Additional Derivative Topics

Transcription

1 BARNMC04_ QXD 2/21/07 1:27 PM Page 216 Aitional Derivative Topics CHAPTER The Constant e an Continuous Compoun Interest 4-2 Derivatives of Eponential an Logarithmic Functions 4-3 Derivatives of Proucts an Quotients 4-4 The Chain Rule I N T R O D U C T I O N In this chapter, we complete our iscussion of erivatives by first looking at the ifferentiation of eponential forms. Net, we will consier the ifferentiation of logarithmic forms. Finally, we will eamine some aitional topics an applications involving all the ifferent types of functions we have encountere thus far, incluing the general chain rule. You will probably fin it helpful to review some of the important properties of the eponential an logarithmic functions given in Chapter 2 before proceeing further. 4-5 Implicit Differentiation 4-6 Relate Rates 4-7 Elasticity of Deman Chapter 4 Review Review Eercise 216

2 BARNMC04_ QXD 2/21/07 1:27 PM Page 217 Section 4-1 The Constant e an Continuous Compoun Interest 217 Section 4-1 THE CONSTANT e AND CONTINUOUS COMPOUND INTEREST The Constant e Continuous Compoun Interest In Chapter 2, both the eponential function with base e an continuous compoun interest were introuce informally. Now, with limit concepts at our isposal, we can give precise efinitions of e an continuous compoun interest. The Constant e The special irrational number e is a particularly suitable base for both eponential an logarithmic functions. The reasons for choosing this number as a base will become clear as we evelop ifferentiation formulas for the eponential function e an the natural logarithmic function ln. In precalculus treatments (Chapter 2), the number e is informally efine as the irrational number that can be approimate by the epression >n24 n by taking n sufficiently large. Now we will use the limit concept to formally efine e as either of the following two limits: DEFINITION The Number e e = lim a1 + 1 n n: q n b e = Á or, alternatively, e = lim s: s2 1>s We will use both of these limit forms. [Note: If s = 1>n, then, as n : q, s : 0. ] The proof that the inicate limits eist an represent an irrational number between 2 an 3 is not easy an is omitte here. I N S I G H T The two limits use to efine e are unlike any we have encountere thus far. Some people reason (incorrectly) that both limits are 1, since 1 + s : 1 as s : 0 an 1 to any power is 1. A calculator eperiment on an orinary scientific calculator with a y key can convince you otherwise. Consier the following table of values for s an f1s2 = 11 + s2 1>s an the graph shown in Figure 1 for s close to 0. f(s) f(s) (1 s) 1/s FIGURE 1 s s approaches O from the left : O ; s approaches O from the right s : 0 ; s2 1>s : e ;

3 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics Compute some of the table values with a calculator yourself, an also try several values of s even closer to 0. Note that the function is iscontinuous at s = 0. Eactly who iscovere e is still ebate. It is name after the great mathematician Leonhar Euler ( ), who compute e to 23 ecimal places, using >n24 n. Continuous Compoun Interest Now we will see how e appears quite naturally in the important application of compoun interest. Let us start with simple interest, move on to compoun interest, an then procee on to continuous compoun interest. On the one han, if a principal P is borrowe at an annual rate r,* then after t years at simple interest, the borrower will owe the lener an amount A given by A = P + Prt = P11 + rt2 Simple interest (1) On the other han, if interest is compoune n times a year, the borrower will owe the lener an amount A given by A = Pa1 + r n b nt Compoun interest (2) where r n is the interest rate per compouning perio an nt is the number of compouning perios. Suppose that P, r, an t in equation (2) are hel fie an n is increase. Will the amount A increase without boun, or will it ten to approach some limiting value? Let us perform a calculator eperiment before we attack the general limit problem. If P = $100, r = 0.06, an t = 2 years, then A = 100a n b 2n We compute A for several values of n in Table 1. The biggest gain appears in the first step; then the gains slow own as n increases. In fact, it appears that A tens to approach $ as n gets larger an larger. TABLE 1 Compouning Frequency Annually 1 $ Semiannually Quarterly Monthly Weekly Daily Hourly 8, n A = 100a n b 2n Eplore & Discuss 1 (A) Suppose that $1,000 is eposite in a savings account that earns 6% simple interest. How much will be in the account after 2 years? (B) Suppose that $1,000 is eposite in a savings account that earns compoun interest at a rate of 6% per year. How much will be in the account after 2 years if interest is compoune annually? Semiannually? Quarterly? Weekly? (C) How frequently must interest be compoune at the 6% rate in orer to have $1,150 in the account after 2 years? * If r is the interest rate written as a ecimal, then 100r% is the rate in percent. For eample, if r = 0.12, we have 100r% = % = 12%. The epressions 0.12 an 12% are therefore equivalent. Unless state otherwise, all formulas in this book use r in ecimal form.

4 BARNMC04_ QXD 2/21/07 1:27 PM Page 219 Section 4-1 The Constant e an Continuous Compoun Interest 219 Now we turn back to the general problem for a moment. Keeping P, r, an t fie in equation (2), we compute the following limit an observe an interesting an useful result: lim Pa1 + r nt n: q n b = P lim a1 + r 1n>r2rt n: q n b = P lim s: s2 1>s 4 rt = P3lim11 + s2 1>s 4 rt s:0 = Pe rt Insert r/r in the eponent an let s = r>n. Note that n : q implies s : 0. Use a limit property.* lim s: s21>s = e The resulting formula is calle the continuous compoun interest formula, a very important an wiely use formula in business an economics. Continuous Compoun Interest where P = principal r = annual nominal interest rate compoune continuously t = time in years A = amount at time t A = Pe rt EXAMPLE 1 Computing Continuously Compoune Interest If $100 is investe at 6% compoune continuously, what amount will be in the account after 2 years? How much interest will be earne? SOLUTION A = Pe rt = 100e % is equivalent to r = L $ (Compare this result with the values calculate in Table 1.) The interest earne is $ $100 = $ MATCHED PROBLEM 1 What amount (to the nearest cent) will an account have after 5 years if $100 is investe at an annual nominal rate of 8% compoune annually? Semiannually? Continuously? EXAMPLE 2 SOLUTION Graphing the Growth of an Investment Recently, Union Savings Bank offere a 5-year certificate of eposit (CD) that earne 5.75% compoune continuously. If $1,000 is investe in one of these CDs, graph the amount in the account as a function of time for a perio of 5 years. We want to graph A = 1,000e t 0 t 5 Using a calculator, we construct a table of values (Table 2). Then we graph the points from the table an join the points with a smooth curve (Fig. 2). * The following new limit property is use: If lim eists, then lim :c 3f124 p = 3lim :c f124 p :c f12, provie that the last epression names a real number. Following common usage, we will often write at 6% compoune continuously, unerstaning that this means at an annual nominal rate of 6% compoune continuously.

5 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics TABLE 2 t A($) 1500 A 0 1, , , , , , t FIGURE 2 I N S I G H T Depening on the omain, the graph of an eponential function can appear to be linear. Table 3 shows the slopes of the line segments connecting the ots in Figure 2. Since these slopes are not ientical, this graph is not the graph of a straight line. TABLE 3 t A m 0 1, , , , , ,333 F 59 F 63 F 66 F 71 F 74 MATCHED PROBLEM 2 If $5,000 is investe in a Union Savings Bank 4-year CD that earns 5.61% compoune continuously, graph the amount in the account as a function of time for a perio of 4 years. EXAMPLE 3 Computing Growth Time How long will it take an investment of $5,000 to grow to $8,000 if it is investe at 5% compoune continuously? SOLUTION Starting with the continous compoun interest formula A = Pe rt, we must solve for t: A = Pe rt 8,000 = 5,000e 0.05t e 0.05t = 1.6 ln e 0.05t = ln t = ln 1.6 ln 1.6 t = 0.05 t L 9.4 years Divie both sies by 5,000 an reverse the equation. Take the natural logarithm of both sies recall that log b b =. FIGURE 3 y 1 = 5,000e 0.05 y 2 = 8,000 Figure 3 shows an alternative metho for solving Eample 3 on a graphing calculator.

6 BARNMC04_ QXD 2/21/07 1:27 PM Page 221 Section 4-1 The Constant e an Continuous Compoun Interest 221 MATCHED PROBLEM 3 How long will it take an investment of $10,000 to grow to $15,000 if it is investe at 9% compoune continuously? EXAMPLE 4 Computing Doubling Time How long will it take money to ouble if it is investe at 6.5% compoune continuously? SOLUTION Starting with the continuous compoun interest formula A = Pe rt, we must solve for t, given A = 2P an r = 0.065: 2P = Pe 0.065t Divie both sies by P an reverse the equation. e 0.065t = 2 Take the natural logarithm of both sies. ln e 0.065t = ln t = ln 2 t = ln t L years MATCHED PROBLEM 4 How long will it take money to triple if it is investe at 5.5% compoune continuously? Eplore & Discuss 2 You are consiering three options for investing $10,000: at 7% compoune annually, at 6% compoune monthly, an at 5% compoune continuously. (A) Which option woul be the best for investing $10,000 for 8 years? (B) How long woul you nee to invest your money for the thir option to be the best? Answers to Matche Problems 1. $146.93; $148.02; $ A = 5,000e t yr yr t A($) A 0 5, , , , , t Eercise 4-1 A Use a calculator to evaluate A to the nearest cent in Problems 1 an A = $1,000e 0.1t for t = 2, 5, an 8 2. A = $5,000e 0.08t for t = 1, 4, an If $6,000 is investe at 10% compoune continuously, graph the amount in the account as a function of time for a perio of 8 years. B 4. If $4,000 is investe at 8% compoune continuously, graph the amount in the account as a function of time for a perio of 6 years. In Problems 5 10, solve for t or r to two ecimal places. 0.1r 5. 2 = e 0.06t 6. 2 = e 0.03t 7. 3 = e 8. 3 = e 0.25t 9. 2 = e 5r = e 10r

7 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics C In Problems 11 an 12, use a calculator to complete each table to five ecimal places n 31 11>n24 n ,000 10, ,000 1,000,000 10,000,000 T p q e = Á s 11 + s2 1>s p p 0 e = Á 13. Use a calculator an a table of values to investigate lim 11 + n21>n n: q Do you think this limit eists? If so, what o you think it is? 14. Use a calculator an a table of values to investigate Do you think this limit eists? If so, what o you think it is? 15. It can be shown that the number e satisfies the inequality a1 + 1 n n b 6 e 6 a1 + 1 n + 1 n b Illustrate this conition by graphing in the same viewing winow, for 1 n It can be shown that y 1 = >n2 n y 2 = L e y 3 = >n2 n + 1 e s = lim a1 + 1 s s:0 + s b lim a1 + s n n: q n b for any real number s. Illustrate this equation graphically for s = 2 by graphing y 1 = >n2 n y 2 = L e 2 n Ú 1 in the same viewing winow, for 1 n 50. Applications 17. Continuous compoun interest. Recently, Provient Bank offere a 10-year CD that earns 5.51% compoune continuously. (A) If $10,000 is investe in this CD, how much will it be worth in 10 years? (B) How long will it take for the account to be worth $15,000? 18. Continuous compoun interest. Provient Bank also offers a 3-year CD that earns 5.28% compoune continuously. (A) If $10,000 is investe in this CD, how much will it be worth in 3 years? (B) How long will it take for the account to be worth $11,000? 19. Present value. A note will pay $20,000 at maturity 10 years from now. How much shoul you be willing to pay for the note now if money is worth 5.2% compoune continuously? 20. Present value. A note will pay $50,000 at maturity 5 years from now. How much shoul you be willing to pay for the note now if money is worth 6.4% compoune continuously? 21. Continuous compoun interest. An investor bought stock for $20,000. Five years later, the stock was sol for $30,000. If interest is compoune continuously, what annual nominal rate of interest i the original $20,000 investment earn? 22. Continuous compoun interest. A family pai $40,000 cash for a house. Fifteen years later, the house was sol for $100,000. If interest is compoune continuously, what annual nominal rate of interest i the original $40,000 investment earn? 23. Present value. Solving A = Pe rt for P, we obtain P = Ae -rt which is the present value of the amount A ue in t years if money earns interest at an annual nominal rate r compoune continuously. (A) Graph P = 10,000e -0.08t, 0 t 50. (B) lim t: q 10,000e -0.08t =? [Guess, using part (A).] [Conclusion: The longer the time until the amount A is ue, the smaller is its present value, as we woul epect.]

