2.8 Nonlinear Models

INTRODUCTION

We finish our discussion of single first-order differential equations by examining some nonlinear mathematical models.

Population Dynamics

If P(t) denotes the size of a population at time t, the model for exponential growth begins with the assumption that dP/dt = kP for some k > 0. In this model the relative, or specific, growth rate defined by

(1)

is assumed to be a constant k. True cases of exponential growth over long periods of time are hard to find, because the limited resources of the environment will at some time exert restrictions on the growth of a population. Thus (1) can be expected to decrease as P increases in size.

The assumption that the rate at which a population grows (or declines) is dependent only on the number present and not on any time-dependent mechanisms such as seasonal phenomena (see Problem 33 in Exercises 1.3) can be stated as

    or     (2)

The differential equation in (2), which is widely assumed in models of animal populations, is called the density-dependent hypothesis.

Logistic Equation

Suppose an environment is capable of sustaining no more than a fixed number of K individuals in its population. The quantity K is called the carrying capacity of the environment. Hence, for the function f in (2) we have f(K) = 0, and we simply let f(0) = r. FIGURE 2.8.1 shows three functions f that satisfy these two conditions. The simplest assumption that we can make is that f(P) is linear—that is, f(P) = c1P + c2. If we use the conditions f(0) = r and f(K) = 0, we find, in turn, c2 = r, c1 = −r/K, and so f takes on the form f(P) = r − (r/K)P. Equation (2) becomes

. (3)

Relabeling constants a = r and b = r/K, the nonlinear equation (3) is the same as

. (4)

A graph. The horizontal axis is labeled P and the vertical axis is labeled function of P. A line begins at a point labeled r on the vertical axis, and goes down and to the right to a point K on the horizontal axis. The two curves begin and end at the same points as the line. One curve above the line goes down and to the right with increasing steepness. The other curve below the line goes down and to the right with decreasing steepness.

FIGURE 2.8.1 Simplest assumption for f(P) is a straight line

Around 1840 the Belgian mathematician/biologist Pierre François Verhulst (1804–1849) was concerned with mathematical models for predicting the human population of various countries. One of the equations he studied was (4), where a > 0, b > 0. Equation (4) came to be known as the logistic equation, and its solution is called the logistic function. The graph of a logistic function is called a logistic curve.

The linear differential equation dP/dt = kP does not provide a very accurate model for population when the population itself is very large. Overcrowded conditions, with the resulting detrimental effects on the environment, such as pollution and excessive and competitive demands for food and fuel, can have an inhibiting effect on population growth. As we shall now see, a solution of (4) that satisfies an initial condition P(0) = P0, where 0 < P0 < a/b, is bounded as t → . If we rewrite (4) as dP/dt = aP − bP2, the nonlinear term −bP2, b > 0, can be interpreted as an “inhibition” or “competition” term. Also, in most applications, the positive constant a is much larger than the constant b.

Logistic curves have proved to be quite accurate in predicting the growth patterns, in a limited space, of certain types of bacteria, protozoa, water fleas (Daphnia), and fruit flies (Drosophila).

Solution of the Logistic Equation

One method of solving (4) is by separation of variables. Decomposing the left side of dP/P(a − bP) = dt into partial fractions and integrating gives

It follows from the last equation that

If P(0) = P0, P0 ≠ a/b, we find c1 = P0/(a − bP0), and so, after substituting and simplifying, the solution becomes

. (5)

Graphs of P(t)

The basic shape of the graph of the logistic function P(t) can be obtained without too much effort. Although the variable t usually represents time and we are seldom concerned with applications in which t < 0, it is nonetheless of some interest to include this interval when displaying the various graphs of P. From (5) we see that

    and    

The dashed line P = a/2b shown in FIGURE 2.8.2 corresponds to the y-coordinate of a point of inflection of the logistic curve. To show this, we differentiate (4) by the Product Rule:

Three graphs. The horizontal axis is labeled t and the vertical axis is labeled P. Two horizontal dashed lines are plotted above the horizontal axis at y equals a over 2 b and at y equals a over b. Graph a. The curve enters the second quadrant, and goes down and to the right with decreasing steepness. It intersects the vertical axis at point P subscript zero and approaches the line y equals a over b. Graph b. The curve enters the second quadrant, and goes up and to the right with increasing steepness away from the horizontal axis. It intersects the vertical axis at point P subscript zero. It continues to go up and to the right up to a point on the horizontal line y equals a over 2 b. It then goes up and to the right with decreasing steepness and approaches y equals a over b. Graph c. The curve enters the second quadrant and goes up and to the right with increasing steepness up to a point on the horizontal line y equals a over 2 b. It then continues to go up and to the right with decreasing steepness, intersects the vertical axis at point P subscript zero, and approaches the line y equals a over b.

