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

M06_BRIG7345_02_SE_C06.1.indd 398 21/10/13 9:20 PM

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.

M06_BRIG7345_02_SE_C06.1.indd 399 21/10/13 11:29 AM

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)

M06_BRIG7345_02_SE_C06.1.indd 400 21/10/13 11:29 AM

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.

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

0.25

0.50

0.75

1.00

“0.25

s(t)

s

0 0.25

0.50

0.75

1.00

“0.25

s

M06_BRIG7345_02_SE_C06.1.indd 401 21/10/13 11:29 AM

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′1×2 dx = 100 + L

t

0 90e-0.1x dx

()* c

N102 N′1×2 = 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 !

90

60

30

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

r)

0

0 t30252015105

N

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′1×2 = 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′1×2 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′1×2 dx = L

15,000

12,000 112 – 0.0002×2 dx Substitute for C′1×2.

= 112x – 0.0001×22 ` 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.

M06_BRIG7345_02_SE_C06.1.indd 406 21/10/13 11:30 AM

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.

t

v

543210

12

16

10

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.

t

v

876543210

10

6

20

14

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?

T

T

T

M06_BRIG7345_02_SE_C06.1.indd 407 21/10/13 3:11 PM

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.

M06_BRIG7345_02_SE_C06.1.indd 408 21/10/13 11:30 AM

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).

t

r

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′1×2 = 2000 – 0.5x 46. C′1×2 = 200 – 0.05x 47. C′1×2 = 300 + 10x – 0.01×2 48. C′1×2 = 3000 – x – 0.001×2 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

3

1

v

0

51.

t51

3

1

v

0

Order now and get 10% discount on all orders above $50 now!!The professional are ready and willing handle your assignment.

ORDER NOW »»