Figure 3.1 Differential calculus is used in Section 3.1 to model the rate at which HIV virions (as depicted) is cleared from the body during drug therapy.

images

Preview

“I loved history, but still loved science, and thought maybe you don't need quite as much calculus to be a biology major.”

Elizabeth Moon, science fiction writer (b. 1945)
Part 2 of an interview with Jayme Lynn Blaschke
November 1999; http://www.sfsite.com/02b/em75.htm

In the previous chapters we have only been able to compute derivatives by directly appealing to the definition of the derivative. As you may have noticed, computing derivatives in this manner quickly becomes tedious. In this chapter we consider the rules and tools that allow us to quickly compute the derivative of any imaginable function. Learning these rules is critical, as expressed in the following admonition (from Colin Adams, Joel Hass, and Abigail Thompson, How to Ace Calculus: The Streetwise Guide, New York: W. H. Freeman, 1998):

“Know these backwards and forwards. They are to calculus what ‘Don't go through a red light’ and ‘Don't run over a pedestrian’ are to driving.”

The first three sections of this chapter provide essential basic rules for calculating derivatives, and the fourth section focuses on the important trigonometric functions. The last three sections expand these tools so that we can apply them to a variety of applications in biology and the life sciences. Applications in this chapter include predicting the growth of a fetal heart and of the Yellowstone bison population. Also, we use differential calculus to investigate the clearance rate of HIV viral particles (see Figure 3.1) from infected humans, how Northwestern crows break whelk shells, dose-response curves in the context of administering drugs and into rates of mortality due to airborne diseases, and Usain Bolt's record-breaking, 100-meter run in the Olympic Games in Beijing.

3.1 Derivatives of Polynomials and Exponentials

As we have seen in Chapters 1 and 2, we can use polynomial and exponential functions to model natural phenomena ranging from the melting of Arctic sea ice to the decay of a drug in the body. To facilitate finding the instantaneous rates of change of these processes, we now derive general rules for computing the derivatives of polynomials and exponentials.

Derivatives of y = xn

In Example 1 of Section 2.7, we proved that

images

and we guessed that images = nxn−1. This powerful result is known as the power rule.

Power Rule

For any real number n ≠ 0,

images

At this point, we are only equipped to prove the power rule for any natural number n. Later, we shall prove the general power rule. The proof when n is a natural number involves the binomial expansion of (a + b)n (if you don't remember this expansion, look it up on the World Wide Web):

images

Now to the proof of the power rule for n a natural number:

Proof. If f(x) = xn, then from the binomial theorem

images

From the definition of derivative we have

images

Note that if n = 0, then f(x) = xn = x0 = 1, so f′(x) = 0 as expected, since 1 is a constant.

Example 1 Using the power rule

Find

images

Solution

images

Derivatives of sums, differences, and scalar multiples

The limit laws from Chapter 2 allow us to quickly compute the derivatives of a sum, difference, or scalar multiple whenever we know the derivatives for f and g. In stating these laws, it is more succinct to use the “prime” rather than full Leibnitz notation.

Elementary Differentiation Rules

Let f and g be differentiable at x. Let c be a constant. Then

Sum (f + g)′(x) = f′(x) + g′(x)

Difference (fg)′(x) = f′(x) − g′(x)

Scalar multiple (cf)′(x) = cf′(x)

In other words, the derivative of a sum is the sum of the derivatives, the derivative of a difference is the difference of the derivatives, and the derivative of a scalar multiple is the scalar multiple of the derivative.

Combining these elementary differentiation rules with the power rule allows us to differentiate any polynomial. Note that throughout this and subsequent chapters, we use the verb differentiate in a technical sense. To differentiate a function is to “take its derivative” using the methods of differential calculus presented in this book. It does not mean that we are trying to distinguish the function from some other function, unless we specifically say so.

Example 2 Using differentiation rules

Let f(x) = x3 + 3x2 + 10.

  1. Find f′. Justify each step of your differentiation.
  2. Determine on what intervals f is increasing and on what intervals f is decreasing.

Solution

images

Hence

images

b. To determine where f is increasing and where f is decreasing, we need to find where f′ > 0 and f′ < 0, respectively. Since

images

we look at the signs of the factors and the product by looking at a number line.

images

images

On the interval (−∞, −2), f is increasing since f′(x) > 0; on (−2, 0), f is decreasing since f′(x) < 0; on (0, ∞), f is increasing again since f′(x) > 0. Graphing y = f(x) confirms these calculations.

Example 3 Growth of a fetal heart

In 1992, a team of cardiologists determined how the left ventricular length L (in centimeters) of the heart in a fetus (Figure 3.2) increases from eighteen weeks until birth. (See J. Tan, N. Silverman, J. Hoffman, M. Villegas, and K. Schmidt, “Cardiac Dimensions Determined by Cross-Sectional Echocardiography in the Normal Human Fetus from 18 Weeks to Term,” American Journal of Cardiology 70(1992): 1459–1497.) The cardiologists used the following function to model the data

images

where t is the age of the fetus (in weeks). Here t = 18 means at the end of week 18.

images

Figure 3.2 Fetal echocardiogram reveals a four-chamber heart correctly oriented in the left chest.

  1. Find L′(t) for 18 ≤ t ≤ 38.
  2. Discuss and interpret the units of L′(t).
  3. During which week between weeks 18 and 38 is the ventricular length growing most rapidly, and what is the associated rate? When is the ventricular length growing most slowly?

Solution

  1. images sum and scalar multiple laws

    images

  2. The units of L′(t) are centimeters per week.L′(t) describes the rate at which the ventricular length is growing.
  3. Since L′(t) is a linear function with negative slope, its largest value on the interval [18, 38] is at t = 18 and its smallest value on this interval is at t = 38. In particular,

    images

    and

    images

    Hence, the ventricular length in the last twenty weeks of pregnancy is increasing most rapidly at the beginning of this twenty week period and growing least rapidly at the time of birth.

In addition to being used to finding derivatives of all polynomials, the power rule and the scalar multiplication rule can be used to find derivatives of all scaling laws.

Example 4 Back to lifting weights

In Example 6 of Section 1.3, we modeled the amount an Olympic weightlifter could lift as

images

where M is the body mass in kilograms of the weightlifter. Find and interpret images at M = 90 kilograms.

Solution To compute the derivative, we note that n = 2/3, and although n is not an integer, the power rule still applies with n − 1 = −images. Thus we obtain

images

Hence, for weightlifters weighing close to 90 kilograms, the rate at which the amount lifted increases with mass of the weightlifter is 2.998 kilograms per kilogram of body mass.

Derivatives of exponentials

Consider the function f(x) = ax for some positive constant a > 0. To find the derivative, we use the definition of the derivative. Let x be a fixed number.

images

Although it is beyond the scope of this book, it can be shown that k = images exists whenever a > 0. In the following example, we estimate the value of k for the case a = 2.

Example 5 Derivative of 2x

Find images by estimating images.

Solution We showed that images = k2x where k = images. To estimate k, we can create the following table with a calculator:

images

Since k ≈ 0.693, images ≈ (0.693)2x. In Example 6, we show that in fact k = ln 2 ≈ 0.69315.

Since f′(x) = kf(x) for an appropriate choice of k whenever f(x) = ax, we can ask this question: Is there a value of a such that k= 1? It turns out that the number e, which we defined in Section 1.4 as e = images (1 + 1/n)n, is the appropriate choice of a. Namely,

images

Except for multiplying by a constant, ex is the only function that remains unchanged under the operation of differentiation; that is, if f(x) = f′(x), then f(x) = aex for some real number a. This fact inspired one mathematician to write: “Who has not been amazed to learn that the function y = ex, like a phoenix rising again from its own ashes, is its own derivative?” (Francois l'Lionnais, Great Currents of Mathematical Thought, vol. 1, New York: Dover Publications, 1962). Armed with the derivative of ex, we can use the rules of differentiation to find the derivative of more general exponential functions f(x) = eax.

Derivative of the Natural Exponential

For any real number a,

images

Further, for b > 0,

images

Proof. If a = 0, then

images

so the statement is true. If a is any nonzero real number, then eax = (ea)x and

images

To find this limit, define Δx = ah. Since h = Δx/a and h → 0 whenever Δx → 0,

images

Thus, we have shown that

images

For the last part, find the value of a such that b = ea and finish the details on your own in Problem 27 in Problem Set 3.1.

Example 6 Déjà Vu

Find the exact value of images.

Solution images = (ln 2)2x. Since ln 2 ≈ 0.693, this agrees with the result obtained in Example 5.

Example 7 Clearance of HIV

Human immunodeficiency virus (shown in Figure 3.1) is a blood-borne pathogen that is typically transmitted through sexual contact or sharing of needles among drug users. HIV attacks the immune system. Understanding how the viral load in the blood of an HIV-infected individual changes with time is critical to treating HIV patients with a “cocktail” of several antiretroviral drugs. Theoretical immunologists have used data from various experiments to model observed changes in the viral load V(t), in particles per milliliter, of an HIV patient undergoing antiretroviral drug therapy for t days.* If no new viral particles are generated by the host, they found that the viral load over time can be modeled by the equation

images

Find V′(t) and interpret it.

Solution

images

The units of V′(t) are particles per milliliter per day. V′(t) describes the rate at which the viral load is changing whenever it is not replenished by new viral particles. Since V′(t) < 0 for all t, the viral load is decreasing.

Note that in HIV-infected patients, new viral particles are produced in the various cells found in the blood and in lymph tissue. In Chapter 6, we account for this second component of the infection process to obtain a more complete model of viral load dynamics within human hosts.

Example 8 Exponential depletion of resources

In Example 4 of Section 1.4, we projected that the U.S. population would contain N(t) = 8.3(1.33)t million individuals t decades after 1815. Suppose the amount of food produced each year, measured in terms of rations (i.e. the amount of food needed to sustain one individual for one year), grew linearly during this same period with the amount given by the equation

images

The number of surplus rations S(t) over this period can be found by taking the difference of the above two functions:

images

Determine at what point in time S(t) starts decreasing.

Solution To find where S changes from increasing to decreasing, we need to determine where S′(t) changes sign from S′(t) > 0 to S′(t) < 0. In particular, we need to find where S′(t) = 0, provided the derivative exists at the point in question. We have

images

If we now solve for the values of t satisfying S′(t) = 0, we obtain

images

images

Figure 3.3 Graph of the number of surplus rations.

Evaluating S′(t) at values of t greater than and less than 1.84, we find that S′(t) > 0 for t < 1.84 and S′(t) < 0 for t > 1.84. Since the units of time are in decades, we see that in the year 1815 + 18.4 ≈ 1833 the surplus of resources will begin to decline. Plotting y = S(t) reveals that at t ≈ 1.84, S(t) takes on its largest value and then begins to decrease, as shown in Figure 3.3.

PROBLEM SET 3.1

Level 1 DRILL PROBLEMS

Differentiate the functions given in Problems 1 to 14. Assume that C is a constant.

images

6. f(x) = 5x3 − 5x2 + 3x − 5

7. f(x) = x5 − 3x2 − 1

8. f(x) = 2x2 − 5x8 + 1

9. s(t) = 4et − 5t + 1

10. f(t) = 5 − e2t

11. f(t) = 5.9(2.25)t

12. f(t) = 82.1(1.85)t

13. g(x) = Cx2 + 5x + e−2x

14. F(x) = 5eCx − 4x2

Determine on what intervals each function given in Problems 15 to 19 is increasing and on what intervals it is decreasing.

15. f(x) = x3x2 + 1

16. g(x) = images

17. f(x) = x5 + 5x4 − 550x3 − 2,000x2 + 60,000x (round to the nearest tenth)

18. g(x) = x3 + 35x2 − 125x − 9,375

19. H(w) = 2wew

20. Let f(x) = x1/2.

  1. Find the derivative using the definition of the derivative.
  2. Apply the power rule with n = 1/2.

21. Let f(x) = x3/2.

  1. Find the derivative using the definition of derivative. Hint: Write x3/2 as ximages and rationalize the numerator.
  2. Apply the power rule with n = 3/2.

Simplify the functions in Problems 22 to 26 and find their derivative whenever it is well defined.

22. g(x) = x2(x3 − 3x)

23. f(x) = images

24. f(x) = (ex − 1)(ex + 1)

25. h(t) = images

26. q(x) = images

Level 2 APPLIED AND THEORY PROBLEMS

27. Prove that for any real number b > 0

images

28. Use the limit laws to prove the sum rule for differentiation:

images

29. Use the limit laws to prove the scalar multiple rule for differentiation:

images

for a constant c.

30. After pouring a mug full of the German beer Erdinger Weissbier (see Section 1.4), German physicist Leike (“Demonstration of Exponential Decay,” pp. 21–26) measured the height of the beer froth at regular time intervals. He estimated the height (in centimeters) of the beer froth as

images

where t is measured in seconds. Find

images

and interpret this quantity.

31. A drug that influences weight gain was tested on eight animals of the same size, age, and sex. Each animal was randomly assigned to a dose level. After two weeks, the difference W in the end and start weight (measured in decagrams) was calculated. The best-fitting quadratic equation to the data was found to be

images

where D is the dose level that ranges from 1 to 8.

  1. Find images and identify its units.
  2. When does weight gain increase with dose level D?

32. Using data from 158 marine species, John Hoenig of the Virginia Institute of Marine Sciences studied how the natural mortality rate M of a species depends on the maximum observed age T (“Empirical Use of Longevity Data to Estimate Mortality Rates,” Fisheries Bulletin 82 (1983): 898–902). Using linear regression, he found

images

where T is measured in years. Find and interpret

images

33. During an outbreak of influenza at a school the number of students who became ill after t days is given by

images

where C is a constant.

  1. If ten people were ill at the beginning of the epidemic (when t = 0), what is C?
  2. At what rate is N(t) increasing when t = 5?

34. A glucose solution is administered intravenously into the blood stream of a patient at a constant rate of r milligrams/hour. As the glucose is being administered, it is converted into other substances and removed from the blood stream. Suppose the concentration of the glucose solution after t hours is given by

images

where k is a constant.

  1. If C0 is the initial concentration of glucose (when t = 0), what is C0 in terms of r and k?
  2. What is the rate at which the concentration of glucose is changing at time t?

35. In Problem 39 in Problem Set 1.3, we found that the length L(cm) of a pumpkin is related to the width W(cm) of a pumpkin by the allometric equation

images

How rapidly is length changing with regard to width for pumpkins of size 5 cm compared with those of 50 cm?

36. At the beginning of Section 1.4, we saw that the population model N(t) = 8.3(1.33)t, where N is in millions and t is in decades starting at t = 0 representing the year 1815, can be used to describe the size of the U.S. population during most of the nineteenth century. Use this relationship to compare the total rate of growth of the U.S. population in 1815 versus the growth rate in 1865 (i.e., five decades later).

37. In Section 1.3, we found that the amount lifted (in kilograms) by an Olympic weightlifter can be predicted by the scaling law

images

where M is the mass of the lifter in kilograms. Find and interpret images.

38. In Example 12 of Section 1.6 (changing the names of the variables from x and y to M and R), we found that the metabolic rate (in kilocalories/day) for animals ranging in size from mice to elephants is given by the function ln R = 0.75 ln M + 4.2 which yields the equation

images

where M is the body mass of the animal in kilograms.