FIGURE 2.8.2 Logistic curves for different initial conditions

From calculus, recall that the points where d2P/dt2 = 0 are possible points of inflection, but P = 0 and P = a/b can obviously be ruled out. Hence P = a/2b is the only possible ordinate value at which the concavity of the graph can change. For 0 < P < a/2b it follows that P″ > 0, and a/2b < P < a/b implies P″ < 0. Thus, as we read from left to right, the graph changes from concave up to concave down at the point corresponding to P = a/2b. When the initial value satisfies 0 < P0 < a/2b, the graph of P(t) assumes the shape of an S, as we see in Figure 2.8.2(b). For a/2b < P0 < a/b the graph is still S-shaped, but the point of inflection occurs at a negative value of t, as shown in Figure 2.8.2(c).

We have already seen equation (4) above in (5) of Section 1.3 in the form dx/dt = kx(n + 1 − x), k > 0. This differential equation provides a reasonable model for describing the spread of an epidemic brought about initially by introducing an infected individual into a static population. The solution x(t) represents the number of individuals infected with the disease at time t.

EXAMPLE 1 Logistic Growth

Suppose a student carrying a flu virus returns to an isolated college campus of 1000 students. If it is assumed that the rate at which the virus spreads is proportional not only to the number x of infected students but also to the number of students not infected, determine the number of infected students after 6 days if it is further observed that after 4 days x(4) = 50.

SOLUTION

Assuming that no one leaves the campus throughout the duration of the disease, we must solve the initial-value problem

By making the identifications a = 1000k and b = k, we have immediately from (5) that

Now, using the information x(4) = 50, we determine k from

We find −1000k = ln = −0.9906. Thus

Finally,

Additional calculated values of x(t) are given in the table in FIGURE 2.8.3(b). ≡

A graph. The horizontal axis is labeled t, and the vertical axis is labeled x. A horizontal dashed line labeled x = 1000 is plotted. A curve begins at the origin and goes up and to the right with increasing steepness, passing through the approximate points (4, 50.7), (5, 124.9), (6, 276.6), (7, 507.9), and (8, 735). It then continues to go up and to the right with decreasing steepness through the approximate points (9, 882.6), and (10, 953.8) and approaches the line x = 1000. The table consists of seven rows and two columns. The column headers are t in days and x or number of students infected. The row entries are as follows. Row 1. Days: 4, number infected: 50. Row 2. Days: 5, number infected: 124. Row 3. Days: 6, number infected: 276. Row 4. Days: 7, number infected: 507. Row 5. Days: 8, number infected: 735. Row 6. Days: 9, number infected: 882. Row 7. Days: 10, number infected: 953.

FIGURE 2.8.3 Number of infected students in Example 1

Modifications of the Logistic Equation

There are many variations of the logistic equation. For example, the differential equations

    and     (6)

could serve, in turn, as models for the population in a fishery where fish are harvested or are restocked at rate h. When h > 0 is a constant, the DEs in (6) can be readily analyzed qualitatively or solved by separation of variables. The equations in (6) could also serve as models of a human population either increased by immigration or decreased by emigration. The rate h in (6) could be a function of time t or may be population dependent; for example, harvesting might be done periodically over time or may be done at a rate proportional to the population P at time t. In the latter instance, the model would look like P′ = P(a − bP) − cP, c > 0. A human population of a community might change due to immigration in such a manner that the contribution due to immigration is large when the population P of the community is itself small, but then the immigration contribution might be small when P is large; a reasonable model for the population of the community is then P′ = P(a − bP) + ce–kP, c > 0, k > 0. Another equation of the form given in (2),

, (7)

is a modification of the logistic equation known as the Gompertz differential equation named after the English self-educated mathematician Benjamin Gompertz (1779–1865). This DE is sometimes used as a model in the study of the growth or decline of population, in the growth of solid tumors, and in certain kinds of actuarial predictions. See Problems 5–10 in Exercises 2.8.

