Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 33 docx

8.3 Derivatives of Sums, Products, Quotients, and Power Functions 301 13. f(x)=|x|−x 14. f(x)=  x 2 , x ≥0 x, x<0 15. The function f(x)=x+|x|is continuous at x = 0 but not differentiable at x =0. Explain, using the definitions of continuity at a point and differentiability at a point. 16. Let f(x)= 1 x 2 +1 . (a) Sketch the graph of f . Do this by graphing x 2 + 1 and looking at the reciprocal. Check your answer with a computer or calculator. (b) Make a rough sketch of f  based on your graph of f . (c) Find f  (x) analytically, using the Quotient Rule. Graph f  . 17. Let f(x)=  x 3 , x ≤1 kx, x>1 (a) What values of k makes f(x)acontinuous function? (b) If k is chosen so that f is continuous at x = 1, is f differentiable there? 18. Find the equation of the tangent line to f(x)=x(x 2 +2) at x = 1. 19. For what value(s) of x is the slope of the tangent line to f(x)= 1 3 x 3 equal to 1? For Problems 20 through 23, find the following: 20. d dx  x +1 x 3 + 3x + 1  21. d dx  π πx + π  22. d dx  2x 2 +x+1 √ 2x  23. d dx  x 2 +5x 2x 10  PART III Exponential, Polynomial, and Rational Functions— with Applications 9 CHAPTER Exponential Functions 9.1 EXPONENTIAL GROWTH: GROWTH AT A RATE PROPORTIONAL TO AMOUNT Look around you at quantities in the world that change. While some functions are linear (the rate of change of output with respect to input is constant), you don’t need to look far to see some very different patterns of growth and decay. Leave a piece of bread on a shelf for too long and mold will grow; not only will the amount of mold increase with time, but the rate at which the mold grows will also increase with time. Or observe the early stages of the spread of infectious disease in a large susceptible population; the rate at which the disease spreads increases as the number of infected people increases. Think about the rate of growth of money in an interest-earning bank account; the more money in the account, the faster additional money will accumulate. For instance, 3% of $1000 is greater than 3% of $100. The growth of bread mold, the spread of disease, and the accumulation of money in a bank account can all be modeled using exponential functions. Similarly, there are quantities that diminish at a rate proportional to the amount present. In particular, radioactive isotopes decay at a rate proportional to the amount present; modeling this decay by exponential functions enables scientists to date various events. We will see that if a quantity grows (or 303 304 CHAPTER 9 Exponential Functions decays) so that its rate of change at any moment is proportional to the amount of the quantity present at that moment, then the quantity grows (or decays) exponentially. ◆ EXAMPLE 9.1 Suppose you are a biologist growing a bacterial culture; you put some bacteria in a petri dish of agar where they will have everything they will need (food, space, etc.) to live and reproduce with abandon. At the start of the experiment, there are 5 bacteria in the petri dish. After an hour, there are 10 bacteria. One hour later, there are 20 bacteria, and the hour after that there are 40. Think about the average rate of change of the bacteria population from one hour to the next. In the first hour, the average rate of change was 5 bacteria per hour, but in the second hour the average rate of change was 10 bacteria per hour, and in the third hour it was 20 bacteria per hour. The bacteria population at time t is clearly not increasing linearly, because the rate of change is not constant. Let’s make a table of the data in order to come up with a model of the number of bacteria as a function of time. We’ll measure time in hours, setting our benchmark time, t = 0, to be when we counted 5 bacteria in the dish. We’ll denote the number of bacteria in the petri dish at time t by B(t). t (in hours) B(t) Change in B since previous hour 0 5 (not applicable) 1105 22010 34020 Let’s begin by looking at the hourly rate of change of the population. Notice that the average rate of growth each hour is equal to the population the previous hour. This is reasonable, as you would expect that the number of new bacteria produced would depend on the number of bacteria already living. 1 In this case it appears that on average each bacterium produces one new bacterium every hour. Looking at B(t) we see that each hour the number of bacteria doubles. We’ll use this observation to figure out an equation for B(t). Let’s rewrite the data in a way that makes the pattern more apparent. t (in hours) B(t) 05 110=5·2 220=5·2·2=5·2 2 340=5·2·2·2=5·2 3 . . . . . . t 5·2 t B(t), the number of bacteria at time t, is given by B(t) = 5 · 2 t . This equation makes sense; 2 we start with 5 bacteria and every hour the population is multiplied by 2. The function 1 Most bacteria reproduce by binary fission; a new cell wall is formed and the one-cell organism splits in two. 2 We should note here that when t is not an integer, this equation may give us a fractional part of a bacterium. For example, B(0.5) = 7.07 As we have seen before, scientists often model discrete phenomena with continuous models. 