C H A P T E R 4. Applications of Differentiation. PDF Created with deskpdf PDF Writer - Trial ::

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1 What What What At Licensed to: Scientists have tried to eplain how rainbows are formed since the time of Aristotle. In the project on page 88, ou will be able to use the principles of differential calculus to eplain the formation, location, and colors of the rainbow. C H A P T E R 4 Applications of Differentiation Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Gregor D. Dimijian, M.D. Licensed to: jsamuels@bmcc.cun.edu 4. M a i m u m a n d M i n i m u m V a l u e s FIGURE Minimum value f(a), maimum value f(d) We have alread investigated some of the applications of derivatives, but now that we know the differentiation rules we are in a better position to pursue the applications of differentiation in greater depth. Here we learn how derivatives affect the shape of a graph of a function and, in particular, how the help us locate maimum and minimum values of functions. Man practical problems require us to minimize a cost or maimize an area or somehow find the best possible outcome of a situation. In particular, we will be able to investigate the optimal shape of a can and to eplain the location of rainbows in the sk. Some of the most important applications of differential calculus are optimization problems, in which we are required to nd the optimal (best) w a of doing something. Here are eamples of such problems that we will solve in this chapter: is the shape of a can that minimizes manufacturing costs? is the maimum acceleration of a space shuttle? (This is an important question to the astronauts who have to withstand the effects of acceleration.) is the radius of a contracted windpipe that epels air most rapidl during a cough? what angle should blood vessels branch so as to minimize the energ epended b the heart in pumping blood? These problems can be reduced to nding the maimum or minimum v alues of a function. Lets rst e plain eactl what we mean b maimum and minimum values. Definition A function f has an absolute maimum (or global maimum) at c if f c f for all in D, where D is the domain of f. The number f c is called the maimum value of f on D. Similarl, f has an absolute minimum at c if f c f for all in D and the number f c is called the minimum value of f on D. The maimum and minimum values of f are called the etreme values of f. Figure shows the graph of a function f with absolute maimum at d and absolute minimum at a. Note that d, f d is the highest point on the graph and a, f a is the lowest point. f(a) a f(d) b c d e Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. 79

2 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu = FIGURE Minimum value, no maimum = FIGURE No minimum, no maimum FIGURE 4 In Figure, if we consider onl values of near b [for instance, if we restrict our attention to the interval a, c], then f b is the largest of those values of f and is called a local maimum value of f. Likewise, f c is called a local minimum value of f because f c f for near c [in the interval b, d, for instance]. The function f also has a local minimum at e. In general, we have the following denition. Definition A function f has a local maimum (or relative maimum) at c if f c f when is near c. [This means that f c f for all in some open interval containing c.] Similarl, f has a local minimum at c if f c f when is near c. EXAMPLE The function f cos takes on its (local and absolute) maimum value of innitel man times, since cos n for an integer n and cos for all. Likewise, cosn is its minimum value, where n is an integer. EXAMPLE If f, then f f because for all. Therefore, f is the absolute (and local) minimum value of f. This corresponds to the fact that the origin is the lowest point on the parabola. (See Figure.) However, there is no highest point on the parabola and so this function has no maimum value. EXAMPLE From the graph of the function f, shown in Figure, we see that this function has neither an absolute maimum value nor an absolute minimum value. In fact, it has no local etreme values either. EXAMPLE 4 The graph of the function f is shown in Figure 4. You can see that f 5 is a local maimum, whereas the absolute maimum is f 7. (This absolute maimum is not a local maimum because it occurs at an endpoint.) Also, f is a local minimum and f 7 is both a local and an absolute minimum. Note that f has neither a local nor an absolute maimum at 4. (_,7) =$-6 +8 (,5) _ 4 5 (,_7) 4 We have seen that some functions have etreme values, whereas others do not. The following theorem gives conditions under which a function is guaranteed to possess etreme values. Licensed to: jsamuels@bmcc.cun.edu FIGURE 5 c d FIGURE 8 {c, f(c)} {d, f(d)} SECTION 4. MAXIMUM AND MINIMUM VALUES 8 The Etreme Value Theorem If f is continuous on a closed interval a, b, then f attains an absolute maimum value f c and an absolute minimum value f d at some numbers c and d in a, b. The Etreme Value Theorem is illustrated in Figure 5. Note that an etreme value can be taken on more than once. Although the Etreme Value Theorem is intuitivel ver plausible, it is difcult to pro ve and so we omit the proof. a c d b a c d=b a c d c b Figures 6 and 7 show that a function need not possess etreme values if either hpothesis (continuit or closed interval) is omitted from the Etreme Value Theorem. FIGURE 6 This function has minimum value f()=, but no maimum value. The function f whose graph is shown in Figure 6 is dened on the closed interv al [, ] but has no maimum value. (Notice that the range of f is [, ). The function takes on values arbitraril close to, but never actuall attains the value.) This does not contradict the Etreme Value Theorem because f is not continuous. [Nonetheless, a discontinuous function could have maimum and minimum values. See Eercise (b).] The function t shown in Figure 7 is continuous on the open interval (, ) but has neither a maimum nor a minimum value. [The range of t is,. The function takes on arbitraril large values.] This does not contradict the Etreme Value Theorem because the interval (, ) is not closed. The Etreme Value Theorem sas that a continuous function on a closed interval has a maimum value and a minimum value, but it does not tell us how to nd these e treme values. We start b looking for local etreme values. Figure 8 shows the graph of a function f with a local maimum at c and a local minimum at d. It appears that at the maimum and minimum points the tangent lines are horizontal and therefore each has slope. We know that the derivative is the slope of the tangent line, so it appears that f c and f d. The following theorem sas that this is alwas true for differentiable functions. FIGURE 7 This continuous function g has no maimum or minimum. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

3 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu Fermat s Theorem is named after Pierre Fermat (6 665), a French lawer who took up mathematics as a hobb. Despite his amateur status, Fermat was one of the two inventors of analtic geometr (Descartes was the other). His methods for finding tangents to curves and maimum and minimum values (before the invention of its and derivatives) made him a forerunner of Newton in the creation of differential calculus. = FIGURE 9 If ƒ=, then fª()= but has no maimum or minimum. 4 Fermat s Theorem If f has a local maimum or minimum at c, and if f c eists, then f c. Proof Suppose, for the sake of deniteness, that f has a local maimum at c. Then, according to Denition, f c f if is sufcientl close to c. This implies that if h is sufcientl close to, with h being positive or negative, then and therefore 5 We can divide both sides of an inequalit b a positive number. Thus, if h and h is sufcientl small, we have Taking the right-hand it of both sides of this inequalit (using Theorem..), we get But since f c eists, we have f c hl and so we have shown that f c. If h, then the direction of the inequalit (5) is reversed when we divide b h: So, taking the left-hand it, we have f c hl hl f c h f c f c h f c h f c h f c h f c h f c h f c h f c h f c h f c h f c f c h hl hl h f c h f c h We have shown that f c and also that f c. Since both of these inequalities must be true, the onl possibilit is that f c. We have proved Fermats Theorem for the case of a local maimum. The case of a local minimum can be proved in a similar manner, or we could use Eercise 76 to deduce it from the case we have just proved (see Eercise 77). The following eamples caution us against reading too much into Fermats Theorem. We cant epect to locate etreme values simpl b setting f and solving for. EXAMPLE 5 If f, then f, so f. But f has no maimum or minimum at, as ou can see from its graph in Figure 9. (Or observe that for but for.) The fact that f simpl means that the curve has a hl f c h f c h Licensed to: jsamuels@bmcc.cun.edu = SECTION 4. MAXIMUM AND MINIMUM VALUES 8 horizontal tangent at,. Instead of having a maimum or minimum at,, the curve crosses its horizontal tangent there. EXAMPLE 6 The function has its (local and absolute) minimum value at, but that value cant be found b setting f because, as was shown in Eample 6 in Section.9, f does not eist. (See Figure.) WARNING Eamples 5 and 6 show that we must be careful when using Fermats Theorem. Eample 5 demonstrates that even when f c there need not be a maimum FIGURE or minimum at c. (In other words, the converse of Fermats Theorem is false in general.) If ƒ=, then f()= is a Furthermore, there ma be an etreme value even when f c does not eist (as in minimum value, but fª() does not eist. Eample 6). Fermats Theorem does suggest that we should at least start looking for etreme values of f at the numbers c where f c or where f c does not eist. Such numbers are given a special name. Figure shows a graph of the function f in Eample 7. It supports our answer because there is a horizontal tangent when.5 and a vertical tangent when..5 _.5 5 _ FIGURE 6 Definition A critical number of a function f is a number c in the domain of f such that either f c or f c does not eist. EXAMPLE 7 Find the critical numbers of f 5 4. SOLUTION The Product Rule gives [The same result could be obtained b rst writing f ] Therefore, f if 8, that is,, and f does not eist when. Thus, the critical numbers are and. In terms of critical numbers, Fermats Theorem can be rephrased as follows (compare Denition 6 with Theorem 4): 7 f f If f has a local maimum or minimum at c, then c is a critical number of f. To nd an absolute maimum or minimum of a continuous function on a closed interval, we note that either it is local [in which case it occurs at a critical number b (7)] or it occurs at an endpoint of the interval. Thus, the following three-step procedure alwas works. The Closed Interval Method To nd the absolute maimum and minimum values of a continuous function f on a closed interval a, b:. Find the values of f at the critical numbers of f in a, b.. Find the values of f at the endpoints of the interval The largest of the values from Steps and is the absolute maimum value; the smallest of these values is the absolute minimum value. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

4 84 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu FIGURE 8 _ FIGURE = (,_) (4,7) π EXAMPLE 8 Find the absolute maimum and minimum values of the function SOLUTION Since f is continuous on [, 4], we can use the Closed Interval Method: Since f eists for all, the onl critical numbers of f occur when f, that is, or. Notice that each of these critical numbers lies in the interval (, 4). The values of f at these critical numbers are The values of f at the endpoints of the interval are Comparing these four numbers, we see that the absolute maimum value is f 4 7 and the absolute minimum value is f. Note that in this eample the absolute maimum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of f is sketched in Figure. If ou have a graphing calculator or a computer with graphing software, it is possible to estimate maimum and minimum values ver easil. But, as the net eample shows, calculus is needed to nd the eact values. EXAMPLE 9 (a) Use a graphing device to estimate the absolute minimum and maimum values of the function f sin,. (b) Use calculus to nd the e act minimum and maimum values. SOLUTION (a) Figure shows a graph of f in the viewing rectangle, b, 8. B moving the cursor close to the maimum point, we see that the -coordinates dont change ver much in the vicinit of the maimum. The absolute maimum value is about 6.97 and it occurs when 5.. Similarl, b moving the cursor close to the minimum point, we see that the absolute minimum value is about.68 and it occurs when.. It is possible to get more accurate estimates b zooming in toward the maimum and minimum points, but instead lets use calculus. (b) The function f sin is continuous on,. Since f cos, we have f when cos and this occurs when or 5. The values of f at these critical points are and f 5 5 f f f 6 f f ( ) 8 f sin s sin 5 f 4 f s Licensed to: jsamuels@bmcc.cun.edu NASA 4. Eercises The values of f at the endpoints are. Eplain the difference between an absolute minimum and a local minimum.. Suppose f is a continuous function dened on a closed interval a, b. f SECTION 4. MAXIMUM AND MINIMUM VALUES 85 Comparing these four numbers and using the Closed Interval Method, we see that the absolute minimum value is f s and the absolute maimum value is f 5 5 s. The values from part (a) serve as a check on our work. EXAMPLE The Hubble Space Telescope was deploed on April 4, 99, b the space shuttle Discover. A model for the velocit of the shuttle during this mission, from liftoff at t until the solid rocket boosters were jettisoned at t 6 s, is given b vt.t.99t.6t.8 (in feet per second). Using this model, estimate the absolute maimum and minimum values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters. SOLUTION We are asked for the etreme values not of the given velocit function, but rather of the acceleration function. So we rst need to differentiate to nd the acceleration: at vt d dt.t.99t.6t.8.96t.858t.6 We now appl the Closed Interval Method to the continuous function a on the interval t 6. Its derivative is The onl critical number occurs when at : Evaluating at at the critical number and at the endpoints, we have a.6 and at.78t.858 t at.5 f 6.8 a So the maimum acceleration is about 6.87 fts and the minimum acceleration is about.5 fts. (a) What theorem guarantees the eistence of an absolute maimum value and an absolute minimum value for f? (b) What steps would ou take to nd those maimum and minimum values? Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

5 86 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu 4 For each of the numbers a, b, c, d, e, r, s, and t, state whether the function whose graph is shown has an absolute maimum or minimum, a local maimum or minimum, or neither a maimum nor a minimum Use the graph to state the absolute and local maimum and minimum values of the function a b c d e r s t a b c d e r s =ƒ =ƒ 7 Sketch the graph of a function f that is continuous on [, 5] and has the given properties. 7. Absolute minimum at, absolute maimum at, local minimum at 4 8. Absolute minimum at, absolute maimum at 5, local maimum at, local minimum at 4 t 9. Absolute maimum at 5, absolute minimum at, local maimum at, local minima at and 4. f has no local maimum or minimum, but and 4 are critical numbers. (a) Sketch the graph of a function that has a local maimum at and is differentiable at. (b) Sketch the graph of a function that has a local maimum at and is continuous but not differentiable at. (c) Sketch the graph of a function that has a local maimum at and is not continuous at.. (a) Sketch the graph of a function on [, ] that has an absolute maimum but no local maimum. (b) Sketch the graph of a function on [, ] that has a local maimum but no absolute maimum.. (a) Sketch the graph of a function on [, ] that has an absolute maimum but no absolute minimum. (b) Sketch the graph of a function on [, ] that is discontinuous but has both an absolute maimum and an absolute minimum. 4. (a) Sketch the graph of a function that has two local maima, one local minimum, and no absolute minimum. (b) Sketch the graph of a function that has three local minima, two local maima, and seven critical numbers. 5 Sketch the graph of f b hand and use our sketch to nd the absolute and local maimum and minimum v alues of f. (Use the graphs and transformations of Sections. and..) 5. f 8, 6. f, 5 7. f, 8. f, 9. f,. f,. f,. f, 5. f t t, t 4. f t t, t 5. f sin, 6. f tan, 4 7. f s 8. f e 9.. f 4 f if if if if Licensed to: jsamuels@bmcc.cun.edu 46 Find the critical numbers of the function. SECTION 4. MAXIMUM AND MINIMUM VALUES 87. f 5 4. f 68. f cos sin,. f 4 4. f 5. st t 4 4t 6t 6. t t 9. tt 5t t 5 4. tt st t 4. F f cos sin 4. G s 44. t 4 tan 45. f ln 46. f e 47 6 Find the absolute maimum and absolute minimum values of f on the given interval. 47. f 5,, 48. f,, 49. f,, 5. f 6 9,, 4 5. f 4,, 5. f,, 5. f,, 54. f 4, 4, f t ts4 t,, 56. f t s t 8 t,, f sin cos,, 58. f cos,, 59. f e,, 6. f ln,, 6. f ln,, 4 6. f e e,, 6. If a and b are positive numbers, nd the maimum v alue of f a b,. ; 64. Use a graph to estimate the critical numbers of correct to one decimal place. f ; (a) Use a graph to estimate the absolute maimum and minimum values of the function to two decimal places. (b) Use calculus to nd the e act maimum and minimum values. 65. f 8, 66. f e, f z z z z 67. f s 69. Between C and C, the volume V (in cubic centimeters) of kg of water at a temperature T is given approimatel b the formula V T.854T.679T Find the temperature at which water has its maimum densit. 7. An object with weight W is dragged along a horizontal plane b a force acting along a rope attached to the object. If the rope makes an angle with the plane, then the magnitude of the force is F where is a positive constant called the coefcient of friction and where. Show that F is minimized when tan. W sin cos 7. A model for the food-price inde (the price of a representative bask et of foods) between 984 and 994 is given b the function It.945t 5.48t 4.656t It.4598t.67t 99. where t is measured in ears since midear 984, so t, and It is measured in 987 dollars and scaled such that I. Estimate the times when food was cheapest and most epensive during the period ; 7. On Ma 7, 99, the space shuttle Endeavour was launched on mission STS-49, the purpose of which was to install a new perigee kick motor in an Intelsat communications satellite. The table gives the velocit data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. Event Time (s) Velocit (fts) Launch Begin roll maneuver 85 End roll maneuver 5 9 Throttle to 89% 447 Throttle to 67% 74 Throttle to 4% 59 5 Maimum dnamic pressure Solid rocket booster separation 5 45 (a) Use a graphing calculator or computer to nd the cubic polnomial that best models the velocit of the shuttle for the time interval t, 5. Then graph this polnomial. (b) Find a model for the acceleration of the shuttle and use it to estimate the maimum and minimum values of the acceleration during the rst 5 seconds. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

