d material draft - nelson€¦ · the rate of change from 2 to 26 is: the function is linear and so...

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lGetting Started, p. #1. The slope between two points can be found by

dividing the change in y by the change in x, .

a)

b)

2. a)

Each successive first difference is 2 times the previ-ous first difference. The function is exponential.b) First differences

Second differences

The second differences are constant so the functionis quadratic.3. a)

and

The zeros are and 2.

b)

Any non-zero number raised to the 0 power is 1, so

c)and

and

Because cannot be a zero. Thezeros are and d)

For and

The zeros are and

4. a) is stretched vertically by a factor of .

b) is stretched vertically by a factor of 2 andtranslated right by 4.c) is stretched vertically by a factor of 3,reflected in the x-axis and translated up 7.d) is stretched vertically by a factor of 5,translated right by 3 and translated down 2.5. a) $1000 is P. 8% or 0.08 is i. is 1.08. n is t.

becomes b) t is 3, or $1259.71c) No, since the interest is compounded each yearyou earn more interest than the previous year. Theinterest earns interest.6. a) is a maximum at so or .

is a minimum at so or

b) The period of is .or t 5 24 s.15°t 5 360°

360°y 5 sinxh(18) 5 1 m.h(18) 5 8 1 7 sin(15° 3 18).

t 5 18.15°t 5 270°270°y 5 sin xh(6) 5 15 m.h(6) 5 8 1 7 sin(15° 3 6).

t 5 615°t 5 90°90°y 5 sinx

A 5 1000(1.08)3A 5 1000(1.08)tA 5 P(1 1 i)n

1 1 i

f(x)

f(x)

f(x)

12

f(x)2270°.290°

cos(2270°) 5 0.cos(290°) 5 02360° # x # 0°,

0 5 cos(x)0 5 2 cos(x)

225°.45°0° # x # 360°, 405°

405° 5 x360° 1 45° 5 x 2 45° 1 45°

225° 5 x180° 1 45° 5 x 2 45° 1 45°

45° 5 x0° 1 45° 5 x 2 45° 1 45°

360° 5 x 2 45°180° 5 x 2 45°,0° 5 x 2 45°,sin(360°) 5 0.sin(180°),sin(0°),

0° # x # 360°0 5 sin(x 2 45°),x 5 0.

1 5 2x0 1 1 5 2x 2 1 1 1

0 5 2x 2 1

232

2 5 x0 1 2 5 x 2 2 1 2

232

5 x

23 5 2x0 2 3 5 2x 1 3 2 3

0 5 x 2 20 5 2x 1 30 5 (2x 1 3)(x 2 2)0 5 2x2 2 x 2 6

35 2 29 5 629 2 23 5 623 2 17 5 617 2 11 5 6

115 2 80 5 3580 2 51 5 2951 2 28 5 2328 2 11 5 1711 2 0 5 11

261 2 (229) 5 232229 2 (213) 5 216213 2 (25) 5 28

25 2 (21) 5 2421 2 1 5 22

5 2 (21)24 2 3

5 267

7 2 35 2 2

543

DyDx

CHAPTER 2Functions: Understanding

Rates of Change

2-1Advanced Functions Solutions Manual

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c)

7.

Lesson 2.1 Determining Average Rate of Change, p. #

1. The average rate of change is equal to the changein y divided by the change in x.a)

Average rate of change

b)


c)


d)


e)


f )


2. a) i) According to the table, the height at is42.00 m and the height at is 27.00 m.

m s

ii) According to the table, the height at is42.00 and is 47.00 m.

m s

b) The flare is gaining height at 15 m s and thenloses height at 5 m s.3. is always increasing at a constant rate and

is decreasing on and increasing onand not constant.

4. a) 1st half hour:

2nd half hour:

3rd half hour:

4th half hour:

5th half hour:

6th half hour:

b) The rate of growth of the crowd at the rally.c) Having a positive rate of growth would indicatethat people were arriving at the rally. Having a negative rate of growth would indicate that peoplewere leaving the rally.

5. a) Day 1:

Day 2:

Day 3:

Day 4:

Day 5:

Day 6:

b) No. Some days the distance travelled was greaterthan others.

1104 2 9586 2 5

5 146 km>day

958 2 739.55 2 4

5 218.5 km>day

739.5 2 5614 2 3

5 178.5 km>day

561 2 3963 2 2

5 165 km>day

396 2 2032 2 1

5 193 km>day

203 2 01 2 0

5 203 km>day

415 2 4323.0 2 2.5

5 234 people>h432 2 4022.5 2 2.0

5 60 people>h402 2 3882.0 2 1.5

5 28 people>h388 2 2451.5 2 1.0

5 286 people>h245 2 1761.0 2 0.5

5 138 people>h176 2 00.5 2 0

5 352 people>h(0, `)

(2`, 0)g(x)f(x)

> >>42 2 47

4 2 35 25

t 5 3t 5 4

>42 2 272 2 1

5 15

t 5 1t 5 2

57.1104 2 72.01 2 2

5 11.04

g(2) 5 75 7.1104

5 16.1604 2 10.05 1 1g(2.01) 5 4(2.01)2 2 5(2.01) 1 1

58.14 2 72.1 2 2

5 11.4

g(2) 5 75 8.145 17.64 2 10.5 1 1

g(2.1) 5 4(2.1)2 2 5(2.1) 1 1

510 2 7

2.25 2 25 12

g(2) 5 75 105 20.25 2 11.25 1 1

g(2.25) 5 4(2.25)2 2 5(2.25) 1 1

513.5 2 72.5 2 2

5 13

g(2) 5 75 13.55 25 2 12.5 1 1

g(2.5) 5 4(2.5)2 2 5(2.5) 1 1

522 2 73 2 2

5 15

g(2) 5 75 225 36 2 15 1 1

g(3) 5 4(3)2 2 5(3) 1 1

545 2 74 2 2

5 19

5 75 16 2 10 1 1

g(2) 5 4(2)2 2 5(2) 1 15 455 64 2 20 1 1

g(4) 5 4(4)2 2 5(4) 1 1

variable;can be positive, negative, or

0 for differentparts of the

same relation

Nonlinear relationsLinear relationsconstant; same as slope of line;positive for linesthat slope up fromleft to right;negative for lines that slopedown from left to right;0 for horizontal lines.

Rates of Change

h(30) 5 15 m.t 5 30. h(30) 5 8 1 7 sin(15° 3 30).

2-2 Chapter 2: Functions: Understanding Rates of Change

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6. The function is To find the average

rate of change find The rate of change from

to is:

The rate of change from 2 to 26 is:

The function is linear and so the rate of change isconstant. That’s why the rate of change for eachinterval is the same.7. For any amount of time up to and including 250the monthly charge is 39 dollars, therefore the rateof change is zero for that interval. After 250minutes the rate of change is a constant 10 centsper minute. The rate is not constant.8. a) Find the ordered pairs for the intervals given.Interval i)– and (0, 5)Interval ii)– and (20, 20)Interval iii)– and (40, 80)Interval iv)– and (0, 5)Use this information to find the change in population over the change in time.

i) or 750 people per year

ii) or 3000 people per year

iii) 12 or 12 000 people per year

iv) or 5250 people per year

b) No; the rate of growth increases as the timeincreases.

c) Assume that the growth continues to follow thispattern and that the population will be 5 120 000people in 2050.9. The function is The average

rate of change is for the interval 10

10. a) The function is

The average rate of change is

i)

$2.60 per sweatshirtii)

$2.00 per sweatshirtiii)

$1.40 per sweatshirtiv)

$0.80 per sweatshirt5 0.8

P(5) 2 P(4)5 2 4

521.15 2 20.35

5 2 4

5 21.15P(5) 5 20.3(5)2 1 3.5(5) 1 11.15

5 1.4

P(4) 2 P(3)4 2 3

520.35 2 18.95

4 2 3

P(3) 5 18.955 20.35

P(4) 5 20.3(4)2 1 3.5(4) 1 11.15

5 2

P(3) 2 P(2)3 2 2

518.95 2 16.95

3 2 2

P(2) 5 16.955 18.95

P(3) 5 20.3(3)2 1 3.5(3) 1 11.15

5 2.6

P(2) 2 P(1)2 2 1

516.95 2 14.35

2 2 1

5 14.35P(1) 5 20.3(1)2 1 3.5(1) 1 11.15

5 16.95P(2) 5 20.3(2)2 1 3.5(2) 1 11.15

P(2) 2 P(1)2 2 1

DP(s)Ds

.

