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Chapter Preview Now that we have some basic techniques for evaluating integrals, we turn our attention to the uses of integration, which are virtually endless. We first illus- trate the general rule that if the rate of change of a quantity is known, then integration can be used to determine the net change or the future value of that quantity over a certain time interval. Next, we explore some rich geometric applications of integration: comput- ing the area of regions bounded by several curves, the volume and surface area of three- dimensional solids, and the length of curves. A variety of physical applications of integration include finding the work done by a variable force and computing the total force exerted by water behind a dam. All of these applications are unified by their use of the slice-and- sum strategy. We end this chapter by revisiting the logarithmic function, exploring the many applications of the exponential function, and introducing hyperbolic functions.

6.1 Velocity and Net Change In previous chapters, we established the relationship between the position and velocity of an object moving along a line. With integration, we can now say much more about this relationship. Once we relate velocity and position through integration, we can make analogous observations about a variety of other practical problems, which include fluid flow, population growth, manufacturing costs, and production and consumption of natu- ral resources. The ideas in this section come directly from the Fundamental Theorem of Calculus, and they are among the most powerful applications of calculus.

Velocity, Position, and Displacement Suppose you are driving along a straight highway and your position relative to a reference point or origin is s1t2 for times t Ú 0. Your displacement over a time interval 3a, b4 is the change in the position s1b2 - s1a2 (Figure 6.1). If s1b2 7 s1a2, then your displacement is positive; when s1b2 6 s1a2, your displacement is negative.

6.1 Velocity and Net Change

6.2 Regions Between Curves

6.3 Volume by Slicing

6.4 Volume by Shells

6.5 Length of Curves

6.6 Surface Area

6.7 Physical Applications

6.8 Logarithmic and Exponential Functions Revisited

6.9 Exponential Models

6.10 Hyperbolic Functions

Applications of Integration

s(b)s ! 0 s(a) s (line of motion)

Position at t ! a Position at t ! b " a

Displacement ! s (b) # s (a) " 0

s(a)s ! 0 s(b) s (line of motion)

Position at t ! b " a Position at t ! a

Displacement ! s (b) # s (a) $ 0Figure 6.1

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6.1 Velocity and Net Change 399

Now assume that v1t2 is the velocity of the object at a particular time t. Recall from Chapter 3 that v1t2 = s′1t2, which means that s is an antiderivative of v. From the Funda- mental Theorem of Calculus, it follows that

L b

a v1t2 dt = Lba s′1t2 dt = s1b2 - s1a2 = displacement.

We see that the definite integral 1ba v1t2 dt is the displacement (change in position) be- tween times t = a and t = b. Equivalently, the displacement over the time interval 3a, b4 is the net area under the velocity curve over 3a, b4 (Figure 6.2a).

Not to be confused with the displacement is the distance traveled over a time interval, which is the total distance traveled by the object, independent of the direction of motion. If the velocity is positive, the object moves in the positive direction and the displacement equals the distance traveled. However, if the velocity changes sign, then the displacement and the distance traveled are not generally equal.

QUICK CHECK 1 A police officer leaves his station on a north-south freeway at 9 a.m., trav- eling north (the positive direction) for 40 mi between 9 a.m. and 10 a.m. From 10 a.m. to 11 a.m., he travels south to a point 20 mi south of the station. What are the distance trav- eled and the displacement between 9 a.m. and 11 a.m.?

To compute the distance traveled, we need the magnitude, but not the sign, of the velocity. The magnitude of the velocity " v1t2 " is called the speed. The distance traveled over a small time interval dt is " v1t2 " dt (speed multiplied by elapsed time). Summing these dis- tances, the distance traveled over the time interval 3a, b4 is the integral of the speed; that is,

distance traveled = L b

a " v1t2 " dt.

As shown in Figure 6.2b, integrating the speed produces the area (not net area) bounded by the velocity curve and the t-axis, which corresponds to the distance traveled. The distance traveled is always nonnegative.Figure 6.2

a b

y ! v(t)

Area ! A1

Area ! A2

Displacement ! A1 " A2 ! ! v(t) dt a

Distance traveled ! A1 # A2 ! ! "v(t)" dt a

y ! "v(t)"

(a)

(b)

DEFINITION Position, Velocity, Displacement, and Distance

1. The position of an object moving along a line at time t, denoted s1t2, is the loca- tion of the object relative to the origin.

2. The velocity of an object at time t is v1t2 = s′1t2. 3. The displacement of the object between t = a and t = b 7 a is

s1b2 - s1a2 = Lba v1t2 dt. 4. The distance traveled by the object between t = a and t = b 7 a is

L b

a " v1t2 " dt,

where " v1t2 " is the speed of the object at time t. QUICK CHECK 2 Describe a possible motion of an object along a line for 0 … t … 5 for which the displacement and the distance traveled are different.

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400 Chapter 6 Applications of Integration

EXAMPLE 1 Displacement from velocity A jogger runs along a straight road with velocity (in mi/hr) v1t2 = 2t2 - 8t + 6, for 0 … t … 3, where t is measured in hours. a. Graph the velocity function over the interval 30, 34. Determine when the jogger moves

in the positive direction and when she moves in the negative direction.

b. Find the displacement of the jogger (in miles) on the time intervals 30, 14, 31, 34, and 30, 34. Interpret these results. c. Find the distance traveled over the interval 30, 34. SOLUTION

a. By solving v1t2 = 2t2 - 8t + 6 = 21t - 121t - 32 = 0, we find that the velocity is zero at t = 1 and t = 3; these values are the t-intercepts of the graph of v, which is an upward-opening parabola with a v-intercept of 6 (Figure 6.3a). The velocity is posi- tive on the interval 0 … t 6 1, which means the jogger moves in the positive s direc- tion. For 1 6 t 6 3, the velocity is negative and the jogger moves in the negative s direction.

b. The displacement (in miles) over the interval 30, 14 is s112 - s102 = L10 v1t2 dt

= L 1

0 12t2 - 8t + 62 dt Substitute for v.

= a2 3

t3 - 4t2 + 6tb ` 1 0

= 8 3

. Evaluate integral.

A similar calculation shows that the displacement over the interval 31, 34 is s132 - s112 = L31 v1t2 dt = - 83.

Over the interval 30, 34, the displacement is 83 + 1-832 = 0, which means the jogger returns to the starting point after three hours.

c. From part (b), we can deduce the total distance traveled by the jogger. On the interval 30, 14, the distance traveled is 83 mi; on the interval 31, 34, the distance traveled is also 8 3 mi. Therefore, the distance traveled on 30, 34 is 163 mi. Alternatively (Figure 6.3b), we can integrate the speed and get the same result:

L 3

0 ! v1t2 ! dt = L10 12t2 - 8t + 62 dt + L31 1-12t2 - 8t + 622 dt Definition of ! v1t2 !

= a 2 3

t3 - 4t2 + 6tb ` 1 0

+ a - 2 3

t3 + 4t2 - 6tb ` 3 1

Evaluate integrals.

= 16 3

. Simplify.

Related Exercises 7–14

Figure 6.3

310

321

!v!

v (t) " 2t2 ! 8t # 6 !v (t)! " !2t2 ! 8t # 6! Displacement

" " v(t) dt " Displacement

" " v(t) dt "

Distance traveled from t " 0 to t " 3

is " !v (t)! dt " Displacement from t " 0to t " 3 is " 0.

Area "

(a) (b)

310

321

!v!

v (t) " 2t2 ! 8t # 6 !v (t)! " !2t2 ! 8t # 6! Displacement

" " v(t) dt " Displacement

" " v(t) dt "

Distance traveled from t " 0 to t " 3

is " !v (t)! dt " Displacement from t " 0to t " 3 is " 0.

Area "

(a) (b)

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6.1 Velocity and Net Change 401

Future Value of the Position Function To find the displacement of an object, we do not need to know its initial position. For ex- ample, whether an object moves from s = -20 to s = -10 or from s = 50 to s = 60, its displacement is 10 units. What happens if we are interested in the actual position of the object at some future time?

Suppose we know the velocity of an object and its initial position s102. The goal is to find the position s1t2 at some future time t Ú 0. The Fundamental Theorem of Calculus gives us the answer directly. Because the position s is an antiderivative of the velocity v, we have

L t

0 v1x2 dx = L t0 s′1x2 dx = s1x2 ` t0 = s1t2 - s102.

Rearranging this expression leads to the following result.

➤ Note that t is the independent variable of the position function. Therefore, another (dummy) variable, in this case x, must be used as the variable of integration.

➤ Theorem 6.1 is a consequence (actually a statement) of the Fundamental Theorem of Calculus.

THEOREM 6.1 Position from Velocity Given the velocity v1t2 of an object moving along a line and its initial position s102, the position function of the object for future times t Ú 0 is

s1t2 = s102 + L t0 v1x2 dx.

position initial displacement

at t position

over 30, t4dee Theorem 6.1 says that to find the position s1t2, we add the displacement over the interval 30, t4 to the initial position s102. QUICK CHECK 3 Is the position s1t2 a number or a function? For fixed times t = a and t = b, is the displacement s1b2 - s1a2 a number or a function?

There are two equivalent ways to determine the position function:

Using antiderivatives (Section 4.9)

Using Theorem 6.1

The latter method is usually more efficient, but either method produces the same result. The following example illustrates both approaches.

EXAMPLE 2 Position from velocity A block hangs at rest from a massless spring at the origin 1s = 02. At t = 0, the block is pulled downward 14 m to its initial position s102 = -14 and released (Figure 6.4). Its velocity 1in m>s2 is given by v1t2 = 14 sin t, for t Ú 0. Assume that the upward direction is positive.

a. Find the position of the block, for t Ú 0. b. Graph the position function, for 0 … t … 3p. c. When does the block move through the origin for the first time? d. When does the block reach its highest point for the first time and what is its position at

that time? When does the block return to its lowest point?Figure 6.4

s(0) ! "E Position of the block at time t ! 0

Position of the block at a later time t

0.25

0.50

0.75

1.00

"0.25

s(t)

0 0.25

0.50

0.75

1.00

"0.25

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406 Chapter 6 Applications of Integration

SOLUTION As shown in Figure 6.9, the growth rate is large when t is small (plenty of food and space) and decreases as t increases. Knowing that the initial population is N102 = 100 cells, we can find the population N1t2 at any future time t Ú 0 using Theorem 6.3:

N1t2 = N102 + L t0 N′1x2 dx = 100 + L

0 90e-0.1x dx

()* c

N102 N′1x2 = 100 + c a 90-0.1 be-0.1x d ` t0 Fundamental Theorem = 1000 - 900e-0.1t. Simplify.

The graph of the population function (Figure 6.10) shows that the population increases, but at a decreasing rate. Note that the initial condition N102 = 100 cells is satisfied and that the population size approaches 1000 cells as t S ∞ .

Figure 6.9

t2010

N !

The growth rate is initially 90 cells/hr and decreases as time increases.

N !(t) " 90e#0.1t

Time (hr)

G ro

w th

ra te

(c el

ls /h

0 t30252015105

1000

800

600

400

200

Time (hr)

C el

l p op

ul at

io n

Max. population ! 1000

N(t) ! 1000 " 900e"0.1t

N(10) " N(5) ! 215 cells N(15) " N(10) ! 130 cells The population is growing at a decreasing rate.

Figure 6.10 Related Exercises 38–44

EXAMPLE 6 Production costs A book publisher estimates that the marginal cost of a particular title (in dollars/book) is given by

C′1x2 = 12 - 0.0002x, where 0 … x … 50,000 is the number of books printed. What is the cost of producing the 12,001st through the 15,000th book?

SOLUTION Recall from Section 3.6 that the cost function C1x2 is the cost required to produce x units of a product. The marginal cost C′1x2 is the approximate cost of produc- ing one additional unit after x units have already been produced. The cost of producing books x = 12,001 through x = 15,000 is the cost of producing 15,000 books minus the cost of producing the first 12,000 books. Therefore, the cost in dollars of producing books 12,001 through 15,000 is

C115,0002 - C112,0002 = L15,00012,000 C′1x2 dx = L

15,000

12,000 112 - 0.0002x2 dx Substitute for C′1x2.

= 112x - 0.0001x22 ` 15,000 12,000

Fundamental Theorem

= 27,900. Simplify. Related Exercises 45–48

➤ Although x is a positive integer (the number of books produced), we treat it as a continuous variable in this example.

QUICK CHECK 6 Is the cost of increas- ing the production from 9000 books to 12,000 books in Example 6 more or less than the cost of increasing the production from 12,000 books to 15,000 books? Explain.

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6.1 Velocity and Net Change 407

SECTION 6.1 EXERCISES Review Questions 1. Explain the meaning of position, displacement, and distance

traveled as they apply to an object moving along a line.

2. Suppose the velocity of an object moving along a line is positive. Are displacement and distance traveled equal? Explain.

3. Given the velocity function v of an object moving along a line, explain how definite integrals can be used to find the displace- ment of the object.

4. Explain how to use definite integrals to find the net change in a quantity, given the rate of change of that quantity.

5. Given the rate of change of a quantity Q and its initial value Q102, explain how to find the value of Q at a future time t Ú 0.

6. What is the result of integrating a population growth rate between times t = a and t = b, where b 7 a?

Basic Skills 7. Displacement and distance from velocity Consider the graph

shown in the figure, which gives the velocity of an object mov- ing along a line. Assume time is measured in hours and distance is measured in miles. The areas of three regions bounded by the velocity curve and the t-axis are also given.

543210

a. On what intervals is the object moving in the positive direction? b. What is the displacement of the object over the interval 30, 34? c. What is the total distance traveled by the object over the

interval 31, 54? d. What is the displacement of the object over the interval 30, 54? e. Describe the position of the object relative to its initial position

after 5 hours.

8. Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.

876543210

a. On what intervals is the object moving in the negative direction? b. What is the displacement of the object over the interval 32, 64? c. How far does the object travel over the interval 30, 64? d. What is the displacement of the object over the interval 30, 84? e. Describe the position of the object relative to its initial position

after 8 hours.

9–14. Displacement from velocity Assume t is time measured in sec- onds and velocities have units of m>s. a. Graph the velocity function over the given interval. Then determine

when the motion is in the positive direction and when it is in the negative direction.

b. Find the displacement over the given interval. c. Find the distance traveled over the given interval.

9. v1t2 = 6 - 2t on 0 … t … 6 10. v1t2 = 10 sin 2t on 0 … t … 2p 11. v1t2 = t2 - 6t + 8 on 0 … t … 5 12. v1t2 = - t2 + 5t - 4 on 0 … t … 5 13. v1t2 = t3 - 5t2 + 6t on 0 … t … 5 14. v1t2 = 50e-2t on 0 … t … 4 15–20. Position from velocity Consider an object moving along a line with the following velocities and initial positions.

a. Graph the velocity function on the given interval and determine when the object is moving in the positive direction and when it is moving in the negative direction.

b. Determine the position function, for t Ú 0, using both the antiderivative method and the Fundamental Theorem of Calculus (Theorem 6.1). Check for agreement between the two methods.

c. Graph the position function on the given interval.

15. v1t2 = sin t on 30, 2p4; s102 = 1 16. v1t2 = - t3 + 3t2 - 2t on 30, 34; s102 = 4 17. v1t2 = 6 - 2t on 30, 54; s102 = 0 18. v1t2 = 3 sin pt on 30, 44; s102 = 1 19. v1t2 = 9 - t2 on 30, 44; s102 = -2 20. v1t2 = 1>1t + 12 on 30, 84; s102 = -4 21. Oscillating motion A mass hanging from a spring is set in

motion, and its ensuing velocity is given by v1t2 = 2p cos pt, for t Ú 0. Assume that the positive direction is upward and that s102 = 0. a. Determine the position function, for t Ú 0. b. Graph the position function on the interval 30, 44. c. At what times does the mass reach its low point the first three

times? d. At what times does the mass reach its high point the first three

times?

22. Cycling distance A cyclist rides down a long straight road at a velocity (in m>min) given by v1t2 = 400 - 20t, for 0 … t … 10, where t is measured in minutes.

a. How far does the cyclist travel in the first 5 min? b. How far does the cyclist travel in the first 10 min? c. How far has the cyclist traveled when her velocity is

250 m>min?

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408 Chapter 6 Applications of Integration

23. Flying into a headwind The velocity (in mi>hr) of an airplane flying into a headwind is given by v1t2 = 30116 - t22, for 0 … t … 3. Assume that s102 = 0 and t is measured in hours. a. Determine and graph the position function, for 0 … t … 3. b. How far does the airplane travel in the first 2 hr? c. How far has the airplane traveled at the instant its velocity

reaches 400 mi>hr? 24. Day hike The velocity (in mi>hr) of a hiker walking along a

straight trail is given by v1t2 = 3 sin2 1pt>22, for 0 … t … 4. Assume that s102 = 0 and t is measured in hours. a. Determine and graph the position function, for 0 … t … 4. 1Hint: sin2 t = 1211 - cos 2t2.2 b. What is the distance traveled by the hiker in the first 15 min of

the hike? c. What is the hiker’s position at t = 3?

25. Piecewise velocity The velocity of a (fast) automobile on a straight highway is given by the function

v1t2 = c 3t if 0 … t 6 2060 if 20 … t 6 45 240 - 4t if t Ú 45,

where t is measured in seconds and v has units of m>s. a. Graph the velocity function, for 0 … t … 70. When is the

velocity a maximum? When is the velocity zero? b. What is the distance traveled by the automobile in the first 30 s? c. What is the distance traveled by the automobile in the first 60 s? d. What is the position of the automobile when t = 75?

26. Probe speed A data collection probe is dropped from a stationary balloon, and it falls with a velocity (in m>s) given by v1t2 = 9.8t, neglecting air resistance. After 10 s, a chute deploys and the probe immediately slows to a constant speed of 10 m>s, which it main- tains until it enters the ocean.

a. Graph the velocity function. b. How far does the probe fall in the first 30 s after it is released? c. If the probe was released from an altitude of 3 km, when does

it enter the ocean?

27–34. Position and velocity from acceleration Find the position and velocity of an object moving along a straight line with the given accel- eration, initial velocity, and initial position.

27. a1t2 = -32, v102 = 70, s102 = 10 28. a1t2 = -32, v102 = 50, s102 = 0 29. a1t2 = -9.8, v102 = 20, s102 = 0 30. a1t2 = e-t, v102 = 60, s102 = 40 31. a1t2 = -0.01t, v102 = 10, s102 = 0 32. a1t2 = 201t + 222, v102 = 20, s102 = 10 33. a1t2 = cos 2t, v102 = 5, s102 = 7 34. a1t2 = 2t1t2 + 122, v102 = 0, s102 = 0 35. Acceleration A drag racer accelerates at a1t2 = 88 ft>s2. Assume

that v102 = 0, s102 = 0, and t is measured in seconds. a. Determine and graph the position function, for t Ú 0. b. How far does the racer travel in the first 4 seconds?

c. At this rate, how long will it take the racer to travel 14 mi? d. How long does it take the racer to travel 300 ft? e. How far has the racer traveled when it reaches a speed of

178 ft>s? 36. Deceleration A car slows down with an acceleration of

a1t2 = -15 ft>s2. Assume that v102 = 60 ft>s, s102 = 0, and t is measured in seconds.

a. Determine and graph the position function, for t Ú 0. b. How far does the car travel in the time it takes to come to rest?

37. Approaching a station At t = 0, a train approaching a station begins decelerating from a speed of 80 mi>hr according to the acceleration function a1t2 = -128011 + 8t2-3, where t Ú 0 is measured in hours. How far does the train travel between t = 0 and t = 0.2? Between t = 0.2 and t = 0.4? The units of accel- eration are mi>hr2.

38. Peak oil extraction The owners of an oil reserve begin extract- ing oil at time t = 0. Based on estimates of the reserves, suppose the projected extraction rate is given by Q′1t2 = 3t2 140 - t22, where 0 … t … 40, Q is measured in millions of barrels, and t is measured in years.

a. When does the peak extraction rate occur? b. How much oil is extracted in the first 10, 20, and 30 years? c. What is the total amount of oil extracted in 40 years? d. Is one-fourth of the total oil extracted in the first one-fourth of

the extraction period? Explain.

39. Oil production An oil refinery produces oil at a variable rate given by

Q′1t2 = c 800 if 0 … t 6 302600 - 60t if 30 … t 6 40 200 if t Ú 40,

where t is measured in days and Q is measured in barrels.

a. How many barrels are produced in the first 35 days? b. How many barrels are produced in the first 50 days? c. Without using integration, determine the number of barrels

produced over the interval 360, 804. 40–43. Population growth 40. Starting with an initial value of P102 = 55, the population of

a prairie dog community grows at a rate of P′1t2 = 20 - t>5 (prairie dogs>month), for 0 … t … 200, where t is measured in months.

a. What is the population 6 months later? b. Find the population P1t2, for 0 … t … 200.

41. When records were first kept 1t = 02, the population of a rural town was 250 people. During the following years, the population grew at a rate of P′1t2 = 3011 + 1t2, where t is measured in years. a. What is the population after 20 years? b. Find the population P1t2 at any time t Ú 0.

42. The population of a community of foxes is observed to fluctuate on a 10-year cycle due to variations in the availability of prey. When population measurements began 1t = 02, the population was 35 foxes. The growth rate in units of foxes>year was observed to be

P′1t2 = 5 + 10 sin pt 5

a. What is the population 15 years later? 35 years later? b. Find the population P1t2 at any time t Ú 0.

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6.1 Velocity and Net Change 409

43. A culture of bacteria in a Petri dish has an initial popula- tion of 1500 cells and grows at a rate 1in cells>day2 of N′1t2 = 100e-0.25t. Assume t is measured in days. a. What is the population after 20 days? After 40 days? b. Find the population N1t2 at any time t Ú 0.

44. Flow rates in the Spokane River The daily discharge of the Spo- kane River as it flows through Spokane, Washington, in April and June is modeled by the functions

r11t2 = 0.25t2 + 37.46t + 722.47 1April2 and r21t2 = 0.90t2 - 69.06t + 2053.12 1June2,

where the discharge is measured in millions of cubic feet per day and t = 1 corresponds to the first day of the month (see figure).

0 15 25205 100

500

1000

1500

2000 r2(t) ! 0.90t

2 " 69.06t # 2053.12

r1(t) ! 0.25t 2 # 37.46t # 722.47

a. Determine the total amount of water that flows through Spokane in April (30 days).

b. Determine the total amount of water that flows through Spokane in June (30 days).

c. The Spokane River flows out of Lake Coeur d’Alene, which contains approximately 0.67 mi3 of water. Determine the per- centage of Lake Coeur d’Alene’s volume that flows through Spokane in April and June.

45–48. Marginal cost Consider the following marginal cost functions.

a. Find the additional cost incurred in dollars when production is increased from 100 units to 150 units.

b. Find the additional cost incurred in dollars when production is increased from 500 units to 550 units.

45. C′1x2 = 2000 - 0.5x 46. C′1x2 = 200 - 0.05x 47. C′1x2 = 300 + 10x - 0.01x2 48. C′1x2 = 3000 - x - 0.001x2 Further Explorations 49. Explain why or why not Determine whether the following

statements are true and give an explanation or counterexample.

a. The distance traveled by an object moving along a line is the same as the displacement of the object.

b. When the velocity is positive on an interval, the displacement and the distance traveled on that interval are equal.

c. Consider a tank that is filled and drained at a flow rate of V′1t2 = 1 - t2>100 1gal>min2, for t Ú 0, where t is measured in minutes. It follows that the volume of water in the tank increases for 10 min and then decreases until the tank is empty.

d. A particular marginal cost function has the property that it is positive and decreasing. The cost of increasing production from A units to 2A units is greater than the cost of increasing production from 2A units to 3A units.

50–51. Velocity graphs The figures show velocity functions for motion along a straight line. Assume the motion begins with an initial position of s102 = 0. Determine the following: a. The displacement between t = 0 and t = 5 b. The distance traveled between t = 0 and t = 5 c. The position at t = 5 d. A piecewise function for s1t2 50.

t51

51.

t51

52–55. Equivalent constant velocity Consider the following veloc- ity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of ________.

52. v1t2 = 2t + 6, for 0 … t … 8 53. v1t2 = 1 - t2>16, for 0 … t … 4 54. v1t2 = 2 sin t, for 0 … t … p 55. v1t2 = t125 - t221>2, for 0 … t … 5 56. Where do they meet? Kelly started at noon 1t = 02 riding a

bike from Niwot to Berthoud, a distance of 20 km, with veloc- ity v1t2 = 15>1t + 122 (decreasing because of fatigue). Sandy started at noon 1t = 02 riding a bike in the opposite direction from Berthoud to Niwot with velocity u1t2 = 20>1t + 122 (also decreasing because of fatigue). Assume distance is measured in kilometers and time is measured in hours.

a. Make a graph of Kelly’s distance from Niwot as a function of time.

b. Make a graph of Sandy’s distance from Berthoud as a function of time.

c. When do they meet? How far has each person traveled when they meet?

d. More generally, if the riders’ speeds are v1t2 = A>1t + 122 and u1t2 = B>1t + 122 and the distance between the towns is D, what conditions on A, B, and D must be met to ensure that the riders will pass each other?

e. Looking ahead: With the velocity functions given in part (d), make a conjecture about the maximum distance each person can ride (given unlimited time).

57. Bike race Theo and Sasha start at the same place on a straight road, riding bikes with the following velocities (measured in mi>hr). Assume t is measured in hours.

Theo: vT1t2 = 10, for t Ú 0 Sasha: vS1t2 = 15t, for 0 … t … 1 and vS1t2 = 15, for t 7 1

a. Graph the velocity functions for both riders. b. If the riders ride for 1 hr, who rides farther? Interpret your

answer geometrically using the graphs of part (a). c. If the riders ride for 2 hr, who rides farther? Interpret your

answer geometrically using the graphs of part (a).

M06_BRIG7345_02_SE_C06.1.indd 409 21/10/13 11:30 AM

414 Chapter 6 Applications of Integration

In cases such as these, we treat y as the independent variable and divide the interval 3c, d4 into n subintervals of width ∆y = 1d - c2>n (Figure 6.17). On the kth subinterval, a point yk

* is selected, and we construct a rectangle that extends from the left curve to the right curve. The kth rectangle has length f 1yk*2 - g1yk*2, so the area of the kth rectangle is 1 f 1yk*2 - g1yk*22∆y. The area of the region is approximated by the sum of the areas of the rectangles. In the limit as n S ∞ and ∆y S 0, the area of the region is given as the definite integral

A = lim nS∞

a n

k = 1 1 f 1yk*2 - g1yk*22∆y = Ldc 1 f 1y2 - g1y22 dy.

➤ This area formula is analogous to the one given on page 412; it is now expressed with respect to the y-axis. In this case, f 1y2 - g1y2 is the length of a rectangle and dy represents its width. We sum (integrate) the areas of the rectangles 1 f 1y2 - g1y22 dy to obtain the area of the region.

Given these points of intersection, we see that the region R1 is bounded by y = -x2 + 3x + 6 and y = -2x on the interval 3-1, 04. Similarly, region R2 is bounded by y = -x2 + 3x + 6 and y = 2x on 30, 34 (Figure 6.15b). Therefore,

A = L 0

-1 11-x2 + 3x + 62 - 1-2x2 2 dx + L30 11-x2 + 3x + 62 - 2x2 dx

(1++++++)++++++* (++++++)+++++1* area of region R1 area of region R2

= L 0

-1 1-x2 + 5x + 62 dx + L30 1-x2 + x + 62 dx Simplify.

= a - x3 3

+ 5 2

x2 + 6xb ` 0 -1

+ a - x3 3

+ 1 2

x2 + 6xb ` 3 0 Fundamental Theorem

= 0 - a 1 3

+ 5 2

- 6b + a -9 + 9 2

+ 18b - 0 = 50 3

. Simplify.

Related Exercises 15–22

Integrating with Respect to y There are occasions when it is convenient to reverse the roles of x and y. Consider the regions shown in Figure 6.16 that are bounded by the graphs of x = f 1y2 and x = g1y2, where f 1y2 Ú g1y2, for c … y … d (which implies that the graph of f lies to the right of the graph of g). The lower and upper boundaries of the regions are y = c and y = d, respectively.

Figure 6.16

x x

x ! g(y) x ! g(y) x ! g(y)

x ! f (y) x ! f (y) x ! f (y)

f (yk*) " g(yk*)

x # g(y) x # f (y)

Area of region: A ! k#1

$ ( f (yk*) " g(yk*)) !y

( f (yk*), yk*)

(g(yk*), yk*)

Area of kth rectangle # ( f (yk*) " g(yk*)) !y

yk*

Figure 6.17

DEFINITION Area of a Region Between Two Curves with Respect to y Suppose that f and g are continuous functions with f 1y2 Ú g1y2 on the interval 3c, d4. The area of the region bounded by the graphs x = f 1y2 and x = g1y2 on 3c, d4 is

A = L d

c 1 f 1y2 - g1y22 dy.

M06_BRIG7345_02_SE_C06.2.indd 414 21/10/13 11:32 AM

6.2 Regions Between Curves 415

EXAMPLE 3 Integrating with respect to y Find the area of the region R bounded by the graphs of y = x3, y = x + 6, and the x-axis.

SOLUTION The area of this region could be found by integrating with respect to x. But this approach requires splitting the region into two pieces (Figure 6.18). Alternatively, we can view y as the independent variable, express the bounding curves as functions of y, and make horizontal slices parallel to the x-axis (Figure 6.19).

