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2.6 Variation and Applications

Find equations of direct variation, inverse variation, and combined variation given values of the variables.
Solve applied problems involving variation.

We now consider applications involving variation.

Direct Variation

The median hourly wage for an elevator and escalator installer/repairer is $35 per hour (Source: U.S. Bureau of Labor Statistics). In 1 hr, $35 is earned; in 2 hr, $70 is earned; in 3 hr, $105 is earned; and so on. This gives rise to a set of ordered pairs:

(1, 35), (2, 70), (3, 105), (4, 140), and so on .

$(1, 35), (2, 70), (3, 105), (4, 140), and so on .$

Note that the ratio of the second coordinate to the first coordinate is the same number for each pair:

\frac{35}{1} = 35, \frac{70}{2} = 35, \frac{105}{3} = 35, \frac{140}{4} = 35, and so on .

$\frac{35}{1} = 35, \frac{70}{2} = 35, \frac{105}{3} = 35, \frac{140}{4} = 35, and so on .$

Whenever a situation produces pairs of numbers in which the ratio is constant, we say that there is direct variation. In this case, the amount earned E varies directly as the time worked t:

\frac{E}{t} = 35 (a constant), or E = 35 t,

$\frac{E}{t} = 35 (a constant), or E = 35 t,$

or, if we use function notation, $E (t) = 35 t$ $E (t) = 35 t$ . This equation is an equation of direct variation. The coefficient, 35, is called the variation constant. In this case, it is the rate of change of earnings with respect to time.

Example 1

Find the variation constant and an equation of variation in which y varies directly as x, and $y = 32$ $y = 32$ when $x = 2$ $x = 2$ .

Solution

We know that (2, 32) is a solution of $y = k x$ $y = k x$ . Thus,

\begin{array}{rcll} y & = & k x \\ 32 & = & k \cdot 2 & Substituting \\ \frac{32}{2} & = & k & Solving for k \\ 16 & = & k . & Simplifying \end{array}

$\begin{array}{rcll} y & = & k x \\ 32 & = & k \cdot 2 & Substituting \\ \frac{32}{2} & = & k & Solving for k \\ 16 & = & k . & Simplifying \end{array}$

The variation constant, 16, is the rate of change of y with respect to x. The equation of variation is $y = 16 x$ $y = 16 x$ .

Now Try Exercise 1.

Example 2

Water from Melting Snow. The number of centimeters W of water produced from melting snow varies directly as S, the number of centimeters of snow. Meteorologists have found that under certain conditions 150 cm of snow will melt to 16.8 cm of water. To how many centimeters of water will 200 cm of snow melt under the same conditions?

Solution

We can express the amount of water as a function of the amount of snow. Thus, $W (S) = k S$ $W (S) = k S$ , where k is the variation constant. We first find k using the given data and then find an equation of variation:

\begin{array}{rcll} W (S) & = & k S & W varies directly as S . \\ W (150) & = & k \cdot 150 & Substituting 150 for S \\ 16.8 & = & k \cdot 150 & Replacing W (150) with 16.8 \\ \frac{16.8}{150} & = & k & Solving for k \\ 0.112 & = & k . & This is the variation constant . \end{array}

$\begin{array}{rcll} W (S) & = & k S & W varies directly as S . \\ W (150) & = & k \cdot 150 & Substituting 150 for S \\ 16.8 & = & k \cdot 150 & Replacing W (150) with 16.8 \\ \frac{16.8}{150} & = & k & Solving for k \\ 0.112 & = & k . & This is the variation constant . \end{array}$

The equation of variation is $W (S) = 0.112 S$ $W (S) = 0.112 S$ .

Next, we use the equation to find how many centimeters of water will result from melting 200 cm of snow:

\begin{array}{rcll} W (S) & = & 0.112 S \\ W (200) & = & 0.112 (200) & Substituting \\ = & 22.4. \end{array}

$\begin{array}{rcll} W (S) & = & 0.112 S \\ W (200) & = & 0.112 (200) & Substituting \\ = & 22.4. \end{array}$

Thus, 200 cm of snow will melt to 22.4 cm of water.