6 88 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu 7. When a foreign object lodged in the trachea (windpipe) forces a person to cough, the diaphragm thrusts upward causing an increase in pressure in the lungs. This is accompanied b a contraction of the trachea, making a narrower channel for the epelled air to o w through. For a given amount of air to escape in a ed time, it must move faster through the narrower channel than the wider one. The greater the velocit of the airstream, the greater the force on the foreign object. X ras show that the radius of the circular tracheal tube contracts to about two-thirds of its normal radius during a cough. According to a mathematical model of coughing, the velocit v of the airstream is related to the radius r of the trachea b the equation vr kr rr where k is a constant and r is the normal radius of the trachea. The restriction on r is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than r is prevented (otherwise the person would suffocate). (a) Determine the value of r in the interval [ r, r] at which v has an absolute maimum. How does this compare with eperimental evidence? å A from Sun to observer APPLIED PROJECT O C å Formation of the primar rainbow B D(å ) r r r The Calculus of Rainbows (b) What is the absolute maimum value of v on the interval? (c) Sketch the graph of v on the interval, r. 74. Show that 5 is a critical number of the function but t does not have a local etreme value at Prove that the function t 5 f 5 has neither a local maimum nor a local minimum. 76. If f has a minimum value at c, show that the function t f has a maimum value at c. 77. Prove Fermats Theorem for the case in which f has a local minimum at c. 78. A cubic function is a polnomial of degree ; that is, it has the form f a b c d, where a. (a) Show that a cubic function can have two, one, or no critical number(s). Give eamples and sketches to illustrate the three possibilities. (b) How man local etreme values can a cubic function have? Rainbows are created when raindrops scatter sunlight. The have fascinated mankind since ancient times and have inspired attempts at scientic e planation since the time of Aristotle. In this project we use the ideas of Descartes and Newton to eplain the shape, location, and colors of rainbows.. The gure sho ws a ra of sunlight entering a spherical raindrop at A. Some of the light is reected, but the line AB shows the path of the part that enters the drop. Notice that the light is refracted toward the normal line AO and in fact Snells Law sas that sin k sin, where is the angle of incidence, is the angle of refraction, and k 4 is the inde of refraction for water. At B some of the light passes through the drop and is refracted into the air, but the line BC shows the part that is reected. (The angle of incidence equals the angle of reection.) When the ra reaches C, part of it is reected, but for the time being we are more interested in the part that leaves the raindrop at C. (Notice that it is refracted awa from the normal line.) The angle of deviation D is the amount of clockwise rotation that the ra has undergone during this three-stage process. Thus Show that the minimum value of the deviation is D 8 and occurs when. The signicance of the minimum de viation is that when we have D, so. This means that man ras with become deviated b approimatel the same amount. It is the concentration of ras coming from near the direction of minimum D D Licensed to: jsamuels@bmcc.cun.edu to observer from Sun D å å A Formation of the secondar rainbow C B APPLIED PROJECT THE CALCULUS OF RAINBOWS 89 deviation that creates the brightness of the primar rainbow. The following gure sho ws that the angle of elevation from the observer up to the highest point on the rainbow is (This angle is called the rainbow angle.) ras from Sun. Problem eplains the location of the primar rainbow, but how do we eplain the colors? Sunlight comprises a range of wavelengths, from the red range through orange, ellow, green, blue, indigo, and violet. As Newton discovered in his prism eperiments of 666, the inde of refraction is different for each color. (The effect is called dispersion.) For red light the refractive inde is k.8 whereas for violet light it is k.45. B repeating the calculation of Problem for these values of k, show that the rainbow angle is about 4. for the red bow and 4.6 for the violet bow. So the rainbow reall consists of seven individual bows corresponding to the seven colors.. Perhaps ou have seen a fainter secondar rainbow above the primar bow. That results from the part of a ra that enters a raindrop and is refracted at A, reected twice (at B and C ), and refracted as it leaves the drop at D (see the gure). This time the deviation angle D is the total amount of counterclockwise rotation that the ra undergoes in this four-stage process. Show that D 6 and D has a minimum value when ras from Sun observer cos k 8 Taking k 4, show that the minimum deviation is about 9 and so the rainbow angle for the secondar rainbow is about 5, as shown in the gure Show that the colors in the secondar rainbow appear in the opposite order from those in the primar rainbow. 8 Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

7 9 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu 4. T h e M e a n V a l u e T h e o r e m Rolle s Theorem was first published in 69 b the French mathematician Michel Rolle (65 79) in a book entitled Méthode pour résoudre les égalitéz. Later, however, he became a vocal critic of the methods of his da and attacked calculus as being a collection of ingenious fallacies. FIGURE a c c b (a) Take cases We will see that man of the results of this chapter depend on one central fact, which is called the Mean Value Theorem. But to arrive at the Mean Value Theorem we rst need the following result. Rolle s Theorem Let f be a function that satises the follo wing three hpotheses:. f is continuous on the closed interval a, b.. f is differentiable on the open interval a, b.. Then there is a number c in a, b such that f c. Before giving the proof lets take a look at the graphs of some tpical functions that satisf the three hpotheses. Figure shows the graphs of four such functions. In each case it appears that there is at least one point c, f c on the graph where the tangent is horizontal and therefore f c. Thus, Rolles Theorem is plausible. a c b Proof f a f b (b) There are three cases: a c c b CASE I f() k, a constant Then f, so the number c can be taken to be an number in a, b. (c) CASE II f() > f(a) for some in (a, b) [as in Figure (b) or (c)] B the Etreme Value Theorem (which we can appl b hpothesis ), f has a maimum value somewhere in a, b. Since f a f b, it must attain this maimum value at a number c in the open interval a, b. Then f has a local maimum at c and, b hpothesis, f is differentiable at c. Therefore, f c b Fermats Theorem. CASE III f() < f(a) for some in (a, b) [as in Figure (c) or (d)] B the Etreme Value Theorem, f has a minimum value in a, b and, since f a f b, it attains this minimum value at a number c in a, b. Again f c b Fermats Theorem. EXAMPLE Lets appl Rolles Theorem to the position function s f t of a moving object. If the object is in the same place at two different instants t a and t b, then f a f b. Rolles Theorem sas that there is some instant of time t c between a and b when f c ; that is, the velocit is. (In particular, ou can see that this is true when a ball is thrown directl upward.) a (d) c b Licensed to: jsamuels@bmcc.cun.edu Figure shows a graph of the function f discussed in Eample. Rolle s Theorem shows that, no matter how much we enlarge the viewing rectangle, we can never find a second -intercept. _ FIGURE _ The Mean Value Theorem is an eample of what is called an eistence theorem. Like the Intermediate Value Theorem, the Etreme Value Theorem, and Rolle s Theorem, it guarantees that there eists a number with a certain propert, but it doesn t tell us how to find the number. SECTION 4. THE MEAN VALUE THEOREM 9 EXAMPLE Prove that the equation has eactl one real root. SOLUTION First we use the Intermediate Value Theorem (.5.) to show that a root eists. Let f. Then f and f. Since f is a polnomial, it is continuous, so the Intermediate Value Theorem states that there is a number c between and such that f c. Thus, the given equation has a root. To show that the equation has no other real root, we use Rolles Theorem and argue b contradiction. Suppose that it had two roots a and b. Then f a f b and, since f is a polnomial, it is differentiable on a, b and continuous on a, b. Thus, b Rolles Theorem, there is a number c between a and b such that f c. But (since ) so f can never be. This gives a contradiction. Therefore, the equation cant have two real roots. Our main use of Rolles Theorem is in proving the following important theorem, which was rst stated b another French mathematician, Joseph-Louis Lagrange. The Mean Value Theorem Let f be a function that satises the follo wing hpotheses:. f is continuous on the closed interval a, b.. f is differentiable on the open interval a, b. Then there is a number c in a, b such that or, equivalentl, Before proving this theorem, we can see that it is reasonable b interpreting it geometricall. Figures and 4 show the points Aa, f a and Bb, f b on the graphs of two differentiable functions. The slope of the secant line AB is A{a,f(a)} P{c,f(c)} B{b,f(b)} a c b f f c f b f a f cb a mab f b f a b a f b f a b a FIGURE FIGURE 4 A for all P P a c c b B Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

8 9 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu A f(b)-f(a) f(a)+ (-a) b-a FIGURE 5 ƒ h() =ƒ The Mean Value Theorem was first formulated b Joseph-Louis Lagrange (76 8), born in Ital of a French father and an Italian mother. He was a child prodig and became a professor in Turin at the tender age of 9. Lagrange made great contributions to number theor, theor of functions, theor of equations, and analtical and celestial mechanics. In particular, he applied calculus to the analsis of the stabilit of the solar sstem. At the invitation of Frederick the Great, he succeeded Euler at the Berlin Academ and, when Frederick died, Lagrange accepted King Louis XVI s invitation to Paris, where he was given apartments in the Louvre. Despite all the trappings of luur and fame, he was a kind and quiet man, living onl for science. B which is the same epression as on the right side of Equation. Since f c is the slope of the tangent line at the point c, f c, the Mean Value Theorem, in the form given b Equation, sas that there is at least one point Pc, f c on the graph where the slope of the tangent line is the same as the slope of the secant line AB. In other words, there is a point P where the tangent line is parallel to the secant line AB. Proof We appl Rolles Theorem to a new function h dened as the dif ference between f and the function whose graph is the secant line AB. Using Equation, we see that the equation of the line AB can be written as or as So, as shown in Figure 5, 4 First we must verif that h satises the three hpotheses of Rolle s Theorem.. The function h is continuous on a, b because it is the sum of f and a rst-de gree polnomial, both of which are continuous.. The function h is differentiable on a, b because both f and the rst-de gree polnomial are differentiable. In fact, we can compute h directl from Equation 4:. (Note that f a and f b f ab a are constants.) Therefore, ha hb. Since h satises the hpotheses of Rolle s Theorem, that theorem sas there is a number c in a, b such that hc. Therefore and so f a f a h f f a h f ha f a f a hb f b f a f b f a f b f a hc f c f c f b f a a b a f b f a a b a f b f a a b a f b f a b a f b f a a a b a f b f a b a b a f b f a b a f b f a b a EXAMPLE To illustrate the Mean Value Theorem with a specic function, lets consider f, a, b. Since f is a polnomial, it is continuous and differentiable Licensed to: jsamuels@bmcc.cun.edu O FIGURE 6 = - B c SECTION 4. THE MEAN VALUE THEOREM 9 for all, so it is certainl continuous on, and differentiable on,. Therefore, b the Mean Value Theorem, there is a number c in, such that Now f 6, f, and f, so this equation becomes which gives c 4, that is, c s. But c must lie in,, so c s. Figure 6 illustrates this calculation: The tangent line at this value of c is parallel to the secant line OB. EXAMPLE 4 If an object moves in a straight line with position function s f t, then the average velocit between t a and t b is and the velocit at t c is f c. Thus, the Mean Value Theorem (in the form of Equation ) tells us that at some time t c between a and b the instantaneous velocit f c is equal to that average velocit. For instance, if a car traveled 8 km in hours, then the speedometer must have read 9 kmh at least once. In general, the Mean Value Theorem can be interpreted as saing that there is a number at which the instantaneous rate of change is equal to the average rate of change over an interval. The main signicance of the Mean Value Theorem is that it enables us to obtain information about a function from information about its derivative. The net eample provides an instance of this principle. EXAMPLE 5 Suppose that f and f 5 for all values of. How large can f possibl be? SOLUTION We are given that f is differentiable (and therefore continuous) everwhere. In particular, we can appl the Mean Value Theorem on the interval,. There eists a number c such that f f f c so We are given that f 5 for all, so in particular we know that f c 5. Multipling both sides of this inequalit b, we have f c, so The largest possible value for f is 7. f f f c 6 c 6c f b f a b a f f f c f c f f c 7 The Mean Value Theorem can be used to establish some of the basic facts of differential calculus. One of these basic facts is the following theorem. Others will be found in the following sections. 5 Theorem If f for all in an interval a, b, then f is constant on a, b. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

