chapter 3 of applied calc

It is impossible to determine how quickly a person is running from a single photograph, since speed is calculated as a change in distance over a change in time. Nevertheless, we may estimate a person’s speed at a particular instant in time by determining the distance traveled over a small interval of time (e.g., one second). A runner’s speed may be classi-fi ed as a rate of change in distance. A key compo-nent of calculus is the study of rates of change.

3.1 Average Rate of Change

Colleges and universities periodically raise their

tuition rates in order to cover rising staffi ng and

facilities costs. As a result, it is often diffi cult for

students to know how much money they should

save to cover future tuition costs. By calculating

the average rate of change in the tuition price

over a period of years, we can estimate projected

increases in tuition costs. In this section, we will

demonstrate how to calculate the average rate of

3.1 Average Rate of Change

3.2 Limits and Instantaneous Rates of Change

3.3 The Derivative as a Slope: Graphical Methods

3.4 The Derivative as a Function: Algebraic Method

3.5 Interpreting the Derivative

Chapter 3

The Derivative

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CH

AP

TE

R

change in the value of a

function over a specifi ed

interval [a, b]. (The interval

notation [a, b] is equivalent

to a # x # b.)

In calculating the difference quo-

tient, we answer the question, “Over

the interval 3a, b 4, on average, how

much does a one-unit increase in the

x-value change the y-value of the

function?”

The Difference Quotient: An Average Rate of Change

The average rate of change of a function y 5 f 1x 2 over an interval 3a, b 4 is

f 1b 2 2 f 1a 2b 2 a

This expression is referred to as the diff erence quotient. For a linear function, the diff erence quotient gives the slope of the line.

3


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48 C h a p t e r 3 : T h e D e r i v a t i v e

EXAMPLE 1 Calculating an Average Rate

of Change from a Table

DThe annual cost of tuition and fees for full-time, resident, undergraduate students majoring in

the Arts and Sciences at the University of Pittsburgh is shown in Table 3.1.

What is the average rate of change in the tuition and fees

from the 2004–2005 academic year to the 2009–2010

academic year? Rounded to the nearest dollar, what do

you estimate the 2012–2013 tuition and fees will be?

SOLUTION

The average rate of change may be calculated by using the difference quotient formula. For the period 2004–2005 to 2009–2010, the interval 3a, b 4 5 30, 5 4. The average rate of change of the tuition is

f 1b 2 2 f 1a 2b 2 a

5f 15 2 2 f 10 2

5 2 0

513,344 2 10,130 dollars

5 years

53214 dollars

5 years

5 642.8 dollars per year

Over the fi ve-year period between 2004–2005 and

2009–2010, tuition increased by $3,214. Although the

annual increase varied from year to year, the average

annual increase was $642.80.

The 2009–2010 tuition and fees cost was $13,344.

We predict that tuition and fees will increase by $642.80

per year in subsequent years. To predict the 2012–2013

tuition and fees cost, we repeatedly increase the annual

cost by $642.80.

2010–2011: 13,344 1 642.80 5 13,986.80

2011–2012: 13,986.80 1 642.80 5 14,629.60

2012–2013: 14,629.60 1 642.80 5 15,272.40

We estimate that the 2012–2013 tuition and fees cost

will be $15,272.40. This value may be calculated more

quickly as follows: 13,344 1 3 1642.80 2 5 15,272.40.

When determining the meaning of an average rate

of change in a real-life problem, it is essential to fi nd

the units of measurement of the result. Fortunately,

the units are easily determined. The units of the rate of

change are the units of the output divided by the units

of the input. In Example 1, the units of the output were

dollars and the units of the input were years. Conse-

quently, the units of the average rate of change were

dollars divided by years, or dollars per year.

EXAMPLE 2 Calculating an Average Rate

of Change from an Equation

DBased on data from 1990–2009, the popula-tion of Washington state may be modeled by

the function P 1 t 2 5 4.933 11.017 2 t, where P is the population in millions of people and t is the num-ber of years since 1990. (Source: Modeled from data at www.census.gov) According to the model, what is the average rate of change in the population between 1990 and 2010?

SOLUTION

We observe that in the model, t is the number of years since 1990. So for the model, t 5 0 represents 1990 and t 5 20 represents 2010. The average rate of change in the population over the interval 30, 20 4 is

P 120 2 2 P 10 2

20 2 05

6.911 2 4.933 million people

20 years

5 0.0989 million people per year

5 98,900 people per year

Between 1990 and 2010, the population of Washington

state increased by an average of 98,900 people per year.

Graphical Interpretation of the Difference QuotientA line connecting any two points on a graph is referred

to as a secant line. Graphically speaking, the differ-

ence quotient for a function y 5 f 1x 2 is the slope of

the secant line connecting (a, f(a)) and 1b, f 1b 2 2 (Fig-

ure 3.1).

Table 3.1

Years (since

2004–2005)

t

Annual Tuition

and Fees (dollars)

f(t)

Change in Tuition

from Prior Year

(dollars)

0 10,130

1 10,736 606

2 11,368 632

3 12,106 738

4 12,832 726

5 13,344 512

Source: University of Pittsburgh


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493 . 1 A v e r a g e R a t e o f C h a n g e

EXAMPLE 3 Finding the Slope

of a Secant Line

DThe graph of the function f 1x 2 5 3x2 2 6x 1 5is shown in Figure 3.2. Calculate the slope of the

secant line of f that passes through (1, 2) and (2, 5).

Figure 3.1 Secant line slope 5f 1b 2 2 f 1a 2

b 2 a

y

x

(a, f (a))

f (b) − f (a)

b − a

(b, f (b))f (b)

Secant line

f (a)

a b

y = f (x)

Figure 3.2

y

x0 1−1

1

2

3

4

5

6

7

8

9

11

10

2

y = 3x2 − 6x + 5

Figure 3.3

Secant line

y

x0 1−1

1

2

3

4

5

6

7

8

9

11

10

2

(2, 5)

(1, 2)

y = 3x2 − 6x + 5

This line is the secant line of the graph between

x 5 1 and x 5 2. The slope of the secant line is given by

m 5f 1b 2 2 f 1a 2

b 2 a

5f 12 2 2 f 11 2

2 2 1

55 2 2

1

5 3

The slope of the secant line is 3. That is, on the interval

[1, 2], a one-unit increase in x results in a three-unit

increase in y, on average.

3.1 Exercises

In Exercises 1–5, calculate the average rate of change of the function over the given interval.

1. f 1x 2 5 2x 2 5 over the interval 33, 5 4 2. v 1x 2 5 2x3 over the interval 321, 1 4 3. v 1m 2 5 m2 2 m over the interval 323, 4 4 4. z 5

ln 1x 2x

over the interval 31, 5 4 5. q 1x 2 5 "x 1 2 over the interval 30, 6 4

SOLUTION

We fi rst plot the points (1, 2) and (2, 5) and draw the line connecting them (Figure 3.3).


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In Exercises 6–10, calculate the average rate of change in the designated quantity over the given interval(s).

6. Air Temperature Temperature between 11:00 A.M. and 3:00 P.M.

Time of Day Temperature (°F)

11:00 A.M. 68

1:00 P.M. 73

3:00 P.M. 75

7. Dow Jones Industrial Average Dow Jones Industrial Average between 2000 and 2007 and between 2002 and 2008.

Year

Dow Jones Industrial

Average Closing Value at

the End of the Year (points)

2000 10,787

2002 8342

2007 13,265

2008 8776

Source: www.census.gov

8. Reading Scores A third-grade student’s reading score between the fi rst and third quarter.

Quarter

Reading Score

(words per minute)

1 69

2 107

3 129

Source: Author’s data

9. Newspaper Subscriptions Daily newspaper subscriptions as the number of cable TV subscribers increased from 50.5 million (in 1990) to 67.7 million (in 2000).

Cable TV Subscribers

(millions)

Daily Newspaper Circulation

(millions)

50.5 62.3

60.9 58.2

66.7 56.0

67.7 55.8


10. Poverty Level Percentage of people below poverty level when the unemployment rate decreased from 5.6 percent (in 1990) to 4.0 percent (in 2000).

Unemployment Rate

(percentage)

People below Poverty Level

(percentage)

5.6 13.5

4.2 11.8

4.0 11.3


In Exercises 11–13, graph each function. Then use the

difference quotient, f 1b 2 2 f 1a 2

b 2 a, to calculate the slope of

the secant line through the points 11, f 11 2 2 and 13, f 13 2 2 . 11. f 1x 2 5 2x

12. f 1x 2 5 5

13. f 1x 2 5 1x 2 2 22In Exercises 14 and 15, use the difference quotient,

f 1b 2 2 f 1a 2b 2 a

, to calculate the slope of the secant line

through the points (1, f(1)) and (3, f(3)) for the given graph of f.

