# Mathematics

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.