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Unformatted text preview: 5E-04(pp 222-231) 1/17/06 2:40 PM Page 222 CHAPTER 4 Scientists have tried to explain how rainbows are formed since the time of Aristotle. In the project on page 232, you will be able to use the principles of differential calculus to explain the formation, location, and colors of the rainbow. A pplications of Differentiation 5E-04(pp 222-231) 1/17/06 2:40 PM Page 223 We have already 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 they help us locate maximum and minimum values of functions. Many practical problems require us to minimize a cost or maximize 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 explain the location of rainbows in the sky. |||| 4.1 Maximum and Minimum Values Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. Here are examples of such problems that we will solve in this chapter: ■ ■ ■ ■ What is the shape of a can that minimizes manufacturing costs? What is the maximum acceleration of a space shuttle? (This is an important question to the astronauts who have to withstand the effects of acceleration.) What is the radius of a contracted windpipe that expels air most rapidly during a cough? At what angle should blood vessels branch so as to minimize the energy expended by the heart in pumping blood? These problems can be reduced to finding the maximum or minimum values of a function. Let’s first explain exactly what we mean by maximum and minimum values. 1 Definition A function f has an absolute maximum (or global maximum) at c f x for all x in D, where D is the domain of f . The number f c is called if f c the maximum value of f on D. Similarly, f has an absolute minimum at c if fc f x for all x in D and the number f c is called the minimum value of f on D. The maximum and minimum values of f are called the extreme values of f . Figure 1 shows the graph of a function f with absolute maximum 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. y f(d) FIGURE 1 Minimum value f(a), maximum value f(d) f(a) a 0 b c d e x 223 5E-04(pp 222-231) 224 ❙❙❙❙ 1/17/06 2:40 PM Page 224 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION In Figure 1, if we consider only values of x 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 x and is called a local maximum value of f . Likewise, f c is called a local minimum value of f because fc f x for x 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 definition. 2 Definition A function f has a local maximum (or relative maximum) at c if fc f x when x is near c. [This means that f c f x for all x in some open interval containing c.] Similarly, f has a local minimum at c if f c f x when x is near c. EXAMPLE 1 The function f x cos x takes on its (local and absolute) maximum value of 1 infinitely many times, since cos 2 n 1 for any integer n and 1 cos x 1 for all x. Likewise, cos 2 n 1 1 is its minimum value, where n is any integer. y y=≈ 0 x 2, then f x f 0 because x 2 0 for all x. Therefore, f 0 0 is the absolute (and local) minimum value of f . This corresponds to the fact that the origin is the lowest point on the parabola y x 2. (See Figure 2.) However, there is no highest point on the parabola and so this function has no maximum value. EXAMPLE 2 If f x x FIGURE 2 Minimum value 0, no maximum x 3, shown in Figure 3, we see that this function has neither an absolute maximum value nor an absolute minimum value. In fact, it has no local extreme values either. E XAMPLE 3 From the graph of the function f x y y=˛ E XAMPLE 4 The graph of the function 0 x fx 3x 4 16 x 3 18 x 2 1 x 4 is shown in Figure 4. You can see that f 1 5 is a local maximum, whereas the absolute maximum is f 1 37. (This absolute maximum is not a local maximum because it occurs at an endpoint.) Also, f 0 0 is a local minimum and f 3 27 is both a local and an absolute minimum. Note that f has neither a local nor an absolute maximum at x 4. FIGURE 3 No minimum, no maximum y (_1, 37) y=3x$-16˛+18≈ (1, 5) _1 1 2 3 4 5 x (3, _27) FIGURE 4 We have seen that some functions have extreme values, whereas others do not. The following theorem gives conditions under which a function is guaranteed to possess extreme values. 5E-04(pp 222-231) 1/17/06 2:40 PM Page 225 S ECTION 4.1 MAXIMUM AND MINIMUM VALUES ❙❙❙❙ 225 3 The Extreme Value Theorem If f is continuous on a closed interval a, b , then f attains an absolute maximum value f c and an absolute minimum value f d at some numbers c and d in a, b . The Extreme Value Theorem is illustrated in Figure 5. Note that an extreme value can be taken on more than once. Although the Extreme Value Theorem is intuitively very plausible, it is difficult to prove and so we omit the proof. y FIGURE 5 0 y a c db 0 x y a c d=b 0 x a c¡ d c™ b x Figures 6 and 7 show that a function need not possess extreme values if either hypothesis (continuity or closed interval) is omitted from the Extreme Value Theorem. y y 3 1 0 1 2 x 0 2 x FIGURE 6 y {c, f (c)} {d, f (d)} 0 FIGURE 8 c d x FIGURE 7 This function has minimum value f(2)=0, but no maximum value. This continuous function g has no maximum or minimum. The function f whose graph is shown in Figure 6 is defined on the closed interval [0, 2] but has no maximum value. (Notice that the range of f is [0, 3). The function takes on values arbitrarily close to 3, but never actually attains the value 3.) This does not contradict the Extreme Value Theorem because f is not continuous. [Nonetheless, a discontinuous function could have maximum and minimum values. See Exercise 13(b).] The function t shown in Figure 7 is continuous on the open interval (0, 2) but has neither a maximum nor a minimum value. [The range of t is 1, . The function takes on arbitrarily large values.] This does not contradict the Extreme Value Theorem because the interval (0, 2) is not closed. The Extreme Value Theorem says that a continuous function on a closed interval has a maximum value and a minimum value, but it does not tell us how to find these extreme values. We start by looking for local extreme values. Figure 8 shows the graph of a function f with a local maximum at c and a local minimum at d. It appears that at the maximum and minimum points the tangent lines are horizontal and therefore each has slope 0. We know that the derivative is the slope of the tangent line, so it appears that f c 0 and f d 0. The following theorem says that this is always true for differentiable functions. 5E-04(pp 222-231) 226 ❙❙❙❙ 1/17/06 2:40 PM Page 226 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION |||| Fermat’s Theorem is named after Pierre Fermat (1601–1665), a French lawyer who took up mathematics as a hobby. Despite his amateur status, Fermat was one of the two inventors of analytic geometry (Descartes was the other). His methods for finding tangents to curves and maximum and minimum values (before the invention of limits and derivatives) made him a forerunner of Newton in the creation of differential calculus. 4 Fermat’s Theorem If f has a local maximum or minimum at c, and if f c exists, then f c 0. Proof Suppose, for the sake of definiteness, that f has a local maximum at c. Then, according to Definition 2, f c f x if x is sufficiently close to c. This implies that if h is sufficiently close to 0, with h being positive or negative, then fc fc h and therefore fc 5 h fc 0 We can divide both sides of an inequality by a positive number. Thus, if h sufficiently small, we have fc h h fc 0 and h is 0 Taking the right-hand limit of both sides of this inequality (using Theorem 2.3.2), we get fc lim h h hl0 fc lim 0 0 hl0 But since f c exists, we have fc lim fc h h hl0 fc lim fc h h hl0 fc and so we have shown that f c 0. If h 0, then the direction of the inequality (5) is reversed when we divide by h : fc h h fc 0 h 0 fc h h So, taking the left-hand limit, we have fc hl0 fc h h fc lim hl0 fc 0 We have shown that f c 0 and also that f c 0. Since both of these inequalities must be true, the only possibility is that f c 0. We have proved Fermat’s Theorem for the case of a local maximum. The case of a local minimum can be proved in a similar manner, or we could use Exercise 70 to deduce it from the case we have just proved (see Exercise 71). y y=˛ 0 lim x The following examples caution us against reading too much into Fermat’s Theorem. 0 and solving for x. We can’t expect to locate extreme values simply by setting f x FIGURE 9 If ƒ=˛, then f ª(0)=0 but ƒ has no maximum or minimum. x 3, then f x 3x 2, so f 0 0. But f has no maximum or minimum at 0, as you can see from its graph in Figure 9. (Or observe that x 3 0 for x 0 but x 3 0 for x 0.) The fact that f 0 0 simply means that the curve y x 3 has a EXAMPLE 5 If f x 5E-04(pp 222-231) 1/17/06 2:40 PM Page 227 S ECTION 4.1 MAXIMUM AND MINIMUM VALUES ❙❙❙❙ 227 horizontal tangent at 0, 0 . Instead of having a maximum or minimum at 0, 0 , the curve crosses its horizontal tangent there. y EXAMPLE 6 The function f x x has its (local and absolute) minimum value at 0, but that value can’t be found by setting f x 0 because, as was shown in Example 6 in Section 3.2, f 0 does not exist. (See Figure 10.) y=| x| 0 | x F IGURE 10 If ƒ=| x |, then f(0)=0 is a minimum value, but f ª(0) does not exist. WARNING Examples 5 and 6 show that we must be careful when using Fermat’s Theorem. Example 5 demonstrates that even when f c 0 there need not be a maximum or minimum at c. (In other words, the converse of Fermat’s Theorem is false in general.) Furthermore, there may be an extreme value even when f c does not exist (as in Example 6). Fermat’s Theorem does suggest that we should at least start looking for extreme values of f at the numbers c where f c 0 or where f c does not exist. Such numbers are given a special name. ■ 6 Definition A critical number of a function f is a number c in the domain of f such that either f c 0 or f c does not exist. |||| Figure 11 shows a graph of the function f in Example 7. It supports our answer because there is a horizontal tangent when x 1.5 and a vertical tangent when x 0. x3 5 4 EXAMPLE 7 Find the critical numbers of f x SOLUTION The Product Rule gives fx 3 5 x 25 4 x3 5 x 1 3.5 34 _0.5 5 _2 FIGURE 11 x. x 5x 2 5x 5 34 x 5x 2 5 x3 5 12 8 x 5x 2 5 [The same result could be obtained by first writing f x 4 x 3 5 x 8 5.] Therefore, 3 fx 0 if 12 8 x 0, that is, x 2 , and f x does not exist when x 0. Thus, the critical numbers are 3 and 0. 2 In terms of critical numbers, Fermat’s Theorem can be rephrased as follows (compare Definition 6 with Theorem 4): 7 If f has a local maximum or minimum at c, then c is a critical number of f. To find an absolute maximum 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 by (7)] or it occurs at an endpoint of the interval. Thus, the following three-step procedure always works. The Closed Interval Method To find the absolute maximum and minimum values of a continuous function f on a closed interval a, b : 1. Find the values of f at the critical numbers of f in a, b . 2. Find the values of f at the endpoints of the interval. 3. The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. 5E-04(pp 222-231) 228 ❙❙❙❙ 1/17/06 2:40 PM Page 228 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION EXAMPLE 8 Find the absolute maximum and minimum values of the function x3 fx SOLUTION Since f is continuous on 3x 2 [ 1 2 y=˛-3≈+1 (4, 17) x3 fx 20 x 4 , 4], we can use the Closed Interval Method: fx y 1 2 1 3x 2 3x 2 1 6x 3x x 2 Since f x exists for all x, the only critical numbers of f occur when f x 0, that is, x 0 or x 2. Notice that each of these critical numbers lies in the interval ( 1 , 4). 2 The values of f at these critical numbers are 15 f0 1 f2 3 10 The values of f at the endpoints of the interval are 5 1 _1 0 _5 f( 2 3 1 2 ) 1 8 f4 17 x 4 Comparing these four numbers, we see that the absolute maximum value is f 4 17 and the absolute minimum value is f 2 3. Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph of f is sketched in Figure 12. (2, _3) FIGURE 12 If you have a graphing calculator or a computer with graphing software, it is possible to estimate maximum and minimum values very easily. But, as the next example shows, calculus is needed to find the exact values. EXAMPLE 9 (a) Use a graphing device to estimate the absolute minimum and maximum values of the function f x x 2 sin x, 0 x 2 . (b) Use calculus to find the exact minimum and maximum values. SOLUTION 8 0 _1 FIGURE 13 2π (a) Figure 13 shows a graph of f in the viewing rectangle 0, 2 by 1, 8 . By moving the cursor close to the maximum point, we see that the y-coordinates don’t change very much in the vicinity of the maximum. The absolute maximum value is about 6.97 and it occurs when x 5.2. Similarly, by moving the cursor close to the minimum point, we see that the absolute minimum value is about 0.68 and it occurs when x 1.0. It is possible to get more accurate estimates by zooming in toward the maximum and minimum points, but instead let’s use calculus. (b) The function f x x 2 sin x is continuous on 0, 2 . Since f x 1 2 cos x , we have f x 0 when cos x 1 and this occurs when x 3 or 5 3. The values 2 of f at these critical points are f and 3 f5 3 3 5 3 2 sin 3 2 sin 5 3 0.684853 s3 3 5 3 s3 6.968039 5E-04(pp 222-231) 1/17/06 2:40 PM Page 229 S ECTION 4.1 MAXIMUM AND MINIMUM VALUES ❙❙❙❙ 229 The values of f at the endpoints are f0 0 and f2 2 6.28 Comparing these four numbers and using the Closed Interval Method, we see that the absolute minimum value is f 3 3 s3 and the absolute maximum value is f5 3 5 3 s3. The values from part (a) serve as a check on our work. EXAMPLE 10 The Hubble Space Telescope was deployed on April 24, 1990, by the space shuttle Discovery. A model for the velocity of the shuttle during this mission, from liftoff at t 0 until the solid rocket boosters were jettisoned at t 126 s, is given by vt 0.001302 t 3 0.09029 t 2 23.61t 3.083 (in feet per second). Using this model, estimate the absolute maximum and minimum values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters. SOLUTION We are asked for the extreme values not of the given velocity function, but rather of the acceleration function. So we first need to differentiate to find the acceleration: at vt d 0.001302 t 3 dt 0.003906 t 2 0.09029 t 2 0.18058 t 23.61t 3.083 23.61 We now apply the Closed Interval Method to the continuous function a on the interval 0 t 126. Its derivative is at 0.007812 t The only critical number occurs when a t t1 0.18058 0: 0.18058 0.007812 23.12 Evaluating a t at the critical number and at the endpoints, we have a0 23.61 a t1 21.52 a 126 62.87 So the maximum acceleration is about 62.87 ft s2 and the minimum acceleration is about 21.52 ft s2. |||| 4.1 Exercises 1. Explain the difference between an absolute minimum and a local minimum. 2. Suppose f is a continuous function defined on a closed interval a, b . (a) What theorem guarantees the existence of an absolute maximum value and an absolute minimum value for f ? (b) What steps would you take to find those maximum and minimum values? 5E-04(pp 222-231) ❙❙❙❙ 230 1/17/06 2:41 PM Page 230 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 3–4 |||| 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 maximum or minimum, a local maximum or minimum, or neither a maximum nor a minimum. 3. 9. Absolute maximum at 5, absolute minimum at 2, local maximum at 3, local minima at 2 and 4 10. f has no local maximum or minimum, but 2 and 4 are critical numbers y ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 11. (a) Sketch the graph of a function that has a local maximum at 2 and is differentiable at 2. (b) Sketch the graph of a function that has a local maximum at 2 and is continuous but not differentiable at 2. (c) Sketch the graph of a function that has a local maximum at 2 and is not continuous at 2. 0 4. a b c d e r s x t 12. (a) Sketch the graph of a function on [ 1, 2] that has an absolute maximum but no local maximum. (b) Sketch the graph of a function on [ 1, 2] that has a local maximum but no absolute maximum. y 13. (a) Sketch the graph of a function on [ 1, 2] that has an absolute maximum but no absolute minimum. (b) Sketch the graph of a function on [ 1, 2] that is discontinuous but has both an absolute maximum and an absolute minimum. t a 0 ■ ■ ■ b cd ■ ■ e ■ ■ r ■ s ■ x ■ ■ one local minimum, and no absolute minimum. (b) Sketch the graph of a function that has three local minima, two local maxima, and seven critical numbers. ■ 5–6 |||| Use the graph to state the absolute and local maximum and minimum values of the function. 5. 14. (a) Sketch the graph of a function that has two local maxima, y 15–30 |||| Sketch the graph of f by hand and use your sketch to find the absolute and local maximum and minimum values of f . (Use the graphs and transformations of Sections 1.2 and 1.3.) 15. f x 8 16. f x y=ƒ 3 17. f x 1 0 18. f x 19. f x x 1 20. f x 6. y 21. f x y=ƒ 3x, x 1 2 x, x 5 2 x 2 2 x 2 2 x 2 2 x 2 x, 0 x, 0 x, 0 x, 0 2 x, 3 x 2 2 22. f x 0 ■ ■ 7–10 ■ ■ ■ ■ ■ ■ ■ ■ Sketch the graph of a function f that is continuous on [1, 5] and has the given properties. |||| 7. Absolute minimum at 2, absolute maximum at 3, ■ 0 t 1 1 t, 0 t 1 sin , 2 26. f ■ 1 t, 25. f x 1, 24. f t 1 x 23. f t 1 1 tan , 27. f x 1 sx 28. f x 1 x3 local maximum at 2, local minimum at 4 x 5 2 4 2 29. f x 1x 2x 4 if 0 if 2 30. f x x2 2 if 1 x 0 if 0 x 1 local minimum at 4 8. Absolute minimum at 1, absolute maximum at 5, 2 ■ ■ ■ x2 ■ ■ x x ■ 2 3 ■ ■ ■ ■ ■ ■ 5E-04(pp 222-231) 1/17/06 2:41 PM Page 231 S ECTION 4.1 MAXIMUM AND MINIMUM VALUES 31– 44 |||| 31. f x 5x 33. f x x3 3x2 4 3 35. s t 3t 37. t x 2x 39. t t 23 5t 41. F x 43. f ■ 6t 45–56 2 x4 5 x 4 3 sx 2 44. t sin ■ st 1 42. G x 2 ■ x ■ ■ 4 ■ ■ V 1 z 13 40. t t 2 x z z2 1 kg of water at a temperature T is given approximately by the formula x x 38. t x 53 2 x2 x3 36. f z 3 t x 34. f x 24 x 2 cos ■ 32. f x 4x 4t 3 1 x 46. f x x 3 47. f x 2x 3 48. f x x3 6x 2 9x 49. f x x4 2x 2 3, 50. f x 51. f x x 12 x 3x ■ ■ 2 1 52. f x x2 x2 4 , 4 53. f t t s4 54. f t 3 st 8 t, 55. f x sin x 56. f x x ■ xa 1 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 velocity data for the shuttle between liftoff and the jettisoning of the solid rocket boosters. 1, 2 0, 8 0, 3 Event , ■ ■ x b, 0 ■ ■ ■ ■ ■ ■ 1. x ; 58. Use a graph to estimate the critical numbers of x3 fx ; 59–62 3x 2 2 correct to one decimal place. Time (s) Velocity (ft s) Launch Begin roll maneuver End roll maneuver Throttle to 89% Throttle to 67% Throttle to 104% Maximum dynamic pressure Solid rocket booster separation 0 10 15 20 32 59 62 125 0 185 319 447 742 1325 1445 4151 |||| (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. (b) Use calculus to find the exact maximum and minimum values. 59. f x x3 60. f x x 4 61. f x x sx 62. f x ■ 99.33 where t is measured in years since midyear 1984, so 0 t 10, and I t is measured in 1987 dollars and scaled such that I 3 100. Estimate the times when food was cheapest and most expensive during the period 1984–1994. 57. If a and b are positive numbers, find the maximum value of fx 0.6270 t 0.06561t 3 ; 66. On May 7, 1992, the space shuttle Endeavour was launched 2 cos x, ■ 0.001438t 4 0.4598t 2 It 4, 4 t2, 0.00009045t 5 It 2, 3 cos x, ■ 1, 4 2, 0, 2 , x2 ■ “basket” of foods) between 1984 and 1994 is given by the function 2, 3 1, 1, 2 1, x 65. A model for the food-price index (the price of a representative 12 x 3 cos where is a positive constant called the coefficient of friction and where 0 2. Show that F is minimized when . tan 0, 3 3x 2 sin 0, 3 5, 1, W F Find the absolute maximum and absolute minimum values of f on the given interval. 3x 2 0.0000679T 3 by 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 t ■ 0.0085043T 2 64. An object with weight W is dragged along a horizontal plane |||| 45. f x 0.06426 T Find the temperature at which water has its maximum density. 23 x 999.87 tan ■ 231 63. Between 0 C and 30 C, the volume V (in cubic centimeters) of Find the critical numbers of the function. 2 ❙❙❙❙ ■ 8x 3x 3 3x 2 x x, 3 0 x 2 x2 cos x ■ 1, 3 (a) Use a graphing calculator or computer to find the cubic polynomial that best models the velocity of the shuttle for the time interval t 0, 125 . Then graph this polynomial. (b) Find a model for the acceleration of the shuttle and use it to estimate the maximum and minimum values of the acceleration during the first 125 seconds. 2 ■ sin x , ■ ■ 0 x ■ 67. When a foreign object lodged in the trachea (windpipe) forces 2 ■ ■ ■ ■ ■ a person to cough, the diaphragm thrusts upward causing an increase in pressure in the lungs. This is accompanied by a 5E-04(pp 232-241) 232 ❙❙❙❙ 1/17/06 2:37 PM Page 232 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION contraction of the trachea, making a narrower channel for the expelled air to flow through. For a given amount of air to escape in a fixed time, it must move faster through the narrower channel than the wider one. The greater the velocity of the airstream, the greater the force on the foreign object. X rays 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 velocity v of the airstream is related to the radius r of the trachea by the equation vr k r0 rr 1 20 2 r r 68. Show that 5 is a critical number of the function tx 2 x 5 3 but t does not have a local extreme value at 5. 69. Prove that the function fx x 101 x 51 x 1 has neither a local maximum nor a local minimum. 70. If f has a minimum value at c, show that the function tx f x has a maximum value at c. 71. Prove Fermat’s Theorem for the case in which f has a local r0 minimum at c. where k is a constant and r0 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 1 r0 is 2 prevented (otherwise the person would suffocate). (a) Determine the value of r in the interval 1 r0 , r0 at which v 2 has an absolute maximum. How does this compare with experimental evidence? (b) What is the absolute maximum value of v on the interval? (c) Sketch the graph of v on the interval 0, r0 . [ 72. A cubic function is a polynomial of degree 3; that is, it has the form fx ax 3 bx 2 cx d where a 0. (a) Show that a cubic function can have two, one, or no critical number(s). Give examples and sketches to illustrate the three possibilities. (b) How many local extreme values can a cubic function have? APPLIED PROJECT The Calculus of Rainbows Rainbows are created when raindrops scatter sunlight. They have fascinated mankind since ancient times and have inspired attempts at scientific explanation since the time of Aristotle. In this project we use the ideas of Descartes and Newton to explain the shape, location, and colors of rainbows. åA from Sun 1. The figure shows a ray of sunlight entering a spherical raindrop at A. Some of the light is ∫ B ∫ O D(å ) ∫ ∫ å to observer C reflected, 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 Snell’s Law says that sin k sin , where is the angle of incidence, is the angle of refraction, and k 4 is the index of 3 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 reflected. (The angle of incidence equals the angle of reflection.) When the ray reaches C, part of it is reflected, but for the time being we are more interested in the part that leaves the raindrop at C. (Notice that it is refracted away from the normal line.) The angle of deviation D is the amount of clockwise rotation that the ray has undergone during this three-stage process. Thus Formation of the primary rainbow D 2 2 4 Show that the minimum value of the deviation is D 138 and occurs when 59.4 . The significance of the minimum deviation is that when 59.4 we have D 0, so D 0. This means that many rays with 59.4 become deviated by approximately the same amount. It is the concentration of rays coming from near the direction of minimum 5E-04(pp 232-241) 1/17/06 2:37 PM Page 233 A PPLIED PROJECT THE CALCULUS OF RAINBOWS ❙❙❙❙ 233 deviation that creates the brightness of the primary rainbow. The following figure shows that the angle of elevation from the observer up to the highest point on the rainbow is 180 138 42 . (This angle is called the rainbow angle.) rays from Sun 138° rays from Sun 42° observer 2. Problem 1 explains the location of the primary rainbow, but how do we explain the colors? Sunlight comprises a range of wavelengths, from the red range through orange, yellow, green, blue, indigo, and violet. As Newton discovered in his prism experiments of 1666, the index of refraction is different for each color. (The effect is called dispersion.) For red light the refractive index is k 1.3318 whereas for violet light it is k 1.3435. By repeating the calculation of Problem 1 for these values of k, show that the rainbow angle is about 42.3 for the red bow and 40.6 for the violet bow. So the rainbow really consists of seven individual bows corresponding to the seven colors. 3. Perhaps you have seen a fainter secondary rainbow above the primary bow. That results from C ∫ D the part of a ray that enters a raindrop and is refracted at A, reflected twice (at B and C ), and refracted as it leaves the drop at D (see the figure). This time the deviation angle D is the total amount of counterclockwise rotation that the ray undergoes in this four-stage process. Show that D 2 6 2 ∫ ∫ å to observer ∫ from Sun ∫ å ∫ and D has a minimum value when B cos A Formation of the secondary rainbow k2 1 8 Taking k 4 , show that the minimum deviation is about 129 and so the rainbow angle for 3 the secondary rainbow is about 51 , as shown in the figure. 42° 51° 4. Show that the colors in the secondary rainbow appear in the opposite order from those in the primary rainbow. 5E-04(pp 232-241) 234 ❙❙❙❙ 1/17/06 2:37 PM Page 234 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION |||| 4.2 The Mean Value Theorem We will see that many 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 first need the following result. Rolle’s Theorem Let f be a function that satisfies the following three hypotheses: |||| Rolle’s Theorem was first published in 1691 by the French mathematician Michel Rolle (1652–1719) in a book entitled Méthode pour résoudre les égalitéz. Later, however, he became a vocal critic of the methods of his day and attacked calculus as being a “collection of ingenious fallacies.” 