Now Try Exercise 17.

Inverse Variation

Suppose a bus is traveling a distance of 20 mi. At a speed of 5 mph, the trip will take 4 hr; at 10 mph, it will take 2 hr; at 20 mph, it will take 1 hr; at 40 mph, it will take $\frac{1}{2}$ $\frac{1}{2}$ hr; and so on. We plot this information on a graph, using speed as the first coordinate and time as the second coordinate to determine a set of ordered pairs:

(5, 4), (10, 2), (20, 1), (40, \frac{1}{2}), and so on .

$(5, 4), (10, 2), (20, 1), (40, \frac{1}{2}), and so on .$

Note that the products of the coordinates are all the same number:

5 \cdot 4 = 20, 10 \cdot 2 = 20, 20 \cdot 1 = 20, 40 \cdot \frac{1}{2} = 20, and so on .

$5 \cdot 4 = 20, 10 \cdot 2 = 20, 20 \cdot 1 = 20, 40 \cdot \frac{1}{2} = 20, and so on .$

Whenever a situation produces pairs of numbers in which the product is constant, we say that there is inverse variation. In this case, the time varies inversely as the speed, or rate:

r t = 20 (a constant), or t = \frac{20}{r},

$r t = 20 (a constant), or t = \frac{20}{r},$

or, if we use function notation, $t (r) = 20 / r$ $t (r) = 20 / r$ . This equation is an equation of inverse variation. The coefficient, 20, is called the variation constant. Note that as the first number increases, the second number decreases.

Example 3

Find the variation constant and an equation of variation in which y varies inversely as x, and $y = 16$ $y = 16$ when $x = 0.3$ $x = 0.3$ .

Solution

We know that (0.3, 16) is a solution of $y = k / x$ $y = k / x$ . We substitute:

\begin{array}{rcll} y & = & \frac{k}{x} \\ 16 & = & \frac{k}{0.3} & Substituting \\ (0.3) 16 & = & k & Solving for k \\ 4.8 & = & k . \end{array}

$\begin{array}{rcll} y & = & \frac{k}{x} \\ 16 & = & \frac{k}{0.3} & Substituting \\ (0.3) 16 & = & k & Solving for k \\ 4.8 & = & k . \end{array}$

The variation constant is 4.8. The equation of variation is $y = 4.8 / x$ $y = 4.8 / x$ .

Now Try Exercise 3.

There are many real-world problems that translate to an equation of inverse variation.

Example 4 Filling a Swimming Pool.

The time t required to fill a swimming pool varies inversely as the rate of flow r of water into the pool. A tank truck can fill a pool in 90 min at a rate of 1500 L/min. How long would it take to fill the pool at a rate of 1800 L/min?

Solution

We can express the amount of time required as a function of the rate of flow. Thus we have $t (r) = k / r$ $t (r) = k / r$ . We first find k using the given information and then find an equation of variation:

\begin{array}{rcll} t (r) & = & \frac{k}{r} & t varies inversely as r . \\ t (1500) & = & \frac{k}{1500} & Substituting 1500 for r \\ 90 & = & \frac{k}{1500} & Replacing t (1500) with 90 \\ 90 \cdot 1500 & = & k & Solving for k \\ 135,000 & = & k . & This is the variation constant . \end{array}

$\begin{array}{rcll} t (r) & = & \frac{k}{r} & t varies inversely as r . \\ t (1500) & = & \frac{k}{1500} & Substituting 1500 for r \\ 90 & = & \frac{k}{1500} & Replacing t (1500) with 90 \\ 90 \cdot 1500 & = & k & Solving for k \\ 135,000 & = & k . & This is the variation constant . \end{array}$

The equation of variation is

t (r) = \frac{135,000}{r} .