P(s) 5 20.30s2 1 3.5s 1 11.5

5 22 m>s5

2105

590 2 10015 2 10

518(15) 2 0.8(15)2 2 (18(10) 2 0.8(10)2)

15 2 10

Dh(t)Dt

5h(15) 2 h(10)

15 2 10

# t # 15.Dh(t)

Dt

h(t) 5 18t 2 0.8t2.

320 2 560 2 0

531560

5 5.25

320 2 6060 2 40

526020

5

80 2 2040 2 20

56020

5 3

20 2 520 2 0

51520

534

(60, 320)(60, 320)

(40, 80)(20, 20)

59624

5 4

5104 2 826 2 2

54(26) 2 4(2)

26 2 2

f(26) 2 f(2)26 2 2

5164

5 4

524 2 86 2 2

54(6) 2 4(2)

6 2 2

f(6) 2 f(2)6 2 2

x 5 6x 5 2

Df(x)Dx

.

f(x) 5 4x.


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b) The rate of change is still positive, but it isdecreasing. This means that the profit is stillincreasing, but at a decreasing rate.c) No; after 6000 sweatshirts are sold, the rate ofchange becomes negative. This means that the profitbegins to decrease after 6000 sweatshirts are sold.11. a)

b) If we were to find the average rate of change ofan interval that is farther in the future, such as2025–2050 instead of 2010–2015, the average rateof change would be greater. The graph indicates thatthe change in y increases as x increases.c) The function is


i)

ii)

iii)

iv)

12. Answers may vary. For example,• The height of a falling object is a real-life examplethat would have a negative rate of change. Thedistance travelled by a runner is an example thatwould have a positive rate of change.• An average rate of change would be useful whenthere are several different rates of change over aspecific interval.• The average rate of change is found by taking thechange in y for the specified interval and dividing itby the change in x over that same interval.13. The car’s starting value is $23 500. After 8 yearsthe car is only worth $8750.The average rate of change in the value of the car is

The value of the car decreases, on average, by$1843.75 per year. As a percentage of the car’s

original value, this is or

decrease.

14. Answers may vary. For example,

15. The average rate of change for the function

is

km L

km L>5 39.5F(50) 5 20.005(50)2 1 0.8(50) 1 12

>5 42F(100) 5 20.005(100)2 1 0.8(100) 1 12

DF(x)Dx

.

7.8%

1843.7523 500

3 100,

5 21843.75.

8750 2 23 5008 2 0

5214 750

8

5 2500 people per year

P(20) 2 P(10)20 2 10

560 000 2 35 000

20 2 10

P(10) 5 35 0005 60 000

P(20) 5 50(20)2 1 1000(20) 1 20 0005 2000 people per year

P(15) 2 P(5)15 2 5

546 250 2 26 250

15 2 5

5 26 250P(5) 5 50(5)2 1 1000(5) 1 20 000

5 46 250P(15) 5 50(15)2 1 1000(15) 1 20 000


P(12) 2 P(2)12 2 2

539 200 2 22 200

12 2 2

5 22 200P(2) 5 50(2)2 1 1000(2) 1 20 000

5 39 200P(12) 5 50(12)2 1 1000(12) 1 20 000


P(10) 2 P(0)10 2 0

535 000 2 20 000

10 2 0

5 20 000P(0) 5 50(0)2 1 1000(0) 1 20 000

5 35 000

P(10) 5 50(10)2 1 1000(10) 1 20 000

DP(t)Dt

.

P(t) 5 50t2 1 1000t 1 20 000.


Visual representation

y

xDx

Dy

Dy——Dx

0

Averagerate ofchange =

Personal example

I record the number of miles I run each week versus theweek number. Then, I cancalculate the average rate ofchange in the distance I runover the course of weeks.

AVERAGE RATE OF CHANGE

Definition in your own words

the change in one quantitydivided by the change in arelated quantity

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lLesson 2.2 Estimating InstantaneousRates of Change from Tables of Values and Equations, p. #



b) As the values of x get closer together on both sidesof 2, the average rate of change gets closer to 20.2. a) Find the average rate of change for intervalsthat approach 2.0 from both sides.

The instantaneous rate of change appears to beapproaching 5.4.b) Find the average rate of change for intervals that approach 2.0.

The instantaneous rate of change is approximately 5.4.c) Answers may vary. For example, The centeredmethod; there is less work involved.3. a) The population at 2.5 months is

b)

c) Use the difference quotient to find the instanta-neous rate of change.

where h is a very small value.

or 50 raccoons per month

d) Part a) asks for the value of at 2.5; part b)asks for the average rate of change over a certaininterval; part c) ask for the instantaneous rate ofchange at 2.5––they are all different values.

P(t)

200.5004 2 2000.01

5 50.04

5 200.5004f(2.51) 5 100 1 30(2.51) 1 4(2.51)2

f(a 1 h) 2 f(a)h

,

200 2 1002.5 2 0

5 40 raccoons per month

5 100P(0) 5 100 1 30(0) 1 4(0)2

5 200P(2.5) 5 100 1 30(2.5) 1 4(2.5)2

P(2.5).

32.38 2 26.982.5 2 1.5

5 5.4

31.4 2 20.63.0 2 1.0

5 5.4

10.3 1 0.52

5 5.4

7.84 1 2.962

5 5.4

32.38 2 30.92.5 2 2.0

5 2.96

31.4 2 30.93.0 2 2.0

5 0.5

30.9 2 26.982.0 2 1.5

5 7.84

30.9 2 20.62.0 2 1

5 10.3

Df(x)Dx

5 20.05f(2) 5 13,f(2.01) 5 13.2,

Df(x)Dx

5 20.5f(2) 5 13,f(2.1) 5 15.05,

Df(x)Dx

5 22.5f(2) 5 13,f(2.5) 5 24.25,

Df(x)Dx

5 25f(2) 5 13,f(3) 5 38,

Df(x)Dx

5 19.95f(1.99) 5 12.8,f(2) 5 13,

Df(x)Dx

5 19.5f(1.9) 5 11.05,f(2) 5 13,

Df(x)Dx

5 17.5f(1.5) 5 4.25,f(2) 5 13,

Df(x)Dx

5 15f(1) 5 22,f(2) 5 13,

Df(x)Dx

.

f(x) 5 5x2 2 7.