1. f is continuous on the closed interval a, b . 2. f is differentiable on the open interval a, b . 3. f a fb Then there is a number c in a, b such that f c 0. Before giving the proof let’s take a look at the graphs of some typical functions that satisfy the three hypotheses. Figure 1 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 0. Thus, Rolle’s Theorem is plausible. y 0 y a c¡ c™ b x 0 y y a (a) c b x (b) 0 a c¡ c™ b x 0 a (c) c b x (d) FIGURE 1 |||| Take cases Proof There are three cases: CASE I f ( x) ■ Then f x k, a constant 0, so the number c can be taken to be any number in a, b . CASE II f ( x) > f ( a) for some x in ( a, b) [as in Figure 1(b) or (c)] By the Extreme Value Theorem (which we can apply by hypothesis 1), f has a maximum value somewhere in a, b . Since f a f b , it must attain this maximum value at a number c in the open interval a, b . Then f has a local maximum at c and, by hypothesis 2, f is differentiable at c. Therefore, f c 0 by Fermat’s Theorem. ■ CASE III f ( x) < f ( a) for some x in ( a, b) [as in Figure 1(c) or (d)] By the Extreme 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 0 by Fermat’s Theorem. ■ EXAMPLE 1 Let’s apply Rolle’s 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 fa f b . Rolle’s Theorem says that there is some instant of time t c between a and b when f c 0; that is, the velocity is 0. (In particular, you can see that this is true when a ball is thrown directly upward.) 5E-04(pp 232-241) 1/17/06 2:37 PM Page 235 S ECTION 4.2 THE MEAN VALUE THEOREM |||| Figure 2 shows a graph of the function fx x 3 x 1 discussed in Example 2. Rolle’s Theorem shows that, no matter how much we enlarge the viewing rectangle, we can never find a second x-intercept. 3 _2 2 EXAMPLE 2 Prove that the equation x 3 FIGURE 2 235 0 has exactly one real root. 1 SOLUTION First we use the Intermediate Value Theorem (2.5.10) to show that a root exists. Let f x x 3 x 1. Then f 0 1 0 and f 1 1 0. Since f is a polynomial, it is continuous, so the Intermediate Value Theorem states that there is a number c between 0 and 1 such that f c 0. Thus, the given equation has a root. To show that the equation has no other real root, we use Rolle’s Theorem and argue by contradiction. Suppose that it had two roots a and b. Then f a 0 f b and, since f is a polynomial, it is differentiable on a, b and continuous on a, b . Thus, by Rolle’s Theorem, there is a number c between a and b such that f c 0. But 3x 2 fx _3 x ❙❙❙❙ 1 1 for all x (since x 2 0 ) so f x can never be 0. This gives a contradiction. Therefore, the equation can’t have two real roots. Our main use of Rolle’s Theorem is in proving the following important theorem, which was first stated by another French mathematician, Joseph-Louis Lagrange. The Mean Value Theorem Let f be a function that satisfies the following hypotheses: 1. f is continuous on the closed interval a, b . 2. f is differentiable on the open interval a, b . |||| The Mean Value Theorem is an example of what is called an existence theorem. Like the Intermediate Value Theorem, the Extreme Value Theorem, and Rolle’s Theorem, it guarantees that there exists a number with a certain property, but it doesn’t tell us how to find the number. Then there is a number c in a, b such that fb b fc 1 fa a or, equivalently, fb 2 fa fcb a Before proving this theorem, we can see that it is reasonable by interpreting it geometrically. Figures 3 and 4 show the points A a, f a and B b, f b on the graphs of two differentiable functions. The slope of the secant line AB is mAB 3 y fb b fa a y P¡ P { c, f(c)} B P™ A A{ a, f(a)} B { b, f(b)} 0 a FIGURE 3 c b x 0 a FIGURE 4 c¡ c™ b x 5E-04(pp 232-241) 236 ❙❙❙❙ 1/17/06 2:37 PM Page 236 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION which is the same expression as on the right side of Equation 1. Since f c is the slope of the tangent line at the point c, f c , the Mean Value Theorem, in the form given by Equation 1, says that there is at least one point P c, 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. y Proof We apply Rolle’s Theorem to a new function h defined as the difference between h(x) A y=ƒ f and the function whose graph is the secant line AB. Using Equation 3, we see that the equation of the line AB can be written as ƒ y 0 x f(a)+ fa fb b fa a x a y B fa fb b fa a x a fa a x x f(b)-f(a) (x-a) b-a F IGURE 5 or as So, as shown in Figure 5, hx 4 fx fb b fa a First we must verify that h satisfies the three hypotheses of Rolle’s Theorem. 1. The function h is continuous on a, b because it is the sum of f and a first-degree polynomial, both of which are continuous. 2. The function h is differentiable on a, b because both f and the first-degree polynomial are differentiable. In fact, we can compute h directly from Equation 4: |||| The Mean Value Theorem was first formulated by Joseph-Louis Lagrange (1736–1813), born in Italy of a French father and an Italian mother. He was a child prodigy and became a professor in Turin at the tender age of 19. Lagrange made great contributions to number theory, theory of functions, theory of equations, and analytical and celestial mechanics. In particular, he applied calculus to the analysis of the stability of the solar system. At the invitation of Frederick the Great, he succeeded Euler at the Berlin Academy 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 luxury and fame, he was a kind and quiet man, living only for science. hx (Note that f a and f b fb b fx fa fa a a are constants.) b ha fa fa fb b fa a a a hb fb fa fb b fa a b a fb 3. fa fb fa Therefore, h a 0 0 hb. Since h satisfies the hypotheses of Rolle’s Theorem, that theorem says there is a number c in a, b such that h c 0. Therefore 0 and so hc fc fc fb b fb b fa a fa a EXAMPLE 3 To illustrate the Mean Value Theorem with a specific function, let’s consider fx x3 x, a 0, b 2. Since f is a polynomial, it is continuous and differentiable 5E-04(pp 232-241) 1/17/06 2:37 PM Page 237 S ECTION 4.2 THE MEAN VALUE THEOREM ❙❙❙❙ 237 for all x, so it is certainly continuous on 0, 2 and differentiable on 0, 2 . Therefore, by the Mean Value Theorem, there is a number c in 0, 2 such that y y=˛- x B f2 Now f 2 6, f 0 f0 fc2 0, and f x 3x 2 1, so this equation becomes 6 O c FIGURE 6 2 x 3c 2 6c 2 12 0 2 which gives c 2 4 , that is, c 2 s3. But c must lie in 0, 2 , so c 2 s3. 3 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 average velocity between t a and t b is fb b f t , then the fa a and the velocity at t c is f c . Thus, the Mean Value Theorem (in the form of Equation 1) tells us that at some time t c between a and b the instantaneous velocity f c is equal to that average velocity. For instance, if a car traveled 180 km in 2 hours, then the speedometer must have read 90 km h at least once. In general, the Mean Value Theorem can be interpreted as saying 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 significance of the Mean Value Theorem is that it enables us to obtain information about a function from information about its derivative. The next example provides an instance of this principle. EXAMPLE 5 Suppose that f 0 3 and f x 5 for all values of x. How large can f 2 possibly be? SOLUTION We are given that f is differentiable (and therefore continuous) everywhere. In particular, we can apply the Mean Value Theorem on the interval 0, 2 . There exists a number c such that f2 f0 fc2 0 so f2 f0 2f c 3 2f c We are given that f x 5 for all x, so in particular we know that f c ing both sides of this inequality by 2, we have 2 f c 10, so f2 3 2f c 3 10 5. Multiply- 7 The largest possible value for f 2 is 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 x 0 for all x in an interval a, b , then f is constant on a, b . Proof Let x 1 and x 2 be any two numbers in a, b with x 1 x 2 . Since f is differentiable on a, b , it must be differentiable on x 1, x 2 and continuous on x 1, x 2 . By 5E-04(pp 232-241) 238 ❙❙❙❙ 1/17/06 2:37 PM Page 238 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION applying the Mean Value Theorem to f on the interval x 1, x 2 , we get a number c such that x 1 c x 2 and f x2 6 Since f x f x1 0 for all x, we have f c f x2 f x1 f c x2 x1 0, and so Equation 6 becomes 0 or f x2 f x1 Therefore, f has the same value at any two numbers x 1 and x 2 in a, b . This means that f is constant on a, b . 7 Corollary If f x on a, b ; that is, f x Proof Let F x fx t x for all x in an interval a, b , then f tx c where c is a constant. t x . Then Fx fx tx 0 for all x in a, b . Thus, by Theorem 5, F is constant; that is, f NOTE ■ t is constant t is constant. Care must be taken in applying Theorem 5. Let fx x x 1 1 if x if x 0 0 The domain of f is D x x 0 and f x 0 for all x in D. But f is obviously not a constant function. This does not contradict Theorem 5 because D is not an interval. Notice that f is constant on the interval 0, and also on the interval ,0 . |||| 4.2 Exercises 1–4 |||| Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle’s Theorem. 1. f x 4x 2. f x x3 3x 2 3. f x sin 2 x, 4. f x ■ x2 x sx ■ ■ 2x 0, 2 1, 1 6, ■ 5, y =ƒ 6, 0 ■ ■ ■ ■ ■ ■ ■ 1 x 2 3. Show that f 1 f 1 but there is no 0. Why does this not number c in 1, 1 such that f c contradict Rolle’s Theorem? 5. Let f x x 1 2. Show that f 0 f 2 but there is no number c in 0, 2 such that f c 0. Why does this not contradict Rolle’s Theorem? 6. Let f x conclusion of the Mean Value Theorem for the interval 0, 8 . y 0, 4 1, 7. Use the graph of f to estimate the values of c that satisfy the ■ 1 0 1 x 8. Use the graph of f given in Exercise 7 to estimate the values of c that satisfy the conclusion of the Mean Value Theorem for the interval 1, 7 . 5E-04(pp 232-241) 1/17/06 2:37 PM Page 239 S ECTION 4.2 THE MEAN VALUE THEOREM ; 9. (a) Graph the function f x x 4 x in the viewing rectangle 0, 10 by 0, 10 . (b) Graph the secant line that passes through the points 1, 5 and 8, 8.5 on the same screen with f . (c) Find the number c that satisfies the conclusion of the Mean Value Theorem for this function f and the interval 1, 8 . Then graph the tangent line at the point c, f c and notice that it is parallel to the secant line. ; 10. (a) In the viewing rectangle 3, 3 by 5, 5 , graph the function f x x 3 2 x and its secant line through the points 2, 4 and 2, 4 . Use the graph to estimate the x-coordinates of the points where the tangent line is parallel to the secant line. (b) Find the exact values of the numbers c that satisfy the conclusion of the Mean Value Theorem for the interval 2, 2 and compare with your answers to part (a). 11–14 |||| Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers c that satisfy the conclusion of the Mean Value Theorem. 11. f x 3x 2 12. f x x3 13. f x 3 sx, 14. f x ■ ■ x 1, 239 22. (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 you generalize parts (a) and (b)? 23. If f 1 10 and f x possibly be? 24. Suppose that 3 18 f8 f2 2 for 1 x 4, how small can f 4 fx 5 for all values of x. Show that 30. 25. Does there exist a function f such that f 0 and f x 1, f 2 4, 2 for all x ? 26. Suppose that f and t are continuous on a, b and differentiable t x for on a, b . Suppose also that f a t a and f x a x b. Prove that f b t b . [Hint: Apply the Mean Value Theorem to the function h f t.] 27. Show that s1 x 1 1 2 x if x 0. 28. Suppose f is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists a number c in b, b such that f c f b b. 1, 1 0, 2 29. Use the Mean Value Theorem to prove the inequality 0, 1 sin a x x 5, 2x ❙❙❙❙ , 2 ■ 30. If f x ■ ■ ■ ■ ■ ■ ■ ■ ■ 15. Let f x x 1 . Show that there is no value of c such that f3 f0 f c 3 0 . Why does this not contradict the Mean Value Theorem? 16. Let f x x 1 x 1 . Show that there is no value of c such that f 2 f0 f c 2 0 . Why does this not contradict the Mean Value Theorem? 17. Show that the equation 1 2x x3 4x 5 0 has exactly one real root. 18. Show that the equation 2 x 1 sin x 0 has exactly one real root. 19. Show that the equation x 3 in the interval sin b a for all a and b b 1, 4 15 x c 0 has at most one root 2, 2 . 20. Show that the equation x 4 that f x 31. Let f x c (c a constant) for all x, use Corollary 7 to show c x d for some constant d. 1 x and tx 1 x 1 if x 1 x 0 if x 0 Show that f x t x for all x in their domains. Can we conclude from Corollary 7 that f t is constant? 32. At 2:00 P.M. a car’s speedometer reads 30 mi h. At 2:10 P.M. it reads 50 mi h. Show that at some time between 2:00 and 2:10 the acceleration is exactly 120 mi h2. 33. Two runners start a race at the same time and finish in a tie. 4x c 0 has at most two real roots. 21. (a) Show that a polynomial of degree 3 has at most three real roots. (b) Show that a polynomial of degree n has at most n real roots. Prove that at some time during the race they have the same speed. [Hint: Consider f t tt h t , where t and h are the position functions of the two runners.] 34. A number a is called a fixed point of a function f if f a Prove that if f x 1 for all real numbers x, then f has at most one fixed point. a. 5E-04(pp 232-241) 240 ❙❙❙❙ 1/17/06 2:37 PM Page 240 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION |||| 4.3 How Derivatives Affect the Shape of a Graph y Many of the applications of calculus depend on our ability to deduce facts about a function f from information concerning its derivatives. Because f x represents the slope of the curve y f x at the point x, f x , it tells us the direction in which the curve proceeds at each point. So it is reasonable to expect that information about f x will provide us with information about f x . D B What Does f Say about f ? A C x 0 FIGURE 1 To see how the derivative of f can tell us where a function is increasing or decreasing, look at Figure 1. (Increasing functions and decreasing functions were defined in Section 1.1.) Between A and B and between C and D, the tangent lines have positive slope and so fx 0. Between B and C, the tangent lines have negative slope and so f x 0. Thus, it appears that f increases when f x is positive and decreases when f x is negative. To prove that this is always the case, we use the Mean Value Theorem. Increasing/Decreasing Test |||| Let’s abbreviate the name of this test to the I/D Test. (a) If f x (b) If f x 0 on an interval, then f is increasing on that interval. 0 on an interval, then f is decreasing on that interval. Proof Resources / Module 3 / Increasing and Decreasing Functions / Increasing-Decreasing Detector (a) Let x 1 and x 2 be any two numbers in the interval with x1 x2 . According to the definition of an increasing function (page 21) we have to show that f x1 f x2 . Because we are given that f x 0, we know that f is differentiable on x1, x2 . So, by the Mean Value Theorem there is a number c between x1 and x2 such that f x2 1 f x1 Now f c 0 by assumption and x 2 Equation 1 is positive, and so f x2 f x1 x1 0 f c x2 x1 0 because x 1 x 2 . Thus, the right side of or f x2 f x1 This shows that f is increasing. Part (b) is proved similarly. EXAMPLE 1 Find where the function f x 3x 4 4x 3 24 x 12 x x 12 x 2 5 is increasing and where it is decreasing. SOLUTION Module 4.3A guides you in determining properties of the derivative f by examining the graphs of a variety of functions f . fx 12 x 3 12 x 2 2x 1 To use the I D Test we have to know where f x 0 and where f x 0. This depends on the signs of the three factors of f x , namely, 12 x, x 2, and x 1. We divide the real line into intervals whose endpoints are the critical numbers 1, 0, and 2 and arrange our work in a chart. A plus sign indicates that the given expression is positive, and a minus sign indicates that it is negative. The last column of the chart gives the conclusion based on the I D Test. For instance, f x 0 for 0 x 2, so f is decreasing on (0, 2). (It would also be true to say that f is decreasing on the closed interval 0, 2 .) 5E-04(pp 232-241) 1/17/06 2:37 PM Page 241 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH 20 _2 Interval 3 _30 1 0 x x x x 12 x x 2 x 1 fx ❙❙❙❙ 241 f 1 decreasing on ( , 1) increasing on ( 1, 0) decreasing on (0, 2) increasing on (2, ) 0 2 2 The graph of f shown in Figure 2 confirms the information in the chart. FIGURE 2 Recall from Section 4.1 that if f has a local maximum or minimum at c, then c must be a critical number of f (by Fermat’s Theorem), but not every critical number gives rise to a maximum or a minimum. We therefore need a test that will tell us whether or not f has a local maximum or minimum at a critical number. You can see from Figure 2 that f 0 5 is a local maximum value of f because f increases on 1, 0 and decreases on 0, 2 . Or, in terms of derivatives, f x 0 for 1 x 0 and f x 0 for 0 x 2. In other words, the sign of f x changes from positive to negative at 0. 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 maximum 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 example, if f is positive on both sides of c or negative on both sides), then f has no local maximum or minimum at c. The First Derivative Test is a consequence of the I D Test. In part (a), for instance, since the sign of f x 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 maximum at c. It is easy to remember the First Derivative Test by visualizing diagrams such as those in Figure 3. y y fª(x)>0 fª(x)<0 fª(x)<0 0 x c fª(x)>0 0 (a) Local maximum x c (b) Local minimum y y fª(x)<0 fª(x)>0 fª(x)<0 fª(x)>0 0 FIGURE 3 c x (c) No maximum or minimum 0 c x (d) No maximum or minimum 5E-04(pp 242-251) 242 ❙❙❙❙ 1/17/06 2:43 PM Page 242 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION EXAMPLE 2 Find the local minimum and maximum values of the function f in Example 1. SOLUTION From the chart in the solution to Example 1 we see that f x changes from negative to positive at 1, so f 1 0 is a local minimum value by the First Derivative Test. Similarly, f changes from negative to positive at 2, so f 2 27 is also a local minimum value. As previously noted, f 0 5 is a local maximum value because f x changes from positive to negative at 0. EXAMPLE 3 Find the local maximum and minimum values of the function tx x 2 sin x 0 x 2 SOLUTION To find the critical numbers of t, we differentiate: tx 1 2 cos x 1 So t x 0 when cos x 3 and 4 3. 2 . The solutions of this equation are 2 Because t is differentiable everywhere, the only critical numbers are 2 3 and 4 3 and so we analyze t in the following table. |||| The + signs in the table come from the fact 1 that t x 0 when cos x 2 . From the graph of y cos x, this is true in the indicated intervals. Interval 2 4 0 3 3 x x x tx 2 4 2 1 2 cos x t 3 3 increasing on (0, 2 3) decreasing on (2 3, 4 3) increasing on (4 3, 2 ) Because t x changes from positive to negative at 2 3, the First Derivative Test tells us that there is a local maximum at 2 3 and the local maximum value is 6 t2 3 2 3 2 sin 2 3 2 3 2 s3 2 Likewise, t x changes from negative to positive at 4 0 2π t4 3 4 3 2 sin 4 3 4 3 2 2 3 s3 3.83 3 and so s3 2 4 3 s3 2.46 FIGURE 4 y=x+2 sin x is a local minimum value. The graph of t in Figure 4 supports our conclusion. What Does f Say about f ? Explore concavity on a roller coaster. Resources / Module 3 / Concavity / Introduction Figure 5 shows the graphs of two increasing functions on a, b . Both graphs join point A to point B but they look different because they bend in different directions. How can we distinguish between these two types 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 . 5E-04(pp 242-251) 1/17/06 2:43 PM Page 243 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH y y B g A A a 0 x b a x b (a) FIGURE 5 (b) y y B B g f A A 0 FIGURE 6 243 B f 0 ❙❙❙❙ 0 x (a) Concave upward x (b) Concave downward 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 . y D B 0a F IGURE 7 b CD P C c CU d CD e CU p CU q x CD Let’s see how the second derivative helps determine the intervals of concavity. Looking at Figure 6(a), you can see that, going from left to right, the slope of the tangent increases. 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 Appendix F with the help of the Mean Value Theorem. Concavity Test (a) If f x (b) If f x 0 for all x in I , then the graph of f is concave upward on I . 0 for all x in I , then the graph of f is concave downward on I . 5E-04(pp 242-251) 244 ❙❙❙❙ 1/17/06 2:43 PM Page 244 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION EXAMPLE 4 Figure 8 shows a population graph for Cyprian honeybees raised in an apiary. 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? P 80 Number of bees (in thousands) 60 40 20 0 3 6 FIGURE 8 9 12 15 t 18 Time (in weeks) SOLUTION By looking at the slope of the curve as t increases, we see that the rate of increase of the population is initially very small, then gets larger until it reaches a maximum at about t 12 weeks, and decreases as the population begins to level off. As the population approaches its maximum value of about 75,000 (called the carrying capacity), the rate of increase, P t , approaches 0. The curve appears to be concave upward on (0, 12) and concave downward on (12, 18). In Example 4, the population curve changed from concave upward to concave downward at approximately the point (12, 38,000). This point is called an inflection point of the curve. The significance of this point is that the rate of population increase has its maximum value there. In general, an inflection point is a point where a curve changes its direction of concavity. Definition A point P on a curve y f x is called an inflection 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 inflection. Notice that if a curve has a tangent at a point of inflection, then the curve crosses its tangent there. In view of the Concavity Test, there is a point of inflection at any point where the second derivative changes sign. EXAMPLE 5 Sketch a possible graph of a function f that satisfies the following conditions: (i f0 0, f2 3, f4 ii fx 0 for 0 x 4, iii fx 0 for x 2, fx 6, fx f0 f4 0 for x 0 for x 0 0 and for x 4 2 SOLUTION Condition (i) tells us that the graph has horizontal tangents at the points 0, 0 and 4, 6 . Condition (ii) says that f is increasing on the interval 0, 4 and decreasing on 5E-04(pp 242-251) 1/17/06 2:43 PM Page 245 S ECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH y 6 (2, 3) 0 dec 2 x 4 inc CD The Second Derivative Test Suppose f is continuous near c. (a) If f c (b) If f c FIGURE 9 y f 0, then f has a local minimum at c. 0, then f has a local maximum at c. For instance, part (a) is true because f x 0 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 10.) x 4 4 x 3 with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. ƒ f(c) c 0 and f c 0 and f c EXAMPLE 6 Discuss the curve y P 0 Another application of the second derivative is the following test for maximum and minimum values. It is a consequence of the Concavity Test. dec CU f ª(c)=0 245 the intervals , 0 and 4, . It follows from the I/D Test that f 0 0 is a local minimum and f 4 6 is a local maximum. Condition (iii) says that the graph is concave upward on the interval , 2 and concave downward on 2, . Because the curve changes from concave upward to concave downward when x 2, the point 2, 3 is an inflection point. We use this information to sketch the graph of f in Figure 9. Notice that we made the curve bend upward when x 2 and bend downward when x 2. (4, 6) 3 ❙❙❙❙ SOLUTION If f x x FIGURE 10 f ·(c)>0, f is concave upward x x4 4 x 3, then fx 4x 3 fx 12 x 2 12 x 2 4x 2 x 3 24 x 12 x x 2 To find the critical numbers we set f x 0 and obtain x 0 and x Second Derivative Test we evaluate f at these critical numbers: f0 0 f3 36 3. To use the 0 0 and f 3 0, f 3 27 is a local minimum. Since f 0 0, the Since f 3 Second Derivative Test gives no information about the critical number 0. But since fx 0 for x 0 and also for 0 x 3, the First Derivative Test tells us that f does not have a local maximum or minimum at 0. [In fact, the expression for f x shows that f decreases to the left of 3 and increases to the right of 3.] Since f x 0 when x 0 or 2, we divide the real line into intervals with these numbers as endpoints and complete the following chart. Interval ( , 0) (0, 2) (2, ) fx 12 x x 2 Concavity upward downward upward The point 0, 0 is an inflection point since the curve changes from concave upward to concave downward there. Also 2, 16 is an inflection point since the curve changes from concave downward to concave upward there. 5E-04(pp 242-251) 246 ❙❙❙❙ 1/17/06 2:43 PM Page 246 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION y Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve in Figure 11. y=x$-4˛ (0, 0) 2 inflection points x 3 NOTE The Second Derivative Test is inconclusive when f c 0. In other words, at such a point there might be a maximum, there might be a minimum, or there might be neither (as in Example 6). This test also fails when f c does not exist. In such cases the First Derivative Test must be used. In fact, even when both tests apply, the First Derivative Test is often the easier one to use. ■ (2, _16) x2 3 6 EXAMPLE 7 Sketch the graph of the function f x (3, _27) fx x Since f x 0 when x numbers are 0, 4, and 6. Interval 0 4 In Module 4.3B you can practice using graphical information about f to determine the shape of the graph of f . |||| Try reproducing the graph in Figure 12 with a graphing calculator or computer. Some machines produce the complete graph, some produce only the portion to the right of the y-axis, and some produce only the portion between x 0 and x 6. For an explanation and cure, see Example 7 in Section 1.4. An equivalent expression that gives the correct graph is x2 13 6 6 x x 6 x 13 . SOLUTION You can use the differentiation rules to check that the first two derivatives are FIGURE 11 y x x x x x 4 x 13 4 6 x x 8 fx 23 x 43 4 and f x does not exist when x x1 3 6 x 23 fx 0 4 6 6 x 53 0 or x 6, the critical f decreasing on ( , 0) increasing on (0, 4) decreasing on (4, 6) decreasing on (6, ) To find the local extreme values we use the First Derivative Test. Since f changes from negative to positive at 0, f 0 0 is a local minimum. Since f changes from positive to negative at 4, f 4 2 5 3 is a local maximum. The sign of f does not change at 6, so there is no minimum or maximum there. (The Second Derivative Test could be used at 4, but not at 0 or 6 since f does not exist at either of these numbers.) Looking at the expression for f x and noting that x 4 3 0 for all x, we have fx 0 for x 0 and for 0 x 6 and f x 0 for x 6. So f is concave downward on , 0 and 0, 6 and concave upward on 6, , and the only inflection point is 6, 0 . The graph is sketched in Figure 12. Note that the curve has vertical tangents at 0, 0 and 6, 0 because f x l as x l 0 and as x l 6. y 4 (4, 2%?