactionto a variety of economic and political events and are unpredictable. This limits theeffectiveness of our model. To allow for this unpredictability, we should do theanalysis for a variety of interest rates ranging from the lowest to the highest that weexpect. This kind of “what if” analysis will allow us to bracket the real outcome.



An interest-bearing account with steady withdrawalsNext let’s look at the balance P(t) in a savings or an investment account from whichthe amount W is withdrawn every year and that pays interest at the rate of r percentper year, compounded continuously. The financial facts provide a formula for therate of change of the balance:

rate of change of P(t) = interest per unit time − withdrawal per unit time

= I − W.(3.3)

I is the interest per year. If r is the annual interest rate, then I = r P . Similarly, Wis the withdrawal measured in dollars per unit time. Hence our model is

d P

dt= r P − W. (3.4)

On the other hand, if we deposit D dollars per year, the equation becomes

P ′ = r P + D. (3.5)

Using the limit quotient definition of derivativeIf that derivation is unclear, there is a slower way to do the derivation that mightbe easier to understand. It uses the limit quotient definition of the derivative. Forexample, instead of leaping directly to equation (3.4), we could compute the changein the balance over a small period of time, 1t . Then

P(t + 1t) − P(t) = interest earned in time 1t − withdrawal in time 1t . (3.6)

The interest earned is approximately

r P(t)1t, (3.7)

since the interest rate r is the percent earned per amount invested per unit time. Onthe other hand, W is the amount withdrawn per unit time, so the amount withdrawnin time 1t is

W1t. (3.8)

Putting these three together, we get

P(t + 1t) − P(t) ≈ r P(t)1t − W1t.

Now we can compute the derivative using the limit quotient definition:

d P

dt= lim

1t→0

P(t + 1t) − P(t)

1t

= lim1t→0

r P(t)1t − W1t

1t

= r P − W.

Thus we reach equation (3.4) using a process that might make more sense.


3.3 Personal Finance 145

Systematic use of dimensions and unitsAnother factor that can often be used to keep confusion at bay during the model-

ing process is the careful and systematic use of dimensions and units. For example,the difference between equations (3.7) and (3.8) can be misleading. In particular,why doesn’t (3.8) read W P1t in complete analogy to (3.7)? The answer is thatr and W have different dimensions. The interest rate r is usually referred to as apercentage. It measures the fraction of a dollar earned per unit time. Since a frac-tion has no dimension and we are measuring time in years, the dimensions of r aresimply years−1. On the other hand, the withdrawal rate W is expressed in terms ofdollars per year. Hence the units of W are dollars × years−1. The left-hand side of(3.6) is measured in dollars. With the dimensions of r and W as found earlier, theexpressions of both (3.7) and (3.8) are also measured in dollars. On the other hand,the expression W P1t is measured in dollars2, which does not match.

The point is that by keeping careful track of the dimensions of the quantitiesbeing used, it is possible to keep from making errors.

Systematic savingsOur next problem is to start planning for retirement. The same process modelssavings towards any goal, such as the education of a child or the purchase of ahouse.

E X A M P L E 3 . 9 u Suppose you are just starting to work and you decide to save $2000 each year.Assuming that you have no savings to begin with and that your savings will earn5% per year, compounded continuously, how much will you accumulate after 30years?

We developed the model for this in equation (3.5). The equation in this case is

P ′ = 0.05 P + 2,

where P is the principal balance in thousands of dollars. This is a linear equation.Let’s work through the solution process carefully.

We first look for an integrating factor, and we know that one is given by

u(t) = e− ∫0.05 dt = e−0.05t .

Hence [e−0.05t P

]′ = e−0.05t[P ′ − 0.05P

] = 2e−0.05t .

Integrating both sides yields

e−0.05t P(t) = −40e−0.05t + C,

orP(t) = −40 + Ce0.05t .

The initial condition becomes

0 = P(0) = −40 + C,



so C = 40 and the solution is

P(t) = 40(e0.05t − 1).

To answer the question, we evaluate this when t = 30 to get

P(30) = 40(e1.5 − 1) = 139.2676.

Our balance is measured in thousands of dollars, so we learn that our balance after30 years is $139,268. u

Planning for retirementThat’s not a bad start for retirement. After all, the amount of money put into savingswas $2000 per year for 30 years, for a total of $60,000. Because of the compoundedinterest, this has grown to more than double that amount.

Now, however, let’s turn the question around. How much money do you needto retire on?

E X A M P L E 3 . 1 0 u After some thought, you have decided that you will need $50,000 each year to liveon after you retire, and that you should plan on living 30 years after your retirement.Assuming that your retirement account will earn 5% interest while you are takingout $50,000 each year, how much money must be in the retirement account whenyou retire?

