answers to even-numbered hw problems

Answers to even-numbered HW problems

S-2 42,943,441.08

Ex 6 a) C(98,230) =$7.04

b) C(1.03, 172) means the cost of operating a car when gas is $1.03 per gallon and you travel 172 miles. It’s value is $5.54.

Ex 8 a) 45 degrees

b) R(30) = 84 degrees

c) 13.66 degrees

d) 10.12 degrees

Ex 16 a) .355 dynes

b) 199.61 dynes

Section 1.1

WebAssign Homework

WebAssign Homework – Section 1-1

Last time, we found a formula for the area of a garden surrounded by 20 yards of chicken wire on three sides.

We said the formula is an example of a function and we used the functional notation A(W) instead of just A.

A(W) = 20W – 2W2

Today, we will refine the concept of a FUNCTION.

Consider the function A(W) = 20W – 2W2

Warm-up

Use a calculator to compute the values of A(4), A( ), A(10), and A(-7).

Round answers to the nearest hundredth (two decimal places).

78


Warm-up

A(4) = 48



78


Warm-up

A( ) = 15.97A(4) = 48 78



78


Warm-up

A( ) = 15.97A(4) = 48 A(10) = 078



78


Warm-up

A( ) = 15.97A(4) = 48 A(10) = 0 A(-7) = -23878



78


Warm-up

A( ) = 15.97A(4) = 48 A(10) = 0 A(-7) = -238

Input valueOutput or Function value

78



78


Warm-up

A( ) = 15.97A(4) = 48 A(10) = 0 A(-7) = -238

We could also write this in table form

78



78


Warm-up

A( ) = 15.97A(4) = 48 A(10) = 0 A(-7) = -238

We could also write this in table form

W 4 10 -7

A(W) 48 15.97 0 -238

78

78

Input values

Output or Function values



78

A function can be thought of as a pairing of the numbers in one set with the numbers in a second set. Sometimes the rule for the pairs is given as an equation or is easy to spot – sometimes it isn’t.

Definition of a function: A function is a correspondence between two sets that assigns to each element of the first set exactly one element of the second set.

The first set of numbers is called the domain.

domain

range.

W A(W)

4 48

15.97

10 0

-7 -238

78

The second set of numbers is called the range.

Definition of a function: A function is a correspondence between two sets that assigns to each element of the domain exactly one element of the range.

We usually refer to the function by the letter name for the range. In this case, we would call it function A.

In this example, the domain is {4, , 10, -7}

and the range is {48, 15.97, 0, -238}

78

Note: the table for the function says the same thing as

A(4) = 48, A( ) = 15.97, A(10) = 0, A(-7) = -238.

78

W A(W)

4 48

15.97

10 0

-7 -238

78

Summary:

In a function, the first components (domain) never repeat.

The second components (range) may repeat.

Example: A(W) = 20W – 2W2

W 4 7/8 10 -7

A(W) 48 15.97 0 -238

A function can be represented in several ways, including: As a formula Example: As a table

As an arrow diagram 4 48 10 0

15.97 -7 -238

Example: 78

As a set of number pairs Example: {(4, 48), ( , 15.97), (10, 0), (-7, -238)}78

For example, W 4 7/8 10 -7

A(W) 48 15.97 0 -238

6

48

Using functions given as tables of values

Year (d) U.S. Population (N) (in millions)

1940 132.1 1950 152.3 1960 179.3 1970 203.3 1980 226.5 1990 248.7 2000 281.4

The population of the United States has increased steadily over the years. Thus, the population (N) of the U.S. is a function of the year (d).

We can use N(d) in place of N.


1940 132.1 1950 152.3 1960 179.3 1970 203.3 1980 226.5 1990 248.7 2000 281.4

N = N(d)

We can still use function notation in this situation.

For example: What is meant by N(1980)?

“the population of the United States in 1980”

How can we use the table to approximate the population of the United States in 1985?

Since 1985 is midway between 1980 and 1990, both of which are in the table, one method is to average the

populations for 1980 and 1990.

N(1980) = 226.5 million

N = N(d)

N(1980) =

N(1990) =

226.5 million

248.7 million

2

7.2485.226 = 237.6 millionAverage population =

A reasonable estimate for the population of the U.S. in 1985 is 237.6 million.


1940 132.1 1950 152.3 1960 179.3 1970 203.3 1980 226.5 1990 248.7 2000 281.4

Therefore, N(1985) 237.6 million

United States Population

July 1, 1987 242,288,918 2,156,031 0.89

July 1, 1986 240,132,887 2,209,092 0.92

July 1, 1985 237,923,795 2,098,893 0.89

July 1, 1984 235,824,902 2,032,908 0.87

July 1, 1983 233,791,994 2,127,536 Source: http://www.npg.org/facts/us_historical_pops.htm

N = N(d)

How would you estimate the population of the United States in 1945?

Averaging assumes that values of the function (in this case, population) increase at a constant (uniform) rate.


1940 132.1 1950 152.3 1960 179.3 1970 203.3 1980 226.5 1990 248.7 2000 281.4


July 1, 1950 152,271,417 3,083,287 2.05July 1, 1949 149,188,130 2,556,828 1.73July 1, 1948 146,631,302 2,505,231 1.72July 1, 1947 144,126,071 2,737,505 1.92July 1, 1946 141,388,566 1,460,401 1.04July 1, 1945 139,928,165 1,530,820 1.10July 1, 1944 138,397,345 1,657,992 1.21July 1, 1943 136,739,353 1,879,800 1.38July 1, 1942 134,859,553 1,457,082 1.09July 1, 1941 133,402,471 1,280,025 0.96July 1, 1940 132,122,446 1,242,728 0.95

The U.S. population grew by less than 8 million from 1940-1945.

