Calculus of a Single Variable 8th Edition [Student ed.] 0883903466, 9780883903469

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2

CHAPTER P

Preparation for Calculus

Section P.1

Graphs and Models

RENÉ DESCARTES (1596–1650) Descartes made many contributions to philosophy, science, and mathematics. The idea of representing points in the plane by pairs of real numbers and representing curves in the plane by equations was described by Descartes .in his book La Géométrie, published in 1637.

• • • • •

Sketch the graph of an equation. Find the intercepts of a graph. Test a graph for symmetry with respect to an axis and the origin. Find the points of intersection of two graphs. Interpret mathematical models for real-life data.

The Graph of an Equation In 1637 the French mathematician René Descartes revolutionized the study of mathematics by joining its two major fields—algebra and geometry. With Descartes’s coordinate plane, geometric concepts could be formulated analytically and algebraic concepts could be viewed graphically. The power of this approach is such that within a century, much of calculus had been developed. The same approach can be followed in your study of calculus. That is, by viewing calculus from multiple perspectives—graphically, analytically, and numerically— you will increase your understanding of core concepts. Consider the equation 3x  y  7. The point 2, 1 is a solution point of the equation because the equation is satisfied (is true) when 2 is substituted for x and 1 is substituted for y. This equation has many other solutions, such as 1, 4 and 0, 7. To find other solutions systematically, solve the original equation for y.

MathBio

y  7  3x

Analytic approach

Then construct a table of values by substituting several values of x. y 8 6 4

(0, 7) (1, 4)

2 −2

3x + y = 7

(2, 1) 2

−4 −6

4

0

1

2

3

4

y

7

4

1

2

5

Numerical approach

From the table, you can see that 0, 7, 1, 4, 2, 1, 3, 2, and 4, 5 are solutions of the original equation 3x  y  7. Like many equations, this equation has an infinite number of solutions. The set of all solution points is the graph of the equation, as shown in Figure P.1.

x 6

(3, −2)

x

8

(4, −5)

Graphical approach: 3x  y  7 Figure P.1

NOTE Even though we refer to the sketch shown in Figure P.1 as the graph of 3x  y  7, it really represents only a portion of the graph. The entire graph would extend beyond the page.

In this course, you will study many sketching techniques. The simplest is point plotting—that is, you plot points until the basic shape of the graph seems apparent. y

EXAMPLE 1

Sketching a Graph by Point Plotting

7

Sketch the graph of y  x 2  2.

6 5

y = x2 − 2

4 3

Solution First construct a table of values. Then plot the points shown in the table.

2 1

x

2

1

0

1

2

3

y

2

1

2

1

2

7

x −4 −3 −2

2

3

. parabola y  x 2  2 The Figure P.2

Editable Graph

4

Finally, connect the points with a smooth curve, as shown in Figure P.2. This graph is a parabola. It is one of the conics you will study in Chapter 10.

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Exploration A

SECTION P.1

3

Graphs and Models

One disadvantage of point plotting is that to get a good idea about the shape of a graph, you may need to plot many points. With only a few points, you could badly misrepresent the graph. For instance, suppose that to sketch the graph of 1 y  30 x39  10x2  x 4

you plotted only five points: 3, 3, 1, 1, 0, 0, 1, 1, and 3, 3, as shown in Figure P.3(a). From these five points, you might conclude that the graph is a line. This, however, is not correct. By plotting several more points, you can see that the graph is more complicated, as shown in Figure P.3(b). y y

(3, 3)

3

1 y = 30 x(39 − 10x 2 + x 4)

3 2 2

(1, 1)

1

1

(0, 0) −3

−2 − 1 (−1, −1) −1 −2

(−3, −3)

−3

x

1

2

3 −3

Plotting only a few points can misrepresent a graph.

−2

x

−1

1

2

3

−1 −2 −3

(a)

(b)

Figure P.3

E X P L O R AT I O N Comparing Graphical and Analytic Approaches Use a graphing utility to graph each equation. In each case, find a viewing window that shows the important characteristics of the graph. a. y  x3  3x 2  2x  5 b. y  x3  3x 2  2x  25 c. y  x3  3x 2  20x  5 d. y  3x3  40x 2  50x  45

Technology has made sketching of graphs easier. Even with technology, however, it is possible to misrepresent a graph badly. For instance, each of the graphing utility screens in Figure P.4 shows a portion of the graph of

TECHNOLOGY

y  x3  x 2  25. From the screen on the left, you might assume that the graph is a line. From the screen on the right, however, you can see that the graph is not a line. So, whether you are sketching a graph by hand or using a graphing utility, you must realize that different “viewing windows” can produce very different views of a graph. In choosing a viewing window, your goal is to show a view of the graph that fits well in the context of the problem.

e. y   x  123

10

f. y  x  2x  4x  6 A purely graphical approach to this problem would involve a simple “guess, check, and revise” strategy. What types of things do you think an analytic approach might involve? For instance, does the graph have symmetry? Does the graph have turns? If so, where are they? As you proceed through Chapters 1, 2, and 3 of this text, you will study many new analytic tools that will help you analyze graphs of equations such as these.

5 −5

−10

5

10

−10

Graphing utility screens of y 

−35

x3



x2

 25

Figure P.4 NOTE In this text, the term graphing utility means either a graphing calculator or computer graphing software such as Maple, Mathematica, Derive, Mathcad, or the TI-89.

4

CHAPTER P

Preparation for Calculus

Intercepts of a Graph Two types of solution points that are especially useful in graphing an equation are those having zero as their x- or y-coordinate. Such points are called intercepts because they are the points at which the graph intersects the x- or y-axis. The point a, 0 is an x-intercept of the graph of an equation if it is a solution point of the equation. To find the x-intercepts of a graph, let y be zero and solve the equation for x. The point 0, b is a y-intercept of the graph of an equation if it is a solution point of the equation. To find the y-intercepts of a graph, let x be zero and solve the equation for y. NOTE Some texts denote the x-intercept as the x-coordinate of the point a, 0 rather than the point itself. Unless it is necessary to make a distinction, we will use the term intercept to mean either the point or the coordinate.

It is possible for a graph to have no intercepts, or it might have several. For instance, consider the four graphs shown in Figure P.5. y

y

y

x

y

x

Three x-intercepts One y-intercept

No x-intercepts One y-intercept

x

One x-intercept Two y-intercepts

x

No intercepts

Figure P.5

EXAMPLE 2

Finding x- and y-intercepts

Find the x- and y-intercepts of the graph of y  x 3  4x. Solution To find the x-intercepts, let y be zero and solve for x.

y

y = x 3 − 4x

x3  4x  0 xx  2x  2  0 x  0, 2, or 2

4 3

(−2, 0) −4 −3

(0, 0) −1 −1 −2 −3

. Intercepts of a graph Figure P.6

Editable Graph

1

(2, 0) 3

x 4

Let y be zero. Factor. Solve for x.

Because this equation has three solutions, you can conclude that the graph has three x-intercepts:

0, 0, 2, 0, and 2, 0.

x-intercepts

To find the y-intercepts, let x be zero. Doing this produces y  0. So, the y-intercept is

0, 0.

y-intercept

(See Figure P.6.)

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Exploration A

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Video

Example 2 uses an analytic approach to finding intercepts. When an analytic approach is not possible, you can use a graphical approach by finding the points at which the graph intersects the axes. Use a graphing utility to approximate the intercepts.

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SECTION P.1

y

Graphs and Models

5

Symmetry of a Graph Knowing the symmetry of a graph before attempting to sketch it is useful because you need only half as many points to sketch the graph. The following three types of symmetry can be used to help sketch the graphs of equations (see Figure P.7).

(x, y)

(−x, y)

x

1. A graph is symmetric with respect to the y-axis if, whenever x, y is a point on the graph, x, y is also a point on the graph. This means that the portion of the graph to the left of the y-axis is a mirror image of the portion to the right of the y-axis. 2. A graph is symmetric with respect to the x-axis if, whenever x, y is a point on the graph, x, y is also a point on the graph. This means that the portion of the graph above the x-axis is a mirror image of the portion below the x-axis. 3. A graph is symmetric with respect to the origin if, whenever x, y is a point on the graph, x, y is also a point on the graph. This means that the graph is unchanged by a rotation of 180 about the origin.

y-axis symmetry

y

(x, y) x

(x, −y)

x-axis symmetry

Tests for Symmetry 1. The graph of an equation in x and y is symmetric with respect to the y-axis if replacing x by x yields an equivalent equation. 2. The graph of an equation in x and y is symmetric with respect to the x-axis if replacing y by y yields an equivalent equation. 3. The graph of an equation in x and y is symmetric with respect to the origin if replacing x by x and y by y yields an equivalent equation.

y

(x, y) x

(−x, −y)

The graph of a polynomial has symmetry with respect to the y-axis if each term has an even exponent (or is a constant). For instance, the graph of

Origin symmetry

y  2x 4  x 2  2 Figure P.7

y-axis symmetry

has symmetry with respect to the y-axis. Similarly, the graph of a polynomial has symmetry with respect to the origin if each term has an odd exponent, as illustrated in Example 3. EXAMPLE 3

Testing for Origin Symmetry

Show that the graph of y

y = 2x − x 3

2

is symmetric with respect to the origin. (1, 1)

1

Solution x

−2

−1

(−1, −1)

1

−1 −2

.. Origin symmetry

y  2x3  x

2

y  2x3  x y  2x3  x y 

2x3

x y  2x3  x

Write original equation. Replace x by x and y by y. Simplify. Equivalent equation

Because the replacements yield an equivalent equation, you can conclude that the graph of y  2x3  x is symmetric with respect to the origin, as shown in Figure P.8.

Figure P.8

Editable Graph

Try It

Exploration A

Video

Video

6

CHAPTER P

Preparation for Calculus

EXAMPLE 4

Using Intercepts and Symmetry to Sketch a Graph

Sketch the graph of x  y 2  1. y

x − y2 = 1

Solution The graph is symmetric with respect to the x-axis because replacing y by y yields an equivalent equation.

(5, 2)

2

(2, 1) 1

(1, 0) x 2

3

4

5

−1

Write original equation. Replace y by y. Equivalent equation

This means that the portion of the graph below the x-axis is a mirror image of the portion above the x-axis. To sketch the graph, first plot the x-intercept and the points above the x-axis. Then reflect in the x-axis to obtain the entire graph, as shown in Figure P.9.

x-intercept

−2

x  y2  1 x  y 2  1 x  y2  1

.

Figure P.9

Try It

Editable Graph

Exploration B

Exploration A

Open Exploration

Graphing utilities are designed so that they most easily graph equations in which y is a function of x (see Section P.3 for a definition of function). To graph other types of equations, you need to split the graph into two or more parts or you need to use a different graphing mode. For instance, to graph the equation in Example 4, you can split it into two parts.

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y1  x  1 y2   x  1

Top portion of graph Bottom portion of graph

Points of Intersection A point of intersection of the graphs of two equations is a point that satisfies both equations. You can find the points of intersection of two graphs by solving their equations simultaneously. y

EXAMPLE 5 2

x−y=1

Find all points of intersection of the graphs of x 2  y  3 and x  y  1.

1

(2, 1) x

−2

−1

1

2

−1

(−1, −2)

Finding Points of Intersection

−2

x2 − y = 3

. points of intersection Two Figure P.10

Editable Graph STUDY TIP You can check the . points of intersection from Example 5 by substituting into both of the original equations or by using the intersect feature of a graphing utility.

Solution Begin by sketching the graphs of both equations on the same rectangular coordinate system, as shown in Figure P.10. Having done this, it appears that the graphs have two points of intersection. You can find these two points, as follows. y  x2  3 yx1 x2  3  x  1 x2  x  2  0 x  2x  1  0 x  2 or 1

Solve first equation for y. Solve second equation for y. Equate y-values. Write in general form. Factor. Solve for x.

The corresponding values of y are obtained by substituting x  2 and x  1 into either of the original equations. Doing this produces two points of intersection:

2, 1 and 1, 2.

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Points of intersection

Exploration A

SECTION P.1

Graphs and Models

7

Mathematical Models Real-life applications of mathematics often use equations as mathematical models. In developing a mathematical model to represent actual data, you should strive for two (often conflicting) goals: accuracy and simplicity. That is, you want the model to be simple enough to be workable, yet accurate enough to produce meaningful results. Section P.4 explores these goals more completely. EXAMPLE 6 The Mauna Loa Observatory in Hawaii has been measuring the increasing . concentration of carbon dioxide in Earth’s atmosphere since 1958.

Comparing Two Mathematical Models

The Mauna Loa Observatory in Hawaii records the carbon dioxide concentration y (in parts per million) in Earth’s atmosphere. The January readings for various years are shown in Figure P.11. In the July 1990 issue of Scientific American, these data were used to predict the carbon dioxide level in Earth’s atmosphere in the year 2035, using the quadratic model

Video

y  316.2  0.70t  0.018t 2

Quadratic model for 1960–1990 data

where t  0 represents 1960, as shown in Figure P.11(a). The data shown in Figure P.11(b) represent the years 1980 through 2002 and can be modeled by y  306.3  1.56t

Linear model for 1980–2002 data

where t  0 represents 1960. What was the prediction given in the Scientific American article in 1990? Given the new data for 1990 through 2002, does this prediction for the year 2035 seem accurate? y

375 370 365 360 355 350 345 340 335 330 325 320 315

CO2 (in parts per million)

CO2 (in parts per million)

y 375 370 365 360 355 350 345 340 335 330 325 320 315

t

t

5 10 15 20 25 30 35 40 45

5 10 15 20 25 30 35 40 45

Year (0 ↔ 1960) (a)

Year (0 ↔ 1960) (b)

Figure P.11

Solution To answer the first question, substitute t  75 (for 2035) into the quadratic model. y  316.2  0.7075  0.018752  469.95

NOTE The models in Example 6 were developed using a procedure called least squares regression (see Section 13.9). The quadratic and linear models have a correlation given by r 2  0.997 and . r 2  0.996, respectively. The closer r 2 is to 1, the “better” the model.

Quadratic model

So, the prediction in the Scientific American article was that the carbon dioxide concentration in Earth’s atmosphere would reach about 470 parts per million in the year 2035. Using the linear model for the 1980–2002 data, the prediction for the year 2035 is y  306.3  1.5675  423.3.

Linear model

So, based on the linear model for 1980–2002, it appears that the 1990 prediction was too high.

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8

CHAPTER P

Preparation for Calculus

E x e r c i s e s f o r S e c t i o n P. 1 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, match the equation with its graph. [Graphs are labeled (a), (b), (c), and (d).] y

(a)

y

(b) 5 4 3

3 2 1

x

x 1 2 3 4 y

(d) 4

2 1

21. y  x 225  x2

22. y  x  1x2  1

23. y 

24. y 

32  x  x

2 1

x

2

2

2

26. y  2x  x 2  1

27. y  x 2  2

28. y  x 2  x



30. y  x3  x

y2

x3

2

2. y  9  x2

3. y  4  x 2

4. y  x 3  x

 4x

31. xy  4

32. xy 2  10

33. y  4  x  3

34. xy  4  x 2  0

2

1. y   12 x  2

35. y 

x2



x 1



x2



x2 1

38. y  x  3

In Exercises 39–56, sketch the graph of the equation. Identify any intercepts and test for symmetry. 39. y  3x  2

1 40. y   2x  2 2 42. y  3 x  1

5. y  32 x  1

6. y  6  2x

1 41. y  2 x  4

7. y  4  x 2

8. y  x  32

43. y  1  x 2

44. y  x 2  3

9. y  x  2

10. y  x  1

45. y  x  3

46. y  2x 2  x

11. y  x  4



12. y  x  2

47. y  x3  2

48. y  x3  4x

1 x1

49. y  xx  2

50. y  9  x2

51. x  y3

52. x  y 2  4

13. y 



36. y 

37. y  x3  x

In Exercises 5–14, sketch the graph of the equation by point plotting.



x 2  3x 3x  12

In Exercises 27–38, test for symmetry with respect to each axis and to the origin.

29.

x 1

20. y 2  x3  4x

25. x 2y  x 2  4y  0

1 y

2

19. y  x 2  x  2

1

1

(c)

In Exercises 19–26, find any intercepts.

2 x

14. y 

2

In Exercises 15 and 16, describe the viewing window that yields the figure. 15. y  x3  3x 2  4

 



53. y 

17. y  5  x

(a) 2, y

(b) x, 3

18. y  x5  5x

(a) 0.5, y

(b) x, 4

54. y 



55. y  6  x

16. y  x  x  10

In Exercises 17 and 18, use a graphing utility to graph the equation. Move the cursor along the curve to approximate the unknown coordinate of each solution point accurate to two decimal places.

1 x

10 x2  1





56. y  6  x

In Exercises 57–60, use a graphing utility to graph the equation. Identify any intercepts and test for symmetry. 57. y 2  x  9

58. x 2  4y 2  4

59. x  3y 2  6

60. 3x  4y 2  8

In Exercises 61–68, find the points of intersection of the graphs of the equations. xy2

62. 2x  3y  13

2x  y  1

5x  3y  1

61. 63.

x2

y6

xy4

64. x  3  y 2 yx1

SECTION P.1

65. x 2  y 2  5

66. x 2  y 2  25

xy1

2x  y  10

where x is the diameter of the wire in mils (0.001 in.). Use a graphing utility to graph the model. If the diameter of the wire is doubled, the resistance is changed by about what factor?

68. y  x3  4x

67. y  x3

y   x  2

yx

Writing About Concepts

In Exercises 69–72, use a graphing utility to find the points of intersection of the graphs. Check your results analytically. 70. y  x 4  2x 2  1

69. y  x3  2x 2  x  1 y  x  3x  1



71. y  x  6



78. The graph has intercepts at x   52, x  2, and x  32.

72. y   2x  3  6

y  x2  4x

In Exercises 77 and 78, write an equation whose graph has the indicated property. (There may be more than one correct answer.) 77. The graph has intercepts at x  2, x  4, and x  6.

y  1  x2

2

9

Graphs and Models

y6x

79. Each table shows solution points for one of the following equations.

73. Modeling Data The table shows the Consumer Price Index (CPI) for selected years. (Source: Bureau of Labor Statistics) Year

1970

1975

1980

1985

1990

1995

2000

CPI

38.8

53.8

82.4

107.6

130.7 152.4 172.2

(i) y  kx  5

(ii) y  x2  k

(iii) y  kx32

(iv) xy  k

Match each equation with the correct table and find k. Explain your reasoning. (a)

(a) Use the regression capabilities of a graphing utility to find a mathematical model of the form y  at 2  bt  c for the data. In the model, y represents the CPI and t represents the year, with t  0 corresponding to 1970. (b) Use a graphing utility to plot the data and graph the model. Compare the data with the model.

(c)

x

1

4

9

y

3

24

81

x

1

4

9

y

36

9

4

(b)

(d)

x

1

4

9

y

7

13

23

x

1

4

9

y

9

6

71

(c) Use the model to predict the CPI for the year 2010. 74. Modeling Data The table shows the average numbers of acres per farm in the United States for selected years. (Source: U.S. Department of Agriculture) Year

1950

1960

1970

1980

1990

2000

Acreage

213

297

374

426

460

434

(a) Use the regression capabilities of a graphing utility to find a mathematical model of the form y  at 2  bt  c for the data. In the model, y represents the average acreage and t represents the year, with t  0 corresponding to 1950. (b) Use a graphing utility to plot the data and graph the model. Compare the data with the model. (c) Use the model to predict the average number of acres per farm in the United States in the year 2010. 75. Break-Even Point Find the sales necessary to break even R  C if the cost C of producing x units is C  5.5x  10,000

Cost equation

and the revenue R for selling x units is R  3.29x.

Revenue equation

76. Copper Wire The resistance y in ohms of 1000 feet of solid copper wire at 77F can be approximated by the model 10,770 y  0.37, x2

5 f x f 100

80. (a) Prove that if a graph is symmetric with respect to the x-axis and to the y-axis, then it is symmetric with respect to the origin. Give an example to show that the converse is not true. (b) Prove that if a graph is symmetric with respect to one axis and to the origin, then it is symmetric with respect to the other axis.

True or False? In Exercises 81–84, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 81. If 1, 2 is a point on a graph that is symmetric with respect to the x-axis, then 1, 2 is also a point on the graph. 82. If 1, 2 is a point on a graph that is symmetric with respect to the y-axis, then 1, 2 is also a point on the graph. 83. If b2  4ac > 0 and a  0, then the graph of y  ax 2  bx  c has two x-intercepts. 84. If b 2  4ac  0 and a  0, then the graph of y  ax 2  bx  c has only one x-intercept. In Exercises 85 and 86, find an equation of the graph that consists of all points x, y having the given distance from the origin. (For a review of the Distance Formula, see Appendix D.) 85. The distance from the origin is twice the distance from 0, 3. 86. The distance from the origin is K K  1 times the distance from 2, 0.

10

CHAPTER P

Preparation for Calculus

Section P.2

Linear Models and Rates of Change • • • • •

y

The Slope of a Line (x2, y2)

y2

y1

Find the slope of a line passing through two points. Write the equation of a line with a given point and slope. Interpret slope as a ratio or as a rate in a real-life application. Sketch the graph of a linear equation in slope-intercept form. Write equations of lines that are parallel or perpendicular to a given line.

The slope of a nonvertical line is a measure of the number of units the line rises (or falls) vertically for each unit of horizontal change from left to right. Consider the two points x1, y1 and x2, y2 on the line in Figure P.12. As you move from left to right along this line, a vertical change of

Δy = y2 − y1

(x1, y1) Δx = x2 − x1 x1

y  y2  y1 Change in y units corresponds to a horizontal change of

x

x2

y  y2  y1  change in y x  x2  x1  change in x

x  x2  x1

Figure P.12

Change in x

units. ( is the Greek uppercase letter delta, and the symbols y and  x are read “delta y” and “delta x.”)

Definition of the Slope of a Line The slope m of the nonvertical line passing through x1, y1 and x2, y2  is y2  y1 y ,  x x2  x1

m .

x1  x2.

Slope is not defined for vertical lines.

Video NOTE

Video

Video

When using the formula for slope, note that

y2  y1   y1  y2 y1  y2   . x2  x1  x1  x2 x1  x2 So, it does not matter in which order you subtract as long as you are consistent and both “subtracted coordinates” come from the same point.

Figure P.13 shows four lines: one has a positive slope, one has a slope of zero, one has a negative slope, and one has an “undefined” slope. In general, the greater the absolute value of the slope of a line, the steeper the line is. For instance, in Figure P.13, the line with a slope of 5 is steeper than the line with a slope of 15. y

y

4

m1 =

y

4

1 5

3

4

m2 = 0

3

y

(0, 4) m3 = −5

3

(−1, 2)

4

(3, 4)

3

2

2

m 4 is undefined.

1

1

(3, 1)

(2, 2)

2

(3, 1) (−2, 0)

1

1 x

−2

−1

1

2

3

−1

If m is positive, then the line rises from left to right. Figure P.13

x

−2

−1

1

2

3

−1

If m is zero, then the line is horizontal.

x

−1

2

−1

(1, −1)

3

4

If m is negative, then the line falls from left to right.

x

−1

1

2

4

−1

If m is undefined, then the line is vertical.

SECTION P.2

E X P L O R AT I O N Investigating Equations of Lines Use a graphing utility to graph each of the linear equations. Which point is common to all seven lines? Which value in the equation determines the slope of each line?

Any two points on a nonvertical line can be used to calculate its slope. This can be verified from the similar triangles shown in Figure P.14. (Recall that the ratios of corresponding sides of similar triangles are equal.) y

(x2*, y2*) (x2, y2)

b. y  4  1x  1 c. y  4 

(x1, y1) (x1*, y1*)

 1

d. y  4  0x  1 e. y  4 

1 2 x

11

Equations of Lines

a. y  4  2x  1  12x

Linear Models and Rates of Change

x

y * − y1* y2 − y1 m= 2 = x2* − x1* x2 − x1

 1

f. y  4  1x  1 g. y  4  2x  1

Any two points on a nonvertical line can be used to determine its slope.

Use your results to write an equation of a line passing through 1, 4 with a slope of m.

Figure P.14

You can write an equation of a nonvertical line if you know the slope of the line and the coordinates of one point on the line. Suppose the slope is m and the point is x1, y1. If x, y is any other point on the line, then y  y1  m. x  x1 This equation, involving the two variables x and y, can be rewritten in the form y  y1  mx  x1, which is called the point-slope equation of a line. Point-Slope Equation of a Line An equation of the line with slope m passing through the point x1, y1 is given by

y

y  y1  mx  x1.

y = 3x − 5

1 x

1

3

Δy = 3

−1 −2 −3

Δx = 1 (1, −2)

−4 −5

The line with a slope of 3 passing through .. the point 1, 2 Figure P.15

Editable Graph

EXAMPLE 1

Finding an Equation of a Line

4

Find an equation of the line that has a slope of 3 and passes through the point 1, 2. Solution y  y1  mx  x1 y  2  3x  1 y  2  3x  3 y  3x  5

Point-slope form Substitute 2 for y1, 1 for x1, and 3 for m. Simplify. Solve for y.

(See Figure P.15.)

Try It

Exploration A

Exploration B

Exploration C

NOTE Remember that only nonvertical lines have a slope. Consequently, vertical lines cannot be written in point-slope form. For instance, the equation of the vertical line passing through the point 1, 2 is x  1.

12

CHAPTER P

Preparation for Calculus

Ratios and Rates of Change The slope of a line can be interpreted as either a ratio or a rate. If the x- and y-axes have the same unit of measure, the slope has no units and is a ratio. If the x- and y-axes have different units of measure, the slope is a rate or rate of change. In your study of calculus, you will encounter applications involving both interpretations of slope.

Population (in millions)

EXAMPLE 2 5 4

355,000

3

10

Population Growth and Engineering Design

a. The population of Kentucky was 3,687,000 in 1990 and 4,042,000 in 2000. Over this 10-year period, the average rate of change of the population was Rate of change 

2



1

1990

2000

Population of Kentucky in census years Figure P.16

4,042,000  3,687,000 2000  1990

 35,500 people per year.

2010

Year

change in population change in years

If Kentucky’s population continues to increase at this same rate for the next 10 years, it will have a 2010 population of 4,397,000 (see Figure P.16). (Source: U.S. Census Bureau) b. In tournament water-ski jumping, the ramp rises to a height of 6 feet on a raft that is 21 feet long, as shown in Figure P.17. The slope of the ski ramp is the ratio of its height (the rise) to the length of its base (the run). Slope of ramp 

rise run



6 feet 21 feet



2 7

Rise is vertical change, run is horizontal change.

In this case, note that the slope is a ratio and has no units.

6 ft

21 ft

.

Dimensions of a water-ski ramp Figure P.17

Try It

Exploration A

Exploration B

The rate of change found in Example 2(a) is an average rate of change. An average rate of change is always calculated over an interval. In this case, the interval is 1990, 2000 . In Chapter 2 you will study another type of rate of change called an instantaneous rate of change.

SECTION P.2

13

Linear Models and Rates of Change

Graphing Linear Models Many problems in analytic geometry can be classified in two basic categories: (1) Given a graph, what is its equation? and (2) Given an equation, what is its graph? The point-slope equation of a line can be used to solve problems in the first category. However, this form is not especially useful for solving problems in the second category. The form that is better suited to sketching the graph of a line is the slopeintercept form of the equation of a line.

The Slope-Intercept Equation of a Line The graph of the linear equation y  mx  b is a line having a slope of m and a y-intercept at 0, b.

.

Video

Video

Sketching Lines in the Plane

EXAMPLE 3

Sketch the graph of each equation. a. y  2x  1

b. y  2

c. 3y  x  6  0

Solution a. Because b  1, the y-intercept is 0, 1. Because the slope is m  2, you know that the line rises two units for each unit it moves to the right, as shown in Figure P.18(a). b. Because b  2, the y-intercept is 0, 2. Because the slope is m  0, you know that the line is horizontal, as shown in Figure P.18(b). c. Begin by writing the equation in slope-intercept form. 3y  x  6  0 3y  x  6 1 y x2 3

Write original equation. Isolate y- term on the left. Slope-intercept form

In this form, you can see that the y-intercept is 0, 2 and the slope is m  13. This means that the line falls one unit for every three units it moves to the right, as shown in Figure P.18(c). y

y

y = 2x + 1

3

3

Δy = 2

2

y 3

y=2

Δx = 3

y = − 13 x + 2

(0, 2)

(0, 1)

Δy = −1

1

1

(0, 2)

Δx = 1 x

1

.

2

(a) m  2; line rises

3

x

x

1

2

3

1

2

(c) m   13 ; line falls

(b) m  0; line is horizontal

Figure P.18

.

Editable Graph

Editable Graph

Try It

3

Editable Graph

Exploration A

4

5

6

14

CHAPTER P

Preparation for Calculus

Because the slope of a vertical line is not defined, its equation cannot be written in the slope-intercept form. However, the equation of any line can be written in the general form Ax  By  C  0

General form of the equation of a line

where A and B are not both zero. For instance, the vertical line given by x  a can be represented by the general form x  a  0. Summary of Equations of Lines 1. 2. 3. 4. 5.

General form: Vertical line: Horizontal line: Point-slope form: Slope-intercept form:

Ax  By  C  0,

A, B  0

xa yb y  y1  mx  x1 y  mx  b

Parallel and Perpendicular Lines The slope of a line is a convenient tool for determining whether two lines are parallel or perpendicular, as shown in Figure P.19. Specifically, nonvertical lines with the same slope are parallel and nonvertical lines whose slopes are negative reciprocals are perpendicular. y

y

m1 = m2 m2 m1 m1

m2

m1 = − m1 x

Parallel lines

2

x

Perpendicular lines

Figure P.19 STUDY TIP In mathematics, the phrase “if and only if” is a way of stating two implications in one statement. For instance, the first statement at the right could be rewritten as the following two implications.

a. If two distinct nonvertical lines are parallel, then their slopes are equal. b. If two distinct nonvertical lines have equal slopes, then they are parallel.

Parallel and Perpendicular Lines 1. Two distinct nonvertical lines are parallel if and only if their slopes are equal—that is, if and only if m1  m2. 2. Two nonvertical lines are perpendicular if and only if their slopes are negative reciprocals of each other—that is, if and only if m1  

1 . m2

SECTION P.2

Linear Models and Rates of Change

15

Finding Parallel and Perpendicular Lines

EXAMPLE 4

Find the general forms of the equations of the lines that pass through the point 2, 1 and are a. parallel to the line 2x  3y  5 (See Figure P.20.)

y 2

3x + 2y = 4

Solution By writing the linear equation 2x  3y  5 in slope-intercept form, y  23 x  53, you can see that the given line has a slope of m  23.

2x − 3y = 5

1

a. The line through 2, 1 that is parallel to the given line also has a slope of 23. y  y1  m x  x1

x 1 −1

4

Lines parallel and perpendicular to . 2x  3y  5 Figure P.20

Point-slope form

y  1  23 x  2 3 y  1  2x  2

(2, −1)

2x − 3y = 7

Substitute. Simplify.

2x  3y  7  0

General form

Note the similarity to the original equation. b. Using the negative reciprocal of the slope of the given line, you can determine that the slope of a line perpendicular to the given line is  32. So, the line through the point 2, 1 that is perpendicular to the given line has the following equation.

Editable Graph

y  y1  mx  x1 y  1   32x  2

Point-slope form Substitute.

2 y  1  3x  2 3x  2y  4  0

. ..

b. perpendicular to the line 2x  3y  5.

Try It

Simplify. General form

Exploration B

Exploration A

Exploration C

Open Exploration

The slope of a line will appear distorted if you use different tick-mark spacing on the x- and y-axes. For instance, the graphing calculator screens in Figures P.21(a) and P.21(b) both show the lines given by y  2x and y  12x  3. Because these lines have slopes that are negative reciprocals, they must be perpendicular. In Figure P.21(a), however, the lines don’t appear to be perpendicular because the tick-mark spacing on the x-axis is not the same as that on the y-axis. In Figure P.21(b), the lines appear perpendicular because the tick-mark spacing on the x-axis is the same as on the y-axis. This type of viewing window is said to have a square setting.

TECHNOLOGY PITFALL

10

−10

6

10

−10

(a) Tick-mark spacing on the x-axis is not the same as tick-mark spacing on the y-axis.

Figure P.21

−9

9

−6

(b) Tick-mark spacing on the x-axis is the same as tick-mark spacing on the y-axis.

16

CHAPTER P

Preparation for Calculus

E x e r c i s e s f o r S e c t i o n P. 2 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–6, estimate the slope of the line from its graph. To print an enlarged copy of the graph, select the MathGraph button. 1.

2.

y 7 6 5 4 3 2 1 x

1 2 3 4 5 6 7

4.

7 6 5

(a) m  400

x

x

1 2 3 4 5 6 7

1 2 3 4 5 6

6.

y

8. 4, 1

(a) 3

(b) 2 (b) 3

(c)  32 (c)

1 3

(d) Undefined (d) 0

In Exercises 9–14, plot the pair of points and find the slope of the line passing through them. 9. 3, 4, 5, 2 11. 2, 1, 2, 5

1 2 3 1 13.  2, 3 ,  4, 6 

10. 1, 2, 2, 4 12. 3, 2, 4, 2 14.

78, 34 , 54,  14 

In Exercises 15–18, use the point on the line and the slope of the line to find three additional points that the line passes through. (There is more than one correct answer.) Point

Slope

8

9

10

11

y

269.7

272.9

276.1

279.3

282.3

285.0

x

Slopes (a) 1

7

1 2 3 4 5 6 7

5 6 7

Point

6

22. Modeling Data The table shows the rate r (in miles per hour) that a vehicle is traveling after t seconds.

In Exercises 7 and 8, sketch the lines through the point with the indicated slopes. Make the sketches on the same set of coordinate axes. 7. 2, 3

t

(b) Use the slope of each line segment to determine the year when the population increased least rapidly.

x 1 2 3

(c) m  0

(a) Plot the data by hand and connect adjacent points with a line segment.

y 70 60 50 40 30 20 10

28 24 20 16 12 8 4

(b) m  100

21. Modeling Data The table shows the populations y (in millions) of the United States for 1996–2001. The variable t represents the time in years, with t  6 corresponding to 1996. (Source: U.S. Bureau of the Census)

y 6 5 4 3 2 1

3 2 1

5.

20. Rate of Change Each of the following is the slope of a line representing daily revenue y in terms of time x in days. Use the slope to interpret any change in daily revenue for a one-day increase in time.

x

1 2 3 4 5 6 7

3.

(a) Find the slope of the conveyor. (b) Suppose the conveyor runs between two floors in a factory. Find the length of the conveyor if the vertical distance between floors is 10 feet.

y 7 6 5 4 3 2 1

y

19. Conveyor Design A moving conveyor is built to rise 1 meter for each 3 meters of horizontal change.

Point

Slope

15. 2, 1

m0

16. 3, 4

m undefined

17. 1, 7

m  3

18. 2, 2

m2

t

5

10

15

20

25

30

r

57

74

85

84

61

43

(a) Plot the data by hand and connect adjacent points with a line segment. (b) Use the slope of each line segment to determine the interval when the vehicle’s rate changed most rapidly. How did the rate change? In Exercises 23–26, find the slope and the y-intercept (if possible) of the line. 23. x  5y  20

24. 6x  5y  15

25. x  4

26. y  1

In Exercises 27–32, find an equation of the line that passes through the point and has the indicated slope. Sketch the line. Point

Slope

27. 0, 3

m

29. 0, 0

m

31. 3, 2

m3

Point

Slope

3 4 2 3

28. 1, 2

m undefined

30. 0, 4

m0

32. 2, 4

m  5

3

SECTION P.2

In Exercises 33–42, find an equation of the line that passes through the points, and sketch the line. 34. 0, 0, 1, 3

35. 2, 1, 0,3

36. 3, 4, 1, 4

37. 2, 8, 5, 0

38. 3, 6, 1, 2

59. 2, 1

39. 5, 1, 5, 8

40. 1, 2, 3, 2

61.

41.

 , 0,  3 4

42.

 ,  7 3 8, 4

5 4,

 14

Point

43. Find an equation of the vertical line with x-intercept at 3. 44. Show that the line with intercepts a, 0 and 0, b has the following equation. x y   1, a b

Line

  3 7 4, 8

63. 2, 5



a  0, b  0

45. x-intercept: 2, 0

46. x-intercept:

0

y-intercept: 0, 2

y-intercept: 0, 3 47. Point on line: 1, 2



 23,

48. Point on line: 3, 4

x-intercept: a, 0

x-intercept: a, 0

y-intercept: 0, a a  0

y-intercept: 0, a a  0

In Exercises 49–56, sketch a graph of the equation. 49. y  3

50. x  4

51. y  2x  1

52. y  3 x  1

53. y  2 

3 2 x

 1

55. 2x  y  3  0

1

54. y  1  3x  4 56. x  2y  6  0

Point 60. 3, 2

xy7

5x  3y  0

62. 6, 4

3x  4y  7

x4

64. 1, 0

y  3

Rate of Change In Exercises 65– 68, you are given the dollar value of a product in 2004 and the rate at which the value of the product is expected to change during the next 5 years. Write a linear equation that gives the dollar value V of the product in terms of the year t. (Let t  0 represent 2000.)

65. $2540

Rate $125 increase per year

66. $156

$4.50 increase per year

67. $20,400

$2000 decrease per year

68. $245,000

$5600 decrease per year

In Exercises 69 and 70, use a graphing utility to graph the parabolas and find their points of intersection. Find an equation of the line through the points of intersection and graph the line in the same viewing window. 69. y  x 2 y  4x 

70. y 

In Exercises 71 and 72, determine whether the points are collinear. (Three points are collinear if they lie on the same line.) 71. 2, 1, 1, 0, 2, 2

72. 0, 4, 7, 6, 5, 11

Writing About Concepts

57. y  x  6, y  x  2

73.

Xmin = -10 Xmax = 10 Xscl = 1 Ymin = -10 Ymax = 10 Yscl = 1

(b)

Xmin = -15 Xmax = 15 Xscl = 1 Ymin = -10 Ymax = 10 Yscl = 1

x 2  4x  3

y  x 2  2x  3

x2

Square Setting In Exercises 57 and 58, use a graphing utility to graph both lines in each viewing window. Compare the graphs. Do the lines appear perpendicular? Are the lines perpendicular? Explain.

(a)

Line

4x  2y  3

2004 Value In Exercises 45–48, use the result of Exercise 44 to write an equation of the line.

17

In Exercises 59– 64, write an equation of the line through the point (a) parallel to the given line and (b) perpendicular to the given line.

33. 0, 0, 2, 6

1 7 2, 2

Linear Models and Rates of Change

In Exercises 73 –75, find the coordinates of the point of intersection of the given segments. Explain your reasoning. (b, c)

(a, 0)

(a, 0)

Perpendicular bisectors 75.

74.

(b, c)

(a, 0)

(a, 0)

Medians

(b, c)

1 58. y  2x  3, y   2 x  1

(a)

Xmin = -5 Xmax = 5 Xscl = 1 Ymin = -5 Ymax = 5 Yscl = 1

(b)

Xmin = -6 Xmax = 6 Xscl = 1 Ymin = -4 Ymax = 4 Yscl = 1

(a, 0)

(a, 0)

Altitudes 76. Show that the points of intersection in Exercises 73, 74, and 75 are collinear.

18

CHAPTER P

Preparation for Calculus

77. Temperature Conversion Find a linear equation that expresses the relationship between the temperature in degrees Celsius C and degrees Fahrenheit F. Use the fact that water freezes at 0C (32F) and boils at 100C (212F). Use the equation to convert 72F to degrees Celsius. 78. Reimbursed Expenses A company reimburses its sales representatives $150 per day for lodging and meals plus 34¢ per mile driven. Write a linear equation giving the daily cost C to the company in terms of x, the number of miles driven. How much does it cost the company if a sales representative drives 137 miles on a given day? 79. Career Choice An employee has two options for positions in a large corporation. One position pays $12.50 per hour plus an additional unit rate of $0.75 per unit produced. The other pays $9.20 per hour plus a unit rate of $1.30. (a) Find linear equations for the hourly wages W in terms of x, the number of units produced per hour, for each option. (b) Use a graphing utility to graph the linear equations and find the point of intersection. (c) Interpret the meaning of the point of intersection of the graphs in part (b). How would you use this information to select the correct option if the goal were to obtain the highest hourly wage? 80. Straight-Line Depreciation A small business purchases a piece of equipment for $875. After 5 years the equipment will be outdated, having no value. (a) Write a linear equation giving the value y of the equipment in terms of the time x, 0  x  5. (b) Find the value of the equipment when x  2. (c) Estimate (to two-decimal-place accuracy) the time when the value of the equipment is $200. 81. Apartment Rental A real estate office handles an apartment complex with 50 units. When the rent is $580 per month, all 50 units are occupied. However, when the rent is $625, the average number of occupied units drops to 47. Assume that the relationship between the monthly rent p and the demand x is linear. (Note: The term demand refers to the number of occupied units.) (a) Write a linear equation giving the demand x in terms of the rent p. (b) Linear extrapolation Use a graphing utility to graph the demand equation and use the trace feature to predict the number of units occupied if the rent is raised to $655. (c) Linear interpolation Predict the number of units occupied if the rent is lowered to $595. Verify graphically. 82. Modeling Data An instructor gives regular 20-point quizzes and 100-point exams in a mathematics course. Average scores for six students, given as ordered pairs x, y where x is the average quiz score and y is the average test score, are 18, 87, 10, 55, 19, 96, 16, 79, 13, 76, and 15, 82. (a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data. (b) Use a graphing utility to plot the points and graph the regression line in the same viewing window.

(c) Use the regression line to predict the average exam score for a student with an average quiz score of 17. (d) Interpret the meaning of the slope of the regression line. (e) The instructor adds 4 points to the average test score of everyone in the class. Describe the changes in the positions of the plotted points and the change in the equation of the line. 83. Tangent Line Find an equation of the line tangent to the circle x2  y2  169 at the point 5, 12. 84. Tangent Line Find an equation of the line tangent to the circle x  12   y  12  25 at the point 4, 3. Distance In Exercises 85–90, find the distance between the point and line, or between the lines, using the formula for the distance between the point x1, y1 and the line Ax  By  C  0. Distance 

Ax1  By1  C A2  B2

85. Point: 0, 0

86. Point: 2, 3

Line: 4x  3y  10 87. Point: 2, 1

Line: 4x  3y  10 88. Point: 6, 2

Line: x  y  2  0 89. Line: x  y  1 Line: x  y  5

Line: x  1 90. Line: 3x  4y  1 Line: 3x  4y  10

91. Show that the distance between the point x1, y1 and the line Ax  By  C  0 is Distance 

Ax1  By1  C. A2  B2

92. Write the distance d between the point 3, 1 and the line y  mx  4 in terms of m. Use a graphing utility to graph the equation. When is the distance 0? Explain the result geometrically. 93. Prove that the diagonals of a rhombus intersect at right angles. (A rhombus is a quadrilateral with sides of equal lengths.) 94. Prove that the figure formed by connecting consecutive midpoints of the sides of any quadrilateral is a parallelogram. 95. Prove that if the points x1, y1 and x2, y2 lie on the same line as xⴱ1, yⴱ1 and xⴱ2, yⴱ2, then yⴱ2  yⴱ1 y2  y1  . x2ⴱ  xⴱ1 x2  x1 Assume x1  x2 and xⴱ1  xⴱ2. 96. Prove that if the slopes of two nonvertical lines are negative reciprocals of each other, then the lines are perpendicular.

True or False? In Exercises 97 and 98, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 97. The lines represented by ax  by  c1 and bx  ay  c2 are perpendicular. Assume a  0 and b  0. 98. It is possible for two lines with positive slopes to be perpendicular to each other.

SECTION P.3

Section P.3

Functions and Their Graphs

19

Functions and Their Graphs • • • • •

Use function notation to represent and evaluate a function. Find the domain and range of a function. Sketch the graph of a function. Identify different types of transformations of functions. Classify functions and recognize combinations of functions.

Functions and Function Notation A relation between two sets X and Y is a set of ordered pairs, each of the form x, y, where x is a member of X and y is a member of Y. A function from X to Y is a relation between X and Y that has the property that any two ordered pairs with the same x-value also have the same y-value. The variable x is the independent variable, and the variable y is the dependent variable. Many real-life situations can be modeled by functions. For instance, the area A of a circle is a function of the circle’s radius r. A  r2

A is a function of r.

In this case r is the independent variable and A is the dependent variable.

X x

Domain

Definition of a Real-Valued Function of a Real Variable f Range y = f(x) Y

A real-valued function f of a real variable Figure P.22

Let X and Y be sets of real numbers. A real-valued function f of a real variable x from X to Y is a correspondence that assigns to each number x in X exactly one number y in Y. The domain of f is the set X. The number y is the image of x under f and is denoted by f x, which is called the value of f at x. The range of f is a subset of Y and consists of all images of numbers in X (see Figure P.22). Functions can be specified in a variety of ways. In this text, however, we will concentrate primarily on functions that are given by equations involving the dependent and independent variables. For instance, the equation x 2  2y  1

FUNCTION NOTATION The word function was first used by Gottfried Wilhelm Leibniz in 1694 as a term to denote any quantity connected with a curve, such as the coordinates of a point on a curve or the slope of a curve. Forty years later, Leonhard Euler used the word “function”to describe any expression made up of a variable and some constants. He introduced the notation y  f x.

Equation in implicit form

defines y, the dependent variable, as a function of x, the independent variable. To evaluate this function (that is, to find the y-value that corresponds to a given x-value), it is convenient to isolate y on the left side of the equation. 1 y  1  x 2 2

Equation in explicit form

Using f as the name of the function, you can write this equation as 1 f x  1  x 2. 2

Function notation

The original equation, x 2  2y  1, implicitly defines y as a function of x. When you solve the equation for y, you are writing the equation in explicit form. Function notation has the advantage of clearly identifying the dependent variable as f x while at the same time telling you that x is the independent variable and that the function itself is “ f.” The symbol f x is read “ f of x.” Function notation allows you to be less wordy. Instead of asking “What is the value of y that corresponds to x  3?” you can ask “What is f 3?”

20

CHAPTER P

Preparation for Calculus

In an equation that defines a function, the role of the variable x is simply that of a placeholder. For instance, the function given by f x  2x 2  4x  1 can be described by the form f 䊏  2䊏  4䊏  1 2

where parentheses are used instead of x. To evaluate f 2, simply place 2 in each set of parentheses. f 2  222  42  1  24  8  1  17

Substitute 2 for

x.

Simplify. Simplify.

NOTE Although f is often used as a convenient function name and x as the independent variable, you can use other symbols. For instance, the following equations all define the same function. f x  x 2  4x  7

Function name is f, independent variable is x.

f t  t 2  4t  7

Function name is f, independent variable is t.

gs  s 2  4s  7

Function name is g, independent variable is s.

EXAMPLE 1

Evaluating a Function

For the function f defined by f x  x 2  7, evaluate each expression. a. f 3a

b. f b  1

c.

f x  x  f x , x  0 x

Solution a. f 3a  3a2  7  9a 2  7

Substitute 3a for x. Simplify.

b. f b  1  b  1  7  b2  2b  1  7  b2  2b  8 2

Substitute b  1 for x. Expand binomial. Simplify.

f x  x  f x x  x  7  x 2  7  x x 2  2xx  x 2  7  x 2  7 x  x 2xx  x 2  x 2

c.

STUDY TIP In calculus, it is important to communicate clearly the domain of a function or expression. For instance, in Example 1(c) the two expressions

f x   x  f x . x x  0



and 2x   x,

are equivalent because  x  0 is excluded from the domain of each expression. Without a stated domain restriction, the two expressions would not be equivalent.

x2x  x x

 2x  x,

Try It

x  0

Exploration A

NOTE The expression in Example 1(c) is called a difference quotient and has a special significance in calculus. You will learn more about this in Chapter 2.

SECTION P.3

Functions and Their Graphs

21

The Domain and Range of a Function .

The domain of a function can be described explicitly, or it may be described implicitly by an equation used to define the function. The implied domain is the set of all real numbers for which the equation is defined, whereas an explicitly defined domain is one that is given along with the function. For example, the function given by

Video Range: y ≥ 0

y

x−1

f(x) =

2

f x 

1 , 4

4 ≤ x ≤ 5

has an explicitly defined domain given by x: 4 ≤ x ≤ 5 . On the other hand, the function given by

1 x 1

2

3

gx 

4

Domain: x ≥ 1

.

x2

(a) The domain of f is 1,  and the range is

0, .

1 x2  4

has an implied domain that is the set x: x  ± 2 .

Finding the Domain and Range of a Function

EXAMPLE 2 Editable Graph

a. The domain of the function f(x) = tan x

y

f x  x  1

3 2

Range

1 x

π

2π

is the set of all x-values for which x  1 ≥ 0, which is the interval 1, . To find the range observe that f x  x  1 is never negative. So, the range is the interval 0, , as indicated in Figure P.23(a). b. The domain of the tangent function, as shown in Figure P.23(b), f x  tan x is the set of all x-values such that x

Domain

. .

(b) The domain of f is all x-values such that  x   n and the range is  , . 2

  n , 2

Exploration A A Function Defined by More than One Equation

EXAMPLE 3

Figure P.23

Domain of tangent function

The range of this function is the set of all real numbers. For a review of the characteristics of this and other trigonometric functions, see Appendix D.

Try It

Editable Graph

n is an integer.

Determine the domain and range of the function.

.

Range: y ≥ 0

y

f(x) =

1 − x,

f x 

x 0 Original graph: Horizontal shift c units to the right: Horizontal shift c units to the left: Vertical shift c units downward: Vertical shift c units upward: Reflection (about the x-axis): Reflection (about the y-axis): Reflection (about the origin):

y  f x y  f x  c y  f x  c y  f x  c y  f x  c y  f x y  f x y  f x

24

CHAPTER P

Preparation for Calculus

Classifications and Combinations of Functions

LEONHARD EULER (1707–1783)

The modern notion of a function is derived from the efforts of many seventeenth- and eighteenth-century mathematicians. Of particular note was Leonhard Euler, to whom we are indebted for the function notation y  f x. By the end of the eighteenth century, mathematicians and scientists had concluded that many real-world phenomena could be represented by mathematical models taken from a collection of functions called elementary functions. Elementary functions fall into three categories.

In addition to making major contributions to almost every branch of mathematics, Euler was one of the first to apply calculus to real-life problems in physics. His extensive published writings include such topics as shipbuilding, acoustics, optics, astronomy, .mechanics, and magnetism.

1. Algebraic functions (polynomial, radical, rational) 2. Trigonometric functions (sine, cosine, tangent, and so on) 3. Exponential and logarithmic functions

MathBio

You can review the trigonometric functions in Appendix D. The other nonalgebraic functions, such as the inverse trigonometric functions and the exponential and logarithmic functions, are introduced in Chapter 5. The most common type of algebraic function is a polynomial function f x  an x n  an1x n1  . . .  a 2 x 2  a 1 x  a 0,

an  0

where the positive integer n is the degree of the polynomial function. The constants ai are coefficients, with an the leading coefficient and a0 the constant term of the polynomial function. It is common practice to use subscript notation for coefficients of general polynomial functions, but for polynomial functions of low degree, the following simpler forms are often used. Zeroth degree: First degree:

f x  a f x  ax  b

Constant function Linear function

Second degree: f x  ax 2  bx  c Third degree: f x  ax3  bx 2  cx  d FOR FURTHER INFORMATION For

Cubic function

Although the graph of a nonconstant polynomial function can have several turns, eventually the graph will rise or fall without bound as x moves to the right or left. Whether the graph of

more on the history of the concept of a function, see the article “Evolution of the Function Concept: A Brief Survey” by. Israel Kleiner in The College Mathematics Journal.

f x  an xn  an1xn1  . . .  a2 x 2  a1x  a0 eventually rises or falls can be determined by the function’s degree (odd or even) and by the leading coefficient an, as indicated in Figure P.29. Note that the dashed portions of the graphs indicate that the Leading Coefficient Test determines only the right and left behavior of the graph.

MathArticle

an > 0

an > 0

an < 0

y

Quadratic function

an < 0

y

y

y

Up to left

Up to right

Up to left

Up to right

Down to left

Down to right

x

Graphs of polynomial functions of even degree

The Leading Coefficient Test for polynomial functions Figure P.29

x

Down to left

x

Down to right

Graphs of polynomial functions of odd degree

x

SECTION P.3

Functions and Their Graphs

25

Just as a rational number can be written as the quotient of two integers, a rational function can be written as the quotient of two polynomials. Specifically, a function f is rational if it has the form

f x 

px , qx

qx  0

where px and qx are polynomials. Polynomial functions and rational functions are examples of algebraic functions. An algebraic function of x is one that can be expressed as a finite number of sums, differences, multiples, quotients, and radicals involving x n. For example, f x  x  1 is algebraic. Functions that are not algebraic are transcendental. For instance, the trigonometric functions are transcendental. Two functions can be combined in various ways to create new functions. For example, given f x  2x  3 and gx  x 2  1, you can form the functions shown.

 f  gx  f x  gx  2x  3  x 2  1  f  gx  f x  gx  2x  3  x 2  1  fgx  f xgx  2x  3x 2  1 f x 2x  3  fgx   2 gx x 1

Difference Product Quotient

You can combine two functions in yet another way, called composition. The resulting function is called a composite function.

f g

Domain of g x

Sum

g(x) g f

Definition of Composite Function f (g(x))

Domain of f

The domain of the composite function f  g Figure P.30

Let f and g be functions. The function given by  f  gx  f gx is called the composite of f with g. The domain of f  g is the set of all x in the domain of g such that gx is in the domain of f (see Figure P.30).

The composite of f with g may not be equal to the composite of g with f. EXAMPLE 4

Finding Composite Functions

Given f x  2x  3 and gx  cos x, find each composite function. a. f  g

b. g  f

Solution a.  f  gx  f gx

Definition of f  g

 f cos x  2cos x  3  2 cos x  3 b. g  f x  g  f x

Substitute cos x for gx. Definition of f x Simplify. Definition of g  f

 g2x  3  cos2x  3 . .

Substitute 2x  3 for f x. Definition of gx

Note that  f  gx   g  f x.

Try It

Exploration A

Exploration B

Exploration C

Exploration D

Exploration E

Open Exploration

26

CHAPTER P

Preparation for Calculus

E X P L O R AT I O N Use a graphing utility to graph each function. Determine whether the function is even, odd, or neither. f x  x 2  x 4 gx  2x 3  1 h x  x 5  2x 3  x

In Section P.1, an x-intercept of a graph was defined to be a point a, 0 at which the graph crosses the x-axis. If the graph represents a function f, the number a is a zero of f. In other words, the zeros of a function f are the solutions of the equation f x  0. For example, the function f x  x  4 has a zero at x  4 because f 4  0. In Section P.1 you also studied different types of symmetry. In the terminology of functions, a function is even if its graph is symmetric with respect to the y-axis, and is odd if its graph is symmetric with respect to the origin. The symmetry tests in Section P.1 yield the following test for even and odd functions.

j x  2  x 6  x 8 k x  x 5  2x 4  x  2

Test for Even and Odd Functions

p x  x 9  3x 5  x 3  x Describe a way to identify a function as odd or even by inspecting the equation.

The function y  f x is even if f x  f x. The function y  f x is odd if f x  f x. NOTE Except for the constant function f x  0, the graph of a function of x cannot have symmetry with respect to the x-axis because it then would fail the Vertical Line Test for the graph of the function.

y

EXAMPLE 5

2

Determine whether each function is even, odd, or neither. Then find the zeros of the function.

1

(−1, 0)

(1, 0) (0, 0)

−2

Even and Odd Functions and Zeros of Functions

1

f (x) = x 3 − x

a. f x  x3  x

b. gx  1  cos x

x

2

Solution

−1

a. This function is odd because f x  x3  x  x3  x   x3  x  f x.

−2

.

The zeros of f are found as shown.

(a) Odd function

x3  x  0 xx 2  1  xx  1x  1  0 x  0, 1, 1

Editable Graph y 3

Let f x  0. Factor. Zeros of f

See Figure P.31(a). b. This function is even because

g(x) = 1 + cos x

g x  1  cosx  1  cos x  g x.

2

cosx  cosx

The zeros of g are found as shown.

1 x π

2π

−1

. .

3π

4π

1  cos x  0

Let gx  0.

cos x  1 x  2n  1, n is an integer.

Subtract 1 from each side. Zeros of g

See Figure P.31(b).

(b) Even function

Editable Graph Figure P.31

Try It

Exploration A

Exploration B

Exploration C

NOTE Each of the functions in Example 5 is either even or odd. However, some functions, such as f x  x 2  x  1, are neither even nor odd.

SECTION P.3

Functions and Their Graphs

27

E x e r c i s e s f o r S e c t i o n P. 3 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, use the graphs of f and g to answer the following. (a) Identify the domains and ranges of f and g. (b) Identify f 2 and g3.

In Exercises 19-24, find the domain of the function. 19. f x  x  1  x 21. gx 

2 1  cos x

23. f x 

x  3

20. f x  x2  3x  2 22. hx 

(c) For what value(s) of x is f x  g x? (d) Estimate the solution(s) of f x  2. (e) Estimate the solutions of g x  0. 1.

2.

y

f

4

g

y

2 4

x 4

2

4

4

f

2

g 2

25. f x 

4

2x  2, 2x  1,

(a) f 1 26. f x 

4

In Exercises 3–12, evaluate (if possible) the function at the given value(s) of the independent variable. Simplify the results. 4. f x  x  3

(a) f 0

(a) f 2

(b) f 3

(b) f 6

(c) f b

(c) f 5

(d) f x  1

(d) f x  x

x2x 2,2,

6. gx  x 2x  4

(a) g0

(a) g4

(c) g2

(c) gc

(d) gt  1

(d) gt  4

(b) g3 

3 (b) g2 

7. f x  cos 2x (a) f 0

8. f x  sin x (a) f 

(b) f  4

(b) f 54

(c) f 3

(c) f 23

9. f x  x3

10. f x  3x  1

f x  x  f x x 11. f x 

1 x  1

f x  f 2 x2

f x  f 1 x1

2

2

f x  f 1 x1

15. f t  sec 17. f x 

1 x

t 4

14. gx  x 2  5 16. ht  cot t 18. gx 

2 x1

1

x2  4

28. f x 

(c) f 2

(d) f t 2  1

(c) f 1

(d) f s 2  2

(c) f 3

(d) f b 2  1

(c) f 5

(d) f 10

x f 1 x > 1

(b) f 0

 x  1, x v 1 x  1, x < 1

(a) f 3

(b) f 1

x  4, x f 5

x  5 , x > 5 2

(a) f 3

(b) f 0

In Exercises 29–36, sketch a graph of the function and find its domain and range. Use a graphing utility to verify your graph. 4 x

29. f x  4  x

30. gx 

31. hx  x  1

1 32. f x  2x3  2

33. f x  9  x 2

34. f x  x  4  x 2

35. gt  2 sin  t

36. h   5 cos

2

Writing About Concepts 37. The graph of the distance that a student drives in a 10-minute trip to school is shown in the figure. Give a verbal description of characteristics of the student’s drive to school.

12. f x  x3  x

In Exercises 13–18, find the domain and range of the function. 13. hx   x  3

27. f x 

s

Distance (in miles)

5. gx  3  x 2

1 2

x < 0 x v 0

(b) f 0

(a) f 2

3. f x  2x  3

24. gx 

sin x 

In Exercises 25 –28, evaluate the function as indicated. Determine its domain and range.

x 2

1

1

10 8

(10, 6)

6 4 2

(4, 2) (6, 2) t

(0, 0) 2 4 6 8 10 Time (in minutes)

28

CHAPTER P

Preparation for Calculus

Writing About Concepts (continued) 38. A student who commutes 27 miles to attend college remembers, after driving a few minutes, that a term paper that is due has been forgotten. Driving faster than usual, the student returns home, picks up the paper, and once again starts toward school. Sketch a possible graph of the student’s distance from home as a function of time.

53. Use the graph of f shown in the figure to sketch the graph of each function. To print an enlarged copy of the graph, select the MathGraph button. (a) f x  3

(b) f x  1

(c) f x  2

(d) f x  4

(e) 3f x

(f)

1 4

3

6

f x

9

f

7

In Exercises 39–42, use the Vertical Line Test to determine whether y is a function of x. To print an enlarged copy of the graph, select the MathGraph button. 39. x  y 2  0

40. x 2  4  y  0 y

y 4

2

1

1

(c) f x  4

(d) f x  1 (f)

1 2

3

(2, 1) 6

f x

2

3

(4, 3)

4

5

x

3 2 1

3

1 2

55. Use the graph of f x  x to sketch the graph of each function. In each case, describe the transformation.

2

(a) y  x  2

x  1, x f 0 x  2, x > 0



42. x 2  y 2  4

y

(b) y   x



(a) hx  sin x 

2 1 x 1

 1 2



(b) hx  sinx  1

57. Given f x  x and gx  x 2  1, evaluate each expression.

1 x

1 1

2

(c) y  x  2

56. Specify a sequence of transformations that will yield each graph of h from the graph of the function f x  sin x.

y

1

4

f

1

2

2

(b) f x  2

2 x

41. y 

(a) f x  4 (e) 2f x

3 1

54. Use the graph of f shown in the figure to sketch the graph of each function. To print an enlarged copy of the graph, select the MathGraph button.

1

(a) f g1

(b) g f 1

(c) g f 0

(d) f g4

(e) f gx

(f) g f x

58. Given f x  sin x and gx  x, evaluate each expression.

2

(a) f g2 In Exercises 43–46, determine whether y is a function of x. 44. x 2  y  4

45. y 2  x 2  1

46. x 2 y  x 2  4y  0

In Exercises 47–52, use the graph of y  f x to match the function with its graph. y

e

6 5

d

3 2

6 5 4 3 2 1 2 3

c

 4 

(d) g f

43. x 2  y 2  4

 12

(b) f g

(c) g f 0

(e) f gx

(f) g f x

In Exercises 59–62, find the composite functions  f  g and  g  f . What is the domain of each composite function? Are the two composite functions equal? 59. f x  x 2

60. f x  x 2  1

gx  x

g

61. f x  f(x)

gx  cos x

3 x

62. f x 

gx  x 2  1

1 x

gx  x  2

x 1 2 3 4 5

7

9 10

b

a

5

47. y  f x  5

48. y  f x  5

49. y  f x  2

50. y  f x  4

51. y  f x  6  2

52. y  f x  1  3

63. Use the graphs of f and g to evaluate each expression. If the result is undefined, explain why. (a)  f  g3 (c) g f 5 (e) g  f 1

(b) g f 2 (d)  f  g3 (f) f g1

y

f

2 2

g x

2

2

4

SECTION P.3

Functions and Their Graphs

29

64. Ripples A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius (in feet) of the outer ripple is given by rt  0.6t, where t is the time in seconds after the pebble strikes the water. The area of the circle is given by the function Ar   r 2. Find and interpret A  rt.

Determine the value of the constant c for each function such that the function fits the data shown in the table.

Think About It In Exercises 65 and 66, F x  f  g  h. Identify functions for f, g, and h. (There are many correct answers.)

80.

65. F x  2x  2

81.

66. F x  4 sin1  x

79.

In Exercises 67–70, determine whether the function is even, odd, or neither. Use a graphing utility to verify your result. 67. f x  x 24  x 2

3 x 68. f x  

69. f x  x cos x

70. f x  sin x

82.

2

Think About It In Exercises 71 and 72, find the coordinates of a second point on the graph of a function f if the given point is on the graph and the function is (a) even and (b) odd. 3 71.  2, 4

72. 4, 9

73. The graphs of f, g, and h are shown in the figure. Decide whether each function is even, odd, or neither. y

y 6

f

4

2 x

4

4

g

Figure for 73

1

0

1

4

y

32

2

0

2

32

x

4

1

0

1

4

y

1

 14

0

1 4

1

x

4

1

0

1

4

y

8

32

Undef.

32

8

x

4

1

0

1

4

y

6

3

0

3

6

83. Graphical Reasoning An electronically controlled thermostat is programmed to lower the temperature during the night automatically (see figure). The temperature T in degrees Celsius is given in terms of t, the time in hours on a 24-hour clock. (a) Approximate T4 and T15. (b) The thermostat is reprogrammed to produce a temperature Ht  Tt  1. How does this change the temperature? Explain.

x

6 4 2

2

4

T

6 24

4

h

4

(c) The thermostat is reprogrammed to produce a temperature Ht  Tt  1. How does this change the temperature? Explain.

4

2

f

x

20

6

16

Figure for 74

12 t

74. The domain of the function f shown in the figure is 6 f x f 6. (a) Complete the graph of f given that f is even. (b) Complete the graph of f given that f is odd. Writing Functions In Exercises 75–78, write an equation for a function that has the given graph.

3

6

9

12 15 18 21 24

84. Water runs into a vase of height 30 centimeters at a constant rate. The vase is full after 5 seconds. Use this information and the shape of the vase shown to answer the questions if d is the depth of the water in centimeters and t is the time in seconds (see figure). (a) Explain why d is a function of t.

75. Line segment connecting 4, 3 and 0, 5

(b) Determine the domain and range of the function.

76. Line segment connecting 1, 2 and 5, 5

(c) Sketch a possible graph of the function.

77. The bottom half of the parabola x  y2  0 78. The bottom half of the circle x2  y2  4 Modeling Data In Exercises 79– 82, match the data with a function from the following list. (i)

f x  cx

(ii) gx 



(iii) hx  c x

cx2

(iv) r x  c/x

30 cm d

30

CHAPTER P

Preparation for Calculus

85. Modeling Data The table shows the average numbers of acres per farm in the United States for selected years. (Source: U.S. Department of Agriculture) Year

1950

1960

1970

1980

1990

2000

Acreage

213

297

374

426

460

434

(b) Use a graphing utility to graph the volume function and approximate the dimensions of the box that yield a maximum volume. (c) Use the table feature of a graphing utility to verify your answer in part (b). (The first two rows of the table are shown.)

(a) Plot the data where A is the acreage and t is the time in years, with t  0 corresponding to 1950. Sketch a freehand curve that approximates the data.

Height, x

Length and Width

Volume, V

1

24  21

1 24  21 2  484

2

24  22

2 24  22 2  800

(b) Use the curve in part (a) to approximate A15. 86. Automobile Aerodynamics The horsepower H required to overcome wind drag on a certain automobile is approximated by Hx  0.002x 2  0.005x  0.029,

10 f x f 100

where x is the speed of the car in miles per hour.

94. Length A right triangle is formed in the first quadrant by the x - and y-axes and a line through the point 3, 2 (see figure). Write the length L of the hypotenuse as a function of x. y

(a) Use a graphing utility to graph H. (b) Rewrite the power function so that x represents the speed in kilometers per hour. Find Hx1.6. 87. Think About It

 

Write the function

4

(0, y)

3

(3, 2)

2 1



f x  x  x  2

(x, 0) x

without using absolute value signs. (For a review of absolute value, see Appendix D.) 88. Writing Use a graphing utility to graph the polynomial functions p1x  x3  x  1 and p2x  x3  x. How many zeros does each function have? Is there a cubic polynomial that has no zeros? Explain. 89. Prove that the function is odd. f x  a2n1 x 2n1  . . .  a3 x 3  a1 x 90. Prove that the function is even. f x  a2n x 2n  a2n2 x 2n2  . . .  a 2 x 2  a0 91. Prove that the product of two even (or two odd) functions is even. 92. Prove that the product of an odd function and an even function is odd. 93. Volume An open box of maximum volume is to be made from a square piece of material 24 centimeters on a side by cutting equal squares from the corners and turning up the sides (see figure).

1

2

3

4

5

6

7

True or False? In Exercises 95–98, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 95. If f a  f b, then a  b. 96. A vertical line can intersect the graph of a function at most once. 97. If f x  f x for all x in the domain of f, then the graph of f is symmetric with respect to the y-axis. 98. If f is a function, then f ax  af x.

Putnam Exam Challenge 99. Let R be the region consisting of the points x, y of the Cartesian plane satisfying both x  y f 1 and y f 1. Sketch the region R and find its area.

 



100. Consider a polynomial f x with real coefficients having the property f  gx  g f x for every polynomial gx with real coefficients. Determine and prove the nature of f x. These problems were composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.

x 24  2x

x

24  2x

x

(a) Write the volume V as a function of x, the length of the corner squares. What is the domain of the function?

SECTION P.4

Section P.4

Fitting Models to Data

31

Fitting Models to Data • Fit a linear model to a real-life data set. • Fit a quadratic model to a real-life data set. • Fit a trigonometric model to a real-life data set.

Fitting a Linear Model to Data

A computer graphics drawing based on the pen and ink drawing of Leonardo da Vinci’s famous study of human proportions, called Vitruvian Man

A basic premise of science is that much of the physical world can be described mathematically and that many physical phenomena are predictable. This scientific outlook was part of the scientific revolution that took place in Europe during the late 1500s. Two early publications connected with this revolution were On the Revolutions of the Heavenly Spheres by the Polish astronomer Nicolaus Copernicus and On the Structure of the Human Body by the Belgian anatomist Andreas Vesalius. Each of these books was published in 1543 and each broke with prior tradition by suggesting the use of a scientific method rather than unquestioned reliance on authority. One basic technique of modern science is gathering data and then describing the data with a mathematical model. For instance, the data given in Example 1 are inspired by Leonardo da Vinci’s famous drawing that indicates that a person’s height and arm span are equal. EXAMPLE 1

Fitting a Linear Model to Data

A class of 28 people collected the following data, which represent their heights x and arm spans y (rounded to the nearest inch).

60, 61, 65, 65, 68, 67, 72, 73, 61, 62, 63, 63, 70, 71, 75, 74, 71, 72, 62, 60, 65, 65, 66, 68, 62, 62, 72, 73, 70, 70, 69, 68, 69, 70, 60, 61, 63, 63, 64, 64, 71, 71, 68, 67, 69, 70, 70, 72, 65, 65, 64, 63, 71, 70, 67, 67 Find a linear model to represent these data. Solution There are different ways to model these data with an equation. The simplest would be to observe that x and y are about the same and list the model as simply y  x. A more careful analysis would be to use a procedure from statistics called linear regression. (You will study this procedure in Section 13.9.) The least squares regression line for these data is

.

Arm span (in inches)

y

y  1.006x  0.23.

76 74 72 70 68 66 64 62 60

Least squares regression line

The graph of the model and the data are shown in Figure P.32. From this model, you can see that a person’s arm span tends to be about the same as his or her height.

Try It x

Height (in inches)

. Linear model and data Figure P.32

Video

Open Exploration

Many scientific and graphing calculators have built-in least squares regression programs. Typically, you enter the data into the calculator and then run the linear regression program. The program usually displays the slope and y-intercept of the best-fitting line and the correlation coefficient r. The correlation coefficient gives a measure of how well the model fits the data. The closer r is to 1, the better the model fits the data. For instance, the correlation coefficient for the model in Example 1 is r  0.97, which indicates that the model is a good fit for the data. If the r-value is positive, the variables have a positive correlation, as in Example 1. If the r-value is negative, the variables have a negative correlation.

TECHNOLOGY 60 62 64 66 68 70 72 74 76

Exploration A

32

CHAPTER P

Preparation for Calculus

Fitting a Quadratic Model to Data A function that gives the height s of a falling object in terms of the time t is called a position function. If air resistance is not considered, the position of a falling object can be modeled by 1 st  2gt 2  v0 t  s0

where g is the acceleration due to gravity, v0 is the initial velocity, and s0 is the initial height. The value of g depends on where the object is dropped. On earth, g is approximately 32 feet per second per second, or 9.8 meters per second per second. To discover the value of g experimentally, you could record the heights of a falling object at several increments, as shown in Example 2. EXAMPLE 2

Fitting a Quadratic Model to Data

A basketball is dropped from a height of about 514 feet. The height of the basketball is recorded 23 times at intervals of about 0.02 second.* The results are shown in the table. Time

0.0

0.02

0.04

0.06

0.08

0.099996

Height

5.23594

5.20353

5.16031

5.0991

5.02707

4.95146

Time

0.119996

0.139992

0.159988

0.179988

0.199984

0.219984

Height

4.85062

4.74979

4.63096

4.50132

4.35728

4.19523

Time

0.23998

0.25993

0.27998

0.299976

0.319972

0.339961

Height

4.02958

3.84593

3.65507

3.44981

3.23375

3.01048

Time

0.359961

0.379951

0.399941

0.419941

0.439941

Height

2.76921

2.52074

2.25786

1.98058

1.63488

Find a model to fit these data. Then use the model to predict the time when the basketball will hit the ground. Solution Begin by drawing a scatter plot of the data, as shown in Figure P.33. From the scatter plot, you can see that the data do not appear to be linear. It does appear, however, that they might be quadratic. To check this, enter the data into a calculator or computer that has a quadratic regression program. You should obtain the model

s

6

Height (in feet)

5 4

s  15.45t 2  1.30t  5.234.

3

Using this model, you can predict the time when the basketball hits the ground by substituting 0 for s and solving the resulting equation for t.

2 1 t 0.1

0.2

0.3

0.4

Time (in seconds)

Scatter plot of data Figure P.33

.

Least squares regression quadratic

0.5

0  15.45t 2  1.30t  5.234 t

1.30 ± 1.30  415.455.234 215.45

Let s  0.

2

t  0.54

Quadratic Formula Choose positive solution.

The solution is about 0.54 second. In other words, the basketball will continue to fall for about 0.1 second more before hitting the ground.

Try It

Exploration A

* Data were collected with a Texas Instruments CBL (Calculator-Based Laboratory) System.

SECTION P.4

Fitting Models to Data

33

Fitting a Trigonometric Model to Data

The plane of Earth’s orbit about the sun and its axis of rotation are not perpendicular. Instead, Earth’s axis is tilted with respect to its orbit. The result is that the amount of daylight received by locations on Earth varies with the time of year. That is, it varies with the position of Earth in its orbit.

What is mathematical modeling? This is one of the questions that is asked in the book Guide to Mathematical Modelling. Here is part of the answer.* 1. Mathematical modeling consists of applying your mathematical skills to obtain useful answers to real problems. 2. Learning to apply mathematical skills is very different from learning mathematics itself. 3. Models are used in a very wide range of applications, some of which do not appear initially to be mathematical in nature. 4. Models often allow quick and cheap evaluation of alternatives, leading to optimal solutions that are not otherwise obvious. 5. There are no precise rules in mathematical modeling and no “correct” answers. 6. Modeling can be learned only by doing. EXAMPLE 3

The number of hours of daylight on Earth depends on the latitude and the time of year. Here are the numbers of minutes of daylight at a location of 20 N latitude on the longest and shortest days of the year: June 21, 801 minutes; December 22, 655 minutes. Use these data to write a model for the amount of daylight d (in minutes) on each day of the year at a location of 20 N latitude. How could you check the accuracy of your model?

d

Daylight (in minutes)

850

365

800

Solution Here is one way to create a model. You can hypothesize that the model is a sine function whose period is 365 days. Using the given data, you can conclude that the amplitude of the graph is 801  6552, or 73. So, one possible model is

73

750

728 700

73

d  728  73 sin

650 t 40

120

200

280

360

440

Day (0 ↔ December 22)

Graph of model Figure P.34

NOTE For more review of trigonometric functions, see Appendix D.

.

Fitting a Trigonometric Model to Data

2 t



 365  2 .

In this model, t represents the number of each day of the year, with December 22 represented by t  0. A graph of this model is shown in Figure P.34. To check the accuracy of this model, we used a weather almanac to find the numbers of minutes of daylight on different days of the year at the location of 20 N latitude. Date Dec 22 Jan 1 Feb 1 Mar 1 Apr 1 May 1 Jun 1 Jun 21 Jul 1 Aug 1 Sep 1 Oct 1 Nov 1 Dec 1

Value of t 0 10 41 69 100 130 161 181 191 222 253 283 314 344

Actual Daylight 655 min 657 min 676 min 705 min 740 min 772 min 796 min 801 min 799 min 782 min 752 min 718 min 685 min 661 min

Daylight Given by Model 655 min 656 min 672 min 701 min 739 min 773 min 796 min 801 min 800 min 785 min 754 min 716 min 681 min 660 min

You can see that the model is fairly accurate.

Try It

Exploration A

* Text from Dilwyn Edwards and Mike Hamson, Guide to Mathematical Modelling (Boca Raton: CRC Press, 1990). Used by permission of the authors.

34

CHAPTER P

Preparation for Calculus

E x e r c i s e s f o r S e c t i o n P. 4 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, a scatter plot of data is given. Determine whether the data can be modeled by a linear function, a quadratic function, or a trigonometric function, or that there appears to be no relationship between x and y. To print an enlarged copy of the graph, select the MathGraph button. 1.

y

2.

y

F

20

40

60

80

100

d

1.4

2.5

4.0

5.3

6.6

(a) Use the regression capabilities of a graphing utility to find a linear model for the data. (b) Use a graphing utility to plot the data and graph the model. How well does the model fit the data? Explain your reasoning. (c) Use the model to estimate the elongation of the spring when a force of 55 newtons is applied.

x

x

3.

y

4.

y

8. Falling Object In an experiment, students measured the speed s (in meters per second) of a falling object t seconds after it was released. The results are shown in the table. t

0

1

2

3

4

s

0

11.0

19.4

29.2

39.4

(a) Use the regression capabilities of a graphing utility to find a linear model for the data. x

x

5. Carcinogens Each ordered pair gives the exposure index x of a carcinogenic substance and the cancer mortality y per 100,000 people in the population.

3.50, 150.1, 3.58, 133.1, 4.42, 132.9, 2.26, 116.7, 2.63, 140.7, 4.85, 165.5, 12.65, 210.7, 7.42, 181.0, 9.35, 213.4 (a) Plot the data. From the graph, do the data appear to be approximately linear? (b) Visually find a linear model for the data. Graph the model. (c) Use the model to approximate y if x  3. 6. Quiz Scores The ordered pairs represent the scores on two consecutive 15-point quizzes for a class of 18 students. 7, 13, 9, 7, 14, 14, 15, 15, 10, 15, 9, 7, 14, 11, 14, 15, 8, 10, 15, 9, 10, 11, 9, 10, 11, 14, 7, 14, 11, 10, 14, 11, 10, 15, 9, 6 (a) Plot the data. From the graph, does the relationship between consecutive scores appear to be approximately linear? (b) If the data appear to be approximately linear, find a linear model for the data. If not, give some possible explanations. 7. Hooke’s Law Hooke’s Law states that the force F required to compress or stretch a spring (within its elastic limits) is proportional to the distance d that the spring is compressed or stretched from its original length. That is, F  kd, where k is a measure of the stiffness of the spring and is called the spring constant. The table shows the elongation d in centimeters of a spring when a force of F newtons is applied.

(b) Use a graphing utility to plot the data and graph the model. How well does the model fit the data? Explain your reasoning. (c) Use the model to estimate the speed of the object after 2.5 seconds. 9. Energy Consumption and Gross National Product The data show the per capita electricity consumptions (in millions of Btu) and the per capita gross national products (in thousands of U.S. dollars) for several countries in 2000. (Source: U.S. Census Bureau) Argentina Chile

(73, 12.05) (68, 9.1)

Bangladesh

(4, 1.59)

Egypt

(32, 3.67)

Greece

(126, 16.86)

Hong Kong

Hungary

(105, 11.99)

India

Mexico

(63, 8.79)

Portugal

(108, 16.99)

South Korea

(167, 17.3)

Spain

(137, 19.26)

Turkey

(47, 7.03)

United Kingdom

(166, 23.55)

Venezuela

(113, 5.74)

Poland

(118, 25.59) (13, 2.34) (95, 9)

(a) Use the regression capabilities of a graphing utility to find a linear model for the data. What is the correlation coefficient? (b) Use a graphing utility to plot the data and graph the model. (c) Interpret the graph in part (b). Use the graph to identify the three countries that differ most from the linear model. (d) Delete the data for the three countries identified in part (c). Fit a linear model to the remaining data and give the correlation coefficient.

SECTION P.4

10. Brinell Hardness The data in the table show the Brinell hardness H of 0.35 carbon steel when hardened and tempered at temperature t (degrees Fahrenheit). (Source: Standard Handbook for Mechanical Engineers) 200

400

600

800

1000

1200

H

534

495

415

352

269

217

13. Health Maintenance Organizations The bar graph shows the numbers of people N (in millions) receiving care in HMOs for the years 1990 through 2002. (Source: Centers for Disease Control) HMO Enrollment N

90

(a) Use the regression capabilities of a graphing utility to find a linear model for the data. (b) Use a graphing utility to plot the data and graph the model. How well does the model fit the data? Explain your reasoning. (c) Use the model to estimate the hardness when t is 500F. 11. Automobile Costs The data in the table show the variable costs for operating an automobile in the United States for several recent years. The functions y1, y2, and y3 represent the costs in cents per mile for gas and oil, maintenance, and tires, respectively. (Source: American Automobile Manufacturers Association)

Enrollment (in millions)

t

35

Fitting Models to Data

76.6

80

76.1

66.8

70 60

50.9 52.5

50 40

81.3 80.9 79.5

33.0 34.0

42.2 36.1 38.4

30 20 10 t 0

1

2

3

4

5

6

7

8

9

10 11 12

Year (0  1990)

(a) Let t be the time in years, with t  0 corresponding to 1990. Use the regression capabilities of a graphing utility to find linear and cubic models for the data. (b) Use a graphing utility to graph the data and the linear and cubic models.

Year

y1

y2

y3

0

5.40

2.10

0.90

1

6.70

2.20

0.90

2

6.00

2.20

0.90

(d) Use a graphing utility to find and graph a quadratic model for the data.

3

6.00

2.40

0.90

(e) Use the linear and cubic models to estimate the number of people receiving care in HMOs in the year 2004.

4

5.60

2.50

1.10

5

6.00

2.60

1.40

6

5.90

2.80

1.40

7

6.60

2.80

1.40

(c) Use the graphs in part (b) to determine which is the better model.

(f) Use a graphing utility to find other models for the data. Which models do you think best represent the data? Explain. 14. Car Performance The time t (in seconds) required to attain a speed of s miles per hour from a standing start for a Dodge Avenger is shown in the table. (Source: Road & Track)

(a) Use the regression capabilities of a graphing utility to find a cubic model for y1 and linear models for y2 and y3. (b) Use a graphing utility to graph y1, y2, y3, and y1  y2  y3 in the same viewing window. Use the model to estimate the total variable cost per mile in year 12. 12. Beam Strength Students in a lab measured the breaking strength S (in pounds) of wood 2 inches thick, x inches high, and 12 inches long. The results are shown in the table. x

4

6

8

10

12

S

2370

5460

10,310

16,250

23,860

(a) Use the regression capabilities of a graphing utility to fit a quadratic model to the data. (b) Use a graphing utility to plot the data and graph the model. (c) Use the model to approximate the breaking strength when x  2.

s

30

40

50

60

70

80

90

t

3.4

5.0

7.0

9.3

12.0

15.8

20.0

(a) Use the regression capabilities of a graphing utility to find a quadratic model for the data. (b) Use a graphing utility to plot the data and graph the model. (c) Use the graph in part (b) to state why the model is not appropriate for determining the times required to attain speeds less than 20 miles per hour. (d) Because the test began from a standing start, add the point 0, 0 to the data. Fit a quadratic model to the revised data and graph the new model. (e) Does the quadratic model more accurately model the behavior of the car for low speeds? Explain.

36

CHAPTER P

Preparation for Calculus

15. Car Performance A V8 car engine is coupled to a dynamometer and the horsepower y is measured at different engine speeds x (in thousands of revolutions per minute). The results are shown in the table.

18. Temperature The table shows the normal daily high temperatures for Honolulu H and Chicago C (in degrees Fahrenheit) for month t, with t  1 corresponding to January. (Source: NOAA) t

1

2

3

4

5

6

H

80.1

80.5

81.6

82.8

84.7

86.5

C

29.0

33.5

45.8

58.6

70.1

79.6

(a) Use the regression capabilities of a graphing utility to find a cubic model for the data.

t

7

8

9

10

11

12

(b) Use a graphing utility to plot the data and graph the model.

H

87.5

88.7

88.5

86.9

84.1

81.2

(c) Use the model to approximate the horsepower when the engine is running at 4500 revolutions per minute.

C

83.7

81.8

74.8

63.3

48.4

34.0

x

1

2

3

4

5

6

y

40

85

140

200

225

245

16. Boiling Temperature The table shows the temperatures T F at which water boils at selected pressures p (pounds per square inch). (Source: Standard Handbook for Mechanical Engineers) p

5

10

14.696 (1 atmosphere)

20

T

162.24

193.21

212.00

227.96

p

30

40

60

80

100

T

250.33

267.25

292.71

312.03

327.81

(a) A model for Honolulu is Ht  84.40  4.28 sin

Find a model for Chicago. (b) Use a graphing utility to graph the data and the model for the temperatures in Honolulu. How well does the model fit? (c) Use a graphing utility to graph the data and the model for the temperatures in Chicago. How well does the model fit? (d) Use the models to estimate the average annual temperature in each city. What term of the model did you use? Explain. (e) What is the period of each model? Is it what you expected? Explain.

(a) Use the regression capabilities of a graphing utility to find a cubic model for the data.

(f) Which city has a greater variability of temperatures throughout the year? Which factor of the models determines this variability? Explain.

(b) Use a graphing utility to plot the data and graph the model. (c) Use the graph to estimate the pressure required for the boiling point of water to exceed 300F. (d) Explain why the model would not be correct for pressures exceeding 100 pounds per square inch. 17. Harmonic Motion The motion of an oscillating weight suspended by a spring was measured by a motion detector. The data collected and the approximate maximum (positive and negative) displacements from equilibrium are shown in the figure. The displacement y is measured in centimeters and the time t is measured in seconds.

6t  3.86.

Writing About Concepts 19. Search for real-life data in a newspaper or magazine. Fit the data to a model. What does your model imply about the data? 20. Describe a possible real-life situation for each data set. Then describe how a model could be used in the real-life setting. (a)

y

(b)

y

(a) Is y a function of t? Explain. (b) Approximate the amplitude and period of the oscillations. (c) Find a model for the data. (d) Use a graphing utility to graph the model in part (c). Compare the result with the data in the figure. y 3

x

(c)

y

x

(d)

y

(0.125, 2.35)

2 1

(0.375, 1.65) t

0.2 1

0.4

0.6

0.8

x

x

REVIEW EXERCISES

37

Review Exercises for Chapter P The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

27. 3, 0,

In Exercises 1–4, find the intercepts (if any). 1. y  2x  3

2. y  x  1x  3

x1 3. y  x2

4. xy  4

5. x2y  x2  4y  0

9.

29. Find equations of the lines passing through 2, 4 and having the following characteristics. 7

6. y  x4  x2  3

In Exercises 7–14, sketch the graph of the equation. 7.

m is undefined.

(a) Slope of 16

In Exercises 5 and 6, check for symmetry with respect to both axes and to the origin.

y  12x  13x  56y

28. 5, 4,

m   23

 3

8. 4x  2y  6

1

10. 0.02x  0.15y  0.25

11. y  7  6x  x 2

12. y  6x  x 2

13. y  5  x

14. y  x  4  4





In Exercises 15 and 16, describe the viewing window of a graphing utility that yields the figure. 15. y  4x2  25

3 x  6 16. y  8

(b) Parallel to the line 5x  3y  3 (c) Passing through the origin (d) Parallel to the y-axis 30. Find equations of the lines passing through 1, 3 and having the following characteristics. 2

(a) Slope of  3 (b) Perpendicular to the line x  y  0 (c) Passing through the point 2, 4 (d) Parallel to the x-axis 31. Rate of Change The purchase price of a new machine is $12,500, and its value will decrease by $850 per year. Use this information to write a linear equation that gives the value V of the machine t years after it is purchased. Find its value at the end of 3 years. 32. Break-Even Analysis A contractor purchases a piece of equipment for $36,500 that costs an average of $9.25 per hour for fuel and maintenance. The equipment operator is paid $13.50 per hour, and customers are charged $30 per hour. (a) Write an equation for the cost C of operating this equipment for t hours.

In Exercises 17 and 18, use a graphing utility to find the point(s) of intersection of the graphs of the equations.

(b) Write an equation for the revenue R derived from t hours of use.

17. 3x  4y  8

18. x  y  1  0

x y5

y  x2  7

(c) Find the break-even point for this equipment by finding the time at which R  C.

19. Think About It Write an equation whose graph has intercepts at x  2 and x  2 and is symmetric with respect to the origin. 20. Think About It For what value of k does the graph of y  kx3 pass through the point? (a) 1, 4

(b) 2, 1

(c) 0, 0

(d) 1, 1

In Exercises 21 and 22, plot the points and find the slope of the line passing through the points. 21.

32, 1, 5, 52 

22. 7, 1, 7, 12

In Exercises 23 and 24, use the concept of slope to find t such that the three points are collinear. 23. 2, 5, 0, t, 1, 1

24. 3, 3, t, 1, 8, 6

In Exercises 25–28, find an equation of the line that passes through the point with the indicated slope. Sketch the line. 25. 0, 5,

3

m2

26. 2, 6,

m0

In Exercises 33–36, sketch the graph of the equation and use the Vertical Line Test to determine whether the equation expresses y as a function of x. 33. x  y 2  0

34. x 2  y  0

35. y  x 2  2x

36. x  9  y 2

37. Evaluate (if possible) the function f x  1x at the specified values of the independent variable, and simplify the results. (a) f 0

(b)

f 1  x  f 1 x

38. Evaluate (if possible) the function at each value of the independent variable. f x 

x 2  2, x < 0

x  2, x  0

(a) f 4

(b) f 0

(c) f 1

39. Find the domain and range of each function. x 2, x 0

and for negative x-values

f (x) = −1

x  1,

lim f x does not exist.

x < 0.

x

.

x→ 0

This means that no matter how close x gets to 0, there will be both positive and negative x-values that yield f x  1 and f x  1. Specifically, if  (the lowercase Greek letter delta) is a positive number, then for x-values satisfying the inequality 0 < x < , you can classify the values of x x as shown.

Figure 1.8

Editable Graph





 , 0

0, 

Negative x-values yield x x  1.

Positive x-values yield x x  1.





. This implies that the limit does not exist.

Try It EXAMPLE 4

Exploration A

Exploration B

Unbounded Behavior

Discuss the existence of the limit lim

x→0

Solution Let f x  1x 2. In Figure 1.9, you can see that as x approaches 0 from either the right or the left, f x increases without bound. This means that by choosing x close enough to 0, you can force f x to be as large as you want. For instance, f x) 1 will be larger than 100 if you choose x that is within 10 of 0. That is,

y

f(x) =

1 x2

4 3



0 < x
100. x2

Similarly, you can force f x to be larger than 1,000,000, as follows.

1

−2

1 . x2

2



0 < x
1,000,000 x2

Because f x is not approaching a real number L as x approaches 0, you can conclude that the limit does not exist.

Try It

Exploration A

Exploration B

SECTION 1.2

1 Discuss the existence of the limit lim sin . x→0 x

1 f (x) = sin x 1

x −1

Solution Let f x  sin1x. In Figure 1.10, you can see that as x approaches 0, f x oscillates between 1 and 1. So, the limit does not exist because no matter how small you choose , it is possible to choose x1 and x2 within  units of 0 such that sin1x1  1 and sin1x2   1, as shown in the table.

1

x .

.

51

Oscillating Behavior

EXAMPLE 5 y

Finding Limits Graphically and Numerically

−1

lim f x does not exist.

x→ 0

Figure 1.10

Editable Graph

2

23

25

27

29

211

x→0

1

1

1

1

1

1

Limit does not exist.

sin 1/x

Try It

Exploration A

Open Exploration

Common Types of Behavior Associated with Nonexistence of a Limit 1. f x approaches a different number from the right side of c than it approaches from the left side. 2. f x increases or decreases without bound as x approaches c. 3. f x oscillates between two fixed values as x approaches c. There are many other interesting functions that have unusual limit behavior. An often cited one is the Dirichlet function f x 

0,1,

if x is rational. if x is irrational.

Because this function has no limit at any real number c, it is not continuous at any real number c. You will study continuity more closely in Section 1.4. When you use a graphing utility to investigate the behavior of a function near the x-value at which you are trying to evaluate a limit, remember that you can’t always trust the pictures that graphing utilities draw. If you use a graphing utility to graph the function in Example 5 over an interval containing 0, you will most likely obtain an incorrect graph such as that shown in Figure 1.11. The reason that a graphing utility can’t show the correct graph is that the graph has infinitely many oscillations over any interval that contains 0.

TECHNOLOGY PITFALL

1.2

PETER GUSTAV DIRICHLET (1805–1859)

.

In the early development of calculus, the definition of a function was much more restricted than it is today, and “functions” such as the Dirichlet function would not have been considered. The modern definition of function was given by the German mathematician Peter Gustav Dirichlet.

−0.25

0.25

− 1.2

MathBio

Incorrect graph of f x  sin1x. Figure 1.11

52

CHAPTER 1

Limits and Their Properties

A Formal Definition of Limit Let’s take another look at the informal description of a limit. If f x becomes arbitrarily close to a single number L as x approaches c from either side, then the limit of f x as x approaches c is L, written as lim f x  L.

x→c

At first glance, this description looks fairly technical. Even so, it is informal because exact meanings have not yet been given to the two phrases “ f x becomes arbitrarily close to L” and “x approaches c.” The first person to assign mathematically rigorous meanings to these two phrases was Augustin-Louis Cauchy. His - definition of limit is the standard used today. In Figure 1.12, let (the lowercase Greek letter epsilon) represent a (small) positive number. Then the phrase “f x becomes arbitrarily close to L” means that f x lies in the interval L  , L  . Using absolute value, you can write this as

L +ε L

(c, L)

 f x  L < .

L−ε

Similarly, the phrase “x approaches c” means that there exists a positive number  such that x lies in either the interval c  , c or the interval c, c  . This fact can be concisely expressed by the double inequality c +δ c c−δ

The - definition of the limit of f x as x approaches c Figure 1.12





0 < x  c < . The first inequality





0 < xc

The distance between x and c is more than 0.

expresses the fact that x  c. The second inequality

x  c < 

x is within  units of c.

says that x is within a distance  of c.

Definition of Limit Let f be a function defined on an open interval containing c (except possibly at c) and let L be a real number. The statement lim f x  L

x→c

means that for each > 0 there exists a  > 0 such that if





0 < x  c < , then

 f x  L < .

FOR FURTHER INFORMATION For

more on the introduction of rigor to calculus, see “Who Gave You the Epsilon? Cauchy and the Origins of Rigorous Calculus” by Judith V. . Grabiner in The American Mathematical Monthly. MathArticle

NOTE

Throughout this text, the expression

lim f x  L

x→c

implies two statements—the limit exists and the limit is L.

Some functions do not have limits as x → c, but those that do cannot have two different limits as x → c. That is, if the limit of a function exists, it is unique (see Exercise 69).

SECTION 1.2

Finding Limits Graphically and Numerically

53

The next three examples should help you develop a better understanding of the - definition of limit.

Finding a  for a Given 

EXAMPLE 6

y = 1.01 y=1 y = 0.99

Given the limit lim 2x  5  1

y

x→3

x = 2.995 x=3 x = 3.005









find  such that 2x  5  1 < 0.01 whenever 0 < x  3 < .

2

Solution In this problem, you are working with a given value of —namely,  0.01. To find an appropriate , notice that

1

x 1

2

3

4

−1





is equivalent to 2 x  3 < 0.01, you can choose   20.01  0.005. This choice works because





0 < x  3 < 0.005

f (x) = 2x − 5

−2

2x  5  1  2x  6  2x  3. Because the inequality 2x  5  1 < 0.01 1 implies that

2x  5  1  2x  3 < 20.005  0.01

. The limit of f x as x approaches 3 is 1.

as shown in Figure 1.13.

Figure 1.13

Try It

Exploration B

Exploration A

NOTE In Example 6, note that 0.005 is the largest value of  that will guarantee 2x  5  1 < 0.01 whenever 0 < x  3 < . Any smaller positive value of  would also work.









In Example 6, you found a -value for a given . This does not prove the existence of the limit. To do that, you must prove that you can find a  for any , as shown in the next example. y=4+ε

Using the - Definition of Limit

EXAMPLE 7

y=4

Use the - definition of limit to prove that

y=4−ε

lim 3x  2  4.

x=2+δ x=2 x=2−δ

y

x→2

Solution You must show that for each > 0, there exists a  > 0 such that 3x  2  4 < whenever 0 < x  2 < . Because your choice of  depends on , you need to establish a connection between the absolute values 3x  2  4 and x  2 .









  3x  2  4  3x  6  3x  2

4

3



So, for a given > 0 you can choose   3. This choice works because 2





0 < x2 <  1

f(x) = 3x − 2

implies that x

1

2

3

4

. The limit of f x as x approaches 2 is 4. Figure 1.14

3

3x  2  4  3x  2 < 33  as shown in Figure 1.14.

Try It

Exploration A



54

CHAPTER 1

Limits and Their Properties

EXAMPLE 8

Using the - Definition of Limit

Use the - definition of limit to prove that

f(x) = x 2

lim x 2  4.

4+ε

x→2

(2 + δ )2

Solution You must show that for each > 0, there exists a  > 0 such that

4

x 2  4 <

(2 − δ )2 4−ε

2+δ 2 2−δ

The . limit of f x as x approaches 2 is 4. Figure 1.15



 To find an appropriate , begin by writing x2  4  x  2x  2. For all x in the interval 1, 3, you know that x  2 < 5. So, letting  be the minimum of 5 and 1, it follows that, whenever 0 < x  2 < , you have whenever 0 < x  2 < .



x2  4  x  2x  2 < 55  as shown in Figure 1.15.

Try It

Exploration A

Throughout this chapter you will use the - definition of limit primarily to prove theorems about limits and to establish the existence or nonexistence of particular types of limits. For finding limits, you will learn techniques that are easier to use than the - definition of limit.

54

CHAPTER 1

Limits and Their Properties

Exercises for Section 1.2 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–8, complete the table and use the result to estimate the limit. Use a graphing utility to graph the function to confirm your result. x2 x2  x  2

1. lim

xq2

x

1.9

1.99

1.999

2.001

2.01

2.1 xq4

x2 x2  4

x

1.9

1.99

1.999

2.001

2.01

2.1

3.9

0.01

0.1

x

x3

x

3.1

f x

3.01

3.1

3.99

3.999

4.001

4.01

4.1

3.01

0.01

0.001

0.001

0.01

0.1

0.01

0.001

0.001

0.01

0.1

f x 0.001

0.001

0.01

0.1

1  x  2

xq3

3.001

x x  1  4 5 x4

8. lim

cos x  1 x

x

0.1

xq0

f x 4. lim

2.999

7. lim sinx x xq0

x 0.1

2.99

f x

x  3  3

x

2.9

x

f x

xq0

1 x  1  1 4 x3

f x 6. lim

2. lim

3. lim

xq3

x

f x

xq2

5. lim

f x 3.001

2.999

2.99

2.9

SECTION 1.2

In Exercises 9–18, use the graph to find the limit (if it exists). If the limit does not exist, explain why. 9. lim 4  x

10. lim x 2  2

x3

y

4

4

3

3

y

(b) lim f x

6 5

xq1

(c) f 4

3 2 1

(d) lim f x

2

xq4

1

1

x

x 1

2

3

1

x 2 1

4

11. lim f x

1

20. (a) f 2

y

(b) lim f x

x1

4 3 2

xq2

f x 

40, x,

x2 x2

f x 

x1,  2,

(c) f 0

x1 x1

2

y

(d) lim f x xq0

y

4

4

3

3

(e) f 2

3

(h) lim f x xq4

x 2 1

4

x  5

13. lim

xq5

1

2

x5

In Exercises 21 and 22, use the graph of f to identify the values of c for which lim f x exists. xqc

1 xq3 x  3

14. lim

y

21.

y 6

y

4 3 2 1

2

(g) f 4

x 2

x 1 2 3 4 5

xq2

1

1

2 1

(f ) lim f x

2 1

1 2 3 4 5 6

2

12. lim f x

x2

55

In Exercises 19 and 20, use the graph of the function f to decide whether the value of the given quantity exists. If it does, find it. If not, explain why. 19. (a) f 1

x1

y

Finding Limits Graphically and Numerically

4

2 1 x 6 7 8 9

x

2

1

2 3 4

x 2

4

2

2

4

2

22.

15. lim sin  x

y 6

16. lim sec x

xq1

4

xq0

y

2

y

x 4

2

1

2

4

6

x 1

2 U 2

x

U 2

In Exercises 23 and 24, sketch the graph of f. Then identify the values of c for which lim f x exists. xqc

17. lim cos xq0

1 x

xq  2

y

1

2 1 x

1

1 1

U 2

U 2

 

x, 23. f x  8  2x, 4,

18. lim tan x

y

2

U

3U 2

x

sin x, 24. f x  1  cos x, cos x,

x f 2 2 < x < 4 x v 4 x < 0 0 f x f  x > 

56

CHAPTER 1

Limits and Their Properties

In Exercises 25 and 26, sketch a graph of a function f that satisfies the given values. (There are many correct answers.) 25. f 0 is undefined.

30. The graph of f x 

26. f 2  0

lim f x  4

f 2  0

f 2  6

lim f x  0

x0











y

lim f x does not exist.

x2



is shown in the figure. Find  such that if 0 < x  2 <  then

 f x  1 < 0.01.

x2

lim f x  3

1 x1

2.0

x2

1.01 1.00 0.99

1.5

27. Modeling Data The cost of a telephone call between two cities is $0.75 for the first minute and $0.50 for each additional minute or fraction thereof. A formula for the cost is given by

1.0

201 2 199 101 99

0.5

Ct  0.75  0.50  t  1

x 1

where t is the time in minutes.

Note: x  greatest integer n such that n  x. For example, 3.2  3 and 1.6  2. (a) Use a graphing utility to graph the cost function for 0 < t  5. (b) Use the graph to complete the table and observe the behavior of the function as t approaches 3.5. Use the graph and the table to find

2

3

4

31. The graph of f x  2 

1 x

is shown in the figure. Find  such that if 0 < x  1 <  then

 f x  1 < 0.1. y

lim C t.

y = 1.1 y=1 y = 0.9

2

t3.5

f

1

t

3

3.3

3.4

C

3.5

3.6

3.7

4 x

?

1

(c) Use the graph to complete the table and observe the behavior of the function as t approaches 3. t

2

2.5

2.9

C

3

3.1

3.5

32. The graph of f x  x 2  1 is shown in the figure. Find  such that if 0 < x  2 <  then f x  3 < 0.2.

4



?



y

Does the limit of Ct as t approaches 3 exist? Explain.

f

4 3

28. Repeat Exercise 27 for

2

Ct  0.35  0.12t  1.

y = 3.2 y=3 y = 2.8

1

29. The graph of f x  x  1 is shown in the figure. Find  such that if 0 < x  2 <  then f x  3 < 0.4.







1

2

3

4

In Exercises 33–36, find the limit L. Then find  > 0 such that f x  L < 0.01 whenever 0 < x  c < .



5 4





33. lim 3x  2

3 2.6

x



y

3.4

2

x2

2



34. lim 4  x4

x 0.5

1.0

1.5

2.0 2.5 1.6 2.4

3.0

x 2



35. lim x 2  3 x2

36. lim x 2  4 x5



SECTION 1.2

Finding Limits Graphically and Numerically

57

In Exercises 37–48, find the limit L. Then use the - definition to prove that the limit is L.

Writing About Concepts (continued)

37. lim x  3

56. Identify three types of behavior associated with the nonexistence of a limit. Illustrate each type with a graph of a function.

x2

38. lim 2x  5 x3

12x  1 2 lim 3x  9 x1

39. lim

x4

40.

57. Jewelry A jeweler resizes a ring so that its inner circumference is 6 centimeters.

41. lim 3

(a) What is the radius of the ring?

42. lim 1

(b) If the ring’s inner circumference can vary between 5.5 centimeters and 6.5 centimeters, how can the radius vary?

x6 x2

3 x 43. lim  x0

44. lim x x4





45. lim x  2 x2

 x3



46. lim x  3

(c) Use the - definition of limit to describe this situation. Identify and . 58. Sports A sporting goods manufacturer designs a golf ball having a volume of 2.48 cubic inches.

47. lim x 2  1

(a) What is the radius of the golf ball?

48. lim x 2  3x

(b) If the ball’s volume can vary between 2.45 cubic inches and 2.51 cubic inches, how can the radius vary?

x1

x3

Writing In Exercises 49– 52, use a graphing utility to graph the function and estimate the limit (if it exists). What is the domain of the function? Can you detect a possible error in determining the domain of a function solely by analyzing the graph generated by a graphing utility? Write a short paragraph about the importance of examining a function analytically as well as graphically. 49. f x 

x  5  3

x4

lim f x)

x4

x3 50. f x  2 x  4x  3 lim f x

x3

x9 51. f x  x  3 lim f x

x9

52. f x 

x3 x2  9

lim f x

x3

Writing About Concepts 53. Write a brief description of the meaning of the notation lim f x  25. x8

54. If f 2  4, can you conclude anything about the limit of f x as x approaches 2? Explain your reasoning. 55. If the limit of f x as x approaches 2 is 4, can you conclude anything about f 2? Explain your reasoning.

(c) Use the - definition of limit to describe this situation. Identify and . 59. Consider the function f x  1  x1 x. Estimate the limit lim 1  x1 x

x0

by evaluating f at x-values near 0. Sketch the graph of f. 60. Consider the function f x 

x  1  x  1. x

Estimate lim

x  1  x  1 x

x0

by evaluating f at x-values near 0. Sketch the graph of f. 61. Graphical Analysis lim

x2

The statement

4 4 x2

x2

means that for each > 0 there corresponds a  > 0 such that if 0 < x  2 < , then









x2  4  4 < . x2



x2  4  4 < 0.001. x2

If  0.001, then



Use a graphing utility to graph each side of this inequality. Use the zoom feature to find an interval 2  , 2   such that the graph of the left side is below the graph of the right side of the inequality.

58

CHAPTER 1

62. Graphical Analysis

Limits and Their Properties

The statement

72. (a) Given that lim 3x  13x  1x2  0.01  0.01

x 2  3x x3 x  3 lim

x0

means that for each > 0 there corresponds a  > 0 such that if 0 < x  3 < , then

prove that there exists an open interval a, b containing 0 such that 3x  13x  1x2  0.01 > 0 for all x  0 in a, b.







x2  3x  3 < . x3





x2  3x  3 < 0.001. x3

(b) Given that lim g x  L, where L > 0, prove that there xc

exists an open interval a, b containing c such that gx > 0 for all x  c in a, b.

If  0.001, then



73. Programming Use the programming capabilities of a graphing utility to write a program for approximating lim f x. xc

Use a graphing utility to graph each side of this inequality. Use the zoom feature to find an interval 3  , 3   such that the graph of the left side is below the graph of the right side of the inequality. True or False? In Exercises 63– 66, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 63. If f is undefined at x  c, then the limit of f x as x approaches c does not exist.

Assume the program will be applied only to functions whose limits exist as x approaches c. Let y1  f x and generate two lists whose entries form the ordered pairs

c ± 0.1 n , f c ± 0.1 n  for n  0, 1, 2, 3, and 4. 74. Programming Use the program you created in Exercise 73 to approximate the limit x 2  x  12 . x4 x4 lim

64. If the limit of f x as x approaches c is 0, then there must exist a number k such that f k < 0.001.

Putnam Exam Challenge

65. If f c  L, then lim f x  L. xc

66. If lim f x  L, then f c  L.

75. Inscribe a rectangle of base b and height h and an isosceles triangle of base b in a circle of radius one as shown. For what value of h do the rectangle and triangle have the same area?

xc

67. Consider the function f x  x. (a) Is lim x  0.5 a true statement? Explain. x0.25

(b) Is lim x  0 a true statement? Explain. x0

68. Writing The definition of limit on page 52 requires that f is a function defined on an open interval containing c, except possibly at c. Why is this requirement necessary? 69. Prove that if the limit of f x as x  c exists, then the limit must be unique. [Hint: Let lim f x  L1 and

xc

lim f x  L 2

xc

and prove that L1  L2.] 70. Consider the line f x  mx  b, where m  0. Use the - definition of limit to prove that lim f x  mc  b. xc

71. Prove that lim f x  L is equivalent to lim f x  L  0. xc

xc

h b

76. A right circular cone has base of radius 1 and height 3. A cube is inscribed in the cone so that one face of the cube is contained in the base of the cone. What is the side-length of the cube? These problems were composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.

SECTION 1.3

Section 1.3

Evaluating Limits Analytically

59

Evaluating Limits Analytically • • • •

.

Evaluate a limit using properties of limits. Develop and use a strategy for finding limits. Evaluate a limit using dividing out and rationalizing techniques. Evaluate a limit using the Squeeze Theorem.

Properties of Limits

Video

In Section 1.2, you learned that the limit of f x as x approaches c does not depend on the value of f at x  c. It may happen, however, that the limit is precisely f c. In such cases, the limit can be evaluated by direct substitution. That is, lim f x  f c.

Substitute c for x.

x→c

Such well-behaved functions are continuous at c. You will examine this concept more closely in Section 1.4. y

f (c) = x

THEOREM 1.1

Some Basic Limits

Let b and c be real numbers and let n be a positive integer.

c+ ε ε =δ

1. lim b  b

f (c) = c

2. lim x  c

x→c

x→c

3. lim x n  c n x→c

ε =δ

c−ε x

c−δ

c

c+δ

Figure 1.16

Proof To prove Property 2 of Theorem 1.1, you need to show that for each > 0 there exists a  > 0 such that x  c < whenever 0 < x  c < . To do this, choose   . The second inequality then implies the first, as shown in Figure 1.16. This completes the proof. (Proofs of the other properties of limits in this section are listed in Appendix A or are discussed in the exercises.)









Evaluating Basic Limits

NOTE When you encounter new notations or symbols in mathematics, be sure . you know how the notations are read. For instance, the limit in Example 1(c) is read as “the limit of x 2 as x approaches 2 is 4.”

EXAMPLE 1

.

The editable graph feature allows you to edit the graph of a function to visually evaluate the limit as x approaches c.

a. lim 3  3 x→2

b. lim x  4

Try It

a.

x→4

x→2

Exploration A

Editable Graph

THEOREM 1.2

c. lim x 2  2 2  4

b.

Editable Graph

c.

Editable Graph

Properties of Limits

Let b and c be real numbers, let n be a positive integer, and let f and g be functions with the following limits. lim f x  L

x→c

and

1. Scalar multiple: 2. Sum or difference: 3. Product: 4. Quotient: 5. Power:

lim g x  K

x→c

lim b f x  bL

x→c

lim f x ± gx  L ± K

x→c

lim f xgx  LK

x→c

lim

x→c

f x L  , gx K

lim f x n  Ln

x→c

provided K  0

60

CHAPTER 1

Limits and Their Properties

EXAMPLE 2

The Limit of a Polynomial

lim 4x 2  3  lim 4x 2  lim 3

x→2

x→2



.