Let P(t) be the balance in your retirement account at time t after retirement. LetP0 denote the balance in your retirement account when you retire. Then P(0) = P0.The problem is to find P0 so that P(30) ≥ 0. Once more we will use a thousanddollars as our unit.

According to the model developed in equation (3.4),

P ′ = 0.05P − 50.

The integrating factor we found in Example 3.9 will also work here. Thus we have

[e−0.05t P

]′ = −50e−0.05t .

Integrating, we gete−0.05t P(t) = 1000e−0.05t + C,

orP(t) = 1000 + Ce0.05t .

If we let P(0) = P0 denote the balance at the time you retire, then we can evaluatethe constant C ,

P0 = P(0) = 1000 + C.

Therefore, C = P0 − 1000, and the solution is

P(t) = 1000 + (P0 − 1000)e0.05t .



Since you want to have that $50,000 each year until you die 30 years afterretiring, you will want P(30) ≥ 0. If you spend your last cent the day you die, youwill want

0 = P(30) = 1000 + (P0 − 1000)e1.5.

We can solve this equation for P0, getting

P0 = 1000(1 − e−1.5) = 776.8698.

Thus you will need to have saved $776,870 before you retire in order to have theretirement you want. u

Saving for retirementWell, that $139,000 we saved in Example 3.9 doesn’t seem so great any more. Youare going to have to do much better than that. But just how are you going to accu-mulate the funds needed to finance a comfortable retirement?

E X A M P L E 3 . 1 1 u After some more thought prompted by the previous example, you decide that youshould put a fixed percentage ρ of your salary into your retirement account. Thequestion is, what value of ρ will achieve our goal?

First, you realize that your current salary of $35,000 per year will not stay atthat level forever, hopefully for no more than one year. We need a model of howyour salary will grow over time. Let’s assume that your salary will grow at 4%per year. That’s only a little more than the inflation rate. This thought leads to thedifferential equation S′ = 0.04S, where S(t) is your annual salary in thousands ofdollars. We are very familiar with the exponential equation, so we easily solve thisequation to get

S(t) = 35e0.04t .

Now you notice that with this model, your salary in year 40 will be over$173,000. This seems excessive, but you are assured by your financial advisorsthat this type of increase over a lifetime is not at all unusual. Remember, we areincluding inflation in our forecasts, and the 4% per year increase we are projectinghardly covers the historical inflation rate. However, there is another concern. If yoursalary at retirement is this large, what kind of income should you plan for in yourretirement years? Clearly the $50,000 in the previous example is too little. Youdecide that $100,000 is a more reasonable figure. This means that the size of yourretirement account at retirement has to be double that found in the previous example.(Why?) You decide to be cautious and aim for a retirement fund of $1,600,000.

We will assume once more that your retirement account will earn an interestrate of 5%. Let P(t) denote the balance in thousands of dollars in your retirementaccount at time t . P will grow between times t and t + 1t from two sources, theinterest on the balance, which is 0.05P(t)1t , and from your investment, which isρ S(t)1t . Hence we have

P(t + 1t) − P(t) ≈ 0.05P(t)1t + ρS(t)1t.



Therefore,

P ′(t) = lim1t→0

P(t + 1t) − P(t)

1t

= 0.05P(t) + ρS(t)

= 0.05P(t) + 35ρe0.04t .

We solve this linear equation as before. Again u(t) = e−0.05t is an integratingfactor, and we get [

e−0.05t P]′ = 35ρe−0.01t .

Integrating, we get

e−0.05t P(t) = 35ρ

∫e−0.01t dt = −3500ρe−0.01t + C,

orP(t) = −3500ρe0.04t + Ce0.05t .

We will again assume that you are starting with no money in your retirementaccount, so this initial condition says

0 = P(0) = −3500ρ + C,

which allows us to conclude that C = 3500ρ, and

P(t) = 3500ρ(e0.05t − e0.04t

).

We can now compute what ρ has to be by comparing this with our goal, whichis that P(40) = 1600. Completing the computation, you find that you must save18.77% of your salary to ensure a comfortable retirement. u

Saving almost 19% of your salary every year seems like a lot. You decide toexplore other strategies.

E X A M P L E 3 . 1 2 u When your salary is low, at the beginning of your career, saving this much of yoursalary will be hard. Perhaps even while you are young you should enjoy more ofthe fruits of your labor. What would happen if you were to start saving at a moremodest rate and slowly increase your savings rate over time? You decide that thepercent to save in year t is

ρ(t) = Rt

40.

With this choice, R will be the savings rate just before you retire after 40 years ofwork. What does R have to be in order to achieve your retirement goal?

The model in this case is similar to what it was in the previous example, but ithas to change to accommodate the variability of your savings rate:

P ′ = 0.05P + ρS

= 0.05P + Rt

40× 35e0.04t

= 0.05P + 0.875Rte0.04t .