5 0.05 km>L>km>hF(100) 2 F(50)

100 2 505

42 2 39.5100 2 50


FollowingInterval

f (x)D xD

Average Rate

of ChangeDf(x)

Dx

2 # x # 3 38 2 13 5 25 3 2 2 5 1 25

2 # x # 2.5 11.25 0.5 22.5

2 # x # 2.1 2.05 0.1 20.5

2 # x # 2.01 0.2005 0.01 20.05

PrecedingInterval

f (x)D xD

Average Rate

of ChangeDf(x)

Dx

1 # x # 2 13 2 (22) 5 15 2 2 1 5 1 15

1.5 # x # 2 8.75 0.5 17.5

1.9 # x # 2 1.95 0.1 19.5

1.99 # x # 2 0.1995 0.01 19.95

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4. Use the difference quotient to find the instanta-neous rate of change.


a)

b)

c)

d)

5. Use the difference quotient to find theinstantaneous rate of change.


The function is

6. Use the difference quotient to find theinstantaneous rate of change.


The function is

per year or

about $11 610 per year



b) Answers may vary. For example, Yes, it makessense. It means that the population in 2000 and 2024are the same, so the average rate of change is 0.c)

The average rate of change during the first 12 yearswas 18 thousand per year. During the second 12 yearsit was thousand per year. The population duringyear 0 is 6 thousand and during year 24 is 6 thousand.d) Because the average rate of change is the sameon each side of 12, we know that the instantaneousrate of change would be 0 at 12.8. Use the difference quotient to find theinstantaneous rate of change.


The function is

When the car turns five, it loses about $960.9. a) The diver will hit the water when The function is

21024.9

524.924.9

t2

210 5 24.9t20 5 10 2 4.9t2

h(t) 5 10 2 4.9t2.h(t) 5 0.

13 207.79 2 13 217.380.01

5 2959

5 13 217.38f(5) 5 18 999(0.93)5

5 13 207.79f(5.01) 5 18 999(0.93)5.01

V(t) 5 18 999(0.93)t.

f(a 1 h) 2 f(a)h

,

218

6 2 22224 2 12

5 218 thousand>year

P(12) 5 222P(24) 5 6

222 2 612 2 0

5 18 thousand>year

P(0) 5 65 222

P(12) 5 21.5(12)2 1 36(12) 1 6

6 2 624 2 0

5 0 people>year

5 6P(0) 5 21.5(0)2 1 36(0) 1 6

5 6P(24) 5 21.5(24)2 1 36(24) 1 6

DyDx

.

P(t) 5 21.5t2 1 36t 1 6.

199 347.13 2 199 231.010.01

5 $11 612

5 199 231.01H(8) 5 125 000(1.06)8

5 199 347.13H(8.01) 5 125 000(1.06)8.01

H(t) 5 125 000(1.06)t.

f(a 1 h) 2 f(a)h

,

28.7295 2 290.01

5 227.05 m>s or 227 m>s5 29

f(3) 5 25(3)2 1 3(3) 1 655 28.7295

f(3.01) 5 25(3.01)2 1 3(3.01) 1 65h(x) 5 25x2 1 3x 1 65.

f(a 1 h) 2 f(a)h

,

380.96 2 3800.01

5 96.06 or 96

5 380f(8) 5 6(8)2 2 4

5 380.9606f(8.01) 5 6(8.01)2 2 4

92.48 2 920.01

5 48.06 or 48

5 92f(4) 5 6(4)2 2 4

5 92.4806f(4.01) 5 6(4.01)2 2 4

23.9994 2 (24)0.01

5 0.06 or 0

5 24f(0) 5 6(0)2 2 4

5 23.9994f(0.01) 5 6(0.01)2 2 4

20 2 19.7622 2 (21.99)

5 223.94 or 224

5 20f(22) 5 6(22)2 2 4

5 19.7606f(21.99) 5 6(21.99)2 2 4

f(a 1 h) 2 f(a)h

,


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The diver enters the water at 1.43 seconds.b) Use the difference quotient to find theinstantaneous rate of change.


The diver is travelling at a rate of m s.10. Use the centered interval method to estimate the instantaneous rate of change at Find values of on either side of 5.

Now, use the difference quotient to find theinstantaneous rate of change.

11. David simply needs to keep track of the totaldistance that he’s travelled and the amount of timethat it has taken him to travel that distance. Dividingthe distance travelled by the time required to travelthat distance will give him his average rate ofspeed.12. a) Use a centered interval to find the instanta-

neous rate of change.

b) Answers may vary. For example, A quadraticmodel for the oven temperature versus time is

Using this model,the instantaneous rate of change at is about

c) Answers may vary. For example, The first rate isusing a larger interval to estimate the instantaneousrate.

d) Answers may vary. For example, The secondestimate is better as it uses a much smaller intervalto estimate the instantaneous rate.13. Answers may vary. For example,

14. a) The formula for finding the area of a circle iswhere r is the radius. The average rate of

change is

The average rate of change is 100 cm2 cm.b) Use the difference quotient to find the instanta-neous rate of change.


cm2 cm or

240 cm2 cm15. The formula for the surface area of a cubegiven the length of a side is where s isthe side length of the cube. Use the differencequotient to find the instantaneous rate of change.


5 54.3606SA 5 6(3.01)2

f(a 1 h) 2 f(a)h

,

V 5 6s2,

>p

>14 402.4001p 2 14 400p

0.018 754.01

5 14 400p

A 5 p(120)25 p(14 402.4001)

A 5 p(120.01)2

f(a 1 h) 2 f(a)h

,

>p

10 000p 2 0100 2 0

5 100p

5 0A 5 p(0)2

5 10 000p

A 5 p(100)2

DADr

.

A 5 pr2,

Method of EstimatingInstantaneous Rateof Change Advantage Disadvantage

series of precedingintervals and followingintervals

accounts fordifferences inthe way thatchange occurson either side ofthe given point

must do twosets ofcalculations

series of centredintervals

accounts forpoints on eitherside of the giveninterval in thesame calculation

to get a preciseanswer, numbersinvolved willneed to haveseveral decimalplaces

difference quotient more precise calculations canbe tedious ormessy

225.5° F>min.x 5 4

y 5 21.96x2 2 9.82x 1 400.71.

305 2 3505 2 3

5 222.5 °F>min

167.669p 2 166.667p

0.015 314.63 or 100p cm3>cm

V(5) 543

p(5)3 5 166.667p

V(5.01) 543

p(5.01)3 5 167.669p

176.868p 2 156.865p

5.1 2 4.95 314.21 or 100p cm3>cm

V(4.9) 543

p(4.9)3 5 156.865p

V(5.1) 543

p(5.1)3 5 176.868p

V(r)r 5 5.

>214

20.160 64 1 0.020 010.01

5 214.063

5 20.020 01h(1.43) 5 10 2 4.9(1.43)2

5 20.160 64h(1.44) 5 10 2 4.9(1.44)2

f(a 1 h) 2 f(a)h

,

1.43 5 t"2.04 5 "t2

2.04 5 t2


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lThe instantaneous rate of change is about 36 cm2 cm.16. The formula for finding the surface area of asphere is Use the difference quotient to find the instantaneous rate of change.


The instantaneous rate of change is about 502.78 cm2 cm or 160 cm2 cm.

Lesson 2.3 Exploring Instantaneous Rates of Change Using Graphs, p. #

1. a) Answers may vary. For example,

The slope is 19.b) Answers may vary. For example,

The slope is 89.

c) Answers may vary. For example,

The slope is 0.25.d)

Because the graph is linear, the slope is the sameeverywhere. The slope is 2.2. a)

b)

c)

d)

0

21

345

y

x

1 2 3 4 5 6–1–2

1 2 3 4 5 60–1

12

y

x

0

5025

75100

x

y

4 8–4–8 –25

0

105

15202525

x

y

4 8–4–8

>p>

1601.6004p 2 1600p

0.018 502.78 cm2>cm

5 1600p

SA 5 4p(20)25 1601.6004p

SA 5 4p(20.01)2

f(a 1 h) 2 f(a)h

,

SA 5 4pr2.

>

54.3606 2 540.01

5 36.06 cm2>cm

5 54SA 5 6(3)2


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3. a) Set A:

The slope of the tangent at is 0.



The slope of the tangent at is 0.Set B:


The slope of the tangent at is about 1.4.