Figure 6.18

20 x

y ! x " 6 y ! x3

(#6, 0)

y ! 0

(2, 8)

Area ! ! ((x " 6) # 0) dx " ! ((x " 6) # x3) dx Figure 6.19

20 x

x ! y1/3

("6, 0)

(2, 8)

Rectangle length ! y1/3 " (y " 6) Rectangle width ! #y

x ! y " 6

Slice the region using horizontal rectangles from y ! 0 to y ! 8.

y ! 0

Area ! ! (y1/3 " (y " 6)) dy

Solving for x in terms of y, the right curve y = x3 becomes x = f 1y2 = y1>3. The left curve y = x + 6 becomes x = g1y2 = y - 6. The intersection point of the curves satisfies the equation y1>3 = y - 6, or y = 1y - 623. Expanding this equation gives the cubic equation

y3 - 18y2 + 107y - 216 = 1y - 821y2 - 10y + 272 = 0, whose only real root is y = 8. As shown in Figure 6.19, the areas of the slices through the region are summed from y = 0 to y = 8. Therefore, the area of the region is given by

L 8

0 1y1>3 - 1y - 622 dy = a 3

4 y4>3 - y2

2 + 6yb ` 8

0 Fundamental Theorem

= a 3 4

# 16 - 32 + 48b - 0 = 28. Simplify. Related Exercises 23–32

QUICK CHECK 3 The region R is bounded by the curve y = 1x, the line y = x - 2, and the x-axis. Express the area of R in terms of (a) integral(s) with respect to x and (b) integral(s) with respect to y.

EXAMPLE 4 Calculus and geometry Find the area of the region R in the first quad- rant bounded by the curves y = x2>3 and y = x - 4 (Figure 6.20). SOLUTION Slicing the region vertically and integrating with respect to x requires two integrals. Slicing the region horizontally requires a single integral with respect to y. The second approach appears to involve less work.

Slicing horizontally, the right bounding curve is x = y + 4 and the left bound- ing curve is x = y3>2. The two curves intersect at 18, 42, so the limits of integration are y = 0 and y = 4. The area of R is

L 4

0 11y + 42 - y3>22 dy = a y2

2 + 4y - 2

5 y5>2b ` 4

0 =

56 5

. s r

right curve left curve

➤ You may use synthetic division or a root finder to factor the cubic polynomial in Example 3. Then the quadratic formula shows that the equation

y2 - 10y + 27 = 0

has no real roots.

Figure 6.20

0 x84

Area of region R

! ! ((y " 4) # y3/2) dy

(8, 4)

y ! x # 4, x ! y " 4

y ! x2/3, x ! y3/2

M06_BRIG7345_02_SE_C06.2.indd 415 21/10/13 11:32 AM

416 Chapter 6 Applications of Integration

Basic Skills 5–8. Finding area Determine the area of the shaded region in the following figures. 5.

y ! x

y ! x2 " 2

y ! x y ! x3

Can this area be found using a different approach? Sometimes it helps to use geometry. Notice that the region R can be formed by taking the entire region under the curve y = x2>3 on the interval 30, 84 and then removing a triangle whose base is the interval 34, 84 (Figure 6.21). The area of the region R1 under the curve y = x2>3 is

L 8

0 x2>3 dx = 3

5 x5>3 ` 8

0 =

96 5

The triangle R2 has a base of length 4 and a height of 4, so its area is 1 2 # 4 # 4 = 8. There-

fore, the area of R is 965 - 8 = 56 5 , which agrees with the first calculation.

➤ To find the point of intersection in Example 4, solve y3>2 = y + 4 by first squaring both sides of the equation.

Figure 6.21

0 84

x 0 84

y ! x2/3

y ! x " 4

Area of region R ! (area of region R1) " (area of region R2)

Related Exercises 33–38

QUICK CHECK 4 An alternative way to determine the area of the region in Example 3 (Figure 6.18) is to compute 18 + 120 1x + 6 - x32 dx. Why?

SECTION 6.2 EXERCISES Review Questions 1. Draw the graphs of two functions f and g that are continuous and

intersect exactly twice on 1- ∞ , ∞2. Explain how to use integra- tion to find the area of the region bounded by the two curves.

2. Draw the graphs of two functions f and g that are continuous and intersect exactly three times on 1- ∞ , ∞2. How is integration used to find the area of the region bounded by the two curves?

3. Make a sketch to show a case in which the area bounded by two curves is most easily found by integrating with respect to x.

4. Make a sketch to show a case in which the area bounded by two curves is most easily found by integrating with respect to y.

M06_BRIG7345_02_SE_C06.2.indd 416 21/10/13 11:32 AM

6.2 Regions Between Curves 417

7. y

y ! 3 " x

y ! 2x

(Hint: Find the intersection point by inspection.)

O x

y ! 4 cos2 x

y ! sec 2 x

9–14. Regions between curves Sketch the region and find its area.

9. The region bounded by y = 21x + 12, y = 31x + 12, and x = 4 10. The region bounded by y = cos x and y = sin x between

x = p>4 and x = 5p>4 11. The region bounded by y = ex, y = e-2x, and x = ln 4

12. The region bounded by y = 2x and y = x2 + 3x - 6

13. The region bounded by y = 2

1 + x2 and y = 1

14. The region bounded by y = 241x and y = 3x2 15–22. Compound regions Sketch each region (if a figure is not given) and then find its total area.

15. The region bounded by y = sin x, y = cos x, and the x-axis between x = 0 and x = p>2

y ! sin x

y ! cos x

16. The regions between y = sin x and y = sin 2x, for 0 … x … p

x!q

y " sin x

y " sin 2x

17. The region bounded by y = x, y = 1>x, y = 0, and x = 2 18. The regions in the first quadrant on the interval 30, 24 bounded by

y = 4x - x2 and y = 4x - 4.

19. The region bounded by y = 2 - ! x ! and y = x2

20. The regions bounded by y = x3 and y = 9x

21. The region bounded by y = ! x - 3 ! and y = x>2 22. The regions bounded by y = x213 - x2 and y = 12 - 4x

23–26. Integrating with respect to y Sketch each region (if a figure is not given) and find its area by integrating with respect to y.

23. The region bounded by y = A x2 + 1, y = 11 - x, and y = 0.

y ! 1 " xy ! " # 12 x

24. The region bounded by x = cos y and x = -sin 2y

1 x

2 "

4 "

x # !sin 2y

x # cos y

25. The region bounded by x = y2 - 3y + 12 and x = -2y2 - 6y + 30

x " !2y2 ! 6y # 30

x " y2 ! 3y # 12

10 3020 x

26. Both regions bounded by x = y3 - 4y2 + 3y and x = y2 - y

x ! y3 " 4y2 # 3y

x ! y2 " y

5 10 x

M06_BRIG7345_02_SE_C06.2.indd 417 21/10/13 11:32 AM

422 Chapter 6 Applications of Integration

QUICK CHECK 1 Why is the volume, as given by the general slicing method, equal to the average value of the area function A on 3a, b4 multiplied by b - a? EXAMPLE 1 Volume of a “parabolic cube” Let R be the region in the first quadrant bounded by the coordinate axes and the curve y = 1 - x2. A solid has a base R, and cross sections through the solid perpendicular to the base and parallel to the y-axis are squares (Figure 6.25a). Find the volume of the solid.

SOLUTION Focus on a cross section through the solid at a point x, where 0 … x … 1. That cross section is a square with sides of length 1 - x2. Therefore, the area of a typical cross section is A1x2 = 11 - x222. Using the general slicing method, the volume of the solid is

V = L 1

0 A1x2 dx General slicing method

= L 1

0 11 - x222 dx Substitute for A1x2.

= L 1

0 11 - 2x2 + x42 dx Expand integrand.

= 8 15

. Evaluate.

The actual solid with a square cross section is shown in Figure 6.25b.

Figure 6.25

x R

y ! 1 " x2

1 " x2

(x, 1 " x2)

Base of solid

Square slice

(a)

y ! 1 " x2

(b)

Related Exercises 7–16

EXAMPLE 2 Volume of a “parabolic hemisphere” A solid has a base that is bounded by the curves y = x2 and y = 2 - x2 in the xy-plane. Cross sections through the solid perpendicular to the base and parallel to the y-axis are semicircular disks. Find the volume of the solid.

SOLUTION Because a typical cross section perpendicular to the x-axis is a semicircular disk (Figure 6.26), the area of a cross section is 12 pr

2, where r is the radius of the cross section.

Figure 6.26

Radius of slice

" ((2 ! x2) ! x2)

" 1 ! x2

y " x2

y " 2 ! x2

Base of solid

Semicircular slice

y 1 2

x 1

M06_BRIG7345_02_SE_C06.3.indd 422 21/10/13 4:45 PM

6.3 Volume by Slicing 423

The key observation is that this radius is one-half of the distance between the upper bound- ing curve y = 2 - x2 and the lower bounding curve y = x2. So the radius at the point x is

r = 1 2

112 - x22 - x22 = 1 - x2. This means that the area of the semicircular cross section at the point x is

A1x2 = 1 2

pr2 = p

2 11 - x222.

The intersection points of the two bounding curves satisfy 2 - x2 = x2, which has solu- tions x = {1. Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

V = L 1

-1 A1x2 dx General slicing method

= L 1

-1

2 11 - x222 dx Substitute for A1x2.

= p

2 L 1

-1 11 - 2x2 + x42 dx Expand integrand.

= 8p 15

. Evaluate.

Related Exercises 7–16

The Disk Method We now consider a specific type of solid known as a solid of revolution. Suppose f is a continuous function with f 1x2 Ú 0 on an interval 3a, b4. Let R be the region bounded by the graph of f , the x-axis, and the lines x = a and x = b (Figure 6.27). Now revolve R around the x-axis. As R revolves once about the x-axis, it sweeps out a three-dimensional solid of revolution (Figure 6.28). The goal is to find the volume of this solid, and it may be done using the general slicing method.

Figure 6.27

y ! f (x)

Figure 6.28

O a

b x

O a b

O a b x

O a b

Revolving the region R generates a solid of revolution.

QUICK CHECK 2 In Example 2, what is the cross-sectional area function A1x2 if cross sections perpendicular to the base are squares rather than semicircles?

M06_BRIG7345_02_SE_C06.3.indd 423 21/10/13 11:32 AM

424 Chapter 6 Applications of Integration

Disk Method about the x-Axis Let f be continuous with f 1x2 Ú 0 on the interval 3a, b4. If the region R bounded by the graph of f , the x-axis, and the lines x = a and x = b is revolved about the x-axis, the volume of the resulting solid of revolution is

V = L b

a p f 1x22 dx.

QUICK CHECK 3 What solid results when the region R is revolved about the x-axis if (a) R is a square with vertices 10, 02, 10, 22, 12, 02, and 12, 22, and (b) R is a triangle with vertices 10, 02, 10, 22, and 12, 02?

With a solid of revolution, the cross-sectional area function has a special form because all cross sections perpendicular to the x-axis are circular disks with radius f 1x2 (Figure 6.29). Therefore, the cross section at the point x, where a … x … b, has area

A1x2 = p1radius22 = pf 1x22. By the general slicing method, the volume of the solid is

V = L b

a A1x2 dx = Lba p f 1x22 dx.

Because each slice through the solid is a circular disk, the resulting method is called the disk method.

Figure 6.29

O x

f (x)

Cross sections of a solid of revolution are circular disks of radius f (x) and area ! f (x)2.

EXAMPLE 3 Disk method at work Let R be the region bounded by the curve f 1x2 = 1x + 122, the x-axis, and the lines x = 0 and x = 2 (Figure 6.30a). Find the volume of the solid of revolution obtained by revolving R about the x-axis.

SOLUTION When the region R is revolved about the x-axis, it generates a solid of revolu- tion (Figure 6.30b). A cross section perpendicular to the x-axis at the point 0 … x … 2 is a circular disk of radius f 1x2. Therefore, a typical cross section has area

A1x2 = p f 1x22 = p11x + 12222. Integrating these cross-sectional areas between x = 0 and x = 2 gives the volume of the solid:

V = L 2

0 A1x2 dx = L20 p 11x + 12222 dx Substitute for A1x2.

= L 2

0 p1x + 124 dx Simplify.

= p u5

5 ` 3 1

= 242 p

5 . Let u = x + 1 and evaluate.

Related Exercises 17–26

Washer Method A slight variation on the disk method enables us to compute the volume of more ex- otic solids of revolution. Suppose that R is the region bounded by the graphs of f and g between x = a and x = b, where f 1x2 Ú g1x2 Ú 0 (Figure 6.31). If R is revolved about the x-axis to generate a solid of revolution, the resulting solid generally has a hole through it.

Once again we apply the general slicing method. In this case, a cross section through the solid perpendicular to the x-axis is a circular washer with an outer radius of R = f 1x2

Figure 6.30

f (x) ! (x " 1)2

Radius ! (x " 1)2

Cross-sectional area ! !((x " 1)2)2

(a)

(b)

M06_BRIG7345_02_SE_C06.3.indd 424 21/10/13 11:32 AM

6.3 Volume by Slicing 425

and an inner radius of r = g1x2, where a … x … b. The area of the cross section is the area of the entire disk minus the area of the hole, or

A1x2 = p1R2 - r22 = p1 f 1x22 - g1x222 (Figure 6.32). The general slicing method gives the area of the solid.

Washer Method about the x-Axis Let f and g be continuous functions with f 1x2 Ú g1x2 Ú 0 on 3a, b4. Let R be the region bounded by y = f 1x2, y = g1x2, and the lines x = a and x = b. When R is revolved about the x-axis, the volume of the resulting solid of revolution is

V = L b

a p1 f 1x22 - g1x222 dx.

Figure 6.31

xa b

b a

x x

y ! f (x)

y ! g(x)

Revolving region R about the x-axis...

... produces a solid with a hole.

O O

QUICK CHECK 4 Show that when g1x2 = 0 in the washer method, the result is the disk method.

EXAMPLE 4 Volume by the washer method The region R is bounded by the graphs of f 1x2 = 1x and g1x2 = x2 between x = 0 and x = 1. What is the volume of the solid that results when R is revolved about the x-axis?

SOLUTION The region R is bounded by the graphs of f and g with f 1x2 Ú g1x2 on 30, 14, so the washer method is applicable (Figure 6.33). The area of a typical cross section at the point x is

A1x2 = p1 f 1x22 - g1x222 = p111x22 - 1x2222 = p1x - x42. Therefore, the volume of the solid is

V = L 1

0 p1x - x42 dx Washer method

= pa x2 2

- x 5

5 b ` 1

0 =

3p 10

. Fundamental Theorem

➤ The washer method is really two applications of the disk method. We compute the volume of the entire solid without the hole (by the disk method) and then subtract the volume of the hole (also computed by the disk method).

Figure 6.32

(x, f (x))

(x, g(x))

Cross-sectional area ! !(f (x)2 " g(x)2)

b a

Cross-sectional area ! !R2 " !r2

M06_BRIG7345_02_SE_C06.3.indd 425 21/10/13 11:33 AM

402 Chapter 6 Applications of Integration

SOLUTION

a. The velocity function (Figure 6.5a) is positive for 0 6 t 6 p, which means the block moves in the positive (upward) direction. At t = p, the block comes to rest momen- tarily; for p 6 t 6 2p, the block moves in the negative (downward) direction. We let s1t2 be the position at time t Ú 0 with the initial position s102 = -14 m. Method 1: Using antiderivatives Because the position is an antiderivative of the velocity, we have

s1t2 = Lv1t2 dt = L 14 sin t dt = - 14 cos t + C. To determine the arbitrary constant C, we substitute the initial condition s102 = -14 into the expression for s1t2:

- 1 4

= - 1 4

cos 0 + C.

Solving for C, we find that C = 0. Therefore, the position for any time t Ú 0 is

s1t2 = - 1 4

cos t.

Method 2: Using Theorem 6.1 Alternatively, we may use the relationship

s1t2 = s102 + L t0 v1x2 dx. Substituting v1x2 = 14 sin x and s102 = -14, the position function is

s1t2 = - 1 4

+ L t

0 1 4

sin x dx

s102 v1x2 = - 1

4 - a 1

4 cos xb ` t

0 Evaluate integral.

= - 1 4

- 1 4

1cos t - 12 Simplify. = - 1

4 cos t. Simplify.

b. The graph of the position function is shown in Figure 6.5b. We see that s102 = -14 m, as prescribed.

c. The block initially moves in the positive s direction (upward), reaching the origin 1s = 02 when s1t2 = -14 cos t = 0. So the block arrives at the origin for the first time when t = p>2.

d. The block moves in the positive direction and reaches its high point for the first time when t = p; the position at that moment is s1p2 = 14 m. The block then reverses direction and moves in the negative (downward) direction, reaching its low point at t = 2p. This motion repeats every 2p seconds.

Related Exercises 15–24

QUICK CHECK 4 Without doing further calculations, what are the displacement and distance traveled by the block in Example 2 over the interval 30, 2p4?

➤ It is worth repeating that to find the displacement, we need to know only the velocity. To find the position, we must know both the velocity and the initial position s102.

Figure 6.5

(a)

(b)

0 t

0.25

!0.25

2!! 3!

0 t

0.25

!0.25

2!! 3!

v(t) " E sin t

Velocity function

Position function

s(t) " !E cos t

M06_BRIG7345_02_SE_C06.1.indd 402 21/10/13 11:30 AM

6.1 Velocity and Net Change 403

EXAMPLE 3 Skydiving Suppose a skydiver leaps from a hovering helicopter and falls in a straight line. He reaches a terminal velocity of 80 m>s at t = 0 and falls for 19 sec- onds, at which time he opens his parachute. The velocity decreases linearly to 6 m>s over a two-second period and then remains constant until he reaches the ground at t = 40 s. The motion is described by the velocity function

v1t2 = c 80 if 0 … t 6 19783 - 37t if 19 … t 6 21 6 if 21 … t … 40.

Determine the height above the ground from which the skydiver jumped.

SOLUTION We let the position of the skydiver increase downward with the origin 1s = 02 corresponding to the position of the helicopter. The velocity is positive, so the distance traveled by the skydiver equals the displacement, which is

L 40

0 0 v1t2 0 dt = L190 80 dt + L2119 1783 - 37t2 dt + L4021 6 dt

= 80t ` 19 0

+ a783t - 37t2 2

b ` 21 19

+ 6t ` 40 21

Fundamental Theorem

= 1720. Evaluate and simplify.

The skydiver jumped from 1720 m above the ground. Notice that the displacement of the skydiver is the area under the velocity curve (Figure 6.6).

➤ The terminal velocity of an object depends on its density, shape, size, and the medium through which it falls. Estimates for human beings in free fall in the lower atmosphere vary from 120 mi>hr 154 m>s2 to 180 mi>hr 180 m>s2.

Figure 6.6

t40302010

Distance traveled ! ! "v(t)" dt ! ! 80 dt " ! (783 # 37t) dt " ! 6 dt

! (783 # 37t) dt 19

! 80 dt 0

v (t) ! 783 # 37t

v (t) ! 6

v (t) ! 80

40! 6 dt

Related Exercises 25–26

QUICK CHECK 5 Suppose (unrealistically) in Example 3 that the velocity of the skydiver is 80 m>s, for 0 … t 6 20, and then it changes instantaneously to 6 m>s, for 20 … t … 40. Sketch the velocity function and, without integrating, find the distance the skydiver falls in 40 s.

Acceleration Because the acceleration of an object moving along a line is given by a1t2 = v′1t2, the relationship between velocity and acceleration is the same as the relationship between

M06_BRIG7345_02_SE_C06.1.indd 403 21/10/13 11:30 AM

404 Chapter 6 Applications of Integration

position and velocity. Given the acceleration of an object, the change in velocity over an interval 3a, b4 is

change in velocity = v1b2 - v1a2 = Lba v′1t2 dt = Lba a1t2 dt. Furthermore, if we know the acceleration and initial velocity v102, then the velocity at future times can also be found.

➤ Theorem 6.2 is a consequence of the Fundamental Theorem of Calculus.

THEOREM 6.2 Velocity from Acceleration Given the acceleration a1t2 of an object moving along a line and its initial velocity v102, the velocity of the object for future times t Ú 0 is

v1t2 = v102 + L t0 a1x2 dx. EXAMPLE 4 Motion in a gravitational field An artillery shell is fired directly upward with an initial velocity of 300 m>s from a point 30 m above the ground (Figure 6.7). Assume that only the force of gravity acts on the shell and it produces an acceleration of 9.8 m>s2. Find the velocity of the shell while it is in the air. SOLUTION We let the positive direction be upward with the origin 1s = 02 correspond- ing to the ground. The initial velocity of the shell is v102 = 300 m>s. The acceleration due to gravity is downward; therefore, a1t2 = -9.8 m>s2. Integrating the acceleration, the velocity is

v1t2 = v102 + L t0 a1x2 dx = 300 + L t0 1-9.82 dx = 300 - 9.8t. 300 m>s -9.8 m>s2 The velocity decreases from its initial value of 300 m>s, reaching zero at the high point of the trajectory when v1t2 = 300 - 9.8t = 0, or at t ≈ 30.6 s (Figure 6.8). At this point, the velocity becomes negative, and the shell begins its descent to Earth.

Knowing the velocity function, you could now find the position function using the methods of Example 3.

Related Exercises 27–37

Net Change and Future Value Everything we have said about velocity, position, and displacement carries over to more general situations. Suppose you are interested in some quantity Q that changes over time; Q may represent the amount of water in a reservoir, the population of a cell culture, or the amount of a resource that is consumed or produced. If you are given the rate Q′ at which Q changes, then integration allows you to calculate either the net change in the quantity Q or the future value of Q.

We argue just as we did for velocity and position: Because Q1t2 is an antiderivative of Q′1t2, the Fundamental Theorem of Calculus tells us that

L b

a Q′1t2 dt = Q1b2 - Q1a2 = net change in Q over 3a, b4.

Geometrically, the net change in Q over the time interval 3a, b4 is the net area under the graph of Q′ over 3a, b4. We interpret the product Q′1t2 dt as a change in Q over a small increment of time. Integrating Q′1t2 accumulates, or adds up, these small changes over the interval 3a, b4. The result is the net change in Q between t = a and t = b. We see that accumulating the rate of change of a quantity over the interval gives the net change in that quantity over the interval.

Figure 6.8

t6020 40

!300

!100

!200

300

200

100

v (t) " 300 ! 9.8t

High point of trajectory at t ! 30.6 s

g ! 9.8 m/s2 v(0) ! 300 m/s

s(0) ! 30 m

Figure 6.7

➤ Note that the units in the integral are consistent. For example, if Q′ has units of gallons>second, and t and x have units of seconds, then Q′1x2 dx has units of 1gallons>second21seconds2 = gallons, which are the units of Q.

M06_BRIG7345_02_SE_C06.1.indd 404 21/10/13 11:30 AM

6.1 Velocity and Net Change 405

Alternatively, suppose we are given both the rate of change Q′ and the initial value Q102. Integrating over the interval 30, t4, where t Ú 0, we have

L t

0 Q′1x2 dx = Q1t2 - Q102.

Rearranging this equation, we write the value of Q at any future time t Ú 0 as

Q1t2 = Q102 + L t0 Q′1x2 dx. e e e

future initial

net change

value value over 30, t4

➤ At the risk of being repetitious, Theorem 6.3 is also a consequence of the Fundamental Theorem of Calculus. We assume that Q′ is an integrable function.

THEOREM 6.3 Net Change and Future Value Suppose a quantity Q changes over time at a known rate Q′. Then the net change in Q between t = a and t = b 7 a is

Q1b2 - Q1a2 = Lba Q′1t2 dt. net change in Q

Given the initial value Q102, the future value of Q at time t Ú 0 is Q1t2 = Q102 + L t0 Q′1x2 dx.

e The correspondences between velocity–displacement problems and more general prob- lems are shown in Table 6.1.

Table 6.1

Velocity–Displacement Problems General Problems

Position s1t2 Quantity Q1t2 (such as volume or population) Velocity: s′1t2 = v1t2 Rate of change: Q′1t2 Displacement: s1b2 - s1a2 = Lba v1t2 dt Net change: Q1b2 - Q1a2 = Lba Q′1t2 dt Future position: s1t2 = s102 + L t0 v1x2 dx Future value of Q: Q1t2 = Q102 + L t0 Q′1x2 dx EXAMPLE 5 Cell growth A culture of cells in a lab has a population of 100 cells when nutrients are added at time t = 0. Suppose the population N1t2 (in cells/hr) increases at a rate given by

N′1t2 = 90e-0.1t. Find N1t2, for t Ú 0.

➤ Although N is a positive integer (the number of cells), we treat it as a continuous variable in this example.

M06_BRIG7345_02_SE_C06.1.indd 405 21/10/13 11:30 AM

410 Chapter 6 Applications of Integration

d. Which rider arrives first at the 10-, 15-, and 20-mile markers of the race? Interpret your answer geometrically using the graphs of part (a).

e. Suppose Sasha gives Theo a head start of 0.2 mi and the riders ride for 20 mi. Who wins the race?

f. Suppose Sasha gives Theo a head start of 0.2 hr and the riders ride for 20 mi. Who wins the race?

58. Two runners At noon 1t = 02, Alicia starts running along a long straight road at 4 mi>hr. Her velocity decreases according to the function v1t2 = 4>1t + 12, for t Ú 0. At noon, Boris also starts running along the same road with a 2-mi head start on Alicia; his velocity is given by u1t2 = 2>1t + 12, for t Ú 0. Assume t is measured in hours.

a. Find the position functions for Alicia and Boris, where s = 0 corresponds to Alicia’s starting point.

b. When, if ever, does Alicia overtake Boris?

59. Running in a wind A strong west wind blows across a circular running track. Abe and Bess start running at the south end of the track, and at the same time, Abe starts running clockwise and Bess starts running counterclockwise. Abe runs with a speed (in units of mi>hr) given by u1w2 = 3 - 2 cos w and Bess runs with a speed given by v1u2 = 3 + 2 cos u, where w and u are the central angles of the runners.

wind

Abe

Bess

start

a. Graph the speed functions u and v, and explain why they describe the runners’ speeds (in light of the wind).

b. Compute the average value of u and v with respect to the central angle.

c. Challenge: If the track has a radius of 110 mi, how long does it take each runner to complete one lap and who wins the race?

Applications 60. Filling a tank A 2000-liter cistern is empty when water be-

gins flowing into it 1at t = 02 at a rate (in L>min) given by Q′1t2 = 31t, where t is measured in minutes. a. How much water flows into the cistern in 1 hour? b. Find and graph the function that gives the amount of water in

the tank at any time t Ú 0. c. When will the tank be full?

61. Depletion of natural resources Suppose that r1t2 = r0 e-kt, with k 7 0, is the rate at which a nation extracts oil, where r0 = 107 barrels>yr is the current rate of extraction. Suppose also that the estimate of the total oil reserve is 2 * 109 barrels. a. Find Q1t2, the total amount of oil extracted by the nation after

t years. b. Evaluate lim

tS∞ Q1t2 and explain the meaning of this limit.

c. Find the minimum decay constant k for which the total oil reserves will last forever.

d. Suppose r0 = 2 * 107 barrels>yr and the decay constant k is the minimum value found in part (c). How long will the total oil reserves last?

62. Snowplow problem With snow on the ground and falling at a constant rate, a snowplow began plowing down a long straight road at noon. The plow traveled twice as far in the first hour as it did in the second hour. At what time did the snow start falling? Assume the plowing rate is inversely proportional to the depth of the snow.

63. Filling a reservoir A reservoir with a capacity of 2500 m3 is filled with a single inflow pipe. The reservoir is empty when the inflow pipe is opened at t = 0. Letting Q1t2 be the amount of water in the reservoir at time t, the flow rate of water into the reservoir 1in m3>hr2 oscillates on a 24-hr cycle (see figure) and is given by

Q′1t2 = 20 a1 + cos pt 12

b .

0 t2412

Time (hr)

Fl ow

ra te

o f w

at er

(m 3 /

hr )

a. How much water flows into the reservoir in the first 2 hr? b. Find and graph the function that gives the amount of water in

the reservoir over the interval 30, t4, where t Ú 0. c. When is the reservoir full?

64. Blood flow A typical human heart pumps 70 mL of blood with each stroke (stroke volume). Assuming a heart rate of 60 beats>min 11 beat>s2, a reasonable model for the outflow rate of the heart is V′1t2 = 7011 + sin 2pt2, where V1t2 is the amount of blood (in milliliters) pumped over the interval 30, t4, V102 = 0, and t is measured in seconds. a. Graph the outflow rate function. b. Verify that the amount of blood pumped over a one-second

interval is 70 mL. c. Find the function that gives the total blood pumped between

t = 0 and a future time t 7 0. d. What is the cardiac output over a period of 1 min? (Use calcu-

lus; then check your answer with algebra.)

65. Air flow in the lungs A simple model (with different parameters for different people) for the flow of air in and out of the lungs is

V′1t2 = -p 2

sin pt 2

where V1t2 (measured in liters) is the volume of air in the lungs at time t Ú 0, t is measured in seconds, and t = 0 corresponds to a time at which the lungs are full and exhalation begins. Only a fraction of the air in the lungs in exchanged with each breath. The amount that is exchanged is called the tidal volume.

a. Find and graph the volume function V assuming that V102 = 6 L.

b. What is the breathing rate in breaths/min? c. What is the tidal volume and what is the total capacity of the

lungs?