# ) 3 2 13 0 FIGURE 12 6 1 2 3 4 5 y=x @ ?#(6-x)! ?# 7x 5E-04(pp 242-251) 1/17/06 2:43 PM Page 247 S ECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH |||| 4.3 1–2 (a) (b) (c) (d) (e) 1. ❙❙❙❙ 247 Exercises |||| Use the given graph of f to find the following. The largest open intervals on which f is increasing. The largest open intervals on which f is decreasing. The largest open intervals on which f is concave upward. The largest open intervals on which f is concave downward. The coordinates of the points of inflection. 7. The graph of the second derivative f of a function f is shown. State the x-coordinates of the inflection points of f . Give reasons for your answers. y y=f·(x) y 0 12 4 6 x 8 4 8. The graph of the first derivative f of a function f is shown. (a) On what intervals is f increasing? Explain. (b) At what values of x does f have a local maximum or minimum? Explain. (c) On what intervals is f concave upward or concave downward? Explain. (d) What are the x-coordinates of the inflection points of f ? Why? 2 0 2. 4 2 6 x 8 y y y=fª(x) 2 0 0 2 4 1 8x 6 3 5 7 9 x _2 9. Sketch the graph of a function whose first and second deriva- tives are always negative. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 10. A graph of a population of yeast cells in a new laboratory cul3. Suppose you are given a formula for a function f . (a) How do you determine where f is increasing or decreasing? (b) How do you determine where the graph of f is concave upward or concave downward? (c) How do you locate inflection points? ture 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 inflection point. 4. (a) State the First Derivative Test. (b) State the Second Derivative Test. Under what circumstances is it inconclusive? What do you do if it fails? 700 600 Number 500 400 of yeast cells 300 200 100 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 x does f have a local maximum or minimum? 5. 6. y y=fª(x) 0 y y=fª(x) ■ ■ 2 ■ 4 ■ 6 ■ x ■ 0 ■ 2 ■ ■ 4 ■ 6 ■ 4 6 8 10 12 14 16 18 Time (in hours) 11–16 0 2 |||| (a) Find the intervals on which f is increasing or decreasing. (b) Find the local maximum and minimum values of f . (c) Find the intervals of concavity and the inflection points. x ■ 11. f x x3 12 x 1 12. f x 5 3x 2 x3 5E-04(pp 242-251) ❙❙❙❙ 248 1/17/06 Page 248 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 13. f x x4 15. f x x 16. f x cos2 x ■ 2:44 PM ■ 2x2 0 x 2 sin x, 0 x x 2 27. y 3 y=fª(x) 3 2 sin x, ■ x2 14. f x 3 ■ ■ ■ 2 2 ■ ■ ■ ■ ■ ■ 0 17–19 |||| Find the local maximum and minimum values of f using both the First and Second Derivative Tests. Which method do you prefer? 17. f x x5 5x 3 19. f x x s1 28. ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 0 0 and f 2 0 and f 6 0 for all x 1, 0 if x 1 or x f2 0 if x 0 if 0 0 if 1 fx fx fx 24. f 1 . 5, what can you say about f ? 0, what can you say about f ? f 29–40 (a) (b) (c) (d) f f f f ■ 27–28 x x x x ■ 0, 2, lim f x f0 f2 0 if 0 x 0 if 2 x 0 if 0 x 0 if 1 x ■ f4 2 or 4 4 or x 1 or 3 3 or x ■ 0 if x fx , xl2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1 or x 29. f x ■ x4 3x 5 35. A x x sx 37. C x x 13 39. f 3 2x 3 33. h x 2 cos 40. f t 2, ■ 0 if x 2 t ■ ; 41–42 3x 2 30. f x 3 3 x cos 2 , ■ x 38. B x 4 cos t, ■ x2 36. G x 5x 3 200 34. h x 6x 2 2 32. t x 12 x 0 2 t ■ ■ x3 3x 8x 3 1 x4 3 4 sx 3x 23 x 2 2 ■ ■ ■ ■ ■ ■ |||| (a) Use a graph of f to estimate the maximum and minimum values. Then find the exact values. (b) Estimate the value of x at which f increases most rapidly. Then find the exact value. 0, f6 x 6, 6, x 5, 5, f x ■ |||| Find the intervals of increase or decrease. Find the local maximum and minimum values. Find the intervals of concavity and the inflection points. Use the information from parts (a)–(c) to sketch the graph. Check your work with a graphing device if you have one. 31. f x fx ■ |||| f4 0, 0 or 2 x 4, x 2 or x 4, x 3, f x 0 if x 1, x3 fx 0 if x 1, 1 if x 2, 2, f x 0, inflection point 0, 1 2 26. f 0 vertical asymptote x 3, f x 0 if 1 0 if x 25. f x ■ , f 1 0, x 0 if 1 0 if 2 x fx fx 2 _2 22–26 |||| Sketch the graph of a function that satisfies all of the given conditions. 23. f 0 8x ■ ■ 21. Suppose f is continuous on fx 6 2 x 4 x 1 3. (b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers? (c) What does the First Derivative Test tell you? 22. f x 4 y=fª(x) 4 20. (a) Find the critical numbers of f x (a) If f 2 (b) If f 6 8x y x x2 6 _2 x 18. f x 4 2 fx ■ ■ ■ ■ 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 x does f have a local maximum or minimum? (c) On what intervals is f concave upward or downward? (d) State the x-coordinate(s) of the point(s) of inflection. (e) Assuming that f 0 0, sketch a graph of f. 41. f x x1 sx 2 1 42. f x x ■ ; 43–44 ■ 2 cos x, ■ ■ 0 ■ x ■ 2 ■ ■ ■ ■ ■ |||| (a) Use a graph of f to give a rough estimate of the intervals of concavity and the coordinates of the points of inflection. (b) Use a graph of f to give better estimates. 43. f x cos x 1 2 cos 2 x, 0 x 2 ■ 5E-04(pp 242-251) 1/17/06 2:44 PM Page 249 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 44. f x ■ C AS ■ x3 x 2 ■ 4 ■ ■ x3 ■ 10x sx ■ 2 249 53. Suppose f is differentiable on an interval I and f x ■ ■ ■ ■ ■ ■ ■ 0 for all numbers x in I except for a single number c. Prove that f is increasing on the entire interval I . ■ 45–46 |||| Estimate the intervals of concavity to one decimal place by using a computer algebra system to compute and graph f . 45. f x ❙❙❙❙ 5 4 ■ x x3 46. f x ■ ■ ■ ■ ■ 1 3 x2 1 x2 ■ 54–56 |||| Assume that all of the functions are twice differentiable and the second derivatives are never 0. 5 4 ■ 54. (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 x f x 2 is concave upward on I . ■ 47. Let K t be a measure of the knowledge you gain by studying 55. (a) If f and t are positive, increasing, concave upward func- K7 for a test for t hours. Which do you think is larger, K 8 or K 3 K 2 ? Is the graph of K concave upward or concave downward? Why? tions on I , show that the product function f t 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, by giving three examples, that f t may be concave upward, concave downward, or linear. Why doesn’t the argument in parts (a) and (b) work in this case? 48. Coffee is being poured into the mug shown in the figure 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 concavity. What is the significance of the inflection point? 56. Suppose f and t are both concave upward on , . Under what condition on f will the composite function h x f tx be concave upward? ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 57. Show that a cubic function (a third-degree polynomial) always has exactly one point of inflection. If its graph has three x-intercepts x 1, x 2, and x 3, show that the x-coordinate of the inflection point is x 1 x 2 x 3 3. ; 58. For what values of c does the polynomial 49. Show that tan x fx tan x x for 0 x 2. [Hint: Show that x is increasing on 0, 2 .] 50. Prove that, for all x 1, 2 sx 59. Prove that if c, f c is a point of inflection of the graph of f 3 1 x ax 3 b x 2 c x d that has a local maximum value of 3 at 2 and a local minimum value of 0 at 1. 51. Find a cubic function f x 52. For what values of a and b does the function fx x3 ax 2 have a local maximum when x when x 1? |||| 4.4 Px x 4 c x 3 x 2 have two inflection points? One inflection point? None? Illustrate by graphing P for several values of c. How does the graph change as c decreases? bx 2 3 and a local minimum 0. and f exists in an open interval that contains c, then f c [Hint: Apply the First Derivative Test and Fermat’s Theorem to the function t f .] x 4, then f 0 inflection point of the graph of f . 60. Show that if f x 61. Show that the function t x 0, but 0, 0 is not an x x has an inflection point at 0, 0 but t 0 does not exist. 62. Suppose that f is continuous and f c fc 0, but fc 0. Does f have a local maximum or minimum at c ? Does f have a point of inflection at c ? Limits at Infinity; Horizontal Asymptotes In Sections 2.2 and 2.4 we investigated infinite limits and vertical asymptotes. There we let x approach a number and the result was that the values of y became arbitrarily large (positive or negative). In this section we let x become arbitrarily large (positive or negative) and see what happens to y. We will find it very useful to consider this so-called end behavior when sketching graphs. 5E-04(pp 242-251) 250 ❙❙❙❙ 1/17/06 2:44 PM Page 250 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION x 0 1 2 3 4 5 10 50 100 1000 Let’s begin by investigating the behavior of the function f defined by fx 1 0 0.600000 0.800000 0.882353 0.923077 0.980198 0.999200 0.999800 0.999998 x2 x2 fx 1 1 as x becomes large. The table at the left gives values of this function correct to six decimal places, and the graph of f has been drawn by a computer in Figure 1. y y=1 0 1 y= x ≈-1 ≈+1 FIGURE 1 As x grows larger and larger you can see that the values of f x get closer and closer to 1. In fact, it seems that we can make the values of f x as close as we like to 1 by taking x sufficiently large. This situation is expressed symbolically by writing lim xl x2 x2 1 1 1 In general, we use the notation lim f x xl L to indicate that the values of f x become closer and closer to L as x becomes larger and larger. 1 Definition Let f be a function defined on some interval a, lim f x xl . Then L means that the values of f x can be made arbitrarily close to L by taking x sufficiently large. Another notation for lim x l f x L is f x lL as xl The symbol does not represent a number. Nonetheless, the expression lim f x xl often read as L is “the limit of f x , as x approaches infinity, is L” or “the limit of f x , as x becomes infinite, is L” or “the limit of f x , as x increases without bound, is L” The meaning of such phrases is given by Definition 1. A more precise definition, similar to the , definition of Section 2.4, is given at the end of this section. 5E-04(pp 242-251) 1/17/06 2:44 PM Page 251 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ❙❙❙❙ 251 Geometric illustrations of Definition 1 are shown in Figure 2. Notice that there are many ways for the graph of f to approach the line y L (which is called a horizontal asymptote) as we look to the far right of each graph. y y y=L y y=ƒ y=L y=ƒ y=ƒ y=L 0 0 x 0 x x FIGURE 2 Examples illustrating lim ƒ=L x` Referring back to Figure 1, we see that for numerically large negative values of x, the values of f x are close to 1. By letting x decrease through negative values without bound, we can make f x as close as we like to 1. This is expressed by writing lim xl x2 x2 1 1 1 The general definition is as follows. 2 Definition Let f be a function defined on some interval lim f x , a . Then L xl means that the values of f x can be made arbitrarily close to L by taking x sufficiently large negative. Again, the symbol is often read as y y=ƒ does not represent a number, but the expression lim f x xl L “the limit of f x , as x approaches negative infinity, is L” Definition 2 is illustrated in Figure 3. Notice that the graph approaches the line y we look to the far left of each graph. y=L 0 L as x 3 y y Definition The line y L is called a horizontal asymptote of the curve f x if either lim f x y=ƒ xl y=L 0 x L x _` lim f x xl For instance, the curve illustrated in Figure 1 has the line y tote because FIGURE 3 Examples illustrating lim ƒ=L or lim xl x2 x2 1 1 1 L 1 as a horizontal asymp- 5E-04(pp 252-261) 252 ❙❙❙❙ 1/17/06 3:26 PM Page 252 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION The curve y f x sketched in Figure 4 has both y 1 and y 2 as horizontal asymptotes because lim f x 1 and lim f x 2 xl xl y 2 y=2 0 y=_1 y=ƒ x _1 F IGURE 4 y EXAMPLE 1 Find the infinite limits, limits at infinity, and asymptotes for the function f whose graph is shown in Figure 5. SOLUTION We see that the values of f x become large as x l 1 from both sides, so 2 lim f x xl 1 0 x 2 Notice that f x becomes large negative as x approaches 2 from the left, but large positive as x approaches 2 from the right. So lim f x and x l2 F IGURE 5 lim f x x l2 Thus, both of the lines x 1 and x 2 are vertical asymptotes. As x becomes large, it appears that f x approaches 4. But as x decreases through negative values, f x approaches 2. So lim f x xl This means that both y EXAMPLE 2 Find lim xl 4 4 and y and lim f x xl 2 2 are horizontal asymptotes. 1 1 and lim . xl x x SOLUTION Observe that when x is large, 1 x is small. For instance, 1 100 y y=Δ 0 FIGURE 6 1 1 lim =0, lim =0 x `x x _` x x 0.01 1 10,000 0.0001 1 1,000,000 0.000001 In fact, by taking x large enough, we can make 1 x as close to 0 as we please. Therefore, according to Definition 1, we have 1 lim 0 xl x Similar reasoning shows that when x is large negative, 1 x is small negative, so we also have 1 lim 0 xl x It follows that the line y 0 (the x-axis) is a horizontal asymptote of the curve y (This is an equilateral hyperbola; see Figure 6.) 1 x. 5E-04(pp 252-261) 1/17/06 3:26 PM Page 253 SECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ❙❙❙❙ 253 Most of the Limit Laws that were given in Section 2.3 also hold for limits at infinity. It can be proved that the Limit Laws listed in Section 2.3 (with the exception of Laws 9 and 10) are also valid if “x l a” is replaced by “x l ” or “ x l .” In particular, if we combine Laws 6 and 11 with the results of Example 2, we obtain the following important rule for calculating limits. 4 Theorem If r 0 is a rational number, then 1 xr lim xl If r 0 0 is a rational number such that x r is defined for all x, then 1 xr lim xl 0 EXAMPLE 3 Evaluate lim xl 3x 2 5x 2 x 4x 2 1 and indicate which properties of limits are used at each stage. SOLUTION As x becomes large, both numerator and denominator become large, so it isn’t obvious what happens to their ratio. We need to do some preliminary algebra. To evaluate the limit at infinity of any rational function, we first divide both the numerator and denominator by the highest power of x that occurs in the denominator. (We may assume that x 0, since we are interested only in large values of x.) In this case the highest power of x in the denominator is x 2, so we have 3x 2 x 4x 2 1 2 1 x 2 x2 lim 5 3x lim x l 5x 2 x x2 4x x2 lim 3 2 4 x 1 x2 lim xl 5x 2 xl y xl y=0.6 0 x lim 3 lim xl 4 lim 3 5 0 0 xl 0 0 xl lim 1 1 x xl lim 5 1 3 xl 5 1 x 4 x 2 x2 1 x2 (by Limit Law 5) 1 x2 1 lim xl x2 2 lim xl 1 x (by 1, 2, and 3) (by 7 and Theorem 4) 3 5 FIGURE 7 y= 3≈-x-2 5≈+4x+1 A similar calculation shows that the limit as x l is also 3 . Figure 7 illustrates the 5 results of these calculations by showing how the graph of the given rational function approaches the horizontal asymptote y 3 . 5 5E-04(pp 252-261) 254 ❙❙❙❙ 1/17/06 3:27 PM Page 254 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION E XAMPLE 4 Find the horizontal and vertical asymptotes of the graph of the function s2x 2 1 3x 5 fx SOLUTION Dividing both numerator and denominator by x and using the properties of limits, we have lim xl s2x 3x 2 1 x2 2 1 lim 5 xl 3 lim 1 x2 2 xl (since sx 2 5 x lim 2 xl 5 x lim 3 xl s2 0 3 50 lim 3 xl x for x 0) 1 x2 1 5 lim xl x lim xl s2 3 Therefore, the line y s 2 3 is a horizontal asymptote of the graph of f . In computing the limit as x l , we must remember that for x 0, we have x x. So when we divide the numerator by x, for x 0 we get sx 2 1 s2x 2 x Therefore s2x 3x lim xl 1 s2x 2 sx 2 1 2 1 2 1 x2 2 1 lim 5 xl 3 y 5 x 3 2 1 x2 1 x2 lim xl 5 lim xl 1 x s2 3 Thus, the line y s 2 3 is also a horizontal asymptote. A vertical asymptote is likely to occur when the denominator, 3x 5, is 0, that is, when x 5 . If x is close to 5 and x 5 , then the denominator is close to 0 and 3x 5 3 3 3 is positive. The numerator s 2 x 2 1 is always positive, so f x is positive. Therefore œ„ 2 y= 3 x œ„ 2 y=_ 3 lim xl 5 3 x= 5 3 If x is close to 5 but x 3 5 3 , then 3x lim FIG URE 8 œ„„„„„„ 2≈+1 y= 3x-5 xl 5 3 The vertical asymptote is x 5 3 5 s2x 2 1 3x 5 0 and so f x is large negative. Thus s2x 2 1 3x 5 . All three asymptotes are shown in Figure 8. 5E-04(pp 252-261) 1/17/06 3:28 PM Page 255 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ( EXAMPLE 5 Compute lim s x 2 xl ❙❙❙❙ 255 x). 1 SOLUTION Because both sx 2 |||| We can think of the given function as having a denominator of 1. 1 and x are large when x is large, it’s difficult to see what happens to their difference, so we use algebra to rewrite the function. We first multiply numerator and denominator by the conjugate radical: lim (s x 2 x) 1 xl lim (s x 2 xl x2 1 sx 2 1 lim xl x) 1 sx 2 sx 2 x2 x lim xl 1 1 x x 1 sx 2 1 x The Squeeze Theorem could be used to show that this limit is 0. But an easier method is to divide numerator and denominator by x. Doing this and using the Limit Laws, we obtain lim (s x 2 1 xl x) lim xl 1 sx 1 2 y lim y=œ„„„„„-x ≈+1 xl 1 x sx 2 1 1 0 x x lim xl 1 x 1 0 0 x FIGURE 9 s1 1 x2 1 0 Figure 9 illustrates this result. EXAMPLE 6 Evaluate lim sin xl |||| The problem-solving strategy for Example 6 is introducing something extra (see page 58). Here, the something extra, the auxiliary aid, is the new variable t. 1 1 x SOLUTION If we let t 1 . x 1 x, then t l 0 as x l . Therefore lim sin xl 1 x lim sin t tl0 0 (See Exercise 69.) EXAMPLE 7 Evaluate lim sin x. xl SOLUTION As x increases, the values of sin x oscillate between 1 and 1 infinitely often and so they don’t approach any definite number. Thus, lim x l sin x does not exist. Infinite Limits at Infinity The notation lim f x xl is used to indicate that the values of f x become large as x becomes large. Similar meanings are attached to the following symbols: lim f x xl lim f x xl lim f x xl 5E-04(pp 252-261) 256 ❙❙❙❙ 1/17/06 3:28 PM Page 256 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION E XAMPLE 8 Find lim x 3 and lim x 3. xl y xl SOLUTION When x becomes large, x 3 also becomes large. For instance, y=˛ 0 10 3 x 100 3 1000 1000 3 1,000,000 1,000,000,000 In fact, we can make x 3 as big as we like by taking x large enough. Therefore, we can write lim x 3 xl Similarly, when x is large negative, so is x 3. Thus lim x 3 xl FIGURE 10 lim x #=`, lim x #=_` x` x _` x 3 in Figure 10. These limit statements can also be seen from the graph of y EXAMPLE 9 Find lim x 2 x. xl | SOLUTION Note that we cannot write lim x 2 lim x 2 x xl lim x xl xl The Limit Laws can’t be applied to infinite limits because can’t be defined). However, we can write lim x 2 xl because both x and x EXAMPLE 10 Find lim xl x lim x x is not a number ( 1 xl 1 become arbitrarily large and so their product does too. x2 3 x . x SOLUTION As in Example 3, we divide the numerator and denominator by the highest power of x in the denominator, which is just x : lim xl because x 1l and 3 x x2 3 1l x x lim xl x 3 x 1 1 1 as x l . The next example shows that by using infinite limits at infinity, together with intercepts, we can get a rough idea of the graph of a polynomial even without computing derivatives. EXAMPLE 11 Sketch the graph of y and its limits as x l and as x l SOLUTION The y-intercept is f 0 found by setting y 0: x 2, x 2 4 x 1 3 x 1 by finding its intercepts . 2413 1 16 and the x-intercepts are 1, 1. Notice that since x 2 4 is positive, the function 5E-04(pp 252-261) 1/17/06 3:28 PM Page 257 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ❙❙❙❙ 257 doesn’t change sign at 2; thus, the graph doesn’t cross the x-axis at 2. The graph crosses the axis at 1 and 1. When x is large positive, all three factors are large, so lim x 2 xl 4 x 1 3 x 1 When x is large negative, the first factor is large positive and the second and third factors are both large negative, so lim x xl 2 4 x 1 3 x 1 Combining this information, we give a rough sketch of the graph in Figure 11. The use of derivatives would enable us to sketch a more accurate graph by giving the precise location of maximum and minimum points and inflection points, but for this particular function the computations would be extremely complex. y 0 _1 1 2 x y=(x-2)$ (x +1)#(x-1) _16 FIGURE 11 Precise Definitions Definition 1 can be stated precisely as follows. 5 Definition Let f be a function defined on some interval a, lim f x xl means that for every . Then L 0 there is a corresponding number N such that fx L whenever x N In words, this says that the values of f x can be made arbitrarily close to L (within a distance , where is any positive number) by taking x sufficiently large (larger than N , where N depends on ). Graphically it says that by choosing x large enough (larger than some number N ) we can make the graph of f lie between the given horizontal lines 5E-04(pp 252-261) 258 ❙❙❙❙ 1/17/06 3:29 PM Page 258 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION yL and y L as in Figure 12. This must be true no matter how small we choose . Figure 13 shows that if a smaller value of is chosen, then a larger value of N may be required. y y=ƒ y=L +∑ ∑ L∑ y=L -∑ ƒ is in here 0 x N FIGURE 12 lim ƒ=L when x is in here x` y y=ƒ y=L+∑ L y=L-∑ 0 FIGURE 13 x N lim ƒ=L x` Similarly, a precise version of Definition 2 is given by Definition 6, which is illustrated in Figure 14. 6 Definition Let f be a function defined on some interval lim f x xl means that for every , a . Then L 0 there is a corresponding number N such that fx L whenever x N y y=ƒ y=L+∑ L y=L-∑ FIGURE 14 x 0 N lim ƒ=L x _` In Example 3 we calculated that lim xl 3x 2 5x 2 x 4x 2 1 3 5 In the next example we use a graphing device to relate this statement to Definition 5 with L 3 and 0.1. 5 5E-04(pp 252-261) 1/17/06 3:29 PM Page 259 SECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ❙❙❙❙ 259 E XAMPLE 12 Use a graph to find a number N such that 3x 2 5x 2 x 4x 2 1 0.6 0.1 whenever x N S OLUTION We rewrite the given inequality as 3x 2 5x 2 0.5 2 1 0.7 We need to determine the values of x for which the given curve lies between the horizontal lines y 0.5 and y 0.7. So we graph the curve and these lines in Figure 15. Then we use the cursor to estimate that the curve crosses the line y 0.5 when x 6.7. To the right of this number the curve stays between the lines y 0.5 and y 0.7. Rounding to be safe, we can say that 1 y=0.7 y=0.5 y= x 4x 3≈-x-2 5≈+4x+1 0 FIGURE 15 3x 2 5x 2 15 In other words, for x 4x 2 1 0.6 0.1 whenever 0.1 we can choose N E XAMPLE 13 Use Definition 5 to prove that lim xl x 7 7 (or any larger number) in Definition 5. 1 x 0. SOLUTION 1. Preliminary analysis of the problem ( guessing a value for N ). Given 0, we want to find N such that 1 x 0 whenever In computing the limit we may assume x 1 x x N 0, in which case 1 x 0 1 x Therefore, we want 1 x that is, whenever N whenever 1 x x x N This suggests that we should take N 1 . 2. Proof (showing that this N works). Given Then 1 x Thus 1 x 0 0 1 x 0, we choose N 1 x whenever 1 N x N 1 . Let x N. 5E-04(pp 252-261) 260 ❙❙❙❙ 1/17/06 3:30 PM Page 260 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Therefore, by Definition 5, lim xl 1 x 0 Figure 16 illustrates the proof by showing some values of of N . y y and the corresponding values y ∑=1 ∑=0.2 0 x N=1 0 ∑=0.1 N=5 x 0 N=10 x FIGURE 16 Finally we note that an infinite limit at infinity can be defined as follows. The geometric illustration is given in Figure 17. y 7 y=M M Definition Let f be a function defined on some interval a, . Then lim f x xl 0 N means that for every positive number M there is a corresponding positive number N such that fx M whenever xN x FIGURE 17 lim ƒ=` Similar definitions apply when the symbol x` |||| 4.4 is replaced by . (See Exercise 70.) Exercises 1. Explain in your own words the meaning of each of the following. (a) lim f x xl 5 (b) lim f x xl (f) The equations of the asymptotes y 3 2. (a) Can the graph of y f x intersect a vertical asymptote? Can it intersect a horizontal asymptote? Illustrate by sketching graphs. (b) How many horizontal asymptotes can the graph of y f x have? Sketch graphs to illustrate the possibilities. 1 x 1 3. For the function f whose graph is given, state the following. (a) lim f x (b) lim f x (c) lim f x (d) lim f x x l2 xl 1 (e) lim f x xl xl 1 xl 4. For the function t whose graph is given, state the following. (a) lim t x (b) lim t x (c) lim t x (d) lim t x xl x l3 xl x l0 5E-04(pp 252-261) 1/17/06 3:31 PM Page 261 S ECTION 4.4 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES 23. lim (sx 2 (e) lim t x xl 2 (f) The equations of the asymptotes sx 2 ax xl bx ) xl 3 26. lim sx xl xl y 27. lim ( x sx ) xl 29. lim x 4 x 2 31. lim xl ■ 2 x 1 7–8 |||| Evaluate the limit and justify each step by indicating the appropriate properties of limits. ■ ■ 9–32 |||| 9. lim xl 11. lim ■ xl 15. lim ul xl ■ ■ ■ xl x2 7 x 2x 2 x3 2x 2 yl 5x 3 12. lim u2 x 14. lim 4 4u 4 5 2 2u 2 tl 1 s9x 6 x xl x3 1 x 19. lim 2 xl 4 sx 17. lim 21. lim (s9x 2 xl ■ 10. lim 3 1 x ■ ■ ■ ■ ■ 1 x ■ ■ ■ ■ ■ ■ ■ x) 1 ; 35–40 |||| Find the horizontal and vertical asymptotes of each curve. Check your work by graphing the curve and estimating the asymptotes. ■ 36. y x x 4 x3 3x 2 10 x 4 sx 4 ■ ■ 4 1 x3 x3 1 x x9 s4x 2 3x 40. F x 1 ■ x2 x2 38. y x 35. y 39. h x ■ x ■ ■ ■ ■ ■ ■ 2 ■ ■ 41. Find a formula for a function f that satisfies the following 1 2x xl s3x 2 8 x 6 s3x 2 3x 1 to estimate the value of lim x l f x to one decimal place. (b) Use a table of values of f x to estimate the limit to four decimal places. (c) Find the exact value of the limit. 37. y 12 x 3 5x 2 1 4x 2 3x 3 8. lim Find the limit. xl 13. lim 4 8 ■ 32. lim x sin ; 34. (a) Use a graph of f x x to estimate the value of lim x l f x correct to two decimal places. (b) Use a table of values of f x to estimate the limit to four decimal places. x 5x x5 x4 by graphing the function f x sx 2 x 1 x. (b) Use a table of values of f x to guess the value of the limit. (c) Prove that your guess is correct. ; 6. (a) Use a graph of xl ■ x4 xl xl by evaluating the function f x x 2 2 x for x 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 50, and 100. Then use a graph of f to support your guess. 