# Solutions for Calculus 11th Larson

Please download to get full document.

View again

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
PDF
187 pages
0 downs
0 views
Share
Description
CLICK HERE TO ACCESS FULL SOLUTIONS MANUALSOLUTIONS MANUAL FOR Calculus11th Edition By Larson ISBN13-9781337275347CLICK HERE TO ACCESS FULL VERSIONChapter 2 …
Transcript
CLICK HERE TO ACCESS FULL SOLUTIONS MANUALSOLUTIONS MANUAL FOR Calculus11th Edition By Larson ISBN13-9781337275347CLICK HERE TO ACCESS FULL VERSIONChapter 2  Differentiation Chapter Comments The material presented in Chapter 2 forms the basis for the remainder of calculus. Much of it needs to be memorized, beginning with the definition of a derivative of a function found on page 103. Students need to have a thorough understanding of the tangent line problem and they need to be able to find an equation of a tangent line. Frequently, students will use the function f ′(x) as the slope of the tangent line. They need to understand that f ′(x) is the formula for the slope and the actual value of the slope can be found by substituting into f ′(x) the appropriate value for x. On pages 105–106 of Section 2.1, you will find a discussion of situations where the derivative fails to exist. These examples (or similar ones) should be discussed in class. As you teach this chapter, vary your notations for the derivative. One time write y′; another time write dydx or f′(x). Terminology is also important. Instead of saying “find the derivative,” sometimes say, “differentiate.” This would be an appropriate time, also, to talk a little about Leibnitz and Newton and the discovery of calculus. Sections 2.2, 2.3, and 2.4 present a number of rules for differentiation. Have your students memorize the Product Rule and the Quotient Rule (Theorems 2.7 and 2.8) in words rather than symbols. Students tend to be lazy when it comes to trigonometry and therefore, you need to impress upon them that the formulas for the derivatives of the six trigonometric functions need to be memorized also. You will probably not have enough time in class to prove every one of these differentiation rules, so choose several to do in class and perhaps assign a few of the other proofs as homework. The Chain Rule, in Section 2.4, will require two days of your class time. Students need a lot of practice with this and the algebra involved in these problems. Many students can find the derivative of f (x) = x2√1 − x2 without much trouble, but simplifying the answer is often difficult for them. Insist that they learn to factor and write the answer without negative exponents. Strive to get the answer in the form given in the back of the book. This will help them later on when the derivative is set equal to zero. Implicit differentiation is often difficult for students. Have students think of y as a function of x and therefore y 3 is [ f (x)] 3. This way they can relate implicit differentiation to the Chain Rule studied in the previous section. Try to get your students to see that related rates, discussed in Section 2.6, are another use of the Chain Rule.Section 2.1  The Derivative and the Tangent Line Problem Section Comments 2.1 The Derivative and the Tangent Line Problem—Find the slope of the tangent line to a curve at a point. Use the limit definition to find the derivative of a function. Understand the relationship between differentiability and continuity.Teaching Tips Ask students what they think “the line tangent to a curve” means. Draw a curve with tangent lines to show a visual picture of tangent lines. For example:16© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.yy = f(x)yy = f(x) xxWhen talking about the tangent line problem, use the suggested example of finding the equation of the tangent line to the parabola y = x2 at the point (1, 1). Compute an approximation of the slope m by choosing a nearby point Q(x, x2) on the parabola and computing the slope mPQ of the secant line PQ. After going over Examples 1–3, return to Example 2 where f (x) = x2 + 1 and note that f′(x) = 2x. How can we find the equation of the line tangent to f and parallel to 4x − y = 0? Because the slope of the line is 4, 2x = 4 x = 2. So, at the point (2, 5), the tangent line is parallel to 4x − y = 0. The equation of the tangent line is y − 5 = 4(x − 2) or y = 4x − 3. Be sure to find the derivatives of various types of functions to show students the different types of techniques for finding derivatives. Some suggested problems are f (x) = 4x3 − 3x2, g(x) = 2(x − 1), and h(x) = √2x + 5.How Do You See It? Exercise Page 108, Exercise 64  The figure shows the graph of g′. y 6 4 2g′ x−6 −44 6 −4 −6(a)  g′(0) = (b)  g′(3) = (c)  What can you conclude about the graph of g knowing that g′(1) = − 83? (d)  What can you conclude about the graph of g knowing that g′(−4) = 73? (e) Is g(6) − g(4) positive or negative? Explain. (f)  Is it possible to find g(2) from the graph? Explain.© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.17Solution (a)  g′(0) = −3 (b)  g′(3) = 0 (c) Because g′(1) = − 83, g is decreasing (falling) at x = 1. (d) Because g′(−4) = 73, g is increasing (rising) at x = −4. (e) Because g′(4) and g′(6) are both positive, g(6) is greater than g(4) and g(6) − g(4) > 0. (f)  No, it is not possible. All you can say is that g is decreasing (falling) at x = 2.Suggested Homework Assignment Pages 107–109:  1, 3, 7, 11, 21–27 odd, 37, 43–47 odd, 53, 57, 61, 77, 87, 93, and 95.Section 2.2  Basic Differentiation Rules and Rates of Change Section Comments 2.2 Basic Differentiation Rules and Rates of Change—Find the derivative of a function using the Constant Rule. Find the derivative of a function using the Power Rule. Find the derivative of a function using the Constant Multiple Rule. Find the derivative of a function using the Sum and Difference Rules. Find the derivatives of the sine function and of the cosine function. Use derivatives to find rates of change.Teaching Tips Start by showing proofs of the Constant Rule and the Power Rule. Students who are mathematics majors need to start seeing proofs early on in their college careers as they will be taking Functions of a Real Variable at some point. 5x2 + x . Show students that before differentiating Go over an example in class like f (x) = x they can rewrite the function as f (x) = 5x + 1. Then they can differentiate to obtain f ′(x) = 5. Use this example to emphasize the prudence of examining the function first before differentiating. Rewriting the function in a simpler, equivalent form can expedite the differentiating process. Give mixed examples of finding derivatives. Some suggested examples are: f (x) = 3x 6 − x 23 + 3 sin x and g(x) =4 3√x+2 − 3 cos x + 7x + π 3. (3x)2This will test students’ understanding of the various differentiation rules of this section.How Do You See It? Exercise Page 119, Exercise 76  Use the graph of f to answer each question. To print an enlarged copy of the graph, go to MathGraphs.com.18© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.yfB C ADE x(a)  Between which two consecutive points is the average rate of change of the function greatest? (b) Is the average rate of change of the function between A and B greater than or less than the instantaneous rate of change at B? (c) Sketch a tangent line to the graph between C and D such that the slope of the tangent line is the same as the average rate of change of the function between C and D. Solution (a)  The slope appears to be steepest between A and B. (b) The average rate of change between A and B is greater than the instantaneous rate of change at B. (c) yf B C ADE xSuggested Homework Assignment Pages 118–120:  1, 3, 5, 7–29 odd, 35, 39–53 odd, 55, 59, 65, 75, 85–89 odd, 91, 95, and 97.Section 2.3 Product and Quotient Rules and Higher-Order Derivatives Section Comments 2.3 Product and Quotient Rules and Higher-Order Derivatives—Find the derivative of a function using the Product Rule. Find the derivative of a function using the Quotient Rule. Find the derivative of a trigonometric function. Find a higher-order derivative of a function.Teaching Tips Some students have difficulty simplifying polynomial and rational expressions. Students should review these concepts by studying Appendices A.2–A.4 and A.7 in Precalculus, 10th edition, by Larson. When teaching the Product and Quotient Rules, give proofs of each rule so that students can see where the rules come from. This will provide mathematics majors a tool for writing proofs, as each proof requires subtracting and adding the same quantity to achieve the desired results. For the Project Rule, emphasize that there are many ways to write the solution. Remind students that there must be one derivative in each term of the solution. Also, the Product Rule can be extended to more that just the product of two functions. Simplification is up to the discretion of the instructor. Examples such as f (x) = (2x2 − 3x)(5x3 + 6) can be done with or without the Product Rule. Show the class both ways.© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.19After the Quotient Rule has been proved to the class, give students the memorization tool of LO d HI – HI d LO. This will give students a way to memorize what goes in the numerator of the Quotient Rule. 2x − 1 4 − (1x) and g(x) = . Save f (x) for the next section x2 + 7x 3 − x2 as this will be a good example for the Chain Rule. g(x) is a good example for first finding the least common denominator. Some examples to use are f (x) =How Do You See It? Exercise Page 132, Exercise 120  The figure shows the graphs of the position, velocity, and acceleration functions of a particle. y 16 12 8 4 −1t 14 5 6 7(a) Copy the graphs of the functions shown. Identify each graph. Explain your reasoning. To print an enlarged copy of the graph, go to MathGraphs.com. (b) On your sketch, identify when the particle speeds up and when it slows down. Explain your reasoning. Solution (a)  ys position function16 12 v velocity functions8v4 t−11456a acceleration function7a(b) The speed of the particle is the absolute value of its velocity. So, the particle’s speed is slowing down on the intervals (0, 43), and (83, 4) and it speeds up on the intervals (43, 83) and (4, 6). 