 4 lim x 2  lim 3

Property 1

 422  3

Example 1

 19

Simplify.

Try It .

Property 2

x→2

x→2

x→2

Exploration A

The editiable graph feature allows you to edit the graph of a function to visually evaluate the limit as x approaches c. Editable Graph In Example 2, note that the limit (as x → 2) of the polynomial function px  4x 2  3 is simply the value of p at x  2. lim px  p2  422  3  19

x→2

This direct substitution property is valid for all polynomial and rational functions with nonzero denominators. THEOREM 1.3

Limits of Polynomial and Rational Functions

If p is a polynomial function and c is a real number, then lim px  pc.

x→c

If r is a rational function given by r x  pxqx and c is a real number such that qc  0, then lim r x  r c 

x→c

EXAMPLE 3

pc . qc

The Limit of a Rational Function

2 Find the limit: lim x  x  2 . x→1 x1

Solution Because the denominator is not 0 when x  1, you can apply Theorem 1.3 to obtain x 2  x  2 12  1  2 4    2. x→1 x1 11 2

.

lim

Try It .

Exploration A

The editiable graph feature allows you to edit the graph of a function to visually evaluate the limit as x approaches c. Editable Graph THE SQUARE ROOT SYMBOL

The first use of a symbol to denote the square root can be traced to the sixteenth century. Mathematicians first used the symbol  , which had only two strokes. This symbol was chosen because it resembled a lowercase r, to .stand for the Latin word radix, meaning root.

Video

Video

Polynomial functions and rational functions are two of the three basic types of algebraic functions. The following theorem deals with the limit of the third type of algebraic function—one that involves a radical. See Appendix A for a proof of this theorem. THEOREM 1.4

The Limit of a Function Involving a Radical

Let n be a positive integer. The following limit is valid for all c if n is odd, and is valid for c > 0 if n is even. n x  n c lim 

x→c

SECTION 1.3

Evaluating Limits Analytically

61

The following theorem greatly expands your ability to evaluate limits because it shows how to analyze the limit of a composite function. See Appendix A for a proof of this theorem.

THEOREM 1.5

The Limit of a Composite Function

If f and g are functions such that lim gx  L and lim f x  f L, then x→c



x→L



lim f g x  f lim gx  f L.

x→c

EXAMPLE 4

x→c

The Limit of a Composite Function

a. Because lim x 2  4  0 2  4  4

x→0

lim x  2

and

x→4

it follows that lim x2  4  4  2.

x→0

b. Because lim 2x 2  10  232  10  8 and

x→3

3 x  2 lim 

x→8

it follows that .

3 3 2x 2  10   8  2. lim 

x→3

Try It .

Exploration A

Open Exploration

The editable graph feature allows you to edit the graph of a function to visually evaluate the limit as x approaches c. a.

Editable Graph

b.

Editable Graph

You have seen that the limits of many algebraic functions can be evaluated by direct substitution. The six basic trigonometric functions also exhibit this desirable quality, as shown in the next theorem (presented without proof). THEOREM 1.6

Limits of Trigonometric Functions

Let c be a real number in the domain of the given trigonometric function. 1. lim sin x  sin c

2. lim cos x  cos c

3. lim tan x  tan c

4. lim cot x  cot c

5. lim sec x  sec c

6. lim csc x  csc c

x→c

x→c

x→c

x→c

x→c

EXAMPLE 5

x→c

Limits of Trigonometric Functions

a. lim tan x  tan0  0 x→0



b. lim x cos x  lim x .

x→ 

x→ 

 lim cos x   cos    x→ 

c. lim sin2 x  lim sin x2  02  0 x→0

Try It

x→0

Exploration A

62

CHAPTER 1

Limits and Their Properties

A Strategy for Finding Limits On the previous three pages, you studied several types of functions whose limits can be evaluated by direct substitution. This knowledge, together with the following theorem, can be used to develop a strategy for finding limits. A proof of this theorem is given in Appendix A. y

3 f(x) = x − 1 x−1

THEOREM 1.7

3

Functions That Agree at All But One Point

Let c be a real number and let f x  gx for all x  c in an open interval containing c. If the limit of gx as x approaches c exists, then the limit of f x also exists and

2

lim f x  lim gx.

x→c

.

x→c

x −2

−1

EXAMPLE 6

1

Editable Graph

Finding the Limit of a Function

Find the limit: lim

x→1

x3  1 . x1

Solution Let f x  x3  1x  1. By factoring and dividing out like factors, you can rewrite f as

y

3

f x 

x  1x2  x  1  x2  x  1  gx, x  1

x  1.

So, for all x-values other than x  1, the functions f and g agree, as shown in Figure 1.17. Because lim gx exists, you can apply Theorem 1.7 to conclude that f and g

2

x→1

have the same limit at x  1. g (x) = x 2 + x + 1 x

.

−2

−1

1

f and g agree at all but one point.

lim

x→1

x  1x 2  x  1 x3  1  lim x1 x→1 x1 x  1x2  x  1  lim x1 x→1  lim x 2  x  1  12  1  1 3

Figure 1.17

Try It

STUDY TIP When applying this strategy for finding a limit, remember that some functions do not have a limit (as x approaches c). For instance, the following limit does not exist.

x3  1 x→1 x  1 lim

Divide out like factors. Apply Theorem 1.7.

x→1

Editable Graph .

.

Factor.

Use direct substitution. Simplify.

Exploration A

Exploration B

Exploration C

Exploration D

A Strategy for Finding Limits 1. Learn to recognize which limits can be evaluated by direct substitution. (These limits are listed in Theorems 1.1 through 1.6.) 2. If the limit of f x as x approaches c cannot be evaluated by direct substitution, try to find a function g that agrees with f for all x other than x  c. [Choose g such that the limit of gx can be evaluated by direct substitution.] 3. Apply Theorem 1.7 to conclude analytically that lim f x  lim gx  gc.

x→c

x→c

4. Use a graph or table to reinforce your conclusion.

SECTION 1.3

Evaluating Limits Analytically

63

Dividing Out and Rationalizing Techniques Two techniques for finding limits analytically are shown in Examples 7 and 8. The first technique involves dividing out common factors, and the second technique involves rationalizing the numerator of a fractional expression.

Dividing Out Technique

EXAMPLE 7

Find the limit: lim

x→3

Solution Although you are taking the limit of a rational function, you cannot apply Theorem 1.3 because the limit of the denominator is 0.

y

−2

lim x 2  x  6  0

x

−1

1

2

x→3

−1

−3

x2  x  6 x→3 x3 lim

−2

f (x) =

x2 + x − 6 x+3

−4

(−3, −5)

x2  x  6 . x3

Direct substitution fails.

lim x  3  0

x→3

Because the limit of the numerator is also 0, the numerator and denominator have a common factor of x  3. So, for all x  3, you can divide out this factor to obtain

−5

. f is undefined when x   3.

f x 

Figure 1.18

x 2  x  6 x  3x  2   x  2  gx, x3 x3

Using Theorem 1.7, it follows that

Editable Graph

x2  x  6  lim x  2 x→3 x3 x→3 lim

 5. NOTE In the solution of Example 7, . be sure you see the usefulness of the Factor Theorem of Algebra. This theorem states that if c is a zero of a polynomial function, x  c is a factor of the polynomial. So, if you apply direct substitution to a rational function and obtain r c 

pc 0  qc 0

you can conclude that x  c must be a common factor to both px and qx.

−3 − δ

−5 + ε −3 + δ

Glitch near (−3, −5)

−5 − ε

Incorrect graph of f Figure 1.19

x  3.

Apply Theorem 1.7. Use direct substitution.

This result is shown graphically in Figure 1.18. Note that the graph of the function f coincides with the graph of the function gx  x  2, except that the graph of f has a gap at the point 3, 5.

Try It

Exploration A

Open Exploration

In Example 7, direct substitution produced the meaningless fractional form 00. An expression such as 00 is called an indeterminate form because you cannot (from the form alone) determine the limit. When you try to evaluate a limit and encounter this form, remember that you must rewrite the fraction so that the new denominator does not have 0 as its limit. One way to do this is to divide out like factors, as shown in Example 7. A second way is to rationalize the numerator, as shown in Example 8. TECHNOLOGY PITFALL

f x 

x2  x  6 x3

and

Because the graphs of gx  x  2

differ only at the point 3, 5, a standard graphing utility setting may not distinguish clearly between these graphs. However, because of the pixel configuration and rounding error of a graphing utility, it may be possible to find screen settings that distinguish between the graphs. Specifically, by repeatedly zooming in near the point 3, 5 on the graph of f, your graphing utility may show glitches or irregularities that do not exist on the actual graph. (See Figure 1.19.) By changing the screen settings on your graphing utility you may obtain the correct graph of f.

64

CHAPTER 1

Limits and Their Properties

Rationalizing Technique

EXAMPLE 8

Find the limit: lim

x  1  1

x

x→0

.

Solution By direct substitution, you obtain the indeterminate form 00. lim x  1  1  0

x→0

lim

x  1  1

Direct substitution fails.

x

x→0

lim x  0

x→0

In this case, you can rewrite the fraction by rationalizing the numerator. x  1  1

y

f(x) =

x  1  1



x  1  1

x x  1  1 x  1  1  xx  1  1 x  xx  1  1 1  , x0 x  1  1

x

1





x +1−1 x



Now, using Theorem 1.7, you can evaluate the limit as shown. x

−1

lim

1

x  1  1

x

x→0

x→0

−1

The . limit of f x as x Figure 1.20

approaches 0 is 12 .

 lim

1 x  1  1



1 11



1 2

A table or a graph can reinforce your conclusion that the limit is 12. (See Figure 1.20.)

Editable Graph x approaches 0 from the left.

0.1

x approaches 0 from the right.

x

0.25

0.01 0.001 0

f x

0.5359 0.5132 0.5013 0.5001 ? f x approaches 0.5.

0.001

0.01

0.1

0.25

0.4999 0.4988 0.4881 0.4721 f x approaches 0.5.

.

Try It

Exploration A

Exploration B

Exploration C

NOTE The rationalizing technique for evaluating limits is based on multiplication by a convenient form of 1. In Example 8, the convenient form is 1

x  1  1 x  1  1

.

SECTION 1.3

Evaluating Limits Analytically

65

The Squeeze Theorem h(x) ≤ f(x) ≤ g (x)

The next theorem concerns the limit of a function that is squeezed between two other functions, each of which has the same limit at a given x-value, as shown in Figure 1.21. (The proof of this theorem is given in Appendix A.)

y

f lies in here.

g

g f

THEOREM 1.8

f

If hx ≤ f x ≤ gx for all x in an open interval containing c, except possibly at c itself, and if

h h

lim hx  L  lim gx

x→c

x

c

.

The Squeeze Theorem

x→c

then lim f x exists and is equal to L. x→c

The Squeeze Theorem Figure 1.21

Video You can see the usefulness of the Squeeze Theorem in the proof of Theorem 1.9. THEOREM 1.9 1. lim

x→0

y

(cos θ , sin θ ) (1, tan θ )

θ

(1, 0)

Two Special Trigonometric Limits

sin x 1 x

2. lim

x→0

1  cos x 0 x

Proof To avoid the confusion of two different uses of x, the proof is presented using the variable , where is an acute positive angle measured in radians. Figure 1.22 shows a circular sector that is squeezed between two triangles.

x

tan θ

1

sin θ θ

θ

A circular sector is used to prove Theorem 1.9. Figure 1.22

θ

1

1

Area of triangle tan

2

≥ ≥

Area of sector

2

1

≥ ≥

Area of triangle sin

2

Multiplying each expression by 2sin produces

1 ≥ ≥ 1 cos sin

and taking reciprocals and reversing the inequalities yields FOR FURTHER INFORMATION For more information on the function f x  sin xx, see the article “The Function sin xx” by William B. . Gearhart and Harris S. Shultz in The College Mathematics Journal. MathArticle

cos ≤

sin

≤ 1.

Because cos  cos   and sin   sin   , you can conclude that this inequality is valid for all nonzero in the open interval  2, 2. Finally, because lim cos  1 and lim 1  1, you can apply the Squeeze Theorem to

→0

→0

conclude that lim sin   1. The proof of the second limit is left as an exercise (see

→0 Exercise 120).

66

CHAPTER 1

Limits and Their Properties

A Limit Involving a Trigonometric Function

EXAMPLE 9

Find the limit: lim

x→0

tan x . x

Solution Direct substitution yields the indeterminate form 00. To solve this problem, you can write tan x as sin xcos x and obtain lim

x→0

f (x) =



tan x sin x  lim x x→0 x

cos1 x.

Now, because

tan x x 4

lim

x→0

sin x 1 x

and

lim

x→0

1 1 cos x

you can obtain − 2

 2

lim

x→0



tan x sin x  lim x→0 x x

 lim cos1 x x→0

 11  1.

−2

The . limit of f x as x approaches 0 is 1.

(See Figure 1.23.)

Figure 1.23

Try It

Editable Graph

Exploration A A Limit Involving a Trigonometric Function

EXAMPLE 10

Find the limit: lim

x→0

sin 4x . x

Solution Direct substitution yields the indeterminate form 00. To solve this problem, you can rewrite the limit as lim

g(x) =

x→0

sin 4x x

lim

x→0

  sin y  4 lim y  y→0

 41  4.

−2

The . limit of gx as x approaches 0 is 4.

Editable Graph

Multiply and divide by 4.

sin 4x sin 4x  4 lim x→0 x 4x

 2

Figure 1.24



Now, by letting y  4x and observing that x → 0 if and only if y → 0, you can write

6

− 2



sin 4x sin 4x .  4 lim x x→0 4x

(See Figure 1.24.)

Try It

Exploration A

Use a graphing utility to confirm the limits in the examples and exercise set. For instance, Figures 1.23 and 1.24 show the graphs of

TECHNOLOGY

f x 

tan x x

and

gx 

sin 4x . x

Note that the first graph appears to contain the point 0, 1 and the second graph appears to contain the point 0, 4, which lends support to the conclusions obtained in Examples 9 and 10.

SECTION 1.3

67

Evaluating Limits Analytically

Exercises for Section 1.3 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, use a graphing utility to graph the function and visually estimate the limits. 1. hx  x 2  5x

12x  3 x9

2. gx 

(a) lim hx

(a) lim gx

(b) lim hx

(b) lim gx

x5

x0

3. f x  x cos x



(b) lim f x

(b) lim f t

t4

x 3

t1

7. lim 2x  1

lim gx  2 1

xc

(a) lim 4f x

(b) lim f x  gx

(b) lim f x  gx

(c) lim f x gx

(c) lim f x gx

xc

xc

xc

f x gx

(d) lim

xc

40. lim f x  27

xc

x3

xc

xc

f x gx

39. lim f x  4

8. lim 3x  2

x0

3

xc

(a) lim 5gx

xc

x2

6x

38. lim f x  2

xc

xc

6. lim x3

x2

x7

In Exercises 37–40, use the information to evaluate the limits.

(d) lim

In Exercises 5–22, find the limit. 5. lim x 4

36. lim sec

lim gx  3



(a) lim f t

x0

4x

xc

4. f t  t t  4

(a) lim f x

x3

37. lim f x  2

x4

x1

35. lim tan

xc

(a) lim f x 3

3 f x (a) lim 

(b) lim f x

(b) lim

xc

xc

11. lim 2x 2  4x  1

12. lim 3x 3  2x 2  4

(c) lim 3f x

f x 18 (c) lim f x 2

1 13. lim x2 x

2 14. lim x3 x  2

(d) lim f x 32

(d) lim f x 23

9. lim x 2  3x

10. lim x 2  1

x3

x1

x3

x1

15. lim

x3 x2  4

16. lim

17. lim

5x x  2

18. lim

x1

x7

xc

x3

x3

xc

2x  3 x5 x  1

x4

19. lim x  1

3 x  4 20. lim  x4

21. lim x  3 2

22. lim 2x  13

x3

x4

xc

xc

xc

xc

In Exercises 41–44, use the graph to determine the limit visually (if it exists). Write a simpler function that agrees with the given function at all but one point. 41. gx 

2x 2  x x

42. hx 

y

y

x0

x

3

2 1

(b) lim gx

24. f x  x  7, gx  x (a) lim f x x3

(c) lim g f x

x4 2

x1

(b) lim gx

(c) lim g f x

x4

x3

25. f x  4  x 2, gx  x  1 (a) lim f x x1

x4

x 2

1

(b) lim gx

(c) lim g f x

x3

x1

(b) lim gx

(a) lim gx

(a) lim hx

(b) lim gx

(b) lim hx

x0

x2 x0

x3  x 43. gx  x1

44. f x 

y

(c) lim g f x

x21

5

1

x1

3 x  6 26. f x  2x 2  3x  1, gx  

(a) lim f x

3

3

1

23. f x  5  x, gx  x3 x1

1

2

In Exercises 23–26, find the limits. (a) lim f x

x 2  3x x

x4

y 2

3

In Exercises 27– 36, find the limit of the trigonometric function.

2

27. lim sin x

28. lim tan x

1

x 29. lim cos x2 3

30. lim sin

31. lim sec 2x

32. lim cos 3x

x x2  x

1 x

x 2

x0

33.

lim sin x

x56

x 

x1

x 2

x 

34.

lim cos x

x53

2 x

2

1

1

2

(a) lim gx

(a) lim f x

(b) lim gx

(b) lim f x

x1

x1

x1

x0

3

68

CHAPTER 1

Limits and Their Properties

In Exercises 45–48, find the limit of the function (if it exists). Write a simpler function that agrees with the given function at all but one point. Use a graphing utility to confirm your result. 45. lim

x1

47. lim

x2

x2  1 x1

46. lim

x1

8 x2

x3

48. lim

2x 2  x  3 x1 1 x1

x3

x1

Graphical, Numerical, and Analytic Analysis In Exercises 79–82, use a graphing utility to graph the function and estimate the limit. Use a table to reinforce your conclusion. Then find the limit by analytic methods. sin 3t t

80. lim

cos x  1 2x2

sin x 2 x x0

82. lim

sin x 3  x

79. lim t0

x0

81. lim

x0

In Exercises 49–62, find the limit (if it exists). x5 2 x5 x  25

50. lim

2x 2 x2 x  4

x2  x  6 x3 x2  9

52. lim

49. lim

51. lim 53. lim

x  5  5

x

x0

55. lim

59. 61. 62.

54. lim

2  x  2

56. lim

Graphical, Numerical, and Analytic Analysis In Exercises 63–66, use a graphing utility to graph the function and estimate the limit. Use a table to reinforce your conclusion. Then find the limit by analytic methods. 63. lim

x  2  2

x

12  x  12 65. lim x x0 x0

x2

x5  32 x2

68. lim

31  cos x x

69. lim

70. lim

cos tan



sin2 x 71. lim x x0

72. lim

tan2 x x

x0

sin x 5x

x0

sin x1  cos x x0 2x2

0

x0

1  cos h2 74. lim  sec  h h0  cos x 1  tan x 75. lim 76. lim x 2 cot x x 4 sin x  cos x sin 3t 77. lim t0 2t sin 2x 2 sin 2x 3x Hint: Find lim . 78. lim 2x 3 sin 3x x0 sin 3x x0



4 x

86. f x  x 2  4x



In Exercises 87 and 88, use the Squeeze Theorem to find lim f x. xc

87. c  0 4  x 2  f x  4  x 2 88. c  a









b  x  a  f x  b  x  a

In Exercises 89–94, use a graphing utility to graph the given function and the equations y  x and y   x in the same viewing window. Using the graphs to observe the Squeeze Theorem visually, find lim f x.





x0

 



89. f x  x cos x

90. f x  x sin x

91. f x  x sin x

92. f x  x cos x



1 x

94. hx  x cos

1 x

95. In the context of finding limits, discuss what is meant by two functions that agree at all but one point. 96. Give an example of two functions that agree at all but one point. 97. What is meant by an indeterminate form? 98. In your own words, explain the Squeeze Theorem.

99. Writing

73. lim



85. f x 

84. f x  x

Writing About Concepts

In Exercises 67–78, determine the limit of the trigonometric function (if it exists). 67. lim

83. f x  2x  3

93. f x  x sin

4  x 64. lim x16 x  16 66. lim

f x  x  f x . x

x x  1  2

x4 x3 x3

13  x  13

1x  4  14 58. lim lim x x x0 x0 2x  x  2x x  x2  x 2 60. lim lim x x x0 x0 2 2 x  x  2x   x  1  x  2x  1 lim x x0 x  x3  x3 lim x x0

x4

57.

x0

x2  5x  4 x4 x2  2x  8 x0

x  5  3

In Exercises 83–86, find lim



f x  x,

Use a graphing utility to graph gx  sin x,

and hx 

sin x x

in the same viewing window. Compare the magnitudes of f x and gx when x is close to 0. Use the comparison to write a short paragraph explaining why lim hx  1.

x0

SECTION 1.3

100. Writing

Use a graphing utility to graph

f x  x, gx  sin2 x, and hx 

sin2 x x

113. lim

x0

x  1 x

sin x 1 x

114. lim

x

lim hx  0.

115. If f x  gx for all real numbers other than x  0, and

x0

Free-Falling Object In Exercises 101 and 102, use the position function st   16t 2  1000, which gives the height (in feet) of an object that has fallen for t seconds from a height of 1000 feet. The velocity at time t  a seconds is given by

ta

69

True or False? In Exercises 113–118, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

in the same viewing window. Compare the magnitudes of f x and gx when x is close to 0. Use the comparison to write a short paragraph explaining why

lim

Evaluating Limits Analytically

lim f x  L, then

lim gx  L.

x0

x0

116. If lim f x  L, then f c  L. xc

117. lim f x  3, where f x 

sa  st . at

x2

3,0,

x  2 x > 2

118. If f x < gx for all x  a, then

101. If a construction worker drops a wrench from a height of 1000 feet, how fast will the wrench be falling after 5 seconds?

lim f x < lim gx.

xa

xa

102. If a construction worker drops a wrench from a height of 1000 feet, when will the wrench hit the ground? At what velocity will the wrench impact the ground?

119. Think About It Find a function f to show that the converse of Exercise 112(b) is not true. [Hint: Find a function f such that lim f x  L but lim f x does not exist.]

Free-Falling Object In Exercises 103 and 104, use the position function st  4.9t 2  150, which gives the height (in meters) of an object that has fallen from a height of 150 meters. The velocity at time t  a seconds is given by

120. Prove the second part of Theorem 1.9 by proving that

xc

lim ta

sa  st . at

lim

x0



103. Find the velocity of the object when t  3.

and

104. At what velocity will the object impact the ground?

gx 

x0

x0

not exist, but lim f x  gx does exist.

xc

1  cos x  0. x

121. Let f x 

105. Find two functions f and g such that lim f x and lim gx do

 

0,1,

0,x,

if x is rational if x is irrational

if x is rational if x is irrational.

Find (if possible) lim f x and lim gx. x0

x0

x0

106. Prove that if lim f x exists and lim f x  gx does not xc

xc

exist, then lim gx does not exist. 108. Prove Property 3 of Theorem 1.1. (You may use Property 3 of Theorem 1.2.)

 



110. Prove that if lim f x  0, then lim f x  0.



111. Prove that if lim f x  0 and gx  M for a fixed number xc

xc



112. (a) Prove that if lim f x  0, then lim f x  0. xc

(Note: This is the converse of Exercise 110.)



  Use the inequality  f x  L   f x  L.

(b) Prove that if lim f x  L, then lim f x  L .

Hint:

xc

(b) Use a graphing utility to graph f. Is the domain of f obvious from the graph? If not, explain. (c) Use the graph of f to approximate lim f x.

xc

(d) Confirm the answer in part (c) analytically. 123. Approximation (a) Find lim

x0

M and all x  c, then lim f xgx  0.

 xc

sec x  1 . x2

x0

109. Prove Property 1 of Theorem 1.2. xc

Consider f x 

(a) Find the domain of f.

xc

107. Prove Property 1 of Theorem 1.1.

xc

122. Graphical Reasoning

1  cos x . x2

(b) Use the result in part (a) to derive the approximation 1 cos x  1  2x 2 for x near 0. (c) Use the result in part (b) to approximate cos0.1. (d) Use a calculator to approximate cos0.1 to four decimal places. Compare the result with part (c). 124. Think About It When using a graphing utility to generate a table to approximate lim sin xx , a student concluded that x0

the limit was 0.01745 rather than 1. Determine the probable cause of the error.

70

CHAPTER 1

Limits and Their Properties

Section 1.4

Continuity and One-Sided Limits • • • •

Determine continuity at a point and continuity on an open interval. Determine one-sided limits and continuity on a closed interval. Use properties of continuity. Understand and use the Intermediate Value Theorem.

Continuity at a Point and on an Open Interval E X P L O R AT I O N Informally, you might say that a function is continuous on an open interval if its graph can be drawn . with a pencil without lifting the pencil from the paper. Use a graphing utility to graph each function on the given interval. From the graphs, which functions would you say are continuous on the interval? Do you think you can trust the results you obtained graphically? Explain your reasoning. Function

Interval

a. y  x2  1

3, 3

b. y 

1 x2

3, 3

c. y 

sin x x

 , 

d. y 

x2  4 x2

3, 3

e. y 

x  1,

2x  4, x ≤ 0 x > 0

3, 3

In mathematics, the term continuous has much the same meaning as it has in everyday usage. Informally, to say that a function f is continuous at x  c means that there is no interruption in the graph of f at c. That is, its graph is unbroken at c and there are no holes, jumps, or gaps. Figure 1.25 identifies three values of x at which the graph of f is not continuous. At all other points in the interval a, b, the graph of f is uninterrupted and continuous.

Animation y

y

y

lim f(x)

f (c) is not defined.

x→c

does not exist.

lim f(x) ≠ f(c) x→c

x

x

a

c

b

a

c

b

x

a

c

b

Three conditions exist for which the graph of f is not continuous at x  c. Figure 1.25

In Figure 1.25, it appears that continuity at x  c can be destroyed by any one of the following conditions. 1. The function is not defined at x  c. 2. The limit of f x does not exist at x  c. 3. The limit of f x exists at x  c, but it is not equal to f c. If none of the above three conditions is true, the function f is called continuous at c, as indicated in the following important definition.

FOR FURTHER INFORMATION For more information on the concept of continuity, see the article “Leibniz and the Spell of the Continuous” by Hardy . in The College Mathematics Grant Journal.

Definition of Continuity Continuity at a Point: conditions are met.

A function f is continuous at c if the following three

1. f c is defined. 2. lim f x exists. x→c

MathArticle

3. lim f x  f c.

.

Continuity on an Open Interval: A function is continuous on an open interval a, b if it is continuous at each point in the interval. A function that is continuous on the entire real line  ,  is everywhere continuous.

x→c

Video

SECTION 1.4

y

Continuity and One-Sided Limits

71

Consider an open interval I that contains a real number c. If a function f is defined on I (except possibly at c), and f is not continuous at c, then f is said to have a discontinuity at c. Discontinuities fall into two categories: removable and nonremovable. A discontinuity at c is called removable if f can be made continuous by appropriately defining (or redefining) f c. For instance, the functions shown in Figure 1.26(a) and (c) have removable discontinuities at c and the function shown in Figure 1.26(b) has a nonremovable discontinuity at c. x

a

c

b

EXAMPLE 1

Continuity of a Function

Discuss the continuity of each function.

(a) Removable discontinuity

a. f x 

y

1 x

b. gx 

x2  1 x1

c. hx 

x  1, x ≤ 0 2  1, x > 0

x

d. y  sin x

Solution

x

a

c

b

(b) Nonremovable discontinuity y

a. The domain of f is all nonzero real numbers. From Theorem 1.3, you can conclude that f is continuous at every x-value in its domain. At x  0, f has a nonremovable discontinuity, as shown in Figure 1.27(a). In other words, there is no way to define f 0 so as to make the function continuous at x  0. b. The domain of g is all real numbers except x  1. From Theorem 1.3, you can conclude that g is continuous at every x-value in its domain. At x  1, the function has a removable discontinuity, as shown in Figure 1.27(b). If g1 is defined as 2, the “newly defined” function is continuous for all real numbers. c. The domain of h is all real numbers. The function h is continuous on  , 0 and 0, , and, because lim hx  1, h is continuous on the entire real line, as shown x→0 in Figure 1.27(c). d. The domain of y is all real numbers. From Theorem 1.6, you can conclude that the function is continuous on its entire domain,  , , as shown in Figure 1.27(d). y

y 3

3 x

a

c

1 f (x) = x

2

(1, 2) 2

2 g (x) = x − 1 x −1

b

Figure 1.26

1

1

(c) Removable discontinuity

x

−1

1

2

3

1

(a) Nonremovable discontinuity at x  0

Editable Graph

Editable Graph y

y = sin x

3 1 2

h (x) =

1

x + 1, x ≤ 0 x 2 + 1, x > 0

x π 2

x −1

Some people may refer to . the function in Example 1(a) as “discontinuous.” We have found that this terminology can be confusing. Rather than saying the function is discontinuous, we . prefer to say that it has a discontinuity at x  0.

3

(b) Removable discontinuity at x  1

y

STUDY TIP

2

−1

−1

.

x

−1

1

2

3

3π 2

−1

−1

(c) Continuous on entire real line

Editable Graph

(d) Continuous on entire real line

Editable Graph

Figure 1.27

Try It

Exploration A

Exploration B

Exploration C

72

CHAPTER 1

Limits and Their Properties

y

One-Sided Limits and Continuity on a Closed Interval To understand continuity on a closed interval, you first need to look at a different type of limit called a one-sided limit. For example, the limit from the right means that x approaches c from values greater than c [see Figure 1.28(a)]. This limit is denoted as

x approaches c from the right. x

cx

One-sided limits are useful in taking limits of functions involving radicals. For instance, if n is an even integer,

(b) Limit from left

Figure 1.28

n lim  x  0.

x→0 

y

EXAMPLE 2

Find the limit of f x  4  x 2 as x approaches 2 from the right.

3

f(x) =

A One-Sided Limit

4 − x2

Solution As shown in Figure 1.29, the limit as x approaches 2 from the right is .

lim 4  x2  0.

1

x→2

Try It

x

−2

−1

1

2

−1

The limit of f x as x approaches  2 from the. right is 0. Figure 1.29

One-sided limits can be used to investigate the behavior of step functions. One common type of step function is the greatest integer function x, defined by x  greatest integer n such that n ≤ x.

Greatest integer function

For instance, 2.5  2 and 2.5  3.

Editable Graph

EXAMPLE 3 y

Exploration A

The Greatest Integer Function

Find the limit of the greatest integer function f x  x as x approaches 0 from the left and from the right.

f(x) = [[x]]

2

Solution As shown in Figure 1.30, the limit as x approaches 0 from the left is given by 1

lim x  1

x

−2

−1

1

2

3

x→0

and the limit as x approaches 0 from the right is given by lim x  0.

−2

. Greatest integer function Figure 1.30

Editable Graph

x→0

The greatest integer function has a discontinuity at zero because the left and right limits at zero are different. By similar reasoning, you can see that the greatest integer function has a discontinuity at any integer n.

Try It

Exploration A

Exploration B

SECTION 1.4

Continuity and One-Sided Limits

73

When the limit from the left is not equal to the limit from the right, the (twosided) limit does not exist. The next theorem makes this more explicit. The proof of this theorem follows directly from the definition of a one-sided limit.

THEOREM 1.10

The Existence of a Limit

Let f be a function and let c and L be real numbers. The limit of f x as x approaches c is L if and only if lim f x  L

x→c

and

lim f x  L.

x→c

The concept of a one-sided limit allows you to extend the definition of continuity to closed intervals. Basically, a function is continuous on a closed interval if it is continuous in the interior of the interval and exhibits one-sided continuity at the endpoints. This is stated formally as follows.

y

Definition of Continuity on a Closed Interval A function f is continuous on the closed interval [a, b] if it is continuous on the open interval a, b and lim f x  f a

x

a

x→a

b

Continuous function on a closed interval Figure 1.31

and

lim f x  f b.

x→b

The function f is continuous from the right at a and continuous from the left at b (see Figure 1.31). Similar definitions can be made to cover continuity on intervals of the form a, b and a, b that are neither open nor closed, or on infinite intervals. For example, the function f x  x is continuous on the infinite interval 0, , and the function gx  2  x is continuous on the infinite interval  , 2 . EXAMPLE 4

Continuity on a Closed Interval

Discuss the continuity of f x  1  x 2. Solution The domain of f is the closed interval 1, 1 . At all points in the open interval 1, 1, the continuity of f follows from Theorems 1.4 and 1.5. Moreover, because

y

1

f (x) =

1 − x2

lim 1  x 2  0  f 1

x→1

Continuous from the right

and x

−1

. f is continuous on  1, 1 . Figure 1.32

Editable Graph

1

lim 1  x 2  0  f 1

x→1

Continuous from the left

you can conclude that f is continuous on the closed interval 1, 1 , as shown in Figure 1.32.

Try It

Exploration A

74

CHAPTER 1

Limits and Their Properties

The next example shows how a one-sided limit can be used to determine the value of absolute zero on the Kelvin scale.

Charles’s Law and Absolute Zero

EXAMPLE 5

On the Kelvin scale, absolute zero is the temperature 0 K. Although temperatures of approximately 0.0001 K have been produced in laboratories, absolute zero has never been attained. In fact, evidence suggests that absolute zero cannot be attained. How did scientists determine that 0 K is the “lower limit” of the temperature of matter? What is absolute zero on the Celsius scale? V

Solution The determination of absolute zero stems from the work of the French physicist Jacques Charles (1746–1823). Charles discovered that the volume of gas at a constant pressure increases linearly with the temperature of the gas. The table illustrates this relationship between volume and temperature. In the table, one mole of hydrogen is held at a constant pressure of one atmosphere. The volume V is measured in liters and the temperature T is measured in degrees Celsius.

30 25

V = 0.08213T + 22.4334 15 10

(−273.15, 0)

− 300

− 200

5 − 100

T

100

The volume of hydrogen gas depends on its. temperature. Figure 1.33

T

40

20

0

20

40

60

80

V

19.1482

20.7908

22.4334

24.0760

25.7186

27.3612

29.0038

The points represented by the table are shown in Figure 1.33. Moreover, by using the points in the table, you can determine that T and V are related by the linear equation V  0.08213T  22.4334

Editable Graph

or

T

V  22.4334 . 0.08213

By reasoning that the volume of the gas can approach 0 (but never equal or go below 0) you can determine that the “least possible temperature” is given by limT  lim

V→0

V→0

V  22.4334 0.08213

0  22.4334 0.08213  273.15. 

.

Use direct substitution.

So, absolute zero on the Kelvin scale 0 K is approximately 273.15 on the Celsius scale.

Try It

Exploration A

The following table shows the temperatures in Example 5, converted to the Fahrenheit scale. Try repeating the solution shown in Example 5 using these temperatures and volumes. Use the result to find the value of absolute zero on the Fahrenheit scale. In 1995, physicists Carl Wieman and Eric Cornell of the University of Colorado at Boulder used lasers and evaporation to produce a supercold gas in which atoms overlap. This gas is called a Bose-Einstein condensate. “We get to within a billionth of a degree of absolute zero,”reported Wieman. (Source: Time magazine, April 10, 2000)

T

40

4

32

68

104

140

176

V

19.1482

20.7908

22.4334

24.0760

25.7186

27.3612

29.0038

NOTE

Charles’s Law for gases (assuming constant pressure) can be stated as

V  RT

Charles’s Law

where V is volume, R is constant, and T is temperature. In the statement of this law, what property must the temperature scale have?

SECTION 1.4

Continuity and One-Sided Limits

75

Properties of Continuity AUGUSTIN-LOUIS CAUCHY (1789–1857)

.

The concept of a continuous function was first introduced by Augustin-Louis Cauchy in 1821. The definition given in his text Cours d’Analyse stated that indefinite small changes in y were the result of indefinite small changes in x. “… f x will be called a continuous function if … the numerical values of the difference f x     f x decrease indefinitely with those of  ….”

MathBio

In Section 1.3, you studied several properties of limits. Each of those properties yields a corresponding property pertaining to the continuity of a function. For instance, Theorem 1.11 follows directly from Theorem 1.2.

THEOREM 1.11

Properties of Continuity

If b is a real number and f and g are continuous at x  c, then the following functions are also continuous at c. 1. Scalar multiple: bf 2. Sum and difference: f ± g 3. Product: fg 4. Quotient:

f , g

if gc  0

The following types of functions are continuous at every point in their domains. px  anxn  an1xn1  . . .  a1x  a0 px 2. Rational functions: rx  , qx  0 qx n x 3. Radical functions: f x   4. Trigonometric functions: sin x, cos x, tan x, cot x, sec x, csc x 1. Polynomial functions:

By combining Theorem 1.11 with this summary, you can conclude that a wide variety of elementary functions are continuous at every point in their domains. EXAMPLE 6

Applying Properties of Continuity

By Theorem 1.11, it follows that each of the following functions is continuous at every point in its domain. .

f x  x  sin x,

Try It

f x  3 tan x,

Exploration A

f x 

x2  1 cos x

Open Exploration

The next theorem, which is a consequence of Theorem 1.5, allows you to determine the continuity of composite functions such as f x  sin 3x,

THEOREM 1.12

f x  x2  1,

1 f x  tan . x

Continuity of a Composite Function

If g is continuous at c and f is continuous at gc, then the composite function given by  f  gx  f gx is continuous at c. One consequence of Theorem 1.12 is that if f and g satisfy the given conditions, you can determine the limit of f gx as x approaches c to be .

lim f gx  f gc.

x→c

Technology

76

CHAPTER 1

Limits and Their Properties

Testing for Continuity

EXAMPLE 7

Describe the interval(s) on which each function is continuous. a. f x  tan x

b. gx 



sin 1 , x  0 x 0, x0

c. hx 



x sin 1 , x  0 x 0, x0

Solution a. The tangent function f x  tan x is undefined at x

  n, 2

n is an integer.

At all other points it is continuous. So, f x  tan x is continuous on the open intervals



. . ., 

3     3 , ,  , , , ,. . . 2 2 2 2 2 2







as shown in Figure 1.34(a). b. Because y  1x is continuous except at x  0 and the sine function is continuous for all real values of x, it follows that y  sin 1x is continuous at all real values except x  0. At x  0, the limit of gx does not exist (see Example 5, Section 1.2). So, g is continuous on the intervals  , 0 and 0, , as shown in Figure 1.34(b). c. This function is similar to that in part (b) except that the oscillations are damped by the factor x. Using the Squeeze Theorem, you obtain



 x ≤ x sin

1 ≤ x, x



x0

and you can conclude that lim hx  0.

x→0

y

So, h is continuous on the entire real line, as shown in Figure 1.34(c).

4 3

y

y

2

y = ⎜x ⎜

1 −π

1 π

1

x

x

−3

−1

1

x

−1

1

−4

f (x) = tan x

. f is continuous on each open interval in its (a) domain. . Editable Graph Figure 1.34

.

−1

−1

g (x) =

1 sin x , x ≠ 0 x=0 0,

(b) g is continuous on  , 0 and 0, .

Editable Graph

Try It

y = − ⎜x ⎜

h (x) =

0,

x=0

(c) h is continuous on the entire real line.

Editable Graph

Exploration A

x sin 1x , x ≠ 0

SECTION 1.4

Continuity and One-Sided Limits

77

The Intermediate Value Theorem Theorem 1.13 is an important theorem concerning the behavior of functions that are continuous on a closed interval.

THEOREM 1.13

Intermediate Value Theorem

If f is continuous on the closed interval a, b and k is any number between f a and f b), then there is at least one number c in a, b such that .

f c  k.

Video NOTE The Intermediate Value Theorem tells you that at least one c exists, but it does not give a method for finding c. Such theorems are called existence theorems.

By referring to a text on advanced calculus, you will find that a proof of this theorem is based on a property of real numbers called completeness. The Intermediate Value Theorem states that for a continuous function f, if x takes on all values between a and b, f x must take on all values between f a and f b. As a simple example of this theorem, consider a person’s height. Suppose that a girl is 5 feet tall on her thirteenth birthday and 5 feet 7 inches tall on her fourteenth birthday. Then, for any height h between 5 feet and 5 feet 7 inches, there must have been a time t when her height was exactly h. This seems reasonable because human growth is continuous and a person’s height does not abruptly change from one value to another. The Intermediate Value Theorem guarantees the existence of at least one number c in the closed interval a, b . There may, of course, be more than one number c such that f c  k, as shown in Figure 1.35. A function that is not continuous does not necessarily exhibit the intermediate value property. For example, the graph of the function shown in Figure 1.36 jumps over the horizontal line given by y  k, and for this function there is no value of c in a, b such that f c  k. y

y

f (a)

f(a)

k k

f (b)

f(b) x

a

c1

c2

c3

b

x

a

b

f is continuous on a, b . [There exist three c’s such that f c  k.]

f is not continuous on a, b . [There are no c’s such that f c  k.]

Figure 1.35

Figure 1.36

The Intermediate Value Theorem often can be used to locate the zeros of a function that is continuous on a closed interval. Specifically, if f is continuous on

a, b and f a and f b differ in sign, the Intermediate Value Theorem guarantees the existence of at least one zero of f in the closed interval a, b .

78

CHAPTER 1

y

Limits and Their Properties

f (x) = x 3 + 2x − 1

An Application of the Intermediate Value Theorem

Use the Intermediate Value Theorem to show that the polynomial function f x  x 3  2x  1 has a zero in the interval 0, 1 .

(1, 2)

2

EXAMPLE 8

Solution Note that f is continuous on the closed interval 0, 1 . Because f 0  0 3  20  1  1 and

1

f 1  13  21  1  2

it follows that f 0 < 0 and f 1 > 0. You can therefore apply the Intermediate Value Theorem to conclude that there must be some c in 0, 1 such that .

(c, 0)

−1

x 1

f c  0

f has a zero in the closed interval 0, 1 .

as shown in Figure 1.37. −1

(0, −1)

f is continuous on 0, 1 with f 0 < 0 and f .1 > 0. Figure 1.37

Editable Graph

Try It

Exploration A

The bisection method for approximating the real zeros of a continuous function is similar to the method used in Example 8. If you know that a zero exists in the closed interval a, b , the zero must lie in the interval a, a  b2 or a  b2, b . From the sign of f  a  b 2, you can determine which interval contains the zero. By repeatedly bisecting the interval, you can “close in” on the zero of the function. You can also use the zoom feature of a graphing utility to approximate the real zeros of a continuous function. By repeatedly zooming in on the point where the graph crosses the x-axis, and adjusting the x-axis scale, you can approximate the zero of the function to any desired accuracy. The zero of x3  2x  1 is approximately 0.453, as shown in Figure 1.38.

TECHNOLOGY

0.2

−0.2

0.013

1

0.4

−0.2

Figure 1.38

−0.012

Zooming in on the zero of f x  x  2x  1 3

0.5

78

CHAPTER 1

Limits and Their Properties

Exercises for Section 1.4 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–6, use the graph to determine the limit, and discuss the continuity of the function.

y

3.

y

4.

c = 2 4

4

(a) lim f x

(b) lim f x

xc

xc

y

1.

(c) lim f x

(3, 1)

xc

x

1

2

3

4

c=3 (2, 2)

4

(3, 0) c=3

2

2

1

6

x 4 3 2 1

1 x

2

1 2

2

x 2

c = 2

(3, 1)

1

3

(2, 2)

y

2.

2

2

(2, 3)

5. 3 2 1

y

6.

y

(4, 2)

4 3

c=4 x

1 3

(1, 2)

1 2 3 4 5 6

(4, 2)

c = 1

2

x

3

(1, 0)

1

SECTION 1.4

In Exercises 7–24, find the limit (if it exists). If it does not exist, explain why. 7. lim

x5 x2  25

8. lim

2x x2  4

x5

x2

9.

10. lim

x 3 2 1

x2  9



15.

16. 17. 18.

3

x  2 x2

30. f t  3  9  t



3  x,

31. f x 

3 

32. gx 

1 x2  4

1 2 x,

2

x  0 x > 0

33. f x  x 2  2x  1 34. f x 

1 x2  1

35. f x  3x  cos x

x 2

19. lim cot x

36. f x  cos

20. lim sec x

37. f x 

x x2  x

21. lim 3x  5

38. f x 

x x2  1

39. f x 

x x2  1

40. f x 

x3 x2  9

41. f x 

x2 x 2  3x  10

42. f x 

x1 x2  x  2

x 

x 2 x4

22. lim2x  x x2

23. lim 2  x  x3

 

x 24. lim 1   x1 2

In Exercises 25–28, discuss the continuity of each function. x2  1 26. f x  x1

1 25. f x  2 x 4

y

y 3 2 1

3 2 1 x 1

3

x

3 2 1 3

1 2

5, 5

3, 3

1, 4

1, 2

In Exercises 33–54, find the x-values (if any) at which f is not continuous. Which of the discontinuities are removable?

  

1 2 3

3

Interval

29. gx  25  x 2

x   x2  x   x  x 2  x x x0 x2 , x  3 2 lim f x, where f x  12  2x x3 , x > 3 3 x2  4x  6, x < 2 lim f x, where f x  x2  4x  2, x  2 x2 x3  1, x < 1 lim f x, where f x  x  1, x  1 x1 x, x  1 lim f x, where f x  1  x, x > 1 x1

3

1 2 2 3

Function

lim 



x

3 2

In Exercises 29–32, discuss the continuity of the function on the closed interval.

1 1  x  x x 13. lim  x x0 14.

1 2

3

x4

x2

3 2 1

3 2 1

x 11. lim x x0 12. lim

y

y

x  2

x4



x, x < 1 x1 28. f x  2, 2x  1, x > 1

1 27. f x  2x  x

x

lim

x3

79

Continuity and One-Sided Limits

3

43. f x 

x  2

44. f x 

x  3

x2 x3

x,x , xx > 11 2x  3, x < 1 46. f x   x , x  1 45. f x 

2

2

80

CHAPTER 1

47. f x 



1 2x

Limits and Their Properties

In Exercises 65–68, use a graphing utility to graph the function. Use the graph to determine any x-values at which the function is not continuous.

 1, x  2

3  x,

x > 2



2x, x  2 48. f x  2 x  4x  1, x > 2

65. f x  x  x

tan  x, 4 49. f x  x,

x < 1 x  1

67. gx 

x

csc  x , 6 50. f x  2,

x  3  2 x  3 > 2

68. f x 



 

51. f x  csc 2x

69. f x 

53. f x  x  1 54. f x  3  x

x2 x 2  4x x  2 56. f x  x4

1

1

x

4

2

2

x 4

72. f x 

4

3

4

x1 x

y 4

4

3

2

2 1 x

4

  

2 4

2 2

4 sin x , x < 0 x a  2x, x  0

1

2

Writing In Exercises 73 and 74, use a graphing utility to graph the function on the interval [4, 4]. Does the graph of the function appear continuous on this interval? Is the function continuous on [4, 4]? Write a short paragraph about the importance of examining a function analytically as well as graphically.

2, x  1 59. f x  ax  b, 1 < x < 3 2, x  3 x2  a2 , xa 60. g x  x  a 8, xa

73. f x 

In Exercises 61– 64, discuss the continuity of the composite function hx  f  gx.

gx  x 2  5

2

(3, 0)

x



1 63. f x  x6

4

y

x3, x  2 57. f x  ax 2, x > 2

g x  x  1

y

1

71. f x  sec



61. f x  x 2

70. f x  xx  3

2

In Exercises 57–60, find the constant a, or the constants a and b, such that the function is continuous on the entire real line.

58. gx 

x x2  1

x

x0

x 2  4x f x  



cos x  1 , x < 0 x 5x, x  0

lim f x.

and

Is the function continuous on the entire real line? Explain. 55.

2x  4, x  3 2  2x, x > 3

y

In Exercises 55 and 56, use a graphing utility to graph the function. From the graph, estimate lim f x

1 x2  x  2

In Exercises 69–72, describe the interval(s) on which the function is continuous.

x 52. f x  tan 2

x0

66. hx 

62. f x 

1 x

g x  x  1 64. f x  sin x g x  x2

sin x x

74. f x 

x3  8 x2

Writing In Exercises 75–78, explain why the function has a zero in the given interval. Interval

Function 75. f x 

77. f x  x 2  2  cos x

1, 2

0, 1

0, 

4 x 78. f x    tan x 8

1, 3

1 4 16 x



x3

3

76. f x  x3  3x  2

 

SECTION 1.4

In Exercises 79–82, use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval [0, 1]. Repeatedly “zoom in” on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places.

Continuity and One-Sided Limits

81

Writing About Concepts (continued) 89. Sketch the graph of any function f such that lim f x  1

lim f x  0.

and

x3

x3

Is the function continuous at x  3? Explain.

79. f x  x3  x  1

90. If the functions f and g are continuous for all real x, is f  g always continuous for all real x? Is fg always continuous for all real x? If either is not continuous, give an example to verify your conclusion.

80. f x  x3  3x  2 81. gt  2 cos t  3t 82. h   1   3 tan

In Exercises 83–86, verify that the Intermediate Value Theorem applies to the indicated interval and find the value of c guaranteed by the theorem. 83. f x  x 2  x  1,

f c  11

0, 5 , 84. f x   6x  8, f c  0

0, 3 , 85. f x  x3  x 2  x  2, f c  4

0, 3 , 2 x x 5 86. f x  f c  6 , ,4 , x1 2

True or False? In Exercises 91–94, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 91. If lim f x  L and f c  L, then f is continuous at c.

x2

xc

92. If f x  gx for x  c and f c  gc, then either f or g is not continuous at c.

 

93. A rational function can have infinitely many x-values at which it is not continuous.



Writing About Concepts 87. State how continuity is destroyed at x  c for each of the following graphs. (a) y (b) y



94. The function f x  x  1 x  1 is continuous on  , . 95. Swimming Pool Every day you dissolve 28 ounces of chlorine in a swimming pool. The graph shows the amount of chlorine f t in the pool after t days. y 140 112 84

c

x

c

x

56 28 t

(c)

y

(d)

1

y

2

3

4

5

6

7

lim f t and lim f t. Estimate and interpret t4 t4 96. Think About It

Describe how the functions

f x  3  x and c

x

c

x

88. Describe the difference between a discontinuity that is removable and one that is nonremovable. In your explanation, give examples of the following descriptions. (a) A function with a nonremovable discontinuity at x  2 (b) A function with a removable discontinuity at x  2 (c) A function that has both of the characteristics described in parts (a) and (b)

gx  3  x differ. 97. Telephone Charges A dial-direct long distance call between two cities costs $1.04 for the first 2 minutes and $0.36 for each additional minute or fraction thereof. Use the greatest integer function to write the cost C of a call in terms of time t (in minutes). Sketch the graph of this function and discuss its continuity.

82

CHAPTER 1

Limits and Their Properties

98. Inventory Management The number of units in inventory in a small company is given by

 t 2 2  t

Nt  25 2

where t is the time in months. Sketch the graph of this function and discuss its continuity. How often must this company replenish its inventory? 99. Déjà Vu At 8:00 A.M. on Saturday a man begins running up the side of a mountain to his weekend campsite (see figure). On Sunday morning at 8:00 A.M. he runs back down the mountain. It takes him 20 minutes to run up, but only 10 minutes to run down. At some point on the way down, he realizes that he passed the same place at exactly the same time on Saturday. Prove that he is correct. [Hint: Let st and r t be the position functions for the runs up and down, and apply the Intermediate Value Theorem to the function f t  st  r t.]

105. Modeling Data After an object falls for t seconds, the speed S (in feet per second) of the object is recorded in the table. t

0

5

10

15

20

25

30

S

0

48.2

53.5

55.2

55.9

56.2

56.3

(a) Create a line graph of the data. (b) Does there appear to be a limiting speed of the object? If there is a limiting speed, identify a possible cause. 106. Creating Models A swimmer crosses a pool of width b by swimming in a straight line from 0, 0 to 2b, b. (See figure.) (a) Let f be a function defined as the y-coordinate of the point on the long side of the pool that is nearest the swimmer at any given time during the swimmer’s path across the pool. Determine the function f and sketch its graph. Is it continuous? Explain. (b) Let g be the minimum distance between the swimmer and the long sides of the pool. Determine the function g and sketch its graph. Is it continuous? Explain. y

(2b, b)

b Not drawn to scale

Saturday 8:00 A.M.

Sunday 8:00 A.M.

x

(0, 0)

100. Volume Use the Intermediate Value Theorem to show that for all spheres with radii in the interval 1, 5 , there is one with a volume of 275 cubic centimeters. 101. Prove that if f is continuous and has no zeros on a, b , then either f x > 0 for all x in a, b or f x < 0 for all x in a, b .

0,1,

if x is rational if x is irrational

103. Show that the function



is continuous only at x  0. (Assume that k is any nonzero real number.) 104. The signum function is defined by



1, x < 0 sgnx  0, x0 1, x > 0.

x  c x > c

108. Prove that for any real number y there exists x in  2, 2 such that tan x  y.

(b) lim sgnx x0

112. (a) Let f1x and f2x be continuous on the closed interval

a, b . If f1a < f2a and f1b > f2b, prove that there exists c between a and b such that f1c  f2c.

(b) Show that there exists c in 0, 2 such that cos x  x. Use a graphing utility to approximate c to three decimal places.

Putnam Exam Challenge

Sketch a graph of sgnx and find the following (if possible). x0

2

111. Discuss the continuity of the function hx  x x.

if x is rational if x is irrational

(a) lim sgnx

x,1  x ,

110. Prove that if lim f c   x  f c, then f is continuous x0 at c.

is not continuous at any real number.

0, f x  kx,

f x 

109. Let f x  x  c2  cx, c > 0. What is the domain of f ? How can you define f at x  0 in order for f to be continuous there?

102. Show that the Dirichlet function f x 

107. Find all values of c such that f is continuous on  , .

(c) lim sgnx x0

113. Prove or disprove: if x and y are real numbers with y  0 and y y  1  x  12, then y y  1  x2. 114. Determine all polynomials Px such that Px2  1  Px2  1 and P0  0. These problems were composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.

SECTION 1.5

Section 1.5

83

Infinite Limits

Infinite Limits • Determine infinite limits from the left and from the right. • Find and sketch the vertical asymptotes of the graph of a function.

Infinite Limits y

Let f be the function given by 3 → ∞, x−2 as x → 2+

6 4 2

x

−6

−4

4

6

f x 

From Figure 1.39 and the table, you can see that f x decreases without bound as x approaches 2 from the left, and f x increases without bound as x approaches 2 from the right. This behavior is denoted as

−2

3 → −∞, −4 x−2 as x → 2− −6

lim

3   x2

f x decreases without bound as x approaches 2 from the left.

lim

3  x2

f x increases without bound as x approaches 2 from the right.

x→2

3 f (x) = x−2

3 . x2

and

f x increases and decreases without bound as x approaches 2.

x→2 

Figure 1.39 x approaches 2 from the right.

x approaches 2 from the left.

x

1.5

1.9

1.99

1.999

2

2.001

2.01

2.1

2.5

f x

6

30

300

3000

?

3000

300

30

6

f x decreases without bound.

f x increases without bound.

A limit in which f x increases or decreases without bound as x approaches c is called an infinite limit.

Definition of Infinite Limits Let f be a function that is defined at every real number in some open interval containing c (except possibly at c itself). The statement lim f x 

x→c

means that for each M > 0 there exists a  > 0 such that f x > M whenever 0 < x  c <  (see Figure 1.40). Similarly, the statement



y

lim f x  

x→c

lim f (x) = ∞

means that for each N < 0 there exists a  > 0 such that f x < N whenever 0 < x  c < . To define the infinite limit from the left, replace 0 < x  c <  by c   < x < c. To define the infinite limit from the right, replace 0 < x  c <  by c < x < c  .

x→c

 

M δ δ

.

c

Infinite limits Figure 1.40



x

 





Video Be sure you see that the equal sign in the statement lim f x  does not mean that the limit exists! On the contrary, it tells you how the limit fails to exist by denoting the unbounded behavior of f x as x approaches c.

84

CHAPTER 1

Limits and Their Properties

E X P L O R AT I O N

Use a graphing utility to graph each function. For each function, analytically find the single real number c that is not in the domain. Then graphically find the limit of f x as x approaches c from the left and from the right. a. f x 

3 x4

b. f x 

1 2x

c. f x 

2 x  3 2

d. f x 

3 x  2 2

EXAMPLE 1

Determining Infinite Limits from a Graph

Use Figure 1.41 to determine the limit of each function as x approaches 1 from the left and from the right. y

y

2

3

1

2 x

2 −2 −3

(a)

−2

f(x) =

1 x−1

2

2

−1 x−1

−1 (x − 1) 2

1

−1

f (x) =

−2

2 −1

x

−2

3

1 (x − 1) 2

(b)

Editable Graph

−1

x

−1

f (x) =

f (x) =

x

1 −2

y

2

3

−1

.

y

2 −1

−2

−2

−3

−3

(c)

Editable Graph

−1

(d)

Editable Graph

Editable Graph

Figure 1.41 Each graph has an asymptote at x  1.

Solution 1 and   x1 1 b. lim  x→1 x  1 2 1 c. lim and  x→1 x  1 1 d. lim   x→1 x  1 2 a. lim x→1

.

Try It

lim

x→1 

1  x1

Limit from each side is .

lim

x→1 

1   x1

Limit from each side is  .

Exploration A

Vertical Asymptotes If it were possible to extend the graphs in Figure 1.41 toward positive and negative infinity, you would see that each graph becomes arbitrarily close to the vertical line x  1. This line is a vertical asymptote of the graph of f. (You will study other types of asymptotes in Sections 3.5 and 3.6.)

NOTE If the graph of a function f has a vertical asymptote at x  c, then f is not continuous at c.

Definition of Vertical Asymptote If f x approaches infinity (or negative infinity) as x approaches c from the right or the left, then the line x  c is a vertical asymptote of the graph of f.

SECTION 1.5

Infinite Limits

85

In Example 1, note that each of the functions is a quotient and that the vertical asymptote occurs at a number where the denominator is 0 (and the numerator is not 0). The next theorem generalizes this observation. (A proof of this theorem is given in Appendix A.)

THEOREM 1.14

Vertical Asymptotes

Let f and g be continuous on an open interval containing c. If f c  0, gc  0, and there exists an open interval containing c such that gx  0 for all x  c in the interval, then the graph of the function given by h x  .

has a vertical asymptote at x  c.

y

f(x) =

1 2(x + 1)

2

Video

1 x

−1

EXAMPLE 2

1 −1

.

a. f x 

(a)

Editable Graph

f x 

4 2 x

−2

2

4

(b)

Editable Graph y

f (x) = cot x

6

2 π

2π

x

−4 −6

. (c)

Editable Graph Functions with vertical asymptotes Figure 1.42

c. f x  cot x

1 2x  1

x2  1 x2  1  x 2  1 x  1x  1

you can see that the denominator is 0 at x  1 and x  1. Moreover, because the numerator is not 0 at these two points, you can apply Theorem 1.14 to conclude that the graph of f has two vertical asymptotes, as shown in Figure 1.42(b). c. By writing the cotangent function in the form f x  cot x 

4

−2π

x2  1 x2  1

is 0 and the numerator is not 0. So, by Theorem 1.14, you can conclude that x  1 is a vertical asymptote, as shown in Figure 1.42(a). b. By factoring the denominator as f x 

.

.

b. f x 

a. When x  1, the denominator of

2

−4

1 2x  1

Solution y

x +1 x2 − 1

Finding Vertical Asymptotes

Determine all vertical asymptotes of the graph of each function.

−2

ff(x) ( =

f x gx

cos x sin x

you can apply Theorem 1.14 to conclude that vertical asymptotes occur at all values of x such that sin x  0 and cos x  0, as shown in Figure 1.42(c). So, the graph of this function has infinitely many vertical asymptotes. These asymptotes occur when x  n, where n is an integer.

Try It

Exploration A

Exploration B

Open Exploration

Theorem 1.14 requires that the value of the numerator at x  c be nonzero. If both the numerator and the denominator are 0 at x  c, you obtain the indeterminate form 00, and you cannot determine the limit behavior at x  c without further investigation, as illustrated in Example 3.

86

CHAPTER 1

Limits and Their Properties

EXAMPLE 3

A Rational Function with Common Factors

Determine all vertical asymptotes of the graph of

f (x) =

f x 

x 2 + 2x − 8 x2 − 4

Solution Begin by simplifying the expression, as shown.

y

4

x 2  2x  8 x2  4 x  4x  2  x  2x  2

f x 

Undefined when x = 2

2 x

−4



2 −2

x 2  2x  8 . x2  4

Vertical asymptote at x = − 2

f x increases and decreases without bound as.x approaches  2. Figure 1.43

x4 , x2

x2

At all x-values other than x  2, the graph of f coincides with the graph of gx  x  4x  2. So, you can apply Theorem 1.14 to g to conclude that there is a vertical asymptote at x  2, as shown in Figure 1.43. From the graph, you can see that

Editable Graph .

lim 

x→2

x 2  2x  8   and x2  4

lim 

x→2

x 2  2x  8  . x2  4

Note that x  2 is not a vertical asymptote.

Try It EXAMPLE 4

Exploration A

Exploration B

Determining Infinite Limits

Find each limit.

f(x) = 6

−4

lim

x→1

x 2 − 3x x−1

and

lim

x→1

x 2  3x x1

Solution Because the denominator is 0 when x  1 (and the numerator is not zero), you know that the graph of 6

−6

f .has a vertical asymptote at x  1. Figure 1.44

x 2  3x x1

f x 

x 2  3x x1

has a vertical asymptote at x  1. This means that each of the given limits is either or  . A graphing utility can help determine the result. From the graph of f shown in Figure 1.44, you can see that the graph approaches from the left of x  1 and approaches  from the right of x  1. So, you can conclude that

Editable Graph

lim

x 2  3x  x1

The limit from the left is infinity.

lim

x2  3x   . x1

The limit from the right is negative infinity.

x→1

and .

x→1

Try It

Exploration A

When using a graphing calculator or graphing software, be careful to interpret correctly the graph of a function with a vertical asymptote—graphing utilities often have difficulty drawing this type of graph.

TECHNOLOGY PITFALL

SECTION 1.5

THEOREM 1.15

Infinite Limits

87

Properties of Infinite Limits

Let c and L be real numbers and let f and g be functions such that lim f x 

lim gx  L.

and

x→c

x→c

1. Sum or difference: lim f x ± gx  x→c

lim f xgx  ,

2. Product:

L > 0

x→c

lim f xgx   ,

x→c

L < 0

gx 0 f x Similar properties hold for one-sided limits and for functions for which the limit of f x as x approaches c is  . 3. Quotient:

lim

x→c

Proof To show that the limit of f x  gx is infinite, choose M > 0. You then need to find  > 0 such that

f x  gx > M

 

 

whenever 0 < x  c < . For simplicity’s sake, you can assume L is positive. Let M1  M  1. Because the limit of f x is infinite, there exists 1 such that f x > M1 whenever 0 < x  c < 1. Also, because the limit of gx is L, there exists  2 such that gx  L < 1 whenever 0 < x  c < 2. By letting  be the smaller of 1 and  2, you can conclude that 0 < x  c <  implies f x > M  1 and gx  L < 1. The second of these two inequalities implies that gx > L  1, and, adding this to the first inequality, you can write

















f x  gx > M  1  L  1  M  L > M. So, you can conclude that lim f x  gx  .

x→c

The proofs of the remaining properties are left as exercises (see Exercise 72).

EXAMPLE 5

Determining Limits

a. Because lim 1  1 and lim x→0



lim 1 

x→0

x→0

1  , you can write x2



1  . x2

Property 1, Theorem 1.15

b. Because lim x 2  1  2 and lim cot  x   , you can write x→1

lim

x→1

x→1

1  0. cot  x

x2

Property 3, Theorem 1.15

c. Because lim 3  3 and lim cot x  , you can write x→0

.

x→0

lim 3 cot x  .

x→0 

Try It

Property 2, Theorem 1.15

Exploration A

88

CHAPTER 1

Limits and Their Properties

Exercises for Section 1.5 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, determine whether f x approaches or  as x approaches 2 from the left and from the right.

 

1. f x  2

x x 4

2. f x 

2

1 x2

y

1 x

2

2

3. f x  tan

1

2 3

4

x 4

4. f x  sec

x 4

3 2 1

x

6

6

2

2

6

Numerical and Graphical Analysis In Exercises 5–8, determine whether f x approaches or  as x approaches 3 from the left and from the right by completing the table. Use a graphing utility to graph the function and confirm your answer. x

3.5

3.1

3.01

3.001

x2  4 x 3  2x 2  x  2

25. f x 

x2  2x  15 x  5x2  x  5

26. ht 

t 2  2t t 4  16

28. g  

tan



2.9

30. f x 

x 2  6x  7 x1

31. f x 

x2  1 x1

32. f x 

sinx  1 x1

In Exercises 33–48, find the limit. 33. lim

x3 x2

34. lim

2x 1x

35. lim

x2 x2  9

36. lim

x2 x 2  16

xq2

lim 

2.5



x2 x 9 2

xq0

x 6. f x  2 x 9 8. f x  sec

x 6

In Exercises 9–28, find the vertical asymptotes (if any) of the graph of the function. 1 9. f x  2 x x2  2 11. hx  2 x x2 13. f x 

x 2  2x  3 x2  x  6

41. lim 1 

1 5. f x  2 x 9

x2 x 4 2

4 10. f x  x  23 2x 12. gx  2 x 1  x 14. f x 

xq1

x2  x 39. lim 2 xq1 x  1x  1

f x

7. f x 

t sin t

x2  1 x1

xq3

2.99

3

29. f x 

37. 2.999

4x 2  4x  24  2x 3  9x 2  18x 24. hx 

xq3

f x x

4x x2  4

t1 15. gt  2 t 1

2s  3 16. hs  2 s  25

17. f x  tan 2x

18. f x  sec  x

 x 2  4x  6x  24

3x 2

x3  1 x1

1 2

1 3 2x

In Exercises 29–32, determine whether the graph of the function has a vertical asymptote or a removable discontinuity at x  1. Graph the function using a graphing utility to confirm your answer.

x 2

20. gx 

23. gx 

27. st 

y

y

6

x4

x

2 2

22. f x 

3 2

4

4 t2

x x2  x  2

21. f x  y

6

19. T t  1 

1 x



xq4

38.

lim

xq 12

40. lim

xq3

44.

45. lim

x csc x

46. lim

47. lim x sec  x xq12

x2 x2

xq0

2 sin x

xq 

6x 2  x  1 4x 2  4x  3



42. lim x 2 

43. lim xq0



lim

xq 2 

xq0

1 x



2 cos x

x2 cot x

48. lim x 2 tan  x xq12

In Exercises 49–52, use a graphing utility to graph the function and determine the one-sided limit. 49. f x 

x2  x  1 x3  1

lim f x

xq1 

51. f x 

1 x 2  25

lim f x

xq5 

50. f x 

x3  1 x x1 2

lim f x

xq1 

52. f x  sec lim f x

xq3 

x 6

SECTION 1.5

Writing About Concepts 53. In your own words, describe the meaning of an infinite limit. Is a real number? 54. In your own words, describe what is meant by an asymptote of a graph. 55. Write a rational function with vertical asymptotes at x  6 and x  2, and with a zero at x  3. 56. Does the graph of every rational function have a vertical asymptote? Explain. 57. Use the graph of the function f (see figure) to sketch the graph of gx  1f x on the interval 2, 3 . To print an enlarged copy of the graph, select the MathGraph button.

89

61. Relativity According to the theory of relativity, the mass m of a particle depends on its velocity v. That is, m

m0 1  v2c2

where m0 is the mass when the particle is at rest and c is the speed of light. Find the limit of the mass as v approaches c  . 62. Rate of Change A 25-foot ladder is leaning against a house (see figure). If the base of the ladder is pulled away from the house at a rate of 2 feet per second, the top will move down the wall at a rate of r

y

2x 625  x2

ft/sec

where x is the distance between the base of the ladder and the house.

2

f x 2 1 1

1

2

3

(a) Find the rate r when x is 7 feet. (b) Find the rate r when x is 15 feet. (c) Find the limit of r as x q 25  .

58. Boyle’s Law For a quantity of gas at a constant temperature, the pressure P is inversely proportional to the volume V. Find the limit of P as V q 0  . 59. Rate of Change A patrol car is parked 50 feet from a long warehouse (see figure). The revolving light on top of the car turns at a rate of 12 revolution per second. The rate at which the light beam moves along the wall is r  50

Infinite Limits

sec2

ft/sec.

(a) Find the rate r when is 6. (b) Find the rate r when is 3. (c) Find the limit of r as q 2  .

25 ft

r

ft 2 sec

63. Average Speed On a trip of d miles to another city, a truck driver’s average speed was x miles per hour. On the return trip the average speed was y miles per hour. The average speed for the round trip was 50 miles per hour. (a) Verify that y 

25x . What is the domain? x  25

(b) Complete the table. x

V 50 ft

30

40

50

60

y Are the values of y different than you expected? Explain. x

60. Illegal Drugs The cost in millions of dollars for a governmental agency to seize x% of an illegal drug is C

528x , 0 f x < 100. 100  x

(c) Find the limit of y as x q 25  and interpret its meaning. 64. Numerical and Graphical Analysis Use a graphing utility to complete the table for each function and graph each function to estimate the limit. What is the value of the limit when the power on x in the denominator is greater than 3?

(a) Find the cost of seizing 25% of the drug.

x

1

(b) Find the cost of seizing 50% of the drug.

f x

(c) Find the cost of seizing 75% of the drug. (d) Find the limit of C as x q 100  and interpret its meaning.

0.5

0.2

0.1

0.01

0.001

(a) lim

x  sin x x

(b) lim

x  sin x x2

(c) lim

x  sin x x3

(d) lim

x  sin x x4

xq0

xq0

xq0

xq0

0.0001

90

CHAPTER 1

Limits and Their Properties

65. Numerical and Graphical Analysis Consider the shaded region outside the sector of a circle of radius 10 meters and inside a right triangle (see figure). (a) Write the area A  f   of the region as a function of . Determine the domain of the function. (b) Use a graphing utility to complete the table and graph the function over the appropriate domain.



0.3

0.6

0.9

1.2

(d) Use a graphing utility to complete the table.



0.3

0.9

1.2

1.5

L (e) Use a graphing utility to graph the function over the appropriate domain. (f) Find

1.5

0.6

f  

lim

 q 2 

L. Use a geometric argument as the basis of

a second method of finding this limit. (g) Find lim L.

(c) Find the limit of A as q 2.

 q0

True or False? In Exercises 67–70, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. V 10 m

66. Numerical and Graphical Reasoning A crossed belt connects a 20-centimeter pulley (10-cm radius) on an electric motor with a 40-centimeter pulley (20-cm radius) on a saw arbor (see figure). The electric motor runs at 1700 revolutions per minute.

20 cm

10 cm

K

67. If px is a polynomial, then the graph of the function given by px has a vertical asymptote at x  1. f x  x1 68. The graph of a rational function has at least one vertical asymptote. 69. The graphs of polynomial functions have no vertical asymptotes. 70. If f has a vertical asymptote at x  0, then f is undefined at x  0. 71. Find functions f and g such that lim f x  and xqc lim gx  but lim f x  gx  0. xqc

xqc

72. Prove the remaining properties of Theorem 1.15. 73. Prove that if lim f x  , then lim xqc

(a) Determine the number of revolutions per minute of the saw. (b) How does crossing the belt affect the saw in relation to the motor? (c) Let L be the total length of the belt. Write L as a function of , where  is measured in radians. What is the domain of the function? (Hint: Add the lengths of the straight sections of the belt and the length of the belt around each pulley.)

74. Prove that if lim

xqc

xqc

1  0. f x

1  0, then lim f x does not exist. f x xqc

Infinite Limits In Exercises 75 and 76, use the - definition of infinite limits to prove the statement. 75. lim xq3

1  x3

76. lim xq4

1   x4

REVIEW EXERCISES

91

Review Exercises for Chapter 1 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, determine whether the problem can be solved using precalculus or if calculus is required. If the problem can be solved using precalculus, solve it. If the problem seems to require calculus, explain your reasoning. Use a graphical or numerical approach to estimate the solution.

x 3  125 x5

19. lim

xq5

x2  4 3 xq2 x  8

20. lim

1  cos x sin x

1. Find the distance between the points 1, 1 and 3, 9 along the curve y  x 2.

21. lim

2. Find the distance between the points 1, 1 and 3, 9 along the line y  4x  3.

22. lim

4x tan x

23. lim

sin 6  x  12 x

In Exercises 3 and 4, complete the table and use the result to estimate the limit. Use a graphing utility to graph the function to confirm your result.

xq0

xq 4

xq0

[Hint: sin    sin cos   cos sin ] 24. lim

xq0

0.1

x

0.01 0.001

0.001

0.01

cos  x  1 x

[Hint: cos    cos cos   sin sin ]

0.1

f x

In Exercises 25 and 26, evaluate the limit given lim f x   4 xqc 2 and lim gx  3. 3

xqc

4x  2  2 xq0 x 4x  2  2  4. lim x xq0 3. lim

25. lim f xgx xqc

26. lim f x  2gx xqc

In Exercises 5 and 6, use the graph to determine each limit. x 2  2x x

5. hx 

6. gx  y

y

3x x2

Numerical, Graphical, and Analytic Analysis In Exercises 27 and 28, consider lim f x.

xq1 

(a) Complete the table to estimate the limit. 8

h

2

x 2 2

2

g

(b) Use a graphing utility to graph the function and use the graph to estimate the limit.

4

4 x

4

4

4

4

8

(c) Rationalize the numerator to find the exact value of the limit analytically. x

(a) lim hx (b) lim hx xq0

xq1

(a) lim gx (b) lim gx xq2

7. lim 3  x xq1

9. lim x 2  3 xq2

8. lim x 10. lim 9

tq4

13. lim

tq2

15. lim

xq4

t2 t2  4

x  2

x4

1x  1  1 x xq0

17. lim

14. lim tq3

16. lim

xq0

18. lim

sq0

1.0001

2x  1  3

x1 3 x 

1 x1

t2

Free-Falling Object In Exercises 29 and 30, use the position function st  4.9t 2  200, which gives the height (in meters) of an object that has fallen from a height of 200 meters. The velocity at time t  a seconds is given by

4  x  2

lim





12. lim 3 y  1 yq4

1.001

Hint: a3  b3  a  ba 2  ab  b2

xq5

In Exercises 11–24, find the limit (if it exists). 11. lim t  2

27. f x  28. f x 

xq9

1.01

f x

xq0

In Exercises 7–10, find the limit L. Then use the - definition to prove that the limit is L.

1.1

9 t3 x

11  s   1 s

tqa

sa  st . at

29. Find the velocity of the object when t  4. 30. At what velocity will the object impact the ground?

92

CHAPTER 1

Limits and Their Properties

In Exercises 31–36, find the limit (if it exists). If the limit does not exist, explain why. 31. lim xq3

51. Let f x 

x  3

x2  4 . Find each limit (if possible). x2





(a) lim f x xq2

x3

(b) lim f x

32. lim x  1

xq2

xq4

x  22, x f 2

2  x, x > 2 1  x, x f 1 34. lim gx, where gx   x  1, x > 1 t  1, t < 1 35. lim ht, where ht   t  1, t v 1 s  4s  2, s f 2 36. lim f s, where f s   s  4s  6, s > 2 33. lim f x, where f x  xq2

(c) lim f x xq2

52. Let f x  xx  1 . (a) Find the domain of f.



(b) Find lim f x.

xq1

xq0

3

(c) Find lim f x.

1 2

tq1

xq1

2

2

sq2

In Exercises 53–56, find the vertical asymptotes (if any) of the graphs of the function.

In Exercises 37–46, determine the intervals on which the function is continuous.

53. gx  1 

37. f x  x  3

55. f x 

38. f x 

3x  x  2 x1 2



41. f x  43. f x 

57. 59.



1 x  2 2

42. f x 

3 x1

44. f x 

x 2

x x 1 x1 2x  2

46. f x  tan 2x

58.

lim

x1 x3  1

60.

xq1 

61. lim xq1

x f 2 x > 2

48. Determine the values of b and c such that the function is continuous on the entire real line. f x 



x  1, x 2  bx  c,

1 < x < 3 x2 v 1





49. Use the Intermediate Value Theorem to show that f x  2x 3  3 has a zero in the interval 1, 2 . 50. Delivery Charges The cost of sending an overnight package from New York to Atlanta is $9.80 for the first pound and $2.50 for each additional pound or fraction thereof. Use the greatest integer function to create a model for the cost C of overnight delivery of a package weighing x pounds. Use a graphing utility to graph the function and discuss its continuity.

lim

xq 12 

x1 x4  1

lim 

x 2  2x  1 x1

62.



64. lim

xq1



xq2

sin 4x 65. lim 5x xq0

x 2x  1

lim

xq1 

x 2  2x  1 x1

1 63. lim x  3 x xq0

67. lim

47. Determine the value of c such that the function is continuous on the entire real line.

xcx3,6,

56. f x  csc  x

2x 2  x  1 x2

xq0

f x 

8 x  10 2

lim 

xq2

5  x, x f 2 2x  3, x > 2

45. f x  csc

4x 4  x2

54. hx 

In Exercises 57–68, find the one-sided limit.

3x 2  x  2 , x  1 x1 39. f x  0, x1 40. f x 

2 x

1 3 x2 

4

sec x 66. lim x xq0

csc 2x x

68. lim xq0

cos 2 x x

69. Environment A utility company burns coal to generate electricity. The cost C in dollars of removing p% of the air pollutants in the stack emissions is C

80,000p , 100  p

0 f p < 100.

Find the cost of removing (a) 15%, (b) 50%, and (c) 90% of the pollutants. (d) Find the limit of C as p q 100. 70. The function f is defined as shown. f x 

tan 2x , x

(a) Find lim

xq0

x0

tan 2x (if it exists). x

(b) Can the function f be defined at x  0 such that it is continuous at x  0?

P.S.

P.S.

93

Problem Solving

Problem Solving

The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

1. Let Px, y be a point on the parabola y  x 2 in the first quadrant. Consider the triangle 䉭PAO formed by P, A0, 1, and the origin O0, 0, and the triangle 䉭PBO formed by P, B1, 0, and the origin. y

3. (a) Find the area of a regular hexagon inscribed in a circle of radius 1. How close is this area to that of the circle? (b) Find the area An of an n-sided regular polygon inscribed in a circle of radius 1. Write your answer as a function of n. (c) Complete the table.

P

A

n

1

6

12

24

48

96

An B O

x

1

(d) What number does An approach as n gets larger and larger? y

(a) Write the perimeter of each triangle in terms of x. 6

(b) Let rx be the ratio of the perimeters of the two triangles,

P(3, 4)

1

Perimeter 䉭PAO . rx  Perimeter 䉭PBO

2 6

Complete the table. 4

x

2

1

0.1

0.01

Perimeter 䉭PAO

Q x

2

6

6

Figure for 3

Perimeter 䉭PBO

2 O

Figure for 4

4. Let P3, 4 be a point on the circle x 2  y 2  25.

r x

(a) What is the slope of the line joining P and O0, 0? (b) Find an equation of the tangent line to the circle at P.

(c) Calculate lim rx. xq0

2. Let Px, y be a point on the parabola y  x 2 in the first quadrant. Consider the triangle 䉭PAO formed by P, A0, 1, and the origin O0, 0, and the triangle 䉭PBO formed by P, B1, 0, and the origin. y

(c) Let Qx, y be another point on the circle in the first quadrant. Find the slope mx of the line joining P and Q in terms of x. (d) Calculate lim mx. How does this number relate to your xq3

answer in part (b)? 5. Let P5, 12 be a point on the circle x 2  y 2  169. y

P

A

1

15

B O

5 x

1

15

5 O

(a) Write the area of each triangle in terms of x.

(a) What is the slope of the line joining P and O0, 0?

Area 䉭PBO . Area 䉭PAO

(b) Find an equation of the tangent line to the circle at P.

Complete the table. 4

x Area 䉭PAO Area 䉭PBO ax

2

1

0.1

0.01

(c) Let Qx, y be another point on the circle in the fourth quadrant. Find the slope mx of the line joining P and Q in terms of x. (d) Calculate lim mx. How does this number relate to your xq5

answer in part (b)? 6. Find the values of the constants a and b such that lim

(c) Calculate lim ax. xq0

Q 15

P(5, 12)

(b) Let ax be the ratio of the areas of the two triangles, ax 

x

5

xq0

a  bx  3

x

 3.

94

CHAPTER 1

Limits and Their Properties

7. Consider the function f x 

3  x13  2

x1

12. To escape Earth’s gravitational field, a rocket must be launched with an initial velocity called the escape velocity. A rocket launched from the surface of Earth has velocity v (in miles per second) given by

.

(a) Find the domain of f. (b) Use a graphing utility to graph the function. f x.

(c) Calculate lim

xq27

v

(d) Calculate lim f x. xq1

8. Determine all values of the constant a such that the following function is continuous for all real numbers. ax , f x  tan x a 2  2,



v 2GM r

2 0

x < 0

9. Consider the graphs of the four functions g1, g2, g3, and g4. y

v g2

2

1

x

3

1

y

2

3

v

y

3

3

g3

2

x

2

x

3

v 10,600 r

2 0

 6.99.

13. For positive numbers a < b, the pulse function is defined as

1 1

 2.17.

2 0

Find the escape velocity for this planet. Is the mass of this planet larger or smaller than that of Earth? (Assume that the mean density of this planet is the same as that of Earth.)

g4

2

1

v 1920 r

(c) A rocket launched from the surface of a planet has velocity v (in miles per second) given by

x

2

1

2

3

For each given condition of the function f, which of the graphs could be the graph of f ?



0, Pa,bx  Hx  a  Hx  b  1, 0, where Hx 

(a) lim f x  3 xq2

1,0,

x < a a f x < b x v b

x v 0 is the Heaviside function. x < 0

(a) Sketch the graph of the pulse function.

(b) f is continuous at 2.

(b) Find the following limits:

(c) lim f x  3 xq2

(i)



1 10. Sketch the graph of the function f x  . x

lim Pa,bx

xqa

(iii) lim Pa,bx xqb

(ii)

lim Pa,bx

xqa

(iv) lim Pa,bx xqb

(a) Evaluate f  , f 3, and f 1.

(c) Discuss the continuity of the pulse function.

(b) Evaluate the limits lim f x, lim f x, lim f x, and xq1 xq1 xq0 lim f x.

(d) Why is

1 4

xq0

Ux 

(c) Discuss the continuity of the function. 11. Sketch the graph of the function f x  x  x. 1 (a) Evaluate f 1, f 0, f 2 , and f 2.7.

(b) Evaluate the limits lim f x, lim f x, and lim1 f x. xq1

 48

Find the escape velocity for the moon.

1 1

2 0

(b) A rocket launched from the surface of the moon has velocity v (in miles per second) given by

3

g1

v 192,000 r

where v0 is the initial velocity, r is the distance from the rocket to the center of Earth, G is the gravitational constant, M is the mass of Earth, and R is the radius of Earth (approximately 4000 miles).

y

2

2GM  R

(a) Find the value of v0 for which you obtain an infinite limit for r as v tends to zero. This value of v0 is the escape velocity for Earth.

x v 0

3



xq1

(c) Discuss the continuity of the function.

xq 2

1 P x b  a a,b

called the unit pulse function? lim f x  L, then 14. Let a be a nonzero constant. Prove that if xq0 lim f ax  L. Show by means of an example that a must be

xq0

nonzero.

96

CHAPTER 2

Differentiation

Section 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.

ISAAC NEWTON (1642–1727) In addition to his work in calculus, Newton made revolutionary contributions to physics, including the Law of Universal Gravitation .and his three laws of motion.

The Tangent Line Problem Calculus grew out of four major problems that European mathematicians were working on during the seventeenth century.

MathBio

1. 2. 3. 4.

y

P

x

The tangent line problem (Section 1.1 and this section) The velocity and acceleration problem (Sections 2.2 and 2.3) The minimum and maximum problem (Section 3.1) The area problem (Sections 1.1 and 4.2)

Each problem involves the notion of a limit, and calculus can be introduced with any of the four problems. A brief introduction to the tangent line problem is given in Section 1.1. Although partial solutions to this problem were given by Pierre de Fermat (1601–1665), René Descartes (1596–1650), Christian Huygens (1629–1695), and Isaac Barrow (1630 –1677), credit for the first general solution is usually given to Isaac Newton (1642–1727) and Gottfried Leibniz (1646–1716). Newton’s work on this problem stemmed from his interest in optics and light refraction. What does it mean to say that a line is tangent to a curve at a point? For a circle, the tangent line at a point P is the line that is perpendicular to the radial line at point P, as shown in Figure 2.1. For a general curve, however, the problem is more difficult. For example, how would you define the tangent lines shown in Figure 2.2? You might say that a line is tangent to a curve at a point P if it touches, but does not cross, the curve at point P. This definition would work for the first curve shown in Figure 2.2, but not for the second. Or you might say that a line is tangent to a curve if the line touches or intersects the curve at exactly one point. This definition would work for a circle but not for more general curves, as the third curve in Figure 2.2 shows. y

y

y

y = f (x)

Tangent line to a circle Figure 2.1

P

P P

x

FOR FURTHER INFORMATION For more information on the crediting of mathematical discoveries to the first “discoverer,” see the article “Mathematical Firsts—Who Done It?” by. Richard H. Williams and Roy D. Mazzagatti in Mathematics Teacher. MathArticle

y = f (x)

y = f (x)

x

Tangent line to a curve at a point Figure 2.2 E X P L O R AT I O N

Identifying a Tangent Line Use a graphing utility to graph the function f x  2x 3  4x 2  3x  5. On the same screen, graph y  x  5, y  2x  5, and y  3x  5. Which of these lines, if any, appears to be tangent to the graph of f at the point 0, 5? Explain your reasoning.

x

SECTION 2.1

y

(c + Δ x , f(c + Δx))

m

(c, f(c)) x

97

Essentially, the problem of finding the tangent line at a point P boils down to the problem of finding the slope of the tangent line at point P. You can approximate this slope using a secant line* through the point of tangency and a second point on the curve, as shown in Figure 2.3. If c, f c is the point of tangency and c  x, f c  x is a second point on the graph of f, the slope of the secant line through the two points is given by substitution into the slope formula

f (c + Δ x) − f (c) = Δy

Δx

The Derivative and the Tangent Line Problem

msec 

The secant line through c, f c and c  x, f c  x

y 2  y1 x 2  x1 f c  x  f c c  x  c

msec 

Figure 2.3

f c  x  f c . x

Change in y Change in x

Slope of secant line

The right-hand side of this equation is a difference quotient. The denominator x is the change in x, and the numerator y  f c  x  f c is the change in y. The beauty of this procedure is that you can obtain more and more accurate approximations of the slope of the tangent line by choosing points closer and closer to the point of tangency, as shown in Figure 2.4.

THE TANGENT LINE PROBLEM In 1637, mathematician René Descartes stated this about the tangent line problem:

(c, f (c))

Δx

Δx

(c, f(c))

“And I dare say that this is not only the most useful and general problem in geometry that I know, but even that I ever desire to know.”

Δy

(c, f (c))

Δy Δx

Δx → 0

Δy

(c, f (c))

Δy

Δx

(c, f (c)) Δy Δx

Δx → 0

Δy

(c, f (c))

(c, f (c))

Δx (c, f(c))

Tangent line

Tangent line

Tangent line approximations Figure 2.4

.

To view a sequence of secant lines approaching a tangent line, select the Animation button.

Animation Definition of Tangent Line with Slope m If f is defined on an open interval containing c, and if the limit lim

x→0

.

y f c  x  f c  lim m x x→0 x

exists, then the line passing through c, f c with slope m is the tangent line to the graph of f at the point c, f c.

Video

Video

The slope of the tangent line to the graph of f at the point c, f c is also called the slope of the graph of f at x  c. * This use of the word secant comes from the Latin secare, meaning to cut, and is not a reference to the trigonometric function of the same name.

98

CHAPTER 2

Differentiation

EXAMPLE 1

The Slope of the Graph of a Linear Function

Find the slope of the graph of f x  2x  3 at the point 2, 1. f (x) = 2x − 3

y

Solution To find the slope of the graph of f when c  2, you can apply the definition of the slope of a tangent line, as shown.

Δx = 1

3

lim

x→0

Δy = 2

2

f 2  x  f 2

22  x  3  22  3  lim x→0 x x  lim

x→0

m=2 1

(2, 1)

4  2x  3  4  3 x

2x x  lim 2  lim

x→0

x 1

2

x→0

3

2

. slope of f at 2, 1 is m  2. The

The slope of f at c, f c  2, 1 is m  2, as shown in Figure 2.5.

Figure 2.5

.Editable Graph

NOTE In Example 1, the limit definition of the slope of f agrees with the definition of the slope of a line as discussed in Section P.2.

Try It

Exploration A

The graph of a linear function has the same slope at any point. This is not true of nonlinear functions, as shown in the following example. y

EXAMPLE 2

Find the slopes of the tangent lines to the graph of

4 3

Tangent line at (−1,2 )

−2

f (x) = x 2 + 1

2

−1

Tangent Lines to the Graph of a Nonlinear Function

Tangent line at (0, 1)

f x  x 2  1 at the points 0, 1 and 1, 2, as shown in Figure 2.6. Solution Let c, f c represent an arbitrary point on the graph of f. Then the slope of the tangent line at c, f c is given by

x 1

2

The slope of f at any point c, f c is m . 2c.

lim

x→0

c  x 2  1  c 2  1 f c  x  f c  lim x→0 x x c 2  2cx  x 2  1  c 2  1 x→0 x 2cx  x 2  lim x→0 x  lim 2c  x  lim

Figure 2.6

Editable Graph

x→0

 2c. So, the slope at any point c, f c on the graph of f is m  2c. At the point 0, 1, the slope is m  20  0, and at 1, 2, the slope is m  21  2. . NOTE

In Example 2, note that c is held constant in the limit process as  x → 0.

Try It

Exploration A

SECTION 2.1

y

lim

x→0

(c, f(c))

x

The graph of f has a vertical tangent line at c, f c. Figure 2.7

99

The definition of a tangent line to a curve does not cover the possibility of a vertical tangent line. For vertical tangent lines, you can use the following definition. If f is continuous at c and

Vertical tangent line

c

The Derivative and the Tangent Line Problem

f c  x  f c  x

or

lim

x→0

f c  x  f c   x

the vertical line x  c passing through c, f c is a vertical tangent line to the graph of f. For example, the function shown in Figure 2.7 has a vertical tangent line at c, f c. If the domain of f is the closed interval a, b , you can extend the definition of a vertical tangent line to include the endpoints by considering continuity and limits from the right for x  a and from the left for x  b.

The Derivative of a Function You have now arrived at a crucial point in the study of calculus. The limit used to define the slope of a tangent line is also used to define one of the two fundamental operations of calculus—differentiation.

Definition of the Derivative of a Function The derivative of f at x is given by fx  lim

x→0

.

f x  x  f x x

provided the limit exists. For all x for which this limit exists, f  is a function of x.

Video Be sure you see that the derivative of a function of x is also a function of x. This “new” function gives the slope of the tangent line to the graph of f at the point x, f x, provided that the graph has a tangent line at this point. The process of finding the derivative of a function is called differentiation. A function is differentiable at x if its derivative exists at x and is differentiable on an open interval a, b if it is differentiable at every point in the interval. In addition to fx, which is read as “ f prime of x,” other notations are used to denote the derivative of y  f x. The most common are fx,

dy , dx

y,

d

f x , dx

Dx y .

Notation for derivatives

The notation dydx is read as “the derivative of y with respect to x” or simply “dy  dx”. Using limit notation, you can write y dy  lim x→0 dx x f x  x  f x  lim x→0 x .

 fx.

History

100

CHAPTER 2

Differentiation

EXAMPLE 3

Finding the Derivative by the Limit Process

Find the derivative of f x  x 3  2x. Solution fx  lim

x→0

 lim

x→0

When using the definition to find a derivative of a function, the key is to rewrite the difference quotient so that x does not occur as a factor of the denominator.

 lim

x→0

STUDY TIP

 lim

x→0

 lim

x→0

 lim

x→0

.

f x  x  f x Definition of derivative x x  x3  2x  x  x3  2x x x3  3x2x  3xx 2  x3  2x  2x  x3  2x x 3x 2x  3xx 2  x3  2x x x 3x 2  3xx  x 2  2 x

3x 2  3xx  x 2  2

 3x 2  2 .

Try It

Exploration A

Exploration C .

Exploration B

Open Exploration

The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph Remember that the derivative of a function f is itself a function, which can be used to find the slope of the tangent line at the point x, f x on the graph of f. EXAMPLE 4

Using the Derivative to Find the Slope at a Point

Find fx for f x  x. Then find the slope of the graph of f at the points 1, 1 and 4, 2. Discuss the behavior of f at 0, 0. Solution Use the procedure for rationalizing numerators, as discussed in Section 1.3. f x  x  f x fx  lim Definition of derivative x→0 x  lim

x  x  x

x

x→0

x→0

y

3

(4, 2) 2

(1, 1)

m=

m= 1 2

(0, 0) 1

x

3

4

The slope of f at x, f x, x > 0, is m .. 1 2x . Figure 2.8

Editable Graph

1 4

f(x) = x

2





1 , 2x

x  x  x



x  x  x x  x  x x x  x  x  lim x→0 x x  x  x  x  lim x→0 x x  x  x  1  lim x→0 x  x  x

 lim



x > 0

At the point 1, 1, the slope is f1  12. At the point 4, 2, the slope is f4  14. See Figure 2.8. At the point 0, 0, the slope is undefined. Moreover, the graph of f has a vertical tangent line at 0, 0.

Try It

Exploration A

Exploration B

Exploration C

SECTION 2.1

The Derivative and the Tangent Line Problem

101

In many applications, it is convenient to use a variable other than x as the independent variable, as shown in Example 5. EXAMPLE 5

Finding the Derivative of a Function

Find the derivative with respect to t for the function y  2t. Solution Considering y  f t, you obtain f t  t  f t dy  lim t→0 dt t 2 2  t  t t  lim t→0 t 2t  2t  t tt  t  lim t→0 t 2t  lim t→0 ttt  t  lim

t→0

.

Definition of derivative

f t  t  2t  t and f t  2t

Combine fractions in numerator. Divide out common factor of t.

2 t t  t

Simplify.

2   2. t

4

.

y=

Evaluate limit as t → 0.

Try It

2 t

Exploration A

Open Exploration

The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph

(1, 2)

A graphing utility can be used to reinforce the result given in Example 5. For instance, using the formula dydt  2t 2, you know that the slope of the graph of y  2t at the point 1, 2 is m  2. This implies that an equation of the tangent line to the graph at 1, 2 is

TECHNOLOGY 0

6 0

y = −2t + 4

At the point 1, 2 the line y   2t  4 is tangent to the graph of y  2 t.

y  2  2t  1 or

Figure 2.9

as shown in Figure 2.9.

y  2t  4

Differentiability and Continuity The following alternative limit form of the derivative is useful in investigating the relationship between differentiability and continuity. The derivative of f at c is

y

(x, f (x))

fc  lim

(c, f (c))

x→c

x−c

f(x) − f (c)

lim

x

f x  f c xc

and

lim

x→c

f x  f c xc

x

As x approaches c, the secant line approaches the tangent line. Figure 2.10

Alternative form of derivative

provided this limit exists (see Figure 2.10). (A proof of the equivalence of this form is given in Appendix A.) Note that the existence of the limit in this alternative form requires that the one-sided limits x→c

c

f x  f c xc

exist and are equal. These one-sided limits are called the derivatives from the left and from the right, respectively. It follows that f is differentiable on the closed interval [a, b] if it is differentiable on a, b and if the derivative from the right at a and the derivative from the left at b both exist.

102

CHAPTER 2

Differentiation

If a function is not continuous at x  c, it is also not differentiable at x  c. For instance, the greatest integer function

y 2

f x  x

1

is not continuous at x  0, and so it is not differentiable at x  0 (see Figure 2.11). You can verify this by observing that

x

−2

−1

1

3

2

f(x) = [[x]] −2

lim

f x  f 0 x  0  lim  x→0 x0 x

Derivative from the left

lim

f x  f 0 x  0  lim  0. x→0 x0 x

Derivative from the right

x→0

The greatest integer function is not differentiable at x  0, because it is not continuous at x  0.

and x→0 

Figure 2.11

Although it is true that differentiability implies continuity (as shown in Theorem 2.1 on the next page), the converse is not true. That is, it is possible for a function to be continuous at x  c and not differentiable at x  c. Examples 6 and 7 illustrate this possibility. EXAMPLE 6 The function

y



f x  x  2

f(x) =⏐x − 2⏐

3

1

3



Derivative from the left





Derivative from the right

x2 0 f x  f 2  lim  1 x→2 x2 x2

lim

x2 0 f x  f 2  lim 1 x→2 x2 x2

and

x 2



lim

x→2

m=1 1



shown in Figure 2.12 is continuous at x  2. But, the one-sided limits

m = −1

2

A Graph with a Sharp Turn

4

f is not differentiable at x  2, because the derivatives from the left and from the right are. not equal. Figure 2.12

x→2

are not equal. So, f is not differentiable at x  2 and the graph of f does not have a tangent line at the point 2, 0.

Try It

Editable Graph

EXAMPLE 7

Exploration A

Open Exploration

A Graph with a Vertical Tangent Line

y

f(x) = x 1/3

The function f x  x13

1

is continuous at x  0, as shown in Figure 2.13. But, because the limit x −2

−1

1

2

lim

x→0

−1

f is not differentiable at x  0, because f has. a vertical tangent at x  0. Figure 2.13

Editable Graph

f x  f 0 x13  0  lim x→0 x0 x 1  lim 23 x→0 x 

is infinite, you can conclude that the tangent line is vertical at x  0. So, f is not differentiable at x  0.

Try It

Exploration A

Exploration B

Exploration C

From Examples 6 and 7, you can see that a function is not differentiable at a point at which its graph has a sharp turn or a vertical tangent.

SECTION 2.1

TECHNOLOGY Some graphing utilities, such as Derive, Maple, Mathcad, Mathematica, and the TI-89, perform symbolic differentiation. Others perform numerical differentiation by finding values of derivatives using the formula f x 

f x  x  f x  x 2x

THEOREM 2.1

The Derivative and the Tangent Line Problem

103

Differentiability Implies Continuity

If f is differentiable at x  c, then f is continuous at x  c. Proof You can prove that f is continuous at x  c by showing that f x approaches f c as x → c. To do this, use the differentiability of f at x  c and consider the following limit.

 f xx  cf c f x  f c   lim x  c lim xc  

lim f x  f c  lim x  c

where x is a small number such as 0.001. Can you see any problems with this definition? For instance, using this definition, what is the value of the derivative of f x  x when x  0?

x→c

x→c

x→c



x→c

 0 f c 0 Because the difference f x  f c approaches zero as x → c, you can conclude that lim f x  f c. So, f is continuous at x  c. x→c

The following statements summarize the relationship between continuity and differentiability. 1. If a function is differentiable at x  c, then it is continuous at x  c. So, differentiability implies continuity. 2. It is possible for a function to be continuous at x  c and not be differentiable at x  c. So, continuity does not imply differentiability.

SECTION 2.1

The Derivative and the Tangent Line Problem

103

Exercises for Section 2.1 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, estimate the slope of the graph at the points x1, y1 and x2, y2. y

1. (a)

y

(b)

In Exercises 3 and 4, use the graph shown in the figure. To print an enlarged copy of the graph, select the MathGraph button. y

(x1, y1) (x2, y2) (x2, y2)

(x1, y1)

x

x

6 5 4 3 2 1

(4, 5)

f

(1, 2) x

1 2 3 4 5 6 y

2. (a)

3. Identify or sketch each of the quantities on the figure.

y

(b)

(a) f 1 and f 4 (x1, y1)

(c) y 

(x2, y2) x

x

(x1, y1)

(x2, y2)

(b) f 4  f 1

f 4  f 1 x  1  f 1 41

4. Insert the proper inequality symbol < or > between the given quantities. (a)

f 4  f 3 f 4  f 1 41 䊏 43

(b)

f 4  f 1 f 1 41 䊏

104

CHAPTER 2

Differentiation

In Exercises 5 –10, find the slope of the tangent line to the graph of the function at the given point. 5. f x  3  2x, 1, 5

6. gx 

7. gx  x 2  4, 1, 3

8. gx  5  x 2,

9. f t  3t  t 2,

0, 0

3 2x

 1, 2, 2

2, 1 2 10. ht  t  3, 2, 7

12. gx  5

13. f x  5x

14. f x  3x  2

15. hs  3  3 s

1 16. f x  9  2x

17. f x  2x 2  x  1

18. f x  1  x 2

19. f x 

20. f x 

2

21. f x 

 12x

x3

1 x1

22. f x 

23. f x  x  1

24. f x 

x3



x

y

x2

4 3 2

1 2 3 4 5

27. f x 

2, 8 1, 1 4 31. f x  x  , 4, 5 x

28. f x 

x 3,

29. f x  x,

x3

 1, 1, 2

30. f x  x  1, 32. f x 

5, 2

1 , 0, 1 x1

Line

33. f x  x 3

3x  y  1  0

34. f x  x 3  2

3x  y  4  0

35. f x 

1

1 36. f x  x  1

x

3 2 1

5 4 3 2 1

f

1 2 3 2 3

42. The tangent line to the graph of y  hx at the point 1, 4 passes through the point 3, 6. Find h1 and h1.

In Exercises 43– 46, sketch the graph of f . Explain how you found your answer. 44.

y 2 1 2

x 1 2

2 3 4

y

4

x

y

x

3 2

3

3 2 1

f

45.

2

4

f

f 6

46.

y 7 6 5 4 3 2 1

2 2

4 5 6

6

38.

x 1 2 3

41. The tangent line to the graph of y  gx at the point 5, 2 passes through the point 9, 0. Find g5 and g 5.

x  2y  7  0

y

f

3 2 1

1 2 3

43.

In Exercises 37–40, the graph of f is given. Select the graph of f . 37.

f

2 3

x  2y  6  0

x

3 2 1

Writing About Concepts

In Exercises 33–36, find an equation of the line that is tangent to the graph of f and parallel to the given line. Function

y

(d)

3 2 1

25. f x  x 2  1, 2, 5

1 2 3

2

y

3 2

x

3 2 1

x

(c)

26. f x  x 2  2x  1, 3, 4

f

f

x

In Exercises 25–32, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results.

x 1 2 3 y

(b)

1

4

f

3 2 1

1 2 3 4 5

5 4 3 2 1

1 x2

5 4 3 2

f

1

(a)

y

40.

5 4 3 2 1

In Exercises 11–24, find the derivative by the limit process. 11. f x  3

y

39.

y 7 6

f

4 3 2 1

f

x 1

1 2 3 4 5 6 7

x 1 2 3 4 5 6 7 8

x 1 2 3

47. Sketch a graph of a function whose derivative is always negative.

SECTION 2.1

Writing About Concepts (continued) 48. Sketch a graph of a function whose derivative is always positive.

60. Graphical Reasoning Use a graphing utility to graph each function and its tangent lines at x  1, x  0, and x  1. Based on the results, determine whether the slopes of tangent lines to the graph of a function at different values of x are always distinct.

In Exercises 49–52, the limit represents f c for a function f and a number c. Find f and c.

5  31  x  2 x0 x 2 x  36 51. lim x6 x6 49. lim

2  x3  8 x0 x 2x  6 52. lim x9 x9 50. lim

(a) f x  x 2

x f x

53. f 0  2;

f x

f  x  3,  < x
0 for x > 0

55. f 0  0; f 0  0; f x > 0 if x  0

57. f x  4x  x 2

(b) g x  x 3

Graphical, Numerical, and Analytic Analysis In Exercises 61 and 62, use a graphing utility to graph f on the interval [2, 2]. Complete the table by graphically estimating the slopes of the graph at the indicated points. Then evaluate the slopes analytically and compare your results with those obtained graphically.

In Exercises 53 –55, identify a function f that has the following characteristics. Then sketch the function. 54. f 0  4; f 0  0;

105

The Derivative and the Tangent Line Problem

65. f x  x4  x

2

1 66. f x  4 x 3

x

1 x 1

3

2

5

6 4 2 4

2

4

(1, 3)

6

The figure shows the graph of g.

59. Graphical Reasoning y

67. f x 

6 4 2

g

4 6 4 6

(a) g0  䊏

1 x

68. f x 

x3  3x 4

Writing In Exercises 69 and 70, consider the functions f and Sx where

x

6 4

Graphical Reasoning In Exercises 67 and 68, use a graphing utility to graph the function and its derivative in the same viewing window. Label the graphs and describe the relationship between them.

Sx x  (b) g3  䊏

(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 g6  g4 positive or negative? Explain. (f) Is it possible to find g 2 from the graph? Explain.

f 2  x  f 2 x  2  f 2. x

(a) Use a graphing utility to graph f and Sx in the same viewing window for x  1, 0.5, and 0.1. (b) Give a written description of the graphs of S for the different values of x in part (a). 69. f x  4  x  3 2

70. f x  x 

1 x

106

CHAPTER 2

Differentiation

In Exercises 71–80, use the alternative form of the derivative to find the derivative at x  c (if it exists). 71. f x  x 2  1, c  2

72. gx  xx  1, c  1

73. f x  x 3  2x 2  1, c  2 74. f x  x 3  2x, c  1



c3

78. gx  x  313, c  3









80. f x  x  4 , c  4

In Exercises 81– 86, describe the x-values at which f is differentiable. 81. f x 

1 x1



12 10 1 x

2

2

2

4

84. f x 

x2 x2  4 5 4 3 2

5 4 3

x

1

4

x

3

86. f x 

x4 x4,,

x  0 x > 0

2

2

y

y

3

4

2

2 4

x 4

4 4

Graphical Analysis In Exercises 87–90, use a graphing utility to find the x-values at which f is differentiable.





88. f x 

89. f x  x25 90. f x 

xx  3x2x,  3x, 3 2

2

x  1 x > 1

95. f x 



x 2  1, 4x  3,

x  2 x > 2

96. f x 



1 2x

 1,

2x ,

x < 2 x  2

97. Graphical Reasoning A line with slope m passes through the point 0, 4 and has the equation y  mx  4.

(b) Graph g and g on the same set of axes.

True or False? In Exercises 99–102, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

2x x1

99. The slope of the tangent line to the differentiable function f at f 2   x  f 2 . the point 2, f 2 is x 100. If a function is continuous at a point, then it is differentiable at that point. 101. If a function has derivatives from both the right and the left at a point, then it is differentiable at that point.

x

1

87. f x  x  3

x  1 x > 1

2

3 4

1 2 3 4 5 6

3

x,x ,

(d) Find f x if f x  x 4. Compare the result with the conjecture in part (c). Is this a proof of your conjecture? Explain.

y

2

94. f x 

(c) Identify a pattern between f and g and their respective derivatives. Use the pattern to make a conjecture about hx if h x  x n, where n is an integer and n  2.

4

y

1

x  1 x > 1

(a) Graph f and f  on the same set of axes. x

4

85. f x  x  1

2

98. Conjecture Consider the functions f x  x 2 and gx  x3.

6 4 2

1

83. f x  x  3 23

xx  11 ,

(b) Use a graphing utility to graph the function d in part (a). Based on the graph, is the function differentiable at every value of m? If not, where is it not differentiable?

y

1

92. f x  1  x 2 3,

(a) Write the distance d between the line and the point 3, 1 as a function of m.



82. f x  x 2  9 y

2



In Exercises 95 and 96, determine whether the function is differentiable at x  2.

77. f x  x  623, c  6 79. hx  x  5 , c  5



91. f x  x  1 93. f x 

75. gx   x , c  0 76. f x  1x,

In Exercises 91–94, find the derivatives from the left and from the right at x  1 (if they exist). Is the function differentiable at x  1?

102. If a function is differentiable at a point, then it is continuous at that point. 103. Let f x 





1 1 x sin , x  0 x 2 sin , x  0 x x . and g x  0, 0, x0 x0

Show that f is continuous, but not differentiable, at x  0. Show that g is differentiable at 0, and find g0. 104. Writing Use a graphing utility to graph the two functions f x  x 2  1 and gx  x  1 in the same viewing window. Use the zoom and trace features to analyze the graphs near the point 0, 1. What do you observe? Which function is differentiable at this point? Write a short paragraph describing the geometric significance of differentiability at a point.



SECTION 2.2

.

Section 2.2

.

Basic Differentiation Rules and Rates of Change

Basic Differentiation Rules and Rates of Change

Video

Video

Video

Video

• • • • • •

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.

The Constant Rule

y

In Section 2.1 you used the limit definition to find derivatives. In this and the next two sections you will be introduced to several “differentiation rules” that allow you to find derivatives without the direct use of the limit definition.

The slope of a horizontal line is 0.

THEOREM 2.2 f (x) = c The derivative of a constant function is 0.

The Constant Rule

The derivative of a constant function is 0. That is, if c is a real number, then d

c  0. dx

x

The Constant Rule

Proof

Figure 2.14

Let f x  c. Then, by the limit definition of the derivative,

d

c  fx dx  lim

f x  x  f x x

 lim

cc x

x→0

NOTE In Figure 2.14, note that the Constant Rule is equivalent to saying that the slope of a horizontal line is 0. This demonstrates the relationship between slope and derivative.

x→0

 lim 0 x→0

 0. EXAMPLE 1

Using the Constant Rule

Function

Derivative

b. f x  0 c. st  3

dy 0 dx fx  0 st  0

d. y  k 2, k is constant

y  0

a. y  7

.

Try It

.

Exploration A

The editable graph feature below allows you to edit the graph of a function. .

107

a.

Editable Graph

b.

Editable Graph

c.

Editable Graph

d.

Editable Graph

E X P L O R AT I O N

Writing a Conjecture Use the definition of the derivative given in Section 2.1 to find the derivative of each function. What patterns do you see? Use your results to write a conjecture about the derivative of f x  x n. a. f x  x1 d. f x  x4

b. f x  x 2 e. f x  x12

c. f x  x 3 f. f x  x1

108

CHAPTER 2

Differentiation

The Power Rule Before proving the next rule, it is important to review the procedure for expanding a binomial.

x  x 2  x 2  2xx  x 2 x  x 3  x 3  3x 2x  3xx2  x3 The general binomial expansion for a positive integer n is

x  x n  x n  nx n1 x 

nn  1x n2 x 2  . . .  x n. 2 x2 is a factor of these terms.

This binomial expansion is used in proving a special case of the Power Rule.

THEOREM 2.3

The Power Rule

If n is a rational number, then the function f x  x n is differentiable and d n

x  nx n1. dx For f to be differentiable at x  0, n must be a number such that x n1 is defined on an interval containing 0.

