8.1 Local Linearity and the Derivative 281 We’ll use the rate at which water is decreasing today to approximate the change in water level over the next two days. dW dt = lim t→0 W t ,so dW dt ≈ W t for t small. W ≈ dW dt t If we let t = 2 and use dW dt t=0 =−115, we have W ≈−115 · 2 =−230. This makes sense; the water is decreasing at a rate of 115 gallons/day so after two days we’d expect it to decrease by about 230 gallons. W(2)≈ G 0 −230 Because W(t) is concave up, we expect the actual water level to be a bit higher than this. ◆ ◆ EXAMPLE 8.2 Approximate √ 16.8. Use a first derivative to get a good approximation. SOLUTION Let’s begin by sketching √ x and getting an off-the-cuff approximation of √ 16.8. This will help us see how a tangent line can be of use. We see that √ 16.8 is a bit larger than 4. y x 16 (16, 4) y = √x Figure 8.3 Question: How do we know this? Answer: 16.8 is close to 16 and √ 16 = 4. Question: How do we know √ 16.8 is a tad more than 4? Answer: We know that √ x is increasing between 16 and 16.8. Question: How can we estimate how much to add to 4 to get a good approximation of √ 16.8? 282 CHAPTER 8 Fruits of Our Labor: Derivatives and Local Linearity Revisited Answer: This depends on the rate at which √ x is increasing near x = 16. This is where the derivative comes into the picture. The derivative of √ x at x =16 gives the rate of increase. d dx √ x x=16 = 1 2 √ x x=16 = 1 2 √ 16 = 1 8 . Here we can adopt one of two equivalent viewpoints. Viewpoint (i): The Derivative as a Tool for Adjustment. We begin with an off-the-cuff approximation of √ 16.8 grounded in a nearby value of √ x that we know with certainty. Knowing the rate of increase of √ x at x = 16 enables us to judge how to adjust our approximation (in this case, 4) to fit a nearby value of x. dy dx ≈ y x for x small (because dy dx =lim x→0 y x ). In this example x = 16.8 − 16 = 0.8. ∆x ∆y 16 16.8 slope = dy dx ∆y ∆x ≈ Figure 8.4 dy dx = 1 8 ≈ y 0.8 ⇒ y ≈ 0.8 8 = 1 10 So y ≈ 0.1 and we have the approximation √ 16.8 ≈ 4 + 0.1 = 4.1. In other words, the “new” y-value is approximately the “old” y-value, 4, plus y , where we approximate y by dy dx ·x. Viewpoint (ii): Tangent Line Approximation. Again we begin with an off-the-cuff ap- proximation grounded in a value of √ x we know with certainty. We choose x = 16. The tangent line to √ x at x = 16 is the best linear approximation of √ x around x = 16, so we use that tangent line to approximate √ 16.8. 16.816 y = √x tangent line at x = 16 slope = 1 8 Figure 8.5 8.1 Local Linearity and the Derivative 283 The y-value corresponding to x = 16.8 on the tangent line is very close to the corre- sponding y-value on the curve. The tangent line has slope 1 8 and passes through the point (16, 4). Its equation is of the form y − y 1 = m(x − x 1 ), from y−y 1 x−x 1 = m where m = 1 8 and (x 1 , y 1 ) = (16, 4). y − 4 = 1 8 · (x − 16) y = 4 + 1 8 · (x − 16) Notice that this just says that the “new” y-value is approximately the “old” y-value, 4, plus the adjustment term—as in the first approach. The y-value on the tangent line when x = 16.8 is y = 4 + 1 8 · (16.8 − 16) = 4.1. Is the approximation √ 16.8 ≈ 4.1 an overestimate or an underestimate? We know that d dx √ x = 1 2 √ x ; thus, as x increases the derivative decreases. Therefore √ x is concave down, the tangent line lies above the graph of the function, and this approximation is a bit too big. Comparing our approximation, √ 16.8 ≈ 4.1, with the calculator approximation, √ 16.8 ≈ 4.09878, we see that the approximation is good. The error is about 0.00122; we are off by about 0.03%. ◆ REMARK If we wanted to further refine our approximation, we could use information supplied by the second derivative to make the required adjustment. 1 Further successive refinements involve higher-order derivatives and give us what is called a Taylor polynomial approximation to the function at x = 16. EXERCISE 8.1 Approximate √ 101 using a tangent line approximation. Compare your answer to that obtained using a calculator. EXERCISE 8.2 Suppose you use a tangent line approximation to estimate √ 4.8. Do you expect the dif- ference between your approximation and the actual answer to be greater or less than the analogous difference we found in Example 8.2? Explain, and then check your reasoning by computation and using a calculator. EXERCISE 8.3 We argued that f(x)= √ x is concave down by reasoning that f is decreasing. Show that f is concave down by showing that f (x) < 0. Use the limit definition of derivative to compute f (x). If you have trouble, look back at Section 5.1 to see how the derivative of f was computed. Answers to Exercises 8.1, 8.2, and 8.3 are supplied at the end of this section. Generalizing Our Method To approximate f () using a tangent line approximation, choose an x near for which the value of f is known. Let’s call this x-value a. Find the tangent line to f at x = a. The slope 1 The larger the value of the second derivative at x =16 the larger the rate of change of the slope near x = 16. 