Chemical Reactions

Suppose that a grams of chemical A are combined with b grams of chemical B. If there are M parts of A and N parts of B formed in the compound and X(t) is the number of grams of chemical C formed, then the numbers of grams of chemicals A and B remaining at any time are, respectively,

    and    

By the law of mass action, the rate of the reaction satisfies

(8)

If we factor out M/(M + N) from the first factor and N/(M + N) from the second and introduce a constant k > 0 of proportionality, (8) has the form

, (9)

where α = a(M + N)/M and β = b(M + N)/N. Recall from (6) of Section 1.3 that a chemical reaction governed by the nonlinear differential equation (9) is said to be a second-order reaction.

EXAMPLE 2 Second-Order Chemical Reaction

A compound C is formed when two chemicals A and B are combined. The resulting reaction between the two chemicals is such that for each gram of A, 4 grams of B are used. It is observed that 30 grams of the compound C are formed in 10 minutes. Determine the amount of C at time t if the rate of the reaction is proportional to the amounts of A and B remaining and if initially there are 50 grams of A and 32 grams of B. How much of the compound C is present at 15 minutes? Interpret the solution as t → .

SOLUTION

Let X(t) denote the number of grams of the compound C present at time t. Clearly X(0) = 0 g and X(10) = 30 g.

If, for example, 2 grams of compound C are present, we must have used, say, a grams of A and b grams of B so that a + b = 2 and b = 4a. Thus we must use a = = 2() g of chemical A and b = = 2() g of B. In general, for X grams of C we must use

The amounts of A and B remaining at any time are then

respectively.

Now we know that the rate at which compound C is formed satisfies

To simplify the subsequent algebra, we factor from the first term and from the second and then introduce the constant of proportionality:

By separation of variables and partial fractions we can write

Integrating gives

    or     (10)

When t = 0, X = 0, so it follows at this point that c2 = . Using X = 30 g at t = 10, we find 210k = ln = 0.1258. With this information we solve the last equation in (10) for X:

(11)

The behavior of X as a function of time is displayed in FIGURE 2.8.4. It is clear from the accompanying table and (11) that X → 40 as t → . This means that 40 grams of compound C are formed, leaving

A graph. The horizontal axis is labeled t, and the vertical axis is labeled x. A horizontal dashed line labeled x = 40 is plotted. The curve begins at the origin and goes up and to the right with decreasing steepness through the approximate points (10, 30), (15, 34.78), (20, 37.25), (25, 38.54), (30, 39.22), and (35, 39.59). It approaches the line x = 40. The table consists of seven rows and two columns. The column headers are t in minutes and x in grams. The row entries are as follows. Row 1. Days: 10, number infected: 30. Row 2. Days: 15, number infected: 34.78. Row 3. Days: 20, number infected: 37.25. Row 4. Days: 25, number infected: 38.54. Row 5. Days: 30, number infected: 39.22. Row 6. Days: 35, number infected: 39.59.

FIGURE 2.8.4 Amount of compound C in Example 2

≡

REMARKS

The indefinite integral can be evaluated in terms of logarithms, the inverse hyperbolic tangent, or the inverse hyperbolic cotangent. For example, of the two results,

            (12)

    ≠ (13)

(12) may be convenient for Problems 19 and 31 in Exercises 2.8, whereas (13) may be preferable in Problem 32.

2.8 Exercises Answers to selected odd-numbered problems begin on page ANS-3.

Logistic Equation

  1. The number N(t) of supermarkets throughout the country that are using a computerized checkout system is described by the initial-value problem

    (a) Use the phase portrait concept of Section 2.1 to predict how many supermarkets are expected to adopt the new procedure over a long period of time. By hand, sketch a solution curve of the given initial-value problem.

    (b) Solve the initial-value problem and then use a graphing utility to verify the solution curve in part (a). How many companies are expected to adopt the new technology when t = 10?