Proof

If n is a positive integer greater than 1, then the binomial expansion produces

x  xn  x n d n

x  lim dx x→0 x nn  1x n2 x 2  . . .  x n  x n 2  lim x x→0 n2 nn  1x x  . . .  x n1  lim nx n1  2 x→0  nx n1  0  . . .  0 x n  nx n1x 





 nx n1. This proves the case for which n is a positive integer greater than 1. You will prove the case for n  1. Example 7 in Section 2.3 proves the case for which n is a negative integer. In Exercise 75 in Section 2.5 you are asked to prove the case for which n is rational. (In Section 5.5, the Power Rule will be extended to cover irrational values of n.)

y 4

y=x

3

When using the Power Rule, the case for which n  1 is best thought of as a separate differentiation rule. That is,

2 1 x 1

2

3

Power Rule when n  1

4

The slope of the line y  x is 1. Figure 2.15

d

x  1. dx

This rule is consistent with the fact that the slope of the line y  x is 1, as shown in Figure 2.15.

SECTION 2.2

EXAMPLE 2

Basic Differentiation Rules and Rates of Change

Using the Power Rule

Function

Derivative

a. f x  x 3

fx)  3x 2 d 13 1 1 gx 

x  x23  23 dx 3 3x dy d 2 2 

x  2x3   3 dx dx x

3 x b. gx  

.

c. y 

109

1 x2

Try It

Exploration A

In Example 2(c), note that before differentiating, 1x 2 was rewritten as x2. Rewriting is the first step in many differentiation problems. Rewrite:

Given: 1 y 2 x

y

f (x) = x 4

Simplify: dy 2  3 dx x

Differentiate: dy  2x3 dx

y  x2

2

EXAMPLE 3 (−1, 1)

1

Finding the Slope of a Graph

Find the slope of the graph of f x  x 4 when

(1, 1)

a. x  1 x

(0, 0)

−1

1

Note that the slope of the graph is negative at the point 1, 1, the slope is zero at the point 0, 0, and the slope is positive at the .. point 1, 1.

c. x  1.

Solution The slope of a graph at a point is the value of the derivative at that point. The derivative of f is fx  4x3. a. When x  1, the slope is f1  413  4. b. When x  0, the slope is f0  403  0. c. When x  1, the slope is f1  413  4.

Slope is negative. Slope is zero. Slope is positive.

See Figure 2.16.

Figure 2.16

Try It

Editable Graph

EXAMPLE 4 f (x) = x 2

3

Point on graph

To find the slope of the graph when x  2, evaluate the derivative, fx  2x, at x  2. m  f2  4

1

x 1

2

y = −4x − 4

The line y   4x  4 is tangent to the .. graph of f x  x 2 at the point  2, 4.

Editable Graph

Finding an Equation of a Tangent Line

2, f 2  2, 4

2

Figure 2.17

Open Exploration

Solution To find the point on the graph of f, evaluate the original function at x  2.

4

−2

Exploration A

Find an equation of the tangent line to the graph of f x  x 2 when x  2.

y

(−2, 4)

b. x  0

Slope of graph at 2, 4

Now, using the point-slope form of the equation of a line, you can write y  y1  mx  x1 y  4  4 x  2 y  4x  4.

Point-slope form Substitute for y1, m, and x1. Simplify.

See Figure 2.17.

Try It

Exploration A

Exploration B

Open Exploration

110

CHAPTER 2

Differentiation

The Constant Multiple Rule THEOREM 2.4

The Constant Multiple Rule

If f is a differentiable function and c is a real number, then cf is also d differentiable and cf x  cfx. dx Proof cf x  x  cf x d

cf x  lim x→0 dx x

Definition of derivative

 f x  xx  f x f x  x  f x  c  lim  x  lim c x→0

Apply Theorem 1.2.

x→0

 cfx Informally, the Constant Multiple Rule states that constants can be factored out of the differentiation process, even if the constants appear in the denominator. d d

cf x  c

dx dx

f x  cfx

d f x d  dx c dx

 

EXAMPLE 5 Function

a. y 

2 x

b. f t 

4t 2 5

c. y  2x 1 3 x2 2 3x e. y   2 d. y  .

Try It

 1c  f x 1 d 1  

f x    fx c dx c Using the Constant Multiple Rule Derivative

d d 2 dy  2x1  2 x1  21x2   2 dx dx dx x d 4 2 4 d 2 4 8 ft 

t  2t  t t  dt 5 5 dt 5 5 dy d 1 1  2x12  2 x12  x12  dx dx 2 x d 1 23 1 2 1 dy  x   x53   53 dx dx 2 2 3 3x

 



  

  d 3 3 3 y    x   1   dx 2 2 2 Exploration A

The Constant Multiple Rule and the Power Rule can be combined into one rule. The combination rule is Dx cx n  cnx n1.

SECTION 2.2

EXAMPLE 6

111

Using Parentheses When Differentiating

Original Function

Rewrite

Differentiate

Simplify

5 2x 3 5 b. y  2x3 7 c. y  2 3x 7 d. y  3x2

5 y  x3 2 5 3 y  x  8 7 2 y  x  3

5 y  3x4 2 5 y  3x4 8 7 y  2x 3

y  

y  63x 2

y  632x

y  126x

Try It

Exploration A

a. y 

.

Basic Differentiation Rules and Rates of Change

15 2x 4 15 y   4 8x 14x y  3

The Sum and Difference Rules THEOREM 2.5

The Sum and Difference Rules

The sum (or difference) of two differentiable functions f and g is itself differentiable. Moreover, the derivative of f  g or f  g is the sum (or difference) of the derivatives of f and g. d

f x  gx  fx  gx dx d

f x  gx  fx  gx dx

Sum Rule

Difference Rule

Proof A proof of the Sum Rule follows from Theorem 1.2. (The Difference Rule can be proved in a similar way.)

f x  x  gx  x  f x  gx d

f x  gx  lim x→0 dx x f x  x  gx  x  f x  gx  lim x→0 x f x  x  f x gx  x  gx  lim  x→0 x x f x  x  f x gx  x  gx  lim  lim x→0 x→0 x x  fx  gx





The Sum and Difference Rules can be extended to any finite number of functions. For instance, if Fx  f x  gx  hx, then Fx  fx  gx  hx. EXAMPLE 7

Using the Sum and Difference Rules

Function

.

a. f x  x 3  4x  5 x4 b. gx    3x 3  2x 2

Try It .

Derivative

fx  3x 2  4 gx  2x 3  9x 2  2

Exploration A

The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph

112

CHAPTER 2

Differentiation

FOR FURTHER INFORMATION For the outline of a geometric proof of the derivatives of the sine and cosine functions, see the article “The Spider’s Spacewalk Derivation of sin and cos ” by Tim . Hesterberg in The College Mathematics Journal.

MathArticle

Derivatives of Sine and Cosine Functions In Section 1.3, you studied the following limits. sin x  1 and x→0 x lim

Derivatives of Sine and Cosine Functions

d

sin x  cos x dx

y′ = 0 y′ = −1 y′ = 1 π

y′ = 1

π

−1

x

2π

2

y′ = 0 y decreasing y increasing

y increasing

d

cos x  sin x dx

Proof

y = sin x

1

x→0

1  cos x 0 x

These two limits can be used to prove differentiation rules for the sine and cosine functions. (The derivatives of the other four trigonometric functions are discussed in Section 2.3.)

THEOREM 2.6

y

lim

d sinx  x  sin x

sin x  lim Definition of derivative x→0 dx x sin x cos x  cos x sin x  sin x  lim x→0 x cos x sin x  sin x1  cos x x sin x 1  cos x  sin x  lim cos x x→0 x x  lim

x→0

y′ positive

y′ negative

y ′ positive

     sin x 1  cos x  cos x lim  sin x lim  x  x

y

π 2

−1

x→0

x

π

 cos x1  sin x0  cos x

2π

y ′ = cos x

The derivative of the sine function is the . function. cosine Figure 2.18

Animation

y = 2 sin x

2

Derivatives Involving Sines and Cosines

Function

3 sin x 2



−

This differentiation rule is shown graphically in Figure 2.18. Note that for each x, the slope of the sine curve is equal to the value of the cosine. The proof of the second rule is left as an exercise (see Exercise 116). EXAMPLE 8

y=

x→0

a. y  2 sin x sin x 1 b. y   sin x 2 2 c. y  x  cos x

Derivative

y  2 cos x 1 cos x y  cos x  2 2 y  1  sin x

A graphing utility can provide insight into the interpretation of a derivative. For instance, Figure 2.19 shows the graphs of

TECHNOLOGY −2

y = sin x

y=

1 sin x 2

d

a sin x  a cos x dx. Figure 2.19

y  a sin x for a  12, 1, 32, and 2. Estimate the slope of each graph at the point 0, 0. Then verify your estimates analytically by evaluating the derivative of each function when x  0.

Try It

Exploration A

Open Exploration

SECTION 2.2

Basic Differentiation Rules and Rates of Change

113

Rates of Change You have seen how the derivative is used to determine slope. The derivative can also be used to determine the rate of change of one variable with respect to another. Applications involving rates of change occur in a wide variety of fields. A few examples are population growth rates, production rates, water flow rates, velocity, and acceleration. A common use for rate of change is to describe the motion of an object moving in a straight line. In such problems, it is customary to use either a horizontal or a vertical line with a designated origin to represent the line of motion. On such lines, movement to the right (or upward) is considered to be in the positive direction, and movement to the left (or downward) is considered to be in the negative direction. The function s that gives the position (relative to the origin) of an object as a function of time t is called a position function. If, over a period of time t, the object changes its position by the amount s  st  t  st, then, by the familiar formula Rate 

distance time

the average velocity is Change in distance s  . Change in time t

EXAMPLE 9

Average velocity

Finding Average Velocity of a Falling Object

If a billiard ball is dropped from a height of 100 feet, its height s at time t is given by the position function s  16t 2  100

Position function

where s is measured in feet and t is measured in seconds. Find the average velocity over each of the following time intervals. a. 1, 2

b. 1, 1.5

c. 1, 1.1

Solution a. For the interval 1, 2 , the object falls from a height of s1  1612  100  84 feet to a height of s2  1622  100  36 feet. The average velocity is s 36  84 48    48 feet per second. t 21 1 b. For the interval 1, 1.5 , the object falls from a height of 84 feet to a height of 64 feet. The average velocity is s 64  84 20    40 feet per second. t 1.5  1 0.5 c. For the interval 1, 1.1 , the object falls from a height of 84 feet to a height of 80.64 feet. The average velocity is s 80.64  84 3.36    33.6 feet per second. t 1.1  1 0.1 .

Note that the average velocities are negative, indicating that the object is moving downward.

Try It

Exploration A

Exploration B

114

CHAPTER 2

Differentiation

s

Suppose that in Example 9 you wanted to find the instantaneous velocity (or simply the velocity) of the object when t  1. Just as you can approximate the slope of the tangent line by calculating the slope of the secant line, you can approximate the velocity at t  1 by calculating the average velocity over a small interval 1, 1  t (see Figure 2.20). By taking the limit as t approaches zero, you obtain the velocity when t  1. Try doing this—you will find that the velocity when t  1 is 32 feet per second. In general, if s  st is the position function for an object moving along a straight line, the velocity of the object at time t is

Tangent line

P

Secant line

t

t1 = 1

t2

The average velocity between t1 and t2 is the slope of the secant line, and the instantaneous velocity at t1 is the slope of the. tangent line. Figure 2.20

Animation

vt  lim

t→0

st  t  st  st. t

Velocity function

In other words, the velocity function is the derivative of the position function. Velocity can be negative, zero, or positive. The speed of an object is the absolute value of its velocity. Speed cannot be negative. The position of a free-falling object (neglecting air resistance) under the influence of gravity can be represented by the equation st 

1 2 gt  v0t  s0 2

Position function

where s0 is the initial height of the object, v0 is the initial velocity of the object, and g is the acceleration due to gravity. On Earth, the value of g is approximately 32 feet per second per second or 9.8 meters per second per second.

.

History EXAMPLE 10

32 ft

Using the Derivative to Find Velocity

At time t  0, a diver jumps from a platform diving board that is 32 feet above the water (see Figure 2.21). The position of the diver is given by st  16t2  16t  32

Position function

where s is measured in feet and t is measured in seconds. a. When does the diver hit the water? b. What is the diver’s velocity at impact? Solution Velocity is positive when an object is rising, . is negative when an object is falling. and Figure 2.21

Animation NOTE In Figure 2.21, note that the diver moves upward for the first halfsecond because the velocity is positive for 0 < t < 12. When the velocity is 0, . diver has reached the maximum the height of the dive.

a. To find the time t when the diver hits the water, let s  0 and solve for t. 16t 2  16t  32  0 16t  1t  2  0 t  1 or 2

Set position function equal to 0. Factor. Solve for t.

Because t ≥ 0, choose the positive value to conclude that the diver hits the water at t  2 seconds. b. The velocity at time t is given by the derivative st  32t  16. So, the velocity at time t  2 is s2  322  16  48 feet per second.

Try It

Exploration A

SECTION 2.2

115

Basic Differentiation Rules and Rates of Change

Exercises for Section 2.2 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, use the graph to estimate the slope of the tangent line to y  xn at the point 1, 1. Verify your answer analytically. To print an enlarged copy of the graph, select the MathGraph button. 1. (a) y  x12

29. y  30. y 

(b) y  x 3

y

Original Function

Rewrite

Differentiate

Simplify

x

x 4 x3

y

2

In Exercises 31–38, find the slope of the graph of the function at the given point. Use the derivative feature of a graphing utility to confirm your results.

2

1

1

(1, 1)

(1, 1)

Function

x 1

x

2

1

2. (a) y  x12

2

(b) y  x1

y 2

(1, 1)

1

2

x 1

3. y  8

4. f x  2

5. y 

6. y 

0,  12 

34. y  3x 3  6

2, 18 0, 1 5, 0 0, 0 , 1

37. f    4 sin 

In Exercises 39–52, find the derivative of the function. 39. f x  x 2  5  3x 2

x8

41. gt  t 2 

1 8. y  8 x

5 x 9. f x  

1 7 33. f x   2  5x 3

38. gt  2  3 cos t

2

In Exercises 3 –24, find the derivative of the function.

1 7. y  7 x

35, 2

36. f x  35  x2

3

x6

3 5t

35. y  2x  1

x 1

1, 3

2

(1, 1)

1

3 x2

32. f t  3 

y

2

31. f x 

Point

43. f x 

4 x 10. gx  

4 t3

40. f x  x 2  3x  3x2 42. f x  x 

x 3  3x 2  4 x2

44. hx 

1 x2

2x 2  3x  1 x

11. f x  x  1

12. gx  3x  1

45. y  xx 2  1

13. f t  2t 2  3t  6

14. y  t 2  2t  3

3 x 47. f x  x  6 

3 x  5 x 48. f x  

15. gx  x 

16. y  8  x

49. hs 

50. f t  t 23  t13  4

2

4x 3

17. st  t 3  2t  4 19. y 

18. f x  2x 3  x 2  3x

 sin  cos

2

22. y  5  sin x

1  3 sin x x

24. y 

5  2 cos x 2x3

In Exercises 25–30, complete the table. Original Function 25. y 

5 2x 2

2 26. y  2 3x 3 27. y  2x 3 28. y 

 3x 2

s 45



s 23

51. f x  6x  5 cos x

52. f x 

20. gt   cos t

1 21. y  x 2  2 cos x

23. y 

3

46. y  3x6x  5x 2

Rewrite

Differentiate

 3 cos x

In Exercises 53–56, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. Function

Simplify

2 3 x 

53. y  x 4  3x 2  2 54. y  x 3  x 55. f x 

2 4 3  x

56. y  x 2  2xx  1

Point 1, 0 1, 2

1, 2 1, 6

116

CHAPTER 2

Differentiation

In Exercises 57–62, determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.

In Exercises 71 and 72, the graphs of a function f and its derivative f are shown on the same set of coordinate axes. Label the graphs as f or f and write a short paragraph stating the criteria used in making the selection. To print an enlarged copy of the graph, select the MathGraph button.

57. y  x 4  8x 2  2 58. y  x 3  x 59. y 

1 x2

60. y  x 2  1 61. y  x  sin x,

Writing About Concepts (continued)

y

71.

0 f x < 2

2 1

3

62. y  3x  2 cos x, 0 f x < 2

x

1

In Exercises 63–66, find k such that the line is tangent to the graph of the function.

y

72.

x 3 2 1

2 1

1 2 3 4

1 2 3

2

Line

Function 63. f x  x  kx

y  4x  9

64. f x  k  x 2

y  4x  7

k 65. f x  x

3 y x3 4

66. f x  kx

yx4

2

Writing About Concepts 67. Use the graph of f to answer each question. To print an enlarged copy of the graph, select the MathGraph button. y

73. Sketch the graphs of y  x 2 and y  x 2  6x  5, and sketch the two lines that are tangent to both graphs. Find equations of these lines. 74. Show that the graphs of the two equations y  x and y  1x have tangent lines that are perpendicular to each other at their point of intersection. 75. Show that the graph of the function f x  3x  sin x  2 does not have a horizontal tangent line.

f

76. Show that the graph of the function

B C A

D

f x  x5  3x3  5x

E 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. 68. Sketch the graph of a function f such that f > 0 for all x and the rate of change of the function is decreasing. In Exercises 69 and 70, the relationship between f and g is given. Explain the relationship between f and g .

does not have a tangent line with a slope of 3. In Exercises 77 and 78, find an equation of the tangent line to the graph of the function f through the point x0, y0 not on the graph. To find the point of tangency x, y on the graph of f , solve the equation f x 

y0  y . x0  x

77. f x  x

x0, y0  4, 0

78. f x 

2 x

x0, y0  5, 0

79. Linear Approximation Use a graphing utility, with a square window setting, to zoom in on the graph of

69. gx  f x  6

f x  4  12 x 2

70. gx  5 f x

to approximate f 1. Use the derivative to find f 1. 80. Linear Approximation Use a graphing utility, with a square window setting, to zoom in on the graph of f x  4x  1 to approximate f 4. Use the derivative to find f 4.

SECTION 2.2

81. Linear Approximation Consider the function f x  x 3 2 with the solution point 4, 8. (a) Use a graphing utility to graph f. Use the zoom feature to obtain successive magnifications of the graph in the neighborhood of the point 4, 8. After zooming in a few times, the graph should appear nearly linear. Use the trace feature to determine the coordinates of a point near 4, 8. Find an equation of the secant line Sx through the two points.

117

Basic Differentiation Rules and Rates of Change

1 , x

91. f x 

1, 2

92. f x  sin x,



0, 6 

Vertical Motion In Exercises 93 and 94, use the position function st  16 t 2  v0 t  s0 for free-falling objects. 93. A silver dollar is dropped from the top of a building that is 1362 feet tall. (a) Determine the position and velocity functions for the coin.

(b) Find the equation of the line T x  f4x  4  f 4

(b) Determine the average velocity on the interval 1, 2 .

tangent to the graph of f passing through the given point. Why are the linear functions S and T nearly the same?

(d) Find the time required for the coin to reach ground level.

3

2

1

0.5

0.1

0

f 4  x T4  x x

0.1

0.5

1

2

3

f 4  x T4  x 82. Linear Approximation Repeat Exercise 81 for the function f x  x 3 where Tx is the line tangent to the graph at the point 1, 1. Explain why the accuracy of the linear approximation decreases more rapidly than in Exercise 81.

95. A projectile is shot upward from the surface of Earth with an initial velocity of 120 meters per second. What is its velocity after 5 seconds? After 10 seconds? 96. To estimate the height of a building, a stone is dropped from the top of the building into a pool of water at ground level. How high is the building if the splash is seen 6.8 seconds after the stone is dropped? Think About It In Exercises 97 and 98, the graph of a position function is shown. It represents the distance in miles that a person drives during a 10-minute trip to work. Make a sketch of the corresponding velocity function. 97.

True or False? In Exercises 83 – 88, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 83. If fx  gx, then f x  gx. 84. If f x  gx  c, then fx  gx. 86. If y  x, then dydx  1. 87. If gx  3 f x, then g x  3fx. 88. If f x  1x n, then f x  1nx n1. In Exercises 89–92, find the average rate of change of the function over the given interval. Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval. 89. f t  2t  7, 1, 2

90. f t  t2  3, 2, 2.1

10 8 6 4 2

98. (10, 6) (4, 2)

(6, 2) t

(0, 0) 2 4 6 8 10 Time (in minutes)

s 10 8 6 4 2

(10, 6) (6, 5) (8, 5) t

(0, 0) 2 4 6 8 10 Time (in minutes)

Think About It In Exercises 99 and 100, the graph of a velocity function is shown. It represents the velocity in miles per hour during a 10-minute drive to work. Make a sketch of the corresponding position function. 99.

v

Velocity (in mph)

85. If y   2, then dydx  2.

s

Distance (in miles)

x

Vertical Motion In Exercises 95 and 96, use the position function st  4.9t 2  v0 t  s0 for free-falling objects.

100.

60 50 40 30 20 10 t 2 4 6 8 10

Time (in minutes)

Velocity (in mph)

(d) Demonstrate the conclusion in part (c) by completing the table.

(e) Find the velocity of the coin at impact. 94. A ball is thrown straight down from the top of a 220-foot building with an initial velocity of 22 feet per second. What is its velocity after 3 seconds? What is its velocity after falling 108 feet?

Distance (in miles)

(c) Use a graphing utility to graph f and T on the same set of coordinate axes. Note that T is a good approximation of f when x is close to 4. What happens to the accuracy of the approximation as you move farther away from the point of tangency?

(c) Find the instantaneous velocities when t  1 and t  2.

v 60 50 40 30 20 10 t 2 4 6 8 10

Time (in minutes)

118

CHAPTER 2

Differentiation

101. Modeling Data The stopping distance of an automobile, on dry, level pavement, traveling at a speed v (kilometers per hour) is the distance R (meters) the car travels during the reaction time of the driver plus the distance B (meters) the car travels after the brakes are applied (see figure). The table shows the results of an experiment. Reaction time

Braking distance

R

B

Driver sees obstacle

Driver applies brakes

Car stops

40

60

80

100

Reaction Time Distance, R

8.3

16.7

25.0

33.3

41.7

Braking Time Distance, B

2.3

1,008,000  6.3Q Q

107. Writing The number of gallons N of regular unleaded gasoline sold by a gasoline station at a price of p dollars per gallon is given by N  f  p. (a) Describe the meaning of f1.479. (b) Is f1.479 usually positive or negative? Explain.

9.0

20.2

35.8

55.9

(a) Use the regression capabilities of a graphing utility to find a linear model for reaction time distance. (b) Use the regression capabilities of a graphing utility to find a quadratic model for braking distance. (c) Determine the polynomial giving the total stopping distance T. (d) Use a graphing utility to graph the functions R, B, and T in the same viewing window. (e) Find the derivative of T and the rates of change of the total stopping distance for v  40, v  80, and v  100. (f) Use the results of this exercise to draw conclusions about the total stopping distance as speed increases. 102. Fuel Cost A car is driven 15,000 miles a year and gets x miles per gallon. Assume that the average fuel cost is $1.55 per gallon. Find the annual cost of fuel C as a function of x and use this function to complete the table. 15

106. Inventory Management The annual inventory cost C for a manufacturer is

where Q is the order size when the inventory is replenished. Find the change in annual cost when Q is increased from 350 to 351, and compare this with the instantaneous rate of change when Q  350.

20

10

1 st   2at 2  c.

C

Speed, v

x

105. Velocity Verify that the average velocity over the time interval t0  t, t0  t is the same as the instantaneous velocity at t  t0 for the position function

20

25

30

35

40

C dC/dx Who would benefit more from a one-mile-per-gallon increase in fuel efficiency—the driver of a car that gets 15 miles per gallon or the driver of a car that gets 35 miles per gallon? Explain. 103. Volume The volume of a cube with sides of length s is given by V  s3. Find the rate of change of the volume with respect to s when s  4 centimeters. 104. Area The area of a square with sides of length s is given by A  s2. Find the rate of change of the area with respect to s when s  4 meters.

108. Newton’s Law of Cooling This law states that the rate of change of the temperature of an object is proportional to the difference between the object’s temperature T and the temperature Ta of the surrounding medium. Write an equation for this law. 109. Find an equation of the parabola y  ax2  bx  c that passes through 0, 1 and is tangent to the line y  x  1 at 1, 0. 110. Let a, b be an arbitrary point on the graph of y  1x, x > 0. Prove that the area of the triangle formed by the tangent line through a, b and the coordinate axes is 2. 111. Find the tangent line(s) to the curve y  x3  9x through the point 1, 9. 112. Find the equation(s) of the tangent line(s) to the parabola y  x 2 through the given point. (a) 0, a

(b) a, 0

Are there any restrictions on the constant a? In Exercises 113 and 114, find a and b such that f is differentiable everywhere.

x  b, cos x, 114. f x   ax  b, 113. f x 

ax3, 2

x f 2 x >2 x < 0 x v 0







115. Where are the functions f1x  sin x and f2x  sin x differentiable? 116. Prove that

d

cos x  sin x. dx

FOR FURTHER INFORMATION For a geometric interpretation of

the derivatives of trigonometric functions, see the article “Sines and Cosines of the Times” by Victor J. Katz in Math Horizons. MathArticle

SECTION 2.3

Section 2.3

Product and Quotient Rules and Higher-Order Derivatives

119

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.

The Product Rule In Section 2.2 you learned that the derivative of the sum of two functions is simply the sum of their derivatives. The rules for the derivatives of the product and quotient of two functions are not as simple.

THEOREM 2.7 NOTE A version of the Product Rule that some people prefer is d

f xg x  f xgx  f xgx. dx

.

The advantage of this form is that it generalizes easily to products involving three or more factors.

The Product Rule

The product of two differentiable functions f and g is itself differentiable. Moreover, the derivative of fg is the first function times the derivative of the second, plus the second function times the derivative of the first. d

f xgx  f xgx  gx fx dx

Video Proof Some mathematical proofs, such as the proof of the Sum Rule, are straightforward. Others involve clever steps that may appear unmotivated to a reader. This proof involves such a step—subtracting and adding the same quantity—which is shown in color.

f x  xgx  x  f xgx d

f xgx  lim dx x→ 0 x f x  xgx  x  f x  xgx  f x  xgx  f xgx x gx  x  gx f x  x  f x  lim f x  x  gx x→ 0 x x gx  x  gx f x  x  f x  lim gx  lim f x  x x→0 x→0 x x gx  x  gx f x   x  f x  lim f x  x  lim  lim gx  lim x→0 x→0 x→0 x→0 x x  f xgx  gxfx  lim

x→0

 









Note that lim f x  x  f x because f is given to be differentiable and therefore THE PRODUCT RULE When Leibniz originally wrote a formula for the Product Rule, he was motivated by the expression

x  dx y  dy  xy from which he subtracted dx dy (as being negligible) and obtained the differential form x dy  y dx. This derivation resulted in the traditional form of the Product Rule. (Source:The History of Mathematics by David M. Burton)

x→ 0

is continuous. The Product Rule can be extended to cover products involving more than two factors. For example, if f, g, and h are differentiable functions of x, then d

f xgxhx  fxgxhx  f xgxhx  f xgxhx. dx For instance, the derivative of y  x2 sin x cos x is dy  2x sin x cos x  x2 cos x cos x  x2 sin xsin x dx  2x sin x cos x  x2cos2x  sin2x.

120

CHAPTER 2

Differentiation

The derivative of a product of two functions is not (in general) given by the product of the derivatives of the two functions. To see this, try comparing the product of the derivatives of f x  3x  2x 2 and gx  5  4x with the derivative in Example 1.

Using the Product Rule

EXAMPLE 1

Find the derivative of hx  3x  2x25  4x. Solution Derivative of second

First

hx  3x  2x2

Derivative of first

Second

d d

5  4x  5  4x 3x  2x2 dx dx

Apply Product Rule.

 3x  2x24  5  4x3  4x  12x  8x2  15  8x  16x2  24x2  4x  15 In Example 1, you have the option of finding the derivative with or without the Product Rule. To find the derivative without the Product Rule, you can write .

Dx 3x  2x 25  4x  Dx 8x 3  2x 2  15x  24x 2  4x  15.

Try It

Exploration A

In the next example, you must use the Product Rule.

Using the Product Rule

EXAMPLE 2

Find the derivative of y  3x2 sin x. Solution d d d

3x2 sin x  3x2 sin x  sin x 3x2 dx dx dx  3x2 cos x  sin x6x  3x2 cos x  6x sin x  3xx cos x  2 sin x

.

Try It .

Exploration A

Technology

The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph

Using the Product Rule

EXAMPLE 3

Find the derivative of y  2x cos x  2 sin x. Solution Product Rule

NOTE In Example 3, notice that you use the Product Rule when both factors of the product are variable, and you use the. Constant Multiple Rule when one of the factors is a constant.

Apply Product Rule.





Constant Multiple Rule





dy d d d

cos x  cos x

2x  2 sin x  2x dx dx dx dx  2xsin x  cos x2  2cos x  2x sin x

Try It

Exploration A

Technology

SECTION 2.3

121

Product and Quotient Rules and Higher-Order Derivatives

The Quotient Rule THEOREM 2.8

The Quotient Rule

The quotient fg of two differentiable functions f and g is itself differentiable at all values of x for which gx  0. Moreover, the derivative of fg is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. d f x gx fx  f xgx  , dx gx

gx 2

 

.

gx  0

Video Proof As with the proof of Theorem 2.7, the key to this proof is subtracting and adding the same quantity. f x f x  x  d f x gx  x gx  lim Definition of derivative x→ 0 dx gx x gx f x  x  f xgx  x  lim x→ 0 xgxgx  x gxf x  x  f xgx  f xgx  f xgx  x  lim x→ 0 xgxg x  x gx f x   x  f x f x gx  x  gx  lim lim x→ 0 x→ 0 x x  lim gxgx  x

 

x→ 0

TECHNOLOGY A graphing utility can be used to compare the graph of a function with the graph of its derivative. For instance, in Figure 2.22, the graph of the function in Example 4 appears to have two points that have horizontal tangent lines. What are the values of y at these two points? y′ =

−5x 2 + 4x + 5 (x 2 + 1) 2

6

.

gx fx  f xgx 

gx 2







x→0

Note that lim gx  x  gx because g is given to be differentiable and therefore x→ 0 is continuous. EXAMPLE 4

Using the Quotient Rule 5x  2 . x2  1

d d

5x  2  5x  2 x 2  1 dx dx x 2  12 x 2  15  5x  22x  x 2  1 2

d 5x  2  dx x 2  1



−4



x 2  1

Apply Quotient Rule.

5x 2  5  10x 2  4x x 2  1 2 5x 2  4x  5  x 2  12 

Graphical comparison of a function and its derivative Figure 2.22

Try It .

f x  x  f x gx  x  gx  f x lim x→0 x x lim gxgx  x

Solution

8

5x − 2 x2 + 1

x→0

Find the derivative of y 

−7

y=





gx lim

Exploration A

Exploration B

The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph

122

CHAPTER 2

Differentiation

Note the use of parentheses in Example 4. A liberal use of parentheses is recommended for all types of differentiation problems. For instance, with the Quotient Rule, it is a good idea to enclose all factors and derivatives in parentheses, and to pay special attention to the subtraction required in the numerator. When differentiation rules were introduced in the preceding section, the need for rewriting before differentiating was emphasized. The next example illustrates this point with the Quotient Rule. EXAMPLE 5

Rewriting Before Differentiating

Find an equation of the tangent line to the graph of f x 

3  1x at 1, 1. x5

Solution Begin by rewriting the function. 3  1x x5 1 x 3 x  xx  5

f x 

f(x) =



1 3− x x+5

y



Multiply numerator and denominator by x.

3x  1 x 2  5x x 2  5x3  3x  12x  5 f  x  x 2  5x2

5



4 3

y=1

(−1, 1) −7 −6 −5 − 4 − 3 −2 −1

Write original function.

3x 2  15x  6x 2  13x  5 x 2  5x 2 3x 2  2x  5  x 2  5x2

2

3

−2 −3 −4 −5

The line y  1 is tangent to the graph of f .x at the point  1, 1. Figure 2.23

.

Quotient Rule



x 1

Rewrite.

Simplify.

To find the slope at 1, 1, evaluate f  1. f  1  0

Slope of graph at 1, 1

Then, using the point-slope form of the equation of a line, you can determine that the equation of the tangent line at 1, 1 is y  1. See Figure 2.23.

Try It

Exploration A

The editable graph feature below allows you to edit the graph of a function. Editable Graph Not every quotient needs to be differentiated by the Quotient Rule. For example, each quotient in the next example can be considered as the product of a constant times a function of x. In such cases it is more convenient to use the Constant Multiple Rule. EXAMPLE 6

Using the Constant Multiple Rule

Original Function

NOTE To see the benefit of using the Constant Multiple Rule for some quotients, try using the Quotient Rule to differentiate the functions in Example . 6—you should obtain the same results, but with more work.

Rewrite

Differentiate

Simplify

a. y 

x 2  3x 6

1 y  x 2  3x 6

1 y  2x  3 6

y 

b. y 

5x 4 8

5 y  x4 8

5 y  4x 3 8

5 y  x 3 2

3 y   3  2x 7 9 y  x2 5

3 y   2 7 9 y  2x3 5

y 

33x  2x 2 7x 9 d. y  2 5x c. y 

Try It

Exploration A

2x  3 6

6 7

y  

18 5x3

SECTION 2.3

Product and Quotient Rules and Higher-Order Derivatives

123

In Section 2.2, the Power Rule was proved only for the case where the exponent n is a positive integer greater than 1. The next example extends the proof to include negative integer exponents.

Proof of the Power Rule (Negative Integer Exponents)

EXAMPLE 7

If n is a negative integer, there exists a positive integer k such that n  k. So, by the Quotient Rule, you can write

 

d n d 1

x  dx dx x k x k 0  1kx k1  x k2

Quotient Rule and Power Rule

0  kx k1 x 2k  kxk1  nx n1. 

n  k

So, the Power Rule Dx x n  nx n1 .

Power Rule

is valid for any integer. In Exercise 75 in Section 2.5, you are asked to prove the case for which n is any rational number.

Try It

Exploration A

Derivatives of Trigonometric Functions Knowing the derivatives of the sine and cosine functions, you can use the Quotient Rule to find the derivatives of the four remaining trigonometric functions. THEOREM 2.9

.

Derivatives of Trigonometric Functions

d

tan x  sec 2 x dx d

sec x  sec x tan x dx

d

cot x  csc2x dx d

csc x  csc x cot x dx

Video Proof Considering tan x  sin xcos x and applying the Quotient Rule, you obtain d cos xcos x  sin xsin x

tan x  dx cos 2 x cos2 x  sin2 x  cos2 x 

Apply Quotient Rule.

1 cos2 x

 sec2 x. The proofs of the other three parts of the theorem are left as an exercise (see Exercise 89).

124

CHAPTER 2

Differentiation

EXAMPLE 8 NOTE Because of trigonometric identities, the derivative of a trigonometric function can take many forms. This presents a challenge when you are trying to match your answers to those given in . the back of the text.

Differentiating Trigonometric Functions

Function

Derivative

a. y  x  tan x

dy  1  sec2 x dx

b. y  x sec x

y  xsec x tan x  sec x1  sec x1  x tan x

Try It EXAMPLE 9

Exploration A

Open Exploration

Different Forms of a Derivative

Differentiate both forms of y 

1  cos x  csc x  cot x. sin x

Solution 1  cos x sin x sin xsin x  1  cos xcos x y  sin2 x sin2 x  cos2 x  cos x  sin2 x

First form: y 



1  cos x sin2 x

Second form: y  csc x  cot x y  csc x cot x  csc2 x To show that the two derivatives are equal, you can write

.

1 1 1  cos x   sin 2 x sin 2 x sin x



x cos sin x 

 csc 2 x  csc x cot x.

Try It

Exploration A

Technology

The summary below shows that much of the work in obtaining a simplified form of a derivative occurs after differentiating. Note that two characteristics of a simplified form are the absence of negative exponents and the combining of like terms. f x After Differentiating

f x After Simplifying

Example 1

3x  2x24  5  4x3  4x

24x2  4x  15

Example 3

2xsin x  cos x2  2cos x

2x sin x

Example 4

x2  15  5x  22x x2  1 2

5x2  4x  5 x2  12

Example 5

x2  5x3  3x  12x  5 x2  5x2

3x2  2x  5 x2  5x2

Example 9

sin xsin x  1  cos xcos x sin2 x

1  cos x sin2 x

SECTION 2.3

Product and Quotient Rules and Higher-Order Derivatives

125

Higher-Order Derivatives Just as you can obtain a velocity function by differentiating a position function, you can obtain an acceleration function by differentiating a velocity function. Another way of looking at this is that you can obtain an acceleration function by differentiating a position function twice. st vt  st at  vt  s t NOTE: The second derivative of f is the derivative of the first derivative of f.

Position function Velocity function Acceleration function

The function given by at is the second derivative of st and is denoted by s t. The second derivative is an example of a higher-order derivative. You can define derivatives of any positive integer order. For instance, the third derivative is the derivative of the second derivative. Higher-order derivatives are denoted as follows.

y,

fx,

dy , dx d 2y , dx 2 d 3y , dx 3

Fourth derivative: y 4,

f 4x,

d4y , dx 4

d4

f x , dx 4

Dx4 y

f nx,

dny , dx n

dn

f x , dx n

Dxn y

y,

fx,

Second derivative: y,

f  x,

First derivative:

Third derivative:

⯗ nth derivative:

EXAMPLE 10

yn,

d

f x , dx d2

f x , dx 2 d3

f x , dx 3

Dx y Dx2 y Dx3 y

Finding the Acceleration Due to Gravity

Because the moon has no atmosphere, a falling object on the moon encounters no air resistance. In 1971, astronaut David Scott demonstrated that a feather and a hammer fall at the same rate on the moon. The position function for each of these falling objects is given by st  0.81t 2  2 where st is the height in meters and t is the time in seconds. What is the ratio of Earth’s gravitational force to the moon’s? THE MOON The moon’s mass is 7.349  1022 kilograms, and Earth’s mass is 5.976  1024 kilograms. The moon’s radius is 1737 kilometers, and Earth’s radius is 6378 kilometers. Because the gravitational force on the surface of a planet is directly proportional to its mass and inversely proportional to the square of its radius, the ratio of the gravitational force on Earth to the gravitational force on the moon is

..

5.976  102463782  6.03. 7.349  102217372

Video

Solution To find the acceleration, differentiate the position function twice. st  0.81t 2  2 st  1.62t s t  1.62

Position function Velocity function Acceleration function

So, the acceleration due to gravity on the moon is 1.62 meters per second per second. Because the acceleration due to gravity on Earth is 9.8 meters per second per second, the ratio of Earth’s gravitational force to the moon’s is Earth’s gravitational force 9.8  Moon’s gravitational force 1.62  6.05.

Try It

Exploration A

126

CHAPTER 2

Differentiation

Exercises for Section 2.3 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–6, use the Product Rule to differentiate the function. 1. gx  x 2  1x 2  2x

2. f x  6x  5x 3  2

3. ht 

4. gs  s4  s 

tt  4

3 

2

2

5. f x  x 3 cos x

6. gx  x sin x

In Exercises 7–12, use the Quotient Rule to differentiate the function. 7. f x  9. hx  11. gx 

x x2  1

8. gt 

3 x 

x3

t2  2 2t  7

s s  1 cos t 12. f t  3 t 10. hs 

1

sin x x2

13. f x  

x3

Value of c  3x

2x 2

 3x  5

14. f x  x 2  2x  1x 3  1 15. f x 

x2  4 x3

4 x3





28. f x  x 4 1 

2x  5 x 31. hs  s3  22

2 x1

3 xx  3 30. f x  

29. f x 

32. hx  x2  12

1 x 33. f x  x3 2

34. gx  x 2

2x  x 1 1

35. f x  3x3  4xx  5x  1 36. f x  x 2  xx 2  1x 2  x  1 37. f x 

x2  c2 , c is a constant x2  c2

38. f x 

c2  x 2 , c is a constant c2  x 2

39. f t  t 2 sin t

40. f      1 cos

c1

cos t 41. f t  t

42. f x 

c1

43. f x  x  tan x

44. y  x  cot x

4 t  8 sec t 45. gt  

46. hs 

c0

sin x x

1  10 csc s s sec x 48. y  x

x1 16. f x  x1

c2

17. f x  x cos x

 c 4

47. y 

 c 6

49. y  csc x  sin x

50. y  x sin x  cos x

sin x 18. f x  x

51. f x  x 2 tan x

52. f x  sin x cos x

Rewrite

Differentiate

4x 32 x

24. y 

3x 2  5 7

3  2x  x 2 x2  1

2

2

57. g  

1  sin

58. f   

sin

1  cos

In Exercises 59–62, evaluate the derivative of the function at the given point. Use a graphing utility to verify your result. Function 59. y 

1  csc x 1  csc x

60. f x  tan x cot x

In Exercises 25–38, find the derivative of the algebraic function. 25. f x 

xx  122x  5 x x3 f x   x  x  1 x 1  2

56.

7 21. y  3 3x

23. y 

54. h   5 sec  tan

cos x

55. gx 

5x 2  3 20. y  4

4 5x 2

x2

In Exercises 55–58, use a computer algebra system to differentiate the function.

Simplify

x 2  2x 19. y  3

22. y 

31  sin x 2 cos x

53. y  2x sin x 

In Exercises 19–24, complete the table without using the Quotient Rule. Function



In Exercises 39–54, find the derivative of the trigonometric function.

In Exercises 13–18, find f x and f c. Function



27. f x  x 1 

26. f x 

x 3  3x  2 x2  1

61. ht 

sec t t

62. f x  sin xsin x  cos x

Point

6 , 3 1, 1

,  1  4 , 1

SECTION 2.3

127

Product and Quotient Rules and Higher-Order Derivatives

In Exercises 63–68, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results.

In Exercises 79 and 80, verify that f x  g x, and explain the relationship between f and g. 79. f x 

3x 5x  4 , gx  x2 x2

63. f x  x3  3x  1x  2,

80. f x 

sin x  3x sin x  2x , gx  x x

64. f x  x  1x 2  2, x , 2, 2 x1

65. f x 

0, 2 66. f x 

4 , 1

67. f x  tan x,

1, 3 x  1 , x  1

2, 31 3 , 2

68. f x  sec x,

In Exercises 81 and 82, use the graphs of f and g. Let f x p x  f xgx and qx  . gx

Famous Curves In Exercises 69–72, find an equation of the tangent line to the graph at the given point. (The graphs in Exercises 69 and 70 are called witches of Agnesi. The graphs in Exercises 71 and 72 are called serpentines.) y

69. 6 4

f (x) =

6

2

x 2

2

y

y

72. 16x f(x) = 2 x + 16

4 3 2 1

4

(2, ) 4 5

x 4

(2,  ) 8 5

x

8

1 2 3 4

f (x) =

8

4x x2 + 6

In Exercises 73–76, determine the point(s) at which the graph of the function has a horizontal tangent line. 73. f x  75. f x 

x1 4x  2 x2

74. f x  76. f x 

x2 x2  1 x4 x2  7

77. Tangent Lines

Find equations of the tangent lines to the x1 graph of f x  that are parallel to the line 2y  x  6. x1 Then graph the function and the tangent lines.

78. Tangent Lines

g

4

g

2

2

4

6

8

x 2

10

2

4

6

8

10

84. Volume The radius of a right circular cylinder is given by 1 t  2 and its height is 2 t, where t is time in seconds and the dimensions are in inches. Find the rate of change of the volume with respect to time. 85. Inventory Replenishment The ordering and transportation cost C for the components used in manufacturing a product is C  100

x2

f

83. Area The length of a rectangle is given by 2t  1 and its height is t, where t is time in seconds and the dimensions are in centimeters. Find the rate of change of the area with respect to time.

4

2

8

8

x 2

2

71.

10

f

8

2

3 2

4

y

10

4

(3, )

4

(b) Find q7.

y

27 x2 + 9

x 2

(b) Find q4.

6

(2, 1) 4

82. (a) Find p4.

y

70. 8 f(x) = 2 x +4

81. (a) Find p1.

Find equations of the tangent lines to the x graph of f x  that pass through the point 1, 5. x1 Then graph the function and the tangent lines.

x  , 200 x x  30  2

x v 1

where C is measured in thousands of dollars and x is the order size in hundreds. Find the rate of change of C with respect to x when (a) x  10, (b) x  15, and (c) x  20. What do these rates of change imply about increasing order size? 86. Boyle’s Law This law states that if the temperature of a gas remains constant, its pressure is inversely proportional to its volume. Use the derivative to show that the rate of change of the pressure is inversely proportional to the square of the volume. 87. Population Growth A population of 500 bacteria is introduced into a culture and grows in number according to the equation



Pt  500 1 

4t 50  t 2



where t is measured in hours. Find the rate at which the population is growing when t  2.

128

CHAPTER 2

Differentiation

88. Gravitational Force Newton’s Law of Universal Gravitation states that the force F between two masses, m1 and m2, is

F

Gm1m2 d2

93. f x  4x32

89. Prove the following differentiation rules. (a)

d

sec x  sec x tan x dx

(c)

d

cot x  csc2 x dx

(b)

d

csc x  csc x cot x dx

94. f x  x  32x2

x x1

95. f x 

where G is a constant and d is the distance between the masses. Find an equation that gives an instantaneous rate of change of F with respect to d. (Assume m1 and m2 represent moving points.)

96. f x 

97. f x  3 sin x

x 2  2x  1 x

98. f x  sec x

In Exercises 99–102, find the given higher-order derivative. 99. fx  x 2,

2 100. f  x  2  , x

f  x

101. fx  2x,

f 4x

fx

102. f 4x  2x  1,

f 6x

Writing About Concepts

90. Rate of Change Determine whether there exist any values of x in the interval 0, 2  such that the rate of change of f x  sec x and the rate of change of gx  csc x are equal. 91. Modeling Data The table shows the numbers n (in thousands) of motor homes sold in the United States and the retail values v (in billions of dollars) of these motor homes for the years 1996 through 2001. The year is represented by t, with t  6 corresponding to 1996. (Source: Recreation Vehicle Industry Association) Year, t

In Exercises 93–98, find the second derivative of the function.

6

7

8

9

10

11

n

247.5

254.5

292.7

321.2

300.1

256.8

v

6.3

6.9

8.4

10.4

9.5

8.6

(a) Use a graphing utility to find cubic models for the number of motor homes sold nt and the total retail value vt of the motor homes. (b) Graph each model found in part (a). (c) Find A  vtnt, then graph A. What does this function represent?

103. Sketch the graph of a differentiable function f such that f 2  0, f < 0 for  < x < 2, and f > 0 for 2 < x < . 104. Sketch the graph of a differentiable function f such that f > 0 and f < 0 for all real numbers x. In Exercises 105–108, use the given information to find f2. g2  3

and

h2  1

g2  2 h 2  4

and

105. f x  2gx  hx

106. f x  4  hx

gx 107. f x  hx

108. f x  gxhx

In Exercises 109 and 110, the graphs of f, f , and f are shown on the same set of coordinate axes. Which is which? Explain your reasoning. To print an enlarged copy of the graph, select the MathGraph button. y

109.

y

110.

2

(d) Interpret At in the context of these data. 92. Satellites When satellites observe Earth, they can scan only part of Earth’s surface. Some satellites have sensors that can measure the angle shown in the figure. Let h represent the satellite’s distance from Earth’s surface and let r represent Earth’s radius.

r

V r

h

2

(b) Find the rate at which h is changing with respect to when

 30. (Assume r  3960 miles.)

1

2

3

1 2

In Exercises 111–114, the graph of f is shown. Sketch the graphs of f and f . To print an enlarged copy of the graph, select the MathGraph button. y

111. (a) Show that h  r csc  1.

x

x

1

4

y

112. 8

f

4

2 x 4 2 2

4

x 8

4

f 4

SECTION 2.3

113.

114.

y

f

4 3 2 1

(a) Use the Product Rule to generate rules for finding f  x, fx, and f 4x.

f

(b) Use the results in part (a) to write a general rule for f nx.

2

U 2

3U 2

1 1 2

4

U 2

U

3U 2

2U

x

115. Acceleration The velocity of an object in meters per second is vt  36  t 2, 0 f t f 6. Find the velocity and acceleration of the object when t  3. What can be said about the speed of the object when the velocity and acceleration have opposite signs? 116. Acceleration An automobile’s velocity starting from rest is 100t v t  2t  15

117. Stopping Distance A car is traveling at a rate of 66 feet per second (45 miles per hour) when the brakes are applied. The position function for the car is st  8.25t 2  66t, where s is measured in feet and t is measured in seconds. Use this function to complete the table, and find the average velocity during each time interval. 0

1

2

3

4

In Exercises 123 and 124, find the derivatives of the function f for n  1, 2, 3, and 4. Use the results to write a general rule for f x in terms of n. 123. f x  x n sin x

124. f x 

cos x xn

Differential Equations In Exercises 125–128, verify that the function satisfies the differential equation. Differential Equation

1 125. y  , x > 0 x

x3 y  2x2 y  0

126. y  2x3  6x  10

y  xy  2y  24x2

127. y  2 sin x  3

y  y  3

128. y  3 cos x  sin x

y  y  0

True or False? In Exercises 129–134, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 129. If y  f xgx, then dydx  fxgx.

st

130. If y  x  1x  2x  3x  4, then d 5ydx 5  0.

vt

131. If fc and gc are zero and hx  f xgx, then hc  0.

at

132. If f x is an nth-degree polynomial, then f n1x  0.

118. Particle Motion The figure shows the graphs of the position, velocity, and acceleration functions of a particle. y 16 12 8 4 1

133. The second derivative represents the rate of change of the first derivative. 134. If the velocity of an object is constant, then its acceleration is zero. 135. Find a second-degree polynomial f x  ax2  bx  c such that its graph has a tangent line with slope 10 at the point 2, 7 and an x-intercept at 1, 0.

t 1

122. Finding a Pattern Develop a general rule for x f x n where f is a differentiable function of x.

Function

where v is measured in feet per second. Find the acceleration at (a) 5 seconds, (b) 10 seconds, and (c) 20 seconds.

t

129

121. Finding a Pattern Consider the function f x  gxhx.

y 4

x

Product and Quotient Rules and Higher-Order Derivatives

4 5 6 7

136. Consider the third-degree polynomial f x  ax3  bx2  cx  d, a  0.

(a) Copy the graphs of the functions shown. Identify each graph. Explain your reasoning. To print an enlarged copy of the graph, select the MathGraph button. (b) On your sketch, identify when the particle speeds up and when it slows down. Explain your reasoning. Finding a Pattern In Exercises 119 and 120, develop a general rule for f nx given f x. 119. f x  x n

1 120. f x  x

Determine conditions for a, b, c, and d if the graph of f has (a) no horizontal tangents, (b) exactly one horizontal tangent, and (c) exactly two horizontal tangents. Give an example for each case.



137. Find the derivative of f x  x x . Does f  0 exist? 138. Think About It Let f and g be functions whose first and second derivatives exist on an interval I. Which of the following formulas is (are) true? (a) fg  f g   fg  fg (b) fg  f g   fg

130

CHAPTER 2

Differentiation

Section 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.

The Chain Rule This text has yet to discuss one of the most powerful differentiation rules—the Chain Rule. This rule deals with composite functions and adds a surprising versatility to the rules discussed in the two previous sections. For example, compare the functions shown below. Those on the left can be differentiated without the Chain Rule, and those on the right are best done with the Chain Rule.

.

Without the Chain Rule

With the Chain Rule

y  x2  1 y  sin x y  3x  2

y  x 2  1 y  sin 6x y  3x  25

y  x  tan x

y  x  tan x2

Basically, the Chain Rule states that if y changes dydu times as fast as u, and u changes dudx times as fast as x, then y changes dydududx times as fast as x.

Video 3

The Derivative of a Composite Function

EXAMPLE 1 Gear 2

Gear 1 Axle 2 Gear 4 1 Axle 1

dy dy  dx du

Gear 3 1

Animation

Axle 3

Solution Because the circumference of the second gear is three times that of the first, the first axle must make three revolutions to turn the second axle once. Similarly, the second axle must make two revolutions to turn the third axle once, and you can write dy 3 du

and

du  2. dx

Combining these two results, you know that the first axle must make six revolutions to turn the third axle once. So, you can write dy  dx  

.

du

 dx .

2

Axle 1: y revolutions per minute Axle 2: u revolutions per minute . 3: x revolutions per minute Axle Figure 2.24

A set of gears is constructed, as shown in Figure 2.24, such that the second and third gears are on the same axle. As the first axle revolves, it drives the second axle, which in turn drives the third axle. Let y, u, and x represent the numbers of revolutions per minute of the first, second, and third axles. Find dydu, dudx, and dydx, and show that

Rate of change of first axle with respect to second axle

dy du



Rate of change of second axle with respect to third axle

du

 dx  3  2  6

Rate of change of first axle with respect to third axle

.

In other words, the rate of change of y with respect to x is the product of the rate of change of y with respect to u and the rate of change of u with respect to x.

Try It

Exploration A

SECTION 2.4

E X P L O R AT I O N Using the Chain Rule Each of the following functions can be differentiated using rules that you studied in Sections 2.2 and 2.3. For each function, find the derivative using those rules. Then find the derivative using the Chain Rule. Compare your results. Which method is simpler? 2 a. 3x  1

b. x  23 c. sin 2x

The Chain Rule

131

Example 1 illustrates a simple case of the Chain Rule. The general rule is stated below.

THEOREM 2.10

The Chain Rule

If y  f u is a differentiable function of u and u  gx is a differentiable function of x, then y  f gx is a differentiable function of x and dy dy  dx du

du

 dx

or, equivalently, d

f gx  fgxg x. dx Proof Let hx  f gx. Then, using the alternative form of the derivative, you need to show that, for x  c, hc  fgcgc. An important consideration in this proof is the behavior of g as x approaches c. A problem occurs if there are values of x, other than c, such that gx  gc. Appendix A shows how to use the differentiability of f and g to overcome this problem. For now, assume that gx  gc for values of x other than c. In the proofs of the Product Rule and the Quotient Rule, the same quantity was added and subtracted to obtain the desired form. This proof uses a similar technique—multiplying and dividing by the same (nonzero) quantity. Note that because g is differentiable, it is also continuous, and it follows that gx → gc as x → c. f gx  f gc x→c xc f gx  f gc  lim x→c gx  gc

hc  lim



f gx  f gc x→c gx  gc  fgcgc



 lim



gx  gc , gx  gc xc



 lim gxx  gcc x→c

When applying the Chain Rule, it is helpful to think of the composite function f  g as having two parts—an inner part and an outer part. Outer function

y  f gx  f u Inner function

The derivative of y  f u is the derivative of the outer function (at the inner function u) times the derivative of the inner function. y  fu  u

132

CHAPTER 2

Differentiation

Decomposition of a Composite Function

EXAMPLE 2 y  f gx

a. y 

1 x1

b. y  sin 2x c. y  3x2  x  1 d. y  tan 2 x

.

Try It

You could also solve the problem in Example 3 without using the Chain Rule by observing that y  x 6  3x 4  3x 2  1 and

. y  6x5  12x3  6x. Verify that this is the same as the derivative in Example 3. Which method would you use to find

.

d 2 x  150? dx

y  f u

ux1

y

u  2x u  3x 2  x  1 u  tan x

y  sin u y  u

1 u

y  u2

Exploration A Using the Chain Rule

EXAMPLE 3 STUDY TIP

u  gx

Find dydx for y  x 2  13. Solution For this function, you can consider the inside function to be u  x 2  1. By the Chain Rule, you obtain dy  3x 2  122x  6xx 2  1 2. dx dy du

Try It

du dx

Exploration A

Exploration B

The editable graph feature below allows you to edit the graph of a function and its derivative. Editable Graph

The General Power Rule The function in Example 3 is an example of one of the most common types of composite functions, y  ux n. The rule for differentiating such functions is called the General Power Rule, and it is a special case of the Chain Rule. THEOREM 2.11

The General Power Rule

If y  ux where u is a differentiable function of x and n is a rational number, then n,

du dy  n ux n1 dx dx or, equivalently, d n

u  nu n1 u. dx

.

Video Proof

Because y  un, you apply the Chain Rule to obtain

 dudx

dy dy  dx du 

d n du

u . du dx

By the (Simple) Power Rule in Section 2.2, you have Du un  nu n1, and it follows that dy du  n ux n1 . dx dx

SECTION 2.4

.

133

Applying the General Power Rule

EXAMPLE 4

Video

The Chain Rule

Find the derivative of f x  3x  2x 23. Solution Let u  3x  2x2. Then f x  3x  2x23  u3 and, by the General Power Rule, the derivative is u

un1

n

d

3x  2x 2 dx  33x  2x 2 23  4x.

fx  33x  2x 22

.

Try It

.

Apply General Power Rule. Differentiate 3x  2x 2.

Exploration A

The editable graph feature below allows you to edit the graph of a function. f(x) =

3

(x 2 − 1) 2

Editable Graph y

Differentiating Functions Involving Radicals

EXAMPLE 5 2

3 Find all points on the graph of f x   x 2  1 2 for which fx  0 and those for which fx does not exist.

Solution Begin by rewriting the function as x

−2

−1

1

2

−1

f x  x 2  123. Then, applying the General Power Rule (with u  x2  1 produces

fx  f ′(x) =

4x 3 3 x2 − 1

The derivative of f is 0 at x  0 and is

.. undefined at x  ± 1. Figure 2.25

Editable Graph



u

un1

n −2

2 2 x  113 2x 3

Apply General Power Rule.

4x . 3 x 1

Write in radical form.

3 2 

So, fx  0 when x  0 and fx does not exist when x  ± 1, as shown in Figure 2.25.

Try It EXAMPLE 6

Exploration A Differentiating Quotients with Constant Numerators

Differentiate gt 

7 . 2t  3 2

Solution Begin by rewriting the function as gt  72t  32. NOTE Try differentiating the function in Example 6 using the Quotient Rule. You should obtain the same result, but using the Quotient Rule is less efficient than using the General Power Rule.

Then, applying the General Power Rule produces un1

n

u

gt  722t  332

Apply General Power Rule.

Constant Multiple Rule

.

 282t  33 28  . 2t  33

Try It

Exploration A

Simplify. Write with positive exponent.

Exploration B

134

CHAPTER 2

Differentiation

Simplifying Derivatives The next three examples illustrate some techniques for simplifying the “raw derivatives” of functions involving products, quotients, and composites.

Simplifying by Factoring Out the Least Powers

EXAMPLE 7

f x  x21  x2  x 21  x 212 d d fx  x 2 1  x 212  1  x 212 x 2 dx dx  x2

 12 1  x 

2 12

 x 31  x 212  2x1  x 212  x1  x 212 x 21  21  x 2 x2  3x 2  1  x 2

.

Try It

f x 

General Power Rule

Factor. Simplify.

Simplifying the Derivative of a Quotient x

Original function

3 x2  4 

x x 2  413 x 2  4131  x13x 2  4232x fx  x 2  423 1 3x 2  4  2x 21  x 2  423 3 x 2  423 x 2  12  3x2  443



Try It



Rewrite.

Quotient Rule

Factor.

Simplify.

Exploration A Simplifying the Derivative of a Power

EXAMPLE 9

3xx  31

2

Original function

2

u

un1

n

3xx  31 dxd  3xx  31 23x  1 x  33  3x  12x   x  3  x  3

y  2

Product Rule

Simplify.



y

Rewrite.

Exploration A

EXAMPLE 8 TECHNOLOGY Symbolic differentiation utilities are capable of differentiating very complicated functions. Often, however, the result is given in unsimplified form. If you have access to such a utility, use it to find the derivatives of the functions given in Examples 7, 8, and 9. Then compare the results with those given .on this page.



2x  1  x 2122x

Original function

2

2

General Power Rule

2

2

2

23x  13x 2  9  6x 2  2x x 2  33 23x  13x 2  2x  9  x 2  33 

.

2

Try It

Exploration A

Quotient Rule

Multiply.

Simplify.

Open Exploration

SECTION 2.4

The Chain Rule

135

Trigonometric Functions and the Chain Rule The “Chain Rule versions” of the derivatives of the six trigonometric functions are as shown.

.

d

sin u  cos u u dx

d

cos u   sin u u dx

d

tan u  sec 2 u u dx d

sec u  sec u tan u u dx

d

cot u   csc 2 u u dx d

csc u   csc u cot u u dx

Technology EXAMPLE 10

Applying the Chain Rule to Trigonometric Functions

u

cos u

d

2x  cos 2x2  2 cos 2x dx y  sinx  1 y  3 sec 2 3x y  cos 2x

a. y  sin 2x .

u

b. y  cosx  1 c. y  tan 3x

Try It

Exploration A

Be sure that you understand the mathematical conventions regarding parentheses and trigonometric functions. For instance, in Example 10(a), sin 2x is written to mean sin2x. EXAMPLE 11 a. b. c. d. .

Parentheses and Trigonometric Functions

y  cos 3x 2  cos3x 2 y  cos 3x 2 y  cos3x2  cos9x 2

y  sin 3x 26x  6x sin 3x 2 y  cos 32x  2x cos 3 y  sin 9x 218x  18x sin 9x 2

y  cos 2 x  cos x 2

y  2cos xsin x  2 cos x sin x 1 sin x y  cos x12sin x   2 2cos x

e. y  cos x  cos x12

Try It

Exploration A

To find the derivative of a function of the form kx  f ghx, you need to apply the Chain Rule twice, as shown in Example 12. EXAMPLE 12

Repeated Application of the Chain Rule

f t  sin3 4t

Original function

 sin 4t3 ft  3sin 4t2

Rewrite.

d

sin 4t dt

d

4t dt  3sin 4t2cos 4t4  12 sin 2 4t cos 4t  3sin 4t2cos 4t

.

Try It

Exploration A

Apply Chain Rule once.

Apply Chain Rule a second time.

Simplify.

136

CHAPTER 2

Differentiation

EXAMPLE 13 y

Tangent Line of a Trigonometric Function

Find an equation of the tangent line to the graph of

f(x) = 2 sin x + cos 2x

f x  2 sin x  cos 2x

2

at the point , 1, as shown in Figure 2.26. Then determine all values of x in the interval 0, 2 at which the graph of f has a horizontal tangent.

( π , 1)

1

x

π 2

π

3π 2

2π

Solution Begin by finding fx. f x  2 sin x  cos 2x

−2

Write original function.

fx  2 cos x  sin 2x2  2 cos x  2 sin 2 x

−3 −4

Apply Chain Rule to cos 2x. Simplify.

To find the equation of the tangent line at , 1, evaluate f. f    2 cos   2 sin 2  2

Figure 2.26

Substitute. Slope of graph at , 1

Now, using the point-slope form of the equation of a line, you can write y  y1  mx  x1 y  1  2x   y  1  2x  2.

Point-slope form Substitute for y1, m, and x1. Equation of tangent line at , 1

  5 3 You can then determine that fx  0 when x  , , , and . So, f has a 6 2 6 2   5 3 horizontal tangent at x  , , , and . 6 2 6 2

STUDY TIP To become skilled at . differentiation, you should memorize each rule. As an aid to memorization, note that the cofunctions (cosine, cotangent, and cosecant) require a negative sign as part of their derivatives.

Try It

Exploration A

This section concludes with a summary of the differentiation rules studied so far.

Summary of Differentiation Rules General Differentiation Rules

Let f, g, and u be differentiable functions of x. Constant Multiple Rule:

Sum or Difference Rule:

d

cf  cf  dx

d

f ± g  f  ± g dx

Product Rule:

Quotient Rule:

d

fg  fg  gf dx

d f gf  fg  dx g g2

Derivatives of Algebraic Functions

Constant Rule:

Simple Power Rule:

d

c  0 dx

d n

x  nxn1, dx

Derivatives of Trigonometric Functions

d

sin x  cos x dx

d

tan x  sec 2 x dx

d

sec x  sec x tan x dx

d

cos x  sin x dx

d

cot x  csc 2 x dx

d

csc x  csc x cot x dx

Chain Rule:

General Power Rule:

d

f u  f u u dx

d n

u  nu n1 u dx

Chain Rule



d

x  1 dx

SECTION 2.4

137

The Chain Rule

Exercises for Section 2.4 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–6, complete the table. y  f gx

u  gx

In Exercises 39 and 40, find the slope of the tangent line to the sine function at the origin. Compare this value with the number of complete cycles in the interval [0, 2]. What can you conclude about the slope of the sine function sin ax at the origin?

y  f u

1. y  6x  54 2. y 

1

y

39. (a)

x  1

y = sin x

2

3. y  x2  1

1  2

5. y  csc 3x 3x 2

7. y  2x  73 9. gx  34  9x

14. gx  x 2  2x  1

15. y 

4 4 2

4 2  9x 16. f x  3 

17. y 

1 x2



x2

18. st 



1 19. f t  t3



2

1 x  2 23. f x  x 2x  24

22. gt  26. y 

x

28. y 

x 2  1

x5 x2  2

  t 30. ht   t  2 1  2v 31. f v   1v 3x  2 32. gx   2x  3 



1 t2  2

24. f x  x3x  93

25. y  x1  x 2

2

1 t 2  3t  1

5 20. y   t  33

21. y 

1 2 2 x 16

 x2

x x 4  4

2

2

1

2

 2



3 2

2

x

2

In Exercises 41–58, find the derivative of the function. 41. y  cos 3x

42. y  sin  x

43. gx  3 tan 4x

44. hx  sec x 2

45. y  sinx2

46. y  cos1  2x2

47. hx  sin 2x cos 2x 49. f x 

cot x sin x

1 1 48. g   sec2  tan2 

50. gv 

cos v csc v

51. y  4 sec2 x

52. gt  5 cos 2  t

53. f    sin 2 2

54. ht  2 cot2 t  2

55. f t  3

56. y  3x  5 cos x2

1 4

 t  1 57. y  x  sin2x2 sec2

3 3 x sin x 58. y  sin 

In Exercises 59–66, evaluate the derivative of the function at the given point. Use a graphing utility to verify your result. Function

3

x  1

x2  1 x1 x

cos  x  1 x

Point

59. st  t  2t  8 2

5 3x 3  4x 60. y  

In Exercises 33–38, use a computer algebra system to find the derivative of the function. Then use the utility to graph the function and its derivative on the same set of coordinate axes. Describe the behavior of the function that corresponds to any zeros of the graph of the derivative.

37. y 

2

2

3



x 

1 4

3

35. y 

1

10. f t  9t  2

3 9x 2  4 13. y  

2

y = sin 2x

2

1

12. gx  5  3x

x



y

(b) y = sin 3x

23

11. f t  1  t

33. y 

 2 2

2

8. y  34  x 25 4

29. gx 

x 2

y

40. (a)

27. y 



2

In Exercises 7–32, find the derivative of the function.

y = sin 2x

2

1

4. y  3 tan x 2 6. y  cos

y

(b)

34. y 



2x x1

36. gx  x  1  x  1 38. y  x 2 tan

1 x

3 x3  4 1 f x  2 x  3x2 3t  2 f t  t1 x1 f x  2x  3 y  37  sec 32x

61. f x  62. 63. 64. 65.

1 66. y   cos x x

2, 4 2, 2

1,  53 4, 161  0, 2 2, 3 0, 36  2 , 2 





138

CHAPTER 2

Differentiation

In Exercises 67–74, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your results. Function

Point

67. f x  3x 2  2 69. y  2x3  12 70. f x  9  x223 71. f x  sin 2x

73. f x  tan x 2

74. y  2 tan3 x

3t

86. f x  sec 2 x

89. f x  cosx2,

,



90. gt  tan 2t,

0, 1  , 3 6

In Exercises 91–94, the graphs of a function f and its derivative f are shown. Label the graphs as f or f and write a short paragraph stating the criteria used in making the selection. To print an enlarged copy of the graph, select the MathGraph button. y

91.

4, 8 4  2t1  t 4 77. s t  , 0, 3 3 2  78. y  t  9 t  2, 2, 10

4 3 2 3

Famous Curves In Exercises 79 and 80, find an equation of the tangent line to the graph at the given point. Then use a graphing utility to graph the function and its tangent line in the same viewing window.

y

4 3 2

3

x

80. Bullet-nose curve 25  x2

y

y

8

f (x) =

x 2  x2

4 3

(3, 4)

2

(1, 1)

1 x

6 4 2 4

2

4

6

x 3 2 1

1

2

3

2

81. Horizontal Tangent Line Determine the point(s) in the interval 0, 2 at which the graph of f x  2 cos x  sin 2x has a horizontal tangent. 82. Horizontal Tangent Line Determine the point(s) at which the x graph of f x  has a horizontal tangent. 2x  1

x 2

2 3 4

4

In Exercises 95 and 96, the relationship between f and g is given. Explain the relationship between f and g . 95. gx  f 3x

2

y

94.

3

4

1 2 3 4

2 3

93.

6

x

x

2

 

f(x) =

y

92.

3 2

76. f x  x 2  x2,

79. Top half of circle



Writing About Concepts

12, 32

2

t2  2t  1

85. f x  sin x 2

 

In Exercises 75–78, (a) use a graphing utility to find the derivative of the function at the given point, (b) find an equation of the tangent line to the graph of the function at the given point, and (c) use the utility to graph the function and its tangent line in the same viewing window. 75. gt 

84. f x 

1 64 87. hx  9 3x  13, 1, 9  1 1 , 0, 88. f x  x  4 2

     

72. y  cos 3x

1 x2

83. f x  2x 2  13

In Exercises 87–90, evaluate the second derivative of the function at the given point. Use a computer algebra system to verify your result.

3, 5 2, 2 1, 1 1, 4 , 0  2 , 4 2  ,1 4  ,2 4

68. f x  13xx 2  5

In Exercises 83–86, find the second derivative of the function.

96. gx  f x 2

97. Given that g5  3, g5  6, h5  3, and h5  2, find f5 (if possible) for each of the following. If it is not possible, state what additional information is required. (a) f x  gxhx (c) f x 

gx hx

(b) f x  ghx (d) f x  gx 3

SECTION 2.4

98. Think About It The table shows some values of the derivative of an unknown function f. Complete the table by finding (if possible) the derivative of each transformation of f. (a) gx  f x  2

(b) hx  2 f x

(c) rx  f 3x

(d) sx  f x  2

x f x

2

1

0

1

2

3

4

2 3

 13

1

2

4

102. Harmonic Motion The displacement from equilibrium of an object in harmonic motion on the end of a spring is 1

1

y  3 cos 12t  4 sin 12t

103. Pendulum A 15-centimeter pendulum moves according to the equation  0.2 cos 8t, where is the angular displacement from the vertical in radians and t is the time in seconds. Determine the maximum angular displacement and the rate of change of when t  3 seconds.

h x

104. Wave Motion A buoy oscillates in simple harmonic motion y  A cos t as waves move past it. The buoy moves a total of 3.5 feet (vertically) from its low point to its high point. It returns to its high point every 10 seconds.

r x s x In Exercises 99 and 100, the graphs of f and g are shown. Let hx  f gx and sx  g f x. Find each derivative, if it exists. If the derivative does not exist, explain why. 99. (a) Find h1.

100. (a) Find h3.

(b) Find s5.

(b) Find s9.

y 10

8

f

4

g

6

g

2

2 x

2

4

6

8

10

x 2

4

6

8

10

101. Doppler Effect The frequency F of a fire truck siren heard by a stationary observer is 132,400 331 ± v

where ± v represents the velocity of the accelerating fire truck in meters per second (see figure). Find the rate of change of F with respect to v when (a) the fire truck is approaching at a velocity of 30 meters per second (use v). (b) the fire truck is moving away at a velocity of 30 meters per second (use v ). 132,400 331 + v

F=

(b) Determine the velocity of the buoy as a function of t.

S  CR 2  r 2

10

f

8

(a) Write an equation describing the motion of the buoy if it is at its high point at t  0. 105. Circulatory System The speed S of blood that is r centimeters from the center of an artery is

y

F=

139

where y is measured in feet and t is the time in seconds. Determine the position and velocity of the object when t  8.

g x

F

The Chain Rule

132,400 331  v

where C is a constant, R is the radius of the artery, and S is measured in centimeters per second. Suppose a drug is administered and the artery begins to dilate at a rate of dRdt. At a constant distance r, find the rate at which S changes with respect to t for C  1.76  105, R  1.2  102, and dRdt  105. 106. Modeling Data The normal daily maximum temperatures T (in degrees Fahrenheit) for Denver, Colorado, are shown in the table. (Source: National Oceanic and Atmospheric Administration) Month

Jan

Feb

Mar

Apr

May

Jun

Temperature

43.2

47.2

53.7

60.9

70.5

82.1

Month

Jul

Aug

Sep

Oct

Nov

Dec

Temperature

88.0

86.0

77.4

66.0

51.5

44.1

(a) Use a graphing utility to plot the data and find a model for the data of the form Tt  a  b sin  t6  c where T is the temperature and t is the time in months, with t  1 corresponding to January. (b) Use a graphing utility to graph the model. How well does the model fit the data? (c) Find T and use a graphing utility to graph the derivative. (d) Based on the graph of the derivative, during what times does the temperature change most rapidly? Most slowly? Do your answers agree with your observations of the temperature changes? Explain.

140

CHAPTER 2

Differentiation

107. Modeling Data The cost of producing x units of a product is C  60x  1350. For one week management determined the number of units produced at the end of t hours during an eight-hour shift. The average values of x for the week are shown in the table. t

0

1

2

3

4

5

6

7

8

x

0

16

60

130

205

271

336

384

392

(a) Use a graphing utility to fit a cubic model to the data. (b) Use the Chain Rule to find dCdt. (c) Explain why the cost function is not increasing at a constant rate during the 8-hour shift. 108. Finding a Pattern Consider the function f x  sin x, where  is a constant. (a) Find the first-, second-, third-, and fourth-order derivatives of the function. (b) Verify that the function and its second derivative satisfy the equation f  x   2 f x  0. (c) Use the results in part (a) to write general rules for the even- and odd-order derivatives f 2kx and f 2k1x. [Hint: 1k is positive if k is even and negative if k is odd.] 109. Conjecture Let f be a differentiable function of period p. (a) Is the function f periodic? Verify your answer. (b) Consider the function gx  f 2x. Is the function g x periodic? Verify your answer. 110. Think About It Let rx  f gx and sx  g f x where f and g are shown in the figure. Find (a) r1 and (b) s4. y 7 6 5 4 3 2 1



u d

u  u , dx u





u  0.

In Exercises 114–117, use the result of Exercise 113 to find the derivative of the function.

    hx  x cos x f x  sin x

114. gx  2x  3

115. f x  x 2  4 116. 117.

Linear and Quadratic Approximations The linear and quadratic approximations of a function f at x  a are P1x  f ax  a  f a and 1 P2x  2 f ax  a 2  f ax  a  f a).

In Exercises 118 and 119, (a) find the specified linear and quadratic approximations of f, (b) use a graphing utility to graph f and the approximations, (c) determine whether P1 or P2 is the better approximation, and (d) state how the accuracy changes as you move farther from x  a. 118. f x  tan

x 4

119. f x  sec 2x

a1

a

 6

True or False? In Exercises 120–122, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 1 120. If y  1  x1 2, then y  21  x1 2.

(6, 6)

121. If f x  sin 22x, then fx  2sin 2xcos 2x.

g

122. If y is a differentiable function of u, u is a differentiable function of v, and v is a differentiable function of x, then

(6, 5)

(2, 4)

113. Let u be a differentiable function of x. Use the fact that u  u 2 to prove that

f x

dy du dv dy  . dx du dv dx

1 2 3 4 5 6 7

111. (a) Find the derivative of the function gx  sin 2 x  cos 2 x in two ways. (b) For f x  sec2 x and gx  tan 2 x, show that fx gx. 112. (a) Show that the derivative of an odd function is even. That is, if f x  f x, then fx  fx. (b) Show that the derivative of an even function is odd. That is, if f x  f x, then fx  fx.

Putnam Exam Challenge 123. Let f x  a1 sin x  a2 sin 2x  . . .  an sin nx, where a1, a2, . . ., an are real numbers and where n is a positive integer. Given that f x  sin x for all real x, prove that a1  2a2  . . .  nan  1.





  



124. Let k be a fixed positive integer. The nth derivative of has the form

1 xk  1

Pnx x k  1n1 where Pnx is a polynomial. Find Pn1. These problems were composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.

SECTION 2.5

Section 2.5

Implicit Differentiation

141

Implicit Differentiation • Distinguish between functions written in implicit form and explicit form. • Use implicit differentiation to find the derivative of a function.

E X P L O R AT I O N Graphing an Implicit Equation How could you use a graphing utility to sketch the graph of the equation x  2y  4y  2? 2

3

Here are two possible approaches. a. Solve the equation for x. Switch the roles of x and y and graph the two resulting equations. The combined graphs will show a 90 rotation of the graph of the original equation. b. Set the graphing utility to parametric mode and graph the equations x   2t 3  4t  2

Implicit and Explicit Functions Up to this point in the text, most functions have been expressed in explicit form. For example, in the equation y  3x 2  5

the variable y is explicitly written as a function of x. Some functions, however, are only implied by an equation. For instance, the function y  1x is defined implicitly by the equation xy  1. Suppose you were asked to find dydx for this equation. You could begin by writing y explicitly as a function of x and then differentiating.

x  2t 3  4t  2 y  t.

.

From either of these two approaches, can you decide whether the graph has a tangent line at the point 0, 1? Explain your reasoning.

Implicit Form

Explicit Form

xy  1

y

1  x1 x

Derivative

1 dy  x2   2 dx x

This strategy works whenever you can solve for the function explicitly. You cannot, however, use this procedure when you are unable to solve for y as a function of x. For instance, how would you find dydx for the equation x 2  2y 3  4y  2

yt and

Explicit form

where it is very difficult to express y as a function of x explicitly? To do this, you can use implicit differentiation. To understand how to find dydx implicitly, you must realize that the differentiation is taking place with respect to x. This means that when you differentiate terms involving x alone, you can differentiate as usual. However, when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x.

Video EXAMPLE 1 a.

Differentiating with Respect to x

d 3

x  3x 2 dx

Variables agree: use Simple Power Rule.

Variables agree un

b.

nu n1 u

dy d 3

y  3y 2 dx dx

Variables disagree: use Chain Rule.

Variables disagree

c. d.

d dy

x  3y  1  3 dx dx d d d

xy 2  x y 2  y 2 x dx dx dx dy  x 2y  y 21 dx



.

 2xy

Try It



dy  y2 dx

Exploration A

Chain Rule:

Product Rule

Chain Rule

Simplify.

d

3y  3y dx

142

CHAPTER 2

Differentiation

Implicit Differentiation Guidelines for Implicit Differentiation 1. Differentiate both sides of the equation with respect to x. 2. Collect all terms involving dydx on the left side of the equation and move all other terms to the right side of the equation. 3. Factor dydx out of the left side of the equation. 4. Solve for dydx. EXAMPLE 2

Implicit Differentiation

Find dydx given that y 3  y 2  5y  x 2  4. Solution NOTE In Example 2, note that implicit differentiation can produce an expression for dydx that contains both x and y.

1. Differentiate both sides of the equation with respect to x. d 3 d

y  y 2  5y  x 2  4 dx dx d 3 d d d d

y  y 2  5y  x 2  4 dx dx dx dx dx dy dy dy 3y 2  2y  5  2x  0 dx dx dx 2. Collect the dydx terms on the left side of the equation and move all other terms to the right side of the equation.

y

3y 2

2

(1, 1)

1

3. Factor dydx out of the left side of the equation.

(2, 0) x

−3

−2

−1

−1 −2

.

−4

1

2

3

(1, −3) y 3 + y 2 − 5y − x 2 = − 4

Point on Graph

Slope of Graph

2, 0 1,  3

 45

x0

0

1, 1

Undefined

1 8

The implicit equation y3  y 2  5y  x 2   4 has the derivative dy 2x  2 . dx 3y  2y  5 Figure 2.27

dy dy dy  2y  5  2x dx dx dx

dy 3y 2  2y  5  2x dx 4. Solve for dydx by dividing by 3y 2  2y  5. 2x dy  dx 3y 2  2y  5

Try It

Exploration A

Video

Video

To see how you can use an implicit derivative, consider the graph shown in Figure 2.27. From the graph, you can see that y is not a function of x. Even so, the derivative found in Example 2 gives a formula for the slope of the tangent line at a point on this graph. The slopes at several points on the graph are shown below the graph. With most graphing utilities, it is easy to graph an equation that explicitly represents y as a function of x. Graphing other equations, however, can require some ingenuity. For instance, to graph the equation given in Example 2, use a graphing utility, set in parametric mode, to graph the parametric representations x  t 3  t 2  5t  4, y  t, and x   t 3  t 2  5t  4, y  t, for 5 ≤ t ≤ 5. How does the result compare with the graph shown in Figure 2.27?

TECHNOLOGY

SECTION 2.5

y

+

y2

=0

(0, 0) x

−1

143

It is meaningless to solve for dydx in an equation that has no solution points. (For example, x 2  y 2  4 has no solution points.) If, however, a segment of a graph can be represented by a differentiable function, dydx will have meaning as the slope at each point on the segment. Recall that a function is not differentiable at (a) points with vertical tangents and (b) points at which the function is not continuous.

1

x2

Implicit Differentiation

1

EXAMPLE 3

−1

Representing a Graph by Differentiable Functions

. If possible, represent y as a differentiable function of x.

(a)

a. x 2  y 2  0

Editable Graph

y=

a. The graph of this equation is a single point. So, it does not define y as a differentiable function of x. See Figure 2.28(a). b. The graph of this equation is the unit circle, centered at 0, 0. The upper semicircle is given by the differentiable function

1 − x2

(−1, 0)

(1, 0) x

−1

1 −1

.

y  1  x 2,

y=−

1 − x2

1 < x < 1

and the lower semicircle is given by the differentiable function y   1  x 2, 1 < x < 1.

(b)

At the points 1, 0 and 1, 0, the slope of the graph is undefined. See Figure 2.28(b). c. The upper half of this parabola is given by the differentiable function

Editable Graph y

y=

y  1  x,

1−x

1

x < 1

and the lower half of this parabola is given by the differentiable function y   1  x, x < 1.

(1, 0) x

.

c. x  y 2  1

Solution

y 1

b. x 2  y 2  1

−1

1

−1

At the point 1, 0, the slope of the graph is undefined. See Figure 2.28(c).

y=−

.

1−x

Try It

(c)

Editable Graph

EXAMPLE 4

Some graph segments can be represented by differentiable functions. Figure 2.28

Exploration B

Exploration A

Finding the Slope of a Graph Implicitly

Determine the slope of the tangent line to the graph of x 2  4y 2  4 at the point 2, 12 . See Figure 2.29. Solution

y 2

x 2 + 4y 2 = 4

x

−1

. . Figure 2.29

1

−2

(

2, − 1 2

)

x 2  4y 2  4 dy 2x  8y  0 dx dy 2x x   dx 8y 4y

Write original equation. Differentiate with respect to x. Solve for

dy . dx

So, at 2, 12 , the slope is dy  2 1   . dx 42 2

Evaluate

dy 1 when x  2 and y   . dx 2

Editable Graph

Try It

Exploration A

Exploration B

Open Exploration

NOTE To see the benefit of implicit differentiation, try doing Example 4 using the explicit function y   124  x 2.

144

CHAPTER 2

Differentiation

EXAMPLE 5

Finding the Slope of a Graph Implicitly

Determine the slope of the graph of 3x 2  y 2 2  100xy at the point 3, 1. Solution d d

3x 2  y 2 2  100xy dx dx dy dy 32x 2  y 2 2x  2y  100 x  y1 dx dx dy dy 12y x 2  y 2  100x  100y  12xx 2  y 2 dx dx dy

12y x 2  y 2  100x  100y  12xx 2  y 2 dx dy 100y  12xx 2  y 2  dx 100x  12yx 2  y 2 25y  3xx 2  y 2  25x  3yx 2  y 2



y 4 3 2 1

(3, 1) x

−4

−2 −1

1

3

4







At the point 3, 1, the slope of the graph is

−4

251  3332  12 25  90 65 13 dy     dx 253  3132  12 75  30 45 9

3(x 2 + y 2) 2 = 100xy

. Lemniscate

as shown in Figure 2.30. This graph is called a lemniscate.

Figure 2.30

Try It

Exploration A

Exploration B

y

EXAMPLE 6

sin y = x

Find dydx implicitly for the equation sin y  x. Then find the largest interval of the form a < y < a on which y is a differentiable function of x (see Figure 2.31).

(1, π2 )

π 2

Determining a Differentiable Function

x

−1

(−1, − π2 )

−π 2

1

− 3π 2

The . derivative is

1 dy  . dx 1  x 2

Figure 2.31

Editable Graph

Solution d d

sin y  x dx dx cos y

dy 1 dx dy 1  dx cos y

The largest interval about the origin for which y is a differentiable function of x is  2 < y < 2. To see this, note that cos y is positive for all y in this interval and is 0 at the endpoints. If you restrict y to the interval  2 < y < 2, you should be able to write dydx explicitly as a function of x. To do this, you can use cos y  1  sin2 y  1  x 2, 

  < y < 2 2

and conclude that .

dy 1  . dx 1  x 2

Try It

Exploration A

SECTION 2.5

ISAAC BARROW (1630–1677)

.

The graph in Figure 2.32 is called the kappa curve because it resembles the Greek letter kappa,  . The general solution for the tangent line to this curve was discovered by the English mathematician Isaac Barrow. Newton was Barrow’s student, and they corresponded frequently regarding their work in the early development of calculus.

Implicit Differentiation

145

With implicit differentiation, the form of the derivative often can be simplified (as in Example 6) by an appropriate use of the original equation. A similar technique can be used to find and simplify higher-order derivatives obtained implicitly.

Finding the Second Derivative Implicitly

EXAMPLE 7

Given x 2  y 2  25, find

d 2y . dx 2

Solution Differentiating each term with respect to x produces MathBio

2x  2y 2y

dy 0 dx dy  2x dx dy 2x x   . dx 2y y

Differentiating a second time with respect to x yields d 2y  y1  xdydx  2 dx y2 y  xxy  y2

.

Quotient Rule Substitute xy for

dy . dx



y 2  x2 y3

Simplify.



25 . y3

Substitute 25 for x 2  y 2.

Try It

Exploration A Finding a Tangent Line to a Graph

EXAMPLE 8

Find the tangent line to the graph given by x 2x 2  y 2  y 2 at the point 22, 22, as shown in Figure 2.32. Solution By rewriting and differentiating implicitly, you obtain x 4  x 2y 2  y 2  0 dy dy 4x 3  x 2 2y  2xy 2  2y 0 dx dx dy 2yx 2  1  2x2x 2  y 2 dx dy x 2x 2  y 2 .  dx y 1  x 2



y

1

( 22 , 22)

At the point 22, 22, the slope is x

−1

1

−1

x 2(x 2 + y 2) = y 2

dy 22 212  12 32   3 dx 12 22 1  12

and the equation of the tangent line at this point is y

. The kappa curve Figure 2.32



2

2



3 x

2

2 y  3x  2.

Try It



Exploration A

146

CHAPTER 2

Differentiation

Exercises for Section 2.5 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

31. Bifolium:

In Exercises 1–16, find dy /dx by implicit differentiation.

32. Folium of Descartes:

    Point: 1, 1 x2

y2 2

1. x 2  y 2  36

2. x 2  y 2  16

3. x12  y12  9

4. x3  y 3  8

5. x3  xy  y 2  4

6. x 2 y  y 2x  2

7. x3y 3  y  x

8. xy  x  2y

2

10. 2 sin x cos y  1

1

9.

x3



3x 2 y



2xy 2

 12

11. sin x  2 cos 2y  1

12. sin  x  cos  y 2  2

13. sin x  x1  tan y

14. cot y  x  y

15. y  sinxy

1 16. x  sec y



17. 19.

9x 2

y2



 16

16y 2

 144



18.

x2

20.

9y 2

y2



2

28.

1

1

1

2

x

2

1

2

33. Parabola

34. Circle

(y  2)2 = 4(x  3)

(x + 1)2 + (y  2)2 = 20 y

8 6 4 2

8

2 4

2

10 12 14

x 8 6

2

y

36. Rotated ellipse 7x 2  6

xy = 1

3xy + 13y 2  16 = 0 y

3 2

3

(1, 1)

1

2 x

3

1

2

(

3, 1)

2

3

3 x 3

Famous Curves In Exercises 29– 32, find the slope of the tangent line to the graph at the given point. 29. Witch of Agnesi:

4

4

35. Rotated hyperbola

 

(3, 4)

4

(4, 0) x

2 4 6 8

1, 1 23 23 x  y  5, 8, 1 x 3  y 3  4xy  1, 2, 1 tanx  y  x, 0, 0  x cos y  1, 2, 3

4

Famous Curves In Exercises 33–40, find an equation of the tangent line to the graph at the given point. To print an enlarged copy of the graph, select the MathGraph button.

9

x2  4 , 2, 0 x2  4

3

2

y

24. x  y3  x3  y 3,

27.

2

2

22. x 2  y 3  0, 1, 1

26.

4

1

21. xy  4, 4, 1

25.

y

x

In Exercises 21–28, find dy/ dx by implicit differentiation and evaluate the derivative at the given point.

23. y 2 

4 8

3

 4x  6y  9  0

x2

Point:  3, 3 

y

In Exercises 17–20, (a) find two explicit functions by solving the equation for y in terms of x, (b) sketch the graph of the equation and label the parts given by the corresponding explicit functions, (c) differentiate the explicit functions, and (d) find dy/ dx and show that the result is equivalent to that of part (c). x2

x3  y 3  6xy  0

4x 2 y

2 3

30. Cissoid: 37. Cruciform

4  xy 2  x3 Point: 2, 2

x 2  4y  8 Point: 2, 1 y

x 2y 2



9x 2

38. Astroid 

4y 2

x 2/3 + y 2/3 = 5

=0

y

y

y 12

6 2

3

4

(4, 2

1

3)

(8, 1) x

x

1 2 x 2

1

1 1

1

2 2

3

6 4 2

2

4

x

6

12

4 12

SECTION 2.5

39. Lemniscate

40. Kappa curve

3(x 2 + y 2)2 = 100(x 2  y 2)

In Exercises 57 and 58, find the points at which the graph of the equation has a vertical or horizontal tangent line.

y 2(x 2 + y 2) = 2x2

y

y

6

3

4

2

(4, 2)

57. 25x 2  16y 2  200x  160y  400  0 58. 4x 2  y 2  8x  4y  4  0 (1, 1)

2 x 6

6

x 3 2

2

4

2

6

3

3

41. (a) Use implicit differentiation to find an equation of the x2 y2 tangent line to the ellipse   1 at 1, 2. 2 8 (b) Show that the equation of the tangent line to the ellipse x x y y y2 x2  2  1 at x0, y0 is 02  02  1. 2 a b a b 42. (a) Use implicit differentiation to find an equation of the x2 y2 tangent line to the hyperbola   1 at 3, 2. 6 8 (b) Show that the equation of the tangent line to the hyperbola x x y y y2 x2  2  1 at x0, y0 is 02  02  1. 2 a b a b In Exercises 43 and 44, find dy/dx implicitly and find the largest interval of the form a < y < a or 0 < y < a such that y is a differentiable function of x. Write dy/dx as a function of x. 43. tan y  x

147

Implicit Differentiation

Orthogonal Trajectories In Exercises 59–62, use a graphing utility to sketch the intersecting graphs of the equations and show that they are orthogonal. [Two graphs are orthogonal if at their point(s) of intersection their tangent lines are perpendicular to each other.] 59. 2x 2  y 2  6 y2

60. y 2  x 3

 4x

2x 2  3y 2  5 62. x3  3 y  1

61. x  y  0

x3y  29  3

x  sin y

Orthogonal Trajectories In Exercises 63 and 64, verify that the two families of curves are orthogonal where C and K are real numbers. Use a graphing utility to graph the two families for two values of C and two values of K. 63. xy  C,

x2  y 2  K

64. x 2  y 2  C 2,

y  Kx

In Exercises 65–68, differentiate (a) with respect to x ( y is a function of x) and (b) with respect to t ( x and y are functions of t). 65. 2y 2  3x 4  0

66. x 2  3xy 2  y 3  10

67. cos  y  3 sin  x  1

68. 4 sin x cos y  1

44. cos y  x

Writing About Concepts In Exercises 45–50, find d 2 y /dx 2 in terms of x and y. 45. x 2  y2  36

46. x 2 y 2  2x  3

47. x 2  y 2  16

48. 1  xy  x  y

49. y 2  x 3

50. y 2  4x

69. Describe the difference between the explicit form of a function and an implicit equation. Give an example of each. 70. In your own words, state the guidelines for implicit differentiation.

In Exercises 51 and 52, use a graphing utility to graph the equation. Find an equation of the tangent line to the graph at the given point and graph the tangent line in the same viewing window. 51. x  y  4, 9, 1

52. y 2 

x1 , x2  1

2, 55 

In Exercises 53 and 54, find equations for the tangent line and normal line to the circle at the given points. (The normal line at a point is perpendicular to the tangent line at the point.) Use a graphing utility to graph the equation, tangent line, and normal line. 53. x 2  y 2  25

54. x 2  y 2  9

55. Show that the normal line at any point on the circle x 2  y 2  r 2 passes through the origin. 56. Two circles of radius 4 are tangent to the graph of y 2  4x at the point 1, 2. Find equations of these two circles.

1671

00

0, 3, 2, 5 

18

4, 3, 3, 4

71. Orthogonal Trajectories The figure below shows the topographic map carried by a group of hikers. The hikers are in a wooded area on top of the hill shown on the map and they decide to follow a path of steepest descent (orthogonal trajectories to the contours on the map). Draw their routes if they start from point A and if they start from point B. If their goal is to reach the road along the top of the map, which starting point should they use? To print an enlarged copy of the graph, select the MathGraph button.

B

1994

A 00

18

148

CHAPTER 2

Differentiation

72. Weather Map The weather map shows several isobars— curves that represent areas of constant air pressure. Three high pressures H and one low pressure L are shown on the map. Given that wind speed is greatest along the orthogonal trajectories of the isobars, use the map to determine the areas having high wind speed.

76. Slope Find all points on the circle x2  y2  25 where the 3 slope is 4. 77. Horizontal Tangent Determine the point(s) at which the graph of y 4  y2  x2 has a horizontal tangent. 78. Tangent Lines Find equations of both tangent lines to the x2 y2 ellipse   1 that passes through the point 4, 0. 4 9 79. Normals to a Parabola The graph shows the normal lines from the point 2, 0 to the graph of the parabola x  y2. How many normal lines are there from the point x0, 0 to the graph 1 1 of the parabola if (a) x0  4, (b) x0  2, and (c) x0  1? For what value of x0 are two of the normal lines perpendicular to each other?

H H

L H

y

73. Consider the equation x 4  44x 2  y 2. (2, 0)

(a) Use a graphing utility to graph the equation.

x

(b) Find and graph the four tangent lines to the curve for y  3.

x = y2

(c) Find the exact coordinates of the point of intersection of the two tangent lines in the first quadrant. 74. Let L be any tangent line to the curve x  y  c. Show that the sum of the x- and y-intercepts of L is c. 75. Prove (Theorem 2.3) that d n

x  nx n1 dx for the case in which n is a rational number. (Hint: Write y  x pq in the form y q  x p and differentiate implicitly. Assume that p and q are integers, where q > 0.)

80. Normal Lines (a) Find an equation of the normal line to the ellipse x2 y2  1 32 8 at the point 4, 2. (b) Use a graphing utility to graph the ellipse and the normal line. (c) At what other point does the normal line intersect the ellipse?

SECTION 2.6

Section 2.6

Related Rates

149

Related Rates • Find a related rate. • Use related rates to solve real-life problems.

r

Finding Related Rates h

You have seen how the Chain Rule can be used to find dydx implicitly. Another important use of the Chain Rule is to find the rates of change of two or more related variables that are changing with respect to time. For example, when water is drained out of a conical tank (see Figure 2.33), the volume V, the radius r, and the height h of the water level are all functions of time t. Knowing that these variables are related by the equation V

r

 2 r h 3

Original equation

you can differentiate implicitly with respect to t to obtain the related-rate equation d d  2 V   rh dt dt 3 h

  dV  dh dr  r  h 2r  dt 3 dt dt  dh dr  2rh .  r 3 dt dt 2

Differentiate with respect to t.

2

From this equation you can see that the rate of change of V is related to the rates of change of both h and r. r

E X P L O R AT I O N

h

Finding a Related Rate In the conical tank shown in Figure 2.33, suppose that the height is changing at a rate of 0.2 foot per minute and the radius is changing at a rate of 0.1 foot per minute. What is the rate of change in the volume when the radius is r  1 foot and the height is h  2 feet? Does the rate of change in the volume depend on the values of r and h? Explain.

EXAMPLE 1 . Volume is related to radius and height. Figure 2.33

Animation

FOR FURTHER INFORMATION To learn more about the history of relatedrate problems, see the article “The Lengthening Shadow: The Story of Related Rates” by Bill Austin, Don .. Barry, and David Berman in Mathematics Magazine.

MathArticle

Two Rates That Are Related

Suppose x and y are both differentiable functions of t and are related by the equation y  x 2  3. Find dydt when x  1, given that dxdt  2 when x  1. Solution Using the Chain Rule, you can differentiate both sides of the equation with respect to t. y  x2  3 d d

y  x 2  3 dt dt dx dy  2x dt dt When x  1 and dxdt  2, you have dy  212  4. dt

Try It

Exploration A

Write original equation. Differentiate with respect to t.

Chain Rule

150

CHAPTER 2

Differentiation

Problem Solving with Related Rates In Example 1, you were given an equation that related the variables x and y and were asked to find the rate of change of y when x  1. Equation: Given rate:

y  x2  3 dx  2 when dt dy dt

Find:

when

x1

x1

In each of the remaining examples in this section, you must create a mathematical model from a verbal description. EXAMPLE 2

Ripples in a Pond

A pebble is dropped into a calm pond, causing ripples in the form of concentric circles, as shown in Figure 2.34. The radius r of the outer ripple is increasing at a constant rate of 1 foot per second. When the radius is 4 feet, at what rate is the total area A of the disturbed water changing? Solution The variables r and A are related by A   r 2. The rate of change of the radius r is drdt  1. Equation:

A  r2

Given rate:

dr 1 dt

Find:

dA dt

when

r4

With this information, you can proceed as in Example 1. d d

A   r 2 dt dt dA dr  2 r dt dt Total area increases as the outer radius increases.

. Figure 2.34

Differentiate with respect to t.

Chain Rule

dA  2 41  8 dt

Substitute 4 for r and 1 for drdt.

When the radius is 4 feet, the area is changing at a rate of 8 square feet per second.

Try It

Exploration A

Video

Video

Guidelines For Solving Related-Rate Problems

NOTE When using these guidelines, be sure you perform Step 3 before Step 4. Substituting the known values of the variables before differentiating will produce an inappropriate derivative.

1. Identify all given quantities and quantities to be determined. Make a sketch and label the quantities. 2. Write an equation involving the variables whose rates of change either are given or are to be determined. 3. Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t. 4. After completing Step 3, substitute into the resulting equation all known values for the variables and their rates of change. Then solve for the required rate of change.

SECTION 2.6

Related Rates

151

The table below lists examples of mathematical models involving rates of change. For instance, the rate of change in the first example is the velocity of a car.

An Inflating Balloon

EXAMPLE 3

Air is being pumped into a spherical balloon (see Figure 2.35) at a rate of 4.5 cubic feet per minute. Find the rate of change of the radius when the radius is 2 feet. Solution Let V be the volume of the balloon and let r be its radius. Because the volume is increasing at a rate of 4.5 cubic feet per minute, you know that at time t the rate of change of the volume is dVdt  92. So, the problem can be stated as shown. Given rate: Find:

dV 9  (constant rate) dt 2 dr when r  2 dt

To find the rate of change of the radius, you must find an equation that relates the radius r to the volume V. Equation:

V

4 r3 3

Volume of a sphere

Differentiating both sides of the equation with respect to t produces dV dr  4 r 2 dt dt

 

dr 1 dV  . dt 4 r 2 dt

Differentiate with respect to t.

Solve for drdt.

Finally, when r  2, the rate of change of the radius is .



dr 1 9  0.09 foot per minute.  dt 16 2

In Example 3, note that the volume is increasing at a constant rate but the radius is increasing at a variable rate. Just because two rates are related does not mean that they are proportional. In this particular case, the radius is growing more and more slowly as t increases. Do you see why?

152

CHAPTER 2

Differentiation

EXAMPLE 4

The Speed of an Airplane Tracked by Radar

An airplane is flying on a flight path that will take it directly over a radar tracking station, as shown in Figure 2.36. If s is decreasing at a rate of 400 miles per hour when s  10 miles, what is the speed of the plane? s

Solution Let x be the horizontal distance from the station, as shown in Figure 2.36. Notice that when s  10, x  10 2  36  8.

6 mi

Given rate: x

Find: Not drawn to scale

An airplane is flying at an altitude of 6 miles, s miles from the station.

dsdt  400 when s  10 dxdt when s  10 and x  8

You can find the velocity of the plane as shown. Equation:

Figure 2.36

x2  62  s2 dx ds 2x  2s dt dt dx s ds  dt x dt dx 10  400 dt 8  500 miles per hour

Pythagorean Theorem Differentiate with respect to t.

 

.

Solve for dxdt.

Substitute for s, x, and dsdt. Simplify.

Because the velocity is 500 miles per hour, the speed is 500 miles per hour.

Try It EXAMPLE 5

Exploration A

Open Exploration

A Changing Angle of Elevation

Find the rate of change in the angle of elevation of the camera shown in Figure 2.37 at 10 seconds after lift-off. Solution Let be the angle of elevation, as shown in Figure 2.37. When t  10, the height s of the rocket is s  50t 2  5010 2  5000 feet. Given rate: Find:

dsdt  100t  velocity of rocket d dt when t  10 and s  5000

Using Figure 2.37, you can relate s and by the equation tan  s2000. tan 

Equation: tan θ = s 2000

s

2000 ft

Animation

 



Not drawn to scale

Figure 2.37

See Figure 2.37.

d

1 ds  dt 2000 dt d

100t  cos 2

dt 2000 2000  s 2  2000 2

sec 2 

θ

A television camera at ground level is filming the lift-off of a space shuttle that is rising vertically according to the position equation s  50t 2, where s is measured in feet and t is measured in seconds. The camera is 2000 feet from . the launch pad.

s 2000

Differentiate with respect to t.

Substitute 100t for dsdt.



2

100t 2000

cos  2000s 2  2000 2

When t  10 and s  5000, you have d

200010010 2   radian per second. dt 50002  20002 29 2 So, when t  10, is changing at a rate of 29 radian per second.

Try It

Exploration A

SECTION 2.6

EXAMPLE 6

Related Rates

153

The Velocity of a Piston

In the engine shown in Figure 2.38, a 7-inch connecting rod is fastened to a crank of radius 3 inches. The crankshaft rotates counterclockwise at a constant rate of 200 revolutions per minute. Find the velocity of the piston when  3.

7

3 θ

θ

.

x

The velocity of a piston is related to the angle of the crankshaft. Figure 2.38

Animation Solution Label the distances as shown in Figure 2.38. Because a complete revolution corresponds to 2 radians, it follows that d dt  2002  400 radians per minute. b

a

θ c

Law of Cosines: b 2  a 2  c 2  2ac cos

Figure 2.39

Given rate:

d

 400 dt

Find:

dx dt

(constant rate)



when

 3

You can use the Law of Cosines (Figure 2.39) to find an equation that relates x and . 7 2  3 2  x 2  23x cos

Equation:

0  2x

6 cos  2x

d

dx dx  6 x sin

 cos

dt dt dt





dx d

 6x sin

dt dt 6x sin

d

dx  dt 6 cos  2x dt

 

When  3, you can solve for x as shown. 7 2  3 2  x 2  23x cos 49  9  x 2  6x

 3

12

0  x 2  3x  40 0  x  8x  5 x8

Choose positive solution.

So, when x  8 and  3, the velocity of the piston is

.

dx 6832 400  dt 612  16 96003  13  4018 inches per minute.

Try It

Exploration A

NOTE Note that the velocity in Example 6 is negative because x represents a distance that is decreasing.

154

CHAPTER 2

Differentiation

Exercises for Section 2.6 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1– 4, assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. Equation

Find

1. y  x

2. y  2x 2  3x

3. xy  4

4. x 2  y 2  25

12. In your own words, state the guidelines for solving relatedrate problems.

Given

(a)

dy when x  4 dt

dx 3 dt

(b)

dx when x  25 dt

dy 2 dt

(a)

dy when x  3 dt

dx 2 dt

(b)

dx when x  1 dt

dy 5 dt

(a)

dy when x  8 dt

dx  10 dt

(b)

dx when x  1 dt

dy  6 dt

(a)

dy when x  3, y  4 dt

dx 8 dt dy  2 dt

dx (b) when x  4, y  3 dt

Writing About Concepts (continued)

13. Find the rate of change of the distance between the origin and a moving point on the graph of y  x2  1 if dxdt  2 centimeters per second. 14. Find the rate of change of the distance between the origin and a moving point on the graph of y  sin x if dxdt  2 centimeters per second. 15. Area The radius r of a circle is increasing at a rate of 3 centimeters per minute. Find the rates of change of the area when (a) r  6 centimeters and (b) r  24 centimeters. 16. Area Let A be the area of a circle of radius r that is changing with respect to time. If drdt is constant, is dAdt constant? Explain. 17. Area The included angle of the two sides of constant equal length s of an isosceles triangle is . 1

(a) Show that the area of the triangle is given by A  2s 2 sin . 1

In Exercises 5–8, a point is moving along the graph of the given function such that dx/dt is 2 centimeters per second. Find dy/dt for the given values of x. 5. y  x 2  1

(a) x  1

(b) x  0

(c) x  1

1 6. y  1  x2

(a) x  2

(b) x  0

(c) x  2

7. y  tan x

 (a) x   3

 (b) x   4

(c) x  0

8. y  sin x

 (a) x  6

 (b) x  4

 (c) x  3

Writing About Concepts In Exercises 9 and 10, 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. y

9.

y

10. 6 5 4 3 2

4

2

f

1 2

3

4

x

3 2 1

(c) Explain why the rate of change of the area of the triangle is not constant even though d dt is constant. 18. Volume The radius r of a sphere is increasing at a rate of 2 inches per minute. (a) Find the rate of change of the volume when r  6 inches and r  24 inches. (b) Explain why the rate of change of the volume of the sphere is not constant even though drdt is constant. 19. Volume A spherical balloon is inflated with gas at the rate of 800 cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is (a) 30 centimeters and (b) 60 centimeters? 20. Volume All edges of a cube are expanding at a rate of 3 centimeters per second. How fast is the volume changing when each edge is (a) 1 centimeter and (b) 10 centimeters? 21. Surface Area The conditions are the same as in Exercise 20. Determine how fast the surface area is changing when each edge is (a) 1 centimeter and (b) 10 centimeters. 1 22. Volume The formula for the volume of a cone is V  3 r 2 h. Find the rate of change of the volume if drdt is 2 inches per minute and h  3r when (a) r  6 inches and (b) r  24 inches.

f

x 1

(b) If is increasing at the rate of 2 radian per minute, find the rates of change of the area when  6 and  3.

1 2 3

11. Consider the linear function y  ax  b. If x changes at a constant rate, does y change at a constant rate? If so, does it change at the same rate as x? Explain.

23. Volume At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 15 feet high?

SECTION 2.6

24. Depth A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep. 25. Depth A swimming pool is 12 meters long, 6 meters wide, 1 meter deep at the shallow end, and 3 meters deep at the deep end (see figure). Water is being pumped into the pool at cubic meter per minute, and there is 1 meter of water at the deep end. (a) What percent of the pool is filled? (b) At what rate is the water level rising?

Figure for 25

Related Rates

155

28. Construction A construction worker pulls a five-meter plank up the side of a building under construction by means of a rope tied to one end of the plank (see figure). Assume the opposite end of the plank follows a path perpendicular to the wall of the building and the worker pulls the rope at a rate of 0.15 meter per second. How fast is the end of the plank sliding along the ground when it is 2.5 meters from the wall of the building? 29. Construction A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position, as shown in the figure. The winch pulls in rope at a rate of meter per second. Find the rate of vertical change and the rate of horizontal change at the end of the pipe when

Figure for 26

26. Depth A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet.

Figure for 29

Figure for 30

(a) If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when is 1 foot deep?

30. Boating A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat (see figure).

(b) If the water is rising at a rate of inch per minute when determine the rate at which water is being pumped into the trough.

(a) The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock?

27. Moving Ladder A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall? (b) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. (c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall.

Figure for 27

(b) Suppose the boat is moving at a constant rate of 4 feet per second. Determine the speed at which the winch pulls in rope when there is a total of 13 feet of rope out. What happens to the speed at which the winch pulls in rope as the boat gets closer to the dock? 31. Air Traffic Control An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other (see figure). One plane is 150 miles from the point moving at 450 miles per hour. The other plane is 200 miles from the point moving at 600 miles per hour. (a) At what rate is the distance between the planes decreasing? (b) How much time does the air traffic controller have to get one of the planes on a different flight path?

Figure for 28

FOR FURTHER INFORMATION For more information on the

mathematics of moving ladders, see the article “The Falling Ladder Paradox” by Paul Scholten and Andrew Simoson in The College Mathematics Journal.

Figure for 31

Figure for 32

156

CHAPTER 2

Differentiation

32. Air Traffic Control An airplane is flying at an altitude of 5 miles and passes directly over a radar antenna (see figure on previous page). When the plane is 10 miles away s  10, the radar detects that the distance s is changing at a rate of 240 miles per hour. What is the speed of the plane? 33. Sports A baseball diamond has the shape of a square with sides 90 feet long (see figure). A player running from second base to third base at a speed of 28 feet per second is 30 feet from third base. At what rate is the player’s distance s from home plate changing? y

2nd 16

1st

8 4

90 ft

x

4

Home

Figure for 33 and 34

39. Evaporation As a spherical raindrop falls, it reaches a layer of dry air and begins to evaporate at a rate that is proportional to its surface area S  4 r 2. Show that the radius of the raindrop decreases at a constant rate. 40. Electricity The combined electrical resistance R of R1 and R2, connected in parallel, is given by 1 1 1   R R1 R2 where R, R1, and R2 are measured in ohms. R1 and R2 are increasing at rates of 1 and 1.5 ohms per second, respectively. At what rate is R changing when R1  50 ohms and R2  75 ohms?

12

3rd

38. Machine Design Repeat Exercise 37 for a position function 3 of xt  35 sin  t. Use the point 10 , 0 for part (c).

8

12 16 20

Figure for 35

34. Sports For the baseball diamond in Exercise 33, suppose the player is running from first to second at a speed of 28 feet per second. Find the rate at which the distance from home plate is changing when the player is 30 feet from second base. 35. Shadow Length A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground (see figure). When he is 10 feet from the base of the light,

41. Adiabatic Expansion When a certain polyatomic gas undergoes adiabatic expansion, its pressure p and volume V satisfy the equation pV 1.3  k, where k is a constant. Find the relationship between the related rates dpdt and dVdt. 42. Roadway Design Cars on a certain roadway travel on a circular arc of radius r. In order not to rely on friction alone to overcome the centrifugal force, the road is banked at an angle of magnitude from the horizontal (see figure). The banking angle must satisfy the equation rg tan  v 2, where v is the velocity of the cars and g  32 feet per second per second is the acceleration due to gravity. Find the relationship between the related rates dvdt and d dt.