284 CHAPTER 8 Fruits of Our Labor: Derivatives and Local Linearity Revisited is f (a) and a point on the line is (a, f(a)).Therefore, the equation of the tangent line is y − f(a)=f (a) · (x − a) or y = f(a)+f (a) · (x − a). Evaluating at x = gives the corresponding y-value on the tangent line. Answers to Selected Exercises Answer to Exercise 8.1 √ 101 ≈ 10 + 1 2 √ 100 · (101 − 100) y ≈ 10 + 1 20 · (1) = 10.5 Answer to Exercise 8.2 The slope of √ x is changing more rapidly at x = 4 than at x = 16. In both cases we are using the tangent line to approximate the value of the function at a distance 0.8 from the known value. Therefore we expect the error to be greater for √ 4.8 than for √ 16.8. √ 4.8 ≈ 2 + 1 2 √ 4 · (4.8 − 4) ≈ 2 + 1 4 · (0.8) = 2.2 The difference between the approximation, 2.2, and the actual value, 2.19089 ,is approximately 0.009. Answer to Exercise 8.3 f (x) = −1 4( √ x) 3 PROBLEMS FOR SECTION 8.1 1. Use the tangent line approximation (linear approximation) of f(x)= √ x at x = 25 to approximate the following. Use the graph of √ x and its tangent line at x =25 to predict the relative accuracy of your approximations. Check the accuracy using a computer or calculator. (a) √ 23 (b) √ 24 (c) √ 24.9 (d) √ 25.1 (e) √ 26 (f) √ 27 2. Suppose we want to use a tangent line approximation of f(x)= √ x at x = a to approximate a particular square root numerically. Which values of a should we choose to approximate each of the following? (a) √ 102 (b) √ 8 (c) √ 18 (d) √ 115.5 3. Use a tangent line approximation to f(x)= 1 x at x = 2 to approximate 1 1.9 . 8.1 Local Linearity and the Derivative 285 4. Approximate √ 98 using the appropriate first derivative to help you. Explain your reasoning. 5. Use the fact that d dx x 1/3 = 1 3 x −2/3 to approximate 3 √ 30. Do you expect your an- swer to be an over-approximation or an under-approximation? Explain. Compare your answer to the approximation supplied by your calculator. 6. A steamboat is traveling down the Mississippi River. It is traveling south, making its way from point St. Paul, Minnesota, to Dubuque, Iowa. At noon it departs St. Paul traveling at 10 mph and is accelerating. It continues to accelerate over the next 10 minutes. Between noon and 12:10 p.m. it has covered 5 miles. Let s(t) be the distance the steamboat has traveled from point St. Paul, where t is measured in minutes. We’ll use noon as benchmark time of t = 0. (a) Determine the sign of s(0), s (0), and s (0). Which of these expressions are you given enough information to specify numerically? (b) Sketch s(t) over the time interval [0, 10]. (c) Find good upper and lower bounds for the distance the boat has traveled between noon and 12:05 p.m. Use a sketch to illustrate your reasoning graphically. 7. Consider the solid right cylinder with a fixed height of 10 inches and a variable radius. Let V(r)be the volume of the cylinder as a function of r, the radius, given in inches. Interpret dV /dr geometrically. Explain why your answer makes sense by looking at V geometrically. 286 CHAPTER 8 Fruits of Our Labor: Derivatives and Local Linearity Revisited Exploratory Problems for Chapter 8 Circles and Spheres You know how to express the area and the circumference of a circle in terms of its radius; you also know how to express the volume and surface area of a sphere in terms of its radius. Having completed the Exploratory Problems for Chapter 7, you know how to take the derivative of both the area function for the circle and the volume function for the sphere. Have you observed the following? A(r) = πr 2 is the area of a circle of radius r. dA dr = 2πr is its circumference. V(r)= 4 3 πr 3 is the volume of a sphere of radius r. dV dr = 4πr 2 is its surface area. These relationships are not coincidental. The problems that follow ask you to make sense out of this observation. (Notice that the area of a square with sides of length s is given by A(s) = s 2 and its perimeter is 4s, not 2s.) r ∆r circle of radius r and radius r + ∆r sphere of radius r and radius r + ∆r ∆ s ∆ s square of side s and side s + ∆ s s s s + ∆ s Figure (a) Figure (b) Figure (c) 1. Let A(r) give the area of a circle as a function of its radius. dA dr = lim r→0 A r where A = A(r + r) − A(r) Geometrically, A corresponds to the shaded area in Figure (a) above. (a) Imagine wrapping a piece of yarn around the outside of a circle of radius r, thereby increasing the radius of the circle by the thickness of the yarn. The thickness of the yarn corresponds to r. Let r be small in comparison to r. Approximate the additional area by clipping the yarn and laying it out straight. Explain why, for r very small, A r ≈ 2πr = circumfer- ence of circle. (b) Now approach the problem more algebraically. Instead of clipping a piece of yarn, look at the algebraic expression for Exploratory Problems for Chapter 8 287 A and show that it is 2πrr + π(r) 2 . Argue that for r very small in comparison to r, we can approximate A by 2πrr and that as r tends toward zero this approximation gets better and better. 2. Investigate the relationship between the derivative of the volume of a sphere and the surface area of a sphere by approximating V , when r is very small in comparison to r.Asanaidin visualization imagine a grapefruit without the skin as the sphere of radius r and the same grapefruit with the skin as the sphere of radius r + r. V corresponds to the volume of the skin of the grapefruit. 3. Let A(s) be the area of a square with sides of length s. The derivative of the area of a square is not its perimeter; it is only half of the perimeter. Explain this by looking at A geometrically and approximating it. (Refer to Figure (c).) 288 CHAPTER 8 Fruits of Our Labor: Derivatives and Local Linearity Revisited 8.2 THE FIRST AND SECOND DERIVATIVES IN CONTEXT: MODELING USING DERIVATIVES We’ve been interpreting the first derivative in context since its introduction but we have not yet used the second derivative extensively in context. Yet, according to a tongue-in- cheek article from The Economist, second-order derivatives were the rage in the 1990s. Below is an excerpt from the article “The Tyranny of Differential Calculus” published in The Economist on April 6, 1991. “The pace of change slows,” said a headline on the Financial Times’s survey of world paints and coatings last week. Growth has been slowing in various countries—slowing quite quickly in some cases. Employers were invited recently to a conference on “techniques for improving performance enhancement.” It’s not enough to enhance your performance, Jones, you must improve your enhancement. Suddenly, everywhere, it is not the rate of change of things that matters, it is the rate of change of the rate of change. Nobody cares much about inflation; only whether it is going up or down. Or rather, whether it is going up fast or down fast Norespectable budget director has talked about the national debt for decades; all talk sternly about the need to reduce the budget deficit, which is, after all, roughly the rate at which the national debt is increasing. Indeed, in recent years it is not the absolute size of the deficit that has mattered so much as the trend: Is the rate of change of the rate of change of the national debt positive or negative? (© 1991 The Economist Newspaper Group, Inc. Reprinted with permission. Further reproduction prohibited. www.economist.com) Consider the comment about the paint industry that reads “Growth has been slowing in various countries.” If we let P(t)be the size of the paint industry at time t, the graph of P(t)is roughly given below. The graph is increasing because the industry is growing; it is concave down to reflect that the growth is slowing. P(t) t Figure 8.6 dP dt > 0 because P is increasing with time; d 2 P dt 2 = d dt dP dt < 0 because the growth is slowing. EXERCISE 8.4 Suppose that in countries A and B the growth of the paint industry is slowing, and at time t ∗ growth is slowing more quickly in country A than in country B. Let P A (t) and P B (t) be the sizes of the paint industries in countries A and B, respectively, at time t. 8.2 The First and Second Derivatives in Context: Modeling Using Derivatives 289 (a) What are the signs of P A (t ∗ ) and P B (t ∗ )? Can you determine which is larger? (b) What are the signs of P A (t ∗ ) and P B (t ∗ )? Can you determine which is larger? (c) What are the signs of P A (t ∗ ) and P B (t ∗ )? Can you determine which is larger? Answers are provided at the end of this section. ◆ EXAMPLE 8.3 When politicians talk about the budget “deficit” they are referring to the difference between the government’s