  2. The number N(t) of people in a community who are exposed to a particular advertisement is governed by the logistic equation. Initially N(0) = 500, and it is observed that N(1) = 1000. Solve for N(t) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000.
  3. A model for the population P(t) in a suburb of a large city is given by the initial-value problem

    ,

    where t is measured in months. What is the limiting value of the population? At what time will the population be equal to one-half of this limiting value?

  4. (a) Census data for the United States between 1790 and 1950 are given in the following table. Construct a logistic population model using the data from 1790, 1850, and 1910.

    Year Population (in millions)
    1790     3.929
    1800     5.308
    1810     7.240
    1820     9.638
    1830   12.866
    1840   17.069
    1850   23.192
    1860   31.433
    1870   38.558
    1880   50.156
    1890   62.948
    1900   75.996
    1910   91.972
    1920 105.711
    1930 122.775
    1940 131.669
    1950 150.697

    (b) Construct a table comparing actual census population with the population predicted by the model in part (a). Compute the error and the percentage error for each entry pair.

Modifications of the Logistic Equation

  1. If a constant number h of fish is harvested from a fishery per unit time, then a model for the population P(t) of the fishery at time t is given by

    ,

    where a, b, h, and P0 are positive constants. Suppose a = 5, b = 1, and h = 4. Show that the differential equation becomes

    Since the DE is autonomous, use the phase portrait concept of Section 2.1 to sketch representative solution curves corresponding to the cases and Determine the long-term behavior of the population in each case.

  2. Solve the IVP in Problem 5. Verify your results of your phase portrait in Problem 5 by using a graphing utility to plot the graph of P(t) with an initial condition taken from each of the three intervals given.
  3. Use the information in Problems 5 and 6 to determine whether the fishery population P(t) becomes extinct in finite time. If so, find that time.
  4. Investigate the harvesting model in Problems 5–7 both qualitatively and analytically in the case a = 5, b = 1, h = . Determine whether the population becomes extinct in finite time. If so, find that time.
  5. Repeat Problem 8 in the case a = 5, b = 1, h = 7.
  6. (a) Suppose a = b = 1 in the Gompertz differential equation (7). Since the DE is autonomous, use the phase portrait concept of Section 2.1 to sketch representative solution curves corresponding to the cases P0 > e and 0 < P0 < e.

    (b) Suppose a = 1, b = −1 in (7). Use a new phase portrait to sketch representative solution curves corresponding to the cases P0 > e– 1 and 0 < P0 < e– 1.

  7. Find an explicit solution of equation (7) subject to P(0) = P0.
  8. The Allee Effect     For an initial population P0, where P0 > K the logistic population model (3) predicts that population cannot sustain itself over time so it decreases but yet never falls below the carrying capacity K of the ecosystem. Moreover, for 0 < P0 < K, the same model predicts that, regardless of how small P0 is, the population increases over time and does not surpass the carrying capacity K. See Figure 2.8.2, where a/b = K. But the American ecologist Warder Clyde Allee (1885–1955) showed that by depleting certain fisheries beyond a certain level, the fishery population never recovers. How would you modify the differential equation (3) to describe a population P that has these same two characteristics of (3) but additionally has a threshold level A, 0 < A < K, below which the population cannot sustain itself and becomes extinct. [Hint: Construct a phase portrait of what you want and then form a DE.]

Chemical Reactions

  1. Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A are used. It is observed that 10 grams of C are formed in 5 minutes. How much is formed in 20 minutes? What is the limiting amount of C after a long time? How much of chemicals A and B remains after a long time?
  2. Solve Problem 13 if 100 grams of chemical A are present initially. At what time is chemical C half-formed?

Additional Nonlinear Models

  1. Leaking Cylindrical Tank     A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3, when friction and contraction of water at the hole are ignored, the height h of water in the tank is described by

    where Aw and Ah are the cross-sectional areas of the water and the hole, respectively.

    (a) Solve for h(t) if the initial height of the water is H. By hand, sketch the graph of h(t) and give its interval I of definition in terms of the symbols Aw, Ah, and H. Use g = 32 ft/s2.

    (b) Suppose the tank is 10 ft high and has radius 2 ft and the circular hole has radius in. If the tank is initially full, how long will it take to empty?

