# Mathematics

398

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

6

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

y

t

t

a b

a

b

y

O

O

y ! v(t)

Area ! A1

Area ! A1

Area ! A2

Area ! A2

Displacement ! A1 ” A2 ! ! v(t) dt a

b

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

b

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

6

4

2

!2

310

v

t

6

4

2

!2

321

!v!

t0

0

1

1

3

0

3

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 ”

Area ”

Area ”

Area ”

(a) (b)

6

4

2

!2

310

v

t

6

4

2

!2

321

!v!

t0

0

1

1

3

0

3

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 ”

Area ”

Area ”

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′1×2 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.