3x2 2x2 30. lim x 2 lim (sx 2 x2 lim x xl 2 7. lim ■ 2x 3 2x2 5 xl x3 x2 x 1 x3 ; 33. (a) Estimate the value of ; 5. Guess the value of the limit fx 28. lim x5 xl 0 261 24. lim cos x 25. lim s x 1 ❙❙❙❙ 16. lim xl 18. lim xl 3x x t2 t 3 lim f x t 22. lim ( x 1 and x x 1 1 1 43. y x 2) f2 0, lim f x x l3 3 and horizontal asymptote y 2x ) ■ x x 1. x ■ 2 x 46. y 1 ■ ■ ■ ■ ■ 2x 2 x2 1 1 44. y x 45. y sx 2 , x l0 43–46 |||| Find the horizontal asymptotes of the curve and use them, together with concavity and intervals of increase and decrease, to sketch the curve. s9x 6 x x3 1 6x 2 lim f x 42. Find a formula for a function that has vertical asymptotes 2 2 , x l3 x2 9x 2 1 s xl xl xl 2 3y 2 5y 2 4y 20. lim (sx 4 3x) conditions: lim f x 0, 5 4 sx ■ ■ 2 1 ■ ■ ■ 5E-04(pp 262-271) ❙❙❙❙ 262 1/17/06 2:48 PM Page 262 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 47–50 |||| Find the limits as x l and as x l . Use this information, together with intercepts, to give a rough sketch of the graph as in Example 11. 47. y x2 x 48. y 2 x31 49. y x 4 50. y ■ 1 ■ 21 5 x xx ■ ■ Then use these sketches to find the following limits. (a) lim x n (b) lim x n x l0 x fx fx fx fx 52. f 2 fx fx fx 4 3 2 ■ 4x x 5 2 ■ 0, f 2 0 if 0 x 0 if 0 x 0 if 1 x f x for all ■ ■ ■ ■ ■ ■ 1, f 0 2, f x 1 or if x 4, lim x l x lim x l ■ ■ ■ ■ ■ ■ ■ (b) What happens to the concentration as t l ? ; 61. Use a graph to find a number N such that 6 x 2 5x 3 2x 2 1 , 0, x ■ ■ xl ■ values as x l and as x l . (a) Describe and compare the end behavior of the functions Px 3x 5 5x 3 2 x and Q x 3x 5 by graphing both functions in the viewing rectangles 2, 2 by 2, 2 and 10, 10 by 10,000, 10,000 . (b) Two functions are said to have the same end behavior if their ratio approaches 1 as x l . Show that P and Q have the same end behavior. N 2 illustrate Definition 5 by finding values of N that correspond to 0.5 and 0.1. ; 63. For the limit lim xl s4x 2 1 x1 2 illustrate Definition 6 by finding values of N that correspond to 0.5 and 0.1. ; 64. For the limit lim xl 2x sx 1 1 illustrate Definition 7 by finding a value of N that corresponds to M 100. 0.0001? (b) Taking r 2 in Theorem 4, we have the statement lim x l 1 x 2 0. Prove this directly using Definition 5. Px Qx if the degree of P is (a) less than the degree of Q and (b) greater than the degree of Q. (ii) n (iv) n x 65. (a) How large do we have to take x so that 1 x 2 57. Let P and Q be polynomials. Find following five cases: (i) n 0 (iii) n 0, n even (v) n 0, n even whenever s4x 2 1 x1 lim ; 56. By the end behavior of a function we mean the behavior of its 58. Make a rough sketch of the curve y 0.2 ; 62. For the limit sin x . x (b) Graph f x sin x x. How many times does the graph cross the asymptote? lim 3 2 xl xl 30 t 200 t Ct 55. (a) Use the Squeeze Theorem to evaluate lim ; 5. 0 for x 0, lim x l t x , , lim x l 0 t x , lim x l 0 t x ■ 3x x2 30 g of salt per liter of water is pumped into the tank at a rate of 25 L min. Show that the concentration of salt after t minutes (in grams per liter) is 0, 0 if x 2, 4, fx 1, f1 0, lim x l 2 f x , lim x l 2 f x , lim x l f x , lim x l f x lim x l 0 f x fx 0 for x 2, f x 0 for x 0 and for 0 ■ for all x 4x 2 fx 60. (a) A tank contains 5000 L of pure water. Brine that contains 0, f 0 1, f x 0 if 0 x 2, 0 if x 2, f x 0 if 0 x 4, 0 if x 4, lim x l f x 0, f x for all x 0, t x tx 1 x 53. f 1 54. t 0 xl 59. Find lim x l f x if x 51–54 |||| Sketch the graph of a function that satisfies all of the given conditions. 51. f 2 (d) lim x n xl x3 3 x l0 (c) lim x n x n (n an integer) for the 0, n odd 0, n odd 66. (a) How large do we have to take x so that 1 s x (b) Taking r 1 2 0.0001? in Theorem 4, we have the statement lim xl 1 sx 0 Prove this directly using Definition 5. 67. Use Definition 6 to prove that lim x l 1x 68. Prove, using Definition 7, that lim x l x 3 0. . 5E-04(pp 262-271) 1/17/06 2:48 PM Page 263 S ECTION 4.5 SUMMARY OF CURVE SKETCHING 69. Prove that 263 70. Formulate a precise definition of lim f x lim f x lim f 1 t xl lim f x xl tl0 lim f 1 t and Then use your definition to prove that xl x3 lim 1 tl0 xl if these limits exist. |||| 4.5 ❙❙❙❙ if these limits exist. Summary of Curve Sketching So far we have been concerned with some particular aspects of curve sketching: domain, range, and symmetry in Chapter 1; limits, continuity, and asymptotes in Chapter 2; derivatives and tangents in Chapter 3; and extreme values, intervals of increase and decrease, concavity, points of inflection, and horizontal asymptotes 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 may 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 you have used a calculator to produce the table of values and corresponding points in Figure 1. y x fx x 22 7 2 4 2 3 1 2 3 4 5 6 20 fx 7 10 11 10 8 8 5 4 3 2 1 0 15 10 _5 _4 _3 _2 _1 5 0 1 2 3 4 5 6 x FIGURE 1 You might then join these points to produce the curve shown in Figure 2, but the cor- | rect graph might be the one shown in Figure 3. You can see the drawbacks of the method of plotting points. Certain essential features of the graph may be missed, such as the maximum and minimum values between 2 and 1 or between 2 and 5. If you just plot points, you don’t know when to stop. (How far should you plot to the left or right?) But the use of calculus ensures that all the important aspects of the curve are illustrated. y y 20 20 15 15 10 10 5 5 _2 _2 0 FIGURE 2 2 4 x 0 FIGURE 3 2 4 x 5E-04(pp 262-271) 264 ❙❙❙❙ 1/17/06 2:48 PM Page 264 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION You might respond: Yes, but what about graphing calculators and computers? Don’t they plot such a huge number of points that the sort of uncertainty demonstrated by Figures 2 and 3 is unlikely to happen? It’s true that modern technology is capable of producing very accurate graphs. But even the best graphing devices have to be used intelligently. We saw in Section 1.4 that it is extremely important to choose an appropriate viewing rectangle to avoid getting a misleading graph. (See especially Examples 1, 3, 4, and 5 in that section.) The use of calculus enables us to discover the most interesting aspects of graphs and in many cases to calculate maximum and minimum points and inflection points exactly instead of approximately. For instance, Figure 4 shows the graph of f x 8 x 3 21x 2 18 x 2. At first glance it seems reasonable: It has the same shape as cubic curves like y x 3, and it appears to have no maximum or minimum point. But if you compute the derivative, you will see that there is a maximum when x 0.75 and a minimum when x 1. Indeed, if we zoom in to this portion of the graph, we see that behavior exhibited in Figure 5. Without calculus, we could easily have overlooked it. 30 _2 y=8˛-21≈+18x+2 8 4 _10 F IGURE 4 y=8˛-21≈+18x+2 0 2 6 FIGURE 5 In the next section we will graph functions by using the interaction between calculus and graphing devices. In this section we draw graphs by first considering the following information. We don’t assume that you have a graphing device, but if you do have one you should use it as a check on your work. Guidelines for Sketching a Curve y 0 x (a) Even function: reflectional symmetry C. Symmetry y 0 (b) Odd function: rotational symmetry FIGURE 6 The following checklist is intended as a guide to sketching a curve y f x by hand. Not every item is relevant to every function. (For instance, a given curve might not have an asymptote or possess symmetry.) But the guidelines provide all the information you need to make a sketch that displays the most important aspects of the function. A. Domain It’s often useful to start by determining the domain D of f , that is, the set of values of x for which f x is defined. B. Intercepts The y-intercept is f 0 and this tells us where the curve intersects the y-axis. To find the x-intercepts, we set y 0 and solve for x. (You can omit this step if the equation is difficult to solve.) x (i) If f x f x for all x in D, that is, the equation of the curve is unchanged when x is replaced by x, then f is an even function and the curve is symmetric about the y-axis. This means that our work is cut in half. If we know what the curve looks like for x 0, then we need only reflect about the y-axis to obtain the complete curve [see Figure 6(a)]. Here are some examples: y x 2, y x 4, y x , and y cos x. (ii) If f x f x for all x in D, then f is an odd function and the curve is symmetric about the origin. Again we can obtain the complete curve if we know what 5E-04(pp 262-271) 1/17/06 2:48 PM Page 265 ❙❙❙❙ S ECTION 4.5 SUMMARY OF CURVE SKETCHING 265 it looks like for x 0. [Rotate 180° about the origin; see Figure 6(b).] Some simple examples of odd functions are y x, y x 3, y x 5, and y sin x. (iii) If f x p f x for all x 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, y sin x has period 2 and y tan x 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). y FIGURE 7 Periodic function: translational symmetry a-p 0 a a+p a+2p x D. Asymptotes (i) Horizontal Asymptotes. Recall from Section 4.4 that if either lim x l f x L or lim x l f x L, then the line y L is a horizontal asymptote of the curve y f x . If it turns out that lim x l f x (or ), then we do not have an asymptote to the right, but that is still useful information for sketching the curve. (ii) Vertical Asymptotes. Recall from Section 2.2 that the line x a is a vertical asymptote if at least one of the following statements is true: 1 lim f x xla lim f x xla E. F. G. In Module 4.5 you can practice using information about f , f , and asymptotes to determine the shape of the graph of f . H. lim f x xla lim f x xla (For rational functions you can locate the vertical asymptotes by equating the denominator to 0 after canceling any common factors. But for other functions this method does not apply.) Furthermore, in sketching the curve it is very useful to know exactly which of the statements in (1) is true. If f a is not defined but a is an endpoint of the domain of f , then you should compute lim x l a f x or lim x l a f x , whether or not this limit is infinite. (iii) Slant Asymptotes. These are discussed at the end of this section. Intervals of Increase or Decrease Use the I/D Test. Compute f x and find the intervals on which f x is positive ( f is increasing) and the intervals on which f x is negative ( f is decreasing). Local Maximum and Minimum Values Find the critical numbers of f [the numbers c where fc 0 or f c does not exist]. Then use the First Derivative Test. If f changes from positive to negative at a critical number c, then f c is a local maximum. If f changes from negative to positive at c, then f c is a local minimum. Although it is usually preferable to use the First Derivative Test, you can use the Second Derivative Test if c is a critical number such that f c 0. Then f c 0 implies that f c is a local minimum, whereas f c 0 implies that f c is a local maximum. Concavity and Points of Inflection Compute f x and use the Concavity Test. The curve is concave upward where f x 0 and concave downward where f x 0. Inflection points occur where the direction of concavity changes. Sketch the Curve Using the information in items A–G, draw the graph. Sketch the asymptotes as dashed lines. Plot the intercepts, maximum and minimum points, and inflection 5E-04(pp 262-271) 266 ❙❙❙❙ 1/17/06 2:48 PM Page 266 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION points. Then make the curve pass through these points, rising and falling according to E, with concavity according to G, and approaching the asymptotes. If additional accuracy is desired near any point, you can compute the value of the derivative there. The tangent indicates the direction in which the curve proceeds. 2x 2 EXAMPLE 1 Use the guidelines to sketch the curve y x 2 1 . A. The domain is x x2 1 0 xx 1 , 1 1, 1 1, B. The x- and y-intercepts are both 0. C. Since f x f x , the function f is even. The curve is symmetric about the y-axis. D. lim xl x 2 lim 1 xl 2 1 x2 1 2 Therefore, the line y 2 is a horizontal asymptote. Since the denominator is 0 when x 1, we compute the following limits: y y=2 lim x l1 0 x lim x=_1 2x 2 xl 1 x=1 2x 2 x2 lim 1 x l1 2x 2 x2 lim 1 xl 1 2x 2 x2 1 2x 2 x2 1 Therefore, the lines x 1 and x 1 are vertical asymptotes. This information about limits and asymptotes enables us to draw the preliminary sketch in Figure 8, showing the parts of the curve near the asymptotes. FIGURE 8 Preliminary sketch |||| We have shown the curve approaching its horizontal asymptote from above in Figure 8. This is confirmed by the intervals of increase and decrease. E. fx 4x x 2 2x 2 2x 1 x 2 1 4x x 1 2 2 2 Since f x 0 when x 0 x 1 and f x 0 when x 0 x 1 , f is increasing on , 1 and 1, 0 and decreasing on 0, 1 and 1, . F. The only critical number is x 0. Since f changes from positive to negative at 0, f0 0 is a local maximum by the First Derivative Test. y G. y=2 4 x2 fx Since 12 x 2 4 1 2 4x 2 x 2 x2 1 4 1 2x 12 x 2 4 x2 1 3 0 for all x, we have 0 x x=_1 x=1 FIGURE 9 Finished sketch of y= 2≈ ≈-1 fx 0 &? x2 1 0 &? x 1 and f x 0 &? x 1. Thus, the curve is concave upward on the intervals , 1 and 1, and concave downward on 1, 1 . It has no point of inflection since 1 and 1 are not in the domain of f . H. Using the information in E–G, we finish the sketch in Figure 9. 5E-04(pp 262-271) 1/17/06 2:49 PM Page 267 S ECTION 4.5 SUMMARY OF CURVE SKETCHING Domain xx 1 0 xx The x- and y-intercepts are both 0. Symmetry: None Since 1 lim xl 1, x2 sx 1 there is no horizontal asymptote. Since s x positive, we have lim xl 1 and so the line x E. G. y= FIGURE 10 0 1 l 0 as x l 1 and f x is always x2 sx 1 1 is a vertical asymptote. 2x sx fx 1 x 2 1 (2 s x x1 1) x 3x 4 2x 132 We see that f x 0 when x 0 (notice that 4 is not in the domain of f ), so the 3 only critical number is 0. Since f x 0 when 1 x 0 and f x 0 when x 0, f is decreasing on 1, 0 and increasing on 0, . F. Since f 0 0 and f changes from negative to positive at 0, f 0 0 is a local (and absolute) minimum by the First Derivative Test. y x=_1 267 x2 . sx 1 EXAMPLE 2 Sketch the graph of f x A. B. C. D. ❙❙❙❙ ≈ œ„„„„ x+1 x fx 2x 1 32 6x 4 4x 3x 2 13 4x 3 x 1 12 3x 2 8x 4x 1 8 52 Note that the denominator is always positive. The numerator is the quadratic 3x 2 8 x 8, which is always positive because its discriminant is b 2 4ac 32, which is negative, and the coefficient of x 2 is positive. Thus, f x 0 for all x in the domain of f , which means that f is concave upward on 1, and there is no point of inflection. H. The curve is sketched in Figure 10. EXAMPLE 3 Sketch the graph of f x A. The domain is . B. The y-intercept is f 0 2 cos x sin 2 x 2 cos x sin 2 x. 2. The x-intercepts occur when 2 cos x 2 sin x cos x 2 cos x 1 sin x 0 that is, when cos x 0 or sin x 1. Thus, in the interval 0, 2 , the x-intercepts are 2 and 3 2. C. f is neither even nor odd, but f x 2 f x for all x and so f is periodic and has period 2 . Thus, in what follows we need to consider only 0 x 2 and then extend the curve by translation in H. D. Asymptotes: None 5E-04(pp 262-271) ❙❙❙❙ 268 1/17/06 2:49 PM Page 268 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION E. fx 2 sin x 2 cos 2 x 2 2 sin2x 21 2 sin2x 2 2 sin x 1 sin x 2 sin x sin x 1 1 Thus, f x 0 when sin x 1 or sin x 1, so in 0, 2 we have x 6, 2 5 6, and 3 2. In determining the sign of f x in the following chart, we use the fact that sin x 1 0 for all x. Interval f fx increasing on 0, 0 x 6 x 5 6 6 decreasing on 6 5 6 x 3 2 increasing on 5 6, 3 3 2 x 2 increasing on 3 2, 2 6, 5 6 2 F. From the chart in E the First Derivative Test says that f 6 3 s3 2 is a local maximum and f 5 6 3 s3 2 is a local minimum, but f has no maximum or minimum at 3 2, only a horizontal tangent. fx G. 2 cos x 4 sin 2 x 2 cos x 1 4 sin x 1 Thus, f x 0 when cos x 0 (so x 2 or 3 2) and when sin x 4. From Figure 11 we see that there are two values of x between 0 and 2 for which 1 sin x x 0 on 2, 1 and 3 2, 2 , 2. Then f 4 . Let’s call them 1 and so f is concave upward there. Also f x 0 on 0, 2 , 1, 3 2 , and 2 , 2 , so f is concave downward there. Inflection points occur when x 2, 1, 3 2, and 2. y y=à x arcsin 1 4 1 2 2 arcsin 1 4 å¡ 0 å™ 1 _4 2π x F IGURE 11 H. The graph of the function restricted to 0 x 2 is shown in Figure 12. Then it is extended, using periodicity, to the complete graph in Figure 13. y y π ”6 , 3œ3 „ 2 ’ 2 2 å¡ 0 π 6 π 2 5π π 6 3π 2 ” FIGURE 12 å™ 2π x _ 3π 2 _π 2 0 π 2 3π 2 5π 2 y=2 Ł x+à 2x 5π 3œ3 „ , _ 2 ’ 6 FIGURE 13 7π 2 9π 2 x 5E-04(pp 262-271) 1/17/06 2:50 PM Page 269 ❙❙❙❙ S ECTION 4.5 SUMMARY OF CURVE SKETCHING 269 Slant Asymptotes y Some curves have asymptotes that are oblique, that is, neither horizontal nor vertical. If y=ƒ lim f x mx xl ƒ-(mx+b) 0 then the line y m x b is called a slant asymptote because the vertical distance between the curve y f x and the line y m x b approaches 0, as in Figure 14. (A similar situation exists if we let x l .) For rational functions, slant asymptotes 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 asymptote can be found by long division as in the following example. y=mx+b 0 b x FIGURE 14 x3 EXAMPLE 4 Sketch the graph of f x A. B. C. D. x 2 1 . ,. The domain is The x- and y-intercepts are both 0. f x , f is odd and its graph is symmetric about the origin. Since f x Since x 2 1 is never 0, there is no vertical asymptote. Since f x l as x l and fxl as x l , there is no horizontal asymptote. But long division gives x3 fx fx So the line y x 2 x x 1 x 2 1 1 x x x x2 1 1 x2 1 l0 as xl x is a slant asymptote. E. 3x 2 x 2 fx Since f x F. Although f 0 x 3 2x 1 x 2 1 x2 x2 3 x2 1 2 2 0 for all x (except 0), f is increasing on ,. 0, f does not change sign at 0, so there is no local maximum or minimum. G. y ˛ y= ≈+1 ”œ„ , 3 ”_œ„ , _ 3 3œ3 „ ’ 4 inflection points 0 when x Interval x x 0 6x x 2 1 2 x Since f x 3œ3 „ ’ 4 x 4x 3 fx 2 0 or x 3 x4 1 3x 2 2 x2 1 2x 4 2x 3 x2 s3, we set up the following chart: x2 x2 1 3 f fx CU on ( s3 s3 ) , s3 x 0 CD on ( s3, 0) 0 x s3 CU on (0, s3 ) x s3 CD on (s3, y=x The points of inflection are ( s3, FIGURE 15 x2 13 3 s3 4), 0, 0 , and (s3, 3 s3 4). H. The graph of f is sketched in Figure 15. ) 5E-04(pp 262-271) ❙❙❙❙ 270 1/17/06 |||| 1. y Exercises 3 9x x 7. y 2x5 5x2 x x1 1 x2 9 x x2 9 x1 x2 13. y 15. y 4x x 3 1 x 2 2 20 x 3 3 x 5 x x 12 x x2 9 x2 x2 9 x2 2 x4 14. y 16. y x 21. y sx 2 1 x2 27. y x x 3x1 29. y x 3 s x2 y L 42. Coulomb’s Law states that the force of attraction between two charged particles is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The figure shows particles with charge 1 located at positions 0 and 2 on a coordinate line and a particle with charge 1 at a position x between them. It follows from Coulomb’s Law that the net force acting on the middle particle is k k Fx 0x2 x2 x 22 1 1 x x 5 xs2 x2 where k is a positive constant. Sketch the graph of the net force function. What does the graph say about the force? +1 sx 2 x5 3 28. y 3 sin x 32. y sin x 33. y 35. y 36. y cos x 37. y sin 2 x ■ ■ 2 sin x, x 1 sin x x x x 2 3 ■ 38. y 2 sin x sin x cos x ■ sin x 40. y ■ ■ ■ 2 ■ x cos x sin x ■ 2x 3 44. y ■ ■ ■ WL 3 x 12EI W L2 2 x 24EI 2x 2 5x 2x 1 49. xy ■ ■ walls. If a constant load W is distributed evenly along its length, the beam takes the shape of the deflection curve W x4 24EI 1 1 ■ 47. y 41. The figure shows a beam of length L embedded in concrete y x x 5x 4 x3 46. y ■ ■ x2 ■ 2 x 2x x2 x2 ■ 3 x 2 ■ ■ ■ 47–52 |||| Use the guidelines of this section to sketch the curve. In guideline D find an equation of the slant asymptote. 2 sin x ■ 4x 3 2x 2 5 2x 2 x 3 2 2 0 x2 x 45. y 3 tan x, 2 39. y 43. y tan x 2x 1 2 2 Find an equation of the slant asymptote. Do not sketch |||| 2 x tan x, 34. y 43–46 +1 x the curve. 1 31. y 1 5x 2 3 _1 0 x 26. y 3 W 0 sx 30. y 9x 3 x 24. y 1 s1 2 sx 22. y x x sx 2 x3 x3 20. y 3 x s5 25. y xx 12. y 3 18. y 19. y 23. y x x 10. y x2 17. y 8x 4 8. y 2 3 5. y 11. y 4. y 2 6. y 15x 4 6x 2 2. y x 2 9. y where E and I are positive constants. (E is Young’s modulus of elasticity and I is the moment of inertia of a cross-section of the beam.) Sketch the graph of the deflection curve. Use the guidelines of this section to sketch the curve. x 3. y Page 270 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION |||| 4.5 1–40 2:50 PM x2 1 x2 x 48. y 50. xy x2 52. y 4 12 2 x x x 1 ■ 2x 3 51. y ■ x2 x ■ 2 ■ 1 1 ■ ■ 53. Show that the curve y y 2 x and y ■ ■ ■ 1 1 ■ 3 2 ■ ■ s 4 x 2 9 has two slant asymptotes: 2 x. Use this fact to help sketch the curve. ■ 5E-04(pp 262-271) 1/17/06 2:50 PM Page 271 S ECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 54. Show that the curve y yx curve. 2 and y b a x and y asymptotes of the hyperbola x 2 a 2 56. Let f x x3 lim |||| 4.6 x 4 1 x in the same manner as in Exercise 56. Then use your results to help sketch the graph of f . 57. Discuss the asymptotic behavior of f x b a x are slant y2 b2 1. 1 x. Show that xl fx x2 271 This shows that the graph of f approaches the graph of y x 2, and we say that the curve y f x is asymptotic to the parabola y x 2. Use this fact to help sketch the graph of f . sx 2 4 x has two slant asymptotes: x 2. Use this fact to help sketch the 55. Show that the lines y ❙❙❙❙ cos x 1 x 2 to sketch its graph without going through the curve-sketching procedure of this section. 58. Use the asymptotic behavior of f x 0 Graphing with Calculus and Calculators |||| If you have not already read Section 1.4, you should do so now. In particular, it explains how to avoid some of the pitfalls of graphing devices by choosing appropriate viewing rectangles. The method we used to sketch curves in the preceding section was a culmination of much of our study of differential calculus. The graph was the final object that we produced. In this section our point of view is completely different. Here we start with a graph produced by a graphing calculator or computer and then we refine 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 technology. The theme is the interaction between calculus and calculators. 2 x 6 3x 5 3x 3 2 x 2. Use the graphs of f and f to estimate all maximum and minimum points and intervals of concavity. EXAMPLE 1 Graph the polynomial f x SOLUTION If we specify a domain but not a range, many graphing devices will deduce a suitable range from the values computed. Figure 1 shows the plot from one such device if we specify that 5 x 5. Although this viewing rectangle is useful for showing that the asymptotic behavior (or end behavior) is the same as for y 2 x 6, it is obviously hiding some finer detail. So we change to the viewing rectangle 3, 2 by 50, 100 shown in Figure 2. 100 41,000 y=ƒ y=ƒ _5 _3 2 5 _50 _1000 FIGURE 2 FIGURE 1 From this graph it appears that there is an absolute minimum value of about 15.33 when x 1.62 (by using the cursor) and f is decreasing on , 1.62 and increasing on 1.62, . Also there appears to be a horizontal tangent at the origin and inflection points when x 0 and when x is somewhere between 2 and 1. Now let’s try to confirm these impressions using calculus. We differentiate and get fx 12 x 5 15x 4 9x 2 4x fx 60 x 4 60 x 3 18 x 4 5E-04(pp 272-281) 272 ❙❙❙❙ 1/17/06 2:54 PM Page 272 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION When we graph f in Figure 3 we see that f x changes from negative to positive when x 1.62; this confirms (by the First Derivative Test) the minimum value that we found earlier. But, perhaps to our surprise, we also notice that f x changes from positive to negative when x 0 and from negative to positive when x 0.35. This means that f has a local maximum at 0 and a local minimum when x 0.35, but these were hidden in Figure 2. Indeed, if we now zoom in toward the origin in Figure 4, we see what we missed before: a local maximum value of 0 when x 0 and a local minimum value of about 0.1 when x 0.35. 20 1 y=ƒ y=fª(x) _1 _3 1 2 _5 _1 FIGURE 3 10 _3 2 y=f·(x) _30 FIGURE 5 FIGURE 4 What about concavity and inflection points? From Figures 2 and 4 there appear to be inflection points when x is a little to the left of 1 and when x is a little to the right of 0. But it’s difficult to determine inflection 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 x 1.23 and from negative to positive when x 0.19. So, correct to two decimal places, f is concave upward on and concave downward on , 1.23 and 0.19, 1.23, 0.19 . The inflection points are 1.23, 10.18 and 0.19, 0.05 . We have discovered that no single graph reveals all the important features of this polynomial. But Figures 2 and 4, when taken together, do provide an accurate picture. EXAMPLE 2 Draw the graph of the function fx x2 7x x2 3 in a viewing rectangle that contains all the important features of the function. Estimate the maximum and minimum values and the intervals of concavity. Then use calculus to find these quantities exactly. SOLUTION Figure 6, produced by a computer with automatic scaling, is a disaster. Some graphing calculators use 10, 10 by 10, 10 as the default viewing rectangle, so let’s try it. We get the graph shown in Figure 7; it’s a major improvement. 3 10 10!* y=ƒ _10 y=ƒ _5 10 5 _10 FIGURE 6 FIGURE 7 5E-04(pp 272-281) 1/17/06 2:54 PM Page 273 S ECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS ❙❙❙❙ 273 The y-axis appears to be a vertical asymptote and indeed it is because lim x2 7x x2 xl0 3 Figure 7 also allows us to estimate the x-intercepts: about 0.5 and 6.5. The exact values are obtained by using the quadratic formula to solve the equation x 2 7x 3 0; we get x ( 7 s37 ) 2. To get a better look at horizontal asymptotes, we change to the viewing rectangle 20, 20 by 5, 10 in Figure 8. It appears that y 1 is the horizontal asymptote and this is easily confirmed: 10 y=ƒ y=1 _20 lim 20 x2 7x x2 xl _5 3 lim xl 7 x 1 3 x2 1 To estimate the minimum value we zoom in to the viewing rectangle 3, 0 by 4, 2 in Figure 9. The cursor indicates that the absolute minimum value is about 3.1 when x 0.9, and we see that the function decreases on , 0.9 and 0, and increases on 0.9, 0 . The exact values are obtained by differentiating: FIGURE 8 2 _3 0 y=ƒ _4 FIGURE 9 fx 7 x2 6 x3 7x 6 x 3 6 6 This shows that f x 0 when x 0 when 7 x 0 and f x 7 and when 6 37 x 0. The exact minimum value is f ( 7 ) 3.08. 