  2. Leaking Cylindrical Tank—Continued     When friction and contraction of the water at the hole are taken into account, the model in Problem 15 becomes

    where 0 < c < 1. How long will it take the tank in Problem 15(b) to empty if c = 0.6? See Problem 13 in Exercises 1.3.

  3. Leaking Conical Tank     A tank in the form of a right-circular cone standing on end, vertex down, is leaking water through a circular hole in its bottom.

    (a) Suppose the tank is 20 ft high and has radius 8 ft and the circular hole has radius 2 in. In Problem 14 in Exercises 1.3 you were asked to show that the differential equation governing the height h of water leaking from a tank is

    In this model, friction and contraction of the water at the hole were taken into account with c = 0.6, and g was taken to be 32 ft/s2. See Figure 1.3.13. If the tank is initially full, how long will it take the tank to empty?

    (b) Suppose the tank has a vertex angle of 60°, and the circular hole has radius 2 in. Determine the differential equation governing the height h of water. Use c = 0.6 and g = 32 ft/s2. If the height of the water is initially 9 ft, how long will it take the tank to empty?

  4. Inverted Conical Tank     Suppose that the conical tank in Problem 17(a) is inverted, as shown in FIGURE 2.8.5, and that water leaks out a circular hole of radius 2 in. in the center of the circular base. Is the time it takes to empty a full tank the same as for the tank with vertex down in Problem 17? Take the friction/contraction coefficient to be c = 0.6 and g = 32 ft/s2.
    A conical tank has a height of 20 feet and a radius of eight feet. Water leaks from a hole at the base. Water in the tank is at a height h, and the cross-sectional area of the water is A superscript w.

    FIGURE 2.8.5 Inverted conical tank in Problem 18

  5. Air Resistance     A differential equation governing the velocity v of a falling mass m subjected to air resistance proportional to the square of the instantaneous velocity is

    where k > 0 is the drag coefficient. The positive direction is downward.

    (a) Solve this equation subject to the initial condition v(0) = v0.

    (b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the DE in Problem 41 in Exercises 2.1.

    (c) If distance s, measured from the point where the mass was released above ground, is related to velocity v by ds/dt = v(t), find an explicit expression for s(t) if s(0) = 0.

  6. How High?—Nonlinear Air Resistance     Consider the 16-pound cannonball shot vertically upward in Problems 38 and 39 in Exercises 2.7 with an initial velocity v0 = 300 ft/s. Determine the maximum height attained by the cannonball if air resistance is assumed to be proportional to the square of the instantaneous velocity. Assume the positive direction is upward and take the drag coefficient to be k = 0.0003. [Hint: Slightly modify the DE in Problem 19.]
  7. That Sinking Feeling     (a) Determine a differential equation for the velocity v(t) of a mass m sinking in water that imparts a resistance proportional to the square of the instantaneous velocity and also exerts an upward buoyant force whose magnitude is given by Archimedes’ principle. See Problem 18 in Exercises 1.3. Assume that the positive direction is downward.

    (b) Solve the differential equation in part (a).

    (c) Determine the limiting, or terminal, velocity of the sinking mass.

  8. Solar Collector     The differential equation

    describes the shape of a plane curve C that will reflect all incoming light beams to the same point and could be a model for the mirror of a reflecting telescope, a satellite antenna, or a solar collector. See Problem 29 in Exercises 1.3. There are several ways of solving this DE.

    (a) Verify that the differential equation is homogeneous (see Section 2.5). Show that the substitution y = ux yields

    Use a CAS (or another judicious substitution) to integrate the left-hand side of the equation. Show that the curve C must be a parabola with focus at the origin and is symmetric with respect to the x-axis.

    (b) Show that the first differential equation can also be solved by means of the substitution u = x2 + y2.

  9. Tsunami     (a) A simple model for the shape of a tsunami is given by

    where W(x) > 0 is the height of the wave expressed as a function of its position relative to a point offshore. By inspection, find all constant solutions of the DE.

    (b) Solve the differential equation in part (a). A CAS may be useful for integration.

    (c) Use a graphing utility to obtain the graphs of all solutions that satisfy the initial condition W(0) = 2.