14.

x2 310

y

1

2

3

4

0

y = f (x)

15.

x2 310

y

3

2

1

−1

y = f (x)


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513 . 2 L i m i t s a n d I n s t a n t a n e o u s R a t e s o f C h a n g e

In Exercises 16–20, calculate the average rate of change in the state populations (in people per year) between 2000 and 2009.

16. Population

Montana

Date Population (in thousands)

2000 902.2

2009 975.0


17. Population

Massachusetts


2000 6349.1

2009 6593.6


18. Population

Missouri


2000 5595.2

2009 5987.6


19. Population

West Virginia


2000 1808.3

2009 1819.8


20. Population

Pennsylvania


2000 12,281.1

2009 12,604.8


3.2 Limits and Instantaneous Rates of Change

In the 2008 Olympic games in Beijing,

Usain Bolt set the world record for the

100-meter dash with a time of 9.69

seconds. How fast was he running (in

meters per second) when the accompa-

nying picture was taken?

From a single photo, it is impossible for us to deter-

mine his speed. However, if we knew how long it took

him to reach various checkpoints during the race, we

could approximate his speed at the fi nish line. In this

section, we will demonstrate how to use the difference

Nicolas Asfouri/AFP/Getty Images


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quotient to estimate and the derivative to calculate an

instantaneous rate of change.Whereas an average rate of change is calculated

over an interval 3a, b 4, an instantaneous rate of change

is calculated at a single value a. For example, the aver-

age highway speed of a person over a 200-mile trip may

have been 59 miles per hour. However, when he passed

a state patrol car exactly 124 miles into the trip, he was

speeding at 84 miles per hour. His average speed on the

interval 30, 200 4 was 59 miles per hour; however, his

instantaneous speed at d 5 124 was 84 miles per hour.

EXAMPLE 1 Estimating an Instantaneous

Rate of Change

DSuppose that a runner in the 100-meter dash recorded the times shown in Table 3.2.

His average speed over the last 10 meters was

Average speed 5100 2 90 meters

9.69 2 8.70 seconds

510 meters

0.99 seconds

5 10.10 meters per second



9.69 2 9.18 seconds

55 meters

0.51 second


And his average speed over the last meter was


9.69 2 9.60 seconds

51 meter

0.09 second


Although each calculation yielded a different result,

all of these difference quotients estimate the runner’s

fi nish-line speed. Which of the estimates do you think

is most accurate?

The last calculation best estimates his fi nish-line

speed because it measures the change in distance over

the smallest interval of time: 0.1 second. (Reducing the

time interval to an even smaller amount of time, say

0.01 second, would further improve the estimate.) We

estimate that the runner’s speed when he crossed the

fi nish line was 11.11 meters per second.

To estimate the instantaneous rate of change of a

function y 5 f sxd at a point (a, f(a)), we calculate the

average rate of change of the function over a very small

interval [a, b]. If we let the variable h represent the dis-

tance between x 5 a and x 5 b, then b 5 a 1 h. Con-

sequently, the difference quotient may be rewritten as

f 1b 2 2 f 1a 2

b 2 a5

f 1a 1 h 2 2 f 1a 21a 1 h 2 2 a

5f 1a 1 h 2 2 f 1a 2

h

Table 3.2

Time (seconds)

(t)

Total Distance

Traveled (meters)

(D(t))

0 0

4.85 50

8.70 90

9.18 95

9.60 99

9.69 100

Estimate his speed when he crossed the fi nish line.

SOLUTION

His average speed over various distances may be

calculated using the difference quotient, D 1b 2 2 D 1a 2

b 2 a.

His average speed over the 100-meter distance was

Average speed 5100 2 0

9.69 2 0 meters

second




9.69 2 4.85 seconds

550 meters

4.84 seconds



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EXAMPLE 2 Estimating an Instantaneous

Rate of Change

DBased on data from 2004–2005 and projections for 2006–2009, the amount of money spent by

consumers on books may be modeled by

b(x) 5 41.01x3 2 416.3x2 1 2999x 1 49,180 million dollars

where x is the number of years since 2004. (Source:

Modeled from Statistical Abstract of the United States,2007, Table 1119) According to the model, how quickly

was consumer spending on books increasing in 2010?

SOLUTION

In 2010, t 5 6. Using the difference quotient

S 16 1 h 2 2 S 16 2h

and selecting increasingly small values of h, we generate

Table 3.3.

We conclude that in 2010 consumer spending on books

is increasing by $2,433 million per year.

LimitsIn Examples 1 and 2, we estimated the instantaneous

rate of change at a point by calculating the average rate

of change over a “short” interval by picking “small”

values of h. The terms short and small are vague.

Numerically, what does “small” mean? Mathematicians

struggled with this dilemma for years before develop-

ing the concept of the limit. We will explore the limit

concept graphically before giving a formal defi nition.

Consider the graph of the function f 1x 2 5 2x2 1 4

over the interval 323, 3 4 (Figure 3.4).

We ask the question, “As x gets close to 2, to what

value does f sxd get close?”

Observe from the graph of f that as the value of xmoves from 0 to 2, the value of f sxd moves from 4 to 0.

We represent this behavior symbolically with the notation

limxS22

f 1x 2 5 0

which is read, “the limit of f sxd as x approaches 2 from the left is 0.” This is commonly referred to as a left-hand

limit because we approach x 5 2 through values to the

left of 2.

W l d h i 2010 di b k

Table 3.3

hS 16 1 h 2 2 S 16 2

h1.000 2795

0.100 2465

0.010 2436

0.001 2433

The Difference Quotient as an Estimate of an Instantaneous Rate of ChangeThe instantaneous rate of change of a function y 5 f(x) at a point (a, f(a)) may be estimated by calculating the diff erence quotient of f at a,

f 1a 1 h 2 2 f 1a 2h

using an h arbitrarily close to 0. (If h 5 0, the diff erence quotient is undefi ned.)

Figure 3.4y

x1−1−2−3 −1

−2

−3

−4

−5

4

3

2

1

2 3

y = f (x)

−3 −2 −1 0 1 2 3

Shut

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tock

Observe from the graph of f that as the value of x

moves from 3 to 2, the value of f sxd moves from 25 to

0. We write

limxS21

f 1x 2 5 0

65639_03_ch03_046-073.indd 5365639_03_ch03_046-073.indd 53 9/29/10 2:15 PM9/29/10 2:15 PM

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which is read, “the limit of f sxd as x approaches 2 from the right is 0.” This is commonly referred to as a right-

hand limit because we approach x 5 2 through values

to the right of 2.

On a number line, 2` lies to the left of 1`. For the

left-hand limit, limxSa2

f 1x 2 , the minus sign is used to indi-

cate that we are approaching x 5 a from the direction

of 2`. For the right-hand limit, limxSa1

f 1x 2 , the plus sign

is used to indicate that we are approaching x 5 a from

the direction of 1`.

The left- and right-hand limit behavior can also be

seen from a table of values for f sxd (Table 3.4).

That is, the left-hand limit of f sxd as x approaches 1 is

1. Similarly, as the value of x moves from 2 to 1, the

value of f s xd moves from 3 to 2. We write

limxS11

f 1x 2 5 2

That is, the right-hand limit of f sxd as x approaches

1 is 2. Since for this function, the left- and right-hand

limits are not equal, we say that “the limit of f sxd as x

approaches 1 does not exist” or, simply, “the limit does

not exist.” This can also be seen from a table of values

for f sxd (Table 3.5).

One of the most powerful features of limits is that

the limit of a function may exist at a point where the

function itself is undefi ned. This feature will be used

often when we introduce the limit defi nition of the

derivative later in this section.

Table 3.4

x f(x)

TLeft of x 5 2

T

0.00 4.000

f(x) gets close to

0 as x nears 2

1.00 3.000

1.90 0.390

1.99 0.040

2.00 0.000

cRight of x 5 2

c

2.01 –0.040f(x) gets close to

0 as x nears 2 2.10 –0.410

3.00 –5.000

Figure 3.5

y = f (x)

y

x1−1−2−3 −1

−2

−3

−4

−5

−6

3

2

1

2 3

Table 3.5

x f(x)

TLeft of x 5 1

T

0.00 2.000f(x) gets close to

1 as x nears 10.90 1.190

0.99 1.020

1.00 1.000

cRight of x 5 1

c

1.01 2.010

f(x) gets close to

2 as x nears 1

1.10 2.100

2.00 3.000

3.00 4.000

When the left- and right-hand limits of f sxdapproach the same fi nite value, we say that “the limit

exists.” In this case, the left- and right-hand limits of

f sxd neared the same value (y 5 0) as x approached 2.