(a) at what rate is the tip of his shadow moving? (b) at what rate is the length of his shadow changing? 36. Shadow Length Repeat Exercise 35 for a man 6 feet tall walking at a rate of 5 feet per second toward a light that is 20 feet above the ground (see figure). y



y

20

r

16

(0, y)

12

1m

8 4

(x, 0)

x

x

4

8

12 16 20

Figure for 36

Figure for 37

37. Machine Design The endpoints of a movable rod of length 1 meter have coordinates x, 0 and 0, y (see figure). The position of the end on the x-axis is xt 

1 t sin 2 6

where t is the time in seconds. (a) Find the time of one complete cycle of the rod. (b) What is the lowest point reached by the end of the rod on the y-axis? (c) Find the speed of the y-axis endpoint when the x-axis endpoint is 14, 0.

43. Angle of Elevation A balloon rises at a rate of 3 meters per second from a point on the ground 30 meters from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 30 meters above the ground. 44. Angle of Elevation A fish is reeled in at a rate of 1 foot per second from a point 10 feet above the water (see figure). At what rate is the angle between the line and the water changing when there is a total of 25 feet of line out?

10 ft

x



SECTION 2.6

45. Angle of Elevation An airplane flies at an altitude of 5 miles toward a point directly over an observer (see figure). The speed of the plane is 600 miles per hour. Find the rates at which the angle of elevation is changing when the angle is (a)  30, (b)  60, and (c)  75.

Related Rates

157

50. Think About It Describe the relationship between the rate of change of y and the rate of change of x in each expression. Assume all variables and derivatives are positive. (a)

dy dx 3 dt dt

(b)

dy dx  xL  x , dt dt

0  x  L

Acceleration In Exercises 51 and 52, find the acceleration of the specified object. (Hint: Recall that if a variable is changing at a constant rate, its acceleration is zero.)

5 mi

51. Find the acceleration of the top of the ladder described in Exercise 27 when the base of the ladder is 7 feet from the wall.

 Not drawn to scale

46. Linear vs. Angular Speed A patrol car is parked 50 feet from a long warehouse (see figure). The revolving light on top of the car turns at a rate of 30 revolutions per minute. How fast is the light beam moving along the wall when the beam makes angles of (a)  30, (b)  60, and (c)  70 with the line perpendicular from the light to the wall?

52. Find the acceleration of the boat in Exercise 30(a) when there is a total of 13 feet of rope out. 53. Modeling Data The table shows the numbers (in millions) of single women (never married) s and married women m in the civilian work force in the United States for the years 1993 through 2001. (Source: U.S. Bureau of Labor Statistics) Year 1993 1994 1995 1996 1997 1998 1999 2000 2001

P





50 ft

30 cm

x

x

Figure for 47

47. Linear vs. Angular Speed A wheel of radius 30 centimeters revolves at a rate of 10 revolutions per second. A dot is painted at a point P on the rim of the wheel (see figure). (a) Find dxdt as a function of . (b) Use a graphing utility to graph the function in part (a). (c) When is the absolute value of the rate of change of x greatest? When is it least? (d) Find dxdt when  30 and  60. 48. Flight Control An airplane is flying in still air with an airspeed of 240 miles per hour. If it is climbing at an angle of 22, find the rate at which it is gaining altitude. 49. Security Camera A security camera is centered 50 feet above a 100-foot hallway (see figure). It is easiest to design the camera with a constant angular rate of rotation, but this results in a variable rate at which the images of the surveillance area are recorded. So, it is desirable to design a system with a variable rate of rotation and a constant rate of movement of the scanning beam along the hallway. Find a model for the variable rate of rotation if dxdt  2 feet per second.





15.0 15.3 15.5 15.8 16.5 17.1 17.6 17.8 18.0

m

32.0 32.9 33.4 33.6 33.8 33.9 34.4 34.6 34.7

(a) Use the regression capabilities of a graphing utility to find a model of the form ms  as3  bs2  cs  d for the data, where t is the time in years, with t  3 corresponding to 1993.

x

Figure for 46

s

(b) Find dmdt. Then use the model to estimate dmdt for t  10 if it is predicted that the number of single women in the work force will increase at the rate of 0.75 million per year. 54. Moving Shadow A ball is dropped from a height of 20 meters, 12 meters away from the top of a 20-meter lamppost (see figure). The ball’s shadow, caused by the light at the top of the lamppost, is moving along the level ground. How fast is the shadow moving 1 second after the ball is released? (Submitted by Dennis Gittinger, St. Philips College, San Antonio, TX)

20 m

y

(0, 50) Shadow



12 m x

100 ft

158

CHAPTER 2

Differentiation

Review Exercises for Chapter 2 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, find the derivative of the function by using the definition of the derivative. 1. f x  x 2  2x  3 3. f x 

.

4. f x 

20. f t  8t 5

21. f x  x  3x 3

22. gs  4s 4  5s 2

2

3 x 23. hx  6x  3

2. f x  x  1

x1 x1

19. ht  3t 4

2 x

25. gt 

24. f x  x12  x12

2 3t 2

27. f    2  3 sin

In Exercises 5 and 6, describe the x-values at which f is differentiable. 5. f x  x  12 3

6. f x 

4x x3

29. f    3 cos 

3

8

2

4

31.

y 12

4

1

y

1

4

4

1

8 1

2





 2

7. Sketch the graph of f x  4  x  2 . (a) Is f continuous at x  2?



x2  4x  2, 1  4x  x2,

x < 2 x  2.

(a) Is f continuous at x  2? (b) Is f differentiable at x  2? Explain. In Exercises 9 and 10, find the slope of the tangent line to the graph of the function at the given point. 2 x 9. gx  x 2  , 3 6 3x  2x 2, 8

1, 65 2,  354

In Exercises 11 and 12, (a) find an equation of the tangent line to the graph of f at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your results. 2 , 0, 2 11. f x  x 3  1, 1, 2 12. f x  x1 In Exercises 13 and 14, use the alternative form of the derivative to find the derivative at x  c (if it exists). 1 , c2 13. gx  x 2x  1, c  2 14. f x  x1 In Exercises 15–30, find the derivative of the function. 15. y  25 17. f x  x

16. y  12 8

18. gx  x12

 2

x

x 1

(b) Is f differentiable at x  2? Explain. 8. Sketch the graph of f x 

y

32.

2

x

8

x 1

10. hx 

sin

4

Writing In Exercises 31 and 32, the figure shows the graphs of a function and its derivative. Label the graphs as f or f and write a short paragraph stating the criteria used in making the selection. To print an enlarged copy of the graph, select the MathGraph button.

y

2

2 3x 2 28. g  4 cos   6 5 sin   2 30. g  3 26. hx 

1

33. Vibrating String When a guitar string is plucked, it vibrates with a frequency of F  200T, where F is measured in vibrations per second and the tension T is measured in pounds. Find the rates of change of F when (a) T  4 and (b) T  9. 34. Vertical Motion A ball is dropped from a height of 100 feet. One second later, another ball is dropped from a height of 75 feet. Which ball hits the ground first? 35. Vertical Motion To estimate the height of a building, a weight is dropped from the top of the building into a pool at ground level. How high is the building if the splash is seen 9.2 seconds after the weight is dropped? 36. Vertical Motion A bomb is dropped from an airplane at an altitude of 14,400 feet. How long will it take for the bomb to reach the ground? (Because of the motion of the plane, the fall will not be vertical, but the time will be the same as that for a vertical fall.) The plane is moving at 600 miles per hour. How far will the bomb move horizontally after it is released from the plane? 37. Projectile Motion A ball thrown follows a path described by y  x  0.02x 2. (a) Sketch a graph of the path. (b) Find the total horizontal distance the ball is thrown. (c) At what x-value does the ball reach its maximum height? (Use the symmetry of the path.) (d) Find an equation that gives the instantaneous rate of change of the height of the ball with respect to the horizontal change. Evaluate the equation at x  0, 10, 25, 30, and 50. (e) What is the instantaneous rate of change of the height when the ball reaches its maximum height?

159

REVIEW EXERCISES

38. Projectile Motion The path of a projectile thrown at an angle of 45 with level ground is yx

32 2 x  v02

where the initial velocity is v0 feet per second.

39. Horizontal Motion The position function of a particle moving along the x-axis is for

 < t
0, is shown below. y

1.0

cos x

x

a

P2x (d) Find the third-degree Taylor polynomial of f x  sin x at x  0. 4. (a) Find an equation of the tangent line to the parabola y  x 2 at the point 2, 4. (b) Find an equation of the normal line to y  x 2 at the point 2, 4. (The normal line is perpendicular to the tangent line.) Where does this line intersect the parabola a second time? (c) Find equations of the tangent line and normal line to y  x 2 at the point 0, 0. (d) Prove that for any point a, b  0, 0 on the parabola y  x 2, the normal line intersects the graph a second time.

(a) Explain how you could use a graphing utility to graph this curve. (b) Use a graphing utility to graph the curve for various values of the constants a and b. Describe how a and b affect the shape of the curve. (c) Determine the points on the curve where the tangent line is horizontal.

162

CHAPTER 2

Differentiation

9. A man 6 feet tall walks at a rate of 5 feet per second toward a streetlight that is 30 feet high (see figure). The man’s 3-foot-tall child follows at the same speed, but 10 feet behind the man. At times, the shadow behind the child is caused by the man, and at other times, by the child. (a) Suppose the man is 90 feet from the streetlight. Show that the man’s shadow extends beyond the child’s shadow. (b) Suppose the man is 60 feet from the streetlight. Show that the child’s shadow extends beyond the man’s shadow. (c) Determine the distance d from the man to the streetlight at which the tips of the two shadows are exactly the same distance from the streetlight. (d) Determine how fast the tip of the shadow is moving as a function of x, the distance between the man and the street light. Discuss the continuity of this shadow speed function.

3



Not drawn to scale

2

Figure for 9

for z in degrees. What is the exact value of this limit? (Hint: 180   radians) (c) Use the limit definition of the derivative to find d sin z dz for z in degrees. (d) Define the new functions Sz  sincz and Cz  coscz, where c  180. Find S90 and C180. Use the Chain Rule to calculate

14. An astronaut standing on the moon throws a rock into the air. The height of the rock is

1

3 ft 10 ft

sin z z

(e) Explain why calculus is made easier by using radians instead of degrees.

(8, 2) 2

6 ft

lim

z0

d Sz. dz

y

30 ft

(b) Use the table to estimate

x 4

6

8

10

27 2 t  27t  6 10

1

s

Figure for 10

where s is measured in feet and t is measured in seconds.

3 x (see figure). 10. A particle is moving along the graph of y   When x  8, the y-component of its position is increasing at the rate of 1 centimeter per second.

(a) How fast is the x-component changing at this moment? (b) How fast is the distance from the origin changing at this moment? (c) How fast is the angle of inclination changing at this moment? 11. Let L be a differentiable function for all x. Prove that if La  b  La  Lb for all a and b, then L x  L 0 for all x. What does the graph of L look like? 12. Let E be a function satisfying E0  E 0  1. Prove that if Ea  b  EaEb for all a and b, then E is differentiable and E x  Ex for all x. Find an example of a function satisfying Ea  b  EaEb. sin x  1 assumes that x is measured x in radians. What happens if you assume that x is measured in degrees instead of radians?

(a) Find expressions for the velocity and acceleration of the rock. (b) Find the time when the rock is at its highest point by finding the time when the velocity is zero. What is the height of the rock at this time? (c) How does the acceleration of the rock compare with the acceleration due to gravity on Earth? 15. If a is the acceleration of an object, the jerk j is defined by j  at. (a) Use this definition to give a physical interpretation of j. (b) Find j for the slowing vehicle in Exercise 117 in Section 2.3 and interpret the result. (c) The figure shows the graph of the position, velocity, acceleration, and jerk functions of a vehicle. Identify each graph and explain your reasoning. y

13. The fundamental limit lim

x0

a b

(a) Set your calculator to degree mode and complete the table.

x

c

z (in degrees) sin z z

0.1

0.01

0.0001

d

164

CHAPTER 3

Applications of Differentiation

Section 3.1

Extrema on an Interval • Understand the definition of extrema of a function on an interval. • Understand the definition of relative extrema of a function on an open interval. • Find extrema on a closed interval.

.

Extrema of a Function

Video y

Maximum

(2, 5)

5

In calculus, much effort is devoted to determining the behavior of a function f on an interval I. Does f have a maximum value on I? Does it have a minimum value? Where is the function increasing? Where is it decreasing? In this chapter you will learn how derivatives can be used to answer these questions. You will also see why these questions are important in real-life applications.

f(x) = x 2 + 1

4

Definition of Extrema

3

Let f be defined on an interval I containing c.

2

Minimum

(0, 1)

x

−1

1

2

3

(a) f is continuous, 1, 2 is closed. y 5

Not a maximum

4

f(x) = x 2 + 1

3 2

Minimum

(0, 1)

x

−1

1

2

y

Maximum

(2, 5)

4

g(x) = 3

x 2 + 1, x ≠ 0 2, x=0

2

Not a minimum x

−1

1

2

3

(c) g is not continuous, 1, 2 is closed. Extrema can occur at interior points or endpoints of an interval. Extrema that occur at the endpoints are called endpoint extrema.

Figure 3.1

A function need not have a minimum or a maximum on an interval. For instance, in Figure 3.1(a) and (b), you can see that the function f x  x 2  1 has both a minimum and a maximum on the closed interval 1, 2 , but does not have a maximum on the open interval 1, 2. Moreover, in Figure 3.1(c), you can see that continuity (or the lack of it) can affect the existence of an extremum on the interval. This suggests the theorem below. (Although the Extreme Value Theorem is intuitively plausible, a proof of this theorem is not within the scope of this text.)

3

(b) f is continuous, 1, 2 is open.

5

1. f c is the minimum of f on I if f c ≤ f x for all x in I. 2. f c is the maximum of f on I if f c ≥ f x for all x in I. The minimum and maximum of a function on an interval are the extreme values, or extrema (the singular form of extrema is extremum), of the function on the interval. The minimum and maximum of a function on an interval are also called the absolute minimum and absolute maximum on the interval.

THEOREM 3.1

The Extreme Value Theorem

If f is continuous on a closed interval a, b , then f has both a minimum and a maximum on the interval.

E X P L O R AT I O N

Finding Minimum and Maximum Values The Extreme Value Theorem (like the Intermediate Value Theorem) is an existence theorem because it tells of the existence of minimum and maximum values but does not show how to find these values. Use the extreme-value capability of a graphing utility to find the minimum and maximum values of each of the following functions. In each case, do you think the x-values are exact or approximate? Explain your reasoning. a. f x  x 2  4x  5 on the closed interval 1, 3 b. f x  x 3  2x 2  3x  2 on the closed interval 1, 3

SECTION 3.1

y

Hill (0, 0)

x

1

2

−2 −3

165

Relative Extrema and Critical Numbers

f(x) = x 3 − 3x 2

−1

Extrema on an Interval

Valley (2, −4)

In Figure 3.2, the graph of f x  x 3  3x 2 has a relative maximum at the point 0, 0 and a relative minimum at the point 2, 4. Informally, you can think of a relative maximum as occurring on a “hill” on the graph, and a relative minimum as occurring in a “valley” on the graph. Such a hill and valley can occur in two ways. If the hill (or valley) is smooth and rounded, the graph has a horizontal tangent line at the high point (or low point). If the hill (or valley) is sharp and peaked, the graph represents a function that is not differentiable at the high point (or low point).

−4

f has a relative maximum at 0, 0 and a relative minimum at 2,  4. Figure 3.2

y

f(x) =

Relative maximum

9(x2 − 3) x3

2

(3, 2) x 2

6

4

Definition of Relative Extrema 1. If there is an open interval containing c on which f c is a maximum, then f c is called a relative maximum of f, or you can say that f has a relative maximum at c, f c. 2. If there is an open interval containing c on which f c is a minimum, then f c is called a relative minimum of f, or you can say that f has a relative minimum at c, f c. The plural of relative maximum is relative maxima, and the plural of relative minimum is relative minima.

−2

Example 1 examines the derivatives of functions at given relative extrema. (Much more is said about finding the relative extrema of a function in Section 3.3.)

−4

. (a) f3  0

Find the value of the derivative at each of the relative extrema shown in Figure 3.3.

y

Solution

f(x) = ⎜x ⎜ 3

a. The derivative of f x 

2 1

Relative minimum

f x 

x

−1

−2

1 −1

2



(0, 0)

. (b) f0 does not exist.

..

x

3π 2

Relative 3π , −1 minimum 2

(

(c) f

)

2   0; f32  0

Editable Graph Figure 3.3

Simplify.

f x  f 0  lim x→0 x0 f x  f 0  lim lim x→0  x→0 x0

( )

−2

99  x 2 . x4

x→0

π , 1 Relative 2 maximum

−1

Differentiate using Quotient Rule.

lim

f(x) = sin x

π 2

x 318x  9x 2  33x 2 x 3 2



y

1

9x 2  3 is x3

At the point 3, 2, the value of the derivative is f3  0 [see Figure 3.3(a)]. b. At x  0, the derivative of f x  x does not exist because the following one-sided limits differ [see Figure 3.3(b)].

Editable Graph

2

The Value of the Derivative at Relative Extrema

EXAMPLE 1

Editable Graph

x  1

Limit from the left



Limit from the right

x x 1 x

c. The derivative of f x  sin x is fx  cos x. At the point 2, 1, the value of the derivative is f2  cos2  0. At the point 32, 1, the value of the derivative is f32  cos32  0 [see Figure 3.3(c)].

Try It

Exploration A

Exploration B

166

CHAPTER 3

Applications of Differentiation

Note in Example 1 that at the relative extrema, the derivative is either zero or does not exist. The x-values at these special points are called critical numbers. Figure 3.4 illustrates the two types of critical numbers.

Definition of a Critical Number Let f be defined at c. If fc  0 or if f is not differentiable at c, then c is a critical number of f. y

y

f ′(c) does not exist. f ′(c) = 0

c

x

Horizontal tangent

c

x

c is a critical number of f. Figure 3.4

THEOREM 3.2

Relative Extrema Occur Only at Critical Numbers

If f has a relative minimum or relative maximum at x  c, then c is a critical number of f.

Proof Case 1: If f is not differentiable at x  c, then, by definition, c is a critical number of f and the theorem is valid. Case 2: If f is differentiable at x  c, then fc must be positive, negative, or 0. Suppose fc is positive. Then fc  lim

x→c

PIERRE DE FERMAT (1601–1665) For Fermat, who was trained as a lawyer, mathematics was more of a hobby than a profession. Nevertheless, Fermat made many contributions to analytic geometry, number theory, calculus, and probability. In letters to friends, he wrote of many of the fundamental ideas of calculus, long before Newton or Leibniz. For instance, the theorem at the right .is sometimes attributed to Fermat.

MathBio

f x  f c > 0 xc

which implies that there exists an interval a, b containing c such that f x  f c > 0, for all x  c in a, b. xc

[See Exercise 72(b), Section 1.2.]

Because this quotient is positive, the signs of the denominator and numerator must agree. This produces the following inequalities for x-values in the interval a, b. Left of c: x < c and f x < f c Right of c: x > c and f x > f c

f c is not a relative minimum f c is not a relative maximum

So, the assumption that f c > 0 contradicts the hypothesis that f c is a relative extremum. Assuming that f c < 0 produces a similar contradiction, you are left with only one possibility—namely, f c  0. So, by definition, c is a critical number of f and the theorem is valid.

SECTION 3.1

Extrema on an Interval

167

Finding Extrema on a Closed Interval Theorem 3.2 states that the relative extrema of a function can occur only at the critical numbers of the function. Knowing this, you can use the following guidelines to find extrema on a closed interval.

Guidelines for Finding Extrema on a Closed Interval To find the extrema of a continuous function f on a closed interval a, b , use the following steps. 1. 2. 3. 4.

.

Find the critical numbers of f in a, b. Evaluate f at each critical number in a, b. Evaluate f at each endpoint of a, b . The least of these values is the minimum. The greatest is the maximum.

Technology The next three examples show how to apply these guidelines. Be sure you see that finding the critical numbers of the function is only part of the procedure. Evaluating the function at the critical numbers and the endpoints is the other part. EXAMPLE 2

Finding Extrema on a Closed Interval

Find the extrema of f x  3x 4  4x 3 on the interval 1, 2 . Solution Begin by differentiating the function. f x  3x 4  4x 3 f  x  12x 3  12x 2

f x  12x 3  12x 2  0

(2, 16) Maximum

x  1  0 x  0, 1

8 4

(0, 0) −1

x

2

−4

Set f x equal to 0.

12x 2

12

(−1, 7)

Differentiate.

To find the critical numbers of f, you must find all x-values for which f x  0 and all x-values for which fx does not exist.

y 16

Write original function.

(1, −1) Minimum

f(x) = 3x 4 − 4x 3

On the closed interval  1, 2 , f has a minimum at 1,  1 and a maximum .. at 2, 16.

Factor. Critical numbers

Because f  is defined for all x, you can conclude that these are the only critical numbers of f. By evaluating f at these two critical numbers and at the endpoints of

1, 2 , you can determine that the maximum is f 2  16 and the minimum is f 1  1, as shown in the table. The graph of f is shown in Figure 3.5. Left Endpoint

Critical Number

Critical Number

Right Endpoint

f 1  7

f 0  0

f 1  1 Minimum

f 2  16 Maximum

Figure 3.5

Editable Graph

Try It

Exploration A

Exploration B

Video

In Figure 3.5, note that the critical number x  0 does not yield a relative minimum or a relative maximum. This tells you that the converse of Theorem 3.2 is not true. In other words, the critical numbers of a function need not produce relative extrema.

168

CHAPTER 3

Applications of Differentiation

y

EXAMPLE 3

(0, 0) Maximum −2

−1

1

x

2

(1, −1)

Finding Extrema on a Closed Interval

Find the extrema of f x  2x  3x 23 on the interval 1, 3 . Solution Begin by differentiating the function.

)3, 6 − 3 3 9 )

f x  2x  3x23 f x  2 

−4

Minimum (−1, −5)

On the closed interval  1, 3 , f has a minimum at  1,  5 and a maximum at.0, 0. Figure 3.6

Editable Graph .

( π2, 3 ) Maximum

(

−1

(0, −1)

π 2

−2 −3

)

π

(

(2π, −1)

7π , − 3 6 2

11π , − 3 6 2

) (

Critical Number

Right Endpoint

f 1  5 Minimum

f 0  0 Maximum

f 1  1

3 9  0.24 f 3  6  3 

Editable Graph

Exploration A Finding Extrema on a Closed Interval

Solution This function is differentiable for all real x, so you can find all critical numbers by differentiating the function and setting f x equal to zero, as shown. f x  2 sin x  cos 2x f x  2 cos x  2 sin 2x  0

)

Figure 3.7

.

x

Write original function. Set f x equal to 0.

2 cos x  4 cos x sin x  0 2cos x1  2 sin x  0

Minima

On the closed interval 0, 2 , f has two minima at 7 6,  32 and 11 6,  32 and a maximum at . 2, 3.

Differentiate.

Find the extrema of f x  2 sin x  cos 2x on the interval 0, 2 .

2

3π , −1 2



Critical Number

EXAMPLE 4

f(x) = 2 sin x − cos 2x

1



Left Endpoint

Try It

y

3

x 13

Write original function.

x 13  1 2 x 13

From this derivative, you can see that the function has two critical numbers in the interval 1, 3 . The number 1 is a critical number because f 1  0, and the number 0 is a critical number because f 0 does not exist. By evaluating f at these two numbers and at the endpoints of the interval, you can conclude that the minimum is f 1  5 and the maximum is f 0  0, as shown in the table. The graph of f is shown in Figure 3.6.

−5

f(x) = 2x − 3x 2/3

4

2

sin 2x  2 cos x sin x Factor.

In the interval 0, 2 , the factor cos x is zero when x  2 and when x  32. The factor 1  2 sin x is zero when x  76 and when x  116. By evaluating f at these four critical numbers and at the endpoints of the interval, you can conclude that the maximum is f 2  3 and the minimum occurs at two points, f 76  32 and f 116  32, as shown in the table. The graph is shown in Figure 3.7.

Left Endpoint

Critical Number

f 0  1

f

2   3

Maximum

Try It

Critical Number f

76   32

Critical Number f

32  1

Minimum

Open Exploration

f

Critical Number

Right Endpoint

116   32

f 2  1

Minimum

SECTION 3.1

169

Extrema on an Interval

Exercises for Section 3.1 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, decide whether each labeled point is an absolute maximum or minimum, a relative maximum or minimum, or neither. 1.

2.

y

y

In Exercises 9–12, approximate the critical numbers of the function shown in the graph. Determine whether the function has a relative maximum, relative minimum, absolute maximum, absolute minimum, or none of these at each critical number on the interval shown.

G

B

B G

C

C

3 x

2

D x

1

4

x

F

F

D

y

10.

5

E

A

y

9.

E

1

1

A

1

x

1

In Exercises 3–8, find the value of the derivative (if it exists) at each indicated extremum. 3. f x 

x2

4. f x  cos

x2  4 y

1

2

3

4

y

11.

x 2

y

1

5 y

12.

5

8

4

6

3

4

2 2

2

1

(0, 0)

1

1 1

2

2

27 2x 2

(

4

)

3

7. f x  x  2

4



8. f x  4  x y

y

2

6

1

4 x

3

2

2

5

23

(2, 0)

1

2

x

2

1 2

x

4

2

2 2

4

6

8

14. gx  x 2x 2  4

15. gt  t4  t, t < 3

16. f x 

17. hx  sin 2 x  cos x

18. f    2 sec  tan

4x x2  1

0 < < 2

4

2x  5 , 0, 5 3

19. f x  23  x, 1, 2

20. f x 

21. f x  x 2  3x, 0, 3

22. f x  x 2  2x  4, 1, 1

3 23. f x  x 3  x 2, 1, 2 2

24. f x  x 3  12x, 0, 4

25. y  3x 23  2x, 1, 1

3 x, 1, 1 26. gx  

2

27. gt 

t , 1, 1 t2  3

28. f x 

29. hs 

1 , 0, 1 s2

30. ht 

(0, 4)

2

1

2

13. f x  x 2x  3

x

1 1

2 2

5

In Exercises 19–36, locate the absolute extrema of the function on the closed interval.

2 (1, 0) 1

2

1

4

0 < x < 2

 2, 2 3 2 3 3

(3, 92 )

3

3

In Exercises 13–18, find any critical numbers of the function.

y

5

2

3

(2, 1)

y

4

2

6. f x  3xx  1

6

1

x

2

1

5. f x  x 

x

x

1 x

2

2

1

(0, 1)





31. y  3  t  3 , 1, 5 1 33. f x  cos  x, 0, 6 35. y 

 

t , 3, 5 t2

32. f x  x, 2, 2   34. gx  sec x,  , 6 3

4 x  tan , 1, 2 x 8

 

36. y  x  2  cos x, 1, 3 2

2x , 2, 2 x2  1





170

CHAPTER 3

Applications of Differentiation

In Exercises 37–40, locate the absolute extrema of the function (if any exist) over each interval.

Writing About Concepts

37. f x  2x  3

38. f x  5  x

(a) 0, 2

(b) 0, 2

(a) 1, 4

(b) 1, 4

In Exercises 53 and 54, graph a function on the interval [2, 5] having the given characteristics.

(c) 0, 2

(d) 0, 2

(c) 1, 4

(d) 1, 4

39. f x 

x2

40. f x  4 

 2x

x2

(a) 1, 2 (b) 1, 3

(a) 2, 2 (b) 2, 0

(c) 0, 2

(c) 2, 2 (d) 1, 2

(d) 1, 4

In Exercises 41–44, sketch the graph of the function. Then locate the absolute extrema of the function over the given interval. 2x  2, 4x2,

 2x , 42. f x   2  3x, 41. f x 

0  x  1 , 0, 3 1 < x  3

53. Absolute maximum at x  2, absolute minimum at x  1, relative maximum at x  3 54. Relative minimum at x  1, critical number at x  0, but no extrema, absolute maximum at x  2, absolute minimum at x  5 In Exercises 55–58, determine from the graph whether f has a minimum in the open interval a, b. 55. (a)

(b)

y

y

1  x < 3 , 1, 5 3  x  5

2

43. f x 

3 , 1, 4 x1

44. f x 

2 , 0, 2 2x

f

a

In Exercises 45 and 46, use a graphing utility to graph the function. Then locate the absolute extrema of the function over the given interval.

x

b

56. (a)

a

x

b

(b)

y

y

45. f x  x 4  2x3  x  1, 1, 3 x 46. f x  x  cos , 2

f

f

f

0, 2

In Exercises 47 and 48, (a) use a computer algebra system to graph the function and approximate any absolute extrema on the given interval. (b) Use the utility to find any critical numbers, and use them to find any absolute extrema not located at the endpoints. Compare the results with those in part (a).

a

x

b

57. (a)

a

x

b

(b)

y

y

47. f x  3.2x 5  5x 3  3.5x, 0, 1 48. f x 

4 x3  x, 3

f

0, 3

In Exercises 49 and 50, use a computer algebra system to find the maximum value of f x on the closed interval. (This value is used in the error estimate for the Trapezoidal Rule, as discussed in Section 4.6.)



49. f x  1  x3, 50. f x 

1 , x2  1



0, 2 1 , 3 2

a

f

x

b

58. (a)

a

x

b

(b)

y

y

 

In Exercises 51 and 52, use a computer algebra system to find the maximum value of f 4x on the closed interval. (This value is used in the error estimate for Simpson’s Rule, as discussed in Section 4.6.) 1 , 1, 1 51. f x  x  1 23, 0, 2 52. f x  2 x 1





f

a

b

f x

a

b

x

SECTION 3.1

59. Power The formula for the power output P of a battery is P  V I  R I 2, where V is the electromotive force in volts, R is the resistance, and I is the current. Find the current (measured in amperes) that corresponds to a maximum value of P in a battery for which V  12 volts and R  0.5 ohm. Assume that a 15-ampere fuse bounds the output in the interval 0  I  15. Could the power output be increased by replacing the 15-ampere fuse with a 20-ampere fuse? Explain. 60. Inventory Cost A retailer has determined that the cost C of ordering and storing x units of a product is 300,000 , 1  x  300. C  2x  x The delivery truck can bring at most 300 units per order. Find the order size that will minimize cost. Could the cost be decreased if the truck were replaced with one that could bring at most 400 units? Explain. 61. Lawn Sprinkler A lawn sprinkler is constructed in such a way that d dt is constant, where ranges between 45 and 135 (see figure). The distance the water travels horizontally is x

v 2 sin 2

, 32

45   135

where v is the speed of the water. Find dxdt and explain why this lawn sprinkler does not water evenly. What part of the lawn receives the most water?  = 105°

y

 = 45°

 



32

v2

x

v2 64

64

FOR FURTHER INFORMATION For more information on the geometric structure of a honeycomb cell, see the article “The Design of Honeycombs” by Anthony L. Peressini in UMAP Module 502, published by COMAP, Inc., Suite 210, 57 Bedford Street, Lexington, MA. True or False? In Exercises 63–66, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 63. The maximum of a function that is continuous on a closed interval can occur at two different values in the interval. 64. If a function is continuous on a closed interval, then it must have a minimum on the interval. 65. If x  c is a critical number of the function f, then it is also a critical number of the function gx  f x  k, where k is a constant. 66. If x  c is a critical number of the function f, then it is also a critical number of the function gx  f x  k, where k is a constant. 67. Let the function f be differentiable on an interval I containing c. If f has a maximum value at x  c, show that f has a minimum value at x  c. 68. Consider the cubic function f x  ax 3  bx2  cx  d where a  0. Show that f can have zero, one, or two critical numbers and give an example of each case.

 = 75°

 = 135°

v2

171

Extrema on an Interval

v2 32

69. Highway Design In order to build a highway, it is necessary to fill a section of a valley where the grades (slopes) of the sides are 9% and 6% (see figure). The top of the filled region will have the shape of a parabolic arc that is tangent to the two slopes at the points A and B. The horizontal distance between the points A and B is 1000 feet.

Water sprinkler: 45°    135° y

Animation FOR FURTHER INFORMATION For more information on the “calculus of lawn sprinklers,” see the article “Design of an Oscillating Sprinkler” by Bart Braden in Mathematics Magazine. MathArticle



3s 2 3  cos

2 sin



where h and s are positive constants and is the angle at which the upper faces meet the altitude of the cell (see figure). Find the angle 6   2 that minimizes the surface area S. 

h

s

9%

grad

e

Highway B de 6% gra

x

Not drawn to scale

62. Honeycomb The surface area of a cell in a honeycomb is S  6hs 

1000 ft A

(a) Find a quadratic function y  ax 2  bx  c, 500  x  500, that describes the top of the filled region. (b) Construct a table giving the depths d of the fill for x  500, 400, 300, 200, 100, 0, 100, 200, 300, 400, and 500. (c) What will be the lowest point on the completed highway? Will it be directly over the point where the two hillsides come together?

172

CHAPTER 3

Applications of Differentiation

Section 3.2

Rolle’s Theorem and the Mean Value Theorem • Understand and use Rolle’s Theorem. • Understand and use the Mean Value Theorem.

Rolle’s Theorem ROLLE’S THEOREM French mathematician Michel Rolle first published the theorem that bears his name in 1691. Before this time, however, Rolle was one of the most vocal critics of calculus, stating that it gave erroneous results and was based on unsound reasoning. Later in life, Rolle came to see the usefulness of calculus.

The Extreme Value Theorem (Section 3.1) states that a continuous function on a closed interval a, b must have both a minimum and a maximum on the interval. Both of these values, however, can occur at the endpoints. Rolle’s Theorem, named after the French mathematician Michel Rolle (1652–1719), gives conditions that guarantee the existence of an extreme value in the interior of a closed interval. E X P L O R AT I O N

Extreme Values in a Closed Interval Sketch a rectangular coordinate plane on a piece of paper. Label the points 1, 3 and 5, 3. Using a pencil or pen, draw the graph of a differentiable function f that starts at 1, 3 and ends at 5, 3. Is there at least one point on the graph for which the derivative is zero? Would it be possible to draw the graph so that there isn’t a point for which the derivative is zero? Explain your reasoning.

THEOREM 3.3 y

Rolle’s Theorem

Let f be continuous on the closed interval a, b and differentiable on the open interval a, b. If

Relative maximum

f a  f b f

then there is at least one number c in a, b such that f c  0. Proof Let f a  d  f b.

d

a

c

b

x

(a) f is continuous on a, b and differentiable on a, b. y

Relative maximum f

d

a

c

b

(b) f is continuous on a, b .

Figure 3.8

x

Case 1: If f x  d for all x in a, b , f is constant on the interval and, by Theorem 2.2, fx  0 for all x in a, b. Case 2: Suppose f x > d for some x in a, b. By the Extreme Value Theorem, you know that f has a maximum at some c in the interval. Moreover, because f c > d, this maximum does not occur at either endpoint. So, f has a maximum in the open interval a, b. This implies that f c is a relative maximum and, by Theorem 3.2, c is a critical number of f. Finally, because f is differentiable at c, you can conclude that fc  0. Case 3: If f x < d for some x in a, b, you can use an argument similar to that in Case 2, but involving the minimum instead of the maximum. From Rolle’s Theorem, you can see that if a function f is continuous on a, b and differentiable on a, b, and if f a  f b, there must be at least one x-value between a and b at which the graph of f has a horizontal tangent, as shown in Figure 3.8(a). If the differentiability requirement is dropped from Rolle’s Theorem, f will still have a critical number in a, b, but it may not yield a horizontal tangent. Such a case is shown in Figure 3.8(b).

SECTION 3.2

EXAMPLE 1

Rolle’s Theorem and the Mean Value Theorem

173

Illustrating Rolle’s Theorem

Find the two x-intercepts of f x  x 2  3x  2

y

and show that f x)  0 at some point between the two x-intercepts.

f(x) = x 2 − 3x + 2 2

Solution Note that f is differentiable on the entire real line. Setting f x equal to 0 produces

1

(1, 0)

(2, 0)

x 3

f ′ ( 32 ) = 0

−1

Horizontal tangent

The x-value for which fx  0 is between . the two x-intercepts. Figure 3.9

x 2  3x  2  0 x  1x  2  0.

Factor.

So, f 1  f 2  0, and from Rolle’s Theorem you know that there exists at least one c in the interval 1, 2 such that f c  0. To find such a c, you can solve the equation f x  2x  3  0

Set fx equal to 0.

and determine that f x  0 when x  1, 2, as shown in Figure 3.9.

Try It

Editable Graph

Set f x equal to 0.

3 2.

Note that the x-value lies in the open interval

Exploration A

Rolle’s Theorem states that if f satisfies the conditions of the theorem, there must be at least one point between a and b at which the derivative is 0. There may of course be more than one such point, as shown in the next example. y

f(−2) = 8

EXAMPLE 2

f(x) = x 4 − 2x 2 f(2) = 8

8 6

2

f ′(0) = 0

x

2

f ′(−1) = 0 −2

Let f x  x 4  2x 2. Find all values of c in the interval 2, 2 such that fc  0. Solution To begin, note that the function satisfies the conditions of Rolle’s Theorem. That is, f is continuous on the interval 2, 2 and differentiable on the interval 2, 2. Moreover, because f 2  f 2  8, you can conclude that there exists at least one c in 2, 2 such that f c  0. Setting the derivative equal to 0 produces

4

−2

Illustrating Rolle’s Theorem

f ′(1) = 0

fx  0 for more than one x-value in the . interval 2, 2. Figure 3.10

f x  4x 3  4x  0 4xx  1x  1  0 x  0, 1, 1.

Factor. x -values for which fx  0

So, in the interval 2, 2, the derivative is zero at three different values of x, as shown in Figure 3.10.

Try It

Editable Graph

Set fx equal to 0.

Exploration A

A graphing utility can be used to indicate whether the points on the graphs in Examples 1 and 2 are relative minima or relative maxima of the functions. When using a graphing utility, however, you should keep in mind that it can give misleading pictures of graphs. For example, use a graphing utility to graph

TECHNOLOGY PITFALL

3

−3

6

−3

Figure 3.11

f x  1  x  1 2 

1 . 1000x  117  1

With most viewing windows, it appears that the function has a maximum of 1 when x  1 (see Figure 3.11). By evaluating the function at x  1, however, you can see that f 1  0. To determine the behavior of this function near x  1, you need to examine the graph analytically to get the complete picture.

174

CHAPTER 3

Applications of Differentiation

The Mean Value Theorem Rolle’s Theorem can be used to prove another theorem—the Mean Value Theorem.

THEOREM 3.4

The Mean Value Theorem

If f is continuous on the closed interval a, b and differentiable on the open interval a, b, then there exists a number c in a, b such that .

y

f c 

Slope of tangent line = f ′(c)

f b  f a . ba

Tangent line

Video Secant line (b, f(b))

y

(a, f(a))

a

c

Proof Refer to Figure 3.12. The equation of the secant line containing the points a, f a and b, f b is

b

x

Figure 3.12



f b  f a x  a  f a. ba



Let gx be the difference between f x and y. Then gx  f x  y  f x 

 f bb  af ax  a  f a.

By evaluating g at a and b, you can see that ga  0  gb. Because f is continuous on a, b it follows that g is also continuous on a, b . Furthermore, because f is differentiable, g is also differentiable, and you can apply Rolle’s Theorem to the function g. So, there exists a number c in a, b such that g c  0, which implies that 0  g c  f c 

f b  f a . ba

So, there exists a number c in a, b such that f  c 

f b  f a . ba

NOTE The “mean” in the Mean Value Theorem refers to the mean (or average) rate of change of f in the interval a, b . JOSEPH-LOUIS LAGRANGE (1736–1813) The Mean Value Theorem was first proved by the famous mathematician Joseph-Louis Lagrange. Born in Italy, Lagrange held a position in the court of Frederick the Great in Berlin for 20 years. Afterward, he moved to France, where he met emperor Napoleon Bonaparte, who is quoted as saying, “Lagrange is the lofty pyramid of the mathe.matical sciences.”

MathBio

Although the Mean Value Theorem can be used directly in problem solving, it is used more often to prove other theorems. In fact, some people consider this to be the most important theorem in calculus—it is closely related to the Fundamental Theorem of Calculus discussed in Chapter 4. For now, you can get an idea of the versatility of this theorem by looking at the results stated in Exercises 77–85 in this section. The Mean Value Theorem has implications for both basic interpretations of the derivative. Geometrically, the theorem guarantees the existence of a tangent line that is parallel to the secant line through the points a, f a and b, f b, as shown in Figure 3.12. Example 3 illustrates this geometric interpretation of the Mean Value Theorem. In terms of rates of change, the Mean Value Theorem implies that there must be a point in the open interval a, b at which the instantaneous rate of change is equal to the average rate of change over the interval a, b . This is illustrated in Example 4.

SECTION 3.2

Given f x  5  4x, find all values of c in the open interval 1, 4 such that

y

Tangent line 4

(2, 3)

Solution The slope of the secant line through 1, f 1 and 4, f 4 is

Secant line

f 4  f 1 4  1   1. 41 41

2

f(x) = 5 − 4x

(1, 1)

Because f satisfies the conditions of the Mean Value Theorem, there exists at least one number c in 1, 4 such that f c  1. Solving the equation f x  1 yields x

1

2

3

f 4  f 1 . 41

f c 

(4, 4)

3

f x 

4

The tangent line at 2, 3 is parallel to the . secant line through 1, 1 and 4, 4. Figure 3.13

4 1 x2

which implies that x  ± 2. So, in the interval 1, 4, you can conclude that c  2, as shown in Figure 3.13.

Try It

Editable Graph

EXAMPLE 4

t = 4 minutes

Not drawn to scale

t=0

At some time t, the instantaneous velocity is . equal to the average velocity over 4 minutes. Figure 3.14

Animation

Exploration A

Open Exploration

Video

Finding an Instantaneous Rate of Change

Two stationary patrol cars equipped with radar are 5 miles apart on a highway, as shown in Figure 3.14. As a truck passes the first patrol car, its speed is clocked at 55 miles per hour. Four minutes later, when the truck passes the second patrol car, its speed is clocked at 50 miles per hour. Prove that the truck must have exceeded the speed limit (of 55 miles per hour) at some time during the 4 minutes.

5 miles

Solution Let t  0 be the time (in hours) when the truck passes the first patrol car. The time when the truck passes the second patrol car is t

4 1  hour. 60 15

By letting st represent the distance (in miles) traveled by the truck, you have 1 s0  0 and s15   5. So, the average velocity of the truck over the five-mile stretch of highway is Average velocity  

.

175

Finding a Tangent Line

EXAMPLE 3

1

Rolle’s Theorem and the Mean Value Theorem

s115  s0 115  0 5  75 miles per hour. 115

Assuming that the position function is differentiable, you can apply the Mean Value Theorem to conclude that the truck must have been traveling at a rate of 75 miles per hour sometime during the 4 minutes.

Try It

Exploration A

A useful alternative form of the Mean Value Theorem is as follows: If f is continuous on a, b and differentiable on a, b, then there exists a number c in a, b such that f b  f a  b  a fc.

Alternative form of Mean Value Theorem

NOTE When doing the exercises for this section, keep in mind that polynomial functions, rational functions, and trigonometric functions are differentiable at all points in their domains.

176

CHAPTER 3

Applications of Differentiation

Exercises for Section 3.2 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, explain why Rolle’s Theorem does not apply to the function even though there exist a and b such that f a  f b.





1. f x  1  x  1

2. f x  cot

y

y

x 2



25. f x  x  1, 1, 1 27. f x  4x  tan  x,

(U , 0) (3U , 0)

1 1 x

(0, 0)

3. f x 

x

3U

1 , x

(a) Verify that f 1  f 2.

4. f x  2  x233,

(b) According to Rolle’s Theorem, what must be the velocity at some time in the interval 1, 2? Find that time.

1, 1

In Exercises 5–8, find the two x-intercepts of the function f and show that f x  0 at some point between the two x-intercepts. 5. f x 

x2

6. f x  xx  3

x2

7. f x  xx  4

30. Reorder Costs The ordering and transportation cost C for components used in a manufacturing process is approximated 1 x by Cx  10  , where C is measured in thousands x x3 of dollars and x is the order size in hundreds.



10.

y

2

(b) According to Rolle’s Theorem, the rate of change of the cost must be 0 for some order size in the interval 3, 6. Find that order size.

(4, 0) 6

2

(1, 0)

x

f (x) = sin 2x

1

(U2 , 0)

(0, 0)

2

In Exercises 31 and 32, copy the graph and sketch the secant line to the graph through the points a, f a and b, f b. Then sketch any tangent lines to the graph for each value of c guaranteed by the Mean Value Theorem. To print an enlarged copy of the graph, select the MathGraph button.

y 2

2

U 4

4



(a) Verify that C3  C6.

8. f x  3xx  1

Rolle’s Theorem In Exercises 9 and 10, the graph of f is shown. Apply Rolle’s Theorem and find all values of c such that f c  0 at some point between the labeled intercepts. 9. f(x) = x2 + 3x  4



26. f x  x  x 13, 0, 1

29. Vertical Motion The height of a ball t seconds after it is thrown upward from a height of 32 feet and with an initial velocity of 48 feet per second is f t  16t 2  48t  32.

2

2



1, 1

U

1

(2, 0)

 14, 14

x x 28. f x   sin , 1, 0 2 6

2

2

In Exercises 25–28, use a graphing utility to graph the function on the closed interval [a, b]. Determine whether Rolle’s Theorem can be applied to f on the interval and, if so, find all values of c in the open interval a, b such that f c  0.

U 2

x

U

y

31.

y

32. f

2

f

In Exercises 11–24, determine whether Rolle’s Theorem can be applied to f on the closed interval [a, b]. If Rolle’s Theorem can be applied, find all values of c in the open interval a, b such that f c  0. 11. f x  x 2  2x, 0, 2

12. f x  x 2  5x  4, 1, 4

13. f x  x  1x  2x  3, 1, 3 14. f x  x  3x  1 2, 1, 3 15. f x  x 23  1, 8, 8





16. f x  3  x  3 , 0, 6

21. f x 

20. f x  cos x, 0, 2

6x   4 sin 2 x, 0,  6

23. f x  tan x, 0, 

 



22. f x  cos 2x, 

  , 12 6

  24. f x  sec x,  , 4 4







x

b

a

x

b

Writing In Exercises 33–36, explain why the Mean Value Theorem does not apply to the function f on the interval [0, 6]. y

33.

x 2  2x  3 x2  1 17. f x  , 1, 3 18. f x  , 1, 1 x2 x 19. f x  sin x, 0, 2

a

y

34.

6

6

5

5

4

4

3

3

2

2

1

1 x 1

35. f x 

2

3

1 x3

4

5

6

x 1

2



3

4



36. f x  x  3

5

6

SECTION 3.2

37. Mean Value Theorem Consider the graph of the function f x  x2  1. (a) Find the equation of the secant line joining the points 1, 2 and 2, 5. (b) Use the Mean Value Theorem to determine a point c in the interval 1, 2 such that the tangent line at c is parallel to the secant line. (c) Find the equation of the tangent line through c. (d) Then use a graphing utility to graph f, the secant line, and the tangent line. y

f (x) = x2 + 1

f (x) = x2  x + 6

(2, 5)

3

(2, 4)

(1, 2) 2

4 2

2

x

Figure for 37

2

Figure for 38

38. Mean Value Theorem Consider the graph of the function f x  x2  x  6. (a) Find the equation of the secant line joining the points 2, 4 and 2, 0. (b) Use the Mean Value Theorem to determine a point c in the interval 2, 2 such that the tangent line at c is parallel to the secant line. (c) Find the equation of the tangent line through c. (d) Then use a graphing utility to graph f, the secant line, and the tangent line. In Exercises 39– 46, determine whether the Mean Value Theorem can be applied to f on the closed interval [a, b]. If the Mean Value Theorem can be applied, find all values of c in the f b  f a open interval a, b such that f c  . ba 39. f x  x 2, 2, 1 40. f x  xx 2  x  2, 1, 1 x1 , x

12, 2

41. f x  x 23, 0, 1

42. f x 

43. f x  2  x, 7, 2

44. f x  x 3, 0, 1

45. f x  sin x, 0,  46. f x  2 sin x  sin 2x, 0,  In Exercises 47– 50, use a graphing utility to (a) graph the function f on the given interval, (b) find and graph the secant line through points on the graph of f at the endpoints of the given interval, and (c) find and graph any tangent lines to the graph of f that are parallel to the secant line. x 1 ,  2, 2 x1



St  200 5 

9 2t



Writing About Concepts

(2, 0)

3 4

47. f x 

52. Sales A company introduces a new product for which the number of units sold S is

(b) During what month does St equal the average value during the first year?

x 1

(b) Use the Mean Value Theorem to verify that at some time during the first 3 seconds of fall the instantaneous velocity equals the average velocity. Find that time.

(a) Find the average value of St during the first year.

4

3 2 1

177

where t is the time in months.

y

5

Rolle’s Theorem and the Mean Value Theorem

48. f x  x  2 sin x,  , 

49. f x  x, 1, 9 50. f x  x 4  4x 3  8x 2  5, 0, 5 51. Vertical Motion The height of an object t seconds after it is dropped from a height of 500 meters is st  4.9t 2  500. (a) Find the average velocity of the object during the first 3 seconds.

53. Let f be continuous on a, b and differentiable on a, b. If there exists c in a, b such that fc  0, does it follow that f a  f b? Explain. 54. Let f be continuous on the closed interval a, b and differentiable on the open interval a, b. Also, suppose that f a  f b and that c is a real number in the interval such that fc  0. Find an interval for the function g over which Rolle’s Theorem can be applied, and find the corresponding critical number of g (k is a constant). (a) gx  f x  k

(b) gx  f x  k

(c) gx  f k x 55. The function f x 

0,1 x,

x0 0 < x f 1

is differentiable on 0, 1 and satisfies f 0  f 1. However, its derivative is never zero on 0, 1. Does this contradict Rolle’s Theorem? Explain. 56. Can you find a function f such that f 2  2, f 2  6, and fx < 1 for all x. Why or why not?

57. Speed A plane begins its takeoff at 2:00 P.M. on a 2500-mile flight. The plane arrives at its destination at 7:30 P.M. Explain why there are at least two times during the flight when the speed of the plane is 400 miles per hour. 58. Temperature When an object is removed from a furnace and placed in an environment with a constant temperature of 90F, its core temperature is 1500F. Five hours later the core temperature is 390F. Explain why there must exist a time in the interval when the temperature is decreasing at a rate of 222F per hour. 59. Velocity Two bicyclists begin a race at 8:00 A.M. They both finish the race 2 hours and 15 minutes later. Prove that at some time during the race, the bicyclists are traveling at the same velocity. 60. Acceleration At 9:13 A.M., a sports car is traveling 35 miles per hour. Two minutes later, the car is traveling 85 miles per hour. Prove that at some time during this two-minute interval, the car’s acceleration is exactly 1500 miles per hour squared.

178

CHAPTER 3

Applications of Differentiation

61. Graphical Reasoning The figure shows two parts of the graph of a continuous differentiable function f on 10, 4 . The derivative f  is also continuous. To print an enlarged copy of the graph, select the MathGraph button. y

4 x 4

4 4 8

(a) Explain why f must have at least one zero in 10, 4 . (b) Explain why f  must also have at least one zero in the interval 10, 4 . What are these zeros called? (c) Make a possible sketch of the function with one zero of f  on the interval 10, 4 . (d) Make a possible sketch of the function with two zeros of f  on the interval 10, 4 . (e) Were the conditions of continuity of f and f necessary to do parts (a) through (d)? Explain. 62. Consider the function f x  3 cos 2

2x.

(a) Use a graphing utility to graph f and f . (b) Is f a continuous function? Is f  a continuous function? (c) Does Rolle’s Theorem apply on the interval 1, 1 ? Does it apply on the interval 1, 2 ? Explain. (d) Evaluate, if possible, lim f x and lim f x. xq3

xq3

Think About It In Exercises 63 and 64, sketch the graph of an arbitrary function f that satisfies the given condition but does not satisfy the conditions of the Mean Value Theorem on the interval [5, 5]. 63. f is continuous on 5, 5 . 64. f is not continuous on 5, 5 . In Exercises 65 and 66, use the Intermediate Value Theorem and Rolle’s Theorem to prove that the equation has exactly one real solution. 65. x 5  x3  x  1  0

66. 2x  2  cos x  0

67. Determine the values a, b, and c such that the function f satisfies the hypotheses of the Mean Value Theorem on the interval 0, 3 .



x  1 1 < x f 0 0 < x f 1 1 < x f 2

Differential Equations In Exercises 69–72, find a function f that has the derivative f x and whose graph passes through the given point. Explain your reasoning.

8

8



a, 2, f x  bx2  c, dx  4,

1, f x  ax  b, x2  4x  c,

x0 0 < x f 1 1 < x f 3

68. Determine the values a, b, c, and d so that the function f satisfies the hypotheses of the Mean Value Theorem on the interval 1, 2 .

69. fx  0, 2, 5

70. fx  4, 0, 1

71. fx  2x, 1, 0

72. fx  2x  3, 1, 0

True or False? In Exercises 73–76, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 73. The Mean Value Theorem can be applied to f x  1x on the interval 1, 1 . 74. If the graph of a function has three x-intercepts, then it must have at least two points at which its tangent line is horizontal. 75. If the graph of a polynomial function has three x-intercepts, then it must have at least two points at which its tangent line is horizontal. 76. If fx  0 for all x in the domain of f, then f is a constant function. 77. Prove that if a > 0 and n is any positive integer, then the polynomial function p x  x 2n1  ax  b cannot have two real roots. 78. Prove that if fx  0 for all x in an interval a, b, then f is constant on a, b. 79. Let px  Ax 2  Bx  C. Prove that for any interval a, b , the value c guaranteed by the Mean Value Theorem is the midpoint of the interval. 80. (a) Let f x  x2 and gx  x3  x2  3x  2. Then f 1  g1 and f 2  g2. Show that there is at least one value c in the interval 1, 2 where the tangent line to f at c, f c is parallel to the tangent line to g at c, gc. Identify c. (b) Let f and g be differentiable functions on a, b where f a  ga and f b  gb. Show that there is at least one value c in the interval a, b where the tangent line to f at c, f c is parallel to the tangent line to g at c, gc. 81. Prove that if f is differentiable on  ,  and fx < 1 for all real numbers, then f has at most one fixed point. A fixed point of a function f is a real number c such that f c  c. 82. Use the result of Exercise 81 to show that f x  12 cos x has at most one fixed point.

 

   

 

83. Prove that cos a  cos b f a  b for all a and b. 84. Prove that sin a  sin b f a  b for all a and b. 85. Let 0 < a < b. Use the Mean Value Theorem to show that b  a
f x2 .

x=b f

ng

Inc

asi

rea

cre

De

sing

.

Video

Constant f ′(x) < 0

f ′(x) = 0

f ′(x) > 0

The derivative is related to the slope of a function. Figure 3.15

x

A function is increasing if, as x moves to the right, its graph moves up, and is decreasing if its graph moves down. For example, the function in Figure 3.15 is decreasing on the interval  , a, is constant on the interval a, b, and is increasing on the interval b, . As shown in Theorem 3.5 below, a positive derivative implies that the function is increasing; a negative derivative implies that the function is decreasing; and a zero derivative on an entire interval implies that the function is constant on that interval.

THEOREM 3.5

Test for Increasing and Decreasing Functions

Let f be a function that is continuous on the closed interval a, b and differentiable on the open interval a, b. 1. If fx > 0 for all x in a, b, then f is increasing on a, b . 2. If fx < 0 for all x in a, b, then f is decreasing on a, b . 3. If fx  0 for all x in a, b, then f is constant on a, b . Proof To prove the first case, assume that fx > 0 for all x in the interval a, b and let x1 < x2 be any two points in the interval. By the Mean Value Theorem, you know that there exists a number c such that x1 < c < x2, and fc 

f x2  f x1 . x2  x1

Because fc > 0 and x2  x1 > 0, you know that f x2  f x1 > 0 which implies that f x1 < f x2. So, f is increasing on the interval. The second case has a similar proof (see Exercise 101), and the third case was given as Exercise 78 in Section 3.2. NOTE The conclusions in the first two cases of Theorem 3.5 are valid even if f x  0 at a finite number of x-values in a, b.

180

CHAPTER 3

Applications of Differentiation

EXAMPLE 1

Intervals on Which f Is Increasing or Decreasing

Find the open intervals on which f x  x 3  32x 2 is increasing or decreasing. Solution Note that f is differentiable on the entire real number line. To determine the critical numbers of f, set f x equal to zero. y

3 f x  x3  x 2 2 2 f x  3x  3x  0 3xx  1  0

f(x) = x 3 − 32 x 2

sing

2

(0, 0)

x

De

−1

1

asing

cre

asi

(

−1

Incre

2

1, − 1

ng

Differentiate and set fx equal to 0. Factor.

x  0, 1

Increa

1

Write original function.

2

)

Because there are no points for which f  does not exist, you can conclude that x  0 and x  1 are the only critical numbers. The table summarizes the testing of the three intervals determined by these two critical numbers. Interval

.

Test Value

Figure 3.16

Editable Graph

Critical numbers

 < x < 0 x  1

Sign of f x

f 1  6 > 0

Conclusion

Increasing

1 < x
0

Decreasing

Increasing

y

Increa

2

1

So, f is increasing on the intervals  , 0 and 1,  and decreasing on the interval 0, 1, as shown in Figure 3.16.

sing

.

Try It

f (x) = x 3 x

−1

1

2

−1

Increa

sing

−2

−2

g easi n Incr

2

1

Constant

ng easi Incr

−2

x 2

3

−x 2, x 1

(b) Not strictly monotonic

Figure 3.17

Video

Example 1 gives you one example of how to find intervals on which a function is increasing or decreasing. The guidelines below summarize the steps followed in the example.

Let f be continuous on the interval a, b. To find the open intervals on which f is increasing or decreasing, use the following steps.

y

−1

Exploration B

Guidelines for Finding Intervals on Which a Function Is Increasing or Decreasing

(a) Strictly monotonic function

−1

Exploration A

1. Locate the critical numbers of f in a, b, and use these numbers to determine test intervals. 2. Determine the sign of fx at one test value in each of the intervals. 3. Use Theorem 3.5 to determine whether f is increasing or decreasing on each interval. These guidelines are also valid if the interval a, b is replaced by an interval of the form  , b, a, , or  , . A function is strictly monotonic on an interval if it is either increasing on the entire interval or decreasing on the entire interval. For instance, the function f x  x 3 is strictly monotonic on the entire real line because it is increasing on the entire real line, as shown in Figure 3.17(a). The function shown in Figure 3.17(b) is not strictly monotonic on the entire real line because it is constant on the interval 0, 1 .

SECTION 3.3

Increasing and Decreasing Functions and the First Derivative Test

181

The First Derivative Test y

After you have determined the intervals on which a function is increasing or decreasing, it is not difficult to locate the relative extrema of the function. For instance, in Figure 3.18 (from Example 1), the function

f(x) = x 3 − 32 x 2

2

3 f x  x 3  x 2 2

1

Relative maximum (0, 0)

x

−1

1

−1

2

(1, − 12 )

Relative minimum

Relative extrema of f

has a relative maximum at the point 0, 0 because f is increasing immediately to the left of x  0 and decreasing immediately to the right of x  0. Similarly, f has a relative minimum at the point 1,  12  because f is decreasing immediately to the left of x  1 and increasing immediately to the right of x  1. The following theorem, called the First Derivative Test, makes this more explicit. THEOREM 3.6

The First Derivative Test

Figure 3.18

Let c be a critical number of a function f that is continuous on an open interval I containing c. If f is differentiable on the interval, except possibly at c, then f c can be classified as follows. 1. If f x changes from negative to positive at c, then f has a relative minimum at c, f c. 2. If f x changes from positive to negative at c, then f has a relative maximum at c, f c. 3. If f x is positive on both sides of c or negative on both sides of c, then f c is neither a relative minimum nor a relative maximum. (+) (−)

(−)

(+) f ′(x) < 0

a

f ′(x) > 0

c

f ′(x) > 0 b

a

Relative minimum

b

Relative maximum (+)

(+)

(−)

(−)

f ′(x) > 0

a

f ′(x) < 0 c

f ′(x) < 0

f ′(x) > 0

c

b

a

f ′(x) < 0

c

b

Neither relative minimum nor relative maximum

Proof Assume that f x changes from negative to positive at c. Then there exist a and b in I such that f x < 0 for all x in a, c and f x > 0 for all x in c, b.

.

By Theorem 3.5, f is decreasing on a, c and increasing on c, b. So, f c is a minimum of f on the open interval a, b and, consequently, a relative minimum of f. This proves the first case of the theorem. The second case can be proved in a similar way (see Exercise 102).

Video

182

CHAPTER 3

Applications of Differentiation

EXAMPLE 2

Applying the First Derivative Test

Find the relative extrema of the function f x  12 x  sin x in the interval 0, 2. Solution Note that f is continuous on the interval 0, 2. To determine the critical numbers of f in this interval, set fx equal to 0. fx 

1  cos x  0 2 cos x  x

Set fx equal to 0.

1 2

 5 , 3 3

Critical numbers

Because there are no points for which f does not exist, you can conclude that x  3 and x  53 are the only critical numbers. The table summarizes the testing of the three intervals determined by these two critical numbers. Interval

Relative maximum

f(x) = 1 x − sin x 2

2

 3

 5 < x < 3 3

 4

5 < x < 2 3

x

x

7 4

Sign of f x

f

4  < 0

f > 0

f

Conclusion

Decreasing

Increasing

Decreasing

3

74 < 0

By applying the First Derivative Test, you can conclude that f has a relative minimum at the point where

1 x −1

x

Test Value

y 4

0 < x
0

f1 < 0

f3 > 0

Conclusion

Decreasing

Increasing

Decreasing

Increasing

Interval

1 −4 −3

4x 3x 2  413

is 0 when x  0 and does not exist when x  ± 2. So, the critical numbers are x  2, x  0, and x  2. The table summarizes the testing of the four intervals determined by these three critical numbers.

6

4

General Power Rule

Test Value

2 < x
0

Conclusion

Decreasing

Increasing

Decreasing

Increasing

x 1

2

3

x-values that are not in the domain of f, as well as critical numbers, determine test . intervals for f.

1

By applying the First Derivative Test, you can conclude that f has one relative minimum at the point 1, 2 and another at the point 1, 2, as shown in Figure 3.22.

Figure 3.22

Editable Graph

Try It

Exploration A

Open Exploration

The most difficult step in applying the First Derivative Test is finding the values for which the derivative is equal to 0. For instance, the values of x for which the derivative of

TECHNOLOGY

f x 

x4  1 x2  1

is equal to zero are x  0 and x  ± 2  1. If you have access to technology that can perform symbolic differentiation and solve equations, use it to apply the First Derivative Test to this function.

SECTION 3.3

If a projectile is propelled from ground level and air resistance is neglected, the object will travel farthest with an initial angle of 45. If, however, the projectile is propelled from a point above ground level, the angle that yields a maximum horizontal distance is not 45 (see Example 5).

EXAMPLE 5

Increasing and Decreasing Functions and the First Derivative Test

185

The Path of a Projectile

Neglecting air resistance, the path of a projectile that is propelled at an angle is y

g sec 2 2 x  tan  x  h, 2v02

0 ≤ ≤

 2

where y is the height, x is the horizontal distance, g is the acceleration due to gravity, v0 is the initial velocity, and h is the initial height. (This equation is derived in Section 12.3.) Let g  32 feet per second per second, v0  24 feet per second, and h  9 feet. What value of will produce a maximum horizontal distance? Solution To find the distance the projectile travels, let y  0, and use the Quadratic Formula to solve for x. g sec2 2 x  tan x  h  0 2v02 32 sec2 2 x  tan x  9  0 2242 

sec2 2 x  tan x  9  0 36 x

tan ± tan2  sec2

sec2 18

x  18 cos sin  sin2  1 ,

x ≥ 0

At this point, you need to find the value of that produces a maximum value of x. Applying the First Derivative Test by hand would be very tedious. Using technology to solve the equation dxd  0, however, eliminates most of the messy computations. The result is that the maximum value of x occurs when

 0.61548 radian, or 35.3. This conclusion is reinforced by sketching the path of the projectile for different values of , as shown in Figure 3.23. Of the three paths shown, note that the distance traveled is greatest for  35. y

θ = 35° θ = 45°

15

10

h=9

θ = 25°

5

x 5

.

15

20

The path of a projectile with initial angle

Figure 3.23

.

10

Simulation

Try It

25

186

CHAPTER 3

Applications of Differentiation

Exercises for Section 3.3 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, use the graph of f to find (a) the largest open interval on which f is increasing, and (b) the largest open interval on which f is decreasing. y

1.

y

2.

10

6

8

4

f

f 2

6

2 2

2 4

6

8

4

10

In Exercises 3–16, identify the open intervals on which the function is increasing or decreasing. 3. f x  x 2  6x  8

4. y   x  12

y 4

x

3

1 1

2

1

2

4

5

y 3 2 x

1

4

25. f x 

2

2

x 7. f x  sin x  2, 0 < x < 2 8. hx  cos , 0 < x < 2 2 y

 5x 5

26. f x  x 4  32x  4

27. f x  x13  1

28. f x  x23  4

29. f x  x  123

30. f x  x  113







31. f x  5  x  5

32. f x  x  3  1

1 33. f x  x  x

x 34. f x  x1

39. f x 

x

4

x2

x2 9

x 2  2x  1 x1

36. f x 

x3 x2

38. f x 

x 2  3x  4 x2

x  cos x 2

40. f x  sin x cos x

41. f x  sin x  cos x

42. f x  x  2 sin x

43. f x 

44. f x  3 sin x  cos x

2x

cos2

45. f x  sin2 x  sin x

46. f x 

sin x 1  cos2 x

y 2

3 2

x

1 x

9. f x 

24. f x  x  22x  1

x5

In Exercises 39–46, consider the function on the interval 0, 2. For each function, (a) find the open interval(s) on which the function is increasing or decreasing, (b) apply the First Derivative Test to identify all relative extrema, and (c) use a graphing utility to confirm your results.

y

4

1

23. f x  x 23  x

37. f x 

6. f x  x 4  2x 2

2

22. f x  x 3  6x 2  15

35. f x 

4

x3  3x 4

2 2

21. f x  2x  3x  12x

3

x

5. y 

1

2

1 1

20. f x   x 2  8x  12

2



y

3

18. f x  x 2  8x  10

19. f x  2x 2  4x  3 3

4

x 2

2

In Exercises 17–38, (a) find the critical numbers of f (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. 17. f x  x 2  6x

x

4

16. f x  cos2 x  cos x, 0 < x < 2

U 2

U

3U 2

2U

1 x2

1

U 2

U

2U

2

10. y 

47. f x  2x9  x 2, x2 x1

12. hx  27x  x

13. y  x16  x 2

14. y  x 

15. y  x  2 cos x,

0 < x < 2

3, 3

48. f x  105  x 2  3x  16 , 0, 5

11. gx  x  2x  8 2

In Exercises 47–52, (a) use a computer algebra system to differentiate the function, (b) sketch the graphs of f and f on the same set of coordinate axes over the given interval, (c) find the critical numbers of f in the open interval, and (d) find the interval(s) on which f is positive and the interval(s) on which it is negative. Compare the behavior of f and the sign of f .

49. f t  t 2 sin t, 0, 2

50. f x 

3

4 x

x 51. f x  3 sin , 3

0, 6

52. f x  2 sin 3x  4 cos 3x, 0, 

x x  cos , 0, 4 2 2

SECTION 3.3

In Exercises 53 and 54, use symmetry, extrema, and zeros to sketch the graph of f . How do the functions f and g differ?

54. f t  cos2 t  sin2 t,

f x > 0 on  , 4

gt  1  2 sin2 t

Think About It In Exercises 55–60, the graph of f is shown in the figure. Sketch a graph of the derivative of f. To print an enlarged copy of the graph, select the MathGraph button. y

55. 4

2 1

1

2

3

69. gx  f x  10

x

2

y

57.

71. Sketch the graph of the arbitrary function f such that

8 6 4 2

f 2 x

2

70. gx  f x  10

y

58.

4 2



f

6 8

> 0, fx undefined, < 0,

x

6 4

4 6 y

59.

4 6

6

4

f

4

f

73. Think About It The function f is differentiable on the interval 1, 1 . The table shows the values of f  for selected values of x. Sketch the graph of f, approximate the critical numbers, and identify the relative extrema.

2 x

2

2

2

x

4

4

2

2

2

4

In Exercises 61–64, use the graph of f to (a) identify the interval(s) on which f is increasing or decreasing, and (b) estimate the values of x at which f has a relative maximum or minimum. 61.

62.

y

y

fe

2

2

x 4

4

2

2

4

2 4

4 6

63.

64.

y 6

0.75

0.50

0.25

f x

10

3.2

0.5

0.8

x

0

0.25

0.50

0.75

1

f x

5.6

3.6

0.2

6.7

20.1

0

6

4

3

2

3.14

0.23

2.45

3.11

0.69

x

23

34

56



f x

3.00

1.37

1.14

2.84

f x fe x

x 2

74. Think About It The function f is differentiable on the interval 0,  . The table shows the values of f for selected values of x. Sketch the graph of f, approximate the critical numbers, and identify the relative extrema.

4

fe

4

2

1

x y

2

2

x

fe 2 x

2

4

x < 4 x  4. x > 4

72. A differentiable function f has one critical number at x  5. Identify the relative extrema of f at the critical number if f4  2.5 and f6  3.

y

60.

6

4

䊏0 g5䊏0 g6䊏0 g0 䊏0 g0 䊏0 g8 䊏0

g0

68. gx  f x

1 1

Sign of gc

67. gx  f x

x

2 1

Supply the appropriate inequality for the indicated value of c.

66. gx  3f x  3

f

1

2

f x > 0 on 6, 

65. gx  f x  5

2

f

f x < 0 on 4, 6

Function

y

56.

Writing About Concepts In Exercises 65–70, assume that f is differentiable for all x. The signs of f are as follows.

x 5  4x 3  3x , gx  xx 2  3) x2  1

53. f x 

187

Increasing and Decreasing Functions and the First Derivative Test

2

2 2

4 4

4

188

CHAPTER 3

Applications of Differentiation

75. Rolling a Ball Bearing A ball bearing is placed on an inclined plane and begins to roll. The angle of elevation of the plane is . The distance (in meters) the ball bearing rolls in t seconds is st  4.9sin t 2. (a) Determine the speed of the ball bearing after t seconds. (b) Complete the table and use it to determine the value of

that produces the maximum speed at a particular time.



0

4

3

2

23

34



s t 76. Numerical, Graphical, and Analytic Analysis The concentration C of a chemical in the bloodstream t hours after injection into muscle tissue is C(t) 

3t , 27  t 3

t v 0.

0

0.5

1

1.5

2

2.5

3

(b) Use a graphing utility to graph the concentration function and use the graph to approximate the time when the concentration is greatest. (c) Use calculus to determine analytically the time when the concentration is greatest. 77. Numerical, Graphical, and Analytic Analysis Consider the functions f x  x and gx  sin x on the interval 0, . (a) Complete the table and make a conjecture about which is the greater function on the interval 0, . 0.5

1

1.5

2

2.5

3

f x gx (b) Use a graphing utility to graph the functions and use the graphs to make a conjecture about which is the greater function on the interval 0, . (c) Prove that f x > gx on the interval 0, . [Hint: Show that hx > 0 where h  f  g.] 78. Numerical, Graphical, and Analytic Analysis Consider the functions f x  x and g x  tan x on the interval 0, 2. (a) Complete the table and make a conjecture about which is the greater function on the interval 0, 2. x f x g x

79. Trachea Contraction Coughing forces the trachea (windpipe) to contract, which affects the velocity v of the air passing through the trachea. The velocity of the air during coughing is v  kR  rr 2, 0 f r < R where k is constant, R is the normal radius of the trachea, and r is the radius during coughing. What radius will produce the maximum air velocity? 80. Profit The profit P (in dollars) made by a fast-food restaurant selling x hamburgers is

0.25

0.5

0.75

1

1.25

1.5

x2  5000, 20,000

0 f x f 35,000.

Find the open intervals on which P is increasing or decreasing. 81. Power The electric power P in watts in a direct-current circuit with two resistors R1 and R 2 connected in parallel is P

Ct

x

(c) Prove that f x < gx on the interval 0, 2. [Hint: Show that hx > 0, where h  g  f.]

P  2.44x 

(a) Complete the table and use it to approximate the time when the concentration is greatest. t

(b) Use a graphing utility to graph the functions and use the graphs to make a conjecture about which is the greater function on the interval 0, 2.

vR1R 2 R1  R 2 2

where v is the voltage. If v and R1 are held constant, what resistance R 2 produces maximum power? 82. Electrical Resistance The resistance R of a certain type of resistor is R  0.001T 4  4T  100 where R is measured in ohms and the temperature T is measured in degrees Celsius. (a) Use a computer algebra system to find dRdT and the critical number of the function. Determine the minimum resistance for this type of resistor. (b) Use a graphing utility to graph the function R and use the graph to approximate the minimum resistance for this type of resistor. 83. Modeling Data The end-of-year assets for the Medicare Hospital Insurance Trust Fund (in billions of dollars) for the years 1995 through 2001 are shown. 1995: 130.3; 1996: 124.9; 1997: 115.6; 1998: 120.4; 1999: 141.4; 2000: 177.5; 2001: 208.7 (Source: U.S. Centers for Medicare and Medicaid Services) (a) Use the regression capabilities of a graphing utility to find a model of the form M  at 2  bt  c for the data. (Let t  5 represent 1995.) (b) Use a graphing utility to plot the data and graph the model. (c) Analytically find the minimum of the model and compare the result with the actual data.

SECTION 3.3

84. Modeling Data The number of bankruptcies (in thousands) for the years 1988 through 2001 are shown. 1988: 594.6; 1989: 643.0; 1990: 725.5; 1991: 880.4;

Increasing and Decreasing Functions and the First Derivative Test

189

93. Relative minima: 0, 0, 4, 0 Relative maximum: 2, 4 94. Relative minimum: 1, 2

1992: 972.5; 1993: 918.7; 1994: 845.3; 1995: 858.1;

Relative maxima: 1, 4, 3, 4

1996: 1042.1; 1997: 1317.0; 1998: 1429.5; 1999: 1392.0; 2000: 1277.0; 2001: 1386.6 (Source: Administrative Office of the U.S. Courts) (a) Use the regression capabilities of a graphing utility to find a model of the form B  at 4  bt 3  ct 2  dt  e for the data. (Let t  8 represent 1988.)

True or False? In Exercises 95–100, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 95. The sum of two increasing functions is increasing. 96. The product of two increasing functions is increasing.

(b) Use a graphing utility to plot the data and graph the model.

97. Every nth-degree polynomial has n  1 critical numbers.

(c) Find the maximum of the model and compare the result with the actual data.

98. An nth-degree polynomial has at most n  1 critical numbers.

Motion Along a Line In Exercises 85– 88, the function st describes the motion of a particle moving along a line. For each function, (a) find the velocity function of the particle at any time t v 0, (b) identify the time interval(s) when the particle is moving in a positive direction, (c) identify the time interval(s) when the particle is moving in a negative direction, and (d) identify the time(s) when the particle changes its direction.

99. There is a relative maximum or minimum at each critical number.

85. st  6t  t 2

103. Let x > 0 and n > 1 be real numbers. Prove that 1  xn > 1  nx.

88. st  t 3  20t 2  128t  280 Motion Along a Line In Exercise 89 and 90, the graph shows the position of a particle moving along a line. Describe how the particle’s position changes with respect to time. 90.

s 28 24 20 16 12 8 4 4 8 12

s 120 100 80 60

t 1 2 3 4 5 6

8

10

Creating Polynomial Functions polynomial function

40 20 t 3

6

9 12 15 18

In Exercises 91– 94, find a

f x  an x n  an1xn1  . . .  a2 x 2  a1x  a 0 that has only the specified extrema. (a) Determine the minimum degree of the function and give the criteria you used in determining the degree. (b) Using the fact that the coordinates of the extrema are solution points of the function, and that the x-coordinates are critical numbers, determine a system of linear equations whose solution yields the coefficients of the required function. (c) Use a graphing utility to solve the system of equations and determine the function. (d) Use a graphing utility to confirm your result graphically. 91. Relative minimum: 0, 0; Relative maximum: 2, 2 92. Relative minimum: 0, 0; Relative maximum: 4, 1000

101. Prove the second case of Theorem 3.5. 102. Prove the second case of Theorem 3.6.

86. st  t 2  7t  10

87. st  t 3  5t 2  4t

89.

100. The relative maxima of the function f are f 1  4 and f 3 10. So, f has at least one minimum for some x in the interval 1, 3.

104. Use the definitions of increasing and decreasing functions to prove that f x  x3 is increasing on  , . 105. Use the definitions of increasing and decreasing functions to prove that f x  1x is decreasing on 0, .

190

CHAPTER 3

Applications of Differentiation

Section 3.4

Concavity and the Second Derivative Test • Determine intervals on which a function is concave upward or concave downward. • Find any points of inflection of the graph of a function. • Apply the Second Derivative Test to find relative extrema of a function.

Concavity You have already seen that locating the intervals in which a function f increases or decreases helps to describe its graph. In this section, you will see how locating the intervals in which f increases or decreases can be used to determine where the graph of f is curving upward or curving downward.

Definition of Concavity Let f be differentiable on an open interval I. The graph of f is concave upward on I if f is increasing on the interval and concave downward on I if f is decreasing on the interval.

The following graphical interpretation of concavity is useful. (See Appendix A for a proof of these results.)

f(x) =

y

1 3 x −x 3 1

Concave m = 0 downward −2

1. Let f be differentiable on an open interval I. If the graph of f is concave upward on I, then the graph of f lies above all of its tangent lines on I. [See Figure 3.24(a).] 2. Let f be differentiable on an open interval I. If the graph of f is concave downward on I, then the graph of f lies below all of its tangent lines on I. [See Figure 3.24(b).]

Concave upward m = −1

−1

y

y

x

Concave upward, f ′ is increasing.

1

m=0

−1

Concave downward, f ′ is decreasing.

y x

1

(a) The graph of f lies above its tangent lines.

(−1, 0) −2

(1, 0)

−1

f ′(x) = x 2 − 1 f ′ is decreasing.

x

(b) The graph of f lies below its tangent lines.

Figure 3.24

1

(0, −1)

f ′ is increasing.

The concavity of f is related to the slope of the derivative. Figure 3.25

x

To find the open intervals on which the graph of a function f is concave upward or downward, you need to find the intervals on which f is increasing or decreasing. For instance, the graph of f x  13x3  x is concave downward on the open interval  , 0 because fx  x2  1 is decreasing there. (See Figure 3.25.) Similarly, the graph of f is concave upward on the interval 0,  because f is increasing on 0, .

SECTION 3.4

191

Concavity and the Second Derivative Test

The following theorem shows how to use the second derivative of a function f to determine intervals on which the graph of f is concave upward or downward. A proof of this theorem follows directly from Theorem 3.5 and the definition of concavity.

THEOREM 3.7

Test for Concavity

Let f be a function whose second derivative exists on an open interval I. 1. If f  x > 0 for all x in I, then the graph of f is concave upward in I. 2. If f  x < 0 for all x in I, then the graph of f is concave downward in I. NOTE A third case of Theorem 3.7 could be that if f  x  0 for all x in I, then f is linear. Note, however, that concavity is not defined for a line. In other words, a straight line is neither concave upward nor concave downward.

To apply Theorem 3.7, locate the x-values at which f  x  0 or f  does not exist. Second, use these x-values to determine test intervals. Finally, test the sign of f  x in each of the test intervals. EXAMPLE 1

Determining Concavity

Determine the open intervals on which the graph of f x 

6 x2  3

is concave upward or downward. y

f(x) =

6 x2 + 3

Solution Begin by observing that f is continuous on the entire real line. Next, find the second derivative of f.

3

f ″ (x) > 0 Concave upward

f ″(x) > 0 Concave upward 1

f ″ (x) < 0 Concave downward x

−2

−1

1

2

−1

From the sign of f  you can determine the . concavity of the graph of f. Figure 3.26

Editable Graph

f x  6x2  31 fx  6x2  322x 12x  2 x  32 x2  3212  12x2x2  32x f  x  x2  34 36x2  1  2 x  3 3

Differentiate. First derivative

Differentiate.

Second derivative

Because f  x  0 when x  ± 1 and f  is defined on the entire real line, you should test f  in the intervals  , 1, 1, 1, and 1, . The results are shown in the table and in Figure 3.26. Interval Test Value Sign of f  x

.

Rewrite original function.

Conclusion

Try It

 < x < 1

1 < x < 1

x  2

x0

x2

f  2 > 0

f  0 < 0

f  2 > 0

Concave upward

Concave downward

Concave upward

Exploration A

Video

1 < x
0

f  0 < 0

f  3 > 0

Concave upward

Concave downward

Concave upward

Try It

Exploration A

2 < x
0

f  1 < 0

f  3 > 0

Concave upward

Concave downward

Concave upward

2 < x
0

Concave upward

In addition to testing for concavity, the second derivative can be used to perform a simple test for relative maxima and minima. The test is based on the fact that if the graph of a function f is concave upward on an open interval containing c, and fc  0, f c must be a relative minimum of f. Similarly, if the graph of a function f is concave downward on an open interval containing c, and fc  0, f c must be a relative maximum of f (see Figure 3.31).

f

.

x

c

If f c  0 and f  c > 0, f c is a relative minimum.

Video THEOREM 3.9

Second Derivative Test

Let f be a function such that fc  0 and the second derivative of f exists on an open interval containing c.

y

f ′′(c) < 0

1. If f  c > 0, then f has a relative minimum at c, f c. 2. If f  c < 0, then f has a relative maximum at c, f c.

Concave downward

If f  c  0, the test fails. That is, f may have a relative maximum at c, a relative minimum at c, f c, or neither. In such cases, you can use the First Derivative Test.

f

x

c

Proof If f  c  0 and f  c > 0, there exists an open interval I containing c for which fx fx  fc  >0 xc xc

If f c  0 and f  c < 0, f c is a relative maximum. Figure 3.31

for all x  c in I. If x < c, then x  c < 0 and fx < 0. Also, if x > c, then x  c > 0 and fx > 0. So, fx changes from negative to positive at c, and the First Derivative Test implies that f c is a relative minimum. A proof of the second case is left to you. EXAMPLE 4

Using the Second Derivative Test

Find the relative extrema for f x  3x 5  5x3. Solution Begin by finding the critical numbers of f. fx  15x 4  15x2  15x21  x2  0 x  1, 0, 1

f(x) = −3x 5 + 5x 3 y

Relative maximum (1, 2)

2

Set fx equal to 0. Critical numbers

Using f  x  60x 3  30x  302x3  x

1

−2

−1

you can apply the Second Derivative Test as shown below. (0, 0) 1

x

2

−1

(−1, −2) Relative minimum

−2

0, 0 is neither a relative minimum nor a relative maximum. .. Figure 3.32

Editable Graph

Point Sign of f x Conclusion

1, 2

1, 2

0, 0

f  1 > 0

f  1 < 0

f  0  0

Relative minimum

Relative maximum

Test fails

Because the Second Derivative Test fails at 0, 0, you can use the First Derivative Test and observe that f increases to the left and right of x  0. So, 0, 0 is neither a relative minimum nor a relative maximum (even though the graph has a horizontal tangent line at this point). The graph of f is shown in Figure 3.32.

Try It

Exploration A

Open Exploration

SECTION 3.4

195

Concavity and the Second Derivative Test

Exercises for Section 3.4 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–10, determine the open intervals on which the graph is concave upward or concave downward. 1. y  x2  x  2



2. y  x3  3x2  2

23. f x  sec x 

y

y

x 21. f x  sin , 0, 4 2

22. f x  2 csc



3

24. f x  sin x  cos x, 0, 2

2

2

25. f x  2 sin x  sin 2x, 0, 2

1

1

2

1

3

4

0, 2

 , 0, 4 2

3

x

3x , 2

26. f x  x  2 cos x, 0, 2

2 1

2

3

4

x

In Exercises 27– 40, find all relative extrema. Use the Second Derivative Test where applicable. 3

3 Generated by Derive

3. f x 

24 x2  12

Generated by Derive

4. f x 

y

37.

2

3 2 1

1 1

1

2

3

3 2

39. 1

1

2

3

x

2 3 Generated by Derive

5. f x 

x2  1 x2  1

Generated by Derive

6. y 

3x 5  40x3  135x 270 6

2

4

1

2

3 2 1

1

2

3

x

6

2

9. y  2x  tan x,

2

2

6

x

Generated by Derive

10. y  x 

44. f x  2x sin x, 0, 2

Writing About Concepts

8. hx  x 5  5x  2

 2 , 2 

In Exercises 41– 44, use a computer algebra system to analyze the function over the given interval. (a) Find the first and second derivatives of the function. (b) Find any relative extrema and points of inflection. (c) Graph f, f , and f on the same set of coordinate axes and state the relationship between the behavior of f and the signs of f and f .

1 1 43. f x  sin x  3 sin 3x  5 sin 5x, 0, 

6

7. gx  3x 2  x3

3

42. f x  x26  x2,  6, 6

4

Generated by Derive

40.

32. f x  x3  9x2  27x

1 6  x 34. gx   8 x  22x  42 f x  x23  3 36. f x  x 2  1 4 x f x  x  38. f x  x x1 f x  cos x  x, 0, 4 f x  2 sin x  cos 2x, 0, 2

x2

41. f x  0.2x2x  33, 1, 4

y

y 3

30. f x   x  52

35.

y

x

29. f x  x  5 33. gx 

3

1

28. f x  x2  3x  8

2

31. f x  x 3  3x 2  3

x2  1 2x  1

3

27. f x  x 4  4x3  2

2 ,  ,  sin x

In Exercises 11–26, find the points of inflection and discuss the concavity of the graph of the function.

45. Consider a function f such that f is increasing. Sketch graphs of f for (a) f < 0 and (b) f > 0. 46. Consider a function f such that f is decreasing. Sketch graphs of f for (a) f < 0 and (b) f > 0. 47. Sketch the graph of a function f that does not have a point of inflection at c, f c even though f  c  0. 48. S represents weekly sales of a product. What can be said of S and S for each of the following statements? (a) The rate of change of sales is increasing.

11. f x  x3  6x2  12x

12. f x  2x3  3x 2  12x  5

(b) Sales are increasing at a slower rate.

13. f x  14 x 4  2x2

14. f x  2x 4  8x  3

(c) The rate of change of sales is constant.

15. f x  xx  43

16. f x  x3x  4

(d) Sales are steady.

17. f x  xx  3

18. f x  xx  1

(e) Sales are declining, but at a slower rate.

x 19. f x  2 x 1

x1 20. f x  x

(f) Sales have bottomed out and have started to rise.

196

CHAPTER 3

Applications of Differentiation

In Exercises 49–52, the graph of f is shown. Graph f, f , and f on the same set of coordinate axes. To print an enlarged copy of the graph, select the MathGraph button. y

49.

y

50. f

2

3 x 60. (a) Graph f x   and identify the inflection point.

f

2

(b) Does f  x exist at the inflection point? Explain.

x

2

x

1

1

1 y

51. 4

1

2

3

4

f

3 x

2

1

y

52.

1

f

1 x

4

1

2

3

4

Think About It In Exercises 53 – 56, sketch the graph of a function f having the given characteristics. 53. f 2  f 4  0

In Exercises 61 and 62, find a, b, c, and d such that the cubic f x  ax3  bx 2  cx  d satisfies the given conditions. 61. Relative maximum: 3, 3

62. Relative maximum: 2, 4

Relative minimum: 5, 1

Relative minimum: 4, 2

Inflection point: 4, 2

Inflection point: 3, 3

63. Aircraft Glide Path A small aircraft starts its descent from an altitude of 1 mile, 4 miles west of the runway (see figure).

2

2

2

(a) Use a graphing utility to graph f for n  1, 2, 3, and 4. Use the graphs to make a conjecture about the relationship between n and any inflection points of the graph of f. (b) Verify your conjecture in part (a).

3

1

59. Conjecture Consider the function f x  x  2n.

(a) Find the cubic f x  ax3  bx2  cx  d on the interval

4, 0 that describes a smooth glide path for the landing. (b) The function in part (a) models the glide path of the plane. When would the plane be descending at the most rapid rate? y

54. f 0  f 2  0

f 3 is defined.

f x > 0 if x < 1

f x < 0 if x < 3

f1  0

f3 does not exist.

fx < 0 if x > 1

fx > 0 if x > 3

f  x < 0

1

x

4

3

2

1

f  x < 0, x  3 55. f 2  f 4  0

56. f 0  f 2  0

fx > 0 if x < 3

fx < 0 if x < 1

f3 does not exist.

f1  0

fx < 0 if x > 3

fx > 0 if x > 1

f  x > 0, x  3

f  x > 0

57. Think About It The figure shows the graph of f . Sketch a graph of f. (The answer is not unique.) To print an enlarged copy of the graph, select the MathGraph button. y 6

f ee

3 2

d

1 1

2

MathArticle 64. Highway Design A section of highway connecting two hillsides with grades of 6% and 4% is to be built between two points that are separated by a horizontal distance of 2000 feet (see figure). At the point where the two hillsides come together, there is a 50-foot difference in elevation.

3

Figure for 57

4

(b) Use a graphing utility to graph the model. (c) Use a graphing utility to graph the derivative of the model.

x 1

of modeling, see the article “How Not to Land at Lake Tahoe!” by Richard Barshinger in The American Mathematical Monthly.

(a) Design a section of highway connecting the hillsides modeled by the function f x  ax3  bx2  cx  d 1000 f x f 1000. At the points A and B, the slope of the model must match the grade of the hillside.

5 4

FOR FURTHER INFORMATION For more information on this type

5

Figure for 58

(d) Determine the grade at the steepest part of the transitional section of the highway.

58. Think About It Water is running into the vase shown in the figure at a constant rate. (a) Graph the depth d of water in the vase as a function of time. (b) Does the function have any extrema? Explain. (c) Interpret the inflection points of the graph of d.

y

Highway A(1000, 60) 6% grad e Not drawn to scale

B(1000, 90) rade 4% g 50 ft

x

SECTION 3.4

65. Beam Deflection The deflection D of a beam of length L is D  2x 4  5Lx3  3L2x2, where x is the distance from one end of the beam. Find the value of x that yields the maximum deflection. 66. Specific Gravity A model for the specific gravity of water S is S

5.755 3 8.521 2 6.540 T  T  T  0.99987, 0 < T < 25 108 106 105

where T is the water temperature in degrees Celsius. (a) Use a computer algebra system to find the coordinates of the maximum value of the function. (b) Sketch a graph of the function over the specified domain. (Use a setting in which 0.996 f S f 1.001.

Concavity and the Second Derivative Test

197

Linear and Quadratic Approximations In Exercises 71–74, use a graphing utility to graph the function. Then graph the linear and quadratic approximations P1x  f a  f ax  a and 1 P2x  f a  f ax  a  2 f ax  a2

in the same viewing window. Compare the values of f, P1 , and P2 and their first derivatives at x  a. How do the approximations change as you move farther away from x  a? Function

Value of a

 4

71. f x  2sin x  cos x

a

67. Average Cost A manufacturer has determined that the total cost C of operating a factory is C  0.5x2  15x  5000, where x is the number of units produced. At what level of production will the average cost per unit be minimized? (The average cost per unit is Cx.)

72. f x  2sin x  cos x

a0

73. f x  1  x

a0

68. Inventory Cost The total cost C for ordering and storing x units is C  2x  300,000x. What order size will produce a minimum cost?

75. Use a graphing utility to graph y  x sin1x. Show that the graph is concave downward to the right of x  1.

(c) Estimate the specific gravity of water when T  20.

69. Sales Growth The annual sales S of a new product are given by 5000t 2 S , 0 f t f 3, where t is time in years. 8  t2 (a) Complete the table. Then use it to estimate when the annual sales are increasing at the greatest rate. t

0.5

1

1.5

2

2.5

74. f x 

77. Prove that every cubic function with three distinct real zeros has a point of inflection whose x-coordinate is the average of the three zeros. 78. Show that the cubic polynomial px  ax3  bx2  cx  d has exactly one point of inflection x0, y0, where x0 

(b) Use a graphing utility to graph the function S. Then use the graph to estimate when the annual sales are increasing at the greatest rate.

a2

76. Show that the point of inflection of f x  x x  62 lies midway between the relative extrema of f.

3

S

x x1

b 3a

and

y0 

2b3 bc   d. 27a2 3a

Use this formula to find the point of inflection of px  x3  3x2  2.

(c) Find the exact time when the annual sales are increasing at the greatest rate.

True or False? In Exercises 79– 82, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

70. Modeling Data The average typing speed S (words per minute) of a typing student after t weeks of lessons is shown in the table.

79. The graph of every cubic polynomial has precisely one point of inflection.

t

5

10

15

20

25

30

S

38

56

79

90

93

94

100t 2 , t > 0. A model for the data is S  65  t 2 (a) Use a graphing utility to plot the data and graph the model. (b) Use the second derivative to determine the concavity of S. Compare the result with the graph in part (a). (c) What is the sign of the first derivative for t > 0? By combining this information with the concavity of the model, what inferences can be made about the typing speed as t increases?

80. The graph of f x  1x is concave downward for x < 0 and concave upward for x > 0, and thus it has a point of inflection at x  0. 81. If fc > 0, then f is concave upward at x  c. 82. If f  2  0, then the graph of f must have a point of inflection at x  2. In Exercises 83 and 84, let f and g represent differentiable functions such that f  0 and g  0. 83. Show that if f and g are concave upward on the interval a, b, then f  g is also concave upward on a, b. 84. Prove that if f and g are positive, increasing, and concave upward on the interval a, b, then fg is also concave upward on a, b.

198

CHAPTER 3

Applications of Differentiation

Section 3.5

Limits at Infinity • Determine (finite) limits at infinity. • Determine the horizontal asymptotes, if any, of the graph of a function. • Determine infinite limits at infinity.

Limits at Infinity y 4

f(x) =

This section discusses the “end behavior” of a function on an infinite interval. Consider the graph of

3x 2 +1

x2

f x  f(x) → 3 as x → −∞

2

f(x) → 3 as x → ∞ x

−4 −3 −2 −1

1

2

3

4

3x 2 x2  1

as shown in Figure 3.33. Graphically, you can see that the values of f x appear to approach 3 as x increases without bound or decreases without bound. You can come to the same conclusions numerically, as shown in the table.

The limit of f x as x approaches  or is 3.

x decreases without bound.

x increases without bound.

Figure 3.33

3

→

f x

→



x

100

10 1

0

1

10

100

→

2.9997

2.97

0

1.5

2.97

2.9997

→3

1.5

f x approaches 3.

f x approaches 3.

The table suggests that the value of f x approaches 3 as x increases without bound x → . Similarly, f x approaches 3 as x decreases without bound x →  . These limits at infinity are denoted by lim f x  3

The statement lim f x  L

NOTE

or lim f x  L, means that the limit x→

Limit at negative infinity

x→

x→

and lim f x  3.

exists and the limit is equal to L.

Limit at positive infinity

x→

To say that a statement is true as x increases without bound means that for some (large) real number M, the statement is true for all x in the interval x: x > M . The following definition uses this concept.

Definition of Limits at Infinity Let L be a real number.

y

lim f(x) = L x→ ∞

ε ε

L

M

f .x is within units of L as x → .

Figure 3.34

x

1. The statement lim f x  L means that for each > 0 there exists an M > 0 x→ such that f x  L < whenever x > M. 2. The statement lim f x  L means that for each > 0 there exists an N < 0 x→ such that f x  L < whenever x < N.









The definition of a limit at infinity is shown in Figure 3.34. In this figure, note that for a given positive number there exists a positive number M such that, for x > M, the graph of f will lie between the horizontal lines given by y  L  and y  L  .

Video

Video

SECTION 3.5

E X P L O R AT I O N Use a graphing utility to graph f x 

2x 2  4x  6 . 3x 2  2x  16

Describe all the important features of the graph. Can you find a single viewing window that shows all of these features clearly? Explain your reasoning. What are the horizontal asymptotes of the graph? How far to the right do you have to move on the graph so that the graph is within 0.001 unit of its horizontal asymptote? Explain your reasoning.

Limits at Infinity

199

Horizontal Asymptotes In Figure 3.34, the graph of f approaches the line y  L as x increases without bound. The line y  L is called a horizontal asymptote of the graph of f.

Definition of a Horizontal Asymptote The line y  L is a horizontal asymptote of the graph of f if lim f x  L

or

x→

lim f x  L.

x→

Note that from this definition, it follows that the graph of a function of x can have at most two horizontal asymptotes—one to the right and one to the left. Limits at infinity have many of the same properties of limits discussed in Section 1.3. For example, if lim f x and lim gx both exist, then x→

x→

lim f x  gx  lim f x  lim gx

x→

x→

x→

and lim f xgx  lim f x lim gx .

x→

x→

x→

Similar properties hold for limits at  . When evaluating limits at infinity, the following theorem is helpful. (A proof of this theorem is given in Appendix A.)

THEOREM 3.10

Limits at Infinity

If r is a positive rational number and c is any real number, then lim

x→

c  0. xr

Furthermore, if x r is defined when x < 0, then lim

x→

c  0. xr

EXAMPLE 1

Finding a Limit at Infinity



Find the limit: lim 5  x→



2 . x2

Solution Using Theorem 3.10, you can write



lim 5 

x→



2 2  lim 5  lim 2 2 x→ x→ x x 50  5.

.

Try It

Exploration A

Property of limits

200

CHAPTER 3

Applications of Differentiation

Finding a Limit at Infinity

EXAMPLE 2

Find the limit: lim

x→

2x  1 . x1

Solution Note that both the numerator and the denominator approach infinity as x approaches infinity. lim 2x  1 →

x→

2x  1 lim x→ x  1

lim x  1 →

x→

NOTE When you encounter an indeterminate form such as the one in Example 2, you should divide the numerator and denominator by the highest power of x in the denominator.

both the numerator and the denominator by x. After dividing, the limit may be evaluated as shown. 2x  1 2x  1 x lim  lim x→ x  1 x→ x  1 x

5

Simplify.

1 x→ x→ x  1 lim 1  lim x→ x→ x

Take limits of numerator and denominator.



x

−1

1 x  lim x→ 1 1 x lim 2  lim

f (x) = 2x − 1 x+1

1 −5 −4 −3 −2

Divide numerator and denominator by x.

2

6

3



, an indeterminate form. To resolve this problem, you can divide

This results in

y

4



1

2

3

20 10

Apply Theorem 3.10.

2 So, the line y  2 is a horizontal asymptote to the right. By taking the limit as x →  , you can see that y  2 is also a horizontal asymptote to the left. The graph of the function is shown in Figure 3.35.

y . 2 is a horizontal asymptote. Figure 3.35

Try It

Editable Graph

Exploration A

Exploration B

You can test the reasonableness of the limit found in Example 2 by evaluating f x for a few large positive values of x. For instance,

3

TECHNOLOGY

f 100  1.9703,

f 1000  1.9970, and

f 10,000  1.9997.

Another way to test the reasonableness of the limit is to use a graphing utility. For instance, in Figure 3.36, the graph of 0

80 0

As x increases, the graph of f moves closer and closer to the line y  2. Figure 3.36

f x 

2x  1 x1

is shown with the horizontal line y  2. Note that as x increases, the graph of f moves closer and closer to its horizontal asymptote.

SECTION 3.5

201

A Comparison of Three Rational Functions

EXAMPLE 3 Find each limit.

MARIA AGNESI (1718–1799)

.

Limits at Infinity

Agnesi was one of a handful of women to receive credit for significant contributions to mathematics before the twentieth century. In her early twenties, she wrote the first text that included both differential and integral calculus. By age 30, she was an honorary member of the faculty at the University of Bologna.

2x  5 x→ 3x 2  1

a. lim

2x 2  5 x→ 3x 2  1

2x 3  5 x→ 3x 2  1

b. lim

c. lim

Solution In each case, attempting to evaluate the limit produces the indeterminate form  . a. Divide both the numerator and the denominator by x 2 .

2x  5x 2 0  0 0 2x  5  lim   0 2 x→ 3x  1 x→ 3  1x 2 30 3 lim

MathBio

b. Divide both the numerator and the denominator by x 2. 2x 2  5 2  5x 2 2  0 2  lim   2 x→ 3x  1 x→ 3  1x 2 30 3 lim

c. Divide both the numerator and the denominator by x 2. 2x 3  5 2x  5x 2  lim  x→ 3x 2  1 x→ 3  1x 2 3 lim

.

You can conclude that the limit does not exist because the numerator increases without bound while the denominator approaches 3.

Try It

.

Exploration A

Open Exploration

Exploration B

Video

Guidelines for Finding Limits at ± of Rational Functions 1. If the degree of the numerator is less than the degree of the denominator, then the limit of the rational function is 0. 2. If the degree of the numerator is equal to the degree of the denominator, then the limit of the rational function is the ratio of the leading coefficients. 3. If the degree of the numerator is greater than the degree of the denominator, then the limit of the rational function does not exist.

y

2

f(x) =

1 x2 + 1

x

−2

−1

lim f(x) = 0

x → −∞

1

2

lim f(x) = 0

x→∞

f has a horizontal asymptote at y  0. Figure 3.37 FOR FURTHER INFORMATION For

more information on the contributions of women to mathematics, see the article “Why Women Succeed in Mathematics” by Mona Fabricant, Sylvia Svitak, and . Patricia Clark Kenschaft in Mathematics Teacher. MathArticle

Use these guidelines to check the results in Example 3. These limits seem reasonable when you consider that for large values of x, the highest-power term of the rational function is the most “influential” in determining the limit. For instance, the limit as x approaches infinity of the function f x 

1 x2  1

is 0 because the denominator overpowers the numerator as x increases or decreases without bound, as shown in Figure 3.37. The function shown in Figure 3.37 is a special case of a type of curve studied by the Italian mathematician Maria Gaetana Agnesi. The general form of this function is f x 

x2

8a 3  4a 2

Witch of Agnesi

and, through a mistranslation of the Italian word vertéré, the curve has come to be known as the Witch of Agnesi. Agnesi’s work with this curve first appeared in a comprehensive text on calculus that was published in 1748.

202

CHAPTER 3

Applications of Differentiation

In Figure 3.37, you can see that the function f x  1x 2  1 approaches the same horizontal asymptote to the right and to the left. This is always true of rational functions. Functions that are not rational, however, may approach different horizontal asymptotes to the right and to the left. This is demonstrated in Example 4. EXAMPLE 4

A Function with Two Horizontal Asymptotes

Find each limit. a. lim

x→

3x  2 2x 2  1

b.

lim

x→

3x  2 2x 2  1

Solution a. For x > 0, you can write x  x 2. So, dividing both the numerator and the denominator by x produces 3x  2 3x  2 x   2x 2  1 2x 2  1 x 2

3



2 x

2x 2  1 x2



3

2 x

2  x1

2

and you can take the limit as follows. 3x  2 lim  lim x→ 2x 2  1 x→

2 x

3

2  x1



30 2  0



3 2

2

y 4

y= 3 , 2 Horizontal asymptote to the right

b. For x < 0, you can write x   x 2. So, dividing both the numerator and the denominator by x produces

x

−6

−4

y=− 3 , 2 Horizontal asymptote to the left

−2

2

−4

f(x) =

4

3x − 2 2x 2 + 1

3x  2 2 2 3 3 3x  2 x x x    2 2x 2  1 2x 2  1 2x  1 1 2 2    x 2 x2 x



and you can take the limit as follows. 3x  2 lim  lim x→ 2x 2  1 x→

Functions that are not rational may have dif. right and left horizontal asymptotes. ferent Figure 3.38

3 

2 x

2  x1



30  2  0



3 2

2

The graph of f x  3x  22x 2  1 is shown in Figure 3.38.

Try It

Editable Graph



Exploration A

2

If you use a graphing utility to help estimate a limit, be sure that you also confirm the estimate analytically—the pictures shown by a graphing utility can be misleading. For instance, Figure 3.39 shows one view of the graph of

TECHNOLOGY PITFALL

−8

8

y −1

The horizontal asymptote appears to be the line y  1 but it is actually the line y  2. Figure 3.39

2x 3  1000x 2  x . x  1000x 2  x  1000 3

From this view, one could be convinced that the graph has y  1 as a horizontal asymptote. An analytical approach shows that the horizontal asymptote is actually y  2. Confirm this by enlarging the viewing window on the graphing utility.

SECTION 3.5

Limits at Infinity

203

In Section 1.3 (Example 9), you saw how the Squeeze Theorem can be used to evaluate limits involving trigonometric functions. This theorem is also valid for limits at infinity. EXAMPLE 5

Limits Involving Trigonometric Functions

Find each limit. a. lim sin x

b. lim

x→

x→

sin x x

y

Solution

y= 1 x

a. As x approaches infinity, the sine function oscillates between 1 and 1. So, this limit does not exist. b. Because 1 ≤ sin x ≤ 1, it follows that for x > 0,

1

f(x) = sinx x x

π



lim sinx x = 0 x→ ∞ −1

1 sin x 1 ≤ ≤ x x x

where lim 1x  0 and lim 1x  0. So, by the Squeeze Theorem, you x→

y=−1 x

x→

can obtain lim

As x increases without bound, f x . approaches 0.

x→

sin x 0 x

as shown in Figure 3.40.

Figure 3.40

Try It

Editable Graph

EXAMPLE 6

Exploration A Oxygen Level in a Pond

Suppose that f t measures the level of oxygen in a pond, where f t  1 is the normal (unpolluted) level and the time t is measured in weeks. When t  0, organic waste is dumped into the pond, and as the waste material oxidizes, the level of oxygen in the pond is f t 

What percent of the normal level of oxygen exists in the pond after 1 week? After 2 weeks? After 10 weeks? What is the limit as t approaches infinity?

f(t)

Solution When t  1, 2, and 10, the levels of oxygen are as shown.

Oxygen level

1.00 0.75 0.50

(10, 0.9)

(2, 0.6)

f 1 

2 t+1 f(t) = t − t2 + 1

(1, 0.5)

t 2

4

6

8

10

Weeks

The level of oxygen in a pond approaches the normal level of 1 as t approaches . Figure 3.41

12  1  1 1   50% 12  1 2

22  2  1 3   60% 22  1 5 2 10  10  1 91 f 10    90.1% 10 2  1 101 f 2 

0.25

.

t2  t  1 . t2  1

1 week

2 weeks

10 weeks

To find the limit as t approaches infinity, divide the numerator and the denominator by t 2 to obtain lim

t→

t2  t  1 1  1t  1t 2 1  0  0  lim   1  100%. 2 t→ t 1 1  1t 2 10

See Figure 3.41.

Try It

Exploration A

204

CHAPTER 3

Applications of Differentiation

Infinite Limits at Infinity Many functions do not approach a finite limit as x increases (or decreases) without bound. For instance, no polynomial function has a finite limit at infinity. The following definition is used to describe the behavior of polynomial and other functions at infinity.

NOTE Determining whether a function has an infinite limit at infinity is useful in analyzing the “end behavior” of its graph. You will see examples of this in Section 3.6 on curve sketching.

Definition of Infinite Limits at Infinity Let f be a function defined on the interval a, . 1. The statement lim f x  means that for each positive number M, there is x→ a corresponding number N > 0 such that f x > M whenever x > N. 2. The statement lim f x   means that for each negative number M, there x→ is a corresponding number N > 0 such that f x < M whenever x > N. Similar definitions can be given for the statements lim f x   .

lim f x  and

x→

x→

EXAMPLE 7

y

Find each limit.

3 2

a. lim x 3

−1

1

2

3

−1

a. As x increases without bound, x 3 also increases without bound. So, you can write lim x 3  . x→

b. As x decreases without bound, x 3 also decreases without bound. So, you can write lim x3   .

−2

x→

−3

.

lim x3

x→

Solution x

−2

b.

x→

f(x) = x 3

1

−3

Finding Infinite Limits at Infinity

The graph of f x  x 3 in Figure 3.42 illustrates these two results. These results agree with the Leading Coefficient Test for polynomial functions as described in Section P.3.

Figure 3.42

Try It

Editable Graph

EXAMPLE 8

Exploration A Finding Infinite Limits at Infinity

Find each limit. 2x 2  4x x→ x  1

a. lim

y

f(x) =

2x 2 − 4x 6 x+1 3 x

−12 −9

−6 −3

−3 −6

. Figure 3.43

Editable Graph

3

6

9

y = 2x − 6

12

b.

2x 2  4x x→ x  1 lim

Solution One way to evaluate each of these limits is to use long division to rewrite the improper rational function as the sum of a polynomial and a rational function. 2x 2  4x 6  lim 2x  6   x→ x  1 x→ x1 2x 2  4x 6 b. lim  lim 2x  6    x→ x  1 x→ x1 a. lim









The statements above can be interpreted as saying that as x approaches ± , the function f x  2x 2  4xx  1 behaves like the function gx  2x  6. In Section 3.6, you will see that this is graphically described by saying that the line y  2x  6 is a slant asymptote of the graph of f, as shown in Figure 3.43.

Try It

Exploration A

SECTION 3.5

Limits at Infinity

Exercises for Section 3.5 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, describe in your own words what the statement means. 1. lim f x  4

2.

x

lim f x  2

(a)

1 x2  1

1 x 3

1

1

2

3

1

1

1

(a) hx 

f x x2

(c) hx 

f x x4

2 3

y

(c)

y

(d)

(a) hx 

f x x

(c) hx 

f x x3

3

3

3 x2  2

(b) hx 

f x x3

(b) hx 

f x x2

16. f x  5x 2  3x  7

x 2

14. f x  4 

x

2 1

8x x 2  3

In Exercises 15 and 16, find lim hx, if possible.

3

3

12. f x 

15. f x  5x 3  3x 2  10

y

(b)

6x 4x 2  5

13. f x  5 

x

In Exercises 3–8, match the function with one of the graphs [(a), (b), (c), (d), (e), or (f)] using horizontal asymptotes as an aid. y

11. f x 

2

In Exercises 17–20, find each limit, if possible.