Again this equation is linear, and again u(t) = e−0.05t is an integrating factor. Weproceed as before. We integrate[

e−0.05t P]′ = 0.875Rte−0.01t ,

to get

e−0.05t P(t) = 0.875R∫

te−0.01t dt

= −87.5R(t + 100)e−0.01t + C,

orP(t) = −87.5R(t + 100)e0.04t + Ce0.05t .

Using P(0) = 0 to compute that C = 8750R, we end up with

P(t) = 87.5R(100e0.05t − (t + 100)e0.04t

).

Finally, we use our goal P(40) = 1600 to compute that R = 0.4021. Thismeans that with this plan, although your savings rate in your early career will besmall, in the last year before your retirement you will have to save over 40% of yoursalary. u

Of course other strategies are possible. In particular you might think aboutexploring ways to increase the income on your investments. Didn’t you hear some-where that over the long term, stocks return more than 8%? We will explore someother cases and strategies in the exercises.

................EXERCISES

1. Suppose that $1200 is invested at a yearly rate of 5%, compoundedcontinuously.

(a) Assuming no additional withdrawals or deposits, how much will be in theaccount after 10 years?

(b) How long will it take the balance to reach $5000?

2. Jamal wishes to invest an unknown sum in an account where interest is com-pounded continuously. Assuming that Jamal makes no additional deposits orwithdrawals, what annual interest rate will allow his initial investment to dou-ble in exactly five years?

3. Alicia opens an account that pays an annual rate of 6% compounded continu-ously with an initial investment of $5000. After that, she deposits an additional$1200 per year. Assuming that no withdrawals are made and she continueswith the same yearly deposit over a period of 10 years, how much will be in theaccount at the end of the 10-year period?

4. On the day of his birth, Jason’s grandmother pledges to make available $50,000on his eighteenth birthday for his college education. She negotiates an accountpaying 6.25% annual interest, compounded continuously, with no initial de-posit, but agrees to deposit a fixed amount each year. What annual depositshould be made to reach her goal?



5. Andre inherits $50,000 from his grandfather’s estate. The money is in an ac-count that pays 5% annual interest, compounded continuously. The terms of theinheritance require that $8000 be withdrawn each year for Andre’s educationalexpenses. If no additional deposits are made, how long will the inheritance lastbefore the funds are completely gone?

6. Clarissa wants to buy a new car. Her loan officer tells her that her annual rateis 8%, compounded continuously, over a four-year term. Clarissa informs herloan officer that she can make equal monthly payments of $225. How much canClarissa afford to borrow?

7. David and Mary would like to purchase a new home. They borrow $100,000at 8% annual interest, compounded continuously. The term of the loan is 30years. What fixed, annual payment will satisfy the terms of their loan?

8. Don and Heidi would like to buy a home. They’ve examined their budget anddetermined that they can afford monthly payments of $1000. If the annualinterest is 7.25%, and the term of the loan is 30 years, what amount can theyafford to borrow?

9. Jose is 25 years old. His current annual salary is $28,000. Over the next 20years, he expects his salary to increase continuously at a rate of 1% per year.He establishes a fund paying 6% annual interest, compounded continuously,with an initial deposit of $2500 and a promise to deposit a fixed percentageof his annual income each year. Find that fixed percentage if Jose wants hisbalance to reach $50,000 at the end of the 20-year period.

10. Adriana opens a savings account with an initial deposit of $1000. The annualrate is 6%, compounded continuously. Adriana pledges that each year her an-nual deposit will exceed that of the previous year by $500. How much will bein the account at the end of the tenth year?

Discrete versus continuous. You may have cast a somewhat skeptical eye at ourfinancial models involving continuous compounding of interest. After all, no onepays off their loans continuously. It would be difficult to imagine how that couldbe accomplished. Payments are made at regular intervals, perhaps yearly, but morelikely monthly. Let’s examine how accurately our continuous models reflect the realworld of finance.

11. The recursive definition,

a(n + 1) = ra(n), a(0) = a0,

is called a first-order difference equation and generates the sequence

a0, ra0, r(ra0), r(r(ra0)), . . . .

A little simplification shows that the nth term of this sequence is

a(n) = a0rn.

Now, suppose that I represents the annual interest rate, but the interest isawarded in discrete packets, m times per year. Then the rate awarded during



each compounding period is I/m. Consequently, if the initial investment isP0, the balance is P0(1 + I/m) at the end of the first compounding period,P0(1 + I/m)2 at the end of the second compounding period, and so on.

(a) Give a first-order difference equation with an initial condition that gener-ates a sequence describing the balance in the account at the end of eachcompounding period.

