390 CHAPTER 6 Rational Expressions - DR. POLLONE'S WEBSITE

390 CHAPTER 6 Rational Expressions

66. A doctor recorded a body-mass index of 47 on a patient's chart. Later, a nurse notices that the doctor recorded the patient's weight as 240 pounds but neglected to record the patient's height. Explain how the nurse can use the information from the chart to find the patient's height. Then find the height.

In physics, when the source of a sound is traveling toward an ob-

server, the relationship between the actual pitch a of the sound and

the pitch h that the observer hears due to the Doppler effect is de-

scribed by the formula h =

a , where s is the speed of the

s

1 - 770

sound source in miles per hour. Use this formula to answer Exer-

cise 67 and 68.

67. An emergency vehicle has a single-tone siren with the pitch of the musical note E. As it approaches an observer standing by the road, the vehicle is traveling 50 mph. Is the pitch that the observer hears due to the Doppler effect lower or higher than the actual pitch? To which musical note is the pitch that the observer hears closest?

Pitch of an Octave of Musical Notes in Hertz (Hz)

Note Middle C D E F G A B

Pitch 261.63 293.66 329.63 349.23 392.00 440.00 493.88

Note: Greater numbers indicate higher pitches (acoustically). (Source: American Standards Association)

68. Suppose an emergency van has a single-tone siren with the pitch of the musical note G. If the van is traveling at 80 mph approaching a standing observer, name the pitch the observer hears (rounded to the nearest tenth) and the musical note closest to that pitch.

In electronics, the relationship among the resistances R1 and R2 of two resistors wired in a parallel circuit and their combined resis-

1 tance R is described by the formula formula to solve Exercises 69 through 71. R

=

1 R1

+

1 .

R2

Use

this

69. If the combined resistance is 2 ohms and one of the two resistances is 3 ohms, find the other resistance.

70. Find the combined resistance of two resistors of 12 ohms each when they are wired in a parallel circuit.

71. The relationship among resistance of two resistors wired in a parallel circuit and their combined resistance may be extended to three resistors of resistances R1, R2, and R3. Write an equation you believe may describe the relationship and use it to find the combined resistance if R1 is 5, R2 is 6, and

R3 is 2.

72.

For the formula

1 x

=

1 y

+

1 z

-

1 w,

find

x

if

y = 2, z = 7,

and w = 6.

6.7 Variation and Problem Solving

OBJECTIVES

1 Solve Problems Involving

Direct Variation.

2 Solve Problems Involving

Inverse Variation.

3 Solve Problems Involving Joint

Variation.

4 Solve Problems Involving

Combined Variation.

OBJECTIVE

1 Solving Problems Involving Direct Variation A very familiar example of direct variation is the relationship of the circumference C of a circle to its radius r. The formula C = 2pr expresses that the circumference is always 2p times the radius. In other words, C is always a constant multiple 12p2 of r. Because it is, we say that C varies directly as r, that C varies directly with r, or that C is directly proportional to r.

C 2pr constant

Section 6.7 Variation and Problem Solving 391

Direct Variation y varies directly as x, or y is directly proportional to x, if there is a nonzero constant k such that

y = kx

The number k is called the constant of variation or the constant of proportionality.

In the above definition, the relationship described between x and y is a linear one. In other words, the graph of y = kx is a line. The slope of the line is k, and the line passes through the origin.

For example, the graph of the direct variation equation C = 2pr is shown. The horizontal axis represents the radius r, and the vertical axis is the circumference C. From the graph, we can read that when the radius is 6 units, the circumference is approximately 38 units. Also, when the circumference is 45 units, the radius is between 7 and 8 units. Notice that as the radius increases, the circumference increases.

C

50

45

40

35

C increases

30

25

20

15

10

5

C 2pr

1 2 3 4 5 6 7 8 9 10 r

as r increases

E X A M P L E 1 Suppose that y varies directly as x. If y is 5 when x is 30, find the constant of variation and the direct variation equation.

Solution Since y varies directly as x, we write y = kx. If y = 5 when x = 30, we have that

y = kx 5 = k1302 1=k 6

Replace y with 5 and x with 30. Solve for k.

1 The constant of variation is .

6

After finding the constant of variation k, the direct variation equation can be written

as

y

=

1 x.

6

PRACTICE

1 Suppose that y varies directly as x. If y is 20 when x is 15, find the constant of variation and the direct variation equation.

E X A M P L E 2 Using Direct Variation and Hooke's Law

Hooke's law states that the distance a spring stretches is directly proportional to the weight attached to the spring. If a 40-pound weight attached to the spring stretches the spring 5 inches, find the distance that a 65-pound weight attached to the spring stretches the spring.

(Continued on next page)

392 CHAPTER 6 Rational Expressions

Solution

1. UNDERSTAND. Read and reread the problem. Notice that we are given that the distance a spring stretches is directly proportional to the weight attached. We let

d = the distance stretched w = the weight attached

The constant of variation is represented by k.