1

1 x 3 2 1

1

2

x

3

1

1

2

3

2 3

3 y

(e)

x2  2 x2  1

(b) lim

x2  2 xq x  1

(c) lim

xq

y

6

5  2x32 3x 32  4

(b) lim

5  2x 32 xq 3x  4

(c) lim

3 2 1

x

3. f x  5. f x  7. f x 

4

1

2

xq

3

2

3x 2 x2  2 x2

(b) lim

x

4. f x 

x 2

2x

6. f x  2 

4 sin x x2  1

8. f x 

x4

x2 1

2x 2  3x  5 x2  1

Numerical and Graphical Analysis In Exercises 9–14, use a graphing utility to complete the table and estimate the limit as x approaches infinity. Then use a graphing utility to graph the function and estimate the limit graphically. x

101

102

103

104

105

106

f x 9. f x 

10. f x 

2x 2 x1

5x32 4x32  1

5x32 xq 4x  1

21. lim

2x  1 3x  2

22. lim

23. lim

x x2  1

24. lim 4 

lim

5x 2 x3

xq

xq

25. 27.

xq

x

lim

xq

lim

x 2  x

2x  1

xq

x 2  x

31. lim

sin 2x x

33. lim

1 2x  sin x

xq

4x  3 2x  1

xq

In Exercises 21–34, find the limit.

29. 100

3  2x 3x  1

5x32 xq 4x 2  1

(c) lim

x 2  2

3  2x 3x 3  1

3  2 x2 xq 3x  1

20. (a) lim

1 2 2

xq

5  2 x 32 xq 3x 2  4

19. (a) lim

2

6 4 2

xq

(c) lim

4

4

18. (a) lim

(b) lim

(f)

8

x2  2 xq x 3  1

17. (a) lim

xq

xq

xq

26. 28.

3x 3  2 9x 3  2x 2  7



lim

xq

lim

xq

3 x



12 x  x4  2

x x 2  1

3x  1 x 2  x x  cos x 32. lim xq x 30.

lim

xq

34. lim cos xq

1 x

205

206

CHAPTER 3

Applications of Differentiation

In Exercises 35–38, use a graphing utility to graph the function and identify any horizontal asymptotes. 35. f x 

x

36. f x 

x1

3x 37. f x  x 2  2

38. f x 

3x  2

52. Sketch a graph of a differentiable function f that satisfies the following conditions and has x  2 as its only critical number.

x2

9x2  2

fx < 0 for x < 2

2x  1

x

1 x

40. lim x tan x

1 x

lim

x

x  x 2  3 

x

45. lim 4x  16x 2  x 

46.

x

x

lim

3x  9x 2  x 

lim

2x  14x  x 

x

x

(b) The graph of f is symmetric to the origin.

102

103

104

105

In Exercises 55–72, sketch the graph of the function using extrema, intercepts, symmetry, and asymptotes. Then use a graphing utility to verify your result. 55. y 

2x 1x

56. y 

x3 x2

57. y 

x x2  4

58. y 

2x 9  x2

x2 9

60. y 

2x 2 x2  4

62. y 

2

Numerical, Graphical, and Analytic Analysis In Exercises 47–50, use a graphing utility to complete the table and estimate the limit as x approaches infinity. Then use a graphing utility to graph the function and estimate the limit. Finally, find the limit analytically and compare your results with the estimates. 101

x

(a) The graph of f is symmetric to the y-axis.

x

44.

100

106

f x

59. y  61. y 

47. f x  x  xx  1

48. f x  x 2  xxx  1

1 49. f x  x sin 2x

x1 50. f x  xx

Writing About Concepts

y 6 4

4

2

2x 1x

2

(b) Use the graphs to estimate lim f x and lim fx. x

(c) Explain the answers you gave in part (b).

x

66. y 

2x 1  x2 1 x

68. y  1 

69. y  3 

2 x

70. y  4 1 



x3 x 2  4

72. y 

1 x2



x x 2

4

In Exercises 73–82, use a computer algebra system to analyze the graph of the function. Label any extrema and/or asymptotes that exist. 1 x2

74. f x 

x2 x2  1

75. f x 

x x2  4

76. f x 

1 x2  x  2

77. f x 

x2 x 2  4x  3

78. f x 

x1 x2  x  1

4

(a) Sketch f.

2x 2 x2  4

3 x2

f

2

x2 9

64. x 2y  4

73. f x  5  x

x2

67. y  2 

71. y 

51. The graph of a function f is shown below. To print an enlarged copy of the graph, select the MathGraph button.

x2

63. xy 2  4 65. y 

2

x

53. Is it possible to sketch a graph of a function that satisfies the conditions of Exercise 52 and has no points of inflection? Explain. 54. If f is a continuous function such that lim f x  5, find,

42. lim 2x  4x 2  1 

43. lim x  x 2  x 

x

x

if possible, lim f x for each specified condition.

In Exercises 41– 46, find the limit. (Hint: Treat the expression as a fraction whose denominator is 1, and rationalize the numerator.) Use a graphing utility to verify your result. 41.

fx > 0 for x > 2

lim f x  lim f x  6

In Exercises 39 and 40, find the limit. Hint: Let x  1/t and find the limit as t  0. 39. lim x sin

Writing About Concepts (continued)

79. f x 

3x

80. gx 

4x 2  1

81. gx  sin

x x 2,

x > 3

82. f x 

2x 3x 2  1

2 sin 2x x

SECTION 3.5

In Exercises 83 and 84, (a) use a graphing utility to graph f and g in the same viewing window, (b) verify algebraically that f and g represent the same function, and (c) zoom out sufficiently far so that the graph appears as a line. What equation does this line appear to have? (Note that the points at which the function is not continuous are not readily seen when you zoom out.) x3  3x 2  2 83. f x  xx  3

x3  2x 2  2 84. f x   2x 2

2 gx  x  xx  3

1 1 gx   x  1  2 2 x

85. Average Cost A business has a cost of C  0.5x  500 for producing x units. The average cost per unit is

Find the limit of C as x approaches infinity. 86. Engine Efficiency The efficiency of an internal combustion engine is



1 Efficiency %  100 1  v1v2c



where v1v2 is the ratio of the uncompressed gas to the compressed gas and c is a positive constant dependent on the engine design. Find the limit of the efficiency as the compression ratio approaches infinity. 87. Physics Newton’s First Law of Motion and Einstein’s Special Theory of Relativity differ concerning a particle’s behavior as its velocity approaches the speed of light, c. Functions N and E represent the predicted velocity, v, with respect to time, t, for a particle accelerated by a constant force. Write a limit statement that describes each theory.

207

89. Modeling Data The table shows the world record times for running 1 mile, where t represents the year, with t  0 corresponding to 1900, and y is the time in minutes and seconds. t

23

33

45

54

y

4:10.4

4:07.6

4:01.3

3:59.4

t

58

66

79

85

99

y

3:54.5

3:51.3

3:48.9

3:46.3

3:43.1

A model for the data is y

C C . x

Limits at Infinity

3.351t 2  42.461t  543.730 t2

where the seconds have been changed to decimal parts of a minute. (a) Use a graphing utility to plot the data and graph the model. (b) Does there appear to be a limiting time for running 1 mile? Explain. 90. Modeling Data The average typing speeds S (words per minute) of a typing student after t weeks of lessons are shown in the table. t

5

10

15

20

25

30

S

28

56

79

90

93

94

A model for the data is S 

100t 2 , 65  t 2

t > 0.

(a) Use a graphing utility to plot the data and graph the model. (b) Does there appear to be a limiting typing speed? Explain.

v

N

91. Modeling Data A heat probe is attached to the heat exchanger of a heating system. The temperature T (degrees Celsius) is recorded t seconds after the furnace is started. The results for the first 2 minutes are recorded in the table.

c E

t

88. Temperature The graph shows the temperature T, in degrees Fahrenheit, of an apple pie t seconds after it is removed from an oven and placed on a cooling rack.

t

0

15

30

45

60

T

25.2

36.9

45.5

51.4

56.0

t

75

90

105

120

T

59.6

62.0

64.0

65.2

T

(0, 425)

(a) Use the regression capabilities of a graphing utility to find a model of the form T1  at 2  bt  c for the data. (b) Use a graphing utility to graph T1. (c) A rational model for the data is T2 

72 t

graphing utility to graph the model.

1451  86t . Use a 58  t

(d) Find T10 and T20. (a) Find lim T. What does this limit represent?

(e) Find lim T2.

(b) Find lim T. What does this limit represent?

(f) Interpret the result in part (e) in the context of the problem. Is it possible to do this type of analysis using T1? Explain.

t0

t

t

208

CHAPTER 3

Applications of Differentiation

92. Modeling Data A container contains 5 liters of a 25% brine solution. The table shows the concentrations C of the mixture after adding x liters of a 75% brine solution to the container. x

0

0.5

1

1.5

2

C

0.25

0.295

0.333

0.365

0.393

x

2.5

3

3.5

4

C

0.417

0.438

0.456

0.472

96. The graph of f x 

y



Not drawn to scale

(a) Find L  lim f x and K  lim f x. (b) Determine x1 and x2 in terms of .

5  3x . Use a 20  4x





x

(e) What is the limiting concentration?

97. Consider lim

x

93. A line with slope m passes through the point 0, 4. (a) Write the distance d between the line and the point 3, 1 as a function of m.

geometrically.

98. Consider

(a) Write the distance d between the line and the point 4, 2 as a function of m. (b) Use a graphing utility to graph the equation in part (a). (c) Find lim dm and m

geometrically. 95. The graph of f x 

lim

m

3x x2  3

. Use the definition of limits at

x

lim

1 0 x2

x

100. lim

x

1 0 x3

102.

lim

2 x

x

0

1 0 x2

103. Prove that if px  an x n  . . .  a1x  a0 and qx  bm x m  . . .  b1x  b0 an  0, bm  0, then

x2

y

x2

99. lim 101.



lim

x

In Exercises 99–102, use the definition of limits at infinity to prove the limit.

dm. Interpret the results

2x2 is shown. 2

. Use the definition of limits at

infinity to find values of N that correspond to (a)  0.5 and (b)  0.1.

dm. Interpret the results

94. A line with slope m passes through the point 0, 2.

3x x2  3

infinity to find values of M that correspond to (a)  0.5 and (b)  0.1.

(b) Use a graphing utility to graph the equation in part (a). lim



(d) Determine N, where N < 0, such that f x  K < for x < N.

(d) Find lim C1 and lim C2. Which model do you think best represents the concentration of the mixture? Explain.



(c) Determine M, where M > 0, such that f x  L < for x > M.

graphing utility to graph C2.

m

x 

x

(c) A rational model for these data is C2 

m

x

x1 

(b) Use a graphing utility to graph C1.

(c) Find lim dm and

f

x2

(a) Use the regression features of a graphing utility to find a model of the form C1  ax 2  bx  c for the data.

x

6x is shown. x2  2

lim

x

f

px ,  qx bm ± ,

n < m n  m. n > m

104. Use the definition of infinite limits at infinity to prove that lim x3  .

x

x1



0, an

x

Not drawn to scale

True or False? In Exercises 105 and 106, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

(a) Find L  lim f x. x

(b) Determine x1 and x2 in terms of .





105. If fx > 0 for all real numbers x, then f increases without bound.





106. If f  x < 0 for all real numbers x, then f decreases without bound.

(c) Determine M, where M > 0, such that f x  L < for x > M. (d) Determine N, where N < 0, such that f x  L < for x < N.

SECTION 3.6

Section 3.6

A Summary of Curve Sketching

209

A Summary of Curve Sketching • Analyze and sketch the graph of a function.

Analyzing the Graph of a Function It would be difficult to overstate the importance of using graphs in mathematics. Descartes’s introduction of analytic geometry contributed significantly to the rapid advances in calculus that began during the mid-seventeenth century. In the words of Lagrange, “As long as algebra and geometry traveled separate paths their advance was slow and their applications limited. But when these two sciences joined company, they drew from each other fresh vitality and thenceforth marched on at a rapid pace toward perfection.” So far, you have studied several concepts that are useful in analyzing the graph of a function.

40

−2

5 −10

200 −10 10

30

• • • • • • • • • • •

x-intercepts and y- intercepts Symmetry Domain and range Continuity Vertical asymptotes Differentiability Relative extrema Concavity Points of inflection Horizontal asymptotes Infinite limits at infinity

(Section P.1) (Section P.1) (Section P.3) (Section 1.4) (Section 1.5) (Section 2.1) (Section 3.1) (Section 3.4) (Section 3.4) (Section 3.5) (Section 3.5)

When you are sketching the graph of a function, either by hand or with a graphing utility, remember that normally you cannot show the entire graph. The decision as to which part of the graph you choose to show is often crucial. For instance, which of the viewing windows in Figure 3.44 better represents the graph of f x  x3  25x2  74x  20?

−1200

Different viewing windows for the graph of f x  x3  25x2  74x  20 Figure 3.44

By seeing both views, it is clear that the second viewing window gives a more complete representation of the graph. But would a third viewing window reveal other interesting portions of the graph? To answer this, you need to use calculus to interpret the first and second derivatives. Here are some guidelines for determining a good viewing window for the graph of a function.

Guidelines for Analyzing the Graph of a Function 1. Determine the domain and range of the function. 2. Determine the intercepts, asymptotes, and symmetry of the graph. 3. Locate the x-values for which fx and f  x either are zero or do not exist. Use the results to determine relative extrema and points of inflection.

NOTE In these guidelines, note the importance of algebra (as well as calculus) for solving the equations f x  0, fx  0, and f  x  0.

210

CHAPTER 3

Applications of Differentiation

Sketching the Graph of a Rational Function

EXAMPLE 1

Analyze and sketch the graph of f x  f(x) =

2(x 2 − 9) x2 − 4

Solution

y

Relative minimum 9 0, 2

( )

4

x

−8

−4

4

(−3, 0)

20x x2  42 203x2  4 Second derivative: f  x  x2  43 x-intercepts: 3, 0, 3, 0 y-intercept: 0, 92  Vertical asymptotes: x  2, x  2 Horizontal asymptote: y  2 Critical number: x  0 Possible points of inflection: None Domain: All real numbers except x  ± 2 Symmetry: With respect to y-axis Test intervals:  , 2, 2, 0, 0, 2, 2,  fx 

First derivative:

Vertical asymptote: x=2

Vertical asymptote: x = −2 Horizontal asymptote: y=2

2x 2  9 . x2  4

8

(3, 0)

Using calculus, you can be certain that you have determined all characteristics of the . of f. graph Figure 3.45

The table shows how the test intervals are used to determine several characteristics of the graph. The graph of f is shown in Figure 3.45.

Editable Graph

f x

f x

f x

Characteristic of Graph





Decreasing, concave downward

Undef.

Undef.

Vertical asymptote





Decreasing, concave upward

0



Relative minimum





Increasing, concave upward

Undef.

Undef.

Vertical asymptote





Increasing, concave downward

 < x < 2

FOR FURTHER INFORMATION For

more information on the use of technology to graph rational functions, see the article “Graphs of Rational Functions for Computer Assisted Calculus” by Stan . and Terry Walters in The College Byrd Mathematics Journal.

x  2

Undef.

2 < x < 0 9 2

x0 0 < x < 2 x2

MathArticle

2 < x
3.

The function z  f x, y has a relative . minimum at  2, 3.

So, a relative minimum of f occurs at 2, 3. The value of the relative minimum is f 2, 3  3, as shown in Figure 13.66.

Figure 13.66

Rotatable Graph

Example 1 shows a relative minimum occurring at one type of critical point—the type for which both fx x, y and fy x, y are 0. The next example concerns a relative maximum that occurs at the other type of critical point—the type for which either fx x, y or fy x, y does not exist.

.

Try It EXAMPLE 2

Exploration A

Exploration B

Open Exploration

Finding a Relative Extremum

Determine the relative extrema of f x, y  1  x 2  y 2 13. Solution Because

Surface: f(x, y) = 1 − (x 2 + y 2)1/3 z

3

(0, 0, 1)

2 4

y

x

fx.x, y and fyx, y are undefined at 0, 0. Figure 13.67

Rotatable Graph

2x 3x 2  y 2 23

Partial with respect to x

fy x, y  

2y 3x 2  y 2 23

Partial with respect to y

and

1

4

fx x, y  

it follows that both partial derivatives exist for all points in the xy-plane except for 0, 0. Moreover, because the partial derivatives cannot both be 0 unless both x and y are 0, you can conclude that (0, 0) is the only critical point. In Figure 13.67, note that f 0, 0 is 1. For all other x, y it is clear that f x, y  1  x 2  y 2 13 < 1. So, f has a relative maximum at 0, 0.

Try It

Exploration A

NOTE In Example 2, fx x, y  0 for every point on the y-axis other than 0, 0. However, because fy x, y is nonzero, these are not critical points. Remember that one of the partials must not exist or both must be 0 in order to yield a critical point.

SECTION 13.8

z

f (x, y) = y 2 − x 2

y

Saddle point at 0, 0, 0: . fx 0, 0  fy 0, 0  0 Figure 13.68

Rotatable Graph

955

The Second Partials Test Theorem 13.16 tells you that to find relative extrema you need only examine values of f x, y at critical points. However, as is true for a function of one variable, the critical points of a function of two variables do not always yield relative maxima or minima. Some critical points yield saddle points, which are neither relative maxima nor relative minima. As an example of a critical point that does not yield a relative extremum, consider the surface given by f x, y  y 2  x 2

x

Extrema of Functions of Two Variables

Hyperbolic paraboloid

as shown in Figure 13.68. At the point 0, 0, both partial derivatives are 0. The function f does not, however, have a relative extremum at this point because in any open disk centered at 0, 0 the function takes on both negative values (along the x-axis) and positive values (along the y-axis). So, the point 0, 0, 0 is a saddle point of the surface. (The term “saddle point” comes from the fact that the surface shown in Figure 13.68 resembles a saddle.) For the functions in Examples 1 and 2, it was relatively easy to determine the relative extrema, because each function was either given, or able to be written, in completed square form. For more complicated functions, algebraic arguments are less convenient and it is better to rely on the analytic means presented in the following Second Partials Test. This is the two-variable counterpart of the Second Derivative Test for functions of one variable. The proof of this theorem is best left to a course in advanced calculus.

THEOREM 13.17

Second Partials Test

Let f have continuous second partial derivatives on an open region containing a point a, b for which fx a, b  0 and

fy a, b  0.

To test for relative extrema of f, consider the quantity d  fxx a, b fyy a, b  fxy a, b 2. 1. 2. 3. 4.

If d > 0 and fxx a, b > 0, then f has a relative minimum at a, b. If d > 0 and fxx a, b < 0, then f has a relative maximum at a, b. If d < 0, then a, b, f a, b is a saddle point. The test is inconclusive if d  0.

NOTE If d > 0, then fxxa, b and fyy a, b must have the same sign. This means that fxxa, b can be replaced by fyy a, b in the first two parts of the test.

A convenient device for remembering the formula for d in the Second Partials Test is given by the 2  2 determinant d



fxx a, b fyx a, b

fxy a, b fyy a, b



where fxy a, b  fyx a, b by Theorem 13.3.

956

CHAPTER 13

Functions of Several Variables

EXAMPLE 3

Find the relative extrema of f x, y  x 3  4xy  2y 2  1.

z 9

Solution Begin by finding the critical points of f. Because fx x, y  3x 2  4y and

8 7

5 4

Saddle point (0, 0, 1)

3

fy x, y  4x  4y

exist for all x and y, the only critical points are those for which both first partial derivatives are 0. To locate these points, let fx x, y and fy x, y be 0 to obtain 3x 2  4y  0 and 4x  4y  0. From the second equation you know that x  y, and, by substitution into the first equation, you obtain two solutions: y  x  0 and 4 y  x  3. Because

6

Relative maximum

Using the Second Partials Test

fxx x, y  6x,

fyy x, y  4, and

fxy x, y  4

it follows that, for the critical point 0, 0, 3

d  fxx 0, 0 fyy 0, 0  fxy 0, 0 2  0  16 < 0

2

and, by the Second Partials Test, you can conclude that 0, 0, 1 is a saddle point of f. Furthermore, for the critical point 43, 43 ,

x

( 43 , 43 )

4

d  fxx 43, 43  fyy 43, 43   fxy 43, 43   84  16

2

y

 16 > 0 ..

and because fxx 43, 43   8 < 0 you can conclude that f has a relative maximum at 43, 43 , as shown in Figure 13.69.

f(x, y) = −x 3 + 4xy − 2y 2 + 1

Figure 13.69

Try It

Rotatable Graph

Exploration A

The Second Partials Test can fail to find relative extrema in two ways. If either of the first partial derivatives does not exist, you cannot use the test. Also, if d  fxx a, b fyy a, b  fxy a, b  0 2

the test fails. In such cases, you can try a sketch or some other approach, as demonstrated in the next example. EXAMPLE 4

Failure of the Second Partials Test

Find the relative extrema of f x, y  x 2 y 2. Solution Because fx x, y  2xy 2 and fy x, y  2x 2y, you know that both partial derivatives are 0 if x  0 or y  0. That is, every point along the x- or y-axis is a critical point. Moreover, because

f(x, y) = x 2 y 2 z

fxx x, y  2y 2,

1

2 x

.

If y = 0, then f(x, y) = 0.

Figure 13.70

Rotatable Graph

fyy x, y  2x 2,

and

fxy x, y  4xy

you know that if either x  0 or y  0, then 2

If x = 0, then f(x, y) = 0.

y

d  fxx x, y fyy x, y  fxy x, y  4x 2 y 2  16x 2 y 2  12x 2 y 2  0. 2

So, the Second Partials Test fails. However, because f x, y  0 for every point along the x- or y-axis and f x, y  x 2 y 2 > 0 for all other points, you can conclude that each of these critical points yields an absolute minimum, as shown in Figure 13.70.

Try It

Exploration A

SECTION 13.8

Extrema of Functions of Two Variables

957

Absolute extrema of a function can occur in two ways. First, some relative extrema also happen to be absolute extrema. For instance, in Example 1, f 2, 3 is an absolute minimum of the function. (On the other hand, the relative maximum found in Example 3 is not an absolute maximum of the function.) Second, absolute extrema can occur at a boundary point of the domain. This is illustrated in Example 5. EXAMPLE 5 z

Surface: f(x, y) = sin xy 1

Find the absolute extrema of the function Absolute minima

Absolute maxima 1

x

.

y

xy = π 2

3

(π , 1) Absolute minima

Figure 13.71

Rotatable Graph

Finding Absolute Extrema

Domain: 0≤x≤π 0≤y≤1

f x, y  sin xy on the closed region given by 0 ≤ x ≤  and 0 ≤ y ≤ 1. Solution From the partial derivatives fx x, y  y cos xy

and

fy x, y  x cos xy

you can see that each point lying on the hyperbola given by xy  2 is a critical point. These points each yield the value f x, y  sin

 1 2

which you know is the absolute maximum, as shown in Figure 13.71. The only other critical point of f lying in the given region is 0, 0. It yields an absolute minimum of 0, because 0 ≤ xy ≤  implies that 0 ≤ sin xy ≤ 1.

.

To locate other absolute extrema, you should consider the four boundaries of the region formed by taking traces with the vertical planes x  0, x  , y  0, and y  1. In doing this, you will find that sin xy  0 at all points on the x-axis, at all points on the y-axis, and at the point , 1. Each of these points yields an absolute minimum for the surface, as shown in Figure 13.71.

Try It

Exploration A

The concepts of relative extrema and critical points can be extended to functions of three or more variables. If all first partial derivatives of w  f x1, x2, x3, . . . , xn exist, it can be shown that a relative maximum or minimum can occur at x1, x2, x3, . . . , xn only if every first partial derivative is 0 at that point. This means that the critical points are obtained by solving the following system of equations. fx1 x1, x2, x3, . . . , xn  0 fx2 x1, x2, x3, . . . , xn  0

⯗

fxn x1, x2, x3, . . . , xn  0 The extension of Theorem 13.17 to three or more variables is also possible, although you will not consider such an extension in this text.

958

CHAPTER 13

Functions of Several Variables

Exercises for Section 13.8 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–6, identify any extrema of the function by recognizing its given form or its form after completing the square. Verify your results by using the partial derivatives to locate any critical points and test for relative extrema. Use a computer algebra system to graph the function and label any extrema.

1 26. f x, y  2xy  2x 4  y 4  1

z 2

1. g x, y  x  1) 2   y  3 2

y

2. g x, y  9  x  3 2   y  2 2

−2

3. f x, y  x 2  y 2  1 4. f x, y   25  x  2 2  y 2 5. f x, y  x 2  y 2  2x  6y  6

3

6. f x, y  x 2  y 2  4x  8y  11

x

In Exercises 7–16, examine the function for relative extrema. 7. f x, y  2x 2  2xy  y 2  2x  3

Rotatable Graph 27. z  ex sin y

8. f x, y  x 2  5y 2  10x  30y  62

z

9. f x, y  5x 2  4xy  y 2  16x  10 10. f x, y  x 2  6xy  10y 2  4y  4

8

11. z  2x 2  3y 2  4x  12y  13

6

12. z  3x 2  2y 2  3x  4y  5

4

13. f x, y 

2

2x 2



y2

3

14. hx, y  x 2  y 213  2

 



15. gx, y  4  x  y



16. f x, y  x  y  2

In Exercises 17–20, use a computer algebra system to graph the surface and locate any relative extrema and saddle points. 17. z 

4x x2  y2  1

19. z  



28. z 

12  x

2



 y 2 e1x



4y 2

y

Rotatable Graph

18. f x, y  y 3  3yx 2  3y 2  3x 2  1 x2

3π

6 x

2 y2

z

2 2 e1x y

2

20. z  exy In Exercises 21–28, examine the function for relative extrema and saddle points. 21. h x, y  x 2  y 2  2x  4y  4

4

22. g x, y  120x  120y  xy  x 2  y 2 23. h x, y  x 2  3xy  y 2

4

24. g x, y  xy

25. f x, y  x 3  3xy  y 3

x

Rotatable Graph

z 7

In Exercises 29 and 30, examine the function for extrema without using the derivative tests and use a computer algebra system to graph the surface. (Hint: By observation, determine if it is possible for z to be negative. When is z equal to 0?) 29. z  4

6 x

Rotatable Graph

y

y

x  y4 x2  y2

30. z 

x 2  y 22 x2  y2

Think About It In Exercises 31–34, determine whether there is a relative maximum, a relative minimum, a saddle point, or insufficient information to determine the nature of the function f x, y at the critical point x0 , y0.

SECTION 13.8

Writing About Concepts 35. Define each of the following for a function of two variables.

Extrema of Functions of Two Variables

In Exercises 45–50, find the critical points and test for relative extrema. List the critical points for which the Second Partials Test fails.

(a) Relative minimum

45. f x, y  x 3  y 3

(b) Relative maximum

46. f x, y  x 3  y 3  6x 2  9y 2  12x  27y  19

(c) Saddle point

47. f x, y  x  1 2 y  4 2

(d) Critical point

48. f x, y  x  1 2   y  2 2

36. State the Second Partials Test for relative extrema and saddle points. In Exercises 37– 40, sketch the graph of an arbitrary function f satisfying the given conditions. State whether the function has any extrema or saddle points. (There are many correct answers.)

38. All of the first and second partial derivatives of f are 0. 39. fx 0, 0  0, fy 0, 0  0

 > 0,



x < 0 > 0, y < 0 , fy x, y x > 0 < 0, y > 0 fxx x, y > 0, fyy x, y < 0, and fxy x, y  0 for all x, y. < 0,

40. fx 2, 1  0,

fy 2, 1  0





> 0, x < 2 > 0, y < 1 , fy x, y < 0, x > 2 < 0, y > 1 fxx x, y < 0, fyy x, y < 0, and fxy x, y  0 for all x, y. fx x, y

41. The figure shows the level curves for an unknown function f x, y. What, if any, information can be given about f at the point A? Explain your reasoning. y

y

A

D

Figure for 41

50. f x, y  x 2  y 223

In Exercises 51 and 52, find the critical points of the function and, from the form of the function, determine whether a relative maximum or a relative minimum occurs at each point. 51. f x, y, z  x 2   y  3 2  z  1 2

In Exercises 53–62, find the absolute extrema of the function over the region R. (In each case, R contains the boundaries.) Use a computer algebra system to confirm your results. 53. f x, y  12  3x  2y R: The triangular region in the xy-plane with vertices 2, 0, 0, 1, and 1, 2 54. f x, y  2x  y2 R: The triangular region in the xy-plane with vertices 2, 0, 0, 1, and 1, 2 55. f x, y  3x 2  2y 2  4y R: The region in the xy-plane bounded by the graphs of y  x 2 and y  4 56. f x, y  2x  2xy  y 2 R: The region in the xy-plane bounded by the graphs of y  x2 and y  1

A

x

x

B

49. f x, y  x 23  y 23

52. f x, y, z  4  x y  1z  2 2

37. fx x, y > 0 and fy x, y < 0 for all x, y.

fx x, y

959

C

Figure for 42

42. The figure shows the level curves for an unknown function f x, y. What, if any, information can be given about f at the points A, B, C, and D? Explain your reasoning.

43. A function f has continuous second partial derivatives on an open region containing the critical point 3, 7. The function has a minimum at 3, 7 and d > 0 for the Second Partials Test. Determine the interval for fxy3, 7 if fxx 3, 7  2 and fyy 3, 7  8. 44. A function f has continuous second partial derivatives on an open region containing the critical point a, b. If fxxa, b and fyya, b have opposite signs, what is implied? Explain.

57. f x, y  x 2  xy,



 

R  x, y : x ≤ 2, y ≤ 1



58. f x, y  x 2  2xy  y 2,

R  x, y : x ≤ 2, y ≤ 1

59. f x, y  x 2  2xy  y 2,

R  x, y : x 2  y 2 ≤ 8

60. f x, y  x 2  4xy  5

R  x, y : 0 ≤ x ≤ 4, 0 ≤ y ≤ x

4xy x 2  1) y 2  1 R  x, y : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 4xy 62. f x, y  2 x  1 y 2  1 R  x, y : x ≥ 0, y ≥ 0, x 2  y 2 ≤ 1 61. f x, y 

True or False? In Exercises 63 and 64, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 63. If f has a relative maximum at x0, y0, z0, then fx x0, y0  fy x0, y0  0. 64. If f is continuous for all x and y and has two relative minima, then f must have at least one relative maximum.

960

CHAPTER 13

Functions of Several Variables

Section 13.9

Applications of Extrema of Functions of Two Variables • Solve optimization problems involving functions of several variables. • Use the method of least squares.

Applied Optimization Problems In this section, you will survey a few of the many applications of extrema of functions of two (or more) variables.

Finding Maximum Volume

EXAMPLE 1 z

(0, 0, 8)

A rectangular box is resting on the xy-plane with one vertex at the origin. The opposite vertex lies in the plane

Plane: 6x + 4y + 3z = 24

6x  4y  3z  24 as shown in Figure 13.72. Find the maximum volume of such a box. Solution Let x, y, and z represent the length, width, and height of the box. Because one vertex of the box lies in the plane 6x  4y  3z  24, you know that z  1324  6x  4y, and you can write the volume xyz of the box as a function of two variables. Vx, y  xy 324  6x  4y 1  324xy  6x 2y  4xy 2 1

x

.

(4, 0, 0)

(0, 6, 0)

y

By setting the first partial derivatives equal to 0 y 1 Vxx, y  3 24y  12xy  4y2  24  12x  4y  0 3 x Vyx, y  13 24x  6x 2  8xy  24  6x  8y  0 3

Figure 13.72

Rotatable Graph

you obtain the critical points 0, 0 and 43, 2. At 0, 0 the volume is 0, so that point does not yield a maximum volume. At the point 43, 2, you can apply the Second Partials Test.

NOTE In many applied problems, the domain of the function to be optimized is a closed bounded region. To find minimum or maximum points, you must not only test critical points, but also consider the values of the function at points on the boundary.

Vxxx, y  4y 8x Vyyx, y  3 1 Vxyx, y  324  12x  8y Because 4 4 4 32 8 64 Vxx3, 2Vyy3, 2  Vxy3, 2  8 9    3   3 > 0 2

2

and Vxx43, 2  8 < 0 you can conclude from the Second Partials Test that the maximum volume is V3, 2  3 243 2  63  2  43 22  64 9 cubic units. 4

.

1

4

4 2

4

Note that the volume is 0 at the boundary points of the triangular domain of V.

Try It

Exploration A

Open Exploration

SECTION 13.9

Applications of Extrema of Functions of Two Variables

961

Applications of extrema in economics and business often involve more than one independent variable. For instance, a company may produce several models of one type of product. The price per unit and profit per unit are usually different for each model. Moreover, the demand for each model is often a function of the prices of the other models (as well as its own price). The next example illustrates an application involving two products. EXAMPLE 2

Finding the Maximum Profit

An electronics manufacturer determines that the profit P (in dollars) obtained by producing x units of a DVD player and y units of a DVD recorder is approximated by the model Px, y  8x  10y  0.001x 2  xy  y 2  10,000. Find the production level that produces a maximum profit. What is the maximum profit? Solution The partial derivatives of the profit function are Pxx, y  8  0.0012x  y FOR FURTHER INFORMATION

For more information on the use of mathematics in economics, see the article “Mathematical Methods of . Economics” by Joel Franklin in The American Mathematical Monthly. MathArticle

and

Pyx, y  10  0.001x  2y.

By setting these partial derivatives equal to 0, you obtain the following system of equations. 8  0.0012x  y  0 10  0.001x  2y  0 After simplifying, this system of linear equations can be written as 2x  y  8000 x  2y  10,000. Solving this system produces x  2000 and y  4000. The second partial derivatives of P are Pxx2000, 4000  0.002 Pyy2000, 4000  0.002 Pxy2000, 4000  0.001. Because Pxx < 0 and Pxx2000, 4000Pyy2000, 4000  Pxy2000, 4000 2  0.0022  0.0012 > 0 you can conclude that the production level of x  2000 units and y  4000 units yields a maximum profit. The maximum profit is

.

P2000, 4000  82000  104000  0.001 20002  20004000  40002  10,000  $18,000.

Try It

Exploration A

NOTE In Example 2, it was assumed that the manufacturing plant is able to produce the required number of units to yield a maximum profit. In actual practice, the production would be bounded by physical constraints. You will study such constrained optimization problems in the next section.

962

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Functions of Several Variables

The Method of Least Squares Many of the examples in this text have involved mathematical models. For instance, Example 2 involves a quadratic model for profit. There are several ways to develop such models; one is called the method of least squares. In constructing a model to represent a particular phenomenon, the goals are simplicity and accuracy. Of course, these goals often conflict. For instance, a simple linear model for the points in Figure 13.73 is y  1.8566x  5.0246. However, Figure 13.74 shows that by choosing the slightly more complicated quadratic model* y  0.1996x 2  0.7281x  1.3749 you can achieve greater accuracy. y = 0.1996x 2 − 0.7281x + 1.3749

y = 1.8566x − 5.0246

y

y

(11, 17)

(11, 17)

15

15

(9, 12)

(9, 12)

10

10

(7, 6)

5

(7, 6)

5

(2, 1)

(5, 2)

(2, 1)

(5, 2)

x 5

x

10

5

Figure 13.73

10

Figure 13.74

As a measure of how well the model y  f x fits the collection of points

x1, y1, x2, y2, x3, y3, . . . , xn, yn

y

you can add the squares of the differences between the actual y-values and the values given by the model to obtain the sum of the squared errors

(x1, y1) d1

y = f(x)

S

n

 f x   y . i

i

2

Sum of the squared errors

i1

d2 (x2, y2)

(x3, y3) d3 x

Sum of the squared errors: S  d12  d22  d32 Figure 13.75

Graphically, S can be interpreted as the sum of the squares of the vertical distances between the graph of f and the given points in the plane, as shown in Figure 13.75. If the model is perfect, then S  0. However, when perfection is not feasible, you can settle for a model that minimizes S. For instance, the sum of the squared errors for the linear model in Figure 13.73 is S  17. Statisticians call the linear model that minimizes S the least squares regression line. The proof that this line actually minimizes S involves the minimizing of a function of two variables. * A method for finding the least squares quadratic model for a collection of data is described in Exercise 39.

SECTION 13.9

ADRIEN-MARIE LEGENDRE (1752–1833)

.

The method of least squares was introduced by the French mathematician Adrien-Marie Legendre. Legendre is best known for his work in geometry. In fact, his text Elements of Geometry was so popular in the United States that it continued to be used for 33 editions, spanning a period of more than 100 years.

Applications of Extrema of Functions of Two Variables

THEOREM 13.18

Least Squares Regression Line

The least squares regression line for x1, y1, x2, y2, . . . , xn, yn is given by f x  ax  b, where n

n a

n

n

xy  xy i i

i1 n



n

x2i 

i1

MathBio

963

i i i1 i1 n 2

b

and

 x 

1 n

  y  a  x . n

n

i

i1

i

i1

i

i1

Proof Let Sa, b represent the sum of the squared errors for the model f x  ax  b and the given set of points. That is, n

 f x   y

Sa, b  

i

i

2

i1 n

 ax  b  y  i

i

2

i1

where the points xi, yi  represent constants. Because S is a function of a and b, you can use the methods discussed in the preceding section to find the minimum value of S. Specifically, the first partial derivatives of S are n

 2x ax  b  y 

Saa, b 

i

i

i

i1

 2a

n

x

 2b

2 i

i1

n

n

 x  2 x y i

i i

i1

i1

n

 2ax  b  y 

Sba, b 

i

i

i1

 2a

n

n

 x  2nb  2  y .

i1

i

i

i1

By setting these two partial derivatives equal to 0, you obtain the values for a and b that are listed in the theorem. It is left to you to apply the Second Partials Test (see Exercise 40) to verify that these values of a and b yield a minimum. If the x-values are symmetrically spaced about the y-axis, then  xi  0 and the formulas for a and b simplify to n

xy

i i

a

i1 n

x

2 i

i1

and b

1 n y. n i1 i



This simplification is often possible with a translation of the x-values. For instance, if the x-values in a data collection consist of the years 2003, 2004, 2005, 2006, and 2007, you could let 2005 be represented by 0.

964

CHAPTER 13

Functions of Several Variables

Finding the Least Squares Regression Line

EXAMPLE 3

Find the least squares regression line for the points 3, 0, 1, 1, 0, 2, and 2, 3. Solution The table shows the calculations involved in finding the least squares regression line using n  4.

TECHNOLOGY Many calculators have “built-in” least squares regression programs. If your calculator has such a program, use it to duplicate the results of Example 3. n



x

y

xy

x2

3

0

0

9

1

1

1

1

0

2

0

0

2

3

6

4

n



xi  2

i1

n



yi  6

i1

xi yi  5

i1

n

x

2 i

 14

i1

Applying Theorem 13.18 produces y

n

n

(2, 3) 8 x + 47 f(x) = 13 26

a

3

n

n

2

−3

−1

x 1

. squares regression line Least . Figure 13.76

Editable Graph

i

2 i



i1 n

i

i1 2

 x 



45  26 8  414  22 13

i

i1

and

(−1, 1) −2

x

i1

1

n

i i

i1

(0, 2) (−3, 0)

n

xy  x y

2

b

1 n

.   y  a  x   14 6  138 2  47 26 n

n

i

i1

i

i1

8 The least squares regression line is f x  13 x  47 26 , as shown in Figure 13.76.

Try It

Exploration A

Exploration B

Exploration C

964

CHAPTER 13

Functions of Several Variables

Exercises for Section 13.9 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, find the minimum distance from the point to the plane 2x  3y  z  12. (Hint: To simplify the computations, minimize the square of the distance.) 1. 0, 0, 0

2. 1, 2, 3

In Exercises 3 and 4, find the minimum distance from the point to the paraboloid z  x 2  y 2. 3. 5, 5, 0

4. 5, 0, 0

9. Maximum Volume The sum of the length and the girth (perimeter of a cross section) of a package carried by a delivery service cannot exceed 108 inches. Find the dimensions of the rectangular package of largest volume that may be sent. 10. Maximum Volume The material for constructing the base of an open box costs 1.5 times as much per unit area as the material for constructing the sides. For a fixed amount of money C, find the dimensions of the box of largest volume that can be made. 11. Maximum Volume The volume of an ellipsoid

In Exercises 5–8, find three positive numbers x, y, and z that satisfy the given conditions.

y2 z2 x2  2 21 2 a b c

5. The sum is 30 and the product is a maximum.

is 4abc3. For a fixed sum a  b  c, show that the ellipsoid of maximum volume is a sphere.

6. The sum is 32 and P  xy 2z is a maximum. 7. The sum is 30 and the sum of the squares is a minimum. 8. The sum is 1 and the sum of the squares is a minimum.

12. Maximum Volume Show that the rectangular box of maximum volume inscribed in a sphere of radius r is a cube.

SECTION 13.9

13. Volume and Surface Area Show that a rectangular box of given volume and minimum surface area is a cube. 14. Maximum Volume Repeat Exercise 9 under the condition that the sum of the perimeters of the two cross sections shown in the figure cannot exceed 144 inches.

965

Applications of Extrema of Functions of Two Variables

21. Minimum Cost A water line is to be built from point P to point S and must pass through regions where construction costs differ (see figure). The cost per kilometer in dollars is 3k from P to Q, 2k from Q to R, and k from R to S. Find x and y such that the total cost C will be minimized. P 2 km

x Q

1 km

R y

S 10 km

Rotatable Graph 15. Area A trough with trapezoidal cross sections is formed by turning up the edges of a 30-inch-wide sheet of aluminum (see figure). Find the cross section of maximum area.

x

x

θ

θ

22. Distance A company has retail outlets located at the points 0, 0, 2, 2, and 2, 2 (see figure). Management plans to build a distribution center located such that the sum of the distances S from the center to the outlets is minimum. From the symmetry of the problem it is clear that the distribution center will be located on the y-axis, and therefore S is a function of the single variable y. Using techniques presented in Chapter 3, find the required value of y. y

30 − 2x

3

(−2, 2)

16. Area

Repeat Exercise 15 for a sheet that is w inches wide.

17. Maximum Revenue A company manufactures two types of sneakers, running shoes and basketball shoes. The total revenue from x1 units of running shoes and x2 units of basketball shoes is R  5x12  8x22  2x1x 2  42x1  102x 2, where x1 and x2 are in thousands of units. Find x1 and x2 so as to maximize the revenue.

y

d2

d3

d1

(0, y)

−3 −2 (0, 0) −2

Figure for 22

4

(2, 2)

2

(−2, 2) d1

x 1

2

(4, 2)

(x, y) d2

d3

(0, 0) 2

−2

x

4

−2

Figure for 23

18. Maximum Revenue A retail outlet sells two types of riding lawn mowers, the prices of which are p1 and p2. Find p1 and p2 so as to maximize total revenue, where R  515p1  805p2  1.5p1 p2  1.5p12 p22.

23. Investigation The retail outlets described in Exercise 22 are located at 0, 0, 4, 2, and 2, 2 (see figure). The location of the distribution center is x, y, and therefore the sum of the distances S is a function of x and y.

19. Maximum Profit A corporation manufactures candles at two locations. The cost of producing x1 units at location 1 is

(a) Write the expression giving the sum of the distances S. Use a computer algebra system to graph S. Does the surface have a minimum?

C1  0.02x12  4x1  500 and the cost of producing x2 units at location 2 is C2  0.05x22  4x 2  275. The candles sell for $15 per unit. Find the quantity that should be produced at each location to maximize the profit P  15x1  x2  C1  C2. 20. Hardy-Weinberg Law Common blood types are determined genetically by three alleles A, B, and O. (An allele is any of a group of possible mutational forms of a gene.) A person whose blood type is AA, BB, or OO is homozygous. A person whose blood type is AB, AO, or BO is heterozygous. The HardyWeinberg Law states that the proportion P of heterozygous individuals in any given population is P p, q, r  2pq  2pr  2qr where p represents the percent of allele A in the population, q represents the percent of allele B in the population, and r represents the percent of allele O in the population. Use the fact that p  q  r  1 to show that the maximum proportion of heterozygous individuals in any population is 23.

(b) Use a computer algebra system to obtain Sx and Sy . Observe that solving the system Sx  0 and Sy  0 is very difficult. So, approximate the location of the distribution center. (c) An initial estimate of the critical point is x1, y1  1, 1. Calculate "S1, 1 with components Sx1, 1 and Sy1, 1. What direction is given by the vector "S1, 1? (d) The second estimate of the critical point is

x 2, y2  x 1  Sxx1, y1t, y1  Syx1, y1t. If these coordinates are substituted into Sx, y, then S becomes a function of the single variable t. Find the value of t that minimizes S. Use this value of t to estimate x 2, y2. (e) Complete two more iterations of the process in part (d) to obtain x4, y4. For this location of the distribution center, what is the sum of the distances to the retail outlets? (f) Explain why "Sx, y was used to approximate the minimum value of S. In what types of problems would you use "Sx, y?

966

CHAPTER 13

Functions of Several Variables

24. Investigation Repeat Exercise 23 for retail outlets located at the points 4, 0, 1, 6, and 12, 2.

Writing About Concepts

Price, x

25. In your own words, state the problem-solving strategy for applied minimum and maximum problems.

Demand, y

26. In your own words, describe the method of least squares for finding mathematical models.

28.

y

y

(2, 3)

3 2

29.

1 −1

(4, 2) (3, 1)

1

(1, 1)

2

(6, 2)

x

3 4

1 2

x

3

375

330

(b) Use the model to estimate the demand when the price is $1.40. 37. Modeling Data An agronomist used four test plots to determine the relationship between the wheat yield y (in bushels per acre) and the amount of fertilizer x (in hundreds of pounds per acre). The results are shown in the table. Fertilizer, x

1.0

1.5

2.0

2.5

Yield, y

32

41

48

53

5

6

(2, 0)

4

38. Modeling Data The table shows the percents x and numbers y (in millions) of women in the work force for selected years. (Source: U.S. Bureau of Labor Statistics)

(4, 1)

(1, 0) (3, 0)

(2, 0)

450

Use the regression capabilities of a graphing utility to find the least squares regression line for the data, and estimate the yield for a fertilizer application of 160 pounds per acre.

(5, 2)

2

(1, 3)

1

3

2

y

30.

2 1

1

−2

(0, 4)

3

(1, 1)

−3 −2 −1

2

y 4

$1.50

x

x −1

(3, 2)

2

(−1, 1) (−3, 0) 1

(0, 1)

(−2, 0) −2

$1.25

4

3

1

$1.00

(a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data.

In Exercises 27–30, (a) find the least squares regression line and (b) calculate S, the sum of the squared errors. Use the regression capabilities of a graphing utility to verify your results. 27.

36. Modeling Data A store manager wants to know the demand y for an energy bar as a function of price x. The daily sales for three different prices of the energy bar are shown in the table.

In Exercises 31–34, find the least squares regression line for the points. Use the regression capabilities of a graphing utility to verify your results. Use the graphing utility to plot the points and graph the regression line.

Year

1965

1970

1975

1980

Percent, x

39.3

43.3

46.3

51.5

Number, y

26.2

31.5

37.5

45.5

Year

1985

1990

1995

2000

Percent, x

54.5

57.5

58.9

59.9

Number, y

51.1

56.8

60.9

66.3

31. 0, 0, 1, 1, 3, 4, 4, 2, 5, 5 32. 1, 0, 3, 3, 5, 6

(a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data.

33. 0, 6, 4, 3, 5, 0, 8, 4, 10, 5 34. 6, 4, 1, 2, 3, 3, 8, 6, 11, 8, 13, 8 35. Modeling Data The ages x (in years) and systolic blood pressures y of seven men are shown in the table. Age, x

16

25

39

45

49

64

70

Systolic Blood Pressure, y

109

122

143

132

199

185

199

(a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data. (b) Use a graphing utility to plot the data and graph the model. (c) Use the model to approximate the change in systolic blood pressure for each one-year increase in age.

(b) According to this model, approximately how many women enter the labor force for each one-point increase in the percent of women in the labor force? 39. Find a system of equations whose solution yields the coefficients a, b, and c for the least squares regression quadratic y  ax 2  bx  c for the points x1, y1, x2, y2, . . . , xn, yn by minimizing the sum Sa, b, c 

n

 y  ax i

2 i

 bxi  c2.

i1

40. Use the Second Partials Test to verify that the formulas for a and b given in Theorem 13.18 yield a minimum.

Hint: Use the fact that n  x ≥   x  . n

n

2 i

i1

2

i

i1

SECTION 13.9

In Exercises 41– 44, use the result of Exercise 39 to find the least squares regression quadratic for the given points. Use the regression capabilities of a graphing utility to confirm your results. Use the graphing utility to plot the points and graph the least squares regression quadratic. 41. 2, 0, 1, 0, 0, 1, 1, 2, 2, 5

Applications of Extrema of Functions of Two Variables

967

47. Modeling Data A meteorologist measures the atmospheric pressure P (in kilograms per square meter) at altitude h (in kilometers). The data are shown below. Altitude, h

0

5

10

15

20

Pressure, P

10,332

5583

2376

1240

517

42. 4, 5, 2, 6, 2, 6, 4, 2 43. 0, 0, 2, 2, 3, 6, 4, 12

44. 0, 10, 1, 9, 2, 6, 3, 0

45. Modeling Data After a new turbocharger for an automobile engine was developed, the following experimental data were obtained for speed y in miles per hour at two-second time intervals x.

(a) Use the regression capabilities of a graphing utility to find a least squares regression line for the points h, ln P. (b) The result in part (a) is an equation of the form ln P  ah  b. Write this logarithmic form in exponential form. (c) Use a graphing utility to plot the original data and graph the exponential model in part (b).

Time, x

0

2

4

6

8

10

(d) If your graphing utility can fit logarithmic models to data, use it to verify the result in part (b).

Speed, y

0

15

30

50

65

70

48. Modeling Data The endpoints of the interval over which distinct vision is possible are called the near point and far point of the eye. With increasing age, these points normally change. The table shows the approximate near points y in inches for various ages x (in years).

(a) Find a least squares regression quadratic for the data. Use a graphing utility to confirm your results. (b) Use a graphing utility to plot the points and graph the model. 46. Modeling Data The table shows the world populations y (in billions) for five different years. (Source: U.S. Bureau of the Census, International Data Base) Year Population, y

1994

1996

1998

2000

2002

5.6

5.8

5.9

6.1

6.2

Let x  4 represent the year 1994. (a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data. (b) Use the regression capabilities of a graphing utility to find the least squares regression quadratic for the data.

Age, x

16

32

44

Near Point, y

3.0

4.7

9.8

50

60

19.7 39.4

(a) Find a rational model for the data by taking the reciprocal of the near points to generate the points x, 1y. Use the regression capabilities of a graphing utility to find a least squares regression line for the revised data. The resulting line has the form 1  ax  b. y Solve for y.

(c) Use a graphing utility to plot the data and graph the models.

(b) Use a graphing utility to plot the data and graph the model.

(d) Use both models to forecast the world population for the year 2010. How do the two models differ as you extrapolate into the future?

(c) Do you think the model can be used to predict the near point for a person who is 70 years old? Explain.

968

CHAPTER 13

Functions of Several Variables

Section 13.10

Lagrange Multipliers • Understand the Method of Lagrange Multipliers. • Use Lagrange multipliers to solve constrained optimization problems. • Use the Method of Lagrange Multipliers with two constraints.

Lagrange Multipliers Many optimization problems have restrictions, or constraints, on the values that can be used to produce the optimal solution. Such constraints tend to complicate optimization problems because the optimal solution can occur at a boundary point of the domain. In this section, you will study an ingenious technique for solving such problems. It is called the Method of Lagrange Multipliers. To see how this technique works, suppose you want to find the rectangle of maximum area that can be inscribed in the ellipse given by x2 y2   1. 32 4 2 Let x, y be the vertex of the rectangle in the first quadrant, as shown in Figure 13.77. Because the rectangle has sides of lengths 2x and 2y, its area is given by f x, y  4xy.

Objective function

You want to find x and y such that f x, y is a maximum. Your choice of x, y is restricted to first-quadrant points that lie on the ellipse y2 x2  2  1. Constraint 2 3 4 Now, consider the constraint equation to be a fixed level curve of gx, y 

x2 y2  . 32 4 2

The level curves of f represent a family of hyperbolas f x, y  4xy  k. In this family, the level curves that meet the given constraint correspond to the hyperbolas that intersect the ellipse. Moreover, to maximize f x, y, you want to find the hyperbola that just barely satisfies the constraint. The level curve that does this is the one that is tangent to the ellipse, as shown in Figure 13.78. Ellipse: x2 y2 + =1 32 42

y

Level curves of f: 4xy = k

y

5

(x, y) 3

2

1 x

−4

−2

−1

−1

1

2

−2 −3

Objective function: f x, y  4xy Figure 13.77

k = 72 k = 56 k = 40 k = 24

3

2

4

1

x

−2 −1 −1

1

2

4

5

6

−2 −3

Constraint: gx, y  Figure 13.78

x2 y2  21 32 4

SECTION 13.10

Lagrange Multipliers

969

To find the appropriate hyperbola, use the fact that two curves are tangent at a point if and only if their gradient vectors are parallel. This means that "f x, y must be a scalar multiple of "gx, y at the point of tangency. In the context of constrained optimization problems, this scalar is denoted by # (the lowercase Greek letter lambda). "f x, y  #"gx, y The scalar # is called a Lagrange multiplier. Theorem 13.19 gives the necessary conditions for the existence of such multipliers.

THEOREM 13.19

Lagrange’s Theorem

Let f and g have continuous first partial derivatives such that f has an extremum at a point x0, y0  on the smooth constraint curve gx, y  c. If "gx0, y0   0, then there is a real number # such that "f x0, y0   #"gx0, y0 .

JOSEPH-LOUIS LAGRANGE (1736–1813) The Method of Lagrange Multipliers is named after the French mathematician JosephLouis Lagrange. Lagrange first introduced the method in his famous paper on mechanics, written when he was just 19 years old.

Proof To begin, represent the smooth curve given by gx, y  c by the vector-valued function rt  xti  ytj,

r t  0

where x and y are continuous on an open interval I. Define the function h as h t  f x t, y t. Then, because f x0, y0  is an extreme value of f, you know that h t0   f xt0 , y t0   f x0, y0  is an extreme value of h. This implies that ht0   0, and, by the Chain Rule, h t0   fxx0, y0  x t0   fy x0, y0 y t0   "f x0, y0   r t0   0.

NOTE Lagrange’s Theorem can be shown to be true for functions of three variables, using a similar argument with level surfaces and Theorem 13.14.

So, "f x0, y0  is orthogonal to r t0 . Moreover, by Theorem 13.12, "gx0, y0  is also orthogonal to r t0 . Consequently, the gradients "f x0, y0  and "gx0, y0  are parallel, and there must exist a scalar # such that "f x0, y0   #"gx0, y0 . The Method of Lagrange Multipliers uses Theorem 13.19 to find the extreme values of a function f subject to a constraint.

Method of Lagrange Multipliers Let f and g satisfy the hypothesis of Lagrange’s Theorem, and let f have a minimum or maximum subject to the constraint gx, y  c. To find the minimum or maximum of f, use the following steps. 1. Simultaneously solve the equations "f x, y  #"gx, y and gx, y  c by solving the following system of equations. fxx, y  #gxx, y NOTE As you will see in Examples 1 and 2, the Method of Lagrange Multipliers requires solving systems of nonlinear equations. This often can require some tricky algebraic manipulation.

fyx, y  #gyx, y gx, y  c 2. Evaluate f at each solution point obtained in the first step. The largest value yields the maximum of f subject to the constraint gx, y  c, and the smallest value yields the minimum of f subject to the constraint gx, y  c.

970

CHAPTER 13

Functions of Several Variables

Constrained Optimization Problems In the problem at the beginning of this section, you wanted to maximize the area of a rectangle that is inscribed in an ellipse. Example 1 shows how to use Lagrange multipliers to solve this problem.

Using a Lagrange Multiplier with One Constraint

EXAMPLE 1

Find the maximum value of f x, y  4xy where x > 0 and y > 0, subject to the constraint x 232   y 242  1. NOTE Example 1 can also be solved using the techniques you learned in Chapter 3. To see how, try to find the maximum value of A  4xy given that y2 x2  2  1. 2 3 4 To begin, solve the second equation for y to obtain 4

y  39  x 2. Then substitute into the first equation to obtain A  4x 439  x 2 . Finally, use the techniques of Chapter 3 to maximize A.

Solution To begin, let gx, y 

x2 y2 2  2  1. 3 4

By equating "f x, y  4yi  4xj and #"gx, y  2# x9 i  #y8 j, you can obtain the following system of equations. 2 4y  # x 9 1 4x  # y 8 2 2 x y  1 32 42

fxx, y  #gxx, y fyx, y  #gyx, y Constraint

From the first equation, you obtain #  18yx, and substitution into the second equation produces 4x 

 

1 18y y 8 x

x2 

9 2 y. 16

Substituting this value for x2 into the third equation produces





1 9 2 1 2 y  y 1 9 16 16

y 2  8.

So, y  ± 22. Because it is required that y > 0, choose the positive value and find that 9 2 y 16 9 9  8  16 2

x2 

x

3 2

.

So, the maximum value of f is .

f

32 , 22  4xy  4 3222   24.

Try It

Exploration A

Exploration B

Note that writing the constraint as gx, y 

x2 y2  21 2 3 4

or

gx, y 

x2 y2  210 2 3 4

does not affect the solution—the constant is eliminated when you form "g.

SECTION 13.10

EXAMPLE 2

Lagrange Multipliers

971

A Business Application

The Cobb-Douglas production function (see Example 5, Section 13.1) for a software manufacturer is given by f x, y  100x 34 y14

Objective function

where x represents the units of labor (at $150 per unit) and y represents the units of capital (at $250 per unit). The total cost of labor and capital is limited to $50,000. Find the maximum production level for this manufacturer. FOR FURTHER INFORMATION For more information on the use of Lagrange multipliers in economics, see the article “Lagrange Multiplier Problems in Economics” by John V. Baxley and . John C. Moorhouse in The American Mathematical Monthly.

MathArticle

Solution From the given function, you have "f x, y  75x14 y 14 i  25x 34 y34 j. The limit on the cost of labor and capital produces the constraint gx, y  150x  250y  50,000.

Constraint

So, #"gx, y  150# i  250# j. This gives rise to the following system of equations. 75x14 y 14  150# 25x 34 y34  250# 150x  250y  50,000

fxx, y  #gxx, y fyx, y  #gyx, y Constraint

By solving for # in the first equation

#

75x14 y14 x14 y14  150 2

and substituting into the second equation, you obtain 25x 34 y34  250

14 y14

x

2



25x  125y.

Multiply by x 14 y 34.

So, x  5y. By substituting into the third equation, you have 1505y  250y  50,000 1000y  50,000 y  50 units of capital x  250 units of labor. So, the maximum production level is .

f 250, 50  100250345014  16,719 product units.

Try It

Exploration A

Economists call the Lagrange multiplier obtained in a production function the marginal productivity of money. For instance, in Example 2 the marginal productivity of money at x  250 and y  50 is

#

x14 y 14 25014 5014   0.334 2 2

which means that for each additional dollar spent on production, an additional 0.334 unit of the product can be produced.

972

CHAPTER 13

Functions of Several Variables

EXAMPLE 3

Lagrange Multipliers and Three Variables

Find the minimum value of f x, y, z  2x 2  y 2  3z 2

Objective function

subject to the constraint 2x  3y  4z  49. Solution Let gx, y, z  2x  3y  4z  49. Then, because "f x, y, z  4xi  2yj  6zk

#"gx, y, z  2# i  3# j  4# k

and

you obtain the following system of equations. 4x  2#

fxx, y, z  #gxx, y, z

2y  3# 6z  4#

fyx, y, z  #gyx, y, z

2x  3y  4z  49

fzx, y, z  #gzx, y, z Constraint

The solution of this system is x  3, y  9, and z  4. So, the optimum value of f is f 3, 9, 4  23 2  92  34 2  147.

Ellipsoid: 2x 2 + y 2 + 3z 2 = 147 z

.

From the original function and constraint, it is clear that f x, y, z has no maximum. So, the optimum value of f determined above is a minimum.

8 y 16 −16 24

Point of tangency (3, −9, −4)

Exploration A

Open Exploration

A graphical interpretation of constrained optimization problems in two variables was given at the beginning of this section. In three variables, the interpretation is similar, except that level surfaces are used instead of level curves. For instance, in Example 3, the level surfaces of f are ellipsoids centered at the origin, and the constraint

x

Plane: 2x − 3y − 4z = 49

.

Try It

2x  3y  4z  49

Figure 13.79

is a plane. The minimum value of f is represented by the ellipsoid that is tangent to the constraint plane, as shown in Figure 13.79.

Rotatable Graph

EXAMPLE 4

Optimization Inside a Region

Find the extreme values of f x, y  x 2  2y 2  2x  3 subject to the constraint x 2  y 2 ≤ 10.

z 40

(−1, −3, 24)

Relative maxima

32

Solution To solve this problem, you can break the constraint into two cases.

(−1, 3, 24)

24 16

Relative minimum (1, 0, 2)

8 2 4

.

x

(

Figure 13.80

Rotatable Graph

Objective function

3

10, 0, 6.675(

4

y

a. For points on the circle x 2  y 2  10, you can use Lagrange multipliers to find that the maximum value of f x, y is 24—this value occurs at 1, 3 and at 1, 3. In a similar way, you can determine that the minimum value of f x, y is approximately 6.675—this value occurs at 10, 0. b. For points inside the circle, you can use the techniques discussed in Section 13.8 to conclude that the function has a relative minimum of 2 at the point 1, 0. By combining these two results, you can conclude that f has a maximum of 24 at 1, ± 3 and a minimum of 2 at 1, 0, as shown in Figure 13.80.

Try It

Exploration A

SECTION 13.10

Lagrange Multipliers

973

The Method of Lagrange Multipliers with Two Constraints For optimization problems involving two constraint functions g and h, you can introduce a second Lagrange multiplier, % (the lowercase Greek letter mu), and then solve the equation "f  #"g  %"h where the gradient vectors are not parallel, as illustrated in Example 5. EXAMPLE 5

Optimization with Two Constraints

Let T x, y, z  20  2x  2y  z 2 represent the temperature at each point on the sphere x 2  y 2  z 2  11. Find the extreme temperatures on the curve formed by the intersection of the plane x  y  z  3 and the sphere. Solution The two constraints are gx, y, z  x 2  y 2  z 2  11

and

hx, y, z  x  y  z  3.

Using "T x, y, z  2i  2j  2zk #"gx, y, z  2# x i  2#y j  2# z k and

%"h x, y, z  % i  % j  % k you can write the following system of equations. 2  2# x  %

Txx, y, z  #gxx, y, z  %hxx, y, z

2  2#y  % 2z  2#z  % 2  y 2  z 2  11 x xyz3

Tyx, y, z  #gyx, y, z  %hyx, y, z Tzx, y, z  #gzx, y, z  %hzx, y, z Constraint 1 Constraint 2

By subtracting the second equation from the first, you can obtain the following system. STUDY TIP The system of equations that arises in the Method of Lagrange Multipliers is not, in general, a linear system, and finding the solution often requires ingenuity.

#x  y  0 2z1  #  %  0 x 2  y 2  z 2  11 xyz3 From the first equation, you can conclude that #  0 or x  y. If #  0, you can show that the critical points are 3, 1, 1 and 1, 3, 1. Try doing this—it takes a little work. If #  0, then x  y and you can show that the critical points occur when x  y  3 ± 23 3 and z  3 $ 43 3. Finally, to find the optimal solutions, compare the temperatures at the four critical points. T 3, 1, 1  T 1, 3, 1  25 3  23 3  23 3  43 91 , ,   30.33 T 3 3 3 3 3  23 3  23 3  43 91 T , ,   30.33 3 3 3 3

 

.

 

So, T  25 is the minimum temperature and T  91 3 is the maximum temperature on the curve.

Try It

Exploration A

974

CHAPTER 13

Functions of Several Variables

Exercises for Section 13.10 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1– 4, identify the constraint and level curves of the objective function shown in the figure. Use the figure to approximate the indicated extrema, assuming that x and y are positive. Use Lagrange multipliers to verify your result. 1. Maximize z  xy

2. Maximize z  xy

Constraint: x  y  10 y

y

12 10

6

17. Minimize f x, y, z  x 2  y 2  z 2 Constraint: x  y  z  1 18. Minimize f x, y  x 2  10x  y 2  14y  70 Constraint: x  y  10

6 2

4 2

x

x 2

4

6

2

8 10 12

Constraint: x  y  4  0

y

4

2

x

2

In Exercises 5 –12, use Lagrange multipliers to find the indicated extrema, assuming that x and y are positive. 5. Minimize f x, y  x 2  y 2 Constraint: x  2y  6  0 6. Maximize f x, y  x  y 2

Constraints: x  y  z  32, Constraints: x  2z  6,

−2

−4

19. Maximize f x, y, z  xyz xyz0

x  y  12

21. Maximize f x, y, z  xy  yz

−2

4

In Exercises 19 –22, use Lagrange multipliers to find the indicated extrema of f subject to two constraints. In each case, assume that x, y, and z are nonnegative.

20. Minimize f x, y, z  x 2  y 2  z 2

c=1 c = 12

x −4

6

Constraint: 2x  4y  5

y

c=8 c=6 c=4 c=2

4

4. Minimize z  x 2  y 2

3. Minimize z  x 2  y 2

Constraint: x  y  z  6  0 Constraint: x  y  z  6  0

c=2 c=4 c=6

4

8

15. Minimize f x, y, z  x 2  y 2  z 2 16. Maximize f x, y, z  xyz

Constraint: 2x  y  4

c = 30 c = 40 c = 50

In Exercises 15 –18, use Lagrange multipliers to find the indicated extrema, assuming that x, y, and z are positive.

Constraints: x  2y  6,

x  3z  0

22. Maximize f x, y, z  xyz Constraints: x 2  z 2  5,

In Exercises 23–26, use Lagrange multipliers to find the minimum distance from the curve or surface to the indicated point. [Hint: In Exercise 23, minimize f x, y  x 2  y 2 subject to the constraint 2x  3y  1.] Point

Curve 23. Line: 2x  3y  1

2

24. Circle: x  4 2  y 2  4

Constraint: 2y  x 2  0 7. Maximize f x, y  2x  2xy  y Constraint: 2x  y  100

Surface 25. Plane: x  y  z  1

8. Minimize f x, y  3x  y  10

x  2y  0

26. Cone: z  x 2  y 2

0, 0 0, 10 Point 2, 1, 1 4, 0, 0

Constraint: x 2y  6 9. Maximize f x, y  6  x 2  y 2 Constraint: x  y  2  0 10. Minimize f x, y  x 2  y 2 Constraint: 2x  4y  15  0 11. Maximize f x, y  e

In Exercises 27 and 28, find the highest point on the curve of intersection of the surfaces. 27. Sphere: x 2  y 2  z 2  36, 28. Cone: x  2

y2



z2

Plane: 2x  y  z  2

 0, Plane: x  2z  4

xy

Writing About Concepts

Constraint: x 2  y 2  8 12. Minimize f x, y  2x  y

29. Explain what is meant by constrained optimization problems.

Constraint: xy  32 In Exercises 13 and 14, use Lagrange multipliers to find any extrema of the function subject to the constraint x 2  y 2 ≤ 1. 13. f x, y  x 2  3xy  y 2

14. f x, y  exy4

30. Explain the Method of Lagrange Multipliers for solving constrained optimization problems.

SECTION 13.10

31. Maximum Volume Use Lagrange multipliers to find the dimensions of the rectangular package of largest volume subject to the constraint that the sum of the length and the girth cannot exceed 108 inches. Compare the answer with that obtained in Exercise 9, Section 13.9. 32. Maximum Volume The material for the base of an open box costs 1.5 times as much per unit area as the material for constructing the sides. Use Lagrange multipliers to find the dimensions of the box of largest volume that can be made for a fixed cost C. Maximize V  xyz subject to 1.5xy  2xz  2yz  C. Compare the answer to that obtained in Exercise 10, Section 13.9. 33. Minimum Cost A cargo container (in the shape of a rectangular solid) must have a volume of 480 cubic feet. The bottom will cost $5 per square foot to construct and the sides and the top will cost $3 per square foot to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has minimum cost. 34. Minimum Surface Area Use Lagrange multipliers to find the dimensions of a right circular cylinder with volume V0 cubic units and minimum surface area. 35. Maximum Volume Use Lagrange multipliers to find the dimensions of a rectangular box of maximum volume that can be inscribed (with edges parallel to the coordinate axes) in the ellipsoid 2

2

2

y z x    1. a2 b2 c2 36. Geometric and Arithmetic Means (a) Use Lagrange multipliers to prove that the product of three positive numbers x, y, and z, whose sum has the constant value S, is a maximum when the three numbers are equal. Use this result to prove that 3  xyz ≤

xyz . 3

n

 x  S, and all x

. . .x, n

i

i

≥ 0. Then prove that

i1 n x x x  1 2 3

P Medium 1

θ1

d1

h y

x Medium 2 a

θ2

Figure for 37

. . .x n . . . x ≤ x1  x 2  x 3  . n n

This shows that the geometric mean is never greater than the arithmetic mean. 37. Refraction of Light When light waves traveling in a transparent medium strike the surface of a second transparent medium, they tend to “bend” in order to follow the path of minimum time. This tendency is called refraction and is described by Snell’s Law of Refraction, sin 1 sin 2  v1 v2 where 1 and 2 are the magnitudes of the angles shown in the figure, and v1 and v2 are the velocities of light in the two media. Use Lagrange multipliers to derive this law using x  y  a.

l

d2 Q

Figure for 38

38. Area and Perimeter A semicircle is on top of a rectangle (see figure). If the area is fixed and the perimeter is a minimum, or if the perimeter is fixed and the area is a maximum, use Lagrange multipliers to verify that the length of the rectangle is twice its height. 39. Hardy-Weinberg Law Use Lagrange multipliers to maximize Pp, q, r  2pq  2pr  2qr subject to p  q  r  1. See Exercise 20 in Section 13.9. 40. Temperature Distribution Let T x, y, z  100  x 2  y 2 represent the temperature at each point on the sphere x 2  y 2  z 2  50. Find the maximum temperature on the curve formed by the intersection of the sphere and the plane x  z  0. Production Level In Exercises 41 and 42, find the maximum production level P if the total cost of labor (at $48 per unit) and capital (at $36 per unit) is limited to $100,000, where x is the number of units of labor and y is the number of units of capital. 41. Px, y  100x 0.25 y 0.75

42. Px, y  100x 0.4y 0.6

Cost In Exercises 43 and 44, find the minimum cost of producing 20,000 units of a product, where x is the number of units of labor (at $48 per unit) and y is the number of units of capital (at $36 per unit). 43. Px, y  100x 0.25 y 0.75

(b) Generalize the result of part (a) to prove that the product x1 x2 x3 . . . xn is a maximum when x1  x 2  x 3 

975

Lagrange Multipliers

44. Px, y  100x 0.6y 0.4

45. Investigation Consider the objective function g, ,   cos  cos  cos  subject to the constraint that , , and  are the angles of a triangle. (a) Use Lagrange multipliers to maximize g. (b) Use the constraint to reduce the function g to a function of two independent variables. Use a computer algebra system to graph the surface represented by g. Identify the maximum values on the graph.

Putnam Exam Challenge 46. A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, what shape will have the greatest volume? This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.

976

CHAPTER 13

Functions of Several Variables

Review Exercises for Chapter 13 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, use the graph to determine whether z is a function of x and y. Explain.

xy x2  y2

21. f x, y, z  z arctan

z

1.

19. gx, y 

2

22. f x, y, z 

20. w  x 2  y 2  z 2 y x 1

1  x 2  y 2  z 2

23. ux, t  cen t sin nx 2

3

4

4

25. Think About It Sketch a graph of a function z  f x, y whose derivative fx is always negative and whose derivative fy is always negative.

y

x

26. Find the slopes of the surface z  x 2 ln y  1 in the x- and y-directions at the point 2, 0, 0.

Rotatable Graph 2.

24. ux, t  c sinakx cos kt

In Exercises 27–30, find all second partial derivatives and verify that the second mixed partials are equal.

z 3

5

x xy

27. f x, y  3x 2  xy  2y 3

28. hx, y 

29. hx, y  x sin y  y cos x

30. gx, y  cosx  2y

x

Laplace’s Equation In Exercises 31–34, show that the function satisfies Laplace’s equation

5 y

 2z  2z  2  0. 2 x y

Rotatable Graph In Exercises 3–6, use a computer algebra system to graph several level curves of the function. 3. f x, y  e

4. f x, y  ln xy

x 2 y 2

5. f x, y  x 2  y 2

6. f x, y 

7. f x, y  ex

y  2

 1 x

8. gx, y  y

9. f x, y, z  x 2  y  z 2, c  1 c0

In Exercises 11–14, find the limit and discuss the continuity of the function (if it exists). xy lim 11. x, y → 1, 1 x 2  y 2 13.

4x 2 y x, y → 0, 0 x 4  y 2 lim

xy lim 12. x, y → 1, 1 x 2  y 2 14.

y  xey x, y → 0, 0 1  x 2

y x2  y2

35. z  x sin

In Exercises 9 and 10, sketch the graph of the level surface f x, y, z  c at the given value of c. 10. f x, y, z  9x 2  y 2  9z 2,

33. z 

32. z  x3  3xy 2 34. z  e x sin y

In Exercises 35 and 36, find the total differential.

x xy

In Exercises 7 and 8, use a computer algebra system to graph the function. 2

31. z  x 2  y 2

2

lim

In Exercises 15–24, find all first partial derivatives. xy xy

15. f x, y  e x cos y

16. f x, y 

17. z  xe y  ye x

18. z  lnx 2  y 2  1

y x

36. z 

xy x 2  y 2

37. Error Analysis The legs of a right triangle are measured to be 1 5 centimeters and 12 centimeters, with a possible error of 2 centimeter. Approximate the maximum possible error in computing the length of the hypotenuse. Approximate the maximum percent error. 38. Error Analysis To determine the height of a tower, the angle of elevation to the top of the tower was measured from a point 1 100 feet ± 2 foot from the base. The angle is measured at 33, with a possible error of 1. Assuming that the ground is horizontal, approximate the maximum error in determining the height of the tower. 39. Volume A right circular cone is measured and the radius and height are found to be 2 inches and 5 inches, respectively. The 1 possible error in measurement is 8 inch. Approximate the maximum possible error in the computation of the volume. 40. Lateral Surface Area Approximate the error in the computation of the lateral surface area of the cone in Exercise 39. The lateral surface area is given by A  rr 2  h 2 .

REVIEW EXERCISES

In Exercises 41– 44, find the indicated derivatives (a) using the appropriate Chain Rule and (b) by substitution before differentiating. dw dt

41. w  lnx 2  y 2, x  2t  3,

x  cos t,

x  r cos t, xy , z

45.



z3

2, 1, 3 4, 4, 9

64. Approximation Consider the following approximations for a function f x, y centered at 0, 0.

!u !u , !r !t

y  r sin t,

Linear approximation:

zt

P1x, y  f 0, 0  fx 0, 0x  fy 0, 0y Quadratic approximation:

y  rt,

z  2r  t

P2x, y  f 0, 0  fx 0, 0x  fy 0, 0y 

In Exercises 45 and 46, differentiate implicitly to find the first partial derivatives of z. x2

Point y 2,

63. Find the angle of inclination of the tangent plane to the surface x 2  y 2  z 2  14 at the point 2, 1, 3.

!w !w , !r !t

x  2r  t,

x2

62. z  25  y 2, y  x

y  sin t

43. u  x 2  y 2  z 2,

44. w 

61. z 

du dt

42. u  y2  x,

In Exercises 61 and 62, find symmetric equations of the tangent line to the curve of intersection of the surfaces at the given point. Surfaces

y4t

977

y  2yz  xz  z  0 2

1 2 fxx0,

0x 2  fxy0, 0xy  12 fyy0, 0y 2

[Note that the linear approximation is the tangent plane to the surface at 0, 0, f 0, 0. (a) Find the linear approximation of f x, y  cos x  sin y centered at 0, 0.

46. xz 2  y sin z  0 In Exercises 47–50, find the directional derivative of the function at P in the direction of v.

(b) Find the quadratic approximation of f x, y  cos x  sin y centered at 0, 0.

47. f x, y  x 2y, 2, 1, v  i  j

(c) If y  0 in the quadratic approximation, you obtain the second-degree Taylor polynomial for what function?

48. f x, y  14 y 2  x 2,

(d) Complete the table.

1, 4, v  2i  j 49. w   xz, 1, 2, 2, v  2i  j  2k 50. w  6x 2  3xy  4y 2z, 1, 0, 1, v  i  j  k y2

In Exercises 51–54, find the gradient of the function and the maximum value of the directional derivative at the given point. 51. z 

y , 1, 1 x2  y2

53. z  ex cos y,

52. z 

 4

  0,

x2 , 2, 1 xy

54. z  x 2 y, 2, 1

In Exercises 55 and 56, use the gradient to find a unit normal vector to the graph of the equation at the given point. 55. 9x 2  4y 2  65, 3, 2 56. 4y sin x  y 2  3,

2 , 1

In Exercises 57–60, find an equation of the tangent plane and parametric equations of the normal line to the surface at the given point. Point

Surface 57. f x, y  x 2 y 58. f x, y  25  y 2 59. z  9  4x  6y  60. z  9  x 2  y 2

x2



y2

2, 1, 4 2, 3, 4 2, 3, 4 1, 2, 2

x

y

0

0

0

0.1

0.2

0.1

0.5

0.3

1

0.5

f x, y

P1x, y

P2x, y

(e) Use a computer algebra system to graph the surfaces z  f x, y, z  P1x, y, and z  P2x, y. How does the accuracy of the approximations change as the distance from 0, 0 increases? In Exercises 65–68, examine the function for relative extrema. Use a computer algebra system to graph the function and confirm your results. 65. f x, y  x 3  3xy  y 2 66. f x, y  2x 2  6xy  9y 2  8x  14 67. f x, y  xy 

1 1  x y

68. z  50x  y  0.1x 3  20x  150 

0.05y3  20.6y  125

978

CHAPTER 13

Functions of Several Variables

Writing In Exercises 69 and 70, write a short paragraph about the surface whose level curves (c-values evenly spaced) are shown. Comment on possible extrema, saddle points, the magnitude of the gradient, etc. y

69.

70.

y

x

x

(a) Use the regression capabilities of a graphing utility to find the least squares regression line for the data. Then use the graphing utility to plot the data and graph the model. (b) Use a graphing utility to plot the points ln t, y. Do these points appear to follow a linear pattern more closely than the plot of the given data in part (a)? (c) Use the regression capabilities of a graphing utility to find the least squares regression line for the points ln t, y and obtain the logarithmic model y  a  b ln t. (d) Use a graphing utility to plot the data and graph the linear and logarithmic models. Which is a better model? Explain. 76. Modeling Data The table shows the drag force y in kilograms for a motor vehicle at indicated speeds x in kilometers per hour.

71. Maximum Profit A corporation manufactures digital cameras at two locations. The cost functions for producing x1 units at location 1 and x2 units at location 2 are C1 

0.05x12

Speed, x

25

50

75

100

125

Drag, y

28

38

54

75

102

 15x1  5400

C2  0.03x 22  15x 2  6100

(a) Use the regression capabilities of a graphing utility to find the least squares regression quadratic for the data.

and the total revenue function is

(b) Use the model to estimate the total drag when the vehicle is moving at 80 kilometers per hour.

R  225  0.4x1  x 2 x1  x 2. Find the production levels at the two locations that will maximize the profit Px1, x2  R  C1  C2. 72. Minimum Cost A manufacturer has an order for 1000 units of wooden benches that can be produced at two locations. Let x1 and x2 be the numbers of units produced at the two locations. The cost function is

Find the number that should be produced at each location to meet the order and minimize cost. 73. Production Level The production function for a candy manufacturer is f x, y  4x  xy  2y where x is the number of units of labor and y is the number of units of capital. Assume that the total amount available for labor and capital is $2000, and that units of labor and capital cost $20 and $4, respectively. Find the maximum production level for this manufacturer. 74. Find the minimum distance from the point 2, 2, 0 to the surface z  x 2  y 2. 75. Modeling Data The data in the table show the yield y (in milligrams) of a chemical reaction after t minutes.

Yield, y Minutes, t Yield, y

77. w  xy  yz  xz Constraint: x  y  z  1 78. z  x 2 y Constraint: x  2y  2

C  0.25x12  10x1  0.15x22  12x 2.

Minutes, t

In Exercises 77 and 78, use Lagrange multipliers to locate and classify any extrema of the function.

1

2

3

4

1.5

7.4

10.2

13.4

5

6

7

8

15.8

16.3

18.2

18.3

79. Minimum Cost A water line is to be built from point P to point S and must pass through regions where construction costs differ (see figure). The cost per kilometer in dollars is 3k from P to Q, 2k from Q to R, and k from R to S. For simplicity, let k  1. Use Lagrange multipliers to find x, y, and z such that the total cost C will be minimized. P 2 km

Q

1 km

R x

y

S z

10 km

80. Investigation Consider the objective function f x, y  ax  by subject to the constraint x 264  y 236  1. Assume that x and y are positive. (a) Use a computer algebra system to graph the constraint. If a  4 and b  3, use the computer algebra system to graph the level curves of the objective function. By trial and error, find the level curve that appears to be tangent to the ellipse. Use the result to approximate the maximum of f subject to the constraint. (b) Repeat part (a) for a  4 and b  9.

P.S.

P.S.

Problem Solving

979

Problem Solving

The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

1. Heron’s Formula states that the area of a triangle with sides of lengths a, b, and c is given by A  ss  as  bs  c where s 

abc , as shown in the figure. 2



4xy , f x, y  x 2  y 2 0,

x, y  0, 0 x, y  0, 0

and the unit vector u 

1 2

i  j.

Does the directional derivative of f at P0, 0 in the direction of u exist? If f 0, 0 were defined as 2 instead of 0, would the directional derivative exist?

b

a

5. Consider the function

c

(a) Use Heron’s Formula to find the area of the triangle with vertices 0, 0, 3, 4, and 6, 0. (b) Show that among all triangles having a fixed perimeter, the triangle with the largest area is an equilateral triangle. (c) Show that among all triangles having a fixed area, the triangle with the smallest perimeter is an equilateral triangle.

6. A heated storage room is shaped like a rectangular box and has a volume of 1000 cubic feet, as shown in the figure. Because warm air rises, the heat loss per unit of area through the ceiling is five times as great as the heat loss through the floor. The heat loss through the four walls is three times as great as the heat loss through the floor. Determine the room dimensions that will minimize heat loss and therefore minimize heating costs. V = xyz = 1000

2. An industrial container is in the shape of a cylinder with hemispherical ends, as shown in the figure. The container must hold 1000 liters of fluid. Determine the radius r and length h that minimize the amount of material used in the construction of the tank.

z y

x

r

Rotatable Graph 7. Repeat Exercise 6 assuming that the heat loss through the walls and ceiling remain the same, but the floor is insulated so that there is no heat loss through the floor.

h

Rotatable Graph 3. Let Px0, y0, z0 be a point in the first octant on the surface xyz  1.

8. Consider a circular plate of radius 1 given by x 2  y 2 ≤ 1, as shown in the figure. The temperature at any point Px, y on the plate is T x, y  2x 2  y 2  y  10. y

(a) Find the equation of the tangent plane to the surface at the point P. (b) Show that the volume of the tetrahedron formed by the three coordinate planes and the tangent plane is constant, independent of the point of tangency (see figure). z

1

x2 + y2 ≤ 1

−1

1

x

−1

3

(a) Sketch the isotherm T x, y  10. To print an enlarged copy of the graph, select the MathGraph button.

P 3

3 x

(b) Find the hottest and coldest points on the plate.

y

9. Consider the Cobb-Douglas production function f x, y  Cxay1a,

Rotatable Graph

(a) Show that f satisfies the equation x

3 x3  1 4. Use a graphing utility to graph the functions f x   and gx  x in the same viewing window.

lim f x  gx  0 and lim f x  gx  0. x→

(b) Find the point on the graph of f that is farthest from the

!f !f y  f. dx dy

(b) Show that f tx, ty  t f x, y. 10. Rewrite Laplace’s equation

(a) Show that x→

0 < a < 1.

coordinates.

!2u !2u !2u   2  0 in cylindrical !x 2 !y 2 !z

980

CHAPTER 13

Functions of Several Variables

11. A projectile is launched at an angle of 45 with the horizontal and with an initial velocity of 64 feet per second. A television camera is located in the plane of the path of the projectile 50 feet behind the launch site (see figure).

15. The figure shows a rectangle that is approximately l  6 centimeters long and h  1 centimeter high. l = 6 cm h = 1 cm

y

(a) Draw a rectangular strip along the rectangular region showing a small increase in length.

(x, y)

(b) Draw a rectangular strip along the rectangular region showing a small increase in height. (−50, 0)

α

45°

x

(a) Find parametric equations for the path of the projectile in terms of the parameter t representing time. (b) Write the angle  that the camera makes with the horizontal in terms of x and y and in terms of t. (c) Use the results of part (b) to find ddt. (d) Use a graphing utility to graph  in terms of t. Is the graph symmetric to the axis of the parabolic arch of the projectile? At what time is the rate of change of  greatest? (e) At what time is the angle  maximum? Does this occur when the projectile is at its greatest height? 12. Consider the distance d between the launch site and the projectile in Exercise 11. (a) Write the distance d in terms of x and y and in terms of the parameter t. (b) Use the results of part (a) to find the rate of change of d. (c) Find the rate of change of the distance when t  2. (d) When is the rate of change of d minimum during the flight of the projectile? Does this occur at the time when the projectile reaches its maximum height? 13. Consider the function f x, y  x 2  y 2ex

2

y 2,



0 <  < .

(a) Use a computer algebra system to graph the function for   1 and   2, and identify any extrema or saddle points. (b) Use a computer algebra system to graph the function for   1 and   2, and identify any extrema or saddle points. (c) Generalize the results in parts (a) and (b) for the function f. 14. Prove that if f is a differentiable function such that "f x0, y0  0 then the tangent plane at x0, y0 is horizontal.

(c) Use the results in parts (a) and (b) to identify the measurement that has more effect on the area A of the rectangle. (d) Verify your answer in part (c) analytically by comparing the value of dA when dl  0.01 and when dh  0.01. 16. Consider converting a point 5 ± 0.05,18 ± 0.05 in polar coordinates to rectangular coordinates x, y. (a) Use a geometric argument to determine whether the accuracy in x is more dependent on the accuracy in r or on the accuracy in . Explain. Verify your answer analytically. (b) Use a geometric argument to determine whether the accuracy in y is more dependent on the accuracy in r or on the accuracy in . Explain. Verify your answer analytically. 17. Let f be a differentiable function of one variable. Show that all tangent planes to the surface z  y f xy intersect in a common point. 18. Consider the ellipse y2 x2  21 2 a b that encloses the circle x2  y2  2x. Find values of a and b that minimize the area of the ellipse. 19. Show that 1 ux, t  sinx  t  sinx  t 2 is a solution to the one-dimensional wave equation !2u !2u  2. !t 2 !x 20. Show that 1 ux, t  f x  ct  f x  ct 2 is a solution to the one-dimensional wave equation ! 2u ! 2u  c2 2. 2 !t !x (This equation describes the small transverse vibration of an elastic string such as those on certain musical instruments.)

982

CHAPTER 14

Multiple Integration

Section 14.1

Iterated Integrals and Area in the Plane • Evaluate an iterated integral. • Use an iterated integral to find the area of a plane region.

Iterated Integrals NOTE In Chapters 14 and 15, you will study several applications of integration involving functions of several variables. Chapter 14 is much like Chapter 7 in that it surveys the use of integration to find plane areas, volumes, surface areas, moments, and centers of mass.

In Chapter 13, you saw that it is meaningful to differentiate functions of several variables with respect to one variable while holding the other variables constant. You can integrate functions of several variables by a similar procedure. For example, if you are given the partial derivative fxx, y  2xy then, by considering y constant, you can integrate with respect to x to obtain f x, y  

  

y

fxx, y dx

Integrate with respect to x.

2xy dx

Hold y constant.

2x dx

Factor out constant y.

 yx 2  C y  x 2 y  C y.

Antiderivative of 2x is x 2. C y is a function of y.

The “constant” of integration, C y, is a function of y. In other words, by integrating with respect to x, you are able to recover f x, y only partially. The total recovery of a function of x and y from its partial derivatives is a topic you will study in Chapter 15. For now, we are more concerned with extending definite integrals to functions of several variables. For instance, by considering y constant, you can apply the Fundamental Theorem of Calculus to evaluate



2y

2y



2xy dx  x 2y

1

x is the variable of integration and y is fixed.

1

 2y2 y  12y  4y 3  y.

Replace x by the limits of integration.

The result is a function of y.

Similarly, you can integrate with respect to y by holding x fixed. Both procedures are summarized as follows.

 

h 2 y

h1 y g x 2

g1x

h2 y



fxx, y dx  f x, y

h1 y

 f h2 y, y  f h1 y, y

With respect to x

 f x, g2x  f x, g1x

With respect to y

g x



fyx, y dy  f x, y

2

g1x

Note that the variable of integration cannot appear in either limit of integration. For instance, it makes no sense to write



x

0

y dx.

SECTION 14.1

983

Integrating with Respect to y

EXAMPLE 1



Iterated Integrals and Area in the Plane

x

Evaluate

2x 2y2  2y dy.

1

Solution Considering x to be constant and integrating with respect to y produces



x

x

2x 2  2y dy   y2 y 1 2 2x 2x 2  x2  1  x 1

 



2x 2y2

1

.

  

Integrate with respect to y.



 3x 2  2x  1.

Try It

Exploration A

Notice in Example 1 that the integral defines a function of x and can itself be integrated, as shown in the next example.

The Integral of an Integral

EXAMPLE 2

  2

Evaluate



x

1

2x 2y2  2y dy dx.

1

Solution Using the result of Example 1, you have

  2

 

x

1

2

2x 2y2  2y dy dx 

1

3x 2  2x  1 dx

1 2





 x3  x 2  x

Integrate with respect to x.

1

 2  1  3.

.

Try It

Exploration A

The integral in Example 2 is an iterated integral. The brackets used in Example 2 are normally not written. Instead, iterated integrals are usually written simply as

 b

y=x

y

R: 1 ≤ x ≤ 2 1≤y≤x

a

2

1

x 1

2

The region of integration for

 2

1

x

f x, y dy dx

1

Figure 14.1

g2x

g1(x

 d

f x, y dy dx

and

c

h2 y

h1 y

f x, y dx dy.

The inside limits of integration can be variable with respect to the outer variable of integration. However, the outside limits of integration must be constant with respect to both variables of integration. After performing the inside integration, you obtain a “standard” definite integral, and the second integration produces a real number. The limits of integration for an iterated integral identify two sets of boundary intervals for the variables. For instance, in Example 2, the outside limits indicate that x lies in the interval 1 ≤ x ≤ 2 and the inside limits indicate that y lies in the interval 1 ≤ y ≤ x. Together, these two intervals determine the region of integration R of the iterated integral, as shown in Figure 14.1. Because an iterated integral is just a special type of definite integral—one in which the integrand is also an integral—you can use the properties of definite integrals to evaluate iterated integrals.

984

CHAPTER 14

Multiple Integration

Area of a Plane Region y

In the remainder of this section, you will take a new look at an old problem—that of finding the area of a plane region. Consider the plane region R bounded by a ≤ x ≤ b and g1x ≤ y ≤ g2x, as shown in Figure 14.2. The area of R is given by the definite integral

Region is bounded by a ≤ x ≤ b and g1(x) ≤ y ≤ g2(x) g2



b

g2x  g1x dx.

Area of R

a

R g1

Δx

Using the Fundamental Theorem of Calculus, you can rewrite the integrand g2x  g1x as a definite integral. Specifically, if you consider x to be fixed and let y vary from g1x to g2x, you can write x

a

b Area =

b

g2(x)

a

g1(x)



g2x

g1x

dy dx

g2x



dy  y

g1x

 g2x  g1x.

Combining these two integrals, you can write the area of the region R as an iterated integral

Vertically simple region

 b

Figure 14.2

a

g2x

g1x

 

b

dy dx 

a b



g2x



y

g1x

dx

Area of R

g2x  g1x dx.

a

Placing a representative rectangle in the region R helps determine both the order and the limits of integration. A vertical rectangle implies the order dy dx, with the inside limits corresponding to the upper and lower bounds of the rectangle, as shown in Figure 14.2. This type of region is called vertically simple, because the outside limits of integration represent the vertical lines x  a and x  b. Similarly, a horizontal rectangle implies the order dx dy, with the inside limits determined by the left and right bounds of the rectangle, as shown in Figure 14.3. This type of region is called horizontally simple, because the outside limits represent the horizontal lines y  c and y  d. The iterated integrals used for these two types of simple regions are summarized as follows.

Region is bounded by c ≤ y ≤ d and h1(y) ≤ x ≤ h2(y) y

d R Δy

Area of a Region in the Plane

c

1. If R is defined by a ≤ x ≤ b and g1x ≤ y ≤ g2x, where g1 and g2 are continuous on a, b , then the area of R is given by h2

h1 Area =

d

h (y) 2

c

h1(y)

x

 b

A

a

dx dy

Horizontally simple region

g2x

dy dx.

Figure 14.2 (vertically simple)

g1x

2. If R is defined by c ≤ y ≤ d and h1 y ≤ x ≤ h2 y, where h1 and h2 are continuous on c, d , then the area of R is given by

Figure 14.3

 d

A

c

h2y

dx dy.

Figure 14.3 (horizontally simple)

h1y

NOTE Be sure you see that the order of integration of these two integrals is different—the order dy dx corresponds to a vertically simple region, and the order dx dy corresponds to a horizontally simple region.

SECTION 14.1

Iterated Integrals and Area in the Plane

985

If all four limits of integration happen to be constants, the region of integration is rectangular, as shown in Example 3. EXAMPLE 3

The Area of a Rectangular Region

Use an iterated integral to represent the area of the rectangle shown in Figure 14.4. y

Solution The region shown in Figure 14.4 is both vertically simple and horizontally simple, so you can use either order of integration. By choosing the order dy dx, you obtain the following.

Rectangular region d

 b

d−c

R

a

 

d

b

dy dx 

c

d

 dx

y

 c

Integrate with respect to y.

c

a b

d  c dx

a

b

a

x



Integrate with respect to x. a

 d  cb  a

b−a

.

b



 d  cx

Notice that this answer is consistent with what you know from geometry.

Figure 14.4

Exploration B

Exploration A EXAMPLE 4

Finding Area by an Iterated Integral

Use an iterated integral to find the area of the region bounded by the graphs of f x  sin x gx  cos x

Sine curve forms upper boundary. Cosine curve forms lower boundary.

between x  4 and x  54. Solution Because f and g are given as functions of x, a vertical representative rectangle is convenient, and you can choose dy dx as the order of integration, as shown in Figure 14.5. The outside limits of integration are 4 ≤ x ≤ 54. Moreover, because the rectangle is bounded above by f x  sin x and below by gx  cos x, you have

5π R: π 4 ≤x≤ 4 cos x ≤ y ≤ sin x

y

Area of R 

y = cos x π 4

π 2

−1

π

3π 2

Δx



x

 y = sin x

Area =

.

5π /4 sin x

π /4

54

sin x

dy dx

cos x 4 54 sin x

y

4 54 4



dx

Integrate with respect to y.

cos x

sin x  cos x dx

cos x

54



 cos x  sin x

dy dx

4

Integrate with respect to x.

 22.

Figure 14.5

Editable Graph

    

Try It

Exploration A

Exploration B

NOTE The region of integration of an iterated integral need not have any straight lines as boundaries. For instance, the region of integration shown in Figure 14.5 is vertically simple even though it has no vertical lines as left and right boundaries. The quality that makes the region vertically simple is that it is bounded above and below by the graphs of functions of x.

986

CHAPTER 14

Multiple Integration

One order of integration will often produce a simpler integration problem than the other order. For instance, try reworking Example 4 with the order dx dy—you may be surprised to see that the task is formidable. However, if you succeed, you will see that the answer is the same. In other words, the order of integration affects the ease of integration, but not the value of the integral. EXAMPLE 5

Comparing Different Orders of Integration

Sketch the region whose area is represented by the integral

 2

y 3

R: 0 ≤ y ≤ 2 y2 ≤ x ≤ 4

2

x = y2

0

dx dy.

y2

Then find another iterated integral using the order dy dx to represent the same area and show that both integrals yield the same value.

(4, 2)

Solution From the given limits of integration, you know that

Δy

1

4

y2 ≤ x ≤ 4 x

1

3

2

4 2 4

−1

Area =

.

0 y2

dx dy

which means that the region R is bounded on the left by the parabola x  y 2 and on the right by the line x  4. Furthermore, because 0 ≤ y ≤ 2

(a)

 2

R: 0 ≤ x ≤ 4 0≤y≤ x

3 2

y=

Outer limits of integration

you know that R is bounded below by the x-axis, as shown in Figure 14.6(a). The value of this integral is

Editable Graph y

Inner limits of integration

0

4

dx dy 

y2



4

x

dy

4  y 2 dy

0



y3 3

 4y 

1

2 Δx 3 4

Area =

dy dx 0 0

(b)

Figure 14.6

x

Integrate with respect to x.

y2

2



1

−1

2

0

(4, 2)

x

 

x

4



2 0



16 . 3

Integrate with respect to y.

To change the order of integration to dy dx, place a vertical rectangle in the region, as shown in Figure 14.6(b). From this you can see that the constant bounds 0 ≤ x ≤ 4 serve as the outer limits of integration. By solving for y in the equation x  y 2, you can conclude that the inner bounds are 0 ≤ y ≤ x. So, the area of the region can also be represented by

 4

0

x

dy dx.

0

By evaluating this integral, you can see that it has the same value as the original integral.

 4

0

x

 

4

dy dx 

0 4

0





x

y

dx

Integrate with respect to y.

0

x dx

0

.



Try It



2 32 x 3

4 0



16 3

Exploration A

Integrate with respect to x.

Exploration B

Open Exploration

SECTION 14.1

Iterated Integrals and Area in the Plane

987

Sometimes it is not possible to calculate the area of a region with a single iterated integral. In these cases you can divide the region into subregions such that the area of each subregion can be calculated by an iterated integral. The total area is then the sum of the iterated integrals. TECHNOLOGY Some computer software can perform symbolic integration for integrals such as those in Example 6. If you have access to such software, use it to evaluate the integrals in the exercises and examples given in this section.

An Area Represented by Two Iterated Integrals

EXAMPLE 6

Find the area of the region R that lies below the parabola y  4x  x 2

Parabola forms upper boundary.

above the x-axis, and above the line y  3x  6.

Line and x-axis form lower boundary.

Solution Begin by dividing R into the two subregions R1 and R2 shown in Figure 14.7. y

y = −3x + 6

4

3

y = 4x − x 2 (1, 3) R1

2

R2

Δx

1

1 2

Area = 1

.

4x − x

−3x + 6

x

Δx

2 2

4

4

4x − x

2

0

dy dx +

2

dy dx

Figure 14.7

Editable Graph In both regions, it is convenient to use vertical rectangles, and you have

  2

Area  

4xx 2

1 3x6 2

 4

dy dx 

2

4xx 2

dy dx

0



4

4x  x 2  3x  6 dx 

1

4x  x 2 dx

2

2

3 4

 2  x3  6x  2x  x3  7 1 8 8 64 15 8  .  14   12    6  32  3 2 3 3 3 2 

.

7x 2

3

2

1

2

The area of the region is 152 square units. Try checking this using the procedure for finding the area between two curves, as presented in Section 7.1.

Try It NOTE In Examples 3 to 6, be sure you see the benefit of sketching the region of integration. You should develop the habit of making sketches to help determine the limits of integration for all iterated integrals in this chapter.

Exploration A

At this point you may be wondering why you would need iterated integrals. After all, you already know how to use conventional integration to find the area of a region in the plane. (For instance, compare the solution of Example 4 in this section with that given in Example 3 in Section 7.1.) The need for iterated integrals will become clear in the next section. In this section, primary attention is given to procedures for finding the limits of integration of the region of an iterated integral, and the following exercise set is designed to develop skill in this important procedure.

988

CHAPTER 14

Multiple Integration

Exercises for Section 14.1 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–10, evaluate the integral.

    

x

1.

2x  y dy

0 2y

3.

1

2.

x

y > 0

4.

y dy x

6.

x  3y  dy

y > 0

8.

1y2

yeyx dy

10.

sin3 x cos y dx

2

11.

                

1

1 x 3

1

x 2  y 2 dy dx 33.

1  cos x dy dx

y

y = 4 − x2

17.

x2



2y 2

 1 dx dy

35. x  y  2, 36. y  x32, 38. xy  9,

23.

sin

0

0

41.

r dr d

0

2

sin dr d

43.

2 10

0

In Exercises 25–28, evaluate the improper iterated integral.

 

1x

25.

1

1

1

  3

y dy dx

26.

0



27.

    4

3r 2

0

1 dx dy xy

y0

x  y  5,

y  x,

y  0,

y0 x9

y  2x, x  2

In Exercises 41– 48, sketch the region R of integration and switch the order of integration.

r dr d

cos

x  0,

x2 y 2  21 a2 b

40. y  x,

2 dx dy 4  y 2

0

4

24.

39.

3y dx dy

0

2

2

y  2x

37. 2x  3y  0,

x  y dx dy

2 cos

0

1

In Exercises 35–40, use an iterated integral to find the area of the region bounded by the graphs of the equations.

10  2x 2  2y 2 dx dy

0

2

x

x 1

64  x 3 dy dx



0

22.

1

−1

0 3y 26y 2 4y2

21.

x2 + y2 = 4

2

0 0 2 2yy 2

20.

5

1

0 y 1 1y2

19.

4

y

34.

y=x+2

3

3

2yex dy dx

1 0 2 2y

18.

2

1

1  x 2 dy dx

4 0 2 4

x

4

2

0 0 4 x2

16.

2≤x≤5

2

2

1 1 1 x

15.

1 x−1

3

0 0 4 x

14.

y= 4

2

1 2  sin x

13.

2

5

3

x  y dy dx

3

1 y

32.

0 0 1 2

12.

x

8

y = 4 − x2

In Exercises 11–24, evaluate the iterated integral. 1

6

4

y

31.

y

0

(3, 1)

x

2

2

x3

1

x  y  dx 2

(1, 1)

2

2

1y2

(3, 3)

(8, 3)

4

2

(1, 3) 3

6 2

x3

y ln x dx, 7. x y e

y

30.

8

y dx

x

x y dy

y

29.

0

2

0 y

9.

In Exercises 29–34, use an iterated integral to find the area of the region.

cos y

y dx, x

4x2

5.

    

x2



0

x2 dy dx 1  y2



28.

0

0

xyex

2 y 2

dx dy

45.

1 1

47.

y

     4

f x, y dx dy

42.

0

0

4x2

2

f x, y dy dx

44.

y

f x, y dx dy

4x 2

f x, y dy dx

0 0 2 ex

0 ln y

f x, y dx dy

46.

1 0 2

0 1

1 x 2

2

f x, y dy dx

48.

2

f x, y dy dx

cos x

0

f x, y dy dx

SECTION 14.1

In Exercises 49–58, sketch the region R whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area.

      1

49.

2

dy dx

50.

0 0 1 1y2

51.

0

53.

dx dy

x

dy dx 

0

0

66.

4x2

67. 68.

dy dx

  9

dy dx

56.

0

3  y

2

58.

dx dy

dy dx

x

4y 2

dx dy

  5

0

50x2

0

  52

y

x 2y 2 dx dy 

5

0

y

( 0, 5

y=

2)

  2

69. 70.

0

42y

x 2 y  xy 2 dx dy

4x2

     2

50y2

71.

x 2 y 2 dx dy

xy dy dx x2  y 2  1

4x 2 xy

e dy dx

0 0 2 2

0

50 − x 2

72.

(5, 5)

73.

5

x 2  y 2 dy dx

0

In Exercises 71–74, use a computer algebra system to approximate the iterated integral.

x 2y 2 dy dx 

x

5

2 dx dy   1 y  1

0 y3 2 4x 24

Think About It In Exercises 59 and 60, give a geometric argument for the given equality. Verify the equality analytically. 59.

sinx  y dx dy

In Exercises 69 and 70, (a) sketch the region of integration, (b) switch the order of integration, and (c) use a computer algebra system to show that both orders yield the same value.

3

2 0

y2

x3  3y 2 dy dx

0 0 x a ax 0

0

2x

0 y 4 y

dy dx

dy dx

dy dx 

4

2

dx dy

6x

x2

1

57.

0 6

   

0 x2 1 2y

4x

2

0 0 2 1

55.

2

  4

0 0 4 x2

54.

52.

989

In Exercises 65–68, use a computer algebra system to evaluate the iterated integral. 65.

4

1 2 2 4x2

1y2

2

  2

Iterated Integrals and Area in the Plane

16  x3  y3 dy dx

0 x 2 1cos

0 0 2 1sin

y=x

74.

0

x

6r 2 cos dr d

15 r dr d

0

5

 2

60.

0

2x

 4

x sin y dy dx 

x2

0

Writing About Concepts

y

75. Explain what is meant by an iterated integral. How is it evaluated?

x sin y dx dy

y2

y

76. Describe regions that are vertically simple and regions that are horizontally simple.

(2, 4)

4

y = 2x

77. Give a geometric description of the region of integration if the inside and outside limits of integration are constants.

3 2

y = x2

78. Explain why it is sometimes an advantage to change the order of integration.

1 x 1

2

In Exercises 61–64, evaluate the iterated integral. (Note that it is necessary to switch the order of integration.)

  2

61.

2

62.

0 x 1 1

63.

0

y

  2

x1  y3 dy dx

2

ey dy dx 2

64.

0

y2

  b

79.

0 x 2 4

sin x 2 dx dy

True or False? In Exercises 79 and 80, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. d

a c 1 x

x sin x dx dy

80.

0

0

  d

f x, y dy dx 

b

f x, y dx dy

c a 1 y

f x, y dy dx 

0

0

f x, y dx dy

990

CHAPTER 14

Multiple Integration

Section 14.2

Double Integrals and Volume • Use a double integral to represent the volume of a solid region. • Use properties of double integrals. • Evaluate a double integral as an iterated integral.

Double Integrals and Volume of a Solid Region Surface: z = f (x, y)

You already know that a definite integral over an interval uses a limit process to assign measure to quantities such as area, volume, arc length, and mass. In this section, you will use a similar process to define the double integral of a function of two variables over a region in the plane. Consider a continuous function f such that f x, y ≥ 0 for all x, y in a region R in the xy-plane. The goal is to find the volume of the solid region lying between the surface given by

z

z  f x, y

y

R

x

. Figure 14.8

Surface lying above the xy-plane

and the xy-plane, as shown in Figure 14.8. You can begin by superimposing a rectangular grid over the region, as shown in Figure 14.9. The rectangles lying entirely within R form an inner partition , whose norm  is defined as the length of the longest diagonal of the n rectangles. Next, choose a point xi, yi in each rectangle and form the rectangular prism whose height is f xi, yi , as shown in Figure 14.10. Because the area of the ith rectangle is  Ai

Rotatable Graph

Area of ith rectangle

it follows that the volume of the ith prism is f xi , yi  Ai

Volume of ith prism

and you can approximate the volume of the solid region by the Riemann sum of the volumes of all n prisms, n

 f x , y  A i

i

i

Riemann sum

i1

as shown in Figure 14.11. This approximation can be improved by tightening the mesh of the grid to form smaller and smaller rectangles, as shown in Example 1. Surface: z = f (x, y)

z

z

z

f(xi , yi)

y

y

y

x

x

The rectangles lying within R form an inner . partition of R.

Rectangular prism whose base has an area of Ai and whose height is f xi, yi 

Volume approximated by rectangular prisms

Figure 14.9

Figure 14.10

Figure 14.11

R

x

Rotatable Graph

Rotatable Graph

Rotatable Graph

SECTION 14.2

EXAMPLE 1

Double Integrals and Volume

991

Approximating the Volume of a Solid

Approximate the volume of the solid lying between the paraboloid 1 1 f x, y  1  x2  y 2 2 2 and the square region R given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Use a partition made up of squares whose sides have a length of 14. Solution Begin by forming the specified partition of R. For this partition, it is convenient to choose the centers of the subregions as the points at which to evaluate f x, y. z

18, 18  38, 18  58, 18  78, 18 

1

18, 38  38, 38  58, 38  78, 38 

18, 58  38, 58  58, 58  78, 58 

18, 78  38, 78  58, 78  78, 78 

1 Because the area of each square is Ai  16 , you can approximate the volume by the sum 16

1 x

y

1

i1

 1  2 x 16

1

i

i1

2

 

1 1  yi 2 2 16

 0.672. This approximation is shown graphically in Figure 14.12. The exact volume of the solid is 23 (see Example 2). You can obtain a better approximation by using a finer 1 partition. For example, with a partition of squares with sides of length 10 , the approximation is 0.668.

Surface: f (x, y) = 1 − 1 x 2 − 1 y 2 2 2

.



f xi yi  Ai 

Figure 14.12

Try It

Rotatable Graph

Exploration A

Some three-dimensional graphing utilities are capable of graphing figures such as that shown in Figure 14.12. For instance, the graph shown in Figure 14.13 was drawn with a computer program. In this graph, note that each of the rectangular prisms lies within the solid region.

TECHNOLOGY

z

In Example 1, note that by using finer partitions, you obtain better approximations of the volume. This observation suggests that you could obtain the exact volume by taking a limit. That is, y x

. Figure 14.13

Rotatable Graph

Volume  lim

n

 f x , y  A . i

→0 i1

i

i

The precise meaning of this limit is that the limit is equal to L if for every > 0 there exists a  > 0 such that



L

n

 f x , y  A i

i1

i



i

<

for all partitions  of the plane region R (that satisfy  < ) and for all possible choices of xi and yi in the ith region. Using the limit of a Riemann sum to define volume is a special case of using the limit to define a double integral. The general case, however, does not require that the function be positive or continuous.

992

CHAPTER 14

Multiple Integration

E X P L O R AT I O N

Definition of Double Integral

The entries in the table represent the depth (in 10-yard units) of earth at the center of each square in the figure below. y



2

3

1

10

9

7

2

7

7

4

3

5

5

4

4

4

5

3

n

 f x , y  A

f x, y d A  lim

→0 i1

R

1

x

If f is defined on a closed, bounded region R in the xy-plane, then the double integral of f over R is given by i

i

i

provided the limit exists. If the limit exists, then f is integrable over R. NOTE Having defined a double integral, you will see that a definite integral is occasionally referred to as a single integral.

Sufficient conditions for the double integral of f on the region R to exist are that R can be written as a union of a finite number of nonoverlapping subregions (see Figure 14.14) that are vertically or horizontally simple and that f is continuous on the region R. A double integral can be used to find the volume of a solid region that lies between the xy-plane and the surface given by z  f x, y.

z 20

30

y

.40

Volume of a Solid Region If f is integrable over a plane region R and f x, y ≥ 0 for all x, y in R, then the volume of the solid region that lies above R and below the graph of f is defined as

x

Rotatable Graph Approximate the number of cubic yards of earth in the first octant. (This exploration was submitted by Robert Vojack, Ridgewood High School, Ridgewood, NJ.)

V



f x, y dA.