9 94 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu Proof Let and be an two numbers in a, b with. Since f is differentiable on a, b, it must be differentiable on, and continuous on,. B appling the Mean Value Theorem to f on the interval,, we get a number c such that c and 6 Since f for all, we have f c, and so Equation 6 becomes Therefore, f has the same value at an two numbers and in a, b. This means that f is constant on a, b. 7 Corollar If f t for all in an interval a, b, then f t is constant on a, b; that is, f t c where c is a constant. Proof Let F f t. Then for all in a, b. Thus, b Theorem 5, F is constant; that is, f t is constant. NOTE Care must be taken in appling Theorem 5. Let The domain of f is D and f for all in D. But f is obviousl not a constant function. This does not contradict Theorem 5 because D is not an interval. Notice that f is constant on the interval, and also on the interval,. EXAMPLE 6 Prove the identit tan cot. SOLUTION Although calculus isnt needed to prove this identit, the proof using calculus is quite simple. If f tan cot, then for all values of. Therefore, f C, a constant. To determine the value of C, we put [because we can evaluate f eactl]. Then Thus, tan cot. f f f c f f F f t f f f if if f C f tan cot 4 4 or Licensed to: jsamuels@bmcc.cun.edu 4. Eercises 4 Verif that the function satises the three hpotheses of Rolles Theorem on the given interval. Then nd all numbers c that satisf the conclusion of Rolles Theorem.. f 4,, 4. f 5,,. f sin,, 4. f s 6, 6, 5. Let f. Show that f f but there is no number c in, such that f c. Wh does this not contradict Rolles Theorem? 6. Let f. Show that f f but there is no number c in, such that f c. Wh does this not contradict Rolles Theorem? 7. Use the graph of f to estimate the values of c that satisf the conclusion of the Mean Value Theorem for the interval, 8. =ƒ 8. Use the graph of f given in Eercise 7 to estimate the values of c that satisf the conclusion of the Mean Value Theorem for the interval, 7. ; 9. (a) Graph the function f 4 in the viewing rectangle, b,. (b) Graph the secant line that passes through the points, 5 and 8, 8.5 on the same screen with f. (c) Find the number c that satises the conclusion of the Mean Value Theorem for this function f and the interval, 8. Then graph the tangent line at the point c, f c and notice that it is parallel to the secant line. ;. (a) In the viewing rectangle, b 5, 5, graph the function f and its secant line through the points, 4 and, 4. Use the graph to estimate the -coordinates of the points where the tangent line is parallel to the secant line. (b) Find the eact values of the numbers c that satisf the conclusion of the Mean Value Theorem for the interval, and compare with our answers to part (a). 4 Verif that the function satises the hpotheses of the Mean Value Theorem on the given interval. Then nd all numbers c that satisf the conclusion of the Mean Value Theorem.. f 5,,. f,,. f e,, 4. f,, 4 5. Let. Show that there is no value of c such that f f f c. Wh does this not contradict the Mean Value Theorem? f 6. Let f. Show that there is no value of c such that f f f c. Wh does this not contradict the Mean Value Theorem? 7. Show that the equation 4 5 has eactl one real root. 8. Show that the equation sin has eactl one real root. 9. Show that the equation 5 c has at most one root in the interval,.. Show that the equation 4 4 c has at most two real roots.. (a) Show that a polnomial of degree has at most three real roots. (b) Show that a polnomial of degree n has at most n real roots.. (a) Suppose that f is differentiable on and has two roots. Show that f has at least one root. (b) Suppose f is twice differentiable on and has three roots. Show that f has at least one real root. (c) Can ou generalize parts (a) and (b)?. If f and f for 4, how small can f 4 possibl be? 4. Suppose that f 5 for all values of. Show that 8 f 8 f. 5. Does there eist a function f such that f, f 4, and f for all? 6. Suppose that f and t are continuous on a, b and differentiable on a, b. Suppose also that f a ta and f t for a b. Prove that f b tb. [Hint: Appl the Mean Value Theorem to the function h f t.] 7. Show that s if. SECTION 4. THE MEAN VALUE THEOREM 95 Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

10 96 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu 8. Suppose f is an odd function and is differentiable everwhere. Prove that for ever positive number b, there eists a number c in b, b such that f c f bb. 9. Use the Mean Value Theorem to prove the inequalit sin a sin b a b. If f c (c a constant) for all, use Corollar 7 to show that f c d for some constant d.. Let f and t for all a and b if if Show that f t for all in their domains. Can we conclude from Corollar 7 that f t is constant? 4. H o w D e r i v a t i v e s A f f e c t t h e S h a p e o f a G r a p h A Let s abbreviate the name of this test to the I/D Test. B FIGURE C D Resources / Module / Increasing and Decreasing Functions / Increasing-Decreasing Detector. Use the method of Eample 6 to prove the identit. Prove the identit sin cos arcsin arctan s 4. At : P.M. a cars speedometer reads mih. At : P.M. it reads 5 mih. Show that at some time between : and : the acceleration is eactl mih. 5. Two runners start a race at the same time and nish in a tie. Prove that at some time during the race the have the same speed. [Hint: Consider f t tt ht, where t and h are the position functions of the two runners.] 6. A number a is called a ed point of a function f if f a a. Prove that if f for all real numbers, then f has at most one ed point. Man of the applications of calculus depend on our abilit to deduce facts about a function f from information concerning its derivatives. Because f represents the slope of the curve f at the point, f, it tells us the direction in which the curve proceeds at each point. So it is reasonable to epect that information about f will provide us with information about f. W h a t D o e s f S a a b o u t f? To see how the derivative of f can tell us where a function is increasing or decreasing, look at Figure. (Increasing functions and decreasing functions were dened in Section..) Between A and B and between C and D, the tangent lines have positive slope and so f. Between B and C, the tangent lines have negative slope and so f. Thus, it appears that f increases when f is positive and decreases when f is negative. To prove that this is alwas the case, we use the Mean Value Theorem. Increasing/Decreasing Test (a) If f on an interval, then f is increasing on that interval. (b) If f on an interval, then f is decreasing on that interval. Proof (a) Let and be an two numbers in the interval with. According to the denition of an increasing function (page ) we have to show that f f. Because we are given that f, we know that f is differentiable on,. So, b the Mean Value Theorem there is a number c between and such that f f f c Now f c b assumption and because. Thus, the right side of Licensed to: jsamuels@bmcc.cun.edu Module 4.A guides ou in determining properties of the derivative f b eamining the graphs of a variet of functions f. FIGURE Equation is positive, and so This shows that f is increasing. Part (b) is proved similarl. EXAMPLE Find where the function f is increasing and where it is decreasing. SOLUTION f f SECTION 4. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 97 f f f 4 To use the ID Test we have to know where f and where f. This depends on the signs of the three factors of f, namel,,, and. We divide the real line into intervals whose endpoints are the critical numbers,, and and arrange our work in a chart. A plus sign indicates that the given epression is positive, and a minus sign indicates that it is negative. The last column of the chart gives the conclusion based on the ID Test. For instance, f for, so f is decreasing on (, ). (It would also be true to sa that f is decreasing on the closed interval,.) Interval f f The graph of f shown in Figure conrms the information in the chart. Recall from Section 4. that if f has a local maimum or minimum at c, then c must be a critical number of f (b Fermats Theorem), but not ever critical number gives rise to a maimum or a minimum. We therefore need a test that will tell us whether or not f has a local maimum or minimum at a critical number. You can see from Figure that f 5 is a local maimum value of f because f increases on, and decreases on,. Or, in terms of derivatives, f for and f for. In other words, the sign of f changes from positive to negative at. This observation is the basis of the following test. The First Derivative Test Suppose that c is a critical number of a continuous function f. (a) If f changes from positive to negative at c, then f has a local maimum at c. (b) If f changes from negative to positive at c, then f has a local minimum at c. (c) If f does not change sign at c (for eample, if f is positive on both sides of c or negative on both sides), then f has no local maimum or minimum at c. The First Derivative Test is a consequence of the ID Test. In part (a), for instance, since the sign of f changes from positive to negative at c, f is increasing to the left of c and decreasing to the right of c. It follows that f has a local maimum at c. or decreasing on (, ) increasing on (, ) decreasing on (, ) increasing on (, ) Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

11 98 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu fª()> c FIGURE 6 (a) Local maimum FIGURE 4 =+sin fª()< The + signs in the table come from the fact that t when cos. From the graph of cos, this is true in the indicated intervals. π It is eas to remember the First Derivative Test b visualizing diagrams such as those in Figure. fª()< fª()> c (b) Local minimum EXAMPLE Find the local minimum and maimum values of the function f in Eample. SOLUTION From the chart in the solution to Eample we see that f changes from negative to positive at, so f is a local minimum value b the First Derivative Test. Similarl, f changes from negative to positive at, so f 7 is also a local minimum value. As previousl noted, f 5 is a local maimum value because f changes from positive to negative at. EXAMPLE Find the local maimum and minimum values of the function SOLUTION To nd the critical numbers of t, we differentiate: So t when cos. The solutions of this equation are and 4. Because t is differentiable everwhere, the onl critical numbers are and 4 and so we analze t in the following table. Interval 4 4 Because t changes from positive to negative at, the First Derivative Test tells us that there is a local maimum at and the local maimum value is t Likewise, t changes from negative to positive at 4 and so t4 4 fª()> c (c) No maimum or minimum t sin fª()> t cos t cos sin s 4 4 sin s c s.8 4 s.46 is a local minimum value. The graph of t in Figure 4 supports our conclusion. fª()< fª()< (d) No maimum or minimum increasing on (, ) decreasing on (, 4) increasing on (4, ) t Licensed to: jsamuels@bmcc.cun.edu Eplore concavit on a roller coaster. Resources / Module / Concavit / Introduction FIGURE 5 FIGURE 6 FIGURE 7 W h a t D o e s f S a a b o u t f? Figure 5 shows the graphs of two increasing functions on a, b. Both graphs join point A to point B but the look different because the bend in different directions. How can we distinguish between these two tpes of behavior? In Figure 6 tangents to these curves have been drawn at several points. In (a) the curve lies above the tangents and f is called concave upward on a, b. In (b) the curve lies below the tangents and t is called concave downward on a, b. Definition If the graph of f lies above all of its tangents on an interval I, then it is called concave upward on I. If the graph of f lies below all of its tangents on I, it is called concave downward on I. Figure 7 shows the graph of a function that is concave upward (abbreviated CU) on the intervals b, c, d, e, and e, p and concave downward (CD) on the intervals a, b, c, d, and p, q. A f a b a b A (a) f (a) Concave upward B B B SECTION 4. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 99 C a b c d e p q CD CU CD CU CU CD Lets see how the second derivative helps determine the intervals of concavit. Looking at Figure 6(a), ou can see that, going from left to right, the slope of the tangent increases. D A A (b) g g (b) Concave downward P B B Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

12 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu FIGURE 8 This means that the derivative f is an increasing function and therefore its derivative f is positive. Likewise, in Figure 6(b) the slope of the tangent decreases from left to right, so f decreases and therefore f is negative. This reasoning can be reversed and suggests that the following theorem is true. A proof is given in Appendi F with the help of the Mean Value Theorem. Concavit Test (a) If f for all in I, then the graph of f is concave upward on I. (b) If f for all in I, then the graph of f is concave downward on I. EXAMPLE 4 Figure 8 shows a population graph for Cprian honebees raised in an apiar. How does the rate of population increase change over time? When is this rate highest? Over what intervals is P concave upward or concave downward? Number of bees (in thousands) P Time (in weeks) SOLUTION B looking at the slope of the curve as t increases, we see that the rate of increase of the population is initiall ver small, then gets larger until it reaches a maimum at about t weeks, and decreases as the population begins to level off. As the population approaches its maimum value of about 75, (called the carring capacit), the rate of increase, Pt, approaches. The curve appears to be concave upward on (, ) and concave downward on (, 8). In Eample 4, the population curve changed from concave upward to concave downward at approimatel the point (, 8,). This point is called an inection point of the curve. The signicance of this point is that the rate of population increase has its maimum value there. In general, an inection point is a point where a curv e changes its direction of concavit. Definition A point P on a curve f is called an inection point if f is continuous there and the curve changes from concave upward to concave downward or from concave downward to concave upward at P. For instance, in Figure 7, B, C, D, and P are the points of inection. Notice that if a curve has a tangent at a point of inection, then the curve crosses its tangent there. In view of the Concavit Test, there is a point of inection at an point where the second derivative changes sign. 8 t Licensed to: jsamuels@bmcc.cun.edu - FIGURE 9 =_ fª(c)= f(c) FIGURE f (c)>, f is concave upward P c ƒ f EXAMPLE 5 Sketch a possible graph of a function f that satises the follo wing conditions: i f on,, ii f on, and,, iii f, l f l Another application of the second derivative is the following test for maimum and minimum values. It is a consequence of the Concavit Test. The Second Derivative Test Suppose f is continuous near c. (a) If f c and f c, then f has a local minimum at c. (b) If f c and f c, then f has a local maimum at c. For instance, part (a) is true because f near c and so f is concave upward near c. This means that the graph of f lies above its horizontal tangent at c and so f has a local minimum at c. (See Figure.) EXAMPLE 6 Discuss the curve 4 4 with respect to concavit, points of inection, and local maima and minima. Use this information to sketch the curve. SOLUTION If f 4 4, then f on, f 4 4 f 4 To nd the critical numbers we set f and obtain and. To use the Second Derivative Test we evaluate f at these critical numbers: f SECTION 4. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH f on, SOLUTION Condition (i) tells us that f is increasing on, and decreasing on,. Condition (ii) sas that f is concave upward on, and,, and concave downward on,. From condition (iii) we know that the graph of f has two horizontal asmptotes: and. We rst dra w the horizontal asmptote as a dashed line (see Figure 9). We then draw the graph of f approaching this asmptote at the far left, increasing to its maimum point at and decreasing toward the -ais at the far right. We also make sure that the graph has inection points when and. Notice that we made the curve bend upward for and, and bend downward when is between and. f 6 Since f and f, f 7 is a local minimum. Since f, the Second Derivative Test gives no information about the critical number. But since f for and also for, the First Derivative Test tells us that f does not have a local maimum or minimum at. [In fact, the epression for f shows that f decreases to the left of and increases to the right of.] Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