revenues and expenditures for a given year. When they speak of “the national debt” they mean the cumulative deficit (or surplus) over all the years of the government’s recorded transactions. Let x(t) be the deficit for year t and let y(t) be the national debt incurred up to and including year t. In our model x(t) and y(t) are both continuous and differentiable. x(t) will be positive if there is a deficit, negative if there is a surplus. (a) There has been considerable talk about a balanced budget. What would a balanced budget mean in terms of x(t)? In terms of y(t)? (b) Politicians also often speak of deficit reduction plans. What would such a reduction mean in terms of x(t) and y(t) and their derivatives? (c) The article from The Economist tells us that the budget deficit is roughly the rate at which the national debt is increasing. What, then, is the calculus-based relationship between x(t) and y(t)? SOLUTION (a) A balanced budget would mean that x(t) = 0 and y(t) is constant. (b) Deficit reduction would mean x(t) is positive and decreasing. x(t) > 0 and x (t) < 0. Deficit reduction would mean y(t) is increasing at a decreasing rate, y (t) > 0 and y (t) < 0. (c) y (t) = x(t). This makes sense both practically speaking and in terms of the answers to parts (a) and (b). ◆ Answers to Selected Exercises Answers to Exercise 8.4 (a) P A (t ∗ ) and P B (t ∗ ) are both positive because the size of the paint industry won’tbe negative. We cannot determine which will be larger. (b) P A (t ∗ ) and P B (t ∗ ) are both positive because both industries are growing. We cannot determine which rate of growth is larger. (c) P A (t ∗ ) and P B (t ∗ ) are both negative, because growth is slowing. Determining which is larger is semantically tricky. Since growth is slowing more rapidly in country A than in country B, P A (t ∗ ) is more negative than P B (t ∗ ). More negative means smaller, so P A (t ∗ ) is smaller than P B (t ∗ ). 290 CHAPTER 8 Fruits of Our Labor: Derivatives and Local Linearity Revisited PROBLEMS FOR SECTION 8.2 1. Water is being poured into a bucket at a steady rate. h(t) gives the height of water at time t. Let t ∗ be the time when the bucket is half full. What can you say about the signs of h (t ∗ ) and h (t ∗ ). Explain your reasoning precisely in plain English. 2. Suppose that the revenue, R, brought in each month by the after-eight shows at a movie theater is a function of the price p of a ticket. Suppose that R is measured in thousands of dollars and that p is measured in dollars. Interpret the following statements in words. (a) R (3.5) = 50 (b) R (7.50) =−15 3. A company is making industrial-size rolls of paper towels. A machine is wrapping paper around a roll at a steady rate. By this we mean that the same number of sheets of paper towels are added to the roll every minute. Let D(t) be the diameter of the roll of paper towels at time t. Determine the sign of the following. Explain your answers using plain English. (a) D(t) (b) D (t) (c) D (t) 4. Let Y(t) be the number of Japanese yen exchangeable for one U.S. dollar, where t is the number of days after January 1, 1996. (a) What is the practical significance of the values of t for which Y (t) is positive? (b) What is the practical significance of the values of t for which Y (t) is negative and Y (t) is negative? (c) What is the meaning of the statement Y (5) = 0.8? (d) Interpret the quantity Y(5)−Y(3) 2 . 8.3 DERIVATIVES OF SUMS, PRODUCTS, QUOTIENTS, AND POWER FUNCTIONS In this section we shift our focus from interpreting and using derivatives to actually comput- ing them. Fruits of our labors with limits will be general procedures for differentiating sums, products, and quotients, and the ability to easily compute the derivative of any function of the form x n , where n is an integer. Sum Rule and Constant Multiple Rule Section 7.4 dealt with principles for working with limits. We restate two of these principles here for reference. . with sides of length s is given by A(s) = s 2 and its perimeter is 4s, not 2s.) r ∆r circle of radius r and radius r + ∆r sphere of radius r and radius r + ∆r ∆ s ∆ s square of side s and side. relationship between x(t) and y(t)? SOLUTION (a) A balanced budget would mean that x(t) = 0 and y(t) is constant. (b) Deficit reduction would mean x(t) is positive and decreasing. x(t) > 0 and x (t) <. differentiating sums, products, and quotients, and the ability to easily compute the derivative of any function of the form x n , where n is an integer. Sum Rule and Constant Multiple Rule Section