8 BARNMC04_ QXD 2/21/07 1:27 PM Page 223 Section 4-1 The Constant e an Continuous Compoun Interest Present value. Referring to Problem 23, in how many years will the $10,000 have to be ue in orer for its present value to be $5,000? 25. Doubling time. How long will it take money to ouble if it is investe at 7% compoune continuously? 26. Doubling time. How long will it take money to ouble if it is investe at 5% compoune continuously? 27. Doubling rate. At what nominal rate compoune continuously must money be investe to ouble in 8 years? 28. Doubling rate. At what nominal rate compoune continuously must money be investe to ouble in 10 years? 29. Growth time. A man with $20,000 to invest ecies to iversify his investments by placing $10,000 in an account that earns 7.2% compoune continuously an $10,000 in an account that earns 8.4% compoune annually. Use graphical approimation methos to etermine how long it will take for his total investment in the two accounts to grow to $35, Growth time. A woman invests $5,000 in an account that earns 8.8% compoune continuously an $7,000 in an account that earns 9.6% compoune annually. Use graphical approimation methos to etermine how long it will take for her total investment in the two accounts to grow to $20, Doubling times (A) Show that the oubling time t (in years) at an annual rate r compoune continuously is given by t = ln 2 r (B) Graph the oubling-time equation from part (A) for 0.02 r Is this restriction on r reasonable? Eplain. (C) Determine the oubling times (in years, to two ecimal places) for r = 5%, 10%, 15%, 20%, 25%, an 30%. 32. Doubling rates (A) Show that the rate r that oubles an investment at continuously compoune interest in t years is given by r = ln 2 t (B) Graph the oubling-rate equation from part (A) for 1 t 20. Is this restriction on t reasonable? Eplain. (C) Determine the oubling rates for t = 2, 4, 6, 8, 10, an 12 years. 33. Raioactive ecay. A mathematical moel for the ecay of raioactive substances is given by where Q 0 = amount of the substance at time t = 0 If the continuous compoun rate of ecay of raium per year is r = , how long will it take a certain amount of raium to ecay to half the original amount? (This perio is the half-life of the substance.) 34. Raioactive ecay. The continuous compoun rate of ecay of carbon-14 per year is r = How long will it take a certain amount of carbon-14 to ecay to half the original amount? (Use the raioactive ecay moel in Problem 33.) 35. Raioactive ecay. A cesium isotope has a half-life of 30 years. What is the continuous compoun rate of ecay? (Use the raioactive ecay moel in Problem 33.) 36. Raioactive ecay. A strontium isotope has a half-life of 90 years. What is the continuous compoun rate of ecay? (Use the raioactive ecay moel in Problem 33.) 37. Worl population. A mathematical moel for worl population growth over short intervals is given by where r = continuous compoun rate of ecay t = time in years Q = amount of the substance at time t P 0 = population at time t = 0 r = continuous compoun rate of growth t = time in years Q = Q 0 e rt P = P 0 e rt P = population at time t How long will it take worl population to ouble if it continues to grow at its current continuous compoun rate of 1.3% per year? 38. U.S. population. How long will it take for the U.S. population to ouble if it continues to grow at a rate of 0.85% per year? 39. Population growth. Some unerevelope nations have population oubling times of 50 years. At what continuous compoun rate is the population growing? (Use the population growth moel in Problem 37.) 40. Population growth. Some evelope nations have population oubling times of 200 years. At what continuous compoun rate is the population growing? (Use the population growth moel in Problem 37.)

9 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics Section 4-2 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS e The Derivative of The Derivative of ln Other Logarithmic an Eponential Functions Eponential an Logarithmic Moels In this section, we obtain formulas for the erivatives of logarithmic an eponential functions. A review of Sections 2 4 an 2 5 may prove helpful. In particular, recall that f12 = e is the eponential function with base e L an the inverse of the function e is the natural logarithm function ln. More generally, if b is a positive real number, b Z 1, then the eponential function b with base b, an the logarithmic function log b with base b, are inverses of each other. The Derivative of e In the process of fining the erivative of e, we will use (without proof) the fact that e h - 1 lim h:0 h = 1 (1) Eplore & Discuss 1 Complete Table 1. TABLE 1 h : 0 ; e h 1 h Do your calculations make it reasonable to conclue that Discuss. We now apply the four-step process (Section 3-4) to the eponential function f12 = e. Step 1. Fin f1 + h2. f1 + h2 = e + h = e e h Step 2. Fin f1 + h2 - f12. f1 + h2 - f12 Step 3. Fin. h f1 + h2 - f12 f1 + h2 - f12 Step 4. Fin f 12 = lim. h:0 h h e h - 1 lim h:0 h f1 + h2 - f12 = e e h - e = e 1e h - 12 = 1? = e 1e h - 12 h See Section 2-4. Factor out e. = e a eh - 1 b h

10 BARNMC04_ QXD 2/21/07 1:27 PM Page 225 Section 4-2 Derivatives of Eponential an Logarithmic Functions 225 Thus, f 12 = lim h:0 f1 + h2 - f12 h = lim e a eh - 1 b h:0 h = e lim a eh - 1 b h:0 h = e # 1 = e e = e Use the limit in (1). The erivative of the eponential function is the eponential function. EXAMPLE 1 Fining Derivatives Fin f 12 for (A) (B) f12 = 5e f12 = -7 e + 2e + e 2 SOLUTIONS (A) f 12 = 5e (B) f 12 = -7e e e Remember that e is a real number, so the power rule (Section 3 5) is use to fin the erivative of e. The erivative of the eponential function e, however, is e. Note that e 2 L is a constant, so its erivative is 0. MATCHED PROBLEM 1 Fin f 12 for (A) (B) f12 = 4e f12 = e e CAUTION: e Z e - 1 e = e The power rule cannot be use to ifferentiate the eponential function. The power rule applies to eponential forms n where the eponent is a constant an the base is a variable. In the eponential form e, the base is a constant an the eponent is a variable. The Derivative of ln We summarize some important facts about logarithmic functions from Section 2 5: SUMMARY Recall that the inverse of an eponential function is calle a logarithmic function. For b 7 0 an b Z 1, Logarithmic form y = log b Domain: 10, q2 Range: 1- q, q2 is equivalent to Eponential form = b y Domain: 1- q, q2 Range: 10, q2

11 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics The graphs of y = log an y = b b are symmetric with respect to the line y =. (See Figure 1.) 10 y y b y 5 y log b FIGURE 1 Of all the possible bases for logarithmic functions, the two most wiely use are log = log 10 Common logarithm (base 10) ln = log e Natural logarithm (base e) We are now reay to use the efinition of the erivative an the four-step process iscusse in Section 3-4 to fin a formula for the erivative of ln. Later we will eten this formula to inclue log b for any base b. Let f12 = ln, 7 0. Step 1. Fin f1 + h2. f1 + h2 = ln1 + h2 ln1 + h2 cannot be simplifie. Step 2. Fin f1 + h2 - f12. f1 + h2 - f12 = ln1 + h2 - ln Use ln A - ln B = ln A B. f1 + h2 - f12 Step 3. Fin. h f1 + h2 - f12 h = + h = ln ln1 + h2 - ln h = 1 h ln + h = # 1 h ln + h Multiply by 1 = > to change form. = 1 c h lna1 + h b Use p ln A = ln A p. = 1 lna1 + h b >h f1 + h2 - f12 Step 4. Fin f 12 = lim. h:0 h f 12 = lim h:0 f1 + h2 - f12 h = lim c 1 h:0 lna1 + h >h b Let s = h>. Note that h : 0 implies s : 0.

12 BARNMC04_ QXD 2/21/07 1:27 PM Page 227 Section 4-2 Derivatives of Eponential an Logarithmic Functions 227 = 1 lim s:0 3ln11 + s21>s 4 Use a new limit property.* = 1 lnc lim 11 + s:0 s21>s = 1 ln e Use the efinition of e. ln e = log e e = 1 = 1 Thus, ln = 1 I N S I G H T In the erivation of the erivative of ln, we use the following properties of logarithms: A ln = ln A - ln B B ln Ap = p ln A We also note that there is no property that simplifies ln1a + B2. (See Theorem 1 in Section 2-5 for a list of properties of logarithms.) EXAMPLE 2 Fining Derivatives Fin y for (A) y = 3e + 5 ln (B) y = 4 - ln 4 SOLUTIONS (A) y =3e + 5 (B) Before taking the erivative, we use a property of logarithms (see Theorem 1, Section 2-5) to rewrite y. y = 4 - ln 4 y = 4-4 ln Use ln M p p ln M. Now take the erivative of both sies. y =4 3-4 MATCHED PROBLEM 2 Fin y for (A) y = ln (B) y = ln 5 + e - ln e 2 Other Logarithmic an Eponential Functions In most applications involving logarithmic or eponential functions, the number e is the preferre base. However, in some situations it is convenient to use a base other than e. Derivatives of y = log an y = b b can be obtaine by epressing these functions in terms of the natural logarithmic an eponential functions. * The following new limit property is use: If lim :c f12 eists an is positive, then lim :c 3ln f124 = ln3lim :c f124.