The slope of the tangent at is about 0.009.Set C:

The slope of the tangent at is

The slope of the tangent at is about

The slope of the tangent at is

The slope of the tangent at is about b) Set: A: All slopes are zero.Set: B: All slopes are positive.Set: C: All slopes are negative.

20.009.x 5 120°

23.x 5 2

20.69.x 5 0

24.x 5 21

x 5 60°

x 5 3

x 5 1

x 5 2

x 5 1

x 5 22

x 5 90°

x 5 3


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4. a) and b)

c) The y-intercept of the tangent line appears to be. Find the slope between the points

and

The slope is 32 .d) Use the data points and

The rate of change is about at e) Answers may vary. For example, The answer inpart d) is the slope of the line connecting two pointson either side of The answer in part c) is theslope of the line tangent to the function at point

The two lines are different and so theirslopes will be different.5. Answers may vary. For example, Similarity: thecalculation; difference: average rate of change is overan interval while instantaneous rate of change is at apoint.

Mid-Chapter Review, p. #1. a)

b) Rate of change is Since we are looking

for the amount of change between each month,will always be 1 month. Therefore, we just need tofind the difference in volume between each month.

or

or

or

or

or

c) Examine each of the answers from the previousexercises. The greatest amount is the greatestamount of change between two months.

The greatest amount of change occurred during m4,between April and May.d) The change in y is the difference between thevolume of water used in each month. The change in x is the difference between the numbers of themonths

or

2. a) The equation models exponential growth. This means that the average rate of change betweenconsecutive years will always increase.b) Use the difference quotient to find theinstantaneous rate of change.


people per year3. a) The average change for a specific interval is

The function is

5 1h(0) 5 25(0)2 1 20(0) 1 1

5 21h(2) 5 25(2)2 1 20(2) 1 1

h(t) 5 25t2 1 20t 1 1.Dh(t)

Dt.

0.095 39 3 10 000 8 950

1.955 627 473 2 1.954 673 5520.01

5 0.095 39

5 1.954 673 552f(10) 5 1.2(1.05)10

5 1.955 627 473f(10.01) 5 1.2(1.05)10.01

f(a 1 h) 2 f(a)h

,

580 m3>month

5.50 2 3.755 2 2

5 0.580 3 1000 m3>month

1.10 , 0.75 , 0.40 , 0.25 , 0.00

400 m3>monthm5: 5.50 2 5.10 5 0.401 000 m3>month1100 m3>monthm4: 5.10 2 4.00 5 1.101 000 m3>month250 m3>monthm3: 4.0 2 3.75 5 0.251 000 m3>month0 m3>monthm2: 3.75 2 3.75 5 0.001 000 m3>month750 m3>monthm1: 3.75 2 3.00 5 0.751 000 m3>month

Dx

Df(x)Dx

.

6

5

4

3

2

1

02 3 510 4 6 7

Volu

me

(1000

m3 )

Month

Water UsageV

m

x 5 5.

x 5 5.

x 5 5.30 °F>min

310 2 2506 2 4

5 30

(4, 250).(6, 310)°F>min

280 2 1205 2 0

5 32

(5, 280).(0, 120)125 °F

450

400

350

300

250

200

150

100

50

020 4 6 8 10 12 14

Tem

pera

ture

(8F)

Time (min)

Oven TemperatureT

t


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lb) ; Answers may vary. For example, Thegraph has its vertex at (2, 21). It appears that a tangent line at this point would be horizontal.

4. Use a centred interval.

So the instantaneous rate of change in the glacier’sposition after 20 days is about 0.9 m day.5. Answers may vary. For example,

6. Answers may vary. For example, Find the valueof y for different values of x on both sides of Put this information in a table.

The slope of the tangent line at is about 12.

7. Examine the graph.

The tangent line appears to be passing through thepoints and Use this information tohelp determine the slope of the tangent line.

The slope of the line is 4.8. The instantaneous rate of change of the functionwhose graph is shown is 4 at 9. a) Answers may vary. For example:

The slope is 0.

x 5 2.

5 4

54 2 02 2 1

m 5Df(x)

Dx

(2, 4).(1, 0)

0

42

68

x

y

1 2 3–2–1–2

(2, 9)

x 5 2.

>

d(20.01) 2 d(19.99)20.01 2 19.99

5 0.9.

5 13.991 001d(19.99) 5 0.01(19.99)2 1 0.5(19.99)

5 14.009 001d(20.01) 5 0.01(20.01)2 1 0.5(20.01)

0

105

152025

t

h(t)

1 2 3 4 5–5–1

(2, 21)

(f(2.01) 2 f(1.99))0.02

8 0

t 5 2

1 2 214 2 2

5 210 m>sh(2) 5 21

5 1h(4) 5 25(4)2 1 20(4) 1 1

21 2 12 2 0

5 10 m>s


approximated by slopes of secant lines

instantaneousrate of change

centred intervals

preceding and following

intervals

differencequotient

slope of tangent line

rate of change

average rateof change

secant line

DyDx

Points Slope of Secant

(2, 9) and (1, 2) 7

(2, 9) and (1.5, 4.375) 9.25

(2, 9) and (1.9, 7.859) 11.41

(2, 9) and (3, 28) 19

(2, 9) and (2.5, 16.625) 15.25

(2, 9) and (2.1, 10.261) 12.61

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b)

The slope is 4.c)

The slope is 5.d)

The slope is 8.

Lesson 2.4 Using Rates of Change to Create a Graphical Model, p. #

1. a) Graph a indicates that as time increases,distance also increases; C.b) Graph b indicates that as time increases, distancedecreases; A.c) Graph c indicates that as time increases, thedistance does not change; B.2. Graph a indicates that distance is increasing at asteady rate over time, meaning that the speed isconstant. However, graph b indicates that distance isdecreasing at a steady rate over time—this alsoindicates that the speed is constant. Graph c indicates that distance does not change so speedis 0, a constant. All 3 are constant speed.3. Draw a graph of Jan’s distance from the sensorover time. Jan is 5 m away from the sensor, whichmeans that her initial position is She thenwalks 4 m towards the sensor for 5 seconds, whichmeans that she will be standing 1 m away from thesensor. Her second position will be She thenwalks 3 meters away for 3 seconds, which meansthat she will be 4 m away from the sensor. Her

third position will be Jan then stops andwaits for 2 seconds, which means she stays 4 maway from the sensor for 2 seconds. Her fourthposition will be Use this information todraw the graph.

4. a) Answers may vary. For example, Draw agraph of Rachel’s distance over time while climbingMt. Fuji. Rachel begins the climb at Level 5 and soher initial position is She walks for 40 minutes at a constant rate to move from Level 5 to Level 6, which means that her second positionwill be It then takes he 90 minutes tomove from Level 6 to Level 7, which means thather third position will be Rachel thendecides to rest for 2 hours, which means that herposition does not change. So her fourth position is After her break, it took Rachel 40 minutes to reach Level 8. Her fifth position is

It took Rachel 45 minutes to go fromLevel 8 to Level 9. Her next position is After the walk from Level 9 to Level 10, Rachelreached the top. This position can be represented as

Use this information to plot the graph.

b) Use the data points from the previous question todetermine Rachel’s average speed during each partof her journey.

02000230026002900320035003800

0 100 200 300 400Time (min)

Elev

atio

n(m

)

Rachel’s Climbh

t

(395, 3740).

(335, 3400).(290, 3100).

(250, 2700).

(130, 2700).

(40, 2400).

(0, 2100).

0

5

10

0 5 10Time (s)

Jan’s Walkd

t

(10, 4).

(8, 4).

(5, 1).

(0, 5).