  1. The average California Condor weighs about 10 kg. Find and interpret images.
  2. The average football player weighs about 100 kg. Find and interpret images.
  3. Compare and discuss the quantities that you found in parts a and b.

39. The number of children newly infected with a particular pathogen that is transmitted through contact with their mothers has been modeled by the function

images

where N(t) is measured in thousands of individuals per year, and t is the number of years since 2000. In epidemiology, N(t) is known as an incidence function.

  1. At what rate is the incidence function N changing with respect to time in the year 2010?
  2. When will the incidence start to decline?

3.2 Product and Quotient Rules

Previously, we saw that the derivative of a sum equals the sum of the derivatives, and the derivative of a difference equals the difference of the derivatives. Armed with these elementary differentiation rules, we might guess that the derivative of a product is the product of the derivatives. The following simple example, however, shows this not the case. Let f(x) = x and g(x) = x2, and consider their product

images

Because f′(x) = 1 and g′(x) = 2x, the product of the derivatives is

images

whereas the actual derivative of p(x) = x3 is p′(x) = 3x2. Hence, our naïve guess is wrong! It is also easy to show that the derivative of a quotient is not the quotient of the derivatives. The goal of this section is to uncover the correct rules for differentiation for products and quotients of functions.

Product rule

To derive a rule for products, we appeal to our geometric intuition by considering areas where Δx > 0 and f(x) and g(x) are assumed to be increasing, positive differentiable functions of x. Note that the algebraic steps stand alone—without considering area or making the assumptions we made in the previous sentence.

Let

images

This product of p can be represented as the area of a rectangle:

images

Next, we find

images

The next step gives us the area of the “inverted L-shaped” region:

images

The key to the proof of the product rule is to rewrite this difference. We can see how to do this by looking at the area of the inverted L-shaped region in another way:

images

Divide both sides by Δx (where Δx ≠ 0):

images

The last step in deriving the product rule is to take the limit as Δx → 0:

images

We have just proved the product rule.

Product Rule

Let f and g be differentiable at x. Then

images

A simple way to remember the product rule is with this mnemonic: “The derivative of the product is the derivative of the first times the second plus the derivative of the second times the first.” Or sing the words of the following ditty aloud or in your mind.

Sing the product rule in time,
One prime two plus one two prime.
Isn't mathematics fun,
One prime two plus two prime one
.

Example 1 Computing with the product rule

Find p′(x) and determine on what intervals p is increasing.

images

Solution

  1. Let f(x) = x and g(x) = ex. Then p(x) = f(x)g(x). By the product rule,

    images

    Since ex > 0 for all x, we have p′(x) > 0 if and only if 1 + x > 0. Hence, p is increasing on the interval (−1, ∞). Indeed, plotting y = p(x) supports this conclusion:

    images

  2. Let f(x) = x2 and g(x) = 2x. Then p(x) = f(x)g(x). Recall that f′(x) = 2x and g′(x) = (ln 2)2x. Hence, by the product rule,

    images

    Since p′ > 0 whenever x > 0 or x < images, p is increasing on these intervals. Indeed, plotting y = p(x) supports this conclusion:

    images

In the next problem, we encounter the notion that many events in biology are not certain but occur with a particular probability. Examples of this is an individual dying over a specified interval of time, a female giving birth to a particular number of individuals (e.g., the litter size of a mouse), or a molecule binding to a receptor site on a cell. You may have already learned, or will learn if you take a course in basic probability theory, that probabilities associated with independent events combine through multiplication. For example, if one flips a coin and the probability of getting heads is p = 1/2, then the probability of getting heads twice in a row is p × p = 1/2 × 1/2 = 1/4. Similarly, if one rolls a fair (honest or unbiased) six-sided die, then the probability of rolling a 3 is p = 1/6 and the probability of rolling an even number is p = 3/6 = 1/2. Thus the probability of getting a 3 on the first roll and an even number on the second roll is 1/6 × 3/6 = 3/36 = 1/12.

Although calculus and probability theory are generally taught as separate courses in college, probability is such an integral part of biological systems that it is useful for us to state and use the following fact for the examples we develop in this text.

Probability of two independent events. If p and q are the probabilities of the events 1 and 2 respectively occurring independent of one another, then pq is the probability that event 1 and event 2 both occur.

In Chapter 7, we revisit probability theory and explore how the integral calculus has become a central tool in the development of this theory.

Example 2 Survival rates

Over time, a zebra in Etosha National Park, Namibia, can die either because it is killed by a predator or because it succumbs to disease (primarily anthrax). Starting at the beginning of each year, suppose f(t) and g(t) respectively represent the probabilities of surviving predation and disease up to week t. Professor Getz's research group (see Figure 3.4) determined that at the height of a typical anthrax season, which occurs around the end of the eleventh week of a typical year (i.e., t = 11), f(11) = 0.965 and g(11) = 0.981. They also determined that the probability of surviving predation and disease at the end of the eleventh week is decreasing by 0.004 and 0.005 per week respectively.

images

Figure 3.4 Dr. Holly Ganz collecting blood from a zebra in Etosha National Park, Namibia, to test if it died of anthrax.

Assume that the events of dying from predation or disease are independent.

  1. Find the probability of surviving to the end of week 11.
  2. Find the rate at which this probability is changing at end of week 11.
  3. Use your answers from parts a and b to estimate the probability of surviving to the end of week 12.

Solution

  1. Since the events of being killed by a predator and dying from disease are assumed to be independent, then the probability p(t) of surviving until week t is given, using the multiplication rule for independent probabilities, by the equation

    images

    Thus, the probability of surviving up to the end of week 11 is given by p(11) = f(11)g(11) = 0.965 × 0.981 = 0.9467 (to 4 decimal places).

  2. The problem statement implies that f′(11) = −0.004 and g(11) = −0.005 (the negative values arise because the probabilities are decreasing). Using the product rule, we get

    images

    The probability of surviving is decreasing at rate of 0.0088 = 0.88% per week at the end of week 11.

  3. Since the probability of surviving is decreasing at a rate of 0.0088 per week at the end of week 11, we can estimate the probability surviving to the end of week 12 by

    images

Example 3 Per capita or intrinsic rate of growth

As we have seen in Section 3.1, single species population models can be of the form

images

where Nn is the population abundance in the nth generation, f(N) is the per capita growth rate of the population density as a function of population N, and g(N) is the growth rate of the whole population as a function of N.

Find an expression in terms of f and N for g′(0). Briefly explain what this expression represents.

Solution Applying the product rule to the relationship g(N) = Nf(N), we have

images

Evaluating at N = 0,

images

Hence, the rate g′(0) at which growth changes at low densities equals the per capita growth rate of the population at low densities.

Quotient rule

Before we derive a quotient rule, we begin with an example for finding the derivative of a reciprocal, which is a special case of a quotient that has 1 in the numerator.

Example 4 Reciprocal rule

Find the derivative of the reciprocal images of a differentiable function f by using the definition of derivative.

Solution Let r(x) = images. Then r(x + h) = images, so using the definition of derivative we find:

images

We restate the result of this example for easy reference.

Reciprocal Rule

Let f be differentiable at x. Then

images

provided that f(x) ≠ 0.

Example 5 Using the reciprocal rule

Find the derivative of g(x) = images.

Solution Let f(x) = x2 + x + 1. Then g(x) = images and f′(x) = 2x + 1. By the reciprocal rule,

images

Example 6 Breaking Whelks

Crows feed on whelks by flying up and dropping the whelks (Figure 3.5) on a hard surface to break them.

images

Figure 3.5 Whelks (Nucella lamellosa, pictured here among other species of marine molluscs) are frequently consumed by Northwestern crows.

Biologists have noticed that Northwestern crows consistently drop whelks from about five meters. As a first step to understanding why this might be the case, we consider data collected by the Canadian scientist Reto Zach in which he repeatedly dropped whelks from various heights to determine how many drops were required to break the whelk. The data are shown in Figure 3.6.

images

Figure 3.6 Data showing how the number of drops to break a whelk depends on the height of the drops.

Data Source: Reto Zach, 1979. Shell Dropping: Decision-Making and Optimal Foraging in Northwestern Crows. Behaviour, 68, pp. 106–117.

A best-fitting curve relating the number of drops, D, to the height, h (in meters), for these data is given by

images

In Chapter 4, we shall use this function to determine the optimal height from which to drop whelks.

Solution

images

At h = 4 meters, the required number of drops decreases at a rate of −2.04 per meter. For instance, if we increased the height by approximately 1 meter, we should expect the number of drops to decrease by approximately 2. This can also be seen on the graph in Figure 3.6 from the fact that at h = 4, D ≈ 8, while at h = 5, D ≈ 8 − 2 = 6.

Combining the reciprocal and product rule, we can find the derivative of a quotient of functions. Let f and g be differentiable functions, and assume that g(x) ≠ 0.

images

Thus, we have derived what is known as the quotient rule.

Quotient Rule

Let f and g be differentiable at x. Then

images

provided g(x) ≠ 0.

There exist a variety of playful mnemonics that can be used to remember the quotient rule. For example, if we replace f by hi and g by lo, then we get the limmrick “lo-dee-hi less hi-dee-lo, draw the line and square below.”

Example 7 Computing with the quotient rule

Find the following derivatives.

images

Solution

  1. Let f(t) = 1 + 2t and g(t) = 3 + 4t. By the quotient rule

    images

  2. Let f(x) = ex and g(x) = 1 + x2. By the quotient rule

    images

Example 8 Dose–response curves revisited

In Example 2 of Section 2.4, a dose–response curve for patients responding to a dose of histamine was given by

images

where x is the natural logarithm of the dosage in millimoles (mmol).

  1. Find images.
  2. Graph images to determine at what logarithmic dosage the response is increasing most rapidly.

Solution

images

b. Graphing images yields

images

Hence, images takes on its largest value at approximately x = −5, and the response increases most rapidly at this logarithmic dose. That is, it increases most rapidly at the dose e−5 ≈ 0.0067 mmol.

PROBLEM SET 3.2

Level 1 DRILL PROBLEMS

Find the derivatives in Problems 1 to 18.

1. p(x) = (3x2 − 1)(7 + 2x3)

2. p(x) = (x2 + 4)(1 − 3x3)

3. q(x) = images

4. q(x) = images

5. f(x) = x2x

6. f(x) = x33x

7. f(x) = (1 + x + x2)ex

8. f(x) = (e3 + e2 + e)x2

9. F(L) = (1 + L + L3 + L4)(LL2)

10. G(M) = (MM3)(1 − 4M)

11. f(x) = (4x + 3)2 Hint: Think (4x + 3)(4x + 3)

12. g(x) = (5 − 2x)2

13. f(x) = images

14. g(t) = images

15. f(p) = images where a is a constant

16. g(m) = images where b is a constant

17. F(x) = images

18. G(x) = images

Find the equation for the tangent line at the prescribed point for each function in Problems 19 to 24.

19. f(x) = (x3 − 2x2)(x + 2) where x = 1

20. G(x) = (x − 5)(x3x) where x = − 1

21. F(x) = images where x = 0

22. f(x) = ex + ex where x = 0

23. F(x) = images

24. g(x) = x ln x where x = 1.

25. a. Differentiate the function

images

b. Factor the function in part a and differentiate by using the product rule. Show that the two answers are the same.

26. a. Use the quotient rule to differentiate

images

b. Rewrite the function in part a as f(x) = x−3(2x − 3) and differentiate by using the product rule.

c. Rewrite the function in part a as f(x) = 2x−2 − 3x−3 and differentiate using the power rule.

d. Show that the answers to parts a, b, and c are all the same.

Level 2 APPLIED AND THEORY PROBLEMS

27. The body mass index (BMI) for an individual who weighs w pounds and is h inches tall is given by

images

A person with a body mass index greater than 30 is considered obese; see the BMI chart.

  1. Consider all adults who are h = 63 inches tall. After setting h = 63 inches so that B is now a function of w only, find images at w = 130 and interpret your results.
  2. Consider all children who weigh 60 lb. After setting w = 60 lb so that B is now a function of h only, find images at h = 54 and interpret your results.

28. In addition to zebra, Professor Getz's group studied springbok (a type of antelope) in Etosha National Park. In a study paralleling Example 2 of this section, they found that the probability of a springbok surviving predation and disease to the end of the eighth week of a typical year is given by, respectively, 94.9% and 98.8%. Suppose they also determined that at the end of this eight weeks the probability of dying from predation and disease is decreasing respectively by 0.6% and 0.25% per week. Assume the events of dying from predation or disease are independent.

  1. Find the probability of surviving to the end of week 8.
  2. Find the rate of change of probability of surviving at the end of week 8.
  3. Estimate the probability of surviving to the end of week 9.

29. Suppose that Professor Getz's group found that the probability of zebra surviving both predation and disease to the end of week 4 was 98.4% and 99.6%, respectively. Suppose they also determined that at the end of week 4 the probability of a zebra dying from predation and disease is decreasing respectively by 0.4% and 0.1% per week (remember when converting percentages to numbers to shift the value by two decimal places). Assume the events of dying from predation or disease are independent.

  1. Find the probability of surviving up to the end of week 4.
  2. Find the rate of change of probability of surviving at the end of week 4.
  3. Estimate the probability of surviving to end of week 5.

30. Consider the generalized Beverton-Holt model of population growth given by

images

where

images

and a > 0, b > 0.

  1. Find g′(N).
  2. Determine what values of b > 0 cause g to be increasing for all N > 0.
  3. When b is outside the range of values found in part b, determine on what interval g is increasing and on what interval g is decreasing.

31. A ligand is a molecule that binds to another molecule or other chemically active structure (e.g., a receptor on a membrane) to form a larger complex. In a study of two ligands (I and II) competing for the same sites on a substrate, ligand II is added to a substrate solution that already contains ligand I. As the concentration of ligand II is increased, the concentration of ligand I bound to the substrate decreases. This one-site ligand competition process is characterized by this equation:

images

where T is the concentration of bound ligand I per milligram of tissue and x is the exponent of the concentration of ligand II in the solution. The constants a and b arise from the relative binding rates of the two ligands and satisfy a > 0 and b > a.

  1. Compute images
  2. Interpret the quantities a, b, and c and sketch the graph.

32. In the 1960s, scientists at Woods Hole Oceanographic Institution measured the uptake rate of glucose by bacterial populations from the coast of Peru (R. R. Vaccaro and H. W. Jannasch, “Variations in Uptake Kinetics for Glucose by Natural Populations in Sea Water,” Limnol. Oceanogr. (1967) 12, 540–542.). In one field experiment, they found that the uptake

images

micrograms per hour where x is micrograms of glucose per liter. Compute and interpret f′(20) and f′(100).

33. In Example 5 of Section 2.4, we found that the predation rate of wolves on moose in North America could be modeled by

images

where x is measured in number of moose per square kilometer. Compute and interpret f′(0.5) and f′(2.0).

34. Cells often use receptors to transport nutrients from outside of the cell membrane to the inner cell. In Example 6 of Section 1.6, we determined that the rate, R, at which nutrients enter the cell depends on the concentration, N, of nutrients outside the cell. The function

images

models the amount of nutrients absorbed in one hour where a and b are positive constants.

  1. Find R when N = b. What does this tell you about b?
  2. Compute and interpret images. When is R increasing? When is R decreasing?