13 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics We begin by fining a relationship between log b an ln for any base b, b 7 0 an b Z 1. Some of you may prefer to remember the process, others the formula. Thus, y = log b b y = ln b y = ln y ln b = ln y = 1 ln b ln Change to eponential form. Take the natural logarithm of both sies. Recall that ln b y = y ln b. Solve for y. b, b 7 0, b Z 1. log b = 1 ln b ln Change-of-base formula for logarithms* (2) Similarly, we can fin a relationship between b an e for any base y = b ln y = ln b ln y = ln b Take the natural logarithm of both sies. Recall that ln b ln b. Take the eponential function of both sies. y = e ln b Thus, b = e ln b Change-of-base formula for eponential functions (3) Differentiating both sies of equation (2) gives log b 1 1 ln ln b ln b a 1 b It can be shown that the erivative of the function e c, where c is a constant, is the function ce c (see Problems in Eercise 4-2 or the more general results of Section 4 4). Therefore, ifferentiating both sies of equation (3), we have b e ln b ln b b ln b For convenient reference, we list the erivative formulas that we have obtaine for eponential an logarithmic functions: DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS For b 7 0, b Z 1, e = e ln = 1 b = b ln b log b = 1 ln b a 1 b EXAMPLE 3 Fining Derivatives Fin g 12 for (A) g12 = 2-3 (B) g12 = log 4 5 * Equation (2) is a special case of the general change-of-base formula for logarithms (which can be erive in the same way): log b = (log a ) (log a b).

14 BARNMC04_ QXD 2/21/07 1:27 PM Page 229 Section 4-2 Derivatives of Eponential an Logarithmic Functions 229 SOLUTIONS (A) g 12 = 2 ln 2-3 ln 3 (B) First use a property of logarithms to rewrite g(). g12 = log Use log b M p 4 5 = p log b M. g12 = 5 log 4 Take the erivative of both sies. g 12 = 5 ln 4 a 1 b MATCHED PROBLEM 3 Fin g 12 for (A) g12 = (B) g12 = log 2-6 log 5 Eplore & Discuss 2 (A) The graphs of f() = log 2 an g() = log 4 are shown in Figure 2. Which graph belongs to which function? 4 y FIGURE 2 (B) Sketch graphs of f () an g (). (C) The function f() is relate to g() in the same way that f () is relate to g (). What is that relationship? Eponential an Logarithmic Moels EXAMPLE 4 SOLUTION Price Deman Moel An Internet store sells blankets mae of the finest Australian wool. If the store sells blankets at a price of $p per blanket, then the price eman equation is p = Fin the rate of change of price with respect to eman when the eman is 800 blankets, an interpret the result. p = ln If = 800, then p = ln L , or - $0.16 When the eman is 800 blankets, the price is ecreasing about $0.16 per blanket. MATCHED PROBLEM 4 The store in Eample 4 also sells a reversible fleece blanket. If the price eman equation for reversible fleece blankets is p = , fin the rate of change of price with respect to eman when the eman is 400 blankets an interpret the result.

15 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics EXAMPLE 5 Cable TV Subscribers A statistician use ata from the U.S. census to construct the moel S1t2 = 21 ln t + 2 where S(t) is the number of cable TV subscribers (in millions) in year t ( t = 0 correspons to 1980). Use this moel to estimate the number of cable TV subscribers in 2010 an the rate of change of the number of subscribers in 2010 (roun both to the nearest tenth of a million). Interpret these results. SOLUTION Since 2010 correspons to t = 30, we must fin S(30) an S S1302 = 21 ln = 73.4 million 1 S 1t2 = 21 t = 21 t S 1302 = 21 = 0.7 million 30 In 2010 there will be approimately 73.4 million subscribers, an this number is growing at the rate of 0.7 million per year. MATCHED PROBLEM 5 A moel for newspaper circulation is C(t) = 83-9 ln t where C(t) is newspaper circulation (in millions) in year t ( t = 0 correspons to 1980). Use this moel to estimate the circulation an the rate of change of circulation in 2010 (roun both to the nearest tenth of a million). Interpret these results. I N S I G H T On most graphing calculators, eponential regression prouces a function of the form y = a # b. Formula (3) enables you to change the base b (chosen by the graphing calculator) to the more familiar base e: y = a # b = a # e ln b On most graphing calculators, logarithmic regression prouces a function of the form y = a + b ln. Formula (2) enables you to write the function in terms of logarithms to any base that you may prefer: y = a + b ln = a + b1ln 2 log Answers to Matche Problems 1. (A) 4e (B) e 2. (A) (B) + e 3. (A) ln 10 (B) a 1 ln 2-6 ln 5 b 1 4. The price is ecreasing at the rate of $0.18 per blanket. 5. The circulation in 2010 is approimately 52.4 million an is ecreasing at the rate of 0.3 million per year.

16 BARNMC04_ QXD 2/21/07 1:27 PM Page 231 Section 4-2 Derivatives of Eponential an Logarithmic Functions 231 Eercise 4-2 A In Problems 1 14, fin f 12. B f12 = -7e f12 = -2 ln f12 = 6 ln f12 = 3-6e 6. f12 = 9e f12 = e + - ln 8. f12 = ln + 2e f12 = ln f12 = ln f12 = 5 - ln f12 = 4 + ln f12 = ln 2 + 4e 14. f12 = ln ln In Problems 15 22, fin the equation of the line tangent to the graph of f at the inicate value of f12 = 5e f12 = 3 + ln ; = 1 f12 = 2 ln ; = 1 f12 = 3e ; = f12 = e + 1; = f12 = ln 3 ; = e 20. f12 = 1 + ln 4 ; = e 21. f12 = 2 + e ; = f12 = 5e ; = A stuent claims that the line tangent to the graph of f12 = e at = 3 passes through the point (2, 0) (see the figure). Is she correct? Will the line tangent at = 4 pass through (3, 0)? Eplain f() Figure for Refer to Problem 23. Does the line tangent to the graph of f12 = e at = 1 pass through the origin? Are there any other lines tangent to the graph of f that pass through the origin? Eplain A stuent claims that the line tangent to the graph of g() = ln at = 3 passes through the origin (see the figure). Is he correct? Will the line tangent at = 4 pass through the origin? Eplain. 1 1 g() Figure for Refer to Problem 25. Does the line tangent to the graph of f() = ln at = e pass through the origin? Are there any other lines tangent to the graph of f that pass through the origin? Eplain. In Problems 27 30, first use appropriate properties of logarithms to rewrite f(), an then fin f y C In Problems 31 42, fin for the inicate function y. 31. y = log y = 3 log y = y = y = 2 - log 36. y = log y = y = y = 3 ln + 2 log y = - log ln 41. y = 2 + e y = e 3-3 In Problems 43 48, use graphical approimation methos to fin the points of intersection of f() an g() (to two ecimal places). 43. f12 = 10 + ln 10 f12 = ln 1 f12 = ln 4 3 f12 = + 5 ln 6 f12 = e ; g12 = 4 [Note that there are three points of intersection an that e is greater than 4 for large values of.] 6

17 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics 44. f12 = e ; g12 = 5 [Note that there are two points of intersection an that e is greater than 5 for large values of.] 45. f12 = 1ln 2 2 ; g12 = 46. f12 = 1ln 2 3 ; g12 = f12 = ln ; g12 = 1>5 48. f12 = ln ; g12 = 1>4 e ch Eplain why lim = c. h:0 h 50. Use the result of Problem 49 an the four-step process to show that if f12 = e c, then f 12 = ce c. Applications 51. Salvage value. The salvage value S (in ollars) of a company airplane after t years is estimate to be given by S1t2 = 300, t What is the rate of epreciation (in ollars per year) after 1 year? 5 years? 10 years? 52. Resale value. The resale value R (in ollars) of a company car after t years is estimate to be given by R1t2 = 20, t What is the rate of epreciation (in ollars per year) after 1 year? 2 years? 3 years? 53. Bacterial growth. A single cholera bacterium ivies every 0.5 hour to prouce two complete cholera bacteria. If we start with a colony of 5,000 bacteria, after t hours there will be A1t2 = 5,000 # 2 2t = 5,000 # 4 t bacteria. Fin A (t), A (1), an A (5), an interpret the results. 54. Bacterial growth. Repeat Problem 53 for a starting colony of 1,000 bacteria such that a single bacterium ivies every 0.25 hour. 55. Bloo pressure. An eperiment was set up to fin a relationship between weight an systolic bloo pressure in normal chilren. Using hospital recors for 5,000 normal chilren, the eperimenters foun that the systolic bloo pressure was given approimately by P() = 17.5(1 + ln ) where P() is measure in millimeters of mercury an is measure in pouns. What is the rate of change of bloo pressure with respect to weight at the 40-poun weight level? At the 90-poun weight level? 56. Refer to Problem 55. Fin the weight (to the nearest poun) at which the rate of change of bloo pressure with respect to weight is 0.3 millimeter of mercury per poun. 57. Psychology: stimulus/response. In psychology, the Weber Fechner law for the response to a stimulus is R = k ln S S 0 where R is the response, S is the stimulus, an S 0 is the lowest level of stimulus that can be etecte. Fin R>S. 58. Psychology: learning. A mathematical moel for the average of a group of people learning to type is given by N1t2 = ln t t Ú 1 where N(t) is the number of wors per minute type after t hours of instruction an practice (2 hours per ay, 5 ays per week). What is the rate of learning after 10 hours of instruction an practice? After 100 hours? Section 4-3 DERIVATIVES OF PRODUCTS AND QUOTIENTS Derivatives of Proucts Derivatives of Quotients The erivative properties iscusse in Section 3-5 ae substantially to our ability to compute an apply erivatives to many practical problems. In this an the net two sections, we a a few more properties that will increase this ability even further. Derivatives of Proucts In Section 3-5, we foun that the erivative of a sum is the sum of the erivatives. Is the erivative of a prouct the prouct of the erivatives?

18 BARNMC04_ QXD 2/21/07 1:27 PM Page 233 Section 4-3 Derivatives of Proucts an Quotients 233 Eplore & Discuss 1 Let F12 = 2, S12 = 3, an f12 = F12S12 = 5. Which of the following is f 12? (A) F 12S 12 (B) F12S 12 (C) F 12S12 (D) F12S 12 + F 12S12 Comparing the various epressions compute in Eplore & Discuss 1, we see that the erivative of a prouct is not the prouct of the erivatives, but appears to involve a slightly more complicate form. Using the efinition of the erivative an the four-step process, we can show that The erivative of the prouct of two functions is the first function times the erivative of the secon function, plus the secon function times the erivative of the first function. That is, THEOREM 1 PRODUCT RULE If an if F 12 an S 12 eist, then Also, f 12 = F12S 12 + S12F 12 y =FS +SF y = f12 = F12S12 y = F S + S F EXAMPLE 1 Differentiating a Prouct Use two ifferent methos to fin f 12 for f12 = SOLUTION Metho 1. Use the prouct rule: f 12 = = = = Metho 2. Multiply first; then take erivatives: f12 = = f 12 = First times erivative of secon, plus secon times erivative of first MATCHED PROBLEM 1 Use two ifferent methos to fin f 12 for f12 = Some of the proucts we encounter can be ifferentiate by either of the methos illustrate in Eample 1. In other situations, the prouct rule must be use. Unless instructe otherwise, you shoul use the prouct rule to ifferentiate all proucts in this section in orer to gain eperience with the use of this important ifferentiation rule.