13 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu =$-4 inection points FIGURE (,) (,_6) (,_7) Tr reproducing the graph in Figure with a graphing calculator or computer. Some machines produce the complete graph, some produce onl the portion to the right of the -ais, and some produce onl the portion between and 6. For an eplanation and cure, see Eample 7 in Section.4. An equivalent epression that gives the correct graph is 4 FIGURE (4,%?#) =@?#(6-)!?# Since f when or, we divide the real line into intervals with these numbers as endpoints and complete the following chart. Interval The point, is an inection point since the curv e changes from concave upward to concave downward there. Also, 6 is an inection point since the curv e changes from concave downward to concave upward there. Using the local minimum, the intervals of concavit, and the inection points, we sketch the curve in Figure. NOTE The Second Derivative Test is inconclusive when f c. In other words, at such a point there might be a maimum, there might be a minimum, or there might be neither (as in Eample 6). This test also fails when f c does not eist. In such cases the First Derivative Test must be used. In fact, even when both tests appl, the First Derivative Test is often the easier one to use. EXAMPLE 7 Sketch the graph of the function f 6. SOLUTION You can use the differentiation rules to check that the rst tw o derivatives are f 4 6 f f Concavit (, ) upward (, ) downward (, ) upward Since f when 4 and f does not eist when or 6, the critical numbers are, 4, and 6. Interval 4 6 f f decreasing on (, ) increasing on (, 4) decreasing on (4, 6) decreasing on (6, ) To nd the local e treme values we use the First Derivative Test. Since f changes from negative to positive at, f is a local minimum. Since f changes from positive to negative at 4, f 4 5 is a local maimum. The sign of f does not change at 6, so there is no minimum or maimum there. (The Second Derivative Test could be used at 4, but not at or 6 since f does not eist at either of these numbers.) Looking at the epression for f and noting that 4 for all, we have f for and for 6 and f for 6. So f is concave downward on, and, 6 and concave upward on 6,, and the onl inection point is 6,. The graph is sketched in Figure. Note that the curve has vertical tangents at, and 6, because f l as l and as l6. EXAMPLE 8 Use the rst and second deri vatives of f e, together with asmptotes, to sketch its graph. Licensed to: jsamuels@bmcc.cun.edu In Module 4.B ou can practice using graphical information about f to determine the shape of the graph of f. = (a) Preinar sketch FIGURE SOLUTION Notice that the domain of f is, so we check for vertical asmptotes b computing the left and right its as l. As l, we know that t l, so and this shows that is a vertical asmptote. As l, we have t l, so As l, we have l and so This shows that is a horizontal asmptote. Now lets compute the derivative. The Chain Rule gives Since e and for all, we have f for all. Thus, f is decreasing on, and on,. There is no critical number, so the function has no maimum or minimum. The second derivative is Since e and 4, we have f when and when f. So the curve is concave downward on (, and concave upward on ( and on. The inection point is ( ), ),, e ). To sketch the graph of f we rst dra w the horizontal asmptote (as a dashed line), together with the parts of the curve near the asmptotes in a preinar sketch [Figure (a)]. These parts reect the information concerning its and the f act that f is decreasing on both, and,. Notice that we have indicated that f l as l even though f does not eist. In Figure (b) we nish the sk etch b incorporating the information concerning concavit and the inection point. In Figure (c) we check our work with a graphing device. inection point f e e 4 = (b) Finished sketch l e tl e t l e tl e t = l e e f e SECTION 4. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH e 4 _ 4 (c) Computer confirmation Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

14 4 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION SECTION 4. HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 5 Licensed to: jsamuels@bmcc.cun.edu 4. Eercises Use the given graph of f to nd the follo wing. (a) The open intervals on which f is increasing. (b) The open intervals on which f is decreasing. (c) The open intervals on which f is concave upward. (d) The open intervals on which f is concave downward. (e) The coordinates of the points of inection.... Suppose ou are given a formula for a function f. (a) How do ou determine where f is increasing or decreasing? (b) How do ou determine where the graph of f is concave upward or concave downward? (c) How do ou locate inection points? 4. (a) State the First Derivative Test. (b) State the Second Derivative Test. Under what circumstances is it inconclusive? What do ou do if it fails? 5 6 The graph of the derivative f of a function f is shown. (a) On what intervals is f increasing or decreasing? (b) At what values of does f have a local maimum or minimum? _ =fª() =fª() The graph of the second derivative f of a function f is shown. State the -coordinates of the inection points of f. Give reasons for our answers. 8. The graph of the rst deri vative f of a function f is shown. (a) On what intervals is f increasing? Eplain. (b) At what values of does f have a local maimum or minimum? Eplain. (c) On what intervals is f concave upward or concave downward? Eplain. (d) What are the -coordinates of the inection points of f? Wh? 9. Sketch the graph of a function whose rst and second deri vatives are alwas negative.. A graph of a population of east cells in a new laborator culture as a function of time is shown. (a) Describe how the rate of population increase varies. (b) When is this rate highest? (c) On what intervals is the population function concave upward or downward? (d) Estimate the coordinates of the inection point. 7 6 Number 5 of 4 east cells =fª() =f () Time (in hours) (a) Find the intervals on which f is increasing or decreasing. (b) Find the local maimum and minimum values of f. (c) Find the intervals of concavit and the inection points.. f. f 5 Licensed to: jsamuels@bmcc.cun.edu. f 4 4. f 5. f sin, 6. f cos sin, 7. f e 8. f e 9. f ln s. f ln Find the local maimum and minimum values of f using both the First and Second Derivative Tests. Which method do ou prefer?. f 5 5. f 4. f s 4. (a) Find the critical numbers of f 4. (b) What does the Second Derivative Test tell ou about the behavior of f at these critical numbers? (c) What does the First Derivative Test tell ou? 5. Suppose f is continuous on,. (a) If f and f 5, what can ou sa about f? (b) If f 6 and f 6, what can ou sa about f? 6 Sketch the graph of a function that satises all of the given conditions. 6. f for all, vertical asmptote, f if or, f if 7. f f f 4, f if or 4, f if or 4, f if, f if or 8. f f, f if, f if, f if, f if, inection point, 9. f if, f if, f,, f if f l. f if, f if, f, f, f f, l f if, f if The graph of the derivative f of a continuous function f is shown. (a) On what intervals is f increasing or decreasing? (b) At what values of does f have a local maimum or minimum? (c) On what intervals is f concave upward or downward? (d) State the -coordinate(s) of the point(s) of inection. (e) Assuming that f, sketch a graph of f =fª() =fª() (a) Find the intervals of increase or decrease. (b) Find the local maimum and minimum values. (c) Find the intervals of concavit and the inection points. (d) Use the information from parts (a) (c) to sketch the graph. Check our work with a graphing device if ou have one.. f 4. f 5. f t h h 9. A s 4. B 4. C 4 4. f ln f cos cos, 44. f t t cos t, t 45 5 (a) Find the vertical and horizontal asmptotes. (b) Find the intervals of increase or decrease. (c) Find the local maimum and minimum values. (d) Find the intervals of concavit and the inection points. (e) Use the information from parts (a) (d) to sketch the graph of f. 45. f 46. f 47. f s 48. f tan, 49. f ln ln 5. f e e Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

15 6 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu 5. f e 5. f lntan ; 5 54 (a) Use a graph of f to estimate the maimum and minimum values. Then nd the e act values. (b) Estimate the value of at which f increases most rapidl. Then nd the e act value f e ; (a) Use a graph of f to give a rough estimate of the intervals of concavit and the coordinates of the points of inection. (b) Use a graph of f to give better estimates. 55. f cos cos, 56. f 4 CAS Estimate the intervals of concavit to one decimal place b using a computer algebra sstem to compute and graph f. 57. f s Let Kt be a measure of the knowledge ou gain b studing for a test for t hours. Which do ou think is larger, K8 K7 or K K? Is the graph of K concave upward or concave downward? Wh? 6. f s Coffee is being poured into the mug shown in the gure at a constant rate (measured in volume per unit time). Sketch a rough graph of the depth of the coffee in the mug as a function of time. Account for the shape of the graph in terms of concavit. What is the signicance of the inection point? ; 6. For the period from 98 to, the percentage of households in the United States with at least one VCR has been modeled b the function Vt 85 5e.5t f 5 4 where the time t is measured in ears since midear 98, so t. Use a graph to estimate the time at which the number of VCRs was increasing most rapidl. Then use derivatives to give a more accurate estimate. 6. The famil of bell-shaped curves occurs in probabilit and statistics, where it is called the normal densit function. The constant is called the mean and the positive constant is called the standard deviation. For simplicit, lets scale the function so as to remove the factor (s) and lets analze the special case where. So we stud the function (a) Find the asmptote, maimum value, and inection points of f. (b) What role does pla in the shape of the curve? ; (c) Illustrate b graphing four members of this famil on the same screen. 6. Find a cubic function f a b c d that has a local maimum value of at and a local minimum value of at. 64. For what values of the numbers a and b does the function f ae b have the maimum value f? s e f e 65. Suppose f is differentiable on an interval I and f for all numbers in I ecept for a single number c. Prove that f is increasing on the entire interval I Assume that all of the functions are twice differentiable and the second derivatives are never. 66. (a) If f and t are concave upward on I, show that f t is concave upward on I. (b) If f is positive and concave upward on I, show that the function t f is concave upward on I. 67. (a) If f and t are positive, increasing, concave upward functions on I, show that the product function ft is concave upward on I. (b) Show that part (a) remains true if f and t are both decreasing. (c) Suppose f is increasing and t is decreasing. Show, b giving three eamples, that ft ma be concave upward, concave downward, or linear. Wh doesnt the argument in parts (a) and (b) work in this case? 68. Suppose f and t are both concave upward on,. Under what condition on f will the composite function h f t be concave upward? 69. Show that tan for. [Hint: Show that f tan is increasing on,.] SECTION 4.4 INDETERMINATE FORMS AND L HOSPITAL S RULE 7 Licensed to: jsamuels@bmcc.cun.edu 7. (a) Show that e for. tion point? None? Illustrate b graphing P for several values (b) Deduce that e for. of c. How does the graph change as c decreases? (c) Use mathematical induction to prove that for and an 7. Prove that if c, f c is a point of inection of the graph of f positive integer n, and f eists in an open interval that contains c, then f c. e n [Hint: Appl the First Derivative Test and Fermats Theorem to the function t f.]! n! 74. Show that if f 4, then f, but, is not an 7. Show that a cubic function (a third-degree polnomial) alwas inection point of the graph of f. has eactl one point of inection. If its graph has three 75. Show that the function t has an inection point at -intercepts,, and, show that the -coordinate of the, but t does not eist. inection point is. 76. Suppose that f is continuous and f c f c, but ; 7. For what values of c does the polnomial f c. Does f have a local maimum or minimum at c? P 4 c have two inection points? One inec- Does f have a point of inection at c? 4.4 I n d e t e r m i n a t e F o r m s a n d L Hospital s R u l e Suppose we are tring to analze the behavior of the function Although F is not dened when, we need to know how F behaves near. In particular, we would like to know the value of the it In computing this it we cant appl Law 5 of its (the it of a quotient is the quotient of the its, see Section.) because the it of the denominator is. In fact, although the it in () eists, its value is not obvious because both numerator and denominator approach and is not dened. In general, if we have a it of the form where both f l and tl as la, then this it ma or ma not eist and is called an indeterminate form of tpe. We met some its of this tpe in Chapter. For rational functions, we can cancel common factors: l l F l la We used a geometric argument to show that ln ln f t sin l l But these methods do not work for its such as (), so in this section we introduce a sstematic method, known as lhospital s Rule, for the evaluation of indeterminate forms. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

16 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu FIGURE a a f g =m (-a) =m (-a) Figure suggests visuall wh l Hospital s Rule might be true. The first graph shows two differentiable functions f and t, each of which approaches as la. If we were to zoom in toward the point a,, the graphs would start to look almost linear. But if the functions actuall were linear, as in the second graph, then their ratio would be which is the ratio of their derivatives. This suggests that la m a m m a m f t la f t Another situation in which a it is not obvious occurs when we look for a horizontal asmptote of F and need to evaluate the it It isnt obvious how to evaluate this it because both numerator and denominator become large as l. There is a struggle between numerator and denominator. If the numerator wins, the it will be ; if the denominator wins, the answer will be. Or there ma be some compromise, in which case the answer ma be some nite positi ve number. In general, if we have a it of the form where both f l (or ) and tl (or ), then the it ma or ma not eist and is called an indeterminate form of tpe. We saw in Section.6 that this tpe of it can be evaluated for certain functions, including rational functions, b dividing numerator and denominator b the highest power of that occurs in the denominator. For instance, This method does not work for its such as (), but lhospitals Rule also applies to this tpe of indeterminate form. L Hospital s Rule Suppose f and t are differentiable and t near a (ecept possibl at a). Suppose that or that l f la f la (In other words, we have an indeterminate form of tpe or.) Then la l la ln f t l and and f f t la t if the it on the right side eists (or is or ). t la t la NOTE LHospitals Rule sas that the it of a quotient of functions is equal to the it of the quotient of their derivatives, provided that the given conditions are satised. It is especiall important to verif the conditions regarding the its of f and t before using lhospitals Rule. NOTE LHospitals Rule is also valid for one-sided its and for its at innit or negative innit; that is, la can be replaced b an of the smbols la, la, l, or l. NOTE For the special case in which f a ta, f and t are continuous, and ta, it is eas to see wh lhospitals Rule is true. In fact, using the alternative form Licensed to: jsamuels@bmcc.cun.edu L Hospital s Rule is named after a French nobleman, the Marquis de l Hospital (66 74), but was discovered b a Swiss mathematician, John Bernoulli ( ). See Eercise 7 for the eample that the Marquis used to illustrate his rule. See the project on page 5 for further historical details. Notice that when using l Hospital s Rule we differentiate the numerator and denominator separatel. We do not use the Quotient Rule. The graph of the function of Eample is shown in Figure. We have noticed previousl that eponential functions grow far more rapidl than power functions, so the result of Eample is not unepected. See also Eercise 67. FIGURE = of the denition of a deri vative, we have SECTION 4.4 INDETERMINATE FORMS AND L HOSPITAL S RULE 9 It is more difcult to pro ve the general version of lhospitals Rule. See Appendi F. ln EXAMPLE Find. l SOLUTION Since we can appl lhospitals Rule: EXAMPLE Calculate. l la l e f f a t ta la l ln ln ln l SOLUTION We have l e and l, so lhospitals Rule gives l Since e l and l as l, the it on the right side is also indeterminate, but a second application of lhospitals Rule gives l e e la la la l l l f f a a t ta a f f a t ta f t d ln d d d d d e e d d l e and f f a a t ta la a l l e l Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