35. In Problem 42 in Problem Set 2.4, we modeled how wolf densities in North America depend on moose densities with the following function

images

where x is number of moose per square kilometer. Determine for what x values f(x) is increasing.

36. In Problem 45 in Problem Set 2.3, two fisheries scientists (T. J. Quinn and R. B. Deriso, “Stock and Recruitment,” Chapter 3 (pp. 86–127) in Quantitative Fish Dynamics, 1999. Oxford University Press, New York, New York) found that the following stock-recruitment function provides a good fit to data pertaining to the southeast Alaska pink salmon fishery:

images

where y is the number of young fish recruited, and x is the number of adult fish involved in recruitment.

  1. Compute images.
  2. Determine for what x values y is increasing and decreasing. Interpret your results.

37. This problem uses the hyperbolic secant function, denoted sech x, which is an important function in mathematics and is defined by the formula

images

We will discuss further in Chapter 4 how two mathematicians, W. O. Kermack and A. G. McKendrick, showed that the weekly mortality rate during the outbreak of the plague in Bombay in 1905–1906 can be reasonably well described by the function

images

where t is measured in weeks. Determine when the mortality rate is increasing and when the mortality rate is decreasing.

38. In a classic paper, V. A. Tucker and K. Schmidt-Koenig modeled the energy expended by a species of Australian parakeet during flight (“Flight of Birds in Relation to Energetics and Wind Directions,” The Auk 88 (1971): 97–107). They used the following function:

images

where v is the bird's velocity (km/h).

  1. Find a formula for the rate of change of energy with respect to v.
  2. At what velocity, v, is the energy expenditure neither increasing nor decreasing? Discuss the importance of this velocity.

    images

    Australian budgerigar (Melopsittacus undulatus)

3.3 Chain Rule and Implicit Differentiation

We now move to the next level in terms of developing tools to differentiate functions that can be regarded as the composite of more elementary functions (discussed in Section 1.5). These tools will give us the power to differentiate functions such as the bell-shaped curve y = ex2, the polynomial y = (1 + 2x + x3)101, and the logarithm function y = ln x.

Chain rule

Suppose we were asked to find the derivative of the function y = (1 + 2x + x3)101. It is not practical to expand this product in order to take the derivative of a polynomial. Instead, we can use a result known as the chain rule. In order to motivate this important rule, let us consider an application.

It is known that the carbon monoxide pollution in the air is changing at the rate of 0.02 ppm (parts per million) for each person in a town whose population is growing at the rate of 1000 people per year. To find the rate at which the level of pollution is increasing with respect to time, we must compute the product

images

We can generalize this commonsense calculation by noting that the pollution level, L, is a function of the population size, P, which itself is a function of time, t. Thus, L as a function of time is (L images P)(t) or, equivalently, L[P(t)]. With this notation, the commonsense calculation becomes:

images

Expressing each of these rates in terms of an appropriate derivative of L[P(t)] in Leibniz form, we obtain the following equation:

images

These observations anticipate the following important result known as the chain rule.

Chain Rule

If y = f(u) is a differentiable function of u, and u, in turn, is a differentiable function of x, then y = f[u(x)] is a differentiable function of x, and its derivative is given by the product

images

Equivalently,

images

Proof. To prove the chain rule, define

images

It should be intuitive that G(h) is continuous at h = 0. (You will be asked to verify this statement in Problem Set 3.3). With this observation in hand, the proof of the chain rule becomes straightforward. By the definition of the derivative.

images

Example 1 Life made easier

Let y = images

Solution View this as the composition of two functions: the “inner” function u(x) = 1 + 2x + x3 and the “outer” function f(u) = u101. Now use the chain rule:

images

In practice, we usually do not write down a function u, but carry out the above process mentally and write

images

Example 2 Escaping parasitism

Parasitoids, usually wasps or flies, are insects whose young develop on and eventually kill their host, typically another insect. Parasitoids have been extremely successful in controlling insect pests, especially in agriculture. To better understand this success, theoreticians have extensively modeled host-parasitoid interactions. A key term in these models is the so-called escape function f(x), which describes the fraction of hosts that escape parasitism when the parasitoid density is x individuals per unit area. If parasitoid attacks are randomly distributed among the hosts, then the escape function is the form f(x) = eax where a is the searching efficiency of the parasitoid.

Suppose that a population of parasitoids attacks alfalfa aphids with a searching efficiency of a = 0.01. If the density of parasitoids is currently 100 wasps per acre and is increasing at a rate of 20 wasps per acre per day, find the rate at which the fraction of aphids escaping parasitism is initially changing.

Solution Let time in days be denoted by the independent variable t, starting with t = 0 at the current time. Since the density of wasps x(t) is changing with time, the fraction of hosts that escape is a composition of two functions f[x(t)]. Hence, by the chain rule

images

Evaluating at the current time t = 0 at which x(0) = 100, we get

images

Thus the fraction of hosts escaping is decreasing at a rate of 0.074 per day at t = 0.

In Example 2, we found the derivative of f(x) = e−0.01x by using the derivative of an exponential (Section 3.1), but we could also use the chain rule. It is worthwhile to restate an extended derivative rule for a natural exponential function:

images

We illustrate this idea with the following example.

Example 3 The bell-shaped curve

Consider the bell-shaped function

images

  1. Find f′ and determine where f is increasing and where it is decreasing.
  2. Plot f and f′ on the same coordinate axes and graphically verify the results from part a.

Solution

  1. Using the extended derivative rule for a natural exponential function with u = −x2, we find

    images

    Since ex2 > 0, the derivative is positive when x < 0, so the function f is increasing on (−∞, 0); and it is negative when x > 0, so the function is decreasing on (0, ∞).

    images

    Figure 3.7 Graph of bell-shaped function (red) and its derivative (blue).

  2. The graph of y = f(x) is shown in red and y = f′(x) in blue in Figure 3.7.

    We see that the derivative function (blue) is positive where the bell-shaped curve (red) is rising, and the derivative function is negative where the bell-shaped curve is falling.

Example 4 Chain rule with graphs

Consider the functions y = f(x) and y = g(x) whose graphs in red and blue, respectively, are as shown here:

images

Find

images

Solution By the chain rule,

images

By inspection, g(−0.5) = 2. To find the derivative of g at −0.5, we note that g is linear on the interval [−1, 0], so the derivative is the slope of the line segment which we find by rise/run = 2/1 = 2, thus, g′(−0.5) = 2.

To find the derivative of f at g(−0.5) = 2, we note that f (red curve) is linear on [0, 2] and has slope m = rise/run 8/2 = 4. Thus, f′[g(−0.5)] = f′(2) = 4.

We conclude that

images

Implicit differentiation

The equation y = images explicitly defines f(x) = images as a function of x for −5 ≤ x ≤ 5. The same function can also be defined implicitly by the equation x2 + y2 = 25, as long as we restrict y by 0 ≤ y ≤ 5 so the vertical line test is satisfied. To find the derivative of the explicit form, we use the chain rule:

images

To obtain the derivative of the same function in its implicit form, we simply differentiate across the equation x2 + y2 = 25, remembering that y is a function of x and using the chain rule:

images

The procedure we have just illustrated is called implicit differentiation.

Example 5 Circular tangents

Consider a circle of radius 5 centered at the origin. Find the equation of the tangent line of this circle at (3, 4).

Solution The equation of this circle is

images

We recognize that this circle is not the graph of a function. However, if we look at a small neighborhood around the point (3, 4), as shown in Figure 3.8, we see that this part of the graph does pass the vertical line test for functions. Thus, the required slope of the tangent line can be found by evaluating the derivative of dy/dx at (3, 4). We have found that

images

images

Figure 3.8 Tangent line (red) to a given circle.

so the slope of the tangent at (3, 4) is

images

Thus, the equation of the tangent line is

images

More generally, given any equation involving x and y, we can differentiate both sides of the equation, use the chain rule, and solve for images. This becomes particularly important when one cannot (or not easily) express y in terms of x explicitly. The next example is of this type and is included because of its historical importance and aesthetic appeal. You may think that such curves have no application to biology; but, for example, the logarithmic spiral (see Figure 1.56) encountered in Project 1C: Golden Ratio (at the end of Chapter 1) describes the growth of the nautilus mollusk.

Example 6 Limaç on of Pascal

The limaçon of Pascal is a famous curve defined by the set of points satisfying

images

The graph is shown in Figure 3.9. This curve was discovered by Étienne Pascal, the father of the more famous Blaise Pascal. The name limaçon comes from the Latin limax, which means “a snail.” Find the equation for the tangent line at the point (0, 1).

Solution To find the slope of the tangent line, we differentiate both sides implicitly and then evaluate at (0, 1).

images

Figure 3.9 Limaçon of Pascal.

images

Hence, the slope of the tangent line is 2 and the tangent line is

images

images

Figure 3.10 Limaçon with tangent line at (0, 1).

The tangent line is shown in Figure 3.10.

Derivatives of logarithms

Implicit differentiation allows us to easily find the derivative of logarithms and power functions.

Derivative of the Natural Logarithm

If y = ln x, then

images

images

In the problem set of this section you are asked to prove the more general statements of this result; namely, if y = ln |x| then, for x ≠ 0, images = images.

The next example deals with the clearance rate—which, as we saw in Example 7 of Section 3.1 in the context of viral particles, is the constant rate of decay—of a drug from the blood stream of an individual.

Example 7 Clearance of acetaminophen

Scientists have estimated that the clearance rate of the drug acetaminophen in the blood stream of an average adult is 0.28 per hour. This means that after an initial dose of acetaminophen at time t = 0, the fraction of acetaminophen in the blood t hours later is e−0.28t.

  1. Find the time, T, it takes for a fraction x of the drug to clear the body.
  2. Find and interpret images.

Solution

  1. Since e−0.28t is the fraction of drug remaining in the body at time t and x is the fraction that has left the body, we need to solve

    images

  2. We use the results of part a to find

    images

    Thus, the time it takes to clear an extra percentage of the drug, given 50% (x = images is 50%) has cleared, is approximately 7.14 × 0.01 = 0.0714 hours or 4 minutes and 17 seconds.

If the base on the logarithm has a base other than the natural base, e, then we use the following result, which you will be asked to verify in the Problem 41 of Problem Set 3.3:

Derivative of a General Logarithm

If b is a positive number (other than 1) and x > 0 then

images

Example 8 Derivative of a log with base 2

For x > 0 differentiate f(x) = x log2 x.

Solution

images

In Section 3.1, we stated the power rule and promised to prove it later in this chapter for all real numbers. We fulfill this promise in the following example.

Example 9 Power law for positive real numbers

Consider y = xn where x > 0 and n is any real number other than 0. Prove that

images

Solution We will prove this for x > 0 by taking the natural logarithm of both sides and then differentiating to find the derivative.

images

You will encounter the proof for x < 0 in the problem set at the end of this section.

Example 10 Modeling problem using the chain rule

Table 3.1 Population as a function of time

Time t Population p(t)
0 4.6696
0.25 4.6717
0.5 4.6779
0.75 4.6884
1 4.7032
1.25 4.7225
1.5 4.7463
1.75 4.7751
2 4.8088
2.25 4.8479
2.5 4.8926
2.75 4.9432
3 5.0000

An environmental study of a certain suburban community suggests that when the population is p thousand people, the amount of carbon monoxide in the air can be modeled by the function

images

where C is measured in parts per million. The population (in millions) at various times (in years) for the last three years is given in Table 3.1.

  1. Decide through visual inspection (plotting each of the two graphs together with the data) which of the following functions model the population data most accurately. (Note how subscripts we apply to y allow us to distinguish between the values predicted by the two models, although once the best model is selected, we will rename the left-hand side p.) Compute the derivatives for each of these models.
    • Linear: yl = 0.109t + 4.618
    • Quadratic: yq = 0.039t2 − 0.009t + 4.672
  2. Using the model you selected in part a, find the rate at which the level of pollution is changing at the end of year three.

You could just use the residual sum-of-squares as discussed in Section 1.2 to decide which is better. Also, see Project 3A at the end of this chapter.

Solution

  1. To determine which of the three proposed models best fits the population data, we plot the graphs shown in Figure 3.11. The rates of change for these models are calculated by finding these derivatives:
    • Linear: yl′ = 0.109 at t = 3, yl = 0.109 ppm
    • Quadratic: yq′ = 0.078t − 0.009 at t = 3, yq = 0.225 ppm

    The quadratic function clearly fits much better than the linear function and predictions differ notably at t = 3. Renaming the right-hand-side of the quadratic function p(t), the function we use in our analysis is the quadratic function, which after renaming the left-hand-side is

    images

    images

    Figure 3.11 Data fitted by the linear (left panel) and quadratic (right panel) functions in part a of Example 10.

  2. By substituting the quadratic population function p(t) selected in part a into the researcher's pollution function C(p), we can represent the level of pollution as C[p(t)], a composite function of time. Applying the chain rule, we find that

    images

    When t = 3, p(3) = 0.039(3)2 − 0.009(3) + 4.672 = 4.996. Therefore, at t = 3 it follows that

    images

Thus, our analysis suggests that after three years, the level of pollution is increasing at the rate of 0.104 parts per million per year.

PROBLEM SET 3.3

Level 1 DRILL PROBLEMS

Use the chain rule to compute the derivative dy/dx for the functions given in Problems 1 to 4.

1. y = u2 + 1; u = 3x − 2

2. y = 2u2u + 5; u = 1 − x2

3. y = images; u = x2 − 9

4. y = u2; u = ln x

Differentiate each function in Problems 5 to 8 with respect to the given variable of the function.

images

Differentiate each function in Problems 9 to 18.

images

Find dy/dx by implicit differentiation in Problems 19 to 25.

19. x2 + y = x3 + y3

20. xy = 25

21. xy(2x + 3y) = 2

images

23. (2x + 3y)2 = 10

24. ln(xy) = e2x

25. exy + ln y2 = x

26. Consider the functions y = f(x) and y = g(x) whose graphs in red and blue, respectively, are shown in Figure 3.12.

images

Figure 3.12 Functions y = f(x) (red) and y = g(x) (blue).

images

27. The graphs of u = g(x) and y = f(u) are shown in Figure 3.13.

images

Figure 3.13 Chain rule with graphs.

  1. Find the approximate value of u at x = 2. What is the slope of the tangent line at that point?
  2. Find the approximate value of y at x = 5. What is the slope of the tangent line at that point?
  3. Find the slope of y = f[g(x)] at x = 2.

28. Let g(x) = f[u(x)], where u(−3) = 5, u(−3) = 2, f(5) = 3, and f (5) = −3. Find an equation for the tangent to the graph of g at the point where x = −3.

29. Let f be a function for which

images

  1. If g(x) = f(3x − 1), what is g′(x)?
  2. If h(x) = fimages, what is h′(x)?

30. The cissoid of Diocles is a curve of the general form represented by the following particular equation

images

as illustrated in Figure 3.14.

images

Figure 3.14 Cissoid of Diocles.

Find the equation of the tangent line to this graph at (3, 3).

31. The folium of Descartes is a curve of the general form represented by the following particular equation

images

as illustrated in Figure 3.15.

images

Figure 3.15 Folium of Descartes.

Find the equation of the tangent line to this graph at (2, 1).