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lc) Answers may vary. For example, use Rachel’saverage rates to make a graph of her speed.During the first 40 minutes of her journey, herspeed was 7.5 m min. This can be represented bya straight line from to Rachel’sspeed during the next 90 minutes is 3.3 m min.This speed can be represented by a straight linefrom to Rachel then rested for2 hours. This can be represented with a straightline from to Rachel travelled ata rate of 10.0 m min for the next 40 minutes. Thisspeed can be represented by a straight line from

to Then she travelled at arate of 6.7 m min for 45 minutes, 4.4 m min for45 minutes, and 9.3 m min for 15 minutes. Thespeeds for these parts or her walk can be repre-sented by the following segments: to

to and to

5. a) Answers may vary. For example, the 2 L plas-tic pop bottle has a uniform shape for the most part.Therefore, as long as the rate of water flowing intothe bottle remains constant, the rate at which theheight is changing will also remain constant.

b) The circumference of the vase changes for anygiven height on the vase. Therefore, the rate ofchange of the height of the water flowing into thevase will vary over time—faster at the very bottomof the vase, slower in the middle and then fasteragain at the top.

6. a) Answers may vary. For example, On a graphthat represents John’s speed, a constant speed wouldbe represented by a straight line, any increase inrate would be represented by a slanted line pointingup, and any decrease in rate would be representedby a slanted line pointing down. John’s speed overhis bike ride could be represented following graph.

b) Answers may vary. For example, The first part ofJohn’s bicycle ride is along a flat road. His heightover this time would be constant. As he travels upthe hill, his height would increase. At the top of thehill, his height would again be constant. As he goesdown the hill, his height would decrease. As he

Time

Spee

d

Speed vs. Times

t

Time

Hei

ght

Water Levelvs. Timeh

t

Time

Hei

ght

Water Levelvs. Time

h

t

0

6

8

2

4

10

0 100 200 300 400 500Time (min)

Spee

d(m

/min

)

Rachel’s Climbs

t

(395, 9.3).(380, 9.3)(380, 4.4),(335, 4.4)(335, 6.7),

(290, 6.7)

> >> (290, 10.0).(250, 10.0)

> (250, 0).(130, 0)

(130, 3.3).(40, 3.3)

>(40, 7.5).(0, 7.5)>

3740 2 3400395 2 335

5 5.7 m>min

3400 2 3100335 2 290

5 6.7 m>min

3100 2 2700290 2 250

5 10.0 m>min

2700 2 2700250 2 130

5 0 m>min

2700 2 2400130 2 40

5 3.3 m>min

2400 2 210040 2 0

5 7.5 m>min


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climbs the second hill his height would againincrease. The graph of his height over time wouldlook something like this.

7. a) Kommy travels 50 m in 45 seconds. This

means that his speed would be m s.

b) During the second part of his swim he travelled50 m in 55 s. This means that his speed would be

m s.

c) The graph of the first length would be steeper,indicating a quicker speed. The graph of the secondlength would be less steep, indicating a slowerspeed.d) Answers may vary. For example, Use the infor-mation for part c) to draw the graph of Kommy’sdistance over time.

e) At Kommy is resting, and so his speedwould be 0.f ) Answers may vary. For example, Kommy’s speedfor the first 45 seconds is 1.11 m s. This would berepresented by the line segment from to

Kommy then rests for 10 s, when hisspeed would be 0. This would be represented by aline segment from to Kommy’sspeed during the second half of his swim is 0.91 m s.This would be represented by a line segment from

to

8. a) A – if the rate at which a speed is increasingincreases, this would be represented by an upwardcurve.b) C – if the rate at which a speed is decreasingdecreases over time, this would be represented by acurve that drops sharply at first and then drops moregradually.c) D – if the rate at which a speed is decreasingincreases, this would be represented by a downwardcurve.d) B – if the rate at which a speed is increasingdecreases, this would be represented by a curve thatrises sharply at first and then rises more gradually.9. Answers may vary. For example, Because the jock-ey is changing the horse’s speed at a non-constantrate—at first slowly and then more quickly—the lineswill have an upward curve when the horse is acceler-ating and a downward curve when decelerating. Thehorse’s speed during the first part of the warm up isconstant, which would be represented by a straightline. She then increases the horse’s speed to a canterand keeps this rate for a while. Draw a graph of thisinformation with speed over time.

10. a) Graph i) shows that distance is decreasing andthen increasing. The first graph shows a personstanding 5 m away from the motion sensor thenmoving to 2 m away. The person then moves back to5 m away from the motion sensor. The person is

Spee

d

s

t

Speed vs. Time

Time

Spee

d(m

/s)

s

t

Speed vs. Time

0

0.5

1.0

1.5

20

0 20 40 60 80 100Time (s)

(110, 0.91).(55, 0.91)

>(55, 0).(45, 0)

(45, 1.11).(0, 1.11)

>t 5 50,

Dis

tanc

e(m

)

d

t

Distance vs. Time

0

20

40

60

80

100

0 20 40 60 80 100Time (s)

>5055

5 0.91

>5045

5 1.11

Time

Hei

ght

Height vs. Timeh

t


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always moving at a constant rate. Graph ii) shows aperson’s initial position being 6 m away from themotion sensor. This person then moves 2 m closer tothe sensor over 2 seconds. Then, he or she rests for asecond and then moves 2 m closer to the sensor over2 more seconds. Finally, this person moves 2 maway from the sensor over 1 second to end up at afinal position of about 4 m away from the sensor.The person is always moving at a constant rate.b) For each graph, determine the point foreach position.Graph A

Graph B

Use these points to find the various speeds.Graph A

so the speed is 1 m s


Graph B




so the speed is 1.5 m s

11. a) Draw a graph of the runner’s speed overtime. The runners positions on the graph will berepresented by the following points:

Plot the points on a graph. Because the runneraccelerates and decelerates at a constant rate, thelines will always be straight.

b) Use the data points on either side of toestimate the instantaneous rate of change at thatpoint. The points are

mi h min

c) The runner’s speed at minute 11 is 10 miles per hour.The runner’s speed at minute 49 is 3 miles per hour.

miles per hour per minute

d) From minute 11 to minute 49, the runnerdecreases his speed by 0.1842 miles per hour eachminute. This is significantly different from what hisprogram requires. His program requires him toincrease and decrease his speed quickly, but theaverage rate of change indicates that he is slowingdown at a slow rate over a long period of time.12. Answers may vary. For example: Walk from

to and stop for 5 s. Then run toContinue walking to and end at

What is the maximum speed and the minimum speed on any interval? Create the speedtime graph from this data.13. Answers may vary. For example: Graphing bothwomen’s speeds on the same graph would meanthat there are two lines on the graph. The firstwoman is decelerating; this means that her linewould have a downward direction. Because she isdecelerating slowly first and then more quickly, theline would also have a downward curve. The secondwoman is accelerating; this means that he line willhave an upward direction. Because she is accelerat-ing quickly at first and then more slowly, the graphwould have a sharp upward curve. The line on thegraph would look something like this:

14. If the original graph showed an increase in rate, itwould mean that the distance travelled during eachsuccessive unit of time would be greater—meaning agraph that curves upward. If the original graphshowed a straight, horizontal line, then it would meanthat the distance travelled during each successive unit

Spee

d

s

t

Speed vs. Time

Time

(25, 0).(25, 5)(15, 30).