17 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu The graph of the function of Eample is shown in Figure. We have discussed previousl the slow growth of logarithms, so it isn t surprising that this ratio approaches as l. See also Eercise 68. _ FIGURE = ln Œ, The graph in Figure 4 gives visual confirmation of the result of Eample 4. If we were to zoom in too far, however, we would get an inaccurate graph because tan is close to when is small. See Eercise 6(d) in Section.. = tan- _ FIGURE 4 ln EXAMPLE Calculate. l s SOLUTION Since ln l and s l as l, lhospitals Rule applies: Notice that the it on the right side is now indeterminate of tpe. But instead of appling lhospitals Rule a second time as we did in Eample, we simplif the epression and see that a second application is unnecessar: tan EXAMPLE 4 Find. [See Eercise 6(d) in Section..] l ln l s l SOLUTION Noting that both tan l and l as l, we use lhospitals Rule: tan sec l l Since the it on the right side is still indeterminate of tpe, we appl lhospitals Rule again: sec sec tan l l 6 Because l sec, we simplif the calculation b writing sec tan l 6 tan l sec l We can evaluate this last it either b using lhospitals Rule a third time or b writing tan as sin cos and making use of our knowledge of trigonometric its. Putting together all the steps, we get sin EXAMPLE 5 Find. l cos tan sec sec tan l l l 6 SOLUTION If we blindl attempted to use lhospitals Rule, we would get l ln l s l ls tan l sec l sin cos cos l sin tan l This is wrong! Although the numerator sin l as l, notice that the denominator cos does not approach, so lhospitals Rule cant be applied here. Licensed to: jsamuels@bmcc.cun.edu Figure 5 shows the graph of the function in Eample 6. Notice that the function is undefined at ; the graph approaches the origin but never quite reaches it. =ln FIGURE 5 SECTION 4.4 INDETERMINATE FORMS AND L HOSPITAL S RULE The required it is, in fact, eas to nd because the function is continuous and the denominator is nonzero at : l Eample 5 shows what can go wrong if ou use lhospitals Rule without thinking. Other its can be found using lhospitals Rule but are more easil found b other methods. (See Eamples and 5 in Section., Eample in Section.6, and the discussion at the beginning of this section.) So when evaluating an it, ou should consider other methods before using lhospitals Rule. I n d e t e r m i n a t e P r o d u c t s If la f and lat (or ), then it isnt clear what the value of la f t, if an, will be. There is a struggle between f and t. If f wins, the answer will be ; if t wins, the answer will be (or ). Or there ma be a compromise where the answer is a nite nonzero number. This kind of it is called an indeterminate form of tpe. We can deal with it b writing the product ft as a quotient: or This converts the given it into an indeterminate form of tpe or so that we can use lhospitals Rule. EXAMPLE 6 Evaluate ln. l ft SOLUTION The given it is indeterminate because, as l, the rst f actor approaches while the second factor ln approaches. Writing, we have l as l, so lhospitals Rule gives ln ln l l l NOTE In solving Eample 6 another possible option would have been to write This gives an indeterminate form of the tpe, but if we appl lhospitals Rule we get a more complicated epression than the one we started with. In general, when we rewrite an indeterminate product, we tr to choose the option that leads to the simpler it. I n d e t e r m i n a t e D i f f e r e n c e s sin cos sin cos f t ln l l ft ln If la f and lat, then the it f t la l is called an indeterminate form of tpe. Again there is a contest between f and t. Will the answer be ( f wins) or will it be ( t wins) or will the compromise on a nite number? To nd out, we tr to con vert the difference into a quotient (for instance, t f Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

18 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu b using a common denominator, or rationalization, or factoring out a common factor) so that we have an indeterminate form of tpe or. EXAMPLE 7 Compute sec tan. l SOLUTION First notice that sec l and tan l as l, so the it is indeterminate. Here we use a common denominator: sec tan sin l l cos cos Note that the use of lhospitals Rule is justied because sin l and cos l as l. I n d e t e r m i n a t e P o w e r s Several indeterminate forms arise from the it. f and t tpe. f and t tpe. f and t tpe Each of these three cases can be treated either b taking the natural logarithm: or b writing the function as an eponential: (Recall that both of these methods were used in differentiating such functions.) In either method we are led to the indeterminate product t ln f, which is of tpe. EXAMPLE 8 Calculate sin 4cot. SOLUTION First notice that as l, we have sin 4l and cot l, so the given it is indeterminate. Let Then la la la l so lhospitals Rule gives la la la sin l cos f t la let f t, then f t e sin 4 cot ln ln sin 4 cot cot ln sin 4 ln sin 4 ln l l tan ln t ln f t ln f l 4 cos 4 sin 4 4 sec So far we have computed the it of ln, but what we want is the it of. To nd this cos l sin Licensed to: jsamuels@bmcc.cun.edu The graph of the function,, is shown in Figure 6. Notice that although is not defined, the values of the function approach as l. This confirms the result of Eample 9. _ FIGURE Eercises 4 Given that we use the fact that e ln : EXAMPLE 9 Find. SECTION 4.4 INDETERMINATE FORMS AND L HOSPITAL S RULE SOLUTION Notice that this it is indeterminate since for an but for an. We could proceed as in Eample 8 or b writing the function as an eponential: In Eample 6 we used lhospitals Rule to show that Therefore which of the following its are indeterminate forms? For those that are not an indeterminate form, evaluate the it where possible.. (a) (c) (e) (b) (d). (a) f p (b) (c). (a) f p (b) (c) la l a l a l a 4. (a) f t (b) f p (c) l a (d) p f (e) p q (f) l a l a h la p p l a q f l a f t la pq p q l a l a l a t h l a p q la la la f p p f hp l a p q l a h p l a l a q sp l sin l 4cot e ln e 4 l l e ln e ln ln l l l e ln e 5 6 Find the it. Use lhospitals Rule where appropriate. If there is a more elementar method, consider using it. If lhospitals Rule doesnt appl, eplain wh. 5. l l 5 cos 9.. l sin e t.. tan p. 4. l tan q ln l 7. ln l 8. 5 t t 9.. t l t. tl t e l. l l tan l sin e t tl t sin l csc e l ln ln l l a b ln sin e sin. 4. l l sinh e l Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

19 4 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu sin l sin 9.. l cos.. l ln e ln. 4. l cos a a a l l s ln 9. cot sin 6 4. l ln tan 44. l l csc l 49. ln 5. l l l 6. cos 6. l 54. ; 6 64 Use a graph to estimate the value of the it. Then use lhospitals Rule to nd the e act value cos l e l l (s ) l l 5 ln 5 ln l tan tan l4 sin l ln l cos m cos n l l l tan 4 s s e l sec e l sin ln l tan sec l 4 tan l csc cot l l ln e l tan l l a ln ln l e l cos 5 l b l 5 ; Illustrate lhospitals Rule b graphing both f t and f t near to see that these ratios have the same it as l. Also calculate the eact value of the it. 65. f e, t f sin, t sec 67. Prove that for an positive integer n. This shows that the eponential function approaches innit f aster than an power of. 68. Prove that l ln l p for an number p. This shows that the logarithmic function approaches more slowl than an power of. 69. If an initial amount A of mone is invested at an interest rate i compounded n times a ear, the value of the investment after t ears is A A i nnt If we let nl, we refer to the continuous compounding of interest. Use lhospitals Rule to show that if interest is compounded continuousl, then the amount after n ears is A Ae it 7. If an object with mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into account, is v mt c e ctm where t is the acceleration due to gravit and c is a positive constant. (In Chapter 9 we will be able to deduce this equation from the assumption that the air resistance is proportional to the speed of the object.) (a) Calculate t l v. What is the meaning of this it? (b) For ed t, use lhospitals Rule to calculate m l v. What can ou conclude about the speed of a ver heav falling object? 7. The rst appearance in print of lhospital s Rule was in the book Analse des Inniment P etits published b the Marquis de lhospital in 696. This was the rst calculus tetbook ever published and the eample that the Marquis used in that book to illustrate his rule was to nd the it of the function e n sa 4 as aa a s 4 a as approaches a, where a. (At that time it was common to write aa instead of a.) Solve this problem. Licensed to: jsamuels@bmcc.cun.edu 7. The gure sho ws a sector of a circle with central angle. Let A be the area of the segment between the chord PR and the arc PR. Let B be the area of the triangle PQR. Find AB. P 7. If f is continuous, f, and f 7, evaluate 74. For what values of a and b is the following equation true? 75. If f is continuous, use lhospitals Rule to show that Thomas Fisher Rare Book Librar l h l O l l Q B( ) f f 5 sin a b f h f h h f Eplain the meaning of this equation with the aid of a diagram. WRITING PROJECT The Internet is another source of information for this project. See the web site and click on Histor of Mathematics. A( ) R WRITING PROJECT THE ORIGINS OF L HOSPITAL S RULE If f is continuous, show that 77. Let ; 78. Let The Origins of L Hospital s Rule hl f h f f h h f e (a) Use the denition of deri vative to compute f. (b) Show that f has derivatives of all orders that are dened on. [Hint: First show b induction that there is a polnomial pn and a nonnegative integer kn such that f n pnf k n for.] f if if (a) Show that f is continuous at. (b) Investigate graphicall whether f is differentiable at b zooming in several times toward the point, on the graph of f. (c) Show that f is not differentiable at. How can ou reconcile this fact with the appearance of the graphs in part (b)? if if f LHospitals Rule was rst published in 696 in the Marquis de lhospital s calculus tetbook Analse des Inniment P etits, but the rule was discovered in 694 b the Swiss mathematician John (Johann) Bernoulli. The eplanation is that these two mathematicians had entered into a curious business arrangement whereb the Marquis de lhospital bought the rights to Bernoullis mathematical discoveries. The details, including a translation of lhospitals letter to Bernoulli proposing the arrangement, can be found in the book b Eves []. Write a report on the historical and mathematical origins of lhospitals Rule. Start b providing brief biographical details of both men (the dictionar edited b Gillispie [] is a good source) and outline the business deal between them. Then give lhospitals statement of his rule, which is found in Struiks sourcebook [4] and more brie in the book of Katz []. Notice that lhospital and Bernoulli formulated the rule geometricall and gave the answer in terms of differentials. Compare their statement with the version of lhospitals Rule given in Section 4.4 and show that the two statements are essentiall the same.. Howard Eves, In Mathematical Circles (Volume : Quadrants III and IV) (Boston: Prindle, Weber and Schmidt, 969), pp... C. C. Gillispie, ed., Dictionar of Scientic Bio graph (New York: Scribners, 974). See the article on Johann Bernoulli b E. A. Fellmann and J. O. Fleckenstein in Volume II and the article on the Marquis de lhospital b Abraham Robinson in Volume VIII.. Victor Katz, A Histor of Mathematics: An Introduction (New York: HarperCollins, 99), p D. J. Struik, ed., A Sourcebook in Mathematics, 8(Princeton, NJ: Princeton Universit Press, 969), pp Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

20 6 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu 4.5 S u m m a r o f C u r v e S k e t c h i n g FIGURE FIGURE _ 5 5 f So far we have been concerned with some particular aspects of curve sketching: domain, range, and smmetr in Chapter ; its, continuit, and asmptotes in Chapter ; derivatives and tangents in Chapters and ; and etreme values, intervals of increase and decrease, concavit, points of inection, and lhospital s Rule in this chapter. It is now time to put all of this information together to sketch graphs that reveal the important features of functions. You ma ask: What is wrong with just using a calculator to plot points and then joining these points with a smooth curve? To see the pitfalls of this approach, suppose ou have used a calculator to produce the table of values and corresponding points in Figure You might then join these points to produce the curve shown in Figure, but the cor- rect graph might be the one shown in Figure. You can see the drawbacks of the method of plotting points. Certain essential features of the graph ma be missed, such as the maimum and minimum values between and or between and 5. If ou just plot points, ou dont know when to stop. (How far should ou plot to the left or right?) But the use of calculus ensures that all the important aspects of the curve are illustrated. 4 f FIGURE 5 _5 _4 4 You might respond: Yes, but what about graphing calculators and computers? Dont the plot such a huge number of points that the sort of uncertaint demonstrated b Figures and is unlikel to happen? Its true that modern technolog is capable of producing ver accurate graphs. But even the best graphing devices have to be used intelligentl. We saw in Section.4 that it is etremel important to choose an appropriate viewing rectangle to avoid getting a misleading graph. (See especiall Eamples,, 4, and 5 in that section.) The use of calculus Licensed to: jsamuels@bmcc.cun.edu = _ 4 FIGURE 4 8 FIGURE 6 _ = FIGURE 5 (a) Even function: reectional smmetr (b) Odd function: rotational smmetr FIGURE 7 Periodic function: translational smmetr enables us to discover the most interesting aspects of graphs and in man cases to calculate maimum and minimum points and inection points eactl instead of approimatel. For instance, Figure 4 shows the graph of f 8 8. At rst glance it seems reasonable: It has the same shape as cubic curves like, and it appears to have no maimum or minimum point. But if ou compute the derivative, ou will see that there is a maimum when.75 and a minimum when. Indeed, if we zoom in to this portion of the graph, we see that behavior ehibited in Figure 5. Without calculus, we could easil have overlooked it. In the net section we will graph functions b using the interaction between calculus and graphing devices. In this section we draw graphs b rst considering the follo wing information. We dont assume that ou have a graphing device, but if ou do have one ou should use it as a check on our work. G u i d e l i n e s f o r S k e t c h i n g a C u r v e The following checklist is intended as a guide to sketching a curve f b hand. Not ever item is relevant to ever function. (For instance, a given curve might not have an asmptote or possess smmetr.) But the guidelines provide all the information ou need to make a sketch that displas the most important aspects of the function. A. Domain Its often useful to start b determining the domain D of f, that is, the set of values of for which f is dened. B. Intercepts The -intercept is f and this tells us where the curve intersects the -ais. To nd the -intercepts, we set and solve for. (You can omit this step if the equation is difcult to solv e.) C. Smmetr (i) If f f for all in D, that is, the equation of the curve is unchanged when is replaced b, then f is an even function and the curve is smmetric about the -ais. This means that our work is cut in half. If we know what the curve looks like for, then we need onl reect about the -ais to obtain the complete curve [see Figure 6(a)]. Here are some eamples:, 4,, and cos. (ii) If f f for all in D, then f is an odd function and the curve is smmetric about the origin. Again we can obtain the complete curve if we know what it looks like for. [Rotate 8 about the origin; see Figure 6(b).] Some simple eamples of odd functions are,, 5, and sin. (iii) If f p f for all in D, where p is a positive constant, then f is called a periodic function and the smallest such number p is called the period. For instance, sin has period and tan has period. If we know what the graph looks like in an interval of length p, then we can use translation to sketch the entire graph (see Figure 7). SECTION 4.5 SUMMARY OF CURVE SKETCHING 7 a-p a a+p a+p D. Asmptotes (i) Horizontal Asmptotes. Recall from Section.6 that if either l f L or l f L, then the line L is a horizontal asmptote of the curve Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