12 Figure 9 also shows that an inflection point occurs somewhere between x 1 and x 2. We could estimate it much more accurately using the graph of the second derivative, but in this case it’s just as easy to find exact values. Since fx 14 x3 18 x4 2 7x 9 x 4 9 0 when x 0 . So f is concave upward on ( 9 , 0) and we see that f x 7x 7 9 0, and concave downward on ( , 7 ). The inflection point is ( 9 , 71 ). 7 27 The analysis using the first two derivatives shows that Figures 7 and 8 display all the major aspects of the curve. E XAMPLE 3 Graph the function f x 10 y=ƒ _10 10 SOLUTION Drawing on our experience with a rational function in Example 2, let’s start by graphing f in the viewing rectangle 10, 10 by 10, 10 . From Figure 10 we have the feeling that we are going to have to zoom in to see some finer detail and also zoom out to see the larger picture. But, as a guide to intelligent zooming, let’s first take a close look at the expression for f x . Because of the factors x 2 2 and x 4 4 in the denominator, we expect x 2 and x 4 to be the vertical asymptotes. Indeed _10 FIGURE 10 x2 x 1 3 . x 22x 44 lim x l2 x2 x 1 3 x 22x 4 4 and lim xl4 x2 x 1 3 x 22x 4 4 5E-04(pp 272-281) 274 ❙❙❙❙ 1/17/06 2:54 PM Page 274 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION To find the horizontal asymptotes we divide numerator and denominator by x 6 : x2 x 1 3 x 22x 4 1 1 x 4 2 2 x 1 1 x 1 3 4 x 4 l0 as xl so the x-axis is the horizontal asymptote. It is also very useful to consider the behavior of the graph near the x-intercepts using an analysis like that in Example 11 in Section 4.4. Since x 2 is positive, f x does not change sign at 0 and so its graph doesn’t cross the x-axis at 0. But, because of the factor x 1 3, the graph does cross the x-axis at 1 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 11. y _1 FIGURE 11 1 2 3 4 x Now that we know what to look for, we zoom in (several times) to produce the graphs in Figures 12 and 13 and zoom out (several times) to get Figure 14. 0.05 0.0001 500 y=ƒ y= _100 1 _1.5 0.5 y=ƒ _0.05 FIGURE 12 _0.0001 FIGURE 13 _1 _10 10 FIGURE 14 We can read from these graphs that the absolute minimum is about 0.02 and occurs when x 20. There is also a local maximum 0.00002 when x 0.3 and a local minimum 211 when x 2.5. These graphs also show three inflection points near 35, 5, and 1 and two between 1 and 0. To estimate the inflection points closely we would need to graph f , but to compute f by hand is an unreasonable chore. If you have a computer algebra system, then it’s easy to do (see Exercise 15). We have seen that, for this particular function, three graphs (Figures 12, 13, and 14) are necessary to convey all the useful information. The only way to display all these features of the function on a single graph is to draw it by hand. Despite the exaggerations and distortions, Figure 11 does manage to summarize the essential nature of the function. EXAMPLE 4 Graph the function f x , locate all maxisin x sin 2 x . For 0 x mum and minimum values, intervals of increase and decrease, and inflection points correct to one decimal place. 5E-04(pp 272-281) 1/17/06 2:55 PM Page 275 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS |||| The family of functions fx sin x sin c x where c is a constant, occurs in applications to frequency modulation (FM) synthesis. A sine wave is modulated by a wave with a different frequency sin c x . The case where c 2 is studied in Example 4. Exercise 19 explores another special case. 1.1 ❙❙❙❙ 275 S OLUTION We first note that f is periodic with period 2 . Also, f is odd and f x 1 for all x. So the choice of a viewing rectangle is not a problem for this function: We start with 0, by 1.1, 1.1 . (See Figure 15.) It appears that there are three local maximum values and two local minimum values in that window. To confirm this and locate them more accurately, we calculate that fx cos x sin 2 x 1 2 cos 2 x and graph both f and f in Figure 16. Using zoom-in and the First Derivative Test, we find the following values to one decimal place. Intervals of increase: π 0, 0.6 , 1.0, 1.6 , 2.1, 2.5 Intervals of decrease: 0 0.6, 1.0 , 1.6, 2.1 , 2.5, Local maximum values: f 0.6 FIGURE 15 1, f 1.6 Local minimum values: _1.1 0.94, f 2.1 f 1.0 1, f 2.5 1 0.94 The second derivative is 1.2 y=ƒ 0 fx π 1 2 cos 2 x 2 sin x sin 2 x 4 sin 2 x cos x sin 2 x Graphing both f and f in Figure 17, we obtain the following approximate values: y=f ª(x) Concave upward on: Concave downward on: FIGURE 16 0, 0.8 , 1.3, 1.8 , 2.3, Inflection points: _1.2 0.8, 1.3 , 1.8, 2.3 0, 0 , 0.8, 0.97 , 1.3, 0.97 , 1.8, 0.97 , 2.3, 0.97 1.2 1.2 f 0 π _2π 2π f· _1.2 FIGURE 17 _1.2 FIGURE 18 Having checked that Figure 15 does indeed represent f accurately for 0 x we can state that the extended graph in Figure 18 represents f accurately for 2 x 2. , Our final example is concerned with families of functions. As discussed in Section 1.4, this means that the functions in the family are related to each other by a formula that contains one or more arbitrary constants. Each value of the constant gives rise to a member of the family and the idea is to see how the graph of the function changes as the constant changes. 5E-04(pp 272-281) 276 ❙❙❙❙ 1/17/06 2:55 PM Page 276 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION _5 4 lim xl _2 c=2 y= 2 1 ≈+2x-2 lim 4 x 1 2x 2 0 c for any value of c, they all have the x-axis as a horizontal asymptote. A vertical asymptote will occur when x 2 2 x c 0. Solving this quadratic equation, we get x 1 s1 c. When c 1, there is no vertical asymptote (as in Figure 19). When c 1, the graph has a single vertical asymptote x 1 because xl 1 _5 c vary as c varies? 2x SOLUTION The graphs in Figures 19 and 20 (the special cases c 2 and c 2) show two very different-looking curves. Before drawing any more graphs, let’s see what members of this family have in common. Since 1 y= ≈+2x+2 F IGURE 19 1 x2 EXAMPLE 5 How does the graph of f x 2 x 1 2x 2 lim 1 xl 1 1 x When c 1, there are two vertical asymptotes: x Now we compute the derivative: fx 2x x2 1 1 2 s1 c (as in Figure 20). 2 2x c 2 _2 FIGURE 20 c=_2 See an animation of Figure 21. Resources / Module 5 / Max and Min / Families of Functions c=_1 FIGURE 21 This shows that f x 0 when x 1 (if c 1), f x 0 when x 1, and fx 0 when x 1. For c 1, this means that f increases on , 1 and decreases on 1, . For c 1, there is an absolute maximum value f1 1 c 1 . For c 1, f 1 1 c 1 is a local maximum value and the intervals of increase and decrease are interrupted at the vertical asymptotes. Figure 21 is a “slide show” displaying five members of the family, all graphed in the viewing rectangle 5, 4 by 2, 2 . As predicted, c 1 is the value at which a transition takes place from two vertical asymptotes to one, and then to none. As c increases from 1, we see that the maximum point becomes lower; this is explained by the fact that 1 c 1 l 0 as c l . As c decreases from 1, the vertical asymptotes become more widely separated because the distance between them is 2 s1 c, which becomes large as c l . Again, the maximum point approaches the x-axis because 1 c 1 l 0 as c l . c=0 c=1 c=2 c=3 The family of functions ƒ=1/(≈+2x+c) There is clearly no inflection point when c fx 2 3x 2 x2 1. For c 6x 2x 4 c 1 we calculate that c 3 and deduce that inflection points occur when x 1 s3 c 1 3. So the inflection points become more spread out as c increases and this seems plausible from the last two parts of Figure 21. 5E-04(pp 272-281) 1/17/06 2:56 PM Page 277 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS |||| 4.6 1–8 |||| Produce graphs of f that reveal all the important aspects of the curve. In particular, you should use graphs of f and f to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points. 4x 4 2. f x x6 3. f x 3 4. f x 5. f x 6. f x sx x 32 x 3 2 95 x 125 x 3 x 5 x3 sin x x 4x ■ 17–18 |||| Use a computer algebra system to graph f and to find f and f . Use graphs of these derivatives to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points of f . sin 2 x , sx 2 1 18. f x 5 cos x 8. f x graphs to estimate the intervals of increase and decrease and concavity of f . 2x 1 4 sx 4 x 1 3 tan x 2 16. If f is the function of Exercise 14, find f and f and use their 17. f x 2x 2 2 x x2 x x 2 4x 1 x CAS CAS 29 2 x ■ 75 x 4 3x 7. f x ■ 89 x 2 15 x 5 4 ■ 4 7 cos x, x 10. f x 8x 3 x ■ ■ ■ ■ ■ ■ ■ ■ ■ x ■ 2 ■ x ■ 2 ■ ■ ■ ■ ■ ■ 13–14 |||| Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to producing graphs (with a graphing device) that display the major features of the curve. Use these graphs to estimate the maximum and minimum values. x 13. f x CAS ■ ■ ■ ■ ■ ■ ■ 4x 3 xx 1 |||| Describe how the graph of f varies as c varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when c changes. You should also identify any transitional values of c at which the basic shape of the curve changes. 20. f x x3 22. f x x 2sc 2 1 23. f x x2 cx 2 cx c 2x 2 1 24. f x 2 x4 21. f x cx 1 x 22 cx 2 4 10x x 1 4 x 23x 1 14. f x ■ ■ 20–25 2 2 sin x, ■ ■ 20 x s9 x 10 11 x x2 12. f x ■ 3 tions f x sin x sin c x that occur in FM synthesis. Here we investigate the function with c 3. Start by graphing f in the viewing rectangle 0, by 1.2, 1.2 . How many local maximum points do you see? The graph has more than are visible to the naked eye. To discover the hidden maximum and minimum points you will need to examine the graph of f very carefully. In fact, it helps to look at the graph of f at the same time. Find all the maximum and minimum values and inflection points. Then graph f in the viewing rectangle 2 , 2 by 1.2, 1.2 and comment on symmetry. 2 ■ 3x 2 11. f x ■ x 19. In Example 4 we considered a member of the family of func- 2 ■ x2 ■ 0 4 9–12 |||| Produce graphs of f that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points, and use calculus to find these quantities exactly. 9. f x 277 Exercises ; 1. f x ❙❙❙❙ ■ ■ ■ 25. f x ■ ■ cx ■ sin x ■ ■ ■ ■ ■ ■ ■ ■ 2 ■ ■ ■ ■ ■ ■ 15. If f is the function considered in Example 3, use a computer algebra system to calculate f and then graph it to confirm that all the maximum and minimum values are as given in the example. Calculate f and use it to estimate the intervals of concavity and inflection points. ■ 26. Investigate the family of curves given by the equation fx x 4 c x 2 x. Start by determining the transitional value of c at which the number of inflection points changes. Then graph several members of the family to see what shapes are possible. There is another transitional value of c at which the number of critical numbers changes. Try to discover it graphically. Then prove what you have discovered. ■ 5E-04(pp 272-281) 278 ❙❙❙❙ 1/17/06 2:56 PM Page 278 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 27. (a) Investigate the family of polynomials given by the equation fx c x 4 2 x 2 1. For what values of c does the curve have minimum points? (b) Show that the minimum and maximum points of every curve in the family lie on the parabola y 1 x 2. Illustrate by graphing this parabola and several members of the family. |||| 4.7 28. (a) Investigate the family of polynomials given by the equation fx 2 x 3 c x 2 2 x. For what values of c does the curve have maximum and minimum points? (b) Show that the minimum and maximum points of every curve in the family lie on the curve y x x 3. Illustrate by graphing this curve and several members of the family. Optimization Problems The methods we have learned in this chapter for finding extreme values have practical applications in many areas of life. A businessperson wants to minimize costs and maximize profits. A traveler wants to minimize transportation time. Fermat’s Principle in optics states that light follows the path that takes the least time. In this section and the next we solve such problems as maximizing areas, volumes, and profits 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 by setting up the function that is to be maximized or minimized. Let’s recall the problem-solving principles discussed on page 58 and adapt them to this situation: Steps in Solving Optimization Problems 1. Understand the Problem The first step is to read the problem carefully until it is clearly 2. 3. 4. 5. 6. understood. Ask yourself: 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 identify the given and required quantities on the diagram. Introduce Notation Assign a symbol to the quantity that is to be maximized or minimized (let’s call it Q for now). Also select symbols a, b, c, . . . , x, y for other unknown quantities and label the diagram with these symbols. It may help to use initials as suggestive symbols—for example, A for area, h for height, t for time. Express Q in terms of some of the other symbols from Step 3. If Q has been expressed as a function of more than one variable in Step 4, use the given information to find relationships (in the form of equations) among these variables. Then use these equations to eliminate all but one of the variables in the expression for Q. Thus, Q will be expressed as a function of one variable x, say, Q f x . Write the domain of this function. Use the methods of Sections 4.1 and 4.3 to find the absolute maximum or minimum value of f . In particular, if the domain of f is a closed interval, then the Closed Interval Method in Section 4.1 can be used. EXAMPLE 1 A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area? |||| Understand the problem |||| Analogy: Try special cases |||| Draw diagrams SOLUTION In order to get a feeling for what is happening in this problem, let’s experiment with some special cases. Figure 1 (not to scale) shows three possible ways of laying out the 2400 ft of fencing. We see that when we try shallow, wide fields or deep, narrow 5E-04(pp 272-281) 1/17/06 2:56 PM Page 279 S ECTION 4.7 OPTIMIZATION PROBLEMS ❙❙❙❙ 279 fields, we get relatively small areas. It seems plausible that there is some intermediate configuration that produces the largest area. 400 1000 2200 700 100 1000 700 1000 100 Area=700 · 1000=700,000 ft@ Area=100 · 2200=220,000 ft@ Area=1000 · 400=400,000 ft@ FIGURE 1 Figure 2 illustrates the general case. We wish to maximize the area A of the rectangle. Let x and y be the depth and width of the rectangle (in feet). Then we express A in terms of x and y: A xy |||| Introduce notation y x A x We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the given information that the total length of the fencing is 2400 ft. Thus 2 x y 2400 From this equation we have y FIGURE 2 A Note that x mize is 0 and x 2 x, which gives x 2400 2x 2400 x 1200 (otherwise A Ax The derivative is A x equation 2400 2400 0). So the function that we wish to maxi- 2x 2 2400 x 2x 2 0 x 1200 4 x, so to find the critical numbers we solve the 2400 4x 0 which gives x 600. The maximum value of A must occur either at this critical number or at an endpoint of the interval. Since A 0 0, A 600 720,000, and A 1200 0, the Closed Interval Method gives the maximum value as A 600 720,000. [Alternatively, we could have observed that A x 4 0 for all x, so A is always concave downward and the local maximum at x 600 must be an absolute maximum.] Thus, the rectangular field should be 600 ft deep and 1200 ft wide. EXAMPLE 2 A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. h r FIGURE 3 SOLUTION Draw the diagram as in Figure 3, 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 cylinder (top, bottom, and sides). From Figure 4 on page 280 we see that the sides are made from a rectangular sheet with dimensions 2 r and h. So the surface area is A 2 r2 2 rh 5E-04(pp 272-281) 280 ❙❙❙❙ 1/17/06 2:56 PM Page 280 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 2πr To eliminate h we use the fact that the volume is given as 1 L, which we take to be 1000 cm3. Thus r 2h 1000 r h which gives h r 2 . Substitution of this into the expression for A gives 1000 2 r2 A Area 2{πr@} Area (2πr)h 1000 r2 2r 2000 r 2 r2 Therefore, the function that we want to minimize is FIGURE 4 2 r2 Ar 2000 r r 0 To find the critical numbers, we differentiate: y Ar y=A(r) 1000 0 10 r FIGURE 5 2000 r2 r3 4 500 r2 3 Then A r 0 when r 3 500, so the only critical number is r s500 . Since the domain of A is 0, , we can’t use the argument of Example 1 concerning 3 endpoints. But we can observe that A r 0 for r s500 and A r 0 for 3 r s500 , so A is decreasing for all r to the left of the critical number and increasing 3 for all r to the right. Thus, r s500 must give rise to an absolute minimum. [Alternatively, we could argue that A r l as r l 0 and A r l as r l , so there must be a minimum value of A r , which must occur at the critical number. See Figure 5.] 3 The value of h corresponding to r s500 is 1000 r2 h |||| In the Applied Project on page 288 we investigate the most economical shape for a can by taking into account other manufacturing costs. 4r 1000 500 23 2 3 500 3 Thus, to minimize the cost of the can, the radius should be s500 should be equal to twice the radius, namely, the diameter. 2r cm and the height NOTE 1 The argument used in Example 2 to justify the absolute minimum is a variant of the First Derivative Test (which applies only to local maximum or minimum values) and is stated here for future reference. ■ Module 4.7 takes you through eight additional optimization problems, including animations of the physical situations. First Derivative Test for Absolute Extreme Values Suppose that c is a critical number of a continuous function f defined on an interval. (a) If f x 0 for all x c and f x 0 for all x c, then f c is the absolute maximum value of f . (b) If f x 0 for all x c and f x 0 for all x c, then f c is the absolute minimum value of f . NOTE 2 An alternative method for solving optimization problems is to use implicit differentiation. Let’s look at Example 2 again to illustrate the method. We work with the same equations ■ A 2 r2 2 rh r 2h 100 5E-04(pp 272-281) 1/17/06 2:57 PM Page 281 S ECTION 4.7 OPTIMIZATION PROBLEMS ❙❙❙❙ 281 but instead of eliminating h, we differentiate both equations implicitly with respect to r : A 4r 2h 2 rh r 2h 2 rh The minimum occurs at a critical number, so we set A equations 2r and subtraction gives 2r h rh h 0 0, or h 2h 0, simplify, and arrive at the rh 0 2r. EXAMPLE 3 Find the point on the parabola y 2 y 0 2 x that is closest to the point 1, 4 . SOLUTION The distance between the point 1, 4 and the point x, y is ¥=2x (1, 4) d 0 1 234 1 d x s( 1 y 2 2 (Alternatively, we could have substituted y of minimizing d, we minimize its square: d2 FIGURE 6 2 y 4 2 y 2 2, so the expression for d (See Figure 6.) But if x, y lies on the parabola, then x becomes (x, y) 1 sx (1 y2 2 fy 1 )2 y 4 2 s 2 x to get d in terms of x alone.) Instead 1)2 y 4 2 (You should convince yourself that the minimum of d occurs at the same point as the minimum of d 2, but d 2 is easier to work with.) Differentiating, we obtain fy C D 8 km B FIGURE 7 1) y 2y 4 y3 8 so f y 0 when y 2. Observe that f y 0 when y 2 and f y 0 when y 2, so by the First Derivative Test for Absolute Extreme Values, the absolute minimum occurs when y 2. (Or we could simply say that because of the geometric nature of the problem, it’s obvious that there is a closest point but not a farthest point.) The corresponding value of x is x y 2 2 2. Thus, the point on y 2 2 x closest to 1, 4 is 2, 2 . 3 km A 2( 1 y 2 2 EXAMPLE 4 A man launches his boat from point A on a bank of a straight river, 3 km wide, and wants to reach point B, 8 km downstream on the opposite bank, as quickly as possible (see Figure 7). He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between C and B and then run to B. If he can row 6 km h and run 8 km h, 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 x be the distance from C to D, then the running distance is DB 8 x and the Pythagorean Theorem gives the rowing distance as AD s x 2 9. We use the equation time distance rate 5E-04(pp 282-291) 282 ❙❙❙❙ 1/17/06 3:22 PM Page 282 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Then the rowing time is s x 2 T as a function of x is 9 6 and the running time is 8 sx 2 6 Tx 9 8 x 8 The domain of this function T is 0, 8 . Notice that if x he rows directly to B. The derivative of T is x Tx Try another problem like this one. Resources / Module 5 / Max and Min / Start of Optimal Lifeguard Thus, using the fact that x Tx x 8, so the total time 6 sx 2 0 he rows to C and if x 8 1 8 9 0, we have 0 &? x 6 sx 2 &? 16 x 2 &? x 1 8 9 9 x2 &? 4 x 9 3 sx 2 &? 7x 2 81 9 9 s7 The only critical number is x 9 s7. To see whether the minimum occurs at this critical number or at an endpoint of the domain 0, 8 , we evaluate T at all three points: T y=T(x) T0 1 0 2 4 x 6 FIGURE 8 1.5 T 9 s7 s7 8 1 1.33 T8 s73 6 1.42 Since the smallest of these values of T occurs when x 9 s7, the absolute minimum value of T must occur there. Figure 8 illustrates this calculation by showing the graph of T . Thus, the man should land the boat at a point 9 s7 km ( 3.4 km) downstream from his starting point. E XAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle of Resources / Module 5 / Max and Min / Start of Max and Min radius r. SOLUTION 1 Let’s take the semicircle to be the upper half of the circle x 2 y 2 r 2 with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the x-axis as shown in Figure 9. Let x, y be the vertex that lies in the first quadrant. Then the rectangle has sides of lengths 2 x and y, so its area is y (x, y) 2x _r FIGURE 9 0 A 2 xy y r x To eliminate y we use the fact that x, y lies on the circle x 2 y sr 2 x 2. Thus A 2 x sr 2 x2 y2 r 2 and so 5E-04(pp 282-291) 1/17/06 3:22 PM Page 283 S ECTION 4.7 OPTIMIZATION PROBLEMS The domain of this function is 0 A 2 sr 2 x ❙❙❙❙ 283 r. Its derivative is 2x 2 x2 sr 2 x 2 2 r 2 2x 2 sr 2 x 2 which is 0 when 2 x 2 r 2, that is, x r s2 (since x 0). This value of x gives a maximum value of A since A 0 0 and A r 0. Therefore, the area of the largest inscribed rectangle is A r s2 2 r s2 r2 r2 2 r2 SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let be the angle shown in Figure 10. Then the area of the rectangle is r 2r cos r sin r 2 2 sin cos r 2 sin 2 We know that sin 2 has a maximum value of 1 and it occurs when 2 2. So A has a maximum value of r 2 and it occurs when 4. Notice that this trigonometric solution doesn’t involve differentiation. In fact, we didn’t need to use calculus at all. ¨ r Ł ¨ FIGURE 10 |||| 4.7 A r à ¨ Exercises 1. Consider the following problem: Find two numbers whose sum is 23 and whose product is a maximum. (a) Make a table of values, like the following one, so that the sum of the numbers in the first two columns is always 23. On the basis of the evidence in your table, estimate the answer to the problem. First number Second number Product 1 2 3 . . . 22 21 20 . . . 22 42 60 . . . (b) Use calculus to solve the problem and compare with your answer to part (a). 2. Find two numbers whose difference is 100 and whose product is a minimum. 3. Find two positive numbers whose product is 100 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 100 m whose area is as large as possible. 6. Find the dimensions of a rectangle with area 1000 m2 whose perimeter is as small as possible. 7. Consider the following problem: A farmer with 750 ft of fenc- ing 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 configurations. Does it appear that there is a maximum area? If so, estimate it. (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols. (c) Write an expression 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 your estimate in part (a). 8. Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have. (a) Draw several diagrams to illustrate the situation, some short boxes with large bases and some tall boxes with small bases. Find the volumes of several such boxes. Does it appear that there is a maximum volume? If so, estimate it. 5E-04(pp 282-291) 284 ❙❙❙❙ 1/17/06 3:22 PM Page 284 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols. (c) Write an expression 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 your estimate in part (a). 9. A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide 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? 10. A box with a square base and open top must have a volume of 32,000 cm3. Find the dimensions of the box that minimize the amount of material used. 11. If 1200 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box. 12. A rectangular storage container with an open top is to have a 3 volume of 10 m . The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container. 13. Do Exercise 12 assuming the container has a lid that is made from the same material as the sides. 14. (a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square. (b) Show that of all the rectangles with a given perimeter, the one with greatest area is a square. 15. Find the point on the line y 4x 7 that is closest to the origin. can be inscribed in a circle of radius r. 24. Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3 cm and 4 cm if two sides of the rectangle lie along the legs. 25. A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible volume of such a cylinder. 26. A right circular cylinder is inscribed in a cone with height h and base radius r. Find the largest possible volume of such a cylinder. 27. A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible surface area of such a cylinder. 28. A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle. See Exercise 52 on page 24.) If the perimeter of the window is 30 ft, find the dimensions of the window so that the greatest possible amount of light is admitted. 29. 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 fixed at 384 cm2, find the dimensions of the poster with the smallest area. 30. A poster is to have an area of 180 in2 with 1-inch margins at the bottom and sides and a 2-inch margin at the top. What dimensions will give the largest printed area? 31. A piece of wire 10 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 maximum? (b) A minimum? 