(b) Find a formula for the nth term of the sequence generated by the first-orderdifference equation created in part (a).

12. Arkady invests $2000 in an account paying 6% annual interest.

(a) If the interest is compounded continuously, and no additional deposits orwithdrawals are made, how much will be in the account at the end of 10years?

(b) If the interest is awarded in discrete annual packets, then you want to usethe discrete formula generated in Exercise 11. In this case, m = 1, so

P(10) = 2000

(1 + 0.06

1

)10

≈ $3,581.70.

Calculate the balance in the account at the end of the 10-year period if theinterest is compounded

(i) semi-annually (twice per year),(ii) monthly (12 times per year),

(iii) daily (365 times per year).(c) Write a small paragraph explaining the point of this problem.

13. The first-order difference equation

a(n + 1) = ra(n) + b, a(0) = a0,

generates the sequence

a0, ra0 + b, r(ra0 + b) + b, r(r(ra0 + b) + b) + b, . . . .

A little simplification shows that the nth term of this sequence is

a(n) = a0rn + b(1 + r + r2 + · · · + rn−1).

(a) Show that

a(n) =(

a0 − b

1 − r

)rn + b

1 − r.

(b) Let I represent the annual interest rate, m the number of compoundingperiods in a year, P0 the initial investment, and d the fixed deposit at theend of each compounding period. Then the balance at the end of eachcompounding period is generated by the first-order difference equation

P(n + 1) =(

1 + I

m

)P(n) + d, P(0) = P0.

Use the result of part (a) to show that the balance at the end of n compound-ing periods is given by

P(n) =(

P0 + md

I

) (1 + I

m

)n

− md

I.



14. Demetrios opens an account with an initial investment of $2000. The annualinterest rate is 5%.

(a) If the interest is compounded continuously and Demetrios makes an addi-tional $1000 deposit every year, what will be the balance at the end of 10years?

(b) If the interest is compounded quarterly (four times per year) and $250 isdeposited at the end of each compounding period, what will be the balanceafter 10 years?

(c) What happens if the interest is compounded daily?

15. Chieh-Hsien purchases a new car and finances $12,000 at an annual rate of 8%for 5 years.

(a) If the interest is compounded continuously, what are his monthly payments?

(b) If the interest is compounded monthly, what are his monthly payments?

3.4 Electrical CircuitsA typical electric circuit involves a voltage source, a resistor, a capacitor, and aninductor. A simple diagram of an electric circuit is given in Figure 1. The voltagesource is denoted by E , the resistor by R, the capacitor by C , and the inductor by L .

The voltage source could be a battery or a generator. A battery provides asteady, constant voltage, while a generator typically produces a variable voltage,usually sinusoidal in nature. The voltage causes electrons, and hence electric charge,to flow through the circuit. The rate at which charge flows is called the current. Wewill denote the source voltage by E = E(t). All voltages are measured in volts,denoted by V, and the currents are measured in amps, denoted by A.

Notice that in Figure 1 we have chosen a direction in which the current flows.This choice is essentially arbitrary, but once chosen it is fixed. An alternating cur-rent by definition flows both ways at different times. When a current flows in thedirection opposite to the chosen direction, it is given negative values.

L

C

R

E

I

I

+

–

Figure 1 An RLC circuit.


3.4 Electrical Circuits 153

The derivation of the differential equation model of the circuit depends on thefollowing experimental laws. The first three laws concern the voltage drop thatoccurs across various components of a typical circuit. We will refer to these as thecomponent laws.

1. Ohm’s law. A resistor impedes the flow of electrical charge. Every electricaldevice, such as a heater, a toaster, or an electrical stove, has some resistance tothe flow of electrical charge. The voltage drop ER across a resistor is propor-tional to the current I ,

ER = RI,

where R is the resistance. The resistance is measured in ohms, denoted by Ä.

2. Faraday’s law. A typical inductor is a coil of wire. Current flowing throughthe coil produces a magnetic field. The magnetic field opposes any change inthe current. The voltage drop EL across an inductor is proportional to rate ofchange of the current,

EL = Ld I

dtwhere L is the inductance. The inductance is measured in henrys, denotedby H.

3. Capacitance law. A capacitor, or condenser, is a device that stores electriccharge. It usually consists of two parallel plates separated by an insulator. Con-sequently, there is little or no current passing through a capacitor. As currentflows up to a capacitor, the charge builds up there. Eventually the charge willbe so large that the current reverses the direction of its flow. The voltage dropEC across a capacitor is proportional to the charge Q on the capacitor,

EC = 1

CQ

where C is the capacitance and Q is the charge on the capacitor. Capacitanceis measured in farads, denoted by F. Charge is measured in coulombs, denotedby C.