We say that “the limit of f sxd as x approaches 2 is 0”

and write

limxS2

f 1x 2 5 0

Sometimes the left- and right-hand limits of a func-

tion at a point are not equal. Consider the graph of the

piecewise function

f 1x 2 5 e2x2 1 2 x # 1

x 1 1 x . 1

on the interval 323, 3 4 (see Figure 3.5).

We ask the question, “As x gets close to 1, to what

value does f sxd get close?”

Observe from the graph that as the value of xmoves from 0 to 1, the value of f sxd moves from 2 to

1. We write

limxS12

f 1x 2 5 1


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The theory surrounding limits is rich and worthy of

study; however, in this text, we are most interested in

the limit of the difference quotient as h approaches zero.

That is,

limhS0

f 1a 1 h 2 2 f 1a 2

hRecall that the difference quotient represents the aver-

age rate of change of f sxd over the interval 3a, a 1 h 4. When we place the limit on the difference quotient, we

are symbolically asking, “As the distance between the two

x-values (a and a 1 h) gets smaller, what happens to the

average rate of change of f sxd on the interval 3a, a 1 h 4?”

The limit of the difference quotient as h approaches zero

(if the limit exists) is the instantaneous rate of change in

f sxd at x 5 a. Note that even though the difference quo-

tient is undefi ned when h 5 0, the limit may still exist.

EXAMPLE 3 Calculating an Instantaneous

Rate of Change

DLet f 1x 2 5 x2. What is the instantaneous rate of change of f sxd when x 5 3?

SOLUTION

We can calculate the instantaneous rate of change by taking the limit of the difference quotient as h approaches zero. The instantaneous rate of change of f 1x 2 at x 5 3 is given by

limhS0

f 13 1 h 2 2 f 13 2h

5 limhS0

13 1 h 22 2 13 22h

5 limhS0

19 1 6h 1 h2 2 2 9

h

5 limhS0

6h 1 h2

h

5 limhS0

h 16 1 h 2h

5 limhS0

16 1 h 2 for h 2 0

Since f ( x ) 5 x2

Since hh

5 1 for h ? 0

As h nears zero, what happens to the value of 6 1 h?

Let’s pick values of h near zero (Table 3.6).

As seen from the table, even though the difference quo-

tient is undefi ned when h 5 0, the value of the simplifi ed

difference quotient, 6 1 h, gets close to 6 as h approaches

zero. In fact, by picking suffi ciently small values of h, we

can get as close to 6 as we would like. So the instanta-

neous rate of change of f sxd when x 5 3 is 6.

Observe that we can attain the same result by plug-

ging in h 5 0 after canceling out the h in the denomina-

tor of the difference quotient. That is,

5 limhS0

16 1 h 2 5 6 1 0

5 6

Throughout the rest of this chapter, we will substi-

tute in h 5 0 after eliminating the h in the denominator

of the difference quotient. This process will simplify our

computations while still giving the correct result.

The limit of the difference quotient as h approaches

zero is used widely throughout calculus and is called

the derivative.

The Limit of a FunctionIf f(x) is defi ned for all values of x near c, then

limxSc

f 1x 2 5 L

means that as x approaches c, f(x) approaches L.

We say that the limit exists if

1. L is a fi nite number and2. Approaching c from the left or right yields

the same value of L.

Table 3.6

h 6 1 h, h 2 0

20.100 5.900

20.010 5.990

20.001 5.999

0.000 Undefi ned

0.001 6.001

0.010 6.010

0.100 6.100

The Derivative of a Function at a PointThe derivative of a function y 5 f 1x 2 at a point (a, f(a)) is

f r 1a 2 5 limhS0

f 1a 1 h 2 2 f 1a 2h

f r 1a 2 is read “f prime of a” and is the instantaneous rate of change of the function f at the point (a, f(a)).


NOT FOR SALE


EXAMPLE 4 Calculating the Derivative

of a Function at a Point

DGiven f 1x 2 5 3x 1 1, fi nd f 9s2d.

SOLUTION

f r 12 2 5 limhS0

f 12 1 h 2 2 f 12 2h

5 limhS0

13 12 1 h 2 1 1 2 2 13 12 2 1 1 2h

5 limhS0

16 1 3h 1 1 2 2 17 2h

5 limhS0

13h 1 7 2 2 17 2h

5 limhS0

3hh

5 limhS0

3

5 3

In this case, the difference quotient turned out to be a

constant value of 3, so taking the limit of the difference

quotient as h approached zero did not alter the value of

the difference quotient.

For linear functions, the slope of the line is the

instantaneous rate of change of the function at any

value of x. Consequently, the derivative of a linear func-

tion will always be a constant value that is equal to the

slope of the line.

EXAMPLE 5 Finding and Interpreting

the Meaning of the Derivative

of a Function at a Point

DThe population of Washington state may be modeled by the function

P 1 t 2 5 4.933 11.017 2 t million people

where t is the number of years since 1990. (Source:

Modeled from www.census.gov data) Find and inter-

pret the meaning of P r 125 2 .SOLUTION

Since t is the number of years since 1990, t 5 25 is the year in 2015. P r 125 2 is the instantaneous rate of change in the population in 2015, given in millions of people per year.

P r 125 2 5 limhS0

P 125 1 h 2 2 P 125 2h

5 limhS0

14.933 11.017 2251h 2 2 14.933 11.017 225 2h

5 limhS0

4.933 1 11.017 2251h 2 11.017 225 2h

5 limhS0

4.933 1 11.017 225 11.017 2h 2 11.017 225 2h

5 limhS0

4.933 11.017 225 1 11.017 2h 2 1 2h

5 limhS0

7.519 1 11.017 2h 2 1 2h

Unlike in Example 3, we are unable to eliminate the

h in the denominator algebraically and calculate the

exact value of P r 125 2 . Nevertheless, by picking a small

value for h (say h 5 0.001), we can estimate the instan-

taneous rate of change.

7.519 1 11.017 20.001 2 1 20.001

5 0.1267 million people per year

5 126.7 thousand people per year

^ 127 thousand people per year

According to the model, the population of Washington

will be increasing by approximately 127,000 people per

year in 2015.

Although we were unable to obtain the exact value

of the derivative of the exponential function in Exam-

ple 5, we will develop the theory in later sections that

will allow us to calculate the exact value of the deriva-

tive of an exponential function.

3.2 Exercises

In Exercises 1–5, use the difference quotient

f 1a 1 h 2 2 f 1a 2h

(with h 5 0.1, h 5 0.01, and h 5 0.001)

to estimate the instantaneous rate of change of the function at the given input value.

1. f 1x 2 5 x2; x 5 2

Since hh

5 1

Since f(2 1 h) 5 (3(2 1 h) 1 1) and f(2) 5 3(2) 1 1

Factor out 4.933

Since 11.017 2251h 5 11.017 225 11.017 2h

Factor out 11.017 225


NOT FOR SALE


2. s 1 t 2 5 216t2 1 64; t 5 2

3. w 1 t 2 5 4t 1 2; t 5 5

4. P 1 t 2 5 5; t 5 25

5. P 1r 2 5 500 11 1 r 22; r 5 0.07

In Exercises 6–10, use the derivative to calculate the instantaneous rate of change of the function at the given input value. (In each exercise, you can eliminate the h algebraically.) Compare your answers to the solutions of Exercises 1–5.

6. f 1x 2 5 x2; x 5 2

7. s 1 t 2 5 216t2 1 64; t 5 2

8. w 1 t 2 5 4t 1 2; t 5 5

9. P 1 t 2 5 5; t 5 25

10. P 1r 2 5 500 11 1 r 22; r 5 0.07

In Exercises 11–15, use the difference quotient (with h 5 0.1, h 5 0.01, and h 5 0.001) to estimate the instantaneous rate of change of the function at the given input value. You may fi nd it helpful to apply the techniques on the Chapter 3 Tech Card.

