Figure 4.1 A great tit is a species of bird whose foraging behavior was studied by biologist Richard Cowie and whose behavior can be predicted by optimal foraging models.

images

Preview

“‘If one way be better than another, that you may be sure is nature's way.’ Aristotle clearly stated the basic premise of optimization in biology, yet it was almost 2,000 years before the power of this idea was appreciated. The essence of optimization is to calculate the most efficient solution to a given problem, and then to test the prediction. The concept has already revolutionized some aspects of biology, but it has the potential for much wider application.”

William Sutherland, on “The best solution” in Nature (2005) 435:569

One of the central ideas in physics, chemistry, and biology is that processes act to optimize some physically or biologically meaningful quantity. For example, from physics we know that light in a vacuum travels along a path that is the shortest distance between two points (taking into account that gravity “bends” space), and from biochemistry we know that proteins fold in a way that minimizes the energy of their constituent amino acid configuration.

Differential calculus is an important tool for analyzing optimization (maximization or minimization). In this chapter we show how optimization applies to various biological problems and processes. Before we do this, however, we study how calculus can be used to construct the graphs of a variety of functions; in particular, we identify where the graph has turning points corresponding to local minimum or maximum values. We then develop procedures for modeling and solving optimization problems, including how to draw the best-fitting line through data plotted on a graph. After considering a number of biological applications, we study how calculus provides insight into dynamic processes such as the growth of populations and the spread of deleterious or mutant genes (e.g., the gene that causes sickle cell anemia) within populations. We end the chapter with an application of difference equations that is at the heart of many numerical methods used by current technologies for finding solutions to nonlinear equations.

4.1 Graphing Using Calculus

In this section, we combine many of the tools that we have studied so far (e.g., limits involving infinity, first and second derivatives) to graph a function. In graphing a function, envision walking along the graph and indicating all the highlights of your walk. For instance, vertical asymptotes are places with such a rapid ascent or descent that they make climbing Mount Everest seem like a stroll in a park. Horizontal asymptotes are places where the landscape levels out into a never-ending plain. Where the derivative is positive, the graph is ascending; and where the derivative is negative, the graph is descending. Switches in the sign of the slope correspond to either hilltops or valley bottoms along your walk. On ascents where the second derivative is positive, the walk is getting harder. On descents where the second derivative is negative, the descent becomes faster.

Properties of graphs

When graphing the function y = f(x) by hand, follow these procedures to find the highlights of the function shape:

Vertical asymptotes: Determine at what points the function is not well defined (e.g., division by zero). At each of these points, say x = a, evaluate the one-sided limits, images f(x) and images f(x), to determine what the graph looks like near x = a. If either of these one-sided limits is +∞ or −∞, then there is a vertical asymptote at this point.

Intervals of increase and decrease: Compute the first derivative f′(x) of f(x) and determine on which intervals f′(x) > 0 and on which intervals f′(x) < 0. On these intervals, f is increasing and decreasing, respectively.

Intervals of concavity: Compute the second derivative f″(x) and determine on which intervals f″(x) > 0 and f″(x) < 0. On these intervals, f is concave up and concave down, respectively.

The x and y intercepts Find the x intercepts (i.e., where f(x) = 0) and the y intercept (i.e., y = f(0)). These points help pin down the placement of the graph.

After identifying these highlights of a function, you are ready to sketch the function.

Example 1 Dropping whelks: a graphical approach

In Example 6 of Section 3.2, we considered how often D(h) a whelk had to be dropped by a crow from a height of h meters before breaking. The function, based on data collected by Reto Zach (“Selection and dropping of whelks, Behaviour,” Vol. 67, pp. 134–148, 1978) is given by

images

  1. Find the horizontal and vertical asymptotes and where D is positive.
  2. Find on what intervals D is increasing and on what intervals D is decreasing.
  3. Find on what intervals D is concave up and on what intervals D is concave down.
  4. Take this information and sketch D(h). Discuss for which h values this function is biologically meaningful.

Solution

  1. There is a vertical asymptote at h = 0.84. Moreover, images D(h) = ±∞ and images D(h) = –∞. Since images D(h) = 1, D has a horizontal asymptote of D = 1.

    D(h) > 0 only if 1 + images > 0, equivalently images > −1. If h > 0.84, the inequality is always satisfied. If h < 0.84, then h − 0.84 is negative and multiplying by h − 0.84 reverses the direction of the inequality, yielding 20.4 < 0.84 − h, equivalently −19.56 > h. Hence, D is positive on the intervals (0.84, ∞) and (−∞, −19.56) as indicated by the third ribbon at the bottom of Figure 4.2.

    images

    Figure 4.2 Graph of the whelk-dropping function D(h) = 1 + images. The ribbons below the graph indicate where D, D′, and D″ are negative (in red) and positive (in green).

  2. Taking the first derivative yields

    images

    Since (h − 0.84)2 is always positive for h ≠ 0.84, we get D′(h) < 0 for all h ≠ 0.84, as indicated by the second ribbon at the bottom of Figure 4.2. Therefore, D is decreasing for all h ≠ 0.84.

  3. Taking the second derivative yields

    images

    which is positive for h > 0.84 and negative for h < 0.84, as indicated by the first ribbon at the bottom of Figure 4.2. Hence, D is concave up for h > 0.84 and concave down for h < 0.84.

  4. Putting all this together yields the graph in Figure 4.2. From the crow's point of view, we require that h > 0 (since the crow does not burrow under ground!). Also, this graph is only meaningful for h > 0.84, since for 0 < h < 0.84 the number of drops predicted is negative. For h > 0.84, the graph we drew is very similar to the graph shown in Figure 3.6 of Chapter 3.

Many functions have no horizontal asymptotes. Nonetheless, understanding the limits as x approaches ±∞ may help graph the function.

Example 2 A “W” shaped curve

Consider the function y = x4 − 2x2.

  1. Find the asymptotes, the intervals where the function is increasing or decreasing, the intervals where the function is concave up or down, the roots (x intercepts), and the y intercept.
  2. Use all the information found in part a to graph the function.

Solution

  1. The function is continuous for all real numbers. Hence, there are no vertical asymptotes. We have images x4 − 2x2= images x2(x2 − 2) = ∞. Hence, there are no horizontal asymptotes, and y gets arbitrarily large as x approaches ±∞. Factoring the function y = x2(x2 − 2) reveals that it is zero at x2 = 0 and x2 = 2; that is, x = 0 and x = ±images are roots of the equation. The function is thus negative over the intervals −images < x < 0 and 0 < x < images, as depicted in the third ribbon below the plot in Figure 4.3. Further, the y intercept is y = 0.

    images

    Figure 4.3 Graph of the function y = x4 − 2x2 with the sign of the function and its first and second derivatives indicated by the green (positive) and red (negative) ribbons below the graph.

    To determine where the function is increasing, we first find the roots of the derivative, which solve the equation

    images

    Hence, the derivative vanishes at x = 0, ±1. Since images = 24 at x = 2, images > 0 on (1, ∞). Since images = −images at x = images, images < 0 on (0, 1), as depicted in the second, ribbon below the plot in Figure 4.3. Since images = images at x = −images, images > 0 on (−1, 0). Since images = −24 at x = −2, images < 0 on (−∞, −1). Therefore, the function is increasing on the intervals (−1, 0) and (1, ∞) and decreasing on the intervals (−∞, −1) and (0, 1), as depicted in Figure 4.3.

    To determine intervals of concave up and concave down, we determine where the second derivative equals zero.

    images

    Since images = −4 at x = 0, y is concave down on images. Since images = 8 at x = ±1, y is concave up on images.

  2. To sketch the graph using the information from part a, we can envision how the graph of the function changes as we move from −∞ to ∞. Since images y = ∞, images < 0 on (−∞, 1), and y = 0 at x = −images, the function decreases from +∞, crosses the x axis at x = −images, and continues to decrease to the value y = −1 at x = −1. Since images > 0 on(−1, 0), the function increases to y = 0 at x = 0. Since images < 0 on (0, 1), the function decreases to y = −1 at x = 1. Since images > 0 on (1, ∞) y at x = images, and images y = ∞, the function increases, crosses the x axis, again at x = images, and approaches +∞ as x approaches +∞. The function changes concavity at the points ±1/images. Hence, the graph has the general characteristics depicted in Figure 4.3.

Example 3 Linear asymptotes

Consider the function y = images.

  1. Find the asymptotes, the intervals where the function is increasing or decreasing, the intervals where the function is concave up or down, and the x and y intercepts.
  2. Use all the information found in part a to graph the function.

Solution

  1. Rewriting the function as y = x + images helps us see that a vertical asymptote exists at x = 0. In fact, images y = +∞ and images y = −∞. Since images y = ∞ and images y = −∞, y has no horizontal asymptotes. Indeed since images goes to zero as x gets large, we expect the graph of y = x + images to approach the line y = x for large x.

    To find the boundaries between the intervals of increase and decrease, we determine where the first derivative equals zero:

    images

    Since images = images at x = ±2 and images = −1 at x = ±1, we find that y is increasing on the intervals (−∞, −images) and (images, ∞) and decreasing on the intervals (−images, 0) and (0, images).

    To determine concavity, we compute the second derivative images, which is positive when x > 0 and negative when x < 0. Hence, y is concave up on (0, ∞) and y is concave down on (−∞, 0).

    There is no y intercept, as the function has a vertical asymptote at x = 0. The roots (x intercepts) must satisfy

    images

    for which there is no real-valued solution. Hence there are no roots.

  2. To graph y = x + images, think about what happens as you move from −∞ to ∞. Since images y = −∞ and images > 0 on (−∞, −images), we find the graph increases from ∞ to y = −2images at x = −images. Since images < 0 on (−images, 0) and images y = −∞, the graph decreases to infinity as x approaches 0 from the left. Since images y = ∞ and images < 0 on (0, images), the graph decreases from ∞ to y = 2images at x = images. Since images > 0 on (images, ∞) and images y = +∞, the graph increases toward +∞ as x approaches +∞. Moreover, the concavity only changes at x = 0. Finally, since images images = 0, it follows that y = x + images behaves like y = x for sufficiently positive or negative values of x. Using this information, we obtain a sketch that looks something like this, where the dotted red line represents the linear asymptote:

    images

Sometimes just using limits and first derivatives is enough to get a good sense of the graph.

Example 4 Tylenol in the blood stream

As a project for a mathematical biology class, three college students developed a model of how acetaminophen (Tylenol) levels diffuse from the stomach and intestines to the blood stream after taking a dose of 1000 mg. Using data from the Federal Drug Administration (FDA), the students found that the concentration of acetaminophen in the blood was modeled by

images

where t is hours after taking the dose. Use information about asymptotes and first derivatives to sketch this function. Discuss the meaning of the graph.

Solution Since C(t) is continuous everywhere, there are no vertical asymptotes. Since e−0.3t and et approach zero as t gets large, images C(t) = 0. Therefore, there is a horizontal asymptote at C = 0. Alternatively, since e−0.3tet = e−0.7t(e0.4t − 1), images e0.4t = 0, and images e−0.7t = ∞, we get images C(t) = −∞.

Taking the first derivative yields

images

We have C′(t) = 0 if and only if

images

Since C′(0) ≈ 20, we have C′(t) > 0 on (−∞, 1.72). Since C′(t) < 0 for very large t, we have C′(t) < 0 on (1.72, ∞). Hence, as t goes from −∞ to 0, the function increases up from −∞ and passes through 0 at t = 0. C(t) increases from t = 0 to t ≈ 1.72, at which point it takes on the value of approximately 12 micrograms/milliliter. For t greater than 1.72, C(t) decreases toward zero as t approaches +∞. Therefore, we can graph the function as follows:

images

This graph is only biologically meaningful for t ≥ 0. It shows that initially there is no drug in the blood stream and then that the concentration of drug increases to a maximum concentration of 12 mcg/ml after 1.7 hours. Hence, the maximum effect of Tylenol is felt after approximately two hours. After reaching the maximum value, the concentration decays to zero.

In the previous example, we note that because the function has a maximum above zero and then approaches 0 as x → ∞, the graph necessarily has a point of inflection where C″(t) switches from being positive to negative. In Problem Set 4.1, we ask you to discuss the sign of the second derivative and to solve for the value of x where this point of inflection occurs.

Graphing families of functions

The shape of some functions, such as f(x) = 3x + a, does not depend in any critical sense on the value of the parameter a. All a does is move the line of slope 3 up and down the xy plane. As we see in this subsection, however, in more complicated functions the value of a parameter can have a surprising effect, and we can use calculus to discover such effects.

Example 5 To infinity and back

Consider the function f(x) = images with the parameter a. Using first derivatives and asymptotes, determine how the shape of this function depends on the parameter a.

Solution Since images = 0, f(x) has a horizontal asymptote y = 0 as x → ±∞. If a < 0, then x2a is positive for all x and there are no vertical asymptotes. If a = 0, then there is a vertical asymptote at x = 0. If a > 0, then there are vertical asymptotes at x = ±images. Computing the first derivative yields

images

Since the denominator of this expression is positive whenever x2a, f′ (x) < 0 for all positive x with x2a and f′(x) > 0 for all negative x with x2a.

These computations suggest there are qualitatively three distinctive graphs. First, consider the case where a < 0. In this case, there are no vertical asymptotes. There are horizontal asymptotes of 0 as x → ±∞. Moreover, f(x) is increasing for negative x and f(x) is decreasing for positive x. Hence, the graph looks something like the figure on the left.

images

Next, consider the case where a = 0. In this case, f(x) = images and there is a vertical asymptote at x = 0. In fact, imagesimages = ∞. There are horizontal asymptotes of y = 0 as x → ±∞. Moreover, f(x) is increasing for negative x and f(x) is decreasing for positive x. Hence the graph, as you by now know, looks something like the figure on the left.

Finally, consider the case where a > 0. In this case, there are vertical asymptotes at x = ±images. In fact, evaluating all the limits as x approaches ±images yields

images

There are horizontal asymptotes of y = 0 as x → ±∞. Moreover, f(x) is increasing for negative x and f(x) is decreasing for positive x. Hence, if we walk along the graph of f from +∞ to −∞, then we initially ascend from 0. The ascent gets exceptionally steep as x approaches − images. As we cross x = −images, we suddenly fall down to y = −∞. After crossing both infinities, we continue to ascend until reaching a maximum of y = −images at x = 0. From there, we descend through −∞ and skyrocket to +∞ at x = images. After this harrowing jump through infinities, we continue with a descent to zero. In other words, our graph looks something like the figure on the left.

images

Example 6 Dose-response curves

Dose-response curves can be used to plot the response of an individual to a dose of a drug or hormone. This response can almost be anything. For instance, the response may be heart rate, dilation of an artery, membrane potential, enzyme activity, or the secretion of a hormone. We previously encountered dose-response curves in Example 2 of Section 2.4 and Problem 31 in Problem Set 3.2. A general form of a dose-response curve is

images

where y is the response of the individual and x is the concentration of the dose of drug or hormone. The parameters a > 0, b > 0, and c > 0 affect the shape of the dose-response curve and often can be used to fit the function to particular data sets. Assuming c = 0, use limits and first derivatives to determine how the shape of this curve depends on the parameters a and b.

Solution Consider f(x) = images. Since f(x) is continuous for all reals, there are no vertical asymptotes. Since

images

there is a horizontal asymptote y = a as x approaches +∞ and a horizontal asymptote y = b as x approaches −∞.

Taking the first derivative of f, we obtain

images

This derivative is negative for all x if b > a, positive for all x if b < a, and zero for all x if b = a.

Hence, the graph of f(x) comes in three flavors. If b > a, then the function decreases from an asymptote of y = b to an asymptote of y = a. If b < a, then the function increases from an asymptote of y = b to an asymptote of y = a. Finally, if b = a, then the function is the constant function y = a. These three graphs are sketched below.

images

Note that when b > a, the graph goes from concave down to concave up and hence passes through an inflection point. Similarly, when a > b, the graph goes from concave up to concave down and hence also passes through an inflection point. You are asked to explore this further in Problem Set 4.1.

Example 7 Stock-recruitment curves

In conservation biology and fisheries management, stock-recruitment curves are used to describe the relationship between the current abundance of a population (i.e., the stock) and the number of juveniles entering the system in the next year (i.e., the recruits). A general class of stock-recruitment curves are given by the functions

images

where N is the current population size, F(N) is the number of recruits in the next generation, and a and b are positive parameters. A useful way to categorize these functions is to consider the relationship between the current population abundance and the average of number recruits per individual, that is,

images

Use limits and first derivatives to determine how the parameter b influences the shape of f(N) for N ≥ 0. Discuss the possible meaning.