(5, 5)(0, 0)

3 2 1049 2 11

52738

5 20.1842

>>10 2 511 2 10

5 5

(11, 10).(10, 5),

t 5 10.5Sp

eed

(mph

)

s

t

Time (min)

2

0

4

6

8

10

0 10 20 30 40 50 60

(16, 10)

(17, 7)(47, 7)

(11, 10)(10, 5)

(0, 5)(49, 3)

(59, 3)

Marathon Training Program

(59, 3).(49, 3),(47, 7),(17, 7),(16, 10),(11, 10),(10, 5),(0, 5),

>3.5 2 26 2 5

5 1.5,

>2 2 45 2 3

5 21,

>4 2 43 2 2

5 0,

>4 2 62 2 0

5 22,

>5 2 26 2 3

5 1,

>2 2 53 2 0

5 21,

(6, 3.5)(5, 2);(3, 4);(2, 4);(0, 6);

(6, 5)(3, 2);(0, 5);

(t, d)


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of time would be greater-meaning a steady increasingstraight line on the second graph. If the originalgraph showed a decrease in rate, it would mean thatthe distance travelled during each successive unite oftime would be less-meaning a line that curves down.

Lesson 2.5 Solving Problems Involving Rates of Change, p. #

1. Answers may vary. For example, Verify that themost economical production level occurs when1500 items are produced by examining the rate ofchange at Because x is in thousands, use

Use the difference quotient to find theinstantaneous rate of change.


When 1500 items are produced, the instantaneous rateof change is zero. Therefore, the most economical pro-duction level occurs when 1500 items are produced.2. The function is Use the difference quotient to find the instantaneousrate of change.


The blood pressure is dropping at a rate of 0 millimeters of mercury per second.3. a) If is a maximum, then the points tothe left of, and very close to the maximum have apositive rate of change. As approaches

from the left, is increasingbecause is a maximum.b) If is a maximum, then the points to theright of, and very close to the maximum have nega-tive rate of change. As moves away from

to the right, is decreasing becauseis a maximum.

4. a) If is a minimum, then the points tothe left of, and very close to the maximum havenegative rate of change. As moves toward

from the left, is decreasingbecause is a minimum.b) If is a minimum, then the points to theright of, and very close to the maximum have apositive rate of change. As moves away from

towards the right, is increasingbecause is a maximum.5. a) The leading coefficient is positive, and so thevalue given will be a minimum. Use the differencequotient to find the instantaneous rate of change.


Find and The function is

The slope is very small, pretty close to zero, and soit can be assumed that is the minimum.b) The leading coefficient is negative, and so thevalue given will be a maximum. Use the differencequotient to find the instantaneous rate of change.


Find and The function is

The number is very close to zero, and so we canassume that the point has an instantaneous rate ofchange is zero and is a maximum.c) The function is Use the differencequotient to find the instantaneous rate of change.

where h is a very small

value. Find and

5 4.999 999f(90.01°) 5 5 sin(90.01°)

f(90.01°).f(90°)

f(a 1 h) 2 f(a)h

,

f(x) 5 5 sin(x).

10.499 994 2 10.50.01

5 20.000 6 or 0

5 10.5f(0.5) 5 26(0.5)2 1 6(0.5) 1 9

5 10.499 994f(0.501) 5 26(0.501)2 1 6(0.501) 1 9

6x 1 9.f(x) 5 26x2 1f(0.501).f(0.5)

f(a 1 h) 2 f(a)h

,

(26, 210.5)

10.5 2 (210.499 995)0.01

5 20.005 or 0

5 210.499 995f(25.99) 5 0.5(25.99)2 1 6(25.99) 1 7.5

5 210.5f(26) 5 0.5(26)2 1 6(26) 1 7.5

6x 1 7.5.f(x) 5 0.5x2 1f(25.99).f(26)

f(a 1 h) 2 f(a)h

,

(a, f(a))y(f(a))(a, f(a))x(a)

(a, f(a))(a, f(a))

y(f(a))(a, f(a))x(a)

(a, f(a))

(a, f(a))y(f(a))(a, f(a))

x(a)

(a, f(a))(a, f(a))

y(f(a))(a, f(a))x(a)

(a, f(a))

119.999 73 2 1200.001

5 0.27 or 0

5 119.999 73P(3.01) 5 220 cos(300° 3 3.001) 1 100

5 120P(3) 5 220 cos(300° 3 3) 1 100

f(a 1 h) 2 f(a)h

,

P(t) 5 220 cos(300°t) 1 100.

1.000 000 3 2 10.01

5 0.000 03

5 1.000 000 3C(1.501) 5 0.3(1.501)2 2 0.9(1.501) 1 1.675

5 1C(1.5) 5 0.3(1.5)2 2 0.9(1.5) 1 1.675

f(a 1 h) 2 f(a)h

,

x 5 1.5.x 5 1500.


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lThe number is very close to zero, and so we canassume that the instantaneous rate of change at thepoint is zero, and so the point is a maximum.d) The function is Use the dif-ference quotient to find the instantaneous rate of change.


Find and

The number is very close to zero, and so we canassume that the instantaneous rate of change at thepoint is zero, and so the point is a maximum.6. Examine the instantaneous rates of change oneither side of the point in question. If the point to theleft of the point in question is negative, then the pointis a minimum. If the point to the left of the point inquestion is positive, then the point is a maximum. Ifthe point to the right of the point in question is posi-tive, then the point is a minimum. If the point to theright of the point in question is negative, then thepoint is a maximum. Use the difference quotient tofind the instantaneous rate of change.


a)Examine which is to the left of

The instantaneous rate of change of is nega-tive and so is a minimum.b)Examine which is to the right of

The instantaneous rate of change of is and so is a maximum.c)Examine which is to the right of

The instantaneous rate of change at ispositive and so is a minimum.d)Examine which is to the left of

The instantaneous rate of change at ispositive, and so is a maximum.e) Examine which is to the right of

The instantaneous rate of change at isand so is a maximum.

f) Examine which is to the left of

The instantaneous rate of change at is 8.97,and so is a maximum.(2, 15)

(1, 10)

10.0897 2 100.01

5 8.97

5 10f(1) 5 2 (1)3 1 12(1) 2 1

5 10.0897f(1.01) 5 2 (1.01)3 1 12(1.01) 2 1

(2, 15).x 5 1,(2, 15)f(x) 5 2x3 1 12x 2 1;

(21, 2)22.9999,(0, 0)

20.029 999 2 00.01

5 22.9999

5 0f(0) 5 (0)3 2 3(0)

5 20.029 999f(0.01) 5 (0.01)3 2 3(0.01)

(21, 2).x 5 0,(21, 2)f(x) 5 x3 2 3x;

(0°, 3)(21°, 2.99)

2.999 55 2 2.999 540.01

5 0.001

5 2.999 54f(21°) 5 3 cos(21°)

5 2.999 55f(20.99°) 5 3 cos(20.99°)

(0°, 3).x 5 21°,(0°, 3)f(x) 5 3 cos x;

(4.5, 220.25)(5, 220)

219.899 2 (220)0.01

5 10.1

5 220f(5) 5 (5)2 2 9(5)

5 219.899f(5.01) 5 (5.01)2 2 9(5.01)

(4.5, 220.25).x 5 5,

(4.5, 220.25)f(x) 5 x2 2 9x;(26, 41.75)22.01,

(25, 40.75)

40.7299 2 40.750.01

5 22.01

5 40.75f(25) 5 2 (25)2 2 12(25) 1 5.75

5 40.7299f(24.99) 5 2 (24.99)2 2 12(24.99) 1 5.75(26, 41.75).

x 5 25,(26, 41.75)f(x) 5 2x2 2 12x 1 5.75;

(2, 1)(1, 2)

1.97 2 20.01

5 23

5 2f(1) 5 (1)2 2 4(1) 1 5

5 1.97f(1.01) 5 (1.01)2 2 4(1.01) 1 5

(2, 1).x 5 1,(2, 1)f(x) 5 x2 2 4x 1 5;

f(a 1 h) 2 f(a)h

,

24.499 999 2 (24.5)0.01

5 0.0001 or 0

5 24.5f(0) 5 24.5 cos(2 ? 0)°

5 24.499 999f(0.01) 5 24.5 cos(2 ? 0.01)°

f(0.01).f(0)

f(a 1 h) 2 f(a)h

,

f(x) 5 24.5 cos(2x).