In addition, there are two laws discovered by Gustav Kirchhoff that are veryimportant for understanding circuits. They are not independent and often, either canbe used to analyze a circuit.

4. Kirchhoff’s voltage law. The sum of the voltage drops around any closed loopin a circuit must be zero.

5. Kirchhoff’s current law. The sum of currents flowing into any junction is zero.

To derive the differential equation that governs the circuit in Figure 1, we startwith Kirchhoff’s voltage law, which states that the sum of the voltage drops acrossthe resistor, the inductor, the capacitor, and the voltage source is zero,

EL + EC + ER − E = 0.

In the preceeding equation, we have started and ended at the positive terminal ofthe voltage source. The voltage drop is positive in the direction of the current and



negative in the opposite direction. We can rearrange this equation to read

EL + EC + ER = E . (4.1)

Using the component laws listed above we obtain

RI + Ld I

dt+ 1

CQ = E . (4.2)

The current I represents moving charge. Therefore

I = d Q

dt. (4.3)

We differentiate equation (4.2) and use (4.3) to eliminate Q from the result to obtain

Rd I

dt+ L

d2 I

dt2+ 1

C

d Q

dt= d E

dt,

or

Ld2 I

dt2+ R

d I

dt+ 1

CI = d E

dt. (4.4)

Initial conditions are usually imposed on the initial charge on the capacitor Q(0)

and the initial current I (0). To solve the second-order equation (4.4), we must knowboth I (0) and I ′(0). I ′(0) is found by inserting I (0) and Q(0) into (4.2) and solvingfor I ′(0).

First-order systems and second-order equations will be the topic of much ofthe rest of this book. However, there are special cases that can be solved usingtechniques learned in Chapter 2.

Suppose there is no capacitor in the circuit. Then equation (4.2) becomes

RI + Ld I

dt= E . (4.5)

This is a linear first-order equation for the current I .

E X A M P L E 4 . 6 u Suppose the electrical circuit has a resistor of R = 1/2 Ä, and an inductor ofL = 1 H. Assume the voltage source is a constant E = 1 V. If the initial current is0 A, find the resulting current.

With the given data (4.5) becomes

d I

dt+ 1

2I = 1.

This equation is both linear and separable. Solving it, we find that

I (t) = 2(1 − e−t/2). u



Next suppose there is no inductor in the circuit. Then equation (4.2) becomesRI + Q/C = E . Using I = d Q/dt, this becomes

Rd Q

dt+ 1

CQ = E . (4.7)

This time we have a first-order equation for the charge on the capacitor.

E X A M P L E 4 . 8 u Suppose the electrical circuit has a resistor of R = 2 Ä and a capacitor of C = 1/5 F.Assume the voltage source is E = cos t V. If the initial current is 0 A, find theresulting current.

With the given data, equation (4.7) becomes

2d Q

dt+ 5Q = cos t.

This is a linear equation for Q. We find that e5t/2 is an integrating factor. Multiply-ing by it, we get

(e5t/2 Q

)′ = e5t/2

(d Q

dt+ 5

2Q

)= 1

2e5t/2 cos t.

Integrating, this becomes

e5t/2 Q(t) = 1

2

∫e5t/2 cos t dt

= 1

29e5t/2 (2 sin t + 5 cos t) + C.

Solving for Q, we get

Q(t) = 1

29(2 sin t + 5 cos t) + Ce−5t/2.

Differentiating, we find the formula for the current:

I (t) = 1

29(2 cos t − 5 sin t) − 5

2Ce−5t/2.

We evaluate the constant C using the initial condition I (0) = 0, and our final answeris

I (t) = 1

29

(2 cos t − 5 sin t − 2e−5t/2

). u

................EXERCISESA resistor (20 Ä) and capacitor (0.1 F) are joined in series with an electromotiveforce (emf) E = E(t), as shown in Figure 2. If there is no charge on the capacitorat time t = 0, find the ensuing charge on the capacitor at time t for the given emf ineach of Exercises 1–6.

R

CE

I

I

+

−

Figure 2 An RC circuit.



1. E(t) = 100 V 2. E(t) = 100e−0.1t V

3. E(t) = 100 sin 2t V 4. E(t) = 100 cos 3t V

5. E(t) = 100 − t V 6. E(t) = 100(1 − e−0.1t) V

An inductor (1 H) and resistor (0.1 Ä) are joined in series with an electromotiveforce (emf) E = E(t), as shown in Figure 3. If there is no current in the circuit attime t = 0, find the ensuing current in the circuit at time t for the given emf in eachof Exercises 7–12.L

R

E

I

I

+

–

Figure 3 An LR circuit.