M06_BRIG7345_02_SE_C06.1.indd 410 21/10/13 11:30 AM

6.1 Velocity and Net Change 411

66. Oscillating growth rates Some species have growth rates that oscillate with an (approximately) constant period P. Consider the growth rate function

N′1t2 = r + A sin 2pt P

where A and r are constants with units of individuals>yr, and t is measured in years. A species becomes extinct if its population ever reaches 0 after t = 0. a. Suppose P = 10, A = 20, and r = 0. If the initial popula-

tion is N102 = 10, does the population ever become extinct? Explain.

b. Suppose P = 10, A = 20, and r = 0. If the initial popula- tion is N102 = 100, does the population ever become extinct? Explain.

c. Suppose P = 10, A = 50, and r = 5. If the initial popula- tion is N102 = 10, does the population ever become extinct? Explain.

d. Suppose P = 10, A = 50, and r = -5. Find the initial population N102 needed to ensure that the population never becomes extinct.

67. Power and energy Power and energy are often used interchange- ably, but they are quite different. Energy is what makes matter move or heat up and is measured in units of joules (J) or Calories (Cal), where 1 Cal = 4184 J. One hour of walking consumes roughly 106 J, or 250 Cal. On the other hand, power is the rate at which energy is used and is measured in watts 1W; 1 W = 1 J>s2. Other useful units of power are kilowatts 11 kW = 103 W2 and megawatts 11 MW = 106 W2. If energy is used at a rate of 1 kW for 1 hr, the total amount of energy used is 1 kilowatt-hour 1kWh2, which is 3.6 * 106 J.

Suppose the power function of a large city over a 24-hr period is given by

P1t2 = E′1t2 = 300 - 200 sin pt 12

where P is measured in megawatts and t = 0 corresponds to 6:00 p.m. (see figure).

0 t2412

400

600

200

Time (hr)

Po w

er (M

W )

!t 12

P(t) ! E "(t) ! 300 # 200 sin

a. How much energy is consumed by this city in a typical 24-hr period? Express the answer in megawatt-hours and in joules.

b. Burning 1 kg of coal produces about 450 kWh of energy. How many kg of coal are required to meet the energy needs of the city for 1 day? For 1 year?

c. Fission of 1 g of uranium-235 1U-2352 produces about 16,000 kWh of energy. How many grams of uranium are needed to meet the energy needs of the city for 1 day? For 1 year?

d. A typical wind turbine can generate electrical power at a rate of about 200 kW. Approximately how many wind turbines are needed to meet the average energy needs of the city?

68. Variable gravity At Earth’s surface, the acceleration due to gravity is approximately g = 9.8 m>s2 (with local variations). However, the acceleration decreases with distance from the sur- face according to Newton’s law of gravitation. At a distance of y meters from Earth’s surface, the acceleration is given by

a1y2 = - g11 + y>R22, where R = 6.4 * 106 m is the radius of Earth.

a. Suppose a projectile is launched upward with an initial velocity of v0 m>s. Let v1t2 be its velocity and y1t2 its height (in meters) above the surface t seconds after the launch. Neglecting forces

such as air resistance, explain why dv dt

= a1y2 and dy dt

= v1t2. b. Use the Chain Rule to show that

dv dt

= 1 2

d dy

1v22. c. Show that the equation of motion for the projectile is

1 2

d dy

1v22 = a1y2, where a1y2 is given previously. d. Integrate both sides of the equation in part (c) with respect to y

using the fact that when y = 0, v = v0. Show that

1 2

1v2 - v202 = g R a 11 + y>R - 1b . e. When the projectile reaches its maximum height, v = 0.

Use this fact to determine that the maximum height is

ymax = Rv0

2gR - v20 .

f. Graph ymax as a function of v0. What is the maximum height when v0 = 500 m>s, 1500 m>s, and 5 km>s?

g. Show that the value of v0 needed to put the projectile into orbit (called the escape velocity) is 12gR.

Additional Exercises 69–72. Another look at the Fundamental Theorem

69. Suppose that f and g have continuous derivatives on an interval 3a, b4. Prove that if f 1a2 = g1a2 and f1b2 = g1b2, then 1ba f ′1x2 dx = 1ba g′1x2 dx.

70. Use Exercise 69 to prove that if two runners start and finish at the same time and place, then regardless of the velocities at which they run, their displacements are equal.

71. Use Exercise 69 to prove that if two trails start at the same place and finish at the same place, then regardless of the ups and downs of the trails, they have the same net change in elevation.

72. Without evaluating integrals, prove that

L 2

d dx

112 sin px22 dx = L20 ddx1x1012 - x232 dx. QUICK CHECK ANSWERS 1. Displacement = -20 mi (20 mi south); distance traveled = 100 mi 2. Suppose the object moves in the posi- tive direction, for 0 … t … 3, and then moves in the negative direction, for 3 6 t … 5. 3. A function; a number 4. Displacement = 0; distance traveled = 1 5. 1720 m 6. The production cost would increase more between 9000 and 12,000 books than between 12,000 and 15,000 books. Graph C′ and look at the area under the curve.

M06_BRIG7345_02_SE_C06.1.indd 411 21/10/13 11:30 AM

412 Chapter 6 Applications of Integration

6.2 Regions Between Curves In this section, the method for finding the area of a region bounded by a single curve is generalized to regions bounded by two or more curves. Consider two functions f and g continuous on an interval 3a, b4 on which f 1x2 Ú g1x2 (Figure 6.11). The goal is to find the area A of the region bounded by the two curves and the vertical lines x = a and x = b.

Once again, we rely on the slice-and-sum strategy (Section 5.2) for finding areas by Riemann sums. The interval 3a, b4 is partitioned into n subintervals using uniformly spaced grid points separated by a distance ∆x = 1b - a2>n (Figure 6.12). On each sub- interval, we build a rectangle extending from the lower curve to the upper curve. On the kth subinterval, a point xk

* is chosen, and the height of the corresponding rectangle is taken to be f 1xk*2 - g1xk*2. Therefore, the area of the kth rectangle is 1 f 1xk*2 - g1xk*22 ∆x (Figure 6.13). Summing the areas of the n rectangles gives an approximation to the area of the region between the curves:

A ≈ a n

k = 1 1 f 1xk*2 - g1xk*22 ∆x.

Figure 6.11

Lower curve y ! g(x)

Upper curve y ! f (x)

Area ! A

Figure 6.12

y ! f (x)

y ! g(x)

a ! x0

"x ! width of each rectangle

! b # a

b ! xn xn#1

x2x1

"x x

Figure 6.13

ba x

f (xk) ! g(xk)

Area of region: A ! k#1

$ ( f (xk) ! g(xk)) "x

Area of kth rectangle # ( f (xk) ! g(xk)) "x

(xk, f (xk))*

* *

(xk, g(xk))

* *

y # g(x)

y # f (x)

DEFINITION Area of a Region Between Two Curves Suppose that f and g are continuous functions with f 1x2 Ú g1x2 on the interval 3a, b4. The area of the region bounded by the graphs of f and g on 3a, b4 is

A = L b

a 1 f 1x2 - g1x22 dx.

As the number of grid points increases, ∆x approaches zero and these sums approach the area of the region between the curves; that is,

A = lim nS∞

a n

k = 1 1 f 1xk*2 - g1xk*22∆x.

The limit of these Riemann sums is a definite integral of the function f - g.

➤ It is helpful to interpret the area formula: f 1x2 - g1x2 is the length of a rectangle and dx represents its width. We sum (integrate) the areas of the rectangles 1 f 1x2 - g1x22 dx to obtain the area of the region.

QUICK CHECK 1 In the area formula for a region between two curves, verify that if the lower curve is g1x2 = 0, the formula becomes the usual formula for the area of the region bounded by y = f 1x2 and the x-axis.

M06_BRIG7345_02_SE_C06.2.indd 412 21/10/13 11:32 AM

6.2 Regions Between Curves 413

EXAMPLE 1 Area between curves Find the area of the region bounded by the graphs of f 1x2 = 5 - x2 and g1x2 = x2 - 3 (Figure 6.14). SOLUTION A key step in the solution of many area problems is finding the intersec- tion points of the boundary curves, which often determine the limits of integration. The intersection points of these two curves satisfy the equation 5 - x2 = x2 - 3. The solutions of this equation are x = -2 and x = 2, which become the lower and upper limits of integration, respectively. The graph of f is the upper curve and the graph of g is the lower curve on the interval 3-2, 24. Therefore, the area of the region is

A = L 2

-2 115 - x22 - 1x2 - 322 dx Substitute for f and g.

s s f 1x2 g1x2

= 2L 2

0 18 - 2x22 dx Simplify and use symmetry.

= 2a8x - 2 3

x3b ` 2 0 Fundamental Theorem

= 64 3

. Simplify.

Notice how the symmetry of the problem simplifies the integration. Additionally, note that the area formula A = 1ba 1 f 1x2 - g1x22 dx is valid even if one or both curves lie below the x-axis, as long as f 1x2 Ú g1x2 on 3a, b4.

Related Exercises 5–14

QUICK CHECK 2 Interpret the area formula when written in the form A = 1ba f 1x2 dx - 1ba g1x2 dx, where f 1x2 Ú g1x2 Ú 0 on 3a, b4. EXAMPLE 2 Compound region Find the area of the region bounded by the graphs of f 1x2 = -x2 + 3x + 6 and g1x2 = ! 2x ! (Figure 6.15a). SOLUTION The lower boundary of the region is bounded by two different branches of the absolute value function. In situations like this, the region is divided into two (or more) subregions whose areas are found independently and then summed; these subregions are labeled R1 and R2 (Figure 6.15b). By the definition of absolute value,

g1x2 = ! 2x ! = b2x if x Ú 0 -2x if x 6 0.

Figure 6.14

Slice the region using vertical rectangles from x ! "2 to x ! 2.

Rectangle height ! f(x) " g(x) Rectangle width !#x

g(x) ! x2 " 3

f(x) ! 5 " x24

("2, 1) (2, 1)

"2 "1 21 x

(x, f(x))

(x, g(x))

Figure 6.15 (a)

!1 321 x

g(x) " !2x!

f (x) " !x2 # 3x # 6

(b)

!1 321

g(x) " !2x!

f (x) " !x2 # 3x # 6

(3, 6)

(!1, 2)

Area " " ((!x2 # 3x # 6) ! (!2x) dx !1

2 0

Area " " ((!x2 # 3x # 6) ! 2x) dx

The left intersection point of f and g satisfies -2x = -x2 + 3x + 6, or x2 - 5x - 6 = 0. Solving for x, we find that 1x + 121x - 62 = 0, which implies x = -1 or x = 6; only the first solution is relevant. The right intersection point of f and g satisfies 2x = -x2 + 3x + 6; you should verify that the relevant solution in this case is x = 3.

➤ The solution x = 6 corresponds to the intersection of the parabola y = -x2 + 3x + 6 and the line y = -2x in the fourth quadrant, not shown in Figure 6.15 because g1x2 = -2x only when x 6 0.

M06_BRIG7345_02_SE_C06.2.indd 413 21/10/13 11:32 AM

418 Chapter 6 Applications of Integration

40–43. Regions between curves Sketch the region and find its area.

40. The region bounded by y = sin x and y = x1x - p2, for 0 … x … p

41. The region bounded by y = 1x - 122 and y = 7x - 19 42. The region bounded by y = 2 and y =

121 - x2 43. The region bounded by y = x2 - 2x + 1 and y = 5x - 9

44–50. Either method Use the most efficient strategy for computing the area of the following regions.

44. The region bounded by x = y1y - 12 and x = -y1y - 12 45. The region bounded by x = y1y - 12 and y = x>3 46. The region bounded by y = x3, y = -x3, and

3y - 7x - 10 = 0

47. The region bounded by y = 1x, y = 2x - 15, and y = 0 48. The region bounded by y = x2 - 4, 4y - 5x - 5 = 0, and

y = 0, for y Ú 0

49. The region in the first quadrant bounded by y = 5 2

- 1 x

and y = x

50. The region in the first quadrant bounded by y = x -1, y = 4x, and y = x>4

51. Comparing areas Let f 1x2 = xp and g1x2 = x1>q, where p 7 1 and q 7 1 are positive integers. Let R1 be the region in the first quadrant between y = f 1x2 and y = x and let R2 be the region in the first quadrant between y = g1x2 and y = x. a. Find the area of R1 and R2 when p = q, and determine which

region has the greater area. b. Find the area of R1 and R2 when p 7 q, and determine which

region has the greater area. c. Find the area of R1 and R2 when p 6 q, and determine which

region has the greater area.

52–55. Complicated regions Find the area of the regions shown in the following figures.

52. y

y ! 4!2x

y ! 2x2

y ! "4x # 6

53.

y ! 9 " x2

y ! "x

y ! 8x

5 2

27–30. Two approaches Express the area of the following shaded regions in terms of (a) one or more integrals with respect to x and (b) one or more integrals with respect to y. You do not need to evaluate the integrals.

27.

y ! x

y ! x2 " 2

28.

y ! x2 " 4x

y ! 2x " 8

29.

x ! 2y

x ! y2 " 3

30.

y ! !x

y ! x3

31–32. Two approaches Find the area of the following regions by (a) integrating with respect to x and (b) integrating with respect to y. Be sure your results agree. Sketch the bounding curves and the region in question.

31. The region bounded by y = 2 - x 2

and x = 2y2

32. The region bounded by x = 2 - y2 and x = ! y !

33–38. Any method Use any method (including geometry) to find the area of the following regions. In each case, sketch the bounding curves and the region in question.

33. The region in the first quadrant bounded by y = x2>3 and y = 4 34. The region in the first quadrant bounded by y = 2 and

y = 2 sin x on the interval 30, p>24 35. The region bounded by y = ex, y = 2e-x + 1, and x = 0

36. The region below the line y = 2 and above the curve y = sec2 x on the interval 30, p>44

37. The region between the line y = x and the curve y = 2x21 - x2 in the first quadrant

38. The region bounded by x = y2 - 4 and y = x>3 Further Explorations 39. Explain why or why not Determine whether the following state-

ments are true and give an explanation or counterexample.

a. The area of the region bounded by y = x and x = y2 can be found only by integrating with respect to x.

b. The area of the region between y = sin x and y = cos x on the interval 30, p>24 is 1p>20 1cos x - sin x2 dx.

c. 110 1x - x22 dx = 110 11y - y2 dy.

M06_BRIG7345_02_SE_C06.2.indd 418 21/10/13 11:32 AM

6.2 Regions Between Curves 419

65. Lorenz curves and the Gini index A Lorenz curve is given by y = L1x2, where 0 … x … 1 represents the lowest fraction of the population of a society in terms of wealth and 0 … y … 1 repre- sents the fraction of the total wealth that is owned by that fraction of the society. For example, the Lorenz curve in the figure shows that L10.52 = 0.2, which means that the lowest 0.5 (50%) of the society owns 0.2 (20%) of the wealth. (See the Guided Project Distribution of Wealth for more on Lorenz curves.)

Line of perfect equality

0 1.00.80.5

1.0

0.6

0.2

Fraction of households

Fr ac

tio n

of to

ta l i

nc om

0.8 of households control 0.6 of the wealth

Lorenz curve, y ! L(x)

0.5 of households control 0.2 of the wealth

a. A Lorenz curve y = L1x2 is accompanied by the line y = x, called the line of perfect equality. Explain why this line is given this name.

b. Explain why a Lorenz curve satisfies the conditions L102 = 0, L112 = 1, L1x2 … x, and L′1x2 Ú 0 on 30, 14.

c. Graph the Lorenz curves L1x2 = xp corresponding to p = 1.1, 1.5, 2, 3, 4. Which value of p corresponds to the most equitable distribution of wealth (closest to the line of perfect equality)? Which value of p corresponds to the least equitable distribution of wealth? Explain.

d. The information in the Lorenz curve is often summarized in a single measure called the Gini index, which is defined as follows. Let A be the area of the region between y = x and y = L1x2 (see figure) and let B be the area of the region between y = L1x2 and the x-axis. Then the Gini index is G =

A A + B. Show that G = 2A = 1 - 2L

L1x2 dx. e. Compute the Gini index for the cases L1x2 = xp and

p = 1.1, 1.5, 2, 3, 4. f. What is the smallest interval 3a, b4 on which values of the

Gini index lie for L1x2 = xp with p Ú 1? Which endpoints of 3a, b4 correspond to the least and most equitable distribution of wealth?

g. Consider the Lorenz curve described by L1x2 = 5x2>6 + x>6. Show that it satisfies the conditions L102 = 0, L112 = 1, and L′1x2 Ú 0 on 30, 14. Find the Gini index for this function.

Additional Exercises 66. Equal area properties for parabolas Consider the parabola

y = x2. Let P, Q, and R be points on the parabola with R between P and Q on the curve. Let /P, /Q, and /R be the lines tangent to the parabola at P, Q, and R, respectively (see figure). Let P′ be the intersection point of /Q and /R, let Q′ be the intersection point of /P and /R, and let R′ be the intersection point of /P and /Q. Prove

T54.

y ! x2

x ! 2 sin2 y

55.

x ! (y " 2)2

y ! 8 " x

56–59. Roots and powers Find the area of the following regions, expressing your results in terms of the positive integer n Ú 2.

56. The region bounded by f 1x2 = x and g1x2 = xn, for x Ú 0 57. The region bounded by f 1x2 = x and g1x2 = x1>n, for x Ú 0 58. The region bounded by f 1x2 = x1>n and g1x2 = xn, for x Ú 0 59. Let An be the area of the region bounded by f 1x2 = x1>n and

g1x2 = xn on the interval 30, 14, where n is a positive integer. Evaluate lim

nS∞ An and interpret the result.

60–63. Bisecting regions For each region R, find the horizontal line y = k that divides R into two subregions of equal area.

60. R is the region bounded by y = 1 - x, the x-axis, and the y-axis.

61. R is the region bounded by y = 1 - # x - 1 # and the x-axis.

62. R is the region bounded by y = 4 - x2 and the x-axis.

63. R is the region bounded by y = 1x and y = x. Applications 64. Geometric probability Suppose a dartboard occupies the square 51x, y2: 0 … # x # … 1, 0 … # y # … 16. A dart is thrown randomly

at the board many times (meaning it is equally likely to land at any point in the square). What fraction of the dart throws land closer to the edge of the board than the center? Equivalently, what is the probability that the dart lands closer to the edge of the board than the center? Proceed as follows.

0 x1 !1

0 x1

a. Argue that by symmetry, it is necessary to consider only one quarter of the board, say the region R: 51x, y2: # x # … y … 16.

b. Find the curve C in this region that is equidistant from the cen- ter of the board and the top edge of the board (see figure).

c. The probability that the dart lands closer to the edge of the board than the center is the ratio of the area of the region R1 above C to the area of the entire region R. Compute this probability.

M06_BRIG7345_02_SE_C06.2.indd 419 21/10/13 11:32 AM

420 Chapter 6 Applications of Integration

71. Area function for a cubic Consider the cubic polynomial f 1x2 = x1x - a21x - b2, where 0 … a … b. a. For a fixed value of b, find the function F 1a2 = 1b0 f 1x2 dx.

For what value of a (which depends on b) is F 1a2 = 0? b. For a fixed value of b, find the function A1a2 that gives the

area of the region bounded by the graph of f and the x-axis between x = 0 and x = b. Graph this function and show that it has a minimum at a = b>2. What is the maximum value of A1a2, and where does it occur (in terms of b)?

72. Differences of even functions Assume f and g are even, integrable functions on 3-a, a4, where a 7 1. Suppose f 1x2 7 g1x2 7 0 on 3-a, a4 and the area bounded by the graphs of f and g on 3-a, a4 is 10. What is the value of

11a0 x1 f 1x22 - g1x222 dx? 73. Roots and powers Consider the functions f 1x2 = xn and

g1x2 = x1>n, where n Ú 2 is a positive integer. a. Graph f and g for n = 2, 3, and 4, for x Ú 0. b. Give a geometric interpretation of the area function

An1x2 = 1 x0 1 f 1s2 - g1s22 ds, for n = 2, 3, 4, cand x 7 0. c. Find the positive root of An1x2 = 0 in terms of n. Does the

root increase or decrease with n?

74. Shifting sines Consider the functions f 1x2 = a sin 2x and g1x2 = 1sin x2>a, where a 7 0 is a real number. a. Graph the two functions on the interval 30, p>24, for a = 12, 1,

and 2. b. Show that the curves have an intersection point x* 1other than

x = 02 on 30, p>24 that satisfies cos x* = 1>12a22, provided a 7 1>12.

c. Find the area of the region between the two curves on 30, x*4 when a = 1.

d. Show that as a S 1>12 +. the area of the region between the two curves on 30, x*4 approaches zero.

QUICK CHECK ANSWERS 1. If g1x2 = 0 and f 1x2 Ú 0, then the area between the curves is 1ba 1 f 1x2 - 02 dx = 1ba f 1x2 dx, which is the area between y = f 1x2 and the x-axis. 2. 1ba f 1x2 dx is the area of the region between the graph of f and the x-axis. 1ba g1x2 dx is the area of the region between the graph of g and the x-axis. The difference of the two integrals is the area of the region between the graphs of f and g. 3. a. 120 1x dx + 142 11x - x + 22 dx b. 120 1y + 2 - y22 dy 4. The area of the triangle to the left of the y-axis is 18. The area of the region to the right of the y-axis is given by the integral.

that Area ∆PQR = 2 # Area ∆P′Q′R′ in the following cases. (In fact, the property holds for any three points on any parabola.) (Source: Mathematics Magazine 81, 2, Apr 2008)

y y ! x2!P

a. P1-a, a22, Q1a, a22, and R10, 02, where a is a positive real number

b. P1-a, a22, Q1b, b22, and R10, 02, where a and b are positive real numbers

c. P1-a, a22, Q1b, b22, and R is any point between P and Q on the curve

67. Minimum area Graph the curves y = 1x + 121x - 22 and y = ax + 1 for various values of a. For what value of a is the area of the region between the two curves a minimum?

68. An area function Graph the curves y = a2x3 and y = 1x for various values of a 7 0. Note how the area A1a2 between the curves varies with a. Find and graph the area function A1a2. For what value of a is A1a2 = 16?

69. Area of a curve defined implicitly Determine the area of the shaded region bounded by the curve x2 = y411 - y32 (see figure).

!1 x1

x2 " y4(1 ! y3)

70. Rewrite first Find the area of the region bounded by the curve

x = 1 2y

- A 14y2 - 1 and the line x = 1 in the first quadrant. (Hint: Express y in terms of x.)

6.3 Volume by Slicing We have seen that integration is used to compute the area of two-dimensional regions bounded by curves. Integrals are also used to find the volume of three-dimensional regions (or solids). Once again, the slice-and-sum method is the key to solving these problems.

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6.3 Volume by Slicing 421

General Slicing Method Consider a solid object that extends in the x-direction from x = a to x = b. Imagine cut- ting through the solid, perpendicular to the x-axis at a particular point x, and suppose the area of the cross section created by the cut is given by a known integrable function A (Figure 6.22).

To find the volume of this solid, we first divide 3a, b4 into n subintervals of length ∆x = 1b - a2>n. The endpoints of the subintervals are the grid points x0 = a, x1, x2, c, xn = b. We now make vertical cuts through the solid perpendicular to the x-axis at each grid point, which produces n slices of thickness ∆x. (Imagine cutting a loaf of bread to create n slices of equal width.) On each subinterval, an arbitrary point xk

* is identified. The kth slice through the solid has a thickness ∆x, and we take A1xk*2 as a representative cross-sectional area of the slice. Therefore, the volume of the kth slice is approximately A1xk*2∆x (Figure 6.23). Summing the volumes of the slices, the approximate volume of the solid is

V ≈ a n

k = 1 A1xk*2∆x.

As the number of slices increases 1n S ∞2 and the thickness of each slice goes to zero 1∆x S 02, the exact volume V is obtained in terms of a definite integral (Figure 6.24): V = lim

nS ∞ a n

k = 1 A1x*k2∆x = Lba A1x2 dx.

Figure 6.22

Cross section with area A(x)

Figure 6.23

Cross-sectional area " A (xk*)

Volume of kth slice ! A(xk*) !x

xk*

Figure 6.24

Increase the number of slices.

n ! !

Volume " lim k"1

# A(xk*)$x

" ! A(x)dx n!!

We summarize the important general slicing method, which is also the basis of other vol- ume formulas to follow.

➤ The factors in this volume integral have meaning: A1x2 is the cross-sectional area of a slice and dx represents its thickness. Summing (integrating) the volumes of the slices A1x2 dx gives the volume of the solid.

General Slicing Method Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on 3a, b4. The volume of the solid is

V = L b

a A1x2 dx.

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426 Chapter 6 Applications of Integration

Related Exercises 27–34

QUICK CHECK 5 Suppose the region in Example 4 is revolved about the line y = -1 in- stead of the x-axis. (a) What is the inner radius of a typical washer? (b) What is the outer radius of a typical washer?

Revolving about the y-Axis Everything you learned about revolving regions about the x-axis applies to revolving re- gions about the y-axis. Consider a region R bounded by the curve x = p1y2 on the right, the curve x = q1y2 on the left, and the horizontal lines y = c and y = d (Figure 6.34a).

To find the volume of the solid generated when R is revolved about the y-axis, we use the general slicing method—now with respect to the y-axis (Figure 6.34b). The area of a typical cross section is A1y2 = p1p1y22 - q1y222, where c … y … d. As before, inte- grating these cross-sectional areas of the solid gives the volume.

➤ Ignoring the factor of p, the integrand in the washer method integral is f 1x22 - g1x22, which is not equal to 1 f 1x2 - g1x222.

Figure 6.33

Interval of integration

g(x) ! x2

g(x)

Area of washer face ! !( f (x)2 " g(x)2) ! !(x " x4)

1 1

f (x) ! x

f (x)

Revolving the region about the x-axis... ... produces a bowl-

shaped solid. y

00 1

Figure 6.34

Interval of integration

q(y)

p(y)

cx ! p(y)

x ! q(y) x ! p(y)

x ! q(y)

Inner radius ! q(y)

Outer radius ! p(y)

(b)(a)

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6.3 Volume by Slicing 427

EXAMPLE 5 Which solid has greater volume? Let R be the region in the first quad- rant bounded by the graphs of x = y3 and x = 4y. Which is greater, the volume of the solid generated when R is revolved about the x-axis or the y-axis?

SOLUTION Solving y3 = 4y, or equivalently, y1y2 - 42 = 0, we find that the bounding curves of R intersect at the points 10, 02 and 18, 22. When the region R (Figure 6.35a) is revolved about the y-axis, it generates a funnel with a curved inner surface (Figure 6.35b). Washer-shaped cross sections perpendicular to the y-axis extend from y = 0 to y = 2. The outer radius of the cross section at the point y is determined by the line x = p1y2 = 4y. The inner radius of the cross section at the point y is determined by the curve x = q1y2 = y3. Applying the washer method, the volume of this solid is

V = L 2

0 p1p1y22 - q1y222 dy Washer method

= L 2

0 p116y2 - y62 dy Substitute for p and q.

= pa 16 3

y3 - y7

7 b ` 2

0 Fundamental Theorem

= 512p

21 . Evaluate.

➤ The disk>washer method about the y-axis is the disk>washer method about the x-axis with x replaced with y.

Disk and Washer Methods about the y-Axis Let p and q be continuous functions with p1y2 Ú q1y2 Ú 0 on 3c, d4. Let R be the region bounded by x = p1y2, x = q1y2, and the lines y = c and y = d. When R is revolved about the y-axis, the volume of the resulting solid of revolution is given by

V = L d

c p1p1y22 - q1y222 dy.

If q1y2 = 0, the disk method results: V = L

c pp1y22 dy.

Figure 6.35 (a)

Interval of integration

Inner radius ! y3

Outer radius ! 4y

q(y) ! y3

(8, 2)

p(y) ! 4y

Area of washer face ! !(p(y)2 " q(y)2) ! !(16y2 " y6)

(b)

y 2

When the region R is revolved about the x-axis, it generates a different funnel (Figure 6.36). Vertical slices through the solid between x = 0 and x = 8 produce wash- ers. The outer radius of the washer at the point x is determined by the curve x = y3, or

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428 Chapter 6 Applications of Integration

y = f 1x2 = x1>3. The inner radius is determined by x = 4y, or y = g1x2 = x>4. The volume of the resulting solid is

V = L 8

0 p1 f 1x22 - g1x222 dx Washer method

= L 8

0 pax2>3 - x2

16 b dx Substitute for f and g.

= pa 3 5

x5>3 - x3 48

b ` 8 0

Fundamental Theorem

= 128p

15 . Evaluate.

We see that revolving the region about the y-axis produces a solid of greater volume.

Figure 6.36

(8, 2)

Interval of integration

Inner radius ! x/4

Outer radius ! x1/3 f (x) ! x1/3

g(x) ! x 4

Area of washer face ! !(f (x)2 " g(x)2)

! !(x2/3 " x 2

16 )

x 8

Related Exercises 35–44

QUICK CHECK 6 The region in the first quadrant bounded by y = x and y = x3 is revolved about the y-axis. Give the integral for the volume of the solid that is generated.

The disk and washer methods may be generalized to handle situations in which a region R is revolved about a line parallel to one of the coordinate axes. The next example discusses three such cases.

EXAMPLE 6 Revolving about other lines Let f 1x2 = 1x + 1 and g1x2 = x2 + 1. a. Find the volume of the solid generated when the region R1 bounded by the graph of f

and the line y = 2 on the interval 30, 14 is revolved about the line y = 2. b. Find the volume of the solid generated when the region R2 bounded by the graphs of f

and g on the interval 30, 14 is revolved about the line y = -1. c. Find the volume of the solid generated when the region R2 bounded by the graphs of f

and g on the interval 30, 14 is revolved about the line x = 2. SOLUTION

a. Figure 6.37a shows the region R1 and the axis of revolution. Applying the disk method, we see that a disk located at a point x has a radius of 2 - f 1x2 = 2 - 11x + 12 = 1 - 1x. Therefore, the volume of the solid generated when R1 is revolved about y = 2 is

L 1

0 p11 - 1x22 dx = pL10 11 - 21x + x2 dx = p6 .