4.999 999 2 50.01

5 20.0001 or 0

5 5f(90°) 5 5 sin(90°)


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7. Use a table to inspect several values of

The height is definitely decreasing after butfrom this data the exact maximum can’t be deter-mined. Examine other values of t to help determinethe maximum.

The maximum appears to be pretty close to 3. 2.8would be a good guess. Verify this using thedifference quotient.


The instantaneous rate of change at isapproximately zero.8. a) i)

The minimum is at approximately

ii)

The maximum is at iii)

The maximum is at approximately iv)

The maximum is at b) i)

ii)

iii)

x 5 6.

x 5 3.25.

x 5 7.6.

x 5 25.

x 5 2.8

10 125.636 054 2 10 125.6360.01

5 0.0054

5 10 125.6361 10 0001 89.67(2.8)f(2.8) 5 216(2.8)2

5 10 125.636 054f(2.801) 5 216(2.801)2 1 89.67(2.801) 1 10 000

f(a 1 h) 2 f(a)h

,

t h(t)

1 10 073.67

2 10 115.34

3 10 125.01

4 10 102.68

5 10 048.35

6 9962.02

t 5 15,

t h(t)

0 10 000

5 10 048.35

15 7745.05

18 6430.06

20 5393.4

25 2241.75

h(t).


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iv)

c) Answers may vary. For example, if and then there is a maximum at x. If

and then there is a mini-mum at x or on either side of the point, the sign ofthe slope of the tangent line changed from either pos-itive to negative (maximum) or negative to positive (minimum).9. a) i) Examine the graph of the equation.

The maximum for the interval appears tobe at or The minimum appears tooccur at or This cannot be verifiedwith the difference quotient because the graph willalways be decreasing. This means that the instanta-neous rate of change for any point on the graph willalways be negative and never be zero.ii) Examine the graph of the function.

b) The minimum appears to be at or and the maximum at or Thiscannot be verified with the difference quotientbecause the graph will always be increasing. Thismeans that the instantaneous rate of change for anypoint on the graph will always be positive and neverbe zero.

10. Answers may vary. For example: examine pointson either side of s to make sure that thediver’s height is increasing before the point anddecreasing afterwards.

The slope to the right of the point is positive.

The function is increasing up to 0.5 s and decreas-ing after 0.5 s––the point is a maximum.11. Answers may vary. For example, yes, this obser-vation is correct. The slope of the tangent at 1.5 s is 0.The slopes of the tangents between 1 s and 1.5 s arenegative, and the slopes of the tangent lines between1.5 s and 2 s are positive. So the minimum of thefunction occurs at 1.5 s.12. Answers may vary. For example, estimate theslope of the tangent line to the curve when bywriting an equation for the slope of an secant lineon the graph if If the slope of the tangent is 0,this will confirm there may be a maximum at If the slopes of tangent lines to the left are positiveand the slopes of tangent lines to the right are nega-tive, this will confirm that a maximum occurs at

13. Answers may vary. For example, Becausegives a maximum value of 1, I know that a

maximum occurs when Solvingthis equation for x will tell me what types of x val-ues will give a maxim. For example, when and

14. Myra is plotting (instantaneous) velocity versustime. The rates of change Myra calculates representacceleration. When Myra’s graph is increasing, thecar is accelerating. When Myra’s graph is decreas-ing, the car is decelerating. When Myra’s graph is

x 5 48°(x 2 3°) 5 45°

(2(x 2 3°)) 5 90°d 5 3,

k 2 2

(k(x 2 d)) 5 90°.sin 90°

x 5 5.

x 5 5.R(x).

x 5 5

11.249 5 2 11.250.01

5 20.05

h(0.5) 5 11.255 11.249 5

h(0.51) 5 25(0.51)2 1 5(0.51) 1 10

11.25 2 11.249 50.01

5 0.05

5 11.25h(0.5) 5 25(0.5)2 1 5(0.5) 1 10

5 11.249 5h(0.49) 5 25(0.49)2 1 5(0.49) 1 10

t 5 0.5

(10, 141.6).x 5 10(0, 35)x 5 0

y

x

15 250

5025

75100125150

5–5–15–25

(5, 44.4).t 5 5(0, 100).x 5 0

0 # t # 5

24 320

25

50y

x

8 16

f(b) . f(x)f(a) . f(x)f(b) , f(x)

f(a) , f(x)


Draft

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l

constant, the velocity of the car is constant; the caris neither accelerating nor decelerating.15. Choose a method and determine the instanta-neous rates of change for the points given. Usetables to examine the relationship between x and theinstantaneous rate of change at x.

The instantaneous rate of change appears to be 2 times the x-coordinate or 2x. Now use a table toexamine the relationship between the points givenand their instantaneous rates of change for the func-tion

The instantaneous rate of change appears to be 3 times square of the x-coordinate or 3x2.

Chapter Review, p. #1. a) Examine the rate of change between eachinterval. If the rate of change is the same for eachinterval, then the data follows a linear relation.

The slope between each interval is the same, and so the relation is linear.

b) Answers may vary. For example,

The graph appears to be linear, and so it wouldappear that my hypothesis is correct.c) The average rate of change from to

d) The cost of one watch is $17.50; this is the slopeof the line on the graph.2. a) Calculate the average rate of change for the inter-val The second point is ; the first is

b) Calculate the average rate of change for theinterval The second point is The firstpoint is

c) The time intervals have the same length. Theamount of change is the same, but with oppositesigns for the two intervals. So the rates of changeare the same for the two intervals, but with oppositesigns.3. a) The company spends $2500 per month inexpenses—this can be represented by 2500m. Theinitial expenses were 10 000. The whole equation is

b) Find the expenses for and 2500(6) 1 10 000 5 25 000

m 5 3.m 5 6E 5 2500m 1 10 000.

1 2 78 2 4

5 21.5 m>s(4, 7).

(8, 1).34, 84.

7 2 14 2 0

5 1.5 m>s(0, 1).(4, 7)30, 44.

420.00 2 350.0025 2 20

5 17.5 per watch

w 5 25.w 5 20

300

350

400

450

500

250

200

150

100

50

050 10 15 20 25 30

Revenue vs. Sales

Reve

nue

($)

Number of watches

w

r

420.00 2 210.0024 2 12

5 17.5

210.00 2 350.0012 2 20

5 17.5

350.00 2 297.5020 2 17

5 17.5

297.50 2 437.5017 2 25

5 17.5

x Rate of Change

22 12

21 3

2 12

3 27

f(x) 5 x3.

x Rate of Change

22 24

21 22

2 4

3 6


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The average rate of change is $2500 per month.c) No, the equation that represents this situation islinear, and the rate of change over time for a linearequation is constant.4. a) Answers may vary. For example, Because theunit of the equation is years, do not choose

and A better choice would beand

b) Answers may vary. For example, The equation isFind and

5. a) Answers may vary. For example: squeezing theinterval.b) Squeezing the interval will be a good method. Usethe interval The equation is

Now use the interval

6. For each point, draw a line tangent to the graphat the point given.a)

The slope of the line appears to be

b)

The slope of the line appears to be 0.c)

The slope of the line appears to be 4.7. Graph the original equation. Find the correspon-ding y for each value of x given. Use this informa-tion to draw a tangent line to the original graph witha graphing calculator.a)

The slope of the line, and therefore the instanta-neous rate of change at is b)

The slope of the line, and therefore the instanta-neous rate of change at is

c)

The slope of the line, and therefore the instanta-neous rate of change at is 0.x 5 20.3,

217.x 5 22,

237.x 5 24,

6 80

42

68

y

x

2 4

0

42

68

y

x2 4 6 8

22.