11. f 1x 2 5 2x23; x 5 3

12. P 1 t 2 5 230 10.9 2 t; t 5 25

13. P 1r 2 5 500 11 1 r 210; r 5 0.07

14. y 5 ln 1x 2 ; x 5 2

15. g 1x 2 5 e3x; x 5 1

In Exercises 16–20, determine the instantaneous rate of change of the function at the indicated input value. (You may fi nd it helpful to apply the techniques on the Chapter 3 Tech Card.) Then explain the real-life meaning of the result.

16. Yogurt Production Based on data from 1997–2005, the amount of yogurt produced in the United States annually may be modeled by

y 1x 2 5 14.99x2 1 62.14x 1 1555 million pounds

where x is the number of years since 1997. (Source: Modeled from Statistical Abstract of the United States, 2007, Table 846) Find and interpret the meaning of S r 110 2 .

17. Heart Disease Death Rate Based on data from 1980–2003, the age-adjusted death rate due to heart disease may be modeled by

R 1 t 2 5 27.597t 1 407.4 deaths per 100,000 people

where t is the number of years since 1980. (Source: Statistical Abstract of the United States, 2006, Table 106) Find and interpret the meaning of D r 155 2 .

18. DVD Player Sales Based on data from 1997–2004, the number of DVD players sold may be modeled by

D 1p 2 5 77,100 10.989 2p thousand DVD players

where p is the average price per DVD player in dollars. (Source: Modeled from Consumer Electronics Association data) Find and interpret the meaning of D r 1100 2 .

19. Theme Park Tickets Based on 2007 ticket prices, the cost of a child’s Disney Park Hopper Bonus Ticket may be modeled by

T 1d 2 5 23.214d2 1 41.19d 1 34.2 dollars

where d is the number of days that the ticket authorizes entrance into Disneyland and Disney California Adventure. (Source: www.disneyland.com) Find and interpret the meaning of T r 14 2 .

20. United States Populations Based on data from 1995–2005, the population of the United States may be modeled by

U 1 t 2 5 298,213 11.009 2 t thousand people

where t is the number of years since 2005. (Source: World Health Statistics 2006, World Health Organization) Find and interpret the meaning of U r 110 2 .

Exercises 21 and 22 deal with the velocity of a free-falling object on earth. The vertical position of a free-falling

object may be modeled by s 1 t 2 5 216t2 1 v0t 1 s0 feet, where v0 is the velocity of the object and s0 is the vertical position of the object at time t 5 0 seconds.

21. Velocity of a Dropped Object A can of soda is dropped from a diving board 40 feet above the bottom of an empty pool. How fast is the can traveling when it reaches the bottom of the pool?

22. Velocity of a Ball A small rubber ball is thrown into the air by a child at a velocity of 20 feet per second. The child releases the ball 4 feet above the ground. What is the velocity of the ball after 1 second?


NOT FOR SALE


3.3 The Derivative as a Slope: Graphical Methods

Based on data from 1990–2003, the

amount of money spent on prescription

drugs (per capita) may be modeled by

P(t) 5 2.889t2 2 2.613t 1 158.7 dollars

where t is the number of years since

1990. (Source: Statistical Abstract of

the United States, 2006, Table 121) To

help project future drug costs, an insur-

ance company wants to know what the

average annual increase in the amount of

money spent on prescription drugs was

from 1990 to 2010 and at what rate

the drug spending will be increasing at

the end of 2010. We will answer these

questions by calculating the average and

instantaneous rates of change in the pre-

scription drug spending.

In Sections 3.1 and 3.2, you learned how to cal-

culate the average rate of change of a function over

an interval and the instantaneous rate of change of a

function at a point. In this section, we will revisit these

concepts from a graphical standpoint. We will also

demonstrate how to use tangent-line approximations

to estimate the value of a function.

As shown in Section 3.1, the difference quotient for-

mula, f 1a 1 h 2 2 f 1a 2

h, gives the average rate of change

in the value of the function between the points (a, f(a))

and 1a 1 h, f 1a 1 h 2 2 . If we let a 5 c 2 h, then the dif-

ference quotient formula becomes f 1c 2 2 f 1c 2 h 2

h. We

will use this modifi ed form of the difference quotient in

our exploration of prescription drug spending, since we

will be approaching t 5 c through values to the left of

t 5 c. Recall from Section 3.1 that, graphically speak-

ing, the difference quotient is the slope of the secant line

connecting the two points on the graph of a function.

According to the prescription drug spending model,

C(0) 5 158.7 and C(20) 5 1262. We’re interested in

the slope of the secant line between (0, 158.7) and

(20, 1262). The slope of this line represents the aver-

age rate of change in the prescription drug spending

between 1990 and 2010 (Figure 3.6).

Figure 3.6

C(t)

C(t) = 2.889t2 − 2.613t + 158.7

t

100

200

300

400

500

600

700

800

900

1000

1100

1200

1300

1400

22201816141210Years (since 1990)

Dru

g sp

endi

ng (

in d

olla

rs)

86420

(20, 2162)

(0, 158.7)

The prescription drug spending graph secant line

between 10, 158.7 2 and (20, 1262) has the slope

m 5C 120 2 2 C 10 2 dollars

20 years

5 1262 2 158.7 dollars

20 years


Between 1990 and 2010, the per capita spending on

prescription drugs increased by $55.17 per year. Does

this mean that from 2010 to 2011, the spending will

increase by about $55.17 ? No. Looking at the graph of

the model, we notice that the prescription drug spend-

ing is rising at an increasing rate as time progresses

(the steeper the graph, the greater the magnitude of the

rate of change). We can approximate the instantaneous

rate of change at the end of 2010 (t 5 20) by calculat-

ing the slope of a secant line through (20, 1262) and a

“nearby” point as shown in Figure 3.7.

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NOT FOR SALE

593 . 3 T h e D e r i v a t i v e a s a S l o p e : G r a p h i c a l M e t h o d s


between (10, 421.5) and (20, 1262) has the slope

m 5C 120 2 2 C 110 2 dollars

10 years

51262 2 421.5 dollars

10 years


As shown in Figure 3.8, the prescription drug

spending graph secant line between 119, 1152 2 and

(20, 1262) has the slope

m 5C 120 2 2 C 119 2 dollars

20 2 19 years

51262 2 1152 dollars

1 year


By zooming in (Figure 3.9), we can see that between

t 5 19 and t 5 20, the secant line and the graph of the

function are extremely close together.

Figure 3.7

C(t)

C

100

200

300

400

500

600

700

800

900

1100

1200

1300

1400

22201816141210Years (since 1990)

Dru

g sp

endi

ng (

in d

olla

rs)

86420

(20, 1262)

(10, 421.5)

Figure 3.8

C(t)

C

t

100

200

300

400

500

600

700

800

900

1000

1100

1200

1300

1400

22201816141210Years (since 1990)

Dru

g sp

endi

ng (

in d

olla

rs)

86420

(20, 1262)

(19, 1152)

Figure 3.9

C(t)

t

1100

1200

1300

212019Years (since 1990)

Dru

g sp

endi

ng (

in d

olla

rs)

18

(20, 1262)

(19, 1152)

C


between (19.9, 1251) and (920, 1262) has the slope

m 5C 120 2 2 C 119.9 2 dollars

20 2 19.9 years

51262 2 1251 dollars

0.1 years


Through (19.9, 1251) and (20, 1262), the slope of the

secant line is $110.0 per year.

Again zooming in, we see that the secant line and

the graph of the function are nearly identical between

t 5 19.9 and t 5 20 (Figure 3.10). In fact, we are

unable to visually distinguish between the two.


NOT FOR SALE


As t gets closer and closer to 20, the slope of the

secant line will approach C r 120 2 . This occurs because

C 120 2 2 C 120 2 h 2h

< C r 120 2 for small values of h.

Since h is the horizontal distance between the points,

h becomes increasingly small as t nears 20. Using the

methods covered in Section 3.2, we determine algebra-

ically that C r 120 2 5 112.9. At the end of 2010, per cap-

ita prescription drug spending was increasing at a rate

of roughly $113 per year. In other words, we anticipate

that per capita prescription drug spending increased by

approximately $113 between 2010 and 2011.

As the preceding example demonstrates, to fi nd the

instantaneous rate of change of a function y 5 f 1x 2 at a

point P 5 1a, f 1a 2 2 , we can fi nd the limit of the slope of

the secant line through P and a nearby point Q as the

point Q gets closer and closer to P. Graphically speak-

ing, we select values Q1, Q2, Q3, . . . , with each con-

secutive value of Qi being a point on the curve that is

closer to the point P than the one before it. The limit of

the slope of the secant lines (imagine the points P and

Q fi nally coinciding) is the line tangent to the curve at

the point P 5 1a, f 1a 2 2 (Figure 3.11).