19 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics EXAMPLE 2 SOLUTION Tangent Lines Let f12 = (A) Fin the equation of the line tangent to the graph of f() at = 3. (B) Fin the value(s) of where the tangent line is horizontal. (A) First, fin f 12: f 12 = = Then, fin f (3) an f 132: f132 = = = -45 f 132 = = = 12 Now, fin the equation of the tangent line at = 3: y - y 1 = m1-1 2 y 1 = f1 1 2 = f132 = -45 y = m = f = f 132 = 12 y = Tangent line at = 3 (B) The tangent line is horizontal at any value of such that f 12 = 0, so f 12 = = = = = 0 The tangent line is horizontal at = 1 an at = 2. = 1, 2 MATCHED PROBLEM 2 Repeat Eample 2 for f12 = I N S I G H T As Eample 2 illustrates, the way we write f 12 epens on what we want to o with it. If we are intereste only in evaluating f 12 at specifie values of, the form in part (A) is sufficient. However, if we want to solve f 12 = 0, we must multiply an collect like terms, as we i in part (B). EXAMPLE 3 Fining Derivatives Fin f 12 for (A) (B) f12 = 2 3 e f12 = 6 4 ln SOLUTIONS (A) f 12 = 2 3 1e 2 +e = 2 3 e + e = 2 2 e (B) f 12 = 6 4 1ln 2 +1ln = ln = ln = ln 2

20 BARNMC04_ QXD 2/21/07 1:27 PM Page 235 Section 4-3 Derivatives of Proucts an Quotients 235 MATCHED PROBLEM 3 Fin f 12 for (A) f12 = 5 8 e (B) f12 = 7 ln Derivatives of Quotients As is the case with a prouct, the erivative of a quotient of two functions is not the quotient of the erivatives of the two functions. Eplore & Discuss 2 Let T12 = 5, B12 = 2, an Which of the following is f 12? T 12 T 12B12 T12B 12 (A) (B) (C) B 12 3B B124 2 (D) The epressions in Eplore & Discuss 2 suggest that the erivative of a quotient leas to a more complicate quotient than you might epect. In general, if T() an B() are any two ifferentiable functions an then T 12B12 3B T12B 12 3B124 2 f12 = T12 B12 = 5 2 = 3 = T 12B12 - T12B 12 3B124 2 f12 = T12 B12 f 12 = B12T 12 - T12B 12 3B124 2 Thus, The erivative of the quotient of two functions is the enominator function times the erivative of the numerator function, minus the numerator function times the erivative of the enominator function, ivie by the enominator function square. THEOREM 2 QUOTIENT RULE If an if T 12 an B 12 eist, then f 12 = y = f12 = T12 B12 B12T 12 - T12B 12 3B124 2 Also, y = BT -TB B 2 y = T B - T B B 2

21 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics EXAMPLE 4 SOLUTION Differentiating Quotients (A) If (A) f12 = 2, fin f f 12 = = = = (B) If y = t2 - t t 3, fin y (C) Fin by using the quotient rule an also by splitting the fraction 2 into two fractions. (B) y = 1t t 2 - t2-1t 2 - t21t t = 1t t t 2 - t213t 2 2 1t = 2t4 - t 3 + 2t - 1-3t 4 + 3t 3 1t = -t4 + 2t 3 + 2t - 1 1t (C) Metho 1. Use the quotient rule: = = Metho 2. Split into two fractions: = The enominator times the erivative of the numerator, minus the numerator times the erivative of the enominator, ivie by the square of the enominator = 6 4 = = = = = 6 3 Comparing methos 1 an 2, we see that it often pays to change an epression algebraically before blinly using a ifferentiation formula. MATCHED PROBLEM 4 Fin (A) (B) (C) f 12 for f12 = y for y = t3-3t t in two ways

22 BARNMC04_ QXD 2/21/07 1:27 PM Page 237 Section 4-3 Derivatives of Proucts an Quotients 237 Eplore & Discuss 3 Eplain why Z is use in the following epression, an then fin the correct erivative: Z EXAMPLE 5 Fining Derivatives Fin f 12 for SOLUTIONS (A) (B) (A) f12 = f12 = 3e 1 + e ln f 12 = 11 + e 213e 2-3e 11 + e e 2 2 = 11 + e 23e - 3e e 11 + e 2 2 (B) = f 12 = 3e 11 + e ln 2-1ln = = # ln ln 22 Multiply by MATCHED PROBLEM 5 Fin f 12 for (A) f 12 = (B) f12 = 3 e ln EXAMPLE 6 Sales Analysis The total sales S (in thousans of games) of a home vieo game t months after the game is introuce are given by S(t) = 125t 2 t (A) Fin S (t). (B) Fin S(10) an S (10). Write a brief verbal interpretation of these results. (C) Use the results from part (B) to estimate the total sales after 11 months.

23 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics SOLUTION (A) S 1t2 = 1t t t 2 1t t = 1t t2-125t 2 12t2 1t = 250t3 + 25,000t - 250t 3 1t = 25,000t 1t (B) S1102 = = 62.5 an S 1102 = 25, = 6.25 The total sales after 10 months are 62,500 games, an sales are increasing at the rate of 6,250 games per month. (C) The total sales will increase by approimately 6,250 games uring the net month. Thus, the estimate total sales after 11 months are 62, ,250 = 68,750 games. MATCHED PROBLEM 6 Refer to Eample 6. Suppose that the total sales S (in thousans of games) t months after the game is introuce are given by S1t2 = 150t t + 3 (A) Fin S 1t2. (B) Fin S(12) an S Write a brief verbal interpretation of these results. (C) Use the results from part (B) to estimate the total sales after 13 months. Answers to Matche Problems (A) (B) 3. (A) y = = -4, = e + e = e (B) 4. (A) (B) (C) 5. (A) 7 # 1 + ln = ln = t t t 3-3t212t2 1t = t4-9t t e e 1e ln ln (B) 11 + ln 2 2 = 11 + ln (A) S 1t2 = 1t (B) S1122 = 120; S 1122 = 2. After 12 months, the total sales are 120,000 games, an sales are increasing at the rate of 2,000 games per month. (C) 122,000 games

24 BARNMC04_ QXD 2/21/07 1:27 PM Page 239 Section 4-3 Derivatives of Proucts an Quotients 239 Eercise 4-3 The answers to most of the problems in this eercise set contain both an unsimplifie form an a simplifie form of the erivative. When checking your work, first check that you applie the rules correctly, an then check that you performe the algebraic simplification correctly. Unless instructe otherwise, when ifferentiating a prouct, use the prouct rule rather than performing the multiplication first. A In Problems 1 26, fin f 12 an simplify. 1. f12 = f12 = f12 = f12 = f12 = f12 = f12 = f12 = f12 = 3e 10. f12 = 2 e 11. f12 = 3 ln f12 = 5 ln f12 = f12 = f12 = f12 = f12 = f12 = f12 = f12 = f12 = 2-4 f12 = B 23. f12 = f12 = ln 26. f12 = 1 + In Problems 27 38, fin h 12, where f() is an unspecifie ifferentiable function. 27. h12 = f h12 = 2 f h12 = 3 f h12 = f e 33. h12 = 34. h12 = 2 f12 f h12 = e f h12 = f h12 = ln 38. f12 h12 = f12 h12 = f12 3 h12 = f12 ln In Problems 39 48, fin the inicate erivatives an simplify. 39. f 12 for f12 = y for y = f12 = 1 - e 1 + e ln e C y 41. for y = 12.5t - t 2 214t t 42. t t t 2-2t y for y = f 12 for f12 = 2-1 w 2-3w w w 2-1 y 46. w for y = w4 - w 3 3w y for y = e 48. In Problems 49 54, fin f 12 an fin the equation of the line tangent to the graph of f at = f12 = f12 = f12 = f12 = f12 = 54. f12 = 1-22 ln 2 In Problems 55 58, fin f 12 an fin the value(s) of where f 12 = f12 = f12 = f12 = 58. f12 = In Problems 59 62, fin f 12 in two ways: by using the prouct or quotient rule an by simplifying first. 59. f12 = f12 = f12 = f12 = In Problems 63 82, fin each inicate erivative an simplify g1w2 = 1w - 52 log 3 w y t for y = 11 + e t 2 ln t f1w2 = 1w w y for y = y for y = 91> > > y for y = log f 12 for f12 = y for y =

25 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics y for y = t 73. g 1t2 if g1t2 = 3t h 1t2 if h1t2 = -0.05t t [10 t log t] 77. t Applications [4 log 5 ] y for y = f 12 for f12 = y for y = y 81. for y = t ln t 82. t e t y u for y = u 2 e u 1 + ln u 83. Sales analysis. The total sales S (in thousans of CDs) of a compact isk are given by 90t 2 S1t2 = t where t is the number of months since the release of the CD. (A) Fin S 1t2. (B) Fin S(10) an S Write a brief verbal interpretation of these results. (C) Use the results from part (B) to estimate the total sales after 11 months. 84. Sales analysis. A communications company has installe a cable television system in a city. The total number N (in thousans) of subscribers t months after the installation of the system is given by N1t2 = 180t t + 4 (A) Fin N 1t2. (B) Fin N(16) an N Write a brief verbal interpretation of these results. (C) Use the results from part (B) to estimate the total number of subscribers after 17 months. 85. Price eman equation. Accoring to classical economic theory, the eman for a quantity in a free market ecreases as the price p increases (see the figure). Suppose that the number of CD players people are willing to buy per week from a retail chain at a price of $p is given by = 4, p + 1 Supply/Deman 2,000 1,600 1, p 70 4, p 1 100p 0.1p Price (ollars) Figure for 85 an 86 (A) Fin p. (B) Fin the eman an the instantaneous rate of change of eman with respect to price when the p price is $40. Write a brief verbal interpretation of these results. (C) Use the results from part (B) to estimate the eman if the price is increase to $ Price supply equation. Also accoring to classical economic theory, the supply of a quantity in a free market increases as the price p increases (see the figure). Suppose that the number of CD players a retail chain is willing to sell per week at a price of $p is given by 100p = 0.1p + 1 (A) Fin p. (B) Fin the supply an the instantaneous rate of change of supply with respect to price when the price is $40. Write a brief verbal interpretation of these results. (C) Use the results from part (B) to estimate the supply if the price is increase to $ Meicine. A rug is injecte into the bloostream of a patient through her right arm. The concentration of the rug (in milligrams per cubic centimeter) in the bloostream of the left arm t hours after the injection is given by C1t2 = 0.14t t (A) Fin C 1t2. (B) Fin C an C 132, an interpret the results. 88. Drug sensitivity. One hour after a ose of milligrams of a particular rug is aministere to a person, the change in boy temperature T(), in egrees Fahrenheit, is given approimately by T12 = 2 a1-9 b 0 7 The rate T 12 at which T changes with respect to the size of the osage,,iscalle the sensitivity of the boy to the osage. (A) Use the prouct rule to fin T 12. (B) Fin T 112, T 132, an T Learning. In the early ays of quantitative learning theory (aroun 1917), L. L. Thurstone foun that a given person successfully accomplishe N() acts after practice acts, as given by N12 = 10 p (A) Fin the instantaneous rate of change of learning, N 12, with respect to the number of practice acts,. (B) Fin N 142 an N 1682.