R

Properties of Double Integrals Double integrals share many properties of single integrals.

THEOREM 14.1

Properties of Double Integrals

Let f and g be continuous over a closed, bounded plane region R, and let c be a constant.

y

R = R1 ∪ R2

1.

    



cf x, y dA  c

R

2. R1

f x, y ± gx, y dA 

R

R2

3. 4.

f x, y dA ≥ 0, f x, y dA ≥

R

Two regions are nonoverlapping if their intersection is a set that has an area of 0. In this figure, the area of the line segment that is common to R1 and R2 is 0. Figure 14.14

5.

R



f x, y d A ±

R

R

x

f x, y dA

R

 

gx, y dA,

R1

gx, y dA

R

if f x, y ≥ 0

R

f x, y dA 



f x, y dA 

if f x, y ≥ gx, y

 R2

two nonoverlapping subregions R1 and R2.

f x, y dA, where R is the union of

SECTION 14.2

Double Integrals and Volume

993

Evaluation of Double Integrals z

(0, 0, 2) 2

Plane: z = 2 − x − 2y

Height: z=2−x

1

(0, 1, 0) (2, 0, 0)

1

Triangular 2 cross section

2

x

Base: y = 2 − x 2

.

Volume:



y

Normally, the first step in evaluating a double integral is to rewrite it as an iterated integral. To show how this is done, a geometric model of a double integral is used as the volume of a solid. Consider the solid region bounded by the plane z  f x, y  2  x  2y and the three coordinate planes, as shown in Figure 14.15. Each vertical cross section taken parallel to the yz-plane is a triangular region whose base has a length of y  2  x2 and whose height is z  2  x. This implies that for a fixed value of x, the area of the triangular cross section is 1 1 2x 2  x2 2  x  . Ax  baseheight  2 2 2 4





By the formula for the volume of a solid with known cross sections (Section 7.2), the volume of the solid is

2

Ax dx

0

 

b

Figure 14.15

Volume 

Ax dx

a 2

2  x2 dx 4 0 2  x3 2 2   . 12 3 0

Rotatable Graph





This procedure works no matter how Ax is obtained. In particular, you can find Ax by integration, as shown in Figure 14.16. That is, you consider x to be constant, and integrate z  2  x  2y from 0 to 2  x2 to obtain z = 2 − x − 2y

Ax 



2x2

2  x  2y dy

0



y=

y=0

2−x 2

2x2



 2  xy  y2

0

2  x .  4 2

Triangular cross section Figure 14.16

Combining these results, you have the iterated integral Volume 



 2

f x, y dA 

0

R

2x2

2  x  2y dy dx.

0

To understand this procedure better, it helps to imagine the integration as two sweeping motions. For the inner integration, a vertical line sweeps out the area of a cross section. For the outer integration, the triangular cross section sweeps out the volume, as shown in Figure 14.17. z

y x

y x

. Integrate with respect to y to obtain the area of the cross section. Figure 14.17

Animation

z

z

x

z

y

x

Integrate with respect to x to obtain the volume of the solid.

y

994

CHAPTER 14

Multiple Integration

The following theorem was proved by the Italian mathematician Guido Fubini (1879–1943). The theorem states that if R is a vertically or horizontally simple region and f is continuous on R, the double integral of f on R is equal to an iterated integral.

THEOREM 14.2

Fubini’s Theorem

Let f be continuous on a plane region R. 1. If R is defined by a ≤ x ≤ b and g1x ≤ y ≤ g2x, where g1 and g2 are continuous on a, b , then





g x

b

f x, y dA 

a

R

2

f x, y dy dx.

g1x

2. If R is defined by c ≤ y ≤ d and h 1 y ≤ x ≤ h 2 y, where h 1 and h 2 are continuous on c, d , then



c

R

EXAMPLE 2

 d

f x, y dA 

h2 y

f x, y dx dy.

h1 y

Evaluating a Double Integral as an Iterated Integral

Evaluate

  R



1 1 1  x2  y 2 dA 2 2

where R is the region given by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. y

Solution Because the region R is a square, it is both vertically and horizontally simple, and you can use either order of integration. Choose dy dx by placing a vertical representative rectangle in the region, as shown in Figure 14.18. This produces the following.

R: 0 ≤ x ≤ 1 0≤y≤1

 

1

R



1 1 1  x2  y 2 dA  2 2 

Δx

     1

0 0 1 0

x

1

1



0

1 1

f (x, y) dA = R

1



f (x, y) dy dx 0 0

2 The . volume of the solid region is 3.



1 1 1  x2  y 2 dy dx 2 2





1 y3 1  x2 y  2 6

1

dx

0



5 1 2  x dx 6 2

 56 x  x6 

3 1 0

2  3

Figure 14.18

Try It

Exploration A

The double integral evaluated in Example 2 represents the volume of the solid region approximated in Example 1. Note that the approximation obtained in Example 1 is quite good 0.672 vs. 23 , even though you used a partition consisting of only 16 squares. The error resulted because the centers of the square subregions were used as the points in the approximation. This is comparable to the Midpoint Rule approximation of a single integral.

SECTION 14.2

Double Integrals and Volume

995

The difficulty of evaluating a single integral a f x dx usually depends on the function f, and not on the interval a, b . This is a major difference between single and double integrals. In the next example, you will integrate a function similar to that in Examples 1 and 2. Notice that a change in the region R produces a much more difficult integration problem. b

E X P L O R AT I O N Volume of a Paraboloid Sector The solid in Example 3 has an elliptical (not a circular) base. Consider the region bounded by the circular paraboloid z  a2  x2  y 2,

Finding Volume by a Double Integral

EXAMPLE 3

a > 0

and the xy-plane. How many ways do you now know for finding the volume of this solid? For instance, you could use the disk method to find the volume as a solid of revolution. Does each method involve integration? z

Find the volume of the solid region bounded by the paraboloid z  4  x2  2y 2 and the xy-plane. Solution By letting z  0, you can see that the base of the region in the xy-plane is the ellipse x2  2y 2  4, as shown in Figure 14.19(a). This plane region is both vertically and horizontally simple, so the order dy dx is appropriate.

4 2 x  ≤ y ≤ 4 2 x  2

Variable bounds for y:



Constant bounds for x:

2 ≤ x ≤ 2

a2

2

The volume is given by

  2

V

−a

.

a

a

y

4x 22

2 4x 22

4  x 2  2y 2 dy dx

2

x



Rotatable Graph

2

4

4  x 2 y 

 

2y 3 3



4x 22

4x 22

See Figure 14.19(b).

dx

2

4  x232 dx 32 2 2 4  16 cos 4 d

32 2 2 64  2 cos4 d

32 0 128 3  32 16  42. 

x  2 sin



 

z

NOTE In Example 3, note the usefulness of Wallis’s Formula to evaluate 2 0 cosn d . You may want to review this formula in Section 8.3.

Surface: f(x, y) = 4 − x 2 − 2y 2

Wallis’s Formula

Base: −2 ≤ x ≤ 2 −

(4 − x 2)/2 ≤ y ≤

(4 − x 2)/2

y

4 2 1

x −1

Δx

1 −1 −2

2

3

.

y

x

(a)

Volume: 2

(4 − x 2)/2

−2 −

(4 − x 2)/2

(4 − x 2 − 2y 2)dy dx

(b)

Rotatable Graph . Figure 14.19 .

Try It Exploration A

Exploration B

Exploration C

Exploration D

996

CHAPTER 14

Multiple Integration

In Examples 2 and 3, the problems could be solved with either order of integration because the regions were both vertically and horizontally simple. Moreover, had you used the order dx dy,you would have obtained integrals of comparable difficulty. There are, however, some occasions in which one order of integration is much more convenient than the other. Example 4 shows such a case.

Comparing Different Orders of Integration

EXAMPLE 4 z

Surface: f )x, y) e

x

Find the volume of the solid region R bounded by the surface

2

f x, y  ex

1

Solution The base of R in the xy-plane is bounded by the lines y  0, x  1, and y  x. The two possible orders of integration are shown in Figure 14.21.

z=0 1

x=1

Surface

and the planes z  0, y  0, y  x, and x  1, as shown in Figure 14.20.

y=0

x

2

1

y

y

y=x

y

R: 0 ≤ x ≤ 1 0≤y≤x

Base is bounded by y  0, y  x, and x . 1.

R: 0 ≤ y ≤ 1 y≤x≤1 (1, 1)

1

Figure 14.20

(1, 1)

1

Rotatable Graph

Δy (1, 0) Δx

1 x

(1, 0)

x

1

x

1

1 1

e−x dy dx 2

0 0

0

e−x dx dy 2

y

Figure 14.21

By setting up the corresponding iterated integrals, you can see that the order dx dy 2 requires the antiderivative ex dx, which is not an elementary function. On the other hand, the order dy dx produces the integral

 1

0

x

 

1

ex dy dx  2

0

x



x 2

e

y dx 0

0 1



xex dx 2

0

1 1 2   ex 2 0 1 1 1  2 e







e1 2e  0.316. 

.

Try It NOTE

Exploration A

Open Exploration

Try using a symbolic integration utility to evaluate the integral in Example 4.

SECTION 14.2

Paraboloid: z = 1 − x2 − y2

z

997

Volume of a Region Bounded by Two Surfaces

EXAMPLE 5

Plane: z=1−y

Double Integrals and Volume

Find the volume of the solid region R bounded above by the paraboloid z  1  x2  y2 and below by the plane z  1  y, as shown in Figure 14.22.

1

Solution Equating z-values, you can determine that the intersection of the two surfaces occurs on the right circular cylinder given by 1

1

y

R: 0 ≤ y ≤ 1 −

y − y2 ≤ x ≤

x2  y  y2.

Because the volume of R is the difference between the volume under the paraboloid and the volume under the plane, you have

x

.

1  y  1  x2  y2

y − y2

   

yy 2

1

Volume 

yy 2 1 yy 2

0

Rotatable Graph 

yy 2

0

y

0

 y  y 2  x 2 dx dy

1



 y  y 2x 

0

 1 2



x − 12

1 2

 

Figure 14.22

.



Try It

4 3

yy 2



dy

yy 2

 y  y232 dy

0

     1 6

x3 3

1

4 3

1 6

 1

1  x2  y2 dx dy 

1 8

1

1  2y  12 32 dy

0

2

cos 4

d

2 2 2

2y  1  sin

cos 4 d

0

16316  32 . Exploration A

Wallis’s Formula

yy 2

yy 2

1  y dx dy

SECTION 14.2

Double Integrals and Volume

997

Exercises for Section 14.2 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

Approximation In Exercises 1– 4, approximate the integral R f x, y d A by dividing the rectangle R with vertices 0, 0, 4, 0, 4, 2, and 0, 2 into eight equal squares and finding the sum 8

 f x , y   A i

i

i

i1

where xi , yi  is the center of the ith square. Evaluate the iterated integral and compare it with the approximation.

    4

1.

0

2. 3.

1 2

4

0

x

f x, y dy dx

0

1

2

3

4

0

32

31

28

23

16

1

31

30

27

22

15

2

28

27

24

19

12

x 2  y 2 dy dx

3

23

22

19

14

7

1 dy dx x  1 y  1

4

16

15

12

7

0

x  y dy dx

0 4 2

0

4

0

x 2 y dy dx

0 0 4 2

0



2

0 0 4 2

4.

5. Approximation The table shows values of a function f over a square region R. Divide the region into 16 equal squares and select xi, yi  to be the point in the ith square closest to the origin. Compare this approximation with that obtained by using the point in the ith square farthest from the origin.

y

998

CHAPTER 14

Multiple Integration

6. Approximation The figure shows the level curves for a function f over a square region R. Approximate the integral using four squares, selecting the midpoint of each square as xi, yi .

 2

0

y

2

f x, y dy dx

In Exercises 21–30, use a double integral to find the volume of the indicated solid. 21.

z

2

4

6

2 4 x

0≤x≤4 0≤y≤2

x

2

In Exercises 7–12, sketch the region R and evaluate the iterated integral R f x, y dA.

1  2x  2y dy dx

8.

0 0 6 3

9. 11.

a a2 x2 1 0

12.

0

10.

0

1 2y

dy 

y1

0

y=x

1

1y

e xy

dx dy

1

0



xy d A

y=x

x

x

y=2

Rotatable Graph 25.

x=2

Rotatable Graph z

26.

2x + 3y + 4z = 12

R

y

2

2

y

2

2

x+y+z=2

z

R: rectangle with vertices 0, 0, 0, 5, 3, 5, 3, 0

14.

4

2

 1

z=4 z=4−x−y

3

x  y dy dx

e xy dx

z

24.

4

x2 y2 dx dy

In Exercises 13–20, set up an integral for both orders of integration, and use the more convenient order to evaluate the integral over the region R. 13.

z

23.

0≤x≤4 0≤y≤2

Rotatable Graph

2

sin2 x cos2 y dy dx

y

2

4 x

Rotatable Graph

0 0 4 y

x  y dx dy

0 y2 a2 x2 a

  

1

y

2

3

8 1

2

z = 6 − 2y

6

1 2

10

   

z

22.

y 2

1

0

1

7.

z=

2



3

sin x sin y d A

R

R: rectangle with vertices , 0, , 0, , 2, , 2

15.

 R

x2

y dA  y2

y

2

2

6 x

x

R: triangle bounded by y  x, y  2x, x  2

16.

y

4



Rotatable Graph

xe y dA

Rotatable Graph

R

R: triangle bounded by y  4  x, y  0, x  0

17.



z

27.

2y ln x dA

R

 R

3

y dA 1  x2

2

R: region bounded by y  0, y  x, x  4

19.



x dA

y=x

x

20.



1

y

1

1

y=1

2

R

R: sector of a circle in the first quadrant bounded by y  25  x2, 3x  4y  0, y  0

z = 4 − y2

4

1

R: region bounded by y  4  x 2, y  4  x

18.

z

28. z = 1 − xy

Rotatable Graph

1

2

y=x

x

y

y=2

Rotatable Graph

x2  y2 d A

R

R: semicircle bounded by y  4  x2, y  0

29. Improper integral

1

z

1

z= z

30. Improper integral

(x +

1)2(y

+

1)2

z = e − (x + y)/2

1

0≤x 0. In Exercises 49–52, evaluate the iterated integral. (Note that it is necessary to switch the order of integration.)

  1

49.

12 2

50.

0 y2 1 arccos y

51.

0

0

   ln 10

ex dx dy

sin x1  sin2 x dx dy 52.

10

ex

0 2

2

0

1 2 2x

1 dy dx ln y

y cos y dy dx

61. Average Production The Cobb-Douglas production function for an automobile manufacturer is f x, y  100x 0.6 y 0.4 where x is the number of units of labor and y is the number of units of capital. Estimate the average production level if the number of units of labor x varies between 200 and 250 and the number of units of capital y varies between 300 and 325. 62. Average Profit A firm’s profit P in marketing two soft drinks is P  192x  576y  x2  5y2  2xy  5000, where x and y represent the numbers of units of the two soft drinks. Use a computer algebra system to evaluate the double integral yielding the average weekly profit if x varies between 40 and 50 units and y varies between 45 and 60 units.

1000

CHAPTER 14

Multiple Integration

Probability A joint density function of the continuous random variables x and y is a function f x, y satisfying the following properties.

(c) P[x, y  R] 



 

(a) f x, y ≥ 0 for all x, y

(b)



 

f x, y d A  1

(a) 200

In Exercises 63–66, show that the function is a joint density function and find the required probability. 63. f x, y 

 0, 0,

65. f x, y 

0,

0 ≤ x ≤ 2, 0 ≤ y ≤ 2 elsewhere P0 ≤ x ≤ 1, 1 ≤ y ≤ 2  x  y,

66. f x, y 

0,

(b) 500

 0

1  x2  y2 dx dy.

0

f x, y d A ≤

R

68. Programming Consider a continuous function f x, y over the rectangular region R with vertices a, c, b, c, a, d, and b, d, where a < b and c < d. Partition the intervals a, b and

c, d into m and n subintervals, so that the subintervals in a given direction are of equal length. Write a program for a graphing utility to compute the sum

  f x , y   A  i

j

i

 a

f x, y d A

77. Let f x  1 et dt. Find the average value of f on the interval

0, 1 .

0

2

70.

0

0

m  4, n  8

 6

71.

4



m  4, n  8

xy

1

dy.



9  x2  y2 dA.

R

80. Determine the region R in the xy-plane that minimizes the value of



x2  y2  4 dA.

82. Use a geometric argument to show that



72.

1

0

9y2

20e

x3



y3

1

m  6, n  4

dx dy

9 . 2

Putnam Exam Challenge

dy dx

2

9  x2  y2 dx dy 

0

x38

0

 4

y cos x dx dy

2

79. Determine the region R in the xy-plane that maximizes the value of

4

m  10, n  20

2

0

Hint: Evaluate  e

 e2x dx. x

2

 2

sin x  y dy dx

ex

0

3





78. Find

2

81. Find 0 arctanx  arctan x dx. (Hint: Convert the integral to a double integral.)

c

Approximation In Exercises 69–72, (a) use a computer algebra system to approximate the iterated integral, and (b) use the program in Exercise 68 to approximate the iterated integral for the given values of m and n. 69.

gx, y d A.

R

d

where xi , yj is the center of a representative rectangle in R.

1

 R

x

i1 j1

(e) 5000

1

V8



67. Approximation The base of a pile of sand at a cement plant is rectangular with approximate dimensions of 20 meters by 30 meters. If the base is placed on the xy-plane with one vertex at the origin, the coordinates on the surface of the pile are 5, 5, 3, 15, 5, 6, 25, 5, 4, 5, 15, 2, 15, 15, 7, and 25, 15, 3. Approximate the volume of sand in the pile.

b

(c) 500 (d) 5

76. If f x, y ≤ gx, y for all x, y in R, and both f and g are continuous over R, then

x ≥ 0, y ≥ 0 elsewhere

m

(e) 1000

R: circle bounded by x2  y2  9

1

0 ≤ x ≤ 3, 3 ≤ y ≤ 6 elsewhere

P0 ≤ x ≤ 1, x ≤ y ≤ 1

n

(d) 125

75. The volume of the sphere x2  y2  z2  1 is given by the integral

P0 ≤ x ≤ 1, 4 ≤ y ≤ 6 exy,

(c) 50

True or False? In Exercises 75 and 76, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

1 4 xy,

1 27 9

(b) 600

74. f x, y  x2  y2 (a) 50

0 ≤ x ≤ 5, 0 ≤ y ≤ 2 elsewhere

P0 ≤ x ≤ 2, 1 ≤ y ≤ 2 64. f x, y 

73. f x, y  4x R: square with vertices 0, 0, 4, 0, 4, 4, 0, 4

f x, y dA

R

1 , 10

Approximation In Exercises 73 and 74, determine which value best approximates the volume of the solid between the xy-plane and the function over the region. (Make your selection on the basis of a sketch of the solid and not by performing any calculations.)

 a

83. Evaluate

0

b

e max b x , a 2 2

2 y2

dy dx, where a and b are positive.

0

1 84. Show that if # > 2 there does not exist a real-valued function u such that for all x in the closed interval 0 ≤ x ≤ 1

ux  1  #x u yu y  x dy. 1

These problems were composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.

SECTION 14.3

Section 14.3

1001

Change of Variables: Polar Coordinates

Change of Variables: Polar Coordinates • Write and evaluate double integrals in polar coordinates.

Double Integrals in Polar Coordinates Some double integrals are much easier to evaluate in polar form than in rectangular form. This is especially true for regions such as circles, cardioids, and rose curves, and for integrands that involve x 2  y 2. In Section 10.4, you learned that the polar coordinates r,  of a point are related to the rectangular coordinates x, y of the point as follows. x  r cos

r 2  x2  y 2

EXAMPLE 1

y  r sin

y and tan  x

and

Using Polar Coordinates to Describe a Region

Use polar coordinates to describe each region shown in Figure 14.23. y

y 4 2 2

x

1

−4

−2

2

4

−2 x

1

2

−4

(b)

(a)

Figure 14.23

Solution a. The region R is a quarter circle of radius 2. It can be described in polar coordinates as R  r, : 0 ≤ r ≤ 2,

π 2

θ

b. The region R consists of all points between the concentric circles of radii 1 and 3. It can be described in polar coordinates as

2

θ

. Δr

1

R (ri, θi) r2

Δθ

0 ≤ ≤ 2 .

r1

R  r, : 1 ≤ r ≤ 3, 0 ≤ ≤ 2 .

Try It

Exploration A

The regions in Example 1 are special cases of polar sectors 0

R  r, : r1 ≤ r ≤ r2, Polar sector Figure 14.24

as shown in Figure 14.24.

1 ≤ ≤ 2

Polar sector

1002

CHAPTER 14

Multiple Integration

π 2

Δθ i (ri, θi)

Ri

g2 Δri

β α

To define a double integral of a continuous function z  f x, y in polar coordinates, consider a region R bounded by the graphs of r  g1  and r  g2  and the lines   and  . Instead of partitioning R into small rectangles, use a partition of small polar sectors. On R, superimpose a polar grid made of rays and circular arcs, as shown in Figure 14.25. The polar sectors Ri lying entirely within R form an inner polar partition , whose norm  is the length of the longest diagonal of the n polar sectors. Consider a specific polar sector Ri, as shown in Figure 14.26. It can be shown (see Exercise 61) that the area of Ri is

g1

 Ai  ri ri  i

0

Polar grid is superimposed over region R. Figure 14.25

Area of R i

where ri  r2  r1 and  i  2  1. This implies that the volume of the solid of height f ri cos i, ri sin i  above Ri is approximately f ri cos i, ri sin i ri ri  i and you have



f x, y dA 

n

 f r cos , r sin r  r  . i

i

i

i

i

i

i

i1

R

The sum on the right can be interpreted as a Riemann sum for f r cos , r sin r. The region R corresponds to a horizontally simple region S in the r -plane, as shown in Figure 14.27. The polar sectors Ri correspond to rectangles Si, and the area Ai of Si is ri  i. So, the right-hand side of the equation corresponds to the double integral



f r cos , r sin r dA.

S

From this, you can apply Theorem 14.2 to write



f x, y dA 

R



 

f r cos , r sin r dA

S  a

g2 

g1  

f r cos , r sin r dr d .

This suggests the following theorem, the proof of which is discussed in Section 14.8. π 2

θ

r = g2(θ )

r = g1(θ )

θ2

β

θ1

Ri Si r1

r2 (ri, θi)

α

(ri, θi) 0

The polar sector Ri is the set of all points r,  such that r1 ≤ r ≤ r2 and 1 ≤ ≤ 2. Figure 14.26

Horizontally simple region S Figure 14.27

r

SECTION 14.3

THEOREM 14.3

Change of Variables: Polar Coordinates

1003

Change of Variables to Polar Form

Let R be a plane region consisting of all points x, y  r cos , r sin  satisfying the conditions 0 ≤ g1  ≤ r ≤ g2 ,  ≤ ≤ , where 0 ≤    ≤ 2. If g1 and g2 are continuous on ,  and f is continuous on R, then



 

f x, y dA 



R

g1  

f r cos , r sin r dr d .

NOTE If z  f x, y is nonnegative on R, then the integral in Theorem 14.3 can be interpreted as the volume of the solid region between the graph of f and the region R.

E X P L O R AT I O N Volume of a Paraboloid Sector In the Exploration feature on page 995, you were asked to summarize the different ways you know for finding the volume of the solid bounded by the paraboloid z  a 2  x 2  y 2,

g2 

The region R is restricted to two basic types, r -simple regions and -simple regions, as shown in Figure 14.28. π 2

g2

θ =β

a > 0 Δθ

and the xy-plane. You now know another way. Use it to find the volume of the solid.

π 2

Fixed bounds for θ : α≤θ ≤β Variable bounds for r: 0 ≤ g1(θ ) ≤ r ≤ g2(θ )

Variable bounds for θ : 0 ≤ h1(r) ≤ θ ≤ h 2(r) h2

Fixed bounds for r: r1 ≤ r ≤ r2

g1

h1

θ =α

Δr

0

r = r1

0

-Simple region

r-Simple region Figure 14.28

EXAMPLE 2

r = r2

Evaluating a Double Polar Integral

Let R be the annular region lying between the two circles x 2  y 2  1 and 2 x 2  y 2  5. Evaluate the integral R x  y dA. R: 1 ≤ r ≤ 5 0 ≤ θ ≤ 2π

π 2

Solution The polar boundaries are 1 ≤ r ≤ 5 and 0 ≤ ≤ 2, as shown in Figure 14.29. Furthermore, x 2  r cos 2 and y  r sin . So, you have



R

x 2  y dA 

R



0 2

3

 

r-Simple region Figure 14.29



     2

5

0 1 2 5 0 1 2 r4 0 2 0 2

4

r 2 cos2  r sin r dr d

r 3 cos2  r 2 sin  dr d

cos2 

6 cos2 

 3  .

5



1

d



55  1 sin d

3

3  3 cos 2 

0



r3 sin

3



55  1 sin d

3

3 sin 2 55  1  cos

2 3



 6.

Try It

Exploration A

2

Exploration B

0

1004

CHAPTER 14

Multiple Integration

In Example 2, be sure to notice the extra factor of r in the integrand. This comes from the formula for the area of a polar sector. In differential notation, you can write dA  r dr d

which indicates that the area of a polar sector increases as you move away from the origin. 16 − x 2 − y 2

Surface: z =

Change of Variables to Polar Coordinates

EXAMPLE 3

z

Use polar coordinates to find the volume of the solid region bounded above by the hemisphere

4

z  16  x 2  y 2

Hemisphere forms upper surface.

and below by the circular region R given by 4 x

4

.

R: x 2 + y 2 ≤ 4

Figure 14.30

Rotatable Graph

y

x2  y 2 ≤ 4

Circular region forms lower surface.

as shown in Figure 14.30. Solution In Figure 14.30, you can see that R has the bounds  4  y 2 ≤ x ≤ 4  y 2,

2 ≤ y ≤ 2

and that 0 ≤ z ≤ 16  x 2  y 2. In polar coordinates, the bounds are 0 ≤ r ≤ 2 and

0 ≤ ≤ 2

with height z  16  x 2  y 2  16  r 2. Consequently, the volume V is given by



V

f x, y dA 

R

   2

0

0



2

1 3

16  r 2 r dr d

2

2



16  r 232

0 2

d

0

1 243  64 d

3 0 2 8   33  8

3 0 



.



Try It

16 8  33   46.979. 3

Exploration A

Exploration B

NOTE To see the benefit of polar coordinates in Example 3, you should try to evaluate the corresponding rectangular iterated integral



4y2

2

2 4y2

16  x 2  y 2 dx dy.

Any computer algebra system that can handle double integrals in rectangular coordinates can also handle double integrals in polar coordinates. The reason this is true is that once you have formed the iterated integral, its value is not changed by using different variables. In other words, if you use a computer algebra system to evaluate

TECHNOLOGY

 2

0

2

16  x 2 x dx dy

0

you should obtain the same value as that obtained in Example 3.

SECTION 14.3

Change of Variables: Polar Coordinates

1005

Just as with rectangular coordinates, the double integral



dA

R

can be used to find the area of a region in the plane.

Finding Areas of Polar Regions

EXAMPLE 4

Use a double integral to find the area enclosed by the graph of r  3 cos 3 . π 2

r = 3 cos 3θ

π π R: − 6 ≤ θ ≤ 6 0 ≤ r ≤ 3 cos 3θ θ=π 6 0 3

θ = −π 6

Solution Let R be one petal of the curve shown in Figure 14.31. This region is r-simple, and the boundaries are as follows. 

  ≤ ≤ 6 6 0 ≤ r ≤ 3 cos 3

Variable bounds on r

So, the area of one petal is 1 A 3

       dA 

R



. Figure 14.31

Editable Graph

.

Fixed bounds on

6

3 cos 3

6 0 6 2

3 cos 3

r 6 2 6

r dr d



9 2



9 4



1 9

 sin 6

4 6



3 . 4

6

6

6

d

0

cos2 3 d

1  cos 6  d



6



6

So, the total area is A  94.

Try It

Exploration B

Exploration A

Open Exploration

As illustrated in Example 4, the area of a region in the plane can be represented by



g2 



A



g1  

r dr d .

If g1   0, you obtain

 

A



g2 

0

r dr d 







r2 2

g2 



0

d 







1 g  2 d

2 2

which agrees with Theorem 10.13. So far in this section, all of the examples of iterated integrals in polar form have been of the form

 



g2 

g1  

f r cos , r sin r dr d

in which the order of integration is with respect to r first. Sometimes you can obtain a simpler integration problem by switching the order of integration, as illustrated in the next example.

1006

CHAPTER 14

Multiple Integration

Changing the Order of Integration

EXAMPLE 5

π 2

θ=π 3

Find the area of the region bounded above by the spiral θ =π 6

r= π 3θ

r

 3

and below by the polar axis, between r  1 and r  2. Solution The region is shown in Figure 14.32. The polar boundaries for the region are

0 1

π R: 0 ≤ θ ≤ 3r 1≤r≤2

region

-Simple . Figure 14.32

Editable Graph

2

1 ≤ r ≤ 2 and

0 ≤ ≤

 . 3r

So, the area of the region can be evaluated as follows.

 2

A

1

3r

0

Try It



2

r d dr 

1

r

3r



dr 

0

Exploration A



2

1

 r dr  3 3

2



1



 3

1006

CHAPTER 14

Multiple Integration

Exercises for Section 14.3 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, the region R for the integral R f x, y dA is shown. State whether you would use rectangular or polar coordinates to evaluate the integral. y

1.

y

7.

4

3

−4

R

2

−2

x 3

In Exercises 9–14, evaluate the double integral R f r,  dA, and sketch the region R.

−4

4

9.

y

4.

R x 2

−2

1

4

13.

R

0

2

3

4

In Exercises 5–8, use polar coordinates to describe the region shown. y

15.

9  r 2 r dr d

12.

0 2

r dr d

14.

r 2 sin cos dr d

rer dr d

2

0 1cos

0

0

  

6

19.

2

4 x 4

8

x −4

−2

2 −2

a2 y2

sin r dr d

0

4

   a

16.

y dx dy

x 2  y 232 dy dx 18.

a2 x2

x dy dx

0 0 2 8y2

x 2  y2 dx dy

0 y 4 4yy2

0 0 2 2xx2 0

−4

0 0 2 3

0 0 3 9x2

17.

y

6.

12

−4

4

In Exercises 15–20, evaluate the iterated integral by converting to polar coordinates.

−1

a

−8

10.

x 1

−4

5.

3r 2 sin dr d

0 2 2 1sin

2

2 −4

11.

     4

6

0 0 2 3

3 4

     2

y

3.

−4

−2

−2 2

4

2

1 1

−2

4

x −6

−4 x

2

R

x

2

4

4

2

4

y

2.

y

8.

xy dy dx

20.

0

0

x 2 dx dy

0

In Exercises 21 and 22, combine the sum of the two iterated integrals into a single iterated integral by converting to polar coordinates. Evaluate the resulting iterated integral.

 2

21.

0

0

  22

x

x 2  y 2 dy dx 

2

8x2

0

x 2  y 2 dy dx

SECTION 14.3

  522

22.

x

0

0

 

25x2

5

xy dy dx 

xy dy dx

522 0

1007

Change of Variables: Polar Coordinates

In Exercises 37–42, use a double integral to find the area of the shaded region. 37.

In Exercises 23–26, use polar coordinates to set up and evaluate the double integral R f x, y dA.

38.

π 2

r=2

r = 6 cos θ

π 2

r=4

23. f x, y  x  y, R: x 2  y 2 ≤ 4, x ≥ 0, y ≥ 0 24. f x, y  ex

2 y 2 2

, R: x 2  y 2 ≤ 25, x ≥ 0

0 1 2 3 4 5

y 25. f x, y  arctan , R: x 2  y 2 ≥ 1, x 2  y 2 ≤ 4, 0 ≤ y ≤ x x

0

7

3

1

26. f x, y  9  x 2  y 2, R: x 2  y 2 ≤ 9, x ≥ 0, y ≥ 0 39. Volume In Exercises 27– 32, use a double integral in polar coordinates to find the volume of the solid bounded by the graphs of the equations.

40.

π 2

π 2

r = 1 + cos θ

27. z  xy, x 2  y 2  1, first octant

0

28. z  x 2  y 2  3, z  0, x 2  y 2  1

0

1 2 3 4

29. z  x 2  y 2, z  0, x 2  y 2  25

r = 2 + sin θ

30. z  lnx  y , z  0, x  y ≥ 1, x  y ≤ 4 2

2

2

2

2

2

31. Inside the hemisphere z  16  x 2  y 2 and inside the cylinder x 2  y 2  4x  0

41.

42.

π 2

32. Inside the hemisphere z  16  x 2  y 2 and outside the cylinder x 2  y 2  1

π 2 r = 3 cos 2θ

r = 2 sin 3θ

0

0

33. Volume Find a such that the volume inside the hemisphere z  16  x 2  y 2 and outside the cylinder x 2  y 2  a 2 is one-half the volume of the hemisphere.

1

3

2

34. Volume Use a double integral in polar coordinates to find the volume of a sphere of radius a. 35. Volume Determine the diameter of a hole that is drilled vertically through the center of the solid bounded by the graphs of the equations z  25ex

2 y 2 4

, z  0,

and

x 2  y 2  16

if one-tenth of the volume of the solid is removed. 36. Machine Design The surfaces of a double-lobed cam are 1 1 modeled by the inequalities 4 ≤ r ≤ 21  cos2  and 9 9 ≤ z ≤ 4x 2  y 2  9 4x 2  y 2  9 where all measurements are in inches. (a) Use a computer algebra system to graph the cam. (b) Use a computer algebra system to approximate the perimeter of the polar curve r

1 2 1



cos2

Writing About Concepts 43. Describe the partition of the region R of integration in the xy-plane when polar coordinates are used to evaluate a double integral. 44. Explain how to change from rectangular coordinates to polar coordinates in a double integral. 45. In your own words, describe r-simple regions and -simple regions. 46. Each figure shows a region of integration for the double integral R f x, y dA. For each region, state whether horizontal representative elements, vertical representative elements, or polar sectors would yield the easiest method for obtaining the limits of integration. Explain your reasoning. (a)

(b)

(c)

y

y

y

.

This is the distance a roller must travel as it runs against the cam through one revolution of the cam. (c) Use a computer algebra system to find the volume of steel in the cam.

R R

R x

x

x

1008

CHAPTER 14

Multiple Integration

47. Think About It Consider the program you wrote to approximate double integrals in rectangular coordinates in Exercise 68, in Section 14.2. If the program is used to approximate the double integral



in polar coordinates, how will you modify f when it is entered into the program? Because the limits of integration are constants, describe the plane region of integration. 48. Approximation Horizontal cross sections of a piece of ice that broke from a glacier are in the shape of a quarter of a circle with a radius of approximately 50 feet. The base is divided into 20 subregions, as shown in the figure. At the center of each subregion, the height of the ice is measured, yielding the following points in cylindrical coordinates.

5, 16 , 7, 15, 16 , 8, 25, 16 , 10, 35, 16 , 12, 45, 16 , 9, 5, 316, 9, 15, 316, 10, 25, 316, 14, 35, 316, 15, 45, 316, 10, 5, 516, 9, 15, 516, 11, 25, 516, 15, 35, 516, 18, 45, 516, 14, 5, 716, 5, 15, 716, 8, 25, 716, 11, 35, 716, 16, 45, 716, 12 (a) Approximate the volume of the solid. (b) Ice weighs approximately 57 pounds per cubic foot. Approximate the weight of the solid. (c) There are 7.48 gallons of water per cubic foot. Approximate the number of gallons of water in the solid.

10 20 30 40 50

4

0

0

dx is





I2 



ex

 





 

2

2





dx

ex



2

2 y 2

ey



2 dy

2

dA

(b) Use the result of part (a) to determine I. FOR FURTHER INFORMATION For more information on this prob2 lem, see the article “Integrating ex Without Polar Coordinates” by William Dunham in Mathematics Teacher.

MathArticle 56. Use the result of Exercise 55 and a change of variables to evaluate each integral. No integration is required.



2







ex dx

(b)



e4x dx 2

f x, y 

Find k such that the function

0,ke

x2 y2,

x ≥ 0, y ≥ 0 elsewhere

59. Think About It Consider the region bounded by the graphs of y  2, y  4, y  x, and y  3x and the double integral R f dA. Determine the limits of integration if the region R is divided into (a) horizontal representative elements, (b) vertical representative elements, and (c) polar sectors.

4

5rer dr d

0

Approximation In Exercises 51 and 52, determine which value best approximates the volume of the solid between the xy-plane and the function over the region. (Make your selection on the basis of a sketch of the solid and not by performing any calculations.)

60. Repeat Exercise 59 for a region R bounded by the graph of the equation x  22  y 2  4. 61. Show that the area A of the polar sector R (see figure) is A  rr , where r  r1  r22 is the average radius of R.

51. f x, y  15  2y; R: semicircle: x 2  y 2  16, y ≥ 0 (b) 200

(c) 300

(d) 200

R

(e) 800

Δr

52. f x, y  xy  2; R: quarter circle: x 2  y 2  9, x ≥ 0, y ≥ 0 (a) 25

22

is a probability density function. r1  r 3 sin  dr d

0

(a) 100

ex

(a) Use polar coordinates to evaluate the improper integral.

58. Probability

5

4

50.





57. Population The population density of a city is approximated 2 2 by the model f x, y  4000e0.01x y , x 2  y 2 ≤ 49, where x and y are measured in miles. Integrate the density function over the indicated circular region to approximate the population of the city.

Approximation In Exercises 49 and 50, use a computer algebra system to approximate the iterated integral.

   



required in the development of the normal probability density function.

(a)

π 8

2

55. Probability The value of the integral I 



3π 8 π 4

49.

53. If R f r,  dA > 0, then f r,  > 0 for all r,  in R. 54. If f r,  is a constant function and the area of the region S is twice that of the region R, then 2 R f r,  dA  S f r,  dA.

f r,  dA

R

π 2

True or False? In Exercises 53 and 54, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

(b) 8

(c) 100

(d) 50

(e) 30

Δθ r1

r2

SECTION 14.4

Section 14.4

Center of Mass and Moments of Inertia

1009

Center of Mass and Moments of Inertia • Find the mass of a planar lamina using a double integral. • Find the center of mass of a planar lamina using double integrals. • Find moments of inertia using double integrals.

Mass y

Section 7.6 discussed several applications of integration involving a lamina of constant density . For example, if the lamina corresponding to the region R, as shown in Figure 14.33, has a constant density , then the mass of the lamina is given by

g2

Mass  A  

R

 

 dA.

dA 

R

g1 x=a

x

x=b

Constant density

R

If not otherwise stated, a lamina is assumed to have a constant density. In this section, however, you will extend the definition of the term lamina to include thin plates of variable density. Double integrals can be used to find the mass of a lamina of variable density, where the density at x, y is given by the density function .

Lamina of constant density  Figure 14.33

Definition of Mass of a Planar Lamina of Variable Density If  is a continuous density function on the lamina corresponding to a plane region R, then the mass m of the lamina is given by m



 x, y dA.

Variable density

R

NOTE Density is normally expressed as mass per unit volume. For a planar lamina, however, density is mass per unit surface area.

Finding the Mass of a Planar Lamina

EXAMPLE 1

Find the mass of the triangular lamina with vertices 0, 0, 0, 3, and 2, 3, given that the density at x, y is x, y  2x  y. y

Solution As shown in Figure 14.34, region R has the boundaries x  0, y  3, and y  3x2 or x  2y3. Therefore, the mass of the lamina is

y=3 3

2

(0, 3)

(2, 3)

m

R



1

(0, 0) 1

 

3



Lamina of variable density .  x, y  2x  y



x2  xy

2y3

dy 0

3

10 9

y 2 dy

0

3

 

10 y3 9 3

0

 10.

Figure 14.34

Editable Graph

2x  y dx dy

0

x

2

2y3

0 0 3

R

x = 2y 3

   3

2x  y dA 

Try It

Exploration A

NOTE In Figure 14.34, note that the planar lamina is shaded so that the darkest shading corresponds to the densest part.

1010

CHAPTER 14

Multiple Integration

y

Finding Mass by Polar Coordinates

EXAMPLE 2 x2 + y2 = 4

Find the mass of the lamina corresponding to the first-quadrant portion of the circle

2

x2  y2  4

(x, y) 1

where the density at the point x, y is proportional to the distance between the point and the origin, as shown in Figure 14.35.

R

Solution At any point x, y, the density of the lamina is x

1

2

. Density at x, y:  x, y  k x2  y 2 Figure 14.35

 x, y  kx  02   y  02  kx2  y2. Because 0 ≤ x ≤ 2 and 0 ≤ y ≤ 4  x2, the mass is given by m

 

kx2  y 2 dA

R 2

Editable Graph 

0

4x2

kx2  y 2 dy dx.

0

To simplify the integration, you can convert to polar coordinates, using the bounds 0 ≤ ≤ 2 and 0 ≤ r ≤ 2. So, the mass is m



kx2  y 2 dA 

R

 

       2

.

Try It

kr 2 dr d

0 0 2 3 2

kr 3

8k 3

2

0

d

d

0

2



8k

3 4k  . 3



kr 2 r dr d

0 0 2 2

0



2

0

Exploration A

On many occasions, this text has mentioned the benefits of computer programs that perform symbolic integration. Even if you use such a program regularly, you should remember that its greatest benefit comes only in the hands of a knowledgeable user. For instance, notice how much simpler the integral in Example 2 becomes when it is converted to polar form.

TECHNOLOGY

Rectangular Form

 2

0

4x2

0

kx2  y 2 dy dx

Polar Form

  2

0

2

kr2 dr d

0

If you have access to software that performs symbolic integration, use it to evaluate both integrals. Some software programs cannot handle the first integral, but any program that can handle double integrals can evaluate the second integral.

SECTION 14.4

Center of Mass and Moments of Inertia

1011

Moments and Center of Mass y

For a lamina of variable density, moments of mass are defined in a manner similar to that used for the uniform density case. For a partition  of a lamina corresponding to a plane region R, consider the ith rectangle Ri of one area Ai , as shown in Figure 14.36. Assume that the mass of Ri is concentrated at one of its interior points xi , yi . The moment of mass of Ri with respect to the x-axis can be approximated by

xi

Ri

(xi, yi)

Mass yi    xi , yi  Ai  yi . yi

Similarly, the moment of mass with respect to the y-axis can be approximated by

Massxi    xi , yi  Ai xi .

x

Mx  mass yi  My  massxi  Figure 14.36

By forming the Riemann sum of all such products and taking the limits as the norm of  approaches 0, you obtain the following definitions of moments of mass with respect to the x- and y-axes.

Moments and Center of Mass of a Variable Density Planar Lamina Let  be a continuous density function on the planar lamina R. The moments of mass with respect to the x- and y-axes are Mx 



y x, y dA and

R

My 



x (x, y dA.

R

If m is the mass of the lamina, then the center of mass is

x, y 

Mm , Mm . y

x

If R represents a simple plane region rather than a lamina, the point x, y is called the centroid of the region. For some planar laminas with a constant density , you can determine the center of mass (or one of its coordinates) using symmetry rather than using integration. For instance, consider the laminas of constant density shown in Figure 14.37. Using symmetry, you can see that y  0 for the first lamina and x  0 for the second lamina. R: 0 ≤ x ≤ 1 − 1 − x2 ≤ y ≤

R: − 1 − y 2 ≤ x ≤ 0≤y≤1

1 − x2

z

z

1

1 −1

−1 −1

−1 1

1

.

1 − y2

x

−1

Rotatable Graph Lamina of constant density and symmetric with respect to the x-axis Figure 14.37

1

y x

1

y

−1

Rotatable Graph Lamina of constant density and symmetric with respect to the y-axis

1012

CHAPTER 14

Multiple Integration

Finding the Center of Mass

EXAMPLE 3

Find the center of mass of the lamina corresponding to the parabolic region

Variable density: y ρ (x, y) = ky y=4−

0 ≤ y ≤ 4  x2

Parabolic region

where the density at the point x, y is proportional to the distance between x, y and the x-axis, as shown in Figure 14.38.

x2 3

(x, y)

Solution Because the lamina is symmetric with respect to the y-axis and

2

 x, y  ky the center of mass lies on the y-axis. So, x  0. To find y, first find the mass of the lamina.

1

−1

1



4x2

2

x −2

Mass 

2

. parabolic region of variable density The

2 0

k 2 k  2

ky dy dx 

Figure 14.38

 

2

4x2



y2

2 2

dx 0

16  8x2  x 4 dx

2

 

k 8x3 x5  16x   2 3 5 64 32   k 32  3 5

Editable Graph





2

2



256k 15

Next, find the moment about the x-axis.

 2

Mx 

4x 2

2 0

 

k 3 k  3

 yky dy dx 

2

2 2

2



4x2

y3

0

dx

64  48x2  12x 4  x 6  dx



12x5 x7 k   64x  16x3  3 5 7 



2

2

4096k 105

So, Variable density: ρ (x, y) = ky

.

z

R: −2 ≤ x ≤ 2 0 ≤ y ≤ 4 − x2

y

and the center of mass is 0, 16 7 .

Center of mass: −2

Mx 4096k105 16   m 256k15 7

(0, 167 )

Try It 1 4

2 x

. Figure 14.39

Rotatable Graph

Exploration A

Open Exploration

y

Although you can think of the moments Mx and My as measuring the tendency to rotate about the x- or y-axis, the calculation of moments is usually an intermediate step toward a more tangible goal. The use of the moments Mx and My is typical—to find the center of mass. Determination of the center of mass is useful in a variety of applications that allow you to treat a lamina as if its mass were concentrated at just one point. Intuitively, you can think of the center of mass as the balancing point of the lamina. For instance, the lamina in Example 3 should balance on the point of a pencil placed at 0, 16 , as shown in Figure 14.39. 7

SECTION 14.4

Center of Mass and Moments of Inertia

1013

Moments of Inertia The moments of Mx and My used in determining the center of mass of a lamina are sometimes called the first moments about the x- and y-axes. In each case, the moment is the product of a mass times a distance. Mx 



 y x, y dA

My 

R



x x, y dA

R

Distance to x -axis

Mass

Distance to y-axis

Mass

You will now look at another type of moment—the second moment, or the moment of inertia of a lamina about a line. In the same way that mass is a measure of the tendency of matter to resist a change in straight-line motion, the moment of inertia about a line is a measure of the tendency of matter to resist a change in rotational motion. For example, if a particle of mass m is a distance d from a fixed line, its moment of inertia about the line is defined as I  md 2  massdistance2. As with moments of mass, you can generalize this concept to obtain the moments of inertia about the x- and y-axes of a lamina of variable density. These second moments are denoted by Ix and Iy , and in each case the moment is the product of a mass times the square of a distance. Ix 



 y 2 x, y dA

Iy 

R

I0 

 

r 2 x, y d A

R

Square of distance to y-axis

Mass

The sum of the moments Ix and Iy is called the polar moment of inertia and is denoted by I0.

Finding the Moment of Inertia

EXAMPLE 4

Find the moment of inertia about the x-axis of the lamina in Example 3. Solution From the definition of moment of inertia, you have

    2

Ix 

4x2

2 0 2

k 4 k  4 

4x2

y4

2 2 2

y2ky dy dx dx

0

256  256x2  96x4  16x6  x8 dx



256x3 96x5 16x7 x9 k 256x     4 3 5 7 9 32,768k  . 315 

.

Mass

x2  y 2 x, y d A

R



x2 x, y dA

R

Square of distance to x -axis

NOTE For a lamina in the xy-plane, I0 represents the moment of inertia of the lamina about the z-axis. The term “polar moment of inertia” stems from the fact that the square of the polar distance r is used in the calculation.



Try It

Exploration A

2



2

1014

CHAPTER 14

Multiple Integration

The moment of inertia I of a revolving lamina can be used to measure its kinetic energy. For example, suppose a planar lamina is revolving about a line with an angular speed of  radians per second, as shown in Figure 14.40. The kinetic energy E of the revolving lamina is E

1 2 I . 2

Kinetic energy for rotational motion

On the other hand, the kinetic energy E of a mass m moving in a straight line at a velocity v is E Planar lamina revolving at  radians per . second

1 mv 2. 2

Kinetic energy for linear motion

So, the kinetic energy of a mass moving in a straight line is proportional to its mass, but the kinetic energy of a mass revolving about an axis is proportional to its moment of inertia. The radius of gyration r of a revolving mass m with moment of inertia I is defined to be

Figure 14.40

Rotatable Graph

r

mI .

Radius of gyration

If the entire mass were located at a distance r from its axis of revolution, it would have the same moment of inertia and, consequently, the same kinetic energy. For instance, the radius of gyration of the lamina in Example 4 about the x-axis is given by y

128   2.469. mI  32,768k315 256k15 21 x

Finding the Radius of Gyration

EXAMPLE 5

Find the radius of gyration about the y-axis for the lamina corresponding to the region R: 0 ≤ y ≤ sin x, 0 ≤ x ≤ , where the density at x, y is given by  x, y  x. Solution The region R is shown in Figure 14.41. By integrating  x, y  x over the region R, you can determine that the mass of the region is . The moment of inertia about the y-axis is

y

2

1

Variable density: ρ (x, y) = x

R: 0 ≤ x ≤ π 0 ≤ y ≤ sin x

Iy 

(x, y)

. Figure 14.41

Editable Graph

π 2

π

x

 

    

sin x

x3 dy dx

0 0 sin x  x3 y 0 

dx

0

x3 sin x dx

0



  3  6 . So, the radius of gyration about the y-axis is

mI   6  

x

y

3

.





 3x2  6sin x  x3  6xcos x

 2  6  1.967.

Try It

Exploration A

0

SECTION 14.4

1015

Center of Mass and Moments of Inertia

Exercises for Section 14.4 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1– 4, find the mass of the lamina described by the inequalities, given that its density is  x, y  xy. (Hint: Some of the integrals are simpler in polar coordinates.) 2. x ≥ 0, 0 ≤ y ≤ 9  x 2

1. 0 ≤ x ≤ 4, 0 ≤ y ≤ 3 3. x ≥ 0, 0 ≤ y ≤ 4  x

In Exercises 5– 8, find the mass and center of mass of the lamina for each density. 5. R: rectangle with vertices 0, 0, a, 0, 0, b, a, b (b)   ky (c)   kx

6. R: rectangle with vertices 0, 0, a, 0, 0, b, a, b (a)   kxy (b)   k x2  y 2 7. R: triangle with vertices 0, 0, b2, h, b, 0 (a)   k

23. y  ex, y  0, x  0, x  2,   ky 24. y  ln x, y  0, x  1, x  e,   kx

2

4. x ≥ 0, 3 ≤ y ≤ 3  9  x 2

(a)   k

In Exercises 23 –26, use a computer algebra system to find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density.

25. r  2 cos 3 , 

26. r  1  cos ,   k In Exercises 27–32, verify the given moment(s) of inertia and find x and y. Assume that each lamina has a density of   1. (These regions are common shapes used in engineering.) 27. Rectangle

28. Right triangle

y

Ix =

(b)   ky (c)   kx

8. R: triangle with vertices 0, 0, 0, a, a, 0 (a)   k

(b)   x2  y2

9. Translations in the Plane Translate the lamina in Exercise 5 to the right five units and determine the resulting center of mass.

Iy =

y

1 bh3 3 1 3 bh 3

1 Ix = 12 bh3 1 3 Iy = 12 bh

h

h b

b

x

29. Circle y

y

a

x

I0 = 12 π a4

I0 = 14 π a4

31. Quarter circle

32. Ellipse y

11. y  a2  x2, y  0

I0 =

(b)   ka  yy

1 π a4 8

b

12. x2  y2  a2, 0 ≤ x, 0 ≤ y (a)   k

(b)   k

x2



a

 a

x

I0 =

y  0, x  2,   kx

14. y 

x3,

15. y 

1 , y  0, x  1, x  1,   k 1  x2

16. xy  4, x  1, x  4,   kx2 17. x  16  y2, x  0,   kx 18. y  9  x2, y  0,   ky2

34. y  a2  x2, y  0,   ky

20. y  cos

x L , y  0, x  0, x  ,   k L 2

36. y  x, y  x2,   kxy



0 ≤ y ≤ x,   k

22. y  a2  x2, y  0, y  x,   kx2  y2

+ b2)

33. y  0, y  b, x  0, x  a,   ky

x , y  0, x  0, x  L,   ky L

21. y 

1 π ab(a2 4

In Exercises 33 – 40, find Ix , Iy , I0 , x, and y for the lamina bounded by the graphs of the equations. Use a computer algebra system to evaluate the double integrals.

19. y  sin

x2,

x

y2

13. y  x, y  0, x  4,   kxy

a2

x

a

y

(a)   k

x

30. Semicircle

10. Conjecture Use the result of Exercise 9 to make a conjecture about the change in the center of mass when a lamina of constant density is translated h units horizontally or k units vertically. Is the conjecture true if the density is not constant? Explain. In Exercises 11–22, find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density or densities. (Hint: Some of the integrals are simpler in polar coordinates.)

  ≤ ≤ , k 6 6

35. y  4  x2, y  0, x > 0,   kx 37. y  x, y  0, x  4,   kxy 38. y  x2, y2  x,   x2  y2 39. y  x2, y2  x,   kx



40. y  x3, y  4x,   k y

1016

CHAPTER 14

Multiple Integration

In Exercises 41– 46, set up the double integral required to find the moment of inertia I, about the given line, of the lamina bounded by the graphs of the equations. Use a computer algebra system to evaluate the double integral. 41. x2  y2  b2,   k, line: x  a a > b 42. y  0, y  2, x  0, x  4,   k, line: x  6 43. y  x, y  0, x  4,   kx, line: x  6 44. y  a  x , y  0,   ky, line: y  a 2

2

45. y  a2  x2, y  0, x ≥ 0,   ka  y, line: y  a 46. y  4  x2, y  0,   k, line: y  2

54. Prove the following Theorem of Pappus: Let R be a region in a plane and let L be a line in the same plane such that L does not intersect the interior of R. If r is the distance between the centroid of R and the line, then the volume V of the solid of revolution formed by revolving R about the line is given by V  2 rA, where A is the area of R. Hydraulics In Exercises 55–58, determine the location of the horizontal axis ya at which a vertical gate in a dam is to be hinged so that there is no moment causing rotation under the indicated loading (see figure). The model for ya is ya  y 

Writing About Concepts The center of mass of the lamina of constant density shown in the figure is  2, 85. In Exercises 47– 50, make a conjecture about how the center of mass x, y will change for the nonconstant density  x, y. Explain. (Make your conjecture without performing any calculations.)

Iy hA

where y is the y-coordinate of the centroid of the gate, Iy is the moment of inertia of the gate about the line y  y, h is the depth of the centroid below the surface, and A is the area of the gate. y

y=L h

y

y=y Iy ya = y − hA

4 3

x 2

(2, 85 )

55.

1

y

2

3

y=L

y=L

x

1

y

56.

d

4





47.  x, y  ky

48.  x, y  k 2  x

49.  x, y  kxy

50.  x, y  k 4  x4  y

51. Give the formulas for finding the moments and center of mass of a variable density planar lamina. 52. Give the formulas for finding the moments of inertia about the x- and y-axes for a variable density planar lamina.

a b

57.

x

y

b

58. b

y=L

x

y

y=L d

53. In your own words, describe what the radius of gyration measures. x

a x

SECTION 14.5

Section 14.5

Surface Area

1017

Surface Area • Use a double integral to find the area of a surface.

Surface Area Surface: z = f(x, y)

At this point you know a great deal about the solid region lying between a surface and a closed and bounded region R in the xy-plane, as shown in Figure 14.42. For example, you know how to find the extrema of f on R (Section 13.8), the area of the base R of the solid (Section 14.1), the volume of the solid (Section 14.2), and the centroid of the base R (Section 14.4). In this section, you will learn how to find the upper surface area of the solid. Later, you will learn how to find the centroid of the solid (Section 14.6) and the lateral surface area (Section 15.2). To begin, consider a surface S given by

z

z  f x, y

y

x

Region R in xy-plane

. Figure 14.42

Rotatable Graph

Surface: z = f (x, y)

ΔTi

z

ΔSi ≈ ΔTi

Surface defined over a region R

defined over a region R. Assume that R is closed and bounded and that f has continuous first partial derivatives. To find the surface area, construct an inner partition of R consisting of n rectangles, where the area of the ith rectangle Ri is Ai  xi yi, as shown in Figure 14.43. In each Ri let xi, yi be the point that is closest to the origin. At the point xi, yi, zi  xi, yi, f xi, yi  on the surface S, construct a tangent plane Ti. The area of the portion of the tangent plane that lies directly above Ri is approximately equal to the area of the surface lying directly above Ri. That is, Ti  Si. So, the surface area of S is given by n

n

 S   T . i

i

i1

i1

To find the area of the parallelogram Ti, note that its sides are given by the vectors u  xi i  fxxi , yi  xi k

R y

and v  yi j  fyxi , yi  yi k.

x

ΔAi

. Figure 14.43

Rotatable Graph

From Theorem 11.8, the area of Ti is given by u



i u  v  xi 0

k fxxi, yi  xi fyxi, yi  yi

j 0 yi





v, where

 fxxi, yi  xi yi i  fyxi, yi  xi yi j  xi yi k  fxxi, yi i  fyxi, yi j  k Ai. So, the area of Ti is u  v   fxxi, yi  2  fyxi, yi  2  1 Ai, and Surface area of S  

n

 S

i

i1 n

 1  f x , y  x

i

i

2

 fyxi, yi  2 Ai.

i1

This suggests the following definition of surface area.

1018

CHAPTER 14

Multiple Integration

Definition of Surface Area If f and its first partial derivatives are continuous on the closed region R in the xy-plane, then the area of the surface S given by z  f x, y over R is given by Surface area 

 

dS

R



1  fxx, y 2  fyx, y 2 dA.

R

As an aid to remembering the double integral for surface area, it is helpful to note its similarity to the integral for arc length.

      b

Length on x-axis:

dx

a b

Arc length in xy-plane:

b

a

Area in xy-plane:

1  fx 2 dx

ds 

a

dA

R

Surface area in space:

dS 

R

1  fxx, y 2  fyx, y 2 dA

R

Like integrals for arc length, integrals for surface area are often very difficult to evaluate. However, one type that is easily evaluated is demonstrated in the next example.

The Surface Area of a Plane Region

EXAMPLE 1 Plane: z=2−x−y

Find the surface area of the portion of the plane

z

z2xy that lies above the circle x 2  y 2 ≤ 1 in the first quadrant, as shown in Figure 14.44.

2

Solution Because fxx, y  1 and fyx, y  1, the surface area is given by S

  

1  fxx, y 2  fyx, y 2 dA

Formula for surface area

R

2

2

y



1  12  12 dA

Substitute.

R

x

.

R: x 2 + y 2 ≤ 1

Figure 14.44

Rotatable Graph



R

 3

3 dA



dA.

R

Note that the last integral is simply 3 times the area of the region R. R is a quarter circle of radius 1, with an area of 14 12 or 4. So, the area of S is S  3 area of R   3 4



.



3 

4

Try It

.

Exploration A

SECTION 14.5

Find the area of the portion of the surface

Surface: f(x, y) = 1 − x 2 + y

f x, y  1  x 2  y that lies above the triangular region with vertices 1, 0, 0, 0, 1, 0, and 0, 1, 0, as shown in Figure 14.45(a).

(0, 1, 2)

2

1019

Finding Surface Area

EXAMPLE 2 z

Surface Area

Solution Because fxx, y  2x and fyx, y  1, you have 1

S



1  fxx, y 2  fyx, y 2 dA 

R

−1

.

y

1

1

R

    1

1x

0

(a)

2  4x 2 dy dx

x1

1



Rotatable Graph

y2  4x2

0 1



y

0 1



0

y=1−x

1

dx x1



1  x2  4x2  x  12  4x2 dx

22  4x 2  2x2  4x 2  dx

Integration tables (Appendix B), Formula 26 and Power Rule 1



2

Try It

y=x−1

−1

1x

2  4x 232 6 0 1  6  ln2  6   6  ln 2  2  1.618. 3

x 1





 x2  4x 2  ln2x  2  4x 2  

R: 0 ≤ x ≤ 1 x−1≤y≤1−x

.

1  4x 2  1 dA.

In Figure 14.45(b), you can see that the bounds for R are 0 ≤ x ≤ 1 and x  1 ≤ y ≤ 1  x. So, the integral becomes S

x



Exploration A

Open Exploration

Change of Variables to Polar Coordinates

EXAMPLE 3 (b)

Find the surface area of the paraboloid z  1  x 2  y 2 that lies above the unit circle, as shown in Figure 14.46.

Figure 14.45

Solution Because fxx, y  2x and fyx, y  2y, you have S

1  fxx, y 2  fyx, y 2 dA 

R

Paraboloid: z = 1 + x2 + y2

S

R: x 2 + y 2 ≤ 1 R

x

Figure 14.46

Rotatable Graph

1  4x 2  4y 2 dA.

R

   2

1

1  4r 2 r dr d

0 0 2

y

0

d

2

55  1

12 0  55  1  6  5.33.

 1

1



1 1  4r 232 12 0 2 55  1  d

12 0 

1



You can convert to polar coordinates by letting x  r cos and y  r sin . Then, because the region R is bounded by 0 ≤ r ≤ 1 and 0 ≤ ≤ 2, you have

z

2

.



Try It



Exploration A

1020

CHAPTER 14

Multiple Integration

Finding Surface Area

EXAMPLE 4

Find the surface area S of the portion of the hemisphere

Hemisphere: 25 −

f(x, y) =

x2

−

y2

f x, y  25  x 2  y 2

z

Hemisphere

that lies above the region R bounded by the circle x 2  y 2 ≤ 9, as shown in Figure 14.47.

5 4 3

Solution The first partial derivatives of f are

2

−4

1

−4

−2

2

1

4 6

−6

2

3

R:

.x

x

fxx, y  4

x2

+

y

5

y2

≤9

25  x 2  y 2

fyx, y 

and

y 25  x 2  y 2

and, from the formula for surface area, you have dS  1  fxx, y 2  fyx, y 2 dA

Figure 14.47



1  



5 dA. 25  x 2  y 2

Rotatable Graph

x 25  x 2  y 2

y

  25  x 2

2



y2



2

dA

So, the surface area is S

 R

5 25  x 2  y 2

dA.

You can convert to polar coordinates by letting x  r cos and y  r sin . Then, because the region R is bounded by 0 ≤ r ≤ 3 and 0 ≤ ≤ 2, you obtain S

   2

0

5 5

.

3

0 2

5

r dr d

25  r 2



 25  r2

0 2

3

0

d

d

0

 10.

Try It

Exploration A

Hemisphere:

25 − x 2 − y2

f(x, y) =

The procedure used in Example 4 can be extended to find the surface area of a sphere by using the region R bounded by the circle x 2  y 2 ≤ a 2, where 0 < a < 5, as shown in Figure 14.48. The surface area of the portion of the hemisphere

z 5

f x, y  25  x 2  y 2 lying above the circular region can be shown to be S a 5 x

. Figure 14.48

Rotatable Graph

a 5

y

R: x 2 + y 2 ≤ a2



  R 2 0

5 dA 25  x 2  y 2 a

0

5 25  r 2

r dr d

 10 5  25  a2 . By taking the limit as a approaches 5 and doubling the result, you obtain a total area of 100. (The surface area of a sphere of radius r is S  4r 2.)

SECTION 14.5

Surface Area

1021

You can use Simpson’s Rule or the Trapezoidal Rule to approximate the value of a double integral, provided you can get through the first integration. This is demonstrated in the next example.

Approximating Surface Area by Simpson’s Rule

EXAMPLE 5

Find the area of the surface of the paraboloid

Paraboloid: f (x, y) = 2 − x 2 − y 2

f x, y  2  x 2  y 2

z

Paraboloid

that lies above the square region bounded by 1 ≤ x ≤ 1 and 1 ≤ y ≤ 1, as shown in Figure 14.49.

2

Solution Using the partial derivatives fxx, y  2x

and

fyx, y  2y

you have a surface area of S

  

1  fxx, y 2  fyx, y 2 dA

R

 y

1 2



R: −1 ≤ x ≤ 1 −1 ≤ y ≤ 1

x

.

1  2x2  2y2 dA

R

1  4x 2  4y 2 dA.

R

In polar coordinates, the line x  1 is given by r cos  1 or r  sec , and you can determine from Figure 14.50 that one-fourth of the region R is bounded by

Figure 14.49

Rotatable Graph 0 ≤ r ≤ sec

and



  ≤ ≤ . 4 4

Letting x  r cos and y  r sin produces y

r = sec θ θ=π 4

1

 x

−1



1

 −1

θ = −π 4

One-fourth of the region R is bounded by   0 ≤ r ≤ sec and  ≤ ≤ . 4 4 . Figure 14.50

    

1 1 S 4 4

1  4x 2  4y 2 dA

R

4

sec

4 0 4

1  4r 2 r dr d

1 12

4

4

sec



1 1  4r232 12 4

0

d

1  4 sec2  32  1 d .

Finally, using Simpson’s Rule with n  10, you can approximate this single integral to be S

1 3



4

4

1  4 sec2 32  1 d

 7.450.

Try It

Exploration A

Most computer programs that are capable of performing symbolic integration for multiple integrals are also capable of performing numerical approximation techniques. If you have access to such software, use it to approximate the value of the integral in Example 5.

TECHNOLOGY

1022

CHAPTER 14

Multiple Integration

Exercises for Section 14.5 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–14, find the area of the surface given by z  f x, y over the region R. (Hint: Some of the integrals are simpler in polar coordinates.) 1. f x, y  2x  2y

R  x, y:



4. f x, y  10  2x  3y

≤ 4

R  x, y:

x2



y2

≤ 9

5. f x, y  9  x 2 R: square with vertices 0, 0, 3, 0, 0, 3, 3, 3 6. f x, y  y 2 R: square with vertices 0, 0, 3, 0, 0, 3, 3, 3 7. f x, y  2  x32 R: rectangle with vertices 0, 0, 0, 4, 3, 4, 3, 0 8. f x, y  2  23 y 32



9. f x, y  ln sec x





10. f x, y  9  x 2  y 2 R  x, y:



y2

≤ 4

R  x, y: 0 ≤ f x, y ≤ 1 R  x, y:



R  x, y:



y2

≤ 16 ≤

b2,

0 < b < a

(e) 36

(d) 55

(e) 500

R: circle bounded by x 2  y 2  9 (b) 150

(c) 9

28. f x, y  25 y52

R  x, y: 0 ≤ x ≤ 4, 0 ≤ y ≤ x 31. f x, y  ex sin y

2

R  x, y: x 2  y 2 ≤ 4

In Exercises 15–18, find the area of the surface. 15. The portion of the plane z  24  3x  2y in the first octant 16. The portion of the paraboloid z  16 

(d) 72

30. f x, y  x 2  3xy  y 2

R  x, y: x  y ≤ a 2

(c) 100

R: square with vertices 1, 1, 1, 1, 1, 1, 1, 1

14. f x, y  a 2  x 2  y 2 2

(b) 200

29. f x, y  x3  3xy  y3

13. f x, y  a 2  x 2  y 2 x2

(a) 16

26. f x, y  14x 2  y 2

In Exercises 29–34, set up a double integral that gives the area of the surface on the graph of f over the region R.

12. f x, y  xy y2

R: square with vertices 0, 0, 4, 0, 4, 4, 0, 4

27. f x, y  e x

11. f x, y  x 2  y 2

x2

25. f x, y  10  12 y 2

In Exercises 27 and 28, use a computer algebra system to approximate the double integral that gives the surface area of the graph of f over the region R  {x, y: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

 R  x, y: 0 ≤ x ≤ , 0 ≤ y ≤ tan x 4 x2

R  x, y: 0 ≤ f x, y ≤ 16

2

Approximation In Exercises 25 and 26, determine which value best approximates the surface area of z  f x, y over the region R. (Make your selection on the basis of a sketch of the surface and not by performing any calculations.)

(a) 100

R  x, y: 0 ≤ x ≤ 2, 0 ≤ y ≤ 2  x



23. f x, y  4  x  y 2

R  x, y: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

R: square with vertices 0, 0, 3, 0, 0, 3, 3, 3 y2

R  x, y: 0 ≤ f x, y

24. f x, y  23x32  cos x

2. f x, y  15  2x  3y

x2

22. f x, y  x 2  y 2

R  x, y: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

R: triangle with vertices 0, 0, 2, 0, 0, 2

3. f x, y  8  2x  2y

21. f x, y  4  x 2  y 2

x2

 y in the first octant 2

32. f x, y  cosx 2  y 2

R  x, y: 0 ≤ x ≤ 4, 0 ≤ y ≤ 10

18. The portion of the cone z  2x 2  y 2 inside the cylinder x2  y 2  4

Writing About Concepts

R: triangle with vertices 0, 0, 1, 0, 1, 1 20. f x, y  2x  y 2 R: triangle with vertices 0, 0, 2, 0, 2, 2



34. f x, y  ex sin y R  x, y: 0 ≤ x ≤ 4, 0 ≤ y ≤ x

19. f x, y  2y  x 2

 2

33. f x, y  exy

17. The portion of the sphere x 2  y 2  z 2  25 inside the cylinder x 2  y 2  9

In Exercises 19–24, write a double integral that represents the surface area of z  f x, y over the region R. Use a computer algebra system to evaluate the double integral.



R  x, y: x 2  y 2 ≤

35. State the double integral definition of the area of a surface S given by z  f x, y over the region R in the xy-plane. 36. Answer the following questions about the surface area S on a surface given by a positive function z  f x, y over a region R in the xy-plane. Explain each answer. (a) Is it possible for S to equal the area of R? (b) Can S be greater than the area of R? (c) Can S be less than the area of R?

SECTION 14.5

37. Find the surface area of the solid of intersection of the cylinders x 2  z 2  1 and y 2  z 2  1 (see figure). z

Surface Area

1023

40. Modeling Data A rancher builds a barn with dimensions 30 feet by 50 feet. The symmetrical shape and selected heights of the roof are shown in the figure. z

z

25

y2

+

z2

=1

(0, 25) (5, 22) (10, 17)

2

(15, 0)

−3

20 3

3 x

y

r

r

x

−2

x2 + z2 = 1

Figure for 37

z = k x 2 + y 2, k > 0

Figure for 38

Rotatable Graph

Rotatable Graph

38. Show that the surface area of the cone z  kx 2  y 2, k > 0 over the circular region x 2  y 2 ≤ r 2 in the xy-plane is r 2k 2  1 (see figure). 39. Building Design A new auditorium is built with a foundation in the shape of one-fourth of a circle of radius 50 feet. So, it forms a region R bounded by the graph of x 2  y 2  502 with x ≥ 0 and y ≥ 0. The following equations are models for the floor and ceiling. Floor: z 

xy 5

Ceiling: z  20 

xy 100

(a) Calculate the volume of the room, which is needed to determine the heating and cooling requirements. (b) Find the surface area of the ceiling.

y

y

50 x

Rotatable Graph (a) Use the regression capabilities of a graphing utility to find a model of the form z  ay 3  by 2  cy  d for the roof line. (b) Use the numerical integration capabilities of a graphing utility and the model in part (a) to approximate the volume of storage space in the barn. (c) Use the numerical integration capabilities of a graphing utility and the model in part (a) to approximate the surface area of the roof. (d) Approximate the arc length of the roof line and find the surface area of the roof by multiplying the arc length by the length of the barn. Compare the results and the integrations with those found in part (c). 41. Product Design A company produces a spherical object of radius 25 centimeters. A hole of radius 4 centimeters is drilled through the center of the object. Find (a) the volume of the object and (b) the outer surface area of the object.

1024

CHAPTER 14

Multiple Integration

Section 14.6

Triple Integrals and Applications • Use a triple integral to find the volume of a solid region. • Find the center of mass and moments of inertia of a solid region.

Triple Integrals z

The procedure used to define a triple integral follows that used for double integrals. Consider a function f of three variables that is continuous over a bounded solid region Q. Then, encompass Q with a network of boxes and form the inner partition consisting of all boxes lying entirely within Q, as shown in Figure 14.51. The volume of the ith box is Vi  xi yi zi .

Volume of ith box

The norm  of the partition is the length of the longest diagonal of the n boxes in the partition. Choose a point xi, yi , z i  in each box and form the Riemann sum y

n

 f x , y , z  V . i

x

i

i

i

i1

Taking the limit as  → 0 leads to the following definition.

. Solid region Q

Definition of Triple Integral

Rotatable Graph

If f is continuous over a bounded solid region Q, then the triple integral of f over Q is defined as

z



f x, y, z dV  lim

n

 f x , y , z  V

→0 i1

i

i

i

i

Q

provided the limit exists. The volume of the solid region Q is given by Volume of Q 



dV.

Q

y x

Volume of Q  .

n

 f x , y , z  V i

i 1

Figure 14.51

i

i

Some of the properties of double integrals in Theorem 14.1 can be restated in terms of triple integrals.

i

1.

  



cf x, y, z dV  c

Q

Rotatable Graph 2.

Q

f x, y, z ± gx, y, z dV 

Q

3.

Q

f x, y, z dV

f x, y, z dV 

 Q1



f x, y, z dV ±

Q

f x, y, z dV 





gx, y, z dV

Q

f x, y, z dV

Q2

In the properties above, Q is the union of two nonoverlapping solid subregions Q1 and Q 2. If the solid region Q is simple, the triple integral  f x, y, z dV can be evaluated with an iterated integral using one of the six possible orders of integration: dx dy dz dy dx dz dz dx dy dx dz dy dy dz dx

dz dy dx.

SECTION 14.6

E X P L O R AT I O N Volume of a Paraboloid Sector On pages 995 and 1003, you were asked to summarize the different ways you know for finding the volume of the solid bounded by the paraboloid z  a2  x2  y2,

1025

The following version of Fubini’s Theorem describes a region that is considered simple with respect to the order dz dy dx. Similar descriptions can be given for the other five orders.

THEOREM 14.4

Evaluation by Iterated Integrals

Let f be continuous on a solid region Q defined by

a > 0

and the xy-plane. You now know one more way. Use it to find the volume of the solid.

Triple Integrals and Applications

a ≤ x ≤ b,

h 1x ≤ y ≤ h 2x, g1x, y ≤ z ≤ g2x, y

where h 1, h 2, g1, and g2 are continuous functions. Then,



z

a

Q

a2

  h2x

b

f x, y, z dV 

h1x

g2x, y

f x, y, z dz dy dx.

g1x, y

To evaluate a triple iterated integral in the order dz dy dx, hold both x and y constant for the innermost integration. Then, hold x constant for the second integration. −a

.

a x

Rotatable Graph

a

y

EXAMPLE 1

Evaluating a Triple Iterated Integral

Evaluate the triple iterated integral

 2

0

x

0

xy

e x y  2z dz dy dx.

0

Solution For the first integration, hold x and y constant and integrate with respect to z.

 2

0

x

0

xy

  2

e x y  2z dz dy dx 

0

x

0 0 2 x



0

xy



e x  yz  z 2

dy dx 0

e x x 2  3xy  2y 2 dy dx

0

For the second integration, hold x constant and integrate with respect to y.

 2

0

x

   2

e xx 2  3xy  2y 2 dy dx 

0

e x x 2y 

0



19 6

3xy 2 2y 3  2 3



x

dx 0

2

x 3e x dx

0

Finally, integrate with respect to x. 19 6



2

0

.

Try It

 



19 x 3 e x  3x 2  6x  6 6 e2 1  19 3  65.797

x3e x dx 

2 0



Exploration A

Exploration B

Example 1 demonstrates the integration order dz dy dx. For other orders, you can follow a similar procedure. For instance, to evaluate a triple iterated integral in the order dx dy dz, hold both y and z constant for the innermost integration and integrate with respect to x. Then, for the second integration, hold z constant and integrate with respect to y. Finally, for the third integration, integrate with respect to z.

1026

CHAPTER 14

Multiple Integration

z

To find the limits for a particular order of integration, it is generally advisable first to determine the innermost limits, which may be functions of the outer two variables. Then, by projecting the solid Q onto the coordinate plane of the outer two variables, you can determine their limits of integration by the methods used for double integrals. For instance, to evaluate

z = g2(x, y)

Q z = g1(x, y) y

x



f x, y, z dz dy dx

Q

first determine the limits for z, and then the integral has the form

Projection onto xy-plane

. region Q lies between two surfaces. Solid

 

g2x, y

g1 x, y

Figure 14.52



f x, y, z dz dy dx.

By projecting the solid Q onto the xy-plane, you can determine the limits for x and y as you did for double integrals, as shown in Figure 14.52.

Rotatable Graph

Using a Triple Integral to Find Volume

EXAMPLE 2

0 ≤ z ≤ 2 4 − x2 − y2 z

Find the volume of the ellipsoid given by 4x 2  4y 2  z 2  16. 4

Solution Because x, y, and z play similar roles in the equation, the order of integration is probably immaterial, and you can arbitrarily choose dz dy dx. Moreover, you can simplify the calculation by considering only the portion of the ellipsoid lying in the first octant, as shown in Figure 14.53. From the order dz dy dx, you first determine the bounds for z. 0 ≤ z ≤ 24  x 2  y 2 2

In Figure 14.54, you can see that the boundaries for x and y are 0 ≤ x ≤ 2 and 0 ≤ y ≤ 4  x 2, so the volume of the ellipsoid is

1

x

2

V

y

          dV

Q 2

24x2 y2

4x2

8

0

8

dy dx

0 0 2 4x2

Integration tables (Appendix B), Formula 37

4  x 2  y 2 dy dx

 16

Rotatable Graph

24x2 y2

z

0

Figure 14.53

dz dy dx

0

4x2

2

Ellipsoid: 4x 2 + 4y 2 + z 2 = 16

.

0

0

0

2

y4  x 2  y 2  4  x 2 arcsin

8 y

0≤x≤2 0 ≤ y ≤ 4 − x2

0



2

0  4  x 2 arcsin1  0  0 dx

8

0

2

2

x2

+

y2

=4

 dx 2

4  x 2

8

0



1

 4 4x 

. x

1

Figure 14.54



y 4  x 2

2

x3 3

2



0

64  . 3

Try It

Exploration A

Exploration B

4x2

0

dx

SECTION 14.6

Triple Integrals and Applications

1027

Example 2 is unusual in that all six possible orders of integration produce integrals of comparable difficulty. Try setting up some other possible orders of integration to find the volume of the ellipsoid. For instance, the order dx dy dz yields the integral





16z22

4

V8

0

0

164y 2 z22

dx dy dz.

0

If you solve this integral, you will obtain the same volume obtained in Example 2. This is always the case—the order of integration does not affect the value of the integral. However, the order of integration often does affect the complexity of the integral. In Example 3, the given order of integration is not convenient, so you can change the order to simplify the problem.

Changing the Order of Integration

EXAMPLE 3

   2

Evaluate

2

0

3

sin  y 2 dz dy dx.

1

x

Solution Note that after one integration in the given order, you would encounter the integral 2 sin y 2 dy, which is not an elementary function. To avoid this problem, change the order of integration to dz dx dy, so that y is the outer variable. The solid region Q is given by

2 ,

0 ≤ x ≤

2 ,

x ≤ y ≤

1 ≤ z ≤ 3

and the projection of Q in the xy-plane yields the bounds Q: 0 ≤ x ≤ z

x≤y≤

π 2 π 2

1≤z≤3

(

3

2

0 ≤ y ≤

0 ≤ x ≤ y.

and

So, you have π , 2

π ,3 2

)

V

         dV

Q



2

0

2



(

2

)

π 2

x

y=x

The volume of the solid region Q is 1.

. Figure 14.55

Rotatable Graph



2

2

y

2

dx dy

1

y

sin y 2 dx dy

0

y



x sin y 2

0

2

π 2

3

0

0

π ,1 2

sin y 2 dz dx dy

z sin y 2

2

π , 2

3

0 1 2 y

0

1

y

dy 0

y sin y 2 dy

0



 cos y 2  1.

2

0

See Figure 14.55.

Try It

Exploration A

1028

CHAPTER 14

Multiple Integration

z

EXAMPLE 4

1

Set up a triple integral for the volume of each solid region.

z = 1 − y2

a. The region in the first octant bounded above by the cylinder z  1  y 2 and lying between the vertical planes x  y  1 and x  y  3

1 y

x= 1−y 3

Δy

Solution

Q: 0 ≤ z ≤ 1 − y 2 1−y≤x≤3−y 0≤y≤1

.

b. The upper hemisphere given by z  1  x 2  y 2 c. The region bounded below by the paraboloid z  x 2  y 2 and above by the sphere x2  y 2  z2  6

x=3−y

x

Determining the Limits of Integration

a. In Figure 14.56, note that the solid is bounded below by the xy-plane z  0 and above by the cylinder z  1  y 2. So,

Figure 14.56

0 ≤ z ≤ 1  y 2.

Rotatable Graph z

Bounds for z

Projecting the region onto the xy-plane produces a parallelogram. Because two sides of the parallelogram are parallel to the x-axis, you have the following bounds: Hemisphere:

1  y ≤ x ≤ 3  y and

1 − x2 − y2

z=

0 ≤ y ≤ 1.

So, the volume of the region is given by 1

    1

V

3y

1y 2

dV 

0

1y

dz dx dy.

0

Q

b. For the upper hemisphere given by z  1  x 2  y 2, you have 1

1

x

y

Circular base: x2 + y2 = 1 Q: 0 ≤ z ≤ −

1−

1 − x2 − y2 y2

≤x≤

1 − y2

In Figure 14.57, note that the projection of the hemisphere onto the xy-plane is the circle given by x 2  y 2  1, and you can use either order dx dy or dy dx. Choosing the first produces

    1

V

Rotatable Graph

dV 

Q

3

1 ≤ y ≤ 1

which implies that the volume of the region is given by

Figure 14.57

z

Bounds for z

 1  y 2 ≤ x ≤ 1  y 2 and

−1 ≤ y ≤ 1

.

0 ≤ z ≤ 1  x 2  y 2.

Sphere: x2 + y2 + z2 = 6

1y2

1x2 y2

dz dx dy.

1 1y 2 0

c. For the region bounded below by the paraboloid z  x 2  y 2 and above by the sphere x 2  y 2  z 2  6, you have x 2  y 2 ≤ z ≤ 6  x 2  y 2.

Paraboloid: z = x2 + y2 −2 2 x

.

2

Q: x 2 + y 2 ≤ z ≤ 6 − x 2 − y 2 − 2 − x2 ≤ y ≤ 2 − x2 − 2≤x≤ 2

Figure 14.58

Rotatable Graph

y

Bounds for z

The sphere and the paraboloid intersect when z  2. Moreover, you can see in Figure 14.58 that the projection of the solid region onto the xy-plane is the circle given by x 2  y 2  2. Using the order dy dx produces  2  x 2 ≤ y ≤ 2  x 2 and

 2 ≤ x ≤ 2

which implies that the volume of the region is given by

    2

V

dV 

Q

Try It

2

2x2

6x2 y2

2x2 x2 y2

Exploration A

dz dy dx.

SECTION 14.6

Triple Integrals and Applications

1029

Center of Mass and Moments of Inertia In the remainder of this section, two important engineering applications of triple integrals are discussed. Consider a solid region Q whose density is given by the density function . The center of mass of a solid region Q of mass m is given by x, y, z, where

E X P L O R AT I O N Sketch the solid (of uniform density) bounded by z  0 and z

1 1  x2  y2

m

where  y ≤ 1. From your sketch, estimate the coordinates of the center of mass of the solid. Now use a computer algebra system to verify your estimate. What do you observe? x2

2

   

 x, y, z dV

Mass of the solid

x x, y, z dV

First moment about yz-plane

y x, y, z dV

First moment about xz-plane

z x, y, z dV

First moment about xy-plane

Q

Myz 

Q

Mxz 

Q

Mxy 

Q

and x NOTE In engineering and physics, the moment of inertia of a mass is used to find the time required for a mass to reach a given speed of rotation about an axis, as shown in Figure 14.59. The greater the moment of inertia, the longer a force must be applied for the mass to reach the given speed.

Myz , m

y

Mxz , m

z

Mxy . m

The quantities Myz, Mxz, and Mxy are called the first moments of the region Q about the yz-, xz-, and xy-planes, respectively. The first moments for solid regions are taken about a plane, whereas the second moments for solids are taken about a line. The second moments (or moments of inertia) about the x-, y-, and z-axes are as follows. Ix 

  

 y 2  z 2 x, y, z dV

Moment of inertia about x-axis

x 2  z 2 x, y, z dV

Moment of inertia about y-axis

x 2  y 2 x, y, z dV

Moment of inertia about z-axis

Q

z

Iy 

Q

y

Iz 

Q

x

For problems requiring the calculation of all three moments, considerable effort can be saved by applying the additive property of triple integrals and writing Ix  Ixz  Ixy,

. Figure 14.59

Rotatable Graph

Iy  Iyz  Ixy, and

where Ixy, Ixz, and Iyz are as follows. Ixy 

  

z 2 x, y, z dV

Q

Ixz 

y 2 x, y, z dV

Q

Iyz 

Q

x 2 x, y, z dV

Iz  Iyz  Ixz

1030

CHAPTER 14

Multiple Integration

Finding the Center of Mass of a Solid Region

EXAMPLE 5 z

Find the center of mass of the unit cube shown in Figure 14.60, given that the density at the point x, y, z is proportional to the square of its distance from the origin. Solution Because the density at x, y, z is proportional to the square of the distance between 0, 0, 0 and x, y, z, you have

1

 x, y, z  kx 2  y 2  z2. (x, y, z)

1

1

y

You can use this density function to find the mass of the cube. Because of the symmetry of the region, any order of integration will produce an integral of comparable difficulty.

        1

m

x

1

0

Variable density: .x, y, z  kx 2  y 2  z2

1

0

0

1

1

k

Figure 14.60

0

1

k

0

x2  y 2 

0

1

k



0

1

x2 

k

0

1



dy dx

0



1 dy dx 3

1 y3 y 3 3

x2 

1



dx

0



2 dx 3 1

 x3  2x3  3

k

z3 3

x 2  y 2z 

0

1

Rotatable Graph

kx 2  y 2  z 2 dz dy dx

k

0

The first moment about the yz-plane is

    1

1

1

xx 2  y 2  z 2 dz dy dx

Myz  k

0

0

1

k

0

1



1

x 2  y 2  z 2 dz dy dx.

x

0

0

0

Note that x can be factored out of the two inner integrals, because it is constant with respect to y and z. After factoring, the two inner integrals are the same as for the mass m. Therefore, you have



1

Myz  k



x x2 

0



2 dx 3

2 1

 x4  x3 

k 

4

0

7k . 12

So, x .

Myz 7k12 7   . m k 12

Finally, from the nature of  and the symmetry of x, y, and z in this solid region, you 7 7 7 have x  y  z, and the center of mass is 12, 12, 12 .

Try It

Open Exploration

SECTION 14.6

Triple Integrals and Applications

1031

Moments of Inertia for a Solid Region

EXAMPLE 6

Find the moments of inertia about the x- and y-axes for the solid region lying between the hemisphere z  4  x 2  y 2 and the xy-plane, given that the density at x, y, z is proportional to the distance between (x, y, z and the xy-plane. Solution The density of the region is given by x, y, z  kz. Considering the symmetry of this problem, you know that Ix  Iy, and you need to compute only one moment, say Ix. From Figure 14.61, choose the order dz dy dx and write

0 ≤ z ≤ 4 − x2 − y2 − 4 − x2 ≤ y ≤ 4 − x2 −2 ≤ x ≤ 2 Hemisphere: z=

Ix 

4 − x2 − y2 z

           

 y 2  z 2 x, y, z dV

Q 2



2

4x2

2 4x2 0 4x2

2

k k 2 2

y



x

Circular base: x2 + y2 = 4

. Variable density:  x, y, z  kz Figure 14.61

 

Rotatable Graph    .

4x2 y 2

k 4

2 4x2 4x2 2 2 4x2 4x2 2

k 4 k 4

4k 5 4k 5

2 4x2 2

2

2

y 2z 2 z 4  2 4

4x2 y 2



dy dx 0

y 24  x 2  y 2 4  x 2  y 2 2  dy dx 2 4



4  x22  y 4 dy dx

4  x 22 y 

2

 y 2  z 2kz dz dy dx

y5 5



4x2

4x2

dx

8 4  x 252 dx 5

2

4  x 252 dx

x  2 sin

0

2

64 cos6 d

0

5 256k  5 32 

Wallis’s Formula

 8k. So, Ix  8k  Iy.

Exploration A In Example 6, notice that the moments of inertia about the x- and y-axes are equal to each other. The moment about the z-axis, however, is different. Does it seem that the moment of inertia about the z-axis should be less than or greater than the moments calculated in Example 6? By performing the calculations, you can determine that Iz 

16 k. 3

This tells you that the solid shown in Figure 14.61 has a greater resistance to rotation about the x- or y-axis than about the z-axis.

1032

CHAPTER 14

Multiple Integration

Exercises for Section 14.6 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–8, evaluate the iterated integral.

      3

2

19.

20. z

z

1

x  y  z dx dy dz

1.

0 0 0 1 1 1

2.

a 2 2 2

x y z dx dy dz

1 1 1 1 x xy

3.

4.

7.

0

0

2

6.

1 1 2

1x

x cos y dz dy dx

8.

0

0

y2 9x2

y3

0 0 4 e2

2zex dy dx dz

1 0 0 4 2

      9

x dz dy dx

0 0 0 4 1 x

5.

z dz dx dy

0 y2

0

a

a

0 1xz

y

x

ln z dy dz dx

z=0

1y

x 2 + y2 + z2 = a2

sin y dz dx dy

12

x

Rotatable Graph

    4x2

9.

2

10.

0

z=4−

4y 2

y dz dy dx

   

11.

4x2

4

0

0

z = 9 − x2

9

4

y=−x+2 x≥0 y≥0 z≥0

2

x 2 sin y dz dy dx z

0 0 1 3 2 2y3 62y3z

12.

x2

x≥0 y≥0 z≥0

In Exercises 11 and 12, use a computer algebra system to approximate the iterated integral. 2

z

x dz dy dx

2x 2 y 2

0

22. z

0

2x2

y

Rotatable Graph

21.

x2

4x2

0

12

0

In Exercises 9 and 10, use a computer algebra system to evaluate the iterated integral. 2

z = 36 − x 2 − y 2

36

x

4 2

y

y = 4 − x2

2 y

x

zex y dx dz dy 2 2

Rotatable Graph

0

In Exercises 13–16, set up a triple integral for the volume of the solid. 13. The solid in the first octant bounded by the coordinate planes and the plane z  4  x  y 14. The solid bounded by z  9  x , z  0, x  0, and y  2x 2

Rotatable Graph

In Exercises 23–26, sketch the solid whose volume is given by the iterated integral and rewrite the integral using the indicated order of integration.

  4

23.

0

15. The solid bounded by the paraboloid z  9  x 2  y 2 and the plane z  0

4x2

0

123x6y4

dz dy dx

0

Rewrite using the order dy dx dz.

  3

16. The solid that is the common interior below the sphere 1 x 2  y 2  z 2  80 and above the paraboloid z  2x 2  y 2

24.

Volume In Exercises 17– 22, use a triple integral to find the volume of the solid shown in the figure.

25.

0

9x2

6xy

dz dy dx

0

0

Rewrite using the order dz dx dy.

 1

0

1

y

1y2

dz dx dy

0

Rewrite using the order dz dy dx.

17.

18. z

26.

z = xy

z=x

0

4 3

  2

z

4

2x

y2 4x2

dz dy dx

0

Rewrite using the order dx dy dz. y

1

In Exercises 27–30, list the six possible orders of integration for the triple integral over the solid region Q

1



x 4

x = 4 − y2

xyz dV.

Q

2

z=0 y

Rotatable Graph

0≤x≤1 0≤y≤1

1

Rotatable Graph

x

27. Q  x, y, z: 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ 3 28. Q  x, y, z: 0 ≤ x ≤ 2, x 2 ≤ y ≤ 4, 0 ≤ z ≤ 2  x 29. Q  x, y, z: x 2  y 2 ≤ 9, 0 ≤ z ≤ 4

SECTION 14.6

In Exercises 31 and 32, the figure shows the region of integration for the given integral. Rewrite the integral as an equivalent iterated integral in the five other orders.

  1

31.

0

1y2

1y

32.

0

0

 3

dz dx dy

0

x

0

9x2

dz dy dx

z

x≥0 y≥0 z≥0

z

z = 9 − x2 1

46. z 

x≥0 y≥0 z≥0

6

y=x

1 , z  0, x  2, x  2, y  0, y  1 y2  1 z

47.

y

x

x = 1 − y2

20 cm x

(0, 0, 4) 5 cm

3

y 3 y

Rotatable Graph

z

48. 12 cm

3

1

44. y  4  x 2, z  y, z  0

45. z  42  x 2  y 2, z  0

9

z=1−y

1

Centroid In Exercises 43–48, find the centroid of the solid region bounded by the graphs of the equations or described by the figure. Use a computer algebra system to evaluate the triple integrals. (Assume uniform density and find the center of mass.) h 43. z  x 2  y 2, z  h r

0

1033

Triple Integrals and Applications

x

x

Rotatable Graph Rotatable Graph

Mass and Center of Mass In Exercises 33–36, find the mass and the indicated coordinates of the center of mass of the solid of given density bounded by the graphs of the equations. 33. Find x using x, y, z  k. Q: 2x  3y  6z  12, x  0, y  0, z  0

Rotatable Graph

Moments of Inertia In Exercises 49–52, find Ix , Iy , and Iz for the solid of given density. Use a computer algebra system to evaluate the triple integrals. 50. (a) x, y, z  k

49. (a)   k

34. Find y using x, y, z  ky.

y

(0, 3, 0)

(5, 0, 0)

(b) x, y, z  kx 2  y 2

(b)   kxyz

Q: 3x  3y  5z  15, x  0, y  0, z  0

z

z

35. Find z using x, y, z  kx.

a a 2

Q: z  4  x, z  0, y  0, y  4, x  0 36. Find y using x, y, z  k. Q:

x y z    1 a, b, c > 0, x  0, y  0, z  0 a b c a

a

Mass and Center of Mass In Exercises 37 and 38, set up the triple integrals for finding the mass and the center of mass of the solid bounded by the graphs of the equations.

x

y

a 2

a 2

y

x

Rotatable Graph

Rotatable Graph

37. x  0, x  b, y  0, y  b, z  0, z  b

x, y, z  kxy

51. (a) x, y, z  k

38. x  0, x  a, y  0, y  b, z  0, z  c

52. (a)   kz (b)   k4  z

(b)   ky

x, y, z  kz

z

z

z=4−x

4

4

Think About It The center of mass of a solid of constant density is shown in the figure. In Exercises 39– 42, make a conjecture about how the center of mass x, y, z  will change for the nonconstant density x, y, z. Explain. 39. x, y, z  kx 40. x, y, z  kz 41. x, y, z  k y  2

z

(2, 0, 85)

x

4

42. x, y, z  kxz 2 y  2 2 3

y

2 y

4 x

3

Rotatable Graph

2

4

4

4

z = 4 − y2

2

1

2

x

Rotatable Graph

y

Rotatable Graph

Moments of Inertia In Exercises 53 and 54, verify the moments of inertia for the solid of uniform density. Use a computer algebra system to evaluate the triple integrals. z

1 53. Ix  12 m3a 2  L2

Iy  Iz 

1 2 2 ma 1 2 12 m3a

a

L

 L2

a

a x

L 2

y

1034

CHAPTER 14

Multiple Integration

1 54. Ix  12 ma 2  b2

Iy  Iz 

1 2 12 mb 1 2 12 ma

z

 c 2

Writing About Concepts (continued) a

c

 c 2

62. Determine whether the moment of inertia about the y-axis of the cylinder in Exercise 53 will increase or decrease for the nonconstant density x, y, z  x 2  z 2 and a  4.

b 2

b c 2

x

a 2

y

Average Value In Exercise 63–66, find the average value of the function over the given solid. The average value of a continuous function f x, y, z over a solid region Q is 1 V



f x, y, z dV

Q

Rotatable Graph

where V is the volume of the solid region Q. Moments of Inertia In Exercises 55 and 56, set up a triple integral that gives the moment of inertia about the z-axis of the solid region Q of density . 55. Q  x, y, z: 1 ≤ x ≤ 1, 1 ≤ y ≤ 1, 0 ≤ z ≤ 1  x

  x 2  y 2  z 2 56. Q  x, y, z: x 2  y 2 ≤ 1, 0 ≤ z ≤ 4  x 2  y 2   kx 2 In Exercises 57 and 58, using the description of the solid region, set up the integral for (a) the mass, (b) the center of mass, and (c) the moment of inertia about the z-axis.

63. f x, y, z  z2  4 over the cube in the first octant bounded by the coordinate planes, and the planes x  1, y  1, and z  1 64. f x, y, z  xyz over the cube in the first octant bounded by the coordinate planes, and the planes x  3, y  3, and z  3 65. f x, y, z  x  y  z over the tetrahedron in the first octant with vertices 0, 0, 0, 2, 0, 0, 0, 2, 0 and 0, 0, 2 66. f x, y, z  x  y over the solid bounded by the sphere x2  y2  z2  2 67. Find the solid region Q where the triple integral



1  2x2  y2  3z2 dV

57. The solid bounded by z  4  x2  y2 and z  0 with density function   kz 58. The solid in the first octant bounded by the coordinate planes and x2  y2  z2  25 with density function   kxy

Q

is a maximum. Use a computer algebra system to approximate the maximum value. What is the exact maximum value? 68. Find the solid region Q where the triple integral

Writing About Concepts



59. Define a triple integral and describe a method of evaluating a triple integral.

is a maximum. Use a computer algebra system to approximate the maximum value. What is the exact maximum value?

1  x2  y2  z2 dV

Q

60. Give the number of possible orders of integration when evaluating a triple integral.

69. Solve for a in the triple integral.

 

(a) Because the solids have the same weight, which has the greater density?

3ay2

1

61. Consider solid A and solid B of equal weight shown below.

0

0

4xy2

dz dx dy 

a

14 15

70. Determine the value of b so that the volume of the ellipsoid

(b) Which solid has the greater moment of inertia? Explain.

x2 

(c) The solids are rolled down an inclined plane. They are started at the same time and at the same height. Which will reach the bottom first? Explain.

is 16.

y2 z2  1 2 b 9

Putnam Exam Challenge 71. Evaluate Axis of revolution

n→

Axis of revolution Solid A

Rotatable Graph

  . . . cos 2n x 1

lim

1

1

2

0

0

0

1



 x2  . . .  xn dx1 dx2 . . . dxn.

This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.

Solid B

SECTION 14.7

Section 14.7

Triple Integrals in Cylindrical and Spherical Coordinates

1035

Triple Integrals in Cylindrical and Spherical Coordinates • Write and evaluate a triple integral in cylindrical coordinates. • Write and evaluate a triple integral in spherical coordinates.

PIERRE SIMON DE LAPLACE (1749–1827)

.

One of the first to use a cylindrical coordinate system was the French mathematician Pierre Simon de Laplace. Laplace has been called the “Newton of France,” and he published many important works in mechanics, differential equations, and probability.

Triple Integrals in Cylindrical Coordinates Many common solid regions such as spheres, ellipsoids, cones, and paraboloids can yield difficult triple integrals in rectangular coordinates. In fact, it is precisely this difficulty that led to the introduction of nonrectangular coordinate systems. In this section, you will learn how to use cylindrical and spherical coordinates to evaluate triple integrals. Recall from Section 11.7 that the rectangular conversion equations for cylindrical coordinates are x  r cos

y  r sin

MathBio

z  z. STUDY TIP An easy way to remember these conversions is to note that the equations for x and y are the same as in polar coordinates and z is unchanged.

In this coordinate system, the simplest solid region is a cylindrical block determined by r1 ≤ r ≤ r2,

1 ≤ ≤ 2, z1 ≤ z ≤ z2

as shown in Figure 14.62. To obtain the cylindrical coordinate form of a triple integral, suppose that Q is a solid region whose projection R onto the xy-plane can be described in polar coordinates. That is,

z

Q  x, y, z: x, y is in R, h1x, y ≤ z ≤ h2x, y and Δzi

R  r, : 1 ≤ ≤ 2,

g1  ≤ r ≤ g2  .

If f is a continuous function on the solid Q, you can write the triple integral of f over Q as θ=π 2

Δri θ =0

riΔ θi

Volume of cylindrical block:

. Vi  ri  ri  i zi Figure 14.62

Rotatable Graph



f x, y, z dV 

Q

  



h2x, y

f x, y, z dz dA

h1x, y

R

where the double integral over R is evaluated in polar coordinates. That is, R is a plane region that is either r-simple or -simple. If R is r-simple, the iterated form of the triple integral in cylindrical form is

 Q

f x, y, z dV 

 

2

1

g2 

g1 

h2r cos , r sin 

h1r cos , r sin 

f r cos , r sin , zr dz dr d .

NOTE This is only one of six possible orders of integration. The other five are dz d dr, dr dz d , dr d dz, d dz dr, and d dr dz.

1036

CHAPTER 14

Multiple Integration

z

To visualize a particular order of integration, it helps to view the iterated integral in terms of three sweeping motions—each adding another dimension to the solid. For instance, in the order dr d dz, the first integration occurs in the r-direction as a point sweeps out a ray. Then, as increases, the line sweeps out a sector. Finally, as z increases, the sector sweeps out a solid wedge, as shown in Figure 14.63.

θ =π 2

θ =0

Integrate with respect to r. E X P L O R AT I O N

z

θ =π 2

θ =0

Integrate with respect to .

−a

and the xy-plane. You now know one more way. Use it to find the volume of the solid. Compare the different methods. What are the advantages and disadvantages of each?

θ =π 2

θ =0

a2

z  a 2  x 2  y 2, a > 0

z

.

z

Volume of a Paraboloid Sector On pages 995, 1003, and 1025, you were asked to summarize the different ways you know for finding the volume of the solid bounded by the paraboloid

a

a

y

x

Rotatable Graph

. Integrate with respect to z. Figure 14.63

Finding Volume by Cylindrical Coordinates

EXAMPLE 1

Animation

Find the volume of the solid region Q cut from the sphere x2  y 2  z 2  4

Sphere: x2 + y2 + z2 = 4

Sphere

by the cylinder r  2 sin , as shown in Figure 14.64.

z 2

Solution Because x 2  y 2  z 2  r 2  z 2  4, the bounds on z are  4  r 2 ≤ z ≤ 4  r 2. R

3 x

Let R be the circular projection of the solid onto the r -plane. Then the bounds on R are 0 ≤ r ≤ 2 sin and 0 ≤ ≤ . So, the volume of Q is

3 y

V

       

0

. Figure 14.64

Cylinder: r = 2 sin θ

2 sin

0 2

2

0

2  

4 3

4r2 2 sin

r dz dr d

2r4  r 2 dr d

0

2

0

Rotatable Graph

4r2

2



2  4  r 232 3

.

0

d

8  8 cos3  d

0

2

32 3

1  cos 1  sin2  d

0

32 sin3

 sin  3 3 16  3  4 9 

2 sin



2



0

 9.644.

Try It

Exploration A

Exploration B

SECTION 14.7

1037

Finding Mass by Cylindrical Coordinates

EXAMPLE 2

0 ≤ z ≤ 16 −

Triple Integrals in Cylindrical and Spherical Coordinates

Find the mass of the ellipsoid Q given by 4x 2  4y 2  z 2  16, lying above the xy-plane. The density at a point in the solid is proportional to the distance between the point and the xy-plane.

4r 2

z

Solution The density function is r, , z  kz. The bounds on z are

4

0 ≤ z ≤ 16  4x 2  4y 2  16  4r 2 where 0 ≤ r ≤ 2 and 0 ≤ ≤ 2, as shown in Figure 14.65. The mass of the solid is m

      2

0 2

0

 x



2 2



y

.

Ellipsoid:

4x 2

+

4y 2

+

z2

= 16

k 2 k 2

kzr dz dr d

0 2

0 0 2 2



164r2

z2r

0

k 2



8r 2  r 4

0

2

dr d

16r  4r 3 dr d

0 0 2

 8k

Figure 14.65

164r2

2

2 0

d

d  16k.

0

Try It

Rotatable Graph

Exploration A

Integration in cylindrical coordinates is useful when factors involving x 2  y 2 appear in the integrand, as illustrated in Example 3.

Finding a Moment of Inertia

EXAMPLE 3 z

Find the moment of inertia about the axis of symmetry of the solid Q bounded by the paraboloid z  x 2  y 2 and the plane z  4, as shown in Figure 14.66. The density at each point is proportional to the distance between the point and the z-axis.

5

Solution Because the z-axis is the axis of symmetry, and x, y, z  kx 2  y 2, it follows that Iz 



kx 2  y 2x 2  y 2 dV.

Q

In cylindrical coordinates, 0 ≤ r ≤ x 2  y 2  z. So, you have

−2 1 2 x

. Figure 14.66

Rotatable Graph

.

      4

1

2

Q: Bounded by z = x2 + y2 z=4

y

Iz  k k k



z

0 0 0 4 2 5 r 0 0 4 2 0

k  5

2

0

r 2rr dr d dz

5

z

0

d dz

z52 d dz 5

4

z52 2 dz

0

2k 2 72 z 5 7

Try It





4 0



512k . 35

Exploration A

1038

CHAPTER 14

Multiple Integration

Triple Integrals in Spherical Coordinates z

Triple integrals involving spheres or cones are often easier to evaluate by converting to spherical coordinates. Recall from Section 11.7 that the rectangular conversion equations for spherical coordinates are

ρi sin φi Δθi

Δ ρi

x   sin  cos

ρi Δ φi

y   sin  sin

z   cos . In this coordinate system, the simplest region is a spherical block determined by

y

, , : 1 ≤  ≤ 2, 1 ≤ ≤ 2, 1 ≤  ≤ 2

x

where 1 ≥ 0, 2  1 ≤ 2, and 0 ≤ 1 ≤ 2 ≤ , as shown in Figure 14.67. If , ,  is a point in the interior of such a block, then the volume of the block can be approximated by V  2 sin     (see Exercise 17 in the Problem Solving exercises for this chapter). Using the usual process involving an inner partition, summation, and a limit, you can develop the following version of a triple integral in spherical coordinates for a continuous function f defined on the solid region Q.

Spherical block: . i  i 2 sin  i  i  i   i V Figure 14.67

Rotatable Graph



f x, y, z dV 

Q



2

1

2

1

2

1

f   sin  cos ,  sin  sin ,  cos 2 sin  d d d .

This formula can be modified for different orders of integration and generalized to include regions with variable boundaries. Like triple integrals in cylindrical coordinates, triple integrals in spherical coordinates are evaluated with iterated integrals. As with cylindrical coordinates, you can visualize a particular order of integration by viewing the iterated integral in terms of three sweeping motions—each adding another dimension to the solid. For instance, the iterated integral

  2

4

0

0

3

 2 sin  d d d

0

(which is used in Example 4) is illustrated in Figure 14.68. Cone: x2 + y2 = z2

Sphere: x2 + y2 + z2 = 9 ρ =3

z

z

z

θ φ

ρ 1 −2 2

1

2

y

x

.

 varies from 0 to 3 with  and held constant.

−2 2

1

2

x

 varies from 0 to  4 with held constant.

y

−2 2

1

2

y

x

varies from 0 to 2 .

Figure 14.68

Animation NOTE The Greek letter  used in spherical coordinates is not related to density. Rather, it is the three-dimensional analog of the r used in polar coordinates. For problems involving spherical coordinates and a density function, this text uses a different symbol to denote density.

SECTION 14.7

1039

Finding Volume in Spherical Coordinates

EXAMPLE 4 Upper nappe of cone: z2 = x2 + y2

Triple Integrals in Cylindrical and Spherical Coordinates

Find the volume of the solid region Q bounded below by the upper nappe of the cone z 2  x 2  y 2 and above by the sphere x 2  y 2  z 2  9, as shown in Figure 14.69.

z

Solution In spherical coordinates, the equation of the sphere is

3

2  x2  y 2  z 2  9

  3.

Furthermore, the sphere and cone intersect when −3

−2 3

2

1

1

x 2  y 2  z 2  z 2  z 2  9 2

3

y

x

Sphere: x2 + y2 + z2 = 9

3 2

and, because z   cos , it follows that

  . 4

3213  cos  .

z

Consequently, you can use the integration order d d d , where 0 ≤  ≤ 3, 0 ≤  ≤ 4, and 0 ≤ ≤ 2. The volume is

Figure 14.69

V

Rotatable Graph

         2

dV 

4

0

0

3

2 sin  d d d

0

Q



2

4

9

9 sin  d d

0

0

2

4

cos 

0

0

9

.

2

1

0

Try It

2

2

d

d  9 2  2   16.563.

Exploration A Finding the Center of Mass of a Solid Region

EXAMPLE 5

Find the center of mass of the solid region Q of uniform density, bounded below by the upper nappe of the cone z 2  x 2  y 2 and above by the sphere x 2  y 2  z 2  9. Solution Because the density is uniform, you can consider the density at the point x, y, z to be k. By symmetry, the center of mass lies on the z-axis, and you need only calculate z  Mxym, where m  kV  9k 2  2  from Example 4. Because z   cos , it follows that Mxy 



    2

3

kz dV  k

Q

0

0

3

2

k

0



k 4

4

3

0

3

0

 cos 2 sin  d d d

0

2

sin2  2

4



3 d d 

0

d d

0

k 2



3

3 d 

0

81k . 8

So, z .

Mxy 92  2  81k8    1.920 m 16 9k 2  2 

and the center of mass is approximately 0, 0, 1.92.

Try It

Exploration A

Open Exploration

1040

CHAPTER 14

Multiple Integration

Exercises for Section 14.7 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

20. Solid inside the sphere x 2  y 2  z 2  4 and above the upper nappe of the cone z 2  x 2  y 2

In Exercises 1–6, evaluate the iterated integral.

            4

1.

2

0 0 2

3.

r cos dr d dz

0 2 cos 2

0 0 2 

4.

0 2

5.

2

0

2r

rz dz dr d

0

Mass In Exercises 21 and 22, use cylindrical coordinates to find the mass of the solid Q.

4r 2

r sin dz dr d

0

21. Q  x, y, z: 0 ≤ z ≤ 9  x  2y, x 2  y 2 ≤ 4

2

e  2 d d d 3

0 0 4 cos 

0

2.

0

0

x, y, z  kx 2  y 2 2 2 22. Q  x, y, z: 0 ≤ z ≤ 12ex y , x 2  y 2 ≤ 4, x ≥ 0, y ≥ 0

 2 sin  d d d

0 0 0 4 4 cos

6.

  4

2

x, y, z  k

 2 sin  cos  d d d

In Exercises 23–28, use cylindrical coordinates to find the indicated characteristic of the cone shown in the figure.

0

In Exercises 7 and 8, use a computer algebra system to evaluate the iterated integral.

   4

7.

2

z

0 0 2

8.

0 

0

re r

sin

0

9.

     

11.



