ESS E NT I A L CAL CUL U S W I T H A P P L ICA TION S Third Edition Wright • Hurd • New Annotated Instructor’s Edition - Not for SaleESS E NT I A L CAL CUL U S W I T H A P P L ICA TION S Third Edition D. Franklin Wright • Spencer P. Hurd • Bill D. NewEditors: Danielle C. Bess, Marvin Glover, Claudia Vance Courseware Developers: Allison Conger, Kyle Gilstrap Manager of Math Content Development: Blair Dunivan Creative Services Manager: Trudy Tronco Designers: Lizbeth Mendoza, Patrick Thompson, Joel Travis Cover Design: Patrick Thompson Composition and Answer Key Assistance: Quant Systems India Pvt. Ltd. A division of Quant Systems, Inc. 546 Long Point Road Mount Pleasant, SC 29464 Copyright © 2023, 2019, 2016, 2008 by Hawkes Learning / Quant Systems, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written consent of the publisher. Library of Congress Control Number: 2022939988 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ISBN: 978-1-64277-549-5 AIE ISBN: 978-1-64277-550-1Table of Contents 0 Algebra Review 1 Functions, Models, and Graphs 2 Limits and the Derivative 3 More about the Derivative iv Table of Contents 4 Applications of the Derivative 5 Exponential and Logarithmic Functions 6 Integration with Applications 7 Additional Integration Topics 8 Multivariable Calculus Table of Contents v 9 Trigonometric Functions 10 Sequences, Taylor Polynomials, and Power Series Answer vi Preface PREFACE Features Did You Know? In the early 1600s, Johannes Kepler (1571–1630) discovered three laws describing the nature of the relationship and relative motion of Earth, the moon, the other known planets, and the sun. Kepler’s first law states that planets travel around the sun in elliptical orbits. His second law states that an imaginary line between a planet and the sun sweeps out equal areas in equal times. His third law helps determine the amount of time it takes a planet to complete one full orbit. Kepler’s Second Law of Planetary Motion illustrates an application of integral calculus to physics. The equality of the two areas occurs because the planet orbits at a higher velocity when it is closer to the sun and at a lower velocity when it is farther away. To determine his second law, Kepler calculated the rate of change of the position of each planet using very accurate data supplied by his mentor, Tycho Brahe (1546–1601). Next, he determined a position function for each planet by using the sum of the areas involved. The process of using rates (derivatives) and areas to get position functions is a consequence of what is presented in Chapter 6 under the imposing title of “The Fundamental Theorem of Calculus.” Integrals form one subcategory of calculus, but what are integrals and integration? When we take a derivative, we find the rate of change of a function. Integration is exactly the opposite. When we take an integral, we transform the rate of change into its original function. For example, we know that velocity is the derivative of position, so we know that position is the integral of velocity. The same relationship is true of velocity and acceleration. In engineering, biology, business, and economics, the process of using data about rates of change of y with respect to x in order to infer the form of a function y = f (x) is now a common approach to problem solving. This chapter will address applications of the integral in these fields. 378 Chapter 6 w Integration with Applications Our emphasis in Chapter 6 is on understanding the process that is the reverse of differentiation by using many basic applications. In Section 6.1, we discuss only a minimum number of basic formulas and rules for integration. In Section 6.2, these formulas are expanded to cover a broader spectrum of functions with a technique called u-substitution. Even here, though, we use only the same few basic formulas of integration. There are two general problems that we will learn how to solve. The first is to determine a possible function f (x) given f ′(x). The second type of problem is to apply this formula to calculate specific values of f (x) that will have geometric or physical meaning in the problem at hand. The many applications will give the opportunity to master the technique while developing an appreciation for the wide extent of calculus applications in business, statistics, and science. The culmination and high point (theoretically!) of the book is in Sections 6.3 and 6.4 where we explain the connection between definite integrals and what is called the Fundamental Theorem of Calculus. In Sections 6.5, 6.6, and 6.7, we apply the theorem to a wide variety of applications. 