The tangent line to a graph f at a point (a, f(a)) is

the line that passes through (a, f(a)) and has slope f r 1a 2 .

EXAMPLE 1 Finding the Equation

of a Tangent Line

DFind the equation of the tangent line to the graph of f 1x 2 5 x2 that passes through (2, 4). Then

graph the tangent line and the graph of f.

SOLUTION

The slope of the tangent line is f 9(2).

f r 12 2 5 limhS0

f 12 1 h 2 2 f 12 2h

5 limhS0

12 1 h 22 2 22

h Since f sxd 5 x2

5 limhS0

4 1 4h 1 h2 2 4

h

5 limhS0

4h 1 h2

h

5 limhS0

h 14 1 h 2h

Figure 3.10

C(t)

C

t

1240

1250

1260

1270

20.12019.9Years (since 1990)

Dru

g sp

endi

ng (

in d

olla

rs)

19.8

(20, 1262)

(19.9, 1251)

Figure 3.11

Q1

Q2

Q3

Q4

P

y

x

Tangent line

f (x)

The Graphical Meaning of the DerivativeThe derivative of a function f at a point (a, f(a)) is the slope of the tangent line to the graph of f at that point. The slope of the tangent line at a point is also referred to as the slope of the curve at that point.


NOT FOR SALE


We already know that

P 14 2 5 8.22. We’ll calcu-

late P 14 1 h 2 and then

substitute the sim-

plifi ed value into

the derivative

formula.

5 limhS0

14 1 h 2 5 4 1 0

5 4

The slope of the tangent line is 4 at the point (2, 4). Using

the slope-intercept form of a line, we have y 5 4x 1 b.

Substituting in the point (2, 4), we get

4 5 4 12 2 1 b

4 5 8 1 b

b 5 24

The equation of the tangent line is y 5 4x 2 4.

We generate a table of values for y 5 4x 2 4 and

f 1x 2 5 x2. Then we graph the results (Figure 3.12).

Our tangent-line estimate 1y 5 3.6 2 was remarkably

close to the actual value of the function 1 f 11.9 2 5 3.61 2 .EXAMPLE 2 Using a Tangent-Line

Approximation for Apple iPod Net Sales

DBased on data from 2005 to 2009, Apple iPod net sales may be modeled by

P 1 t 2 5 20.58t2 1 3.2t 1 4.7 billion dollars

where t is the number of years since the end of 2005.

(Source: Model based on data from Apple Computer

Corporation) According to the model, iPod net sales

were $8.22 billion in 2009.

Use the model to predict how quickly iPod net sales

were changing at the end of 2009. Then use a tangent line

to estimate iPod net sales at the end of 2010. Compare

the estimate to the actual value predicted by the model.

Figure 3.12

f (x) = x2

y = 4x − 4

(2, 4)

y

x1−1

−6−4−2

810121416

642

2 3 4

x y f(x)

–1 –8 1

0 –4 0

1 0 1

2 4 4

3 8 9

4 12 16

Figure 3.13

C(t)

t

3.2

3.4

3.6

3.8

4.0

4.2

4.4

4.6

4.8

2.22.12.01.91.8

(2,4)

y = 4x−4

y = x2

Tangent-Line ApproximationsIn Example 1 you may have noticed that the tangent line

lies very near to the graph of f for values of x near a. In

fact, if we zoom in to the region immediately surround-

ing (a, f(a)), the graph of f and the tangent line to the

graph of f at (a, f(a)) appear nearly identical. For values

of x near 2, the tangent-line y-value is a good approxi-

mation of the actual function value (see Figure 3.13).

Because it is frequently easier to calculate the values

of the tangent line than the values of the function, some-

times the tangent line is used to estimate the value of the

function. For example, suppose we wanted to estimate

f 11.9 2 given f 1x 2 5 x2. Since x 5 1.9 is near x 5 2, we

may use the equation of the tangent line, y 5 4x 2 4,

to estimate f 11.9 2 . That is, f 11.9 2 < 4 11.9 2 2 4 5 3.6.

The actual value is

f 11.9 2 5 11.9 22 5 3.61

Shut

ters

tock

SOLUTION

Since t 5 4 corresponds with the year 2009, the instantaneous rate of change in iPod net sales at the end of 2009 is given by

P r 14 2 5 limhS0

P 14 1 h 2 2 P 14 2 million dollars

h years


NOT FOR SALE


P 14 1 h 2 5 20.58 14 1 h 22 1 3.2 14 1 h 2 1 4.7

5 20.58 116 1 8h 1 h2 2 1 12.8 1 3.2h 1 4.7

5 29.28 2 4.64h 2 0.58h2 1 3.2h 1 17.5

5 20.58h2 2 1.44h 1 8.22

P r 14 2 5 limhS0

P 14 1 h 2 2 P 14 2h

5 limhS0

120.58h2 2 1.44h 1 8.22 2 2 8.22

h

5 limhS0

20.58h2 2 1.44hh

5 limhS0

h 120.58h 2 1.44 2h

5 limhS0

120.58h 2 1.44 2 5 20.58 10 2 2 1.44

5 21.44 billion dollars

year

At the end of 2009, iPod net sales were decreasing at

a rate of $1.44 billion per year. That is, between 2009

and 2010, the iPod net sales were expected to decrease

by approximately $1.44 billion.

The point-slope form of the tangent line is

y 2 y1 5 21.44 1x 2 x1 2 . Using the point 14, 8.22 2 , we

determine

y 2 8.22 5 21.44 1x 2 4 2This is the tangent-line equation. At t 5 5, we have

y 2 8.22 5 21.44 15 2 4 2 y 5 8.22 2 1.44

5 6.78

Using the tangent-line equation, we estimate that iPod

net sales are about $6.8 billion in 2010. According to

the model, the actual number of iPod net sales was

somewhat less.

P 15 2 5 20.58 15 22 1 3.2 15 2 1 4.7

5 6.2

The model indicates that iPod net sales were $6.2 bil-

lion in 2010.

Why was there a discrepancy between the two esti-

mates in Example 2? Look at the graph of the model

and the tangent line (Figure 3.14).

Although both functions were equal at t 5 4, by the

time t reached 5, the value of the model fell below the

tangent-line estimate by about $0.6 billion. Although

tangent-line estimates of a function’s value are not

exact, they are often good enough for their intended

purpose.

Numerical DerivativesOften we encounter real-life data in tables or charts. Is

it possible to calculate a derivative from a table of data?

We’ll investigate this question by looking at a table of

data for f 1x 2 5 x2 (see Table 3.7). For this function,

f r 12 2 5 4.

Table 3.7

x f(x)

0 0

1 1

2 4

3 9

4 16

Figure 3.14

C(t)

t

1

2

4

3

5

6

7

8

11

9

10

12

654321Years (since 2005)

iPod

net

sal

es (

bill

ion

doll

ars)

y = −1.44(x – 4) + 8.22

y = −0.58x2 + 3.2x + 4.7

We can estimate f 9(2) by calculating the slope of the

secant line through points whose x-values are equidis-

tant from x 5 2. That is,

65639_03_ch03_046-073.indd 6265639_03_ch03_046-073.indd 62 9/29/10 1:52 PM9/29/10 1:52 PM

NOT FOR SALE


When estimating a derivative numerically, we typi-

cally select the two closest data points that are horizon-

tally equidistant from our point of interest. Doing so

often yields a line that is parallel or near-parallel to the

tangent line. Picking two points that are equidistant from

the point of interest will tend to give the best estimate

of the derivative and can be used as long as the point of

interest is not an endpoint. (If the point of interest is an

endpoint, we fi nd the slope of the secant line between the

endpoint and the next closest point.) If we assume that

each output in a table of data represents f 1a 2 for a corre-

sponding input a, then we can symbolically represent the

process of numerically estimating a derivative as follows.

If a is the largest domain value in the data set, then

f r 1a 2 <f 1a 2 2 f 1a 2 h 21a 2 2 1a 2 h 2

<f 1a 2 2 f 1a 2 h 2

h

EXAMPLE 3 Estimating the Derivative

from a Table of Data

DUse Table 3.8 to estimate how quickly home sales prices in the southern United States were

increasing at the end of 2006. (That is, estimate the slope of the tangent line at (6, 208.2).)

SOLUTION

We’ll estimate the slope of the tangent line at (6, 208.2) by calculating the slope of the secant line between (4, 181.1) and (8, 203.7).