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6.3 Volume by Slicing 429

3. The region bounded by the curves y = 2x and y = x2 is revolved about the x-axis. Give an integral for the volume of the solid that is generated.

4. The region bounded by the curves y = 2x and y = x2 is revolved about the y-axis. Give an integral for the volume of the solid that is generated.

b. When the graph of f is revolved about y = -1, it sweeps out a solid of revolution whose radius at a point x is f 1x2 + 1 = 1x + 2. Similarly, when the graph of g is revolved about y = -1, it sweeps out a solid of revolution whose radius at a point x is g1x2 + 1 = x2 + 2 (Figure 6.37b). Using the washer method, the volume of the solid generated when R2 is revolved about y = -1 is

L 1

0 p111x + 222 - 1x2 + 2222 dx

= pL 1

0 1-x4 - 4x2 + x + 41x2 dx

= 49p 30

c. When the region R2 is revolved about the line x = 2, we use the washer method and integrate in the y-direction. First note that the graph of f is described by y = 1x + 1, or equivalently, x = 1y - 122, for y Ú 1. Also, the graph of g is described by y = x2 + 1, or equivalently, x = 1y - 1 for y Ú 1 (Figure 6.37c). When the graph of f is revolved about the line x = 2, the radius of a typical disk at a point y is 2 - 1y - 122. Similarly, when the graph of g is revolved about x = 2, the radius of a typical disk at a point y is 2 - 1y - 1. Finally, observe that the extent of the region R2 in the y-direction is the interval 1 … y … 2.

Applying the washer method, simplifying the integrand, and integrating powers of y, the volume of the solid of revolution is

L 2

1 p112 - 1y - 12222 - 12 - 1y - 1222 dy = 31p

30 .

Figure 6.37

Radius: ! 2 " ( x # 1) ! 1 " x

Interval of integration

R1 2

y ! x # 1

(a)

y " x2 # 1

y " !1

Outer radius " x # 2

y " x # 1

Inner radius " x2 # 2

Interval of integration

(b)

R2 x ! (y " 1)2

x ! 2

Interval of integration y

x ! y " 1

Inner radius ! 2 " y " 1

Outer radius ! 2 " (y " 1)2

(c)

Related Exercises 45–52

SECTION 6.3 EXERCISES Review Questions 1. Suppose a cut is made through a solid object perpendicular to the

x-axis at a particular point x. Explain the meaning of A1x2. 2. A solid has a circular base and cross sections perpendicular to the

base are squares. What method should be used to find the volume of the solid?

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430 Chapter 6 Applications of Integration

11. The solid with a semicircular base of radius 5 whose cross sec- tions perpendicular to the base and parallel to the diameter are squares

12. The solid whose base is the region bounded by y = x2 and the line y = 1, and whose cross sections perpendicular to the base and parallel to the x-axis are squares

square cross section

base

y ! x2

13. The solid whose base is the triangle with vertices 10, 02, 12, 02, and 10, 22, and whose cross sections perpendicular to the base and parallel to the y-axis are semicircles

14. The pyramid with a square base 4 m on a side and a height of 2 m (Use calculus.)

15. The tetrahedron (pyramid with four triangular faces), all of whose edges have length 4

16. A circular cylinder of radius r and height h whose axis is at an angle of p>4 to the base

h r

circular base

17–26. Disk method Let R be the region bounded by the following curves. Use the disk method to find the volume of the solid generated when R is revolved about the x-axis.

17. y = 2x, y = 0, x = 3 (Verify that your answer agrees with the volume formula for a cone.)

y ! 2x

(3, 6)

5. Why is the disk method a special case of the general slicing method?

6. The region R bounded by the graph of y = f 1x2 Ú 0 and the x-axis on 3a, b4 is revolved about the line y = -2 to form a solid of revolution whose cross sections are washers. What are the inner and outer radii of the washer at a point x in 3a, b4?

Basic Skills 7–16. General slicing method Use the general slicing method to find the volume of the following solids.

7. The solid whose base is the region bounded by the curves y = x2 and y = 2 - x2, and whose cross sections through the solid per- pendicular to the x-axis are squares

x y

8. The solid whose base is the region bounded by the semicircle y = 21 - x2 and the x-axis, and whose cross sections through the solid perpendicular to the x-axis are squares

y x !1 ! x2 y "

y x

9. The solid whose base is the region bounded by the curve y = 1cos x and the x-axis on 3-p>2, p>24, and whose cross sections through the solid perpendicular to the x-axis are isosceles right triangles with a horizontal leg in the xy-plane and a vertical leg above the x-axis

x y ! cos x

10. The solid with a circular base of radius 5 whose cross sections perpen- dicular to the base and parallel to the x-axis are equilateral triangles

equilateral triangles

circular base

y ! 2 " x2

y ! x2

x y

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6.3 Volume by Slicing 431

28. y = x, y = 24 x y ! x

y ! 4 x

29. y = ex>2, y = e-x>2, x = ln 2, x = ln 3 y ! ex/2

y ! e"x/2

xln 2O ln 3

30. y = x, y = x + 2, x = 0, x = 4

y ! x " 2

y ! x

31. y = x + 3, y = x2 + 1

32. y = 1sin x, y = 1, x = 0 33. y = sin x, y = 1sin x, for 0 … x … p>2 34. y = ! x ! , y = 2 - x2

35–40. Disks ,washers about the y-axis Let R be the region bounded by the following curves. Use the disk or washer method to find the vol- ume of the solid generated when R is revolved about the y-axis.

35. y = x, y = 2x, y = 6

0 x

y ! x

y ! 2x

y ! 6

18. y = 2 - 2x, y = 0, x = 0 (Verify that your answer agrees with the volume formula for a cone.)

y ! 2 " 2x

19. y = e-x, y = 0, x = 0, x = ln 4

ln 4

y ! e"x

20. y = cos x on 30, p>24, y = 0, x = 0 1Recall that cos2 x = 1 211 + cos 2x2.2

y ! cos x

21. y = sin x on 30, p4, y = 0 (Recall that sin2 x = 1211 - cos 2x2.2 22. y = 225 - x2, y = 0 (Verify that your answer agrees with the

volume formula for a sphere.)

23. y = 124 1 - x2, y = 0, x = 0, and x = 12

24. y = sec x, y = 0, x = 0, and x = p4

25. y = 121 + x2, y = 0, x = -1, and x = 1

26. y = 124 1 - x2, y = 0, x = -12, and x = 12

27–34. Washer method Let R be the region bounded by the following curves. Use the washer method to find the volume of the solid gener- ated when R is revolved about the x-axis.

27. y = x, y = 21x R

y ! 2!x

y ! x

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434 Chapter 6 Applications of Integration

We divide 3a, b4 into n subintervals of length ∆x = 1b - a2>n and identify an arbi- trary point xk

* on the kth subinterval, for k = 1, c, n. Now observe the rectangle built on the kth subinterval with a height of f 1xk*2 and a width ∆x (Figure 6.40). As it revolves about the y-axis, this rectangle sweeps out a thin cylindrical shell.

6.4 Volume by Shells You can solve many challenging volume problems using the disk>washer method. There are, however, some volume problems that are difficult to solve with this method. For this reason, we extend our discussion of volume problems to the shell method, which—like the disk>washer method—is used to compute the volume of solids of revolution. Cylindrical Shells Let R be a region bounded by the graph of f, the x-axis, and the lines x = a and x = b, where 0 … a 6 b and f 1x2 Ú 0 on 3a, b4. When R is revolved about the y-axis, a solid is generated (Figure 6.39) whose volume is computed with the slice-and-sum strategy.

➤ Why another method? Suppose R is the region in the first quadrant bounded by the graph of y = x2 - x3 and the x-axis (Figure 6.38). When R is revolved about the y-axis, the resulting solid has a volume that is difficult to compute using the washer method. The volume is much easier to compute using the shell method.

Figure 6.38

Figure 6.39

Figure 6.40

y ! x2 " x3

0 1

0.14

0.07

y y

Revolving region R about the y-axis...

... produces a solid.

a x

f (xk*)

xk*

Revolving the kth rectangle about the y-axis...

... produces a cylindrical shell with height f (xk*) and thickness !x.

b a

When the kth cylindrical shell is unwrapped (Figure 6.41), it approximates a thin rect- angular slab. The approximate length of the slab is the circumference of a circle with ra- dius xk

*, which is 2pxk *. The height of the slab is the height of the original rectangle f 1xk*2

and its thickness is ∆x; therefore, the volume of the kth shell is approximately

2pxk * # f 1xk*2 # ∆x = 2pxk* f 1xk*2∆x.

length height thickness

Summing the volumes of the n cylindrical shells gives an approximation to the volume of the entire solid:

V ≈ a n

k = 1 2pxk

* f 1xk*2∆x.

(1)1* (1)1* ()*

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6.4 Volume by Shells 435

As n increases and as ∆x approaches 0 (Figure 6.42), we obtain the exact volume of the solid as a definite integral:

V = lim nS∞

a n

k = 1 2p xk

* f 1xk*2∆x = Lba 2px f 1x2 dx. shell shell circumference thickness

Figure 6.41

f (xk*)

Circumference ! 2! " radius ! 2!xk*

Radius ! xk*

Height ! f (xk*)

Thickness ! #x

Length ! 2! " radius ! 2!xk*

shell height()*

(1)1* ()*

Increase the number of shells.

Volume ! lim k!1

" 2!xk* f (xk*)#x

! ! 2!x f (x) dx

n ! $

n!$

Before doing examples, we generalize this method as we did for the disk method. Suppose that the region R is bounded by two curves, y = f 1x2 and y = g1x2, where f 1x2 Ú g1x) on 3a, b4 (Figure 6.43). What is the volume of the solid generated when R is revolved about the y-axis?

➤ Rather than memorizing, think of the meaning of the factors in this formula: f 1x2 is the height of a single cylindrical shell, 2px is the circumference of the shell, and dx corresponds to the thickness of a shell. Therefore, 2px f 1x2 dx represents the volume of a single shell, and we sum the volumes from x = a to x = b. Notice that the integrand for the shell method is the function A1x2 that gives the surface area of the shell of radius x, for a … x … b.

Figure 6.42

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436 Chapter 6 Applications of Integration

The situation is similar to the case we just considered. A typical rectangle in R sweeps out a cylindrical shell, but now the height of the kth shell is f 1xk*2 - g1xk*2, for k = 1, c, n. As before, we take the radius of the kth shell to be xk*, which means the volume of the kth shell is approximated by 2pxk

*1 f 1xk*2 - g1xk*22∆x (Figure 6.43). Sum- ming the volumes of all the shells gives an approximation to the volume of the entire solid:

V ≈ a n

k = 1 2pxk

* 1 f 1xk*2 - g1xk*22∆x. shell shell circumference height

Taking the limit as n S ∞ (which implies that ∆x S 0), the exact volume is the definite integral

V = lim nS∞

a n

k = 1 2pxk

*1 f 1xk*2 - g1xk*22∆x = Lba 2px1 f 1x2 - g1x22 dx. We now have the formula for the shell method.

➤ An analogous formula for the shell method when R is revolved about the x-axis is obtained by reversing the roles of x and y:

V = L d

c 2py1 p 1y2 - q1y22 dy.

We assume R is bounded by the curves x = p1y2 and x = q1y2, where p1y2 Ú q1y2 on 3c, d4.

(+++1)+++1*(1)1*

y y

xa b

y " f (x)

y " g(x)

Shell height " f (xk*) # g(xk*)

Shell radius " xk*

xk*

Volume of kth shell ! 2!xk*( f (xk*) # g(xk*))!x

xk*

Figure 6.43

EXAMPLE 1 A sine bowl Let R be the region bounded by the graph of f 1x2 = sin x2, the x-axis, and the vertical line x = 1p>2 (Figure 6.44). Find the volume of the solid generated when R is revolved about the y-axis.

SOLUTION Revolving R about the y-axis produces a bowl-shaped region (Figure 6.45). The radius of a typical cylindrical shell is x and its height is f 1x2 = sin x2. Therefore, the volume by the shell method is

V = L b

a 2px f 1x2 dx = L1p>20 2px sin x2 dx.rr

shell circumference

shell height

Volume by the Shell Method Let f and g be continuous functions with f 1x2 Ú g1x2 on 3a, b4. If R is the region bounded by the curves y = f 1x2 and y = g1x2 between the lines x = a and x = b, the volume of the solid generated when R is revolved about the y-axis is

V = L b

a 2px1 f 1x2 - g1x22 dx.

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6.4 Volume by Shells 437

Now we make the change of variables u = x2, which means that du = 2x dx. The lower limit x = 0 becomes u = 0 and the upper limit x = 1p>2 becomes u = p>2. The volume of the solid is

V = L 1p>2

0 2px sin x2 dx = p L

p>2 0

sin u du u = x2, du = 2x dx

= p1-cos u2 ` p>2 0

Fundamental Theorem

= p10 - 1-122 = p. Simplify. Related Exercises 5–14

QUICK CHECK 1 The triangle bounded by the x-axis, the line y = 2x, and the line x = 1 is revolved about the y-axis. Give an integral that equals the volume of the resulting solid using the shell method.

EXAMPLE 2 Shells about the x-axis Let R be the region in the first quadrant bounded by the graph of y = 1x - 2 and the line y = 2. Find the volume of the solid generated when R is revolved about the x-axis.

SOLUTION The revolution is about the x-axis, so the integration in the shell method is with respect to y. A typical shell runs parallel to the x-axis and has radius y, where 0 … y … 2; the shells extend from the y-axis to the curve y = 1x - 2 (Figure 6.46). Solving y = 1x - 2 for x, we have x = y2 + 2, which is the height of the shell at the point y (Figure 6.47). Integrating with respect to y, the volume of the solid is

V = L 2

0 2py 1y2 + 22 dy = 2pL20 1y3 + 2y2 dy = 16p.

shell shell circumference height

➤ When computing volumes using the shell method, it is best to sketch the region R in the xy-plane and draw a slice through the region that generates a typical shell.

Figure 6.44

Figure 6.45

Interval of integration

f (x) ! sin x2

Height ! sin x 2

! /2

Shell radius ! x 1

f (x) ! sin x2

!! /2

y Shell circumference ! 2!x

Shell radius ! x

Shell height ! sin x2

Figure 6.46

ceInterval of

in te

gr at

io n

x2 4 6

y " 2 (6, 2)

y " x ! 2 x " y2 # 2

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438 Chapter 6 Applications of Integration

Related Exercises 15–26

EXAMPLE 3 Volume of a drilled sphere A cylindrical hole with radius r is drilled symmetrically through the center of a sphere with radius a, where 0 … r … a. What is the volume of the remaining material?

SOLUTION The y-axis is chosen to coincide with the axis of the cylindrical hole. We let R be the region in the xy-plane bounded above by f 1x2 = 2a2 - x2, the upper half of a circle of radius a, and bounded below by g1x2 = - 2a2 - x2, the lower half of a circle of radius a, for r … x … a (Figure 6.48a). Slices are taken perpendicular to the x-axis from x = r to x = a. When a slice is revolved about the y-axis, it sweeps out a cylindri- cal shell that is concentric with the hole through the sphere (Figure 6.48b). The radius of a typical shell is x and its height is f 1x2 - g1x2 = 22a2 - x2. Therefore, the volume of the material that remains in the sphere is

V = L a

r 2px122a2 - x22 dx

= -2pL 0

a2 - r2 1u du u = a2 - x2, du = -2x dx

= 2pa 2 3

u3>2b ` a2 - r2 0

Fundamental Theorem

= 4p 3

1a2 - r223>2. Simplify.

Shell height ! y2 " 2

Shell radius ! y

x ! y2 " 2

Shell circumference ! 2!y

Figure 6.47

Figure 6.48

x a

r r

f (x) ! a2 " x2

g(x) ! " a2 " x2

Shell height ! f (x) " g(x)

Shell radius ! x

(a) (b)

➤ In Example 2, we could use the disk> washer method to compute the volume, but notice that this approach requires splitting the region into two subregions. A better approach is to use the shell method and integrate along the y-axis.

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6.4 Volume by Shells 439

It is important to check the result by examining special cases. In the case that r = a (the radius of the hole equals the radius of the sphere), our calculation gives a volume of 0, which is correct. In the case that r = 0 (no hole in the sphere), our calculation gives the correct volume of a sphere, 43 pa

3. Related Exercises 27–32

EXAMPLE 4 Revolving about other lines Let R be the region bounded by the curve y = 1x, the line y = 1, and the y-axis (Figure 6.49a). a. Use the shell method to find the volume of the solid generated when R is revolved

about the line x = -12 (Figure 6.49b). b. Use the disk/washer method to find the volume of the solid generated when R is

revolved about the line y = 1 (Figure 6.49c).

SOLUTION

a. Using the shell method, we must imagine taking slices through R parallel to the y-axis. A typical slice through R at a point x, where 0 … x … 1, has length 1 - 1x. When that slice is revolved about the line x = -12, it sweeps out a cylindrical shell with a radius of x + 12 and a height of 1 - 1x (Figure 6.50). A slight modification of the standard shell method gives the volume of the solid:

L 1

0 2pax + 1

2 b11 - 1x2 dx = 2pL10 ax - x3>2 + 12 - x1>22 b dx Expand integrand.

= 2pa 1 2

x2 - 2 5

x5>2 + 1 2

x - 1 3

x3>2b ` 1 0

= 8p 15

. Evaluate integral.

(a)

y ! 1

y ! x

x (b)

1 x ! "

"2 1

y ! 1

1 (c)

y ! x

Shell radius ! x " #12

Interval of integration

y ! 1

Shell height ! 1 # x

1x ! # 2

Figure 6.49

Figure 6.50

shell circumference

shell height

b. Using the disk/washer method, we take slices through R parallel to the y-axis. Consider a typical slice at a point x, where 0 … x … 1. Its length, now measured with respect to the line y = 1, is 1 - 1x. When that slice is revolved about the line y = 1, it sweeps out a disk of radius 1 - 1x (Figure 6.51). By the disk/washer method, the volume of the solid is

L 1

0 p11 - 1x22 dx = pL10 11 - 21x + x2 dx Expand integrand.

= pax - 4 3

x3>2 + 1 2

x2b ` 1 0 Evaluate integral.

= p

6 .

Related Exercises 33–40

➤ If we instead revolved about the y-axis 1x = 02, the radius of the shell would be x. Because we are revolving about the line x = -12, the radius of the shell is x +

1 2.

➤ The disk/washer method can also be used for part (a), and the shell method can also be used for part (b).

Figure 6.51

Interval of integration

y ! 1

Disk radius ! 1 " x

1 xx

y ! x

radius of disk

M06_BRIG7345_02_SE_C06.4.indd 439 31/10/13 9:44 AM

440 Chapter 6 Applications of Integration

QUICK CHECK 2 Write the volume integral in Example 4b in the case that R is revolved about the line y = -5.

Restoring Order After working with slices, disks, washers, and shells, you may feel somewhat over- whelmed. How do you choose a method, and which method is best?

Notice that the disk method is just a special case of the washer method. So for solids of revolution, the choice is between the washer method and the shell method. In principle, either method can be used. In practice, one method usually produces an integral that is easier to evaluate than the other method. The following table summarizes these methods.

SUMMARY Disk ,Washer and Shell Methods Integration with respect to x

y ! f (x)

y ! g(x) O a

y ! f (x)

y ! g(x) O a

Disk ,washer method about the x-axis Disks>washers are perpendicular to the x-axis.

L b

a p1 f 1x22 - g1x222 dx

Shell method about the y-axis Shells are parallel to the y-axis.

L b

a 2px1 f 1x2 - g1x22 dx

Integration with respect to y

x ! p(y)

x ! q(y)

O x

x ! p(y)

x ! q(y)

Disk ,washer method about the y-axis Disks>washers are perpendicular to the y-axis.

L d

c p1p1y22 - q1y222 dy

Shell method about the x-axis Shells are parallel to the x-axis.

L d

c 2py1p1y2 - q1y22 dy

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6.4 Volume by Shells 441

EXAMPLE 5 Volume by which method? The region R is bounded by the graphs of f 1x2 = 2x - x2 and g1x2 = x on the interval 30, 14 (Figure 6.52). Use the washer method and the shell method to find the volume of the solid formed when R is revolved about the x-axis.

SOLUTION Solving f 1x2 = g1x2, we find that the curves intersect at the points 10, 02 and 11, 12. Using the washer method, the upper bounding curve is the graph of f , the lower bounding curve is the graph of g, and a typical washer is perpendicular to the x-axis (Figure 6.53). Therefore, the volume is

V = L 1

0 p112x - x222 - x22 dx Washer method

= pL 1

0 1x4 - 4x3 + 3x22 dx Expand integrand.

= p a x5 5

- x4 + x3b ` 1 0

= p

5 . Evaluate integral.

The shell method requires expressing the bounding curves in the form x = p1y2 for the right curve and x = q1y2 for the left curve. The right curve is x = y. Solving y = 2x - x2 for x, we find that x = 1 - 11 - y describes the left curve. A typical shell is parallel to the x-axis (Figure 6.54). Therefore, the volume is

V = L 1

0 2py 1y - 11 - 11 - y22 dy.

This integral equals p5 , but it is more difficult to evaluate than the integral required by the washer method. In this case, the washer method is preferable. Of course, the shell method may be preferable for other problems.

➤ To solve y = 2x - x2 for x, write the equation as x2 - 2x + y = 0 and complete the square or use the quadratic formula.

Figure 6.52

f (x) ! 2x " x2 (1, 1)

g(x) ! x

p1y2 q1y2f5

Related Exercises 41–48

QUICK CHECK 3 Suppose the region in Example 5 is revolved about the y-axis. Which method (washer or shell) leads to an easier integral?

Figure 6.53

y ! x

(1, 1)

(Outer radius)2 ! (2x " x2)2

(Inner radius)2 ! x2

y ! 2x " x2

O x

f (x) g(x)

Figure 6.54

x ! y

(1, 1)

Shell height ! y " (1 " !1 " y) Shell radius ! y

x ! 1 " !1 " y

The following example shows that while two methods may be used on the same problem, one of them may be preferable.

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442 Chapter 6 Applications of Integration

Review Questions 1. Assume f and g are continuous with f 1x2 Ú g 1x2 on 3a, b4. The

region bounded by the graphs of f and g and the lines x = a and x = b is revolved about the y-axis. Write the integral given by the shell method that equals the volume of the resulting solid.

2. Fill in the blanks: A region R is revolved about the y-axis. The vol- ume of the resulting solid could (in principle) be found using the disk>washer method and integrating with respect to ________ or using the shell method and integrating with respect to ________.

3. Fill in the blanks: A region R is revolved about the x-axis. The vol- ume of the resulting solid could (in principle) be found using the disk>washer method and integrating with respect to ________ or using the shell method and integrating with respect to ________.

4. Are shell method integrals easier to evaluate than washer method integrals? Explain.

Basic Skills 5–14. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about the y-axis.

5. y = x - x2, y = 0

y ! x " x2

1 x

6. y = -x2 + 4x + 2, y = x2 - 6x + 10

y ! x2 " 6x # 10

y ! "x2 # 4x # 2

0 41 x

7. y = 11 + x22-1, y = 0, x = 0, and x = 2

0 x2

y ! 1 1 " x2

8. y = 6 - x, y = 0, x = 2, and x = 4

0 x42

y ! 6 " x

9. y = 3x, y = 3, and x = 0 (Use integration and check your answer using the volume formula for a cone.)

0 x1

y ! 3x

10. y = 1 - x2, x = 0, and y = 0, in the first quadrant

11. y = x3 - x8 + 1, y = 1

12. y = 1x, y = 0, and x = 1 13. y = cos x2, y = 0, for 0 … x … 1p>2 14. y = 24 - 2x2, y = 0, and x = 0, in the first quadrant 15–26. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about the x-axis.

15. y = 1x, y = 0, and x = 4

y ! !x

16. y = 8, y = 2x + 2, x = 0, and x = 2

0 x2

y ! 2x " 2

SECTION 6.4 EXERCISES

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6.5 Length of Curves 445

72. Ellipsoids An ellipse centered at the origin is described by the equation x2>a2 + y2>b2 = 1. If an ellipse R is revolved about either axis, the resulting solid is an ellipsoid.

a. Find the volume of the ellipsoid generated when R is revolved about the x-axis (in terms of a and b).

b. Find the volume of the ellipsoid generated when R is revolved about the y-axis (in terms of a and b).

c. Should the results of parts (a) and (b) agree? Explain.

73. Change of variables Suppose f 1x2 7 0 for all x and 140 f 1x2 dx = 10. Let R be the region in the first quadrant bounded by the coordinate axes, y = f 1x22, and x = 2. Find the volume of the solid generated by revolving R about the y-axis.

74. Equal integrals Without evaluating integrals, explain the following equalities. (Hint: Draw pictures.)

a. pL 4

0 18 - 2x22 dx = 2pL80 ya4 - y2 b dy

b. L 2

0 125 - 1x2 + 1222 dx = 2L51 y1y - 1 dy

75. Volumes without calculus Solve the following problems with and without calculus. A good picture helps.

a. A cube with side length r is inscribed in a sphere, which is inscribed in a right circular cone, which is inscribed in a right circular cylinder. The side length (slant height) of the cone is equal to its diameter. What is the volume of the cylinder?

b. A cube is inscribed in a right circular cone with a radius of 1 and a height of 3. What is the volume of the cube?

c. A cylindrical hole 10 in long is drilled symmetrically through the center of a sphere. How much material is left in the sphere? (Enough information is given.)

QUICK CHECK ANSWERS

1. 110 2px12x2 dx 2. V = 110 p136 - 11x + 5222 dx 3. The shell method is easier.

6.5 Length of Curves The space station orbits Earth in an elliptical path. How far does it travel in one orbit? A baseball slugger launches a home run into the upper deck and the sportscaster claims it landed 480 feet from home plate. But how far did the ball actually travel along its flight path? These questions deal with the length of trajectories or, more generally, with arc length. As you will see, their answers can be found by integration.

There are two common ways to formulate problems about arc length: The curve may be given explicitly in the form y = f 1x2 or it may be defined parametrically. In this sec- tion, we deal with the first case. Parametric curves are introduced in Section 10.1, and the associated arc length problem is discussed in Section 11.8.

Arc Length for y = f 1x 2 Suppose a curve is given by y = f 1x2, where f is a function with a continuous first derivative on the interval 3a, b4. The goal is to determine how far you would travel if you walked along the curve from 1a, f 1a22 to 1b, f 1b22. This distance is the arc length, which we denote L.

As shown in Figure 6.55, we divide 3a, b4 into n subintervals of length ∆x = 1b - a2>n, where xk is the right endpoint of the kth subinterval, for k = 1, c, n. Join- ing the corresponding points on the curve by line segments, we obtain a polygonal line with n line segments. If n is large and ∆x is small, the length of the polygonal line is a good approximation to the length of the actual curve. The strategy is to find the length of the polygonal line and then let n increase, while ∆x goes to zero, to get the exact length of the curve.

➤ More generally, we may choose any point in the kth subinterval and ∆x may vary from one subinterval to the next. Using right endpoints, as we do here, simplifies the discussion and leads to the same result.

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446 Chapter 6 Applications of Integration

y ! f (x)

x1x0 ! a x2 xk"1 xn"1 xn ! bxk… …O x

The length of the red polygonal line from (a, f (a)) to (b, f (b)) approximates L.

(xk"1, f (xk"1)) (xk, f (xk))

|#yk | #x

Figure 6.55

Consider the kth subinterval 3xk - 1, xk4 and the line segment between the points 1xk - 1, f 1xk - 122 and 1xk, f 1xk22. We let the change in the y-coordinate between these points be

∆yk = f 1xk2 - f 1xk - 12. The kth line segment is the hypotenuse of a right triangle with sides of length ∆x and " ∆yk " = " f 1xk2 - f 1xk - 12 " . The length of each line segment is21∆x22 + " ∆yk "2, for k = 1, 2, c, n. Summing these lengths, we obtain the length of the polygonal line, which approximates the length L of the curve:

L ≈ a n

k = 1 21∆x22 + " ∆yk "2.

In previous applications of the integral, we would, at this point, take the limit as n S ∞ and ∆x S 0 to obtain a definite integral. However, because of the presence of the ∆yk term, we must complete one additional step before taking a limit. Notice that the slope of the line segment on the kth subinterval is ∆yk>∆x (rise over run). By the Mean Value Theorem (see the margin figure and Section 4.6), this slope equals f ′1x*k2 for some point xk

* on the kth subinterval. Therefore,

L ≈ a n

k = 1 21∆x22 + " ∆yk "2

= a n

k = 1 B1∆x22a1 + a ∆yk∆x b2b Factor out 1∆x22. = a

k = 1 B1 + a ∆yk∆x b2 ∆x Bring ∆x out of the square root. = a

k = 1 21 + f ′1x*k22 ∆x. Mean Value Theorem

Now we have a Riemann sum. As n increases and as ∆x approaches zero, the sum approaches a definite integral, which is also the exact length of the curve. We have

L = lim nS∞

a n

k = 1 21 + f ′1x*k22 ∆x = Lba 21 + f ′1x22 dx.

➤ Notice that ∆x is the same for each subinterval, but ∆yk depends on the subinterval.

xxkxk!1

!"yk!