  10. Evaporation     An outdoor decorative pond in the shape of a hemispherical tank is to be filled with water pumped into the tank through an inlet in its bottom. Suppose that the radius of the tank is R = 10 ft, that water is pumped in at a rate of π ft3/min, and that the tank is initially empty. See FIGURE 2.8.6. As the tank fills, it loses water through evaporation. Assume that the rate of evaporation is proportional to the area A of the surface of the water and that the constant of proportionality is k = 0.01.

    (a) The rate of change dV/dt of the volume of the water at time t is a net rate. Use this net rate to determine a differential equation for the height h of the water at time t. The volume of the water shown in the figure is V = πRh2 − πh3, where R = 10. Express the area of the surface of the water A = πr2 in terms of h.

    (b) Solve the differential equation in part (a). Graph the solution.

    The first illustration represents a hemispherical tank. The volume of water is V and the surface of the water is A. The input is water pumped from the bottom at a rate of pi cubic feet per minute. The output is evaporation at a rate proportional to the area A of the surface. The second illustration represents the cross-section of the tank. The tank has a radius R. The height of the water is r, and it is at a distance h from the top of the tank.

    FIGURE 2.8.6 Pond in Problem 24

    (c) If there were no evaporation, how long would it take the tank to fill?

    (d) With evaporation, what is the depth of the water at the time found in part (c)? Will the tank ever be filled? Prove your assertion.

  11. Sawing Wood     A long uniform piece of wood (cross sections are the same) is cut through perpendicular to its length by a vertical saw blade. See FIGURE 2.8.7. If the friction between the sides of the saw blade and the wood through which the blade passes is ignored, then it can be assumed that the rate at which the saw blade moves through the piece of wood is inversely proportional to the width of the wood in contact with its cutting edge. As the blade advances through the wood (moving left to right), the width of a cross section changes as a nonnegative continuous function w. If a cross section of the wood is described as a region in the xy-plane defined over an interval [a, b],, then as shown in Figure 2.8.7(c) the position x of the saw blade is a function of time t and the vertical cut made by the blade can be represented by a vertical line segment. The length of this vertical line is the width w(x) of the wood at that point. Thus the position x(t) of the saw blade and the rate dx/dt at which it moves to the right are related to w(x) by

    (14)

    Here k represents the number of square units of the material removed by the saw blade per unit time. In the problems that follow, we assume that the saw can be programmed so that k = 1. Find an explicit solution x(t) of the initial-value problem (14) when a cross section of the piece of wood is triangular and is bounded by the graphs of y = x, x = 1, and y = 0 How long does it take the saw to cut through this piece of wood?

    The first illustration represents a man standing on top of a log and sawing it across the width. The cutting edge of the saw vertical to the log. The second illustration represents a log with a cut made perpendicular to the length. The third illustration is a graph. Two vertical lines are drawn at points a and b located on the horizontal axis. The curve begins at a point on the vertical line passing through a. It goes up and to the right to a high point, then goes down and to the left to a point on the vertical line passing through b. The region under the curve is labeled cross-section. A vertical dashed line at a point x between a and b is labeled cutting edge of saw blade moving left to right.

    FIGURE 2.8.7 Sawing a log in Problem 25

  12. (a) Find an implicit solution of the initial-value problem (14) in Problem 25 when the piece of wood is a circular log. Assume a cross section is a circle of radius 2 centered at [Hint: To save time, see formula 33 in the table of integrals given on the right page inside the front cover.]

    (b) Solve the implicit solution obtained in part (b) for time t as a function of x. Graph the function With the aid of the graph, approximate the time that it takes the saw to cut through this piece of wood. Then find the exact value of this time.

  13. Solve the initial-value problem (14) in Problem 25 when a cross section of a uniform piece of wood is the triangular region given in FIGURE 2.8.8. Assume again that k = 1. How long does it take to cut through this piece of wood?
    A graph is plotted on the x y plane. It begins at the origin and goes up and to the right to a high point at (square root of 2 over 2, square root of 2 over 2). It then goes down to the right to (square root of 2, 0).