Figure 3.15

f (x) = x2

f ' (2) = 4

m = 4

y

6

8

4

2

16

14

12

10

x1−1 2 3 4 5 60

(1, 1)

(3, 9)

(2, 4)

FunctionSecantTangent

Numerical Estimate of the DerivativeThe derivative of a function f at a point (a, f(a)) may be approximated from a table by

f r 1a 2 <f 1a 1 h 2 2 f 1a 2 h 21a 1 h 2 2 1a 2 h 2

<f 1a 1 h 2 2 f 1a 2 h 2

2h

where h is the horizontal distance between a and a 1 h.

Table 3.8

Median Sales Price of a New One-Family House in the Southern United States

Years (since 2000)

(t)

Price (thousands of dollars)

(P)

0 148.0

2 163.4

4 181.1

6 208.2

8 203.7

Source: Statistical Abstract of the United States, 2010, Table 940

f r 12 2 <f 13 2 2 f 11 2

3 2 1

59 2 1

2

58

2

5 4

In this case, our estimate was equal to the tangent-

line slope. Let’s look at the situation graphically (see

Figure 3.15).

If a is the smallest domain value in the data set, then

f r 1a 2 <f 1a 1 h 2 2 f 1a 21a 1 h 2 2 1a 2

<f 1a 1 h 2 2 f 1a 2

hIn both cases, h is the distance between a and the next

closest domain value.


NOT FOR SALE


f r 16 2 <f 18 2 2 f 14 2 thousand dollars

8 2 4 years

<203.7 2 181.1

4

< 5.65 thousand dollars per year

We estimate that home prices in the southern United

States were increasing at a rate of about 5.7 thousand

dollars per year at the end of 2006. (Since the original

data were accurate to one decimal place, we rounded

our fi nal result to one decimal place.)

3.3 Exercises

In Exercises 1–10, determine the equation of the tangent line of the function at the given point. Then graph the tangent line and the function together.

1. f 1x 2 5 x2 2 4x; 11, 23 2 2. f 1x 2 5 2x2 1 6; 12, 2 2 3. g 1x 2 5 x2 1 2x 1 1; 10, 1 2 4. g 1x 2 5 x2 2 4; 13, 5 2 5. g 1x 2 5 x2 2 4x 2 5; 14, 25 2 6. h 1x 2 5 x3; 12, 8 2 7. h 1x 2 5 x3; 10, 0 2 8. h 1x 2 5 x3; 11, 1 2 9. f 1x 2 5 1x 2 3 22; 13, 0 2 10. f 1x 2 5 1x 1 2 22; 121, 1 2In Exercises 11–15, answer the questions by calculating the slope of the tangent line and the tangent-line equation, as appropriate.

11. Median Price of a New Home Based on data from 2003 to 2008, the median price of a home in the western region of the United States may be modeled by

W 1 t 2 5 29.2t2 1 55t 1 253 thousand dollars

where t is the number of years since 2003. (Source: Modeled from Statistical Abstract of the United States, 2010, Table 940) How quickly was the median sales price changing in 2009? What was the estimated median sales price in 2010? (Use a tangent-line approximation.)

12. New One-Family Home Size Based on data from 2000 to 2008, the average number of square feet in new one-family homes may be modeled by

H 1 t 2 5 20.0023t2 1 3447t 1 2265 square feet

where t is the number of years after 2000. (Source: Modeled from Statistical Abstract of the United States, 2010, Table 936) At what rate was home size changing in 2008? What was the estimated size of a new home in 2009? (Use a tangent-line approximation.)

13. Student-to-Teacher Ratio Based on data from 1995 to 2007, the student-to-teacher ratio at private elementary and secondary schools may be modeled by

R(t) 5 0.00639t2 2 0.284t 1 15.7 students per teacher

where t is the number of years since 1995. (Source: Modeled from Statistical Abstract of the United States, 2010, Table 245) According to the model, how quickly was the student-to-teacher ratio changing in 2008? What was the estimated student-to-teacher ratio for 2009? (Use a tangent-line approximation.)

14. Cassette Tape Shipment Value Based on data from 1990–2004, the value of music cassette tapes shipped may be modeled by

V(s) 5 20.00393s2 1 9.74s 2 63.0 million dollars

where s represents the number of music cassette tapes shipped (in millions). (Source: Modeled from Statistical Abstract of the United States, 2006, Table 1131) According to the model, at what rate is the value of the cassette tapes shipped changing when the number of cassette tapes shipped is 250 million? What was the estimated shipment value when 251 million cassette tapes were shipped? (Use a tangent-line approximation.)

15. College Attendance Based on data from 2000–2002 and Census Bureau projections for 2003–2013, private college enrollment may be modeled by

P 1x 2 5 0.340x 2 457 thousand students

where x is the number of students (in thousands) enrolled in public colleges. (Source: Modeled from Statistical Abstract of the United States, 2006, Table 204)


NOT FOR SALE


Calculate the rate of change in private college enrollment when 12,752 thousand public college students are enrolled. Use a tangent-line approximation to estimate the number of students enrolled in private colleges when there are 12,753 thousand public college students.

In Exercises 16–20, estimate the specifi ed derivative by using the data in the table. Then interpret the result.

16. Worker Wages

Average Annual Salary per Worker: Motion Picture and Sound Recording Industries

Years (since 2000)

tAnnual Salary (dollars)

W 1 t 20 55,355

1 54,776

2 54,877

3 55,991

4 60,424

5 62,051

6 65,764

7 67,055


Estimate W r 16 2 . 17. Ice Cream Prices

Ice Cream Retail Price

Years (since 2004)

t

Price of 1/2 Gallon

of Ice Cream

I 1 t 20 3.85

1 3.69

2 3.90

3 4.08

4 4.28


Estimate I r 13 2 .

18. Undergraduate Tuition at the University of Pittsburgh

Full-Time Resident Tuition and Fees

Years(since 2004–2005)

t

Annual Tuition and Fees (dollars)

f 1 t 20 10,130

1 10,736

2 11,368

3 12,106

4 12,832

5 13,344

Source: University of Pittsburgh

Estimate f r 14 2 and f r 15 2 . What can you conclude about the rate of increase in tuition and fees at the University of Pittsburgh?

19. Steak Prices

Boneless Sirloin Steak Prices

Years (since 2004)

tPrice per Pound (dollars)

S 1 t 20 6.09

1 5.93

2 5.79

3 5.91

4 6.07


Estimate S r 13 2 . 20. Bread Prices

Whole Wheat Bread Prices

Years (since 2004)

tPrice per Loaf (dollars)

B 1 t 20 1.30

1 1.29

2 1.62

3 1.81

4 1.95


Estimate B r 12 2 .


NOT FOR SALE


3.4 The Derivative as a Function: Algebraic Method

Based on data from 2000 to 2006, the

cumulative number of homicides resulting

from a romantic triangle may be modeled

by R(t) 5 21.57t2 1 119t 1 125 homi-

cides between the start of 2000 and the

end of year t, where t is the number of

years since 2000. (Source: Modeled

from Crime in the United States 2006,

Uniform Crime Report, FBI) At what rate

was the cumulative number of homicides

increasing at the end of 2004, 2005,

and 2006? Although we could calculate

the derivative at t 5 4, t 5 5, and t 5 6,

we can save time by fi nding the derivative

function and then substituting in the vari-

ous values of t.

In this section, we will introduce the derivative

function and show how to fi nd it algebraically. Your

skill in fi nding the derivative at a point will prove to

be especially useful in this section. We’ll begin with a

simple example before returning to the romantic tri-

angle problem.

EXAMPLE 1 Calculating the Derivative

of a Function at Multiple Points

DDetermine the instantaneous rate of change of f 1x 2 5 3x2 at 11, 3 2 , 13, 27 2 , and 110, 300 2 .

SOLUTION

Although we could calculate f r 11 2 , f r 13 2 , and f r 110 2individually, it will be more effi cient to fi nd the derivative function itself and then substitute in the different values of x.