$t (r) = \frac{135,000}{r} .$

Next, we use the equation to find the time that it would take to fill the pool at a rate of 1800 L/min:

\begin{array}{rcll} t (r) & = & \frac{135,000}{r} \\ t (1800) & = & \frac{135,000}{1800} & Substituting \\ t & = & 75. \end{array}

$\begin{array}{rcll} t (r) & = & \frac{135,000}{r} \\ t (1800) & = & \frac{135,000}{1800} & Substituting \\ t & = & 75. \end{array}$

Thus it would take 75 min to fill the pool at a rate of 1800 L/min.

Now Try Exercise 15.

Let’s summarize the procedure for solving variation problems.

Combined Variation

We now look at other kinds of variation.

There are other types of combined variation as well. Consider the formula for the volume of a right circular cylinder, $V = π r^{2} h$ $V = π r^{2} h$ , in which V, r, and h are variables and $π$ $π$ is a constant. We say that V varies jointly as h and the square of r. In this formula, $π$ $π$ is the variation constant.

Example 5

Find an equation of variation in which y varies directly as the square of x, and $y = 12$ $y = 12$ when $x = 2$ $x = 2$ .

Solution

We write an equation of variation and find k:

\begin{array}{rcll} y & = & k x^{2} \\ 12 & = & k \cdot 2^{2} & Substituting \\ 12 & = & k \cdot 4 \\ 3 & = & k . \end{array}

$\begin{array}{rcll} y & = & k x^{2} \\ 12 & = & k \cdot 2^{2} & Substituting \\ 12 & = & k \cdot 4 \\ 3 & = & k . \end{array}$

Thus, $y = 3 x^{2}$ $y = 3 x^{2}$ .

Now Try Exercise 27.

Example 6

Find an equation of variation in which y varies jointly as x and z, and $y = 42$ $y = 42$ when $x = 2$ $x = 2$ and $z = 3$ $z = 3$ .

Solution

We have

\begin{array}{rcll} y & = & k x z \\ 42 & = & k \cdot 2 \cdot 3 & Substituting \\ 42 & = & k \cdot 6 \\ 7 & = & k . \end{array}

$\begin{array}{rcll} y & = & k x z \\ 42 & = & k \cdot 2 \cdot 3 & Substituting \\ 42 & = & k \cdot 6 \\ 7 & = & k . \end{array}$

Thus, $y = 7 x z$ $y = 7 x z$ .

Now Try Exercise 29.

Example 7

Find an equation of variation in which y varies jointly as x and z and inversely as the square of w, and $y = 105$ $y = 105$ when $x = 3, z = 20$ $x = 3, z = 20$ , and $w = 2$ $w = 2$ .

Solution

We have

\begin{array}{rcll} y & = & k \cdot \frac{x z}{w^{2}} \\ 105 & = & k \cdot \frac{3 \cdot 20}{2^{2}} & Substituting \\ 105 & = & k \cdot 15 \\ 7 & = & k . \end{array}

$\begin{array}{rcll} y & = & k \cdot \frac{x z}{w^{2}} \\ 105 & = & k \cdot \frac{3 \cdot 20}{2^{2}} & Substituting \\ 105 & = & k \cdot 15 \\ 7 & = & k . \end{array}$

Thus, $y = 7 \frac{x z}{w^{2}}$ $y = 7 \frac{x z}{w^{2}}$ , or $y = \frac{7 x z}{w^{2}}$ $y = \frac{7 x z}{w^{2}}$ .

Now Try Exercise 33.

Example 8

Volume of a Tree The volume of wood V in a tree varies jointly as the height h and the square of the girth g. (Girth is distance around.) If the volume of a redwood tree is $216 m^{3}$ $216 m^{3}$ when the height is 30 m and the girth is 1.5 m, what is the height of a tree whose volume is $344 m^{3}$ $344 m^{3}$ and whose girth is 1.6 m?