    FIGURE 2.8.8 Triangular cross section in Problem 27

Computer Lab Assignments

  1. Regression Line Read the documentation for your CAS on scatter plots (or scatter diagrams) and least-squares linear fit. The straight line that best fits a set of data points is called a regression line or a least squares line. Your task is to construct a logistic model for the population of the United States, defining f(P) in (2) as an equation of a regression line based on the population data in the table in Problem 4. One way of doing this is to approximate the left-hand side of the first equation in (2) using the forward difference quotient in place of dP/dt:

    (a) Make a table of the values t, P(t), and Q(t) using t = 0, 10, 20, . . ., 160, and h = 10. For example, the first line of the table should contain t = 0, P(0), and Q(0). With P(0) = 3.929 and P(10) = 5.308,

    Note that Q(160) depends on the 1960 census population P(170). Look up this value.

    (b) Use a CAS to obtain a scatter plot of the data (P(t), Q(t)) computed in part (a). Also use a CAS to find an equation of the regression line and to superimpose its graph on the scatter plot.

    (c) Construct a logistic model dP/dt = Pf (P), where f(P) is the equation of the regression line found in part (b).

    (d) Solve the model in part (c) using the initial condition P(0) = 3.929.

    (e) Use a CAS to obtain another scatter plot, this time of the ordered pairs (t, P(t)) from your table in part (a). Use your CAS to superimpose the graph of the solution in part (d) on the scatter plot.

    (f) Look up the U.S. census data for 1970, 1980, and 1990. What population does the logistic model in part (c) predict for these years? What does the model predict for the U.S. population P(t) as t → ?

  2. Immigration Model (a) In Examples 3 and 4 of Section 2.1, we saw that any solution P(t) of (4) possesses the asymptotic behavior P(t) → a/b as t → for P0 > a/b and for 0 < P0 < a/b; as a consequence, the equilibrium solution P = a/b is called an attractor. Use a root-finding application of a CAS (or a graphic calculator) to approximate the equilibrium solution of the immigration model

    (b) Use a graphing utility to graph the function F(P) = P(1 − P) + 0.3e-P. Explain how this graph can be used to determine whether the number found in part (a) is an attractor.

    (c) Use a numerical solver to compare the solution curves for the IVPs

    ,

    for P0 = 0.2 and P0 = 1.2 with the solution curves for the IVPs

    ,

    for P0 = 0.2 and P0 = 1.2. Superimpose all curves on the same coordinate axes but, if possible, use a different color for the curves of the second initial-value problem. Over a long period of time, what percentage increase does the immigration model predict in the population compared to the logistic model?

  3. What Goes Up . . . In Problem 20 let ta be the time it takes the cannonball to attain its maximum height and let td be the time it takes the cannonball to fall from the maximum height to the ground. Compare the value of ta with the value of td and compare the magnitude of the impact velocity vi with the initial velocity v0. See Problem 53 in Exercises 2.7. A root-finding application of a CAS may be useful here. [Hint: Use the model in Problem 19 when the cannonball is falling.]
  4. Skydiving A skydiver is equipped with a stopwatch and an altimeter. She opens her parachute 25 seconds after exiting a plane flying at an altitude of 20,000 ft and observes that her altitude is 14,800 ft. Assume that air resistance is proportional to the square of the instantaneous velocity, her initial velocity upon leaving the plane is zero, and g = 32 ft/s2.

    (a) Find the distance s(t), measured from the plane, that the skydiver has traveled during free fall in time t. [Hint: The constant of proportionality k in the model given in Problem 19 is not specified. Use the expression for terminal velocity vt obtained in part (b) of Problem 19 to eliminate k from the IVP. Then eventually solve for vt.]

    (b) How far does the skydiver fall and what is her velocity at t = 15 s?

  5. Hitting Bottom A helicopter hovers 500 feet above a large open tank full of liquid (not water). A dense compact object weighing 160 pounds is dropped (released from rest) from the helicopter into the liquid. Assume that air resistance is proportional to instantaneous velocity v while the object is in the air and that viscous damping is proportional to v2 after the object has entered the liquid. For air, take k = , and for the liquid, k = 0.1. Assume that the positive direction is downward. If the tank is 75 feet high, determine the time and the impact velocity when the object hits the bottom of the tank. [Hint: Think in terms of two distinct IVPs. If you use (13), be careful in removing the absolute value sign. You might compare the velocity when the object hits the liquid—the initial velocity for the second problem—with the terminal velocity vt of the object falling through the liquid.]