16t= 8 312 8 4vspeed t−413567−8 − 12 − 16t= 4t=43Suggested Homework Assignment Pages 129–132:  1, 3, 9, 13, 19, 23, 29–55 odd, 59, 61, 63, 75, 77, 91–107 odd, 111, 113, 117, and 131–135 odd.20© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Section 2.4  The Chain Rule Section Comments 2.4 The Chain Rule—Find the derivative of a composite function using the Chain Rule. Find the derivative of a function using the General Power Rule. Simplify the derivative of a function using algebra. Find the derivative of a trigonometric function using the Chain Rule.Teaching Tips Begin this section by asking students to consider finding the derivative of F(x) = √x2 + 1. F is a composite function. Letting y = f (u) = √u and u = g(x) = x2 + 1, then y = F(x) = f (g(x)) or F = f ∘ g. When stating the Chain Rule, be sure to state it using function notation and using Leibniz notation as students will see both forms when studying other courses with other texts. Following the definition, be sure to prove the Chain Rule as done on page 134. Be sure to give examples that involve all rules discussed so far. Some examples include: 3 + sin(2x) 2 2 f (x) = (sin(6x))4, g(x) = , and h(x) = x− ∙ [8x + cos(x2 + 1)]3. 3 x √ x+3()(√)You can use Exercise 98 on page 141 to review the following concepts: •  Product Rule •  Chain Rule •  Quotient Rule •  General Power RuleStudents need to understand these rules because they are the foundation of the study of differentiation. Use the solution to show students how to solve each problem. As you apply each rule, give the definition of the rule verbally. Note that part (b) is not possible because we are not given g′(3). Solution (a)  f (x) = g(x)h(x) f′(x) = g(x)h′(x) + g′(x)h(x) f′(5) = (−3)(−2) + (6)(3) = 24(b)  f (x) = g(h(x)) f′(x) = g′(h(x))h′(x) f′(5) = g′(3)(−2) = −2g′(3)Not possible. You need g′(3) to find f′(5). g(x) (c)  f (x) = h(x) f ′(x) =h(x)g′(x) − g(x)h′(x) [h(x)]2 f ′(x) =(3)(6) − (−3)(−2) 12 4 = = (3)2 9 3© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.21(d)  f (x) = [g(x)]3 f′(x) = 3[g(x)]2g′(x) f′(5) = 3(−3)2(6) = 162How Do You See It? Exercise Page 142, Exercise 106  The cost C (in dollars) of producing x units of a product is C = 60x + 1350. For one week, management determined that the number of units produced x at the end of t hours can be modeled by x = −1.6t3 + 19t2 − 0.5t − 1. The graph shows the cost C in terms of the time t. Cost of Producing a Product CCost (in dollars)25,000 20,000 15,000 10,000 5,000 t 12345678Time (in hours)(a) Using the graph, which is greater, the rate of change of the cost after 1 hour or the rate of change of the cost after 4 hours? (b)  Explain why the cost function is not increasing at a constant rate during the eight-hour shift. Solution (a)  According to the graph, C′(4) > C′(1). (b)  Answers will vary.Suggested Homework Assignment Pages 140–143:  1–53 odd, 63, 67, 75, 81, 83, 91, 97, 121, and 123.Section 2.5  Implicit Differentiation Section Comments 2.5 Implicit Differentiation—Distinguish between functions written in implicit form and explicit form. Use implicit differentiation to find the derivative of a function.Teaching Tips Material learned in this section will be vital for students to have for related rates. Be sure to ask dy students to find when x = c. dx You can use the exercise below to review the following concepts: •  Finding derivatives when the variables agree and when they disagree •  Using implicit differentiation to find the derivative of a function22© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Determine if the statement is true. If it is false, explain why and correct it. For each statement, assume y is a function of x. (a) d cos(x 2) = −2x sin(x 2) dx(b) d cos( y2) = 2y sin( y2) dy(c) d cos( y2) = −2y sin( y2) dxImplicit differentiation is often difficult for students, so as you review this concept remind students to think of y as a function of x. Part (a) is true, and part (b) can be corrected as shown below. Part dy (c) requires implicit differentiation. Note that the result can also be written as −2y sin( y2) . dx Solution (a) True (b) False.d cos( y2) = −2y sin( y2). dy(c) False.d cos( y2) = −2yy′ sin( y2). dxA good way to teach students how to understand the differentiation of a mix of variables in part (c) is to let g = y. Then g′ = y′. So, d d cos( y2) = cos( g2) dx dx = −sin ( g2) ∙ 2gg′ = −sin( y2) ∙ 2yy′How Do You See It? Exercise Page 151, Exercise 