Solution

We first find k using the first set of data. Then we solve for h using the second set of data.

\begin{array}{rcll} V & = & k h g^{2} \\ 216 & = & k \cdot 30 \cdot {1.5}^{2} \\ 216 & = & k \cdot 30 \cdot 2.25 \\ 216 & = & k \cdot 67.5 \\ 3.2 & = & k \end{array}

$\begin{array}{rcll} V & = & k h g^{2} \\ 216 & = & k \cdot 30 \cdot {1.5}^{2} \\ 216 & = & k \cdot 30 \cdot 2.25 \\ 216 & = & k \cdot 67.5 \\ 3.2 & = & k \end{array}$

Thus the equation of variation is $V = 3.2 h g^{2}$ $V = 3.2 h g^{2}$ . We substitute the second set of data into the equation:

\begin{array}{rcll} 344 & = & 3.2 \cdot h \cdot {1.6}^{2} \\ 344 & = & 3.2 \cdot h \cdot 2.56 \\ 344 & = & 8.192 \cdot h \\ 42 & \approx & h . \end{array}

$\begin{array}{rcll} 344 & = & 3.2 \cdot h \cdot {1.6}^{2} \\ 344 & = & 3.2 \cdot h \cdot 2.56 \\ 344 & = & 8.192 \cdot h \\ 42 & \approx & h . \end{array}$

The height of the tree is about 42 m.

Now Try Exercise 35.

2.6 Exercise Set

Find the variation constant and an equation of variation for the given situation.

1. y varies directly as x, and $y = 54$ $y = 54$ when $x = 12$ $x = 12$
2. y varies directly as x, and $y = 0.1$ $y = 0.1$ when $x = 0.2$ $x = 0.2$
3. y varies inversely as x, and $y = 3$ $y = 3$ when $x = 12$ $x = 12$
4. y varies inversely as x, and $y = 12$ $y = 12$ when $x = 5$ $x = 5$
5. y varies directly as x, and $y = 1$ $y = 1$ when $x = \frac{1}{4}$ $x = \frac{1}{4}$
6. y varies inversely as x, and $y = 0.1$ $y = 0.1$ when $x = 0.5$ $x = 0.5$
7. y varies inversely as x, and $y = 32$ $y = 32$ when $x = \frac{1}{8}$ $x = \frac{1}{8}$
8. y varies directly as x, and $y = 3$ $y = 3$ when $x = 33$ $x = 33$
9. y varies directly as x, and $y = \frac{3}{4}$ $y = \frac{3}{4}$ when $x = 2$ $x = 2$
10. y varies inversely as x, and $y = \frac{1}{5}$ $y = \frac{1}{5}$ when $x = 35$ $x = 35$
11. y varies inversely as x, and $y = 1.8$ $y = 1.8$ when $x = 0.3$ $x = 0.3$
12. y varies directly as x, and $y = 0.9$ $y = 0.9$ when $x = 0.4$ $x = 0.4$
13. Child’s Allowance. The Harrisons decide to give their children a weekly allowance that is directly proportional to each child’s age. Their 6-year-old daughter receives an allowance of $5.50. What is their 9-year-old son’s allowance?
14. Sales Tax. The amount of sales tax paid on a product is directly proportional to the purchase price. In Iowa, the sales tax on a Nook Glowlight^TM that sells for $119 is $7.14. What is the sales tax on an e-book that sells for $21?
15. Rate of Travel. The time t required to drive a fixed distance varies inversely as the speed r. It takes 5 hr at a speed of $80 k m / h$ $80 k m / h$ to drive a fixed distance. How long will it take to drive the same distance at a speed of $70 k m / h$ $70 k m / h$ ?
16. Beam Weight. The weight W that a horizontal beam can support varies inversely as the length L of the beam. Suppose an 8-m beam can support 1200 kg. How many kilograms can a 14-m beam support?
17. Fat Intake. The maximum number of grams of fat that should be in a diet varies directly as a person’s weight. A person weighing 120 lb should have no more than 60 g of fat per day. What is the maximum daily fat intake for a person weighing 180 lb?
18. House of Representatives. The number of representatives N that each state has varies directly as the number of people P living in the state. If California, with 38,040,000 residents, has 53 representatives, how many representatives does Texas, with a population of 26,060,000, have?
19. Work Rate. The time T required to do a job varies inversely as the number of people P working. It takes 5 hr for 7 bricklayers to build a park wall. How long will it take 10 bricklayers to complete the job?
20. Pumping Rate. The time t required to empty a tank varies inversely as the rate r of pumping. If a pump can empty a tank in 45 min at the rate of $600 k L / m i n$ $600 k L / m i n$ , how long will it take the pump to empty the same tank at the rate of $1000 k L / m i n$ $1000 k L / m i n$ ?
21. Hooke’s Law. Hooke’s law states that the distance d that a spring will stretch varies directly as the mass m of an object hanging from the spring. If a 3-kg mass stretches a spring 40 cm, how far will a 5-kg mass stretch the spring?
22. Relative Aperture. The relative aperture, or f-stop, of a 23.5-mm diameter lens is directly proportional to the focal length F of the lens. If a 150-mm focal length has an f-stop of 6.3, find the f-stop of a 23.5-mm diameter lens with a focal length of 80 mm.
23. Musical Pitch. The pitch P of a musical tone varies inversely as its wavelength W. One tone has a pitch of 330 vibrations per second and a wavelength of 3.2 ft. Find the wavelength of another tone that has a pitch of 550 vibrations per second.
24. Weight on Mars. The weight M of an object on Mars varies directly as its weight E on Earth. A person who weighs 95 lb on Earth weighs 35.9 lb on Mars. How much would a 100-lb person weigh on Mars?