7. E(t) = 1 V 8. E(t) = e−0.1t V

9. E(t) = 5 sin 2π t V 10. E(t) = 4 cos 3t V

11. E(t) = 10 − 2t V 12. E(t) = 10(1 − e−0.05t) V

13. Solve the general initial value problem modeling the RC circuit,

Rd Q

dt+ 1

CQ = E, Q(0) = 0,

where E is a constant source of emf.

14. Solve the general initial value problem modeling the LR circuit,

Ld I

dt+ RI = E, I (0) = I0,

where E is a constant source of emf.

15. A resistor (20 Ä) and capacitor (1 F) are linked in series with an electromotiveforce (emf) E = E(t) in a RC circuit (see Figure 2). If the emf is given asE(t) = 10e−0.01t and the charge on the capacitor is zero at time t = 0, find themaximum charge on the capacitor and the time that it will occur.

16. An inductor (10 H) and resistor (1 Ä) are linked in series with an electromotiveforce (emf) E = E(t) in an LR circuit (see Figure 3). If the emf is given asE(t) = 5e−0.05t and the current is zero at time t = 0, find the maximum currentin the circuit and the time that it will occur.

17. A resistor (10 Ä) and capacitor (1 F) are linked in series with an electromotiveforce (see Figure 2). Initially, there is no charge on the capacitor. The emfproduces a constant voltage difference of 100 V for the first five seconds, afterwhich it is switched off.

(a) Use a numerical solver to approximate the charge on the capacitor at theend of 10 seconds.

(b) Set up and solve a differential equation modeling this circuit, then use theresulting formula to calculate the charge on the capacitor at the end of 10seconds.

18. An RC circuit (see Figure 2) contains a resistor (10 Ä) and a capacitor (0.2 F)linked in series with an electromotive force E(t) = 10 sin 2π t .

(a) Use a numerical solver to sketch the graph of the charge on the capac-itor versus time for each of the initial conditions Q(0) = −1, −0.75,

−0.5, . . . , 1 on the time interval [0, 15]. What appears to happen to each



of the solutions with the passage of time? Explain what might be meant bya steady-state response.

(b) What is the frequency of the emf? What appears to be the frequency ofthe steady-state response? Hint: In the function f (t) = sin ωt , recall thatω is called the angular frequency, the period is given by T = 2π/ω, andthe frequency is the reciprocal of the period. Think of the frequency as thenumber of cycles per second, a unit known as Hertz (Hz).

19. An LR circuit (see Figure 3) contains an inductor (1 H) and a resistor (0.5 Ä)linked in series with an electromotive force E(t) = cos π t . Use a numericalsolver to sketch the graph of the current in the circuit versus time for each of theinitial conditions I (0) = −1, −0.75, −0.5, . . . , 1 on the time interval [0, 20].What is the frequency of the emf? What appears to be the frequency of thesteady-state response (see Exercise 18)?

20. An LR circuit, modeled by the initial value problem

I ′ + I = cos 2t, I (0) = 0,

has solution

I = 2

5sin 2t + 1

5cos 2t − 1

5e−t ,

which is made up of a transient response, −(1/5)e−t , and a steady-state re-sponse, (2/5) sin 2t + (1/5) cos 2t .

(a) On one plot, sketch the graph of the solution in black, the transient responsein red, and the steady-state response in blue.

(b) Create a second plot that contains the steady-state response in blue. Useyour numerical solver to superimpose the solutions of I ′ + I = cos 2ton your plot for the initial conditions I (0) = −1, −0.5, . . . , 1. Explain thebehavior of the solutions in terms of the steady-state and transient response.

21. In the circuit modeled by the differential equation

RI + 1

CQ = E sin ωt,

show that, regardless of initial condition, the eventual charge on the capacitoris given by the expression

EC

1 + R2C2ω2(sin ωt − RCω cos ωt).

22. In the lab, measuring the voltage drop across the capacitor of an RC circuit inFigure 2 is easier than measuring the charge buildup on the capacitor. Showthat

Rd Q

dt+ 1

CQ = E cos ωt

is equivalent to the differential equation

RCdV

dt+ V = E cos ωt,



where V is the voltage drop across the capacitor. Show that, regardless of theinitial voltage drop across the capacitor, the eventual voltage is given by

V = E

1 + R2C2ω2(RCω sin ωt + cos ωt).

23. Consider the circuit modeled by the differential equation

L I ′ + 1

CQ = 0, (4.9)

with initial conditions Q(0) = Q0 and I (0) = 0. Note that the substitutionI ′ = Q′′ is not much use, because we have not yet learned how to solve second-order differential equations. However, you can use the chain rule to show thatI ′ = I (d I/d Q) and this substitution will make equation (4.9) separable. Solvethe resulting equation for I in terms of Q, then solve this result for Q in termsof t . Finally, find the current as a function of t .