0

42

68y

x

2 4 6 8

0.004 19 2 (20.004 19)12.001 2 11.999

5 4.19 cm>s2sin(120°(12.001)) 5 0.004 192sin(120°(11.999)) 5 20.004 19

11.999 # t # 12.001.

0.0419 2 (20.0419)12.01 2 11.99

5 4.19 cm>s2sin(120°(12.01)) 5 0.04192sin(120°(11.99)) 5 20.0419

y 5 2sin(120°t).11.99 # t # 12.01.

600.558 280 1 621.912 9762

5 611.24

4372.515 625 – 4222.376 0554.0 – 3.75

5 600.558 280

V(4.0) 5 4372.515 625V(3.75) 5 2500(1.15)3.75 5 4222.376 055

4527.993 869 2 4372.515 6254.25 2 4.0

5 621.912 976

5 4427.993 869V(4.25) 5 2500(1.15)4.25

5 4372.515 625V(4.0) 5 2500(1.15)4.0

V(4.2).V(4.0)V(t) 5 2500(1.15)t.

4.0 # t # 4.25.3.75 # t # 4.04 # t # 5.3 # t # 4

25 000 2 17 5006 2 3

5 2500

2500(3) 1 10 000 5 17 500


Draft

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d)

The slope of the line, and therefore the instanta-neous rate of change at is 23.

8.

9. a) Answers may vary. For example:

b) Find the average rate of change in the bicyclerider’s speed on the interval The speed at

was 10 km h. The speed at was 5 km h.

The average speed is

c) From to the rate of change ofspeed is d) The speed is decreasing at a constant rate from

to So find the average rate of changeon any interval between those two numbers and itwill be the same as the instantaneous rate of changeat

10. The roller coaster moves at a slow steady speedbetween A and B. At B it begins to accelerate as itmoves down to C. Going uphill from C to D it decel-erates. At D it starts to move down and accelerates toE, where the speed starts to decrease until, where itmaintains a slower speed to G, the end of the track.

11. Graph each function using a graphing calculatorto determine whether each function is a maximumor a minimum.a)

The graph shows that is a minimum.b)

The graph shows that is a maximum.c)

The graph shows that is a maximum.d)

The graph shows that is a minimum.(45°, 21)

(17, 653)

(23, 5)

(5, 218)

Spee

d

Time

s

tA B

C

D

E

F G

0 2 516 2 10

5 20.83 km>h>sx 5 12.

t 5 16.t 5 10

20.33 km>h.(12, 3.33),(7, 5)

5 2 107 2 0

5257

.

>t 5 7>t 5 00 # t # 7.

Spee

d(k

m/

h)

Speed vs. Time

0

2

4

6

8

10

0 4 8 12 16Time (s)

s

t

50

40

30

20

10

020 4 6 8 10

Hei

ght

(cm

)

Time (s)

h

t

Height vs. Time

x 5 2,


Draft

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e)

The graph shows that is a minimum.f)

The graph shows that is a maximum.

12. a)i)

The slope is ii)

b) For each of the points given, the value of hwould be equal to zero. Substitute 0 in for h to findthe instantaneous rate of change for each point.i)ii)13. a) To the left of a maximum, the instantaneousrates of change are positive. To the right, the instan-taneous rates of change are negative.b) To the left of a minimum, the instantaneous ratesof change are negative. To the right, the instanta-neous rates of change are positive.

Chapter Self-Test, p. #1. a)

b) At s the speed is approximately 25 knots.At the speed is approximately 3 knots.

At the boats speed is 25 knots.

The two different average rates of change indicatethat the boat was increasing its speed from to

at a rate of 11 kn min and moving at a con-stant speed from to c) Because the rate of change is constant over theinterval, the instantaneous rate of change at would be the same as it was over the interval

11 kn min.2. a) The slope of the secant line between (5, 70)

and (50, 25) would be

b) The hot cocoa is cooling by on average.c) Examine the graph to and draw a line tangent tothe graph at the point (30, 35).

1 °C>min

25 2 7050 2 5

5 21.

>6 # t # 8.

t 5 7

t 5 13.t 5 8>t 5 8

t 5 6

25 2 2513 2 8

5 0 kn>min

t 5 13

25 2 38 2 6

5 11 kn>min

t 5 6t 5 8

30

35

40

45

50

25

20

15

10

5

050 10 15 20 25

Speed vs. Time

Spee

d(k

nots

)Time (min)

t

s

(23, 0)

(19, 45)(14, 45)

(13, 25)

(8, 25)

(6, 3)(0, 0)

(1, 3)

m 5 24(0) 2 48 5 248m 5 0 2 26 5 226

4h2 2 48h 1 68 2 6821 1 h 2 (21)

5 24h 2 48

5 685 24 1 56 1 16

g(21) 5 24(21)2 2 56(21) 1 165 24h2 2 48h 1 685 24 1 8h 2 4h2 1 56 2 56h 1 165 24(1 2 2h 1 h2) 1 56 2 56h 1 16

g(21 1 h) 5 24(21 1 h)2 2 56(21 1 h) 1 16a 5 21g(x) 5 24x2 2 56x 1 16;

m 5 h 2 26.

256 2 26h 1 h2 2 (256)2 1 h 2 2

5 h 2 26

5 256f(2) 5 (2)2 2 30(2)

5 256 2 26h 1 h25 22 1 2(2)h 1 h2 2 30(2) 2 30h

f(2 1 h) 5 (2 1 h)2 2 30(2 1 h)f(x) 5 x2 2 30x

a23,95b

(225°, 24)


Draft

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lThe slope of the tangent line is d) The hot cocoa is cooling by after 30 min.e) The rate of decrease decreases over the interval,until it is nearly 0 and constant.3. a) Calculate both and

The average rate of change is $310 per dollar spent.b) Use the different quotient to estimate the instan-taneous rate of change.

The instantaneous rate of change is approximatelyper dollar spent.

c) The sign for part a) means that the company isincreasing its profit when it spends between $8000and $10 000 on advertising. The negative signmeans that the company’s profit is decreasing whenit spends $50 000 on advertising.

4. Use a graphing calculator to help estimate theinstantaneous rate of change at each of the pointsgiven.a)

The instantaneous rate of change when is The instantaneous rate of change when is 0. The point is a minimum.The instantaneous rate of change when is 7.b)

The instantaneous rate of change when is 4.5.The instantaneous rate of change when is The instantaneous rate of change when is 0.This point is a maximum.

x 5 124.5.

x 5 4

x 5 22

k(x)

–4

1214

6

x

3 4 50

42

–2–4

68

10

1–1–2–3–5–6 2

p 5 1

p 5 20.7521.

p 5 21

h(p)

–1

p

3 4 50

21

–1–2–3

34567

1–2–3–4–5 2

2$100

4948.9995 2 49500.01

5 2100.05

5 4950P(50) 5 25(50)2 1 400(50) 2 2550

5 4948.9995P(50.01) 5 25(50.01)2 1 400(50.01) 2 2550

950 2 33010 2 8

5 310

5 330P(8) 5 25(8)2 1 400(8) 2 2550

5 950P(10) 5 25(10)2 1 400(10) 2 2550

P(8).P(10)

0.75 °C>min20.75.

Tem

pera

ture

(deg

rees

cels

ius)

Time (min)

90

80

70

60

50

40

30

20

10

0200 40 60 80 100


d material draft - nelson€¦ · the rate of change from 2 to 26 is: the function is linear and so...

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