6.1 THE INDEFINITE INTEGRAL The Antiderivative We have developed several formulas for differentiation and used those formulas to find rates of change, slopes of tangent lines, marginal cost, maxima and minima, and so on. Now we want to consider reversing the process of differentiation. That is, we want to antidifferentiate. If the management of a company knows the current rate of change of cost (i.e., they know the marginal cost function), they can use this knowledge to find the related cost function and proceed to make a reasonably accurate budget projection. With special instruments, scientists can measure the speed of particles along straight lines. That is, they know the derivative of the position function (or velocity), and with this information, they can find the function that will tell the location of the particle at any particular time. In such situations, a derivative is known and the objective is to find the function (or functions) with this derivative. Antiderivative If F and f are two functions and F ′(x) = f (x) for all x in the domain of f, then F is an antiderivative of f. Note: We also say that F(x) is an antiderivative of f (x). For example, suppose that P(x) = 4x2, Q(x) = 4x2 + 5, and R(x) = 4x2 − 100. In each case, we have P ′(x) = Q ′(x) = R ′(x) = 8x. Objectives • Apply the definition of the antiderivative. • Use formulas of integration to find integrals. • Use integration to solve applications. 270 Chapter 4 w Applications of the Derivative 4.2 THE SECOND DERIVATIVE TEST We have seen how the first derivative of a function can be used to locate local extrema and how the second derivative can be used to analyze concavity and locate points of inflection. Now we will show how the second derivative, if it exists, provides a relatively simple test for local extrema. Second Derivative Test for Local Extrema Suppose that f is a function, f ′ and f ″ exist on the interval (a, b), c is in (a, b), and f ′(c) = 0. 1. If f ″(c) > 0, then f (c) is a local minimum. (See Figure 1(A).) 2. If f ″(c) < 0, then f (c) is a local maximum. (See Figure 1(B).) 3. If f ″(c) = 0, then the test fails to give any information about local extrema. xacb y y = f (x) local minimum f ″(c) > 0 f ′(c) = 0 At a local minimum, x = c, f (x) must be concave upward ( f ″(c) > 0). (A) xacb y y = f (x) local maximum f ″(c) < 0 f ′(c) = 0 At a local maximum, x = c, f (x) must be concave downward ( f ″(c) < 0). (B) FIGURE 1 In the pursuit of fully understanding calculus, it is very important that the student is capable of graphing a few functions successfully by using calculus, without the aid of a calculator. The next example will demonstrate this. Example 1: Using the Second Derivative Test Let f (x) = x4 − 18x2. a. Find the local extrema. b. Find the points of inflection. Objectives • Use the Second Derivative Test to find the local extrema of a function. • Solve applications involving points of diminishing returns. { NOTE The principle being applied in the Second Derivative Test is simple. If f ″(c) is defined at a local maximum (c, f (c)), a curve is concave down. If f ″(c) is defined at a local minimum (c, f (c)), a curve is concave up. Of course, the primary value of the Second Derivative Test occurs only when the algebra formula is available for analysis. Continuity at x = a The function f is continuous at x = a if and only if limlim. xaxa fxfafx Finding Absolute Extrema Assume that f is a continuous function on the interval [a, b]. To find the absolute extrema of f on [a, b], perform the following steps. 1. Find all the critical values for f in [a, b]. That is, find all c in [a, b] where a. f ′ (c) = 0 or b. f ′(c) is undefined. 2. Evaluate f (a), f (b), and f (c) for all critical values c. 3. The largest value found in Step 2 is the absolute maximum. The smallest value found in Step 2 is the absolute minimum. vi Preface Did You Know? Each chapter begins with a list of sections and an interesting part of math history related to the chapter at hand. Introduction Before the first section of each chapter, introductory paragraphs provide an overview of the chapter and its purpose. Objectives The objectives provide students with a clear and concise list of skills presented in each section. They are helpful for both reference and class preparation. Definitions, Theorems, Formulas, Properties, and Procedures Definitions, theorems, and formulas are clearly set apart in highly visible green boxes for easy reference, and properties and procedures are similarly set apart in distinctive blue boxes. All formally identified terms appear in bold print when first defined, and other useful terms appear in italic font. Preface Preface vii Notes The green note boxes in the margins help clarify subtle details and provide problem-solving tips. Cautions Many common errors are pointed out, along with how to correct them. These are set apart in red boxes. Examples Examples are presented in a step-by-step manner that is easy for students to follow. Each example has a title indicating the problem-solving skill being presented. Examples make use of tables, diagrams, and graphs for additional clarity where applicable. Technology Instructions Technology notes and screenshots are included throughout the text to highlight ways that graphing calculators can help solve problems or explain concepts. Step-by-step instructions for using a TI-84 Plus are given in many cases. { NOTE We have used the letter u throughout this section. This particular letter is commonly used in most calculus courses, and the substitution technique is sometimes known as the