Find an equation of variation for the given situation.

25. y varies inversely as the square of x, and $y = 0.15$ $y = 0.15$ when $x = 0.1$ $x = 0.1$
26. y varies inversely as the square of x, and $y = 6$ $y = 6$ when $x = 3$ $x = 3$
27. y varies directly as the square of x, and $y = 0.15$ $y = 0.15$ when $x = 0.1$ $x = 0.1$
28. y varies directly as the square of x, and $y = 6$ $y = 6$ when $x = 3$ $x = 3$
29. y varies jointly as x and z, and $y = 56$ $y = 56$ when $x = 7$ $x = 7$ and $z = 8$ $z = 8$
30. y varies directly as x and inversely as z, and $y = 4$ $y = 4$ when $x = 12$ $x = 12$ and $z = 15$ $z = 15$
31. y varies jointly as x and the square of z, and $y = 105$ $y = 105$ when $x = 14$ $x = 14$ and $z = 5$ $z = 5$
32. y varies jointly as x and z and inversely as w, and $y = \frac{3}{2}$ $y = \frac{3}{2}$ when $x = 2, z = 3$ $x = 2, z = 3$ , and $w = 4$ $w = 4$
33. y varies jointly as x and z and inversely as the product of w and p, and $y = \frac{3}{28}$ $y = \frac{3}{28}$ when $x = 3$ $x = 3$ , $z = 10$ $z = 10$ , $w = 7$ $w = 7$ , and $p = 8$ $p = 8$
34. y varies jointly as x and z and inversely as the square of w, and $y = \frac{12}{5}$ $y = \frac{12}{5}$ when $x = 16$ $x = 16$ , $z = 3$ $z = 3$ , and $w = 5$ $w = 5$
35. Intensity of Light. The intensity I of light from a light bulb varies inversely as the square of the distance d from the bulb. Suppose that I is $90 W / m^{2}$ $90 W / m^{2}$ (watts per square meter) when the distance is 5 m. How much farther would it be to a point where the intensity is $40 W / m^{2}$ $40 W / m^{2}$ ?
36. Atmospheric Drag. Wind resistance, or atmospheric drag, tends to slow down moving objects. Atmospheric drag varies jointly as an object’s surface area A and velocity v. If a car traveling at a speed of 40 mph with a surface area of $37.8 {ft}^{2}$ $37.8 {ft}^{2}$ experiences a drag of 222 N (Newtons), how fast must a car with $51 {ft}^{2}$ $51 {ft}^{2}$ of surface area travel in order to experience a drag force of 430 N?
37. Stopping Distance of a Car. The stopping distance d of a car after the brakes have been applied varies directly as the square of the speed r. If a car traveling 60 mph can stop in 200 ft, how fast can a car travel and still stop in 72 ft?
38. Weight of an Astronaut. The weight W of an object varies inversely as the square of the distance d from the center of the earth. At sea level (3978 mi from the center of the earth), an astronaut weighs 220 lb. Find his weight when he is 200 mi above the surface of the earth.
39. Earned-Run Average. A pitcher’s earned-run average E varies directly as the number R of earned runs allowed and inversely as the number I of innings pitched. In 2013, Jon Lester of the Boston Red Sox had an earned-run average of 3.75. He gave up 89 earned runs in 213.1 innings. How many earned runs would he have given up had he pitched 235 innings with the same average? Round to the nearest whole number.
40. Boyle’s Law. The volume V of a given mass of a gas varies directly as the temperature T and inversely as the pressure P. If $V = 231 {cm}^{3}$ $V = 231 {cm}^{3}$ when $T = 42 °$ $T = 42 °$ and $P = 20 kg / {cm}^{2}$ $P = 20 kg / {cm}^{2}$ , what is the volume when $T = 30 °$ $T = 30 °$ and $P = 15 kg / {cm}^{2}$ $P = 15 kg / {cm}^{2}$ ?