21 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu In Module 4.5 ou can practice using information about f, f, and asmptotes to determine the shape of the graph of f. f. If it turns out that l f (or ), then we do not have an asmptote to the right, but that is still useful information for sketching the curve. (ii) Vertical Asmptotes. Recall from Section. that the line a is a vertical asmptote if at least one of the following statements is true: (For rational functions ou can locate the vertical asmptotes b equating the denominator to after canceling an common factors. But for other functions this method does not appl.) Furthermore, in sketching the curve it is ver useful to know eactl which of the statements in () is true. If f a is not dened b ut a is an endpoint of the domain of f, then ou should compute la f or la f, whether or not this it is innite. (iii) Slant Asmptotes. These are discussed at the end of this section. E. Intervals of Increase or Decrease Use the I/D Test. Compute f and nd the interv als on which f is positive ( f is increasing) and the intervals on which f is negative ( f is decreasing). F. Local Maimum and Minimum Values Find the critical numbers of f [the numbers c where f c or f c does not eist]. Then use the First Derivative Test. If f changes from positive to negative at a critical number c, then f c is a local maimum. If f changes from negative to positive at c, then f c is a local minimum. Although it is usuall preferable to use the First Derivative Test, ou can use the Second Derivative Test if c is a critical number such that f c. Then f c implies that f c is a local minimum, whereas f c implies that f c is a local maimum. G. Concavit and Points of Inflection Compute f and use the Concavit Test. The curve is concave upward where f and concave downward where f. Inection points occur where the direction of concavit changes. H. Sketch the Curve Using the information in items A G, draw the graph. Sketch the asmptotes as dashed lines. Plot the intercepts, maimum and minimum points, and inection points. Then make the curve pass through these points, rising and falling according to E, with concavit according to G, and approaching the asmptotes. If additional accurac is desired near an point, ou can compute the value of the derivative there. The tangent indicates the direction in which the curve proceeds. EXAMPLE Use the guidelines to sketch the curve. A. The domain is B. The - and -intercepts are both. C. Since f f, the function f is even. The curve is smmetric about the -ais. D. f f la la f f la la,,, l l Therefore, the line is a horizontal asmptote. Licensed to: jsamuels@bmcc.cun.edu = FIGURE 8 Preinar sketch We have shown the curve approaching its horizontal asmptote from above in Figure 8. This is confirmed b the intervals of increase and decrease. = =_ =_ = FIGURE 9 Finished sketch of = - = E. SECTION 4.5 SUMMARY OF CURVE SKETCHING 9 Since the denominator is when, we compute the following its: Therefore, the lines and are vertical asmptotes. This information about its and asmptotes enables us to draw the preinar sketch in Figure 8, showing the parts of the curve near the asmptotes. Since f when and f when, f is increasing on, and, and decreasing on, and,. F. The onl critical number is. Since f changes from positive to negative at, f is a local maimum b the First Derivative Test. G. Since 4 for all, we have and. Thus, the curve is concave upward on the intervals, and, and concave downward on,. It has no point of inection since and are not in the domain of f. H. Using the information in E G, we nish the sk etch in Figure 9. EXAMPLE Sketch the graph of f. s A. Domain, B. The - and -intercepts are both. C. Smmetr: None D. Since E. l l f f f &? f 4 4 &? l there is no horizontal asmptote. Since s l as l and f is alwas positive, we have l s and so the line is a vertical asmptote. s &? f s (s ) l l We see that f when (notice that 4 is not in the domain of f ), so the onl critical number is. Since f when and f when, f is decreasing on, and increasing on,. F. Since f and f changes from negative to positive at, f is a local (and absolute) minimum b the First Derivative Test. 4 Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

22 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu =_ FIGURE FIGURE (_,_/e) = œ + = G. Note that the denominator is alwas positive. The numerator is the quadratic 8 8, which is alwas positive because its discriminant is b 4ac, which is negative, and the coefcient of is positive. Thus, f for all in the domain of f, which means that f is concave upward on, and there is no point of inection. H. The curve is sketched in Figure. EXAMPLE Sketch the graph of f e. A. The domain is. B. The - and -intercepts are both. C. Smmetr: None D. Because both and e become large as l, we have l e. As l, however, e l and so we have an indeterminate product that requires the use of lhospitals Rule: E. Thus, the -ais is a horizontal asmptote. Since e is alwas positive, we see that f when, and f when. So f is increasing on, and decreasing on,. F. Because f and f changes from negative to positive at, f e is a local (and absolute) minimum. G. Since f if and f if, f is concave upward on, and concave downward on,. The inection point is, e. H. We use this information to sketch the curve in Figure. EXAMPLE 4 Sketch the graph of f cos sin. A. The domain is. B. The -intercept is f. The -intercepts occur when that is, when cos or sin. Thus, in the interval,, the -intercepts are and. C. f is neither even nor odd, but f f for all and so f is periodic and has period. Thus, in what follows we need to consider onl and then etend the curve b translation in H. D. Asmptotes: None E. f cos sin cos sin cos cos sin e l l e e l e l f e e e f e e e f sin cos sin sin sin sin sin sin Thus, f when sin or sin, so in, we have 6, 56, and. In determining the sign of f in the following chart, we use the Licensed to: jsamuels@bmcc.cun.edu arcsin 4 arcsin 4 FIGURE π œ, 6 å å π π 5π π π π 6 6 5π œ,_ 6 FIGURE fact that sin for all. SECTION 4.5 SUMMARY OF CURVE SKETCHING F. From the chart in E the First Derivative Test sas that f 6 s is a local maimum and f 56 s is a local minimum, but f has no maimum or minimum at, onl a horizontal tangent. G. Thus, f when cos (so or ) and when sin 4. From Figure we see that there are two values of between and for which sin 4. Lets call them and. Then f on, and,, so f is concave upward there. Also f on,,,, and,, so f is concave downward there. Inection points occur when,,, and. = H. The graph of the function restricted to is shown in Figure. Then it is etended, using periodicit, to the complete graph in Figure 4. _ π FIGURE 4 EXAMPLE 5 Sketch the graph of ln4. A. The domain is _ 4 Interval f cos 4 sin cos 4 sin _ π 4 4, B. The -intercept is f ln 4. To nd the -intercept we set π f π ln4 increasing on, 6 decreasing on 6, 56 å å 5π =Ł+ f increasing on 56, increasing on, 7π π 9π Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

23 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu =_ FIGURE 5 =ln(4 - ) FIGURE 6 (,ln4) {_œ,} =ƒ ƒ-(m+b) {œ,} =m+b = We know that ln loge (since e ), so we have 4? and therefore the -intercepts are s. C. Since f f, f is even and the curve is smmetric about the -ais. D. We look for vertical asmptotes at the endpoints of the domain. Since 4 l as l and also as l, we have E. Thus, the lines and are vertical asmptotes. Since f when and f when, f is increasing on, and decreasing on,. F. The onl critical number is. Since f changes from positive to negative at, f ln 4 is a local maimum b the First Derivative Test. G. Since f for all, the curve is concave downward on, and has no inection point. H. Using this information, we sketch the curve in Figure 5. S l a n t A s m p t o t e s l ln4 f f Some curves have asmptotes that are oblique, that is, neither horizontal nor vertical. If then the line m b is called a slant asmptote because the vertical distance between the curve f and the line m b approaches, as in Figure 6. (A similar situation eists if we let l.) For rational functions, slant asmptotes occur when the degree of the numerator is one more than the degree of the denominator. In such a case the equation of the slant asmptote can be found b long division as in the following eample. EXAMPLE 6 Sketch the graph of f. A. The domain is,. B. The - and -intercepts are both. C. Since f f, f is odd and its graph is smmetric about the origin. D. Since is never, there is no vertical asmptote. Since f l as l and f l as l, there is no horizontal asmptote. But long division gives f f l So the line is a slant asmptote. 4 f m b l l ln4 as l Licensed to: jsamuels@bmcc.cun.edu œ _œ,_ 4 = FIGURE Eercises = + œ œ, 4 inflection points E. SECTION 4.5 SUMMARY OF CURVE SKETCHING Since f for all (ecept ), f is increasing on,. F. Although f, f does not change sign at, so there is no local maimum or minimum. G. f f Since f when or s, we set up the following chart: The points of inection are (s, s4),,, and (s, s4). H. The graph of f is sketched in Figure 7. Interval f f s s s s 5 Use the guidelines of this section to sketch the curve s5. s. s s s s s CU on (, s) CD on (s, ) CU on (, s) CD on (s, ) s. s. sin sin. sin tan. tan, 4. tan, 5. sin, 6. cos sin 7. sin sin 8. sin 9. sin 4. cos cos sin 4. e 4. e e 4. ln 44. e 45. e 46. ln 47. lnsin 48. ln 49. e 5. e e 5. e 5. tan 4 e Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

24 4 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu 5. The gure sho ws a beam of length L embedded in concrete walls. If a constant load W is distributed evenl along its length, the beam takes the shape of the deection curv e where E and I are positive constants. ( E is Youngs modulus of elasticit and I is the moment of inertia of a cross-section of the beam.) Sketch the graph of the deection curv e. 54. Coulombs Law states that the force of attraction between two charged particles is directl proportional to the product of the charges and inversel proportional to the square of the distance between them. The gure sho ws particles with charge located at positions and on a coordinate line and a particle with charge at a position between them. It follows from Coulombs Law that the net force acting on the middle particle is where k is a positive constant. Sketch the graph of the net force function. What does the graph sa about the force? Find an equation of the slant asmptote. Do not sketch the curve. 55. W 4EI 4 WL EI WL 4EI F k + k _ W L Graphing with Calculus and Calculators If ou have not alread read Section.4, ou should do so now. In particular, it eplains how to avoid some of the pitfalls of graphing devices b choosing appropriate viewing rectangles Use the guidelines of this section to sketch the curve. In guideline D nd an equation of the slant asmptote e Show that the curve tan has two slant asmptotes: and. Use this fact to help sketch the curve. 66. Show that the curve s 4 has two slant asmptotes: and. Use this fact to help sketch the curve. 67. Show that the lines ba and ba are slant asmptotes of the hperbola a b. 68. Let f. Show that f l 5 4 This shows that the graph of f approaches the graph of, and we sa that the curve f is asmptotic to the parabola. Use this fact to help sketch the graph of f. 69. Discuss the asmptotic behavior of f 4 in the same manner as in Eercise 68. Then use our results to help sketch the graph of f. 7. Use the asmptotic behavior of f cos to sketch its graph without going through the curve-sketching procedure of this section. The method we used to sketch curves in the preceding section was a culmination of much of our stud of differential calculus. The graph was the nal object that we produced. In this section our point of view is completel different. Here we start with a graph produced b a graphing calculator or computer and then we rene it. We use calculus to make sure that we reveal all the important aspects of the curve. And with the use of graphing devices we can tackle curves that would be far too complicated to consider without technolog. The theme is the interaction between calculus and calculators. Licensed to: jsamuels@bmcc.cun.edu 4, _5 5 _ FIGURE =ƒ _ FIGURE =ƒ _5 _ FIGURE 5 =f () _ SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 5 EXAMPLE Graph the polnomial f 6 5. Use the graphs of f and f to estimate all maimum and minimum points and intervals of concavit. SOLUTION If we specif a domain but not a range, man graphing devices will deduce a suitable range from the values computed. Figure shows the plot from one such device if we specif that 5 5. Although this viewing rectangle is useful for showing that the asmptotic behavior (or end behavior) is the same as for 6, it is obviousl hiding some ner detail. So we change to the vie wing rectangle, b 5, shown in Figure. From this graph it appears that there is an absolute minimum value of about 5. when.6 (b using the cursor) and f is decreasing on,.6 and increasing on.6,. Also there appears to be a horizontal tangent at the origin and inection points when and when is somewhere between and. Now lets tr to conrm these impressions using calculus. We differentiate and get When we graph f in Figure we see that f changes from negative to positive when.6; this conrms (b the First Deri vative Test) the minimum value that we found earlier. But, perhaps to our surprise, we also notice that f changes from positive to negative when and from negative to positive when.5. This means that f has a local maimum at and a local minimum when.5, but these were hidden in Figure. Indeed, if we now zoom in toward the origin in Figure 4, we see what we missed before: a local maimum value of when and a local minimum value of about. when.5. =fª() _ FIGURE _5 What about concavit and inection points? From Figures and 4 there appear to be inection points when is a little to the left of and when is a little to the right of. But its difcult to determine inection points from the graph of f, so we graph the second derivative f in Figure 5. We see that f changes from positive to negative when. and from negative to positive when.9. So, correct to two decimal places, f is concave upward on,. and.9, and concave downward on.,.9. The inection points are.,.8 and.9,.5. We have discovered that no single graph reveals all the important features of this polnomial. But Figures and 4, when taken together, do provide an accurate picture. EXAMPLE Draw the graph of the function f f _ FIGURE 4 f 7 in a viewing rectangle that contains all the important features of the function. Estimate _ =ƒ Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

25 6 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu _ FIGURE 9 = =ƒ _ FIGURE 8 _5 =ƒ _4 the maimum and minimum values and the intervals of concavit. Then use calculus to nd these quantities e actl. SOLUTION Figure 6, produced b a computer with automatic scaling, is a disaster. Some graphing calculators use, b, as the default viewing rectangle, so lets tr it. We get the graph shown in Figure 7; its a major improvement. =ƒ!* _5 5 FIGURE 6 The -ais appears to be a vertical asmptote and indeed it is because Figure 7 also allows us to estimate the -intercepts: about.5 and 6.5. The eact values are obtained b using the quadratic formula to solve the equation 7 ; we get (7 s7). To get a better look at horizontal asmptotes, we change to the viewing rectangle, b 5, in Figure 8. It appears that is the horizontal asmptote and this is easil conrmed: l l 7 7 To estimate the minimum value we zoom in to the viewing rectangle, b 4, in Figure 9. The cursor indicates that the absolute minimum value is about. when.9, and we see that the function decreases on,.9 and, and increases on.9,. The eact values are obtained b differentiating: f 7 6 l This shows that f when 6 and when 6 7 f and when. The eact minimum value is f ( 6 7) Figure 9 also shows that an inection point occurs some where between and. We could estimate it much more accuratel using the graph of the second derivative, but in this case its just as eas to nd e act values. Since f we see that f when 9. So is concave upward on ( 9 7 f 7, ) and, and concave downward on (, 9 7). The inection point is ( 9 7, 7 7). The analsis using the rst tw o derivatives shows that Figures 7 and 8 displa all the major aspects of the curve. _ FIGURE 7 _ =ƒ Licensed to: jsamuels@bmcc.cun.edu.5 _ FIGURE =ƒ =ƒ _ FIGURE 4 FIGURE _.5 EXAMPLE Graph the function f. 4 4 SOLUTION Drawing on our eperience with a rational function in Eample, lets start b graphing f in the viewing rectangle, b,. From Figure we have the feeling that we are going to have to zoom in to see some ner detail and also zoom out to see the larger picture. But, as a guide to intelligent zooming, lets rst tak e a close look at the epression for f. Because of the factors and 4 4 in the denominator, we epect and 4 to be the vertical asmptotes. Indeed l To nd the horizontal asmptotes we di vide numerator and denominator b : so the -ais is the horizontal asmptote. It is also ver useful to consider the behavior of the graph near the -intercepts using an analsis like that in Eample in Section.6. Since is positive, f does not change sign at and so its graph doesnt cross the -ais at. But, because of the factor, the graph does cross the -ais at and has a horizontal tangent there. Putting all this information together, but without using derivatives, we see that the curve has to look something like the one in Figure. Now that we know what to look for, we zoom in (several times) to produce the graphs in Figures and and zoom out (several times) to get Figure 4. =ƒ. _.5.5 FIGURE _. SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 7 and 4 4 l We can read from these graphs that the absolute minimum is about. and occurs when. There is also a local maimum. when. and a local minimum when.5. These graphs also show three inection points near 5, 5, and and two between and. To estimate the inection points closel we would need to graph f, but to compute f b hand is an unreasonable chore. If ou have a computer algebra sstem, then its eas to do (see Eercise 7). We have seen that, for this particular function, three graphs (Figures,, and 4) are necessar to conve all the useful information. The onl wa to displa all these features of the function on a single graph is to draw it b hand. Despite the eaggerations and distortions, Figure does manage to summarize the essential nature of the function. l4 5 as =ƒ FIGURE l 6 Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