0

0

 2

2

r dz dr d

10.

0

3

0

r dz dr d

0

 2 sin  d d d

 2 sin  d d d

2

        2

0

15.

4

16.

16x2 y2

4x2

0

a

a2 x2

1x2

0

3

x dz dy dx

28. Moment of Inertia Assume that the density of the cone is

a

 x, y, z  kx 2  y 2

x 2  y 2  z 2 dz dy dx

0

and find the moment of inertia about the z-axis.

Volume In Exercises 17–20, use cylindrical coordinates to find the volume of the solid. 17. Solid inside both x 2  y 2  z 2  a 2 and x  a22  y2  a22 18. Solid inside

x2



26. Center of Mass Find the center of mass of the cone assuming that its density at any point is proportional to the distance between the point and the base. Use a computer algebra system to evaluate the triple integral.

Iz  10 mr02.

a a2 x2 y2

1x2 y2

25. Center of Mass Find the center of mass of the cone assuming that its density at any point is proportional to the distance between the point and the axis of the cone. Use a computer algebra system to evaluate the triple integral.

27. Moment of Inertia Assume that the cone has uniform density and show that the moment of inertia about the z-axis is

x 2  y 2 dz dy dx

0

a2 x2

0

x dz dy dx

x2 y2

a

1

Rotatable Graph

24. Centroid Find the centroid of the cone.

5

2 4x2

14.

x

23. Volume Find the volume of the cone.

0

4x2

2

y

3r 2

In Exercises 13–16, convert the integral from rectangular coordinates to both cylindrical and spherical coordinates, and evaluate the simplest iterated integral. 13.

r0

4

6

2

(

h

2 cos  2 d d d

0

2

0

12.

er

0

2

(

0

3

0

z = h 1 − rr 0

d dr dz

In Exercises 9–12, sketch the solid region whose volume is given by the iterated integral, and evaluate the iterated integral. 2

z

y2



z2

 16 and outside z 

x 2



y2

19. Solid bounded by the graphs of the sphere r 2  z 2  a 2 and the cylinder r  a cos

Moment of Inertia In Exercises 29 and 30, use cylindrical coordinates to verify the given formula for the moment of inertia of the solid of uniform density. 1 29. Cylindrical shell: Iz  2 ma 2  b2

0 < a ≤ r ≤ b,

0 ≤ z ≤ h 3

30. Right circular cylinder: Iz  2ma 2 r  2a sin , 0 ≤ z ≤ h Use a computer algebra system to evaluate the triple integral.

SECTION 14.7

Triple Integrals in Cylindrical and Spherical Coordinates

1041

Volume In Exercises 31 and 32, use spherical coordinates to find the volume of the solid.

Writing About Concepts

31. The torus given by   4 sin  (Use a computer algebra system to evaluate the triple integral.)

39. Give the equations for the coordinate conversion from rectangular to cylindrical coordinates and vice versa.

32. The solid between the spheres x 2  y 2  z 2  a 2 and x 2  y 2  z 2  b2, b > a, and inside the cone z 2  x 2  y 2

40. Give the equations for the coordinate conversion from rectangular to spherical coordinates and vice versa.

Mass In Exercises 33 and 34, use spherical coordinates to find the mass of the sphere x 2  y 2  z 2  a 2 with the given density. 33. The density at any point is proportional to the distance between the point and the origin. 34. The density at any point is proportional to the distance of the point from the z-axis. Center of Mass In Exercises 35 and 36, use spherical coordinates to find the center of mass of the solid of uniform density.

41. Give the iterated form of the triple integral  f x, y, z dV Q in cylindrical form. 42. Give the iterated form of the triple integral  f x, y, z dV Q in spherical form. 43. Describe the surface whose equation is a coordinate equal to a constant for each of the coordinates in (a) the cylindrical coordinate system and (b) the spherical coordinate system. 44. When evaluating a triple integral with constant limits of integration in the cylindrical coordinate system, you are integrating over a part of what solid? What is the solid when you are in spherical coordinates?

35. Hemispherical solid of radius r 36. Solid lying between two concentric hemispheres of radii r and R, where r < R

45. Find the “volume” of the “four-dimensional sphere” x 2  y 2  z 2  w2  a 2

Moment of Inertia In Exercises 37 and 38, use spherical coordinates to find the moment of inertia about the z-axis of the solid of uniform density. 37. Solid bounded by the hemisphere   cos , 4 ≤  ≤ 2, and the cone   4 38. Solid lying between two concentric hemispheres of radii r and R, where r < R

by evaluating

 a

a2 x2

16

0

0



a2 x2 y2

0



a2 x2 y2 z2

dw dz dy dx.

0

46. Use spherical coordinates to show that







x2 y2 z2

x2  y2  z2 e

  

dx dy dz  2.

1042

CHAPTER 14

Multiple Integration

Section 14.8

Change of Variables: Jacobians • Understand the concept of a Jacobian. • Use a Jacobian to change variables in a double integral.

CARL GUSTAV JACOBI (1804–1851) The Jacobian is named after the German mathematician Carl Gustav Jacobi. Jacobi is known for his work in many areas of mathematics, but his interest in integration stemmed from the problem of finding the circumference .of an ellipse.

Jacobians For the single integral



b

f x dx

a

you can change variables by letting x  gu, so that dx  g u du, and obtain



b

a

MathBio



d

f x dx 

f  g ug u du

c

where a  gc and b  gd . Note that the change-of-variables process introduces an additional factor g u into the integrand. This also occurs in the case of double integrals



f x, y dA 

R





f  gu, v, h u, v

S



!x !y !y !x du dv  !u !v !u !v Jacobian

where the change of variables x  gu, v and y  hu, v introduces a factor called the Jacobian of x and y with respect to u and v. In defining the Jacobian, it is convenient to use the following determinant notation.

Definition of the Jacobian If x  gu, v and y  hu, v, then the Jacobian of x and y with respect to u and v, denoted by !x, y!u, v, is

 

!x !x, y !u  !u, v !y !u

!x !v !x !y !y !x   . !u !v !u !v !y !v

The Jacobian for Rectangular-to-Polar Conversion

EXAMPLE 1

Find the Jacobian for the change of variables defined by x  r cos

and

 

y  r sin .

Solution From the definition of a Jacobian, you obtain

.

!x !x !x, y !r !

 !r,  !y !y !r !

cos r sin

 sin

r cos

2  r cos  r sin2

 r.

Try It



Exploration A

Exploration B

SECTION 14.8

θ



θ =β

r=a

S

α

f x, y dA 

R

r=b

 

f r cos , r sin  r dr d , r > 0

 

S



θ =α r

b

y

!x, y dr d

!r, 

f r cos , r sin 

S

a

1043

Example 1 points out that the change of variables from rectangular to polar coordinates for a double integral can be written as

T(r, θ ) = (r cos θ, r sin θ)

β

Change of Variables: Jacobians

where S is the region in the r -plane that corresponds to the region R in the xy-plane, as shown in Figure 14.70. This formula is similar to that found on page 1003. In general, a change of variables is given by a one-to-one transformation T from a region S in the uv-plane to a region R in the xy-plane, to be given by T u, v  x, y   gu, v, h u, v

θ =β

r=b

where g and h have continuous first partial derivatives in the region S. Note that the point u, v lies in S and the point x, y lies in R. In most cases, you are hunting for a transformation in which the region S is simpler than the region R.

R

r=a

θ =α x

x  2y  0,

Figure 14.70

x+ 4

3

y= 1

3

x

y=

x+

(− , )

y −2

=−

4 3

R

8 3

8 3

1

4 3

x

−2

1

2

3

2 3

1 3

xy1

Figure 14.71 v u=1

v=0 −1

2

The four boundaries for R in the xy-plane give rise to the following bounds for S in the uv-plane. Bounds in the uv-Plane

u=4 (4, 0) u 3

−1

v  4

The region S is shown in Figure 14.72. Note that the transformation T maps the vertices of the region S onto the vertices of the region R. For instance, T 1, 0  13 21  0 , 13 1  0   23, 13  1

−3

v = −4

. Region S in the uv-plane

u4 v0

T 4, 0  3 24  0 , 3 4  0   83, 43 

S

(1, −4)

u1

x  2y  0 x  2y  4

Region R in the xy-plane

Figure 14.72

and

1 y  u  v. 3

and

xy1 xy4

−2

(1 , 0)

1 x  2u  v 3

Bounds in the xy-Plane

(, )

−1

−5

x  y  4,

Solution To begin, let u  x  y and v  x  2y. Solving this system of equations for x and y produces T u, v  x, y, where

4

( , ) 2y = 0 x− (, )

2

−2

x  2y  4,

as shown in Figure 14.71. Find a transformation T from a region S to R such that S is a rectangular region (with sides parallel to the u- or v-axis). y

5 3

Finding a Change of Variables to Simplify a Region

Let R be the region bounded by the lines

S is the region in the r -plane that corresponds to R in the xy-plane.

2

EXAMPLE 2

(4, −4)

1

T 4, 4  13 24  4 , 13 4  4   43, 83  1 1 T 1, 4  3 21  4 , 3 1  4  2 5   3, 3  .

Try It

Exploration A

1044

CHAPTER 14

Multiple Integration

Change of Variables for Double Integrals THEOREM 14.5

Change of Variables for Double Integrals

Let R and S be regions in the xy- and uv-planes that are related by the equations x  gu, v and y  hu, v such that each point in R is the image of a unique point in S. If f is continuous on R, g and h have continuous partial derivatives on S, and !x, y!u, v is nonzero on S, then



f x, y dx dy 

R

v

(u, v + Δv)

 

f  g u, v, hu, v

S

!x, y du dv. !u, v

Proof Consider the case in which S is a rectangular region in the uv-plane with vertices u, v, u  u, v, u  u, v  v, and u, v  v, as shown in Figure 14.73. The images of these vertices in the xy-plane are shown in Figure 14.74. If u and v are small, the continuity of g and h implies that R is approximately a parallelogram determined by the vectors MN and MQ . So, the area of R is

(u + Δu, v + Δv)

S

\

\

A   MN

(u + Δu, v)

(u, v)



u



\

\

MQ .

Moreover, for small u and v, the partial derivatives of g and h with respect to u can be approximated by

Area of S  u v u > 0, v > 0

guu, v 

gu  u, v  gu, v u

huu, v 

hu  u, v  hu, v . u

Figure 14.73

and y

Q

P

Consequently, \

R M = (x, y)

N x

x = g(u, v) y = h(u, v)

The vertices in the xy-plane are M  g u, v, h u, v, N  g u  u, v, h u  u, v, P  g u  u, v  v , hu  u, v  v , and Q  g u, v  v , h u, v  v . Figure 14.74

MN  gu  u, v  gu, v i  hu  u, v  hu, v j  guu, v u i  huu, v u j !x !y  ui  uj. !u !u



\

Similarly, you can approximate MQ by i

j

k

!x !y vi  vj, which implies that !v !v

 

!x !u !y 0  !x !u u !v !y v 0 !v It follows that, in Jacobian notation, !x MN  MQ  !u u !x v !v \

\

\

A  MN



\

MQ  

 

!y !u u vk. !y !v

!x, y u v. !u, v

Because this approximation improves as u and v approach 0, the limiting case can be written as \

dA  MN



\

MQ  

 

!x, y du dv. !u, v

SECTION 14.8

Change of Variables: Jacobians

1045

The next two examples show how a change of variables can simplify the integration process. The simplification can occur in various ways. You can make a change of variables to simplify either the region R or the integrand f x, y, or both. EXAMPLE 3 y

x+ y=

x

=−

Let R be the region bounded by the lines

4

x  2y  0,

y=

4

x+

3

y −2

1

2

x−

R

= 2y

0

1 x

−2

1 −1 −2

Figure 14.75

2

Using a Change of Variables to Simplify a Region

x  2y  4,

x  y  4,

and

xy1

as shown in Figure 14.75. Evaluate the double integral



3xy dA.

R

3

Solution From Example 2, you can use the following change of variables. 1 x  2u  v 3

and

1 y  u  v 3

The partial derivatives of x and y are !x 2  , !u 3

!x 1  , !v 3

!y 1  , !u 3

1 !y  !v 3

and

  

which implies that the Jacobian is !x !x !x, y !u !v  !u, v !y !y !u !v 2 1 3 3  1 1 3 3 2 1   9 9 1  . 3 So, by Theorem 14.5, you obtain



3xy dA 

R

     S 4



1

4

1  9 1  9

.

Try It



1 4



1



0

uv 2 v 3  du 2 3 4 64 8u 2  8u  du 3 2u 2v 

8u 3

1  4u 2 9 3 164  . 9 

 

1 1 !x, y 2u  v u  v dv du 3 3 !u, v 0 1 2u 2  uv  v 2 dv du 9 4 3

 64  u 3

4

1

Open Exploration

1046

CHAPTER 14

Multiple Integration

Using a Change of Variables to Simplify an Integrand

EXAMPLE 4

Let R be the region bounded by the square with vertices 0, 1, 1, 2, 2, 1, and 1, 0. Evaluate the integral



x  y 2 sin2x  y dA.

R

Solution Note that the sides of R lie on the lines x  y  1, x  y  1, x  y  3, and x  y  1, as shown in Figure 14.76. Letting u  x  y and v  x  y, you can determine the bounds for region S in the uv-plane to be

−1

y

1

x−

y=

3

1 ≤ u ≤ 3

x−

y=

(1, 2)

2

as shown in Figure 14.77. Solving for x and y in terms of u and v produces x+

R

(0, 1)

1 x  u  v 2

y=

(2, 1)

3 2

x

3

x+

(1, 0)

1 ≤ v ≤ 1

and

1 y  u  v. 2

and

y= 1

The partial derivatives of x and y are !x 1  , !u 2

Region R in the xy-plane Figure 14.76

!x 1  , !v 2

!y 1  , !u 2

and

  

1 !y  !v 2

which implies that the Jacobian is v

1

u=1

(3, 1)

(1, 1) v=1

!x !x, y !u  !u, v !y !u

u=3

S u 1

−1

2

3

v = −1 (1, −1)

(3, −1)

!x !v !y !v

1 2  1 2

1 2 1  2

By Theorem 14.5, it follows that



x  y 2 sin2x  y dA 

R

1 1 1    . 4 4 2

    1

Figure 14.77

1 1

12 du dv u v  dv 3

u 2 sin2 v

1 1 1

1  2

Region S in the uv-plane

3

sin2

3 3 1

13 sin2v dv  3 1 1 13  1  cos 2v dv 6 1 1 13 1  v  sin 2v 6 2 1

  13 1 1   2  sin 2  sin2 6 2 2 

.

13 2  sin 2 6

 2.363.

Try It

Exploration A

In each of the change-of-variables examples in this section, the region S has been a rectangle with sides parallel to the u- or v-axis. Occasionally, a change of variables can be used for other types of regions. For instance, letting T u, v  x, 12 y changes the circular region u 2  v 2  1 to the elliptical region x 2   y 24  1.

SECTION 14.8

Exercises for Section 14.8 The symbol

.

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–8, find the Jacobian x, y /u, v for the indicated change of variables. 1. x 

1  2u

 v, y 

1 2 u

y

y

(−1, 1)

6

 v

4

2. x  au  bv, y  cu  dv

(3, 3)

2

−2

5. x  u cos  v sin , y  u sin  v cos

6. x  u  a, y  v  a 7. x 

sin v, y 

cos v

15.

16.

y

4

2

R

(2, 2)

2

(3, 0)

(4, 1) x

3

2

(0, 0) 2 3 4 5 6

In Exercises 11–16, use the indicated change of variables to evaluate the double integral. 4x2  y 2 dA

12.

R



 v

x

 v

y

(0, 1)

2



18.

(1, 2) (0, 1)

20.

(2, 1) x

1

1 −1

(0, −1)

y x  y dA

R

4 x

y sin xy dA

(1, 0) 1

14.

yv

17. f x, y  x  yexy

19.

(1, 0)

−1

y

In Exercises 17–22, use a change of variables to find the volume of the solid region lying below the surface z  f x, y and above the plane region R.

1 2 u  v  12u  v

y

y

13.

60xy dA

R

(−1, 0)

1 y , x

R

1

x

1

y  2x,

R: region lying between the graphs of xy  1, xy  4, y  1, y4

(6, 3)

3

x



u x , v

6 5

y

uv

R

(2, 3)

1 2 u 1 2 u

uv, y 

1 y  x, 4

y  13 u  v

y



Figure for 14

R: first-quadrant region lying between the graphs of

10. x  13 4u  v

y  3v

11.

−1

exy2 dA

x

9. x  3u  2v

(0, 0)



x

R

In Exercises 9 and 10, sketch the image S in the uv-plane of the region R in the xy-plane using the given transformations.

1

8

Figure for 13

u 8. x  , y  u  v v

3

(0, 0) (4, 0) 6

(0, 0) 1

−1

x

4. x  uv  2u, y  uv

eu

(1, 1)

(7, 3)

3. x  u  v 2, y  u  v

eu

1047

Change of Variables: Jacobians



x 2

21.

4x  yexy dA

R

xuv

x  12u  v

yu

y  2 u  v 1

22.

R: region bounded by the square with vertices 4, 0, 6, 2, 4, 4, 2, 2 f x, y  x  y2 sin2x  y R: region bounded by the square with vertices , 0, 32, 2, , , 2, 2 f x, y  x  yx  4y R: region bounded by the parallelogram with vertices 0, 0, 1, 1, 5, 0, 4, 1 f x, y  3x  2y2y  x32 R: region bounded by the parallelogram with vertices 0, 0, 2, 3, 2, 5, 4, 2 f x, y  x  y R: region bounded by the triangle with vertices 0, 0, a, 0, 0, a, where a > 0 xy f x, y  1  x 2y 2 R: region bounded by the graphs of xy  1, xy  4, x  1, x  4 Hint: Let x  u, y  vu.

1048

CHAPTER 14

Multiple Integration

23. Consider the region R in the xy-plane bounded by the ellipse x2 a2



y2 b2

1

and the transformations x  au and y  bv. (a) Sketch the graph of the region R and its image S under the given transformation. (b) Find

!x, y . !u, v

(c) Find the area of the ellipse. 24. Use the result of Exercise 23 to find the volume of each domeshaped solid lying below the surface z  f x, y and above the elliptical region R. (Hint: After making the change of variables given by the results in Exercise 23, make a second change of variables to polar coordinates.) x2 y2  ≤ 1 (a) f x, y  16  x 2  y 2; R: 16 9



 (b) f x, y  A cos 2



In Exercises 27–30, find the Jacobian x, y, z/u, v, w for the indicated change of variables. If x  f u, v, w , y  gu, v, w, and z  hu, v, w, then the Jacobian of x, y, and z with respect to u, v, and w is

 

x u y x, y, z  u u, v, w z u

x v y v z v

x w y . w z w

27. x  u1  v, y  uv1  w, z  uvw 28. x  4u  v, y  4v  w, z  u  w 29. Spherical Coordinates x   sin  cos , y   sin  sin , z   cos  30. Cylindrical Coordinates x  r cos , y  r sin , z  z

x2 y2 x2 y2  2 ≤ 1 2  2 ; R: 2 a b a b

Writing About Concepts 25. State the definition of the Jacobian. 26. Describe how to use the Jacobian to change variables in double integrals.

Putnam Exam Challenge 31. Let A be the area of the region in the first quadrant bounded by the line y  12 x, the x-axis, and the ellipse 19 x2  y2  1. Find the positive number m such that A is equal to the area of the region in the first quadrant bounded by the line y  mx, the y-axis, and the ellipse 19 x2  y2  1. This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.

1048

CHAPTER 14

Multiple Integration

Review Exercises for Chapter 14 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

10. Region bounded by the graphs of y  6x  x 2 and y  x 2  2x

In Exercises 1 and 2, evaluate the integral.



x2

1.



2y

x ln y dy

2.

x 2  y 2 dx

y

1

In Exercises 3–6, evaluate the iterated integral. Change the coordinate system when convenient.

  1

3.

0

3

5.

0

1x

  2

3x  2y dy dx

4.

0

0

9x2

0

6.

0

12. Region bounded by the graphs of x  y 2  1, x  0, y  0, and y  2 13. Region bounded by the graphs of x  y  3 and x  y 2  1 14. Region bounded by the graphs of x  y and x  2y  y 2

2x

x 2  2y dy dx

x2

3

4x dy dx

11. Region enclosed by the graph of y 2  x 2  x 4

2 4y2

Think About It In Exercises 15 and 16, give a geometric argument for the given equality. Verify the equality analytically.

dx dy

2 4y2



x  y dx dy 





1

15.

0

Area



22y 2

2

f x, y dA

2

16.

0

for both orders of integration. Compute the area of R by letting f x, y  1 and integrating. 7. Triangle: vertices 0, 0, 3, 0, 0, 1 8. Triangle: vertices 0, 0, 3, 0, 2, 2 9. The larger area between the graphs of x 2  y 2  25 and x  3

5y

3y2

3

e xy dx dy 

0

0 22

2 2x3

0

x2

x  y dy dx 

0

2y

In Exercises 7–14, write the limits for the double integral

R

  

8x22

x  y dy dx

0

 5

e xy dy dx 

3

5x

e xy dy dx

0

Volume In Exercises 17 and 18, use a multiple integral and a convenient coordinate system to find the volume of the solid. 17. Solid bounded by the graphs of z  x 2  y  4, z  0, y  0, x  0, and x  4

REVIEW EXERCISES

18. Solid bounded by the graphs of z  x  y, z  0, x  0, x  3, and y  x Approximation In Exercises 19 and 20, determine which value best approximates the volume of the solid between the xy-plane and the function over the region. (Make your selection on the basis of a sketch of the solid and not by performing any calculations.) 19. f x, y  x  y 9 2

(b) 5

20. f x, y 

(c) 13

(e) 100

(d) 100

10x 2y 2 (b) 15

(c)

2 3

(d) 3

(e) 15

Probability In Exercises 21 and 22, find k such that the function is a joint density function and find the required probability, where

 d

Pa ≤ x ≤ b, c ≤ y ≤ d 

c

21. f x, y 



kxyexy, 0,

29. Solid bounded by the graphs of z  0 and z  h, outside the cylinder x 2  y 2  1 and inside the hyperboloid x2  y2  z2  1 30. Solid that remains after drilling a hole of radius b through the center of a sphere of radius R b < R

x 2  y 22  9x 2  y 2. (a) Convert the equation to polar coordinates. Use a graphing utility to graph the equation.

R: circle bounded by x 2  y 2  1 (a) 

Volume In Exercises 29 and 30, use a multiple integral and a convenient coordinate system to find the volume of the solid.

31. Consider the region R in the xy-plane bounded by the graph of the equation

R: triangle with vertices 0, 0, 3, 0, 3, 3 (a)

1049

b

f x, y dx dy.

a

(b) Use a double integral to find the area of the region R. (c) Use a computer algebra system to determine the volume of the solid over the region R and beneath the hemisphere z  9  x 2  y 2 . 32. Combine the sum of the two iterated integrals into a single iterated integral by converting to polar coordinates. Evaluate the resulting iterated integral.

  813

x ≥ 0, y ≥ 0 elsewhere

3x2

0

0

  4

xy dy dx 

813

16x2

xy dy dx

0

P0 ≤ x ≤ 1, 0 ≤ y ≤ 1 22. f x, y 

kxy, 0,

0 ≤ x ≤ 1, 0 ≤ y ≤ x elsewhere

Mass and Center of Mass In Exercises 33 and 34, find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density or densities. Use a computer algebra system to evaluate the multiple integrals.

P0 ≤ x ≤ 0.5, 0 ≤ y ≤ 0.25 True or False? In Exercises 23 – 26, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

 b

23.

a

d





b

f xg y dy dx 

c

d

f x dx

a



g y dy

c

24. If f is continuous over R1 and R2, and



dA 

R1



dA

R2

then

  

f x, y dA 

R1 1

25.

1

26.

0

0



f x, y dA.

 1

cosx 2  y 2 dx dy  4

0

1

cos x 2  y 2 dx dy

0

 h

0



In Exercises 35 and 36, find Ix, Iy, I0, x, and y for the lamina bounded by the graphs of the equations. Use a computer algebra system to evaluate the double integrals.

Surface Area In Exercises 37 and 38, find the area of the surface given by z  f x, y over the region R. 37. f x, y  16  x 2  y 2 38. f x, y  16  x  y 2 R  x, y: 0 ≤ x ≤ 2, 0 ≤ y ≤ x

x

0



h x x2 2   2 ,   k, first quadrant 2 L L

R  x, y: x 2  y 2 ≤ 16

 1 dx dy < 1  x2  y2 4

In Exercises 27 and 28, evaluate the iterated integral by converting to polar coordinates. 27.

34. y 

(b)   kx 2  y 2

36. y  4  x 2, y  0, x > 0,   ky

1

1

(a)   kxy

35. y  0, y  b, x  0, x  a,   kx

R2

1 1

33. y  2x, y  2x 3, first quadrant

 4

x 2  y 2 dy dx

28.

0

16y2

0

x 2  y 2 dx dy

Use a computer algebra system to evaluate the integral. 39. Surface Area Find the area of the surface of the cylinder f x, y  9  y 2 that lies above the triangle bounded by the graphs of the equations y  x, y  x, and y  3.

1050

CHAPTER 14

Multiple Integration

40. Surface Area The roof over the stage of an open air theater at a theme park is modeled by



f x, y  25 1  ex

2 y 21000

y x 1000  2

cos 2

2

55. Investigation Consider a spherical segment of height h from a sphere of radius a, where h ≤ a, and constant density x, y, z  k (see figure).

where the stage is a semicircle bounded by the graphs of y  50 2  x 2 and y  0.

h

(a) Use a computer algebra system to graph the surface. (b) Use a computer algebra system to approximate the number of square feet of roofing required to cover the surface. In Exercises 41– 44, evaluate the iterated integral.

       9x2

3

41.

3 9x2

42.

x2 y2

2 4x2 a

43.

5

0

(b) Find the centroid of the solid.



x2



 dz dy dx

y2

0

c

25x2 y2

0

0

   

1x2 y2

1x2

1 1x2  2

0

(d) Find lim z. h→0

0

25x2

1

46.

(c) Use the result of part (b) to find the centroid of a hemisphere of radius a.

0

4x2

0

1x2 y2

(e) Find Iz.

1 dz dy dx 1  x2  y2  z2

In Exercises 45 and 46, use a computer algebra system to evaluate the iterated integral. 45.

(a) Find the volume of the solid.

x 2  y 2  z 2 dx dy dz

0

44.

b

x 2  y 2 dz dy dx

x 2y22

4x2

2

Rotatable Graph

9

x 2  y 2 dz dy dx

(f) Use the result of part (e) to find Iz for a hemisphere. 56. Moment of Inertia Find the moment of inertia about the z2 z-axis of the ellipsoid x 2  y 2  2  1, where a > 0. a In Exercises 57 and 58, give a geometric interpretation of the iterated integral.

xyz dz dy dx

57.

0 

0

Volume In Exercises 47 and 48, use a multiple integral to find the volume of the solid. 47. Solid inside the graphs of r  2 cos and r 2  z 2  4 48. Solid inside the graphs of r 2  z  16, z  0, and r  2 sin

Center of Mass In Exercises 49–52, find the center of mass of the solid of uniform density bounded by the graphs of the equations. 49. Solid inside the hemisphere   cos , 4 ≤  ≤ 2, and outside the cone   4 50. Wedge:

x2



y2



a 2,

z  cy c > 0, y ≥ 0, z ≥ 0

58.

53. The solid of uniform density inside the paraboloid z  16  x 2  y 2, and outside the cylinder x 2  y 2  9, z ≥ 0. 54. x 2  y 2  z 2  a 2, density proportional to the distance from the center

0 2

 2 sin  d d d

0 1r 2

r dz dr d

0

59. x  u  3v,

y  2u  3v

60. x  u2  v2,

y  u2  v2

In Exercises 61 and 62, use the indicated change of variables to evaluate the double integral. 61.



lnx  y dA

62.

R

 R

1 x  u  v, 2

2

Moment of Inertia In Exercises 53 and 54, find the moment of inertia Iz of the solid of given density.

6 sin 

In Exercises 59 and 60, find the Jacobian x, y/u, v for the indicated change of variables.

52. x  y  z  25, z  4 (the larger solid) 2



0

0

51. x 2  y 2  z 2  a 2, first octant 2

    2

4x2 y2

1 y  u  v 2

x dA 1  x2y2

x  u,

y

y

v u

y 6

4

(2, 3)

x=1

5

3

4

R

2

(1, 2) 1

3

(3, 2)

2 1

(2, 1)

xy = 5 R

x=5

x

1

2

3

4

x

1 xy = 1 4

5

6

P.S.

P.S.

Problem Solving

1051

Problem Solving

The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

1. (a) Find the volume of the solid of intersection of the three cylinders x 2  z 2  1, y 2  z 2  1, and x 2  y 2  1 (see figure). z

verify that −3 3

y

3

x

3

3

y





2 2 dv du  . 2 2 2u v 9

prove that −3

Rotatable Graph (b) Use the Monte Carlo Method (see Section 4.2 exercises) to confirm the answer in part (a). (Hint: Generate random points inside the cube of volume 8 centered at the origin.) 2. Let a, b, c, and d be positive real numbers. The first octant of the plane ax  by  cz  d is shown in the figure. Show that the surface area of this portion of the plane is equal to AR a 2  b 2  c 2 c



1  2 n1 n



 1

0

1

0

1 dx dy. 1  xy

yx xy and v  to 2 2 1 2 1 dx dy  I1  I2  . 1  xy 6 0

(g) Use the change of variables u 

x

−3

22 u 2

1 sin

2 

 tan . cos

2

(f) Use the formula for the sum of an infinite geometric series to

3

−3

 

u 2

2

(e) Prove that I2 

z

3

(d) Prove the trigonometric identity



1 2  n n1



 1

0

4. Consider a circular lawn with a radius of 10 feet, as shown in the figure. Assume that a sprinkler distributes water in a radial fashion according to the formula f r 

r r2  16 160

(measured in cubic feet of water per hour per square foot of lawn), where r is the distance in feet from the sprinkler. Find the amount of water that is distributed in 1 hour in the following two annular regions. A  r, : 4 ≤ r ≤ 5, 0 ≤ ≤ 2}

where AR is the area of the triangular region R in the xy-plane, as shown in the figure.

B  r, : 9 ≤ r ≤ 10, 0 ≤ ≤ 2} Is the distribution of water uniform? Determine the amount of water the entire lawn receives in 1 hour.

z

1 ft B

A 4 ft

R

y

x

Rotatable Graph 3. Derive Euler’s famous result that was mentioned in Section 9.3, 1 2 , by completing each step. 2  6 n1 n



(a) Prove that



1 dv v   C. arctan 2  u 2  v 2 2  u 2 2  u 2

y

y = 13 x 2

  22

(b) Prove that I1 

5. The figure shows the region R bounded by the curves y  x, x2 x2 y  2x, y  , and y  . Use the change of variables 3 4 x  u13 v23 and y  u23 v13 to find the area of the region R.

0

u

2 2 by 2 2 dv du  2  u  v 18 u

using the substitution u  2 sin . (c) Prove that

   2

I2 

22

4

2

arctan

6

y=

2x

y=

x

R

u 2

u 2

y = 14 x 2

2 dv du 2  u2  v2

1  sin

d

cos

by using the substitution u  2 sin .

x

1052

CHAPTER 14

Multiple Integration

6. The figure shows a solid bounded below by the plane z  2 and above by the sphere x 2  y 2  z 2  8.

A(b)

z 4

x2 + y2 + z2 = 8 V(b)

A(a) V(a) −2 2

y

2

x

Figure for 12

Rotatable Graph

Rotatable Graph (a) Find the volume of the solid using cylindrical coordinates. (b) Find the volume of the solid using spherical coordinates. 7. Sketch the solid whose volume is given by the sum of the iterated integrals

  6

0

3

y

  6

dx dy dz 

z2 z2

0

12z2 6y

3

dx dy dz.

13. The angle between a plane P and the xy-plane is , where 0 ≤ < 2. The projection of a rectangular region in P onto the xy-plane is a rectangle whose sides have lengths  x and y, as shown in the figure. Prove that the area of the rectangular region in P is sec x y.

z2

Area: sec θ Δx Δy

Then write the volume as a single iterated integral in the order dy dz dx.

 1

8. Prove that lim

n→

0

P

1

xn

yn

dx dy  0.

0

In Exercises 9 and 10, evaluate the integral. (Hint: See Exercise 55 in Section 14.3.)

 

9.

θ Δx

x 2ex dx 2

Area in xy-plane: ΔxΔy

0

1

10.

ln

0

1 dx x

Rotatable Graph

11. Consider the function f x, y 

kexya, 0,



Δy

x ≥ 0, y ≥ 0 elsewhere.

Find the relationship between the positive constants a and k such that f is a joint density function of the continuous random variables x and y. 12. From 1963 to 1986, the volume of the Great Salt Lake approximately tripled while its top surface area approximately doubled. Read the article “Relations between Surface Area and Volume in Lakes” by Daniel Cass and Gerald Wildenberg in The College Mathematics Journal. Then give examples of solids that have “water levels” a and b such that Vb  3Va and Ab  2Aa (see figure), where V is volume and A is area.

14. Use the result of Exercise 13 to order the planes in ascending order of their surface areas for a fixed region R in the xy-plane. Explain your ordering without doing any calculations. (a) z1  2  x (b) z2  5 (c) z3  10  5x  9y (d) z4  3  x  2y





15. Evaluate the integral

0

0

16. Evaluate the integrals

 1

0

1

0

xy dx dy and x  y3

1 dx dy. 1  x2  y22

 1

0

1

0

xy dy dx. x  y3

Are the results the same? Why or why not? 17. Show that the volume of a spherical block can be approximated by V  2 sin     .

1054

CHAPTER 15

Vector Analysis

Section 15.1

Vector Fields • • • •

Understand the concept of a vector field. Determine whether a vector field is conservative. Find the curl of a vector field. Find the divergence of a vector field.

Vector Fields In Chapter 12, you studied vector-valued functions—functions that assign a vector to a real number. There you saw that vector-valued functions of real numbers are useful in representing curves and motion along a curve. In this chapter, you will study two other types of vector-valued functions—functions that assign a vector to a point in the plane or a point in space. Such functions are called vector fields, and they are useful in representing various types of force fields and velocity fields.

Definition of Vector Field Let M and N be functions of two variables x and y, defined on a plane region R. The function F defined by Fx, y  Mi  Nj NOTE Although a vector field consists of infinitely many vectors, you can get a good idea of what the vector field looks like by sketching several representative vectors Fx, y whose initial points are x, y.

Plane

is called a vector field over R. Let M, N, and P be functions of three variables x, y, and z, defined on a solid region Q in space. The function F defined by Fx, y, z  Mi  Nj  Pk

Space

is called a vector field over Q. From this definition you can see that the gradient is one example of a vector field. For example, if f x, y  x 2  y 2 then the gradient of f "f x, y  fxx, yi  fyx, yj  2xi  2yj

Vector field in the plane

is a vector field in the plane. From Chapter 13, the graphical interpretation of this field is a family of vectors, each of which points in the direction of maximum increase along the surface given by z  f x, y. For this particular function, the surface is a paraboloid and the gradient tells you that the direction of maximum increase along the surface is the direction given by the ray from the origin through the point x, y. Similarly, if f x, y, z  x 2  y 2  z 2 then the gradient of f "f x, y, z  fxx, y, zi  fyx, y, zj  fzx, y, zk  2xi  2yj  2zk

Vector field in space

is a vector field in space. A vector field is continuous at a point if each of its component functions M, N, and P is continuous at that point.

SECTION 15.1

Vector Fields

1055

Some common physical examples of vector fields are velocity fields, gravitational fields, and electric force fields. Velocity field

1. Velocity fields describe the motions of systems of particles in the plane or in space. For instance, Figure 15.1 shows the vector field determined by a wheel rotating on an axle. Notice that the velocity vectors are determined by the locations of their initial points—the farther a point is from the axle, the greater its velocity. Velocity fields are also determined by the flow of liquids through a container or by the flow of air currents around a moving object, as shown in Figure 15.2. 2. Gravitational fields are defined by Newton’s Law of Gravitation, which states that the force of attraction exerted on a particle of mass m 1 located at x, y, z by a particle of mass m 2 located at 0, 0, 0 is given by Fx, y, z 

Rotating wheel

Gm 1m 2 u x2  y 2  z2

where G is the gravitational constant and u is the unit vector in the direction from the origin to x, y, z. In Figure 15.3, you can see that the gravitational field F has the properties that Fx, y, z always points toward the origin, and that the magnitude of Fx, y, z is the same at all points equidistant from the origin. A vector field with these two properties is called a central force field. Using the position vector

Figure 15.1

r  xi  yj  zk for the point x, y, z, you can write the gravitational field F as Gm 1m 2 r r2 r Gm 1m 2  u. r2

 

Fx, y, z  Air flow vector field Figure 15.2

3. Electric force fields are defined by Coulomb’s Law, which states that the force exerted on a particle with electric charge q1 located at x, y, z by a particle with electric charge q2 located at 0, 0, 0 is given by

z

(x, y, z)

Fx, y, z  y

cq1q2 u r2

where r  xi  yj  zk, u  rr, and c is a constant that depends on the choice of units for r, q1, and q2. Note that an electric force field has the same form as a gravitational field. That is,

x

m1 is located at (x, y, z). m2 is located at (0, 0, 0).

Gravitational force field Figure 15.3

Fx, y, z 

k u. r2

Such a force field is called an inverse square field.

Definition of Inverse Square Field Let rt  xti  ytj  ztk be a position vector. The vector field F is an inverse square field if Fx, y, z 

k u r2

where k is a real number and u  rr is a unit vector in the direction of r.

1056

CHAPTER 15

Vector Analysis

Because vector fields consist of infinitely many vectors, it is not possible to create a sketch of the entire field. Instead, when you sketch a vector field, your goal is to sketch representative vectors that help you visualize the field. EXAMPLE 1

Sketching a Vector Field

Sketch some vectors in the vector field given by Fx, y  yi  xj. Solution You could plot vectors at several random points in the plane. However, it is more enlightening to plot vectors of equal magnitude. This corresponds to finding level curves in scalar fields. In this case, vectors of equal magnitude lie on circles. F  c  y2  c x2  y 2  c2

y 3 2

Equation of circle

To begin making the sketch, choose a value for c and plot several vectors on the resulting circle. For instance, the following vectors occur on the unit circle.

1 x 1

3

Vector field: F(x, y) = −yi + xj

.

Vectors of length c

x 2

Point

Vector

1, 0 0, 1 1, 0 0, 1

F1, 0  j F0, 1  i F1, 0  j F0, 1  i

These and several other vectors in the vector field are shown in Figure 15.4. Note in the figure that this vector field is similar to that given by the rotating wheel shown in Figure 15.1.

Figure 15.4

Try It EXAMPLE 2

Exploration A Sketching a Vector Field

Sketch some vectors in the vector field given by Fx, y  2xi  yj. Solution For this vector field, vectors of equal length lie on ellipses given by

y

F  2x2   y2  c

4

which implies that

c=2 c=1

−4

−3

4x 2  y 2  c 2. x

−2

2

3

For c  1, sketch several vectors 2xi  yj of magnitude 1 at points on the ellipse given by 4x 2  y 2  1.

−4

4x 2  y 2  4.

. Vector field: F(x, y) = 2xi + yj

Figure 15.5

For c  2, sketch several vectors 2xi  yj of magnitude 2 at points on the ellipse given by

These vectors are shown in Figure 15.5.

Try It

Exploration A

SECTION 15.1

EXAMPLE 3 z

Vector Fields

1057

Sketching a Velocity Field

Sketch some vectors in the velocity field given by vx, y, z  16  x 2  y 2k

16

where x 2  y 2 ≤ 16. Solution You can imagine that v describes the velocity of a liquid flowing through a tube of radius 4. Vectors near the z-axis are longer than those near the edge of the tube. For instance, at the point 0, 0, 0, the velocity vector is v0, 0, 0  16k, whereas at the point 0, 3, 0, the velocity vector is v0, 3, 0  7k. Figure 15.6 shows these and several other vectors for the velocity field. From the figure, you can see that the speed of the liquid is greater near the center of the tube than near the edges of the tube.

.

Try It

Exploration A

Conservative Vector Fields

4

4

x

.

Velocity field: v(x, y, z) = (16 − x 2 − y 2)k

Figure 15.6

y

Notice in Figure 15.5 that all the vectors appear to be normal to the level curve from which they emanate. Because this is a property of gradients, it is natural to ask whether the vector field given by Fx, y  2xi  yj is the gradient for some differentiable function f. The answer is that some vector fields can be represented as the gradients of differentiable functions and some cannot—those that can are called conservative vector fields.

Definition of Conservative Vector Field A vector field F is called conservative if there exists a differentiable function f such that F  "f. The function f is called the potential function for F.

Rotatable Graph EXAMPLE 4

Conservative Vector Fields

a. The vector field given by Fx, y  2xi  yj is conservative. To see this, consider the potential function f x, y  x 2  12 y 2. Because "f  2xi  yj  F it follows that F is conservative. b. Every inverse square field is conservative. To see this, let Fx, y, z 

k u and r2

f x, y, z 

k x 2  y 2  z 2

where u  rr. Because kx ky kz i 2 j 2 k x2  y2  z232 x  y2  z232 x  y2  z232 xi  yj  zk k  2 x  y 2  z 2 x 2  y 2  z 2 k r  r2 r k u  r2

"f 



.

it follows that F is conservative.

Try It

Exploration A



1058

CHAPTER 15

Vector Analysis

As can be seen in Example 4(b), many important vector fields, including gravitational fields and electric force fields, are conservative. Most of the terminology in this chapter comes from physics. For example, the term “conservative” is derived from the classic physical law regarding the conservation of energy. This law states that the sum of the kinetic energy and the potential energy of a particle moving in a conservative force field is constant. (The kinetic energy of a particle is the energy due to its motion, and the potential energy is the energy due to its position in the force field.) The following important theorem gives a necessary and sufficient condition for a vector field in the plane to be conservative.

THEOREM 15.1

Test for Conservative Vector Field in the Plane

Let M and N have continuous first partial derivatives on an open disk R. The vector field given by Fx, y  Mi  Nj is conservative if and only if !N !M  . !x !y Proof To prove that the given condition is necessary for F to be conservative, suppose there exists a potential function f such that Fx, y  "f x, y  Mi  Nj. Then you have fxx, y  M

fxyx, y 

!M !y

fy x, y  N

fyx x, y 

!N !x

and, by the equivalence of the mixed partials fxy and fyx, you can conclude that !N!x  !M!y for all x, y in R. The sufficiency of the condition is proved in Section 15.4. NOTE Theorem 15.1 requires that the domain of F be an open disk. If R is simply an open region, the given condition is necessary but not sufficient to produce a conservative vector field.

EXAMPLE 5

Testing for Conservative Vector Fields in the Plane

Decide whether the vector field given by F is conservative. a. Fx, y  x 2yi  xyj

b. Fx, y  2xi  yj

Solution a. The vector field given by Fx, y  x 2yi  xyj is not conservative because !M !  x 2y  x 2 and !y !y

!N !  xy  y. !x !x

b. The vector field given by Fx, y  2xi  yj is conservative because .

!M !  2x  0 !y !y

Try It

and

!N !  y  0. !x !x

Exploration A

SECTION 15.1

Vector Fields

1059

Theorem 15.1 tells you whether a vector field is conservative. It does not tell you how to find a potential function of F. The problem is comparable to antidifferentiation. Sometimes you will be able to find a potential function by simple inspection. For instance, in Example 4 you observed that f x, y  x 2 

1 2 y 2

has the property that "f x, y  2xi  yj.

Finding a Potential Function for Fx, y

EXAMPLE 6

Find a potential function for Fx, y  2xyi  x 2  yj. Solution From Theorem 15.1 it follows that F is conservative because !

2xy  2x !y

and

! 2

x  y  2x. !x

If f is a function whose gradient is equal to Fx, y, then "f x, y  2xyi  x 2  yj which implies that fxx, y  2xy and fyx, y  x 2  y. To reconstruct the function f from these two partial derivatives, integrate fxx, y with respect to x and fyx, y with respect to y, as follows. f x, y  f x, y 

 

fxx, y dx  fyx, y dy 

 

2xy dx  x 2y  g y

x 2  y dy  x 2 y 

y2  hx 2

Notice that g y is constant with respect to x and hx) is constant with respect to y. To find a single expression that represents f x, y, let g y  

y2 2

and

hx  K.

Then, you can write f x, y  x 2 y  g y  K y2  K.  x2 y  2 .

You can check this result by forming the gradient of f. You will see that it is equal to the original function F.

Try It

Exploration A

Exploration B

NOTE Notice that the solution in Example 6 is comparable to that given by an indefinite integral. That is, the solution represents a family of potential functions, any two of which differ by a constant. To find a unique solution, you would have to be given an initial condition satisfied by the potential function.

1060

CHAPTER 15

Vector Analysis

Curl of a Vector Field Theorem 15.1 has a counterpart for vector fields in space. Before stating that result, the definition of the curl of a vector field in space is given.

Definition of Curl of a Vector Field The curl of Fx, y, z  Mi  Nj  Pk is curl Fx, y, z  "  Fx, y, z !P !N !P !M !N !M   i  j  k. !y !z !x !z !x !y



NOTE

 

 



If curl F  0, then F is said to be irrotational.

The cross product notation used for curl comes from viewing the gradient "f as the result of the differential operator " acting on the function f. In this context, you can use the following determinant form as an aid in remembering the formula for curl.

 

curl Fx, y, z  "  Fx, y, z i j k 

! ! !x !y

! !z

M N P !P !N !P !M !N !M  i  j  k  !y !z !x !z !x !y



EXAMPLE 7

 

 



Finding the Curl of a Vector Field

Find curl F for the vector field given by Fx, y, z  2xyi  x 2  z 2j  2yzk. Is F irrotational? Solution The curl of F is given by



curl Fx, y, z  "

Fx, y, z i j 

k



! ! !  !x !z !y 2xy x 2  z 2 2yz ! ! ! !y !z !x  i x2  z2 2yz 2xy



 

! ! !z j  !x 2yz 2xy  2z  2zi  0  0j  2x  2xk .

 0.

Because curl F  0, F is irrotational.

Try It

Open Exploration



! !y x z 2

k

2

SECTION 15.1

Vector Fields

1061

Later in this chapter, you will assign a physical interpretation to the curl of a vector field. But for now, the primary use of curl is shown in the following test for conservative vector fields in space. The test states that for a vector field whose domain is all of three-dimensional space (or an open sphere), the curl is 0 at every point in the domain if and only if F is conservative. The proof is similar to that given for Theorem 15.1.

THEOREM 15.2

Test for Conservative Vector Field in Space

Suppose that M, N, and P have continuous first partial derivatives in an open sphere Q in space. The vector field given by Fx, y, z  Mi  Nj  Pk is conservative if and only if curl Fx, y, z  0. That is, F is conservative if and only if !P !N  , !y !z

!P !M  , and !x !z

!N !M  . !x !y

From Theorem 15.2, you can see that the vector field given in Example 7 is conservative because curl Fx, y, z  0. Try showing that the vector field Fx, y, z  x 3y 2zi  x 2zj  x 2yk is not conservative—you can do this by showing that its curl is curl Fx, y, z  x3y 2  2xyj  2xz  2x 3yzk  0. For vector fields in space that pass the test for being conservative, you can find a potential function by following the same pattern used in the plane (as demonstrated in Example 6). EXAMPLE 8 NOTE Examples 6 and 8 are illustrations of a type of problem called recovering a function from its gradient. If you go on to take a course in differential equations, you will study other methods for solving this type of problem. One popular method gives an interplay between successive “partial integrations” and partial differentiations.

Finding a Potential Function for Fx, y, z

Find a potential function for Fx, y, z  2xyi  x 2  z 2j  2yzk. Solution From Example 7, you know that the vector field given by F is conservative. If f is a function such that Fx, y, z  "f x, y, z, then fxx, y, z  2xy,

fyx, y, z  x 2  z 2, and

fzx, y, z  2yz

and integrating with respect to x, y, and z separately produces f x, y, z  f x, y, z  f x, y, z 

  

M dx  N dy  P dz 

  

2xy dx  x 2 y  g y, z

x 2  z 2 dy  x 2 y  yz 2  hx, z

2yz dz  yz 2  kx, y.

Comparing these three versions of f x, y, z, you can conclude that g y, z  yz 2  K, .

hx, z  K, and

So, f x, y, z is given by f x, y, z  x 2 y  yz 2  K.

Try It

Exploration A

kx, y  x 2 y  K.

1062

CHAPTER 15

Vector Analysis

Divergence of a Vector Field NOTE Divergence can be viewed as a type of derivative of F in that, for vector fields representing velocities of moving particles, the divergence measures the rate of particle flow per unit volume at a point. In hydrodynamics (the study of fluid motion), a velocity field that is divergence free is called incompressible. In the study of electricity and magnetism, a vector field that is divergence free is called solenoidal.

You have seen that the curl of a vector field F is itself a vector field. Another important function defined on a vector field is divergence, which is a scalar function.

Definition of Divergence of a Vector Field The divergence of Fx, y  Mi  Nj is div Fx, y  "  Fx, y 

!M !N  . !x !y

Plane

The divergence of Fx, y, z  Mi  Nj  Pk is div Fx, y, z  "

 Fx, y, z 

!M !N !P   . !x !y !z

Space

If div F  0, then F is said to be divergence free. The dot product notation used for divergence comes from considering " as a differential operator, as follows. "  Fx, y, z  

EXAMPLE 9

 !x!  i  !y!  j  !z! k  Mi  Nj  Pk !M !N !P   !x !y !z

Finding the Divergence of a Vector Field

Find the divergence at 2, 1, 1 for the vector field Fx, y, z  x3y 2zi  x 2zj  x 2yk. Solution The divergence of F is div Fx, y, z  .

! 3 2 ! !

x y z  x 2z  x 2y  3x 2y 2z. !x !y !z

At the point 2, 1, 1, the divergence is div F2, 1, 1  322121  12.

Try It

Exploration A

There are many important properties of the divergence and curl of a vector field F (see Exercises 77– 83). One that is used often is described in Theorem 15.3. You are asked to prove this theorem in Exercise 84.

THEOREM 15.3

Relationship Between Divergence and Curl

If Fx, y, z  Mi  Nj  Pk is a vector field and M, N, and P have continuous second partial derivatives, then div curl F  0.

SECTION 15.1

Vector Fields

1063

Exercises for Section 15.1 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–6, match the vector field with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] y

(a)

y

(b)

4

6

x

x −6

4

6

In Exercises 21–26, find the gradient vector field for the scalar function. (That is, find the conservative vector field for the potential function.) 21. f x, y  5x 2  3xy  10y 2

22. f x, y  sin 3x cos 4y

23. f x, y, z  z  ye x

24. f x, y, z 

2

25. gx, y, z  xy lnx  y

z xz y   z x y

26. gx, y, z  x arcsin yz

In Exercises 27–30, verify that the vector field is conservative. −6 y

(c)

y

(d) 5

5

x

x

5

5

(f)

y

28. Fx, y 

1  yi  xj x2

29. Fx, y  sin y i  x cos yj

30. Fx, y 

1  yi  xj xy

In Exercises 31–34, determine if the vector field is conservative. 31. Fx, y  5y 23yi  xj 32. Fx, y 

−5

(e)

27. Fx, y  12x yi  6x 2  yj

y

33. Fx, y 

3

34. Fx, y 

2 1

x 5

2

−1

3

−3

2. Fx, y  y i

3. Fx, y  x i  3yj

4. Fx, y  y i  xj

5. Fx, y  x, sin y!

6. Fx, y 

 12xy, 14x 2!

In Exercises 7–16, sketch several representative vectors in the vector field. 8. Fx, y  2i 10. Fx, y  x i  yj

11. Fx, y, z  3yj

12. Fx, y  x i

13. Fx, y  4x i  yj

14. Fx, y  x 2  y 2i  j

15. Fx, y, z  i  j  k

16. Fx, y, z  x i  yj  zk

In Exercises 17–20, use a computer algebra system to graph several representative vectors in the vector field. 1 17. Fx, y  82xyi  y 2j

18. Fx, y  2y  3xi  2y  3xj x i  yj  zk 19. Fx, y, z  x 2  y 2  z 2 20. Fx, y, z  x i  yj  zk

2 2xy e  yi  xj y2 1 1  x 2y 2

 yi  xj

In Exercises 35–42, determine whether the vector field is conservative. If it is, find a potential function for the vector field. 35. Fx, y  2xyi  x 2j

1. Fx, y  x j

9. Fx, y  x i  yj

x i  yj

x −3 − 2

−5

7. Fx, y  i  j

1 x 2  y 2

36. Fx, y 

1  yi  2xj y2

37. Fx, y  xe x y 2yi  xj 2

38. Fx, y  3x 2y 2 i  2x 3 yj 39. Fx, y 

xi  yj x2  y 2

40. Fx, y 

2y x2 i 2j x y

41. Fx, y  e x cos yi  sin yj 42. Fx, y 

2xi  2yj x 2  y 22

In Exercises 43–46, find curl F for the vector field at the given point. Vector Field 43. Fx, y, z  xyzi  yj  zk 44. Fx, y, z  x 2zi  2xzj  yzk 45. Fx, y, z  ex sin yi  e x cos yj 46. Fx, y, z  exyz i  j  k

Point 1, 2, 1 2, 1, 3 0, 0, 3 3, 2, 0

1064

CHAPTER 15

Vector Analysis

In Exercises 47–50, use a computer algebra system to find the curl F for the vector field.



x i  lnx 2  y 2 j  k 47. Fx, y, z  arctan y 48. Fx, y, z 

In Exercises 71 and 72, find curl curl F      F. 71. Fx, y, z  xyzi  yj  zk 72. Fx, y, z  x 2zi  2xz j  yzk In Exercises 73 and 74, find div F  G.

xz xy yz i j k yz xz xy

49. Fx, y, z  sinx  yi  sin y  zj  sinz  xk 50. Fx, y, z  x 2  y 2  z 2 i  j  k In Exercises 51–56, determine whether the vector field F is conservative. If it is, find a potential function for the vector field.

73. Fx, y, z  i  2 xj  3yk

74. Fx, y, z  x i  zk

Gx, y, z  x i  yj  zk

Gx, y, z  x 2i  yj  z 2k

In Exercises 75 and 76, find div curl F      F. 75. Fx, y, z  xyzi  yj  zk 76. Fx, y, z  x 2zi  2xz j  yzk

51. Fx, y, z  sin yi  x cos yj  k 52. Fx, y, z  ez  yi  xj  k

In Exercises 77–84, prove the property for vector fields F and G and scalar function f. (Assume that the required partial derivatives are continuous.)

53. Fx, y, z  ez  yi  xj  xyk 54. Fx, y, z  y 2z3i  2xyz 3j  3xy 2z 2 k 55. Fx, y, z 

x 1 i  2 j  2z  1 k y y

77. curl F  G  curl F  curl G

56. Fx, y, z 

x y i 2 jk x2  y 2 x  y2

79. divF  G  div F  div G

78. curl"f   "  "f   0 80. divF  G  curl F  G  F  curl G

In Exercises 57– 60, find the divergence of the vector field F.

81. "  "f  "  F  "  "  F

57. Fx, y, z 

82. "   f F  f "  F  "f   F

6x 2 i



xy 2j

83. div f F  f div F  "f  F

58. Fx, y, z  xe x i  ye y j 59. Fx, y, z  sin x i  cos yj 

84. divcurl F  0 (Theorem 15.3)

z 2k

60. Fx, y, z  lnx 2  y 2i  xyj  ln y 2  z 2k In Exercises 61– 64, find the divergence of the vector field F at the given point. Point

Vector Field 61. Fx, y, z  xyzi  yj  zk

1, 2, 1 2, 1, 3 0, 0, 3 3, 2, 1

62. Fx, y, z  x 2z i  2xzj  yzk 63. Fx, y, z  e x sin yi  e x cos yj 64. Fx, y, z  lnxyzi  j  k

In Exercises 85– 88, let Fx, y, z  xi  yj  zk, and let f x, y, z  Fx, y, z .





85. Show that "ln f   87. Show that "f  nf n

F . f2

86. Show that "

 1f    fF . 3

n2F.

88. The Laplacian is the differential operator "2  "  " 

!2 !2 !2 2  2  !x !y !z2

and Laplace’s equation is

Writing About Concepts

"2w 

65. Define a vector field in the plane and in space. Give some physical examples of vector fields. 66. What is a conservative vector field and how do you test for it in the plane and in space? 67. Define the curl of a vector field. 68. Define the divergence of a vector field in the plane and in space.

!2w !2w !2w  2  2  0. !x 2 !y !z

Any function that satisfies this equation is called harmonic. Show that the function 1f is harmonic. True or False? In Exercises 89–92, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. 89. If Fx, y  4xi  y2 j, then Fx, y  → 0 as x, y → 0, 0. 90. If Fx, y  4xi  y2j and x, y is on the positive y-axis, then the vector points in the negative y-direction.

In Exercises 69 and 70, find curl F  G. 69. Fx, y, z  i  2xj  3yk Gx, y, z  x i  yj  zk

70. Fx, y, z  x i  zk Gx, y, z 

x 2 i  yj  z 2k

91. If f is a scalar field, then curl f is a meaningful expression. 92. If F is a vector field and curl F  0, then F is irrotational but not conservative.

SECTION 15.2

Section 15.2

Line Integrals

1065

Line Integrals • • • •

Understand and use the concept of a piecewise smooth curve. Write and evaluate a line integral. Write and evaluate a line integral of a vector field. Write and evaluate a line integral in differential form.

Piecewise Smooth Curves JOSIAH WILLARD GIBBS (1839–1903) Many physicists and mathematicians have contributed to the theory and applications described in this chapter—Newton, Gauss, Laplace, Hamilton, and Maxwell, among others. However, the use of vector analysis to describe these results is attributed primarily to the American mathematical physicist Josiah Willard Gibbs.

A classic property of gravitational fields is that, subject to certain physical constraints, the work done by gravity on an object moving between two points in the field is independent of the path taken by the object. One of the constraints is that the path must be a piecewise smooth curve. Recall that a plane curve C given by rt  xti  ytj, a ≤ t ≤ b is smooth if dx dt

dy dt

and

are continuous on [a, b] and not simultaneously 0 on a, b. Similarly, a space curve C given by rt  xti  ytj  ztk, a ≤ t ≤ b is smooth if dx , dt

dy , dt

and

dz dt

are continuous on [a, b] and not simultaneously 0 on a, b. A curve C is piecewise smooth if the interval [a, b] can be partitioned into a finite number of subintervals, on each of which C is smooth. EXAMPLE 1 z

Find a piecewise smooth parametrization of the graph of C shown in Figure 15.7. C = C1 + C2 + C3

1

(0, 0, 0) 1 x

C1 (1, 2, 0)

Figure 15.7

Finding a Piecewise Smooth Parametrization

(1, 2, 1) C3 C2

Solution Because C consists of three line segments C1, C2, and C3, you can construct a smooth parametrization for each segment and piece them together by making the last t-value in Ci correspond to the first t-value in Ci1, as follows.

(0, 2, 0) y

C1: xt  0,

yt  2t,

zt  0,

C2: xt  t  1, C3: xt  1,

yt  2, yt  2,

zt  0, zt  t  2,

0 ≤ t ≤ 1 1 ≤ t ≤ 2 2 ≤ t ≤ 3

So, C is given by

.



2tj, rt  t  1i  2j, i  2j  t  2k,

0 ≤ t ≤ 1 1 ≤ t ≤ 2. 2 ≤ t ≤ 3

Because C1, C2, and C3 are smooth, it follows that C is piecewise smooth.

Try It

Exploration A

Recall that parametrization of a curve induces an orientation to the curve. For instance, in Example 1, the curve is oriented such that the positive direction is from 0, 0, 0, following the curve to 1, 2, 1. Try finding a parametrization that induces the opposite orientation.

1066

CHAPTER 15

Vector Analysis

Line Integrals Up to this point in the text, you have studied various types of integrals. For a single integral



b

f x dx

Integrate over interval [a, b].

a

you integrated over the interval [a, b]. Similarly, for a double integral



f x, y dA

Integrate over region R.

R

you integrated over the region R in the plane. In this section, you will study a new type of integral called a line integral



f x, y ds

Integrate over curve C.

C

for which you integrate over a piecewise smooth curve C. (The terminology is somewhat unfortunate—this type of integral might be better described as a “curve integral.”) To introduce the concept of a line integral, consider the mass of a wire of finite length, given by a curve C in space. The density (mass per unit length) of the wire at the point x, y, z is given by f x, y, z. Partition the curve C by the points P0, P1, . . . , Pn z

producing n subarcs, as shown in Figure 15.8. The length of the ith subarc is given by si. Next, choose a point xi, yi, zi  in each subarc. If the length of each subarc is small, the total mass of the wire can be approximated by the sum

(xi , yi , zi) P0

P1 P2

Pi − 1 C Δsi

Pi P n−1

Mass of wire 

Pn

n

 f x , y , z  s . i

i

i

i

i1

x

Partitioning of curve C Figure 15.8

y

If you let  denote the length of the longest subarc and let  approach 0, it seems reasonable that the limit of this sum approaches the mass of the wire. This leads to the following definition.

Definition of Line Integral If f is defined in a region containing a smooth curve C of finite length, then the line integral of f along C is given by



f x, y ds  lim



f x, y, z ds  lim

C

or

C

n

 f x , y  s

→0 i1

i

i

Plane

i

n

 f x , y , z  s

→0 i1

i

i

i

i

Space

provided this limit exists.

As with the integrals discussed in Chapter 14, evaluation of a line integral is best accomplished by converting to a definite integral. It can be shown that if f is continuous, the limit given above exists and is the same for all smooth parametrizations of C.

SECTION 15.2

Line Integrals

1067

To evaluate a line integral over a plane curve C given by rt  xti  ytj, use the fact that ds  rt dt   xt 2  yt 2 dt. A similar formula holds for a space curve, as indicated in Theorem 15.4.

THEOREM 15.4

Evaluation of a Line Integral as a Definite Integral

Let f be continuous in a region containing a smooth curve C. If C is given by rt  xti  ytj, where a ≤ t ≤ b, then





b

f x, y ds 

f xt, yt xt 2  yt 2 dt.

a

C

If C is given by rt  xti  ytj  ztk, where a ≤ t ≤ b, then





b

f x, y, z ds 

f xt, yt, zt xt 2  yt 2  zt 2 dt.

a

C

Note that if f x, y, z  1, the line integral gives the arc length of the curve C, as defined in Section 12.5. That is,



EXAMPLE 2



C

(1, 2, 1)

1 x

Evaluating a Line Integral

Evaluate

1

(0, 0, 0)

rt dt  length of curve C.

a

C

z



b

1 ds 

x2  y  3z ds

C

1 2

y

Figure 15.9

where C is the line segment shown in Figure 15.9. Solution Begin by writing a parametric form of the equation of a line: x  t,

y  2t, and

z  t,

0 ≤ t ≤ 1.

Therefore, xt  1, yt  2, and zt  1, which implies that  xt 2  yt 2  zt 2  12  22  12  6.

So, the line integral takes the following form.





1

x2  y  3z ds 

C

t 2  2t  3t6 dt



0

1

NOTE The value of the line integral in Example 2 does not depend on the parametrization of the line segment C (any smooth parametrization will produce the same value). To convince . yourself of this, try some other parametrizations, such as x  1  2t, 1 y  2  4t, z  1  2t,  2 ≤ t ≤ 0, or x  t, y  2t, z  t, 1 ≤ t ≤ 0.

 6

t 2  t dt

0

 6



56  6

Try It



t3 t2  3 2

Exploration A

1 0

1068

CHAPTER 15

Vector Analysis

Suppose C is a path composed of smooth curves C1, C2, . . . , Cn. If f is continuous on C, it can be shown that



f x, y ds 

C



f x, y ds 

C1



f x, y ds  . . . 

C2



f x, y ds.

Cn

This property is used in Example 3. EXAMPLE 3 y

Evaluate

(0, 0)

Solution Begin by integrating up the line y  x, using the following parametrization.

(1, 1)

y=x

C1: x  t, y  t,

 xt 2  yt 2  2

C2

and you have x 1

Figure 15.10

0 ≤ t ≤ 1

For this curve, rt  ti  tj, which implies that xt  1 and yt  1. So,

y = x2

C1

x ds, where C is the piecewise smooth curve shown in Figure 15.10.

C

C = C1 + C2 1



Evaluating a Line Integral Over a Path





1

x ds 

t2 dt 

0

C1

2

2

1



t2

0

2



2

.

Next, integrate down the parabola y  x2, using the parametrization C2: x  1  t, y  1  t2,

0 ≤ t ≤ 1.

For this curve, rt  1  ti  1  t2j, which implies that xt  1 and yt  21  t. So,  xt 2  yt 2  1  41  t2

and you have





1

x ds 

  Consequently, .



1  t1  4 1  t2 dt

0

C2

x ds 

C

1



0

1 32 5  1. 12



C1

Try It



1 2

1  41  t2 32 8 3

x ds 



x ds 

C2

2

2



1 32 5  1  1.56. 12

Exploration A

For parametrizations given by rt  xti  ytj  ztk, it is helpful to remember the form of ds as ds  rt dt   xt 2  yt 2  zt 2 dt. This is demonstrated in Example 4.

SECTION 15.2



1069

Evaluating a Line Integral

EXAMPLE 4 Evaluate

Line Integrals

x  2 ds, where C is the curve represented by

C

4 32 1 t j  t 2 k, 0 ≤ t ≤ 2. 3 2

rt  ti 

Solution Because rt  i  2t12j  tk, and rt   xt 2  yt 2  zt 2  1  4t  t 2 it follows that



  2

x  2 ds 

t  21  4t  t 2 dt

0

C

2

1  2t  21  4t  t 212 dt 2 0 2 1  1  4t  t 232 3 0





.



1 1313  1 3

 15.29.

Try It

Exploration A

The next example shows how a line integral can be used to find the mass of a spring whose density varies. In Figure 15.11, note that the density of this spring increases as the spring spirals up the z-axis.

Finding the Mass of a Spring

EXAMPLE 5 z

Density: ρ (x, y, z) = 1 + z

Find the mass of a spring in the shape of the circular helix rt 

1 cos ti  sin tj  tk, 2

0 ≤ t ≤ 6

where the density of the spring is x, y, z  1  z, as shown in Figure 15.11. Solution Because rt 

1 2

sin t2  cos t2  12  1

it follows that the mass of the spring is Mass 



C



6

1  t 2 dt t  t  22 3  6  1  2 

1  z ds 

0

2

6 0

2 x

..

2

y

r(t) = 1 (cos ti + sin tj + tk) 2

Figure 15.11

Rotatable Graph

 144.47. The mass of the spring is approximately 144.47.

Try It

Exploration A

1070

CHAPTER 15

Vector Analysis

Line Integrals of Vector Fields

Inverse square force field F

One of the most important physical applications of line integrals is that of finding the work done on an object moving in a force field. For example, Figure 15.12 shows an inverse square force field similar to the gravitational field of the sun. Note that the magnitude of the force along a circular path about the center is constant, whereas the magnitude of the force along a parabolic path varies from point to point. To see how a line integral can be used to find work done in a force field F, consider an object moving along a path C in the field, as shown in Figure 15.13. To determine the work done by the force, you need consider only that part of the force that is acting in the same direction as that in which the object is moving (or the opposite direction). This means that at each point on C, you can consider the projection F  T of the force vector F onto the unit tangent vector T. On a small subarc of length si, the increment of work is Wi  forcedistance  Fxi, yi, zi   Txi, yi, zi  si where xi, yi, z i  is a point in the ith subarc. Consequently, the total work done is given by the following integral. W



C

Vectors along a parabolic path in the force field F

Fx, y, z  Tx, y, z ds z

z

Figure 15.12

T has the direction of F.

C F

z

C

(F • T)T T

F

(F • T)T C

T

y

y

T

y

(F • T)T x

x

x

At each point on C, the force in the direction of motion is F Figure 15.13

 TT.

This line integral appears in other contexts and is the basis of the following definition of the line integral of a vector field. Note in the definition that rt rt dt rt  F  rt dt  F  dr.

F  T ds  F 

Definition of Line Integral of a Vector Field Let F be a continuous vector field defined on a smooth curve C given by rt, a ≤ t ≤ b. The line integral of F on C is given by



C

F  dr 



C

F  T ds 



b

a

Fxt, yt), zt  rt dt.

SECTION 15.2

1071

Work Done by a Force

EXAMPLE 6 z

Line Integrals

Find the work done by the force field

(−1, 0, 3π )

1 1 1 Fx, y, z   xi  yj  k 2 2 4

3π

Force field F

on a particle as it moves along the helix given by rt  cos ti  sin tj  tk

−2

π

−2

−1

−1

from the point 1, 0, 0 to 1, 0, 3, as shown in Figure 15.14.

(1, 0, 0)

Solution Because

1

2

2

x

Space curve C

rt  xti  ytj  ztk  cos ti  sin tj  tk

y

Figure 15.14

it follows that xt  cos t, yt  sin t, and zt  t. So, the force field can be written as 1 1 1 Fxt, yt, zt   cos ti  sin tj  k. 2 2 4 To find the work done by the force field in moving a particle along the curve C, use the fact that rt  sin ti  cos tj  k and write the following. W

      F

C b



dr

Fxt, yt, zt  rt dt

a 3



1 1 1  cos ti  sin tj  k  sin ti  cos tj  k dt 2 2 4 0 3 1 1 1 dt  sin t cos t  sin t cos t  2 2 4 0 3 1  dt 4 0 

z

.

3



1 t 4 3  4



0

Try It

y x Generated by Mathematica

Figure 15.15



Open Exploration

NOTE In Example 6, note that the x- and y-components of the force field end up contributing nothing to the total work. This occurs because in this particular example the z-component of the force field is the only portion of the force that is acting in the same (or opposite) direction in which the particle is moving (see Figure 15.15).

The computer-generated view of the force field in Example 6 shown in Figure 15.15 indicates that each vector in the force field points toward the z-axis.

TECHNOLOGY

1072

CHAPTER 15

Vector Analysis

For line integrals of vector functions, the orientation of the curve C is important. If the orientation of the curve is reversed, the unit tangent vector Tt is changed to Tt, and you obtain



C

F  dr  

EXAMPLE 7



C

F  dr.

Orientation and Parametrization of a Curve

Let Fx, y  yi  x2j and evaluate the line integral C F  dr for each parabolic curve shown in Figure 15.16.

C1: r1(t) = (4 − t)i + (4t − t 2)j C2: r2(t) = ti + (4t − t 2)j

a. C1: r1t  4  ti  4t  t 2 j, 0 ≤ t ≤ 3 b. C2: r2t  ti  4t  t 2j, 1 ≤ t ≤ 4

y

4

Solution a. Because r1 t  i  4  2tj and

(1, 3) 3

2

C2

Fxt, yt  4t  t 2i  4  t2j

C1

the line integral is

1

(4, 0) x

1

2

3



C1

F  dr 

4



Figure 15.16

  

4t  t 2  64  64t  20t 2  2t 3 dt



t4  7t 3  34t 2  64t 2

3

4t  t 2i  4  t2j  i  4  2tj dt

0 3

0 3



2t 3  21t 2  68t  64 dt

0

  NOTE Although the value of the line integral in Example 7 depends on the orientation of C, it does not depend on the parametrization of C. To see this, let C3 be represented by r3  t  2i  4  j t2

where 1 ≤ t ≤ 2. The graph of this curve is the same parabolic segment shown in Figure 15.16. Does the value of the line integral over C3 agree with the value over C1 or C2? Why or why not?



3 0

69  . 2 b. Because r2 t  i  4  2tj and Fxt, yt  4t  t 2i  t 2j the line integral is



C2

F  dr  

  

4t  t 2  4t 2  2t 3 dt



t4  t 3  2t 2 2

4

4t  t 2 i  t 2j  i  4  2tj dt

1 4

1 4



2t 3  3t 2  4t dt

1

   .



4 1

69 . 2

The answer in part (b) is the negative of that in part (a) because C1 and C2 represent opposite orientations of the same parabolic segment.

Try It

Exploration A

SECTION 15.2

Line Integrals

1073

Line Integrals in Differential Form A second commonly used form of line integrals is derived from the vector field notation used in the preceding section. If F is a vector field of the form Fx, y  Mi  Nj, and C is given by rt  xti  ytj, then F  dr is often written as M dx  N dy.



F  dr 

C



     F

C b

a b



Mi  Nj  xti  ytj dt M

a



dr dt dt



dx dy N dt dt dt

M dx  N dy

C

This differential form can be extended to three variables. The parentheses are often omitted, as follows.



M dx  N dy



and

C

M dx  N dy  P dz

C

Notice how this differential notation is used in Example 8. y

Evaluating a Line Integral in Differential Form

EXAMPLE 8

4

Let C be the circle of radius 3 given by rt  3 cos ti  3 sin tj, 0 ≤ t ≤ 2

2

x −4

−2

2

4

−2



y3 dx  x3  3xy2 dy.

C

−4

.

as shown in Figure 15.17. Evaluate the line integral

r(t) = 3 cos ti + 3 sin tj

Figure 15.17

Solution Because x  3 cos t and y  3 sin t, you have dx  3 sin t dt and dy  3 cos t dt. So, the line integral is



M dx  N dy

C

Editable Graph

 

 

y3 dx  x3  3xy2 dy

C 2

NOTE The orientation of C affects the value of the differential form of a line integral. Specifically, if C has the orientation opposite to that of C, then



C

M dx  N dy  



M dx  N dy.

C

. So, of the three line integral forms presented in this section, the orientation of C does not affect the form C f x, y ds, but it does affect the vector form and the differential form.

27 sin3 t3 sin t  27 cos3 t  81 cos t sin2 t3 cos t dt

  

0

 81  81  81

2

0 2 0 2 0



 81

cos4 t  sin4 t  3 cos2 t sin2 t dt

cos t  sin t  34 sin 2t dt cos 2t  341  2cos 4t dt 2

2

sin 2t 3 3 sin 4t  t 2 8 32

2

2



0

243  . 4

Try It

Exploration A

1074

CHAPTER 15

Vector Analysis

For curves represented by y  gx, a ≤ x ≤ b, you can let x  t and obtain the parametric form xt

and

y  gt, a ≤ t ≤ b.

Because dx  dt for this form, you have the option of evaluating the line integral in the variable x or t. This is demonstrated in Example 9. EXAMPLE 9 y

Evaluating a Line Integral in Differential Form

Evaluate



C: y = 4x − x 2 4

y dx  x2 dy

C

(1, 3)

3

where C is the parabolic arc given by y  4x  x2 from 4, 0 to 1, 3, as shown in Figure 15.18.

2

Solution Rather than converting to the parameter t, you can simply retain the variable x and write

1

(4, 0) x

.

1

2

Figure 15.18

Editable Graph

3

4

dy  4  2x dx.

y  4x  x2

Then, in the direction from 4, 0 to 1, 3, the line integral is



 

1

y dx  x2 dy 

4x  x2 dx  x24  2x dx

4 1

C



4x  3x2  2x3 dx

4



.

 2x2  x3 

Try It

x4 2



1 4



69 . 2

See Example 7.

Exploration A E X P L O R AT I O N

Finding Lateral Surface Area The figure below shows a piece of tin that has been cut from a circular cylinder. The base of the circular cylinder is modeled by x2  y2  9. At any point x, y on the base, the height of the object is given by f x, y  1  cos

x . 4

Explain how to use a line integral to find the surface area of the piece of tin. z

2

1 −2

x

.

1 + cos π x 4

−1

3 3

x2 + y2 = 9

Rotatable Graph

y

(x, y)

SECTION 15.2

1075

Line Integrals

Exercises for Section 15.2 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–6, find a piecewise smooth parametrization of the path C. 1.

y

2.

x 2 + y2 = 9

y

x 2 y2 + =1 16 9

4 2

2

1 1

x −2

2

2 −2

−2

C

3.

In Exercises 15–18, evaluate

 C

x −2 −1

14. C: counterclockwise around the circle x2  y2  4 from 2, 0 to 0, 2

4. (3, 3)

3

16. C: line from 0, 0 to 3, 9 17. C: counterclockwise around the triangle with vertices 0, 0, 1, 0, and 0, 1

y 5

18. C: counterclockwise around the square with vertices 0, 0, 2, 0, 2, 2, and 0, 2

(5, 4)

4 2

along the given path. 15. C: line from 0, 0 to 1, 1

C

−4

y

x  4y  ds

3

C

C

2

1

In Exercises 19 and 20, evaluate

1 x

x 1

5.

2

1

3

6.

y

y=

3

5



2x  y2  z ds

C

along the path C shown in the figure. (2, 4)

4

(1, 1)

4

y

C

x

1

2

19.

20. z

z

3

C 2

y=x

y = x2 (1, 0, 1)

1

x 1

x

1

2

3

C

4

C

7.



x  y ds

8.

C



1 y x

4 xy ds

C: rt  ti  2  tj

0 ≤ t ≤ 2

x2  y2  z2 ds

10.

C



0 ≤ t ≤ 2 8xyz ds

Mass In Exercises 21 and 22, find the total mass of two turns of a spring with density  in the shape of the circular helix

C

C: rt  sin ti  cos tj  8tk 0 ≤ t ≤ 2

C: rt  12ti  5tj  3k 0 ≤ t ≤ 2

In Exercises 11–14, evaluate



y

1

(1, 1, 1)

(0, 1, 0)

C

C: rt  4ti  3tj

9.



x

(0, 0, 0)

(0, 0, 0)

(1, 0, 0)

In Exercises 7–10, evaluate the line integral along the given path.

(0, 1, 1)

1

1

x2  y2 ds

C

rt  3 cos ti  3 sin tj  2tk. 1 21.  x, y, z  2x2  y2  z2

22.  x, y, z  z Mass In Exercises 23–26, find the total mass of the wire with density .

along the given path.

23. rt  cos ti  sin tj, x, y  x  y, 0 ≤ t ≤ 

11. C: x-axis from x  0 to x  3

3 24. rt  t2 i  2tj, x, y  y, 4

12. C: y-axis from y  1 to y  10 13. C: counterclockwise around the circle x2  y2  1 from 1, 0 to 0, 1

0 ≤ t ≤ 1

25. rt  t2 i  2tj  tk, x, y, z  kz k > 0,

1 ≤ t ≤ 3

26. rt  2 cos ti  2 sin tj  3tk, x, y, z  k  z k > 0, 0 ≤ t ≤ 2

1076

CHAPTER 15

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In Exercises 27–32, evaluate



C

y

F  dr

1

where C is represented by rt.

C

27. Fx, y  xyi  yj C: rt  4ti  tj,

x

0 ≤ t ≤ 1

1

28. Fx, y  xyi  yj C: rt  4 cos ti  4 sin tj,

0 ≤ t ≤ 2

29. Fx, y  3x i  4yj C: rt  2 cos ti  2 sin tj,

0 ≤ t ≤ 2

30. Fx, y  3x i  4yj C: rt  ti  4  t2 j, 31. Fx, y, z 

x2 yi

2 ≤ t ≤ 2

Figure for 36 37. Fx, y  2xi  yj C: counterclockwise around the triangle with vertices 0, 0, 1, 0, and 1, 1 y

 x  zj  x yzk

C: rt  ti  t 2j  2k, 0 ≤ t ≤ 1

(1, 1) 1

32. Fx, y, z  x2i  y2j  z2k C: rt  2 sin ti  2 cos tj  12 t 2k, 0 ≤ t ≤ 

C

In Exercises 33 and 34, use a computer algebra system to evaluate the integral



C

F  dr

x 1

38. Fx, y  yi  xj

where C is represented by rt.

C: counterclockwise along the semicircle y  4  x2 from 2, 0 to 2, 0

33. Fx, y, z  x2zi  6yj  yz2k

y

C: rt  ti  t2j  ln tk, 1 ≤ t ≤ 3

3

xi  yj  zk x2  y2  z2 C: rt  ti  tj  et k, 0 ≤ t ≤ 2

34. Fx, y, z 

Work In Exercises 35– 40, find the work done by the force field F on a particle moving along the given path. 35. Fx, y  x i  2yj C: y  x3 from 0, 0 to 2, 8 y

C 1 x −2

−1

1

2

−1

39. Fx, y, z  x i  yj  5zk C: rt  2 cos ti  2 sin tj  tk, 0 ≤ t ≤ 2 z

(2, 8)

8

2π

6 4

C

π

C

2

−3

−3

x 2

4

6

8

36. Fx, y  x2 i  xyj C: x  cos3 t, y  sin3 t from 1, 0 to 0, 1

x

3

3

y

SECTION 15.2

40. Fx, y, z  yzi  xzj  xyk z

49.

2 5

 

x  3y2 dy

50.

C

3

x

1077

In Exercises 49–52, evaluate the line integral along the path C given by x  2t, y  10t, where 0 ≤ t ≤ 1.

C: line from 0, 0, 0 to 5, 3, 2

C

Line Integrals

51.

1

 

x  3y2 dx

C

xy dx  y dy

52.

C

3y  x dx  y2 dy

C

In Exercises 53–60, evaluate the integral

3



y

2x  y dx  x  3y dy

C

along the path C. 41. Work Find the work done by a person weighing 150 pounds walking exactly one revolution up a circular helical staircase of radius 3 feet if the person rises 10 feet.

53. C: x-axis from x  0 to x  5

42. Work A particle moves along the path y  from the point 0, 0 to the point 1, 1. The force field F is measured at five points along the path and the results are shown in the table. Use Simpson’s Rule or a graphing utility to approximate the work done by the force field.

55. C: line segments from 0, 0 to 3, 0 and 3, 0 to 3, 3

x2

x, y

0, 0

14, 161 

12, 14 

34, 169 

1, 1

F x, y

5, 0!

3.5, 1!

2, 2!

1.5, 3!

1, 5!

In Exercises 43 and 44, evaluate C F  dr for each curve. Discuss the orientation of the curve and its effect on the value of the integral.

54. C: y-axis from y  0 to y  2 56. C: line segments from 0, 0 to 0, 3 and 0, 3 to 2, 3 57. C: arc on y  1  x 2 from 0, 1 to 1, 0 58. C: arc on y  x32 from 0, 0 to 4, 8 59. C: parabolic path x  t, y  2t 2, from 0, 0 to 2, 8 60. C: elliptic path x  4 sin t, y  3 cos t, from 0, 3 to 4, 0 Lateral Surface Area In Exercises 61–68, find the area of the lateral surface (see figure) over the curve C in the xy-plane and under the surface z  f x, y, where Lateral surface area 



f x, y ds.

C

43. Fx, y  x 2 i  xyj

z

(a) r1t  2ti  t  1j,

1 ≤ t ≤ 3

Surface: z = f(x, y)

(b) r2t  23  ti  2  tj, 0 ≤ t ≤ 2 44. Fx, y  x 2 yi  xy32j (a) r1t  t  1 i  t 2j,

0 ≤ t ≤ 2

(b) r2t  1  2 cos ti  4 cos2 tj,

0 ≤ t ≤ 2

Lateral surface

In Exercises 45– 48, demonstrate the property that



C

(xi, yi)

F  dr  0

x

P

Q

y

Δsi

regardless of the initial and terminal points of C, if the tangent vector r t is orthogonal to the force field F.

C: Curve in xy-plane

Rotatable Graph

45. Fx, y  yi  xj C: rt  t i  2tj

61. f x, y  h, C: line from 0, 0 to 3, 4

46. Fx, y  3yi  xj

62. f x, y  y, C: line from 0, 0 to 4, 4)

C: rt  t i  t 3j



47. Fx, y  x3  2x2i  x  C: rt  t i  t2j 48. Fx, y  xi  yj C: rt  3 sin ti  3 cos tj



y j 2

63. f x, y  xy,

C: x2  y2  1 from 1, 0 to 0, 1

64. f x, y  x  y, C: x2  y2  1 from 1, 0 to 0, 1 65. f x, y  h, C: y  1  x2 from 1, 0 to 0, 1 66. f x, y  y  1, C: y  1  x2 from 1, 0 to 0, 1 67. f x, y  xy, 68. f x, y 

x2

C: y  1  x2 from 1, 0 to 0, 1

 y2  4, C: x2  y2  4

1078

CHAPTER 15

Vector Analysis

69. Engine Design A tractor engine has a steel component with a circular base modeled by the vector-valued function rt  2 cos t i  2 sin tj. Its height is given by z  1  y2. (All measurements of the component are given in centimeters.) (a) Find the lateral surface area of the component. (b) The component is in the form of a shell of thickness 0.2 centimeter. Use the result of part (a) to approximate the amount of steel used in its manufacture.

76. Investigation Determine the value of c such that the work done by the force field Fx, y  15 4  x2yi  xyj on an object moving along the parabolic path y  c1  x2 between the points 1, 0 and 1, 0 is a minimum. Compare the result with the work required to move the object along the straight-line path connecting the points.

(c) Draw a sketch of the component. 70. Building Design The ceiling of a building has a height above the floor given by z  20  14x, and one of the walls follows a path modeled by y  x 32. Find the surface area of the wall if 0 ≤ x ≤ 40. (All measurements are given in feet.)

Writing About Concepts 77. Define a line integral of a function f along a smooth curve C in the plane and in space. How do you evaluate the line integral as a definite integral?

Moments of Inertia Consider a wire of density x, y given by the space curve

78. Define a line integral of a continuous vector field F on a smooth curve C. How do you evaluate the line integral as a definite integral?

C: rt  xti  ytj,

79. Order the surfaces in ascending order of the lateral surface area under the surface and over the curve y  x from 0, 0 to 4, 2 in the xy-plane. Explain your ordering without doing any calculations.

a ≤ t ≤ b.

The moments of inertia about the x- and y-axes are given by Ix 

 

y2 x, y ds

C

Iy 

x2 x, y ds.

C

In Exercises 71 and 72, find the moments of inertia for the wire of density .

(a) z1  2  x

(b) z2  5  x

(c) z3  2

(d) z4  10  x  2y

80. For each of the following, determine whether the work done in moving an object from the first to the second point through the force field shown in the figure is positive, negative, or zero. Explain your answer.

71. A wire lies along rt  a cos ti  a sin tj, 0 ≤ t ≤ 2 and a > 0, with density x, y  1.

(a) From 3, 3 to 3, 3

72. A wire lies along rt  a cos ti  a sin tj, 0 ≤ t ≤ 2 and a > 0, with density x, y  y.

(c) From 5, 0 to 0, 3

y

(b) From 3, 0 to 0, 3

x

Approximation In Exercises 73 and 74, determine which value best approximates the lateral surface area over the curve C in the xy-plane and under the surface z  f x, y. (Make your selection on the basis of a sketch of the surface and not by performing any calculations.) 73. f x, y  e xy

True or False? In Exercises 81–84, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

C: line from 0, 0 to 2, 2 (a) 54

(b) 25

(c) 250

(d) 75

(e) 100

74. f x, y  y C: y  x2 from 0, 0 to 2, 4 (a) 2

(b) 4

(c) 8

(d) 16

75. Investigation The top outer edge of a solid with vertical sides and resting on the xy-plane is modeled by rt  3 cos t i  3 sin tj  1  sin2 2tk, where all measurements are in centimeters. The intersection of the plane y  b 3 < b < 3 with the top of the solid is a horizontal line. (a) Use a computer algebra system to graph the solid. (b) Use a computer algebra system to approximate the lateral surface area of the solid. (c) Find (if possible) the volume of the solid.

81. If C is given by xt  t, yt  t, 0 ≤ t ≤ 1, then





1

xy ds 

t 2 dt.

0

C

82. If C2  C1, then



C1

f x, y ds 



f x, y ds  0.

C2

83. The vector functions r1  t i  t 2j, 0 ≤ t ≤ 1, and r2  1  ti  1  t2j, 0 ≤ t ≤ 1, define the same curve. 84. If



F  T ds  0, then F and T are orthogonal.

C

85. Work Consider a particle that moves through the force field Fx, y   y  xi  xyj from the point 0, 0 to the point 0, 1 along the curve x  kt1  t, y  t. Find the value of k such that the work done by the force field is 1.

SECTION 15.3

Section 15.3

Conservative Vector Fields and Independence of Path

1079

Conservative Vector Fields and Independence of Path • Understand and use the Fundamental Theorem of Line Integrals. • Understand the concept of independence of path. • Understand the concept of conservation of energy.

Fundamental Theorem of Line Integrals y

The discussion in the preceding section pointed out that in a gravitational field the work done by gravity on an object moving between two points in the field is independent of the path taken by the object. In this section, you will study an important generalization of this result—it is called the Fundamental Theorem of Line Integrals. To begin, an example is presented in which the line integral of a conservative vector field is evaluated over three different paths.

(1, 1)

1

C1

x

(0, 0)

1

C1: y = x

Line Integral of a Conservative Vector Field

EXAMPLE 1

Find the work done by the force field

(a)

1 1 Fx, y  xyi  x 2j 2 4

y

on a particle that moves from 0, 0 to 1, 1 along each path, as shown in Figure 15.19. a. C1: y  x

(1, 1)

1

b. C2: x  y 2

c. C3: y  x 3

Solution

C2

a. Let rt  ti  tj for 0 ≤ t ≤ 1, so that x

(0, 0)

1

C2: x = y2

dr  i  j dt

and

Then, the work done is

(b)

W



C1

y

F  dr 



1

0

1 1 Fx, y  t 2 i  t 2j. 2 4 1



3 2 1 t dt  t 3 4 4

1  . 4

0

b. Let rt  ti  t j for 0 ≤ t ≤ 1, so that 1



dr  i 

(1, 1)

2t  1

j dt

Then, the work done is C3

W x

(0, 0)

1

C3: y = x 3 (c)

Figure 15.19



C2



1

0

1 1 Fx, y  t 32 i  t 2j. 2 4 1



5 32 1 t dt  t 52 8 4

0

1  . 4

c. Let rt  12ti  18t 3j for 0 ≤ t ≤ 2, so that dr 

12 i  38 t j dt 2

and

Fx, y 

1 4 1 t i  t 2j. 32 16

Then, the work done is W



C3

.

F  dr 

and

F  dr 



2

0

2



5 4 1 5 t dt  t 128 128

0

1  . 4

So, the work done by a conservative vector field is the same for all paths.

Try It

Exploration A

1080

CHAPTER 15

Vector Analysis

In Example 1, note that the vector field Fx, y  12xyi  14x 2j is conservative because Fx, y  "f x, y, where f x, y  14x 2y. In such cases, the following theorem states that the value of C F  dr is given by



C

F  dr  f x1, y1  f x0, y0 

1 0 4



1. 4

THEOREM 15.5

Fundamental Theorem of Line Integrals

Let C be a piecewise smooth curve lying in an open region R and given by NOTE Notice how the Fundamental Theorem of Line Integrals is similar to the Fundamental Theorem of Calculus (Section 4.4), which states that



rt  xti  ytj, a ≤ t ≤ b. If Fx, y  Mi  Nj is conservative in R, and M and N are continuous in R, then



b

f x dx  Fb  Fa

C

a

where Fx  f x.



F  dr 

C

"f  dr  f xb, yb  f xa, ya

where f is a potential function of F. That is, Fx, y  "f x, y. Proof A proof is provided only for a smooth curve. For piecewise smooth curves, the procedure is carried out separately on each smooth portion. Because Fx, y  "f x, y  fxx, yi  fyx, yj, it follows that



C

F  dr  

   b

dr dt dt dx dy fxx, y  fyx, y dt dt dt

F

a b a



and, by the Chain Rule (Theorem 13.6), you have



C



b

d

f xt, yt dt dt a  f xb, yb  f xa, y a.

F  dr 

The last step is an application of the Fundamental Theorem of Calculus. In space, the Fundamental Theorem of Line Integrals takes the following form. Let C be a piecewise smooth curve lying in an open region Q and given by rt  xti  ytj  ztk, a ≤ t ≤ b. If Fx, y, z  Mi  Nj  Pk is conservative and M, N, and P are continuous, then



C

F  dr 



C

"f  dr

 f xb, yb, zb  f xa, ya, za

where Fx, y, z  "f x, y, z. The Fundamental Theorem of Line Integrals states that if the vector field F is conservative, then the line integral between any two points is simply the difference in the values of the potential function f at these points.

SECTION 15.3

EXAMPLE 2 F(x, y) = 2xyi + (x 2 − y)j

Evaluate

(−1, 4)

Using the Fundamental Theorem of Line Integrals

F  dr, where C is a piecewise smooth curve from 1, 4 to 1, 2 and

4

as shown in Figure 15.20. Solution From Example 6 in Section 15.1, you know that F is the gradient of f where

(1, 2)

2

f x, y  x 2y  1

x

−2

1081

Fx, y  2xyi  x 2  yj

3

C



C

y

Conservative Vector Fields and Independence of Path

−1

1

2

Using the Fundamental Theorem of Line Integrals, C F  dr.

y2  K. 2

Consequently, F is conservative, and by the Fundamental Theorem of Line Integrals, it follows that



C

F  dr  f 1, 2  f 1, 4



Figure 15.20

 122 

 

22 42  124  2 2



 4. .

Note that it is unnecessary to include a constant K as part of f, because it is canceled by subtraction.

Try It EXAMPLE 3 Evaluate

F(x, y, z) = 2xyi + (x 2 + z 2)j + 2yzk



C

z

Exploration A Using the Fundamental Theorem of Line Integrals

F  dr, where C is a piecewise smooth curve from 1, 1, 0 to 0, 2, 3 and

Fx, y, z  2xyi  x 2  z 2j  2yzk

3

as shown in Figure 15.21. (0, 2, 3)

Solution From Example 8 in Section 15.1, you know that F is the gradient of f where f x, y, z  x 2y  yz 2  K. Consequently, F is conservative, and by the Fundamental Theorem of Line Integrals, it follows that

2

1

C



1

C

F  dr  f 0, 2, 3  f 1, 1, 0  022  232  121  10 2  17.

2

. x

(1, 1, 0)

2

y

Using the Fundamental Theorem of Line Integrals, C F  dr. Figure 15.21

Try It

Exploration A

In Examples 2 and 3, be sure you see that the value of the line integral is the same for any smooth curve C that has the given initial and terminal points. For instance, in Example 3, try evaluating the line integral for the curve given by rt  1  t i  1  t j  3tk. You should obtain



C

F  dr 



1

0

 17.

30t 2  16t  1 dt

1082

CHAPTER 15

Vector Analysis

Independence of Path A C R1

R2 B

R1 is connected.

R2 is not connected.

From the Fundamental Theorem of Line Integrals it is clear that if F is continuous and conservative in an open region R, the value of C F  dr is the same for every piecewise smooth curve C from one fixed point in R to another fixed point in R. This result is described by saying that the line integral C F  dr is independent of path in the region R. A region in the plane (or in space) is connected if any two points in the region can be joined by a piecewise smooth curve lying entirely within the region, as shown in Figure 15.22. In open regions that are connected, the path independence of C F  dr is equivalent to the condition that F is conservative.

Figure 15.22

THEOREM 15.6

Independence of Path and Conservative Vector Fields

If F is continuous on an open connected region, then the line integral



C

F  dr

is independent of path if and only if F is conservative.

(x1, y)

(x, y)

C2 C4

C1 C3 (x0, y0 )

Figure 15.23

(x, y1)

Proof If F is conservative, then, by the Fundamental Theorem of Line Integrals, the line integral is independent of path. Now establish the converse for a plane region R. Let Fx, y  Mi  Nj, and let x0, y0 be a fixed point in R. If x, y is any point in R, choose a piecewise smooth curve C running from x0, y0 to x, y, and define f by f x, y 



C

F  dr 



M dx  N dy.

C

The existence of C in R is guaranteed by the fact that R is connected. You can show that f is a potential function of F by considering two different paths between x0, y0 and x, y. For the first path, choose x1, y in R such that x  x1. This is possible because R is open. Then choose C1 and C2, as shown in Figure 15.23. Using the independence of path, it follows that f x, y 

 

M dx  N dy

C



M dx  N dy 

C1



M dx  N dy.

C2

Because the first integral does not depend on x, and because dy  0 in the second integral, you have f x, y  g y 



M dx

C2

and it follows that the partial derivative of f with respect to x is fxx, y  M. For the second path, choose a point x, y1 . Using reasoning similar to that used for the first path, you can conclude that fyx, y  N. Therefore, "f x, y  fxx, y i  fyx, y j  Mi  Nj  Fx, y and it follows that F is conservative.

SECTION 15.3

EXAMPLE 4

Conservative Vector Fields and Independence of Path

1083

Finding Work in a Conservative Force Field

For the force field given by Fx, y, z  e x cos yi  e x sin yj  2k show that C F  dr is independent of path, and calculate the work done by F on an object moving along a curve C from 0, 2, 1 to 1, , 3. Solution Writing the force field in the form Fx, y, z  Mi  Nj  Pk, you have M  e x cos y, N  e x sin y, and P  2, and it follows that !N !P 0 !y !z !P !M 0 !x !z !M !N  e x sin y  . !x !y So, F is conservative. If f is a potential function of F, then fxx, y, z  e x cos y fyx, y, z  e x sin y fzx, y, z  2. By integrating with respect to x, y, and z separately, you obtain f x, y, z  f x, y, z  f x, y, z 

  

fxx, y, z dx  fyx, y, z dy  fzx, y, z dz 

  

e x cos y dx  e x cos y  g y, z e x sin y dy  e x cos y  hx, z 2 dz  2z  kx, y.

By comparing these three versions of f x, y, z, you can conclude that f x, y, z  e x cos y  2z  K. Therefore, the work done by F along any curve C from 0, 2, 1 to 1, , 3 is W



C

F  dr



.

1, , 3



 e x cos y  2z

0, 2, 1

 e  6  0  2  4  e.

Try It

Exploration A

How much work would be done if the object in Example 4 moved from the point 0, 2, 1 to 1, , 3 and then back to the starting point 0, 2, 1? The Fundamental Theorem of Line Integrals states that there is zero work done. Remember that, by definition, work can be negative. So, by the time the object gets back to its starting point, the amount of work that registers positively is canceled out by the amount of work that registers negatively.

1084

CHAPTER 15

Vector Analysis

A curve C given by rt for a ≤ t ≤ b is closed if r a  r b. By the Fundamental Theorem of Line Integrals, you can conclude that if F is continuous and conservative on an open region R, then the line integral over every closed curve C is 0.

THEOREM 15.7 NOTE Theorem 15.7 gives you options for evaluating a line integral involving a conservative vector field. You can use a potential function, or it might be more convenient to choose a particularly simple path, such as a straight line.

Let Fx, y, z  Mi  Nj  Pk have continuous first partial derivatives in an open connected region R, and let C be a piecewise smooth curve in R. The following conditions are equivalent. 1. F is conservative. That is, F  "f for some function f. 2.

 

C

3.

C

F  dr is independent of path. F  dr  0 for every closed curve C in R.

Evaluating a Line Integral

EXAMPLE 5 Evaluate

C1: r(t) = (1 − cos t)i + sin tj



C1

y

Equivalent Conditions

F  dr, where

Fx, y   y 3  1i  3xy 2  1j and C1 is the semicircular path from 0, 0 to 2, 0, as shown in Figure 15.24.

1

C1

Solution You have the following three options.

C2 (0, 0)

(2, 0) 1

C2: r(t) = ti

Figure 15.24

2

x

a. You can use the method presented in the preceding section to evaluate the line integral along the given curve. To do this, you can use the parametrization rt  1  cos t i  sin tj, where 0 ≤ t ≤ . For this parametrization, it follows that dr  rt dt  sin ti  cos tj dt, and



C1

F  dr 





sin t  sin4 t  cos t  3 sin2 t cos t  3 sin2 t cos2 t dt.

0

This integral should dampen your enthusiasm for this option. b. You can try to find a potential function and evaluate the line integral by the Fundamental Theorem of Line Integrals. Using the technique demonstrated in Example 4, you can find the potential function to be f x, y  xy 3  x  y  K, and, by the Fundamental Theorem, W



C1

F  dr  f 2, 0  f 0, 0  2.

c. Knowing that F is conservative, you have a third option. Because the value of the line integral is independent of path, you can replace the semicircular path with a simpler path. Suppose you choose the straight-line path C2 from 0, 0 to 2, 0. Then, rt  ti, where 0 ≤ t ≤ 2. So, dr  i dt and Fx, y   y 3  1 i  3xy 2  1 j  i  j, so that

.



F  dr 

C1



C2

F  dr 



2



1 dt  t

0

2 0

 2.

Of the three options, obviously the third one is the easiest.

Try It

Exploration A

Open Exploration

SECTION 15.3

Conservative Vector Fields and Independence of Path

1085

Conservation of Energy MICHAEL FARADAY (1791–1867) Several philosophers of science have considered Faraday’s Law of Conservation of Energy to be the greatest generalization ever conceived by humankind. Many physicists have contributed to our knowledge of this law. Two early and influential ones were James Prescott Joule (1818–1889) and Hermann Ludwig Helmholtz (1821–1894).

In 1840, the English physicist Michael Faraday wrote, “Nowhere is there a pure creation or production of power without a corresponding exhaustion of something to supply it.” This statement represents the first formulation of one of the most important laws of physics—the Law of Conservation of Energy. In modern terminology, the law is stated as follows: In a conservative force field, the sum of the potential and kinetic energies of an object remains constant from point to point. You can use the Fundamental Theorem of Line Integrals to derive this law. From physics, the kinetic energy of a particle of mass m and speed v is k  12 mv 2. The potential energy p of a particle at point x, y, z in a conservative vector field F is defined as px, y, z  f x, y, z, where f is the potential function for F. Consequently, the work done by F along a smooth curve C from A to B is W



C



F  dr  f x, y, z

B A



B

 px, y, z

A

 pA  pB as shown in Figure 15.25. In other words, work W is equal to the difference in the potential energies of A and B. Now, suppose that rt is the position vector for a particle moving along C from A  ra to B  rb. At any time t, the particle’s velocity, acceleration, and speed are vt  rt, at  r t, and vt  vt , respectively. So, by Newton’s Second Law of Motion, F  mat  mvt, and the work done by F is W



C

F  dr  

  

b

a b a b



a



m 2

F  rt dt F  vt dt 



b

a

mvt  vt dt

m vt  vt dt

 

b

a b

d

vt  vt dt dt

m d 

vt 2 dt 2 a dt b m  vt 2 2 a

y

A

  m   vt  2

F

b

2

C

a

1 1  m vb 2  m va 2 2 2  kB  kA.

B x

The work done by F along C is W



C

F

 dr  pA  pB .

Figure 15.25

Equating these two results for W produces pA  pB  kB  kA pA  kA  pB  kB which implies that the sum of the potential and kinetic energies remains constant from point to point.

1086

CHAPTER 15

Vector Analysis

Exercises for Section 15.3 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, show that the value of C F  d r is the same for each parametric representation of C.

14. Fx, y  xy 2 i  2x 2y j 1 (a) r1t  t i  j, t

1. Fx, y  x i  xy j 2

(a) r1t  t i  t 2 j,

1 (b) r2t  t  1 i  3t  3 j,

0 ≤ t ≤ 1

(b) r2   sin i  sin2 j,

0 ≤ ≤

 2

15.



y 2 dx  2xy dy y

(a)

y=

4

0 ≤ w ≤ 2

C2

(4, 4) C1

2

 (a) r1   sec i  tan j, 0 ≤ ≤ 3

(−1, 0)

1

−1

x

4. Fx, y  y i  x 2 j

3

2

4

y

(c)

y

(d)

y=

C3

(−1, 1)

e3

(1, 1)

(−1, 0)

x

5. Fx, y  e xsin y i  cos yj (−1, −1)

16.

9. Fx, y, z 

 2xyz j 



y

y

(b)

C2

C1

2

x

(4, 1)

−1

(0, 0)

F  dr.

−1

x 1

(Hint: If F is conservative, the integration may be easier on an alternative path.) (a) r1t  t i  t 2 j, 0 ≤ t ≤ 1

6

(b) r2t  t i  t 3 j, 0 ≤ t ≤ 1

4

12. Fx, y  ye xy i  xe xy j

(a) r1t  t i  t j, 0 ≤ t ≤ 1



C4 x

−1

C3 x 2

−1

(0, −1)

2xy dx  x 2  y 2 dy

C

y2 x2   1 from 5, 0 to 0, 4 25 16

t 2 j,

0 ≤ t ≤ 1

(a) C: ellipse

(c) r3t  t i  t 3 j,

0 ≤ t ≤ 1

(b) C: parabola y  4  x 2 from 2, 0 to 0, 4

(b) r2t  t i 

1 − y2

1

y = ex

1

17.

x=

(0, 1)

(0, 1)

(b) The closed path consisting of line segments from 0, 3 to 0, 0, and then from 0, 0 to 3, 0

y

(2, e 2)

2

(a) r1t  t i  t  3 j, 0 ≤ t ≤ 3

(0, −1)

4

(d)

8

11. Fx, y  2xy i  x j

3

2

y

(c)

2

1 − y2

1

(2, 3) 3

1

13. Fx, y  y i  x j

x=

(0, 1)

4

In Exercises 11–24, find the value of the line integral

C

−1

(1, −1)

2x  3y  1 dx  3x  y  5 dy

(a)

xy 2 k

10. Fx, y, z  sin yz i  xz cos yz j  xy sin yz k



x

1

C

xy 8. Fx, y, z  y ln z i  x ln z j  k z y 2z i

(1, 0)

−1

6. Fx, y  15x 2y 2 i  10x 3yj

1 − x2

C4

In Exercises 5–10, determine whether or not the vector field is conservative.

1  y i  xj y2

x

1

(0, 0) 1

(a) r1t  2  t i  3  t j, 0 ≤ t ≤ 3

(1, 0)

−1

(b) r2t  t  1 i  t j, 0 ≤ t ≤ 3

7. Fx, y 

1 − x2

3

3. Fx, y  y i  x j

(b) r2w  2  ln w i  3  ln w j, 1 ≤ w ≤

y

(b) (3, 4)

0 ≤ t ≤ 4

(b) r2w  w 2 i  w j,

0 ≤ t ≤ 2

C

2. Fx, y  x 2  y 2 i  x j (a) r1t  t i  t j,

1 ≤ t ≤ 3

SECTION 15.3

18.



x 2  y 2 dx  2xy dy

Conservative Vector Fields and Independence of Path

31.

C

(a) C: line segment from 0, 0, 0 to 1, 1, 1

0 ≤ t ≤ 2

(b) r2t  2 cos t i  2 sin t j,

0 ≤ t ≤

(b) C: line segments from 0, 0, 0 to 0, 0, 1 to 1, 1, 1

 2

(c) C: line segments from 0, 0, 0 to 1, 0, 0 to 1, 1, 0 to 1, 1, 1

19. Fx, y, z  yz i  xz j  xy k (a) r1t  t i  2 j  t k, 0 ≤ t ≤ 4 (b) r2t  t 2 i  tj  t 2 k,

32. Repeat Exercise 31 using the integral

0 ≤ t ≤ 2

20. Fx, y, z  i  z j  y k (b) r2t  1  2t i   2t k, 0 ≤ t ≤ 1

33.

(a) r1t  t i 

22. Fx, y, z  y i  x j  3xz 2 k 0 ≤ t ≤ 1

23. Fx, y, z  ez y i  x j  xy k (a) r1t  4 cos t i  4 sin t j  3k, 0 ≤ t ≤  0 ≤ t ≤ 1

24. Fx, y, z  y sin z i  x sin z j  xy cos xk (a) r1t  t 2 i  t 2 j,

0 ≤ t ≤ 2

(b) r2t  4t i  4tj, 0 ≤ t ≤ 1 In Exercises 25 –34, evaluate the line integral using the Fundamental Theorem of Line Integrals. Use a computer algebra system to verify your results.

C

 y i  x j  dr



C

2x  y i  2x  y j  dr

C: smooth curve from 2, 2 to 4, 3

27.



cos x sin y dx  sin x cos y dy

C

C: smooth curve from 0,   to 28.



y dx  x dy x2  y 2



e x sin y dx  e x cos y dy

C

32, 2 

C: smooth curve from 1, 1 to 23, 2

29.



6x dx  4z dy  4y  20z dz

C: smooth curve from 0, 0, 0 to 4, 3, 1 Work In Exercises 35 and 36, find the work done by the force field F in moving an object from P to Q. 35. Fx, y  9x 2y 2 i  6x3y  1 j; P0, 0, Q5, 9 36. Fx, y 

2x x2 i  2 j; P3, 2, Q1, 4 y y

37. Work A stone weighing 1 pound is attached to the end of a two-foot string and is whirled horizontally with one end held fixed. It makes 1 revolution per second. Find the work done by the force F that keeps the stone moving in a circular path. [Hint: Use Force  (mass)(centripetal acceleration).] 38. Work If Fx, y, z  a1i  a 2 j  a3 k is a constant force vector field, show that the work done in moving a particle along any path from P to Q is W  F  PQ . \

C: smooth curve from 0, 0 to 3, 8

26.

34.

2 , 3, 4

C

(a) r1t  cos t i  sin t j  t k, 0 ≤ t ≤  (b) r2t  1  2t i   t k,

sin x dx  z dy  y dz

C: smooth curve from 0, 0, 0 to

 k, 0 ≤ t ≤ 1

(b) r2t  t i  tj   2t  12 k, 0 ≤ t ≤ 1

(b) r2t  4  8t i  3k,

zy dx  xz dy  xy dz.

C

21. Fx, y, z  2y  x i  x 2  z j  2y  4z k t 2j

 

C

(a) r1t  cos t i  sin t j  t 2 k, 0 ≤ t ≤ 



 y  2z dx  x  3z dy  2x  3y dz

C

(a) r1t  t 3 i  t 2 j,

25.



1087

39. Work To allow a means of escape for workers in a hazardous job 50 meters above ground level, a slide wire is installed. It runs from their position to a point on the ground 50 meters from the base of the installation where they are located. Show that the work done by the gravitational force field for a 150-pound man moving the length of the slide wire is the same for each path. (a) rt  t i  50  t j (b) rt  t i  5050  t2j 1

40. Work Can you find a path for the slide wire in Exercise 39 such that the work done by the gravitational force field would differ from the amounts of work done for the two paths given? Explain why or why not.

C

C: cycloid x   sin , y  1  cos from 0, 0 to 2, 0



2x 2y 30. 2  y2 2 dx  x 2  y 2 2 dy  x C C: circle x  42   y  5 2  9 clockwise from 7, 5 to 1, 5

Writing About Concepts 41. State the Fundamental Theorem of Line Integrals. 42. What does it mean that a line integral is independent of path? State the method for determining if a line integral is independent of path.

1088

CHAPTER 15

Vector Analysis

Writing About Concepts (continued)

In Exercises 45 and 46, consider the force field shown in the figure. Is the force field conservative? Explain why or why not.