32. Answer Exercise 31 if one piece is bent into a square and the other into a circle. 33. A cylindrical can without a top is made to contain V cm3 of 16. Find the point on the line 6x point 23. Find the dimensions of the isosceles triangle of largest area that y 9 that is closest to the 3, 1 . 17. Find the points on the ellipse 4 x 2 y2 4 that are farthest away from the point 1, 0 . ; 18. Find, correct to two decimal places, the coordinates of the point on the curve y tan x that is closest to the point 1, 1 . 19. Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius r. 20. Find the area of the largest rectangle that can be inscribed in the ellipse x 2 a 2 y2 b2 1. liquid. Find the dimensions that will minimize the cost of the metal to make the can. 34. 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? 35. A cone-shaped drinking cup is made from a circular piece of paper of radius R by cutting out a sector and joining the edges CA and CB. Find the maximum capacity of such a cup. A B R 21. 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. 22. Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola y 8 x 2. C 5E-04(pp 282-291) 1/17/06 3:23 PM Page 285 S ECTION 4.7 OPTIMIZATION PROBLEMS 36. A cone-shaped paper drinking cup is to be made to hold 27 cm3 speed of 20 km h. Another boat has been heading due east at 15 km h and reaches the same dock at 3:00 P.M. At what time were the two boats closest together? 37. A cone with height h is inscribed in a larger cone with height H so that its vertex is at the center of the base of the larger cone. Show that the inner cone has maximum volume when h 1 H. 3 41. Solve the problem in Example 4 if the river is 5 km wide and point B is only 5 km downstream from A. 42. A woman at a point A on the shore of a circular lake with 38. For a fish swimming at a speed v relative to the water, the energy expenditure per unit time is proportional to v 3. It is radius 2 mi wants to arrive at the point C diametrically opposite A on the other side of the lake in the shortest possible time. She can walk at the rate of 4 mi h and row a boat at 2 mi h. How should she proceed? believed that migrating fish try to minimize the total energy required to swim a fixed distance. If the fish are swimming against a current u u v , then the time required to swim a distance L is L v u and the total energy E required to swim the distance is given by av 3 285 40. A boat leaves a dock at 2:00 P.M. and travels due south at a of water. Find the height and radius of the cup that will use the smallest amount of paper. Ev ❙❙❙❙ B L v u where a is the proportionality constant. A (a) Determine the value of v that minimizes E. (b) Sketch the graph of E. ¨ 2 C 2 Note: This result has been verified experimentally; migrating fish swim against a current at a speed 50% greater than the current speed. 43. The illumination of an object by a light source is directly 39. In a beehive, each cell is a regular hexagonal prism, open at proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 ft apart, where should an object be placed on the line between the sources so as to receive the least illumination? one end with a trihedral angle at the other end. It is believed that bees form their cells in such a way as to minimize the surface area for a given volume, thus using the least amount of wax in cell construction. Examination of these cells has shown that the measure of the apex angle is amazingly consistent. Based on the geometry of the cell, it can be shown that the surface area S is given by S 6sh off the least area from the first quadrant. (3s 2s3 2) csc 32 2 s cot where s, the length of the sides of the hexagon, and h, the height, are constants. (a) Calculate dS d . (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 by more than 2 . trihedral angle ¨ rear of cell 45. Let a and b be positive numbers. Find the length of the shortest line segment that is cut off by the first quadrant and passes through the point a, b . 46. At which points on the curve y 1 tangent line have the largest slope? b front of cell 40 x 3 3 x 5 does the 47. Show that of all the isosceles triangles with a given perimeter, the one with the greatest area is equilateral. CAS 48. The frame for a kite is to be made from six pieces of wood. The four exterior pieces have been cut with the lengths indicated in the figure. To maximize the area of the kite, how long should the diagonal pieces be? a h s 44. Find an equation of the line through the point 3, 5 that cuts b a b 5E-04(pp 282-291) 286 ❙❙❙❙ 1/17/06 3:23 PM Page 286 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION ; 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 figure). Express L as a function of x AP and use the graphs of L and dL d x to estimate the minimum value. the figure. Show that the shortest length of such a rope occurs when 1 2. P S A P 5m Q 2m B ¨™ ¨¡ R T 3m 53. The upper right-hand corner of a piece of paper, 12 in. by 8 in., C D 50. 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 very low speeds the engine runs inefficiently, so initially c decreases as the speed increases. But at high speeds the fuel consumption increases. You can see that c v is minimized for this car when v 30 mi h. However, for fuel efficiency, what must be minimized is not the consumption in gallons per hour but rather the fuel consumption in gallons per mile. Let’s call this consumption G. Using the graph, estimate the speed at which G has its minimum value. as in the figure, is folded over to the bottom edge. How would you fold it so as to minimize the length of the fold? In other words, how would you choose x to minimize y ? 12 y x 8 c 54. A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-angled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner? 0 20 40 6 √ 60 51. Let v1 be the velocity of light in air and v2 the velocity of light in water. According to Fermat’s Principle, a ray of light will travel from a point A in the air to a point B in the water by a path ACB that minimizes the time taken. Show that sin sin 1 v2 9 v1 2 ¨ 55. An observer stands at a point P, one unit away from a track. where 1 (the angle of incidence) and 2 (the angle of refraction) are as shown. This equation is known as Snell’s Law. A Two runners start at the point S in the figure and run along the track. One runner runs three times as fast as the other. Find the maximum value of the observer’s angle of sight between the runners. [Hint: Maximize tan .] ¨¡ P C ¨ ¨™ 1 B 52. Two vertical poles PQ and ST are secured by a rope PRS going from the top of the first pole to a point R on the ground between the poles and then to the top of the second pole as in S 5E-04(pp 282-291) 1/17/06 3:23 PM Page 287 S ECTION 4.7 OPTIMIZATION PROBLEMS 56. A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle . How should be chosen so that the gutter will carry the maximum amount of water? ¨ 10 cm 10 cm 10 cm 57. Where should the point P be chosen on the line segment AB so as to maximize the angle ? B C L r4 where L is the length of the blood vessel, r is the radius, and C is a positive constant determined by the viscosity of the blood. (Poiseuille established this law experimentally, but it also follows from Equation 9.4.2.) The figure shows a main blood vessel with radius r1 branching at an angle into a smaller vessel with radius r2. C 2 ¨ P 287 pumping the blood. In particular, this energy is reduced when the resistance of the blood is lowered. One of Poiseuille’s Laws gives the resistance R of the blood as R ¨ ❙❙❙❙ r™ 3 b vascular branching A 5 A r¡ ¨ B 58. A painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer (as in the figure). How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle subtended at his eye by the painting?) a (a) Use Poiseuille’s Law to show that the total resistance of the blood along the path ABC is R C a b cot r4 1 b csc r4 2 h ¨ d where a and b are the distances shown in the figure. (b) Prove that this resistance is minimized when cos 59. Find the maximum area of a rectangle that can be circum- scribed about a given rectangle with length L and width W . ¨ L W 60. The blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that convey blood from the heart to the organs and back to the heart. This system should work so as to minimize the energy expended by the heart in r4 2 r4 1 (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. 5E-04(pp 282-291) 288 ❙❙❙❙ 1/17/06 3:23 PM Page 288 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 61. Ornithologists have determined that some species of birds tend to avoid flights over large bodies of water during daylight hours. It is believed that more energy is required to fly over water than land because air generally rises over land and falls over water during the day. A bird with these tendencies is released from an island that is 5 km from the nearest point B on a straight shoreline, flies to a point C on the shoreline, and then flies along the shoreline to its nesting area D. Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points B and D are 13 km apart. (a) In general, if it takes 1.4 times as much energy to fly over water as land, to what point C should the bird fly in order to minimize the total energy expended in returning to its nesting area? (b) Let W and L denote the energy (in joules) per kilometer flown over water and land, respectively. What would a large value of the ratio W L mean in terms of the bird’s flight? What would a small value mean? Determine the ratio W L corresponding to the minimum expenditure of energy. (c) What should the value of W L be in order for the bird to fly directly to its nesting area D? What should the value of W L be for the bird to fly to B and then along the shore to D ? (d) If the ornithologists observe that birds of a certain species reach the shore at a point 4 km from B, how many times more energy does it take a bird to fly over water than land? ; 62. Two light sources of identical strength are placed 10 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 figure). We want to locate P on so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. (a) Find an expression for the intensity I x at the point P. (b) If d 5 m, use graphs of I x and I x to show that the intensity is minimized when x 5 m, that is, when P is at the midpoint of . (c) If d 10 m, show that the intensity (perhaps surprisingly) is not minimized at the midpoint. (d) Somewhere between d 5 m and d 10 m there is a transitional value of d at which the point of minimal illumination abruptly changes. Estimate this value of d by graphical methods. Then find the exact value of d. island x P d 5 km C B 13 km D nest 10 m APPLIED PROJECT The Shape of a Can h r In this project we investigate the most economical shape for a can. We first interpret this to mean that the volume V of a cylindrical can is given and we need to find the height h and radius r that minimize the cost of the metal to make the can (see the figure). If we disregard any waste metal in the manufacturing process, then the problem is to minimize the surface area of the cylinder. We solved this problem in Example 2 in Section 4.7 and we found that h 2r; that is, the height should be the same as the diameter. But if you go to your cupboard or your supermarket with a ruler, you will discover that the height is usually greater than the diameter and the ratio h r varies from 2 up to about 3.8. Let’s see if we can explain this phenomenon. 1. The material for the cans is cut from sheets of metal. The cylindrical sides are formed by bending rectangles; these rectangles are cut from the sheet with little or no waste. But if the top and bottom discs are cut from squares of side 2r (as in the figure), this leaves considerable waste metal, which may be recycled but has little or no value to the can makers. If this is the case, show that the amount of metal used is minimized when Discs cut from squares h r 8 2.55 5E-04(pp 282-291) 1/17/06 3:23 PM Page 289 S ECTION 4.8 APPLICATIONS TO BUSINESS AND ECONOMICS ❙❙❙❙ 289 2. A more efficient packing of the discs is obtained by dividing the metal sheet into hexagons and cutting the circular lids and bases from the hexagons (see the figure). Show that if this strategy is adopted, then h r 4 s3 2.21 3. The values of h r that we found in Problems 1 and 2 are a little closer to the ones that Discs cut from hexagons actually occur on supermarket shelves, but they still don’t account for everything. If we look more closely at some real cans, we see that the lid and the base are formed from discs with radius larger than r that are bent over the ends of the can. If we allow for this we would increase h r. More significantly, in addition to the cost of the metal we need to incorporate the manufacturing of the can into the cost. Let’s assume that most of the expense is incurred in joining the sides to the rims of the cans. If we cut the discs from hexagons as in Problem 2, then the total cost is proportional to 4 s3 r 2 2 rh k4 r h where k is the reciprocal of the length that can be joined for the cost of one unit area of metal. Show that this expression is minimized when 3 sV k 3 h r 2 hr hr 4 s3 3 ; 4. Plot sV k as a function of x h r and use your graph to argue that when a can is large or joining is cheap, we should make h r approximately 2.21 (as in Problem 2). But when the can is small or joining is costly, h r should be substantially larger. 5. Our analysis shows that large cans should be almost square but small cans should be tall and thin. Take a look at the relative shapes of the cans in a supermarket. Is our conclusion usually true in practice? Are there exceptions? Can you suggest reasons why small cans are not always tall and thin? |||| 4.8 Applications to Business and Economics y inflection point c (x)=slope y=C(x) C (x) 0 x FIGURE 1 Cost function In Section 3.4 we introduced the idea of marginal cost. Recall that if C x , the cost function, is the cost of producing x units of a certain product, then the marginal cost is the rate of change of C with respect to x. In other words, the marginal cost function is the derivative, C x , of the cost function. The graph of a typical cost function is shown in Figure 1. The marginal cost C x is the slope of the tangent to the cost curve at x, C x . Notice that the cost curve is initially concave downward (the marginal cost is decreasing) because of economies of scale (more efficient use of the fixed costs of production). But eventually there is an inflection point and the cost curve becomes concave upward (the marginal cost is increasing), perhaps because of overtime costs or the inefficiencies of a large-scale operation. The average cost function x 1 cx Cx x represents the cost per unit when x units are produced. We sketch a typical average cost 5E-04(pp 282-291) 290 ❙❙❙❙ 1/17/06 3:24 PM Page 290 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION y function in Figure 2 by noting that C x x is the slope of the line that joins the origin to the point x, C x in Figure 1. It appears that there will be an absolute minimum. To find it we locate the critical point of c by using the Quotient Rule to differentiate Equation 1: y=c(x) xC x cx Now c x x Cx 0 and this gives Cx 0 0 when x C x Cx 2 Cx x x FIGURE 2 Average cost function cx Therefore: If the average cost is a minimum, then marginal cost average cost This principle is plausible because if our marginal cost is smaller than our average cost, then we should produce more, thereby lowering our average cost. Similarly, if our marginal cost is larger than our average cost, then we should produce less in order to lower our average cost. |||| See Example 8 in Section 3.4 for an explanation of why it is reasonable to model a cost function by a polynomial. EXAMPLE 1 A company estimates that the cost (in dollars) of producing x items is Cx 2600 2 x 0.001x 2. (a) Find the cost, average cost, and marginal cost of producing 1000 items, 2000 items, and 3000 items. (b) At what production level will the average cost be lowest, and what is this minimum average cost? SOLUTION (a) The average cost function is Cx x 2600 x Cx cx 2 2 0.001x The marginal cost function is 0.002 x We use these expressions to fill in the following table, giving the cost, average cost, and marginal cost (in dollars, or dollars per item, rounded to the nearest cent). x Cx cx Cx 1000 2000 3000 5,600.00 10,600.00 17,600.00 5.60 5.30 5.87 4.00 6.00 8.00 (b) To minimize the average cost we must have marginal cost Cx 2 0.002 x average cost cx 2600 x 2 0.001x 5E-04(pp 282-291) 1/17/06 3:24 PM Page 291 SECTION 4.8 APPLICATIONS TO BUSINESS AND ECONOMICS |||| Figure 3 shows the graphs of the marginal cost function C and average cost function c in Example 1. Notice that c has its minimum value when the two graphs intersect. 0.001x 2600 x so x2 2600 0.001 and x c 0 3000 291 This equation simplifies to 10 Cª ❙❙❙❙ 2,600,000 s2,600,000 1612 To see that this production level actually gives a minimum, we note that cx 5200 x 3 0, so c is concave upward on its entire domain. The minimum average cost is FIGURE 3 c 1612 2600 1612 2 0.001 1612 $5.22 i tem Now let’s consider marketing. Let p x be the price per unit that the company can charge if it sells x units. Then p is called the demand function (or price function) and we would expect it to be a decreasing function of x. If x units are sold and the price per unit is p x , then the total revenue is Rx xp x and R is called the revenue function (or sales function). The derivative R of the revenue function is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold. If x units are sold, then the total profit is Px Rx Cx and P is called the profit function. The marginal profit function is P , the derivative of the profit function. In order to maximize profit we look for the critical numbers of P, that is, the numbers where the marginal profit is 0. But if Px then Rx Rx Cx 0 Cx Therefore: If the profit is a maximum, then marginal revenue marginal cost To ensure that this condition gives a maximum, we could use the Second Derivative Test. Note that Px when Rx Rx Cx 0 Cx and this condition says that the rate of increase of marginal revenue is less than the rate of 5E-04(pp 292-301) 292 ❙❙❙❙ 1/17/06 3:16 PM Page 292 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION increase of marginal cost. Thus, the profit will be a maximum when Rx Cx and Rx Cx E XAMPLE 2 Determine the production level that will maximize the profit for a company with cost and demand functions Cx 84 0.01x 2 1.26 x 0.00007x 3 and px 3.5 0.01x SOLUTION The revenue function is Rx xp x 3.5x 0.01x 2 so the marginal revenue function is Rx 3.5 0.02 x and the marginal cost function is Cx |||| Figure 4 shows the graphs of the revenue and cost functions in Example 2. The company makes a profit when R C and the profit is a maximum when x 103. Notice that the curves have parallel tangents at this production level because marginal revenue equals marginal cost. 1.26 0.02 x 0.00021x 2 Thus, marginal revenue is equal to marginal cost when 3.5 0.02 x 1.26 0.02 x 0.00021x 2 Solving, we get 320 x R 2.24 0.00021 103 To check that this gives a maximum, we compute the second derivatives: C Rx 0 FIGURE 4 160 0.02 Thus, R x C x for all x mize the profit. Cx 0.02 0.00042 x 0. Therefore, a production level of 103 units will maxi- EXAMPLE 3 A store has been selling 200 DVD players a week at $350 each. A market survey indicates that for each $10 rebate offered to buyers, the number of players sold will increase by 20 a week. Find the demand function and the revenue function. How large a rebate should the store offer to maximize its revenue? SOLUTION If x is the number of DVD players sold per week, then the weekly increase in sales is x 200. For each increase of 20 players sold, the price is decreased by $10. So 1 for each additional player sold, the decrease in price will be 20 10 and the demand function is px 350 10 20 x 200 450 1 2 x The revenue function is Rx xp x 450 x 1 2 x2 Since R x 450 x, we see that R x 0 when x 450. This value of x gives an absolute maximum by the First Derivative Test (or simply by observing that the graph of 5E-04(pp 292-301) 1/17/06 3:17 PM Page 293 S ECTION 4.8 APPLICATIONS TO BUSINESS AND ECONOMICS ❙❙❙❙ 293 R is a parabola that opens downward). The corresponding price is p 450 and the rebate is 350 225 offer a rebate of $125. |||| 4.8 1 2 450 450 225 125. Therefore, to maximize revenue the store should Exercises (b) Sketch a graph of the profit function. (c) Sketch a graph of the marginal profit function. 1. A manufacturer keeps precise records of the cost C x of making x items and produces the graph of the cost function shown in the figure. (a) Explain why C 0 0. (b) What is the significance of the inflection point? (c) Use the graph of C to sketch the graph of the marginal cost function. y y=R(x) y=C (x) C x 0 5–8 |||| For each cost function (given in dollars), find (a) the cost, average cost, and marginal cost at a production level of 1000 units; (b) the production level that will minimize the average cost; and (c) the minimum average cost. x 0 5. C x 40,000 300x x2 6. C x 25,000 120x 0.1x 2 7. C x 16,000 200 x 4x 3 ; 8. C x 10,000 340 x 0.3 x 2 2. The graph of a cost function C is given. (a) Draw a careful sketch of the marginal cost function. (b) Use the geometric interpretation of the average cost c x as a slope (see Figure 1) to draw a careful sketch of the average cost function. (c) Estimate the value of x for which c x is a minimum. How are the average cost and the marginal cost related at that value of x ? C 400 ■ ■ ■ 3700 10. C x 339 ■ 4 6 ■ ■ ■ ■ ■ ■ ■ |||| A cost function is given. (a) Find the average cost and marginal cost functions. (b) Use graphs of the functions in part (a) to estimate the production level that minimizes the average cost. (c) Use calculus to find the minimum average cost. (d) Find the minimum value of the marginal cost. 9. C x 2 ■ 0.0001 x 3 ; 9–10 200 0 ■ 2 ■ ■ 0.04 x 2 5x 0.09x 2 25x ■ 0.0003x 3 ■ 0.0004 x 3 ■ ■ ■ ■ ■ ■ ■ x 11–14 3. The average cost of producing x units of a commodity is cx 21.4 0.002 x. Find the marginal cost at a production level of 1000 units. In practical terms, what is the meaning of your answer? 4. The figure shows graphs of the cost and revenue functions reported by a manufacturer. (a) Identify on the graph the value of x for which the profit is maximized. |||| For the given cost and demand functions, find the production level that will maximize profit. 11. C x 680 4x 0.01x 2, px 12. C x 680 4x 0.01x 2, px 13. C x 14. C x px ■ ■ 1450 36 x x 2 ■ ■ 12 3 0.001x , 16,000 500x 1700 7x ■ 12 1.6 x ■ 2 x 500 px 60 0.01x 3 0.004 x , ■ ■ ■ ■ ■ ■ 5E-04(pp 292-301) 294 ❙❙❙❙ 1/17/06 3:17 PM Page 294 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 15–16 |||| Find the production level at which the marginal cost function starts to increase. 15. C x 16. C x ■ ■ 0.001x 3 0.0002 x ■ ■ 0.3x 3 2 0.25x ■ 6x 2 ■ 900 4x ■ 1500 ■ ■ ■ ■ ■ ; 17. The cost, in dollars, of producing x yards of a certain fabric is C x 1200 12 x 0.1x 2 0.0005x 3 and the company finds that if it sells x yards, it can charge p x 29 0.00021x dollars per yard for the fabric. (a) Graph the cost and revenue functions and use the graphs to estimate the production level for maximum profit. (b) Use calculus to find the production level for maximum profit. 18. An aircraft manufacturer wants to determine the best selling ; price for a new airplane. The company estimates that the initial cost of designing the airplane and setting up the factories in which to build it will be 500 million dollars. The additional cost of manufacturing each plane can be modeled by the function m x 20 x 5x 3 4 0.01x 2, where x is the number of aircraft produced and m is the manufacturing cost, in millions of dollars. The company estimates that if it charges a price p (in millions of dollars) for each plane, it will be able to sell xp 320 7.7p planes. (a) Find the cost, demand, and revenue functions. (b) Find the production level and the associated selling price of the aircraft that maximizes profit. 19. A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at $10, the average attendance had been 27,000. When ticket prices were lowered to $8, the average attendance rose to 33,000. (a) Find the demand function, assuming that it is linear. (b) How should ticket prices be set to maximize revenue? 20. During the summer months Terry makes and sells necklaces on the beach. Last summer he sold the necklaces for $10 each and his sales averaged 20 per day. When he increased the price by $1, he found that he lost two sales per day. (a) Find the demand function, assuming that it is linear. (b) If the material for each necklace costs Terry $6, what should the selling price be to maximize his profit? |||| 4.9 21. A manufacturer has been selling 1000 television sets a week at $450 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will increase by 100 per week. (a) Find the demand function. (b) How large a rebate should the company offer the buyer in order to maximize its revenue? (c) If its weekly cost function is C x 68,000 150 x, how should the manufacturer set the size of the rebate in order to maximize its profit? 22. The manager of a 100-unit apartment complex knows from experience that all units will be occupied if the rent is $800 per month. A market survey suggests that, on average, one additional unit will remain vacant for each $10 increase in rent. What rent should the manager charge to maximize revenue? 23. Store managers want an optimal inventory policy. Overstocking results in excessive storage and interest costs, whereas a small inventory means added costs for reordering and delivery. A supermarket manager estimates that a total of 800 cases of soup will be sold at a steady rate during the coming year and it costs $4 to store a case for a year. If he places several orders per year, each consisting of x cases, then there will be an average of 1 x cases in stock at any time and so storage costs for 2 the year are ( 1 x) 4 2 x dollars. He also estimates that the 2 handling cost for each delivery is $100. What is the optimal reorder quantity x that minimizes costs? 