Project 3.5 The Spruce Budworm

In 1978, Ludwig, Jones, and Holling published a paper detailing the interactionbetween the spruce budworm and the balsam fir forests in eastern Canada. Theybegan by assuming that the budworm population B obeys the logistic model definedby

d B

dt= rB B

(1 − B

K B

),

where rB is the intrinsic growth rate of the budworm and K B is the carrying capacity,which is assumed to depend upon the amount of foliage available.

The growth of the budworm is affected by predation, mainly in the form of birdsand parasites. Thus, the equation governing the budworm population becomes

d B

dt= rB B

(1 − B

K B

)− p(B),

where the rate of predation p(B) is assumed to depend on the budworm population.

1. Ludwig et al. made the assumption that

p(B) = β B2

α2 + B2

where α and β are positive. Sketch the graph of p(B) versus B, assuming apositive budworm population. What is the limiting rate of predation? At whatvalue of B does p(B) reach one-half of its limiting rate of predation? Explainhow the consumption of budworms by avian predators is limited by “saturation”and is nicely modeled by the function p(B).


Project 3.5 The Spruce Budworm 159

The analysis of the model

d B

dt= rB B

(1 − B

K B

)− β B2

α2 + B2(5.1)

is complicated by the presence of four different parameters, rB , K B , α, and β. Lud-wig introduces dimensionless variables at this point, one immediate benefit beingthe reduction of the number of parameters.

2. Argue that B and α must have the same units and that µ = B/α is a dimen-sionless quantity. Divide both sides of equation (5.1) by β, then substitute thedimensionless variable µ = B/α to show that

α

β

dµ

dt= αrB

βµ

(1 − αµ

K B

)− µ2

1 + µ2. (5.2)

Argue that

τ = β

αt, R = αrB

β, and Q = K B

α

are dimensionless variables and their insertion into equation (5.2) produces theequation

dµ

dτ= Rµ

(1 − µ

Q

)− µ2

1 + µ2. (5.3)

3. Ludwig continues his qualitative analysis of the budworm model by calculatingthe equilibrium points. Of course, this is accomplished by setting the right-handside of equation (5.3) equal to zero. Factoring,

µ

[R

(1 − µ

Q

)− µ

1 + µ2

]= 0, (5.4)

it is easily seen that µ = 0 is an equilibrium point. Argue that µ = 0 is anunstable equilibrium.

4. The remaining equilibrium points are calculated by setting the remaining factor

R

Qµ

Figure 1 Three equilibriumpoints.

of equation (5.4) equal to zero. This eventually leads to a rather intractablesolution of a cubic polynomial, so Ludwig adopted an alternate strategy, that ofdetermining where

R

(1 − µ

Q

)= µ

1 + µ2. (5.5)

The analysis continues by plotting graphs of the left- and right-hand sides ofequation (5.5) versus µ and looking for points of intersection. Fix R. If Q issufficiently small, there is only one point of intersection. However, for largevalues of Q, you can have one, two, or even three points of intersection, de-pending on the value of R, as shown in Figure 1. If you gradually decrease Rin Figure 1, two points of intersection will coalesce into one, then disappear,



leaving only one point of intersection. The same thing will happen if you in-crease R. Further, because µ > 0, the sign of dµ/dt is determined solely bythe sign of

R

(1 − µ

Q

)− µ

1 + µ2,

which in turn is easily determined by noting where the graph of R(1 − µ/Q)

lies above or below the graph of µ/(1 + µ2).Sketch all possible ways that the graphs of R(1 − µ/Q) and µ/(1 + µ2)

can intersect. In each case,

• create a phase line below the graph where equilibrium points and directionof flow are indicated, and

• write a description of the ecological significance of the given phase line.Why are some scenarios more/less dangerous than others?

You’ve seen that the positions where two equilibrium points coalesce into one,then disappear, are located where the curves in Figure 1 intersect tangentially.Therefore, at these positions of bifurcation,

R

(1 − µ

Q

)= µ

1 + µ2(5.6)

and

d

dµ

{R

(1 − µ

Q

)}= d

dµ

{µ

1 + µ2

}. (5.7)

5. After performing the differentiation in equation (5.7), substitute the result intoequation (5.6) to show that

R = 2µ3

(1 + µ2)2. (5.8)

After that, show that

Q = 2µ3

µ2 − 1. (5.9)

Taken together, these equations define parametrically (in terms of µ) the bi-furcation curve in the (RQ) parameter space. Sketch the bifurcation curve in(RQ) parameter space, then label the regions that have one, two, and three equi-librium points (not including µ = 0 in your count). Discuss the ecologicalsignificance of these regions. Which region leads to a stable, manageable bud-worm population? Which region leads to a serious outbreak of the budwormpopulation?