Solution Notice that if 0 < b < 1, then there is a vertical asymptote at N = 0 and images f(N) = ∞. If b ≥ 1, there is no vertical asymptote. To determine the horizontal asymptote as N → ∞, notice that the power of the numerator is less than the power of the higher order term in the denominator. Hence, images f(N) = 0.

The first derivative of f(N) is given by

images

Hence, if b ≤ 1, f′(N) < 0 for all N > 0. However, if b > 1, then f′(N) > 0 for 0 ≤ N ≤ (b − 1)1/b and f′(N) < 0 otherwise. Therefore, we get three types of graphs, depending on whether b < 1, b = 1, or b > 1:

images

For b ≤ 1, the number of recruits constantly decreases with stock levels. One interpretation of this fact is at higher population densities, there are fewer resources per individuals and, consequently, fewer recruits produced per individual. For b > 1, the number of recruits per individual initially increases and then decreases. One possible explanation is that for b > 1, individuals have difficulty finding mates for reproduction. Therefore, at low densities as densities increase, the chance of finding mates increases and the number of recruits produced per individual increases. However, as the population density increases too much (i.e., beyond (b − 1)1/b), the advantage of finding mates is outweighed by the limited resources available per individual. Consequently, at higher densities, the number of recruits per individual decreases. For b > 1, the population exhibits what ecologists call depensation or a strong Allee effect. Working with data from 128 species, Ran Myers and colleagues used F(N) to evaluate to what extent fish populations exhibit depensation and discussed the implications for populations to recover from environmental disturbances. (See R. A. Myers, N. J. Barrowman, J. A. Hutchings, and A. A. J. Rosenberg, “Population Dynamics of Exploited Fish Stocks at Low Population Levels,” Science 269 (1995): 1106–1108.)

PROBLEM SET 4.1

Level 1 DRILL PROBLEMS

In Problems 1 to 14, graph the functions by hand by finding asymptotes and using first and second derivatives. Compare your graphs to what you get using technology.

images

images

images

In Problems 15 to 20, graph the families of functions by finding asymptotes and using first and second derivatives. In particular, determine how the graph of the functions depends on the parameter a > 0.

images

In Problems 21 and 22, sketch the graph of a function with the given properties.

21. x = 2, x = −2 are vertical asymptotes f is increasing for 0 < x < 2 and x > 2 f is decreasing for x < −2 and −2 < x < 0 graph is concave down on (−∞, −2) and (2, ∞) intercepts are (−1, 0), (−3, 0), (3, 0) and (1, 0)

22. y = 1, y = −1 are horizontal asymptotes f is increasing for x < −images and for x > images f is decreasing for −1 < x < 1 graph is concave down for x < −1 and for 0 < x < 1 graph is concave up for x > 1 and for −1 < x < 0

Level 2 APPLIED AND THEORY PROBLEMS

23. Consider the graph of y = ax2 + bx + c for constants a, b, and c. Use second derivatives to determine what happens to the graph as a changes.

24. Consider the graph of y = images. Use limits and first derivatives to determine how the shape of this curve depends on the parameter a.

25. In Example 6, we saw that the dose-response curve

images

has asymptotes y = a and y = b, respectively, as x approaches ±∞. Find the second derivative y″(x) and discuss its properties including an equation that can be used to identify any points of inflection, if they exist.

26. In Example 4 of Section 2.7, we consider patterns of local species richness of ants along an elevational gradient. A function that best fits the data is

images

where x is elevation measured in kilometers and S is the number of species. Plot this function using information about first derivatives.

27. In Example 6 of Section 1.5, we develop the Michaelis-Menton model for the rate at which an organism consumes its resource. For bacterial populations in the ocean, this model was given by

images

where x is the concentration of glucose (micrograms per liter) in the environment. Use asymptotes and first derivatives to sketch this function by hand.

28. In Example 5 of Section 2.4 we found that the rate at which wolves kill moose can be modeled by

images

where x is measured in number of moose per square kilometer. Use asymptotes and first derivatives to sketch this function.

29. In Problem 42 in Problem Section 2.4, we examined how wolf densities in North America depend on moose densities. We found that the following function provides a good fit to the data:

images

where x is number of moose per square kilometer.

  1. Find the horizontal and vertical asymptotes.
  2. Determine on which intervals f is increasing and decreasing.
  3. Determine on which intervals f is concave up and concave down.
  4. Use the information from parts ac to sketch the graph of f(x).

30. Two mathematicians, W. O. Kermack and A. G. McKendrick, showed that the weekly mortality rate during the outbreak of the plague in Bombay (1905–1906) is reasonably well described by the function

images

where t is measured in weeks. Sketch this function using information about asymptotes and first derivatives. Recall that

images

31. In Example 4 we modeled the rate at which acetaminophen diffuses from the stomach and intestines to the blood stream using this equation:

images

Calculate the second derivative and discuss its behavior. Identify if the function C(t) has any points of inflection for t > 0.

32. In an experiment, a microbiologist introduces a toxin into a bacterial colony growing in an agar dish. The data on the area of the dish covered by living colony members at time t minutes after the introduction of the toxin are given by this equation:

images

Sketch the graph of A(t) showing its salient features.

33. Let f be a function that represents the weight of a fish at age t. Write a function that satisfies the following properties.

  • The weight of the fish at birth must be positive.
  • As the fish ages, the weight increases at decreasing rate.
  • No fish can grow bigger than 2 kg.

34. Aerobic rate is the rate of a person's oxygen consumption and is sometimes modeled by the function A defined by

images

for x ≥ 10. Graph this function.

35. As a project for their mathematical biology class, three college students developed a model of how acetaminophen levels varied in the blood stream of a child after taking a dose of 325 milligrams. Using FDA data, they found that

images

where t is hours after taking the dose.

  1. Use information about asymptotes and first derivatives to sketch this function.
  2. Discuss the meaning of your graph. In particular, address when the maximum concentration is achieved and what the maximum concentration is.

36. A naturalist at an animal sanctuary determined that the function

images

provides a good measure of the number of animals in the sanctuary that are x years old. Sketch the graph of f for x > 0.

4.2 Getting Extreme

When viewing the graph of a function as a landscape of hilltops and valley bottoms, each top and each bottom corresponds to a place where the function in question has an extremum. Methods to identify extrema play an important role in applications. For instance, if the function of interest represents how profits due to harvesting a crop depend on the amount of seeds planted, then the farmer would like to know how many seeds per acre yield the greatest profits. In other words, the farmer would like to identify the largest hilltop of the function. Alternatively, if a northwestern crow minimizes the amount of energy required to break whelk shells, then the optimal behavior for a crow corresponds to the deepest valley of a function. In this section, we develop methods to find these hilltops and valleys.

Local extrema

Local Maxima and Minima

Let f(x) be a function of x. We say that f has a local maximum at x = a if

images

for all x near a. We say that f has a local minimum at x = a if

images

for all x near a. We say f has a local extremum at x = a if there is a local maximum or a local minimum at x = a.

Example 1 Finding Extrema

Estimate for what x values, y = f(x), as graphed below, has local maxima and local minima on the domain D = {x: −3 ≤ x ≤ 3}.

images

Solution There are local minima at x ≈ −1.75, x = 0, and x = 3. There are local maxima at x = −3, x ≈ −1, and x = 2.

The previous example suggests that either extrema occur at end points of the domain, points where f is not differentiable, or points where the derivative of f equals zero. The following theorem verifies these observations.

Theorem 4.1 Fermat's theorem

If f is defined on (a, b) and has a local extremum at c images (a, b), then either f′(c) = 0 or f′(c) is not defined.

Proof. Suppose that f is defined on (a, b) and has a local extremum at c images (a, b). This extremum is either a local maximum or a local minimum. Suppose that this extremum is a local minimum. Then we have f(c) ≤ f(x) for all x near c. Equivalently, f(c) ≤ f(c + h) for all h sufficiently small. Taking a difference quotient yields that

images

for all sufficiently small positive h and

images

for all sufficiently small negative h. Assume f′(c) exists. Then, taking one-sided limits yields

images

and

images

Therefore, f′(c) = 0. The case of a local maximum can be proved similarly, which you see in Problem 31 in Problem Set 4.2.

Fermat's theorem shows that we can find possible local maxima and local minima by finding points where f′(x) = 0 or f′ is not defined. Such points have a special name.

Critical Points and Values

If f′(c) = 0 or f′(c) is not defined, then c is a critical point for f. The value of f at a critical point is called a critical value.

Although all local extrema are critical values, not all critical values are local extrema. Consider, for example, y = x3 plotted on the left for −1 ≤ x ≤ 1.

images

The derivative is images = 3x2. Hence, x = 0 is the only critical point. However as y = x3 increases over all the reals, x = 0 is neither a local maximum nor a local minimum.

Example 2 Finding and classifying critical points

Find the critical points of y = x3 − 3x2 − 4 and determine whether these critical points are local maxima, local minima, or neither.

Solution We have images = 3x2 − 6x = 3x(x − 2). Hence, images = 0 at x = 0 and x = 2. These are the critical points of y. To determine whether these critical points correspond to local maxima, local minima, or neither, we can consider how the sign of the derivative varies over the real line. Since images < 0 for 0 < x < 2 and images > 0 for x > 2, the function decreases over the interval (0, 2) and increases on (2, ∞). Hence, there is a local minimum of y = −8 at x = 2. Alternatively, since images > 0 for x < 0 and images < 0 for 0 < x < 2, the function increases until x = 0 and then decreases. Thus, there is a local maximum y = −4 at x = 0. Graphing y = x3 − 3x2 − 4 with technology in the plot on the left corroborates these statements.

images

Example 2 illustrates one method for identifying local maxima and local minima; this method is called the first derivative test.

First Derivative Test

Assume f′(c) = 0 and f is differentiable near x = c.

  • If the sign of f′ changes from positive to negative at x = c (i.e., f changes from increasing to decreasing), then f has a local maximum at x = c.
  • If the sign of f′ changes from negative to positive (i.e., f changes from decreasing to increasing) at x = c, then f has a local maximum at x = c.

Example 3 Thermodilution

Cardiac output can be determined by thermodilution as illustrated in Figure 4.4. Let's say a doctor injects 10 milliliters (ml) of a cold dextrose solution into a vein entering the heart. As the cold solution mixes with the blood in the heart, the temperature variations in the blood leaving the heart are measured. A typical temperature variation curve (i.e., degrees below normal temperature), may be described by the function T(t) = 0.2t2et degrees Celsius, where t is measured in seconds. Find the critical points and classify them. Discuss the meaning of your results.

images

images

Figure 4.4 Thermodilution involves injecting a cold dextrose solution in the vena cava of the heart and measuring the temperature of the blood leaving the heart (e.g., from the aorta).

Solution Taking the derivative yields

images

We have T′(t) = 0 at t = 0 and t = 2. Hence, t = 0 and t = 2 are the critical points of T. To apply the first derivative test, we need to determine the sign of T′ on the intervals (−∞, 0), (0, 2), and (2, ∞). Since T′ is continuous everywhere, it can only have sign changes at the points t = 0, 2. Therefore, it suffices to check the sign of T′ at one point in each of the intervals. Since T′(−1) = −0.6e1 < 0, T′ is negative on (−∞, 0). Since T′(1) = 0.2e−1 > 0, T′ is positive on (0, 2). Since T′(3) = −0.6e−3 < 0, T′ is negative on (2, ∞). Since at t = 0 the sign of T′ changes from negative to positive, we have a local minimum at t = 0. Since at t = 2 the sign of T′ changes from positive to negative, at t = 2 we have a local maximum. Hence, the temperature of blood leaving the heart drops T(2) ≈ 0.11 degrees Celsius after two seconds, before returning to its normal temperature.

Another possibility for identifying local maxima and local minima is using the second derivative. Suppose f has a critical point at x = a and has second-order derivatives at x = a. In Section 3.6, we saw that a second-order approximation of f(x) is given by

images

Since a is a critical point, f′(a) = 0 and the second-order approximation reduces to

images

images

Figure 4.5 The second derivative test determines whether a critical point x = a where f′(a) = 0 is a local minimum (a) or a local maximum (b).

Provided that f″(a) ≠ 0, the graph of this approximation is given by a parabola whose vertex is at x = a. Furthermore, if f″ (a) > 0, then this parabola is facing up, which suggests that there is a local minimum at x = a as shown in Figure 4.5a. Alternatively, if f″(a) < 0, then this parabola is facing down, which suggests that there is a local maximum at x = a as shown in Figure 4.5b. What is suggested by the second approximation can be proven, thereby providing an alternative test for classifying extrema.

Second Derivative Test

Let f have first and second derivatives at x = a. Assume that f′(a) = 0.

Local maximum If f″(a) < 0, then there is a local maximum at x = a.

Local minimum If f″(a) > 0, then there is a local minimum at x = a.

Inconclusive If f″(a) = 0, then we can not determine whether the critical point is a maximum or minimum from the second derivative.

Example 4 Using the second derivative test

Find and classify the critical points of y = −x3 + 6x2 + 2 using the second derivative test.

Solution Computing the first and second derivatives of y = −x3 + 6x2 + 2 yields

images

This derivative always exists, so the critical points correspond to the solutions of −3x(x − 4) = 0. Hence, they are given by x = 0 and x = 4. Evaluating the second derivatives at x = 0 and x = 4 yields

images

Hence, there is a local minimum at x = 0 and a local maximum at x = 4. Graphing the function y = −x3 + 6x2 + 2 demonstrates these conclusions in the graph on the left.

images

Global extrema

Often in applied problems, we want to find the largest or small value of a function on its domain.

Global Extrema

Let f be a function with domain D.

f has a global minimum at x = a if

images

f has a global maximum at x = a if

images

A global maximum or a global minimum is called a global extremum.

Example 5 Finding global extrema

Consider the function whose graph is illustrated on the left. Find the global extrema of this function.

Solution The global maximum of y = 9 occurs at x = −2 and the global minimum of approximately 0.1 occurs approximately at x = −0.6.

images

Example 5 illustrates that global extrema may occur at critical points or end points for a continuous function on a closed interval. Thus, we have the following procedure for finding global extrema.

Closed Interval Method

Let f be a continuous function defined on the closed interval [a, b]. To find the global extrema of f, do the following:

Find critical points on the interval (a, b).

Evaluate f at all critical points and at end points a and b.

Identify the global extrema The largest value of f at a critical point or end point is the global maximum of f. The smallest value of f at a critical point or end point is the global minimum of f.

Example 6 Using the closed interval method

Find the global extrema of f(x) = images on the interval [−3, 6].

Solution Taking the derivative of f yields

images

The critical points are x = 3 and x = −2. Evaluating f at the critical points and end points yields f(−3) = 8.5, f(−2) = 11images, f(3) = −9.5, and f(6) = 22. Therefore, the global maximum of 22 occurs at the end point x = 6. The global minimum of −9.5 occurs at the critical point x = 3. Plotting this function on the left demonstrates our findings.

images

Example 7 Getting extreme with carbon dioxide

In Example 4 of Section 1.2, we examined how CO2 concentrations in parts per million (ppm) varied from 1974 to 1985 at the Mauna Loa Observatory in Hawaii. Using linear and periodic functions, we found that the following function gives an excellent fit to the data:

images

where x is months after April 1974. Using the closed interval method, find the global maximum and minimum CO2 levels in the one-year interval [0, 12].

Solution To find the critical points of f(x), we differentiate

images

Although we can solve for the critical points by hand by recalling properties of inverse sine, we circumvent such analysis by using a root finder on a graphing calculator. Finding all the roots of f′(x) on the interval [0, 12] yields x = 0.149 and x = 5.851. Evaluating f to two decimal points at these critical points and the end points yields

images

Hence, the global minimum CO2 level occurred at x = 5.851 (sometime in late October 1974). The global maximum occurred at x = 12 (in April 1975). Plotting the function over the interval [0, 12] demonstrates these extremes:

images

Example 8 Search period of the codling moth

After a codling moth (Cydia pomonella, see Figure 4.6) larva hatches from its egg case, it goes looking for an apple in which to burrow. The period between hatching and finding the apple is called the search period. For individual larvae, this period will vary, but the average time s that it takes is known to be a function of temperature T in degrees Celsius. A good fit to the available data illustrated in Figure 4.7 is provided by the equation

images

Use the tools of calculus to find the largest and smallest values of s(T) over the range 20 ≤ T ≤ 30.

images

Figure 4.6 Codling moth adult.