The U.S. population grew by more than 12 million from 1945-1950.

Source: http://www.npg.org/facts/us_historical_pops.htm

The rate of growth was faster in the second half of the decade.

If we had averaged, million people2.1422

3.1521.132

N = N(d)

How would you estimate the population of the United States in 1945?

Averaging assumes that values of the function (in this case, population) increase at a constant (uniform) rate.


1940 132.1 1950 152.3 1960 179.3 1970 203.3 1980 226.5 1990 248.7 2000 281.4

Averaging would not be an appropriate method for estimating the United States population in 1945 because historical evidence suggests the population did not rise at a uniform rate between 1940 and 1950. The population increased far more rapidly from 1945 to 1950 because of the post World War II Baby Boom.

N = N(d)

What if we wanted an estimate of the population in 1983?

To find the average rate of change in population over an interval, divide the change (difference) in population by the number of years in the interval (change in years).

In our example: 22.210

2.22

10

5.2267.248

19801990

)1980(N)1990(N

So the average rate of change for the ten year period from 1980 to 1990 is 2.22 million people per year.

To do this, we will find the average yearly change in population between 1980 and 1990 (called the average rate of change).

This means that, on average, the population increased by 2.22 million per year from 1980 to 1990.

Therefore, our estimate for the population in 1983 is

226.5 + 3(2.22) = 233.16 million


1940 132.1 1950 152.3 1960 179.3 1970 203.3 1980 226.5 1990 248.7 2000 281.4


226.5 + 3(2.22) = 233.16 million


Jul 1, 1990 249.62 million

Jul 1, 1989 246.82 million

Jul 1, 1988 244.50 million

Jul 1, 1987 242.29 million

Jul 1, 1986 240.13 million

Jul 1, 1985 237.92 million

Jul 1, 1984 235.83 million

Jul 1, 1983 233.79 million

Jul 1, 1982 231.66 million

Jul 1, 1981 229.47 million

Jul 1, 1980 227.23 million http://www.multpl.com/united-states-population/table


July 1, 1950 152,271,417 3,083,287 2.05July 1, 1949 149,188,130 2,556,828 1.73July 1, 1948 146,631,302 2,505,231 1.72July 1, 1947 144,126,071 2,737,505 1.92July 1, 1946 141,388,566 1,460,401 1.04July 1, 1945 139,928,165 1,530,820 1.10July 1, 1944 138,397,345 1,657,992 1.21July 1, 1943 136,739,353 1,879,800 1.38July 1, 1942 134,859,553 1,457,082 1.09July 1, 1941 133,402,471 1,280,025 0.96July 1, 1940 132,122,446 1,242,728 0.95

The U.S. population grew by less than 8 million from 1940-1945.

The U.S. population grew by more than 12 million from 1945-1950.

Source: http://www.npg.org/facts/us_historical_pops.htm

Average rate of change from 1940-1945:

5

1.1329.1391.56 million people per year

Average rate of change from 1945-1950:

5

9.1393.1522.48 million people per year

N = N(d)

To find the average rate of change in population over an interval, divide the change (difference) in population by the number of years in the interval (change in years).

22.210

2.22

10

5.2267.248

19801990

)1980(N)1990(N

So the average rate of change for the ten year period from 1980 to 1990 is 2.22 million people per year.

This means that, on average, the population increased by 2.22 million per year from 1980 to 1990.


226.5 + 3(2.22) = 233.16 million


1940 132.1 1950 152.3 1960 179.3 1970 203.3 1980 226.5 1990 248.7 2000 281.4

P = Number of Breeding cccPairs of Bald Eagles

Bald Eagles were once very common throughout most of the United States. Their population numbers have been estimated at 300,000 to 500,000 birds in the early 1700s. Their population fell to threatened levels in the continental U.S. of less than 10,000 nesting pairs by the 1950s.

Bald eagles were officially declared an endangered species in 1967 in all areas of the United States

Shown is a table indicating the estimated number of breeding pairs of bald eagles in the lower 48 states in various years.

Answer each of the following questions about the function P = P(y), where y is the year and P is the number of breeding pairs of bald eagles.

1940

1952 1956 1981 1986 1992 2000

9,551 4,750 3,210 1,188 1,875 3,399 6,471

y = Year

Practice

1. Explain the meaning of P(1956) and give its value.

2. Estimate the value of P(1989).

3. Find the average yearly rate of change from 1992 to 2000.

4. Use your answer to question 3 to estimate the number of breeding pairs of bald eagles in 1994.

P = Number of Breeding cccPairs of Bald Eagles

1940

1952 1956 1981 1986 1992 2000

9,551 4,750 3,210 1,188 1,875 3,399 6,471

y = Year

P(1989) is approximately 2,637.

The average yearly rate of change from 1992 to 2000 was 384 breeding pairs per year.

P(1956) represents the number of breeding pairs of bald eagles in the U.S. in 1956.

P(1956) = 3,210

There were approximately 4,167 breeding pairs of bald eagles in the U.S. in 1994. [3,399 + 2(384)]

8

33996471

Homework:

Read section 1.2 (omit pages 46–51)

Do the following problems:

Page 52 # S-1, S-3, S-8, S-11, S-15

Pages 53-56 # 1, 7, 8, 10

Not part c Not parts g and h

answers to even-numbered hw problems

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