Slope # "yk "x

Slope # f $(xk*)

(xk*, f (xk*))

xk*

Mean Value Theorem

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6.5 Length of Curves 447

DEFINITION Arc Length for y = f 1x 2 Let f have a continuous first derivative on the interval 3a, b4. The length of the curve from 1a, f 1a22 to 1b, f 1b22 is

L = L b

a 21 + f ′1x22 dx.

➤ Note that 1 + f ′1x22 is positive, so the square root in the integrand is defined whenever f ′ exists. To ensure that 21 + f ′1x22 is integrable on 3a, b4, we require that f ′ be continuous.

QUICK CHECK 1 What does the arc length formula give for the length of the line y = x between x = 0 and x = a, where a Ú 0?

EXAMPLE 1 Arc length Find the length of the curve f 1x2 = x3>2 between x = 0 and x = 4 (Figure 6.56).

SOLUTION Notice that f ′1x2 = 32 x1>2, which is continuous on the interval 30, 44. Using the arc length formula, we have

L = L b

a 21 + f ′1x22 dx = L40 B1 + a 32 x1>2b2 dx Substitute for f ′1x2.

= L 4

0 A1 + 94 x dx Simplify. =

4 9 L

1 2u du u = 1 + 9x

4 , du =

9 4

= 4 9

a 2 3

u3>2b ` 1

Fundamental Theorem

= 8 27

1103>2 - 12. Simplify. The length of the curve is 827 1103>2 - 12 ≈ 9.1 units.

Related Exercises 3–16

EXAMPLE 2 Arc length of an exponential curve Find the length of the curve f 1x2 = 2ex + 18 e-x on the interval 30, ln 24. SOLUTION We first calculate f ′1x2 = 2ex - 18 e-x and f ′1x22 = 4e2x - 12 + 164 e-2x. The length of the curve on the interval 30, ln 24 is

L = L ln 2

0 21 + f ′1x22 dx = L ln 20 31 + 14e2x - 12 + 164 e-2x2 dx

= L ln 2

0 34e2x + 12 + 164 e-2x dx Simplify.

= L ln 2

0 312ex + 18 e-x22 dx Factor.

= L ln 2

0 12ex + 18 e-x2 dx Simplify.

= 12ex - 18 e-x2 ` ln 2 0

= 33 16

. Evaluate the integral.

Related Exercises 3–16

Figure 6.56

x4321

y ! x3/2

(4, 8)

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448 Chapter 6 Applications of Integration

EXAMPLE 3 Circumference of a circle Confirm that the circumference of a circle of radius a is 2pa.

SOLUTION The upper half of a circle of radius a centered at 10, 02 is given by the func- tion f 1x2 = 2a2 - x2 for ! x ! … a (Figure 6.57). So we might consider using the arc length formula on the interval 3-a, a4 to find the length of a semicircle. However, the circle has vertical tangent lines at x = {a and f ′1{a2 is undefined, which prevents us from using the arc length formula. An alternative approach is to use symmetry and avoid the points x = {a. For example, let’s compute the length of one-eighth of the circle on the interval 30, a>124 (Figure 6.57).

We first determine that f ′1x2 = - x2a2 - x2 , which is continuous on 30, a>124. The length of one-eighth of the circle is

L a>12

0 21 + f ′1x22 dx = La>120 C1 + a - x2a2 - x2 b2 dx

= L a>12

0 A a2a2 - x2 dx Simplify. = aL

a>12 0

dx2a2 - x2 Simplify; a 7 0. = a sin-1

x a ` a>12 0

Integrate.

= a asin-1 112 - 0b Evaluate. =

pa 4

. Simplify.

It follows that the circumference of the full circle is 81pa>42 = 2pa units. Related Exercises 3–16

EXAMPLE 4 Looking ahead Consider the segment of the parabola f 1x2 = x2 on the interval 30, 24. a. Write the integral for the length of the curve. b. Use a calculator to evaluate the integral.

SOLUTION

a. Noting that f ′1x2 = 2x, the arc length integral is L

0 21 + f ′1x22 dx = L20 21 + 4x2 dx.

b. Using integration techniques presented so far, this integral cannot be evaluated (the required method is given in Section 7.4). This is typical of arc length integrals—even simple functions can lead to arc length integrals that are difficult to evaluate analyti- cally. Without an analytical method, we may use numerical integration to approximate the value of a definite integral (Section 7.7). Many calculators have built-in functions for this purpose. For this integral, the approximate arc length is

L 2

0 21 + 4x2 dx ≈ 4.647.

Related Exercises 17–26

➤ The arc length integral for the semicircle on 3-a, a4 is an example of an improper integral, a topic considered in Section 7.8.

➤ When relying on technology, it is a good idea to check whether an answer is plausible. In Example 4, we found that the arc length of y = x2 on 30, 24 is approximately 4.647. The straight- line distance between 10, 02 and 12, 42 is 220 ≈ 4.472, so our answer is reasonable.

Figure 6.57

!a x

! 4

One-eighth of the circley

y " !a2 ! x2

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6.5 Length of Curves 449

Arc Length for x = g 1 y2 Sometimes it is advantageous to describe a curve as a function of y—that is, x = g1y2. The arc length formula in this case is derived exactly as in the case of y = f 1x2, switching the roles of x and y. The result is the following arc length formula.

DEFINITION Arc Length for x = g 1 y 2 Let x = g1y2 have a continuous first derivative on the interval 3c, d4. The length of the curve from 1g1c2, c2 to 1g1d2, d2 is

L = L d

c 21 + g′1y22 dy.

EXAMPLE 5 Arc length Find the length of the curve y = f 1x2 = x2>3 between x = 0 and x = 8 (Figure 6.58).

SOLUTION The derivative of f 1x2 = x2>3 is f ′1x2 = 23 x-1>3, which is undefined at x = 0. Therefore, the arc length formula with respect to x cannot be used, yet the curve certainly appears to have a well-defined length.

The key is to describe the curve with y as the independent variable. Solving y = x2>3 for x, we have x = g1y2 = {y3>2. Notice that when x = 8, y = 82>3 = 4, which says that we should use the positive branch of {y3>2. Therefore, finding the length of the curve y = f 1x2 = x2>3 from x = 0 to x = 8 is equivalent to finding the length of the curve x = g1y2 = y3>2 from y = 0 to y = 4. This is precisely the problem solved in Example 1. The arc length is 8271103>2 - 12 ≈ 9.1 units.

Related Exercises 27–30

EXAMPLE 6 Ingenuity required Find the length of the curve y = f 1x2 = ln 1x + 2x2 - 12 on the interval 31,124 (Figure 6.59). SOLUTION Calculating f ′ shows that the graph of f has a vertical tangent line at x = 1. Therefore, the integrand in the arc length integral is undefined at x = 1. An alternative strategy is to express the function in the form x = g1y2 and evalu- ate the arc length integral with respect to y. Noting that x Ú 1 and y Ú 0, we solve y = ln 1x + 2x2 - 12 for x in the following steps:

ey = x + 2x2 - 1 Exponentiate both sides. ey - x = 2x2 - 1 Subtract x from both sides.

e2y - 2eyx = -1 Square both sides and cancel x2.

x = e2y + 1

2ey =

ey + e-y

2 . Solve for x.

We conclude that the given curve is also described by the function

x = g1y2 = ey + e-y 2

. The interval 1 … x … 12 corresponds to the interval 0 … y … ln 112 + 12 (Figure 6.59). Note that g′1y2 = ey - e-y

2 is continuous on 30, ln 112 + 124. The arc length is

L ln 112 + 12

0 21 + g′1y22 dy = L ln 112 + 120 B1 + a ey - e-y2 b2dy Substitute for g′1y2.

= 1 2 L

ln 112 + 12 0

1ey + e-y2 dy Expand and simplify. =

1 2

1ey - e-y2 ` ln 112 + 12 0

= 1. Fundamental Theorem

Related Exercises 27–30

➤ The function 12 1ey + e-y2 is the hyperbolic cosine, denoted cosh y. The function 12 1ey - e-y2 is the hyperbolic sine, denoted sinh y. See Section 6.10.

x8642

y ! x2/3

x ! y3/2

(8, 4)

Figure 6.58

1.5

0.5

ey ! e"y

y # ln(x ! !x2 " 1)

x #

ln(!2 ! 1)

Figure 6.59

QUICK CHECK 2 What does the arc length formula give for the length of the line x = y between y = c and y = d, where d Ú c? Is the result consistent with the result given by the Pythagorean theorem?

QUICK CHECK 3 Write the integral for the length of the curve x = sin y on the interval 0 … y … p.

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6.6 Surface Area 451

for more on hyperbolic functions). Estimate the length of the Gateway Arch.

630 ft

Additional Exercises 39. Lengths of related curves Suppose the graph of f on the interval 3a, b4 has length L, where f ′ is continuous on 3a, b4. Evaluate the

following integrals in terms of L.

a. L b>2

a>2 21 + f ′12x22 dx b. Lb>ca>c 21 + f ′1cx22 dx if c ≠ 0 40. Lengths of symmetric curves Suppose a curve is described

by y = f 1x2 on the interval 3-b, b4, where f ′ is continuous on 3-b, b4. Show that if f is symmetric about the origin ( f is odd) or f is symmetric about the y-axis ( f is even), then the length of the curve y = f 1x2 from x = -b to x = b is twice the length of the curve from x = 0 to x = b. Use a geometric argument and prove it using integration.

41. A family of exponential functions

a. Show that the arc length integral for the function f 1x2 = Aeax + 1

4Aa2 e-ax, where a 7 0 and A 7 0, may be

integrated using methods you already know. b. Verify that the arc length of the curve y = f 1x2 on the interval 30, ln 24 is

A12a - 12 - 1 4a2 A

12-a - 12. 42. Bernoulli’s “parabolas” Johann Bernoulli (1667–1748) evalu-

ated the arc length of curves of the form y = x12n + 12>2n, where n is a positive integer, on the interval 30, a4. a. Write the arc length integral.

b. Make the change of variables u2 = 1 + a2n + 1 2n

b2x1>n to obtain a new integral with respect to u.

c. Use the Binomial Theorem to expand this integrand and evalu- ate the integral.

d. The case n = 1 1y = x3>22 was done in Example 1. With a = 1, compute the arc length in the cases n = 2 and n = 3. Does the arc length increase or decrease with n?

e. Graph the arc length of the curves for a = 1 as a function of n.

QUICK CHECK ANSWERS 1. 12a (The length of the line segment joining the points) 2. 121d - c2 (The length of the line segment joining the points) 3. L = 1p0 21 + cos2 y dy

6.6 Surface Area In Sections 6.3 and 6.4, we introduced solids of revolution and presented methods for computing the volume of such solids. We now consider a related problem: computing the area of the surface of a solid of revolution. Surface area calculations are important in aero- dynamics (computing the lift on an airplane wing) and biology (computing transport rates across cell membranes), to name just two applications. Here is an interesting observation: A surface area problem is “between” a volume problem (which is three-dimensional) and an arc length problem (which is one-dimensional). For this reason, you will see ideas that appear in both volume and arc length calculations as we develop the surface area integral.

Some Preliminary Calculations Consider a curve y = f 1x2 on an interval 3a, b4, where f is a nonnegative function with a continuous first derivative on 3a, b4. Now imagine revolving the curve about the x-axis to generate a surface of revolution (Figure 6.60). Our objective is to find the area of this surface.

Figure 6.60

y ! f (x) y

xa b

b a

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452 Chapter 6 Applications of Integration

Before tackling this problem, we consider a preliminary problem upon which we build a general surface area formula. First consider the graph of f 1x2 = rx>h on the interval 30, h4, where h 7 0 and r 7 0. When this line segment is revolved about the x-axis, it generates the surface of a cone of radius r and height h (Figure 6.61). A formula from geometry states that the surface area of a right circular cone of radius r and height h (excluding the base) is pr2r2 + h2 = pr/, where / is the slant height of the cone (the length of the slanted “edge” of the cone).

➤ One way to derive the formula for the surface area of a cone (not including the base) is to cut the cone on a line from its base to its vertex. When the cone is unfolded, it forms a sector of a circular disk of radius / with a curved edge of length 2pr. This sector is a fraction 2pr 2p/

= r /

of a full circular disk of radius /.

So the area of the sector, which is also the surface area of the cone, is p/2 # r

/ = pr/.

Figure 6.62

f (x) ! cx

ba x

(b, f (b))

Surface area of large cone "

ba x

(a, f (a))

ba x

Surface area of small cone Surface area of frustrum

Surface area Sb Surface area S ! Sa " SbSurface area Sa

Figure 6.61

y ! x

r r h

x h

Surface area ! !rl ! !r!r2 " h2

l !!r2 " h2 r

QUICK CHECK 1 Which is greater, the surface area of a cone of height 10 and radius 20 or the surface area of a cone of height 20 and radius 10 (excluding the bases)?

With this result, we can solve a preliminary problem that will be useful. Consider the linear function f 1x2 = cx on the interval 3a, b4, where 0 6 a 6 b and c 7 0. When this line segment is revolved about the x-axis, it generates a frustum of a cone (a cone whose top has been sliced off ). The goal is to find S, the surface area of the frustum. Figure 6.62 shows that S is the difference between the surface area Sb of the cone that extends over the interval 30, b4 and the surface area Sa of the cone that extends over the interval 30, a4.

Notice that the radius of the cone on 30, b4 is r = f 1b2 = cb, and its height is h = b. Therefore, this cone has surface area

Sb = pr2r2 + h2 = p1bc221bc22 + b2 = pb2c2c2 + 1. Similarly, the cone on 30, a4 has radius r = f 1a2 = ca and height h = a, so its surface area is

Sa = p1ac221ac22 + a2 = pa2c2c2 + 1. The difference of the surface areas Sb - Sa is the surface area S of the frustum on 3a, b4:

S = Sb - Sa = pb2c2c2 + 1 - pa2c2c2 + 1 = pc1b2 - a222c2 + 1.

Curved edge length ! 2!r

S ! ! ( f (b) " f (a)) Surface area of frustrum:

! ! (r2 " r1)

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6.6 Surface Area 453

Figure 6.63

y ! f (x) y

a b

x1x0 x2 xk"1 xn"1 xnxk

Figure 6.64

! (!x )2 " |!yk |2

x1 x2 xk#1 xn

|!yk | $ | f (xk ) # f (xk#1)| !x

xn#1

A slightly different form of this surface area formula will be useful. Observe that the line segment from 1a, f 1a22 to 1b, f 1b22 (which is the slant height of the frustum in Fig- ure 6.62) has length

/ = 21b - a22 + 1bc - ac22 = 1b - a22c2 + 1. Therefore, the surface area of the frustum can also be written

S = pc1b2 - a222c2 + 1 = pc1b + a21b - a22c2 + 1 Factor b2 - a2. = p¢cb + ca≤1b - a22c2 + 1 Expand c1b + a2.

6 (++++)++++*

f 1b2 f 1a2 / = p1 f 1b2 + f 1a22/.

This result can be generalized to any linear function g1x2 = cx + d that is positive on the interval 3a, b4. That is, the surface area of the frustum generated by revolving the line segment between 1a, g1a22 and 1b, g1b22 about the x-axis is given by p1g1b2 + g1a22/ (Exercise 36).

QUICK CHECK 2 What is the surface area of the frustum of a cone generated when the graph of f 1x2 = 3x on the interval 32, 54 is revolved about the x-axis? Surface Area Formula With the surface area formula for a frustum of a cone, we now derive a general area for- mula for a surface of revolution. We assume the surface is generated by revolving the graph of a positive, differentiable function f on the interval 3a, b4 about the x-axis. We begin by subdividing the interval 3a, b4 into n subintervals of equal length ∆x = b - a

n .

The grid points in this partition are

x0 = a, x1, x2, c, xn - 1, xn = b. Now consider the kth subinterval 3xk - 1, xk4 and the line segment between the points 1xk - 1, f 1xk - 122 and 1xk, f 1xk22 (Figure 6.63). We let the change in the y-coordinates between these points be ∆yk = f 1xk2 - f 1xk - 12.

When this line segment is revolved about the x-axis, it generates a frustum of a cone (Fig- ure 6.64). The slant height of this frustum is the length of the hypotenuse of a right triangle whose sides have lengths ∆x and " ∆yk " . Therefore, the slant height of the kth frustum is21∆x22 + " ∆yk "2 = 21∆x22 + 1∆yk22

M06_BRIG7345_02_SE_C06.6.indd 453 21/10/13 4:47 PM

454 Chapter 6 Applications of Integration

and its surface area is

Sk = p1 f 1xk2 + f 1xk - 12221∆x22 + 1∆yk22. It follows that the area S of the entire surface of revolution is approximately the sum of the surface areas of the individual frustums Sk, for k = 1, c, n; that is,

S ≈ a n

k = 1 Sk = a

k = 1 p1 f 1xk2 + f 1xk - 12221∆x22 + 1∆yk22.

We would like to identify this sum as a Riemann sum. However, one more step is required to put it in the correct form. We apply the Mean Value Theorem on the kth subinterval 3xk - 1, xk4 and observe that

f 1xk2 - f 1xk - 12 ∆x

= f ′1xk*2, for some number xk

* in the interval 1xk - 1, xk2, for k = 1, c, n. It follows that ∆yk = f 1xk2 - f 1xk - 12 = f ′1xk*2∆x.

We now replace ∆yk with f ′1xk*2∆x in the expression for the approximate surface area. The result is

S ≈ a n

k = 1 Sk = a

k = 1 p1 f 1xk2 + f 1xk - 12221∆x22 + 1∆yk 22

= a n

k = 1 p1 f 1xk2 + f 1xk - 12221∆x2 211 + f ′1xk*222 Mean Value Theorem

= a n

k = 1 p1 f 1xk2 + f 1xk - 12221 + f ′1xk*22 ∆x. Factor out ∆x.

When ∆x is small, we have xk - 1 ≈ xk ≈ xk*, and by the continuity of f, it follows that f 1xk - 12 ≈ f 1xk2 ≈ f 1xk*2, for k = 1, c, n. These observations allow us to write

S ≈ a n

k = 1 p1 f 1xk*2 + f 1xk*2221 + f ′1xk*22 ∆x

= a n

k = 1 2p f 1xk*221 + f ′1xk*22 ∆x.

This approximation to S, which has the form of a Riemann sum, improves as the number of subintervals increases and as the length of the subintervals approaches 0. Specifically, as n S ∞ and as ∆x S 0, we obtain an integral for the surface area:

S = lim nS∞

a n

k = 1 2p f 1x*k221 + f ′1x*k22 ∆x

= L b

a 2p f 1x221 + f ′1x22 dx.

➤ Notice that f is assumed to be differentiable on 3a, b4; therefore, it satisfies the conditions of the Mean Value Theorem. Recall that a similar argument was used to derive the arc length formula in Section 6.5.

DEFINITION Area of a Surface of Revolution Let f be a nonnegative function with a continuous first derivative on the interval 3a, b4. The area of the surface generated when the graph of f on the interval 3a, b4 is revolved about the x-axis is

S = L b

a 2p f 1x221 + f ′1x22 dx.

QUICK CHECK 3 Let f 1x2 = c, where c 7 0. What surface is generated when the graph of f on 3a, b4 is revolved about the x-axis? Without using calculus, what is the area of the surface? M06_BRIG7345_02_SE_C06.6.indd 454 21/10/13 11:35 AM

6.6 Surface Area 455

Figure 6.65

1 3

y ! 2 x

(b)

!a a!h x

y " a2 ! x2

EXAMPLE 1 Using the surface area formula The graph of f 1x2 = 21x on the interval 31, 34 is revolved about the x-axis. What is the area of the surface generated (Figure 6.65)?

SOLUTION Noting that f ′1x2 = 11x, the surface area formula gives S = L

a 2p f 1x221 + f ′1x22 dx

= 2pL 3

1 21xA1 + 1x dx Substitute for f and f ′.

= 4pL 3

1 2x + 1 dx Simplify.

= 8p 3 1x + 123>2 ` 3

1 =

16p 3

14 - 222. Integrate and simplify. Related Exercises 5–14

EXAMPLE 2 Surface area of a spherical cap A spherical cap is produced when a sphere of radius a is sliced by a horizontal plane that is a vertical distance h below the north pole of the sphere, where 0 … h … 2a (Figure 6.66a). We take the spherical cap to be that part of the sphere above the plane, so that h is the depth of the cap. Show that the area of a spherical cap of depth h cut from a sphere of radius a is 2pah.

SOLUTION To generate the spherical surface, we revolve the curve f 1x2 = 2a2 - x2 on the interval 3-a, a4 about the x-axis (Figure 6.66b). The spherical cap of height h cor- responds to that part of the sphere on the interval 3a -h, a4, for 0 … h … 2a. Noting that f ′1x2 = -x1a2 - x22-1>2, the surface area of the spherical cap of height h is

S = L b

a 2pf 1x221 + f ′1x22 dx

= 2pL a

a - h 2a2 - x221 + 1-x1a2 - x22-1>222 dx Substitute for f and f ′.

= 2pL a

a - h 2a2 - x2B a2a2 - x2 dx Simplify.

= 2pL a

a - h a dx = 2pah. Simplify and integrate.

It is worthwhile to check this result with three special cases. With h = 2a, we have a complete sphere, so S = 4pa2. The case h = a corresponds to a hemispherical cap, so S = 14pa22>2 = 2pa2. The case h = 0 corresponds to no spherical cap, so S = 0.

Related Exercises 5–14

➤ Notice that f is not differentiable at {a. Nevertheless, in this case, the surface area integral can be evaluated using methods you know.

➤ The surface area of a sphere of radius a is 4pa2.

Figure 6.66

h h

(a)

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456 Chapter 6 Applications of Integration

EXAMPLE 3 Painting a funnel The curved surface of a funnel is generated by

revolving the graph of y = f 1x2 = x3 + 1 12x

on the interval 31, 24 about the x-axis (Figure 6.67). Approximately what volume of paint is needed to cover the outside of the funnel with a layer of paint 0.05 cm thick? Assume that x and y are measured in centimeters.

SOLUTION Note that f ′1x2 = 3x2 - 1 12x2

. Therefore, the surface area of the funnel

in cm2 is

S = L b

a 2pf 1x221 + f ′1x22 dx

= 2pL 2

1 ax3 + 1

12x bB1 + a3x2 - 112x2 b2 dx Substitute for f and f ′.

= 2pL 2

1 ax3 + 1

12x bB a3x2 + 112x2 b2 dx Expand and factor under square root.

= 2pL 2

1 ax3 + 1

12x b a3x2 + 1

12x2 b dx Simplify.

= 12,289

192 p. Evaluate integral.

Because the paint layer is 0.05 cm thick, the approximate volume of paint needed isa 12,289p 192

cm2b10.05 cm2 ≈ 10.1 cm3. Related Exercises 15–16

The derivation that led to the surface area integral may be used when a curve is revolved about the y-axis (rather than the x-axis). The result is the same integral with x replaced with y. For example, if the curve x = g1y2 on the interval 3c, d4 is revolved about the y-axis, the area of the surface generated is

S = L d

c 2pg1y221 + g′1y22 dy.

To use this integral, we must first describe the given curve as a differentiable function of y.

EXAMPLE 4 Change of perspective Consider the function y = ln a x + 2x2 - 1 2

b . Find the area of the surface generated when the part of the curve between the points 154, 02 and 1178 , ln 22 is revolved about the y-axis (Figure 6.68). SOLUTION We solve for x in terms of y in the following steps:

y = ln a x + 2x2 - 1 2

b ey =

x + 2x2 - 1 2

Exponentiate both sides.

2ey - x = 2x2 - 1 Rearrange terms. 4e2y - 4xey + x2 = x2 - 1 Square both sides.

x = g1y2 = ey + 1 4

e-y. Solve for x.Figure 6.68

ln 2

5 4

17 8

y ! ln( )! x2 " 1x # 2

Figure 6.67

1 2

8 y ! x3 " 112x

11111 x

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6.7 Physical Applications 459

Figure 6.69 x ! a x ! b

Figure 6.70

x ! a x ! b

Mass of kth subinterval: mk ! ! (xk*)"x

xk*

any linear function g1x2 = cx + d that is positive on the interval 3a, b4, where / is the slant height of the frustum. 37. Scaling surface area Let f be a nonnegative function with a con-

tinuous first derivative on 3a, b4 and suppose that g1x2 = c f 1x2 and h1x2 = f 1cx2, where c 7 0. When the curve y = f 1x2 on 3a, b4 is revolved about the x-axis, the area of the resulting surface is A. Evaluate the following integrals in terms of A and c.

a. L b

a 2pg1x22c2 + g′1x22 dx b. Lb>ca>c 2ph1x22c2 + h′1x22 dx

38. Surface plus cylinder Suppose f is a nonnegative function with a continuous first derivative on 3a, b4. Let L equal the length of the graph of f on 3a, b4 and let S be the area of the surface generated by revolving the graph of f on 3a, b4 about the x-axis. For a positive constant C, assume the curve y = f 1x2 + C is revolved about the x-axis. Show that the area of the resulting surface equals the sum of S and the surface area of a right circular cylinder of radius C and height L.

QUICK CHECK ANSWERS

1. The surface area of the first cone 120015p2 is twice as great as the surface area of the second cone 110015p2. 2. The surface area is 63110p. 3. The surface is a cylinder of radius c and height b - a. The area of the curved surface is 2pc1b - a2.

the volume of the object is important. Animals typically generate heat at a rate proportional to their volume and lose heat at a rate pro- portional to their surface area. Therefore, animals with a low SAV ratio tend to retain heat, whereas animals with a high SAV ratio (such as children and hummingbirds) lose heat relatively quickly.

a. What is the SAV ratio of a cube with side lengths a? b. What is the SAV ratio of a ball with radius a? c. Use the result of Exercise 34 to find the SAV ratio of an ellip-

soid whose long axis has length 2a13 4, for a Ú 1, and whose other two axes have half the length of the long axis. (This scal- ing is used so that, for a given value of a, the volumes of the ellipsoid and the ball of radius a are equal.) The volume of a

general ellipsoid is V = 4p 3

ABC, where the axes have lengths 2A, 2B, and 2C.

d. Graph the SAV ratio of the ball of radius a Ú 1 as a function of a (part (b)) and graph the SAV ratio of the ellipsoid de- scribed in (part (c)) on the same set of axes. Which object has the smaller SAV ratio?

e. Among all ellipsoids of a fixed volume, which one would you choose for the shape of an animal if the goal is to minimize heat loss?

Additional Exercises 36. Surface area of a frustum Show that the surface area of the frus-

tum of a cone generated by revolving the line segment between 1a, g1a22 and 1b, g1b22 about the x-axis is p1g1b2 + g1a22/, for 6.7 Physical Applications We continue this chapter on applications of integration with several problems from phys- ics and engineering. The physical themes in these problems are mass, work, force, and pressure. The common mathematical theme is the use of the slice-and-sum strategy, which always leads to a definite integral.

Density and Mass Density is the concentration of mass in an object and is usually measured in units of mass per volume 1for example, g>cm32. An object with uniform density satisfies the basic relationship

mass = density # volume. When the density of an object varies, this formula no longer holds, and we must appeal to calculus.

In this section, we introduce mass calculations for thin objects that can be viewed as line segments (such as wires or thin bars). The bar shown in Figure 6.69 has a density r that varies along its length. For one-dimensional objects, we use linear density with units of mass per length 1for example, g>cm2. What is the mass of such an object? QUICK CHECK 1 In Figure 6.69, suppose a = 0, b = 3, and the density of the rod in g>cm is r1x2 = 14 - x2. (a) Where is the rod lightest and heaviest? (b) What is the density at the middle of the bar?

We begin by dividing the bar, represented by the interval a … x … b, into n sub- intervals of equal length ∆x = 1b - a2>n (Figure 6.70). Let xk* be any point in the kth subinterval, for k = 1, c, n. The mass of the kth segment of the bar mk is approxi- mately the density at xk

* multiplied by the length of the interval, or mk ≈ r1xk*2∆x. So the approximate mass of the entire bar is

a n

k = 1 mk ≈ a

k = 1 r1xk*2∆x.

➤ In Chapter 13, we return to mass calculations for two- and three- dimensional objects (plates and solids).

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460 Chapter 6 Applications of Integration

The exact mass is obtained by taking the limit as n S ∞ and as ∆x S 0, which produces a definite integral.

➤ Note that the units of the integral work out as they should: r has units of mass per length and dx has units of length, so r1x2 dx has units of mass.

➤ Another interpretation of the mass integral is that mass equals the average value of the density multiplied by the length of the bar b - a.

Figure 6.71

O xba

y ! F(x)

force

Force varies on [a, b]

DEFINITION Mass of a One-Dimensional Object Suppose a thin bar or wire is represented by the interval a … x … b with a density function r (with units of mass per length). The mass of the object is

m = L b

a r1x2 dx.

EXAMPLE 1 Mass from variable density A thin 2-m bar, represented by the interval 0 … x … 2, is made of an alloy whose density in units of kg>m is given by r1x2 =11 + x22. What is the mass of the bar? SOLUTION The mass of the bar in kilograms is

m = L b

a r1x2 dx = L20 11 + x22 dx = ax + x33 b ` 20 = 143 .

Related Exercises 9–16

QUICK CHECK 2 A thin bar occupies the interval 0 … x … 2 and has a density in kg>m of r1x2 = 11 + x22. Using the minimum value of the density, what is a lower bound for the mass of the object? Using the maximum value of the density, what is an upper bound for the mass of the object?