Solution The polynomial p(T) = −0.03T2 + 1.67T − 13.65 has roots at T ≈ 9.95 and T ≈ 45.7. Hence, p(T) ≠ 0 for 20 ≤ T ≤ 30 and s(T) = images is continuous on this interval. Using the quotient rule, the derivative is

images

which satisfies s′(T) = 0 only if 0.06T = 1.67, equivalently T ≈ 27.83°C. Evaluating s(T) at this critical point and at the end points we obtain

images

Hence, on the interval 20 ≤ T ≤ 30, s(T) has a minimum at T ≈ 27.83 and a maximum at T = 20.

images

Figure 4.7 Codling moth search period.

Data Source: P. L. Shaffer, and H. J. Gold, “A Simulation Model of Population Dynamics of the Codling Moth, Cydia pomonella,” Ecological Modeling 30 (1985): 247–274.

In many problems we need to find the global extrema on open intervals, half-closed intervals, or intervals involving infinity. For each of these cases, we deal with the limits as we approach the end points of the intervals, as illustrated with the open interval method. (The other cases appear as exercises in Problem Set 4.2.)

Open Interval Method

Let f be a continuous function defined on the open interval (a, b). Assume the limits images are well defined. Here we allow L and M to be ±∞, a to be −∞, and b to be +∞. To find the global extrema of f on (a, b), do the following:

Find critical points Find all the critical points on the interval (a, b).

Evaluate at critical points Evaluate f at all critical points.

Identify the extrema If L or M is greater than f evaluated at any critical point, then f has no global maximum on (a, b). Alternatively, if f evaluated at a critical point x = c is greater than or equal to L, M, and f evaluated at any other critical point, then f(c) is the global maximum. If L or M is less than f evaluated at any critical point, then f has no global minimum on (a, b). Alternatively, if f evaluated at a critical point x = c is less than or equal to L, M, and f evaluated at any other critical point, then f(c) is the global minimum.

Example 9 Using the open interval method

Use the open interval method to find the global extrema of the following functions on the indicated intervals.

images

Solution

  1. We have f (x) = images is continuous on (1, 2). Note f′(x) exists for all x on (1, 2). Moreover,

    images

    Hence, f(x) has no global maximum on (1, 2). Solving for the critical points on (1, 2), we get

    images

    Since f has only one critical point and f(1.5) = 4 is less than images and images, the global minimum is 4 and occurs at x = 1.5.

  2. Since 1 + x2 is positive for all x, f(x) = images is continuous on (−∞, ∞). Taking limits at infinity, we get

    images

    Solving for the critical points, we get

    images

    Since f(1) = images and f(1) = −images are greater than 0 and less than 0, respectively, these correspond to the global minimum and global maximum.

PROBLEM SET 4.2

Level 1 DRILL PROBLEMS

In Problems 1 to 4, identify the local and global extrema.

images

images

In Problems 5 to 12, find the critical points and use the first derivative test to classify them.

images

In Problems 13 to 16, find the critical points and use the second derivative test to classify them.

images

In Problems 17 to 20, use the closed interval method to find the global extrema on the indicated intervals.

17. f(x) = x2 − 4x + 2 on [0, 3]

18. f(x) = x3 − 12x + 2 on [−3, 3]

19. f(x) = x + images on [0.1, 10]

20. f(x) = xex on [0, 100]

In Problems 21 to 24, use the open interval method to find the global extrema on the indicated intervals.

21. f(x) = x2 − 4x + 2 on (−∞, ∞)

22. f(x) = x3 − 12x + 2 on (0, ∞)

23. f(x) = x + images on (0, ∞)

24. f(x) = xex on (−∞, ∞)

25. Let f be continuous on the half-open interval [a, b) with b possibly equal to +∞. Devise a method to find the global extrema of f on this interval.

26. Let f be continuous on the half-open interval (a, b] with a possibly equal to −∞. Devise a method to find the global extrema of f on this interval.

In Problems 27 to 30, use the half-open interval methods you developed in Problems 25 and 26 to find the global extrema on the indicated intervals.

27. f(x) = x2 − 4x + 2 on [0, ∞)

28. f(x) = x3 − 12x + 2 on [1, 10)

29. f(x) = x + images on [−1, ∞)

30. f(x) = xex on (−∞, −1]

Level 2 APPLIED AND THEORY PROBLEMS

31. Let f be defined on (a, b) and c images (a, b). Prove that if x = c is a local maximum and f is differentiable at x = c, then f′(c) = 0.

32. In Example 4 of Section 1.2, we examined how CO2 concentrations (in ppm) have varied from 1974 to 1985. Using linear and periodic functions, we found that the following function gives an excellent fit to the data:

images

where x is months after April 1974. Using the closed interval method, find the global maximum and minimum CO2 levels on the interval [12, 24].

33. In the previous problem, use the closed interval method and find the global maximum and minimum CO2 levels between April 2000 and April 2001.

34. A close relative of the codling moth is the pea moth, Cydia nigricana, which is a pest of cultivated and garden peas in several European countries. If its search period in one of the regions where it is a pest is given by the function

images

then graph s(T) using information about the first derivative over the domain 20 ≤ T ≤ 30. Be sure that your graph indicates the largest and smallest value of s over this interval.

35. In Problem 30 of Problem Set 4.1, we saw that the weekly mortality rate during the outbreak of the plague in Bombay (1905–1906) can be reasonably well described by the function

images

where t is measured in weeks. Find the global maximum of this function. Recall that

images

36. Some species of plants (e.g., bamboo) flower once and then die. A well-known formula for the average growth rate r of a semelparous species (a species that breeds only once) that breeds at age x is

images

where s(x) represents the proportion of plants that survive from germination to age x, n(x) is the number of seeds produced at age x, and p is the proportion of seeds that germinate.

  1. Find the age of reproduction that maximizes r in terms of the parameters a, b, c, and p where

    images

    and

    images

    0 < c < 1.

  2. Sketch the graph of y = r(x) for the case where a = 0.2, b = 3, c = 0.8, and p = 0.5.

37. The production of blood cells plays an important role in medical research involving leukemia and other so-called dynamical diseases. In 1977, a mathematical model was developed by A. Lasota. (See W. B. Gearhart and M. Martelli, “A Blood Cell Population Model, Dynamical Diseases, and Chaos,” in UMAP Modules 1990:Tools for Teaching [Arlington, MA: Consortium for Mathematics and Its Applications, 1991].) The model involved the cell production function

images

where A, s, and r are positive constants and x is the number of granulocytes (a type of white blood cell) present.

  1. Find the granulocyte level x that maximizes the production function P. How do you know it is a maximum?
  2. Graph this function.

38. When you cough, the radius of your trachea (windpipe) decreases, thereby affecting the speed of the air in the trachea. If r is the normal radius of the trachea, the relationship between the speed S of the air and the radius r of the trachea during a cough is given by a function of the form

images

where a is a positive constant. (Philip M. Tuchinsky, “The Human Cough,” UMAP Modules 1976: Tools for Teaching [Lexington, MA: Consortium for Mathematics and Its Applications, 1977].) Find the radius r for which the speed of the air is the greatest.

39. Research indicates that the power P required by a bird to maintain flight is given by the formula

images

where v is the relative speed of the bird, w is its weight, ρ is the density of air, and S and A are constants associated with the bird's size and shape. (See C. J. Pennycuick, “The Mechanics of Bird Migration,” IBIS III (1969): 525–556.) What speed will minimize the power? You may assume that w, ρ, S, and A are all positive and constant.

40. An epidemic spreads through a community in such a way that t weeks after its outbreak, the number of residents who have been infected is given by a function of the form

images

where A is the total number of susceptible residents. Show that the epidemic is spreading most rapidly when half the susceptible residents have been infected.

4.3 Optimization in Biology

One of the most important applications of calculus to biology in particular, and science and technology in general, involves finding the extrema of functions. Consider, for example, the problem of determining the most effective treatment regimen for a malignant tumor using chemo- or radiation therapy. Such treatments are toxic to the body, so physicians want to prescribe the minimum dosage that will do the job. Typically, a single treatment will not destroy the tumor. Instead, the tumor will initially shrink in size after treatment and sometime later will begin to regrow. Ideally, therapy should be readministered immediately—when the tumor is at its smallest—before this regrowth phase. Using calculus in conjunction with tumor growth data, we can estimate the time when therapy should be reapplied. As another example, think of a farmer planting a corn crop. The farmer might be interested in knowing the planting density of seeds that would maximize his or her profit. To find out, the farmer could formulate a function that describes how profits depend on planting density, and maximizing this function. In this section, we consider these problems as well as the behavior of dogs fetching balls, sustainable harvesting of arctic fin whales, and vascular branching. More examples appear in the problem set. We end this section with applications to finding best-fitting functions to empirically collected data.

Steps to solve an optimization problem

Optimization problems typically require that we develop an appropriate model for the problem being considered, then analyze this model to find the optimal solution for the problem. To tackle these problems successfully, use the following steps:

  1. Read, understand, and visualize. Take the time to carefully read the problem so that you fully understand what is being asked. In particular, ask yourself: What am I trying to maximize or minimize? What information am I given? Is it sufficient to solve the problem at hand? When appropriate, draw a picture or figure that summarizes the problem.
  2. Identify key variables and quantities. Ask yourself: What are the important quantities in the problem? What quantity is being optimized? This is the dependent variable. Which of the variables is the one whose value I can control to obtain my optimal solution? This is the independent variable. What additional quantities presented in the problem do I need to obtain the sought after relationship between the dependent and independent variable? Associate units with each of these variables.
  3. Write the function. In this step, you need to determine how the dependent variable is determined by the quantities that you identified in the previous step. Think carefully about this crucial step. Make sure that units on both sides of your equation agree.
  4. Optimize. Determine whether you need to minimize or maximize the function and over what interval you need to perform the optimization. To find the optimal value, it suffices to find the critical points and evaluate the function at these critical points and at the end points of the interval. Whichever of these values is the largest (smallest) yields the maximum (minimum).
  5. Interpret your answer. Interpret the results of your optimization. Ask yourself whether your answer makes sense. If not, check your work.

In this next example, we demonstrate how these steps are applied to the problem of determining the density of seedlings that a farmer needs to plant to optimize profits from a particular crop. In the remaining examples in this section, we do not stress the steps involved, but these steps are implicit in our approach. Remember to you these steps whenever you get stuck in solving an optimization problem. The examples relate to behavior, physiology, resource management, and fitting functions to data. You won't necessarily have time to study them all, so you can pick and choose those of greatest interest to you.

Example 1 Maximum economic yield

In an article titled “A ‘Cookbook’ Approach for Determining the ‘Point of Maximum Economic Return,’” Gaspar and colleagues lay out procedures for farmers to conduct field trials to determine the impact of different planting densities on crop yields. In an experiment to assess the impact of planting densities on yield Y of a corn hybrid, they fitted the following function to the data depicted in Figure 4.8.

images

where x is thousands of seeds planted per acre.

In a particular year, suppose that the price a farmer can obtain for his corn is $1.50 per bushel and that the cost of seed is $3 per thousand seeds. Use this information to determine the density of seeds (i.e., seeds per acre) that maximizes the farmer's net profit per acre.

images

Figure 4.8 A yield curve (red) fitted to data (blue points) obtained from a corn hybrid experiment (since the curve does not pass through the point (0, 0), it should not be used to estimate yield at densities close to zero).

Data Source: Gaspar, P.E., S. Paszkiewicz, P. Carter, M. McLeod, T. Doerge, and S. Butzen. 1999. Corn hybrid response to plant populations. Pioneer Hi-Bred International Inc. Northern Agronomic Research Summary. pp. 29–40.

Solution We follow the five steps for solving an optimization problem.

  1. From the statement of the problem, we realize that the farmer wants to maximize his net profit, measured in dollars per acre, rather than his yield, measured in bushels per acre. The planting density of seeds is the quantity that can be varied.
  2. The independent variable is x, with units 1,000 seeds per acre; the dependent variable is net profit P(x), with units dollars per acre.
  3. Since P(x) is determined by the total revenue R(x) generated from the crops each season minus the cost C(x) of the seeds each season, it follows that

    images

    where R(x) is the yield Y(x) in bushels per acre multiplied by the price p = 1.5 in dollars per bushel of corn; that is,

    images

    On the other hand, the cost per acre is $3 for each thousand seeds, so that for x thousand seeds per acre the cost is

    images

    Hence, we can write the function

    images

    images

    that we want to maximize with respect to x on the interval [0, ∞). If we plot P(x) verus x, then the red line in the figure on the left indicates optimum planting density.

  4. The point at which P(x) is maximized can be found by solving for the critical points of P(x). Since

    images

    it follows that x = images = 27.617 is the unique critical point. Since the graph of P(x) is a downward facing parabola, the global maximum occurs at x = 27.617.

  5. Our interpretation of 27.617 is that the farmer should plant approximately 27.6 thousand seeds per acre and, in doing so, should obtain a net profit of P(27.6) ≈ $155 per acre.

In many problems in population biology, a variable x is used to represent the density (or number) of individuals in a population and the function G(x) is used to represent the population growth rate. For instance, for the discrete logistic equation presented in Example 7 of Section 2.5, we modeled the growth using the function.

images

where r > 0 has the interpretation of the maximum per capita growth rate and K > 0 has the interpretation of the environmental carrying capacity. Here, we use the function G(x) to determine the optimal rate at which a population of whales should be harvested, if not illegal to do so.

The Arctic fin whale Balaenoptera physalus, which at fifty to seventy tons for adults of both sexes is second in size to the blue whale, was a highly desirable catch during the whaling heydays of the nineteenth and twentieth centuries. As many as 30,000 individuals were slaughtered each year from 1935 to 1965. This level of exploitation could not be sustained for long, so today population levels are estimated to be an order of magnitude (i.e., a factor of 10) below historical highs of around a half million individuals. Some individuals are still taken each year for purposes of subsistence by aboriginal people in Greenland. A moratorium on whale hunting is needed, however, to allow this species to recover to levels where the populations can be safely exploited on a sustainable basis.

images

Figure 4.9 Arctic fin whale.

Example 2 Sustainable exploitation of the Arctic fin whale

Assume the Arctic fin whale growth rate is modeled by the logistic function G(x) = rx(1 − x/K) with r = 0.08 (i.e., an 8% annual growth rate when the whale densities are low) and K = 500,000 (i.e., prior to exploitation, the Arctic fin whale population was estimated to be around half a million individuals). If the population is harvested at a constant rate of H individuals per year for an extended period of time, then this harvesting rate is sustainable if there exists a positive number x of whales at which the growth rate G(x) equals the harvesting rate H. That is, it is possible for the growth to keep pace with the loss from harvesting. Determine the maximal sustainable harvesting rate.

Solution According to the statement of the problem, a harvesting rate H is sustainable if there is a positive x such that

images

Hence, maximizing a sustainable harvesting rate is equivalent to finding x > 0 which maximizes G(x). Since the graph of G is a downward facing parabola, we can find its maximum by taking the derivative of G, setting it equal to zero, and solving for x:

images

Hence, the maximum sustainable yield occurs at a harvesting rate of H = G(250,000) = 10,000 whales per year at which the whale population consists of 250,000 individuals. This maximum sustainable harvesting rate of 10,000 whales per year is three times smaller than the harvesting rate in the early twentieth century. Hence, the model reaffirms the statement that harvesting at 30,000 whales per year in the early twentieth century was not sustainable and may explain why the current population sizes are an order of magnitude lower than half a million.

Sometimes when solving a problem it is useful to sketch a figure, as illustrated in the next example.

Example 3 Do dogs know calculus?

Professor Tim Pennings from Hope College wanted to determine whether his dog, Elvis, fetched balls thrown into Lake Michigan in an optimal way. (See T. Pennings, “Do Dogs Know Calculus?” College Mathematics Journal 34(2003):178–182.) Standing along the shoreline with Elvis at his side, Tim would throw the ball into the water. Elvis could choose to swim out directly from where Tim was standing to get the ball, hence taking a minimal-distance trajectory. Alternatively, he could run along the shore before he jumped into the water and swam to the ball. Because Elvis can only swim at an average speed of 0.91 meters per second whereas he can run at an average speed of 6.4 meters per second, it is likely that he ran for some distance along the shore. But how far along the shore should Elvis run?

Professor Pennings performed an experiment to assess what strategy Elvis was playing by throwing the ball repeatedly into the water and keeping track of where Elvis entered the water. For one throw, the ball landed 6 meters from the shore, as illustrated in Figure 4.10a.

images

Figure 4.10 How will Elvis fetch a ball that landed 6 meters from shore?