26 BARNMC04_ QXD 2/21/07 1:27 PM Page 241 Section 4-4 The Chain Rule 241 Section 4-4 THE CHAIN RULE Composite Functions General Power Rule The Chain Rule The wor chain in the name chain rule comes from the fact that a function forme by composition involves a chain of functions that is, a function of a function. The chain rule enables us to compute the erivative of a composite function in terms of the erivatives of the functions making up the composition. In this section, we review composite functions, introuce the chain rule by means of the special case known as the general power rule, an then iscuss the chain rule itself. Composite Functions The function m12 = is a combination of a quaratic function an a cubic function. To see this more clearly, let y = f1u2 = u 3 an u = g12 = Then we can epress y as a function of as follows: y = f1u2 = f3g124 = = m12 The function m is sai to be the composite of the two functions f an g. In general, we have the following: DEFINITION Composite Functions A function m is a composite of functions f an g if m12 = f3g124 The omain of m is the set of all numbers such that is in the omain of g an g() is in the omain of f. EXAMPLE 1 Composite Functions Let f1u2 = e u an g12 = -3. Fin f [g()] an g[ f(u)]. SOLUTION f 3g124 = f1-32 = e -3 g3f1u24 = g1e u 2 = -3e u MATCHED PROBLEM 1 Let f1u2 = 2u an g12 = e. Fin f[g()] an g[ f(u)]. EXAMPLE 2 Composite Functions Write each function as a composition of two simpler functions. (A) y = 100e 0.04 (B) y = 24-2 SOLUTION (A) Let y = f1u2 = 100e u u = g12 = 0.04 Check: y = f3g124 = 100e g12 = 100e 0.04

27 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics (B) Let y = f1u2 = 2u u = g12 = 4-2 Check: y = f3g124 = 2g12 = 24-2 MATCHED PROBLEM 2 Write each function as a composition of two simpler functions. (A) y = 50e -2 (B) y = I N S I G H T There can be more than one way to epress a function as a composition of simpler functions. Choosing y = f1u2 = 100u an u = g12 = e 0.04 in Eample 2A prouces the same result: y = f3g124 = 100g12 = 100e 0.04 Since we will be using composition as a means to an en (fining a erivative), usually it will not matter what choices you make for the functions in the composition. General Power Rule We have alreay mae etensive use of the power rule, (1) n = n n - 1 Now we want to generalize this rule so that we can ifferentiate composite functions of the form 3u124 n, where u() is a ifferentiable function. Is rule (1) still vali if we replace with a function u()? Eplore & Discuss 1 Let u12 = 2 2 an f12 = 3u124 3 = 8 6. Which of the following is f 12? (A) 33u124 2 (B) 33u (C) 33u124 2 u 12 The calculations in Eplore & Discuss 1 show that we cannot generalize the power rule simply by replacing with u() in equation (1). How can we fin a formula for the erivative of 3u124 n, where u() is an arbitrary ifferentiable function? Let s begin by consiering the erivatives of 3u124 2 an 3u124 3 to see if a general pattern emerges. Since 3u124 2 = u12u12, we use the prouct rule to write 3u1242 = 3u12u124 (2) Because 3u124 3 = 3u124 2 u12, we now use the prouct rule an the result in equation (2) to write 3u1243 = 53u1242 u126 = 3u124 2 u12 + u12 3u1242 = 3u124 2 u 12 + u1232u12u 124 = 33u124 2 u 12 = u12u 12 + u12u 12 = 2u12u 12 Use equation (2) to substitute for 3u1242.

28 BARNMC04_ QXD 2/21/07 1:27 PM Page 243 Continuing in this fashion, we can show that Section 4-4 The Chain Rule 243 3u124n = n3u124 n - 1 u 12 n a positive integer (3) Using more avance techniques, we can establish formula (3) for all real numbers n. Thus, we have the general power rule: THEOREM 1 GENERAL POWER RULE If u() is a ifferentiable function, n is any real number, an then This rule is often written more compactly as y =nu n - 1 u or y = f12 = 3u124 n f 12 = n3u124 n - 1 u 12 un = nu n - 1 u where u = u12 EXAMPLE 3 Using the General Power Rule Fin the inicate erivatives: (A) f 12 if f12 = (B) y if y = (C) t 1t 2 + t (D) if SOLUTION (A) f12 = Let u = 3 + 1, n = 4. f 12 = nu n - 1 u = u = 3 = (B) (C) h w y = y = = = t 1t 2 + t h1w2 = 23 - w Let u = , n = 7. nu n - 1 u u = 32 = t 1t2 + t = -31t 2 + t t 2 + t + 42 = -31t 2 + t t t + 12 = 1t 2 + t Let u = t 2 + t + 4, n = -3. nu n - 1 u t u t = 2t + 1

29 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics (D) h1w2 = 23 - w = 13 - w2 1>2 h w = w2-1> w2 = w2-1> Let u = 3 - w, n = 1 2 nu n - 1 u w u w = = - or w2 1> w MATCHED PROBLEM 3 Fin the inicate erivatives: (A) h 12 if h12 = (B) y if y = g (C) (D) if g1w2 = 24 - w t 1t w Notice that we use two steps to ifferentiate each function in Eample 3. First, we applie the general power rule; then we foun u. As you gain eperience with the general power rule, you may want to combine these two steps. If you o this, be certain to multiply by u. For eample, = Z Correct u> = 5 4 is missing I N S I G H T If we let u12 =, then u> = 1, an the general power rule reuces to the (orinary) power rule iscusse in Section 3-5. Compare the following: n = n n - 1 un = nu n - 1 u un Z nu n - 1 Yes power rule Yes general power rule Unless u12 = + k, so that u> = 1 The Chain Rule We have use the general power rule to fin erivatives of composite functions of the form f(g()), where f(u) = u n is a power function. But what if f is not a power function? Then a more general rule, the chain rule, enables us to compute the erivatives of many composite functions of the form f(g()). Suppose that is a composite of f an g, where y = m12 = f3g124 y = f1u2 an u = g12

30 BARNMC04_ QXD 2/21/07 1:27 PM Page 245 Section 4-4 The Chain Rule 245 We woul like to epress the erivative y> in terms of the erivatives of f an g. From the efinition of a erivative (see Section 3-4), we have y = lim m1 + h2 - m12 h:0 h f3g1 + h24 - f3g124 = lim h:0 h f3g1 + h24 - f3g124 = lim c h:0 h f3g1 + h24 - f3g124 = lim c # h:0 g1 + h2 - g12 # Substitute m1 + h2 = f3g1 + h24 an m12 = f3g124. g1 + h2 - g12 Multiply by 1 = g1 + h2 - g12. g1 + h2 - g12 g1 + h2 - g12 g1 + h2 - g12 h (4) We recognize the secon factor in equation (4) as the ifference quotient for g(). To interpret the first factor as the ifference quotient for f(u), we let k = g1 + h2 - g12. Since u = g12, we can write u + k = g12 + g1 + h2 - g12 = g1 + h2 Substituting in equation (1), we now have y (5) = lim f1u + k2 - f1u2 g1 + h2 - g12 c # h:0 k h If we assume that k = 3g1 + h2 - g124 : 0 as h : 0, we can fin the limit of each ifference quotient in equation (5): y = c lim f1u + k2 - f1u2 g1 + h2 - g12 clim k:0 k h:0 h = f 1u2g 12 = y u u This result is correct uner rather general conitions an is calle the chain rule, but our erivation is superficial, because it ignores a number of hien problems. Since a formal proof of the chain rule is beyon the scope of this book, we simply state it as follows: THEOREM 2 CHAIN RULE If y = f1u2 an u = g12 efine the composite function y = m12 = f3g124 then y = y u provie that y u an u u eist or, equivalently, m 12 = f 3g124g 12 provie that f 3g124 an g 12 eist 2 EXAMPLE 4 Using the Chain Rule Fin y u, u, an y (epress y as a function of ) for (A) y = u 3>2 an u = (B) y = e u an u = (C) y = ln u an u =

31 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics SOLUTION (A) y u = 3 2 u1>2 an u = 6 y = y u u Basic erivative rules Chain rule = 3 2 u1>2 162 = >2 Since u = (B) y u = eu an u = 62 y = y u u = e u = 6 2 e Basic erivative rules Chain rule Since u = (C) y u = 1 u an u = 2-4 y = y u u = 1 u = Basic erivative rules Chain rule Since u = MATCHED PROBLEM 4 Fin y u, u, an y (epress y as a function of ) for (A) y = u -5 an u = (B) y = e u an u = (C) y = ln u an u = Eplore & Discuss 2 Let m12 = f3g124. Use the chain rule an the graphs in Figures 1 an 2 to fin (A) f(4) (B) g(6) (C) m(6) (D) f (4) (E) g (6) (F) m (6) y u 30 y f(u) 6 u (6, 4) 10 (4, 12) 2 y 2u 4 u g() 5 10 u 5 10 FIGURE 1 FIGURE 2 The chain rule can be etene to compositions of three or more functions. For eample, if y = f1w2, w = g1u2, an u = h12, then y = y w u w u

32 BARNMC04_ QXD 2/21/07 1:27 PM Page 247 Section 4-4 The Chain Rule 247 EXAMPLE 5 Using the Chain Rule For y = h12 = e 1 + 1ln 22, fin y. SOLUTION Note that h is of the form y = e w, where w = 1 + u 2 an u = ln. Thus, y = y w u w u = e w 12u2a 1 b = e 1 + u2 12u2a 1 b Since w = 1 + u 2 = e 1 + 1ln ln 2a 1 b Since u = ln = 2 + 1ln 22 1ln 2e1 MATCHED PROBLEM 5 For y = h12 = 3ln11 + e 24 3, fin y. The chain rule generalizes basic erivative rules. We list three general erivative rules here for convenient reference [the first, equation (6), is the general power rule of Theorem 1]. General Derivative Rules 3f124n = n3f124 n - 1 f 12 ln3f124 = 1 f12 f 12 ef12 = e f12 f 12 (6) (7) (8) Unless irecte otherwise, you now have a choice between the chain rule an the general erivative rules. However, practicing with the chain rule will help prepare you for concepts that appear later in the tet. Eamples 4 an 5 illustrate the chain rule metho, an the net eample illustrates the general erivative rules metho. EXAMPLE 6 Using General Derivative Rules (A) e2 = e 2 2 Using equation (5) = e = 2e 2 (B) 1 ln = Using equation (4) 1 = = (C) = e e e2 Using equation (3) = e e 2 2 Using equation (5) = e e = 6e e 2 2 2