u-substitution technique. A rule of thumb that you may have observed while using this technique is to substitute u for 1. the exponent in an exponential function, 2. an expression in parentheses, 3. a denominator, or 4. an expression under a radical sign. ¡ CAUTION There are no rules to simplify ln ln ln . MN M N and For instance, lnlnlnlnln.235236 Also, ln ln .lnln.. 10 5 143 10 5 20 69 176 Chapter 3 w Functions and Their Graphs p = S(x) Supply function: price per unit at which the supplier will supply x units of an item. p = D(x) Demand function: price per unit at which consumers will buy x units of an item. The equilibrium point, denoted as (x E, p E), is the point where the price is such that the production level (supply) is equal to the purchase level (demand). (See Figure 1.) x 5 10 15 20 p 3 6 9 12 p = 0.2x + 3 x 5 10 15 20 p 3 6 9 12 p = −0.4x + 12 (15, 6) x 5 10 15 20 p 3 6 9 12 (x E, p E) p = S(x) Supply Function (A) p = D(x) Demand Function (B) S(x) = D(x) Equilibrium Point (C) FIGURE 1 In general, supply and demand functions are not linear functions as they are shown to be in Figure 1. However, supply functions are increasing functions, and demand functions are decreasing functions. Suppliers will happily provide more products as prices increase, and consumers will happily buy more products as prices decrease. Example 5: Equilibrium Point Suppose that the supply function for a particular product is p = S(x) = x2 + x + 3 and the demand function p = D(x) = (x − 5)2 where x represents thousands of units and p represents thousands of dollars. Find the equilibrium point (x E, p E). Solution We solve for x E by setting S(x) = D(x). Then we substitute this value for x into either S(x) or D(x) to find p E. SxDx xxx xxxx x x E EE EE EE EE E E () = () ++ =− () ++ =−+ = = 2 2 22 35 31025 1122 2 ppS xS EE = () = () =+ += 2 2239 2 The equilibrium point occurs where x = 2000 units and p = $9000. Units (in thousands) 1 2 3 4 5 Price (in thousands of dollars) 2 4 6 8 10 12 14 16 (2, 9) Equilibrium point p x (x E, p E) p = S(x) = x2 + x + 3 p = D(x) = (x − 5)2 466 Chapter 7 w Additional Integration Topics Improper Integrals Using Technology Once a convergent improper integral has been set up, it is possible to use a graphing calculator such as the TI-84 Plus to do the calculating. For example, let us integrate 2 3 1 xe dxx . The following methods are recommended. Method 1 Step 1: Graph the function in a suitable window, say [−1, 3] by [−1, 1]. The displayed portion of the graph is large enough to easily see that there is very little area beyond x = 3. (See Figure 3.) FIGURE 3 Step 2: Press 2nd trace and select item 7: ∫f(x)dx (this is the integration symbol). (See Figure 4.) At the prompt, type the lower limit 1 and enter. At the next prompt, type the upper limit 3 . The area is shaded and the decimal answer 0.0439809 appears at the bottom of the screen (see Figure 5). This is an approximation to the actual value, but when using this method, the upper and lower limits must be in the range of x-values plotted on the screen. FIGURE 4 FIGURE 5 Method 2 Step 1: Press mode and select CLASSIC . Step 2: From the home screen, press math and select item 9:fnInt( (function integral). (See Figure 6.) FIGURE 6 296 Chapter 4 w Applications of the Derivative Example 1: Minimizing Inventory Costs A furniture dealer sells 500 desks per year. The desks take up floor space and warehouse space, and the dealer estimates his storage costs at $6 per desk. The distributor charges the dealer a $60 fee for each order. How many times per year and in what lot size should the dealer order to minimize inventory costs? Solution Using the given information, we can determine a function for the inventory costs, C(x). Let x = lot size. Then 500 x is the number of orders per year, and x 2 is the average inventory. Cx x storage cost per itemcost per order 2 500 xx Cx x x 6 2 60 500 The number of desks ordered is between 1 and 500. At the extremes, one order for 500 desks would cost C 500 500 500 6 2 60 500 1560$, and 500 orders for one desk at a time would cost C 1 1 1 6 2 60 500 30 003$,. Now we need to differentiate C(x) so we can determine the local minima. Cx x x xx Cxx 6 2 60 500 330 000 330 000 1 2 , , 33 30 000 2 , x Rewrite C(x) using exponents. We set C ′(x) = 0 and solve for x. 3 30 000 0 330 000 10 000 100 2 2 2 , , , x x x x Note that x = −100 is not in the interval [1, 500], so it must be disregarded. Substituting x = 100 into the equation for the number of orders per year, we get 500500 5 100 x orders per year. Ɨ TECH TRAINING To evaluate the function C(x) in Example 1 with a TI-84 Plus calculator, perform the following steps: 1. Enter the function C(x) that was found into Y1 . 2. Now, on the main screen, press vars , scroll right to Y-VARS , select 1:Function , and then 1:Y1 to display Y1 on the main screen. 3. With your cursor after Y1 , type an opening parenthesis and then the x-value we are evaluating, 500. Type a closing parenthesis. 4. Press enter to calculate the corresponding y-value. 5. This process can be repeated for the next x-value, 1.Next >