What path would Elvis take if he were to minimize that amount of time it took him to retrieve the ball?

Solution Let us begin by sketching a figure that indicates a hypothetical path Elvis could take (see Figure 4.10b). In this drawing, 15 − x is the distance Elvis runs along the shore. Assuming that Elvis wants to minimize the time to getting to the ball, we need to write a function that describes how the amount of time to get to the ball depends on x. Since he is running at a speed of 6.4 meters per second and runs a distance of 15 − x meters along the shore, we get that the time he spends running on the shore is

images

By the Pythagorean theorem, the distance Elvis swims to the ball is images. Hence, the time he spends swimming is

images

Hence, the total time T it takes him to get to the ball as a function of x is given by

images

We want to understand the graph of this function for 0 ≤ x ≤ 15, so let us take the derivative of T.

images

To find the critical points, we need to solve T′(x) = 0. Doing so yields

images

Hence, on the interval [0, 15], T′ vanishes only at x = 0.86. Since T′(2) ≈ 0.19 > 0, we see that T is increasing on the interval (0.86, 15]. On the other hand, since T′(0) ≈ −0.16, T is decreasing on [0, 0.86). Thus, the minimum time is achieved at x = 0.86. Therefore, Elvis should run 14.1 meters along the shore before jumping into the water.

So what was the outcome of Professor Pennings' experiment? When Professor Pennings measured the point at which Elvis entered the water, he found that Elvis ran 15 − x = 14.1 meters along the shore (i.e., x = 0.9). Does this dog know calculus? Well, he could have been lucky on this one throw. So Professor Pennings performed thirty-five throws, with the ball landing different distances d from the shoreline. Professor Pennings measured the point x where Elvis entered the water on each throw. In the problems at the end of this section, you will be asked to show that the optimal place to enter the water as a function of the distance d the ball lands from the shore is

images

A scatter plot of the data and the line is shown in Figure 4.11. This figure illustrates that Elvis is on average acting pretty optimal, which is quite remarkable.

images

Figure 4.11 Scatter plot of distance of ball from shore (in the horizontal direction) and Elvis's point of entry 15 − d in the water (in the vertical direction). The best-fitting line x = 0.144d passes through the center of the scatter plot.

The next example concerns treating tumors with chemotherapy (Figure 4.12), which was mentioned in the introduction to this section. Refer to Problems 44 and 45 in Problem Set 1.4 and Problem Set 44 in Problem Set 1.5 for further insights into modeling tumor growth.

images

Figure 4.12 Chemotherapy is a treatment for cancer patients in which powerful drugs are used to kill cancer cells. Often it is used in conjunction with other treatments, including radiation therapy and surgery.

Example 4 Tumor regrowth

In an experimental study performed at Dartmouth College (E. Demidenko. Mixed Models: Theory and Applications. Wiley 2004), two groups of mice with tumors were treated with the chemotherapeutic drug cisplatin. Before the therapy, the tumor consisted of proliferating cells (also known as clonogenic cells) that grew exponentially with a doubling time of approximately 2.9 days. Each of the mice was given a dose of 10 mg/kg of cisplatin. At the time of the therapy, the average tumor size was approximately 0.5 cm3. After treatment, 99% of the proliferating cells became quiescent cells (also known as nonproliferating or resting cells). These quiescent cells do not divide and decay with a half-life of approximately 5.7 days.

  1. Write a function V(t) that represents the volume of the tumor t days after therapy. The tumor volume includes the volume of the proliferating cells and the quiescent cells.
  2. Determine at what point in time the tumor starts to regrow and therapy should be readministered.

Solution

  1. If P(t) and Q(t) represent the respective volumes of proliferating and quiescent cells in a tumor, then the total volume V(t) of the tumor is given by

    images

    Assume the proliferating cells are increasing at an exponential rate and have an initial volume of P(0) = 0.01 × 0.5 = 0.005 cm3 (i.e., 1% of the previous untreated average size). Hence, P(t) = 0.005eat where we need to solve for a. Since the doubling time is 2.9 days, we can solve for a as follows:

    images

    Hence, P(t) = 0.005e0.24t.

    Similarly, we have Q(0) = 0.99(0.5) = 0.495 and Q(t) = 0.495ebt where we have to solve for b. Since the half-life of quiescent cells is 5.7 days, we can solve for b as follows:

    images

    Hence, Q(t) = 0.495e−0.12t.

    Thus, it follows from the first equation that

    images

  2. To determine when V(t) is increasing or decreasing, we need to compute its derivative:

    images

    Since V′(0) ≈ −0.0582, the volume of the tumor is initially decreasing after therapy. To see when V′(t) changes sign, we solve

    images

    Hence, after 10.84 days, the tumor begins to regrow and therapy should be read-ministered. Indeed, this prediction is supported by the data shown in the left panel of Figure 4.13. (The data in the right panel are examined in Problem Set 4.3.)

    images

    Figure 4.13 Plots of tumor size (log scale; each line comes from a different mouse) before and after a treatment of 10 mg/kg (left panel) and 25 mg/kg (right panel) of the drug cisplatin.

    From E. Demidenko, Mixed Models: Theory and Applications, John Wiley & Sons, 2004. p. 544. Used with permission.

In the next example, we consider the vascular system, which consists of arteries and veins that branch in different directions to pump blood through all parts of the body. Ideally, the body is designed to minimize the amount of energy it expends in pumping blood. According to one of Poiseuille's laws, the resistance blood experiences by traveling down the center of a blood vessel with radius r and length L is proportional to

images

Without loss of generality, we assume that this proportionality constant equals one, and use this law to determine optimal branching angles in the vascular system of animals.

Example 5 Vascular branching

Consider a blood vessel that branches as illustrated below:

images

where a and b are positive constants. Given a and b, determine the angle θ which minimizes the total resistance in the blood flow from the point A to the point C.

Solution We want to minimize the total resistance along the blood vessel from A to C. Let B be the point where the vessel branches. We need to determine the resistance from A to B and the resistance from B to C. To determine the resistance along the blood vessel from A to B, we need to determine how the distance from A to B depends on θ. Using the right triangle as shown below

images

we get that the length from B to D is given by b cot θ. Hence, the distance from A to B is

images

and, as the radius of the vessel from A to B is 3, the resistance from A to B is

images

Using the same right triangle, we get that the distance from B to C is b csc θ. Hence, the resistance from B to C is

images

as the radius of the vessel from B to C is 2. Adding the resistance from A to B to the resistance from B to C, we get that the resistance from A to C is given by

images

To minimize the resistance on the half-open interval images, we need to determine the critical points along this interval as follows:

images

Hence, the optimal angle is given by ≈ 1.37 radians. Equivalently, since one radian is approximately 57.3 degrees, θ ≈ 78.5 degrees.

Regression

Regression, loosely speaking, is the process of drawing a graph through a set of data, where the graph represents the information in the data after process and measurement errors have been “averaged out.” The term regression dates back to Sir Francis Galton's work on measuring heritable traits in humans, such as height, and his publication of this work in 1886 under the title “Regression Towards Mediocrity in Hereditary Status” (Journal of the Anthropological Institute of Great Britain and Ireland, 15 (1886): 246–263). To simplify our notation in developing some of the ideas associated with regression theory, we introduce the summation notation, a notation that we revisit in Chapter 5 when developing integral calculus.

Summation Notation

The sum of a sequence of n real numbers a1, a2, a3,···, an is

images

The index i does not have to start at i = 1, but could start at any integer value less than or equal to n.

The following three properties of this summation notation are easily verified.

Properties of Summations

Let a1, a2, a3,..., an, b1, b2, b3,..., bn, and c be real numbers. Then

images

In our development of regression, we use the following definition that we informally introduced in Section 1.2 and depicted in Figure 1.14 for the case of linear regression. Here we provide a definition for any function f(x), not just the linear function f(x) = mx + c.

Residual Sum-of-Squares

The residual sum-of-squares S of the function y = f(x) from a data set

images

is the residuals

images

squared and then summed to obtain

images

If we interpret the residuals ei as “errors” from the true value, then S is the sum of squared errors. To find a best-fitting function to a data set, regression aims to minimize the sum of these squared errors.

Example 6 Best line through the origin

In Example 7 of Section 2.2, we determined the slope a of how the rate (cells/hour) at which copepods feed on diatoms increases as a function of diatom density (cells/milliliter). In the first two rows of data in Table 4.1 we have extracted eleven representative points from these data, as diatom density increases from 0 to 200 cells/milliliter.

Table 4.1 Diatom densities x (cells/milliliter) and copepod feeding rates y (cells/hour)

images

  1. Find the slope a of the line f(x) = ax through the origin that minimizes the residual sum-of-squares through the data. This is the so-called best-fitting or regression line.
  2. Plot these data and the regression line on the same graph.

Solution

  1. From the definition of the residual sum-of-squares with f(x) = ax and properties of summations, we have

    images

    S is a quadratic function of a. Since the coefficient of a2 is positive, the graph of S(a) is an upward facing parabola and, hence, has a unique minimum. To minimize S(a) with respect to a simply requires finding where its derivative S′(a) equals 0. Differentiating S yields

    images

    Setting S′(a) = 0 and solving yields

    images

    Using the values of the sums, as given in Table 4.1, we obtain that

    images

    is the value of a that minimizes S(a).

  2. The line f(x) = 5.86x and the data are plotted below

    images

Recall Example 5 of Section 1.7. We plotted the data in Figure 1.46 of that example on the decline of the Glued gene in a species of fruit fly. The data, which exhibit an exponential decline from 0.5 to 0.0036 over six generations, are reported in Table 4.2. To model how rapidly the Glued gene is lost from the experimental fruit fly population in each generation, we used the function 0.5ert for some appropriately chosen constant r > 0. Taking the logarithm of this equation yields the linear function ln 0.5 − rt of t. In the next example, we use this observation to find the best-fitting line to the logarithmically transformed data.

Table 4.2 Changes in the frequency x of the Glued gene in a species of fruit fly over seven (generation 0 is also experimentally generated) experimental generations t(values to two significant digits)

images

Example 7 Regression and lethal genes

  1. Find the value of the decay rate parameter r > 0 of the function ln 0.5 − rt that minimizes the residual sum-of-squares of this function with respect to the data (1, ln x1), (2, ln x2),..., (6, ln x6) from Table 4.2.
  2. Provide a semi-log plot of these data and the best-fitting, log-transformed, exponential decay function on the same graph.

Solution

  1. Following the approach of Example 6, we want to minimize

    images

    Differentiating S with respect to r yields

    images

    Setting S′(r) = 0 and solving for r yields

    images

    Calculating relevant sums from Table 4.2, we obtain the following table.

    images

    Thus, from the above equation for r we have

    images

    We conclude that the best estimate of the rate at which the Glued gene is purged from the fruit fly population during the experiment is very close to 33% per generation.

  2. The log-transformed data ln xt and the best-fitting function ln 0.5 − rt are shown below.

    images

PROBLEM SET 4.3

Level 1 DRILL PROBLEMS

1. In Example 1, the selling price of corn was $1.50 per bushel and the seeds cost $3 per thousand seeds. Determine the density of seeds that maximize profit if the selling price and cost are both doubled.

2. In Example 1, determine the density of seeds that maximize profit if the selling price is $5 per bushel and the seeds cost $2 per thousand seeds.

3. In Example 1, determine the density of seeds that maximize profit if the selling price is $2.20 per bushel and the seeds cost $2.50 per thousand seeds.

4. In Example 2, we estimated that the maximum per capita growth rate of the Arctic fin whale is r = 0.08. Suppose a better estimate is r = 0.1. Determine the maximum sustainable harvesting rate for this value of r.

5. In Example 2, we estimated that the long-term abundance of the Arctic fin whales in the absence of harvesting the Arctic fin whale is K = 500,000. Suppose a better estimate is K = 400,000. Determine the maximum sustainable harvesting rate for this value of K.

6. In a species of fish the growth rate function is given by G(x) = 1.4x(1 − x/K), where K = 5 million metric tons (i.e., the population of fish is measured in metric tons rather than number of individuals). If the harvest rate is a function of the harvesting effort h and the total amount of fish x, that is H = hx, find the harvesting effort value h that corresponds to the maximum sustainable yield.

7. In a species of fish, the growth rate function is given by G(x) = 2.1x(1 − x/K), where K = 8 million metric tons. If the harvest rate is H = hx, find the harvesting effort value h that corresponds to the maximum sustainable yield.

In Problems 8 to 12, find the optimal angle for the following vascular branching problems, as considered in Example 5.

8. A larger artery has a radius of 0.05 mm, and a smaller artery of radius 0.025 mm branches from the larger artery with branching angle θ.

9. A larger artery has a radius of 0.06 mm, and a smaller artery of radius 0.04 mm branches from the larger artery with branching angle θ.

10. The radius of the main blood vessel is r1 = 2 and the radius of the branching vessel is r2 = 1.

11. The radius of the main blood vessel is r1 = 4 and the radius of the branching vessel is r2 = 3.

12. In a general case, the radius of the main blood vessel is r1 and the radius of the branching vessel is r2. Assume that r1 > r2.

13. In Example 3, calculate at what point x along the shore Elvis should enter the water if the distance of the ball from the shore is 20 meters rather than 6.

14. In Example 3, calculate at what point x along the shore Elvis should enter the water if the distance of the ball from the shore is 10 meters rather than 6.

15. In Example 3, calculate at what point x along the shore Elvis should enter the water if the distance of the ball from the shore is d meters.

In Problems 16 to 22, calculate the residual sum-of-squares of the listed function through this data set:

images

Draw a graph of the function and the data on the same plot. In Problems 21 to 22, plot the residual sum-of-squares graph as a function of the free parameter and selected range for this parameter. Note that you are not being asked to fit the exponential functions in the relevant examples below on a semi-log plot, but to use the function itself directly in the calculations of the residuals with respect to the data.

16. f(x) = 2x

17. f(x) = 2x + 1

18. f(x) = 3x − 1

19. f(x) = 0.4e0.8x

20. f(x) = 0.2e0.9x

21. f(x) = mx, 0 ≤ m ≤ 4

22. f(x) = 0.2erx, 0 ≤ r ≤ 1

Level 2 APPLIED AND THEORY PROBLEMS

23. Find a general formula for which Example 3 is a specific case that describes how to calculate at what point x along the shore Elvis should enter the water if the distance of the ball from the shore is d meters (rather than 6) and the point on the shore to which this distance d holds is k meters (rather than 15) from where Tim is standing.

24. In a species of fish, the growth rate function is given by G(x) = 1.5x(1 − x/K), where K = 6,000 metric tons (i.e., the population of fish, x, is measured in metric tons rather than number of individuals). The price a fisher can get is p = $600 per metric ton. If the amount the fisher can harvest is determined by the function H = hx, where each unit of h costs the fisher c = $100, what is the maximum amount of money the fisher can expect to make on a sustainable basis? (Hint: The fisher's sustainable income is given by pHch, where H is a sustainable harvesting rate.)

25. In the tumor growth study described in Example 4, the tumor consisted of proliferating cells (clonogenic cells) that grew exponentially with a doubling time of approximately 2.9 days. Suppose that each mouse was given a dose of 25 mg/kg of cisplatin per treatment with the following results: At the time of the therapy, the average tumor size was approximately 0.44 cm3. After treatment, 99.73% of the proliferating cells became quiescent cells and decayed with a half-life of approximately 6.24 days.

  1. Write a function V(t) that represents the size of the tumor (proliferating plus quiescent cells) t days after therapy.
  2. Determine at what point in time the tumor starts to regrow and therapy should be readministered.
  3. Compare your answer to the data figure in Example 4.

26. In a follow-up study to the tumor growth study described in Example 4, mice were infected with a relatively aggressive line of proliferating clonogenic cells that grew exponentially with a doubling time of approximately 1.8 days. Each mouse was given a dose of 20 mg/kg of cisplatin per treatment with the following results: At the time of the therapy, the average tumor size was approximately 0.6cm3. After treatment, 99.10% of the proliferating cells became quiescent cells and decayed with a half-life of approximately 4.4 days.

  1. Write a function V(t) that represents the size of the tumor (proliferating plus quiescent cells) t days after therapy.
  2. Determine at what point in time the tumor starts to regrow and therapy should be readministered.

27. In certain tissues, cells exist in the shape of circular cylinders. Suppose such a cylinder has radius r and height h. If the volume is fixed (say, at v), find the value of r that minimizes the total surface area (S = 2πr2 + 2πrh) of the cell.