We begin with the derivative formula; however,

instead of substituting a specifi c value for a, we replace

a with the variable x.

f r 1x 2 5 limhS0

f 1x 1 h 2 2 f 1x 2h

5 limhS0

13 1x 1 h 22 2 2 13x2 2h

Since f 1x 2 5 3x2

5 limhS0

13 1x2 1 2hx 1 h2 2 2 2 13x2 2h

5 limhS0

13x2 1 6hx 1 3h2 2 2 13x2 2h

5 limhS0

6hx 1 3h2

h

5 limhS0

h 16x 1 3h 2

h

5 limhS0

16x 1 3h 2 5 6x 1 3 10 2 5 6x

The result f r 1x 2 5 6x is the derivative function for

f 1x 2 5 3x2. It can be used to calculate the instanta-

neous rate of change of f at any point (a, f(a)).

f r 11 2 5 6 11 2 5 6

The instantaneous rate of change of f at 11, 3 2 is 6.

f r 13 2 5 6 13 2 5 18


f r 110 2 5 6 110 2 5 60


As demonstrated in Example 1, the techniques used

to fi nd the derivative function are virtually identical to

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NOT FOR SALE

673 . 4 T h e D e r i v a t i v e a s a F u n c t i o n : A l g e b r a i c M e t h o d

the procedures used to fi nd the derivative at a point.

However, knowing the derivative function allows us to

calculate the derivative at a number of different points

more quickly than calculating the derivative at each

point separately. The derivative function for a function

f is called the derivative of f.

The Derivative FunctionThe derivative function of a function f is given by

f r 1x 2 5 limhS0

f 1x 1 h 2 2 f 1x 2h

if the limit exists.

EXAMPLE 2 Calculating the

Instantaneous Rate of Change

of a Function at Multiple Points

DBased on data from 2000 to 2006, the cumu-lative number of homicides resulting from a

romantic triangle between the start of 2000 and the end

of year t may be modeled by

R 1 t 2 5 21.57t2 1 119t 1 125 homicides

where t is the number of years since 2000. (Source:

Modeled from Crime in the United States 2006, Uni-

form Crime Report, FBI) At what rate was the cumula-

tive number of homicides increasing at the end of 2004,

2005, and 2006?

SOLUTION

We will begin by fi nding the derivative R r 1 t 2 .R r 1 t 2 5 lim

hS0

R 1 t 1 h 2 2 R 1 t 2h

Because of the complex nature of R 1 t 2 , we will fi rst cal-

culate R 1 t 1 h 2 and then substitute the result into the

derivative formula.

R(t 1 h) 5 21.57(t 1 h)2 1 119(t 1 h) 1 125

5 21.57t2 (t2 1 2ht 1 h2) 1 119t 1 119h 1 125

5 21.57t2 2 3.14ht 2 1.57h2 1 119t 1 119h 1 125

We already know that R 1 t 2 5 21.57t2 1 119t 1 125.

Substituting both of these quantities into the derivative

formula yields

R r 1 t 2 5 limhS0

R 1 t 1 h 2 2 R 1 t 2h

5 limhS0

23.14ht 2 1.57h2 1 119hh

5 limhS0

h 123.14t 2 1.57h 1 119 2h

5 limhS0

123.14t 2 1.57h 1 119 2 5 23.14t 2 1.57 10 2 1 119

5 23.14t 1 119

So R r 1 t 2 5 23.14t 1 119 homicides per year. We can

now compute the instantaneous rate of change in the

cumulative number of homicides in 2004, 2005, and 2006.

R r 14 2 5 23.14 14 2 1 119

5 106.4

< 106 homicides per year

R r 15 2 5 23.14 15 2 1 119

5 103.3


R r 16 2 5 23.14 16 2 1 119

5 100.2


The cumulative number of homicides resulting

from a romantic triangle was increasing at a rate of 106

homicides per year in 2004, 103 homicides per year in

2005, and 100 homicides per year in 2006. Accord-

ing to the model, although the cumulative number of

homicides continued to increase, the rate at which these

homicides were increasing slowed between the end of

2004 and the end of 2006.

EXAMPLE 3 Finding the Derivative

of a Function

DFind the derivative of g 1 t 2 5 2t3 2 4t 1 3.

SOLUTION

We must fi nd g r 1 t 2 5 limhS0

g 1 t 1 h 2 2 g 1 t 2

h. We’ll fi rst

fi nd g 1 t 1 h 2 and then substitute the result into the derivative formula.

Since hh

5 1 for h 2 0


NOT FOR SALE


g 1 t 1 h 2 5 2 1 t 1 h 23 2 4 1 t 1 h 2 1 3

5 2 1 t3 1 3t2h 1 3th2 1 h3 2 2 4t 2 4h 1 3

5 2t3 1 6t2h 1 6th2 1 2h3 2 4t 2 4h 1 3

5 g 1 t 2 1 6t2h 1 6th2 1 2h3 2 4h

g r 1 t 2 5 limhS0

g 1 t 1 h 2 2 g 1 t 2h

5 limhS0

16t2h 1 6th2 1 2h3 2 4h 2h

5 limhS0

h 16t2 1 6th 1 2h2 2 4 2h

5 limhS0

16t2 1 6th 1 2h2 2 4 2 5 6t2 1 6t 10 2 1 2 10 22 2 4

5 6t2 2 4

The derivative of g 1 t 2 5 2t3 2 4t 1 3 is g r 1 t 2 5 6t2 2 4.

Estimating DerivativesFor polynomial functions, all terms in the numerator of

the derivative formula without an h will cancel out. This

allows us always to eliminate the h in the denominator.

However, with some other types of functions, the h in

the denominator cannot be eliminated algebraically. In

this case, we can estimate the derivative function by sub-

stituting in a small positive value (e.g., 0.001) for h. The

closer the value of h is to zero, the more accurate the

estimate of the derivative will be.

EXAMPLE 4 Calculating the

Instantaneous Rate of Change of

a Function at Multiple Points

DThe per capita consumption of bot-tled water in the United States may

be modeled by

W 5 2.593 11.106 2 t gallons

where t is the number of years since the end of 1980. (Source: Mod-

eled from Statistical Abstract of the United States, 2001, Table

204)

Determine how quickly

bottled water consumption

was increasing at the end of

2002, 2004, and 2006.

SOLUTION

We must evaluate W r 1 t 2 at t 5 22, 24, and 26.

W r 1 t 2 5 limhS0

W 1 t 1 h 2 2 W 1 t 2h

5 limhS0

12.593 11.106 2 t1h 2 2 12.593 11.106 2 t 2h

5 limhS0

12.593 11.106 2 t 11.106 2h 2 2 12.593 11.106 2 t 2h

5 limhS0

2.593 11.106 2 t 1 11.106 2h 2 1 2h

5 2.593 11.106 2 t # limhS0

1 11.106 2h 2 1 2h

We can move the expression 2.593 11.106 2 t to the other

side of the limit because it does not contain an h. Since

limhS0

1 11.106 2h 2 1 2h

<1 11.106 20.001 2 1 2

0.001

< 0.1008

we have

W r 1 t 2 5 2.593 11.106 2 t # limhS0

1 11.106 2h 2 1 2h

< 2.593 11.106 2 t # 10.1008 2 < 0.2614 11.106 2 t

We will now evaluate the derivative function at

t 5 22, 24, and 26.

W r 122 2 < 0.2614 11.106 222

< 2.398

According to the model, bottled water consumption was

increasing by 2.398 gallons per year at the end of 2002.

W r 124 2 < 0.2614 11.106 224

< 2.934

Bottled water consumption was increasing by 2.934

gallons per year at the end of 2004.

W r 126 2 < 0.2614 11.106 226

< 3.589

Bottled water consumption was increasing at a rate of

3.589 gallons per year at the end of 2006.

Since 11.106 2 t1h 5 11.106 2 t 11.106 2h

Factor out 2.593(1.106)t

Shut

ters

tock


NOT FOR SALE

693 . 4 T h e D e r i v a t i v e a s a F u n c t i o n : A l g e b r a i c M e t h o d

3.4 Exercises

In Exercises 1–5, fi nd the derivative of the function.

1. f 1x 2 5 x2 2 4x

2. g 1x 2 5 x2 1 2x 1 1

3. g 1x 2 5 x2 2 4x 2 5

4. j 1x 2 5 x3 1 2

5. f 1 t 2 5 1 t 2 3 22In Exercises 6–10, fi nd the slope of the tangent line of the function at x 5 1, x 5 3, and x 5 5.

6. g 1x 2 5 2x2 1 x 2 1

7. f 1x 2 5 x2 2 2x

8. j 1x 2 5 25

9. W 1x 2 5 24x 1 9

10. S 1x 2 5 3x2 2 2x 1 1

In Exercises 11–13, estimate the derivative of the function. When you are unable to eliminate the h in the denominator of the derivative formula algebraically, use h 5 0.001.