26 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu The famil of functions f sin sin c where c is a constant, occurs in applications to frequenc modulation (FM) snthesis. A sine wave is modulated b a wave with a different frequenc sin c. The case where c is studied in Eample 4. Eercise 5 eplores another special case.. π _. FIGURE 5. π _. FIGURE 6 =ƒ =fª() EXAMPLE 4 Graph the function f sin sin. For, locate all maimum and minimum values, intervals of increase and decrease, and inection points correct to one decimal place. SOLUTION We rst note that f is periodic with period. Also, f is odd and for all. So the choice of a viewing rectangle is not a problem for this function: We start with, b.,.. (See Figure 5.) It appears that there are three local maimum values and two local minimum values in that window. To conrm this and locate them more accuratel, we calculate that and graph both f and f in Figure 6. Using zoom-in and the First Derivative Test, we nd the follo wing values to one decimal place. The second derivative is Graphing both f and f in Figure 7, we obtain the following approimate values:. π _. Concave upward on:.8,.,.8,. Concave downward on:,.8,.,.8,., Inflection points:,,.8,.97,.,.97,.8,.97,.,.97 f FIGURE 7 Having checked that Figure 5 does indeed represent f accuratel for, we can state that the etended graph in Figure 8 represents f accuratel for. f cos sin cos Intervals of increase:,.6,.,.6,.,.5 Intervals of decrease:.6,.,.6,.,.5, Local maimum values: f.6, f.6, f.5 Local minimum values: f..94, f..94 f cos sin sin 4 sin cos sin f _π FIGURE 8 Our nal e ample is concerned with families of functions. As discussed in Section.4, this means that the functions in the famil are related to each other b a formula that contains one or more arbitrar constants. Each value of the constant gives rise to a member of the famil and the idea is to see how the graph of the function changes as the constant changes.. _. f π Licensed to: jsamuels@bmcc.cun.edu _5 4 = ++ FIGURE 9 c= FIGURE c= _5 4 FIGURE _ c=_ See an animation of Figure. Resources / Module 5 / Ma and Min / Families of Functions = +- SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 9 EXAMPLE 5 How does the graph of f c var as c varies? SOLUTION The graphs in Figures 9 and (the special cases c and c ) show two ver different-looking curves. Before drawing an more graphs, lets see what members of this famil have in common. Since c= The famil of functions ƒ=/( ++c) for an value of c, the all have the -ais as a horizontal asmptote. A vertical asmptote will occur when c. Solving this quadratic equation, we get s c. When c, there is no vertical asmptote (as in Figure 9). When c, the graph has a single vertical asmptote because l When c, there are two vertical asmptotes: s c (as in Figure ). Now we compute the derivative: This shows that f when (if c ), f when, and f when. For c, this means that f increases on, and decreases on,. For c, there is an absolute maimum value f c. For c, f c is a local maimum value and the intervals of increase and decrease are interrupted at the vertical asmptotes. Figure is a slide sho w displaing v e members of the famil, all graphed in the viewing rectangle 5, 4 b,. As predicted, c is the value at which a transition takes place from two vertical asmptotes to one, and then to none. As c increases from, we see that the maimum point becomes lower; this is eplained b the fact that c l as cl. As c decreases from, the vertical asmptotes become more widel separated because the distance between them is s c, which becomes large as cl. Again, the maimum point approaches the -ais because c l as cl. c= l c l f c There is clearl no inection point when c. For c we calculate that f 6 4 c c and deduce that inection points occur when sc. So the inection points become more spread out as c increases and this seems plausible from the last two parts of Figure. c= c= Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

27 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu 4.6 ; Eercises 8 Produce graphs of f that reveal all the important aspects of CAS 7. If f is the function considered in Eample, use a computer the curve. In particular, ou should use graphs of f and f to estimate the intervals of increase and decrease, etreme values, inter- that all the maimum and minimum values are as given in the algebra sstem to calculate f and then graph it to conrm vals of concavit, and inection points. eample. Calculate f and use it to estimate the intervals of. f concavit and inection points.. f CAS 8. If f is the function of Eercise 6, nd f and f and use their graphs to estimate the intervals of increase and decrease and. f s 5 concavit of f. 4. f 4 5. f 4 6. f tan 5 cos 7. f 4 7 cos, f e 9 9 Produce graphs of f that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease, etreme values, intervals of concavit, and inection points, and use calculus to nd these quantities e actl. 9. f 8. f. f s9. f sin, 4 (a) Graph the function. (b) Use lhospitals Rule to eplain the behavior as l. (c) Estimate the minimum value and intervals of concavit. Then use calculus to nd the e act values.. f ln 4. f e 5 6 Sketch the graph b hand using asmptotes and intercepts, but not derivatives. Then use our sketch as a guide to producing graphs (with a graphing device) that displa the major features of the curve. Use these graphs to estimate the maimum and minimum values. 5. f f CAS 9 Use a computer algebra sstem to graph f and to nd f and f. Use graphs of these derivatives to estimate the intervals of increase and decrease, etreme values, intervals of concavit, and inection points of f. 9. f sin, s. f s 4 4 e. f e. f e tan CAS 4 (a) Graph the function. (b) Eplain the shape of the graph b computing the it as l or as l. (c) Estimate the maimum and minimum values and then use calculus to nd the e act values. (d) Use a graph of f to estimate the -coordinates of the inection points.. f 4. f sin sin 5. In Eample 4 we considered a member of the famil of functions f sin sin c that occur in FM snthesis. Here we investigate the function with c. Start b graphing f in the viewing rectangle, b.,.. How man local maimum points do ou see? The graph has more than are visible to the naked ee. To discover the hidden maimum and minimum points ou will need to eamine the graph of f ver carefull. In fact, it helps to look at the graph of f at the same time. Find all the maimum and minimum values and inection points. Then graph f in the viewing rectangle, b.,. and comment on smmetr. 6 Describe how the graph of f varies as c varies. Graph several members of the famil to illustrate the trends that ou discover. In particular, ou should investigate how maimum and Licensed to: jsamuels@bmcc.cun.edu minimum points and inection points mo ve when c changes. You should also identif an transitional values of c at which the basic shape of the curve changes. 6. f c 7. f 4 c 8. f sc 9. f e c. f ln c. f c c. f. f c sin c 4. The famil of functions f t Ce at e bt, where a, b, and C are positive numbers and b a, has been used to model the concentration of a drug injected into the blood at time t. Graph several members of this famil. What do the have in common? For ed values of C and a, discover graphicall what happens as b increases. Then use calculus to prove what ou have discovered. 5. Investigate the famil of curves given b f e c, where c is a real number. Start b computing the its as l. Identif an transitional values of c where the basic shape 4.7 O p t i m i z a t i o n P r o b l e m s SECTION 4.7 OPTIMIZATION PROBLEMS changes. What happens to the maimum or minimum points and inection points as c changes? Illustrate b graphing several members of the famil. 6. Investigate the famil of curves given b the equation f 4 c. Start b determining the transitional value of c at which the number of inection points changes. Then graph several members of the famil to see what shapes are possible. There is another transitional value of c at which the number of critical numbers changes. Tr to discover it graphicall. Then prove what ou have discovered. 7. (a) Investigate the famil of polnomials given b the equation f c 4. For what values of c does the curve have minimum points? (b) Show that the minimum and maimum points of ever curve in the famil lie on the parabola. Illustrate b graphing this parabola and several members of the famil. 8. (a) Investigate the famil of polnomials given b the equation f c. For what values of c does the curve have maimum and minimum points? (b) Show that the minimum and maimum points of ever curve in the famil lie on the curve. Illustrate b graphing this curve and several members of the famil. The methods we have learned in this chapter for nding e treme values have practical applications in man areas of life. A businessperson wants to minimize costs and maimize prots. A traveler wants to minimize transportation time. Fermats Principle in optics states that light follows the path that takes the least time. In this section and the net we solve such problems as maimizing areas, volumes, and prots and minimizing distances, times, and costs. In solving such practical problems the greatest challenge is often to convert the word problem into a mathematical optimization problem b setting up the function that is to be maimized or minimized. Lets recall the problem-solving principles discussed on page 8 and adapt them to this situation: Steps in Solving Optimization Problems. Understand the Problem The rst step is to read the problem carefull until it is clearl understood. Ask ourself: What is the unknown? What are the given quantities? What are the given conditions?. Draw a Diagram In most problems it is useful to draw a diagram and identif the given and required quantities on the diagram.. Introduce Notation Assign a smbol to the quantit that is to be maimized or minimized (lets call it Q for now). Also select smbols a, b, c,...,, for other unknown quantities and label the diagram with these smbols. It ma help to use initials as suggestive smbolsfor e ample, A for area, h for height, t for time. 4. Epress Q in terms of some of the other smbols from Step. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

28 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu FIGURE Area= =, ft@ FIGURE Understand the problem Analog: Tr special cases Draw diagrams Introduce notation A 5. If Q has been epressed as a function of more than one variable in Step 4, use the given information to nd relationships (in the form of equations) among these variables. Then use these equations to einate all but one of the variables in the epression for Q. Thus, Q will be epressed as a function of one variable, sa, Q f. Write the domain of this function. 6. Use the methods of Sections 4. and 4. to nd the absolute maimum or minimum value of f. In particular, if the domain of f is a closed interval, then the Closed Interval Method in Section 4. can be used. EXAMPLE A farmer has 4 ft of fencing and wants to fence off a rectangular eld that borders a straight river. He needs no fence along the river. What are the dimensions of the eld that has the lar gest area? SOLUTION In order to get a feeling for what is happening in this problem, lets eperiment with some special cases. Figure (not to scale) shows three possible was of laing out the 4 ft of fencing. We see that when we tr shallow, wide elds or deep, narro w elds, we get relati vel small areas. It seems plausible that there is some intermediate conguration that produces the lar gest area. 7 7 Area=7 =7, ft@ Figure illustrates the general case. We wish to maimize the area A of the rectangle. Let and be the depth and width of the rectangle (in feet). Then we epress A in terms of and : A We want to epress A as a function of just one variable, so we einate b epressing it in terms of. To do this we use the given information that the total length of the fencing is 4 ft. Thus 4 From this equation we have 4, which gives A 4 4 Note that and (otherwise A ). So the function that we wish to maimize is A 4 The derivative is A 4 4, so to nd the critical numbers we solv e the equation Area= 4=4, ft@ Licensed to: jsamuels@bmcc.cun.edu FIGURE r Area {πr@} FIGURE 4 πr Area (πr)h =A(r) FIGURE 5 r In the Applied Project on page 4 we investigate the most economical shape for a can b taking into account other manufacturing costs. h r h SECTION 4.7 OPTIMIZATION PROBLEMS which gives 6. The maimum value of A must occur either at this critical number or at an endpoint of the interval. Since A, A6 7,, and A, the Closed Interval Method gives the maimum value as A6 7,. [Alternativel, we could have observed that A 4 for all, so A is alwas concave downward and the local maimum at 6 must be an absolute maimum.] Thus, the rectangular eld should be 6 ft deep and ft wide. EXAMPLE A clindrical can is to be made to hold L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. SOLUTION Draw the diagram as in Figure, where r is the radius and h the height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the clinder (top, bottom, and sides). From Figure 4 we see that the sides are made from a rectangular sheet with dimensions r and h. So the surface area is A r rh To einate h we use the fact that the volume is given as L, which we take to be cm. Thus r h which gives h r. Substitution of this into the epression for A gives A r r r r r Therefore, the function that we want to minimize is Ar r r To nd the critical numbers, we dif ferentiate: r Ar 4r 4r 5 r r Then Ar when r 5, so the onl critical number is r s 5. Since the domain of A is,, we cant use the argument of Eample concerning endpoints. But we can observe that Ar for r s 5 and Ar for r s 5, so A is decreasing for all r to the left of the critical number and increasing for all r to the right. Thus, r s 5 must give rise to an absolute minimum. [Alternativel, we could argue that Arl as rl and Arl as rl, so there must be a minimum value of Ar, which must occur at the critical number. See Figure 5.] The value of h corresponding to r s 5 is h r 5 5 r Thus, to minimize the cost of the can, the radius should be s 5 cm and the height should be equal to twice the radius, namel, the diameter. NOTE The argument used in Eample to justif the absolute minimum is a variant of the First Derivative Test (which applies onl to local maimum or minimum values) and is stated here for future reference. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

29 4 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu FIGURE 6 Module 4.7 takes ou through eight additional optimization problems, including animations of the phsical situations. (,4) (,) 4 = First Derivative Test for Absolute Etreme Values Suppose that c is a critical number of a continuous function f dened on an interv al. (a) If f for all c and f for all c, then f c is the absolute maimum value of f. (b) If f for all c and f for all c, then f c is the absolute minimum value of f. NOTE An alternative method for solving optimization problems is to use implicit differentiation. Lets look at Eample again to illustrate the method. We work with the same equations but instead of einating h, we differentiate both equations implicitl with respect to r: The minimum occurs at a critical number, so we set A, simplif, and arrive at the equations and subtraction gives r h, or h r. EXAMPLE Find the point on the parabola that is closest to the point, 4. SOLUTION The distance between the point, 4 and the point, is (See Figure 6.) But if, lies on the parabola, then, so the epression for d becomes (Alternativel, we could have substituted s to get d in terms of alone.) Instead of minimizing d, we minimize its square: (You should convince ourself that the minimum of d occurs at the same point as the minimum of, but d is easier to work with.) Differentiating, we obtain d A r rh A 4r h rh r h rh r h h rh d s 4 d s( ) 4 rh r h d f ( ) 4 f ( ) 4 8 so f when. Observe that f when and f when, so b the First Derivative Test for Absolute Etreme Values, the absolute minimum occurs when. (Or we could simpl sa that because of the geometric nature of the problem, its obvious that there is a closest point but not a farthest point.) The corresponding value of is. Thus, the point on closest to, 4 is,. Licensed to: jsamuels@bmcc.cun.edu A km FIGURE 7 C D B 8 km Tr another problem like this one. Resources / Module 5 / Ma and Min / Start of Optimal Lifeguard T FIGURE =T() SECTION 4.7 OPTIMIZATION PROBLEMS 5 EXAMPLE 4 A man launches his boat from point A on a bank of a straight river, km wide, and wants to reach point B, 8 km downstream on the opposite bank, as quickl as possible (see Figure 7). He could row his boat directl across the river to point C and then run to B, or he could row directl to B, or he could row to some point D between C and B and then run to B. If he can row 6 kmh and run 8 kmh, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared with the speed at which the man rows.) SOLUTION If we let be the distance from C to D, then the running distance is DB 8 and the Pthagorean Theorem gives the rowing distance as. We use the equation AD s 9 Then the rowing time is s 96 and the running time is 8 8, so the total time T as a function of is The domain of this function T is, 8. Notice that if he rows to C and if 8 he rows directl to B. The derivative of T is Thus, using the fact that, we have T &? &? &? T s 9 6 T 6s time distance rate 9 s s The onl critical number is 9s7. To see whether the minimum occurs at this critical number or at an endpoint of the domain, 8, we evaluate T at all three points: T.5 T 9 T8 s7.4 s7 s Since the smallest of these values of T occurs when 9s7, the absolute minimum value of T must occur there. Figure 8 illustrates this calculation b showing the graph of T. Thus, the man should land the boat at a point 9s7 km (.4 km) downstream from his starting point. &? &? 4 s 9 Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