Work Work can be described as the change in energy when a force causes a displacement of an object. When you carry a basket of laundry up a flight of stairs or push a stalled car, you apply a force that results in the displacement of an object, and work is done. If a constant force F displaces an object a distance d in the direction of the force, the work done is the force multiplied by the distance:

work = force # distance. It is easiest to use metric units for force and work. A newton 1N2 is the force required to give a 1-kg mass an acceleration of 1 m>s2. A joule 1J2 is 1 newton-meter 1N@m2, the work done by a 1-N force over a distance of 1 m.

Calculus enters the picture with variable forces. Suppose an object is moved along the x-axis by a variable force F that is directed along the x-axis (Figure 6.71). How much work is done in moving the object between x = a and x = b? Once again, we use the slice-and-sum strategy.

The interval 3a, b4 is divided into n subintervals of equal length ∆x = 1b - a2>n. We let xk

* be any point in the kth subinterval, for k = 1, c, n. On that subinterval, the force is approximately constant with a value of F 1xk*2. Therefore, the work done in mov- ing the object across the kth subinterval is approximately F 1xk*2∆x 1force # distance2. Summing the work done over each of the n subintervals, the total work over the interval 3a, b4 is approximately

W ≈ a n

k = 1 F 1xk*2∆x.

This approximation becomes exact when we take the limit as n S ∞ and ∆x S 0. The total work done is the integral of the force over the interval 3a, b4 (or, equivalently, the net area under the force curve in Figure 6.71).

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6.7 Physical Applications 461

An application of force and work that is easy to visualize is the stretching and com- pression of a spring. Suppose an object is attached to a spring on a frictionless horizontal surface; the object slides back and forth under the influence of the spring. We say that the spring is at equilibrium when it is neither compressed nor stretched. It is convenient to let x be the position of the object, where x = 0 is the equilibrium position (Figure 6.72).

Figure 6.72

Compressed

Equilibrium

Stretched

x ! 0

x " 0

F " 0

x ! 0

x ! 0x # 0

F # 0

Figure 6.73

F(x) ! kx

Spring is compressed.

Spring is stretched.

DEFINITION Work The work done by a variable force F moving an object along a line from x = a to x = b in the direction of the force is

W = L b

a F 1x2 dx.

QUICK CHECK 3 Explain why the sum of the work over n subintervals is only an approxi- mation of the total work.

According to Hooke’s law, the force required to keep the spring in a compressed or stretched position x units from the equilibrium position is F 1x2 = kx, where the positive spring constant k measures the stiffness of the spring. Note that to stretch the spring to a position x 7 0, a force F 7 0 (in the positive direction) is required. To compress the spring to a position x 6 0, a force F 6 0 (in the negative direction) is required (Figure 6.73). In other words, the force required to displace the spring is always in the direction of the displacement.

EXAMPLE 2 Compressing a spring Suppose a force of 10 N is required to stretch a spring 0.1 m from its equilibrium position and hold it in that position.

a. Assuming that the spring obeys Hooke’s law, find the spring constant k. b. How much work is needed to compress the spring 0.5 m from its equilibrium position? c. How much work is needed to stretch the spring 0.25 m from its equilibrium position? d. How much additional work is required to stretch the spring 0.25 m if it has already

been stretched 0.1 m from its equilibrium position?

SOLUTION

a. The fact that a force of 10 N is required to keep the spring stretched at x = 0.1 m means (by Hooke’s law) that F 10.12 = k10.1 m2 = 10 N. Solving for the spring constant, we find that k = 100 N>m. Therefore, Hooke’s law for this spring is F 1x2 = 100x.

b. The work in joules required to compress the spring from x = 0 to x = -0.5 is

W = L b

a F 1x2 dx = L -0.50 100x dx = 50x2 ` -0.50 = 12.5.

➤ Hooke’s law was proposed by the English scientist Robert Hooke (1635–1703), who also coined the biological term cell. Hooke’s law works well for springs made of many common materials. However, some springs obey more complicated spring laws (see Exercise 51).

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462 Chapter 6 Applications of Integration

Figure 6.74

yn " b

yn#1

yk#1

yk*

y0 " 0

h # yk*

y " h (outflow level)

Work to lift kth layer ! A(yk*) !y !g $ (h # yk*)

Work to lift all layers

! k"1

% A(yk*) !y !g $ (h # yk*)

Volume of kth layer ! A(yk*) !y

Force on kth layer ! A(yk*) !y $ ! $ g

The kth layer occupying the interval 3yk - 1, yk4, for k = 1, c, n, is approximately yk* units above the bottom of the tank, where yk

* is any point in 3yk - 1, yk4. The cross-sectional area of the kth layer at yk

*, denoted A1yk*2, is determined by the shape of the tank; the solution depends on being able to find A for all values of y. Because the volume of the kth layer is approximately A1yk*2∆y, the force on the kth layer (its weight) is

mass t

Fk = mg ≈ A1yk*2∆y # r # g. t 6

volume density

To reach the level y = h, the kth layer is lifted an approximate distance 1h - yk*2 (Figure 6.74). So the work in lifting the kth layer to a height h is approximately

Wk = A1yk*2∆yrg # 1h - yk*2. force distance

Summing the work required to lift all the layers to a height h, the total work is

W ≈ a n

k = 1 Wk = a

k = 1 A1yk*2rg1h - yk*2∆y.

This approximation becomes more accurate as the width of the layers ∆y tends to zero and the number of layers tends to infinity. In this limit, we obtain a definite integral from y = 0 to y = b. The total work required to empty the tank is

W = lim nS ∞ a

k = 1 A1yk*2rg1h - yk*2∆y = Lb0 rgA1y21h - y2 dy.

This derivation assumes that the bottom of the tank is at y = 0, in which case the distance that the slice at level y must be lifted is D1y2 = h - y. If you choose a different location for the origin, the function D will be different. Here is a general procedure for any choice of origin.

➤ The choice of a coordinate system is somewhat arbitrary and may depend on the geometry of the problem. You can let the y-axis point upward or downward, and there are usually several logical choices for the location of y = 0. You should experiment with different coordinate systems.

c. The work in joules required to stretch the spring from x = 0 to x = 0.25 is

W = L b

a F 1x2 dx = L0.250 100x dx = 50x2 ` 0.250 = 3.125.

d. The work in joules required to stretch the spring from x = 0.1 to x = 0.35 is

W = L b

a F 1x2 dx = L0.350.1 100x dx = 50x2 ` 0.350.1 = 5.625.

Comparing parts (c) and (d), we see that more work is required to stretch the spring 0.25 m starting at x = 0.1 than starting at x = 0. Related Exercises 17–26

Lifting Problems Another common work problem arises when the motion is vertical and the force is the gravitational force. The gravitational force exerted on an object with mass m is F = mg, where g ≈ 9.8 m>s2 is the acceleration due to gravity near the surface of Earth. The work in joules required to lift an object of mass m a vertical distance of y meters is

work = force # distance = mgy. This type of problem becomes interesting when the object being lifted is a body of

water, a rope, or a chain. In these situations, different parts of the object are lifted different distances—so integration is necessary. Here is a typical situation and the strategy used.

Suppose a fluid such as water is pumped out of a tank to a height h above the bottom of the tank. How much work is required, assuming the tank is full of water? Three key observations lead to the solution.

Water from different levels of the tank is lifted different vertical distances, requiring different amounts of work.

Two equal volumes of water from the same horizontal plane are lifted the same distance and require the same amount of work.

A volume V of water has mass rV, where r = 1 g>cm3 = 1000 kg>m3 is the density of water.

To solve this problem, we let the y-axis point upward with y = 0 at the bottom of the tank. The body of water that must be lifted extends from y = 0 to y = b (which may be the top of the tank). The level to which the water must be raised is y = h, where h Ú b (Figure 6.74). We now slice the water into n horizontal layers, each having thickness ∆y.

➤ Notice again that the units in the integral are consistent. If F has units of N and x has units of m, then W has units of F dx, or N@m, which are the units of work 11 N@m = 1 J2.

QUICK CHECK 4 In Example 2, explain why more work is needed in part (d) than in part (c), even though the displacement is the same.

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6.7 Physical Applications 463

PROCEDURE Solving Lifting Problems 1. Draw a y-axis in the vertical direction (parallel to gravity) and choose a con-

venient origin. Assume the interval 3a, b4 corresponds to the vertical extent of the fluid.

2. For a … y … b, find the cross-sectional area A1y2 of the horizontal slices and the distance D1y2 the slices must be lifted.

3. The work required to lift the water is

W = L b

a rgA1y2D1y2 dy.

The kth layer occupying the interval 3yk - 1, yk4, for k = 1, c, n, is approximately yk* units above the bottom of the tank, where yk

* is any point in 3yk - 1, yk4. The cross-sectional area of the kth layer at yk

mass t

Fk = mg ≈ A1yk*2∆y # r # g. t 6 volume density

To reach the level y = h, the kth layer is lifted an approximate distance 1h - yk*2 (Figure 6.74). So the work in lifting the kth layer to a height h is approximately

Wk = A1yk*2∆yrg # 1h - yk*2. force distance

Summing the work required to lift all the layers to a height h, the total work is

W ≈ a n

k = 1 Wk = a

k = 1 A1yk*2rg1h - yk*2∆y.

W = lim nS ∞ a

k = 1 A1yk*2rg1h - yk*2∆y = Lb0 rgA1y21h - y2 dy.

t t

We now use this procedure to solve two pumping problems.

EXAMPLE 3 Pumping water How much work is needed to pump all the water out of a cylindrical tank with a height of 10 m and a radius of 5 m? The water is pumped to an outflow pipe 15 m above the bottom of the tank.

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464 Chapter 6 Applications of Integration

SOLUTION In this problem, we choose a different origin by letting y = 0 and y = -5 correspond to the center and the bottom of the tank, respectively. For -5 … y … 0, a hori- zontal layer of gasoline located at a depth y is a rectangle with a length of 10 and width of 2225 - y2 (Figure 6.76). Therefore, the cross-sectional area of the layer at depth y is

A1y2 = 20225 - y2. ➤ Again, there are several choices for the

location of the origin. The location in this example makes A1y2 easy to compute.

SOLUTION Figure 6.75 shows the cylindrical tank filled to capacity and the outflow 15 m above the bottom of the tank. We let y = 0 represent the bottom of the tank and y = 10 represent the top of the tank. In this case, all horizontal slices are circular disks of radius r = 5 m. Therefore, for 0 … y … 10, the cross-sectional area is

A1y2 = pr2 = p52 = 25p. Note that the water is pumped to a level h = 15 m above the bottom of the tank, so the lifting distance is D1y2 = 15 - y. The resulting work integral is

W = L 10

0 rgA1y2 D1y2 dy = 25prgL100 115 - y2 dy.

25p 15 - y

Substituting r = 1000 kg>m3 and g = 9.8 m>s2, the total work in joules is W = 25prgL

0 115 - y2 dy

= 25p11000219.82a15y - 1 2

y2b ` 10 0

r g

≈ 7.7 * 107.

The work required to pump the water out of the tank is approximately 77 million joules. Related Exercises 27–37

QUICK CHECK 5 In the previous example, how would the integral change if the outflow pipe were at the top of the tank?

EXAMPLE 4 Pumping gasoline A cylindrical tank with a length of 10 m and a radius of 5 m is on its side and half-full of gasoline (Figure 6.76). How much work is required to empty the tank through an outlet pipe at the top of the tank? 1The density of gasoline is r ≈ 737 kg>m3.2

➤ Recall that g ≈ 9.8 m>s2. You should verify that the units are consistent in this calculation: The units of r, g, A1y2, D1y2, and dy are kg>m3, m>s2, m2, m, and m, respectively. The resulting units of W are kg m2>s2, or J. A more convenient unit for large amounts of work and energy is the kilowatt-hour, which is 3.6 million joules.

e e eFigure 6.75

r ! 5

y ! 15 (outflow level)

y ! 10 (tank top)

y ! water layer level

y ! 0

(bottom)

15 " y ! distance water layer is lifted

The area of water layer is A(y) ! !(52).

!r2

Figure 6.76

length ! 10

D(y) ! 5 " y

y ! "5

y ! 0

2. ... the width of the gasoline layer is 2 25 " y2.

3. The area of the gasoline layer is A(y) ! 10 # 2 25 " y2.

1. Because this width is 25 " y2...

5!5

The equation of the right side of the circle is x " 25 ! y2.

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6.7 Physical Applications 465

Figure 6.77

w(y)a ! yk*

y " 0

y " a

Pressure on strip ! !g(a ! yk*) Force on strip ! !g(a ! yk*) # area of strip ! !g(a ! yk*) w(yk*) $y

$y yk*

yk!1

The distance the layer at level y must be lifted to reach the top of the tank is D1y2 = 5 - y, where 5 … D1y2 … 10. The resulting work integral is

W = 73719.82L0-5 20225 - y2 15 - y2 dy = 144,452L0-5 225 - y2 15 - y2 dy. r g A1y2 D1y2 This integral is evaluated by splitting the integrand into two pieces and recognizing that one piece is the area of a quarter circle of radius 5:

L 0

-5 225 - y2 15 - y2 dy = 5L0-5 225 - y2 dy - L0-5y225 - y2 dy

area of quarter circle let u = 25 - y2; du = -2y dy

= 5 # 25p 4

+ 1 2 L

0 1u du

= 125p

4 + 1

3 u3>2 `

= 375p + 500

12 .

Multiplying this result by 20 rg = 144,452, we find that the work required is approxi- mately 20.2 million joules. Related Exercises 27–37

Force and Pressure Another application of integration deals with the force exerted on a surface by a body of water. Again, we need a few physical principles.

Pressure is a force per unit area, measured in units such as newtons per square meter 1N>m22. For example, the pressure of the atmosphere on the surface of Earth is about 14 lb>in2 1approximately 100 kilopascals, or 105 N>m22. As another example, if you stood on the bottom of a swimming pool, you would feel pressure due to the weight (force) of the column of water above your head. If your head is flat and has surface area A m2 and it is h meters below the surface, then the column of water above your head has volume Ah m3. That column of water exerts a force (its weight)

F = mass # acceleration = volume # density # g = Ahrg, mass

where r is the density of water and g is the acceleration due to gravity. Therefore, the pressure on your head is the force divided by the surface area of your head:

pressure = force

A =

Ahrg

A = rgh.

This pressure is called hydrostatic pressure (meaning the pressure of water at rest), and it has the following important property: It has the same magnitude in all directions. Specifically, the hydrostatic pressure on a vertical wall of the swimming pool at a depth h is also rgh. This is the only fact needed to find the total force on vertical walls such as dams. We assume that the water completely covers the face of the dam.

The first step in finding the force on the face of the dam is to introduce a coordinate system. We choose a y-axis pointing upward with y = 0 corresponding to the base of the dam and y = a corresponding to the top of the dam (Figure 6.77). Because the pressure varies with depth (y-direction), the dam is sliced horizontally into n strips of equal thick- ness ∆y. The kth strip corresponds to the interval 3yk - 1, yk4, and we let yk* be any point in that interval. The depth of that strip is approximately h = a - yk*, so the hydrostatic pres- sure on that strip is approximately rg1a - yk*2.

The crux of any dam problem is finding the width of the strips as a function of y, which we denote w1y2. Each dam has its own width function; however, once the width function is known, the solution follows directly. The approximate area of the kth strip is

e e g c i i

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466 Chapter 6 Applications of Integration

EXAMPLE 5 Force on a dam A large vertical dam in the shape of a symmetric trapezoid has a height of 30 m, a width of 20 m at its base, and a width of 40 m at the top (Figure 6.78). What is the total force on the face of the dam when the reservoir is full?

SOLUTION We place the origin at the center of the base of the dam (Figure 6.79). The right slanted edge of the dam is a segment of the line that passes through the points 110, 02 and 120, 302. An equation of that line is

y - 0 = 30 10

1x - 102 or y = 3x - 30 or x = 1 3

1y + 302. Notice that at a depth of y, where 0 … y … 30, the width of the dam is

w1y2 = 2x = 2 3

1y + 302.➤ You should check the width function: w102 = 20 (the width of the dam at its base) and w1302 = 40 (the width of the dam at its top).

its width multiplied by its thickness, or w1yk*2∆y. The force on the kth strip (which is the area of the strip multiplied by the pressure) is approximately

Fk = rg1a - yk*2w1yk*2∆y. pressure area of strip

Summing the forces over the n strips, the total force is

F ≈ a n

k = 1 Fk = a

k = 1 rg1a - yk*2w1yk*2∆y.

To find the exact force, we let the thickness of the strips tend to zero and the number of strips tend to infinity, which produces a definite integral. The limits of integration cor- respond to the base 1y = 02 and top 1y = a2 of the dam. Therefore, the total force on the dam is

F = lim nS ∞ a

k = 1 rg1a - yk*2w1yk*2∆y = La0 rg1a - y2w1y2 dy.

➤ We have chosen y = 0 to be the base of the dam. Depending on the geometry of the problem, it may be more convenient (less computation) to let y = 0 be at the top of the dam. Experiment with different choices.

d d PROCEDURE Solving Force Problems 1. Draw a y-axis on the face of the dam in the vertical direction and choose a

convenient origin (often taken to be the base of the dam).

2. Find the width function w1y2 for each value of y on the face of the dam. 3. If the base of the dam is at y = 0 and the top of the dam is at y = a, then the

total force on the dam is

F = L a

0 rg1b - y2w1y2 dy.

depth width

Figure 6.78

Figure 6.79

!30 0

(20, 30)

y " 3x ! 30

x " W(y # 30)

w(y) " (y # 30)

(10, 0)

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6.8 Logarithmic and Exponential Functions Revisited 471

6.8 Logarithmic and Exponential Functions Revisited

Earlier in the text, we made several claims about exponential and logarithmic functions, but we did not prove them. (For example, these functions are continuous and differen- tiable on their domains.) Our objective in this section is to place these important functions on a solid foundation by presenting a more rigorous development of their properties.

Before embarking on this program, we offer a roadmap to help guide you through the section. We carry out the following four steps.

1. We first define the natural logarithm function in terms of an integral, and then derive the properties of ln x directly from this new definition.

60. Orientation and force A plate shaped like an equilateral triangle 1 m on a side is placed on a vertical wall 1 m below the surface of a pool filled with water. On which plate in the figure is the force greater? Try to anticipate the answer and then compute the force on each plate.

surface

1 m

surface

1 m

1 m 1 m

61. Orientation and force A square plate 1 m on a side is placed on a vertical wall 1 m below the surface of a pool filled with water. On which plate in the figure is the force greater? Try to anticipate the answer and then compute the force on each plate.

surface

1 m

surface

1 m

62. A calorie-free milkshake? Suppose a cylindrical glass with a diameter of 112 m and a height of

1 10 m is filled to the brim with a

400-Cal milkshake. If you have a straw that is 1.1 m long (so the top of the straw is 1 m above the top of the glass), do you burn off all the calories in the milkshake in drinking it? Assume that the density of the milkshake is 1 g>cm3 11 Cal = 4184 J2.

63. Critical depth A large tank has a plastic window on one wall that is designed to withstand a force of 90,000 N. The square window is 2 m on a side, and its lower edge is 1 m from the bottom of the tank.

a. If the tank is filled to a depth of 4 m, will the window withstand the resulting force?

b. What is the maximum depth to which the tank can be filled without the window failing?

64. Buoyancy Archimedes’ principle says that the buoyant force ex- erted on an object that is (partially or totally) submerged in water is equal to the weight of the water displaced by the object (see figure). Let rw = 1 g>cm3 = 1000 kg>m3 be the density of water and let r be the density of an object in water. Let f = r>rw. If 0 6 f … 1, then the object floats with a fraction f of its volume submerged; if f 7 1, then the object sinks.

buoyant force

volume of displaced water

Consider a cubical box with sides 2 m long floating in water with one-half of its volume submerged 1r = rw>22. Find the force required to fully submerge the box (so its top surface is at the water level).

(See the Guided Project Buoyancy and Archimedes’ Principle for further explorations of buoyancy problems.)

QUICK CHECK ANSWERS

1. a. The bar is heaviest at the left end and lightest at the right end. b. r = 2.5 g>cm. 2. Minimum mass = 2 kg; maximum mass = 10 kg 3. We assume that the force is constant over each subinterval, when, in fact, it varies over each subinterval. 4. The restoring force of the spring in- creases as the spring is stretched 1 f 1x2 = 100x2. Greater restoring forces are encountered on the interval 30.1, 0.354 than on the interval 30, 0.254. 5. The factor 115 - y2 in the integral is replaced with 110 - y2.

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472 Chapter 6 Applications of Integration

All the familiar geometric and algebraic properties of the natural logarithmic function fol- low directly from this new integral definition.

Properties of the Natural Logarithm Domain, range, and sign Because the natural logarithm is defined as a definite integral, its value is the net area under the curve y = 1>t between t = 1 and t = x. The integrand is undefined at t = 0, so the domain of ln x is 10, ∞2. On the interval 11, ∞2, ln x is positive because the net area of the region under the curve is positive (Figure 6.80a). On 10, 12, we have L x1 1t dt = - L1x 1t dt, which implies ln x is negative (Figure 6.80b). As expected, when x = 1, we have ln 1 = L

1 t dt = 0. The net area interpretation of ln x

also implies that the range of ln x is 1- ∞ , ∞2 (see Exercise 72 for an outline of a proof).

2. The natural exponential function ex is introduced as the inverse of ln x, and the prop- erties of ex are developed by appealing to this inverse relationship. We also present derivative and integral formulas associated with these functions.

3. Next, we define the general exponential function bx in terms of ex, and the general logarithmic function logb x in terms of ln x. The derivative and integral results stated in Section 3.9 follow immediately.

4. Finally, we revisit the General Power Rule (Section 3.9) and we also derive a limit that can be used to approximate e.

Step 1: The Natural Logarithm Our aim is to develop the properties of the natural logarithm using definite integrals. It all begins with the following definition.

DEFINITION The Natural Logarithm

The natural logarithm of a number x 7 0, is ln x = L x

1 t dt.

Figure 6.80

1 (a)

x dt t

! 0.

If x ! 1,

net area " ln x " !

If 0 # x # 1,

net area " ln x " ! 1

x dt t

# 0.

1 t

y "

1 (b)

1 t

y "

Derivative The derivative of the natural logarithm follows immediately from its defini- tion and the Fundamental Theorem of Calculus:

d dx

1ln x2 = d dx L

dt t

= 1 x

, for x 7 0.

➤ By the Fundamental Theorem of Calculus,

d dx L

a f 1t2 dt = f 1x2.

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6.8 Logarithmic and Exponential Functions Revisited 473

We have two important consequences.

Because its derivative is defined for x 7 0, ln x is a differentiable function for x 7 0, which means it is continuous on its domain (Theorem 3.1).

Because 1>x 7 0 for x 7 0, ln x is strictly increasing and one-to-one on its domain; therefore, it has a well-defined inverse.

The Chain Rule allows us to extend the derivative property to all nonzero real numbers (Exercise 70). By differentiating ln 1-x2 for x 6 0, we find that

d dx

1ln ! x !2 = 1 x

, for x ≠ 0.

More generally, by the Chain Rule,

d dx

1ln ! u1x2 !2 = 1 u1x2 u′1x2 = u′1x2u1x2 .

Graph of ln x As noted before, ln x is continuous and strictly increasing for x 7 0. The

second derivative, d2

dx2 1ln x2 = - 1

x2 , is negative for x 7 0, which implies the graph of

ln x is concave down for x 7 0. As demonstrated in Exercise 72,

lim xS∞

ln x = ∞ , and lim xS0+

ln x = - ∞ .

This information, coupled with the fact that ln 1 = 0, gives the graph of y = ln x (Figure 6.81).

Logarithm of a product The familiar logarithm property

ln xy = ln x + ln y, for x 7 0, y 7 0,

may be proved using the integral definition:

ln xy = L xy

dt t

Definition of ln xy

= L x

dt t

+ L xy

dt t

Additive property of integrals

= L x

dt t

+ L y

du u

Substitute u = t>x in second integral. = ln x + ln y. Definition of ln

Logarithm of a quotient Assuming x 7 0 and y 7 0, the product property and a bit of algebra give

ln x = ln ay # x y b = ln y + ln x

y .

Solving for ln 1x>y2, we have ln

x y

= ln x - ln y,

which is the quotient property for logarithms.

QUICK CHECK 1 What is the domain of ln ! x !?

Figure 6.81

y ! ln x

(1, 0)

lim ln x ! "

lim ln x ! #"

x!"

x!0 $

dx2 (ln x) % 0 !& concave down

d dx

(ln x) & 0 !& ln x is increasing for x > 0

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474 Chapter 6 Applications of Integration

EXAMPLE 1 Integrals with ln x Evaluate L 4

x2 + 9 dx.

SOLUTION

L 4

x2 + 9 dx =

1 2 L

du u

Let u = x2 + 9; du = 2x dx.

= 1 2

ln ! u ! ` 9

Fundamental Theorem

= 1 2

1ln 25 - ln 92 Evaluate. = ln

5 3

Properties of logarithms

Related Exercises 7–20

Logarithm of a power Assuming x 7 0 and p is rational, we have

ln xp = L xp

dt t

Definition of ln xp

= pL x

du u

Let t = up; dt = pup - 1du.

= p ln x. By definition, ln x = L x

du u

This argument relies on the Power Rule 1dt = pup - 1 du2, which we proved only for ratio- nal exponents. Later in this section, we prove that ln xp = p ln x, for all real values of p.

Integrals Because d dx

1ln ! x !2 = 1 x

, we have

L 1 x

dx = ln ! x ! + C.

We have shown that the familiar properties of ln x follow from its new integral definition.

THEOREM 6.4 Properties of the Natural Logarithm

1. The domain and range of ln x are 10, ∞2 and 1- ∞ , ∞2, respectively. 2. ln xy = ln x + ln y, for x 7 0 and y 7 0 3. ln 1x>y2 = ln x - ln y, for x 7 0 and y 7 0 4. ln xp = p ln x, for x 7 0 and p a rational number

5. d dx 1ln ! x !2 = 1

x , for x ≠ 0

6. d dx

1ln ! u1x2 !2 = u′1x2 u1x2 , for u1x2 ≠ 0

7. L 1 x

dx = ln ! x ! + C

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6.8 Logarithmic and Exponential Functions Revisited 475

The number e has the property that the area of the region bounded by the graph of

y = 1 t and the t-axis on the interval 31, e4 is 1 (Figure 6.83). Note that ln 2 6 1 and

ln 3 7 1 (Exercise 73). Because ln x is continuous on its domain, the Intermediate Value Theorem ensures that there is a number e with 2 6 e 6 3 such that ln e = 1.

Step 2: The Exponential Function We have established that f 1x2 = ln x is a continuous, increasing function on the interval 10, ∞2. Therefore, it is one-to-one and its inverse function exists on 10, ∞2. We denote the inverse function f -11x2 = exp 1x2. Its graph is obtained by reflecting the graph of f 1x2 = ln x about the line y = x (Figure 6.82). The domain of exp 1x2 is 1- ∞ , ∞2 be- cause the range of ln x is 1- ∞ , ∞2, and the range of exp 1x2 is 10, ∞2 because the domain of ln x is 10, ∞2.

The usual relationships between a function and its inverse also hold:

y = exp 1x2 if and only if x = ln y; exp 1ln x2 = x, for x 7 0, and ln 1exp 1x22 = x, for all x.

We now appeal to the properties of ln x and use the inverse relations between ln x and exp 1x2 to show that exp 1x2 satisfies the expected properties of any exponential func- tion. For example, if x1 = ln y1 and x2 = ln y2, then it follows that y1 = exp 1x12, y2 = exp 1x22, and

exp 1x1 + x22 = exp 1ln y1 + ln y22 Substitute x1 = ln y1, x2 = ln y2. ln y1y2

= exp 1ln y1y22 Properties of logarithms = y1y2 Inverse property of exp 1x2 and ln x = exp 1x12 exp 1x22. y1 = exp 1x12, y2 = exp 1x22

Therefore, exp 1x2 satisfies the property of exponential functions bx1 + x2 = bx1bx2. Similar arguments show that exp 1x2 satisfies other characteristic properties of exponential func- tions ( Exercise 71):

exp 102 = 1, exp 1x1 - x22 = exp 1x12exp 1x22, and

1exp 1x22p = exp 1px2, for rational numbers p. Suspecting that exp 1x2 is an exponential function, the next task is to identify its base. Let’s consider the real number exp (1), and with a bit of forethought, call it e. The inverse relationship between ln x and exp 1x2 implies that

if e = exp 112, then ln e = ln 1exp 1122 = 1. Using the fact that ln e = 1 and the integral definition of ln x, we now formally define e.

DEFINITION The Number e

The number e is the real number that satisfies ln e = L e

dt t

= 1.

Figure 6.82

42 x

y ! x

y ! ln x domain (0, ") range (#", ")

y ! exp(x) domain (#", ") range (0, ")

Figure 6.83

t3e21

1 t

y !

ln 2 " 1 ln 3 # 1

dt tln e ! ! ! 1

area ! ln e ! 1 1

➤ We give a limit definition that provides a good approximation to e at the end of this section.

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476 Chapter 6 Applications of Integration

We may now dispense with the notation exp 1x2 and use ex as the inverse of ln x. The usual inverse relationships between ex and ln x hold, and the properties of exp 1x2 can now be written for ex.

DEFINITION The Exponential Function For any real number x, y = ex = exp 1x2, where x = ln y.

We can now show that indeed exp 1x2 is the exponential function ex. Assume that p is a rational number and note that ep 7 0. By property 4 of Theorem 6.4 we have

ln ep = p ln e = p. 1

Using the inverse relationship between ln x and exp 1x2, we also know that ln exp 1p2 = p.