9.1 Exponential Growth: Growth at a Rate Proportional to Amount 305 B(t) = 5 · 2 t is an example of an exponential function. In the expression b a the “a”atthe top right is called the exponent and the “b” is called the base. Figure 9.1 is a graph of the number of bacteria versus t. Figure 9.2 illustrates the average rate of change of population over a one-hour time period by the slope of the secant line through (t, B(t)) and (t + 1, B(t + 1)) for two different values of t. t B 5 10 15 20 25 30 35 1234 Figure 9.1 1 2 3 5 10 15 20 t B 25 30 35 Figure 9.2 Average rate of change of B over [t, t + 1] = B(t + 1) − B(t) (t + 1) − t = 5 · 2 t+1 − 5 · 2 t 1 = 5 · 2 t (2 − 1) = 5 · 2 t = B(t). ◆ Generally the rate of growth of a bacteria population under favorable conditions, while not equal to the population, is proportional 3 to the population, where the constant of proportionality depends on the organism. Below we give a real-life example. ◆ EXAMPLE 9.2 Escherichia coli is a bacterium that lives in the large intestine and helps the body with digestion. If E. coli enter the abdomen through a perforation in the intestine their presence can cause peritonitis, a potentially fatal disease if untreated. In order to understand the progress of the disease it is necessary to understand the growth of a population of E. coli. 3 Recall that y is proportional to x means y = kx for some constant k, called the constant of proportionality. 306 CHAPTER 9 Exponential Functions Microbiologists have found that under favorable conditions a population of E. coli doubles every 20 minutes. Such rapid growth rates are typical of bacteria. Suppose that at time t = 0, t measured in minutes, a person has 10 E. coli in his abdomen. How many E. coli are in this unfortunate individual’s abdomen 12 hours later if the condition is untreated and there are no limiting factors? SOLUTION Let’s begin by figuring out how many E. coli will be in his abdomen t minutes later if the condition is left untreated. Let B(t) = the number of bacteria in his abdomen at time t.We’ll make a table and search for a pattern. t (in minutes) B(t) 010 20 10 · 2 40 (10 · 2) · 2 = 10 · 2 · 2 = 10 · 2 2 60 (10 · 2 2 ) · 2 = 10 · 2 3 80 (10 · 2 3 ) · 2 = 10 · 2 4 . . . . . . t 10 · 2 t/20 Notice that on the right-hand side, 2 is raised to a power that is always 1/20 of t. At t = 20, for example, 2 is raised to the (1/20) · 20 = 1; at t = 60, 2 is raised to the (1/20) · 60 = 3, and so on. So B(t) = 10 · 2 t/20 . Using this equation, we find that 12 hours later (12 hr · 60 min/hr = 720 min later) there are 10 · 2 720/20 = 10 · 2 36 ≈ 10 · (6.87 · 10 10 ) = 6.87 · 10 11 bacteria. The number of E. coli in the poor fellow’s abdomen has grown from a mere 10 to about 687,000,000,000, or 687 billion after only 12 hours. This example shows why it is important to treat bacterial infections without delay. 4 ◆ EXERCISE 9.1 Let B(t) = 10 · 2 t/20 . Show analytically that the average rate of change of B(t) over the interval [t , t + 20] is B(t) 20 . That is, compute the average rate of change of B over the interval [t, t + 20] as we did at the end of Example 9.1. If you have trouble with the algebra involved, read Section 9.2 and then try again. Let’s determine growth equations for other patterns of population growth; you’ll see that we could have arrived at the last two growth equations bypassing the construction of a table of values. Let P 0 denote the size of a population at some benchmark time t = 0, t in years, and let P(t)be the population at time t. For each of the examples below, try to arrive at the growth equation yourself; cover up the right-hand side of the page with a sheet of paper. (Note: The answers can be expressed in a myriad of ways. If your answer is not the one given, determine whether or not it is equivalent.) 4 Another example of disease caused by bacterial infection is an equine disease manifested by the appearance of huge welts on a horse’s skin. If treated immediately the disease is completely curable, but left untreated it can be fatal. When a groom at the Belmont Racetrack near New York informed his trainer that his horse had “gotten eaten real badly by mosquitos last night,” the frenzy that ensued and the urgency with which the veterinarian worked made it clear that the groom had misdiagnosed the cause of the welts. 