43. Consider the force field shown in the figure.

45.

y

y

46.

y

x

x

x

−5

True or False? In Exercises 47–50, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. −5

(a) Give a verbal argument that the force field is not conservative because you can identify two paths that require different amounts of work to move an object from 4, 0 to 3, 4. Identify two paths and state which requires the greater amount of work. To print an enlarged copy of the graph, select the MathGraph button. (b) Give a verbal argument that the force field is not conservative because you can find a closed curve C such that



C

F  dr  0.

47. If C1, C2, and C3 have the same initial and terminal points and C1 F  dr1  C2 F  dr2, then C1 F  dr1  C3 F  dr3. 48. If F  y i  x j and C is given by rt  4 sin t i  3 cos t j, 0 ≤ t ≤ , then C F  dr  0. 49. If F is conservative in a region R bounded by a simple closed path and C lies within R, then C F  dr is independent of path. 50. If F  M i  N j and !M!x  !N!y, then F is conservative. 51. A function f is called harmonic if f is harmonic, then

 C

! 2f ! 2f  2  0. Prove that if 2 !x !y

!f !f dx  dy  0 !y !x



where C is a smooth closed curve in the plane.

44. Wind Speed and Direction The map shows the jet stream wind speed vectors over the United States for March 19, 2004. In planning a flight from Dallas to Atlanta in a small plane at an altitude of 5000 feet, is the amount of fuel required independent of the flight path? Is the vector field conservative? Explain.

52. Kinetic and Potential Energy The kinetic energy of an object moving through a conservative force field is decreasing at a rate of 10 units per minute. At what rate is the potential energy changing? 53. Let Fx, y 

y x i 2 j. x2  y 2 x  y2

(a) Show that !N !M  !x !y where M

y x . and N  2 x2  y 2 x  y2

(b) If rt  cos t i  sin t j for 0 ≤ t ≤ , find C F  dr. Dallas

Atlanta

(c) If rt  cos t i  sin t j for 0 ≤ t ≤ , find C F  dr.

(d) If rt  cos t i  sin t j for 0 ≤ t ≤ 2, find C F  dr. Why doesn’t this contradict Theorem 15.7?



(e) Show that " arctan



x  F. y

SECTION 15.4

Section 15.4

Green’s Theorem

1089

Green’s Theorem • Use Green’s Theorem to evaluate a line integral. • Use alternative forms of Green’s Theorem.

r(a) = r(b)

Green’s Theorem In this section, you will study Green’s Theorem, named after the English mathematician George Green (1793–1841). This theorem states that the value of a double integral over a simply connected plane region R is determined by the value of a line integral around the boundary of R. A curve C given by rt  xti  ytj, where a ≤ t ≤ b, is simple if it does not cross itself—that is, rc  rd for all c and d in the open interval a, b. A plane region R is simply connected if its boundary consists of one simple closed curve, as shown in Figure 15.26.

R1

Simply connected R3 R2

THEOREM 15.8

Green’s Theorem

Let R be a simply connected region with a piecewise smooth boundary C, oriented counterclockwise (that is, C is traversed once so that the region R always lies to the left). If M and N have continuous partial derivatives in an open region containing R, then

Not simply connected

Figure 15.26



M dx  N dy 

C

R

C2: y = f2(x)



M dx 

C

M dx 





M dx

C2

b



C1: y = f1(x)

b

Mx, f1x  Mx, f2x dx

On the other hand,

 R

C ′1: x = g1(y)

!M dA  !y

   b

a

b



d



R

R is horizontally simple.

f1x

!M dy dx !y



f2x f1 x

dx

Mx, f2x  Mx, f1x dx.

a

C′2: x = g2(y)

C ′ = C ′1 + C′2

f2x

Mx, y

a b

c

Mx, f2x dx

b

a

x

R is vertically simple.



a

Mx, f1x dx 

a b

 a C = C1 + C2

  

C1

R

Figure 15.27

!N !M  dA. !x !y

Proof A proof is given only for a region that is both vertically simple and horizontally simple, as shown in Figure 15.27.

y

y

 

x

Consequently,



C

M dx  

 R

!M dA. !y

Similarly, you can use g1 y and g2 y to show that C N dy  R !N!x dA. By adding the integrals C M dx and C N dy, you obtain the conclusion stated in the theorem.

1090

CHAPTER 15

Vector Analysis

EXAMPLE 1 y

Using Green’s Theorem

Use Green’s Theorem to evaluate the line integral



C = C1 + C2 (1, 1)

y=x

1

C

where C is the path from 0, 0 to 1, 1 along the graph of y  x3 and from 1, 1 to 0, 0 along the graph of y  x, as shown in Figure 15.28.

C1 C2

y = x3 x

(0, 0)

y 3 dx  x3  3xy 2 dy

1

C is simple and closed, and the region R always lies to the left of C. Figure 15.28

Solution Because M  y 3 and N  x 3  3xy 2, it follows that !N  3x 2  3y 2 and !x

!M  3y 2. !y

Applying Green’s Theorem, you then have



y 3 dx  x 3  3xy 2 dy 

C

      

!N !M  dA !x !y

R 1





x

3x2  3y 2  3y 2 dy dx

0 x3 1 x



3x 2 dy dx

0 x3 1



3x 2y

0 1



x

dx x3

3x 3  3x5 dx

0





3x 4 x 6  4 2

1



0

1  . 4

.

Try It GEORGE GREEN (1793–1841) Green, a self-educated miller’s son, first published the theorem that bears his name in 1828 in an essay on electricity and magnetism. At that time there was almost no mathematical theory to explain electrical phenomena. “Considering how desirable it was that a power of universal agency, like electricity, should, as far as possible, be submitted to calculation, . . . I was induced to try whether it would be possible to discover any general relations existing between this function and the quantities of electricity in the bodies producing it.”

Exploration A

Green’s Theorem cannot be applied to every line integral. Among other restrictions stated in Theorem 15.8, the curve C must be simple and closed. When Green’s Theorem does apply, however, it can save time. To see this, try using the techniques described in Section 15.2 to evaluate the line integral in Example 1. To do this, you would need to write the line integral as



C

y 3 dx  x 3  3xy 2 dy 



y 3 dx  x 3  3xy 2 dy 

C1



y 3 dx  x 3  3xy 2 dy

C2

where C1 is the cubic path given by rt  ti  t 3j from t  0 to t  1, and C2 is the line segment given by rt  1  ti  1  tj from t  0 to t  1.

SECTION 15.4

EXAMPLE 2

Green’s Theorem

1091

Using Green’s Theorem to Calculate Work

While subject to the force

F(x, y) = y 3 i + (x 3 + 3xy 2)j

Fx, y  y 3i  x3  3xy 2j

y

a particle travels once around the circle of radius 3 shown in Figure 15.29. Use Green’s Theorem to find the work done by F.

C 2

Solution From Example 1, you know by Green’s Theorem that

1



x

−2

−1

1

2

y 3 dx  x 3  3xy 2 dy 

C

−1



3x 2 dA.

R

In polar coordinates, using x  r cos and dA  r dr d , the work done is

−2

r=3

W



3x 2 dA 

R

Figure 15.29

     2

0

3

3

2

3r cos 2 r dr d

0

3

r 3 cos2 dr d

0 0 2 4



3 r cos2 d

4 0 0 2 81 3 cos2 d

4 0

3



.

2

243 8

1  cos 2  d

0



sin 2

243

 8 2



243 . 4

Try It



2



0

Exploration B

Exploration A

When evaluating line integrals over closed curves, remember that for conservative vector fields (those for which !N!x  !M!y), the value of the line integral is 0. This is easily seen from the statement of Green’s Theorem:



M dx  N dy 

C

y

R

EXAMPLE 3 C

 

!N !M  dA  0. !x !y



Green’s Theorem and Conservative Vector Fields

Evaluate the line integral



y 3 dx  3xy 2 dy

C

x

C is closed. Figure 15.30

.

where C is the path shown in Figure 15.30. Solution From this line integral, M  y 3 and N  3xy 2. So, !N!x  3y 2 and !M!y  3y 2. This implies that the vector field F  Mi  Nj is conservative, and because C is closed, you can conclude that



y 3 dx  3xy 2 dy  0.

C

Try It

Exploration A

1092

CHAPTER 15

Vector Analysis

Using Green’s Theorem for a Piecewise Smooth Curve

EXAMPLE 4 y

Evaluate



(0, 3) C

arctan x  y 2 dx  e y  x2 dy

C

R

where C is the path enclosing the annular region shown in Figure 15.31. x

(−3, 0)

(−1, 0)

(1, 0)

C is piecewise smooth. Figure 15.31

(3, 0)

Solution In polar coordinates, R is given by 1 ≤ r ≤ 3 for 0 ≤ ≤ . Moreover, !N !M   2x  2y  2r cos  r sin . !x !y So, by Green’s Theorem,



arctan x  y 2 dx  ey  x 2 dy 

C

  

   

2x  y dA

R 

3

2r cos  sin r dr d

0 1 

2cos  sin 

0 



0

.

Try It



3 1

d



52 cos  sin  d

3





52 sin  cos

3



104 . 3

Exploration A

r3 3





0

Open Exploration

In Examples 1, 2, and 4, Green’s Theorem was used to evaluate line integrals as double integrals. You can also use the theorem to evaluate double integrals as line integrals. One useful application occurs when !N!x  !M!y  1.



M dx  N dy 

C

   R



!N !M  dA !x !y



1 dA

R

!N !M  1 !x !y

 area of region R Among the many choices for M and N satisfying the stated condition, the choice of M  y2 and N  x2 produces the following line integral for the area of region R.

THEOREM 15.9

Line Integral for Area

If R is a plane region bounded by a piecewise smooth simple closed curve C, oriented counterclockwise, then the area of R is given by A

1 2



C

x dy  y dx.

SECTION 15.4

Green’s Theorem

1093

Finding Area by a Line Integral

EXAMPLE 5

Use a line integral to find the area of the ellipse y2 x2  2  1. 2 a b y

x2 a2

+

y2 b2

Solution Using Figure 15.32, you can induce a counterclockwise orientation to the elliptical path by letting

=1

x  a cos t

and

y  b sin t, 0 ≤ t ≤ 2.

So, the area is

b

a x

A

R

1 2



 

2

1

a cos tb cos t dt  b sin ta sin t dt 2 0 2 ab  cos 2 t  sin2 t dt 2 0

x dy  y dx 

C



Figure 15.32

.

2



ab t 2

0

  ab.

Try It

Exploration A

Green’s Theorem can be extended to cover some regions that are not simply connected. This is demonstrated in the next example. y 2

C1

R

−3

EXAMPLE 6

C1: Ellipse C2: Circle

Let R be the region inside the ellipse x 29   y 24  1 and outside the circle x 2  y 2  1. Evaluate the line integral

C3 C2

3

C4

Green’s Theorem Extended to a Region with a Hole

x



2xy dx  x 2  2x dy

C

where C  C1  C2 is the boundary of R, as shown in Figure 15.33. −2

C3: y = 0, 1 ≤ x ≤ 3 C4: y = 0, 1 ≤ x ≤ 3

Figure 15.33

Solution To begin, you can introduce the line segments C3 and C4, as shown in Figure 15.33. Note that because the curves C3 and C4 have opposite orientations, the line integrals over them cancel. Furthermore, you can apply Green’s Theorem to the region R using the boundary C1  C4  C2  C3 to obtain



2xy dx  x 2  2x dy 

C

    R



!N !M  dA !x !y



2x  2  2x dA

R

2

dA

R

 2area of R  2ab   r 2  2  32   12  10.

.

Try It

Exploration A

1094

CHAPTER 15

Vector Analysis

In Section 15.1, a necessary and sufficient condition for conservative vector fields was listed. There, only one direction of the proof was shown. You can now outline the other direction, using Green’s Theorem. Let Fx, y  Mi  Nj be defined on an open disk R. You want to show that if M and N have continuous first partial derivatives and !M !N  !y !x then F is conservative. Suppose that C is a closed path forming the boundary of a connected region lying in R. Then, using the fact that !M!y  !N!x, you can apply Green’s Theorem to conclude that



C

F  dr  

  

M dx  N dy

C

R

!N !M dA  !x !y



 0. This, in turn, is equivalent to showing that F is conservative (see Theorem 15.7).

Alternative Forms of Green’s Theorem This section concludes with the derivation of two vector forms of Green’s Theorem for regions in the plane. The extension of these vector forms to three dimensions is the basis for the discussion in the remaining sections of this chapter. If F is a vector field in the plane, you can write Fx, y, z  Mi  Nj  0k

 

so that the curl of F, as described in Section 15.1, is given by i j k ! ! ! curl F  "  F  !x !y !z M N 0 !M !N !M !N i j  k.  !z !z !x !y





Consequently,



curl F  k   

!N !M !N !M i j  k !z !z !x !y



 k

!N !M  . !x !y

With appropriate conditions on F, C, and R, you can write Green’s Theorem in the vector form



C

F  dr  

   R

R

!N !M  dA !x !y



curl F  k dA.

First alternative form

The extension of this vector form of Green’s Theorem to surfaces in space produces Stokes’s Theorem, discussed in Section 15.8.

SECTION 15.4

Green’s Theorem

1095

For the second vector form of Green’s Theorem, assume the same conditions for F, C, and R. Using the arc length parameter s for C, you have rs  xsi  ysj. So, a unit tangent vector T to curve C is given by rs  T  xsi  ysj.

n C

T θ

N = −n

T  cos i  sin j   n  cos  i  sin  j 2 2   sin i  cos j N  sin i  cos j









From Figure 15.34 you can see that the outward unit normal vector N can then be written as N  ysi  xsj. Consequently, for Fx, y  Mi  Nj, you can apply Green’s Theorem to obtain



C

F  N ds  

Figure 15.34

       b

a b

Mi  Nj   ysi  xsj ds M

a





dy dx ds N ds ds

M dy  N dx

C



N dx  M dy

C



R



!M !N  dA !x !y



Green’s Theorem

div F dA.

R

Therefore,



C

F  N ds 



div F dA.

Second alternative form

R

The extension of this form to three dimensions is called the Divergence Theorem, discussed in Section 15.7. The physical interpretations of divergence and curl will be discussed in Sections 15.7 and 15.8.

SECTION 15.4

Green’s Theorem

1095

Exercises for Section 15.4 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, verify Green’s Theorem by evaluating both integrals



y2 dx  x 2 dy 

C

  R

N M  dA x y



1. C: square with vertices 0, 0, 4, 0, 4, 4, 0, 4 2. C: triangle with vertices 0, 0, 4, 0, 4, 4 3. C: boundary of the region lying between the graphs of y  x and y  x 24 4. C: circle given by



y2

1

xe y dx  e x dy 

C

for the given path.

  R



 y  x dx  2x  y dy

C

for the given path. 7. C: boundary of the region lying between the graphs of y  x and y  x 2  x 8. C: x  2 cos , y  sin

In Exercises 5 and 6, verify Green’s Theorem by using a computer algebra system to evaluate both integrals



6. C: boundary of the region lying between the graphs of y  x and y  x3 in the first quadrant In Exercises 7–10, use Green’s Theorem to evaluate the integral

for the given path.

x2

5. C: circle given by x 2  y 2  4

N M dA  x y



9. C: boundary of the region lying inside the rectangle bounded by x  5, x  5, y  3, and y  3, and outside the square bounded by x  1, x  1, y  1, and y  1 10. C: boundary of the region lying inside the semicircle y  25  x 2 and outside the semicircle y  9  x 2

1096

CHAPTER 15

Vector Analysis

In Exercises 11–20, use Green’s Theorem to evaluate the line integral. 11.



25. R: region bounded by the graph of x 2  y 2  a 2

2xy dx  x  y dy

C

C: boundary of the region lying between the graphs of y  0 and y  4  x 2

12.



y 2 dx  xy dy

C

C: boundary of the region lying between the graphs of y  0, y  x, and x  9

13.



x 2  y 2 dx  2xy dy

14.

C



x 2  y 2 dx  2xy dy

C: r  1  cos



y 2 arctan dx  lnx 2  y 2 dy x C

C: x  4  2 cos , y  4  sin

16.



26. R: triangle bounded by the graphs of x  0, 3x  2y  0, and x  2y  8 27. R: region bounded by the graphs of y  2x  1 and y  4  x2 28. R: region inside the loop of the folium of Descartes bounded by the graph of x

t3

3t , 1

y

3t 2 1

t3

C

C: x 2  y 2  a 2

15.

Area In Exercises 25–28, use a line integral to find the area of the region R.

Writing About Concepts 29. State Green’s Theorem. 30. Give the line integral for the area of a region R bounded by a piecewise smooth simple curve C.

e x cos 2y dx  2e x sin 2y dy

C

C: x 2  y 2  a 2

17.



sin x cos y dx  xy  cos x sin y dy

C

C: boundary of the region lying between the graphs of y  x and y  x

18.



ex 2  y dx  ey 2

22

 x dy

In Exercises 31 and 32, use Green’s Theorem to verify the line integral formulas. 31. The centroid of the region having area A bounded by the simple closed path C is x

C

C: boundary of the region lying between the graphs of the circle x  6 cos , y  6 sin and the ellipse x  3 cos , y  2 sin

19.



xy dx  x  y dy

C

C: boundary of the region lying between the graphs of x 2  y 2  1 and x 2  y 2  9

20.



3x 2e y dx  ey dy

C

C: boundary of the region lying between the squares with vertices 1, 1, 1, 1, 1, 1, and 1, 1, and 2, 2, 2, 2, 2, 2, and 2, 2 Work In Exercises 21–24, use Green’s Theorem to calculate the work done by the force F on a particle that is moving counterclockwise around the closed path C. 21. Fx, y  xyi  x  yj C: x 2  y 2  4 22. Fx, y  e x  3yi  ey  6xj C: r  2 cos

23. Fx, y  x 32  3yi  6x  5y j C: boundary of the triangle with vertices 0, 0, 5, 0, and 0, 5 24. Fx, y  3x 2  yi  4xy 2 j C: boundary of the region lying between the graphs of y  x, y  0, and x  9

1 2A



y

x 2 dy,

C

1 2A



y 2 dx.

C

32. The area of a plane region bounded by the simple closed path C 1

given in polar coordinates is A  2

r C

2

d .

Centroid In Exercises 33–36, use a computer algebra system and the results of Exercise 31 to find the centroid of the region. 33. R: region bounded by the graphs of y  0 and y  4  x 2 34. R: region bounded by the graphs of y  a 2  x 2 and y  0 35. R: region bounded by the graphs of y  x3 and y  x, 0 ≤ x ≤ 1 36. R: triangle with vertices a, 0, a, 0, and b, c, where a ≤ b ≤ a Area In Exercises 37–40, use a computer algebra system and the results of Exercise 32 to find the area of the region bounded by the graph of the polar equation. 37. r  a1  cos 

38. r  a cos 3

39. r  1  2 cos (inner loop)

40. r 

41. Think About It I



C

3 2  cos

Let

y dx  x dy x2  y 2

where C is a circle oriented counterclockwise. Show that I  0 if C does not contain the origin. What is I if C contains the origin?

SECTION 15.4

42. (a) Let C be the line segment joining x1, y1 and x2, y2. Show that



y dx  x dy  x1 y2  x2 y1.

C

(b) Let x1, y1, x2, y2, . . . , xn, yn  be the vertices of a polygon. Prove that the area enclosed is 1 2 x1y2

 x2 y1  x2 y3  x3 y2  . . . 

xn1 yn  xn yn1  xn y1  x1 yn . Area In Exercises 43 and 44, use the result of Exercise 42(b) to find the area enclosed by the polygon with the given vertices.

Green’s Theorem

In Exercises 46 and 47, prove the identity where R is a simply connected region with boundary C. Assume that the required partial derivatives of the scalar functions f and g are continuous. The expressions DN f and DN g are the derivatives in the direction of the outward normal vector N of C, and are defined by DN f  f  N, and DN g  g  N. 46. Green’s first identity:

 R

 f " 2g  "f  "g dA 



[Hint: Use the second alternative form of Green’s Theorem and the property div  f G  f div G  "f  G. 47. Green’s second identity:



45. Investigation Consider the line integral

(Hint: Use Exercise 46 twice.)



y n dx  x n dy

C

where C is the boundary of the region lying between the graphs of y  a 2  x 2 a > 0 and y  0. (a) Use a computer algebra system to verify Green’s Theorem for n, an odd integer from 1 through 7. (b) Use a computer algebra system to verify Green’s Theorem for n, an even integer from 2 through 8. (c) For n an odd integer, make a conjecture about the value of the integral.

f DNg ds

C

44. Hexagon: 0, 0, 2, 0, 3, 2, 2, 4, 0, 3, 1, 1

43. Pentagon: 0, 0, 2, 0, 3, 2, 1, 4, 1, 1

1097

 f "2g  g"2f  dA 

R



 f DNg  g DN f  ds

C

48. Use Green’s Theorem to prove that



f x dx  g y dy  0

C

if f and g are differentiable functions and C is a piecewise smooth simple closed path. 49. Let F  Mi  Nj, where M and N have continuous first partial derivatives in a simply connected region R. Prove that if C is simple, smooth, and closed, and Nx  My , then



C

F  dr  0.

1098

CHAPTER 15

Section 15.5

Vector Analysis

Parametric Surfaces • • • •

Understand the definition of and sketch a parametric surface. Find a set of parametric equations to represent a surface. Find a normal vector and a tangent plane to a parametric surface. Find the area of a parametric surface.

Parametric Surfaces You already know how to represent a curve in the plane or in space by a set of parametric equations—or, equivalently, by a vector-valued function. rt  xti  ytj rt  xti  ytj  ztk

Plane curve Space curve

In this section, you will learn how to represent a surface in space by a set of parametric equations—or by a vector-valued function. For curves, note that the vector-valued function r is a function of a single parameter t. For surfaces, the vector-valued function is a function of two parameters u and v.

Definition of Parametric Surface Let x, y, and z be functions of u and v that are continuous on a domain D in the uv-plane. The set of points x, y, z given by ru, v  xu, vi  yu, vj  zu, vk

Parametric surface

is called a parametric surface. The equations x  xu, v,

y  yu, v, and

z  zu, v

Parametric equations

are the parametric equations for the surface.

If S is a parametric surface given by the vector-valued function r, then S is traced out by the position vector ru, v as the point u, v moves throughout the domain D, as shown in Figure 15.35. v

z

D

S (u, v)

r(u, v)

y

.

u

Figure 15.35

x

Rotatable Graph

Some computer algebra systems are capable of graphing surfaces that are represented parametrically. If you have access to such software, use it to graph some of the surfaces in the examples and exercises in this section.

TECHNOLOGY

SECTION 15.5

z

EXAMPLE 1

Parametric Surfaces

1099

Sketching a Parametric Surface

3

Identify and sketch the parametric surface S given by ru, v  3 cos ui  3 sin uj  vk where 0 ≤ u ≤ 2 and 0 ≤ v ≤ 4. 4

Solution Because x  3 cos u and y  3 sin u, you know that for each point x, y, z on the surface, x and y are related by the equation x2  y2  32. In other words, each cross section of S taken parallel to the xy-plane is a circle of radius 3, centered on the z-axis. Because z  v, where 0 ≤ v ≤ 4, you can see that the surface is a right circular cylinder of height 4. The radius of the cylinder is 3, and the z-axis forms the axis of the cylinder, as shown in Figure 15.36.

y

.

x

Figure 15.36

Try It

Rotatable Graph

Exploration A

As with parametric representations of curves, parametric representations of surfaces are not unique. That is, there are many other sets of parametric equations that could be used to represent the surface shown in Figure 15.36. EXAMPLE 2

z

c3

Identify and sketch the parametric surface S given by

c2 d1

c4

Sketching a Parametric Surface

ru, v  sin u cos vi  sin u sin vj  cos uk where 0 ≤ u ≤  and 0 ≤ v ≤ 2.

c1

d2

d3

x

d4

Solution To identify the surface, you can try to use trigonometric identities to eliminate the parameters. After some experimentation, you can discover that y

x2  y2  z2  sin u cos v2  sin u sin v2  cos u2  sin2 u cos2 v  sin2 u sin2 v  cos2 u  sin2 ucos2 v  sin2 v  cos2 u  sin2 u  cos2 u  1.

. Figure 15.37

Rotatable Graph

So, each point on S lies on the unit sphere, centered at the origin, as shown in Figure 15.37. For fixed u  di , ru, v traces out latitude circles x2  y2  sin2 di,

.

0 ≤ di ≤ 

that are parallel to the xy-plane, and for fixed v  ci , ru, v traces out longitude (or meridian) half-circles.

Try It

Exploration A

NOTE To convince yourself further that the vector-valued function in Example 2 traces out the entire unit sphere, recall that the parametric equations x   sin  cos ,

y   sin  sin , and

z   cos 

where 0 ≤ ≤ 2 and 0 ≤  ≤ , describe the conversion from spherical to rectangular coordinates, as discussed in Section 11.7.

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CHAPTER 15

Vector Analysis

Finding Parametric Equations for Surfaces In Examples 1 and 2, you were asked to identify the surface described by a given set of parametric equations. The reverse problem—that of writing a set of parametric equations for a given surface—is generally more difficult. One type of surface for which this problem is straightforward, however, is a surface that is given by z  f x, y. You can parametrize such a surface as rx, y  xi  yj  f x, yk.

z

Representing a Surface Parametrically

EXAMPLE 3

3

Write a set of parametric equations for the cone given by 2

z  x2  y2 as shown in Figure 15.38. Solution Because this surface is given in the form z  f x, y, you can let x and y be the parameters. Then the cone is represented by the vector-valued function

−2 1 2

x

1 2

.

rx, y  xi  yj  x2  y2 k

y

where x, y varies over the entire xy-plane.

Figure 15.38

Try It

Rotatable Graph

Exploration A

A second type of surface that is easily represented parametrically is a surface of revolution. For instance, to represent the surface formed by revolving the graph of y  f x, a ≤ x ≤ b, about the x-axis, use x  u,

.

y  f u cos v, and

z  f u sin v

where a ≤ u ≤ b and 0 ≤ v ≤ 2. Select the animation button to see an example.

Animation Representing a Surface of Revolution Parametrically

EXAMPLE 4

Write a set of parametric equations for the surface of revolution obtained by revolving

z

1 f x  , x

1 1

y

1 ≤ x ≤ 10

about the x-axis. Solution Use the parameters u and v as described above to write x  u,

10 x

.

Figure 15.39

Rotatable Graph

y  f u cos v 

1 cos v, and u

z  f u sin v 

1 sin v u

where 1 ≤ u ≤ 10 and 0 ≤ v ≤ 2. The resulting surface is a portion of Gabriel’s Horn, as shown in Figure 15.39.

Try It

Exploration A

Open Exploration

The surface of revolution in Example 4 is formed by revolving the graph of y  f x about the x-axis. For other types of surfaces of revolution, a similar parametrization can be used. For instance, to parametrize the surface formed by revolving the graph of x  f z about the z-axis, you can use z  u,

x  f u cos v, and

y  f u sin v.

SECTION 15.5

Parametric Surfaces

1101

Normal Vectors and Tangent Planes Let S be a parametric surface given by ru, v  xu, vi  yu, vj  zu, vk over an open region D such that x, y, and z have continuous partial derivatives on D. The partial derivatives of r with respect to u and v are defined as ru 

!x !y !z u, vi  u, vj  u, vk !u !u !u

rv 

!x !y !z u, vi  u, vj  u, vk. !v !v !v

and

Each of these partial derivatives is a vector-valued function that can be interpreted geometrically in terms of tangent vectors. For instance, if v  v0 is held constant, then ru, v0  is a vector-valued function of a single parameter and defines a curve C1 that lies on the surface S. The tangent vector to C1 at the point xu0, v0 , yu0, v0 , zu0, v0  is given by ruu0, v0   N

as shown in Figure 15.40. In a similar way, if u  u0 is held constant, then ru0, v  is a vector-valued function of a single parameter and defines a curve C2 that lies on the surface S. The tangent vector to C2 at the point xu0, v0 , yu0, v0 , zu0, v0  is given by

z

(x0, y0, z0) rv C2

ru

rvu0, v0  

C1 S

x y

. Figure 15.40

!x !y !z u , v i  u0, v0 j  u0, v0 k !u 0 0 !u !u

!x !y !z u , v i  u0, v0 j  u0, v0 k. !v 0 0 !v !v

If the normal vector ru  rv is not 0 for any u, v in D, the surface S is called smooth and will have a tangent plane. Informally, a smooth surface is one that has no sharp points or cusps. For instance, spheres, ellipsoids, and paraboloids are smooth, whereas the cone given in Example 3 is not smooth.

Rotatable Graph Normal Vector to a Smooth Parametric Surface Let S be a smooth parametric surface ru, v  xu, vi  yu, vj  zu, vk defined over an open region D in the uv-plane. Let u0, v0  be a point in D. A normal vector at the point

x0, y0, z0   xu0, v0 , yu0, v0 , zu0, v0  is given by

  i

!x N  ruu0, v0   rvu0, v0   !u !x !v

j

!y !u !y !v

k

!z !u . !z !v

NOTE Figure 15.40 shows the normal vector ru  rv . The vector rv  ru is also normal to S and points in the opposite direction.

1102

CHAPTER 15

Vector Analysis

EXAMPLE 5

Finding a Tangent Plane to a Parametric Surface

Find an equation of the tangent plane to the paraboloid given by ru, v  ui  vj  u2  v2k z

at the point (1, 2, 5). Solution The point in the uv-plane that is mapped to the point x, y, z  1, 2, 5 is u, v  1, 2. The partial derivatives of r are

7 6

ru  i  2uk and

(1, 2, 5)

rv  j  2vk.

 

The normal vector is given by ru

−3

 rv

i  1 0

j 0 1

k 2u  2ui  2vj  k 2v

which implies that the normal vector at 1, 2, 5 is ru equation of the tangent plane at 1, 2, 5 is −2

−1

1

2

2

y

3

3

x

.

 rv

 2i  4j  k. So, an

2x  1  4 y  2  z  5  0 2x  4y  z  5. The tangent plane is shown in Figure 15.41.

Figure 15.41

Try It

Rotatable Graph

Exploration A

Area of a Parametric Surface

v

To define the area of a parametric surface, you can use a development that is similar to that given in Section 14.5. Begin by constructing an inner partition of D consisting of n rectangles, where the area of the ith rectangle Di is Ai  ui vi , as shown in Figure 15.42. In each Di let ui, vi  be the point that is closest to the origin. At the point xi, yi, zi   xui, vi , yui, vi , zui, vi  on the surface S, construct a tangent plane Ti . The area of the portion of S that corresponds to Di, Ti , can be approximated by a parallelogram in the tangent plane. That is, Ti  Si . So, the surface of S is given by  Si   Ti . The area of the parallelogram in the tangent plane is

Di Δvi Δui

ui ru  vi rv   ru

(ui, vi)

u

 rv 

ui vi

which leads to the following definition.

Area of a Parametric Surface

z

Δvi rv

Let S be a smooth parametric surface ru, v  xu, vi  yu, vj  zu, vk

S

defined over an open region D in the uv-plane. If each point on the surface S corresponds to exactly one point in the domain D, then the surface area of S is given by Δui ru

Surface area 

   dS 

S

y x

Figure 15.42

where ru 

ru

 rv

dA

D

!y !z !y !z !x !x i j k and rv  i j k. !u !u !u !v !v !v

SECTION 15.5

Parametric Surfaces

1103

For a surface S given by z  f x, y, this formula for surface area corresponds to that given in Section 14.5. To see this, you can parametrize the surface using the vectorvalued function rx, y  xi  yj  f x, yk defined over the region R in the xy-plane. Using rx  i  fxx, yk and



you have

i rx  ry  1 0

j 0 1

ry  j  fyx, yk



k fxx, y  fxx, yi  fyx, yj  k fyx, y

and rx  ry    fxx, y  fyx, y 2  1. This implies that the surface area of S is 2

Surface area 

 

rx  ry  dA

R



1  fxx, y 2  fyx, y 2 dA.

R

Finding Surface Area

EXAMPLE 6 NOTE The surface in Example 6 does not quite fulfill the hypothesis that each point on the surface corresponds to exactly one point in D. For this surface, ru, 0  ru, 2 for any fixed value of u. However, because the overlap consists of only a semicircle (which has no area), you can still apply the formula for the area of a parametric surface.

Find the surface area of the unit sphere given by ru, v  sin u cos vi  sin u sin vj  cos uk where the domain D is given by 0 ≤ u ≤  and 0 ≤ v ≤ 2. Solution Begin by calculating ru and rv. ru  cos u cos vi  cos u sin vj  sin uk rv  sin u sin vi  sin u cos vj



The cross product of these two vectors is ru

 rv

i j k  cos u cos v cos u sin v sin u sin u sin v sin u cos v 0



 sin2 u cos vi  sin2 u sin vj  sin u cos uk which implies that ru

 rv 

 sin2 u cos v2  sin2 u sin v2  sin u cos u2  sin4 u  sin2 u cos2 u  sin2 u  sin u.

sin u > 0 for 0 ≤ u ≤ 

Finally, the surface area of the sphere is A



ru

 rv 

dA 

D

0

 .

  2

2



sin u du dv

0

2 dv

0

 4 .

Try It

Exploration A

1104

CHAPTER 15

Vector Analysis

Finding Surface Area

EXAMPLE 7

E X P L O R AT I O N For the torus in Example 7, describe the function ru, v for fixed u. Then describe the function ru, v for fixed v.

Find the surface area of the torus given by ru, v  2  cos u cos vi  2  cos u sin vj  sin uk where the domain D is given by 0 ≤ u ≤ 2 and 0 ≤ v ≤ 2. (See Figure 15.43.)

z

Solution Begin by calculating ru and rv . ru  sin u cos vi  sin u sin vj  cos uk rv   2  cos u sin vi  2  cos u cos vj





The cross product of these two vectors is

y x

.

Figure 15.43

Rotatable Graph

i j k ru  rv  sin u cos v sin u sin v cos u  2  cos u sin v 2  cos u cos v 0   2  cos u cos v cos ui  sin v cos uj  sin uk which implies that ru

 rv 

 2  cos ucos v cos u2  sin v cos u2  sin2 u  2  cos ucos2 ucos2 v  sin2 v  sin2 u  2  cos ucos2 u  sin2 u  2  cos u.

Finally, the surface area of the torus is A



ru

 rv 

dA 

D

 .

  2

2

2  cos u du dv

0 0 2

4 dv

0

 8 2.

Try It

Exploration A

Exploration B

If the surface S is a surface of revolution, you can show that the formula for surface area given in Section 7.4 is equivalent to the formula given in this section. For instance, suppose f is a nonnegative function such that f is continuous over the interval a, b . Let S be the surface of revolution formed by revolving the graph of f, where a ≤ x ≤ b, about the x-axis. From Section 7.4, you know that the surface area is given by



b

Surface area  2

f x1  fx 2 dx.

a

To represent S parametrically, let x  u, y  f u cos v, and z  f u sin v, where a ≤ u ≤ b and 0 ≤ v ≤ 2. Then, ru, v  ui  f u cos vj  f u sin vk. Try showing that the formula Surface area 



ru

 rv 

dA

D

is equivalent to the formula given above (see Exercise 52).

SECTION 15.5

Parametric Surfaces

1105

Exercises for Section 15.5 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, match the vector-valued function with its graph. [The graphs are labeled (a), (b), (c), and (d).] z

(a)

4

4

x

0 ≤ v ≤ 2

0 ≤ u ≤ 3,

2

2

0 ≤ v ≤ 2

0 ≤ u ≤ 2,

11. su, v  u cos vi  u sin vj  u2k

z

(b)

10. su, v  u cos vi  u2j  u sin vk

12. su, v  4u cos vi  4u sin vj  u2k 0 ≤ v ≤ 2

0 ≤ u ≤ 2,

y 2

x

2

y

In Exercises 13–18, use a computer algebra system to graph the surface represented by the vector-valued function. 13. ru, v  2u cos vi  2u sin vj  u 4k

Rotatable Graph

0 ≤ v ≤ 2

0 ≤ u ≤ 1,

Rotatable Graph

14. ru, v  2 cos v cos ui  4 cos v sin uj  sin vk (c)

z

(d)

z

0 ≤ u ≤ 2,

4

0 ≤ v ≤ 2

0 ≤ u ≤ 2, 2

−4

y

4

4

x

0 ≤ v ≤ 2

15. ru, v  2 sinh u cos vi  sinh u sin vj  cosh uk

2

16. ru, v  2u cos vi  2u sin vj  vk

2 y

0 ≤ v ≤ 3

0 ≤ u ≤ 1,

17. ru, v  u  sin ucos vi  1  cos usin vj  uk

x

0 ≤ u ≤ ,

Rotatable Graph

Rotatable Graph

1. ru, v  ui  vj  uvk 2. ru, v  u cos vi  u sin vj  uk 3. ru, v  2 cos v cos ui  2 cos v sin uj  2 sin vk 4. ru, v  4 cos ui  4 sin uj  vk In Exercises 5– 8, find the rectangular equation for the surface by eliminating the parameters from the vector-valued function. Identify the surface and sketch its graph. 5. ru, v  ui  vj 

v k 2

1 6. ru, v  2u cos vi  2u sin vj  2 u2 k

7. ru, v  2 cos ui  vj  2 sin uk

0 ≤ v ≤ 2

18. ru, v  cos3 u cos vi  sin3 u sin vj  uk 0 ≤ u ≤

 , 0 ≤ v ≤ 2 2

In Exercises 19–26, find a vector-valued function whose graph is the indicated surface. 19. The plane z  y

20. The plane x  y  z  6

21. The cylinder x2  y2  16 22. The cylinder 4x2  y2  16 23. The cylinder z  x2 24. The ellipsoid

x2 y2 z2   1 9 4 1

25. The part of the plane z  4 that lies inside the cylinder x2  y2  9

8. ru, v  3 cos v cos ui  3 cos v sin uj  5 sin vk

26. The part of the paraboloid z  x2  y2 that lies inside the cylinder x2  y2  9

Think About It In Exercises 9–12, determine how the graph of the surface su, v differs from the graph of ru, v  u cos vi  u sin vj  u2k (see figure) where 0 ≤ u ≤ 2 and 0 ≤ v ≤ 2. (It is not necessary to graph s.)

Surface of Revolution In Exercises 27–30, write a set of parametric equations for the surface of revolution obtained by revolving the graph of the function about the given axis.

x 27. y  , 2

4

r(u, v)

x

2

2

Rotatable Graph

0 ≤ x ≤ 6

y

x-axis

28. y  x 32,

0 ≤ x ≤ 4

x-axis

29. x  sin z,

0 ≤ z ≤ 

z-axis

30. z  4  y2,

−2

−2

Axis of Revolution

Function

z

0 ≤ y ≤ 2

y-axis

1106

CHAPTER 15

Vector Analysis

Tangent Plane In Exercises 31–34, find an equation of the tangent plane to the surface represented by the vector-valued function at the given point.

Area In Exercises 35–42, find the area of the surface over the given region. Use a computer algebra system to verify your results.

31. ru, v  u  vi  u  vj  vk, 1, 1, 1

35. The part of the plane

z

ru, v  2ui 

2

(1, −1, 1)

−4

−2

where 0 ≤ u ≤ 2 and 0 ≤ v ≤ 1 36. The part of the paraboloid ru, v  4u cos v i  4u sin vj  u2 k, where 0 ≤ u ≤ 2 and 0 ≤ v ≤ 2

2

4

v v j k 2 2

2

x

4

37. The part of the cylinder ru, v  a cos ui  a sin uj  vk, where 0 ≤ u ≤ 2 and 0 ≤ v ≤ b

y

−2

38. The sphere ru, v  a sin u cos v i  a sin u sin vj a cos uk, where 0 ≤ u ≤  and 0 ≤ v ≤ 2

Rotatable Graph 32. ru, v  ui  vj  uv k,

39. The part of the cone ru, v  au cos vi  au sin vj  uk, where 0 ≤ u ≤ b and 0 ≤ v ≤ 2

1, 1, 1

40. The torus ru, v  a  b cos vcos ui  a  b cos vsin uj  b sin vk, where a > b, 0 ≤ u ≤ 2, and 0 ≤ v ≤ 2

z 2

41. The surface of revolution ru, v  u cos vi  u sin vj  uk, where 0 ≤ u ≤ 4 and 0 ≤ v ≤ 2

1 (1, 1, 1)

42. The surface of revolution ru, v  sin u cos vi  uj  sin u sin vk, where 0 ≤ u ≤  and 0 ≤ v ≤ 2

y 2

Writing About Concepts 1

43. Define a parametric surface.

2 x

44. Give the double integral that yields the surface area of a parametric surface over an open region D.

Rotatable Graph 33. ru, v  2u cos vi  3u sin vj  u2 k, 0, 6, 4 z 6 5

(0, 6, 4)

45. The four figures are graphs of the surface ru, v  ui  sin u cos vj  sin u sin vk,  0 ≤ u ≤ , 0 ≤ v ≤ 2. 2 Match each of the four graphs with the point in space from which the surface is viewed. The four points are 10, 0, 0, 10, 10, 0, 0, 10, 0, and 10, 10, 10. z

(a)

z

(b)

−6 4

2

y 2

x

4

6

y

x

y

Rotatable Graph Rotatable Graph 34. ru, v  2u cosh vi  2u sinh vj  12 u2 k, 4, 0, 2 z

z

(c)

(d)

z

4

x

(−4, 0, 2) 2

x

6

4

−2

2 4

y

Rotatable Graph

−4

−6

y

SECTION 15.5

46. Use a computer algebra system to graph three views of the graph of the vector-valued function

1107

Parametric Surfaces

53. Open-Ended Project The parametric equations x  3  sin u 7  cos3u  2v  2 cos3u  v

ru, v  u cos vi  u sin vj  vk, 0 ≤ u ≤ , 0 ≤ v ≤ 

y  3  cos u 7  cos3u  2v  2 cos3u  v

from the points 10, 0, 0, 0, 0, 10, and 10, 10, 10.

z  sin3u  2v  2 sin3u  v

47. Investigation Use a computer algebra system to graph the torus

where   ≤ u ≤  and   ≤ v ≤ , represent the surface shown below. Try to create your own parametric surface using a computer algebra system.

ru, v  a  b cos v cos ui 

a  b cos v sin uj  b sin vk for each set of values of a and b, where 0 ≤ u ≤ 2 and 0 ≤ v ≤ 2. Use the results to describe the effects of a and b on the shape of the torus. (a) a  4,

b1

(b) a  4,

b2

(c) a  8,

b1

(d) a  8,

b3

48. Investigation Consider the function in Exercise 16. (a) Sketch a graph of the function where u is held constant at u  1. Identify the graph. (b) Sketch a graph of the function where v is held constant at v  23. Identify the graph. (c) Assume that a surface is represented by the vector-valued function r  ru, v. What generalization can you make about the graph of the function if one of the parameters is held constant? 49. Surface Area given by

The surface of the dome on a new museum is

Rotatable Graph 54. Möbius Strip The surface shown in the figure is called a Möbius Strip and can be represented by the parametric equations



ru, v  20 sin u cos vi  20 sin u sin vj  20 cos uk

x  a  u cos

where 0 ≤ u ≤ 3 and 0 ≤ v ≤ 2 and r is in meters. Find the surface area of the dome.







v v v cos v, y  a  u cos sin v, z  u sin 2 2 2

where 1 ≤ u ≤ 1, 0 ≤ v ≤ 2, and a  3. Try to graph other Möbius strips for different values of a using a computer algebra system.

50. Find a vector-valued function for the hyperboloid x2  y2  z2  1

z

and determine the tangent plane at 1, 0, 0.

2

51. Graph and find the area of one turn of the spiral ramp −3

ru, v  u cos vi  u sin vj  2vk where 0 ≤ u ≤ 3, and 0 ≤ v ≤ 2. 52. Let f be a nonnegative function such that f is continuous over the interval a, b . Let S be the surface of revolution formed by revolving the graph of f, where a ≤ x ≤ b, about the x-axis. Let x  u, y  f u cos v, and z  f usin v, where a ≤ u ≤ b and 0 ≤ v ≤ 2. Then, S is represented parametrically by ru, v  ui  f u cos vj  f u sin vk. Show that the following formulas are equivalent.

 

b

Surface area  2

f x1  fx 2 dx

a

Surface area 

D

ru  rv  dA

x

4

2

−4

−1

1 3 −2

Rotatable Graph

y

1108

CHAPTER 15

Vector Analysis

Section 15.6

Surface Integrals • • • •

Evaluate a surface integral as a double integral. Evaluate a surface integral for a parametric surface. Determine the orientation of a surface. Understand the concept of a flux integral.

Surface Integrals The remainder of this chapter deals primarily with surface integrals. You will first consider surfaces given by z  gx, y. Later in this section you will consider more general surfaces given in parametric form. Let S be a surface given by z  gx, y and let R be its projection onto the xy-plane, as shown in Figure 15.44. Suppose that g, gx, and gy are continuous at all points in R and that f is defined on S. Employing the procedure used to find surface area in Section 14.5, evaluate f at xi, yi , z i  and form the sum

z

S: z = g(x, y)

n

(xi , yi, zi)

 f x , y , z  S i

i

i

i

i1

x

(xi, yi)

where Si  1  gxxi , yi  2  gyxi , yi  2 Ai. Provided the limit of the above sum as  approaches 0 exists, the surface integral of f over S is defined as

R

Scalar function f assigns a number to each . of S. point Figure 15.44

y



f x, y, z dS  lim

n

 f x , y , z  S .

 →0 i1

S

i

i

i

i

This integral can be evaluated by a double integral.

Rotatable Graph THEOREM 15.10

Evaluating a Surface Integral

Let S be a surface with equation z  gx, y and let R be its projection onto the xy-plane. If g, gx , and gy are continuous on R and f is continuous on S, then the surface integral of f over S is



f x, y, z dS 

S



f x, y, gx, y1  gxx, y 2  gyx, y 2 dA.

R

For surfaces described by functions of x and z (or y and z), you can make the following adjustments to Theorem 15.10. If S is the graph of y  gx, z and R is its projection onto the xz-plane, then



f x, y, z dS 

S



f x, gx, z, z1  gxx, z 2  gzx, z 2 dA.

R

If S is the graph of x  g y, z and R is its projection onto the yz-plane, then

 S

f x, y, z dS 



f g y, z, y, z1  gy y, z 2  gz y, z 2 dA.

R

If f x, y, z  1, the surface integral over S yields the surface area of S. For instance, suppose the surface S is the plane given by z  x, where 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. The surface area of S is 2 square units. Try verifying that S f x, y, z dS  2.

SECTION 15.6

EXAMPLE 1

Surface Integrals

1109

Evaluating a Surface Integral

Evaluate the surface integral



 y 2  2yz dS

S

where S is the first-octant portion of the plane 2x  y  2z  6. Solution Begin by writing S as 1 z  6  2x  y 2 1 gx, y  6  2x  y. 2 Using the partial derivatives gxx, y  1 and gyx, y   12, you can write

1  1  41  32 .

1  gxx, y 2  gyx, y 2  z

Using Figure 15.45 and Theorem 15.10, you obtain

z = 12 (6 − 2x − y)

(0, 0, 3)



 y 2  2yz dS 

S

    

f x, y, gx, y1  gxx, y 2  gyx, y 2 dA

R

S

 (0, 6, 0)

R

23x

3

x

y

(3, 0, 0)

3

y = 2(3 − x)

.

y 2  2y

126  2x  y32 dA

y3  x dy dx

0 0 3

3  x3 dx

6

Figure 15.45

0



3   3  x4 2

Rotatable Graph

3 0

243 .  2

.

Try It z

(0, 0, 3) z=

Exploration A

An alternative solution to Example 1 would be to project S onto the yz-plane, as shown in Figure 15.46. Then, x  126  y  2z, and

6−y 2

1  gyy, z 2  gzy, z 2 

S (0, 6, 0) x

y

(3, 0, 0) x=

. Figure 15.46

1 2

(6 − y − 2z)

So, the surface integral is



 y 2  2yz dS 

S

   R 6



0

Rotatable Graph

Exploration B

1  14  1  32 .

f  g y, z, y, z1  gy y, z 2  gz y, z 2 dA

6y2

0 6

 y 2  2yz

32 dz dy

3 36y  y 3 dy  8 0 243 .  2 Try reworking Example 1 by projecting S onto the xz-plane.

1110

CHAPTER 15

Vector Analysis

In Example 1, you could have projected the surface S onto any one of the three coordinate planes. In Example 2, S is a portion of a cylinder centered about the x-axis, and you can project it onto either the xz-plane or the xy-plane.

Evaluating a Surface Integral

EXAMPLE 2 z

Evaluate the surface integral

R: 0 ≤ x ≤ 4 0≤y≤3

3



x  z dS

S

where S is the first-octant portion of the cylinder y 2  z 2  9 between x  0 and x  4, as shown in Figure 15.47.

3

2

1 3

4 x

.

S:

y2

+

z2

y

Solution Project S onto the xy-plane, so that z  gx, y  9  y 2, and obtain 1  gxx, y 2  gyx, y 2 

 



3 . 9  y 2

=9

Figure 15.47

Rotatable Graph

1

y 9  y 2



2

Theorem 15.10 does not apply directly because gy is not continuous when y  3. However, you can apply the theorem for 0 ≤ b < 3 and then take the limit as b approaches 3, as follows.



      b

x  z dS  lim b→3

S

0

4

0 b 4

x  9  y 2 

3 dx dy 9  y 2



x  1 dx dy b→3 9  y 2 0 0 b 4 x2  lim 3  x dy 2 b→3 0 0 29  y b 8  lim 3  4 dy b→3 9  y 2 0 b y  lim 3 4y  8 arcsin b→3 3 0  lim 3





  b  lim 34b  8 arcsin  3   36  24   2 b→3 

.

 36  12

Try It

Exploration A

Exploration B

Open Exploration

Some computer algebra systems are capable of evaluating improper integrals. If you have access to such computer software, use it to evaluate the improper integral

TECHNOLOGY

  3

0

4

0

x  9  y 2 

3 9  y 2

dx dy.

Do you obtain the same result as in Example 2?

SECTION 15.6

Surface Integrals

1111

You have already seen that if the function f defined on the surface S is simply f x, y, z  1, the surface integral yields the surface area of S.



Area of surface 

1 dS

S

On the other hand, if S is a lamina of variable density and x, y, z is the density at the point x, y, z, then the mass of the lamina is given by Mass of lamina 



x, y, z dS.

S

Finding the Mass of a Surface Lamina

EXAMPLE 3 z

A cone-shaped surface lamina S is given by

Cone: z=4−2

x2 + y2

4

z  4  2x 2  y 2,

0 ≤ z ≤ 4

as shown in Figure 15.48. At each point on S, the density is proportional to the distance between the point and the z-axis. Find the mass m of the lamina.

3

Solution Projecting S onto the xy-plane produces S: z  4  2x2  y2  gx, y, 0 ≤ z ≤ 4 R: x 2  y 2 ≤ 4

2

with a density of x, y, z  kx 2  y 2. Using a surface integral, you can find the mass to be

1

1

2

x

.

m

1

2

R: x 2 + y 2 = 4

       

x, y, z dS

S

y



kx 2  y 21  gxx, y 2  gyx, y 2 dA

R

Figure 15.48

R

Rotatable Graph

k k 

2

2

2



x2

4y 2 dA  y2

5x 2  y 2 dA

R 2 0

5k

3

2



5r r dr d

0 2

0

Polar coordinates

2



r3

0

d

2

85k d

3 0 2 85k 165k . 

 3 3 0 

.

1  x 4x y

x 2  y 2

k



Try It

Exploration A

Exploration B

Use a computer algebra system to confirm the result shown in Example 3. The computer algebra system Derive evaluated the integral as follows.

TECHNOLOGY

 2

4y 2

k

2 4y 2

5x 2  y 2 dx dy 

165k 3

1112

CHAPTER 15

Vector Analysis

Parametric Surfaces and Surface Integrals For a surface S given by the vector-valued function ru, v  xu, v i  yu, vj  zu, vk

Parametric surface

defined over a region D in the uv-plane, you can show that the surface integral of f x, y, z over S is given by





f x, y, z dS 

S

f xu, v, yu, v, zu, v ruu, v  rvu, v  dA.

D

Note the similarity to a line integral over a space curve C.





b

f x, y, z ds 

f xt, yt, zt r t  dt

Line integral

a

C

NOTE Notice that ds and dS can be written as ds   r t  dt and dS   ruu, v  rvu, v  dA.

EXAMPLE 4 z

Evaluating a Surface Integral

Example 2 demonstrated an evaluation of the surface integral



3

x  z dS

S

where S is the first-octant portion of the cylinder y 2  z 2  9 between x  0 and x  4 (see Figure 15.49). Reevaluate this integral in parametric form. Solution In parametric form, the surface is given by

1 2 3

3

y

4 x Generated by Mathematica

Figure 15.49

rx,   xi  3 cos j  3 sin k where 0 ≤ x ≤ 4 and 0 ≤ ≤ 2. To evaluate the surface integral in parametric form, begin by calculating the following. rx  i r  3 sin j  3 cos k i j k rx  r  1 0 0  3 cos j  3 sin k 0 3 sin 3 cos

rx  r   9 cos 2  9 sin 2  3





So, the surface integral can be evaluated as follows.



   4

x  3 sin 3 dA 

2

3x  9 sin  d dx

0 0 4

D



0 4



0

3x  9 cos

3 x  9 dx 2

  9x

3 2 x 4  12  36 

.

Try It



Exploration A

4 0

2



dx 0

SECTION 15.6

Surface Integrals

1113

Orientation of a Surface Unit normal vectors are used to induce an orientation to a surface S in space. A surface is called orientable if a unit normal vector N can be defined at every nonboundary point of S in such a way that the normal vectors vary continuously over the surface S. If this is possible, S is called an oriented surface. An orientable surface S has two distinct sides. So, when you orient a surface, you are selecting one of the two possible unit normal vectors. If S is a closed surface such as a sphere, it is customary to choose the unit normal vector N to be the one that points outward from the sphere. Most common surfaces, such as spheres, paraboloids, ellipses, and planes, are orientable. (See Exercise 43 for an example of a surface that is not orientable.) Moreover, for an orientable surface, the gradient vector provides a convenient way to find a unit normal vector. That is, for an orientable surface S given by z  gx, y

Orientable surface

let Gx, y, z  z  gx, y.

S: z = g(x, y)

Then, S can be oriented by either the unit normal vector

N = ∇G ⏐⏐∇G⏐⏐

z

N 

S

"Gx, y, z "Gx, y, z  gxx, yi  gyx, yj  k 1  gxx, y 2  gyx, y 2

Upward unit normal

or the unit normal vector N

y

.

Upward direction

x



S is oriented in an upward direction.

"Gx, y, z "Gx, y, z  gxx, yi  gyx, yj  k 1  gxx, y 2  gyx, y 2

as shown in Figure 15.50. If the smooth orientable surface S is given in parametric form by

Rotatable Graph

r u, v  xu, v i  yu, v j  zu, v k

Parametric surface

the unit normal vectors are given by

S: z = g(x, y) z

Downward unit normal

N = − ∇G ⏐⏐∇G⏐⏐

N

ru  rv ru  rv 

N

rv  ru . rv  ru 

and S

y

.

x

NOTE Suppose that the orientable surface is given by y  gx, z or x  g y, z. Then you can use the gradient vector

Downward direction

S is oriented in a downward direction.

"Gx, y, z  gxx, zi  j  gzx, zk

Gx, y, z  y  gx, z

"Gx, y, z  i  gy y, zj  gz y, zk

Gx, y, z  x  g y, z

or

Rotatable Graph Figure 15.50

to orient the surface.

1114

CHAPTER 15

Vector Analysis

Flux Integrals One of the principal applications involving the vector form of a surface integral relates to the flow of a fluid through a surface S. Suppose an oriented surface S is submerged in a fluid having a continuous velocity field F. Let S be the area of a small patch of the surface S over which F is nearly constant. Then the amount of fluid crossing this region per unit of time is approximated by the volume of the column of height F  N, as shown in Figure 15.51. That is,

z

N

F

F·N ΔS

V  heightarea of base  F  N S. y x

The velocity field F indicates the direction of the. fluid flow. Figure 15.51

Rotatable Graph

Consequently, the volume of fluid crossing the surface S per unit of time (called the flux of F across S) is given by the surface integral in the following definition.

Definition of Flux Integral Let Fx, y, z  Mi  Nj  Pk, where M, N, and P have continuous first partial derivatives on the surface S oriented by a unit normal vector N. The flux integral of F across S is given by

 S

F  N dS.

Geometrically, a flux integral is the surface integral over S of the normal component of F. If x, y, z is the density of the fluid at x, y, z, the flux integral

 S

 F  N dS

represents the mass of the fluid flowing across S per unit of time. To evaluate a flux integral for a surface given by z  gx, y, let Gx, y, z  z  gx, y. Then, N dS can be written as follows. "Gx, y, z dS "Gx, y, z  "Gx, y, z  gx  2   gy2  1 dA   gx 2   gy2  1  "Gx, y, z dA

N dS 

THEOREM 15.11

Evaluating a Flux Integral

Let S be an oriented surface given by z  gx, y and let R be its projection onto the xy-plane.

  S

S

F  N dS  F  N dS 

  R

R

F  gxx, yi  gyx, yj  k dA

Oriented upward

F  gxx, yi  gyx, yj  k dA

Oriented downward

For the first integral, the surface is oriented upward, and for the second integral, the surface is oriented downward.

SECTION 15.6

EXAMPLE 5 z

Surface Integrals

1115

Using a Flux Integral to Find the Rate of Mass Flow

Let S be the portion of the paraboloid z  gx, y  4  x 2  y 2

8

lying above the xy-plane, oriented by an upward unit normal vector, as shown in Figure 15.52. A fluid of constant density  is flowing through the surface S according to the vector field

6

Fx, y, z  xi  yj  zk. Find the rate of mass flow through S. Solution Begin by computing the partial derivatives of g. −4

gxx, y  2x 4

.

x

Figure 15.52

4

y

and gyx, y  2y The rate of mass flow through the surface S is

Rotatable Graph

 S

 F  N dS   

       F

xi  yj  4  x 2  y 2 k  2xi  2yj  k dA

R



gxx, yi  gyx, yj  k dA

R

2x 2  2y 2  4  x 2  y 2 dA

R

  

4  x 2  y 2 dA

R 2

2

4  r 2r dr d

Polar coordinates

0 0 2

12 d

0

.

 24.

Try It

Exploration A

For an oriented surface S given by the vector-valued function ru, v  xu, vi  yu, vj  zu, vk

Parametric surface

defined over a region D in the uv-plane, you can define the flux integral of F across S as

 S

F  N dS  

  D

D

F

r

ru

F  ru

 rv

u  rv 

 rv

 r

u

 rv 

dA

dA.

Note the similarity of this integral to the line integral



C

F  dr 



C

F  T ds.

A summary of formulas for line and surface integrals is presented on page 1117.

1116

CHAPTER 15

Vector Analysis

Finding the Flux of an Inverse Square Field

EXAMPLE 6

Find the flux over the sphere S given by

S: x 2 + y 2 + z 2 = a2

x2  y 2  z2  a2

z

N

Sphere S

where F is an inverse square field given by Fx, y, z 

a

N

N

kq r kqr  r 2 r r 3

Inverse square field F

and r  xi  yj  z k. Assume S is oriented outward, as shown in Figure 15.53. a x

R: x 2 + y 2 ≤ a 2

.

a

y

Solution The sphere is given by r u, v  xu, v i  yu, v j  zu, v k  a sin u cos vi  a sin u sin vj  a cos uk

N

where 0 ≤ u ≤  and 0 ≤ v ≤ 2. The partial derivatives of r are ruu, v  a cos u cos v i  a cos u sin vj  a sin uk

Figure 15.53

Rotatable Graph

and rvu, v  a sin u sin vi  a sin u cos vj



which implies that the normal vector ru ru

 rv

i  a cos u cos v a sin u sin v

 rv

is

j a cos u sin v a sin u cos v

k a sin u 0



 a 2sin 2 u cos vi  sin 2 u sin vj  sin u cos uk. Now, using Fx, y, z 

kqr r3

 kq 

xi  yj  zk xi  yj  zk3

kq a sin u cos vi  a sin u sin vj  a cos uk a3

it follows that F  ru

 rv



kq

a sin u cos vi  a sin u sin vj  a cos uk  a3 a2sin2 u cos vi  sin2 u sin vj  sin u cos uk

 kqsin3 u cos2 v  sin3 u sin2 v  sin u cos2 u  kq sin u. Finally, the flux over the sphere S is given by

 S

F  N dS  

.

 

kq sin u dA

D 2 0



kq sin u du dv

0

 4 kq.

Try It

Exploration A

SECTION 15.6

Surface Integrals

1117

The result in Example 6 shows that the flux across a sphere S in an inverse square field is independent of the radius of S. In particular, if E is an electric field, the result in Example 6, along with Coulomb’s Law, yields one of the basic laws of electrostatics, known as Gauss’s Law:

 S

E  N dS  4 kq

Gauss’s Law

where q is a point charge located at the center of the sphere and k is the Coulomb constant. Gauss’s Law is valid for more general closed surfaces that enclose the origin, and relates the flux out of the surface to the total charge q inside the surface. This section concludes with a summary of different forms of line integrals and surface integrals.

Summary of Line and Surface Integrals Line Integrals

ds  r t  dt   x t 2  yt 2  zt 2 dt

 

C

C



b

f x, y, z ds  F  dr  

 

C b

a

f xt, yt, zt ds

Scalar form

a

F  T ds Fxt, yt, zt  rt dt

Vector form

Surface Integrals z  gx, y

dS  1  gxx, y 2  gyx, y 2 dA

  S

  

f x, y, z dS 

S

f x, y, gx, y1  gxx, y 2  gyx, y 2 dA

Scalar form

R

F  N dS 

F

gxx, y i  gyx, y j  k dA

Vector form (upward normal)

R

Surface Integrals  parametric form

dS  ruu, v  rvu, v  dA

  S

  

f x, y, z dS 

S

f xu, v, yu, v, zu, v dS

Scalar form

D

F  N dS 

F

D

ru  rv dA

Vector form

1118

CHAPTER 15

Vector Analysis

Exercises for Section 15.6 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, evaluate



x  2y  z dS.

In Exercises 17–22, evaluate

S

4. S: z 

17. f x, y, z  x 2  y 2  z 2

0 ≤ y ≤ 4

2. S: z  15  2x  3y, 0 ≤ x ≤ 2, 0 ≤ x ≤ 1,

In Exercises 5 and 6, evaluate

18. f x, y, z 

0 ≤ y ≤ x



S: z  x  2,

0 ≤ y ≤ 4

x2  y2 ≤ 1

2 32 , 3x

f x, y, z dS.

S

1. S: z  4  x, 0 ≤ x ≤ 4, 3. S: z  10,



x2  y 2 ≤ 1

xy z

S: z  x 2  y 2,

4 ≤ x 2  y 2 ≤ 16

19. f x, y, z  x  y 2  z 2 2

xy dS.

S: z  x 2  y 2,

S

x2  y 2 ≤ 4

5. S: z  6  x  2y, first octant

20. f x, y, z  x  y 2  z 2

6. S: z  h, 0 ≤ x ≤ 2,

x  12  y 2 ≤ 1  z2 21. f x, y, z   S: x 2  y 2  9, 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, 0 ≤ z ≤ 9 22. f x, y, z  x 2  y 2  z 2 S: x 2  y 2  9, 0 ≤ x ≤ 3, 0 ≤ z ≤ x

2

0 ≤ y ≤ 4  x 2

S: z  x 2  y 2, x2

In Exercises 7 and 8, use a computer algebra system to evaluate



xy dS.

S

7. S: z  9  x 2, 1 8. S: z  2 xy,

0 ≤ x ≤ 2, 0 ≤ y ≤ x

0 ≤ x ≤ 4,

In Exercises 23–28, find the flux of F through S,

0 ≤ y ≤ 4

In Exercises 9 and 10, use a computer algebra system to evaluate



0 ≤ y ≤ 2

F  N dS

S: 2x  3y  z  6, first octant

11. S: 2x  3y  6z  12, first octant, x, y, z  x 2  y 2 12. S: z  a 2  x 2  y 2, In Exercises 13–16, evaluate

 x, y, z  kz



25. Fx, y, z  x i  yj  zk S: z  9  x 2  y 2,

z ≥ 0

26. Fx, y, z  x i  yj  zk S: x 2  y 2  z 2  36, first octant 27. Fx, y, z  4 i  3j  5k S: z  x 2  y 2,

f x, y dS.

x2  y 2 ≤ 4

28. Fx, y, z  x i  yj  2zk

S

S: z  a 2  x 2  y 2

13. f x, y  y  5 v S: ru, v  u i  vj  k, 2

S: x  y  z  1, first octant 24. Fx, y, z  x i  yj

Mass In Exercises 11 and 12, find the mass of the surface lamina S of density .

0 ≤ u ≤ 1,

14. f x, y  x  y S: ru, v  2 cos u i  2 sin u j  v k

 , 0 ≤ v ≤ 2 2

15. f x, y  xy

0 ≤ v ≤ 2

In Exercises 29 and 30, find the flux of F over the closed surface. (Let N be the outward unit normal vector of the surface.) 29. Fx, y, z  4xy i  z 2 j  yzk S: unit cube bounded by x  0, x  1, y  0, y  1, z  0, z1 30. Fx, y, z  x  y i  yj  zk

S: ru, v  2 cos u i  2 sin u j  v k

 0 ≤ u ≤ , 0 ≤ v ≤ 2 2 16. f x, y  x  y S: ru, v  4u cos v i  4u sin v j  3u k 0 ≤ u ≤ 4,

S

23. Fx, y, z  3z i  4j  yk 0 ≤ x ≤ 2,

 1 10. S: z  cos x, 0 ≤ x ≤ , 0 ≤ y ≤ x 2 2

0 ≤ u ≤



where N is the upward unit normal vector to S.

x 2  2xy dS.

S

9. S: z  10  x 2  y 2,

y2

0 ≤ v ≤ 

S: z  1  x 2  y 2,

z0

SECTION 15.6

Writing About Concepts 31. Define a surface integral of the scalar function f over a surface z  gx, y. Explain how to evaluate the surface integral. 32. Describe an orientable surface. 33. Define a flux integral and explain how it is evaluated. 34. Is the surface shown in the figure orientable? Explain.

Surface Integrals

1119

Flow Rate In Exercises 41 and 42, use a computer algebra system to find the rate of mass flow of a fluid of density  through the surface S oriented upward if the velocity field is given by Fx, y, z  0.5zk. 41. S: z  16  x 2  y 2, 42. S: z  16 

x2



z ≥ 0

y2

43. Investigation (a) Use a computer algebra system to graph the vector-valued function ru, v  4  v sin u cos2ui  4  v sin u sin2uj  v cos uk, 0 ≤ u ≤ ,

1 ≤ v ≤ 1.

This surface is called a Möbius strip. (b) Explain why this surface is not orientable. Double twist

(c) Use a computer algebra system to graph the space curve represented by ru, 0. Identify the curve. (d) Construct a Möbius strip by cutting a strip of paper, making a single twist, and pasting the ends together.

Rotatable Graph

(e) Cut the Möbius strip along the space curve graphed in part (c), and describe the result. 35. Electrical Charge Let E  yz i  xz j  xy k be an electrostatic field. Use Gauss’s Law to find the total charge enclosed by the closed surface consisting of the hemisphere z  1  x 2  y 2 and its circular base in the xy-plane. 36. Electrical Charge Let E  x i  y j  2z k be an electrostatic field. Use Gauss’s Law to find the total charge enclosed by the closed surface consisting of the hemisphere z  1  x 2  y 2 and its circular base in the xy-plane. Moment of Inertia In Exercises 37 and 38, use the following formulas for the moments of inertia about the coordinate axes of a surface lamina of density . Ix 

  

 y 2  z 2 x, y, z dS

S

Iy 

x 2  z 2 x, y, z dS

S

Iz 

x 2  y 2 x, y, z dS

S

37. Verify that the moment of inertia of a conical shell of uniform density about its axis is 12ma 2, where m is the mass and a is the radius and height. 38. Verify that the moment of inertia of a spherical shell of uniform density about its diameter is 23ma 2, where m is the mass and a is the radius. Moment of Inertia In Exercises 39 and 40, find Iz for the given lamina with uniform density of 1. Use a computer algebra system to verify your results. 39. x 2  y 2  a 2, 40. z  x 2  y 2,

0 ≤ z ≤ h 0 ≤ z ≤ h

1120

CHAPTER 15

Vector Analysis

Section 15.7

Divergence Theorem • Understand and use the Divergence Theorem. • Use the Divergence Theorem to calculate flux.

CARL FRIEDRICH GAUSS (1777–1855) The Divergence Theorem is also called Gauss’s Theorem, after the famous German mathematician Carl Friedrich Gauss. Gauss is recognized, with Newton and Archimedes, as one of the three greatest mathematicians in history. One of his many contributions to mathematics was made at the age of 22, when, as part of his doctoral dissertation, he proved .the Fundamental Theorem of Algebra.

MathBio

Divergence Theorem Recall from Section 15.4 that an alternative form of Green’s Theorem is



C

F  N ds  

   R

!M !N  dA !x !y



div F dA.

R

In an analogous way, the Divergence Theorem gives the relationship between a triple integral over a solid region Q and a surface integral over the surface of Q. In the statement of the theorem, the surface S is closed in the sense that it forms the complete boundary of the solid Q. Regions bounded by spheres, ellipsoids, cubes, tetrahedrons, or combinations of these surfaces are typical examples of closed surfaces. Assume that Q is a solid region on which a triple integral can be evaluated, and that the closed surface S is oriented by outward unit normal vectors, as shown in Figure 15.54. With these restrictions on S and Q, the Divergence Theorem is as follows. z

S1: Oriented by upward unit normal vector N S2: Oriented by downward unit normal vector

S1

S2 y

N x

. Figure 15.54

Rotatable Graph

THEOREM 15.12

The Divergence Theorem

Let Q be a solid region bounded by a closed surface S oriented by a unit normal vector directed outward from Q. If F is a vector field whose component functions have continuous partial derivatives in Q, then

 S

F  N dS 



div F dV.

Q

NOTE As noted at the left above, the Divergence Theorem is sometimes called Gauss’s Theorem. It is also sometimes called Ostrogradsky’s Theorem, after the Russian mathematician Michel Ostrogradsky (1801–1861).

SECTION 15.7

Proof

Divergence Theorem

1121

If you let Fx, y, z  Mi  Nj  Pk, the theorem takes the form

 S

F  N dS  

   

Mi N  Nj  N  Pk  N dS

S

!M !N !P   dV. !x !y !z



Q

You can prove this by verifying that the following three equations are valid.

   S

S

S

Mi  N dS 

   Q

Nj  N dS 

z

Q

Pk  N dS 

z  g1x, y

S3 S1

N (downward) R S1: z = g1(x, y)

. Figure 15.55

Rotatable Graph

Upper surface

and lower surface

S2

x

!P dV !z

Q

z  g2x, y

N (horizontal)

!N dV !y

Because the verifications of the three equations are similar, only the third is discussed. Restrict the proof to a simple solid region with upper surface

S2: z = g2(x, y)

N (upward)

!M dV !x

y

Lower surface

whose projections onto the xy-plane coincide and form region R. If Q has a lateral surface like S3 in Figure 15.55, then a normal vector is horizontal, which implies that Pk  N  0. Consequently, you have

 S

Pk

 N dS 



Pk

S1

 N dS 



Pk

S2

 N dS  0.

On the upper surface S2, the outward normal vector is upward, whereas on the lower surface S1, the outward normal vector is downward. So, by Theorem 15.11, you have the following.

 S1

 S2

Pk  N dS 

   

Px, y, g1x, yk 

R



!g1

!g1

 !x i  !y j  k dA

Px, y, g1x, y dA

R

Pk  N dS  

R



Px, y, g2x, yk  

!g2 !g i  2 j  k dA !x !y

Px, y, g2x, y dA

R

Adding these results, you obtain

 S

Pk  N dS  

    

Px, y, g2x, y  Px, y, g1x, y dA

R

g2x, y

R



g1x, y

!P dV. !z

Q



!P dz dA !z



1122

CHAPTER 15

Vector Analysis

Using the Divergence Theorem

EXAMPLE 1

Let Q be the solid region bounded by the coordinate planes and the plane 2x  2y  z  6, and let F  xi  y 2j  zk. Find



F  N dS

S

where S is the surface of Q. z

Solution From Figure 15.56, you can see that Q is bounded by four subsurfaces. So, you would need four surface integrals to evaluate



6

S2: yz-plane S1: xz-plane

F  N dS.

S

However, by the Divergence Theorem, you need only one triple integral. Because div F  S4

4

 1  2y  1  2  2y

3

x

you have 3

S4: 2x + 2y + z = 6

.

!M !N !P   !x !y !z

4

S3: xy-plane

y



F  N dS 

S

      

div F dV

Q

Figure 15.56

3



3y

62x2y

2  2y dz dx dy

0 0 0 3 3y

Rotatable Graph 

0 0 3 3y



62x2y



2z  2yz

dx dy 0

12  4x  8y  4xy  4y 2 dx dy

0 0 3





12x  2x 2  8xy  2x 2y  4xy 2

0 3



3y

dy

0

18  6y  10y 2  2y 3 dy

0



 18y  3y 2 

10y 3 y 4  3 2



3 0

63  . 2

.

Try It

Exploration A

If you have access to a computer algebra system that can evaluate triple-iterated integrals, use it to verify the result in Example 1. When you are using such a utility, note that the first step is to convert the triple integral to an iterated integral—this step must be done by hand. To give yourself some practice with this important step, find the limits of integration for the following iterated integrals. Then use a computer to verify that the value is the same as that obtained in Example 1.

TECHNOLOGY

 ?

?

?

?

?

?

 ?

2  2y dy dz dx,

?

?

?

?

?

2  2y dx dy dz

SECTION 15.7

z

1123

Verifying the Divergence Theorem

EXAMPLE 2 S2: z = 4 − x 2 − y 2

Divergence Theorem

Let Q be the solid region between the paraboloid z  4  x2  y 2

4

and the xy-plane. Verify the Divergence Theorem for

N2

Fx, y, z  2zi  xj  y 2k.

x

S1: z = 0

.

2

2

N1 = −k R: x 2 + y2 ≤ 4

Figure 15.57

Rotatable Graph

y

Solution From Figure 15.57 you can see that the outward normal vector for the surface S1 is N1  k, whereas the outward normal vector for the surface S2 is N2 

2xi  2yj  k . 4x 2  4y 2  1

So, by Theorem 15.11, you have

 S

F  N dS 

           F



F

k dS 

R

   

4xz  2xy  y 2 dA

2

4y 2

2

4y 2 4y 2

2 4y 2 2 4y 2 2 4y 2

2

4y 2

2

y 2 dx dy 

4xz  2xy  y 2 dx dy

2 4y 2

4xz  2xy dx dy

4x4  x 2  y 2  2xy dx dy 16x  4x 3  4xy 2  2xy dx dy

2



2xi  2yj  k dS

F

R

4y 2

2 4y 2

2

N2 dS

S2

y 2 dA 

2

F

S2

S1



     

N1 dS 

S1



8x2  x 4  2x 2y 2  x 2y

4y 2

4y 2

dy

2



0 dy

2

 0. On the other hand, because div F 

! ! !

2z  x  y 2  0  0  0  0 !x !y !z

you can apply the Divergence Theorem to obtain the equivalent result

 S

F  N dS 

div F dV

Q



.

 

0 dV  0.

Q

Try It

Exploration A

1124

CHAPTER 15

Vector Analysis

z

EXAMPLE 3

9

Plane: x+z=6

Using the Divergence Theorem

Let Q be the solid bounded by the cylinder x 2  y 2  4, the plane x  z  6, and the xy-plane, as shown in Figure 15.58. Find

8



7 6

S

F  N dS

where S is the surface of Q and Fx, y, z  x 2  sin zi  xy  cos zj  eyk. Solution Direct evaluation of this surface integral would be difficult. However, by the Divergence Theorem, you can evaluate the integral as follows. 2

2



y

x

S

F  N dS 

div F dV

Q

Cylinder: x 2 + y2 = 4

.

      



2x  x  0 dV

Q

Figure 15.58



Rotatable Graph

3x dV

Q

 

2

6r cos

3r cos r dz dr d

0 0 0 2 2

18r 2 cos  3r 3 cos 2  dr d

0

0



2

2

48 cos  12 cos 2 d

0





 48 sin  6 

1 sin 2

2

2



0

 12 Notice that cylindrical coordinates with x  r cos and dV  r dz dr d were used to evaluate the triple integral.

. z

S2

Try It

N2

Even though the Divergence Theorem was stated for a simple solid region Q bounded by a closed surface, the theorem is also valid for regions that are the finite unions of simple solid regions. For example, let Q be the solid bounded by the closed surfaces S1 and S2, as shown in Figure 15.59. To apply the Divergence Theorem to this solid, let S  S1  S2. The normal vector N to S is given by N1 on S1 and by N2 on S2. So, you can write

−N1 S1

x

Exploration A

y



div F dV 

. Figure 15.59

Rotatable Graph

      F

N dS

F

N1 dS 

S

Q



S1



F

S1

    F

N2 dS

S2

N1 dS 

F

S2

N2 dS.

SECTION 15.7

z

Divergence Theorem

1125

Flux and the Divergence Theorem To help understand the Divergence Theorem, consider the two sides of the equation

Sα

Solid region Q

S

(x0, y0, z0)

.

y

x

Figure 15.60

Rotatable Graph



F  N dS 



div F dV.

Q

You know from Section 15.6 that the flux integral on the left determines the total fluid flow across the surface S per unit of time. This can be approximated by summing the fluid flow across small patches of the surface. The triple integral on the right measures this same fluid flow across S, but from a very different perspective—namely, by calculating the flow of fluid into (or out of) small cubes of volume Vi. The flux of the ith cube is approximately Flux of ith cube  div Fxi, yi, zi Vi for some point xi, yi, zi in the ith cube. Note that for a cube in the interior of Q, the gain (or loss) of fluid through any one of its six sides is offset by a corresponding loss (or gain) through one of the sides of an adjacent cube. After summing over all the cubes in Q, the only fluid flow that is not canceled by adjoining cubes is that on the outside edges of the cubes on the boundary. So, the sum n

 div Fx , y , z  V i

.

i

i

i

i1

Rotatable Graph (a) Source: div F > 0

approximates the total flux into (or out of) Q, and therefore through the surface S. To see what is meant by the divergence of F at a point, consider V to be the volume of a small sphere S of radius  and center x0, y0, z0, contained in region Q, as shown in Figure 15.60. Applying the Divergence Theorem to S produces Flux of F across S 

 

div F dV

Q

 div Fx0, y0, z0, V where Q is the interior of S. Consequently, you have div Fx0, y0, z0  .

Rotatable Graph (b) Sink: div F < 0

flux of F across S V

and, by taking the limit as  → 0, you obtain the divergence of F at the point x0, y0, z0. flux of F across S V  flux per unit volume at x0, y0, z0

div Fx0, y0, z0  lim

 →0

The point x0, y0, z0 in a vector field is classified as a source, a sink, or incompressible, as follows. 1. Source, if div F > 0 2. Sink, if div F < 0

See Figure 15.61(a).

3. Incompressible, if div F  0

See Figure 15.61(c).

See Figure 15.61(b).

.

Rotatable Graph (c) Incompressible: div F  0

Figure 15.61

NOTE In hydrodynamics, a source is a point at which additional fluid is considered as being introduced to the region occupied by the fluid. A sink is a point at which fluid is considered as being removed.

1126

CHAPTER 15

Vector Analysis

EXAMPLE 4

Calculating Flux by the Divergence Theorem

Let Q be the region bounded by the sphere x 2  y 2  z 2  4. Find the outward flux of the vector field Fx, y, z  2x3 i  2y 3j  2z3 k through the sphere. Solution By the Divergence Theorem, you have Flux across S 

      

N dS 

F

S





div F dV

Q

6x 2  y 2  z 2 dV

Q 2

6

0



0

2

6

0



2

4 sin  d d d

Spherical coordinates

0

24 sin  d d

0 2

 12

24 d

0

 24 .



Try It

325

768 . 5

Exploration A

Exploration B

Open Exploration

1126

CHAPTER 15

Vector Analysis

Exercises for Section 15.7 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–4, verify the Divergence Theorem by evaluating

 S

3. Fx, y, z  2x  yi  2y  zj  zk S: surface bounded by the plane 2x  4y  2z  12 and the coordinate planes

F  N dS

4. Fx, y, z  xyi  zj  x  yk

as a surface integral and as a triple integral.

S: surface bounded by the planes y  4 and z  4  x and the coordinate planes

1. Fx, y, z  2x i  2yj  z 2k S: cube bounded by the planes x  0, x  a, y  0, y  a, z  0, z  a

z

z

6

4

2. Fx, y, z  2x i  2yj  z 2k S: cylinder x 2  y 2  4, 0 ≤ z ≤ h z

z

a

4 h

6

4

x

3 y

x

Figure for 3 a

a

Rotatable Graph

y

x x

Figure for 1

Rotatable Graph

2

Figure for 2

Rotatable Graph

2

y

Figure for 4

Rotatable Graph

y

SECTION 15.7

In Exercises 5–16, use the Divergence Theorem to evaluate

 S

5. Fx, y, z  x 2i  y 2j  z 2k

  S

verify that



9. Fx, y, z  x i  yj  zk S: x 2  y 2  z 2  4

S

10. Fx, y, z  xyz j  9, z  0, z  4

25. Given the vector field Fx, y, z  x i  yj  zk

12. Fx, y, z  xy 2  cos zi  x 2y  sin zj  e z k

verify that



S: z  12x 2  y 2, z  8 13. Fx, y, z  x 3i  x 2yj  x 2ey k

S

S: z  4  y, z  0, x  0, x  6, y  0

26. Given the vector field Fx, y, z  x i  yj  zk

S: x 2  y 2  z 2  9

verify that

16. Fx, y, z  2x i  yj  zk S: z  4  x 2  y 2, z  0

1  F

In Exercises 17 and 18, evaluate

 S

curl F  N dS

where S is the closed surface of the solid bounded by the graphs of x  4, and z  9  y 2, and the coordinate planes. 17. Fx, y, z  4xy  z 2i  2x 2  6yzj  2xzk 18. Fx, y, z  xy cos z i  yz sin xj  xyzk

Writing About Concepts

F  N dS  3V

where V is the volume of the solid bounded by the closed surface S.

14. Fx, y, z  xe z i  ye z j  e z k S: z  4  y, z  0, x  0, x  6, y  0

F  N dS  0

where V is the volume of the solid bounded by the closed surface S.

S: x2  y2  9, z  0, z  4

15. Fx, y, z  xyi  4yj  xzk

curl F  N dS  0

Fx, y, z  a1i  a2 j  a3k

S: z  a 2  x 2  y 2, z  0



 S

F  N dS 

3  F



dV.

Q

In Exercises 27 and 28, prove the identity, assuming that Q, S, and N meet the conditions of the Divergence Theorem and that the required partial derivatives of the scalar functions f and g are continuous. The expressions DN f and DN g are the derivatives in the direction of the vector N and are defined by DN f  f  N, 27.



DN g  g  N.

 f "2g  "f  "gdV 

Q



f DNg dS

S

[Hint: Use div  f G  f div G  "f  G.]

19. State the Divergence Theorem. 20. How do you determine if a point x0, y0, z0 in a vector field is a source, a sink, or incompressible?

z dx dy.

S

for any closed surface S.

8. Fx, y, z  xy i  yz j  yzk

S:



24. For the constant vector field given by

S: z  a 2  x 2  y 2, z  0

11. Fx, y, z  x i  y 2j  zk

y dz dx 

S

23. Verify that

S: x  0, x  a, y  0, y  a, z  0, z  a

y2



22. Verify the result of Exercise 21 for the cube bounded by x  0, x  a, y  0, y  a, z  0, and z  a.

S: x  0, x  a, y  0, y  a, z  0, z  a

x2

x dy dz 

S

6. Fx, y, z  x 2z2 i  2yj  3xyzk 7. Fx, y, z  x 2 i  2xyj  xyz 2 k

1127

21. Use the Divergence Theorem to verify that the volume of the solid bounded by a surface S is

F  N dS

and find the outward flux of F through the surface of the solid bounded by the graphs of the equations. Use a computer algebra system to verify your results.

Divergence Theorem

28.

 Q

 f " 2g  g" 2f  dV 

 S

(Hint: Use Exercise 27 twice.)

 f DNg  g DN f  dS

1128

CHAPTER 15

Vector Analysis

Section 15.8

Stokes’s Theorem • Understand and use Stokes’s Theorem. • Use curl to analyze the motion of a rotating liquid.

GEORGE GABRIEL STOKES (1819–1903) Stokes became a Lucasian professor of mathematics at Cambridge in 1849. Five years later, he published the theorem that bears his name .as a prize examination question there.

MathBio

Stokes’s Theorem A second higher-dimension analog of Green’s Theorem is called Stokes’s Theorem, after the English mathematical physicist George Gabriel Stokes. Stokes was part of a group of English mathematical physicists referred to as the Cambridge School, which included William Thomson (Lord Kelvin) and James Clerk Maxwell. In addition to making contributions to physics, Stokes worked with infinite series and differential equations, as well as with the integration results presented in this section. Stokes’s Theorem gives the relationship between a surface integral over an oriented surface S and a line integral along a closed space curve C forming the boundary of S, as shown in Figure 15.62. The positive direction along C is counterclockwise relative to the normal vector N. That is, if you imagine grasping the normal vector N with your right hand, with your thumb pointing in the direction of N, your fingers will point in the positive direction C, as shown in Figure 15.63.

z

N

N

Surface S

C

S y

R

. .

x

C

Direction along C is counterclockwise relative to N.

Figure 15.62

Rotatable Graph

Figure 15.63

Rotatable Graph

THEOREM 15.13

Stokes’s Theorem

Let S be an oriented surface with unit normal vector N, bounded by a piecewise smooth simple closed curve C with a positive orientation. If F is a vector field whose component functions have continuous partial derivatives on an open region containing S and C, then



C

F  dr 

 S

curl F  N dS.

NOTE The line integral may be written in the differential form C M dx  N dy  P dz or in the vector form C F  T ds.

SECTION 15.8

Stokes’s Theorem

1129

Using Stokes’s Theorem

EXAMPLE 1

Let C be the oriented triangle lying in the plane 2x  2y  z  6, as shown in Figure 15.64. Evaluate

z



6

S: 2x + 2y + z = 6

C

F  dr

where Fx, y, z  y 2 i  zj  xk. C2

C3

R 3

.

x

C1 x+y=3

Figure 15.64

 

Solution Using Stokes’s Theorem, begin by finding the curl of F. N (upward)

3

i ! curl F  !x y 2 y

k !  i  j  2yk !z x

Considering z  6  2x  2y  gx, y, you can use Theorem 15.11 for an upward normal vector to obtain



C

Rotatable Graph

j ! !y z

F  dr  

    

curl F  N dS

S

R



R 3



i  j  2yk  gxx, y i  gyx, y j  k dA i  j  2yk  2i  2j  k dA

3y

2y  4 dx dy

0 0 3



2y 2  10y  12 dy

0



 

.



2y 3  5y 2  12y 3

3 0

 9.

Try It

Exploration A

Try evaluating the line integral in Example 1 directly, without using Stokes’s Theorem. One way to do this would be to consider C as the union of C1, C2, and C3, as follows. C1: r1t  3  t i  tj,

0 ≤ t ≤ 3 C2: r2t  6  t j  2t  6 k, 3 ≤ t ≤ 6 C3: r3t  t  6 i  18  2t k, 6 ≤ t ≤ 9 The value of the line integral is



C

F  dr  

 

C1 3

F  r1 t dt 





C2

F  r2 t dt 

6

t 2 dt 

0

999  9.

3





C3

F  r3 t dt

9

2t  6 dt 

6

2t  12 dt

1130

CHAPTER 15

Vector Analysis

Verifying Stokes’s Theorem

EXAMPLE 2 S: z = 4 − x 2 − y 2

Verify Stokes’s Theorem for Fx, y, z  2zi  xj  y 2 k, where S is the surface of the paraboloid z  4  x 2  y 2 and C is the trace of S in the xy-plane, as shown in Figure 15.65.

z 4

S N (upward)

−3

R 3 x

. Figure 15.65

 

Solution As a surface integral, you have z  gx, y  4  x 2  y 2 and

C

3

R: x 2 + y 2 ≤ 4

y

i ! curl F  !x 2z

j ! !y x

k !  2yi  2j  k. !z y2

By Theorem 15.11 for an upward normal vector N, you obtain

 S

curl F  N dS 

Rotatable Graph



    

2yi  2j  k  2xi  2yj  k dA

R 2

4y 2

2 4y 2

4xy  4y  1 dx dy

2





2x 2 y  4y  1x

2

4y 2

dy

4y 2

2



2

24y  14  y 2 dy

2



2

8y4  y 2  24  y 2  dy



8 y   4  y 2 32  y4  y 2  4 arcsin 3 2  4. As a line integral, you can parametrize C by rt  2 cos ti  2 sin tj  0k,

0 ≤ t ≤ 2.

For Fx, y, z  2zi  xj  y k, you obtain 2



C

F  dr   

    

M dx  N dy  P dz

C

2z dx  x dy  y 2 dz

C 2

0  2 cos t 2 cos t  0 dt

0



2

4 cos 2 t dt

0

2

2

1  cos 2t dt

0



2 t

.



1 sin 2t 2

2 0

 4.

Try It

Exploration A

Open Exploration



2 2

SECTION 15.8

Stokes’s Theorem

1131

Physical Interpretation of Curl T F

α (x, y, z)

F T

Cα

F N

N

Stokes’s Theorem provides insight into a physical interpretation of curl. In a vector field F, let S be a small circular disk of radius , centered at x, y, z and with boundary C , as shown in Figure 15.66. At each point on the circle C , F has a normal component F  N and a tangential component F  T. The more closely F and T are aligned, the greater the value of F  T. So, a fluid tends to move along the circle rather than across it. Consequently, you say that the line integral around C measures the circulation of F around C . That is,



Disk Sα

Figure 15.66

C

curl F N S

F  T ds  circulation of F around C .

Now consider a small disk S to be centered at some point x, y, z on the surface S, as shown in Figure 15.67. On such a small disk, curl F is nearly constant, because it varies little from its value at x, y, z. Moreover, curl F  N is also nearly constant on S, because all unit normals to S are about the same. Consequently, Stokes’s Theorem yields



(x, y, z) Sα

C

F  T ds 

 S

curl F  N dS

 curl F  N

.

dS

S

 curl F  N  2.

Figure 15.67

Rotatable Graph



So,

curl F  N 



C

F  T ds

 2 circulation of F around C  area of disk S  rate of circulation.

Assuming conditions are such that the approximation improves for smaller and smaller disks  → 0, it follows that 1  →0  2

curl F  N  lim



C

F  T ds

which is referred to as the rotation of F about N. That is, curl Fx, y, z  N  rotation of F about N at x, y, z. In this case, the rotation of F is maximum when curl F and N have the same direction. Normally, this tendency to rotate will vary from point to point on the surface S, and Stokes’s Theorem

 S

curl F  N dS 

Surface integral



C

F  dr

Line integral

says that the collective measure of this rotational tendency taken over the entire surface S (surface integral) is equal to the tendency of a fluid to circulate around the boundary C (line integral).

1132

CHAPTER 15

Vector Analysis

An Application of Curl

EXAMPLE 3 z

A liquid is swirling around in a cylindrical container of radius 2, so that its motion is described by the velocity field Fx, y, z  yx 2  y 2 i  xx 2  y 2 j as shown in Figure 15.68. Find

 S

curl F  N dS

where S is the upper surface of the cylindrical container. 2 x 2

.



Solution The curl of F is given by

y

i ! curl F  !x yx 2  y 2

Figure 15.68

Rotatable Graph

Letting N  k, you have

 S

curl F  N dS  

j ! !y xx 2  y 2

    

3x 2  y 2 dA

R 2

2

3r r dr d

0

0



2

.

2

d

r3

0





k !  3x 2  y 2 k. !z 0

2

0

8 d

0

 16.

Try It

Exploration A

NOTE If curl F  0 throughout region Q, the rotation of F about each unit normal N is 0. That is, F is irrotational. From earlier work, you know that this is a characteristic of conservative vector fields.

Summary of Integration Formulas Fundamental Theorem of Calculus:



Fundamental Theorem of Line Integrals:



b

F x dx  Fb  Fa

a

C

Green's Theorem:

 

M dx  N dy 

C

C

F  N ds 



  R

!N !M  dA  !x !y

S



C

F  T ds 



C

F  dr 



C

 R

" f  dr  f xb, y b  f xa, y a

curl F  kdA

div F dA

R

Divergence Theorem:





F  dr 

F  N dS 

 Q

Stokes's Theorem:

div F dV



C

F  dr 

 S

curl F  N dS

SECTION 15.8

Stokes’s Theorem

1133

Exercises for Section 15.8 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1–6, find the curl of the vector field F.

20. Fx, y, z  xyz i  y j  z k,

S: the first-octant portion of z  x 2 over x 2  y 2  a 2

1. Fx, y, z  2y  z i  xyz j  e z k 2. Fx, y, z  x 2 i  y 2 j  x 2 k

Motion of a Liquid In Exercises 21 and 22, the motion of a liquid in a cylindrical container of radius 1 is described by the velocity field Fx, y, z. Find

3. Fx, y, z  2z i  4x 2 j  arctan x k 4. Fx, y, z  x sin y i  y cos x j  yz 2 k 5. Fx, y, z  e x

2 y 2

i  ey

2 z 2



j  xyz k

6. Fx, y, z  arcsin y i  1  x 2 j  y 2 k

S

In Exercises 7–10, verify Stokes’s Theorem by evaluating



C

F  T ds 



C

curl F  N dS

where S is the upper surface of the cylindrical container. 21. Fx, y, z  i  j  2k

F  dr

22. Fx, y, z  z i  yk

Writing About Concepts

as a line integral and as a double integral. 7. Fx, y, z  y  z i  x  z j  x  y k S: z  1  x 2  y 2

23. State Stokes’s Theorem. 24. Give a physical interpretation of curl.

8. Fx, y, z  y  z i  x  z j  x  y k S: z  4  x 2  y 2,

x2  y2 ≤ a2

z ≥ 0

9. Fx, y, z  xyz i  y j  zk

25. According to Stokes’s Theorem, what can you conclude about the circulation in a field whose curl is 0? Explain your reasoning.

S: 3x  4y  2z  12, first octant 10. Fx, y, z  z 2 i  x 2 j  y 2 k S: z  y 2,

0 ≤ x ≤ a,

26. Let f and g be scalar functions with continuous partial derivatives, and let C and S satisfy the conditions of Stokes’s Theorem. Verify each identity.

0 ≤ y ≤ a

In Exercises 11–20, use Stokes’s Theorem to evaluate C F  dr. Use a computer algebra system to verify your results. In each case, C is oriented counterclockwise as viewed from above.

C

C: triangle with vertices 0, 0, 0, 0, 2, 0, 1, 1, 1 x 12. Fx, y, z  arctan i  lnx 2  y 2 j  k y C: triangle with vertices 0, 0, 0, 1, 1, 1, 0, 0, 2 13. Fx, y, z 



x2 j

S: z  4  x 2  y 2,



y2 k

 S

 f " f   dr  0

 " f  " g  N dS (c)



C

 f " g  g" f   dr  0

28. Let C be a constant vector. Let S be an oriented surface with a unit normal vector N, bounded by a smooth curve C. Prove that

S

C

1

 N dS  2



C

C  r  dr.

z ≥ 0

15. Fx, y, z  z i  y j  xz k 2

Putnam Exam Challenge

S: z  4  x 2  y 2 16. Fx, y, z  x 2 i  z 2 j  xyz k

29. Let Gx, y 

S: z  4  x 2  y 2 x 17. Fx, y, z  lnx 2  y 2 i  arctan j  k y S: z  9  2x  3y over one petal of r  2 sin 2 in the first octant 18. Fx, y, z  yz i  2  3y j  x 2  y 2 k,

x2  y2 ≤ 16

S: the first-octant portion of x 2  z 2  16 over x 2  y 2  16 19. Fx, y, z  xyz i  y j  z k S: z  x 2,

 f "g  dr 

27. Demonstrate the results of Exercise 26 for the functions f x, y, z  xyz and gx, y, z  z. Let S be the hemisphere z  4  x 2  y 2.



z ≥ 0

14. Fx, y, z  4xz i  y j  4xy k S: z  9  x 2  y 2,

 

C

(b)

11. Fx, y, z  2y i  3z j  x k

z2 i

(a)

0 ≤ x ≤ a,

0 ≤ y ≤ a

N is the downward unit normal to the surface.

x

2

y x , ,0 .  4y2 x2  4y2



Prove or disprove that there is a vector-valued function F x, y, z  Mx, y, z, Nx, y, z, Px, y, z with the following properties. (i) M, N, P have continuous partial derivatives for all x, y, z  0, 0, 0; (ii) Curl F  0 for all x, y, z  0, 0, 0; (iii) F x, y, 0  Gx, y. This problem was composed by the Committee on the Putnam Prize Competition. © The Mathematical Association of America. All rights reserved.

1134

CHAPTER 15

Vector Analysis

Review Exercises for Chapter 15 The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

In Exercises 1 and 2, sketch several representative vectors in the vector field. Use a computer algebra system to verify your results. 1. Fx, y, z  x i  j  2 k

2. Fx, y  i  2y j

23.



x 2  y 2 ds

C

C: r t  cos t  t sin t i  sin t  t cos t j, 0 ≤ t ≤ 2

24.



x ds

C

In Exercises 3 and 4, find the gradient vector field for the scalar function. 3. f x, y, z  8x 2  xy  z 2

4. f x, y, z  x 2e yz

C: r t  t  sin t i  1  cos t j, 0 ≤ t ≤ 2

25.



2x  y dx  x  3y dy

C

(a) C: line segment from 0, 0 to 2, 3 In Exercises 5–12, determine if the vector field is conservative. If it is, find a potential function for the vector field. 5. Fx, y 

1 y i 2j y x

6. Fx, y  

26.

8. Fx, y   9.

11. 12.

2x  y dx  x  3y dy

C: r t  cos t  t sin t i  sin t  t sin t j, 0 ≤ t ≤ 2

y 1 i j x2 x

1  cos 2x j Fx, y, z  4x y  z i    6y j  2z k Fx, y, z  4xy  z 2 i  2x 2  6yz j  2 xz k yz i  xz j  xy k Fx, y, z  y 2z 2 Fx, y, z  sin z  y i  x j  k 2y 3



C

7. Fx, y  6xy 2  3x 2 i  6x 2 y  3y 2  7 j

10.

(b) C: counterclockwise around the circle x  3 cos t, y  3 sin t

sin 2x i 

3y 2

2x 2

In Exercises 27 and 28, use a computer algebra system to evaluate the line integral over the given path. 27.



2x  y ds

28.

C



x 2  y 2  z 2 ds

C

r t  a cos 3 t i  a sin 3 t j,

r t  t i  t 2 j  t 32k,

0 ≤ t ≤ 2

0 ≤ t ≤ 4

In Exercises 13–20, find (a) the divergence of the vector field F and (b) the curl of the vector field F.

Lateral Surface Area In Exercises 29 and 30, find the lateral surface area over the curve C in the xy-plane and under the surface z  f x, y.

13. Fx, y, z  x 2 i  y 2 j  z 2 k

29. f x, y  5  sinx  y

14. Fx, y, z  xy 2 j  z x 2 k 15. Fx, y, z  cos y  y cos x i  sin x  x sin y j  xyz k 16. Fx, y, z  3x  y i   y  2z j  z  3x k

C: y  3x from 0, 0 to 2, 6 30. f x, y  12  x  y C: y  x 2 from 0, 0 to 2, 4

17. Fx, y, z  arcsin x i  xy 2 j  yz 2 k 18. Fx, y, z  x 2  y i  x  sin2y j 19. Fx, y, z  lnx 2  y 2 i  lnx 2  y 2 j  z k 20. Fx, y, z 

z z i  j  z2 k x y

In Exercises 21–26, evaluate the line integral along the given path(s). 21.



x 2  y 2 ds

C

(a) C: line segment from 1, 1 to 2, 2 (b) C:

22.



  16, one revolution counterclockwise, starting at 4, 0 x2

y2

xy ds

C

(a) C: line segment from 0, 0 to 5, 4 (b) C: counterclockwise around the triangle with vertices 0, 0, 4, 0, 0, 2

In Exercises 31–36, evaluate



C

F  dr.

31. Fx, y  xy i  x 2 j C: r t  t 2 i  t 3 j, 0 ≤ t ≤ 1 32. Fx, y  x  y i  x  y j C: r t  4 cos t i  3 sin t j, 0 ≤ t ≤ 2 33. Fx, y, z  x i  y j  z k C: r t  2 cos t i  2 sin t j  t k, 0 ≤ t ≤ 2 34. Fx, y, z  2y  z i  z  x j  x  y k C: curve of intersection of x 2  z 2  4 and y 2  z 2  4 from 2, 2, 0 to 0, 0, 2 35. Fx, y, z   y  z i  z  x j  x  y k C: curve of intersection of z  x 2  y 2 and x  y  0 from 2, 2, 8 to 2, 2, 8 36. Fx, y, z  x 2  z i   y 2  z j  x k C: curve of intersection of z  x 2 and x 2  y 2  4 from 0, 2, 0 to 0, 2, 0

REVIEW EXERCISES

In Exercises 37 and 38, use a computer algebra system to evaluate the line integral. 37.



C

x 2  y 2 dx  2xy dy

C: x 2  y 2  a 2

C





C

xy dx  x 2  y 2 dy

C: y  x 2 from 0, 0 to 2, 4 and y  2x from 2, 4 to 0, 0

38.

48.

1135

49.



xy dx  x 2 dy

C

C: boundary of the region between the graphs of y  x 2 and yx

F  dr

Fx, y  2x  y i  2y  x j C: r t  2 cos t  2t sin t i  2 sin t  2t cos t j, 0 ≤ t ≤ 

50.



y 2 dx  x 43 dy

C

C: x 23  y 23  1

39. Work Find the work done by the force field F  x i  y j along the path y  x 32 from 0, 0 to 4, 8.

In Exercises 51 and 52, use a computer algebra system to graph the surface represented by the vector-valued function.

40. Work A 20-ton aircraft climbs 2000 feet while making a 90 turn in a circular arc of radius 10 miles. Find the work done by the engines.

51. r u, v  sec u cos v i  1  2 tan u sin v j  2u k

In Exercises 41 and 42, evaluate the integral using the Fundamental Theorem of Line Integrals.

52. r u, v  eu4 cos v i  eu4 sin v j 

41.



C



53. Investigation Consider the surface represented by the vectorvalued function

C: smooth curve from 0, 0, 1 to 4, 4, 4

Use a computer algebra system to do the following.



y2

dx  2xy dy.

C

(a) C: r t  1  3t i  1  t j, 0 ≤ t ≤ 1 (b) C: r t  t i  t j,

1 ≤ t ≤ 4

(c) Use the Fundamental Theorem of Line Integrals, where C is a smooth curve from 1, 1 to 4, 2. 44. Area and Centroid Consider the region bounded by the x-axis and one arch of the cycloid with parametric equations x  a  sin  and y  a1  cos . Use line integrals to find (a) the area of the region and (b) the centroid of the region. In Exercises 45 –50, use Green’s Theorem to evaluate the line integral.



C



xy dx  x 2  y 2 dy

C

C: boundary of the square with vertices 0, 0, 0, 2, 2, 0, 2, 2



xy 2 dx  x 2 y dy

C

C: x  4 cos t,

(a) Graph the surface for 0 ≤ u ≤ 2 and  (b) Graph the surface for 0 ≤ u ≤ 2 and (c) Graph the surface for 0 ≤ u ≤

  ≤ v ≤ . 2 2

  ≤ v ≤ . 4 2

  and 0 ≤ v ≤ . 4 2

(d) Graph and identify the space curve for 0 ≤ u ≤ 2 and  v . 4 (e) Approximate the area of the surface graphed in part (b). (f) Approximate the area of the surface graphed in part (c). 54. Evaluate the surface integral



z dS over the surface S:

S

r u, v  u  v i  u  v j  sin v k

y dx  2x dy

C: boundary of the square with vertices 0, 0, 0, 2, 2, 0, 2, 2

47.

0 ≤ v ≤ 2

ru, v  3 cos v cos u i  3 cos v sin u j  sin v k.

43. Evaluate the line integral

46.

u k 6

1 dz z

y dx  x dy 

C

45.

 , 0 ≤ v ≤ 2 3

0 ≤ u ≤ 4,

2xyz dx  x 2z dy  x 2y dz

C: smooth curve from 0, 0, 0 to 1, 4, 3

42.

0 ≤ u ≤

y  2 sin t

where 0 ≤ u ≤ 2 and 0 ≤ v ≤ . 55. Use a computer algebra system to graph the surface S and approximate the surface integral

  x  y dS S

where S is the surface S: r u, v  u cos v i  u sin v j  u  12  u k over 0 ≤ u ≤ 2 and 0 ≤ v ≤ 2.

1136

CHAPTER 15

Vector Analysis

56. Mass A cone-shaped surface lamina S is given by

58. Fx, y, z  x i  y j  z k

z  aa  x  y , 0 ≤ z ≤ a . 2

2

Q: solid region bounded by the coordinate planes and the plane 2x  3y  4z  12

2

At each point on S, the density is proportional to the distance between the point and the z-axis. (a) Sketch the cone-shaped surface. (b) Find the mass m of the lamina.

In Exercises 59 and 60, verify Stokes’s Theorem by evaluating

 F  dr C

In Exercises 57 and 58, verify the Divergence Theorem by evaluating

  F  N dS S

as a surface integral and as a triple integral. 57. Fx, y, z  x 2 i  xy j  z k Q: solid region bounded by the coordinate planes and the plane 2x  3y  4z  12

as a line integral and as a double integral. 59. Fx, y, z  cos y  y cos x i  sin x  x sin y j  xyz k S: portion of z  y 2 over the square in the xy-plane with vertices 0, 0, a, 0, a, a, 0, a N is the upward unit normal vector to the surface. 60. Fx, y, z  x  z i  y  z j  x 2 k S: first-octant portion of the plane 3x  y  2z  12 61. Prove that it is not possible for a vector field with twicedifferentiable components to have a curl of xi  yj  zk.

P.S.

P.S.

Problem Solving

1137

Problem Solving

The symbol

indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.

Click on

to view the complete solution of the exercise.

Click on

to print an enlarged copy of the graph.

1. Heat flows from areas of higher temperature to areas of lower temperature in the direction of greatest change. As a result, measuring heat flux involves the gradient of the temperature. The flux depends on the area of the surface. It is the normal direction to the surface that is important, because heat that flows in directions tangential to the surface will give no heat loss. So, assume that the heat flux across a portion of the surface of area S is given by H  k"T  N dS, where T is the temperature, N is the unit normal vector to the surface in the direction of the heat flow, and k is the thermal diffusivity of the material. The heat flux across the surface S is given by H

 S

N

1

S

Rotatable Graph 1

1

x

y

Figure for 2 3. Consider a wire of density x, y, z given by the space curve C: rt  xti  ytj  ztk, a ≤ t ≤ b.

 k"T  N dS.

The moments of inertia about the x-, y-, and z-axes are given by

Consider a single heat source located at the origin with temperature Tx, y, z 

z

Ix 

  

 y 2  z 2x, y, z ds

C

25 . x  y 2  z 2

Iy 

2

x 2  z 2x, y, z ds

C

(a) Calculate the heat flux across the surface



1 1 S  x, y, z: z  1  x 2,  ≤ x ≤ , 0 ≤ y ≤ 1 2 2



as shown in the figure.

Iz 

x 2  y 2x, y, z ds.

C

Find the moments of inertia for a wire of uniform density   1 in the shape of the helix rt  3 cos ti  3 sin tj  2tk, 0 ≤ t ≤ 2 (see figure).

z

r(t) = 3 cos t i + 3 sin t j + 2t k 2

N

z

S

1

12

Rotatable Graph

10 8 1

1

6

y

4

x

2

(b) Repeat the calculation in part (a) using the parametrization

 2 x  cos u, y  v, z  sin u, ≤ u ≤ , 0 ≤ v ≤ 1. 3 3 2. Consider a single heat source located at the origin with temperature Tx, y, z 

25 . x 2  y 2  z 2

2

1 4. Find the moments of inertia for the wire of density   1t given by the curve C: rt 

t2 22 t 32 i  tj  k, 2 3

z

r(t) =

(a) Calculate the heat flux across the surface S  x, y, z: z  1  x 2  y 2, x 2  y 2 ≤ 1

2

as shown in the figure.

1

(b) Repeat the calculation in part (a) using the parametrization  x  sin u cos v, y  sin u sin v, z  cos u, 0 ≤ u ≤ , 2 0 ≤ v ≤ 2.

y

2

x

1 x

1

3/2 t2 i + t j + 2 2t k 3 2

2

y

0 ≤ t ≤ 1 (see figure).

1138

CHAPTER 15

Vector Analysis

5. Use a line integral to find the area bounded by one arch of the cycloid x   a  sin ,

y   a1  cos ,

0 ≤ ≤ 2

8. The force field Fx, y  3x 2 y 2 i  2x 3 y j is shown in the figure below. Three particles move from the point 1, 1 to the point 2, 4 along different paths. Explain why the work done is the same for each particle, and find the value of the work.

as shown in the figure.

y

y

6 2a 5 4 3 2 x

1

2π a

x

6. Use a line integral to find the area bounded by the two loops of the eight curve xt 

1 sin 2t, 2

yt  sin t, 0 ≤ t ≤ 2

2

3

y

4

5

6

9. Let S be a smooth oriented surface with normal vector N, bounded by a smooth simple closed curve C. Let v be a constant vector, and prove that



as shown in the figure.

S

2v  N dS 



v  r  dr.

C

x2 y2  2  1 compare with the 2 a b magnitude of the work done by the force field

1

10. How does the area of the ellipse x

−1

1

1 1 Fx, y   yi  xj 2 2

1 −1

on a particle that moves once around the ellipse (see figure)? y

7. The force field Fx, y  x  yi  x 2  1j acts on an object moving from the point 0, 0 to the point 0, 1, as shown in the figure.

1

y x

1

−1

1

−1

x 1

(a) Find the work done if the object moves along the path x  0, 0 ≤ y ≤ 1. (b) Find the work done if the object moves along the path x  y  y 2, 0 ≤ y ≤ 1. (c) Suppose the object moves along the path x  c y  y 2, 0 ≤ y ≤ 1, c > 0. Find the value of the constant c that minimizes the work.

11. A cross section of Earth’s magnetic field can be represented as a vector field in which the center of Earth is located at the origin and the positive y-axis points in the direction of the magnetic north pole. The equation for this field is Fx, y  Mx, yi  Nx, yj 

m

3xyi  2y 2  x 2j x 2  y 252

where m is the magnetic moment of Earth. Show that this vector field is conservative.