2. TRANSLATE. Because d is directly proportional to w, we write

d = kw

d 65 lb

3. SOLVE. When a weight of 40 pounds is attached, the spring stretches 5 inches. That is, when w = 40, d = 5.

d = kw

5 = k1402 Replace d with 5 and w with 40.

1=k 8

Solve for k.

Now when we replace k with 1 in the equation d = kw, we have 8

d = 1w 8

To find the stretch when a weight of 65 pounds is attached, we replace w with 65 to find d.

d = 1 1652 8

=

65

=

1 8

or

8.125

88

4. INTERPRET.

Check: Check the proposed solution of 8.125 inches in the original problem.

State: The spring stetches 8.125 inches when a 65-pound weight is attached.

PRACTICE

2 Use Hooke's law as stated in Example 2. If a 36-pound weight attached to a spring stretches the spring 9 inches, find the distance that a 75-pound weight attached to the spring stretches the spring.

OBJECTIVE

2 Solving Problems Involving Inverse Variation

When y is proportional to the reciprocal of another variable x, we say that y varies inversely as x, or that y is inversely proportional to x. An example of the inverse variation relationship is the relationship between the pressure that a gas exerts and the volume of its container. As the volume of a container decreases, the pressure of the gas it contains increases.

Inverse Variation

y varies inversely as x, or y is inversely proportional to x, if there is a nonzero con-

stant k such that

y

=

k x

The number k is called the constant of variation or the constant of proportionality.

Section 6.7 Variation and Problem Solving 393

Notice

that

y

=

k x

is

a

rational

equation. Its

graph

for

k

7

0

and

x

7

0

is

shown.

From the graph, we can see that as x increases, y decreases.

y

y decreases

y

k x

,

k

0,

x

0

x

as x increases

E X A M P L E 3 Suppose that u varies inversely as w. If u is 3 when w is 5, find the

constant of variation and the inverse variation equation.

Solution Since u varies and we solve for k.

inversely

as

w,

we

have

u

=

k w.

We

let

u

=

3

and

w

=

5,

u

=

k w

3 = k Let u = 3 and w = 5. 5

15 = k Multiply both sides by 5.

The constant of variation k is 15. This gives the inverse variation equation

u

=

15 w

PRACTICE

3 Suppose that b varies inversely as a. If b is 5 when a is 9, find the constant of variation and the inverse variation equation.

E X A M P L E 4 Using Inverse Variation and Boyle's Law

Boyle's law says that if the temperature stays the same, the pressure P of a gas is inversely proportional to the volume V. If a cylinder in a steam engine has a pressure of 960 kilopascals when the volume is 1.4 cubic meters, find the pressure when the volume increases to 2.5 cubic meters.

Solution

1. UNDERSTAND. Read and reread the problem. Notice that we are given that the pressure of a gas is inversely proportional to the volume. We will let P = the pressure and V = the volume. The constant of variation is represented by k.

2. TRANSLATE. Because P is inversely proportional to V, we write

P

=

k V

When P = 960 kilopascals, the volume V = 1.4 cubic meters. We use this information to find k.

960 = k Let P = 960 and V = 1.4. 1.4

1344 = k Multiply both sides by 1.4.

Thus, the value of k is 1344. Replacing k with 1344 in the variation equation, we have

P

=

1344 V

Next we find P when V is 2.5 cubic meters.

(Continued on next page)

394 CHAPTER 6 Rational Expressions

3. SOLVE.

P = 1344 Let V = 2.5 . 2.5

= 537.6

4. INTERPRET.

Check: Check the proposed solution in the original problem.

State: When the volume is 2.5 cubic meters, the pressure is 537.6 kilopascals.

PRACTICE

4 Use Boyle's law as stated in Example 4. When P = 350 kilopascals and V = 2.8 cubic meters, find the pressure when the volume decreases to 1.5 cubic meters.

OBJECTIVE

3 Solving Problems Involving Joint Variation Sometimes the ratio of a variable to the product of many other variables is constant. For example, the ratio of distance traveled to the product of speed and time traveled is always 1.

d = 1 or d = rt rt Such a relationship is called joint variation.

Joint Variation

If the ratio of a variable y to the product of two or more variables is constant, then y varies jointly as, or is jointly proportional to, the other variables. If

y = kxz

then the number k is the constant of variation or the constant of proportionality.

CONCEPT CHECK

Which type of variation is represented by the equation xy = 8? Explain.

a. Direct variation

b. Inverse variation

c. Joint variation

E X A M P L E 5 Expressing Surface Area The lateral surface area of a cylinder varies jointly as its radius and height. Express this surface area S in terms of radius r and height h.

r

h

Answer to Concept Check: b; answers may vary

Solution Because the surface area varies jointly as the radius r and the height h, we equate S to a constant multiple of r and h.

S = krh

In the equation, S = krh, it can be determined that the constant k is 2p, and we then have the formula S = 2prh . (The lateral surface area formula does not include the areas of the two circular bases.)

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