70  Use the graph to answer the questions. yy 3 − 9y 2 + 27y + 5x 2 = 474 2 −2x 2(a) Which is greater, the slope of the tangent line at x = −3 or the slope of the tangent line at x = −1? (b)  Estimate the point(s) where the graph has a vertical tangent line. (c)  Estimate the point(s) where the graph has a horizontal tangent line. Solution (a)  The slope is greater at x = −3. (b)  The graph has vertical tangent lines at about (−2, 3) and (2, 3). (c)  The graph has a horizontal tangent line at about (0, 6). © 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.23Suggested Homework Assignment Pages 149–150:  1–17 odd, 25–35 odd, 53, and 61.Section 2.6  Related Rates Section Comments 2.6 Related Rates—Find a related rate. Use related rates to solve real-life problems.Teaching Tips Begin this lesson with a quick review of implicit differentiation with an implicit function in terms of x and y differentiated with respect to time. Follow this with an example similar to Example 1 on page 152, outlining the step-by-step procedure at the top of page 153 along with the guidelines at the bottom of page 153. Be sure to tell students, that for every related rate problem, to write down the given information, the equation needed, and the unknown quantity. A suggested problem to work out with the students is as follows: A ladder 10 feet long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 foot per second, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 feet from the wall? Be sure to go over a related rate problem similar to Example 5 on page 155 so that students are exposed to working with related rate problems involving trigonometric functions.How Do You See It? Exercise Page 159, Exercise 34  Using the graph of f, (a) determine whether dydt is positive or negative given that dxdt is negative, and (b) determine whether dxdt is positive or negative given that dydt is positive. Explain. y(i)        (ii) y 6 5 4 3 242f1fx 1234−3 −2 −1x1 2 3Solution dx dy (i) (a) negative ⇒ positive dt dt dy dx (b) positive ⇒ negative dt dt (ii) (a) dx dy negative ⇒ negative dt dtdy dx (b) positive ⇒ positive dt dtSuggested Homework Assignment Pages 157–160:  1, 7, 11, 13, 15, 17, 21, 25, 29, and 41.24© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.Chapter 2  Project Timing a Handoff You are a competitive bicyclist. During a race, you bike at a constant velocity of k meters per second. A chase car waits for you at the ten-mile mark of a course. When you cross the ten-mile mark, the car immediately accelerates to catch you. The position function of the chase car is given 5 3 2 by the equation s(t) = 15 4 t − 12 t , for 0 ≤ t ≤ 6, where t is the time in seconds and s is the distance traveled in meters. When the car catches you, you and the car are traveling at the same velocity, and the driver hands you a cup of water while you continue to bike at k meters per second. Exercises   1.  Write an equation that represents your position s (in meters) at time t (in seconds).   2. Use your answer to Exercise 1 and the given information to write an equation that represents the velocity k at which the chase car catches you in terms of t.   3.  Find the velocity function of the car.   4. Use your answers to Exercises 2 and 3 to find how many seconds it takes the chase car to catch you.   5.  What is your velocity when the car catches you?   6. Use a graphing utility to graph the chase car’s position function and your position function in the same viewing window.   7. Find the point of intersection of the two graphs in Exercise 6. What does this point represent in the context of the problem?   8. Describe the graphs in Exercise 6 at the point of intersection. Why is this important for a successful handoff?  9. Suppose you bike at a constant velocity of 9 meters per second and the chase car’s position function is unchanged. (a) Use a graphing utility to graph the chase car’s position function and your position function in the same viewing window. (b) In this scenario, how many times will the chase car be in the same position as you after the 10-mile mark? (c) In this scenario, would the driver of the car be able to successfully handoff a cup of water to you? Explain.10. Suppose you bike at a constant velocity of 8 meters per second and the chase car’s position function is unchanged. (a) Use a graphing utility to graph the chase car’s position function and your position function in the same viewing window. (b) In this scenario, how many times will the chase car be in the same position as you after the ten-mile mark? (c) In this scenario, why might it be difficult for the driver of the chase car to successfully handoff a cu
Related Search
Advertisements
Similar documents

### Solutions for Calculus 8th Stewart

View more...
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us. Thanks to everyone for your continued support.

No, Thanks