33 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics MATCHED PROBLEM 6 Fin (A) + 2 (B) ln e32 (C) 12 + e Answers to Matche Problems 1. f3g124 = 2e, g3f1u24 = e 2u 2. (A) f1u2 = 50e u, u = -2 (B) f1u2 = 2 3 u, u = [Note: There are other correct answers.] 3. (A) (B) (C) (D) 4. (A) (B) (C) 15(5 + 2) ( 4-5) 4-4t>(t 2 + 4) 3-1>(224 - w) y =-5u-4, u = 62, y =-302 ( ) -6 y u = eu, u = 123, y = 123 e y u = 1 u, u y = 2 + 9, = e [ln(1 + e )] e 6. (A) (B) 6e (C) - 8e -2 (2 + e -2 ) 3 Eercise 4-4 For many of the problems in this eercise set, the answers in the back of the book inclue both an unsimplifie form an a simplifie form. When checking your work, first check that you applie the rules correctly, an then check that you performe the algebraic simplification correctly. A In Problems 1 4, fin f[g()]. 1. f1u2 = u 3 ; g12 = f1u2 = u 4 ; g12 = f1u2 = e u ; g12 = g1u2 = e u ; g12 = 3 3 In Problems 5 8, write each composite function in the form y = f1u2 an u = g y = y = y = e y = e In Problems 9 16, replace the? with an epression that will make the inicate equation vali = ? = ? = ? = ? = e e e4-2 = e 4-2?? ln(4 + 1) = ln = - 3?? In Problems 17 36, fin f 12 an simplify. 17. f12 = (2 + 5) f12 = (3-7) 19. f12 = (5-2) f12 = (9-5) 21. f12 = ( ) f12 = (6-0.5) 23. f12 = ( ) f12 = (5-3) f12 = e f12 = 6e 27. f12 = 3e f12 = e f12 = (2-5) 1> >2 f12 = (4 + 3) 31. f12 = ( 4 + 1) f12 = ( 5 + 2) f12 = 3 ln(1 + 2 ) 34. f12 = 2 ln( ) 35. f12 = (1 + ln ) f12 = ( - 2 ln ) 4 In Problems 37 42, fin f () an the equation of the line tangent to the graph of f at the inicate value of. Fin the value(s) of where the tangent line is horizontal. 37. f12 = ; = f12 = ; = 1

34 BARNMC04_ QXD 2/21/07 1:27 PM Page 249 Section 4-4 The Chain Rule f12 = >2 ; = 3 f12 = >2 ; = 4 In Problems 69 74, fin f 12 an fin the value(s) of where the tangent line is horizontal. 69. f12 = f12 = f12 = 5e ; = f12 = ln f12 = 72. f12 = - 1 2; = B In Problems 43 62, fin the inicate erivative an simplify. 73. f12 = f12 = y if y = Suppose a stuent reasons that the functions 44. y if y = f12 = ln an g12 = 4 ln must have the same erivative, since he has entere f(), g(), 45. t 21t2 + 3t2-3 f 12, an g 12 into a graphing calculator, but only three graphs appear (see the figure). Is his reasoning correct? 46. t 31t3 + t Are f 12 an g 12 the same function? Eplain h w if h1w2 = 2w g w if g1w2 = w - 7 g 12 if g12 = 4e 3 e h 12 if h12 = ln(1 + ) [4 ln(1 + 4 )] 53. F 1t2 if F1t2 = (e t2 + 1 ) G 1t2 if G1t2 = (1 - e 2t ) y if y = ln( 2 + 3) 3>2 56. y if y = [ln( 2 + 3)] 3> w 1w w 1w y 59. if y = y 60. if y = a b f 1t2 if f1t2 = 2t 2-3t g 1t2 if g1t2 = 2 3 t - t 2 In Problems 63 68, fin f 12 an fin the equation of the line tangent to the graph of f at the inicate value of. 63. f12 = ; = f12 = ; = f12 = ; = f12 = ; = f12 = 2ln ; = e 68. f12 = e 1 ; = 1 (A) Figure for (B) 76. Suppose a stuent reasons that the functions f12 = ln an g12 = >3 must have the same erivative, since she has entere f(), g(), f 12, an g 12 into a graphing calculator, but only three graphs appear (see the figure). Is her reasoning correct? Are f 12 an g 12 the same function? Eplain. 1 (A) (B) Figure for 76 C In Problems 77 92, fin each erivative an simplify log 2(3 2-1) log 3( ) (2-1)3 ( 2 + 3) log(3-1) log 5(5 2-1 ) 10ln B

35 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics Applications 93. Cost function. The total cost (in hunres of ollars) of proucing calculators per ay is C12 = (see the figure). Cost (hunre ollars) C() Prouction Figure for 93 (A) Fin C 12. (B) Fin C 1242 an C 1422, an interpret the results. 94. Cost function. The total cost (in hunres of ollars) of proucing cameras per week is C12 = (A) Fin C 12. (B) Fin C 1152 an C 1242, an interpret the results. 95. Price supply equation. The number of stereo speakers a retail chain is willing to sell per week at a price of $p is given by = 802p p 100 (see the figure). (A) Fin p. (B) Fin the supply an the instantaneous rate of change of supply with respect to price when the price is $75. Write a brief verbal interpretation of these results. Supply/Deman , p p p Price (ollars) Figure for 95 an Price eman equation. The number of stereo speakers people are willing to buy per week from a retail chain at a price of $p is given by = 1, p p 100 (see the previous figure). (A) Fin p. (B) Fin the eman an the instantaneous rate of change of eman with respect to price when the price is $75. Write a brief verbal interpretation of these results. 97. Drug concentration. The concentration of a rug in the bloostream t hours after injection is given approimately by C1t2 = 4.35e -t 0 t 5 where C(t) is concentration in milligrams per milliliter. (A) What is the rate of change of concentration after 1 hour? After 4 hours? (B) Graph C. 98. Water pollution. The use of ioine crystals is a popular way of making small quantities of nonpotable water safe to rink. Crystals place in a 1-ounce bottle of water will issolve until the solution is saturate. After saturation, half of the solution is poure into a quart container of nonpotable water, an after about an hour, the water is usually safe to rink. The half-empty 1-ounce bottle is then refille, to be use again in the same way. Suppose that the concentration of ioine in the 1-ounce bottle t minutes after the crystals are introuce can be approimate by C1t2 = e -t 2 t Ú 0 where C(t) is the concentration of ioine in micrograms per milliliter. (A) What is the rate of change of the concentration after 1 minute? After 4 minutes? (B) Graph C for 0 t Bloo pressure an age. A research group using hospital recors evelope the following approimate mathematical moel relating systolic bloo pressure an age: P12 = ln Here, P() is pressure, measure in millimeters of mercury, an is age in years. What is the rate of change of pressure at the en of 10 years? At the en of 30 years? At the en of 60 years? 100. Biology. A yeast culture at room temperature (68 F) is place in a refrigerator maintaining a constant temperature of 38 F. After t hours, the temperature T of the culture is given approimately by T = 30e -0.58t + 38 t Ú 0 What is the rate of change of temperature of the culture at the en of 1 hour? At the en of 4 hours? 101. Learning. In 1930, L. L. Thurstone evelope the following formula to inicate how learning time T epens on the length of a list n: T = f1n2 = c k n2n - a Here, a, c, an k are empirical constants. Suppose that, for a particular person, the time T (in minutes) require to learn a list of length n is T = f1n2 = 2n2n - 2 (A) Fin T n. (B) Fin f 1112 an f 1272, an interpret the results.

36 BARNMC04_ QXD 2/21/07 1:27 PM Page 251 Section 4-5 Implicit Differentiation 251 Section 4-5 IMPLICIT DIFFERENTIATION Special Function Notation Implicit Differentiation Special Function Notation The equation y = (1) efines a function f with y as a epenent variable an as an inepenent variable. Using function notation, we woul write In orer to reuce to a minimum the number of symbols involve in a iscussion, we will often write equation (1) in the form where y is both a epenent variable an a function symbol. This is a convenient notation, an no harm is one as long as one is aware of the ouble role of y. Other eamples are This type of notation will simplify much of the iscussion an work that follows. Until now, we have consiere functions involving only one inepenent variable. There is no reason to stop there: The concept can be generalize to functions involving two or more inepenent variables, an this will be one in etail in Chapter 8. For now, we will borrow the notation for a function involving two inepenent variables. For eample, specifies a function F involving two inepenent variables. Implicit Differentiation y = f12 or f12 = y = = y12 = 2t 2-3t + 1 = 1t2 z = 2u 2-3u = z1u2 1 r = = r1s2 1s 2 2>3-3s2 F1, y2 = 2-2y + 3y 2-5 Consier the equation y - 2 = 0 (2) an the equation obtaine by solving equation (2) for y in terms of, y = (3) Both equations efine the same function with as the inepenent variable an y as the epenent variable. On the one han, we can write, for equation (3), y = f12 where f12 = (4) an we have an eplicit (irectly state) rule that enables us to etermine y for each value of. On the other han, the y in equation (2) is the same y as in equation (3), an equation (2) implicitly gives (implies, though oes not irectly epress) y as a function of. Thus, we say that equations (3) an (4) efine the function f eplicitly an equation (2) efines f implicitly. The irect use of an equation that efines a function implicitly to fin the erivative of the epenent variable with respect to the inepenent variable is calle

37 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics implicit ifferentiation. Let us ifferentiate equation (2) implicitly an equation (3) irectly, an compare results. Starting with y - 2 = 0 we think of y as a function of that is, y = y12 an write y12-2 = 0 Then we ifferentiate both sies with respect to : y = y12-2 = y -0 = 0 Since y is a function of, but is not eplicitly given, we simply write y12 = y to inicate its erivative. Now we solve for y : y =-6 Note that we get the same result if we start with equation (3) an ifferentiate irectly: y = y =-6 Why are we intereste in implicit ifferentiation? In general, why o we not solve for y in terms of an ifferentiate irectly? The answer is that there are many equations of the form F1, y2 = 0 (5) that are either ifficult or impossible to solve for y eplicitly in terms of (try it for 2 y 5-3y + 5 = 0 or for e y - y = 3, for eample). But it can be shown that, uner fairly general conitions on F, equation (5) will efine one or more functions in which y is a epenent variable an is an inepenent variable. To fin y uner these conitions, we ifferentiate equation (5) implicitly. Eplore & Discuss 1 (A) How many tangent lines are there to the graph in Figure 1 when = 0? When = 1? When = 2? When = 4? When = 6? y FIGURE 1 (B) Sketch the tangent lines referre to in part (A), an estimate each of their slopes. (C) Eplain why the graph in Figure 1 is not the graph of a function.

38 BARNMC04_ QXD 2/21/07 1:27 PM Page 253 Section 4-5 Implicit Differentiation 253 EXAMPLE 1 Differentiating Implicitly Given F1, y2 = 2 + y 2-25 = 0 fin y an the slope of the graph at = 3. (6) y 2 y 2 25 FIGURE 2 SOLUTION We start with the graph of 2 + y 2-25 = 0 (a circle, as shown in Fig. 2) so that we can interpret our results geometrically. From the graph, it is clear that equation (6) oes not efine a function. But with a suitable restriction on the variables, equation (6) can efine two or more functions. For eample, the upper half an the lower half of the circle each efine a function. On each half-circle, a point that correspons to = 3 is foun by substituting = 3 into equation (6) an solving for y: y 2-25 = y 2 = 25 y 2 = 16 y = ;4 Thus, the point (3, 4) is on the upper half-circle, an the point 13, -42 is on the lower half-circle. We will use these results in a moment. We now ifferentiate equation (6) implicitly, treating y as a function of [i.e., y = y12]: 2 + y 2-25 = y = y = y = y y 12-0 = yy =0 y =- 2 2y y =- y Use the chain rule. Solve for y in terms of an y. Leave the answer in terms of an y. 2 y y y 4 y 4 (3, 4) 0 FIGURE 3 4 (3, 4) Slope! Slope! We have foun y without first solving 2 + y 2-25 = 0 for y in terms of. An by leaving y in terms of an y, we can use y =->y to fin y for any point on the graph of 2 + y 2-25 = 0 (ecept where y = 0). In particular, for = 3, we foun that (3, 4) an 13, -42 are on the graph; thus, the slope of the graph at (3, 4) is y ƒ 13, 42 = The slope of the graph at (3, 4) an the slope at 13, -42 is y ƒ 13, -42 = = 3 4 The slope of the graph at 13, -42 The symbol y ƒ 1a, b2 is use to inicate that we are evaluating y at = a an y = b. The results are interprete geometrically in Figure 3 on the original graph. In Eample 1, the fact that y is given in terms of both an y is not a great isavantage. We have only to make certain that when we want to evaluate y œ for a particular value of an y, say, 1 0, y 0 2, the orere pair must satisfy the original equation.

