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Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Rational Expressions, Equations, and Functions6
Rational Expressions, Equations, and Functions
6.1 Rational Expressions and Functions:
Multiplying and Dividing
6.2 Rational Expressions and Functions:
Adding and Subtracting
6.3 Complex Rational Expressions
6.4 Rational Equations
6.5 Solving Applications Using Rational
Equations
6.6 Division of Polynomials
6.7 Synthetic Division
6.8 Formulas, Applications, and Variation
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3. Formulas, Applications, and Variation
6.8
Formulas, Applications, and Variation
Formulas
Direct Variation
Inverse Variation
Joint and Combined Variation
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4. Formulas
Formulas occur frequently as mathematical models. Many formulas contain rational expressions, and to solve such formulas for a specified letter, we proceed as when solving rational equations.
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5. Example
In a hydraulic system, a fluid is confined to two connecting chambers. The pressure in each chamber is the same and is given by finding the force exerted (F) divided by the surface area (A). Therefore, we know
Solve for A2.
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6. Solution
Multiplying both sides by the LCD
Simplifying by removing factors
Dividing both sides by F1
This formula can be used to calculate A2 whenever A1, F2, and F1 are known.
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7. To Solve a Rational Equation for a Specified Variable
If necessary, multiply both sides by the LCD to clear fractions.
Multiply, as needed, to remove parentheses.
Get all terms with the specified variable alone on one side.
Factor out the specified variable if it is in no more than one term.
Multiply or divide on both sides to isolate the specified variable.
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8. Variation
To extend our study of formulas and functions, we now examine three real-world situations: direct variation, inverse variation, and combined variation.
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9. Direct Variation
A mass transit driver earns $17 per hour. In 1 hr, $17 is earned. In 2 hr, $34 is earned. In 3 hr, $51 is earned, and so on. This gives rise to a set of ordered pairs: (1, 22), (2, 34), (3, 51), (4, 68), and so on.
Note that the ratio of earnings E to time t is 17/1 in every case.
If a situation gives rise to pairs of numbers in which the ratio is constant, we say that there is direct variation. Here earnings vary directly as the time:
We have E/t = 17, so E = 17t, or using function notation, E(t) = 17t.
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10. Direct Variation
When a situation gives rise to a linear function of the form f (x) = kx, or y = kx, where k is a nonzero constant, we say that there is direct variation, that y varies directly as x, or that y is proportional to x. The number k is called the variation constant, or constant of proportionality.
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11. Example
Find the variation constant and an equation of variation if y varies directly as x, and y = 15 when x = 3.
Solution
We know that (3, 15) is a solution of y = kx. Therefore,
Substituting
Solving for k
The variation constant is 5. The equation of variation is y = 5x. The notation y(x) = 5x or f (x) = 5x is also used.
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12. Inverse Variation
To see what we mean by inverse variation, suppose it takes one person 8 hours to paint the baseball fields for the local park district. If two people do the job, it will take only 4 hours. If three people paint the fields, it will take only 2 and 2/3 hours, and so on. This gives rise to pairs of numbers, all have the same product: (1, 8), (2, 4), (3, 8/3), (4, 2), and so on.
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13. Inverse Variation
Note that the product of each pair of numbers is 8. Whenever a situation gives rise to pairs of numbers for which the product is constant, we say that there is inverse variation. Since pt = 8, the time t, in hours, required for the fields to be painted by p people is given by t = 8/p or, using function notation, t(p) = 8/p.
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14. Inverse Variation
When a situation gives rise to a rational function of the form f (x) = k/x, or y = k/x, where k is a nonzero constant, we say that there is inverse variation, that is y varies inversely as x, or that y is inversely proportional to x. The number k is called the variation constant, or constant of proportionality.
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15. Example
The time, t, required to empty a tank varies inversely as the rate, r, of pumping. If a pump can empty a tank in 90 minutes at the rate of 1080 kL/min, how long will it take the pump to empty the same tank at the rate of 1500 kL/min?
Solution
1. Familiarize. Because of the phrase “ varies inversely as the rate, r, of pumping,” we express the amount of time needed to empty the tank as a function of the rate: t(r) = k/r
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16. Solution (continued)
2. Translate. We use the given information to solve for k. Then we use that result to write the equation of variation.
Using function notation
Replacing r with 1080
Replacing t(90) with 1080
Solving for k, the variation constant
The equation of variation is t(r) = 97,200/r. This is the translation.
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17. Solution (continued)
3. Carry out. To find out how long it would take to pump out the tank at a rate of 1500 mL/min, we calculate t(1500).
t = 64.8 when r = 1500
4. Check. We could now recheck each step. Note that, as expected, as the rate goes up, the time it takes goes down.
5. State. If the pump is emptying the tank at a rate of 1500 mL/min, then it will take 64.8 minutes to empty the entire tank.
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18. Joint and Combined Variation
When a variable varies directly with more than one other variable, we say that there is joint variation. For example, in the formula for the volume of a right circular cylinder, V = πr2h, we say that V varies jointly as h and the square of r.
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19. Joint Variation
y varies jointly as x and z if, for some nonzero constant k, y = kxz.
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20. Example
Find an equation of variation if a varies jointly as b and c, and a = 48 when b = 4 and c = 2.
Solution
We have a = kbc, so
The variation constant is 6.
The equation of variation is a = 6bc.
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21. Joint variation is one form of combined variation
Joint variation is one form of combined variation. In general, when a variable varies directly and/or inversely, at the same time, with more than one other variable, there is combined variation.
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22. Example
Find an equation of variation if y varies jointly as x and z and inversely as the product of w and p, and y = 60 when x = 24, z = 5, w = 2, and p = 3.
Solution
The equation if variation is of the form
so, substituting, we have:
Thus,
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