28. Farmers regularly use fertilizers to enhance the productivity of their crops. Determining the appropriate amount of fertilizer to use requires balancing the costs of fertilization with the increases in yield. In a 2004 study, Baker and colleagues studied the relationship between nitrogen fertilization and yield of hard red spring wheat. (See Dustin A. Baker, Douglas L. Young, David R. Huggins, and William L. Pan, “Economically Optimal Nitrogen Fertilization for Yield and Protein in Hard Red Spring Wheat,” Agronomy Journal 96 (2004): 116–123.) For conventional tillage practices in eastern Washington in the late 1980s, they found that the grain yield (in kilograms per hectare) as a function of nitrogen (in kilograms per hectare) is well approximated by

images

These researchers suggested that a high selling price for wheat would be $191.1/kg and low cost for nitrogen would be $0.49/kg. Determine the amount of nitrogen that maximizes profits per hectare.

29. Baker and colleagues suggested that a low selling price for wheat would be $139.65/kg and a high cost for nitrogen would be $0.71/kg. Using the same yield function as in the previous problem, determine the amount of nitrogen that maximizes profits per hectare.

30. If the effects of density dependence in a whale population set in less rapidly closer to the final carrying capacity, K, than the logistic equation used in Example 2, then the equation should be replaced by an asymmetric growth model

images

for some α images (0, 1). For the case α = 0.5, calculate the stock level x that provides the maximum sustainable.

31. If the effects of density dependence in a whale population set in less rapidly or more rapidly closer to the final carrying capacity, K, than the logistic equation used in Example 2, then the equation should be replaced by an asymmetric growth model

images

For α > 0, r > 0, and K > 0, calculate the stock level x that provides the maximum sustainable yield. Discuss whether rapid onset of density dependence (i.e., large α) or gradual onset of density dependence (i.e., small α) leads to larger sustainable yields.

32. During the winter, a species of bird migrates from the coast of a mainland to an island 500 miles southeast. If the energy the bird requires to fly one mile over the water is twice more than the amount of energy it requires to fly over the land, determine what path the species should fly to minimize the amount of energy used.

33. The Statue of Liberty is 92 meters high, including the 46 meter pedestal upon which it stands. How far from the base should an individual stand to ensure that the view angle, θ, is maximized?

images

34. In the northeastern part of Sweet Water County, a large dam is being constructed on the Shuga River to produce hydroelectricity (i.e., the generation of electricity through water pressure). An important part of this project is running power lines from the power stations at the downstream side of the dam to various parts of the county, including Pickle City, the largest city in the county. On the recommendation of a number of other counties, Sweet Water County officials have hired you as a consultant to resolve cost issues for running these power lines.

County officials have informed you that the Shuga River runs due east, and on its southern side lies an expanse of federally protected wetlands. Pickle City lies several miles to the south of these wetlands, as shown in the map below.

images

The federally protected wetlands are divided into two regions. In the western region, county officials expect that due to federal regulations it will cost 40% more to run conduit here than it does through non-wetland ground. The eastern region of the wetlands is a habitat for the endangered Brown Barbaloots. Consequently, federal law prevents the county from running conduits through this region.

As the county officials intend to submit a budget proposal for the project to the county council in the next week, they would like you to determine the path from the power station to downtown Pickle City that minimizes the cost of installing the conduit.

35. An oil spill has fouled 200 miles of Pacific shoreline. The oil company responsible has been given fourteen days to clean up the shoreline, after which a fine will be levied in the amount of $10,000/day. The local cleanup crew can scrub five miles of beach per day at a cost of $500/day. Additional crews can be brought in at a cost of $18,000, plus $800/day for each crew. Determine how many additional crews should be brought in to minimize the total cost to the company and how much the cleanup will cost.

36. Consider a spherical cell with radius r. Assume that the cell gains energy at a rate proportional to its surface area (i.e., nutrients diffusing in from outside of the cell) and that the cell loses energy at a rate proportional to its volume (i.e., all parts of the cell are using energy). If the cell is trying to maximize its net gain of energy, determine the optimal radius of the cell. Note: Your final expression will depend on your proportionality constants.

37. Consider a cylindrical cell with radius r and height r/2. Assume that the cell gains energy at a rate proportional to its surface area (i.e., nutrients diffusing in from outside of the cell) and that the cell loses energy at a rate proportional to its volume (i.e., all parts of the cell are using energy). If the cell is trying to maximize its net gain of energy, determine the optimal value of r. Note: Your final expression will depend on your proportionality constants.

38. A dune buggy is in the desert at a point A located 40 km from a point B, which lies on a long, straight road, as shown in Figure 4.14. The driver can travel at 45 km/hour on the desert and 75 km/hour on the road. The driver will win a prize if she arrives at the finish line at point D, 50km from B, in 85 minutes or less. Set up and analyze a model to help her decide on a route to minimize the time of travel. Does she win the prize?

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Figure 4.14 Path traveled by a dune buggy.

39. The question of whether an optimal body size exists for different kinds of animals is one that is of great interest to biologists. The reproductive power P of an individual can be modeled, following the ideas of ecologist James H. Brown (see his book Macro-ecology, University of Chicago Press, 1995), as the harmonic mean of two limiting rates. The harmonic mean of two numbers a and b is the reciprocals of the average of the inverses of the two numbers: 1/(1/a + 1/b) = ab/(a + b). The two rates are a per-unit-mass rate R1 at which individuals acquire resources, and a per-unit-mass rate R2 at which individuals convert those resources into new individuals; that is,

images

Assuming both R1 and R2 are the following allometric functions of body mass measure in kilograms

images

find the body mass M that maximizes the reproductive power P, and show that this extremum is a maximum for the case b1 = 0.75 and b2 = −0.25.

40. Suppose that we express the two rate functions in Problem 39 using the general form R1 = c1Mb1 and R2 = c2Mb2. Show in this case that the maximum body size is given by the expression

images

41. In Example 6, we determined the slope a of the linear component of a type I functional response for the rate (cells/hour) at which copepods feed on diatoms as a function of diatom density (cells/milliliter). Suppose now that only the following partial set of data (Table 4.3) is available for estimating the parameter a.

Table 4.3 Diatom densities x (cells/milliliter) and copepod feeding rates y (cells/hour)

images

  1. Find the slope a of the line f(x) = ax through the origin that minimizes the residual sum-of-squares through the data.
  2. Plot these data and the best-fit line on the same graph.

42. In Table 1.4, the CO2 concentrations at the Mauna Loa Observatory in Hawaii are listed by month, starting at month 1 (May 1974) until month 140 (December 1985). The data for the month of May, for the twelve periods spanning May 1974 to May 1985, are shown in Table 4.4.

Table 4.4 CO2x(cell/milliliter) in year t

images

  1. Find the slope m of the regression line f(x) = mx + 333.2 through the point (t, xt) = (0, 333.2) that minimizes the residual sum-of-squares through the remaining eleven data points.
  2. Plot these data and the regression line on the same graph.

43. In Example 7, we found the value of the decay rate parameter r > 0 using linear regression to fit a semi-log plot of the frequency of Glued genes in an experimental fruit fly population. This experiment was repeated by the same group of researchers and the following data were obtained:

images

Use these data to find the value r of the best-fitting exponential decay function xt = 0.59e−rt through the starting point (t, ln xt) = (0, −0.53). Present the solution on a semi-log plot together with a plot of the data.

44. In Section 1.4, we presented the following data on the growth of the United States from 1815 until 1895.

Year Population (in millions)
1815 8.3
1825 11.0
1835 14.7
1845 19.7
1855 26.7
1865 35.2
1875 44.4
1885 55.9
1895 68.9

Use linear regression on a semi-log transformation of the above data to find the best estimate of the annual growth rate r > 0 in the populationmodel x(t) = 8.3ert (million individuals) for t images [0, 80] (years), with t = 0 corresponding to the year 1815.

4.4 Decisions and Optimization

Optimal decisions

The behavior of animals has been honed by natural selection to maximize the reproductive potential of individuals. Thus, from an individual point of view, individuals should act in ways that maximize the number of offspring they rear to sexual maturity. This number is referred to as an individual's fitness. From a genetic point of view, a gene encoding for a behavior that maximizes an individual's fitness will have a greater representation in the gene pool of future generations than a gene that encodes for a behavior that is detrimental to an individual's fitness (e.g., a gene that causes an individual to be excessively reckless, making it likely that the individual will die before reaching sexual maturity). Theories of optimal behavior are based on the premise that organisms maximize their fitness by behaving in a certain way. Using models, researchers can develop hypotheses about these optimal behaviors. These hypotheses can be tested experimentally or through comparative studies.

In our first example in this section, we obtain insights into the reason why Northwestern crows consistently drop whelks from a specific height to break them open. If the crows fly too low, the shells require too many drops to get them to break open. If they fly too high, the crows waste energy. Assuming that crows have evolved to minimize energy expenditures, scientists might be interested in testing this hypothesis by formulating a suitable function to minimize. This function would characterize the number of of drops, and hence work, required to break open a whelk as a function of the height from which the shell is dropped. In addition to modeling the dropping behavior of Northwestern crows, this section investigates optimal foraging in a patchy environment, optimal timing of seed production, and optimal time to harvest crops. In addition to these examples, we present a key theorem called the marginal value theorem, which has applications to problems maximizing or minimizing average rates of change.

The Northwestern crow, illustrated in Figure 4.15, feeds on whelks, a type of mollusk. To get the meat from inside the whelk's shell, individual crows lift whelks into the air and drop them onto a rock to break open the shell. The biologist Reto Zach (see Example 6 of Section 3.2) observed that individual crows typically drop the shells from a height of five meters. This led him to ask this question: Does the height from which Northwestern crows drop whelk shells minimize the amount energy required to open a shell?

images

Figure 4.15 A Northwestern crow.

Example 1 Northwestern crows and whelks

After collecting data by dropping whelks from different heights, Zach found that, on average, the number of drops required to break a whelk dropped from h meters is modeled by the function

images

This relationship implies images D(h) = ∞, which in turn implies that if h ≤ 0.84, the shell will never open. Use this equation

images

to find the optimal height from which a whelk should be dropped to minimize the amount of work required to break a whelk shell.

Solution Since work is force times distance, the amount of work required to drop a shell of fixed weight is in proportion to the total height the crow flies when breaking a whelk. The total height is given by the number of drops times the height of the drop. In other words, up to a proportionality constant, the average amount of work that it will take a crow to break open a whelk shell is

images

To determine where this function takes on its smallest value, we need to understand the graph of the function. It has a vertical asymptote at h = 0.84. Taking the derivative yields

images

Since the denominator is positive wherever h ≠ 0.84, we only need to understand when the numerator is positive or negative. Solving −16.4304 − 1.68h + h2 = 0 for h yields h ≈ −3.3 meters and h ≈ 4.98 meters. Since this quadratic corresponds to an upward facing parabola, we get that the numerator of W′(h) is positive when h > 4.98 and negative on the interval (0.84, 4.98). Hence, W(h) decreases on the interval (0.84, 4.98) and increases on the interval (4.98, ∞). The height h ≈ 4.98 is a global minimum for h > 0.84.

Hence, the height that minimizes the amount of work is approximately five meters, the height observed by Reto Zach!

The next example explores the optimal time for a plant to produce seeds. This is just one in a class of optimal time-to-reproduction problems. Others include the optimal time for a honey bee colony to swarm and the optimal time for salmon to return from the ocean to lay eggs upriver.

Example 2 Optimal time for producing seeds

A particular plant is known to have the following growth and seed production characteristics. At time of planting (t = 0), the seedling has a mass of 5 grams. At time (t > 0) days after planting, the seedling has grown into a plant that weighs w(t) = 5 + 400tt2 grams. The plant has a gene that can be manipulated to control the age t at which the plant matures. The number of seeds S(t) produced by a plant maturing at age t is

images

A farmer asks the geneticists to genetically engineer a plant line that accounts for the fact that on his farm, because of losses from pests, drought, and disease, only a proportion

images

of germinating seeds develop and survive to age t. What age of maturity should the geneticist select for the plants to maximize the seed production of the mature crop for the farmer?

Solution For every N seeds that the farmer plants on his land at time t = 0, N × P(t) will mature at time t > 0. The total yield from these plants is then

images

Since N is just a scaling factor that depends on the number of acres that farmer plants, we can set it to any convenient value such as N = 1. To find the germination time that maximizes this yield, we need to understand the first derivative:

images

Thus, Y′(t) exists for t > 0 and the derivative vanishes at solutions to the equation

images

We can use technology or the quadratic formula to obtain the roots

images

Since Y′(0) > 0 and Y′(200) < 0, we get that Y increases on the interval (0, 123.6) and decreases on the interval (123.6, ∞). Hence, Y(t) is maximized at t ≈ 123.6. We verify this directly by plotting Y as a function of time, as illustrated in Figure 4.16. The vertical line indicates the optimal maturation time t* = 123 days.

images

Figure 4.16 The red curve is the number of seeds S(t) produced by a plant that survives to age t. The blue curve, which has its maximum value at t*, is the expected number of seeds that plant will produce after taking into account the probability it may die before starting to seed.

Optimal foraging and marginal value

Very often, food is not distributed homogeneously over the environment; rather, it occurs in discrete patches in the environment. For fruit bats, a patch may correspond to a fruit tree or a stand of fruit trees. For a hummingbird, which feeds on the nectar of flowers, a patch may correspond to a single flower or a field of flowers. In optimal foraging theory, we want to know how long an animal should continue to collect resources in a patch when it has the choice of traveling to another resource-rich patch. The question of when to leave a patch as resources in the patch are being depleted is known as the optimal residence time problem.

Example 3 Optimal foraging in a multi-patch environment

images

Figure 4.17 House martin parent feeding its young

House martins make sorties from their nests to collect food to bring back to their young. In an experiment carried out in the early 1980s, two British scientists, D. M. Bryant and A. K. Turner, found that the travel time of martins from a particular nest to nearby foraging areas ranged from half a minute to several minutes, and the weight of the load of insects the martins collected and brought back to their nest to feed their chicks (see Figure 4.17) varied between 20 and 100 mg. (See Central place foraging by swallows (Hirundinidae): The question of load size. Animal Behavior 30 (1982): 845–856.) On an average foraging bout, Bryant and Turner observed that these martins collected insects at the rate of (roughly) 10 mg/minute from time of departure from the nest. Assume one of these martins encounters a patch three minutes after leaving its nest, and its cumulative load of insects after foraging for t minutes is given by the function

images

If the martin is trying to maximize the average load accumulated per minute since leaving its nest, then what is the optimal time for the martin to quit foraging in this patch?

Solution Since it takes three minutes for the martin to reach the patch, the average load accumulated per minute after t minutes in the patch is R(t) = B(t)/(t + 3). To determine the best time to leave the patch, we need to understand the graph of R(t) for t ≥ 0. Taking the first derivative of R we get

images

We have images = 0 when t2 = 18. Equivalently, t = ±images ≈ ±4.24 minutes. Only the positive solution is relevant. Since R′(0) > 0 and R′(images) < 0, R is increasing on the interval (0, images) and decreasing on the interval (images, ∞). Hence, the maximum is achieved at t = images at which

images

which exceeds the background average rate of 10 mg/minute. This conclusion is reaffirmed by graphing R(t) as follows:

images

In Example 3, the average rate of change was being maximized over a time interval. A fundamental result for problems of this type is the marginal value theorem.

Theorem 4.2 Marginal value theorem

Let V(t) be a function defined on an interval [a, ∞). If V(t) represents the accumulated value of the resource by time ta, then the average rate of resource accumulation over time ta is given by

images

If a maximum or minimum of A(t) occurs at t = b > a and V is differentiable at t = b, then b satisfies the equation

images

In other words, this maximum or minimum occurs at a time where the average rate of change equals the instantaneous rate of change.

Proof. Since V achieves a maximum or minimum at t = b > a, we have A′(b) = 0. Computing the derivative yields

images

Setting A′(b) = 0 and multiplying both sides of the equation by (ba)2, we get

images

Equivalently, V′(b) = images.

Example 4 Optimal foraging of great tits

In a classic paper on animal behavior, biologist Richard Cowie studied the foraging behavior of great tits by constructing experimental trees in an aviary (see Figure 4.18). On these experimental trees, food was placed in plastic containers in a manner that would allow Cowie to manipulate the average travel time T between food containers. Through these experiments, Cowie estimated that the energy gained by a bird after eating from a container for t ≥ 0 seconds is

images

Assuming the great tits are maximizing their average energy gain, do the following:

images

Figure 4.18 Experimental tree in Cowie's experiments.