32. Another version of the folium of Descartes is given by the equation

images

as illustrated in Figure 3.16

images

Figure 3.16 Folium of Descartes.

Find at what points the tangent line is horizontal.

Level 2 APPLIED AND THEORY PROBLEMS

33. Arteriosclerosis develops when plaque forms in the arterial walls, blocking the flow of blood; this, in turn, often leads to heart attack or stroke (Figure 3.17). Model the cross-section of an artery as a circle with radius R centimeters, and assume that plaque is deposited in such a way that when the patient is t years old, the plaque is p(t) cm thick, where

images

Find the rate at which the cross-sectional area covered by plaque is changing with respect to time in a sixty-year-old patient.

images

Figure 3.17 Blood flow in a healthy artery (top) and in an artery narrowed by arteriosclerosis (bottom).

34. In Problem 38 in Problem Set 3.2, the energy expended by a species of Australian parakeet during flight was modeled by the function

images

where v is the bird's velocity (km/h). Assume at one point in time (say t = 0) in its flight, a parakeet's velocity is 25 km/hour and its velocity is increasing at a rate of 2 km/hour2. Find the instantaneous rate of change of this parakeet's energy use at t = 0. Be sure to identify the correct units for your answer.

35. In Example 5 of Section 2.4, we found that the predation rate of wolves in North America could be modeled by

images

where x is measured in number of moose per square kilometer. If the current moose density is x = 0.5 and is increasing at a rate of 0.1 per year, determine the rate at which the predation rate is increasing.

36. In Problem 42 in Problem Set 2.4, we modeled how wolf densities in North America depend on moose densities with the following function

images

where x is the number of moose per square kilometer. If the current moose density is x = 2.0 and decreasing at a rate of 0.2 per year, determine the current rate of change of the wolf densities.

37. The proportion of a species of aphid that escapes parasitism is

images

where x is the density of parasitoids. If the density of parasitoids is currently 10 wasps per acre and decreasing at a rate 20 wasps per acre per day, find at what rate the likelihood of escaping parasitism is changing.

38. In Example 10 we saw that an environmental study of a certain suburban community suggested the following relationship between the population level p (thousand people) and the amount of carbon monoxide C (ppm) in the air:

images

Now suppose that the population p (in thousands) at time t is modeled by the function

images

Use the chain rule to find the rate at which the pollution level is changing after three years and compare this with the estimate obtained in Example 10.

39. The bicorn (also called the cocked hat) is a quartic curve studied by mathematician James Sylvester (1814–1897) in 1864. As illustrated in Figure 3.18, it is given by the set of points that satisfy the equation

images

Find the formulas for the two tangent lines at x = 1/2.

images

Figure 3.18 Bicorn curve.

40. Prove that if f is differentiable at u(a) and u is continuous at a, then

images

is continuous at h = 0.

41. Prove that images

42. Prove that

images

for y = xn where x < 0 and n is any non-zero number for which y is defined.

43. The average height of boys in the United States as a function of age is shown in Figure 3.19. This relationship is well described by the linear function

images

images

Figure 3.19 Heights in inches of boys and girls in the United States as a function of time in months.

where t is measured in months. The relationship between height and weight can be modeled by the power function

images

Find the instantaneous rates at which height and weight are changing at age ten years.

44. The average height of girls in the United States as a function of age is shown in Figure 3.19. This relationship is well-described by the linear function

images

where t is measured in months. The relationship between height and weight can be modeled by the power function

images

Find the instantaneous rates at which height and weight are changing at age ten years.

3.4 Derivatives of Trigonometric Functions

Many physical and biological processes change periodically over time and consequently are represented by a periodic function. A powerful result in mathematical analysis proves that periodic functions can be represented as a sum of sines and cosines; this is called the Fourier series representation. In this section, we find the derivative of these fundamental functions and their functional relatives—tangent, secant, cotangent, and cosecant. Recall from Chapter 1 that we assume the trigonometric functions are functions of real numbers or of angles measured in radians. We make this assumption because the trigonometric differentiation formulas rely on limit formulas that become more complicated if degrees, rather than radians, are used to measure angles.

Derivatives of sine and cosine

Before stating the theorem regarding the derivatives of the sine and cosine functions, we look at the graph of images for small values of h. In particular, for h = 0.01, the expression images has the following graph:

images

From this graph, it appears that the derivative of f(x) = sin x is f′(x) = cosx. This relationship appears to match up, as illustrated in Figure 3.20: sine is increasing on intervals where cosine is positive and has turning points where cosine is zero.

images

Figure 3.20 Graphs of cosine (blue) and sine (red).

Before verifying this assertion, we need two important limits. The first limit is presented numerically in Example 1. You will be asked to give analytical derivations of the limits in Problems 29 and 30 in Problem Set 3.4.

Two Important Trigonometric Limits

images

Example 1 Numerical approach to one of the trigonometric limits

Find images numerically and graphically using technology.

Solution Note that

images

so the left- and right-hand limits should be the same. Thus, for the numerical approach, we develop a table for x values approaching 0 from the right.

images

The numbers in this table appear to be approaching 1 as x tends toward 0 from the right (x > 0). Thus, we might infer from the table that

images

Figure 3.21 Graph of y = images near the point x = 0

images

Plotting sin x/x near 0 in Figure 3.21 reaffirms our tabular approach to finding the limit.

We can now state the derivative rule for the sine and cosine functions.

Derivative Rules for Sine and Cosine

The functions sin x and cos x are differentiable for all x and

images

Proof. We will prove the first derivative formula using the trigonometric identity

images

and the definition of derivative. For a fixed x

images

To find the derivative of cosine, we use the trigonometric identities

images

and the chain rule

images

Example 2 Derivatives involving sine and cosine functions

Differentiate the functions

images

Solution

  1. Setting u = 2x and y = f(u) = sin u, the chain rule implies that

    images

  2. By the product rule,

    images

images

Example 3 Rate of change of CO2

In Section 1.2, we initially approximated the concentration of CO2 (in ppm) at the Mauna Loa Observatory in Hawaii with the function

images

Find h′(3) and compare it to the approximation found in Example 2 of Section 2.1.

Solution

images

Evaluating at t = 3 yields

images

This agrees with the numerical solution of Example 2 in Section 2.1.

Periodic fluctuations in biological systems can be internally or externally driven, as the next two examples illustrate.

Example 4 Circadian rhythms

Circadian rhythms are roughly twenty-four-hour cycles in physiological processes of an organism that are primarily driven by the individual's circadian clock. Examples in mammals include heart and breathing patterns and daily fluctuations in body temperature and hormones. In humans, the circadian clock driving these periodic patterns is formed by a cluster of neurons found in the hypothalamus, a part of the brain. However, many brainless organisms, ranging from yeast cells to plants, also exhibit circadian rhythms.

A simple model of circadian rhythm for body temperature in an organism is

images

where t is measured in hours. Assume that 0 ≤ C ≤ 6.

  1. Sketch T(t) and discuss the meaning of A, B, and C.
  2. Find and sketch T′(t). Discuss what this plot tells you about the circadian rhythm of body temperature.

Solution

  1. Using what we studied about trigonometric functions in Chapter 1, we have that T(t) oscillates around the value A with an amplitude of B and a period of 24 hours. The −C term shifts the graph to the right by C hours. Therefore, we get the following plot of T(t):

    images

    Therefore, the body temperature of A° Celsius occurs at hours C and C + 12 each day and the maximum temperature A + B° Celsius occurs at hour C + 6 each day.

  2. Taking the derivative of T yields

    images

    Plotting the derivative using what we learned in Chapter 1 gives

    images

    The maximum instantaneous rate of change of body temperature occurs at hour C each day, while the minimum rate of change occurs at hour C + 12 each day. Since T′(t) > 0 on [0, C + 6) and T(t) < 0 on (C + 6, C + 18), body temperature is increasing between hours 0 and C + 6 and decreasing between hours C + 6 and C + 18.

Example 5 Periodic populations

Many populations live in environments that change in a periodic fashion with time (e.g., diurnal and seasonal cycles). To understand the dynamics of an algal population growing in a climate chamber programmed to have a particular light-dark cycle, a research scientist conducted an experiment in which she found that the algae abundance (in cells per liter) over time t in hours could be modeled by the function

images

  1. Verify that N(t) satisfies the relationship

    images

    Explain what this relationship means. Based on the assumption that the growth rate of the population is proportional to the intensity of light, at what times are the light intensity greatest?

  2. Determine at what times the population is increasing and at what times the population is decreasing.

Solution

  1. Taking the derivative of N(t) with the chain rule, we get

    images

    Hence, by the definition of N(t), we have

    images

    We can interpret cos t as the per capita growth rate of the population. This per capita growth is greatest (i.e., equals one) at t = 0, ±2π, ±4π,... Hence, at these moments of time the light intensity must be the greatest.

  2. The population increases when N′(t) > 0. This occurs when cos t > 0, in other words, when t is in the intervals (0, π/2), (3π/2, 5π/2),.... On the complementary intervals, the population is decreasing; that is, on these intervals, those algal cells that are dying more quickly than dividing.

Derivatives of other trigonometric functions

If you know the derivatives of sine and cosine and the basic rules of differentiation, then all the other trigonometric derivatives follow.

Derivative Rules for Trigonometric Functions

The six basic trigonometric functions sin x, cos x, tan x, csc x, sec x, and cot x are all differentiable wherever they are defined and

images

All the additional derivative rules are proved by using the quotient rule along with formulas for the derivative of sine and cosine. Here we will obtain the derivative of the tangent function and leave the rest to Problem Set 3.4.

images

Notice that the derivatives of all “co” trig functions except for cosine have the “co-trig” derivative form of their corresponding trigonometric partners, but with a sign change. Thus, for example, because the derivative of tangent is secant squared, this rule implies that the derivative of cotangent is the negative of cosecant squared.

Example 6 Derivative of a product of trigonometric functions

Differentiate f(x) = sec x tan x.

Solution

images

PROBLEM SET 3.4

Level 1 DRILL PROBLEMS

Differentiate the functions given in Problems 1 to 20.

1. f(x) = sin x + cos x

2. g(x) = 2 sin x + tan x

3. y = sin 2x

4. y = cos 2x

5. f(t) = t2 + cos t + cos images

6. g(t) = 2 sec t + 3 tan t − tan images

7. y = ex sin x

8. y = tan x2

9. f(θ) = sin2 θ

10. g(θ) = cos2 θ

11. y = cos x101

12. y = (cos x)101

13. p(t) = (t2 + 2) sin t

14. y = x sec x

images

18. y = sin(2t3 + 1)

19. y = ln(sin x + cos x)

20. y = ln(sec x + tan x)

Use the given trigonometric identity in parentheses and the basic rules of differentiation to find the derivatives of the functions given in Problems 21 to 24.

images

Level 2 APPLIED AND THEORY PROBLEMS

29. Consider three areas as shown in Figure 3.22.

images

Figure 3.22 Triangles and a unit circle with subtended angle h.

  1. What is the area g(h) of the blue-shaded triangle?
  2. What is the area f(h) of the pink-shaded sector? Hint: The area of a sector of a circle of radius r and central angle θ measured in radians is images.
  3. What is the area k(h) of the green-shaded triangle?
  4. Use the fact that g(h) ≤ f(h) ≤ k(h) for small h > 0 to prove that

    images

    by beginning with the inequality

    images

30. Prove

images

Hint: Multiply by 1 written as images and use a fundamental trigonometric identity.

31. A researcher studying a certain species of fish in a northern lake models the population after t months of the study by the function

images

At what rate is the population changing after two months? Is the population growing or declining at this time?

32. In an experiment with algae, a research scientist manipulated the light in growth chambers so that

images

described the population density (in cells per liter) as a function of t (in hours).

  1. Find a function r(t) such that

    images

  2. Under the assumption that the growth rate of the population is proportional to the intensity of light, what is the period of the light fluctuations in the chamber?
  3. Determine at what times (if any) the population is decreasing in abundance.

33. In a more ambitious experiment with algae, a research scientist manipulated the light in algal tanks so that

images

describes the population density (in cells per liter) as a function of t (in hours).

  1. Find a function r(t) such that

    images

  2. Under the assumption that the growth rate of the population is proportional to the intensity of light, what is the period of the light fluctuations in the tank?
  3. Determine at what times (if any) the population is decreasing in abundance.

34. In Example 4 of Section 1.5, we modeled the tides for Toms Cove in Assateague Beach, Virginia, on August 19, 2004 with the function

images

where H is the height of the tide (in feet) and t is the time (in hours after midnight). Find and interpret images.

35. The temperature in a swimming pool is given by

images

where t is measured in hours since midnight.

  1. Find T′(12) and interpret.
  2. During what hours is the temperature of the pool increasing?

36. The human heart goes through cycles of contraction and relaxation (called systoles). During these cycles, blood pressure goes up and down repeatedly; as the heart contracts, pressure rises, and as the heart relaxes (for a split second), the pressure drops. Consider the following approximate function for the blood pressure of a patient:

images

where t is measured in minutes. Find and interpret P′(t).

3.5 Linear Approximation

We have seen that the tangent line is the line that just touches a curve at a single point. In this section, we will discover that the tangent line can be used to provide a reasonable approximation to a curve. Using these linear approximations we will be able to make projections about the size of a bison population (Figure 3.23) estimate images, and estimate the effects of measurement error.

images

Figure 3.23 The North American plain's bison (Bison bison) is one of two subspecies of bison that roamed the Great Plains. Their numbers have dropped from tens of millions to thousands over the last 200 years.

Approximating with the tangent line

We begin with an example that illustrates how well a tangent line can approximate a curve.

Example 1 Zooming in at a point

Consider the function y = ln x.

  1. Find the tangent line at x = 1.
  2. Graph y = ln x and the tangent line over the intervals [0.1, 2], [0.5, 1.5], [0.9, 1.1]. Discuss what you find.

Solution

  1. Since

    images

    we get that the tangent line is the line of slope 1 through the point (1, 0)—that is, the equation

    images

    as we claimed in Section 2.1.

  2. The graphs y = ln x (in blue) and y = x − 1 (in red) on the intervals [0.1, 2], [0.5, 1.5], and [0.9, 1.1] are shown in Figure 3.24. This figure illustrates that as we zoom into the point (1, 0), the tangent line provides a better and better approximation of our original function.

    images

    Figure 3.24 Zooming in on the graphs of y = ln x and y = x − 1 about the point (1, 0)

The difference between a tangent line and the associated curve becomes more and more negligible as you zoom into the point of contact. Thus, it seems quite reasonable to approximate the function with the tangent line in a neighborhood of the point at which this tangent is constructed. This is called the linearization of a function.

Linear Approximation

If f is differentiable at x = a, then the linear approximation of f around a is given by

images

for x near a.

With linear approximations, we can make predictions about the future, as illustrated in the next example. We use data on the abundance of the North American bison in Yellowstone National Park going back as early as 1902. (Estimates of bison population levels in Yellowstone from 1902–1931 can be found at http://www.seattlecentral.edu/qelp/Data.html.) Annual abundances for bison in Yellowstone for the twenty-nine-year period 1902 to 1931 are shown in Figure 3.25. These data suggest that the bison population was recovering in the first part of the twentieth century after years of intense hunting in the nineteenth century.

images

Figure 3.25 Bison abundance in Yellowstone National Park.