Skill Maintenance

Vocabulary Reinforcement

In each of Exercises 41–45, fill in the blank with the correct term. Some of the given choices will not be used.

even function
odd function
constant function
composite function
direct variation
inverse variation
relative maximum
relative minimum
solution
zero
perpendicular
parallel

41. Nonvertical lines are _______________ if and only if they have the same slope and different y-intercepts. [1.4]
42. An input c of a function f is a(n) _______________ of the function if $f (c) = 0$ $f (c) = 0$ . [1.5]
43. For a function f for which $f (c)$ $f (c)$ exists, $f (c)$ $f (c)$ is a(n) _______________ if $f (c)$ $f (c)$ is the lowest point in some open interval. [2.1]
44. If the graph of a function is symmetric with respect to the origin, then f is a(n) _______________. [2.4]
45. An equation $y = k / x$ $y = k / x$ is an equation of _______________. [2.6]

Synthesis

46. In each of the following equations, state whether y varies directly as x, inversely as x, or neither directly nor inversely as x.
1. $7 x y = 14$ $7 x y = 14$
2. $x - 2 y = 12$ $x - 2 y = 12$
3. $- 2 x + 3 y = 0$ $- 2 x + 3 y = 0$
4. $x = \frac{3}{4} y$ $x = \frac{3}{4} y$
5. $\frac{x}{y} = 2$ $\frac{x}{y} = 2$
47. Volume and Cost. An 18-oz jar of peanut butter in the shape of a right circular cylinder is 5 in. high and 3 in. in diameter and sells for $2.89. In the same store, a 22-oz jar of the same brand is $5 \frac{1}{4}$ $5 \frac{1}{4}$ in. high and $3 \frac{1}{4}$ $3 \frac{1}{4}$ in. in diameter. If the cost is directly proportional to volume, what should the price of the larger jar be? If the cost is directly proportional to weight, what should the price of the larger jar be?
48. Describe in words the variation given by the equation

$Q = \frac{k p^{2}}{q^{3} .}$ $Q = \frac{k p^{2}}{q^{3} .}$
49. Area of a Circle. The area of a circle varies directly as the square of the length of a diameter. What is the variation constant?

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Table of Contents for 2.6 Variation and Applications

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Table of Contents for
2.6 Variation and Applications