  1. Use the marginal value theorem to determine the relationship between the average travel time T and the optimal residence time t in a patch.
  2. Solve for T in terms of the optimal residence time and plot it.
  3. Discuss your findings with respect to data collected by Cowie, as depicted in Figure 4.19.

Solution

  1. Assume that at t = 0 the bird arrives at a food container. Since it takes T seconds to get to a container, we are interested in the time interval [−T, ∞) where t = −T corresponds to the moment that the bird begins traveling to the container. Since we assume there is no energy gain during the flight, we define E(t) = 0 for t ≤ 0.

    images

    Figure 4.19 Amount of time spent by great tits in a foraging patch is plotted as the solid line through the data (solid points with measurement error bars) as a function of the amount of time taken by individuals to reach the patch. The dotted line is discussed in the solution to part c of Example 4.

    Data Source: R. Cowie, “Optimal Foraging in Great Tits (Parus major),” Nature 268 (1977): 137–139.

    Clearly, the maximum cannot occur during the interval [−T, 0]. By the marginal value theorem with a = −T, the time t at which the maximum occurs must satisfy

    images

  2. Solving for the average travel time T in terms of the optimal residence time t yields

    images

    Plotting T as a function of t yields this graph:

    images

  3. The graph plotted in part b implies that as the average travel time T between patches increases, the optimal residence time t within a patch should increase at a greater than linear rate (i.e., it is concave up). Notice that in the graph of Cowie's data in Figure 4.19 the axes are switched, so that the inverse of our function T = f(t) is plotted as the dashed line. While five of the twelve data points are very close to the dashed line, the remaining seven data points lie significantly above it. In other words, for these seven experiments, the birds were spending more time in the patches than predicted by the model. One possible explanation for this discrepancy is that the model does not account for the energetic costs of traveling. Cowie adjusted the model to account for these energetic costs and the resulting prediction is plotted as a solid curve in Figure 4.19. In Problem 4 of this section's problem set, you will be asked to redo the analysis in this example in a way that accounts for these energetic costs.

The marginal value theorem has a simple graphical interpretation, which is explored in the next example.

Example 5 Optimal time to harvest

Over a six-decade period, a forestry company has collected data on the profit P(t) of stands of trees harvested at various ages t (units are decades). Initially, P(t) is zero because the costs required to bring in the heavy equipment needed to harvest the trees exceeds the value of the harvest itself, and the company will not harvest before making a profit. Once the trees reach a certain size, a profit is possible and it steadily increases as the stand of trees ages. The company found that the function that best fit its data has the following graph:

images

where the profit P is measured in thousands of U.S. dollars and t is measured in decades.

The company wants to maximize the profits it makes per unit time. Write the function A(t) that the company wants to maximize and illustrate where the maximum occurs graphically.

Solution The company wants to maximize the average rate of accumulation of profit P(t), which is A(t) = P(t)/t. Since P(0) = 0, we can apply the marginal value theorem, which asserts that the average rate of change P(a)/a equal the instantaneous rate of change P′(a), where t = a is the optimal harvesting time. Graphically, this corresponds to the line from (0, 0) to (a, P(a)) being tangent to the graph of P(t) at t = a. To find the optimal value a, we take a ruler and place one end at (0, 0) and rotate the ruler until the line determined by its edge is tangent to the graph of P(t). Doing so yields the following plot:

images

where the optimal time to harvest is approximately a ≈ 1.2 decades, about 12 years.

Our final example in this section introduces the concept of discounting when optimizing a sustainable stream of revenue calculated for all time in the future. Discounting arises if someone promises to pay you D dollars next year, and the current interest rate compounded continuously is r%; then this person should be willing to receive Der/100 dollars now. Namely, if this person took the Der/100 dollars now and invested it, then a year later the investment would yield er/100 dollars for each dollar invested. Hence, a year later the person would have er/100Der/100 = D dollars. As a result of this reasoning, economists use the discount factor e−δt, where δ = r/100 to reduce D dollars needed at time t in the future to their current value De−δt dollars now.

Example 6 Optimal rotation period for a plantation

In the mid-nineteenth century, a German forester by the name of Faustmann developed a theory for the optimal rotation period of a plantation. He calculated that if one planted a stand and harvested it every T years, and received the same value V(T) each time, then the sum of all the discounted amounts (i.e., the sum of V(T)e−δT obtained after T years, V(T)e−δ2T obtained after 2T years, V(T)e−δ3T obtained after 3T years, and so on for all time into the future) constitutes the so-called present value P(T) of the stand given by the formula

images

Now continue his analysis by doing the following:

  1. Using his formula for P(T), find a general expression for the optimal stand rotation period T* that is defined to be the value of T > 0 that maximizes the present value P(T) of the stand.
  2. What does the expression in part a imply as δ → 0?
  3. Use your technology to find the optimal rotation period when

    images

    and the discount rate is δ = 0.1.

Solution

  1. The optimal rotation period T* is an extremum of P(T). If T* > 0, then, T* will satisfy the equation P′(T) = 0 where

    images

    Therefore, if δ > 0, P′(T) = 0 implies that

    images

  2. Using l'Hôpital's rule to calculate the limit as δ approaches 0, we obtain that the optimal rotation period T* satisfies the equation

    images

    This equation implies that T* maximizes the average profit accumulation rate over each harvesting period in the limit δ = 0.

  3. From part a and the specific form for V(T), the optimal rotation period when δ = 0.1 is the solution to

    images

PROBLEM SET 4.4

Level 1 DRILL PROBLEMS

In Problems 1 to 6, the amount of energy a hummingbird gains after remaining in a patch for t seconds is given. For each problem, find how long a hummingbird should stay in a patch if it wants to maximize its average energy intake rate.

1. The travel time between patches is 15 seconds and

images

2. The travel time between patches is 5 seconds and

images

3. The travel time between patches is 10 seconds and

images

4. The travel time between patches is 5 seconds and

images

5. The travel time between patches is 5 seconds and

images

6. The travel time between patches is 10 seconds and

images

In Problems 7 to 10, rework Example 5 with the given graphs.

images

images

Assume the house martins in Example 3 can choose between two patches. In Problems 11 to 16, the time to fly to a patch and the energy yield as a function of patch residence time (t minutes) are given for two patches. If an individual can visit only one patch and wants to maximize the average amount of calories it receives, then which patch of each pair should it choose?

11. B(t) = images calories with a travel time of 2 minutes or B(t) = images calories with a travel time of 3 minutes.

12. B(t) = images calories with a travel time of 1 minute or B(t) = images calories with a travel time of 2 minutes.

13. B(t) = images calories with a travel time of 3 minutes or B(t) = images calories with a travel time of 2 minutes.

14. B(t) = images calories with a travel time of 2 minutes or B(t) = images calories with a travel time of 3 minutes.

15. B(t) = images calories with a travel time of 2 minutes or B(t) images calories with a travel time of 3 minutes.

16. B(t) = images calories with a travel time of 2 minutes or B(t) = images calories with a travel time of 15 seconds.

In Example 5 (optimal time to harvest), assume that the profit function P(t) has the form specified in Problems 17 to 22. For these profit functions, find the optimal age at which to harvest the stands of trees to maximize profit where t is measured in decades.

images

Level 2 APPLIED AND THEORY PROBLEMS

27. At the National Council of Teachers of Mathematics (NCTM) illuminations website, students are encouraged to collect data on how many drops are required to break a blanched peanut in two pieces. The sample data provided at this website are shown in the following graph.

images

The data can be modeled by the function

images

where h is the height in centimeters. Suppose that the “peanut hummingbird” collects peanuts and wants to minimize the amount of work required to break a peanut into two halves. Determine the height which minimizes the amount of work to break open the peanuts.

28. In Example 4, we found how the optimal residence time in a patch for a great tit depended on the travel time between patches. Although our prediction described the data reasonably well, more than half of the data points lay above the optimal curve. Cowie proposed that part of the reason for this result was that the birds expend energy traveling between patches and searching for food within a patch. In this problem, determine how these expenditures of energy influence the optimal residence time. Let

images

denote the amount of energy gained by a bird after residing in a patch for t seconds. Assume that the bird requires T seconds to travel the patch. Cowie found that great tits expend approximately 0.697 calories per second while traveling between patches and expend approximately 0.155 calories per second while searching for food in a patch.

  1. Write a function V(t) that represents the average gain in energy in a patch after residing there for t ≥ 0 seconds.
  2. Use the marginal value theorem to find an expression relating the optimal residence time t to the travel time T.
  3. Compare your solution to the solution found in Example 4.

29. Suppose the crop developed by plant geneticists, as discussed in Example 2 — that is, the weight of the crop t days after planting satisfies the growth equation w (t) = 5 + 400tt2—is grown in a location that is relatively pest free, so that the proportion of germinating seeds is

images

The crop, however, must be harvested on or before the first frosty day of fall. Suppose the crop has relative value 1 when harvested at its optimum time of maturity, as represented by the day T on which the yield Y = a w(t)P(t) is maximized and that this value is reduced by 10te%, where te is the number of days prior to T that harvest actually occurs, so that the relative crop value is 0 if harvested at time T − 10 or earlier.

  1. What is the value of T?
  2. If the expected number of days ts in the growing season—that is the number of frost-free days plus 1—is equally likely to fall on any day from day 165 to day 190, then what is the expected value of the harvest in any year?

30. When a codling moth larva hatches from its egg, it goes looking for an apple. The period between hatching and finding an apple is called the search period. The search period S seems to be a function of the temperature, as shown in Table 4.5.

Table 4.5 Search period for the codling moth

Temperature S, in days
20°C 0.129
21°C 0.122
22°C 0.116
23°C 0.112
24°C 0.109
25°C 0.106
26°C 0.105
27°C 0.104
28°C 0.104
29°C 0.105
30°C 0.106

Source: P. L. Shaffer and H. J. Gold, 1985. “A simulation model of population dynamics of the codling moth, Cydia pomonella” Ecological Modeling 30:247–274.

Following the lead of Shaffer and Gold (see Section 4.2, Example 8), find 1/S for each data value and then use technology to fit a quadratic function to the data. Find the largest and smallest value of this fitted function S.

31. In a plantation of a particular species of trees, a forest economist estimated the number of board feet that can be harvested as a function of the age of the plantation. Data are given in Table 4.6. By using your technology to fit a quadratic function to the data, estimate at what age the plantation should be harvested to maximize the yield of board per acre.

Table 4.6 Harvest yield for a lumber crop

Age (years) Yield (board feet per acre)
15 6013
20 7021
25 8793
30 9411
35 9786
40 9958
45 9921
50 9766

32. By using your technology to fit a cubic equation to the data in Problem 31, find the age in [15, 50] at which the plantation represented by the data should be harvested to maximize the yield.

33. By using your technology to fit a quartic equation to the data in Problem 31, find the age in [15, 50] at which the plantation represented by this data should be harvested to maximize the yield.

34. By using your technology to fit a quintic equation to the data in Problem 31, find the age in [15, 50] at which the plantation represented by the data should be harvested to maximize the yield.

4.5 Linearization and Difference Equations

As we have seen in earlier chapters, difference equations xn+1 = f(xn) are useful for describing biological dynamics. The simplest dynamics occur at an equilibrium, because by definition, these equilibria are the solutions of the difference equation that remain constant for all time. Specifically, if a given first value x0 satisfies x0 = f(x0), then our difference equation implies x1 = x0. Repeated application results in xn+1 = xn = ··· = x0 for all n > 0.

While equilibria may be easily identified, as discussed in Section 1.7, by solving the equation x = f(x), their biological relevance depends on their stability. Many biological systems, when perturbed, naturally return to their equilibrium state. The temperature of the human body is a case in point. If a person's temperature is perturbed because of an infection, it returns to its equilibrium value of 98.6°F once he or she is well again. Not all equilibria, however, are stable. If we stand up a six-month-old child, the child may stay upright for a second or two, but until the child is around a year old, he or she will soon fall over. Standing vertically without feedback control from muscles constantly moving to correct the tendency to fall over is an unstable situation.

Thus, when a biological system is perturbed away from equilibrium, it may do one of two things. A system may return to the equilibrium state, in which case the equilibrium is considered stable. Alternatively, even if the perturbation is small, a system may continue to drift away from the equilibrium. In this case, the equilibrium is unstable. In this section, we make the notion of stability precise and provide a simple algebraic method for checking stability—a method that relies on linearizing the difference equation near the equilibrium. These ideas and methods are applied to models of population growth and population genetics.

We conclude the section by considering another application of linearization and difference equations, namely, numerically solving for the roots of a nonlinear equation. This numerical method is an important alternative to the bisection method presented in Example 10 of Section 2.3.

Equilibrium stability

We begin with the following example, which motivates the notion of a stable equilibrium.

Example 1 Logistic equation

In Example 7 of Section 2.5, we introduced the discrete logistic equation, which is a simple model of population growth. If xn denotes the population density (e.g., average number of individuals per acre) in the nth generation, then the model is given by

images

where r is the per capita growth rate at low densities and K is the carrying capacity of the population. The only two equilibrium solutions are x = 0 and x = K. For K = 100 and r = 0.5, 1.5, 3.0, simulate the model for the initial condition x0 = 99. Discuss what you find.

Solution Simulating the model with x0 = 99 for twenty-five iterations yields the following figures:

images

When r = 0.5, the population density gradually increases from the density 99 to the equilibrium density 100. When r = 1.5, the population density exhibits oscillations that dampen toward the equilibrium density 100. When r = 3, the population exhibits irregular oscillations that never approach the equilibrium density 100, despite having started near this equilibrium density.

Example 1 illustrates that some solutions starting near the equilibria approach the equilibrium, while other solutions starting near an equilibrium move away. These observations suggest the following definitions.

Equilibrium Stability

An equilibrium to xn+1 = f(xn), that is a solution satisfying x* = f(x*), is:

stable provided that there exists an open interval (a, b) containing x* such that images xn = x* and x1 lies in (a, b); whenever x0 lies in (a, b);

unstable provided that there is an interval (a, b) containing x* such that all solutions xn eventually leave (a, b) whenever x0 lies in (a, b) but x0x*;

Note: In the definition of stability, the second condition that “x1 lies in (a, b) whenever x0 lies in (a, b)” is equivalent to the image of (a, b) under f lies in (a, b).

Stated more simply, stability of an equilibrium means that if the solution starts near the equilibrium, then it remains nears the equilibrium and asymptotically approaches the equilibrium. Alternatively, solutions starting near (but not at) an unstable equilibrium eventually move further away from the unstable equilibrium.

Example 2 Stability the hard way

Find the equilibria of the following difference equations and verify stability using the definitions of stable and unstable.

images

Solution

  1. The equilibria are given by solutions of x = x/2. Hence, the only equilibrium is x = 0. Given any x0, and using the methods developed in Section 1.7, it follows that xn = images. Therefore, given any a > 0, we get that images xn = 0 for any x0 in (−a, a). In addition, the image of (−a, a) under f is images. Therefore, x* is stable.
  2. The equilibria of xn+1 = images must sastisfy x = x2. Hence, the equilibria are given by x = 0 and x = 1. For any x0, we have that x1 = images. Hence, xn = images. If x0 lies in the interval (−1, 1), then images = 0. Moreover, the image of (−1, 1) under the function f(x) = x2 is [0, 1), which lies in (−1, 1). Hence, 0 is a stable equilibrium for this difference equation. For any x0 > 1 or x0 < −1, xn = images approaches +∞ as n approaches ∞. Hence, for any initial condition near 1, the solution moves away from 1 so that the equilibrium 1 is unstable.

Example 3 Stability of linear difference equations

Consider the linear difference equation

images

For this difference equation, the origin, x = 0, is always an equilibrium. Determine for which r values, the origin is stable or unstable.

Solution The solution of this difference equation is given by xn = rnx0. Suppose x0 ≠ 0. If |r| < 1, then |xn| = |r|n|x0| is decreasing to zero at a geometric rate. Therefore, if |r| < 1, then x = 0 is stable. Alternatively, if |r| > 1, then |xn| = |r|n|x0| is increasing without bound. Hence, x = 0 is unstable when |r| > 1. If r = 1, then xn = x0 for all n. Hence, xn neither approaches or moves away from 0, so that 0 is neither stable or unstable when r = 1. Similarly, if r = −1, you can show that x = 0 is neither stable nor unstable.

Stability via linearization

Stability of an equilibrium can be verified directly using the definition, but this method can be challenging. To make things easier, we take advantage of linearization and our work in Example 3.