Equating these two expressions for p, we conclude that ln e p = ln exp 1p2. Because ln x is a one-to-one function, it follows that

ep = exp 1p2, for rational numbers p, and we conclude that exp (x) is the exponential function with base e.

We already know how to evaluate ex when x is rational. For example, e3 = e # e # e, e-2 =

1 e # e, and e1>2 = 1e. But how do we evaluate ex when x is irrational? We proceed

as follows. The function x = ln y is defined for y 7 0 and its range is all real numbers. Therefore, the domain of its inverse y = exp 1x2 is all real numbers; that is, exp 1x2 is de- fined for all real numbers. We now define ex to be exp 1x2 when x is irrational.

THEOREM 6.5 Properties of ex

The exponential function ex satisfies the following properties, all of which result from the integral definition of ln x. Let x and y be real numbers.

1. ex + y = exey

2. ex - y = ex>ey 3. 1ex2p = exp, where p is a rational number 4. ln 1ex2 = x 5. eln x = x, for x 7 0

➤ The restriction on p in property 3 will be lifted shortly.

QUICK CHECK 2 Simplify eln 2x, ln 1e2x2, e2 ln x, and ln 12ex2. Derivatives and Integrals By Theorem 3.23 (derivatives of inverse functions), the

derivative of the exponential function exists for all x. To compute d dx 1ex2, we observe that

ln 1ex2 = x and then differentiate both sides with respect to x: d dx

1ln ex2 = d dx

1x2 1

1 ex

d dx

1ex2 = 1 d dx

1ln u1x22 = u′1x2 u1x2 (Chain Rule)

d dx

1ex2 = ex. Solve for d dx

1ex2.

e c

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6.8 Logarithmic and Exponential Functions Revisited 477

Once again, we obtain the remarkable result that the exponential function is its own deriva- tive. It follows that ex is its own antiderivative up to a constant; that is,

Lex dx = ex + C. Extending these results using the Chain Rule, we have the following theorem.

EXAMPLE 2 Integrals with ex Evaluate L ex

1 + ex dx.

SOLUTION The change of variables u = 1 + ex implies du = ex dx:

L 1

1 + ex e x dx = L

1 u

du u = 1 + ex, du = ex dx

u du

= ln ! u ! + C Antiderivative of u-1

= ln 11 + ex2 + C. Replace u with 1 + ex. Note that the absolute value may be removed from ln ! u ! because 1 + ex 7 0, for all x.

Related Exercises 21–26

Step 3: General Logarithmic and Exponential Functions We now turn to exponential and logarithmic functions with a general positive base b. The first step is to define the exponential function bx for positive bases with b ≠ 1 and for all real numbers x. We use property 3 of Theorem 6.5 and the fact that b = eln b. If x is a rational number, then

bx = 1eln b2x = ex ln b; b

this important relationship expresses bx in terms of ex. Because ex is defined for all real x, we use this relationship to define bx for all real x.

DEFINITION Exponential Functions with General Bases Let b be a positive real number with b ≠ 1. Then for all real x,

bx = ex ln b.

QUICK CHECK 3 What is the slope of the curve y = ex at x = ln 2? What is the area of the region bounded by the graph of y = ex and the x-axis between x = 0 and x = ln 2?

THEOREM 6.6 Derivative and Integral of the Exponential Function For real numbers x,

d dx

1eu1x22 = eu1x2u′1x2 and Lex dx = ex + C.

This definition fills the gap in property 4 of Theorem 6.4 1ln xp = p ln x2. We use the definition of bx to write

xp = ep ln x, for x 7 0 and p real.

Taking the natural logarithm of both sides and using the inverse relationship between ex and ln x, we find that

ln xp = ln ep ln x = p ln x, for x 7 0 and p real.

In this way, we extend property 4 of Theorem 6.4 to real powers. Just as bx is defined in terms of ex, logarithms with base b 7 1 and b ≠ 1 may be

expressed in terms of ln x. All that is needed is the change of base formula (Section 1.3)

logb x = ln x ln b

➤ Knowing that ln xp = p ln x for real p, we can also extend property 3 of Theorem 6.5 to real numbers. For real x and y, we take the natural logarithm of both sides of z = 1ex2y, which gives ln z = y ln ex = xy, or z = exy. Therefore, 1ex2y = exy.

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6.8 Logarithmic and Exponential Functions Revisited 477

Once again, we obtain the remarkable result that the exponential function is its own deriva- tive. It follows that ex is its own antiderivative up to a constant; that is,

Lex dx = ex + C. Extending these results using the Chain Rule, we have the following theorem.

EXAMPLE 2 Integrals with ex Evaluate L ex

1 + ex dx.

SOLUTION The change of variables u = 1 + ex implies du = ex dx:

L 1

1 + ex e x dx = L

1 u

du u = 1 + ex, du = ex dx

u du

= ln ! u ! + C Antiderivative of u-1

= ln 11 + ex2 + C. Replace u with 1 + ex. Note that the absolute value may be removed from ln ! u ! because 1 + ex 7 0, for all x.

Related Exercises 21–26

bx = 1eln b2x = ex ln b; b

this important relationship expresses bx in terms of ex. Because ex is defined for all real x, we use this relationship to define bx for all real x.

DEFINITION Exponential Functions with General Bases Let b be a positive real number with b ≠ 1. Then for all real x,

bx = ex ln b.

QUICK CHECK 3 What is the slope of the curve y = ex at x = ln 2? What is the area of the region bounded by the graph of y = ex and the x-axis between x = 0 and x = ln 2?

THEOREM 6.6 Derivative and Integral of the Exponential Function For real numbers x,

d dx

1eu1x22 = eu1x2u′1x2 and Lex dx = ex + C.

This definition fills the gap in property 4 of Theorem 6.4 1ln xp = p ln x2. We use the definition of bx to write

xp = ep ln x, for x 7 0 and p real.

Taking the natural logarithm of both sides and using the inverse relationship between ex and ln x, we find that

ln xp = ln ep ln x = p ln x, for x 7 0 and p real.

In this way, we extend property 4 of Theorem 6.4 to real powers. Just as bx is defined in terms of ex, logarithms with base b 7 1 and b ≠ 1 may be

expressed in terms of ln x. All that is needed is the change of base formula (Section 1.3)

logb x = ln x ln b

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478 Chapter 6 Applications of Integration

Theorems 3.18 and 3.20 give us the derivative results for exponential and logarithmic functions with a general base b 7 0. Extending those results with the Chain Rule, we have the following derivatives and integrals.

SUMMARY Derivatives and Integrals with Other Bases Let b 7 0 and b ≠ 1. Then

d dx

1logb 0 u1x2 0 2 = u′1x2u1x2 ln b, for u1x2 ≠ 0 and ddx 1bu1x22 = 1ln b2bu1x2u′1x2. For b 7 0 and b ≠ 1, Lbx dx =

1 ln b

bx + C.

QUICK CHECK 4 Verify that the derivative and integral results for a general base b reduce to the expected results when b = e.

EXAMPLE 3 Integrals involving exponentials with other bases Evaluate the follow- ing integrals.

a. Lx 3x 2 dx b. L

6-1x1x dx SOLUTION

a. Lx 3x 2 dx =

1 2 L3u du u = x2, du = 2x dx

= 1 2

ln 3 3u + C Integrate.

= 1

2 ln 3 3x

2 + C Substitute u = x2.

b. L 4

6-1x1x dx = -2L -2-1 6u du u = - 1x, du = - 121x dx = - 2

ln 6 6u `

-1

-2 Fundamental Theorem

= 5

18 ln 6 Simplify.

Related Exercises 27–32

Step 4: General Power Rule With the identity xp = ep ln x, we can state and prove the final version of the Power Rule. In Chapter 3, we showed that

d dx

1xp2 = pxp - 1 when p is a rational number. This result is extended to all real values of p by differentiat- ing xp = ep ln x:

d dx

1xp2 = d dx

1ep ln x2 xp = ep ln x = ep ln x

p x

Chain Rule xp

= xp p x

ep ln x = xp

= pxp - 1. Simplify.

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6.8 Logarithmic and Exponential Functions Revisited 479

EXAMPLE 4 Derivative of a tower function Evaluate the derivative of f 1x2 = x2x. SOLUTION We use the inverse relationship eln x = x to write x2x = eln 1x2x2 = e2x ln x. It follows that

d dx

1x2x2 = d dx

1e2x ln x2 = e2x ln x

d dx

12x ln x2 d dx

1eu1x22 = eu1x2u′1x2 x2x

= x2x a2 ln x + 2x # 1 x b Product Rule

= 2x2x11 + ln x2. Simplify. Related Exercises 33–40

Computing e We have shown that the number e serves as a base for both ln x and ex, but how do we approximate its value? Recall that the derivative of ln x at x = 1 is 1. By the definition of the derivative, it follows that

1 = d dx

1ln x2 ` x = 1

= lim hS0

ln 11 + h2 - ln 1

h Derivative of ln x at x = 1

= lim hS0

ln 11 + h2

h ln 1 = 0

= lim hS0

ln 11 + h21>h. p ln x = ln xp The natural logarithm is continuous for x 7 0, so it is permissible to interchange the order of lim

hS0 and the evaluation of ln 11 + h21>h. The result is that

ln 1 lim hS0

11 + h21>h2 = 1. e

Observe that the limit within the brackets is e because ln e = 1 and only one number satisfies this equation. Therefore, we have isolated e as a limit:

e = lim hS0

11 + h21>h. It is evident from Table 6.2 that 11 + h21>h S 2.718282c as h S 0. The value of this limit is e, and it has been computed to millions of digits. A better approximation,

e ≈ 2.718281828459045,

is obtained by methods introduced in Chapter 9.

➤ Because d dx

1ln x2 = 1 x

d dx

1ln x2 ` x = 1

= 1 1

= 1.

➤ We rely on Theorem 2.12 of Section 2.6 here. If f is continuous at g1a2 and lim

xSa g1x2 exists, then

lim xSa

f 1g1x22 = f 1 lim xSa

g1x22.

THEOREM 6.7 General Power Rule For any real number p,

d dx

1xp2 = pxp - 1 and d dx

1u1x2p2 = pu1x2p - 1u′1x2.

Table 6.2

h 11 + h 21,h h 11 + h 21,h 10-1 2.593742 -10-1 2.867972 10-2 2.704814 -10-2 2.731999 10-3 2.716924 -10-3 2.719642 10-4 2.718146 -10-4 2.718418 10-5 2.718268 -10-5 2.718295 10-6 2.718280 -10-6 2.718283 10-7 2.718282 -10-7 2.718282

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482 Chapter 6 Applications of Integration

6.9 Exponential Models The uses of exponential functions are wide-ranging. In this section, you will see them ap- plied to problems in finance, medicine, ecology, biology, economics, pharmacokinetics, anthropology, and physics.

Exponential Growth Exponential growth functions have the form y1t2 = Cekt, where C is a constant and the rate constant k is positive (Figure 6.84). If we start with this function and take its derivative, we find that

y′1t2 = d dt

1Cekt2 = C # kekt = k1Cekt2; y

that is, y′1t2 = ky. Here is the first insight about exponential functions: Their rate of change is proportional to their value. If y represents a population, then y′1t2 is the growth rate with units such as people/month or cells/hr. We see that the larger the population, the faster its growth.

Another way to talk about growth rates is to use the relative growth rate, which is the growth rate divided by the current value of that quantity, or y′1t2>y1t2. For example, if y is a population, the relative growth rate is the fraction or percentage by which the population grows each unit of time. Examples of relative growth rates are 5% per year or a factor of 1.2 per month. Therefore, when the equation y′1t2 = ky is written in the form y′1t2>y = k, it has another interpretation. It says a quantity that grows exponentially has a constant relative growth rate. Constant relative or percentage change is the hallmark of exponential growth.

EXAMPLE 1 Linear versus exponential growth Suppose the population of the town of Pine is given by P1t2 = 1500 + 125t, while the population of the town of Spruce is given by S1t2 = 1500e0.1t, where t Ú 0 is measured in years. Find the growth rate and the relative growth rate of each town.

SOLUTION Note that Pine grows according to a linear function, while Spruce

grows exponentially (Figure 6.85). The growth rate of Pine is dP dt

= 125 people/year,

which is constant for all times. The growth rate of Spruce is

dS dt

= 0.111500e0.1t2 = 0.1S1t2, S1t2 showing that the growth rate is proportional to the population. The relative growth rate

of Pine is 1 P

dP dt

= 125

1500 + 125t, which decreases in time. The relative growth rate of

Spruce is

1 S

dS dt

= 0.1 # 1500e0.1t

1500e0.1t = 0.1,

which is constant for all times. In summary, the linear population function has a constant absolute growth rate and the exponential population function has a constant relative growth rate.

Related Exercises 9–10

QUICK CHECK 1 Population A increases at a constant rate of 4%>yr. Population B increases at a constant rate of 500 people/yr. Which population exhibits exponential growth? What kind of growth is exhibited by the other population?

➤ The derivative dy

dt is the absolute growth

rate but is usually simply called the growth rate.

➤ A consumer price index that increases at a constant rate of 4% per year increases exponentially. A currency that is devalued at a constant rate of 3% per month decreases exponentially. By contrast, linear growth is characterized by constant absolute growth rates, such as 500 people per year or $400 per month.

Figure 6.84

!1 1

y " 2e2t

y " e2t

y " 0.1e10t

y " et

4000

3000

2000

1000

t1284

Spruce: S(t) ! 1500e0.1t

exponential growth

Pine: P(t) ! 1500 " 125t linear growth

Figure 6.85

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6.9 Exponential Models 483

The rate constant k in y1t2 = Cekt determines the growth rate of the exponential function. We adopt the convention that k 7 0; then it is clear that y1t2 = Cekt describes exponential growth and y1t2 = Ce-kt describes exponential decay, to be discussed shortly. For problems that involve time, the units of k are time-1; for example, if t is measured in months, the units of k are month-1. In this way, the exponent kt is dimen- sionless (without units).

Unless there is good reason to do otherwise, it is customary to take t = 0 as the refer- ence point for time. Notice that with y1t2 = Cekt, we have y102 = C. Therefore, C has a simple meaning: It is the initial value of the quantity of interest, which we denote y0. In the examples that follow, two pieces of information are typically given: the initial value and clues for determining the rate constant k. The initial value and the rate constant deter- mine an exponential growth function completely.

➤ The unit time-1 is read per unit time. For example, month-1 is read per month.

Exponential Growth Functions Exponential growth is described by functions of the form y1t2 = y0ekt. The initial value of y at t = 0 is y102 = y0, and the rate constant k 7 0 determines the rate of growth. Exponential growth is characterized by a constant relative growth rate.

Because exponential growth is characterized by a constant relative growth rate, the time required for a quantity to double (a 100% increase) is constant. Therefore, one way to describe an exponentially growing quantity is to give its doubling time. To compute the time it takes the function y1t2 = y0ekt to double in value, say from y0 to 2y0, we find the value of t that satisfies

y1t2 = 2y0 or y0ekt = 2y0. Canceling y0 from the equation y0e

kt = 2y0 leaves the equation ekt = 2. Taking logarithms

of both sides, we have ln ekt = ln 2, or kt = ln 2, which has the solution t = ln 2

k . We

denote this doubling time T2 so that T2 = ln 2

k . If y increases exponentially, the time it takes

to double from 100 to 200 is the same as the time it takes to double from 1000 to 2000.

➤ Note that the initial value y0 appears on both sides of this equation. It may be canceled, meaning that the doubling time is independent of the initial condition: The doubling time is constant for all t.

QUICK CHECK 2 Verify that the time needed for y1t2 = y0ekt to double from y0 to 2y0 is the same as the time needed to double from 2y0 to 4y0.

EXAMPLE 2 World population Human population growth rates vary geographically and fluctuate over time. The overall growth rate for world population peaked at an annual rate of 2.1% per year in the 1960s. Assume a world population of 6.0 billion in 1999 1t = 02 and 6.9 billion in 2009 1t = 102. a. Find an exponential growth function for the world population that fits the two data

points.

b. Find the doubling time for the world population using the model in part (a). c. Find the (absolute) growth rate y′1t2 and graph it, for 0 … t … 50. d. How fast was the population growing in 2014 1t = 152?

DEFINITION Doubling Time The quantity described by the function y1t2 = y0ekt, for k 7 0, has a constant dou- bling time of T2 =

ln 2 k

, with the same units as t.

➤ World population

1804 1 billion 1927 2 billion 1960 3 billion 1974 4 billion 1987 5 billion 1999 6 billion 2011 7 billion 2050 9 billion (proj.)

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484 Chapter 6 Applications of Integration

SOLUTION

a. Let y1t2 be world population measured in billions of people t years after 1999. We use the growth function y1t2 = y0ekt, where y0 and k must be determined. The initial value is y0 = 6 (billion). To determine the rate constant k, we use the fact that y1102 = 6.9. Substituting t = 10 into the growth function with y0 = 6 implies

y1102 = 6e10k = 6.9. Solving for k yields the rate constant k =

ln 16.9>62 10

≈ 0.013976 ≈ 0.014 year-1.

Therefore, the growth function is

y1t2 = 6e0.014t. b. The doubling time of the population is

T2 = ln 2

k ≈

ln 2 0.014

≈ 50 years.

c. Working with the growth function y1t2 = 6e0.014t, we find that y′1t2 = 6 10.0142e0.014t = 0.084e0.014t,

which has units of billions of people/year. As shown in Figure 6.86 the growth rate itself increases exponentially.

d. In 2014 1t = 152, the growth rate was y′1152 = 0.084e10.01421152 ≈ 0.104 billion people>year,

or roughly 104 million people/year. Related Exercises 11–20

QUICK CHECK 3 Assume y1t2 = 100e0.05t. By (exactly) what percentage does y increase when t increases by 1 unit?

A Financial Model Exponential functions are used in many financial applications, sev- eral of which are explored in the exercises. For now, consider a simple savings account in which an initial deposit earns interest that is reinvested in the account. Interest payments are made on a regular basis (for example, annually, monthly, daily), or interest may be compounded continuously. In all cases, the balance in the account increases exponentially at a rate that can be determined from the advertised annual percentage yield (or APY) of the account. Assuming that no additional deposits are made, the balance in the account is given by the exponential growth function y1t2 = y0ekt, where y0 is the initial deposit, t is measured in years, and k is determined by the annual percentage yield.

EXAMPLE 3 Compounding The APY of a savings account is the percentage increase in the balance over the course of a year. Suppose you deposit $500 in a savings account that has an APY of 6.18% per year. Assume that the interest rate remains constant and that no additional deposits or withdrawals are made. How long will it take the balance to reach $2500?

SOLUTION Because the balance grows by a fixed percentage every year, it grows expo- nentially. Letting y1t2 be the balance t years after the initial deposit of y0 = $500, we have y1t2 = y0ekt, where the rate constant k must be determined. Note that if the initial balance is y0, one year later the balance is 6.18% more, or

y112 = 1.0618 y0 = y0ek. Solving for k, we find that the rate constant is

k = ln 1.0618 ≈ 0.060 yr-1.

➤ It is a common mistake to assume that if the annual growth rate is 1.4% per year, then k = 1.4% = 0.014 year-1. The rate constant k must be calculated, as it is in Example 2, to give k = 0.013976. For larger growth rates, the difference between k and the actual growth rate is greater.

Figure 6.86

0.1

0.2

t4020

y!(t) " 0.084e0.014t

Years (after 1999)

G ro

w th

ra te

(b ill

io ns

o f p

eo pl

e/ ye

ar )

➤ Converted to a daily rate (dividing by 365), the world population in 2014 increased at a rate of roughly 284,000 people per day.

➤ The concept of continuous compounding was introduced in Exercise 109 of Section 4.7.

➤ If the balance increases by 6.18% in one year, it increases by a factor of 1.0618 in one year.

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6.9 Exponential Models 485

Therefore, the balance at any time t Ú 0 is y1t2 = 500e0.060t. To determine the time required for the balance to reach $2500, we solve the equation

y1t2 = 500e0.060t = 2500. Dividing by 500 and taking the natural logarithm of both sides yields

0.060t = ln 5.

The balance reaches $2500 in t = 1ln 52>0.060 ≈ 26.8 yr. Related Exercises 11–20

Resource Consumption Among the many resources that people use, energy is cer- tainly one of the most important. The basic unit of energy is the joule (J), roughly the energy needed to lift a 0.1-kg object (say an orange) 1 m. The rate at which energy is con- sumed is called power. The basic unit of power is the watt (W), where 1 W = 1 J>s. If you turn on a 100-W lightbulb for 1 min, the bulb consumes energy at a rate of 100 J>s, and it uses a total of 100 J>s # 60 s = 6000 J of energy.

A more useful measure of energy for large quantities is the kilowatt-hour (kWh). A kilowatt is 1000 W or 1000 J>s. So if you consume energy at the rate of 1 kW for 1 hr (3600 s), you use a total of 1000 J>s # 3600 s = 3.6 * 106 J, which is 1 kWh. A person running for one hour consumes roughly 1 kWh of energy. A typical house uses on the order of 1000 kWh of energy in a month.

Assume that the total energy used (by a person, machine, or city) is given by the function E1t2. Because the power P1t2 is the rate at which energy is used, we have P1t2 = E′1t2. Using the ideas of Section 6.1, the total amount of energy used between the times t = a and t = b is

total energy used = L b

a E′1t2 dt = Lba P1t2 dt.

We see that energy is the area under the power curve. With this background, we can inves- tigate a situation in which the rate of energy consumption increases exponentially.

EXAMPLE 4 Energy consumption At the beginning of 2010, the rate of energy consumption for the city of Denver was 7000 megawatts (MW), where 1 MW = 106 W. That rate is expected to increase at an annual growth rate of 2% per year.

a. Find the function that gives the power or rate of energy consumption for all times after the beginning of 2010.

b. Find the total amount of energy used during 2014. c. Find the function that gives the total (cumulative) amount of energy used by the city

between 2010 and any time t Ú 0.

SOLUTION

a. Let t Ú 0 be the number of years after the beginning of 2010 and let P1t2 be the power function that gives the rate of energy consumption at time t. Because P increases at a constant rate of 2% per year, it increases exponentially. Therefore, P1t2 = P0ekt, where P0 = 7000 MW. We determine k as before by setting t = 1; after one year the power is

P112 = P0ek = 1.02P0. Canceling P0 and solving for k, we find that k = ln 1.02 ≈ 0.0198. Therefore, the power function (Figure 6.87) is

P1t2 = 7000e0.0198t, for t Ú 0.

➤ In one year, the power function increases by 2% or by a factor of 1.02.

Figure 6.87

8000

4000

t302010

Power ! rate of energy consumption

Years after 2010

Po w

er (M

W )

P(t) ! 7000e0.0198t

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486 Chapter 6 Applications of Integration

b. The entire year 2014 corresponds to the interval 4 … t … 5. Substituting P1t2 = 7000e0.0198t, the total energy used in 2014 was

L 5

4 P1t2 dt = L54 7000e0.0198t dt Substitute for P1t2.

= 7000

0.0198 e0.0198t `

Fundamental Theorem

≈ 7652. Evaluate.

Because the units of P are MW and t is measured in years, the units of energy are MW-yr. To convert to MWh, we multiply by 8760 hr>yr to get the total energy of about 6.7 * 107 MWh (or 6.7 * 1010 kWh).

c. The total energy used between t = 0 and any future time t is given by the future value formula (Section 6.1):

E1t2 = E102 + L t0 E′1s2 ds = E102 + L t0 P1s2 ds. Assuming t = 0 corresponds to the beginning of 2010, we take E102 = 0. Substitut- ing again for the power function P, the total energy (in MW-yr) at time t is

E1t2 = E102 + L t0 P1s2 ds = 0 + L

0 7000e0.0198s ds Substitute for P1s2 and E102.

= 7000

0.0198 e0.0198s `

Fundamental Theorem

≈ 353,5351e0.0198t - 12. Evaluate. As shown in Figure 6.88, when the rate of energy consumption increases exponentially, the total amount of energy consumed also increases exponentially.

Related Exercises 11–20

Exponential Decay Everything you have learned about exponential growth carries over directly to exponen- tial decay. A function that decreases exponentially has the form y1t2 = y0e-kt, where y0 = y102 is the initial value and k 7 0 is the rate constant.

Exponential decay is characterized by a constant relative decay rate and by a constant half-life. For example, radioactive plutonium has a half-life of 24,000 years. An initial sample of 1 mg decays to 0.5 mg after 24,000 years and to 0.25 mg after 48,000 years. To compute the half-life, we determine the time required for the quantity y1t2 = y0e-kt to reach one-half of its current value; that is, we solve y0e

-kt = y0>2 for t. Canceling y0 and taking logarithms of both sides, we find that

e-kt = 1 2 1 -kt = ln 1

2 = - ln 2 1 t = ln 2

k .

The half-life is given by the same formula as the doubling time.

Figure 6.88

0 t302010

200,000

100,000

Years after 2010

Energy consumed t years after 2010

E ne

rg y

(M W

-y r)

y ! E(t)

QUICK CHECK 4 If a quantity decreases by a factor of 8 every 30 years, what is its half-life?

Exponential Decay Functions Exponential decay is described by functions of the form y1t2 = y0e-kt. The initial value of y is y102 = y0, and the rate constant k 7 0 determines the rate of decay. Exponential decay is characterized by a constant relative decay rate. The constant

half-life is T1>2 = ln 2k , with the same units as t.

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6.9 Exponential Models 487

Radiometric Dating A powerful method for estimating the age of ancient objects (for example, fossils, bones, meteorites, and cave paintings) relies on the radioactive decay of certain elements. A common version of radiometric dating uses the carbon isotope C@14, which is present in all living matter. When a living organism dies, it ceases to replace C@14, and the C@14 that is present decays with a half-life of about T1>2 = 5730 years. Comparing the C@14 in a living organism to the amount in a dead sample provides an estimate of its age.

EXAMPLE 5 Radiometric dating Researchers determine that a fossilized bone has 30% of the C@14 of a live bone. Estimate the age of the bone. Assume a half-life for C@14 of 5730 years.

SOLUTION The exponential decay function y1t2 = y0e-kt represents the amount of C-14 in the bone t years after its owner died. By the half-life formula, T1>2 = 1ln 22>k. Substituting T1>2 = 5730 yr, the rate constant is

k = ln 2 T1>2 = ln 25730 yr ≈ 0.000121 yr-1.

Assume that the amount of C@14 in a living bone is y0. Over t years, the amount of C@14 in the fossilized bone decays to 30% of its initial value, or 0.3y0. Using the decay function, we have

0.3y0 = y0e-0.000121t.

Solving for t, the age of the bone (in years) is

t = ln 0.3

-0.000121 ≈ 9950.

Related Exercises 21–26

Pharmacokinetics Pharmacokinetics describes the processes by which drugs are assimilated by the body. The elimination of most drugs from the body may be modeled by an exponential decay function with a known half-life (alcohol is a notable exception). The simplest models assume that an entire drug dose is immediately absorbed into the blood. This assumption is a bit of an idealization; more refined mathematical models account for the absorption process.

EXAMPLE 6 Pharmacokinetics An exponential decay function y1t2 = y0e-kt models the amount of drug in the blood t hr after an initial dose of y0 = 100 mg is administered. Assume the half-life of the drug is 16 hours.

a. Find the exponential decay function that governs the amount of drug in the blood. b. How much time is required for the drug to reach 1% of the initial dose (1 mg)? c. If a second 100-mg dose is given 12 hr after the first dose, how much time is required

for the drug level to reach 1 mg?

SOLUTION

a. Knowing that the half-life is 16 hr, the rate constant is

k = ln 2 T1>2 = ln 216 hr ≈ 0.0433 hr-1.

Therefore, the decay function is y1t2 = 100e-0.0433t. b. The time required for the drug to reach 1 mg is the solution of

100e-0.0433t = 1.

Solving for t, we have

t = ln 0.01

-0.0433 hr-1 ≈ 106 hr.

It takes more than 4 days for the drug to be reduced to 1% of the initial dose.

➤ Half-lives of common drugs

Penicillin 1 hr Amoxicillin 1 hr Nicotine 2 hr Morphine 3 hr Tetracycline 9 hr Digitalis 33 hr Phenobarbitol 2–6 days

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6.10 Hyperbolic Functions 491

6.10 Hyperbolic Functions In this section, we introduce a new family of functions called the hyperbolic functions, which are closely related to both trigonometric functions and exponential functions. Hyperbolic functions find widespread use in applied problems in fluid dynamics, projectile motion, architecture, and electrical engineering, to name just a few areas. Hyperbolic functions are also important in the development of many theoretical results in mathematics.

Relationship Between Trigonometric and Hyperbolic Functions The trigonometric functions defined in Chapter 1 are based on relationships involving a circle—for this reason, trigonometric functions are also known as circular functions. Specifically, cos t and sin t are equal to the x- and y-coordinates, respectively, of the point P1x, y2 on the unit circle that corresponds to an angle of t radians (Figure 6.90). We can also regard t as the length of the arc from 11, 02 to the point P1x, y2.

Figure 6.90

x2 ! y2 " 1 1

x Area " t/2

Three ways to interpret t : t = angle in radians t = arc length t = 2 · (area of sector)

P(x, y) " (cos t, sin t)

Arc length " t

There is yet another way to interpret the number t, and it is this third interpretation that links the trigonometric and hyperbolic functions. Observe that t is twice the area of the circular sector in Figure 6.90. The functions cos t and sin t are still defined as the x- and y-coordinates of the point P, but now we associate P with a sector whose area is one-half of t.