30 6 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu _r FIGURE 9 FIGURE Resources / Module 5 / Ma and Min / Start of Ma and Min 4.7 Eercises r r Ł (,) r r. Consider the following problem: Find two numbers whose sum is and whose product is a maimum. (a) Make a table of values, like the following one, so that the sum of the numbers in the rst tw o columns is alwas. On the basis of the evidence in our table, estimate the answer to the problem. First number Second number Product EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle of radius r. SOLUTION Lets take the semicircle to be the upper half of the circle r with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the -ais as shown in Figure 9. Let, be the verte that lies in the rst quadrant. Then the rectangle has sides of lengths and, so its area is A To einate we use the fact that, lies on the circle r and so sr. Thus A sr The domain of this function is r. Its derivative is which is when r, that is, rs (since ). This value of gives a maimum value of A since A and Ar. Therefore, the area of the largest inscribed rectangle is As r sr r r r SOLUTION A simpler solution is possible if we think of using an angle as a variable. Let be the angle shown in Figure. Then the area of the rectangle is A sr sr A r cos r sin r sin cos r sin We know that sin has a maimum value of and it occurs when. So A has a maimum value of r and it occurs when. Notice that this trigonometric solution doesnt involve differentiation. In fact, we didnt need to use calculus at all. 4 r sr (b) Use calculus to solve the problem and compare with our answer to part (a).. Find two numbers whose difference is and whose product is a minimum.. Find two positive numbers whose product is and whose sum is a minimum. 4. Find a positive number such that the sum of the number and its reciprocal is as small as possible. 5. Find the dimensions of a rectangle with perimeter m whose area is as large as possible. 6. Find the dimensions of a rectangle with area m whose perimeter is as small as possible. Licensed to: jsamuels@bmcc.cun.edu 7. Consider the following problem: A farmer with 75 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens? (a) Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these congurations. Does it appear that there is a maimum area? If so, estimate it. (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with our smbols. (c) Write an epression for the total area. (d) Use the given information to write an equation that relates the variables. (e) Use part (d) to write the total area as a function of one variable. (f) Finish solving the problem and compare the answer with our estimate in part (a). 8. Consider the following problem: A bo with an open top is to be constructed from a square piece of cardboard, ft wide, b cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a bo can have. (a) Draw several diagrams to illustrate the situation, some short boes with large bases and some tall boes with small bases. Find the volumes of several such boes. Does it appear that there is a maimum volume? If so, estimate it. (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with our smbols. (c) Write an epression for the volume. (d) Use the given information to write an equation that relates the variables. (e) Use part (d) to write the volume as a function of one variable. (f) Finish solving the problem and compare the answer with our estimate in part (a). 9. A farmer wants to fence an area of.5 million square feet in a rectangular eld and then di vide it in half with a fence parallel to one of the sides of the rectangle. How can he do this so as to minimize the cost of the fence?. A bo with a square base and open top must have a volume of, cm. Find the dimensions of the bo that minimize the amount of material used.. If cm of material is available to make a bo with a square base and an open top, nd the lar gest possible volume of the bo.. A rectangular storage container with an open top is to have a volume of m. The length of its base is twice the width. Material for the base costs $ per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.. Do Eercise assuming the container has a lid that is made from the same material as the sides. 4. (a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square. SECTION 4.7 OPTIMIZATION PROBLEMS 7 (b) Show that of all the rectangles with a given perimeter, the one with greatest area is a square. 5. Find the point on the line 4 7 that is closest to the origin. 6. Find the point on the line 6 9 that is closest to the point,. 7. Find the points on the ellipse 4 4 that are farthest awa from the point,. ; 8. Find, correct to two decimal places, the coordinates of the point on the curve tan that is closest to the point,. 9. Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius r.. Find the area of the largest rectangle that can be inscribed in the ellipse a b.. Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on the base of the triangle.. Find the dimensions of the rectangle of largest area that has its base on the -ais and its other two vertices above the -ais and ling on the parabola 8.. Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r. 4. Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths cm and 4 cm if two sides of the rectangle lie along the legs. 5. A right circular clinder is inscribed in a sphere of radius r. Find the largest possible volume of such a clinder. 6. A right circular clinder is inscribed in a cone with height h and base radius r. Find the largest possible volume of such a clinder. 7. A right circular clinder is inscribed in a sphere of radius r. Find the largest possible surface area of such a clinder. 8. A Norman window has the shape of a rectangle surmounted b a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle. See Eercise 5 on page 4.) If the perimeter of the window is ft, nd the dimensions of the window so that the greatest possible amount of light is admitted. 9. The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is ed at 84 cm, nd the dimensions of the poster with the smallest area.. A poster is to have an area of 8 in with -inch margins at the bottom and sides and a -inch margin at the top. What dimensions will give the largest printed area?. A piece of wire m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maimum? (b) A minimum? Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

31 8 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu. Answer Eercise if one piece is bent into a square and the other into a circle.. A clindrical can without a top is made to contain V cm of liquid. Find the dimensions that will minimize the cost of the metal to make the can. 4. A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building? 5. A cone-shaped drinking cup is made from a circular piece of paper of radius R b cutting out a sector and joining the edges CA and CB. Find the maimum capacit of such a cup. A 6. A cone-shaped paper drinking cup is to be made to hold 7 cm of water. Find the height and radius of the cup that will use the smallest amount of paper. 7. A cone with height h is inscribed in a larger cone with height H so that its verte is at the center of the base of the larger cone. Show that the inner cone has maimum volume when h H. 8. For a sh swimming at a speed v relative to the water, the energ ependiture per unit time is proportional to v. It is believed that migrating sh tr to minimize the total ener g required to swim a ed distance. If the sh are swimming against a current u u v, then the time required to swim a distance L is Lv u and the total energ E required to swim the distance is given b Ev av L v u where a is the proportionalit constant. (a) Determine the value of v that minimizes E. (b) Sketch the graph of E. C Note: This result has been veried e perimentall; migrating sh swim against a current at a speed 5% greater than the current speed. 9. In a beehive, each cell is a regular heagonal prism, open at one end with a trihedral angle at the other end. It is believed that bees form their cells in such a wa as to minimize the surface area for a given volume, thus using the least amount of wa in cell construction. Eamination of these cells has shown that the measure of the ape angle is amazingl consistent. Based on the geometr of the cell, it can be shown that the R B surface area S is given b where s, the length of the sides of the heagon, and h, the height, are constants. (a) Calculate. (b) What angle should the bees prefer? (c) Determine the minimum surface area of the cell (in terms of s and h). Note: Actual measurements of the angle in beehives have been made, and the measures of these angles seldom differ from the calculated value b more than. 4. A boat leaves a dock at : P.M. and travels due south at a speed of kmh. Another boat has been heading due east at 5 kmh and reaches the same dock at : P.M. At what time were the two boats closest together? 4. Solve the problem in Eample 4 if the river is 5 km wide and point B is onl 5 km downstream from A. 4. A woman at a point A on the shore of a circular lake with radius mi wants to arrive at the point C diametricall opposite A on the other side of the lake in the shortest possible time. She can walk at the rate of 4 mih and row a boat at mih. How should she proceed? 4. S 6sh s cot (s s) csc dsd rear of cell A b The illumination of an object b a light source is directl proportional to the strength of the source and inversel proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed ft apart, where should an object be placed on the line between the sources so as to receive the least illumination? 44. Find an equation of the line through the point, 5 that cuts off the least area from the rst quadrant. s h trihedral angle front of cell B C SECTION 4.7 OPTIMIZATION PROBLEMS 9 Licensed to: jsamuels@bmcc.cun.edu 45. Let a and b be positive numbers. Find the length of the shortest line segment that is cut off b the rst quadrant and passes travel from a point A in the air to a point B in the water b a path ACB that minimizes the time taken. Show that through the point a, b. sin v 46. At which points on the curve 4 5 does the sin v tangent line have the largest slope? where (the angle of incidence) and (the angle of refraction) 47. Show that of all the isosceles triangles with a given perimeter, the one with the greatest area is equilateral. are as shown. This equation is known as Snells Law. A CAS 48. The frame for a kite is to be made from si pieces of wood. The four eterior pieces have been cut with the lengths indicated in the gure. To maimize the area of the kite, how long C should the diagonal pieces be? ; 49. A point P needs to be located somewhere on the line AD so that the total length L of cables linking P to the points A, B, and C is minimized (see the gure). Epress L as a function of AP and use the graphs of L and dld to estimate the minimum value. A B 5. The graph shows the fuel consumption c of a car (measured in gallons per hour) as a function of the speed v of the car. At ver low speeds the engine runs inefcientl, so initiall c decreases as the speed increases. But at high speeds the fuel consumption increases. You can see that cv is minimized for this car when v mih. However, for fuel efcienc, what must be minimized is not the consumption in gallons per hour but rather the fuel consumption in gallons per mile. Lets call this consumption G. Using the graph, estimate the speed at which G has its minimum value. c a a P m m D Let v be the velocit of light in air and v the velocit of light in water. According to Fermats Principle, a ra of light will b b C 5 m 5. Two vertical poles PQ and ST are secured b a rope PRS going from the top of the rst pole to a point R on the ground between the poles and then to the top of the second pole as in the gure. Sho w that the shortest length of such a rope occurs when. 5. The upper right-hand corner of a piece of paper, in. b 8 in., as in the gure, is folded o ver to the bottom edge. How would ou fold it so as to minimize the length of the fold? In other words, how would ou choose to minimize? A steel pipe is being carried down a hallwa 9 ft wide. At the end of the hall there is a right-angled turn into a narrower hallwa 6 ft wide. What is the length of the longest pipe that can be carried horizontall around the corner? P Q R T 9 B 6 S Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.

32 4 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Licensed to: jsamuels@bmcc.cun.edu 55. An observer stands at a point P, one unit awa from a track. Two runners start at the point S in the gure and run along the track. One runner runs three times as fast as the other. Find the maimum value of the observers angle of sight between the runners. [Hint: Maimize tan.] P S 56. A rain gutter is to be constructed from a metal sheet of width cm b bending up one-third of the sheet on each side through an angle. How should be chosen so that the gutter will carr the maimum amount of water? cm cm cm 57. Where should the point P be chosen on the line segment AB so as to maimize the angle? B P A A painting in an art galler has height h and is hung so that its lower edge is a distance d above the ee of an observer (as in the gure). Ho w far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maimize the angle subtended at his ee b the painting?) h d 59. Find the maimum area of a rectangle that can be circumscribed about a given rectangle with length L and width W. 6. The blood vascular sstem consists of blood vessels (arteries, arterioles, capillaries, and veins) that conve blood from the heart to the organs and back to the heart. This sstem should work so as to minimize the energ epended b the heart in pumping the blood. In particular, this energ is reduced when the resistance of the blood is lowered. One of Poiseuilles Laws gives the resistance R of the blood as where L is the length of the blood vessel, r is the radius, and C is a positive constant determined b the viscosit of the blood. (Poiseuille established this law eperimentall, but it also follows from Equation 8.4..) The gure sho ws a main blood vessel with radius r branching at an angle into a smaller vessel with radius r. A vascular branching r B W L R C L r 4 (a) Use Poiseuilles Law to show that the total resistance of the blood along the path ABC is R C a b cot b csc r 4 r 4 where a and b are the distances shown in the gure. (b) Prove that this resistance is minimized when a cos r 4 r 4 r C b Licensed to: jsamuels@bmcc.cun.edu (c) Find the optimal branching angle (correct to the nearest degree) when the radius of the smaller blood vessel is twothirds the radius of the larger vessel. Manfred Kage / Peter Arnold 6. Ornithologists have determined that some species of birds tend to avoid ights o ver large bodies of water during dalight hours. It is believed that more energ is required to o ver water than land because air generall rises over land and falls over water during the da. A bird with these tendencies is released from an island that is 5 km from the nearest point B on a straight shoreline, ies to a point C on the shoreline, and then ies along the shoreline to its nesting area D. Assume that the bird instinctivel chooses a path that will minimize its energ ependiture. Points B and D are km apart. (a) In general, if it takes.4 times as much energ to o ver water as land, to what point C should the bird in order to minimize the total energ epended in returning to its nesting area? (b) Let W and L denote the energ (in joules) per kilometer o wn over water and land, respectivel. What would a large value of the ratio WL mean in terms of the birds ight? What would a small value mean? Determine the ratio WL corresponding to the minimum ependiture of energ. (c) What should the value of WL be in order for the bird to directl to its nesting area D? What should the value of WL be for the bird to to B and then along the shore to D? APPLIED PROJECT r h The Shape of a Can APPLIED PROJECT THE SHAPE OF A CAN 4 (d) If the ornithologists observe that birds of a certain species reach the shore at a point 4 km from B, how man times more energ does it take a bird to o ver water than land? 5 km island B ; 6. Two light sources of identical strength are placed m apart. An object is to be placed at a point P on a line parallel to the line joining the light sources and at a distance d meters from it (see the gure). We want to locate P on so that the intensit of illumination is minimized. We need to use the fact that the intensit of illumination for a single source is directl proportional to the strength of the source and inversel proportional to the square of the distance from the source. (a) Find an epression for the intensit I at the point P. (b) If d 5 m, use graphs of I and I to show that the intensit is minimized when 5 m, that is, when P is at the midpoint of. (c) If d m, show that the intensit (perhaps surprisingl) is not minimized at the midpoint. (d) Somewhere between d 5 m and d m there is a transitional value of d at which the point of minimal illumination abruptl changes. Estimate this value of d b graphical methods. Then nd the e act value of d. d C km In this project we investigate the most economical shape for a can. We rst interpret this to mean that the volume V of a clindrical can is given and we need to nd the height h and radius r that minimize the cost of the metal to make the can (see the gure). If we disre gard an waste metal in the manufacturing process, then the problem is to minimize the surface area of the clinder. We solved this problem in Eample in Section 4.7 and we found that h r; that is, the height should be the same as the diameter. But if ou go to our cupboard or our supermarket with a ruler, ou will discover that the height is usuall greater than the diameter and the ratio hr varies from up to about.8. Lets see if we can eplain this phenomenon. P m D nest Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part. Copright 5 Thomson Learning, Inc. All Rights Reserved. Ma not be copied, scanned, or duplicated, in whole or in part.