In the next example, we use the data only from 1908 to 1915. However, a project involving all the data is outlined at the end of the chapter.

Example 2 Predicting bison abundance

Suppose it is January 1910, and you are the Yellowstone National Park manager.

  1. Use a linear function to extrapolate what the abundance of bison might be in 1910 through 1915, given that you know the bison abundance in 1908 and 1909 are 95 and 118, respectively.
  2. Compare your estimates to the actual population size in 1910 to 1915 as given in the last column in Table 3.2.

Table 3.2 Estimate and measured abundance of Bison in Yellowstone National Park, 1910–1915

images

Solution

  1. Let N(t) denote the number of bison in year t, and for the sake of simplicity, set t = 0 at the year 1900. If we approximate N(t) at t = 8 by a linear function we get

    images

    To approximate N′(8), we can use

    images

    Hence,

    images

    for t “near” 8. Hence, our approximation yields the predictions shown in Table 3.2, with actual abundance in the last column.

  2. As we can see in Figure 3.25, our estimates are pretty good; nevertheless, they underestimate the actual population size more and more as time moves on. This is consistent with our expectation that population growth might be exponential.

Using linear approximation, we can estimate the value of a function at points near known values of the function. As we will see, however, linear approximations of nonlinear functions generally get increasingly worse as we move away from the point of approximation.

Example 3 Approximating images

Consider the function f(x) = images.

  1. Find the linear approximation of f at x = 9.
  2. Use the linear approximation found in part a to approximate images. Compare this approximation to a calculator approximation.
  3. How well does this same approximation work for images?

Solution

  1. f′(x) = images and the linear approximation for images at x = 9 is

    images

    for x near 9.

  2. If we now apply the approximation in part a to find images, we obtain

    images

    This is fairly close to the calculator approximation

    images

    So the error is 0.004 (to three decimal places)

  3. Similarly,

    images

    Since we know the answer is 4, the approximation now has an error of more than 1.1.

The next example shows that the linear approximation of sin x in the neighborhood of 0 is very simple, but it fails badly as the approximation is pushed too far beyond 0.

Example 4 Approximating sin x

Consider y = sin x

  1. Find the linear approximation of sin x at x = 0.
  2. Plot the difference between y = sin x and its linear approximation on the intervals [−1, 1], [− 0.5, 0.5], and [−0.1, 0.1]. Discuss the meaning of these plots.
  3. Approximate sin 2, sin 1, and sin 0.25 with the linear approximation from part a. Compare your approximations to calculator approximations.

Solution

  1. Since images = cos 0 = 1, we get the linear approximation at 0:

    images

  2. The graphs of sin xx on the intervals [−1, 1], [−0.5, 0.5] and [−0.1, 0.1] are illustrated in Figure 3.26. These figures illustrate that the difference between sin x and x gets smaller and smaller as you zoom around the point x = 0. Hence, y = x is a better and better approximation for sin x as x approaches 0.

    images

    Figure 3.26 Graphs of sin xx zooming onto the point (0, 0)

images

Development of organisms from immature forms, like eggs or seeds, to adult forms depends on a series of biochemical reactions. These biochemical reactions are often temperature sensitive, breaking down at too low or too high temperatures. For plants and insects whose internal temperatures are largely determined by the temperature of their environment, it follows that developmental rates depend on the temperature of the environment. This temperature dependence can be measured experimentally, as the following example illustrates.

Example 5 Developmental rates of the spider mite destroyer

Roy and colleagues estimated how developmental rates of a beetle (Stethorus punctillum) varied with temperature. They modeled the rate at which the fourth stage of development is completed with the following function

images

where T is measured in degrees Celsius. This model and the data used to parameterize the model are shown in Figure 3.27. This figure suggests that the developmental rate is approximately linear in T for temperatures near 20°C.

images

Figure 3.27 Developmental rate (percent completed per day) for a stage of development of the beetle Stethorus punctillum (also known as spider mite destroyer).

Data Source: Michèle Roy, Jacques Brodeur, and Conrad Cloutier. “Relationship between temperature and developmental rate of Stethorus punctillum (Coleoptera: Coccinellidae) and its prey Tetranychus mcdanieli (Acari: Tetranychidae),” Environmental Entomology 31 (2002), 177–187.

  1. Find a linear approximation to D(T) near the value T = 20.
  2. Plot the linear approximation found in part a. How does this compare to the data?

Solution

  1. To find the linear approximation at T = 20, we need to compute D′(20). Using the product rule and the chain rule, we get

    images

    Evaluating this expression at T = 20 yields

    images

    Using the point slope formula for a line, we get the linear approximation

    images

    Equivalently,

    images

  2. Plotting the linear approximation (red line) from part a against D(T) and the data yields

    images

    The linear approximation works quite well, and it even appears to provide a better estimate for the lower developmental threshold where the developmental rate equals zero at T = 12.

Error analysis

When a scientist makes a measurement, it is always subject to some measurement error. Hence, we have

measurement = actual value + measurement error

For instance, when the clearance rate of acetaminophen is given as 0.28 per hour, this estimate is the average of a series of measurements, each of which may vary by 0.05 or so. Consequently, when we estimate the half-life of acetaminophen, or the amount in the blood stream several hours after taking the drug, it is important to understand how small variations in the estimate 0.28 influence half-life estimates.

Example 6 Professor Getz's headache

Professor Getz takes 1000 mg of acetaminophen to combat a headache.

  1. Solve for the half-life T of the drug as a function of the clearance rate x per hour.
  2. Determine the half-life of the acetaminophen, assuming that x = 0.28 is a good estimate of the clearance rate.
  3. What is the derivative of the half-life T with respect to x and its value at x = 0.28?
  4. Use linear approximation to estimate the change ΔT in the estimated half-life if the estimate x = 0.28 is off by Δx. Interpret this result.

Solution

  1. Let A(t) denote the amount of acetaminophen in the body at time t hours. Since the clearance rate is x, we have

    images

    The half-life is the time t = T such that

    images

    Hence, the half-life as a function of x is

    images

  2. Evaluating T at x = 0.28 yields T(0.28) = 2.47553 hours.
  3. Differentiating the half-life function derived in part a yields

    images

    from which we can calculate T′(0.28) = −8.84116.

  4. If x = 0.28 + Δx where Δx can be viewed as a small measurement error, then by linear approximation we have

    images

    Thus,

    images

    Hence, for a measurement error of Δx per hour, half-life changes by approximately −8.84116 Δx hours. For instance, if the measurement error in the clearance rate is Δx = 0.05, then our estimate of the half-life decreases approximately by 8.84116 · 0.05 = 0.4421 hours. Thus the estimate of the half-life, T, is quite sensitive to the estimate of the clearance rate.

Example 6 illustrates how an error in the measurement of the independent variable x propagates to an error in the dependent variable y—a process called error propagation.

Error Estimates and Sensitivity

Suppose y = f(x) is a quantity of interest and x = a is the true value of x. If there is an error of Δx in measuring x = a, then the approximate resulting error in y is given by

images

f′(a) is called the sensitivity of y to x at x = a. The greater the sensitivity the greater the propagation of error.

Consider another example.

Example 7 Estimating metabolic rates

In Example 12 of Section 1.6 (below we changed the names of the variables from x and y to M and R), we discovered that the mouse-to-elephant curve describing how the metabolic rate R (in kilocalories/day) depends on body mass M (in kilograms) is approximately given by

images

Thus, taking exponentials yields the equation

images

  1. Estimate the metabolic rate of a California condor weighing 10 kg.
  2. Determine the sensitivity of your estimate to the measurement of 10 kg. Discuss how a small error ΔM propagates to an error ΔR in your estimate for R.

Solution

  1. For the 10 kg condor, we get R = e4.2 × 100.75 ≈ 375 kilocalories/day.
  2. The sensitivity of this estimate to our estimate for the condor weight is

    images

    Hence, ΔR ≈ 28.13 ΔM. For example, an error of ΔM = 0.1 kg yields an error of ΔR = 2.81 kilocalories/day in estimating R.

Elasticity

Often scientists are more interested in the percent error and not the absolute error. For example, a scientist may ask this question: How does a 10% error in the measurement of the clearance rate result in a percentage error in the estimate of the half-life? If x = a is the true value of the independent variable and there is a measurement error of Δx, then the percent error in x is

images

With an error of Δx in the independent variable, we get an error of Δy = f(a + Δx) − f(a) in y. Hence, the percent error in y is

images

The ratio of the percentage error in y over the percentage error in x is given by

images

and can be approximated by

images

This quantity is used quite commonly in the analysis of biological models and, consequently, has a special name: elasticity.

Elasticity

Let y = f(x) be a function that is differentiable at x = a. The elasticity of f with respect to x at a is

images

We can interpret E as stating that for a 1% error in the measurement of x = a, there is a E% error in the measurement of y.

Example 8 Elasticity of metabolic rates

Let us revisit Example 7 where we estimated the metabolic rate of a California condor weighing 10 kg.

  1. Find the elasticity of your estimate of the metabolic rate to the estimate of the condor's weight.
  2. Interpret your elasticity in terms of 10% error in the estimate of the condor's weight.

Solution

  1. To compute the elasticity, recall we found that R(10) ≈ 375 kilocalories/day and R′(10) ≈ 28.13. Hence, the elasticity at M = 10 is

    images

  2. Since the elasticity is 0.75, a 10% measurement error in the weight of the condor would result in approximately a 7.5% error in the estimate of the metabolic rate.

Using elasticity, we can estimate with what accuracy we need to measure an independent variable to ensure a certain accuracy in the estimate of a dependent variable.

Example 9 Determining measurement accuracy

The body mass index (BMI) for individual weighing w pounds and h inches tall is given by

images

  1. Determine the elasticity of B with respect to the variable h for a particular weight class w (i.e., we regard w as constant, so that B is a function of the variable h alone).
  2. Estimate how accurate your height measurement needs to be to guarantee less than a ±5% error in your BMI measurement.

Solution

  1. To compute the elasticity, we first need the derivative

    images

    Hence, the elasticity is

    images

    Note that this answer does not depend on w or h but is a pure number! Think about why this is the case.

  2. Since the elasticity is −2, an x% error in h results in a −2x% error in our estimate for BMI. To ensure that our error is no greater than ±5%, we need to ensure that the error in the measurement of the height is no greater than ±2.5%.

PROBLEM SET 3.5

Level 1 DRILL PROBLEMS

In Problems 1 to 6 find the linear approximation of y = f(x) at the specified point, and use technology to graph the function and its linear approximation to determine whether the linear approximation tends to overestimate or underestimate y = f(x) near the specified point.

images

In Problems 7 to 12, estimate the indicated quantity using a linear approximation around the given point x and compare to the true value obtained using technology.

images

Find the sensitivity of y = f(x) at the point specified in Problems 13 to 18, and use it to estimate Δy for the given measurement error Δx.

images

Find the elasticity of y = f(x) at the point specified in Problems 19 to 24 and use it to estimate the percent error in y for the given percent error in x.

images

images

Level 2 APPLIED AND THEORY PROBLEMS

25. In Example 5, we saw that Roy and colleagues estimated how developmental rates of a beetle (Stethorus punctillum) varied with temperature. They modeled the rate at which egg development is completed with the following function

images

where T is measured in degrees Celsius. This model and the data used to parameterize the model are shown in Figure 3.28. This figure suggests that the developmental rate is approximately linear in T for temperatures near 20°C.

images

Figure 3.28 Developmental rate (percent development completed per day) for egg development of the beetle Stethorus punctillum(spider mite destroyer).

  1. Find a linear approximation to D(T) near the value T = 20.
  2. Sketch the linear approximation on top of the graph in Figure 3.28. How does this compare to the data?

26. In Example 5, we saw that Roy and colleagues estimated how developmental rates of a beetle (Stethorus punctillum) varied with temperature. They modeled the rate at which the first stage of larval development is completed with the following function

images

where T is measured in degrees Celsius. This model and the data used to parameterize the model are shown in Figure 3.29. This figure suggests that the developmental rate is approximately linear in T for temperatures near 20°C.

images

Figure 3.29 Developmental rate (percent completed per day) for the first stage of larval development for the beetle Stethorus punctillum (spider mite destroyer).

  1. Find a linear approximation to D(T) near the value T = 20.
  2. Sketch the linear approximation on top of the graph in Figure 3.29. How does this compare to the data?

27. If your measurement of the radius of a circle is accurate to within 3%, approximately how accurate (to the nearest percentage) is your calculation to the area A when the radius is r = 12 cm? (Recall the formula A = πr2).

28. Suppose a 12-ounce can of Coke® (i.e., Coca-Cola) has a height of 4.75 inches. If your measurement of the radius has an accuracy to within 1%, how accurate is your measurement for the volume? Check your answer by examining a Coke can.

29. An environmental study suggests that t years from now, the average level of carbon monoxide in the air will be

images

parts per million (ppm). By approximately how much will the carbon monoxide level change during the next six months?

30. A certain cell is modeled as a sphere. If the formulas S = 4πr2 and V = imagesπr3 are used to compute the surface area and volume of the sphere, respectively, estimate the effect of S and V produced by a 1% increase in the radius r.

31. In a model developed by John Helms (“Environmental Control of Net Photosynthesis in Naturally Grown Pinus Ponderosa Nets,” Ecology (Winter 1972, p. 92), the water evaporation E(T) for a ponderosa pine is modeled by

images

where T (degrees Celsius) is the surrounding air temperature.

  1. Compute the elasticity of E(T) at T = 30.
  2. If the temperature is increased by 5% from 30°C, estimate the corresponding percentage change in E(T).

32. In a healthy person of height x inches, the average pulse rate in beats per minute is modeled by the formula

images

  1. Compute the sensitivity of P at x = 60.
  2. Use your answer to part a to estimate the change in pulse rate that corresponds to a height change from 59 to 60 inches.
  3. Compute the elasticity of P. Does it depend on x?
  4. Determine how accurate the measurement of x needs to be to ensure the estimate for P has an error of less than 10%.

33. In Example 6, we showed that the half-life, T, of a drug with clearance rate x is given by

images

Suppose that the true value of the clearance rate of some drug is given by x = a.

  1. Find the elasticity of T with respect to x.
  2. If you want to estimate the half-life of this drug within an error of 2%, how accurately do you have to measure the clearance rate of the drug?

34. A drug is injected into a patient's blood stream. The concentration of the drug in the blood stream t hours after the drug is injected is modeled by the formula

images

where C is measured in milligrams per cubic centimeter.

  1. Compute the sensitivity of C at t = 30.
  2. Use your answer to part a to estimate the change in concentration over the time period from 30 to 35 minutes after injection.

35. According to Poiseuille's law, the speed of blood flowing along the central axis of an artery of radius R is modeled by the formula

images

where c is a constant. What percentage error (rounded to the nearest percent) will you make in the calculation of S(R) from this formula if you make a 1% error in the measurement of R?

36. The gross U.S. federal debt (in trillions of dollars) from 2000 to 2004 is given in the following table

Year Gross federal debt
2000 5.629
2001 5.770
2002 6.198
2003 6.760
2004 7.355

Data source: From the historical tables of the Office of Management and Budget, 2006, as downloaded from www.whitehouse.gov/omb/budget/Historicals.