6. Pick an (R, Q) in each region determined by the bifurcation curve in Exercise 5and use your numerical solver to sketch the fate of µ versus τ for various initialpopulation sizes. [Use equation (5.3)].


Project 3.6 Social Security—Now or Later 161

If you wish to pursue this model in greater depth, here are some resources youwill find useful.

• Ludwig, Jones, and Holling. Qualitative Analysis of Insect Outbreak Systesm:The Spruce Budworm and Forest, Journal of Animal Ecology (1978), 47, 315–332.

• Murray, J. Mathematical Biology (1993), Springer-Verlag (New York).• Strogatz, S. Nonlinear Dynamics and Chaos (1994), Addison-Wesley (Read-

ing, Mass.)

Project 3.6 Social Security—Now or Later

In the spring of 2000, a change in the Social Security law allowed individuals overtheir full retirement age7 to continue to work full time and still receive full socialsecurity retirement benefits. However, they could choose to delay the start of theirretirement benefits until a later time, in which case the monthly benefit would belarger. In May 2000, all working individuals over full retirement age had an im-portant decision to make, Should I take social security now, or later? Anyone whocontinues to work after their full retirement age will face the same question.

Many people faced with this decision find it an easy choice to make. If thegovernment will give you money, take it. That is not bad logic, but others will wantto examine the situation a little more deeply. They ask the question, Which waymaximizes my financial situation?

It is not easy to see what this question means, but here is one way to approachthe problem. It is based on the idea of setting up an account in which the socialsecurity benefits are deposited, which pays a certain fixed rate of return, and fromwhich taxes are paid on the benefits and on the return. Of course such an accountwould be purely hypothetical. No one would actually do this. However, if wepostulate two such accounts, one with balance P1(t) set up at full retirement age,and another with balance P2(t) set up at the true retirement age, say 70, we cancompare the balances over time.

Of course, P2(t) = 0 for t < 70 while P1(t) is increasing. Then starting att = 70, P2 will start increasing at a faster rate than P1 because of the larger benefits.The question is, At what age will P2 catch up with P1? This “catch-up” age mighthelp someone make the decision on when to start receiving social security. If thecatch-up age is low, then it might be a good idea to delay taking social security. Onthe other hand, if the catch-up age is high, then the opposite strategy might be best.Of course, it is up to each person to say what is high and what is low.

To carry out the modeling of the account balances it is necessary to accumulatesome data.

• The most important data is the monthly benefit amounts for retirement at thefull retirement age and 70. The correct figures depend in a fairly complicated

7 As a result of the same change in the Social Security law, the full retirement age was also changed. Itused to be 65 for everyone, but now it varies depending on a person’s birth date. For individuals born in1937 or earlier it is still 65, but for younger people it increases gradually to the point where for peopleborn in 1967 or later it is 67.



way on one’s entire income history. However, it is possible to make esti-mates. Use your Internet browser to visit the Social Security Administration(http://www.ssa.gov/). There you will find calculators that will compute themonthly benefit for you, once you provide it with income data. The point isto find the monthly payment amounts starting at both full retirement age andat 70 based on the same income data. Perhaps the easiest thing to do is to findthe maximum benefits for each age. Let A1 be the annual benefit starting atfull retirement age, and A2 that starting at 70. Furthermore let Td be the timein years between full retirement age and 70.

• It will be necessary to deduct the taxes paid on each of the accounts. Consultwith IRS documents to discover what portion of the benefits is taxable. Inaddition taxes will have to be paid on the income into the accounts. Make anassumption about the marginal tax rate, which we will denote by ρ.

• The only other item of information needed is r , the rate of return on the invest-ment accounts. Since this is uncertain, the computation should be done for arange of return rates varying from 0 to a maximum that will be computed.

With the data you have collected, perform the following tasks.

1. Along the lines of the derivations in Section 3.3, construct a differential equa-tion model for the growth of the two accounts. Assume that all payments aremade continuously. Be sure to account for taxes and return on investments.

2. Solve the differential equations exactly.

3. Show that if

r > rmax = 1

(1 − ρ)Tdln(A2/A1), (6.1)

then P1(t) > P2(t) for all t . In other words, if the return on investment is large,the account started at the full retirement age is always larger than the one startedat 70.

4. Show that for 0 < r < rmax the catch-up time T is

T = 1

(1 − ρ)rln

(A2 − A1

A2e−(1−ρ)rTd − A1

).

Plot the catch-up time T versus the investment rate r with the data you got fromthe Social Security Administration.

5. Redo the whole exercise with r = 0. In other words, assume that there is noreturn on investment. Show that the catch-up time is

T = Td A2

A2 − A1.

6. What would you advise an older relative or friend who asks you if he or sheshould delay receiving social security?


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