9.1 Exponential Growth: Growth at a Rate Proportional to Amount 307 ◆ EXAMPLE 9.3 (a) If the population doubles every year, then P(t)= P 0 ·2 t . (b) If the population triples every year, then P(t)= P 0 ·3 t . (c) If the population is cut in half every year, then P(t)= P 0 ( 1 2 ) t . (d) If the population increases by 30% every year, then P(t)= P 0 (1.30) t . (e) If the population decreases by 10% every year, then P(t)= P 0 (1−0.10) t = P 0 (0.90) t . (f) If the population doubles every 5 years, then P(t)= P =P 0 ·2 t/5 . (g) If the population triples every 10 years, then P(t)= P 0 ·3 t/10 . (h) If the population triples every half year, then P(t)= P 0 ·3 2t . (i) If the population increases by 10% every 3 years, then P(t)=P 0 (1.1) t/3 . (j) If the population is cut in half every 3 years, then P(t)= P 0 ( 1 2 ) t/3 . SELECTED EXPLANATIONS (d) To increase A by 30% gives A + 30%A = A + 30 100 A = A + 0.30A = A(1 + .030).In other words, to increase A by 30% is to get 130% of A, or 1.30A. Therefore, each year the population is 130% of the previous year’s population. (e) To decrease A by 10% is to be left with 90% of A, or 0.90A. Therefore each year the population is 90% of the previous year’s population. (f) The population must be multiplied by 2 not each year, but every 5 years. We want P(5) to be P 0 · 2 and P(10) to be P 0 · 2 2 . In the expression P(t)=P 0 ·2 t/5 , each time t goes up by 5 the exponent increases by 1. (h) Notice that when t = 1 2 we have P( 1 2 ) = P 0 · 3 as specified. After one year the popu- lation triples twice: P(1)=P 0 ·3 2 . ◆ COMMON ERROR Suppose a population doubles every three years. The temptation to write P(t)=P 0 ·2 3t can be thwarted by seeing what happens when t = 3. Using this incorrect equation we get P(3)=P 0 ·2 3·3 =P 0 ·2 9 , and the error becomes glaringly clear. We can model the situation correctly using P(t)=P 0 ·2 t/3 . PROBLEMS FOR SECTION 9.1 1. The number of bacteria in a certain culture is known to triple every day. Suppose that at noon today there are 200 bacteria. (a) Construct a table of values to find a function that gives the number of bacteria after t days. (b) Approximately what was the population count at noon yesterday? At noon 4 days ago? (c) From now on, suppose the population at noon today is called B 0 rather than being specifically 200. Find a function that gives the number of bacteria after t days. (d) Express the number of bacteria as a function of w, where w is time measured in weeks. (e) How many bacteria will be present at noon one week from today? 308 CHAPTER 9 Exponential Functions 2. As the story goes, a long time ago there was a Persian ruler who enjoyed the game of chess so much that in order to demonstrate his gratitude he offered its inventor any reward the man wanted. When the man was called in front of the ruler he requested 1 grain of rice for the first square of the chessboard, 2 for the second, 4 for the third, 8 for the fourth, and so on until rice was allocated for all 64 squares of the chessboard. At first the ruler thought that the request was quite modest—he would have been happy to give the man jewels instead of rice. Eventually he realized that the request was not modest at all and—according to some versions of this story—he ordered the man beheaded. What was all the fuss about? (This classic story is so popular that it has even been used on the TV show “I Love Lucy.”) (a) How many grains of rice were allocated to the 64th square? (b) If we assume that a grain of rice is 0.02 g, approximately how many grams of rice are allocated to the 64th square? Compare this with the annual world production of rice (≈ 4x10 8 tons in 1980). Note: 1 ton is 907.18 kilograms. For general edification, a table of commonly used prefixes is supplied below. Symbol Factor G giga- 1 000 000 000 = 10 9 M mega- 1 000 000 = 10 6 k kilo- 1 000 = 10 3 m milli- 0.001 = 10 −3 n nano- 0.000000 001 = 10 −9 (c) Challenge problem (optional): Find the total mass of the rice allocated to all 64 squares. Note: If you have to add 64 numbers without any shortcut, please skip this part of the problem! (Facts about rice production are taken from J.C. Newby, Mathematics for the Biological Sciences, Oxford University Press, 1980. p. 53. The story itself is an old standard.) 3. Consider two strains of bacteria, one (E. coli) whose population doubles every 20 minutes and another, strain X, whose population doubles every 15 minutes. Suppose that at present the number of E. coli is 600 and the number of bacteria of strain X is 100. (a) Express the number of E. coli as a function of h, the number of hours from now. (b) Express the number of strain X bacteria as a function of h, the number of hours from now. (c) After approximately how many hours will the populations be equal in number? 4. In the 1980s the small town of Old Bethpage, New York, made the front page of the New York Times magazine section as an illustration of what was termed a “dying suburb.” In Old Bethpage schools are being converted to nursing homes as the population ages and the baby boomers move out. Suppose that the number of school-age children in 1980 was C 0 , and was decreasing at a rate of 6% per year. Let’s assume that the number of school-age children continues to drop at a rate of 6% each year. Let C(t) be the number of school-age children in Old Bethpage t years after 1980. 