  1. Plot the data and the linear approximation of the data at t = 0 (2000). Discuss the quality of this approximation.
  2. Use a linear approximation to estimate the federal debt in 2010. Look up the actual gross federal debt to see how well the approximation worked.

37. In Example 7 of Section 2.6, the function

images

was used to fit data on the extent of the Arctic sea ice S(million square kilometers) as a function of years t since 1980 (corresponding to t = 0).

  1. What is the sensitivity of S in 1980?
  2. How has the sensitivity changed from 1980 to 2000? Does this make sense? If not, why not?

38. Consider a power function f(x) = axb with a > 0 and b ≠ 0. Show that the elasticity of f(x) is independent of the value x. What does it depend on? Use this answer to quickly solve the following problems.

  1. If there is a 5% error in estimating the mass M of a weightlifter, what is approximately the percent error in estimating the lift L = 20.15M2/3kg of the weightlifter?
  2. If there is a 10% error in estimating the mass M of an organism, then what is approximately the percent error in estimating the metabolic rate RM3/4kcal/hour of the organism?
  3. If there is a 2% error in measuring the weight W of a person, what is approximately the percent error in estimating the body mass index BW of the person?

39. In Problem 31 in Problem Set 3.2, we used the following function (here we replace c with the constant k = ec−10 and x with ln y)

images

to model T—the concentration of bound ligand I per milligram of tissue—in terms of y representing the concentration of a second ligand II in the solution.

  1. What is the elasticity of T with respect to changes in y?
  2. If there is a 10% error in estimating the concentration y of ligand II, then what is the error in calculating the level of ligand I as a function of y for the case a = 1, b = 2, and k = 1.

40. In Example 2 of Section 2.4, we used the function

images

to represent the percent of patients exhibiting an above normal temperature response to dose x millimoles (mM) of a particular histamine.

  1. What is the elasticity of R with respect to changes in x?
  2. If there is a 5% error in estimating the dose x when x = −5 (mM), then what is the percent error in calculating the response R?

3.6 Higher Derivatives and Approximations

The derivative of a function can be interpreted as the instantaneous rate of change, and it yields linear approximations to the function. Since the derivative of a function is also a function, this latter function also has a derivative. What does this derivative of a derivative represent? How useful is it? The goal of this section is to answer these questions—and considerably more.

Second derivatives

The second derivative of a function f is the derivative of f′ and is denoted f″. In other words,

images

Equivalently, we write

images

or if y = f(x),

images

Note that images is regarded as the symbol for the “operation of taking the second derivative of a function with respect to its argument x.”

Example 1 Finding second derivatives

Find f″(x) for the given functions.

images

Solution

images

What do these second derivatives represent? Consider the following definition.

Concave Up/Concave Down

If the graph of a function f lies above all its tangents on an interval I, then it is said to be concave up on I. If the graph of f lies below all of its tangents on I, it is said to be concave down.

images

Since f″ is the derivative of f′, the mean value theorem implies that if f″ > 0 on an interval, then f′ is increasing on this interval. What does this mean? In terms of tangent lines, this means that the slope of the tangent line is increasing in the interval. Hence, f is “bending upward” or, equivalently, is concave up on this interval. Alternatively, if f″ < 0, then the slope of the tangent line is decreasing and f is “bending downward” or, equivalently, is concave down.

Concavity

Let f be a function whose first and second derivatives are defined at x = a. If f″(a) < 0, then y = f(x) is concave down near x = a. If f″(a) > 0, then y = f(x) is concave up near x = a.

Example 2 Identifying concavities

Identify the concavities of the function defined by the given graphs. In other words, determine where the graphs are concave up and where they are concave down.

images

Solution

  1. The easiest way to proceed is to place a straight edge (e.g., ruler, pencil) on the graph; keeping it tangent to the curve, move from left to right. Whenever the straight edge is rotating in a counterclockwise fashion, the slope of the tangent line is increasing and f″ > 0. Hence, the function is concave up for these values of x. Alternatively, whenever the straight edge is rotating in a clockwise fashion, the slope of the tangent line is decreasing and f″ < 0. Hence, the function is concave down for these value of x.

    For the graph in part a of the example, there is a clockwise rotation from x = −1 to x = −0.5 and from x = 0 to x = 0.5. Hence, the function is concave down on (−1, −0.5) and (0, 0.5). Alternatively, there is a counterclockwise rotation from x = −0.5 to x = 0 and from x = 0.5 to x = 1. Hence, the function is concave up on (−0.5, 0) and (0.5, 1).

  2. Using the approach described in part a, we find that this graph is concave down over (−1, 1) and concave up on (1, 4).

A point on a continuous graph that separates a concave downward portion of a curve from a concave upward portion is called an inflection point. The following example illustrates this idea using data from mortality due to airborne diseases.

Example 3 Sigmoidal decay in deaths due to airborne diseases

In a study of deaths in the United States, Ausubel and colleagues found that deaths from aerially transmitted diseases as a fraction of all deaths could be very well described by the sigmoidal function shown in Figure 3.30. Determine where this function is concave up and down. Find the point of inflection. Discuss what these changes in concavity mean.

images

Figure 3.30 Fraction of deaths as a function of time (in years) after 1880.

Data: J. H. Ausubel, P. S. Meyer, and I. K. Wernick, “Death and the Human Environment: The United States in the 20th Century,” Technology in Society 23(2) (2001): 131–146. Reprinted with permission.

Solution To estimate the intervals of concavity, we can place a ruler as a tangent to the curve and slowly move it from the left side to the right side. Doing so, we notice that the ruler is rotating clockwise from x = 0 to x ≈ 40 and rotating counterclockwise from x ≈ 40 to x = 120. Hence, the point of inflection appears to be located at x = 40, and the fraction of deaths due to airborne diseases is decreasing at a faster and faster rate from 1880 (x = 0) to 1920 (x = 40); that is, the curve is concave down. The fraction of deaths is decreasing at a slower and slower rate from 1920 (x = 40) to 1980 (x = 100); that is, the curve is concave up.

The curve in Example 3 is an example of a sigmoidal curve, a curve that is monotonic, with horizontal asymptotes at positive and negative infinity and a single point of inflection. Another example of a sigmoidal curve is Figure 1.28 in Section 1.4; this is a model of population growth in the United States where growth is initially exponential and then levels off asymptotically. The following graphs show both forms of sigmoidal curves.

images

Graphs of increasing (in blue) and decreasing (in red) sigmoidal functions.

Changing rates of change

Recall from Section 2.1 the interpretation of the derivative of a function as the rate of change of that function with respect to increasing values of the function's argument. If we go back to the origins of the calculus, we see that velocity was interpreted by Newton as the instantaneous rate of change over time of the position of a particle in space, and acceleration was interpreted as the instantaneous rate of change over time of velocity. These interpretations led Newton to formulate his famous second law on the relationship between the force acting on an object, the mass of the object, and its resulting acceleration.

Average and Instantaneous Acceleration

Let v(t) be the velocity of an object at time t in a predetermined direction. The average acceleration of an object from time t to time t + h is

images

while the instantaneous acceleration of an object at time t is

images

The field of kinematics—that is, the motion of points or bodies through space—and the application of calculus to physics are developed in physics courses. But no calculus book is complete without at least one example dealing with acceleration, and here we focus on a biological one.

Example 4 When does Usain Bolt slow down?

Example 3 of Section 2.1 provides data on times for the 10-meter splits of Usain Bolt's record-breaking 100-meter win in the 2008 Beijing Olympic Games. Use these data to answer the following questions.

  1. Use technology to fit a fifth-order polynomial s(t) to these data that specifies the distance covered by Bolt as a function of time, beginning with the moment he leaves the starting blocks until the end of this race.
  2. Calculate and plot the first and second derivatives of s(t) to obtain graphs of Bolt's instantaneous velocity and acceleration during the race.
  3. From the plots in part b, determine the time ts during the race that Bolt switches from speeding up to slowing down, and calculate his velocity vmax, which is a maximum, at this time.
  4. What is Bolt's average deceleration (negative of acceleration) from time ts until the end of the race?

Solution

  1. From Table 2.1 in Example 3 of Section 2.1, and taking into account that it takes 0.17 second for Usain Bolt to leave his starting block, we can construct the following table of how long it takes Usain Bolt to reach each of the successive 10-meter marks along the race.

    images

    From these data we can use technology to obtain the following function s(t) that specifies distance covered as time progress from leaving the blocks at t = 0 to finishing the race at t = 9.52, which with the time it takes for Usain Bolt to leave blocks, gives him a final race time of 9.52+0.17 = 9.69 seconds (each term is specified to five significant figures):

    images

    This function is illustrated in the right panel of Figure 3.31.

    images

    Figure 3.31 Distance covered as a function of time (seconds) by Usain Bolt in his record-breaking 100-meter win in the 2008 Beijing Olympic Games. Right panel: data and five-degree polynomial fit; middle panel: corresponding velocity profile; left panel: acceleration profile. Maximum velocity vmax = 12.28 is achieved (dotted vertical line) at t = 7.17 seconds, when the acceleration curve passes through zero.

  2. Taking the first derivative, we obtain the velocity function (middle panel in Figure 3.31)

    images

    and taking the second derivative we obtain the acceleration function (left panel in Figure 3.31)

    images

  3. The function s″ (t) has a root at ts = 7.17. We see from Figure 3.31 that s″ (t) is decreasing and goes from positive on [0, 7.16] to negative on [7.18, 9.52]. Thus, the velocity and acceleration plots show that Bolt accelerates, first strongly, and then more slowly until ts = 7.17, at which time he reaches his maximum instantaneous velocity of vmax = 12.28 meters/second (m/s).
  4. The average acceleration from time ts = 7.17 to the end of the race at t = 9.52 is, by definition with v(t) = s′ (t),

    images

    so taking the negative sign into account, the average deceleration is 0.83 m/s2.

The reason for choosing a fifth-order polynomial rather than a fourth or sixth in Example 4 is that the runners accelerate strongly at the beginning and tend to decelerate as they tire at the end. Thus, acceleration is initially positive and eventually negative. Since this type of behavior is best modeled by odd-ordered rather than even-ordered polynomials, we choose an odd-ordered polynomial for distance, as it becomes odd again after taking two derivatives. We choose a fifth-order instead of third-order polynomial as it does a better job modeling the intermediate, close to zero acceleration segment that we see occur approximately around seconds 4 to 7 during the course of the race (see Problem 53 in Problem Set 3.6).

Example 5 Declining rates

A recent news article reported that SAT scores are declining at a slower rate. Use calculus to describe this report.

Solution The key statement is that “SAT scores are declining at a slower rate.” If S(t) denotes the average SAT score as a function of time, then the phrase “SAT scores are declining” means that S′(t) < 0 and the phrase “at a slower rate” means that S″(t) > 0 as the rate S′(t) is increasing. In other words, SAT scores are decreasing, yet concave up and so “leveling off”!

Using our interpretations of the first and second derivative, we should be able to identify the graph of one from the other.

Example 6 Finding f, f′, and f″

The graphs of y = f(x), y = f′(x), and y = f″(x) are shown in Figure 3.32. Identify f, f′, and f″.

images

Figure 3.32 Graph of a function and its first and second derivatives.

Solution To identify f, f′, f″, we can start with one graph—say, the black one— and determine when it is rising and falling. We can see that it is falling roughly on the intervals [− 2, − 1.25] and [0, 1.25] and rising on the complementary intervals. Since the blue curve is negative on [−2, −1.25] and [0, 1.25] and positive on the complementary intervals, the blue curve may be the graph of the derivative of function defined by the black curve. On the other hand, the black curve appears to be negative where the red curve is falling and positive where the red curve is rising. Hence, the black curve appears to be the graph of the derivative of the function defined by the red curve. Therefore, we conclude that y = f(x) is defined by the red curve, y = f′(x) is defined by the black curve, and y = f″(x) is defined by the blue curve.

Second-order approximations

In Section 3.5, we approximated functions with their tangent lines. While a good start, these approximations can be improved upon by using first and second derivatives.

Example 7 Stripping away the tangent line

Consider y = e2x.

  1. Find the tangent line at x = 0.
  2. Compute images and determine whether the linear approximation overestimates or underestimates y2x near x = 0.
  3. Plot the difference between y = e2x and its tangent line. Discuss what you notice.

Solution

  1. Since images = 2e0 = 2, the tangent line has slope 2 and passes through the point (0, 1)—that is, the line

    images

  2. images, which is 4 at x = 0. Since the second derivative is positive, images is increasing near x = 0, and we would expect the tangent line to underestimate (i.e., lie under) y = e2x near x = 0. Indeed, graphing the function y = e22x (blue curve) and y = 2x + 1 (red curve) confirms this prediction.

    images

  3. Plotting y = e2x − 2x − 1 yields a function that looks like a parabola.

    images

The preceding example suggests that function obtained by taking the difference between the original function and its derivative is approximately parabolic. If we want to approximate this function by a quadratic function, how do we find the best quadratic approximation? To answer this question, we develop the following procedure for finding the best quadratic approximation to a function f around the point x = 0. To find

images

near x = 0, we require

images

To have the first derivatives of f(x) and the approximation a + bx +cx2 agree at x = 0, we can take derivatives of both sides

images

At x = 0, we want f′(0) = b, so we define b = f′(x). Finally, to have their second derivatives agree at x = 0, we differentiate one more time:

images

This leads us to define c = images. This gives a quadratic (second-order or parabolic) approximation at x = 0:

images

Let us see how well this approximation works.

Example 8 Quadratic approximation

Find the quadratic approximation to y = e2x at x = 0. Plot y = e2x, its linear approximation, and its quadratic approximation.

Solution Let f(x) = e2x, so from the previous example f (x) = 2e2x and f″(x) = 4e2x. The linear approximation is

images

Figure 3.33 Graph of a function f(x) = e2x(in black) along with its linear (in red) and quadratic approximations (in blue).

images

The quadratic approximation is

images

The graphs of y = e2x, y = 2x + 1, and y = 1 + 2x + 2x2 are shown in Figure 3.33. The quadratic approximation does a significantly better job of approximating the function y = e2x.

In cases where the linear approximation is a horizontal line, the quadratic approximation is the first approximation to give real information about the concavity of the curve in question.

Example 9 Approximating the cosine

  1. Find the linear and quadratic approximations of y = cos x at x = 0.
  2. Use the quadratic approximation to estimate cos 1, cos 0.5, and cos 0.1. Compare your approximations to the answers given by a calculator.

images

Figure 3.34 Graph of a function f(x) = cos x (in black) along with its linear (in red) and quadratic approximations (in blue).

Solution

  1. Let f (x) = cos x, so f′(0) = −sin 0 = 0 and f″(0) = −cos 0 = −1. The linear approximation is

    images

    The quadratic approximation is

    images

    The graph is a downward facing parabola. The graphs of y = cos x, y = 1, and y = 1− imagesx2 are shown in Figure 3.34.

  2. We compare the quadratic and calculator approximations.

images

The approximations get better and better as you get closer and closer to x = 0.

More generally, we may wish to approximate a function near a point x = a with a parabola. As an exercise, you can verify that by forcing the quadratic approximation and the function to agree up to the second derivative at x = a, you obtain the following second-order approximation of f at x = a.