11. P 1x 2 5 3x

12. C 1x 2 5 23 # 4x

13. R 1x 2 5 5.042 # 10.98 2xIn Exercises 14–18, use the derivative function to answer the questions.

14. Median Price of a New Home Based on data from 2003 to 2008, the median price of a home in the western region of the United States may be modeled by

W 1 t 2 5 29.2t2 1 55t 1 253 thousand dollars

where t is the number of years since 2003. (Source: Modeled from Statistical Abstract of the United States, 2010, Table 940) According to the model, was the median sales price changing more quickly at the end of 2006 or the end of 2008?

15. Median Price of a New Home Based on data from 2003 to 2008, the median price of a home in the United States may be modeled by

U 1 t 2 5 25.1t2 1 33t 1 194 thousand dollars

where t is the number of years since 2003. (Source: Modeled from Statistical Abstract of the United States, 2010, Table 940) According to the model, was the median sales price changing more quickly at the end of 2007 or the end of 2009?

16. Walmart Net Sales Based on data from 1996–2006,

the net sales of Walmart may be modeled by

s 1 t 2 5 0.8636t2 1 14.39t 1 84.72 billion dollars

where t is the number of years since 1996. (Source: Modeled from Walmart Annual Report, 2006, pp.18–19) According to the model, how much more rapidly were net sales increasing in 2006 compared to 2005?

17. Prescription Drug Spending Based on data from 1990–2003, per capita prescription drug spending may be modeled by

P 1 t 2 5 2.889t2 2 2.613t 1 158.7 dollars

where t is the number of years since 1990. (Source: Statistical Abstract of the United States, 2006, Table 121) In what year was prescription drug spending increasing twice as fast as it was increasing in 2000?

18. Oil Production Based on data from 1985–2004, the difference between U.S. oil fi eld production and net oil imports may be modeled by

b 1 t 2 5 4.29t2 2 278t 1 2250 million barrels

where t is the number of years since 1985. (Source: Modeled from Statistical Abstract of the United States, 2007, Table 881) At what rate was the difference in U.S. oil fi eld production and net oil imports changing in 2000 and 2005, according to the model?

Exercises 19 and 20 are intended to challenge your understanding.

19. Given f r 1x 2 5 3x and g 1x 2 5 x2 1 3 1 f 1x 2 , fi nd g r 1x 2 .

20. Given f 1x 2 5 x3 2 3x, determine where f r 1x 2 5 0.


NOT FOR SALE


3.5 Interpreting the Derivative

Many of us feel inundated by the adver-

tisements we are sent through the mail.

Don’t expect this to let up anytime soon:

Spending on direct-mail advertising has

risen every year since 1990. The amount

of money spent on direct-mail advertising

may be modeled by

A(t) 5 70.54t2 1 1488t 1 22,828

million dollars

where t is the number of years since

1990. (Source: Modeled from Statisti-

cal Abstract of the United States, 2001,

Table 1272) According to the model,

A(9) 5 41,934 and A9(9) 5 2,758. But

what does this mean? In this section, we

will discuss how to interpret the mean-

ing of a derivative in the context of a

real-life problem.

Recall that the units of the deriv-

ative are the units of the output

divided by the units of the input.

In this case, the units of A r are

millions of dollars

year or millions of

dollars per year. Note that t 5 9 cor-

responds to the year 1999. We

conclude that in 1999, 41,934

million dollars were spent on

direct-mail advertising, and

spending was increasing by

2,758 million dollars per year.

In other words, according to the

model, 41,934 million dollars were

spent on direct-mail advertising in

1999 and spending increased by about

EXAMPLE 1 Interpreting the Meaning

of the Derivative

DBased on data from 1940 to 2000, the average monthly Social Security benefi t

for men may be modeled by

P 1 t 2 5 0.3486t2 2 5.505t 1 35.13 dollars

where t is the number of years since 1940.

(Source: Modeled from Social Security

Administration data)

Interpret the meaning of P 160 2 5 959.79

and P r 160 2 5 36.33.

SOLUTION

Since t is the number of years since 1940, t 5 60

corresponds to 2000. P 160 2 5 959.79 means that (according to the model) the average Social Security

benefi t for men in 2000 was $959.79.

The units of the derivative are dollars

year.

P9 s60d 5 36.33 means that in 2000, the average Social

Security benefi t for men was increasing by $36.33 per

year. In other words, the average Social Security ben-

efi t for men was expected to increase by about $36.33

between 2000 and 2001.

Interpreting the Derivative

Let f(x) be a function. The meaning of f r 1a 2 5 c may be written in either of the following two ways:

• When x 5 a, the value of the function f is increasing (decreasing) at a rate of c units of output per unit of input.

• The value of the function f will increase (decrease) by about c units of output between a units of input and a 1 1 units of input.

Shut

terst

ock

2,758 million dollars between 1999 and 2000. The terms

about, approximately, or roughly must be used when

using the second interpretation of the derivative, since we

are using a tangent-line approximation to estimate the

increase over the next year. (For a graphical discussion of

tangent-line approximations, refer to Section 3.3.)


NOT FOR SALE

713 . 5 I n t e r p r e t i n g t h e D e r i v a t i v e

EXAMPLE 2 Using a Tangent-Line

Approximation to Estimate

a Function Value

DThe average weight of a boy between 2 and 13 years of age may be modeled by

W 1a 2 5 0.215a2 1 2.993a 1 23.78 pounds

where a is the age of the boy in years. (Source: Modeled

from www.babybag.com data)

Interpret the meaning of W 110 2 5 75.21 and

W r 110 2 5 7.29. Then use a tangent-line approxima-

tion to estimate W 111 2 .SOLUTION

Since a is the age of the boy, a 5 10 corresponds to a 10-year-old boy. W 110 2 5 75.21 means that the average weight of a 10-year-old boy is 75.21 pounds.

The units of the derivative are pounds

year of his age.

W r 110 2 5 7.29 means that the average weight of a

10-year-old boy is increasing by 7.29 pounds per year

of his age. In other words, the average weight of a boy

will increase by about 7.29 pounds between his 10th

and 11th years.

We use a tangent-line approximation to estimate

W 111 2 . W 111 2 < W 110 2 1 W r 110 2

< 75.21 1 7.29

< 82.5

We estimate that the average weight of an 11-year-old

boy is 82.5 pounds.

EXAMPLE 3 Interpreting the Meaning

of the Derivative

DBased on data from 1992–2005, the number of television sets in U.S. homes may be modeled by

T 1x 2 5 0.06680x3 2 1.276x2 1 12.46x 1 190.7 million sets

where x is the number of years since 1992. (Source: Sta-tistical Abstract of the United States, 2007, Table 1111)

Interpret the meaning of T 112 2 5 271.9 and

T r 112 2 5 10.7.

SOLUTION

Since t is the number of years, t 5 12 corresponds with 2004. T 112 2 5 271.9 means that in 2004, there were 271.9 million television sets in U.S. homes.

The units of the derivative are television sets

year.

T r 112 2 5 10.7 means that in 2004, the number of tele-

vision sets in U.S. homes was increasing at a rate of

10.7 million sets per year. In other words, the number

of television sets in homes was expected to increase by

about 10.7 million sets between 2004 and 2005.

3.5 Exercises

In Exercises 1–18, interpret the real-life meaning of the indicated values. Answer additional questions as appropriate.

1. Body Weight The weight of a girl between 2 and 13 years of age may be modeled by

W 1a 2 5 0.289a2 1 2.464a 1 23.10 pounds

where a is the age of the girl. (Source: Modeled from www.babybag.com data) Interpret the meaning of W 110 2 5 76.64 and W r 110 2 5 8.24. Then estimate W 111 2 .

2. Body Weight Compare the results of Example 2 and Exercise 1. Were boys or girls expected to gain more weight between their 10th and 11th years? Explain.

3. Carbon Monoxide Pollution Based on data from

1990–2003, carbon monoxide pollution may be

modeled by

P 1 t 2 5 20.248t 1 5.99 parts per million

where t is the number of years since 1990. (Source: Statistical Abstract of the United States, 2006, Table 359) Interpret the meaning of P 115 2 5 2.27 and P r 115 2 5 20.248.

4. Kazakhstan Population Based on data from

1995–2005, the population of Kazakhstan may be

modeled by

K 1 t 2 5 14,825 10.993 2 t thousand people

where t is the number of years since 2005. (Source: World Health Statistics 2006, World Health Organization) Interpret the meaning of K 110 2 5 13,819 and P r 110 2 5 297.08.

5. India Population Based on data from 1995–2005, the population of the India may be modeled by

I 1 t 2 5 1,103,371 11.015 2 t thousand people


NOT FOR SALE

chapter 3 of applied calc

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