39 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics I N S I G H T In the solution for Eample 1, notice that the erivative of with respect to is 2yy, not just 2y. When first learning about implicit ifferentiation, it is a goo iea to replace y with y() in the original equation to emphasize that y is a function of. y 2 MATCHED PROBLEM 1 Graph 2 + y = 0, fin y by implicit ifferentiation, an fin the slope of the graph when = 5. EXAMPLE 2 Differentiating Implicitly Fin the equation(s) of the tangent line(s) to the graph of y - y = 0 (7) at the point(s) where = 1. SOLUTION We first fin y when = 1: Thus, there are two points on the graph of (7) where = 1, namely, 11, -12 an (1, 2). We net fin the slope of the graph at these two points by ifferentiating equation (7) implicitly: y - y = 0 y - y = 0 y -1 # 2yy +y = 0 y -2yy -y = 0 Now fin the slope at each point: y - y = 0 y - 112y = 0 y - y = 0 y 2 - y - 2 = 0 1y - 221y + 12 = 0 y -2yy =y y2y =y 2-2 y = y y y = -1, 2 y ƒ 11, -12 = = = -1 3 = -1 3 y ƒ 11,22 = = = 2-3 = -2 3 Use the prouct rule an the chain rule for y2. Solve for y by getting all terms involving y on one sie. Equation of tangent line at 11, -12: Equation of tangent line at (1, 2): y - y 1 = m1-1 2 y - y 1 = m1-1 2 y + 1 = -1 y - 2 = y + 1 = y = y - 2 = y =

40 BARNMC04_ QXD 2/21/07 1:27 PM Page 255 Section 4-5 Implicit Differentiation 255 MATCHED PROBLEM 2 Repeat Eample 2 for 2 + y 2 - y - 7 = 0 at = 1. Eplore & Discuss 2 The slopes of the tangent lines to y 2 + 3y + 4 = 9 when = 0 can be foun either (1) by ifferentiating the equation implicitly or (2) by solving for y eplicitly in terms of (using the quaratic formula) an then computing the erivative. Which of the two methos is more efficient? Eplain. EXAMPLE 3 Differentiating Implicitly Fin for = 1t2 efine implicitly by t ln = e t - 1 an evaluate at 1t, 2 = 10, 12. SOLUTION It is important to remember that is the epenent variable an t is the inepenent variable. Therefore, we ifferentiate both sies of the equation with respect to t (using prouct an chain rules where appropriate) an then solve for : t ln = e t - 1 t 1t ln 2 = t 1et 2 - t 1 t + ln = et + e t # t + # ln = # e t + # e t t + ln = 2 e t + e t t -e = 2 e t - ln 1t - e t 2 = 2 e t - ln = 2 e t - ln t - e t Differentiate implicitly with respect to t. Clear fractions. Z 0 Solve for. Factor out. Now we evaluate at 1t, 2 = 10, 12, as requeste: ƒ 10,12 = 1122 e 0-1 ln 1 0-1e 0 = 1-1 = -1 MATCHED PROBLEM 3 Fin for = 1t2 efine implicitly by an evaluate at 1t, 2 = 11, ln t = te ƒ ƒ ƒ Answers to Matche Problems 1. y =->y.when = 5, y = ;12; thus, y 15,122 = - 5 an y 15,-122 = te - y = y = t ln t - t 2 e ; 11,02 = -1 2y - ; y = , y =

41 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics Eercise 4-5 A In Problems 1 4, fin y in two ways: (A) Differentiate the given equation implicitly an then solve for y. (B) Solve the given equation for y an then ifferentiate irectly y + 9 = y B In Problems 5 22, use implicit ifferentiation to fin y an evaluate y at the inicate point In Problems 23 an 24, fin for = 1t2 efine implicitly by the given equation. Evaluate at the inicate point y - 4 = y - 18 = y - 2 = 0 y = 0; 11, y - 1 = 0; 11, y 3-3 = 0; 12, 12 y = 0; 1-2, 22 y 2 + 2y + 3 = 0; 1-1, 12 y 2 - y - 4 = 0; 10, 12 y - 6 = 0; 12, 32 3y = 0; 12, 12 2y + y + 2 = 0; 1-1, 22 2y + y - 1 = 0; 1-1, 12 2 y = 0; 12, y = 0; 1-1, 32 e y = 2 + y 2 ; 11, y = 4e y ; 12, y = ln y; 11, 12 ln y = 2y 2 - ; 12, 12 ln y + 2y = 2 3 ; 11, 12 e y - y = 2-2; 12, t 2 + t = 0; 1-2, t 2-4 = 0; 1-3, -22 Problems 25 an 26 refer to the equation an graph shown in the figure. 25. Use implicit ifferentiation to fin the slopes of the tangent lines at the points on the graph where = 1.6. Check your answers by visually estimating the slopes on the graph in the figure. 26. Fin the slopes of the tangent lines at the points on the graph where = 0.2. Check your answers by visually estimating the slopes on the graph in the figure. In Problems 27 30, fin the equation(s) of the tangent line(s) to the graphs of the inicate equations at the point(s) with the given value of. 27. y = 0; = y + 1 = 0; = y 2 - y - 6 = 0; = y 2 - y - 2 = 0; = If e y = 1, fin y in two ways, first by ifferentiating implicitly an then by solving for y eplicitly in terms of. Which metho o you prefer? Eplain. 32. Eplain the ifficulty that arises in solving 3 + y + e y = 1 for y as an eplicit function of. Fin the slope of the tangent line to the graph of the equation at the point (0, 1). C In Problems 33 40, fin y an the slope of the tangent line to the graph of each equation at the inicate point y2 3 + y = + 7; 12, y = y; 1-3, y2 3 = 2y 2-3; 11, y2 4 - y 3 = 8; 1-1, y = 0; 12, y >2-2 = 0; 14, ln1y2 = y 2-1; 11, e y - 2 = y + 1; 10, Fin the equation(s) of the tangent line(s) at the point(s) on the graph of the equation y 3 - y - 3 = 2 where = 1. Roun all approimate values to two ecimal places. 42. Refer to the equation in Problem 41. Fin the equation(s) of the tangent line(s) at the point(s) on the graph where y = -1. Roun all approimate values to two ecimal places. 1 ( 1) 2 (y 1) 2 1 2

42 BARNMC04_ QXD 2/21/07 1:27 PM Page 257 Section 4-6 Relate Rates 257 Applications For the eman equations in Problems 43 46, fin the rate of change of p with respect to by ifferentiating implicitly ( is the number of items that can be sol at a price of $p) = p 2-2p + 1,000 = p 3-3p = 210,000 - p = 2 3 1,500 - p Biophysics. In biophysics, the equation 1L + m21v + n2 = k is calle the funamental equation of muscle contraction, where m, n, an k are constants an V is the velocity of the shortening of muscle fibers for a muscle subjecte to a loa L. Fin L V by implicit ifferentiation. 48. Biophysics. In Problem 47, fin V L by implicit ifferentiation. Section 4-6 RELATED RATES The workers in a union are concerne that the rate at which wages are increasing is lagging behin the rate of increase in the company s profits. An automobile ealer wants to preict how baly an anticipate increase in interest rates will ecrease his rate of sales. An investor is stuying the connection between the rate of increase in the Dow Jones average an the rate of increase in the gross omestic prouct over the past 50 years. In each of these situations, there are two quantities wages an profits in the first instance, for eample that are changing with respect to time. We woul like to iscover the precise relationship between the rates of increase (or ecrease) of the two quantities. We will begin our iscussion of such relate rates by consiering some familiar situations in which the two quantities are istances an the two rates are velocities. EXAMPLE 1 Relate Rates an Motion A 26-foot laer is place against a wall (Fig. 1). If the top of the laer is sliing own the wall at 2 feet per secon, at what rate is the bottom of the laer moving away from the wall when the bottom of the laer is 10 feet away from the wall? y FIGURE 1 SOLUTION 26 ft Many people reason that since the laer is of constant length, the bottom of the laer will move away from the wall at the rate that the top of the laer is moving own the wall. This is not the case, as we will see. At any moment in time, let be the istance of the bottom of the laer from the wall an let y be the istance of the top of the laer from the groun (see Fig. 1). Both an y are changing with respect to time an can be thought of as functions of time; that is, = 1t2 an y = y1t2. Furthermore, an y are relate by the Pythagorean relationship: 2 + y 2 = 26 2 (1) Differentiating equation (1) implicitly with respect to time t an using the chain rule where appropriate, we obtain 2 (2) t + 2y y t = 0 The rates t an y t are relate by equation (2); hence, this type of problem is referre to as a relate-rates problem. Now, our problem is to fin t when = 10 feet, given that y>t = -2 (y is ecreasing at a constant rate of 2 feet per secon). We have all the quantities we nee in equation (2) to solve for t, ecept y. When = 10, y can be foun from equation (1): y 2 = 26 2 y = = 24 feet

43 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics Substitute y>t = -2, = 10, an y = 24 into (2); then solve for t: t = 0 t = = 4.8 feet per secon Thus, the bottom of the laer is moving away from the wall at a rate of 4.8 feet per secon. I N S I G H T In the solution to Eample 1, we use equation (1) in two ways: first, to fin an equation relating y t an t, an secon, to fin the value of y when = 10. These steps must be one in this orer; substituting = 10 an then ifferentiating oes not prouce any useful results: 2 + y 2 = y 2 = yy =0 y =0 Substituting 10 for has the effect of stopping the laer. The rate of change of a stationary object is always 0, but that is not the rate of change of the moving laer. MATCHED PROBLEM 1 Again, a 26-foot laer is place against a wall (Fig. 1). If the bottom of the laer is moving away from the wall at 3 feet per secon, at what rate is the top moving own when the top of the laer is 24 feet up the wall? Eplore & Discuss 1 (A) For which values of an y in Eample 1 is t equal to 2 (i.e., the same rate at which the laer is sliing own the wall)? (B) When is t greater than 2? Less than 2? DEFINITION Suggestions for Solving Relate-Rates Problems Step 1. Sketch a figure if helpful. Step 2. Ientify all relevant variables, incluing those whose rates are given an those whose rates are to be foun. Step 3. Epress all given rates an rates to be foun as erivatives. Step 4. Fin an equation connecting the variables ientifie in step 2. Step 5. Implicitly ifferentiate the equation foun in step 4, using the chain rule where appropriate, an substitute in all given values. Step 6. Solve for the erivative that will give the unknown rate. EXAMPLE 2 Relate Rates an Motion Suppose that two motorboats leave from the same point at the same time. If one travels north at 15 miles per hour an the other travels east at 20 miles per hour, how fast will the istance between them be changing after 2 hours?