Second-Order Approximation

Let f have a first and second derivative defined at x = a. The second-order approximation of f around x = a is given by

images

Example 10 Professor Getz's headache continues

Recall in Example 6 of Section 3.5 that we found that the half-life for a drug as a function of the clearance rate x per hour is given by

images

For a dose of 1000 mg of acetaminophen, we estimated a clearance rate of approximately 0.28 per hour.

  1. Compute the first- and second-order approximation of T(x) at x = 0.28.
  2. Plot both approximations together with the function T(x).
  3. Discuss whether an error analysis using the sensitivity T′(0.28) overestimates or underestimates the propagation of error from estimating x to estimating T(x).

Solution

  1. Computing the first and second derivatives of T(x) at x = 0.28 gives

    images

    Therefore, the first-order approximation is given by

    images

    and the second-order approximation is given by

    images

  2. Plotting T(x) (in black), the first-order approximation (in blue), and the second-order approximation (in red) yields

    images

  3. Since the sensitivity analysis using T′(0.28) is based on the linear approximation of T(x) near x = 0.28 and T is concave up, the sensitivity analysis underestimates the propagation of error from estimating x to estimating T(x). This underestimation corresponds to the tangent line (in black) in part a lying under the graph of T(x).

Even higher derivatives

Once we have taken second derivatives, there is no reason to stop. We can attempt to take the derivative of the second derivative, and the derivative of the resulting derivatives until the function obtained is 0; or we can continue indefinitely (e.g., see parts b and c of the next example). These higher derivatives and the associated notations are defined as follows:

images

Example 11 Higher derivatives

Find the following higher derivatives.

images

Solution

  1. If y = 1 + x + x3, then

    images

  2. If f(x) = e2x, then

    images

  3. At first this problem might seem insane. Take 101 derivatives of a function? However, if we proceed calmly, a pattern will quickly emerge that will dispel this insanity. If f(x) = sin x, then

    f(x) = cos x

    f(x) = −sin x

    f′″(x) = − cos x

    f(4)(x) = sin x

We are back to where we started, and the derivatives cycle in a fixed pattern of period four; that is, repetition occurs every four derivatives so that f(1)(x) = f(5)(x) = f(9)(x)= ···cos x, f(2)(x) = f(6)(x) = f(10)(x)= ··· −sin x, and so on. Hence, f(100)(x) = sin x and f(101)(x) = images sin x = cos x.

We conclude with an application of higher derivatives from politics.

Example 12 Presidential proclamation

In the fall of 1972, President Nixon announced that the rate of increase of inflation was decreasing. This was the first time a sitting president used the third derivative to advance his case for reelection.

So reported Hugo Rossi in “Mathematics Is an Edifice, Not a Toolbox” (Notices of the AMS, 43, October 1996). Discuss how a third derivative was being used by President Nixon.

Solution Let V denote the “value of a dollar” at time t in years. Inflation means that the value of a dollar is decreasing, so images < 0. If inflation is increasing, then the value of a dollar is decreasing at a faster rate; that is, images < 0. Finally, if the rate of increase of inflation is decreasing, we get images > 0. Hence, at the level of the third derivative, things were looking less bleak for the value of the dollar!

PROBLEM SET 3.6

Level 1 DRILL PROBLEMS

Find the higher derivatives indicated in Problems 1 to 12.

images

images

In Problems 13 to 18 you are given the function s (t) that specifies the position of an object on a line as a function of time t. Find expressions for the velocity and acceleration as a function of time for t ≥0. Solve for t when the acceleration is 0; use technology where needed to find the values and when multiple values exist, use the closest to t = 0.

images

In Problems 19 to 24 find the linear approximations of the given functions f(x) around x = a. Using second-order derivatives, determine whether the linear approximation tends to overestimate or underestimate f(x) near x = a.

images

In Problems 25 to 34 determine on what intervals f is increasing, decreasing, concave up, concave down, and find the points of inflection.

images

Find the first- and second-order approximations of y = f(x) around x = a in Problems 35 to 40. Use technology to plot the function and its approximations near x = a.

images

In Problems 41 to 44, identify y = f(x), y = f′(x), and y = f(x).

images

45. Sketch the graph of a function with all of the following properties:

images

46. Sketch the graph of a function with all of the following properties:

images

Level 2 APPLIED AND THEORY PROBLEMS

47. The slogan of a particular home improvement company is “Improving Home Improvement.” Explain the role of derivatives in this slogan.

48. A politician claims that “Under a new law, prices would rise slower than if the law were not passed.” Explain the role of higher derivatives in this statement.

49. At the website http://www.nlreg.com/aids.htm, you can find the following figure that graphs the number of new cases of AIDS since 1980.

  1. Estimate where the function is concave up and concave down.
  2. Describe in words what these changes in the concavity mean for the AIDS epidemic.

    images

50. In Example 2 of Section 2.4, a dose-response curve for patients responding to a dose of histamine is given by

images

where x is the natural logarithm of the dose in millimoles (mmol).

  1. Compute images
  2. Determine for what dose ranges R is concave up and concave down. Interpret your results.

51. images

images

One of the most famous women in the history of mathematics is Maria Gaetana Agnesi.

She was born in Milan, the first of twenty-one children. Her first publication was at age nine, when she wrote a Latin discourse defending higher education for women. Her most important work was a now classic calculus textbook published in 1748. Agnesi is primarily remembered for a curve defined by the equation

images

for a positive constant a. The curve was named versiera (from the Italian verb to turn) by Agnesi, but John Colson, an Englishman who translated her work, confused the word versiera with the word avversiera, which means “wife of the devil” in Italian; the curve has ever since been called the “witch of Agnesi.” This was particularly unfortunate because Colson wanted Agnesi's work to serve as a model for budding young mathematicians, especially young women. Graph this curve, find the points of inflection (if any), and discuss its concavity.

52. The spruce budworm is a moth whose larvae eat the leaves of coniferous trees. These insects suffer predation by birds. Ludwig and colleagues* suggested as model for the per capita predation rate, p(x):

images

where b is the maximum predation rate and a is the number of budworms at which the predation rate is half its maximum rate. What is the concavity of this curve, and is there a point of inflection?

53. In Example 4 of this section, we fitted a fifth-order polynomial to the data on the distance covered as a function of time during Usain Bolt's record-breaking 100-meter win in the 2008 Beijing Olympic Games. Use technology to fit a third-order polynomial to these data; then differentiate this polynomial to obtain the corresponding velocity and acceleration functions of time. Plot these three functions (displacement, velocity, acceleration) and estimate the time ts at which Bolt switches from acceleration to deceleration during the course of the race and his velocity vmax, which is a maximum, at this switching point. What is the average deceleration from the switching point to the end of the race? Compare your values of ts, vmax, and average deceleration with those obtained in Example 4.

54. In Example 4 of this section, we fitted a fifth-order polynomial to the data on the distance covered as a function of time during Usain Bolt's record-breaking 100-meter win in the 2008 Beijing Olympic Games. Use technology to fit a seventh-order polynomial to these data; then differentiate this polynomial to obtain the corresponding velocity and acceleration functions of time. Plot these three functions (displacement, velocity, acceleration) and estimate the time ts at which Bolt switches from acceleration to deceleration during the course of the race and his velocityvmax, which is a maximum, at this switching point. What is the average deceleration from the switching point to the end of the race? Compare your values of ts, vmax, and average deceleration with those obtained in Example 4.

55. In Example 4 of this section, we fitted a fifth-order polynomial to the data on the distance covered as a function of time during Usain Bolt's record-breaking 100-meter win in the 2008 Beijing Olympic Games. Use Ben Johnson's split times given in Table 2.1 in Example 3 of Section 2.1 to obtain data relating distance and time of Johnson's performance during the 1988 Seoul Olympics 100-meter final. Fit a fifth-order polynomial to the data; then differentiate this polynomial to obtain the corresponding velocity and acceleration functions of time. Plot these three functions (displacement, velocity, acceleration) and estimate the time ts at which Johnson switches from acceleration to deceleration during the course of the race. What is Johnson's maximum velocity, which occurs at ts, and what is his average deceleration from time ts until the end of the race? Compare your results to Bolt's performance analyzed in Example 4.

56. Let f be a function that is twice differentiable on an interval I containing the point x = a. If there exists a K > 0 such that |f″(x)|≤K for all x in I, then show that

images

for all x in I. This result gives the error of the first-order approximation. Hint: Pick any point ba in I. Define

G(x) = f(x) − f(a) − f′(a) (xa) − C(xa)2

where C is chosen such that G(b) = 0. Differentiate G and apply the mean value theorem to G and f′.

57. Let f be a function with first- and second-order derivatives at x = a. Consider a quadratic of the form q(x) = b + c(xa) + d(xa)2. Show that f(a) = q(a), f′(a) = q′(a), and f″(a) = q″(a) if and only if b = f(a), c = f′(a) and d = f″(a)/2.

3.7 l'Hôpital's Rule

In models of tumor or population growth, spread of rumors, and risk of being infected, one may encounter limits of the form

images

where limxa f(x) and limxa g(x) are both zero or both infinite. Such limits are called 0/0 indeterminate form and ∞/∞ indeterminate form, respectively, because their value cannot be determined without further analysis. In this section, we study an approach to handling these limits and explore some applications.

The 0/0 and ∞/∞ indeterminate forms

In 1694, French mathematician Guillaume de l'Hôpital (see Historical Quest in Problem Set 3.7) found a useful method for evaluating limits involving indeterminate forms. He considered the special case where f(a) = g(a) = 0, f and g are differentiable at x = a, and g′(a) ≠ 0. Under these assumptions, we have

images

The following example illustrates how replacing the original limit with a limit involving derivatives can be helpful.

Example 1 Using l'Hôpital's argument

Evaluate the following limits.

images

Solution

  1. This limit is of indeterminate form because sin x and x both approach 0 as x →0. l'Hôpital's rule applies because both sin x and x are differentiable at x = 0. Thus

    images

  2. For this example, f(x) = x7 − 128 and g(x) = x3 − 8 and the limit is of the form 0/0. Since f′(2) = 7 · 26 and g′(2) = 3 · 22 ≠ 0, we can apply l'Hôpital's argument to obtain

    images

This approach to computing limits is known as l'Hôpital's rule. The general statement of this rule is given in the following theorem.

Theorem 3.1 l'Hôpital's rule

Let f and g be differentiable functions on an open interval containing a (except possibly at a itself). Suppose images produces an indeterminate form images and that

images

where L is either a finite number, −∞, or ∞. Then

images

The theorem also applies to one-sided limits and to limits at infinity where x → ∞ and x → − ∞.

When we use l'Hôpital's rule, we use the symbol images as shown in the following example.

Example 2 Using l'Hôpital's rule

Find the following limits.

images

Solution

  1. Since sin(3 − 3) = 3 − 3 = 0, this limit is of an 0/0 indeterminate form. Since both sin(x − 3) and x − 3 are differentiable near x = 3, we can apply l'Hôpital's rule as follows:

    images

  2. Since images x + ex = ∞ and images 2x + 1 = ∞, this limit is an ∞/∞ indeterminate form. Since both x + ex and 2x + 1 are differentiable for large x, we can apply l'Hôpital's rule as follows:

    images

  3. Since cos π/2 = 1 − sin π/2 = 0, this limit is of a 0/0 indeterminate form. Since both cos x and 1 − sin x are differentiable near x = π/2, we can apply l'Hôpital's rule as follows:

    images

  4. Since images images = ∞ and images ln x = ∞, this limit is an ∞/∞ indeterminate form. Since both images and ln x are differentiable for large x, we can apply l'Hôpital's rule as follows:

    images

We consider two applications of l'Hôpital's rule to models of population growth.

Example 3 Exponential versus arithmetic growth

In An Essay on the Principle of Population, first published anonymously in 1798, but later attributed to Thomas Malthus, we find the following text:

“Population, when unchecked, increases in a geometrical ratio. Subsistence increases only in an arithmetical ratio. A slight acquaintance with numbers will shew [sic] the immensity of the first power in comparison of the second.”

While Example 4 of Section 1.4 explored a special case of this observation, l'Hôpital's rule allows us to fully appreciate the observation of Malthus. Let P(t) = P0ct for some c > 1 and P0 > 0 represent the size of a population at time t, and let F(t) = a + bt for some a > 0 and b > 0 represent the total amount of food available at time t. Find

images

and discuss its implications.

Solution Since both P0ct and a + bt approach infinity as t approaches infinity, we obtain

images

Hence, as time marches on, the amount of food per individual approaches nothing.

What do tumor growth, sales of mobile phones, spread of rumor or infection, and population growth have in common? They all can be modeled mathematically by specifying how the growth rate of the tumor, rumor, or population depends on its current size, frequency, or abundance. The next example looks an important family of growth functions introduced by F. J. Richards in a 1959 article that appeared in the Journal of Experimental Botany[2: 290–301].

Example 4 Generalized logistic growth function and indeterminate form 0/0

Whether it be the spread of a rumor, tumor growth, or population growth, an important class of models describing the rate at which the size N of a population, or a tumor for that matter, changes is

images

where r > 0, v > 0, and K > 0 are positive constants that influence the shape of the growth function.

  1. For what population sizes N does the population grow?
  2. Let r = 1 and K = 100. Plot G(N) on the interval [0, 110] for v = 2, 1, 0.1, 0.01. Discuss what you find.
  3. For K > N > 0, r = 1, and K = 100, find

    images

    and sketch the resulting curve.

Solution

  1. Since r, v, and K are positive, the population growth rate G(N) is positive if

    images

    Therefore, the population grows whenever its population size is positive and less than K.

  2. Plotting G(N) for the different v values yields the dashed colored curves in Figure 3.35. Smaller v values correspond to higher curves. Hence, as v gets smaller, the growth rate gets larger. Moreover, the growth function G appears to be approaching a limiting function as v approaches zero.

    images

    Figure 3.35 Generalized logistic growth function G(N) = N(1 − (N/100)v)/v for v = 2 (in red), 1 (in green), 0.5 (in purple), 0.25 (in cyan), and 0.1 (in magenta). The limiting Gompertz growth function G(N) = −N log (N/100) is shown as a solid black curve.

  3. For 0 < N < K, G(N) is positive and the limit

    images

    is of indeterminate form 0/0 as r N(1 − (N/K)v) andvequal zero at v = 0. Therefore, taking derivatives with respect to v gives

    images

    for N > 0. This limiting growth function −rN ln(N/K) is known as the Gompertz growth function, which has been used extensively to model tumor growth. Figure 3.35 plots this function (in black) for K = 100 and r = 1. As anticipated in part b, the plots of the generalized logistic growth functions approach the Gompertz growth function as v approaches zero.

Sometimes repeated applications of l'Hôpital's rule are necessary to get anywhere.

Example 5 Applying l'Hôpital's rule twice

Evaluate images.

Solution If we consider the limit of the numerator and denominator separately, we obtain ∞/∞. However, if we apply l'Hôpital's rule twice we obtain

images

Note that L'Hôpital's rule is not the only way to solve the above example. We could have divided both the numerator and denominator by x2 to obtain

images

Most examples in this section, however, do not yield to this simple procedure, so l'Hôpital's rule must be used. Before applying l'Hôpital's rule, however, we must check that the conditions of Theorem 3.1 apply. If they do not hold, then the analysis is not valid, as illustrated by the next two examples.

Example 6 Limit is not an indeterminate form

Evaluate images.