9.2 Exponential: The Bare Bones 309 (a) Find C(t). (b) Express the number of school-age children in Old Bethpage in 1994 as a percentage of the 1980 population. (c) Use your calculator to estimate the year in which the population of school-age children in Old Bethpage will be half of its size in 1980. 5. In the middle of the 1994–95 academic year, in the middle of the week, in the middle of the day, there was a bank robbery and subsequent shootout in the middle of Harvard Square. Throughout the afternoon the news spread by word-of-mouth. Suppose that at the time of the occurrence 30 people know the story. Every 15 minutes each person who knows the news passes it along to one other person. Let N(t) be the number of people who know at time t. (a) Make a table with time in one column and N(t) in the other. Identify the pattern and write N as a function of t. (b) If you’ve written your equation for N(t) with t in minutes, convert to hours. If you’ve done it in hours, convert to minutes. Make a table to check your answers. (It’s easy to make a mistake the first time you do this.) 9.2 EXPONENTIAL: THE BARE BONES You are going to be using exponential functions a great deal, so you need to be loose and limber with exponential algebra. You need to know exponential functions like the back of your hand. You will want to be able to picture an exponential function in your mind, with your eyes closed and your calculator behind your back. So without further ado, let’s start from ground zero and rapidly work our way up. The Algebra of Exponents If n is a positive integer, then b n is b multiplied by itself n times. In fact, b a can be defined for any real exponent a and positive base b. How to do this will become clearer as we proceed. The Basics of Exponential Algebra Let x be a positive number. Then ExponentLaws i) x a x b = x a+b ii) x a x b = x a−b iii) (x a ) b = x ab These statements should make sense intuitively when you think of a and b as positive integers. (For (ii) think of a>bfor starters.) In actuality we will define x a so that these statements are true for any exponents a and b. You have undoubtedly seen these laws before; you’ll need to be able to recall and apply them without problem. Practice will aid application 310 CHAPTER 9 Exponential Functions and concrete examples can supplement recall. Suppose, for instance, that you can’t recall whether x a x b is x a+b or x ab . Work through a simple concrete example to clarify your thoughts. Try an example where a and b are small positive integers, as done below. 5 x 2 x 3 = (xx)(xxx) =x 5 ,sox a x b must be x a+b , not x ab . If the three exponent rules are to work for rational exponents, then the following must be true. x 0 = 1 x −k = 1 x k x 1/n = n √ x, i.e., the nth root of x These results can be deduced from the first three laws as follows. 1 = x b x b = x b−b = x 0 ,sox 0 =1 1 x k = x 0 x k = x 0−k = x −k ,sox −k = 1 x k (x 1/n ) n = x n/n = x,sox 1/n is a number that, when raised to the nth power, gives x. Therefore, x 1/n = n √ x. Notice the following: 1 x −k = (x −k ) −1 = x k . x k/n =(x k ) 1/n = n  x k and x k/n =(x 1/n ) k =( n √ x) k ; consequently, n  x k = ( n √ x) k . At this point, we have defined b a for any rational exponent a and positive number b.In fact, we can define b t for any real number t. For now we’ll deal with irrational exponents simply by approximating the irrational number by a sequence of rational ones. A more satisfactory definition can be given after taking up logarithmic functions. Suppose s is irrational, r and t are rational, and r<s<t.Then b r <b s <b t .Wedefine b s so as to make the graph of b x continuous. Why do we restrict ourselves to b positive? If b =0, then b t is undefined for t ≤ 0. If b is negative, then b t is not a real number for many values of t. For example, (−4) 1/2 = √ −4 is not a real number; there is no real number whose square is −4. Try graphing y = (−8) t on your graphing calculator or computer and see what happens. We know that (−8) 1/3 = 3 √ −8 =−2and (−8) −1 = 1 −8 , but how does your calculator respond to the command to graph (−8) t ? How does it respond if you ask for √ −8, that is, (−8) 1/2 ? Different calculators respond differently; can you make sense out of your calculator’s response? Keep the following points in mind: An exponent refers only to the number immediately below it unless otherwise indicated by parentheses. 5 Make sure you choose a and b so that a + b = ab; otherwise you have defeated the purpose of your experiment! . In order to understand the progress of the disease it is necessary to understand the growth of a population of E. coli. 3 Recall that y is proportional to x means y = kx for some constant k, called. great deal, so you need to be loose and limber with exponential algebra. You need to know exponential functions like the back of your hand. You will want to be able to picture an exponential function. in order to demonstrate his gratitude he offered its inventor any reward the man wanted. When the man was called in front of the ruler he requested 1 grain of rice for the first square of the chessboard,