Suppose x* is an equilibrium of xn+1 = f(xn) and f is differentiable at x*. If we approximate f by its tangent line at x = x*, we get

images

Using this linear approximation and setting r = f′(x*), we get

images

Equivalently,

images

Using the change of variables yn = xnx*, we get

images

Example 3 suggests that if |r| < 1 and x0 is sufficiently close to x*, then we expect that yn approaches zero at a geometric rate. Equivalently, since we defined yn = xnx*, it follows that xn approaches x* at a geometric rate. Alternatively, if |r| > 1 and x0 is sufficiently close to x*, then we expect that yn increases initially at a geometric rate. Equivalently, xn initially moves away from x* at a geometric rate. As it turns out, all of these statements hold provided that xn is sufficiently close to x*, as the following theorem states.

Theorem 4.3 Stability via linearization theorem

If xn+1 = f(xn) has an equilibrium at x = x* and r = f′(x*) exists, then x* is stable if |r| < 1 and unstable if |r| > 1.

Theorem 4.3 is inconclusive about stability if |f′(x*)| = 1.

Example 4 Logistic revisited

Consider the logistic difference equation

images

with r > 0. Determine for which r values x* = 100 is stable.

Solution Let f(x) = x + rx images. To determine whether an equilibrium is stable or not, we need to compute

images

and evaluate at x = 100

images

For stability, we need that |1 − r| < 1. Equivalently,

images

Hence, the equilibrium x* = 100 is stable provided that 0 < r < 2 and unstable provided that r > 2. This conclusion is consistent with the simulations in Example 1. Indeed, for r = 0.5 and r = 1.5, the simulations approached the equilibrium x* = 100. However, for r = 3, the simulation oscillated irregularly and never converged to any density.

Example 5 Stability of insect population dynamics

Biology professor T. S. Bellows investigated the ability of several different difference equation models to describe the population dynamics of various insect species. (See T. S. Bellows, “The Descriptive Properties of Some Models for Density Dependence.” The Journal of Animal Ecology 50(1) (Feb. 1981), 139–156) He found that the so-called generalized Beverton-Holt model provided the best mathematical description for the insect species that he studied. If xn denotes the population density in the nth generation, then the model is of the form

images

where r is the intrinsic fitness of the population and b measures the abruptness of density dependence. For three insect species, Bellows made the following parameter estimates:

  • Budworm moth: r = 3.5 and b = 2.7
  • Colorado potato beetle: r = 75 and b = 4.8
  • Meadow plant bug: r = 2.2 and b = 1.4

These insects are shown in Figure 4.20.

images

Figure 4.20 Photos of the meadow plant bug (left), Colorado potato beetle (center), and the budworm moth (right).

  1. Use these parameter estimates to determine which population, according to the model, supports a stable equilibrium.
  2. For the species that do not support a stable equilibrium, simulate their dynamics.

Solution

  1. To begin with, we need to find the equilibria of the model that must satisfy x = images. Equivalently, x = 0 or

    images

    Hence, for the budworm moth, the equilibria are given by

    images

    For the Colorado potato beetle, the equilibria are given by

    images

    For the meadow plant bug, the equilibria are given by

    images

    Let f(x) = images. To determine the stability of these equilibria, we need to evaluate the derivative

    images

    at the equilibria. Since f′(0) = r and r > 1 for all three species, 0 is an unstable equilibrium for all three species.

    For the budworm moth, f′(1.4) ≈ −0.93. Since | −0.93| = 0.93 < 1, the equilibrium x ≈ 1.4 is stable for the budworm moth model. For the Colorado potato beetle, f′(2.45) ≈ −3.75. Since | −3.75| > 1, the equilibrium x ≈ 2.45 is unstable for the Colorado potato beetle model. Therefore, the Colorado potato beetle model has no stable equilibria. For the meadow plant bug, f′(1.14) ≈ 0.24. Since 0.24 < 1, the equilibrium x ≈ 1.14 is stable for the meadow plant bug model.

  2. Since all of the equilibria for the Colorado potato beetle model are unstable, we can ask the question: What is the long-term behavior of a nonequilibrium solution? Simulating the model with x0 = 2.4 (a value “close to” the equilibrium 2.45) yields the following numerical solution.

    images

    This figure suggests that the Colorado potato beetle is subject to episodic population outbreaks, which is a characteristic associated with agricultural insect pests.

Stability of monotone difference equations

A special, and important, class of difference equations xn+1 = f(xn) was introduced in Section 2.5. For these difference equations, f is a continuous and increasing function over some domain of interest. Within this domain, solutions to this difference equation are monotone (i.e., either increasing for all n or decreasing for all n). As a consequence of this monotonicity, it is possible to provide a simple graphical approach to stability for these difference equations.

Theorem 4.4 Stability of monotone difference equations theorem

Let f be a continuous, increasing function on an interval(a, b). Let x* be an equilibrium for xn+1 = f(xn) that lies in(a, b). Then x* is:

Stable if f(x) > x for x in (a, x*) and f(x) < x for x in (x*, b). In particular, images xn = x* whenever x0 lies in(a, b).

Unstable if f(x) < x for x in(a, x*) and f(x) > x for x in(x*, b). In particular, xn leaves (a, b) for some n whenever x0 lies in (a, x*) or (x*, b).

Combined with the monotone convergence theorem in Section 2.5, this stability theorem allows us to determine the fate of solutions to difference equations where f is a continuous, increasing function.

Example 6 Graphical approach to stability

Consider the difference equation

images

where the graph of f is given by

images

Assuming f is increasing on [0, 1], identify the equilibria and determine their stability.

Solution The equilibria correspond to points where the graph of y = x intersects with the graph of y = f(x). These intersections occur at x = 0, images, and 1. Inspection of the graph of f yields that f(x) > x for x in (−0.2, 0) and (0.5, 1). Alternatively, f(x) < x for x in (0, 0.5) and (1, 1.2). Applying the stability theorem for monotone difference equations implies that 0 and 1 are stable, while 0.5 is unstable. Moreover, xn converges to 0 whenever x0 lies in (−0.2, 0.5) and xn converges to 1 whenever x0 lies in (0.5, 1.2).

We can verify this stability with cobwebbing. Cobwebbing with an x0 slightly above 0.5 and an x0 slightly below 0.5 leads to the following figures:

images

As we saw in Example 5 of Section 1.7, we can construct models that allow us to trace the fate of alleles that code for genes affecting the biological fitness (i.e., the ability to survive and reproduce) of individuals. Recalling our discussion in Section 1.7 regarding genetic models of diploid organisms, we consider an allele that codes for a particular trait. If the frequency of this allele in the population is x images [0, 1], then a well-known model describing how the frequency of this trait changes from generation n to generation n + 1 in a very large (essentially infinite) randomly mating population is

images

where w1 and w2 are the fitness of individuals who, respectively, have two or no copies, relative to individuals who have only one copy, of the allele in question. Referring back to the equation given in Example 5 of Section 1.7 regarding the spread of a deleterious mutant allele, we see that the equation is the same as the above equation for the special case w1 = 0 and w2 = 1. This case is equivalent to the statement that the allele a in question is recessive (heterozygous Aa individuals are not affected) and lethal (aa individuals die before reproducing). Despite the drastic effect of this lethal allele a, we found that it can take a very long time for it to be eliminated from the population. In the next example, we consider a variant of this model in which the allele that is lethal in the homozygous state actually confers a benefit on an individual when combined with the other allele (i.e., when in the heterozygous state).

Example 7 Fate of the sickle cell allele

In areas of the world where malaria occurs, it is known that individuals who have one sickle cell allele are more resistant to malaria than those who do not have the allele. On the other hand, individuals who have two sickle cell alleles suffer from sickle cell anemia, which can cause premature death. Let x denote the frequency of the allele that does not cause sickle cell anemia. Assume when malaria is prevalent that individuals not protected by the sickle cell allele will, on average, have 10% fewer progeny than individuals who have one sickle cell allele—that is, w1 = 0.9. For the sake of simplicity, we assume that individuals with sickle anemia die before they reproduce (even though, in reality, this assumption is too extreme)—that is, w2 = 0.

  1. Write and simplify the difference equation, xn+1 = f(xn), under the assumption that x ≠ 0.
  2. Verify that f is increasing on the interval (0, 1].
  3. Find the equilibria on the interval (0, 1] and determine their stability.
  4. Interpret your results.

Solution

  1. Under the assumption that w1 = 0.9, w2 = 0, and x ≠ 0, we get

    images

  2. To verify that f(x) is increasing on the interval, we compute the derivative of f(x):

    images

    Hence, f′(x) > 0 on (0, 1] and f is increasing on this interval.

  3. To find the equilibria in (0, 1], we solve x = f(x):

    images

    Hence, the equilibria are given by x = 1 and x = 1/1.1 ≈ 0.91.

    To determine their stability, we can use the derivative calculated in part b. We have f′(1) = images ≈ 1.11. Hence, x = 1 is unstable. Alternatively, f′(1/1.1) = 0.9 so that x = 0.91 is stable. In fact, these calculations imply that f(x) < x on the interval (0.91, 1) and f(x) > x on the interval (0, 0.91). Hence, the stability theorem for monotone difference equations implies that images xn = 0.91 whenever x0 lies in (0, 1).

  4. The results imply that as long as both alleles are present in the population, they will persist, and the frequency of the sickle cell anemia allele will approach a value of 1 − 1/1.1 ≈ 0.09. Hence, under the assumptions made, we expect approximately 9% of this population to have the sickle cell allele.

Newton's method

The final application of linearization to difference equations is to illustrate the inner workings of an algorithm called Newton's method. This method is used to find the roots of nonlinear algebraic equations of the form f(x) = 0 that are too difficult or impossible to solve algebraically. The algorithm is based on the Newton-Raphson difference equation, which we now describe. Suppose our initial guess for the solution of f(x) = 0 is x = x0. Assuming that this guess is not the solution, we need to find an improved guess for the root. Since the nonlinear function in question is too hard to manipulate by hand, we consider the linear approximation of y = f(x) at x = x0:

images

To get our next guess, x1, for the solution to f(x) = 0, we set x = x1 and y = 0 in the linear approximation and solve for x1:

images

To get the next guess, x2, we can proceed similarly to get the equation

images

Proceeding inductively, we get the following difference equation:

images

This difference equation is illustrated in Figure 4.21. In this figure, r is a root of the equation f(x) = 0.

images

images

Figure 4.21 Graphical representation of Newton's method.

One of the key requirements of the method is to start with a reasonable guess x0 for the root x* because the closer x0 is to x* the more likely the solution will converge to x*. The following theorem implies that if the sequence converges, then it converges to a root.

Theorem 4.5 Newton's method

Let f(x) be a continuously differentiable function with f′(x) ≠ 0. Any solution to

images

will approach a limit that is a root of the equation or else will not have a finite limit.

When applying Newton's method, we choose a number images > 0 that determines the allowable tolerance for estimated solutions. Given an appropriate initial guess, x0, we iteratively compute xn until |f(xn)| < images. This procedure is shown in the following flowchart.

Example 8 Time to tumor regrowth

In Example 4 of Section 4.3, we considered the growth of a tumor in a mouse after the mouse was given a drug treatment. To model the volume of the tumor, we used the function (renaming the variable x rather than t)

images

where x is measured in days after the drug was administered. Using Newton's method, solve for the time x > 0 at which the tumor regrows to its starting volume of 0.5cm3. Find this time within a tolerance of 0.01.

Solution We want to find a root of

images

To use Newton's method, we need to compute the first derivative:

images

We will see what happens if we start with x0 = 20, although other start values close to the anticipated solution can be chosen. To find x1, we compute

images

Since |f(x1)| ≈ 0.02 > 0.01, our stopping criterion has not been met and we compute

images

Since f(18.732) ≈ 0.0004, the stopping criterion has been met and our answer is x = 18.732.

Implementation of Newton's method for finding roots is widespread, as a quick search of the Web will reveal. Several websites will turn up that allow you to input a function, an initial condition, and the number of iterations, and you will obtain the corresponding sequence from Newton's method.

Newton's method may not converge to a solution, as shown by the following example.

Example 9 Nonconvergence of Newton's method

Consider the function f(x) = ex − 2x. Use Newton's method with x0 = 1 to find a solution to f(x) = 0. Discuss what you find.

Solution Note that f′(x) = ex − 2 so that

images

If we let x0 = 1, then we find

images

images

Figure 4.22 Graph of y = ex − 2x.

Note that the values simply alternate, and the method does not converge to a solution. The graph in Figure 4.22 shows why there can be no solution: the graph does not intersect with the x-axis and hence does not have any roots.

PROBLEM SET 4.5

Level 1 DRILL PROBLEMS

In Problems 1 to 4, find the equilibria of xn+1 = f(xn) and determine their stability using cobwebbing.

images

images

In Problems 5 to 10, find the equilibria of the difference equation. Moreover, use the definitions of an unstable equilibrium and a stable equilibrium to determine their stability.

images

In Problems 11 to 16, find the equilibria of the difference equation. Moreover, use linearization to determine their stability.

images

17. Consider the following alternative formulation of the logistic difference equation, which is different from the formulation presented in Example 1: xn+1 = rxn(1 − xn/100) with r > 0.

  1. Find the equilibria.
  2. Determine under what conditions the origin is stable.
  3. Determine under what conditions the nonzero equilibrium is positive.
  4. Determine under what conditions the nonzero equilibrium is stable.

18. Consider the logistic difference equation xn+1 = rxn(1 − xn/50) with r > 0.

  1. Find the equilibria.
  2. Determine under what conditions the origin is stable.
  3. Determine under what conditions the nonzero equilibrium is positive.
  4. Determine under what conditions the nonzero equilibrium is stable.

19. Consider the Beverton-Holt difference equation xn+1 = images with r > 0.

  1. Find the equilibria.
  2. Determine under what conditions the origin is stable.
  3. Determine under what conditions the nonzero equilibrium is positive.
  4. Determine under what conditions the nonzero equilibrium is stable.

20. Consider the Beverton-Holt difference equation xn+1 = images with r > 0.

  1. Find the equilibria.
  2. Determine under what conditions the origin is stable.
  3. Determine under what conditions the nonzero equilibrium is positive.
  4. Determine under what conditions the nonzero equilibrium is stable.

Following the approach laid out in Example 7 (i.e., graphing and using Theorem 4.4) investigate the fate of an allele in a large, randomly mating population when the fitnesses of individuals with two and zero copies of the allele relative to those that have one copy are given in Problems 21 to 24.

21. w1 = 1/2 and w2 = 1/2

22. w1 = 2 and w2 = 1

23. w1 = 1/2 and w2 = 2

24. w1 = 2 and w2 = 2

Use Newton's method to estimate a root of the equations in Problems 25 to 32. Use x0 as a starting value and iterate twenty times.

25. x2 − 2 = 0, x0 = 1

26. x2 + 2 = 0, x0 = 1

27. x3x + 1 = 0, x0 = −1

28. x4 + 2x − 1 = 1, x0 = 1

29. cos x = x, x0 = 1

30. sin x + 0.1 = x2, x0 = 0

31. ex − 5x = 0, x0 = 0

32. ex + x = 0, x0 = −1

33. Let f(x) = −2x4 + 3x2 + images

  1. Show that the equation f(x) = 0 has at least two solutions.
  2. Use x0 = 2 and Newton's method to estimate a root of the equation f(x) = 0.
  3. Show that Newton's method fails if you choose x0 = images as the initial estimate.

34. Let f(x) = x6x5 + x3 − 3

  1. Show that the equation f(x) = 0 has at least two solutions.
  2. Use x0 = 2 and Newton's method to find a root of the equation f(x) = 0.
  3. Show that Newton's method fails if you choose x0 = 0 as the initial estimate.

Level 2 APPLIED AND THEORY PROBLEMS

35. For the beetle species Lasioderma serricorne, Bellows found that the fraction f(x) of eggs surviving as a function of their initial density x is well described by

images

A graph of this function and the corresponding data are shown below:

images

If we assume that each adult produces two eggs, then the dynamics of the population is given by

images

  1. Find the equilibria and determine their stability.
  2. Simulate the model with x0 = 0.1

36. For the flour beetle species Tribolium castaneum, Bellows found that the fraction f(x) of eggs surviving as a function of their initial density x is well described by

images

A graph of this function and the corresponding data are shown below:

images

If we assume that each adult produces r eggs, then the dynamics of the population is given by

images

  1. Find the equilibria and determine their stability for r = 2, 4, 6.
  2. Simulate the model with x0 = 0.1 for r = 2, 4, 6.