Solution You must always remember to check that you have an indeterminate form before applying l'Hôpital's rule. The limit is

images

If you apply l'Hôpital's rule in Example 6, you obtain the WRONG answer:

images

Example 7 Conditions of l'Hôpital's rule are not satisfied

Evaluate images.

Solution This limit has the indeterminate form ∞/∞. If you try to apply l'Hôpital's rule, you find

images

The limit on the right does not exist, because both sin x and cos x oscillate between −1 and 1 as x → ∞. Recall that l'Hôpital's rule applies only if images is finite or is ±∞. This does not mean that the limit of the original expression does not exist or that we cannot find it; it simply means that we cannot apply l'Hôpital's rule. To find this limit, factor out an x from the numerator and denominator and proceed as follows:

images

Other indeterminate forms

Remember that l'Hôpital's rule itself applies only to the indeterminate forms 0/0 and ∞/∞. Other indeterminate forms, such as 1, 00, ∞0, ∞ − ∞, and 0 · ∞, can often be manipulated algebraically, or by taking logarithms, into one of the standard forms 0/0 or ∞/∞, and then evaluated using l'Hôpital's rule. In a case where we have taken the logarithm to obtain one of the standard forms, we need to remember to transform back by applying exponentiation to our solution.

Example 8 Limit of the form 00

Find images xsinx.

Solution This is a 00 indeterminate form. From the graph shown in Figure 3.36, it looks as though the desired limit is 1.

images

Figure 3.36 Graph of xsin x.

We can verify this conjecture analytically. We proceed as with the previous example by using properties of logarithms.

images

taking inverse of logarithms L = e0 = 1

Example 9 Escaping infection and the indeterminate form 0−∞

In models of host-pathogen and host-parasite interactions, the fraction of hosts escaping parasitism is often given by a negative-binomial escape function

images

where P is the density of the parasites or pathogens, a > 0 is the rate at which hosts encounter parasites or pathogens, and k > 0 is a clumping parameter. Small values of k correspond to parasites or pathogens being highly aggregated in the environment and large values of k correspond to parasites or pathogens being more evenly distributed across the environment. Assume a = 0.1.

  1. For k = 0.1, 1, 5, 10, plot f(P) over the interval [0, 10]. What effect does k have on the risk of being parasitized or infected?
  2. For P > 0, find images f(P).

Solution

  1. Plots of f(P) for k = 0.1, 1, 5, 10 shown in Figure 3.37 suggest that as k increases the likelihood of escaping parasitism or infection goes down. Hence, as k increases and parasites or pathogens are more evenly distributed across the environment, the risk of parasitism or infection goes up.
  2. For P > 0, 1 + aP/k approaches 0 and −k approaches −∞ as k approaches ∞. Therefore we have an indeterminate form of 0−∞. To turn this problem to an indeterminate form of 0/0, we take the ln of f(P) which yields

    images

    images

    Figure 3.37 Negative binomial escape function f(P) = (1 + 0.1P/k)k for k = 0.1 (in red), 1 (in green), 5 (in purple), and 10 (in cyan). The limiting Poisson escape function f(P) = exp(− 0.1P) is shown as a solid black curve.

    Since ln(1 + aP/k) and −1/k approach zero as k approaches ∞, we can apply l'Hôpital's rule.

    images

    Therefore, we get images ln f(P) = −aP and by exponentiating

    images

    for P > 0. This limiting escape function is known as the Poisson escape function that corresponds to parasitism or infection events occurring randomly among all hosts. This limiting function is plotted in black in Figure 3.37 and illustrates that the greatest risk of infection or parasitism occurs in this limiting case.

Example 10 Finding a horizontal asymptote and the indeterminate form ∞0

Find the horizontal asymptote of the graph f(x) = x1/x for x > 0.

Solution To determine if the graph of f has a horizontal asymptote for x > 0, we evaluate

images

This limit is indeterminate of the form ∞0. To evaluate it, we take the natural logarithm and proceed as follows:

images

Thus, y = 1 is a horizontal asymptote for the graph of y = x1/x, as shown in Figure 3.38.

images

Figure 3.38 Graph of y = x1/x with horizontal asymptote.

We saw in Figure 3.38 that the graph of f(x) = x1/x approaches the line y = 1 asymptotically as x → ∞, but how does f(x) behave as x → 0+? That is, what is

images

It may seem that to answer this question, we need to apply l'Hôpital's rule again, but this limit has the form 0, which is simply 0 and is not indeterminate at all. Other forms that may appear to be indeterminate, but really are not, are 0/∞, ∞ · ∞, ∞ + ∞ and − ∞ − ∞.

PROBLEM SET 3.7

Level 1 DRILL PROBLEMS

1. An incorrect use of l'Hôpital's rule is illustrated in the following limit computations. In each case, explain what is wrong and find the correct value of the limit.

images

2. Sometimes l'Hôpital's rule leads nowhere. For example, observe what happens when the rule is applied to

images

Use any method you wish to evaluate this limit.

Find the limits, if possible, in Problems 3 to 18.

images

images

In Problems 19 to 22, use l'Hôpital's rule to determine all horizontal asymptotes to the graph of the given function. You are NOT required to sketch the graph.

images

Verify the statements in Problems 23 to 25.

images

Level 2 APPLIED AND THEORY PROBLEMS

26. Fisheries scientists have found that a Ricker stock-recruitment relationship, which has the form

images

where y is a measure (also called an index) of the number of individuals recruited to the fishery each year (typically one-year-olds), and x is an index of the spawning stock biomass (sometimes measured in terms of eggs produced), provides a reasonable fit to various species. Consider the case where the parameter values are a = 5.9 and b = 0.0018.

  1. What is the value of the recruitment index as x → ∞?
  2. What is the maximum value of the recruitment index and at what spawning stock index value does it occur?
  3. Over what range of spawning stock index values is the recruitment function concave up and over what values is it concave down?
  4. Use the information obtained in parts a, b, and c to sketch this function.

27. An agronomist experimenting with a new breed of giant potato has found that individual tubers x months after planting have a biomass in kilograms given by the equation y(x) = 2e−1/(5x) for x > 0.

  1. Calculate the rate of growth of the tuber over time and determine what happens to this rate in the limit as x → 0 and x → ∞.
  2. Find the time after planting when the growth rate of the tuber is maximized.
  3. Show that the growth rate is positive for all x > 0 and determine the regions over which the growth is accelerating and decelerating.
  4. Sketch the biomass of the potato, as well as its growth rate, indicating the important points and regions calculated in parts a, b, and c.

28. Determine which function, f(x) = xn with n > 0 or g(x) = eax with a > 0, grows faster at ∞ by computing images.

29. Determine which function, f(x) = xn with n > 0 or g(x) = ln x, grows faster at ∞ by computing images.

30. Consider a drug in the body whose current concentration is 1 mg/liter. In this problem, you investigate the meaning of exponential decay of the drug.

  1. If one-half of the drug particles cleared the body after one hour, determine the concentration of the drug that remains after one hour.
  2. If one-quarter of the drug particles cleared the body every half an hour, determine the concentration of the drug that remains after one hour.
  3. If one-twentieth of the drug particles cleared the body every six minutes, determine the concentration of the drug that remains after one hour.
  4. If images of the drug particles cleared the body every 1/nth of an hour, determine the concentration cn of the drug that remains after one hour.
  5. Find images cn.

31. images The French mathematician Guillaume de l'Hôpital (1661–1704) is best known today for the rule that bears his name, but that rule was discovered by l'Hôpital's teacher, Johann Bernoulli. Not only did l'Hôpital neglect to cite his sources in his book, but there is also evidence that he paid Bernoulli for his results and for keeping their arrangements for payment confidential. In a letter dated March 17, 1694, he asked Bernoulli “to communicate to me your discoveries”—with the request not to mention them to others: “it would not please me if they were made public.” (See D. J. Stuik, A Source Book in Mathematics, 1200–1800, Cambridge, MA: Harvard University Press, 1969, 313–316.) L'Hôpital's argument, which was originally given without using functional notation, can easily be reproduced:

images

First, place some conditions on the functions f and g that will make this argument true. Next, supply reasons for this argument, and give necessary conditions for the functions f and g.

32. Consider the general logistic growth function

images

from Example 4.

  1. For v > 0, find the density N > 0 that maximizes G(N); that is, solve G′(N) = 0. The answer will depend on the parameters r, v, and K.
  2. Take your answer from part a and compute its limit as v → 0.
  3. Find density N > 0 that maximizes−rN log(N/K) and compare your answer to what you found in part b.

CHAPTER 3 REVIEW QUESTIONS

  1. Find images for the following expressions.

    images

  2. Approximate 651/3 and determine whether this approximation overestimates or underestimates the true answer. Justify your answers by using derivatives.
  3. Find images where y = x2(2x − 3)3.
  4. Use the definition of derivative to calculate, showing all details, images(x − 3x2).
  5. Find the first- and second-order approximations to y = ex2 at x = 0. Graph the function and its approximations.
  6. In Figure 3.39, which graph represents the function and which graph the derivative?

    images

    Figure 3.39 A function and its derivative.

  7. Sketch the graph of a function with the following properties:

    images

    What can you say about the derivative of f when x = 1?

  8. The developmental rate of insects and plants as a function of temperature T can be modeled by the Briere model

    images

    where TL is the lower developmental threshold below which an individual does not develop, TU is the upper developmental threshold above which an individual does not develop, and a is a proportionality constant. Find a linear approximation to D(T) at T = TL, discuss what it means, and discuss where it breaks down.

  9. Let f be a function defined by

    images

    Determine where the function is increasing, where it is decreasing, and where the graph is concave up and where it is concave down.

  10. Suppose the proportion of insect hosts escaping parasitism depends on the parasitoid density, d, and is modeled by the function f(d) = e−.05d. Does the proportion escaping parasitism increase or decrease with parasitoid density? What is the concavity of this curve, and is there a point of inflection?
  11. Determine the concavity and inflection points and use l'Hôpital's rule to find the horizontal asymptotes of the graph of

    images

  12. Suppose the concentration in the blood at time t of a drug injected into the body is modeled by

    images

    Use l'Hôpital's rule to find the horizontal asymptote. Find the time when C′(t) = 0. Graph this curve and verify that the largest concentration occurs at the solution to C′(t) = 0.

  13. Say you want to estimate the height of a tall tree. To do so, you cannot simply drop a tape measure from the top of the tree. However, you can determine the height by using a sextant to determine the angle θ between the ground and the tip of the tree at a distance of 100 feet from the base of the tree.
    1. Find the height of the tree, H, as a function of θ.
    2. If you measure an angle θ = 1.1 radians, determine the height of the tree.
    3. Determine the elasticity of the height in part b to θ. Discuss how a 10% error in measuring θ influences the estimate for the height of the tree when θ = 1.1.
  14. A bacterial colony is estimated to have a population of P thousand individuals, where

    images

    and t is the number of hours after a toxin is introduced.

    1. At what rate is the population changing when t = 0 and t = 1?
    2. Is the rate increasing or decreasing at t = 0 and t = 1?
    3. At what time does the population begin to decrease?
  15. As we saw in Example 6 of Section 1.5, scientists at Woods Hole Oceanographic Institution measured the uptake rate of glucose by bacterial populations from the coast of Peru. In one field experiment, they found that the uptake rate can be modeled by f(x) = images micrograms per hour, where x is micrograms of glucose per liter. If the current uptake rate is twelve, determine the current level of glucose and, hence, determine the rate at which this uptake rate is itself changing per unit increase in glucose.
  16. The gross U.S. federal debt (in trillions of dollars) is plotted below.

    images

    Regarding this debt, President Ronald Reagan stated in 1979 that the United States is “going deeper into debt at a faster rate than we ever have before.” Discuss the role of higher-order derivatives in the graph of federal debt from 1950 to the end of the graph in the context of President Reagan's statement.

  17. The figure eight curve shown below is defined implicitly by the equation x4 = x2y2.

    images

    Find all the points on this curve that have horizontal tangents.

  18. Consider the functions y = f(x) (in blue) and y = g(x) (in red) whose graphs are shown below.

    images

    Find images f(g(x)) at x = −1 and x = 1.

  19. Find the tangent line of y = (1 + sin 2x)100 at x = π/2.
  20. We modeled the concentration of carbon dioxide (in ppm) at the Mauna Loa Observatory of Hawaii with the function

    images

    where t is months since May 1974. Find for what months h is increasing and decreasing over the interval [0, 12].

GROUP PROJECTS

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

Project 3 Modeling North American Bison Population

If we look closely at the data plotted in Example 2 of Section 3.5, on the abundance of North American bison in Yellowstone National Park from 1902 to 1931, we get the distinct impression that the data can be represented much better by two linear functions than by one: the first representing data from 1902 to 1915 and the second representing the data from 1915 to 1931. We can fit these two functions by eye, or we can work more precisely using the concept of a sum-of-squares measure to gauge how well the functions fit the data. This concept requires that we know the actual values of the data points. Specifically, if we have n data points, indexed by i = 1,..., n, then we need to know the values (xi, yi) for each data point. For the bison, these data are specified in Table 3.3. (Note that the data for some years are missing. This is not a problem if we just ignore these missing points when indexing the data.)

Table 3.3 Population for the North American bison

images

For the bison data, consider piecing together two linear functions so that they both meet at the point (x12, y12) = (1915,270). Since both lines pass through this point, they must both satisfy the equation

images

for some constant c. If the line fitted to the 1902–1915 date is specified by a constant c1 and the line fitted to the 1915–1931 data is specified by a constant c2, then the the actual function fitted to the data is y = f(x) where

images

The question now is to find the values of c1 and c2 that provide the best fit of the function f(x) to the data in the sense of minimizing the sum-of-squares measure, denoted by S, of the fit.

Before we do this, recall, that for any sequence of n points a1, a2, a3,···, an, the sum of these points can be written as:

images

(Also note that i does not have to start at i = 1, but could start at any integer value less than or equal to n).

Returning to our problem, if we define the value of this measure to be S, where

images

then we can plot the value of S for different choices of c1 and c2. This is best done by considering separately the sums

images

and

images

  1. By calculating S1 for a range of values of c1 and S2 for a range of values of c2 and then plotting the results, find to two significant figures for the values of c1 and c2 that minimize the sum S = S1 + S2. (Find these by “playing around” with the functions until you find the appropriate intervals over which to plot the two sums.) This is a graphical approach to finding the best-fitting function f(x) defined above.
  2. Can you think of a way that you might use differential calculus to solve this problem analytically? Once you find a way to do this, then solve the problem analytically and compare this analytical solution with your graphical solution.
  3. What advantages does the analytical solution have over the graphical solution and vice versa?

* A. S. Perelson, A. U. Neumann, M. Markowitz, J. M. Leonard, D. D. Ho, “HIV-1 Dynamics In Vivo: Virion Clearance Rate, Infected Cell Lifespan, and Viral Generation Time,” Science 271 (1996): 1582–1586. Also, A. S. Perelson and P. W. Nelson, “Mathematical Analysis of HIV-1 Dynamics In Vivo,” SIAM Review 41(1999):3–44.

* D. Ludwig, D. D. Hones, and C. S. Holling, “Qualitative Analysis of Insect Outbreak Systems: The Spruce Budworm and Forest,” Journal of Animal Ecology 47(1978): 315–332.

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