The hyperbolic cosine and hyperbolic sine are defined in an analogous fashion us- ing the hyperbola x2 - y2 = 1 instead of the circle x2 + y2 = 1. Consider the region bounded by the x-axis, the right branch of the unit hyperbola x2 - y2 = 1, and a line seg- ment from the origin to a point P1x, y2 on the hyperbola (Figure 6.91); let t equal twice the area of this region.

The hyperbolic cosine of t, denoted cosh t, is the x-coordinate of P and the hyperbolic sine of t, denoted sinh t, is the y-coordinate of P. Expressing x and y in terms of t leads to the standard definitions of the hyperbolic functions. We accomplish this task by writing t, which is an area, as an integral that depends on the coordinates of P. In Exercise 112, we ask you to carry out the calculations to show that

x = cosh t = et + e-t

2 and y = sinh t =

et - e-t

2 .

Everything that follows in this section is based on these two definitions.

Definitions, Identities, and Graphs of the Hyperbolic Functions Once the hyperbolic cosine and hyperbolic sine are defined, the four remaining hyperbolic functions follow in a manner analogous to the trigonometric functions.

➤ Recall that the area of a circular sector of radius r and angle u is A = 12 r2u. With r = 1 and u = t, we have A = 12t, which implies t = 2A.

Figure 6.91

x2 ! y2 " 1

P(x, y) " (cosh t, sinh t)

t " twice the area of hyperbolic sector

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492 Chapter 6 Applications of Integration

The hyperbolic functions satisfy many important identities. Let’s begin with the fun- damental identity for hyperbolic functions, which is analogous to the familiar trigonomet- ric identity cos2 x + sin2 x = 1:

cosh2 x - sinh2 x = 1.

This identity is verified by appealing to the definitions:

cosh2 x - sinh2 x = a ex + e-x 2

b2 - a ex - e-x 2

b2 Definition of cosh x and sinh x =

e2x + 2 + e-2x - 1e2x - 2 + e-2x2 4

Expand and combine fractions.

= 4 4

= 1. Simplify.

EXAMPLE 1 Deriving hyperbolic identities

a. Use the fundamental identity cosh2 x - sinh2 x = 1 to prove that 1 - tanh2 x = sech2 x. b. Derive the identity sinh 2x = 2 sinh x cosh x.

SOLUTION

a. Dividing both sides of the fundamental identity cosh2 x - sinh2 x = 1 by cosh2 x leads to the desired result:

cosh2 x - sinh2 x = 1 Fundamental identity

cosh2 x cosh2 x

- sinh 2 x

cosh2 x =

1 cosh2 x

Divide both sides by cosh2 x. (+)+* (+)+* tanh2 x sech2 x

1 - tanh2 x = sech2 x. Identify functions.

b. Using the definition of the hyperbolic sine, we have

sinh 2x = e2x - e-2x

2 Definition of sinh

= 1ex - e-x21ex + e-x2

2 Factor; difference of perfect squares

= 2 sinh x cosh x. Identify functions. Related Exercises 11–18

The identities in Example 1 are just two of many useful hyperbolic identities, some of which we list next.

➤ There is no universally accepted pronunciation of the names of the hyperbolic functions. In the United States, cohsh x (long oh sound) and sinch x are common choices for cosh x and sinh x. The pronunciations tanch x, cotanch x, seech x or sech x, and coseech x or cosech x are used for the other functions. International pronunciations vary as well.

➤ The fundamental identity for hyperbolic functions can also be understood in terms of the geometric definition of the hyperbolic functions. Because the point P1cosh t, sinh t2 is on the hyperbola x2 - y2 = 1, the coordinates of P satisfy the equation of the hyperbola, which leads immediately to

cosh2 t - sinh2 t = 1.

DEFINITION Hyperbolic Functions Hyperbolic cosine Hyperbolic sine

cosh x = ex + e-x

2 sinh x =

ex - e-x

Hyperbolic tangent Hyperbolic cotangent

tanh x = sinh x cosh x

= ex - e-x

ex + e-x coth x = cosh x sinh x

= ex + e-x

ex - e-x

Hyperbolic secant Hyperbolic cosecant

sech x = 1

cosh x =

2 ex + e-x csch x =

1 sinh x

= 2

ex - e-x

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6.10 Hyperbolic Functions 493

Graphs of the hyperbolic functions are relatively easy to produce be- cause they are based on the familiar graphs of ex and e-x. Recall that lim

xS ∞ e-x = 0 and that lim

xS- ∞ ex = 0. With these facts in mind, we see that

the graph of cosh x (Figure 6.92) approaches the graph of y = 1 2

ex as x S ∞

because cosh x = ex + e-x

2 ≈

2 for large values of x. A similar argument

shows that as x S -∞ , cosh x approaches y = 1 2

e-x. Note also that cosh x is an even function:

cosh1-x2 = e-x + e-1-x2 2

= ex + e-x

2 = cosh x.

Finally, cosh 0 = e0 + e0

2 = 1, so its y-intercept is 10, 12. The behavior of

sinh x, an odd function also shown in Figure 6.92, can be explained in much the same way.

Hyperbolic Identities cosh2 x - sinh2 x = 1 cosh1-x2 = cosh x 1 - tanh2 x = sech2 x sinh1-x2 = -sinh x coth2 x - 1 = csch2 x tanh1-x2 = - tanh x cosh1x + y2 = cosh x cosh y + sinh x sinh y sinh1x + y2 = sinh x cosh y + cosh x sinh y cosh 2x = cosh2 x + sinh2 x sinh 2x = 2 sinh x cosh x

cosh2 x = cosh 2x + 1

2 sinh2 x =

cosh 2x - 1 2

Figure 6.92

1 2

x!2 !1 1 2

y " !ex 1 2

y " !e!x

1 2

y " !!e!x

y " cosh x domain (!#, #) range [1, #)

(0, 1)

cosh x is an even function

y " sinh x domain (!#, #) range (!#, #)

sinh x is an odd function

Figure 6.93

!2 !1

(0, 1)

x1 2

y " tanh x domain (!#, #) range (!1, 1)

y " coth x domain x ! 0 range |y| $ 1

!2 !1

x1 2

y " sech x domain (!#, #) range (0, 1]

y " csch x domain x ! 0 range y ! 0

QUICK CHECK 1 Use the definition of the hyperbolic sine to show that sinh x is an odd function.

The graphs of the other four hyperbolic functions are shown in Figure 6.93. As a con- sequence of their definitions, we see that the domain of cosh x, sinh x, tanh x, and sech x is 1- ∞ , ∞2, whereas the domain of coth x and csch x is the set of all real numbers excluding 0.

QUICK CHECK 2 Explain why the graph of tanh x has the horizontal asymptotes y = 1 and y = -1.

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494 Chapter 6 Applications of Integration

Derivatives and Integrals of Hyperbolic Functions Because the hyperbolic functions are defined in terms of ex and e-x, computing their derivatives is straightforward. The derivatives of the hyperbolic functions are given in Theorem 6.8—reversing these formulas produces corresponding integral formulas.

➤ The identities, derivative formulas, and integral formulas for the hyperbolic functions are similar to the corresponding formulas for the trigonometric functions, which makes them easy to remember. However, be aware of some subtle differences in the signs associated with these formulas. For instance,

d>dx1cos x2 = -sin x, whereas

d>dx1cosh x2 = sinh x.

THEOREM 6.8 Derivative and Integral Formulas

1. d dx 1cosh x2 = sinh x 1 Lsinh x dx = cosh x + C

2. d dx 1sinh x2 = cosh x 1 Lcosh x dx = sinh x + C

3. d dx 1tanh x2 = sech2 x 1 Lsech2 x dx = tanh x + C

4. d dx 1coth x2 = -csch2 x 1 Lcsch2 x dx = -coth x + C

5. d dx 1sech x2 = -sech x tanh x 1 Lsech x tanh x dx = -sech x + C

6. d dx 1csch x2 = -csch x coth x 1 Lcsch x coth x dx = -csch x + C

Proof: Using the definitions of cosh x and sinh x, we have

d dx

1cosh x2 = d dx

a ex + e-x 2

b = ex - e-x 2

= sinh x and

d dx

1sinh x2 = d dx

a ex - e-x 2

b = ex + e-x 2

= cosh x.

To prove formula (3), we begin with tanh x = sinh x>cosh x and then apply the Quotient Rule:

d dx

1tanh x2 = d dx

a sinh x cosh x

b Definition of tanh x =

cosh x1cosh x2 - sinh x1sinh x2 cosh2 x

Quotient Rule

= 1

cosh2 x cosh2 x - sinh2 x = 1

= sech2 x. sech x = 1>cosh x The proofs of the remaining derivative formulas are assigned in Exercises 19–21. The

integral formulas are a direct consequence of their corresponding derivative formulas.

EXAMPLE 2 Derivatives and integrals of hyperbolic functions Evaluate the follow- ing derivatives and integrals.

a. d dx

1sech 3x2 b. d2 dx2

1sech 3x2 c. L

csch2 1x1x dx d. L ln 30 sinh3 x cosh x dx

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6.10 Hyperbolic Functions 495

SOLUTION

a. Combining formula (5) of Theorem 6.8 with the Chain Rule gives

d dx

1sech 3x2 = -3 sech 3x tanh 3x. b. Applying the Product Rule and Chain Rule to the result of part (a), we have

dx2 1sech 3x2 = d

dx 1-3 sech 3x tanh 3x2

= d dx

1-3 sech 3x2 # tanh 3x + 1-3 sech 3x2 # d dx

1tanh 3x2 Product Rule v u

9 sech 3x tanh 3x 3 sech2 3x

= 9 sech 3x tanh2 3x - 9 sech3 3x Chain Rule = 9 sech 3x1tanh2 3x - sech2 3x2. Simplify.

c. The integrand suggests the substitution u = 1x: L

csch2 1x1x dx = 2Lcsch2 u du Let u = 1x ; du = 121x dx. = -2 coth u + C Formula (4), Theorem 6.8 = -2 coth 1x + C. u = 1x

d. The derivative formula d>dx1sinh x2 = cosh x suggests the substitution u = sinh x: L

ln 3

0 sinh3 x cosh x dx = L

4>3 0

u3 du. Let u = sinh x; du = cosh x dx.

The new limits of integration are determined by the calculations

x = 0 1 u = sinh 0 = 0 and

x = ln 3 1 u = sinh1ln 32 = eln 3 - e-ln 3 2

= 3 - 1>3

2 =

4 3

We now evaluate the integral in the variable u:

L 4>3

0 u3 du =

1 4

u4 ` 4>3 0

= 1 4

aa 4 3 b4 - 04b = 64

81 .

Related Exercises 19–40

QUICK CHECK 3 Find both the derivative and indefinite integral of f 1x2 = 4 cosh 2x. Theorem 6.9 presents integral formulas for the four hyperbolic functions not covered

in Theorem 6.8.

THEOREM 6.9 Integrals of Hyperbolic Functions

1. L tanh x dx = ln cosh x + C 2. Lcoth x dx = ln ! sinh x ! + C 3. Lsech x dx = tan-11sinh x2 + C 4. Lcsch x dx = ln ! tanh1x>22 ! + C

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496 Chapter 6 Applications of Integration

Proof: Formula (1) is derived by first writing tanh x in terms of sinh x and cosh x:

L tanh x dx = L sinh x cosh x

dx Definition of tanh x

= L 1 u

du Let u = cosh x; du = sinh x dx.

= ln ! u ! + C Evaluate integral. = ln cosh x + C. u = cosh x 7 0

Formula (2) is derived in a similar fashion (Exercise 44). The more challenging proofs of formulas (3) and (4) are considered in Exercises 107 and 108.

EXAMPLE 3 Integrals involving hyperbolic functions Determine the indefinite integral 1x coth 1x22 dx. SOLUTION The integrand suggests the substitution u = x2:

Lx coth x2 dx = 1 2 Lcoth u du. Let u = x2; du = 2x dx.

= 1 2

ln ! sinh u ! + C Evaluate integral; use Theorem 6.9.

= 1 2

ln1sinh x22 + C. u = x2; sinh x2 Ú 0 Related Exercises 41–44

QUICK CHECK 4 Determine the indefinite integral 1csch 2x dx. Inverse Hyperbolic Functions

At present, we don’t have the tools for evaluating an integral such as L dx2x2 + 4.

By studying inverse hyperbolic functions, we can discover new integration formulas. Inverse hyperbolic functions are also useful for solving equations involving hyperbolic functions.

Figures 6.92 and 6.93 show that the functions sinh x, tanh x, coth x, and csch x are all one-to-one on their respective domains. This observation implies that each of these func- tions has a well-defined inverse. However, the function y = cosh x is not one-to-one on 1- ∞ , ∞2, so its inverse, denoted y = cosh-1 x, exists only if we restrict the domain of cosh x. Specifically, when y = cosh x is restricted to the interval 30, ∞2, it is one-to-one, and its inverse is defined as follows:

y = cosh-1 x if and only if x = cosh y, for x Ú 1 and 0 … y 6 ∞ .

Figure 6.94a shows the graph of y = cosh-1 x, obtained by reflecting the graph of y = cosh x on 30, ∞2 over the line y = x. The definitions and graphs of the other five inverse hyperbolic functions are also shown in Figure 6.94. Notice that the domain of y = sech x (Figure 6.94d) must be restricted to 30, ∞2 to ensure the existence of its inverse.

Because hyperbolic functions are defined in terms of exponential functions, we can find explicit formulas for their inverses in terms of logarithms. For example, let’s start with the definition of the inverse hyperbolic sine. For all real x and y, we have

y = sinh-1 x 3 x = sinh y.

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6.10 Hyperbolic Functions 497

Following the procedure outlined in Section 1.3, we solve

x = sinh y = ey - e-y

for y to give a formula for sinh-1 x:

x = ey - e-y

2 1 ey - 2x - e-y = 0 Rearrange equation.

1 1ey22 - 2xey - 1 = 0. Multiply by ey. At this stage, we recognize a quadratic equation in ey and solve for ey using the quadratic formula, with a = 1, b = -2x, and c = -1:

ey = 2x { 24x2 + 4

2 = x { 2x2 + 1 = x + 2x2 + 1.

choose positive root

Because ey 7 0 and 2x2 + 1 7 x, the positive root must be chosen. We now solve for y by taking the natural logarithm of both sides:

ey = x + 2x2 + 1 1 y = ln1x + 2x2 + 12. Therefore, the formula we seek is sinh-1 x = ln1x + 2x2 + 12.

A similar procedure can be carried out for the other inverse hyperbolic functions (Exercise 110). Theorem 6.10 lists the results of these calculations.

y ! cosh"1 x

y ! cosh x

y ! cosh"1 x ⇔ x ! cosh y for x # 1 and 0 $ y % &

(a)

1 2

y ! sinh x

y ! sinh"1 x

y ! sinh"1 x ⇔ x ! sinh y for "# $ x $ # and "# $ y $ #

(b)

x21"1"2

y ! tanh"1 x ⇔ x ! tanh y for "1 # x # 1 and "$ # y # $

(c)

y ! tanh x

y ! tanh"1 x y

x321 "1

"1"2"3

Figure 6.94

y ! sech"1 x

y ! sech x

y ! sech"1 x ⇔ x ! sech y for 0 # x $ 1 and 0 $ y # %

(d)

x321

y ! csch"1 x

y ! csch x

y ! csch"1 x ⇔ x ! csch y for x ! 0 and y ! 0

(e)

x21"1"2

y ! coth"1 x ⇔ x ! coth y for |x| # 1 and y ! 0

( f )

x21"1"2

y ! coth"1 x

y ! coth x

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498 Chapter 6 Applications of Integration

Notice that the formulas in Theorem 6.10 for the inverse hyperbolic secant, cosecant, and cotangent are given in terms of the inverses of their corresponding reciprocal func- tions. Justification for these formulas follows from the definitions of the inverse functions. For example, from the definition of csch-1 x, we have

y = csch-1 x 3 x = csch y 3 1>x = sinh y. Applying the inverse hyperbolic sine to both sides of 1>x = sinh y yields

sinh-111>x2 = sinh-11sinh y2 or y = csch-1 x = sinh-111>x2. u

Similar derivations yield the other two formulas.

EXAMPLE 4 Points of intersection Find the points at which the curves y = cosh x and y = 53 intersect (Figure 6.95).

SOLUTION The x-coordinates of the points of intersection satisfy the equation cosh x = 53, which is solved by applying cosh-1 to both sides of the equation. However, evaluating cosh-11cosh x2 requires care—in Exercise 105, you are asked to show that cosh-11cosh x2 = ! x ! . With this fact, the points of intersection can be found:

cosh x = 53 Set equations equal to one another. cosh-11cosh x2 = cosh-1 53 Apply cosh-1 to both sides.

! x ! = ln153 + 215322 - 12 Simplify; use Theorem 6.10. x = { ln 3. Simplify.

The points of intersection lie on the line y = 53, so the points are 1- ln 3, 532 and 1ln 3, 532. Related Exercises 45–46

Derivatives of the Inverse Hyperbolic Functions and Related Integral Formulas The derivatives of the inverse hyperbolic functions can be computed directly from the logarithmic formulas given in Theorem 6.10. However, it is more efficient to use the definitions in Figure 6.94.

Recall that the inverse hyperbolic sine is defined by

y = sinh-1 x 3 x = sinh y.

We differentiate both sides of x = sinh y with respect to x and solve for dy>dx: x = sinh y y = sinh-1 x 3 x = sinh y

1 = 1cosh y2 dy dx

Use implicit differentiation.

dx =

1 cosh y

Solve for dy>dx. dy

dx =

{2sinh2 y + 1 cosh2 y - sinh2 y = 1 dy

dx =

12x2 + 1. x = sinh y

Figure 6.95

1 y ! cosh x

5y ! 3

THEOREM 6.10 Inverses of the Hyperbolic Functions Expressed as Logarithms

cosh-1 x = ln 1x + 2x2 - 12 1x Ú 12 sech-1 x = cosh-1 1 x

10 6 x … 12 sinh-1 x = ln 1x + 2x2 + 12 csch-1 x = sinh-1 1

x 1x ≠ 02

tanh-1 x = 1 2

ln a 1 + x 1 - x b 1 ! x ! 6 12 coth-1 x = tanh-1 1x 1 ! x ! 7 12

➤ Most calculators allow for the direct evaluation of the hyperbolic sine, cosine, and tangent, along with their inverses, but are not programmed to evaluate sech-1 x, csch-1 x, and coth-1 x. The formulas in Theorem 6.10 are useful for evaluating these functions on a calculator.

QUICK CHECK 5 Use the results of Example 4 to write an integral for the area of the region bounded by y = cosh x and y = 53 (Figure 6.95) and then evaluate the integral.

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6.10 Hyperbolic Functions 499

The restrictions associated with the formulas in Theorem 6.11 are a direct conse- quence of the domains of the inverse functions (Figure 6.94). Note that the derivative of both tanh-1 x and coth-1 x is 1>11 - x22, although this result is valid on different domains 1 ! x ! 6 1 for tanh-1 x and ! x ! 7 1 for coth-1 x2. These facts have a bearing on formula (3) in the next theorem, which is a reversal of the derivative formulas in Theorem 6.11. Here we list integral results, where a is a positive constant; each formula can be verified by differentiation.

THEOREM 6.11 Derivatives of the Inverse Hyperbolic Functions

d dx

1cosh-1 x2 = 12x2 - 1 1x 7 12 ddx 1sinh-1x2 = 12x2 + 1 d dx

1tanh-1 x2 = 1 1 - x2

1 ! x ! 6 12 d dx

1coth-1 x2 = 1 1 - x2

1 ! x ! 7 12 d dx

1sech-1 x2 = - 1 x21 - x2 10 6 x 6 12 ddx 1csch-1 x2 = - 1! x ! 21 + x2 1x ≠ 02

THEOREM 6.12 Integral Formulas

1. L dx2x2 - a2 = cosh-1 xa + C, for x 7 a

2. L dx2x2 + a2 = sinh-1 xa + C, for all x

3. L dx

a2 - x2 = µ 1a tanh-1 xa + C, for ! x ! 6 a

1 a

coth-1 x a

+ C, for ! x ! 7 a

4. L dx

x2a2 - x2 = - 1a sech-1 xa + C, for 0 6 x 6 a 5. L

x2a2 + x2 = - 1a csch-1 ! x !a + C, for x ≠ 0 EXAMPLE 5 Derivatives of inverse hyperbolic functions Compute dy>dx for each function.

a. y = tanh-1 3x b. y = x2 sinh-1 x

SOLUTION

a. Using the Chain Rule, we have

dx =

d dx

1tanh-1 3x2 = 1 1 - 13x22 # 3 = 31 - 9x2.

b. dy

dx = 2x sinh-1 x + x2 # 12x2 + 1 Product Rule; Theorem 6.11

= x a22x2 + 1 # sinh-1 x + x2x2 + 1 b Simplify. Related Exercises 47–52

➤ The integrals in Theorem 6.12 appear again in Chapter 7 in terms of logarithms and with fewer restrictions on the variable of integration.

➤ The function 3>11 - 9x22 in the solution to Example 5a is defined for all x ≠ {1>3. However, the derivative formula dy>dx = 3>11 - 9x22 is valid only on -1>3 6 x 6 1>3 because tanh-1 3x is defined only on -1>3 6 x 6 1>3. The result of computing d>dx1coth-1 3x2 is the same, but valid on 1- ∞ , -1>32 ∪ 11>3, ∞2.

In the last step, the positive root is chosen because cosh y 7 0 for all y. The derivatives of the other inverse hyperbolic functions, listed in Theorem 6.11, are

derived in a similar way (Exercise 106).

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500 Chapter 6 Applications of Integration

EXAMPLE 6 Integral computations

a. Compute the area of the region bounded by y = 1>2x2 + 16 over the interval 30, 34. b. Evaluate L

dx1x14 - x2. SOLUTION

a. The region in question is shown in Figure 6.96, and its area is given by

L 3

dx2x2 + 16. Using formula (2) in Theorem 6.12 with a = 4, we have L

dx2x2 + 16 = sinh-1 x4 ` 30 Theorem 6.12 = sinh-1

3 4

- sinh-1 0 Evaluate.

= sinh-1 3 4

. sinh-10 = 0

A calculator gives an approximate result of sinh-113>42 ≈ 0.693. The exact result can be written in terms of logarithms using Theorem 6.10:

sinh-113>42 = ln13>4 + 213>422 + 12 = ln 2. b. The integral doesn’t match any of the formulas in Theorem 6.12, so we use the substi-

tution u = 1x: L

dx1x14 - x2 = 2L53 du4 - u2. Let u = 1x; du = dx21x . The new integral now matches formula (3), with a = 2. We conclude that

2L 5

4 - u2 = 2 # 1

2 coth-1

u 2 ` 5 3

L dx

a2 - x2 =

1 a

coth-1 x a

+ C

= coth-1 5 2

- coth-1 3 2

. Evaluate.

The antiderivative involving coth-1 x was chosen because the interval of integration 13 … u … 52 satisfies " u " 7 a = 2. Theorem 6.10 is used to express the result in numerical form in case your calculator cannot evaluate coth-1 x:

coth-1 5 2

- coth-1 3 2

= tanh-1 2 5

- tanh-1 2 3

≈ -0.381.

Related Exercises 53–64

QUICK CHECK 6 Evaluate L 1

4 - u2 .

Applications of Hyperbolic Functions This section concludes with a brief look at two applied problems associated with hyper- bolic functions. Additional applications are presented in the exercises.

The Catenary When a free-hanging rope or flexible cable supporting only its own weight is attached to two points of equal height, it takes the shape of a curve known as a catenary. You can see catenaries in telephone wires, ropes strung across chasms for Tyrolean traverses (Example 7), and spider webs.

The equation for a general catenary is y = a cosh1x>a2, where a ≠ 0 is a real num- ber. When a 6 0, the curve is called an inverted catenary, sometimes used in the design of arches. Figure 6.97 illustrates catenaries for several values of a.

Figure 6.96

x2 ! 16 y "

0.25

321#1

y ! acosh(x/a) for several values of a " 0

When a # 0, inverted catenaries result (reflections across the x-axis)

a ! 0.5

a ! 1.0

a ! 2.0

a ! 1.5

a ! 3.0

x41 2 3

$1$2$3$4

Figure 6.97

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6.10 Hyperbolic Functions 501

EXAMPLE 7 Length of a catenary A climber anchors a rope at two points of equal height, separated by a distance of 100 ft, in order to perform a Tyrolean traverse. The rope follows the catenary f 1x2 = 200 cosh1x>2002 over the interval 3-50, 504 (Figure 6.98). Find the length of the rope between the two anchor points.

➤ A Tyrolean traverse is used to pass over difficult terrain, such as a chasm between two cliffs or a raging river. A rope is strung between two anchor points, the climber clips onto the rope and then traverses the gap by pulling on the rope. y

100

200

x50!50

y " 200 cosh (x /200)

Figure 6.98

SOLUTION Recall from Section 6.5 that the arc length of the curve y = f 1x2 over the interval 3a, b4 is L = Lba 21 + f ′1x22 dx. Also note that

f ′1x2 = 200 sinh a x 200

b # 1 200

= sinh x

200 .

Therefore, the length of the rope is

L = L 50

-50 B1 + sinh2a x200 b dx Arc length formula = 2L

0 B1 + sinh2a x200 b dx Use symmetry. = 400L

1>4 0

21 + sinh2 u du Let u = x 200

= 400L 1>4

0 cosh u du 1 + sinh2 u = cosh2 u

= 400 sinh u ` 1>4 0

Evaluate integral.

= 400asinh 1 4

- sinh 0b Simplify. ≈ 101 ft. Evaluate. Related Exercises 65–68

Velocity of a Wave To describe the characteristics of a traveling wave, research- ers formulate a wave equation that reflects the known (or hypothesized) properties of the wave and that often takes the form of a differential equation (Section 7.9). Solving a wave equation produces additional information about the wave, and it turns out that hyper- bolic functions may arise naturally in this context.

EXAMPLE 8 Velocity of an ocean wave The velocity v (in meters/second) of an ideal- ized surface wave traveling on the ocean is modeled by the equation

v = B gl2p tanha 2pdl b ,

➤ Using the principles of vector analysis introduced in Chapter 11, the tension in the rope and the forces acting upon the anchors in Example 7 can be computed. This is crucial information for anyone setting up a Tyrolean traverse; the sag angle (Exercise 68) figures into these calculations. Similar calculations are important for catenary lifelines used in construction and for rigging camera shots in Hollywood movies.

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502 Chapter 6 Applications of Integration

5. Express sinh-1 x in terms of logarithms.

6. What is the domain of sech-1 x? How is sech-1 x defined in terms of the inverse hyperbolic cosine?

7. A calculator has a built-in sinh-1 x function, but no csch-1 x func- tion. How do you evaluate csch-1 5 on such a calculator?

8. On what interval is the formula d>dx1tanh-1 x2 = 1>1x2 - 12 valid?

where g = 9.8 m>s2 is the acceleration due to gravity, l is the wavelength measured in meters from crest to crest, and d is the depth of the undisturbed water, also measured in meters (Figure 6.99).

a. A sea kayaker observes several waves that pass beneath her kayak, and she estimates that l = 12 m and v = 4 m>s. How deep is the water in which she is kayaking?

b. The deep-water equation for wave velocity is v = A gl2p, which is an approximation to the velocity formula given above. Waves are said to be in deep water if the depth-

to-wavelength ratio d>l is greater than 1 2

. Explain why v = A gl2p is a good approxi- mation when d>l 7 1

2 .

SOLUTION

a. We substitute l = 12 and v = 4 into the velocity equation and solve for d.

4 = B g # 122p tanha 2pd12 b 1 16 = 6gp tanhapd6 b Square both sides. 1 8p

3g = tanhapd

6 b Multiply by p

6g .

In order to extract d from the argument of tanh, we apply tanh-1 to both sides of the equation and then use the property tanh-11tanh x2 = x, for all x.

tanh-1a 8p 3g

b = tanh-1 a tanhapd 6 b b Apply tanh-1 to both sides.

tanh-1a 8p 29.4

b = pd 6

Simplify; 3g = 29.4.

d = 6 p

tanh-1a 8p 29.4

b ≈ 2.4 m Solve for d. Therefore, the kayaker is in water that is about 2.4 m deep.

b. Recall that y = tanh x is an increasing function 1dy>dx = sech2 x 7 02 whose values approach 1 as x S ∞ . Also notice that when d

l =

1 2

, tanha 2pd l

b = tanh p ≈ 0.996, which is nearly equal to 1. These facts imply that whenever

d l

7 1 2

, we can replace

tanha 2pd l

b with 1 in the velocity formula, resulting in the deep-water velocity function v = A gl2p.

Related Exercises 69–72

➤ In fluid dynamics, water depth is often discussed in terms of the depth-to- wavelength ratio d>l, not the actual depth of the water. Three classifications are generally used:

shallow water: d>l 6 0.05 intermediate depth: 0.05 6 d>l 6 0.5 deep water: d>l 7 0.5

Figure 6.99

x31 2

!1!2!3

tanh x ! 1 as x ! "

y # tanh x

QUICK CHECK 7 Explain why longer waves travel faster than shorter waves in deep water.

SECTION 6.10 EXERCISES Review Questions

1. State the definition of the hyperbolic cosine and hyperbolic sine functions.

2. Sketch the graphs of y = cosh x, y = sinh x, and y = tanh x (include asymptotes), and state whether each function is even, odd, or neither.

3. What is the fundamental identity for hyperbolic functions?

4. How are the derivative formulas for the hyperbolic functions and the trigonometric functions alike? How are they different?

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