37. Consider the genetic model

images

Show that the following statements are true:

  1. This model has three equilibrium solutions: images
  2. p = 1 is stable and p = 0 is unstable when w1 > 1 > w2 > − 1.
  3. (Harder problem) p* as defined in part a is the only stable equilibrium when w1 < 1 and w2 < 1 (a condition known as heterozygote superiority).
  4. (Harder problem) p* as defined in part a is the only unstable equilibrium when w1 > 1 and w2 > 1 (a condition known as inbreeding depression).

38. It can be shown that the volume of a spherical cap is given by

images

where R is the radius of the sphere and H is the height of the cap, as shown in Figure 4.23.

images

Figure 4.23 Spherical segment is the portion of a sphere between two parallel planes

If V = 8 and R = 2, use Newton's method to estimate the corresponding H.

39. images

images

The Greek geometer Archimedes is acknowledged to be one of the greatest mathematicians of all time. Ten treatises (as well as traces of some lost works) of Archimedes have survived the rigors of time and are masterpieces of mathematical exposition. In one of these works, On the Sphere and Cylinder, Archimedes asks where a sphere should be cut in order to divide it into two pieces whose volumes have a given ratio.

Show that if a plane at distance d from the center of a sphere with R = 1 divides the sphere into two parts, one with volume five times that of the other, then

images

where H = 1 − d. Find d by using the Newton-Raphson method to estimate H. (Hint: see Problem 38.)

40. Suppose the plane in Problem 39 is located so that it divides the sphere in the ratio of 1:3. Find an equation for d, and estimate the value of d using Newton's method.

41. In Example 4 of Section 4.3, we considered the growth of a tumor in a mouse after the mouse was given a drug treatment. To model the volume of the tumor, we used the function

images

where x is measured in days after the drug was administered. Using Newton's method, solve within a tolerance of 0.01 for the time x at which the tumor volume has doubled in volume. For an initial guess, use x = 25 days.

42. In Example 4 of Section 4.3, we considered the growth of a tumor in a mouse after the mouse was given a drug treatment. To model the volume of the tumor, we used the function

images

where x is measured in days after the drug was administered. Using Newton's method, solve within a tolerance of 0.01 for the time x at which the tumor volume has quadrupled in volume. For an initial guess, use x = 30 days.

43. In Problem 25 in Problem Set 4.3, you found that the volume of a tumor for mice under a different drug regimen was

images

where x is days after treatment. Using Newton's method, solve within a tolerance of 0.01 for the time x at which tumor volume has regrown to its original volume. For an initial guess, use x = 20 days.

44. In Problem 25 in Problem Set 4.3, you found that the volume of a tumor for mice under a different drug regimen was

images

where x is days after treatment. Using Newton's method, solve within a tolerance of 0.01 for the time x at which tumor volume has doubled in volume. For an initial guess, use x = 25 days.

45. Show that for different initial values Newton's method converges to a unique solution for the function

images

but yet converges to one of three solutions for the function

images

Why is this the case?

46. According to an online article in the New Scientist (Catherine Brahic, “Carbon Emissions Rising Faster Than Ever,” p. 9), recent research suggests that stabilizing carbon dioxide concentrations in the atmosphere at 450 parts per million (ppm) could limit global warming to 2°C. In Section 1.2, we modeled carbon dioxide concentrations in the atmosphere with the following function (which we now present to higher precision to make more transparent the numerical details of the convergence process):

images

where x is months after April 1974. In Example 11 of Section 2.3, we used the bisection method to estimate the first time that the model predicts carbon dioxide levels of 450 ppm. Use Newton's method to estimate this time with a stopping value of images = 0.001.

47. Repeat Problem 46 except estimate the first time until reaching 400 ppm.

CHAPTER 4 REVIEW QUESTIONS

  1. Use the first derivative test and the second derivative test to find and classify all the extrema of g(x) = x3 − 3x − 4.
  2. Find the global maximum and global minimum of f(x) = images on [0, 6].
  3. Using asymptotes and first derivatives, graph f(x) = images by hand and then check it using a calculator.
  4. Consider the family of curves

    images

    Using calculus, graph the curves for the given values of b.

    1. b = 0
    2. b = 0.05
    3. b = 0.01
    4. b = −0.05
    5. b = −0.1
  5. The canopy height (in meters) of a tropical grass may be modeled by (for 0 ≤ t ≤ 30)

    images

    where t is the number of days after mowing.

    1. Sketch the graph of h(t).
    2. When was the canopy height growing most rapidly? Least rapidly?
  6. Find the value of r in the function f(x) = erx that provides the best fit through a semi-log plot of the points{(0, 1), (1, 0.6), (2, 0.4), (3, 0.3)}.
  7. A travel company plans to sponsor a tour to Africa. There will be accommodations for no more than forty people, and the tour will be canceled if no more than ten people book reservations. Based on past experience, the manager determines that if n people book the tour, the profit (in dollars) may be modeled by the function

    images

    What is the maximum profit?

  8. Two towns, A and B, are 12.0 miles apart and are located 5.0 and 3.0 miles, respectively, from a long, straight highway, as shown in Figure 4.24.

    images

    Figure 4.24 Highway construction project.

    A construction company has a contract to build a road from A to the highway and then to B. Analyze a model to determine the length (to the nearest tenth of a mile) of the shortest road that meets these requirements.

  9. A particular plant is known to have the following growth and seed production characteristics: At time of planting (t = 0), the seedling has a mass of 3 grams. At time t > 0 days after planting, the seedling has grown into a plant that weighs

    images

    grams. The plant has a gene that can be manipulated to control the age t at which the plant matures. At maturity the number of seeds S(t) produced by the plant is given by

    images

    A farmer asks a geneticist to genetically engineer a plant line that accounts for the fact that on his farm, because of losses from pests, drought, and disease, a proportion

    images

    of germinating seeds can be expected to develop and survive to age t as plants. What age of maturity should the geneticist select for the plants to maximize the seed production of the mature crop for the farmer?

  10. If the value of a forest stand (units are dollars per square meter) is given by the function

    images

    where t represents the number of years after the stand has been clear-cut, and the discount is 0.02 per year, solve for the optimal rotation period, assuming that V(t) applies every time the stand is clear-cut.

  11. On a particular island in the tropics, scientists have determined that individuals who are not protected by the sickle cell allele will have, on average, 20% fewer progeny than individuals who have one sickle cell allele, while individuals who have two sickle cell alleles will not reproduce.
    1. Write a difference equation for the proportion xn of non-sickle-cell alleles in the population in generation n, assuming that initially x0 > 0, under assumptions of random mating and random segregation of alleles.
    2. Find the equilibria proportions for x on the interval (0, 1] and determine if they are stable.
    3. Interpret your results.
  12. Let C(t) denote the concentration in the blood at time t of a drug injected into the body intramuscularly. In a now classic paper by E. Heinz (“Probleme bei der Diffusion kleiner Substanzmengen inner-halb des menschlichen Körpers,” Biochem. Z, 319 (1949): 482–492), the concentration was modeled by the function

    images

    where a, b (with b > a), and k are positive constants that depend on the drug. At what time does the largest concentration occur? What happens to the concentration as t → +∞?

  13. Consider a bird that has arrived at a wooded patch with two trees. If the bird spends x minutes foraging for insects on the first tree, she gains E1(x) = 200(1 − ex) Calories from insects. If the bird spends x minutes on the second tree, she gains E2(x) = 100(1 − ex) Calories of insects. Assuming the bird has five minutes to spend in the patch, determine the time she should spend on each tree to optimize her energy intake.
  14. For the flour beetle species Tribolium confusum, Bellows found that the fraction f(x) of eggs surviving as a function of their initial density x is well described by

    images

    A graph of this function and the corresponding data are shown below:

    images

    If we assume that each adult produces r eggs, then the dynamics of the population is given by

    images

    1. Find the equilibria and determine their stability for r = 2, 4, 6.
    2. Simulate the model with x0 = 0.1 for r = 2, 4, 6.
  15. A model for the population growth rate of Eastern Pacific yellowfin tuna is given by

    images

    where N is measured in thousands of tons, t is measured in years, and θ > 0 is a parameter that determines the strength of density dependence. Find the population size at the maximum sustainable harvesting rate. Discuss what effectθhas on this population size.

  16. The energy gain from nectar for a hummingbird flying at t = 0 to a flower is shown in the graph below:

    images

    Ignoring the cost of flying constantly, find the time at which the hummingbird's energy gain per unit time is maximal and, consequently, the bird should fly to another flower. Discuss what effect including a fixed energetic cost per unit time would have on your answer.

  17. A simple model of a biochemical switch is given by

    images

    where xn is the concentration of a biochemical at time n.

    1. Find the equilibria of this model.
    2. Verify that f(x) = images is an increasing function of x for x > 0.
    3. Discuss what you can say about all nonnegative solutions to this difference equation.
  18. Public awareness of a new drug is modeled by

    images

    where t is the number of months after FDA approval and P(t) is the fraction of people who are aware of the drug and its possible uses.

    1. Find the critical points for P(t).
    2. Sketch the graph of P(t).
    3. At what time, t, during the time interval 0 ≤ t ≤ 36 is P(t) the largest?
  19. Suppose that systolic blood pressure of a patient t years old is modeled by

    images

    for 0 ≤ t ≤ 60, where P(t) is measured in millimeters of mercury.

    1. Sketch the graph of P(t).
    2. At what rate is P(t) increasing at age t?
  20. During the time period 1905–1920, hunters virtually wiped out all large predators on the Kaibab Plateau near the Grand Canyon in northern Arizona. This, in turn, resulted in a rapid increase in the deer population P(t), until food supplies were exhausted and famine led to a steep decline in P(t). A study of this ecological disaster determined that during the time period 1905–1920, the rate of change of the population, P′(t), could be modeled by the function

    images

    0 ≤ t ≤ 20, where t is the number of years after the base year of 1905.

    1. In what year during this period was the deer population the largest?
    2. In what year did the rate of growth P′(t) begin to decline?

GROUP PROJECTS

Seeing a project through on your own, or working in a small group to complete a project, teaches important skills. The following projects provide opportunities to develop such skills.

Project 4A Optimal swimming patterns

In getting from one spot to another, fish have to contend with drag forces and gravity. Drag forces are much greater when a fish is swimming than when it is merely gliding. To reduce the amount of time spent swimming, fish that are heavier than water engage in burst swimming in which they alternate between gliding and swimming upward. This burst swimming leads to a vertical zigzag motion of the fish in the water as shown below:

images

where a is the angle of the upward burst and b is the angle of the downward glide.

In this project, you will investigate the optimal swimming pattern under the following assumptions:

  • Throughout its swim, the fish maintains a constant speed s to the right.
  • The forces acting on the fish are its weight W relative to the water and drag forces.
  • The drag on the gliding fish is D and the drag on the swimming fish is kD where k ≥ 1.
  • The fish has sufficient top/bottom surface area (e.g., a skate) that frictional forces perpendicular to the top/bottom of the fish cancel the component of the gravitational force that is perpendicular to the top/bottom of the fish.
  • The energy expended by the fish in swimming is proportional to the force it exerts in moving.

Under these assumptions, your project should do the following:

  • Find the ratio of energy in the burst mode to the energy for continuous horizontal swimming from A to B.
  • It has been found empirically that tan a ≈ 0.2. Given this information, find the optimal value of b for the fish.
  • Determine how much energy the fish saves by swimming with this b instead of swimming horizontally.
  • Determine how sensitive the amount of energy used is to b, and how sensitive the optimal b is to the estimate of a.

Project 4B Stability and bifurcation diagrams

Consider the normalized version of the logistic model introduced in the first example in Section 4.5; that is, we set K = 1 or, equivalently, interpret the units of x in terms of multiples of K to obtain the equation

images

Now explore the behavior of this equation as follows:

  1. Solve for the equilibrium solutions as a function of r and determine the stability properties of these equilibria for r images [0, 5]. You will notice that as r increases, an equilibrium solution jumps at some point rb from being stable on one side of rb to unstable on the other side of rb. The value rb is called a bifurcation point.
  2. Plot the equilibria in the rx plane (r is the horizontal axis spanning [0, 5]) using a solid line to denote where the nontrivial equilibrium solution images to x = f(x) is stable and a dotted line where it is unstable.
  3. Now consider the equilibria of the iterated logistic map xn+2 = f(f(xn)) by constructing (see Section 1.6) the composite map (f images f)(x). Use the terminology f2 ≡ (f images f). Find all the equilibria of f2(x) as a function of r and plot them on the same rx plane as above, but this time plot only the nontrivial stable solutions using a solid line (if you plot where they are unstable, your diagram will become too busy). Note that the equation x = f2(x) has many more solutions than the equation x = f(x): It has both all the solutions to equation x = f(x) (demonstrate this) and additional solutions that come in pairs, say x* and x**, such that the sequence {x*, x**, x*, x**, ...} is a two-cycle solution of the original equation x = f(x) (demonstrate this). Further, if for a particular value of r, x* and x** are stable equilibrium solutions of x = f2(x), then the two-cycle {x*, x**, x*, x**, ...} is a stable attractor of the equation xn+1 = f(xn). By this, we mean that for any initial condition x0 starting close to x* or x**, the resulting sequence generated by our original equation will oscillate between two values that get closer and closer to x* and x** as time progresses.
  4. You have now reached the limit of what you can probably do analytically. Research the literature (a good source is J. D. Murray's book Mathematical Biology: I, An Introduction, 3rd ed., New York: Springer-Verlag, 2001). Then discuss what happens as r increases on [0, 5], focusing in terms of bifurcation values at which stable equilibrium solutions of the logistic equation are replaced by stable two-cycles, as well as stable n-cycles for n > 2.
  5. If you have command of an appropriate technology, use it to summarize graphically your discussion in what is called a bifurcation diagram. (Instructions on how to do this are available in textbooks and on the Web, so locate a set of instructions and see if you can follow them.)

Project 4C Economic production versus ecological welfare

Economic activities, such as extraction and processing of raw materials and the manufacture of finished goods, always result in some damage to the ecosystem. Because of pollution and the destruction of natural habitats, such activities may even severely degrade the ecosystem's delivery of clean water and clean air. Activities may also compromise the ecosystem's ability to produce food and provide a place for relaxation and recreation. In this project, you are asked to use optimization to explore the trade-off between economic production and ecological welfare. (This problem follows Problem II.5 in J. Harte, Consider a Cylindrical Cow. Sausalito, University Science Books, 2001.)

images

Figure 4.25 Industrial pollution.

Suppose the level of economic activity is measured by a variable X, the value of goods and services produced by this activity (also known as economic output) is measured by a variable Y, and the value of ecosystem services is measured by an environmental quality variable Z.

A very simple model of human welfare W is based on these assumptions:

  • Welfare is proportional to both economic output Y and environmental quality Z.
  • Economic output Y is itself proportional to economic activity X and and environmental quality Z. (The first assumption is self-evident and the second arises from the notion that it is much more difficult to produce the same unit of economic output in a poor environment where resources are depleted than in a pristine environment where resources are plentiful.)
  • The environment declines from a pristine level linearly with activity X.

These assumptions are equivalent to the following mathematical statement: For positive constants a, b, c, and Z0, our variables satisfy the equations

images

  1. Demonstrate that human welfare W is maximized at X* = Z0/3c and has the maximum value

    images

  2. Show that the value of economic activity images that maximizes production Y is 1.5 times larger than X*; that is, ρ = images/X* = 3/2. Further, show that if images is the welfare obtained when production is maximized, then the “cost of greed” (defined to be the ratio γ = images/W*) is γ = 27/32. Discuss the implications of the fact that ρ > 1 and γ < 1.
  3. Assume that the economic production level Y has the more general Cobb-Douglas form Y = bXα Zβ than assumed in Equation 4.1, where α and β are nonnegative, empirically determined constants with values that depend on the economic sector under consideration. If, in addition, welfare has the general form W = aXμ Yν, then find the values of X that maximize both economic output and welfare. Calculate the ratios ρ and γ for this more general case. What do you conclude?
  4. Show in the case of the given equations for W, Y, and Z that the level of economic output that maximizes welfare-per-unit-output—that is, the ratio W/Y—is X = 0. Does this hold true for the more general case when α, β, μ, and ν are not necessarily 1?
  5. Look through the literature and see how many Cobb-Douglas functions you can find and what values of α and β are associated with various sectors of the world economy. Also, see if you can find a real problem where most of the parameters a, b, c, Z0, α, β, μ, and ν are known. Describe the problem and the values of the parameters. (If one of or more of α, β, μ, and ν are not known, then set them equal to 1, and it is fine if relative rather than global values of the other constants are known or guessed at.) Now calculate the optimum production levels images and X* with respect to economic output and welfare, respectively, and elaborate in anyway you think appropriate.
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