44 BARNMC04_ QXD 2/21/07 1:27 PM Page 259 Section 4-6 Relate Rates 259 y N z SOLUTION First, raw a picture, as shown in Figure 2. All variables,, y, an z, are changing with time. Hence, they can be thought of as functions of time: = 1t2, y = y1t2, an z = z1t2, given implicitly. It now makes sense to take erivatives of each variable with respect to time. From the Pythagorean theorem, z 2 = 2 + y 2 (3) FIGURE 2 E We also know that t = 20 miles per hour an y t = 15 miles per hour We woul like to fin z t at the en of 2 hours that is, when = 40 miles an y = 30 miles. To o this, we ifferentiate both sies of equation (3) with respect to t an solve for z t: We have everything we nee ecept z. From equation (3), when = 40 an y = 30, we fin z to be 50. Substituting the known quantities into equation (4), we obtain z t z 2z t = 2 t + 2y y t z t = = 25 miles per hour Thus, the boats will be separating at a rate of 25 miles per hour. (4) MATCHED PROBLEM 2 Repeat Eample 2 for the situation at the en of 3 hours. EXAMPLE 3 SOLUTION y ( 3, 4) y 2 25 FIGURE 3 Relate Rates an Motion Suppose a point is moving along the graph of 2 + y 2 = 25 (Fig. 3). When the point is at 1-3, 42, its coorinate is increasing at the rate of 0.4 unit per secon. How fast is the y coorinate changing at that moment? Since both an y are changing with respect to time, we can think of each as a function of time, namely, = 1t2 an y = y1t2 but restricte so that 2 + y 2 = 25 (5) Our problem now is to fin y t, given = -3, y = 4, an >t = 0.4. Implicitly ifferentiating both sies of equation (5) with respect to t, we have 2 + y 2 = 25 2 t + 2y y t = 0 t + y y t = 0 y t = 0 y = 0.3 unit per secon t Divie both sies by 2. Substitute = -3, y = 4, an >t = 0.4, an solve for y>t.

45 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics MATCHED PROBLEM 3 A point is moving on the graph of y 3 = 2. When the point is at 1-8, 42, its y coorinate is ecreasing by 2 units per secon. How fast is the coorinate changing at that moment? EXAMPLE 4 SOLUTION Relate Rates an Business Suppose that for a company manufacturing transistor raios, the cost, revenue, an profit equations are given by C = 5, R = P = R - C Cost equation Revenue equation Profit equation where the prouction output in 1 week is raios. If prouction is increasing at the rate of 500 raios per week when prouction is 2,000 raios, fin the rate of increase in (A) Cost (B) Revenue (C) Profit If prouction is a function of time (it must be, since it is changing with respect to time), then C, R, an P must also be functions of time. These functions are implicitly (rather than eplicitly) given. Letting t represent time in weeks, we ifferentiate both sies of each of the preceing three equations with respect to t an then substitute = 2,000 an >t = 500 to fin the esire rates. (A) C = 5, Think: C = C1t2 an = 1t2. (B) (C) C t C t Since >t = 500 when = 2,000, Cost is increasing at a rate of $1,000 per week. R = R t R t R t Since >t = 500 when = 2,000, Differentiate both sies with respect to t. Revenue is increasing at a rate of $3,000 per week. P = R - C P t = t 15, t 122 = t = 2 t = t t = 10 t t = t = R t R t = , = $3,000 per week - C t C t = $3,000 - $1,000 = $2,000 per week = = $1,000 per week Results from parts (A) an (B) Profit is increasing at a rate of $2,000 per week.

46 BARNMC04_ QXD 2/21/07 1:27 PM Page 261 Section 4-6 Relate Rates 261 MATCHED PROBLEM 4 Repeat Eample 4 for a prouction level of 6,000 raios per week. Eplore & Discuss 2 (A) In Eample 4, suppose that 1t2 = 500t Fin the time an prouction level at which the profit is maimize. (B) Suppose that 1t2 = t t Fin the time an prouction level at which the profit is maimize. (C) Eplain why it is unnecessary to know a formula for (t) in orer to etermine the prouction level at which the profit is maimize. Answers to Matche Problems y>t = ft>sec z>t = 25 mi>hr >t = 6 units>sec 4. (A) (B) C>t = $1,000>wk R>t = -$1,000>wk (C) P>t = -$2,000>wk Eercise 4-6 A In Problems 1 6, assume that = 1t2 an y = y1t2. Fin the inicate rate, given the other information. B 1. y = 2 + 2; >t = 3 when = 5; fin y t 2. y = 3-3; >t = -2 when = 2; fin y t y 2 = 1; y>t = -4 when = -0.6 an y = 0.8; fin t y 2 = 4; y>t = 5 when = 1.2 an y = -1.6; fin t y + y 2 = 11; >t = 2 when = 1 an y = 2; fin y t y - y 2 = 7; y>t = -1 when = 2 an y = -1; fin t 7. A point is moving on the graph of y = 36. When the point is at (4, 9), its coorinate is increasing by 4 units per secon. How fast is the y coorinate changing at that moment? 8. A point is moving on the graph of y 2 = 36. When the point is at (3, 0), its y coorinate is ecreasing by 2 units per secon. How fast is its coorinate changing at that moment? 9. A boat is being pulle towar a ock as inicate in the figure. If the rope is being pulle in at 3 feet per secon, how fast is the istance between the ock an the boat ecreasing when it is 30 feet from the ock? Rope Figure for 9 an 10 4 ft 10. Refer to Problem 9. Suppose that the istance between the boat an the ock is ecreasing by 3.05 feet per secon. How fast is the rope being pulle in when the boat is 10 feet from the ock? 11. A rock thrown into a still pon causes a circular ripple. If the raius of the ripple is increasing by 2 feet per secon, how fast is the area changing when the raius is 10 feet? [Use A = pr 2, p L ] 12. Refer to Problem 11. How fast is the circumference of a circular ripple changing when the raius is 10 feet? [Use C = 2pR, p L ] 13. The raius of a spherical balloon is increasing at the rate of 3 centimeters per minute. How fast is the volume changing when the raius is 10 centimeters? [Use V = 4 3 pr3, p L ] 14. Refer to Problem 13. How fast is the surface area of the sphere increasing when the raius is 10 centimeters? [Use S = 4pR 2, p L ] 15. Boyle s law for enclose gases states that if the volume is kept constant, the pressure P an temperature T are relate by the equation P T = k where k is a constant. If the temperature is increasing at 3 kelvins per hour, what is the rate of change of pressure when the temperature is 250 kelvins an the pressure is 500 pouns per square inch? 16. Boyle s law for enclose gases states that if the temperature is kept constant, the pressure P an volume V of a gas are relate by the equation VP = k where k is a constant. If the volume is ecreasing by 5 cubic inches per secon, what is the rate of change of pressure when the volume is 1,000 cubic inches an the pressure is 40 pouns per square inch?

47 BARNMC04_ QXD 2/21/07 1:27 PM Page CHAPTER 4 Aitional Derivative Topics C 17. A 10-foot laer is place against a vertical wall. Suppose the bottom of the laer slies away from the wall at a constant rate of 3 feet per secon. How fast is the top of the laer sliing own the wall (negative rate) when the bottom is 6 feet from the wall? [Hint: Use the Pythagorean theorem, a 2 + b 2 = c 2, where c is the length of the hypotenuse of a right triangle an a an b are the lengths of the two shorter sies.] 18. A weather balloon is rising vertically at the rate of 5 meters per secon. An observer is staning on the groun 300 meters from the point where the balloon was release. At what rate is the istance between the observer an the balloon changing when the balloon is 400 meters high? 19. A streetlight is on top of a 20-foot pole. A person who is 5 feet tall walks away from the pole at the rate of 5 feet per secon. At what rate is the tip of the person s shaow moving away from the pole when he is 20 feet from the pole? 20. Refer to Problem 19. At what rate is the person s shaow growing when he is 20 feet from the pole? 21. Helium is pumpe into a spherical balloon at a constant rate of 4 cubic feet per secon. How fast is the raius increasing after 1 minute? After 2 minutes? Is there any time at which the raius is increasing at a rate of 100 feet per secon? Eplain. 22. A point is moving along the ais at a constant rate of 5 units per secon. At which point is its istance from (0, 1) increasing at a rate of 2 units per secon? At 4 units per secon? At 5 units per secon? At 10 units per secon? Eplain. 23. A point is moving on the graph of y = e in such a way that its coorinate is always increasing at a rate of 3 units per secon. How fast is the y coorinate changing when the point crosses the ais? 24. A point is moving on the graph of 3 + y 2 = 1 in such a way that its y coorinate is always increasing at a rate of 2 units per secon. At which point(s) is the coorinate increasing at a rate of 1 unit per secon? Applications 25. Cost, revenue, an profit rates. Suppose that for a company manufacturing calculators, the cost, revenue, an profit equations are given by C = 90, R = P = R - C where the prouction output in 1 week is calculators. If prouction is increasing at a rate of 500 calculators per week when prouction output is 6,000 calculators, fin the rate of increase (ecrease) in (A) Cost (B) Revenue (C) Profit 26. Cost, revenue, an profit rates. Repeat Problem 25 for C = 72, R = P = R - C where prouction is increasing at a rate of 500 calculators per week at a prouction level of 1,500 calculators. 27. Avertising. A retail store estimates that weekly sales s an weekly avertising costs (both in ollars) are relate by s = 60,000-40,000e The current weekly avertising costs are $2,000, an these costs are increasing at the rate of $300 per week. Fin the current rate of change of sales. 28. Avertising. Repeat Problem 27 for s = 50,000-20,000e Price eman. The price p (in ollars) an eman for a prouct are relate by p + 50p 2 = 80,000 (A) If the price is increasing at a rate of $2 per month when the price is $30, fin the rate of change of the eman. (B) If the eman is ecreasing at a rate of 6 units per month when the eman is 150 units, fin the rate of change of the price. 30. Price eman. Repeat Problem 29 for 2 + 2p + 25p 2 = 74, Pollution. An oil tanker agroun on a reef is leaking oil that forms a circular oil slick about 0.1 foot thick (see the figure). To estimate the rate V t (in cubic feet per minute) at which the oil is leaking from the tanker, it was foun that the raius of the slick was increasing at 0.32 foot per minute (R t = 0.32) when the raius R was 500 feet. Fin V t, using p L Tanker Oil slick R Figure for 31 A pr 2 V 0.1 A 32. Learning. A person who is new on an assembly line performs an operation in T minutes after performances of the operation, as given by T = 6 a b If t = 6 operations per hour, where t is time in hours, fin T t after 36 performances of the operation.