24. Suppose a person has an amount A of spending money deposited to her savings account each month, where it earns interest at a monthly rate R. Assume she spends the entire amount throughout the month at a steady rate. When she makes cash withdrawals from the account, she incurs a transaction cost T (a combination of bank fees and the cost of her time). She would save money by making fewer withdrawals, but the more money she leaves in the account the more interest she earns. Suppose she makes n cash withdrawals for the same amount during the month. Then her average cash balance at any given time is A 2n . (Why?) Find the value of n that minimizes total costs (transaction costs and lost interest), and then show that the optimal average cash balance is sAT 2R . Newton ’s Method Resources / Module 5 / Newton’s Method / Start of Newton’s Method Suppose that a car dealer offers to sell you a car for $18,000 or for payments of $375 per month for five years. You would like to know what monthly interest rate the dealer is, in effect, charging you. To find the answer, you have to solve the equation 1 48 x 1 x 60 1 x 60 1 0 (The details are explained in Exercise 39.) How would you solve such an equation? For a quadratic equation ax 2 b x c 0 there is a well-known formula for the roots. For third- and fourth-degree equations there are also formulas for the roots, but they are extremely complicated. If f is a polynomial of degree 5 or higher, there is no such formula 5E-04(pp 292-301) 1/17/06 3:17 PM Page 295 SECTION 4.9 NEWTON’S METHOD 0.15 0 0.012 _0.05 FIGURE 1 |||| Try to solve Equation 1 using the numerical rootfinder on your calculator or computer. Some machines are not able to solve it. Others are successful but require you to specify a starting point for the search. y { x ¡, f (x ¡)} y=ƒ L 0 x™ x ¡ r x f x1 f x1 x Since the x-intercept of L is x 2 , we set y 0 If f x 1 { x ¡, f (x ¡)} r x™ x¡ x1 x x2 f x1 f x1 f x2 f x2 If we keep repeating this process, we obtain a sequence of approximations x 1, x 2, x 3, x 4, . . . as shown in Figure 3. In general, if the nth approximation is x n and f x n 0, then the next approximation is given by FIGURE 3 2 |||| Sequences were briefly introduced in A Preview of Calculus on page 6. A more thorough discussion starts in Section 12.1. x1 0, we can solve this equation for x 2 : x3 { x ™, f (x ™)} x£ f x1 x2 We use x2 as a second approximation to r. Next we repeat this procedure with x 1 replaced by x 2 , using the tangent line at x 2 , f x 2 . This gives a third approximation: y x¢ x1 0 and obtain f x1 x2 0 295 (see the note on page 187). Likewise, there is no formula that will enable us to find the exact roots of a transcendental equation such as cos x x. We can find an approximate solution to Equation 1 by plotting the left side of the equation. Using a graphing device, and after experimenting with viewing rectangles, we produce the graph in Figure 1. We see that in addition to the solution x 0, which doesn’t interest us, there is a solution between 0.007 and 0.008. Zooming in shows that the root is approximately 0.0076. If we need more accuracy we could zoom in repeatedly, but that becomes tiresome. A faster alternative is to use a numerical rootfinder on a calculator or computer algebra system. If we do so, we find that the root, correct to nine decimal places, is 0.007628603. How do those numerical rootfinders work? They use a variety of methods, but most of them make some use of Newton’s method, also called the Newton-Raphson method. We will explain how this method works, partly to show what happens inside a calculator or computer, and partly as an application of the idea of linear approximation. The geometry behind Newton’s method is shown in Figure 2, where the root that we are trying to find is labeled r. We start with a first approximation x 1, which is obtained by guessing, or from a rough sketch of the graph of f , or from a computer-generated graph of f. Consider the tangent line L to the curve y f x at the point x 1, f x 1 and look at the x-intercept of L, labeled x 2. The idea behind Newton’s method is that the tangent line is close to the curve and so its x-intercept, x2 , is close to the x-intercept of the curve (namely, the root r that we are seeking). Because the tangent is a line, we can easily find its x-intercept. To find a formula for x2 in terms of x1 we use the fact that the slope of L is f x1 , so its equation is y FIGURE 2 ❙❙❙❙ xn xn 1 f xn f xn If the numbers x n become closer and closer to r as n becomes large, then we say that the sequence converges to r and we write lim x n nl r 5E-04(pp 292-301) 296 ❙❙❙❙ 1/17/06 3:18 PM Page 296 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION | Although the sequence of successive approximations converges to the desired root for functions of the type illustrated in Figure 3, in certain circumstances the sequence may not converge. For example, consider the situation shown in Figure 4. You can see that x 2 is a worse approximation than x 1. This is likely to be the case when f x 1 is close to 0. It might even happen that an approximation (such as x 3 in Figure 4) falls outside the domain of f . Then Newton’s method fails and a better initial approximation x 1 should be chosen. See Exercises 29–32 for specific examples in which Newton’s method works very slowly or does not work at all. y x™ 0 x£ x¡ r x EXAMPLE 1 Starting with x 1 equation x 3 2 x 5 0. FIGURE 4 2, find the third approximation x 3 to the root of the SOLUTION We apply Newton’s method with x3 fx 2x 5 and 3x 2 fx 2 Newton himself used this equation to illustrate his method and he chose x 1 2 after some experimentation because f 1 6, f 2 1, and f 3 16. Equation 2 becomes 3 xn 2xn 5 xn 1 xn 2 3x n 2 With n 1 we have |||| Figure 5 shows the geometry behind the first step in Newton’s method in Example 1. Since f 2 10, the tangent line to y x 3 2 x 5 at 2, 1 has equation y 10 x 21 so its x-intercept is x 2 2.1. x2 x3 1 x1 23 2 2x1 5 3x 2 2 1 22 5 322 2 2.1 1 Then with n 1.8 2.2 x3 x™ x2 2.1 y=10x-21 _2 FIGURE 5 2 we obtain x3 2 2x2 5 3x 2 2 2 2.1 3 2 2.1 3 2.1 2 2 It turns out that this third approximation x 3 5 2.0946 2.0946 is accurate to four decimal places. Suppose that we want to achieve a given accuracy, say to eight decimal places, using Newton’s method. How do we know when to stop? The rule of thumb that is generally used is that we can stop when successive approximations x n and x n 1 agree to eight decimal places. (A precise statement concerning accuracy in Newton’s method will be given in Exercises 12.12.) Notice that the procedure in going from n to n 1 is the same for all values of n. (It is called an iterative process.) This means that Newton’s method is particularly convenient for use with a programmable calculator or a computer. 6 EXAMPLE 2 Use Newton’s method to find s2 correct to eight decimal places. 6 SOLUTION First we observe that finding s2 is equivalent to finding the positive root of the equation x6 2 0 5E-04(pp 292-301) 1/17/06 3:18 PM Page 297 SECTION 4.9 NEWTON’S METHOD so we take f x becomes x6 If we choose x 1 297 6 x 5 and Formula 2 (Newton’s method) 2. Then f x xn ❙❙❙❙ 1 x6 2 n 5 6x n xn 1 as the initial approximation, then we obtain x2 1.16666667 x3 1.12644368 x4 1.12249707 x5 1.12246205 x6 1.12246205 Since x 5 and x 6 agree to eight decimal places, we conclude that 6 s2 1.12246205 to eight decimal places. EXAMPLE 3 Find, correct to six decimal places, the root of the equation cos x x. SOLUTION We first rewrite the equation in standard form: cos x Therefore, we let f x xn y y=x y=cos x 1 π 2 x π x cos x 1 x. Then f x xn cos x n x n sin x n 1 0 sin x xn cos x n sin x n xn 1 In order to guess a suitable value for x 1 we sketch the graphs of y cos x and y x in Figure 6. It appears that they intersect at a point whose x-coordinate is somewhat less than 1, so let’s take x 1 1 as a convenient first approximation. Then, remembering to put our calculator in radian mode, we get x2 0.75036387 x3 0.73911289 x4 0.73908513 x5 FIGURE 6 1, so Formula 2 becomes 0.73908513 Since x 4 and x 5 agree to six decimal places (eight, in fact), we conclude that the root of the equation, correct to six decimal places, is 0.739085. Instead of using the rough sketch in Figure 6 to get a starting approximation for Newton’s method in Example 3, we could have used the more accurate graph that a calculator or computer provides. Figure 7 suggests that we use x1 0.75 as the initial approximation. Then Newton’s method gives 1 y=cos x x2 0 FIGURE 7 1 0.73911114 x3 y=x 0.73908513 x4 0.73908513 and so we obtain the same answer as before, but with one fewer step. 5E-04(pp 292-301) 298 ❙❙❙❙ 1/17/06 3:19 PM Page 298 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION You might wonder why we bother at all with Newton’s method if a graphing device is available. Isn’t it easier to zoom in repeatedly and find the roots as we did in Section 1.4? If only one or two decimal places of accuracy are required, then indeed Newton’s method is inappropriate and a graphing device suffices. But if six or eight decimal places are required, then repeated zooming becomes tiresome. It is usually faster and more efficient to use a computer and Newton’s method in tandem—the graphing device to get started and Newton’s method to finish. |||| 4.9 Exercises 6. x 3 ton’s method is used to approximate the root r of the equation fx 0 with initial approximation x 1 1. (a) Draw the tangent lines that are used to find x 2 and x 3, and estimate the numerical values of x 2 and x 3. (b) Would x 1 5 be a better first approximation? Explain. x2 1 0, x1 7. x 4 20 0, x1 2 8. x 1. The figure shows the graph of a function f . Suppose that New- 5 ■ 2 ■ 0, ■ x1 1 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ; 9. Use Newton’s method with initial approximation x1 1 to find x 2 , the second approximation to the root of the equation x 3 x 3 0. Explain how the method works by first graphing the function and its tangent line at 1, 1 . y ; 10. Use Newton’s method with initial approximation x1 1 to find x 2 , the second approximation to the root of the equation x 4 x 1 0. Explain how the method works by first graphing the function and its tangent line at 1, 1 . 1 0 r 1 s x 11–12 2. Follow the instructions for Exercise 1(a) but use x 1 9 as the starting approximation for finding the root s. |||| Use Newton’s method to approximate the given number correct to eight decimal places. ■ 3. Suppose the line y 5x 4 is tangent to the curve y f x when x 3. If Newton’s method is used to locate a root of the equation f x 0 and the initial approximation is x1 3, find the second approximation x2. 4. For each initial approximation, determine graphically what happens if Newton’s method is used for the function whose graph is shown. (a) x1 0 (b) x1 1 (c) x1 3 (d) x1 4 (e) x1 5 7 12. s1000 3 11. s30 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 13–16 |||| Use Newton’s method to approximate the indicated root of the equation correct to six decimal places. 13. The root of 2 x 3 14. The root of x 4 6x 2 x 3x 4 0 in the interval 1, 2 x2 15. The positive root of sin x 16. The positive root of 2 cos x ■ ■ ■ ■ ■ 0 in the interval 2, 3 1 x4 ■ ■ ■ ■ ■ ■ ■ y 17–22 |||| Use Newton’s method to find all roots of the equation correct to six decimal places. 17. x 4 0 1 3 5 x 1 3 19. s x x2 21. cos x ■ 5–8 |||| Use Newton’s method with the specified initial approximation x 1 to find x 3 , the third approximation to the root of the given equation. (Give your answer to four decimal places.) 5. x 3 2x 4 0, x1 1 18. x 5 1 ■ 2 3 x2 22. tan x sx ■ 5x 20. sx x ■ ■ ■ ■ ■ ; 23–26 s1 ■ x2 ■ ■ ■ |||| Use Newton’s method to find all the roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations. 23. x 5 x4 5x 3 x2 4x 3 0 5E-04(pp 292-301) 1/17/06 3:19 PM Page 299 S ECTION 4.9 NEWTON’S METHOD 24. x 2 4 2 25. x s 2 ■ 4 x2 ■ x ■ x 299 34. Use Newton’s method to find the absolute minimum value of x2 2 ❙❙❙❙ 26. 3 sin x 1 ■ ■ ■ ■ 2 ■ ■ ■ ■ a 0 to derive the following square-root algorithm (used by the ancient Babylonians to compute sa ) : 1 xn 2 1 28. (a) Apply Newton’s method to the equation 1 x a 0 to derive the following reciprocal algorithm: 1 2 ax n 2xn x3 point of the curve y cos x correct to six decimal places. 36. Of the infinitely many lines that are tangent to the curve y sin x and pass through the origin, there is one that has the largest slope. Use Newton’s method to find the slope of that line correct to six decimal places. 37. A grain silo consists of a cylindrical main section, with height a xn (b) Use part (a) to compute s1000 correct to six decimal places. xn sin x correct to six decimal places. 35. Use Newton’s method to find the coordinates of the inflection 2x ■ 27. (a) Apply Newton’s method to the equation x 2 xn x2 the function f x 1 30 ft, and a hemispherical roof. In order to achieve a total volume of 15,000 ft 3 (including the part inside the roof section), what would the radius of the silo have to be? 38. In the figure, the length of the chord AB is 4 cm and the length of the arc AB is 5 cm. Find the central angle , in radians, correct to four decimal places. Then give the answer to the nearest degree. 5 cm (This algorithm enables a computer to find reciprocals without actually dividing.) (b) Use part (a) to compute 1 1.6984 correct to six decimal places. 4 cm A B ¨ 29. Explain why Newton’s method doesn’t work for finding the root of the equation x 3 tion is chosen to be x 1 3x 1. 6 0 if the initial approxima- 30. (a) Use Newton’s method with x 1 ; 1 to find the root of the equation x 3 x 1 correct to six decimal places. (b) Solve the equation in part (a) using x 1 0.6 as the initial approximation. (c) Solve the equation in part (a) using x 1 0.57. (You definitely need a programmable calculator for this part.) (d) Graph f x x 3 x 1 and its tangent lines at x1 1, 0.6, and 0.57 to explain why Newton’s method is so sensitive to the value of the initial approximation. 39. A car dealer sells a new car for $18,000. He also offers to sell the same car for payments of $375 per month for five years. What monthly interest rate is this dealer charging? To solve this problem you will need to use the formula for the present value A of an annuity consisting of n equal payments of size R with interest rate i per time period: 3 tion sx 0 with any initial approximation x 1 your explanation with a sketch. 0. Illustrate 48 x 1 fx 0 0 x 60 1 3x 4 28 x 3 6x 2 correct to two decimal places.. x 60 1 24 x 0 the point 1, 0 . (The unit here is the distance between the centers of Earth and the Sun, called an astronomical unit: y L¢ Earth 2 function f x 28 x 6x 24 x correct to three 3x decimal places. (b) Find the absolute minimum value of the function fx n 40. The figure shows the Sun located at the origin and Earth at 33. (a) Use Newton’s method to find the critical numbers of the 3 i Use Newton’s method to solve this equation. then the root of the equation f x 0 is x 0. Explain why Newton’s method fails to find the root no matter which initial approximation x 1 0 is used. Illustrate your explanation with a sketch. 4 1 Replacing i by x, show that 32. If if x sx s x if x R 1 i A 31. Explain why Newton’s method fails when applied to the equa- 1 x Sun L∞ L¡ 7 L£ L™ x 5E-04(pp 292-301) 300 ❙❙❙❙ 1/17/06 3:20 PM Page 300 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 1 AU 1.496 10 8 km.) There are five locations L 1 , L 2 , L 3 , L 4 , and L 5 in this plane of rotation of Earth about the Sun where a satellite remains motionless with respect to Earth because the forces acting on the satellite (including the gravitational attractions of Earth and the Sun) balance each other. These locations are called libration points. (A solar research satellite has been placed at one of these libration points.) If m1 is the mass of the Sun, m 2 is the mass of Earth, and r m 2 m1 m 2 , it turns out that the x-coordinate of L 1 is |||| 4.10 the unique root of the fifth-degree equation x5 2 r x4 21 px rx 2r x 3 1 r 1 1 r x2 0 and the x-coordinate of L 2 is the root of the equation 2rx 2 px 0 6 Using the value r 3.04042 10 , find the locations of the libration points (a) L 1 and (b) L 2. Antiderivatives A physicist who knows the velocity of a particle might wish to know its position at a given time. An engineer who can measure the variable rate at which water is leaking from a tank wants to know the amount leaked over a certain time period. A biologist who knows the rate at which a bacteria population is increasing might want to deduce what the size of the population will be at some future time. In each case, the problem is to find a function F whose derivative is a known function f. If such a function F exists, it is called an antiderivative of f. Resources / Module 6 / Antiderivatives / Start of Antiderivatives Definition A function F is called an antiderivative of f on an interval I if Fx y ˛ y= 3 +3 f x for all x in I . For instance, let f x x 2. It isn’t difficult to discover an antiderivative of f if we keep 13 the Power Rule in mind. In fact, if F x x 2 f x . But the function 3 x , then F x 13 2 Gx 100 also satisfies G x x . Therefore, both F and G are antiderivatives 3x 13 of f . Indeed, any function of the form H x C, where C is a constant, is an anti3x derivative of f . The question arises: Are there any others? To answer this question, recall that in Section 4.2 we used the Mean Value Theorem to prove that if two functions have identical derivatives on an interval, then they must differ by a constant (Corollary 4.2.7). Thus, if F and G are any two antiderivatives of f , then Fx ˛ y= 3 +2 ˛ y= 3 +1 fx Gx so G x Fx C, where C is a constant. We can write this as G x have the following result. Fx C, so we y= ˛ 0 x 3 1 Theorem If F is an antiderivative of f on an interval I , then the most general antiderivative of f on I is Fx C ˛ y= 3 -1 ˛ y= 3 -2 where C is an arbitrary constant. FIGURE 1 Members of the family of antiderivatives of ƒ=≈ Going back to the function f x x 2, we see that the general antiderivative of f is x 3 C. By assigning specific values to the constant C, we obtain a family of functions whose graphs are vertical translates of one another (see Figure 1). 3 EXAMPLE 1 Find the most general antiderivative of each of the following functions. (a) f x sin x (b) f x x n, n 0 (c) f x x 3 5E-04(pp 292-301) 1/17/06 3:20 PM Page 301 SECTION 4.10 ANTIDERIVATIVES ❙❙❙❙ 301 S OLUTION (a) If F x cos x, then F x sin x, so an antiderivative of sin x is Theorem 1, the most general antiderivative is G x cos x C. xn 1 n1 d dx (b) n 1 xn n1 xn x n is Thus, the general antiderivative of f x xn 1 n1 Fx This is valid for n cos x. By 0 because then f x C x n is defined on an interval. (c) If we put n 3 in part (b) we get the particular antiderivative F x x2 2 3 by the same calculation. But notice that f x x is not defined at x 0. Thus, Theorem 1 tells us only that the general antiderivative of f is x 2 2 C on any interval that does not contain 0. So the general antiderivative of f x 1 x 3 is 1 2x 2 1 2x 2 Fx C1 if x 0 C2 if x 0 As in Example 1, every differentiation formula, when read from right to left, gives rise to an antidifferentiation formula. In Table 2 we list some particular antiderivatives. Each formula in the table is true because the derivative of the function in the right column appears in the left column. In particular, the first formula says that the antiderivative of a constant times a function is the constant times the antiderivative of the function. The second formula says that the antiderivative of a sum is the sum of the antiderivatives. (We use the notation F f, G t.) 2 Table of Antidifferentiation Formulas Function Particular antiderivative cf x |||| To obtain the most general antiderivative from the particular ones in Table 2, we have to add a constant (or constants), as in Example 1. fx xn n cF x tx Particular antiderivative cos x Fx sin x sin x Gx cos x 2 sec x tan x sec x tan x xn 1 n1 1 Function sec x EXAMPLE 2 Find all functions t such that tx 4 sin x 2x 5 sx x SOLUTION We first rewrite the given function as follows: 2x 5 sx x x Thus, we want to find an antiderivative of tx 4 sin x tx 4 sin x 4 sin x 2x 4 x 12 2x 4 1 sx 5E-04(pp 302-310) 302 ❙❙❙❙ 1/17/06 3:10 PM Page 302 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION Using the formulas in Table 2 together with Theorem 1, we obtain tx 4 cos x x5 5 2 2 5 4 cos x x1 2 C 1 2 x5 2sx C In applications of calculus it is very common to have a situation as in Example 2, where it is required to find a function, given knowledge about its derivatives. An equation that involves the derivatives of a function is called a differential equation. These will be studied in some detail in Chapter 10, but for the present we can solve some elementary differential equations. The general solution of a differential equation involves an arbitrary constant (or constants) as in Example 2. However, there may be some extra conditions given that will determine the constants and therefore uniquely specify the solution. EXAMPLE 3 Find f if f x x s x and f 1 2. SOLUTION The general antiderivative of x3 2 fx is x5 2 fx To determine C we use the fact that f 1 2 5 2 8 5 2 5 12 x 2 x C C 2 , so the particular solution is 2x 5 2 5 fx EXAMPLE 4 Find f if f x5 2 2: f1 Solving for C, we get C 2 5 C 5 2 6x 8 4, f 0 SOLUTION The general antiderivative of f x 4, and f 1 12 x 2 6x 4 is x3 x2 6 4 x C 4 x 3 3x 2 3 2 Using the antidifferentiation rules once more, we find that fx fx 4 12 x4 4 3 x3 3 4 x2 2 Cx D x4 x3 To determine C and D we use the given conditions that f 0 f0 0 D 4, we have D 4. Since f1 we have C 1 1 2 C 1. 4x C 2x 2 Cx 4 and f 1 4 1. Since 1 3x D 4 3. Therefore, the required function is fx x4 x3 2x 2 The Geometry of Antiderivatives If we are given the graph of a function f , it seems reasonable that we should be able to sketch the graph of an antiderivative F . Suppose, for instance, that we are given that 5E-04(pp 302-310) 1/17/06 3:10 PM Page 303 S ECTION 4.10 ANTIDERIVATIVES ❙❙❙❙ 303 F0 1. Then we have a place to start, the point 0, 1 , and the direction in which we move our pencil is given at each stage by the derivative F x f x . In the next example we use the principles of this chapter to show how to graph F even when we don’t have a formula for f . This would be the case, for instance, when f x is determined by experimental data. y EXAMPLE 5 The graph of a function f is given in Figure 2. Make a rough sketch of an antiderivative F , given that F 0 2. y=ƒ 0 1 2 3 4 x FIGURE 2 y y=F(x) SOLUTION We are guided by the fact that the slope of y F x is f x . We start at the point 0, 2 and draw F as an initially decreasing function since f x is negative when 0 x 1. Notice that f 1 f3 0, so F has horizontal tangents when x 1 and x 3. For 1 x 3, f x is positive and so F is increasing. We see that F has a local minimum when x 1 and a local maximum when x 3. For x 3, f x is negative and so F is decreasing on 3, . Since f x l 0 as x l , the graph of F becomes flatter as x l . Also notice that F x f x changes from positive to negative at x 2 and from negative to positive at x 4, so F has inflection points when x 2 and x 4. We use this information to sketch the graph of the antiderivative in Figure 3. 2 s1 x 3 x, sketch the graph of the antiderivative F that satisfies the initial condition F 1 0. EXAMPLE 6 If f x 1 0 FIGURE 3 1 x SOLUTION We could try all day to think of a formula for an antiderivative of f and still be unsuccessful. A second possibility would be to draw the graph of f first and then use it to graph F as in Example 5. That would work, but instead let’s create a more accurate graph by using what is called a direction field. Since f 0 1, the graph of F has slope 1 when x 0. So we draw several short tangent segments with slope 1, all centered at x 0. We do the same for several other values of x and the result is shown in Figure 4. It is called a direction field because each segment indicates the direction in which the curve y F x proceeds at that point. y 4 y 4 3 3 2 2 1 1 _1 1 2 3x _1 _1 1 2 3x _1 FIGURE 4 FIGURE 5 A direction field for ƒ=œ„„„„„-x. 1+˛ The slope of the line segments above x=a is f(a). The graph of an antiderivative follows the direction field. Now we use the direction field to sketch the graph of F . Because of the initial condition F 1 0, we start at the point 1, 0 and draw the graph so that it follows the directions of the tangent segments. The result is pictured in Figure 5. Any other antiderivative would be obtained by shifting the graph of F upward or downward. 5E-04(pp 302-310) 304 ❙❙❙❙ 1/17/06 3:10 PM Page 304 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION R ectilinear Motion Antidifferentiation is particularly useful in analyzing the motion of an object moving in a straight line. Recall that if the object has position function s f t , then the velocity funcs t . This means that the position function is an antiderivative of the veloction is v t v t , so the velocity function is ity function. Likewise, the acceleration function is a t an antiderivative of the acceleration. If the acceleration and the initial values s 0 and v 0 are known, then the position function can be found by antidifferentiating twice. EXAMPLE 7 A particle moves in a straight line and has acceleration given by 6t 4. Its initial velocity is v 0 6 cm s and its initial displacement is at s0 9 cm. Find its position function s t . SOLUTION Since v t at 6t vt Note that v 0 4, antidifferentiation gives 6 t2 2 4t C. But we are given that v 0 4t 6, so C 3t 2 vt Since v t 3t 2 C 4t C 6 and 6 s t , s is the antiderivative of v : st This gives s 0 function is 3 t3 3 4 t2 2 6t D D. We are given that s 0 t3 st t3 2t 2 9, so D 2t 2 6t 6t D 9 and the required position 9 An object near the surface of the earth is subject to a gravitational force that produces a downward acceleration denoted by t. For motion close to the ground we may assume that t is constant, its value being about 9.8 m s2 (or 32 ft s2 ). EXAMPLE 8 A ball is thrown upward with a speed of 48 ft s from the edge of a cliff 432 ft above the ground. Find its height above the ground t seconds later. When does it reach its maximum height? When does it hit the ground? SOLUTION The motion is vertical and we choose the positive direction to be upward. At time t the distance above the ground is s t and the velocity v t is decreasing. Therefore, the acceleration must be negative and we have at dv dt 32 Taking antiderivatives, we have vt 32 t C To determine C we use the given information that v 0 vt 32 t 48 48. This gives 48 0 C, so 5E-04(pp 302-310) 1/17/06 3:11 PM Page 305 ❙❙❙❙ S ECTION 4.10 ANTIDERIVATIVES The maximum height is reached when v t antidifferentiate again and obtain 16t 2 st Using the fact that s 0 48t 432, we have 432 v t , we D 0 16t 2 st |||| Figure 6 shows the position function of the ball in Example 8. The graph corroborates the conclusions we reached: The ball reaches its maximum height after 1.5 s and hits the ground after 6.9 s. 0, that is, after 1.5 s. Since s t D and so 48t 432 The expression for s t is valid until the ball hits the ground. This happens when s t that is, when 16t 2 48t 305 432 0 27 0, 0 500 t2 or, equivalently, 3t Using the quadratic formula to solve this equation, we get 3 t 8 0 We reject the solution with the minus sign since it gives a negative value for t. Therefore, the ball hits the ground after 3(1 s13 ) 2 6.9 s. FIGURE 6 |||| 4.10 Exercises 1–16 |||| Find the most general antiderivative of the function. (Check your answer by differentiation.) 1. f x 6x 2 3. f x 1 8x x3 5. f x 5x 1 4 7. f x 6 sx 9. f x 2. f x 5x 5 3x 7 15. f t ■ 17–36 |||| 21. f x t 4 sx 3 12. f x 5 sin x ■ 3 cos t 16. f sec t tan t ■ x2 3 14. f t u2 ■ ■ 6 ■ ■ ■ 3x , f1 f 24 x 2 2x 10, 4 sin cos , f0 3 s t, f4 20, 2 34. f x 20 x 35. f x 2 36. f 18. f x 2 1 45 20. f x 22. f t t 60t 23. f x 1 24. f x 8x 3 6 x, f0 12 x 8 3, f1 st x6 x 12 x cos x, sin x, f0 4, 4 1 f0 4 7 f2 15 f0 f0 3 f0 f4 1, 1, 3 f1 2, 3, f0 f0 5, 9, 2 f 0 f1 f0 2, 8, f f1 2 5 0 1, f0 ■ ■ 1 cos x x 2 x3 12 x, 3 t 1 40 x 3, 6x 3 2 2 33. f x ■ f1 sec 2 t, 32. f t 13 Find f . 12 x 2 0, 31. f 7 sec2 6x x 10 30. f x 4 sin t ■ 3 x 4, f1 29. f x 2x 6 2x 2 2 cos t 5x , 28. f x 3 sx 4 4x 3 x6 5 2x 27. f t 8 sx 6 26. f x 5x 3 3x 1.7 8. f x 3 su ■ 17. f x 19. f u4 25. f x 4 x 10 2x 10. t x 4 st ■ x 20 6. f x 6 sx x3 13. h x 7x 3 4 x2 4 4. f x 3 10 x9 11. f u 3 s13 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ 37. Given that the graph of f passes through the point 1, 6 and that the slope of its tangent line at x, f x is 2 x 38. Find a function f such that f x 6 ■ is tangent to the graph of f . 3 1, find f 2 . x and the line x y 0 5E-04(pp 302-310) 306 ❙❙❙❙ 1/17/06 3:12 PM Page 306 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 39–40 |||| The graph of a function f is shown. Which graph is an antiderivative of f and why? 39. 40. y f |||| Draw a graph of f and use it to make a rough sketch of the antiderivative that passes through the origin. y 45. f x a sin x 2 , 46. f x f b a ; 45–46 1 x4 ■ ■ b c 47. ■ ■ ■ 4 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ sketch of an antiderivative F , given that F 0 48. y ■ ■ ■ ■ y 2 41. The graph of a function is shown in the figure. Make a rough 10 8 1 0. 6 y 0 1 4 3x 2 2 _1 0 _2 _2 0 ■ 47–48 |||| A direction field is given for a function. Use it to draw the antiderivative F that satisfies F 0 2. c ■ x x x ■ 0 4 12 x 8 x 2 ■ ■ 49–50 ■ |||| fies F 0 ■ ■ 49 f x sin x , x 50. f x √ x tan x, ■ ■ ■ ■ ■ ■ ■ Use a direction field to graph the antiderivative that satis0. 42. The graph of the velocity function of a car is shown in the fig- ure. Sketch the graph of the position function. ■ ■ ■ 0 x 2 2 ■ ■ x 2 ■ ■ ■ ■ ■ ■ ■ 51. A function is defined by the following experimental data. Use a t 0 direction field to sketch the graph of its antiderivative if the initial condition is F 0 0. x 43. The graph of f is shown in the figure. Sketch the graph of f if f is continuous and f 0 1. 0.2 0.4 0.6 0.8 1.0 1.2 1.4 fx 0 0.2 0.5 0.8 1.0 0.6 0.2 0 1.6 0.1 1 x 2 and use it to sketch several members of the family of antiderivatives. (b) Compute the general antiderivative explicitly and sketch several particular antiderivatives. Compare with your sketch in part (a). 52. (a) Draw a direction field for the function f x y 2 y=fª(x) 1 0 _1 0 1 2 x 53–58 |||| A particle is moving with the given data. Find the position of the particle. 53. v t ; 44. (a) Use a graphing device to graph f x 2 x 3 sx. (b) Starting with the graph in part (a), sketch a rough graph of the antiderivative F that satisfies F 0 1. (c) Use the rules of this section to find an expression for F x . (d) Graph F using the expression in part (c). Compare with your sketch in part (b). sin t 54. v t cos t, s0 1.5 st, s4 10 55. a t t s0 1, v 0 56. a t cos t sin t, s0 57. a t 10 sin t 2, 3 cos t, 0 3 0, v 0 s0 0, 5 s2 12 5E-04(pp 302-310) 1/17/06 3:12 PM Page 307 S ECTION 4.10 ANTIDERIVATIVES 58. a t ■ ■ 10 ■ 3 t 2, ■ s0 ■ 3t ■ 0, s 2 ■ ■ ■ ■ ■ 59. A stone is dropped from the upper observation deck (the Space Deck) of the CN Tower, 450 m above the ground. (a) Find the distance of the stone above ground level at time t. (b) How long does it take the stone to reach the ground? (c) With what velocity does it strike the ground? (d) If the stone is thrown downward with a speed of 5 m s, how long does it take to reach the ground? 60. Show that for motion in a straight line with constant acceleration a, initial velocity v0 , and initial displacement s0 , the dis- placement after time t is 1 2 s at 2 v0 t per second from a point s0 meters above the ground. Show that vt v2 0 19.6 s t x 1 s x, in grams per centimeter, where x is measured in centimeters from one end of the rod. Find the mass of the rod. 67. Since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A raindrop has an initial downward velocity of 10 m s and its downward acceleration is a 9 0 0.9t if 0 if t t 10 10 If the raindrop is initially 500 m above the ground, how long does it take to fall? 68. A car is traveling at 50 mi h when the brakes are fully applied, s0 61. An object is projected upward with initial velocity v0 meters 2 307 66. The linear density of a rod of length 1 m is given by 10 ■ ❙❙❙❙ producing a constant deceleration of 22 ft s2. What is the distance covered before the car comes to a stop? 69. What constant acceleration is required to increase the speed of a car from 30 mi h to 50 mi h in 5 s? s0 70. A car braked with a constant deceleration of 16 ft s2, produc62. Two balls are thrown upward from the edge of the cliff in Example 8. The first is thrown with a speed of 48 ft s and the other is thrown a second later with a speed of 24 ft s. Do the balls ever pass each other? 63. A stone was dropped off a cliff and hit the ground with a speed of 120 ft s. What is the height of the cliff? length L and linear density , then the board takes on the shape of a curve y f x , where mt L x 1 2 tL x 2 and E and I are positive constants that depend on the material of the board and t 0 is the acceleration due to gravity. (a) Find an expression for the shape of the curve. (b) Use f L to estimate the distance below the horizontal at the end of the board. y 0 71. A car is traveling at 100 km h when the driver sees an accident 80 m ahead and slams on the brakes. What constant deceleration is required to stop the car in time to avoid a pileup? 72. A model rocket is fired vertically upward from rest. Its acceler- 64. If a diver of mass m stands at the end of a diving board with EI y ing skid marks measuring 200 ft before coming to a stop. How fast was the car traveling when the brakes were first applied? x 65. A company estimates that the marginal cost (in dollars per item) of producing x items is 1.92 0.002 x. If the cost of producing one item is $562, find the cost of producing 100 items. ation for the first three seconds is a t 60 t, at which time the fuel is exhausted and it becomes a freely “falling” body. Fourteen seconds later, the rocket’s parachute opens, and the (downward) velocity slows linearly to 18 ft s in 5 s. The rocket then “floats” to the ground at that rate. (a) Determine the position function s and the velocity function v (for all times t). Sketch the graphs of s and v. (b) At what time does the rocket reach its maximum height, and what is that height? (c) At what time does the rocket land? 73. A high-speed bullet train accelerates and decelerates at the rate of 4 ft s2. Its maximum cruising speed is 90 mi h. (a) What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes? (b) Suppose that the train starts from rest and must come to a complete stop in 15 minutes. What is the maximum distance it can travel under these conditions? (c) Find the minimum time that the train takes to travel between two consecutive stations that are 45 miles apart. (d) The trip from one station to the next takes 37.5 minutes. How far apart are the stations? 5E-04(pp 302-310) ❙❙❙❙ 308 |||| 1/17/06 3:12 PM Page 308 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 4 Review ■ CONCEPT CHECK 1. Explain the difference between an absolute maximum and a local maximum. Illustrate with a sketch. 9. (a) Given an initial approximation x1 to a root of the equation (b) State the Mean Value Theorem and give a geometric interpretation. 5. (a) State the Increasing /Decreasing Test. (b) State the Concavity Test. 6. (a) State the First Derivative Test. (b) State the Second Derivative Test. (c) What are the relative advantages and disadvantages of these tests? 7. Explain the meaning of each of the following statements. L ■ 0, then f has a local maximum or minimum at c. 2. If f has an absolute minimum value at c, then f c 0. 3. If f is continuous on a, b , then f attains an absolute maxi- mum value f c and an absolute minimum value f d at some numbers c and d in a, b . 1 f 1 , then there is a number 0. 5. If f x 0 for 1 6. If f 2 0, then 2, f 2 is an inflection point of the f x. curve y 7. If f x 0 x t x for 0 1. x 6, then f is decreasing on (1, 6). x 1, then f x 8. There exists a function f such that f 1 fx fx 0 for all x. 10. (a) What is an antiderivative of a function f ? (b) Suppose F1 and F2 are both antiderivatives of f on an interval I . How are F1 and F2 related? ■ 10. There exists a function f such that f x fx 11. If f and t are increasing on an interval I , then f t is increas- ing on I . 12. If f and t are increasing on an interval I , then f t is increas- ing on I . 13. If f and t are increasing on an interval I , then f t is increasing on I . 14. If f and t are positive increasing functions on an interval I , then f t is increasing on I . 0 on I , then t x 16. The most general antiderivative of f x 0, and Fx 0, f x 0, and 1 f x is decreasing on I . t x for 2, f 3 0, f x 0 for all x. 15. If f is increasing and f x 1 for all x. 9. There exists a function f such that f x fx 0, explain geometrically, with a diagram, how the second approximation x2 in Newton’s method is obtained. (b) Write an expression for x2 in terms of x1, f x 1 , and f x 1 . (c) Write an expression for x n 1 in terms of x n , f x n , and f xn . (d) Under what circumstances is Newton’s method likely to fail or to work very slowly? TRUE-FALSE QUIZ Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. 1 and f c L. need calculus to graph a function? 4. (a) State Rolle’s Theorem. c such that c f x has the horizontal asymptote y 8. If you have a graphing calculator or computer, why do you (b) Define a critical number of f . 4. If f is differentiable and f L xl 3. (a) State Fermat’s Theorem. 1. If f c xl (d) The curve y (b) Explain how the Closed Interval Method works. xl (b) lim f x (c) lim f x 2. (a) What does the Extreme Value Theorem say? (a) lim f x ■ 1 x x 2 is C 0, and 17. If f x exists and is nonzero for all x, then f 1 f 0. 5E-04(pp 302-310) 1/17/06 3:14 PM Page 309 C HAPTER 4 REVIEW ■ EXERCISES |||| Find the local and absolute extreme values of the function on the given interval. 10 2. f x x 3. f x sx, x x x2 x2 4. f x 5. f x x 6. f x x 3, 27x y y=f ª(x) 0, 4 _2 0 _1 2, 0 , 2 x 3, 1 2 ■ 7–12 ■ |||| cos x, 17–28 0, |||| 5 ■ ■ 8. lim tl 9. lim xl t 1t t 2 2t ■ ■ ■ ■ xl ■ ■ 3x ■ x 2 x3 6x 2 15 x 4 x4 3x 3 3x 2 x 2x ■ 2 t 1 xl x3 1 22. y 1 x x 1 24. y x s1 x x x s2 26. y sx sin2x x 3 sx 2 cos x 4 sin x 12. lim xl sx 2 x) ■ 25. y 27. y x2 10. lim x x2 2 3 23. y 1 8 xx 2 1 20. y 1 21. y s4x 2 1 3x 1 11. lim (s 4 x 2 7 x3 17. y ■ Find the limit. 3 6 Use the guidelines of Section 4.5 to sketch the curve. 19. y ■ 3x 4 x 5 7. lim xl 6x 4 2 x 2 1 ■ 4 0, 2 18. y ■ 3 2, 1 sin 2 x, sin x ■ 0, 4 1 309 (d) Sketch a possible graph of f . 1–6 1. f x ❙❙❙❙ ■ ■ 28. y ■ ■ ■ ■ 4x ■ tan x, ■ 2 ■ ■ x 2 ■ ■ ■ ■ ■ ■ ■ ■ ; 29–32 13–15 |||| Produce graphs of f that reveal all the important aspects of the curve. Use graphs of f and f to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points. In Exercise 29 use calculus to find these quantities exactly. Sketch the graph of a function that satisfies the given conditions. |||| 13. f 0 0, f 2 f1 f9 0, lim x l f x 0, lim x l 6 f x , fx 0 on , 2 , 1, 6 , and 9, , fx 0 on 2, 1 and 6, 9 , fx 0 on , 0 and 12, , fx 0 on 0, 6 and 6, 12 14. f 0 fx fx ■ ■ 31. f x 32. f x 0, f is continuous and even, 1 if 1 2 x if 0 x 1, f x 1 if x 3 15. f is odd, fx fx 29. f x fx 0 for x 0 for x ■ ■ 0 for 0 x 2, 2, f x 0 for 0 3, lim x l f x ■ ■ ■ ■ ■ x ■ x2 1 x3 3x 6 5x 5 x4 2 sin x cos x, ■ ■ 3 sx 30. f x 0 ■ 5x 3 x ■ 1 2x 2 x 2 2 ■ ■ ■ ■ ■ 3, 33. Show that the equation x 101 x 51 x 1 0 has exactly one real root. x 2 3, 34. Suppose that f is continuous on 0, 4 , f 0 2 ■ ■ ■ ■ 16. The figure shows the graph of the derivative f of a function f . (a) On what intervals is f increasing or decreasing? (b) For what values of x does f have a local maximum or minimum? (c) Sketch the graph of f . fx 5 for all x in 0, 4 . Show that 9 1, and f4 21. 35. By applying the Mean Value Theorem to the function fx x 1 5 on the interval 32, 33 , show that 2 5 s33 2.0125 36. For what values of the constants a and b is 1, 6 a point of inflection of the curve y x3 ax 2 bx 1? ■ 5E-04(pp 302-310) 310 ❙❙❙❙ 1/17/06 3:14 PM Page 310 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION f x 2 , where f is twice differentiable for all x, fx 0 for all x 0, and f is concave downward on and concave upward on 0, . (a) At what numbers does t have an extreme value? (b) Discuss the concavity of t. 50. Use Newton’s method to find all roots of the equation 37. Let t x 38. Find two positive integers such that the sum of the first number and four times the second number is 1000 and the product of the numbers is as large as possible. 39. Show that the shortest distance from the point x 1, y1 to the x2 sin x ,0 51. Use Newton’s method to find the absolute maximum value of the function f t places. By C cos t 0 53–58 40. Find the point on the hyperbola x y |||| Find f . sx 5 55. f t 8 that is closest to the point 3, 0 . 41. Find the smallest possible area of an isosceles triangle that is circumscribed about a circle of radius r. 42. Find the volume of the largest circular cone that can be x sin x, 2 . Use Newton’s method when necessary. x 2t 0 is A x 1 B y1 C sA2 B 2 t 2 correct to eight decimal t 52. Use the guidelines in Section 4.5 to sketch the curve y 53. f x straight line Ax 1 correct to six decimal places. 3x u 56. f u ■ su 1 58. f x 2x ■ 54. f x 3 sin t, 2 u 57. f x ■ 4 5 sx , 3x ■ 5 f1 3 48 x 2, 6x 3 f0 3 sec2x 8x f0 2 4x ■ 1, 5, ■ f0 f0 ■ 2 2, ■ f1 ■ 0 ■ ■ ■ inscribed in a sphere of radius r. 43. In ABC, D lies on AB, CD AB, AD BD 4 cm, 5 cm. Where should a point P be chosen on CD and CD so that the sum PA PB PC is a minimum? 44. Do Exercise 43 when CD 2 cm. K L C C L where K and C are known positive constants. What is the length of the wave that gives the minimum velocity? 46. A metal storage tank with volume V is to be constructed in the shape of a right circular cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal? 47. A hockey team plays in an arena with a seating capacity of 15,000 spectators. With the ticket price set at $12, average attendance at a game has been 11,000. A market survey indicates that for each dollar the ticket price is lowered, average attendance will increase by 1000. How should the owners of the team set the ticket price to maximize their revenue from ticket sales? ; 48. A manufacturer determines that the cost of making x units of a commodity is C x 1800 25x 0.2 x 2 0.001x 3 and the demand function is p x 48.2 0.03x. (a) Graph the cost and revenue functions and use the graphs to estimate the production level for maximum profit. (b) Use calculus to find the production level for maximum profit. (c) Estimate the production level that minimizes the average cost. 49. Use Newton’s method to find the root of the equation x5 x4 3x2 3x six decimal places. 2 , and use that graph to sketch fx x 2 sin x 2 , 0 x the antiderivative F of f that satisfies the initial condition F0 0. x 4 x 3 c x 2. In particular you should determine the transitional value of c at which the number of critical numbers changes and the transitional value at which the number of inflection points changes. Illustrate the various possible shapes with graphs. ; 60. Investigate the family of curves given by f x 45. The velocity of a wave of length L in deep water is v ; 59. Use a graphing device to draw a graph of the function 0 in the interval 1, 2 correct to 61. A canister is dropped from a helicopter 500 m above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of 100 m s. Will it burst? 62. In an automobile race along a straight road, car A passed car B twice. Prove that at some time during the race their accelerations were equal. State the assumptions that you make. 63. A light is to be placed atop a pole of height h feet to illuminate a busy traffic circle, which has a radius of 40 ft. The intensity of illumination I at any point P on the circle is directly proportional to the cosine of the angle (see the figure) and inversely proportional to the square of the distance d from the source. (a) How tall should the light pole be to maximize I ? (b) Suppose that the light pole is h feet tall and that a woman is walking away from the base of the pole at the rate of 4 ft s. At what rate is the intensity of the light at the point on her back 4 ft above the ground decreasing when she reaches the outer edge of the traffic circle? ¨ h d 40 5E-04(pp 311-313) 1/17/06 3:00 PM Page 311 PROBLEMS PLUS One of the most important principles of problem solving is analogy (see page 58). If you are having trouble getting started on a problem, it is sometimes helpful to start by solving a similar, but simpler, problem. The following example illustrates the principle. Cover up the solution and try solving it yourself first. EXAMPLE If x, y, and z are positive numbers, prove that x2 1 y2 1 z2 x yz 1 8 SOLUTION It may be difficult to get started on this problem. (Some students have tackled it by multiplying out the numerator, but that just creates a mess.) Let’s try to think of a similar, simpler problem. When several variables are involved, it’s often helpful to think of an analogous problem with fewer variables. In the present case we can reduce the number of variables from three to one and prove the analogous inequality x2 1 1 2 x for x 0 In fact, if we are able to prove (1), then the desired inequality follows because x2 1 y 2 1 z2 x yz x2 1 y2 1 x z2 1 1 z y 222 8 The key to proving (1) is to recognize that it is a disguised version of a minimum problem. If we let fx x2 1 x What have we learned from the solution to this example? |||| To solve a problem involving several variables, it might help to solve a similar problem with just one variable. |||| When trying to prove an inequality, it might help to think of it as a maximum or minimum problem. 1 1 x x x x 0 1 x 2 , so f x 0 when x 1. Also, f x 0 for 0 1. Therefore, the absolute minimum value of f is f 1 then f x 1 fx 0 for x means that Look Back x2 x 1 and 2. This for all positive values of x 2 and, as previously mentioned, the given inequality follows by multiplication. The inequality in (1) could also be proved without calculus. In fact, if x 0, we have x2 1 2 &? x2 &? x x 1 1 2 2 x &? x2 2x 1 0 0 Because the last inequality is obviously true, the first one is true too. 311 5E-04(pp 311-313) 1/17/06 3:01 PM P RO B L E M S Page 312 1. Show that sin x s2 for all x. cos x 2. Show that x 2 y 2 4 x2 4 y2 16 for all numbers x and y such that x 3. Let a and b be positive numbers. Show that not both of the numbers a 1 2 and y b and b 1 2. a 1 can be greater than 4. 1 x 2 at which the tangent line cuts from the first quadrant the triangle with the smallest area. 4. Find the point on the parabola y 5. Find the highest and lowest points on the curve x 2 y2 xy 12. 6. Water is flowing at a constant rate into a spherical tank. Let V t be the volume of water in the tank and H t be the height of the water in the tank at time t. (a) What are the meanings of V t and H t ? Are these derivatives positive, negative, or zero? (b) Is V t positive, negative, or zero? Explain. (c) Let t1, t 2, and t 3 be the times when the tank is one-quarter full, half full, and three-quarters full, respectively. Are the values H t1 , H t 2 , and H t 3 positive, negative, or zero? Why? 7. Find the absolute maximum value of the function 1 x 1 fx 1 8. Find a function f such that f x 1 2 1 1 ,f 0 2 0, and f x 0 for all x, or prove that such a function cannot exist. y m x b intersects the parabola y x 2 in points A and B (see the figure). Find the point P on the arc AOB of the parabola that maximizes the area of the triangle PAB. 9. The line y y=≈ B 10. Sketch the graph of a function f such that f x fx A 0 for x 1, and lim x l fx 0 for all x, f x 0. x 0 for x 1, 11. Determine the values of the number a for which the function f has no critical number: y=mx+b fx O P x a2 a 6 cos 2 x 2x cos 1 12. Sketch the region in the plane consisting of all points x, y such that x 2 xy FIGURE FOR PROBLEM 9 a y x2 y2 13. Let ABC be a triangle with BAC 120 and AB AC (a) Express the length of the angle bisector AD in terms of x (b) Find the largest possible value of AD . C 1. AB . 14. (a) Let ABC be a triangle with right angle A and hypotenuse a BC . (See the figure.) If the inscribed circle touches the hypotenuse at D, show that D CD 1 2 ( BC AC AB ) 1 2 C, express the radius r of the inscribed circle in terms of a and . (b) If (c) If a is fixed and varies, find the maximum value of r. A FIGURE FOR PROBLEM 14 B 15. A triangle with sides a, b, and c varies with time t, but its area never changes. Let be the angle opposite the side of length a and suppose always remains acute. (a) Express d dt in terms of b, c, , db dt, and dc dt. (b) Express da dt in terms of the quantities in part (a). 16. ABCD is a square piece of paper with sides of length 1 m. A quarter-circle is drawn from B to D with center A. The piece of paper is folded along EF , with E on AB and F on AD, so that A falls on the quarter-circle. Determine the maximum and minimum areas that the triangle AEF could have. 312 5E-04(pp 311-313) 1/17/06 3:01 PM Page 313 17. The speeds of sound c1 in an upper layer and c2 in a lower layer of rock and the thickness h of the upper layer can be determined by seismic exploration if the speed of sound in the lower layer is greater than the speed in the upper layer. A dynamite charge is detonated at a point P and the transmitted signals are recorded at a point Q, which is a distance D from P. The first signal to arrive at Q travels along the surface and takes T1 seconds. The next signal travels from P to a point R, from R to S in the lower layer, and then to Q, taking T2 seconds. The third signal is reflected off the lower layer at the midpoint O of RS and takes T3 seconds to reach Q. (a) Express T1, T2, and T3 in terms of D, h, c1 , c2 , and . (b) Show that T2 is a minimum when sin c1 c2. (c) Suppose that D 1 km, T1 0.26 s, T2 0.32 s, T3 0.34 s. Find c1 , c2 , and h. P Q D Speed of sound=c¡ h ¨ ¨ R S O Speed of sound=c™ Note: Geophysicists use this technique when studying the structure of Earth’s crust, whether searching for oil or examining fault lines. 18. For what values of c is there a straight line that intersects the curve y cx 3 12 x 2 5x 2 in four distinct points? d B x4 E C x r F D FIGURE FOR PROBLEM 19 19. One of the problems posed by the Marquis de l’Hospital in his calculus textbook Analyse des Infiniment Petits concerns a pulley that is attached to the ceiling of a room at a point C by a rope of length r. At another point B on the ceiling, at a distance d from C (where d r), a rope of length is attached and passed through the pulley at F and connected to a weight W . The weight is released and comes to rest at its equilibrium position D. As l’Hospital argued, this happens when the distance ED is maximized. Show that when the system reaches equilibrium, the value of x is r (r 4d sr 2 8d 2 ) Notice that this expression is independent of both W and . 20. Given a sphere with radius r, find the height of a pyramid of minimum volume whose base is a square and whose base and triangular faces are all tangent to the sphere. What if the base of the pyramid is a regular n-gon (a polygon with n equal sides and angles)? (Use the fact that the volume of a pyramid is 1 Ah, where A is the area of the base.) 3 21. Assume that a snowball melts so that its volume decreases at a rate proportional to its surface area. If it takes three hours for the snowball to decrease to half its original volume, how much longer will it take for the snowball to melt completely? 22. A hemispherical bubble is placed on a spherical bubble of radius 1. A smaller hemispherical FIGURE FOR PROBLEM 22 bubble is then placed on the first one. This process is continued until n chambers, including the sphere, are formed. (The figure shows the case n 4.) Use mathematical induction to prove that the maximum height of any bubble tower with n chambers is 1 sn. 313 ...
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This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas at Austin.

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