What you should learn

To solve problems involving direct and inverse variation

NCTM Curriculm Standards 2,, 4, 6 - 9

 

In doing this the teacher wants to make sure that the following words are incorporated into the introductory lesson:

Direct Variation

Constant of Variation

Inverse Variation

Introduction: Do you find yourself singing in the shower? If you are like the average American, you have enough time to sing several songs. The national average for time spent in the shower is 12.2 minutes. A standard shower head uses about 6 gallons of water per minute. That menas, you use 73.2 gallons of water for each shower you take. Taking one shower per day for one year would use 26,718 gallons of water!

The number of gallons of water used depends directly on the amount of time spent in the shower. The table below shows the number of gallons of water used y as a function of time in the shower x.

The relationship between the number of minutes in the shower and the gallons of water used is shown by the equation y = 6x. This type of equation is called a direct variation. We say that y varies directly as x or y is directly proportional to x. This means that as x increases in value, y increases in value, or as x decreases in value, y decreases in value.

Definition of Direct Variation: A direct variation is described by an equation of the form y = kx, where k0.

 

In the equation y = kx, k is called the constant of variation. To find the constant of variation, divide each side by x.

Exercise 1: Julio's wages vary directly as the number of hours that he works. If his wages for 5 hours are $29.75, how much will they be for 30 hours?

First, find Julio's hourly pay. Let x = number of hours Julio works, and let y = Julio's pay. Find the value of k in the equation y = kx. The value of k is the amount of money Julio is paid per hour.

k = y/x

k = 29.75/5

k = 5.95

Julio is paid $5.95 per hour.

Next find out how much Julio's wage will be for 30 hours.

y = kx

y = 5.95(30)

y = 178.50

Thus, Julio's wages will be $178.50 fo r30 hours of work.

 

 

Using the table in the application at the beginning of the lesson, many proportions can be formed. Two examples are shown below.

Two general forms for proportions like these can be derived from the equation y = kx. Let (x1, y1) be a solution for y = kx. Let a second solution be (x2, y2). Then y1 = kx1 and y2 = kx2.

Another proportion can be derived from the proportion.

You can use either of these forms to solve problems involving direct proportion.

Exercise 2: If y varies directly as x, and y = 28 when x = 7, find x when y = 52.

Use y1/y2 = x1/x2 to solve the problem.

28/53 = 7/x2

28x2 = 52(7)

x2 = 13

Thus, x = 13 when y = 52.

 

 

The reversse of direct variation is inverse variation. We say that y varies inversely as x. This means that as x increases in value, y decreases in value, or as y decreases in value, x increases in value. For example, the more miles you drive, the less gasoline you have in the tank.

Definition of Inverse Variation: An inverse variation is described by an equation of the form xy = k, where k0.

Exercise 3: The length of a violin string varies inversely as the frequency of its vibrations. A violin string 10 inches long vibrates at a frequency of 512 cycles per second. Find the frequency of an 8-inch string.

Let l represent the length in inches and f represents the frequency in cycles per second. Find the value of k.

lf = k

(10)(512) = k

5120 = k

Next, find the frequency, in cycles per second, of the 8-inch string.

lf = k

8 * f = 5120

f = 5120/8 = 640

The frequency of an 8-inch string is 640 cycles per second.

 

 

Let (x1, y1) be a solution of an inverse variation, xy = k. Let (x2, y2) be a second solution. Then x1y1 = k and x2y2 = k

The equation x1y1 = x2y2 is called the product rule for inverse variations. Study how it can be used to form a proportion.

You can use either the product rule or the proportion rule to solve problems involving inverse variation.

Exercise 4: If y varies inversely as x, and y = 5 when x = 15, find x when y = 3.

If you have observed people on a seesaw, you may have noticed that the heavier person must sit closer to the fulcrum (pivot point) to balance the seesaw. A seesaw is a type of lever, and all lever problems involve inverse variation.

Suppose weights w1 and w2 are placed on a lever at distances d1 and d2, respectively, from the fulcrum. The lever is balanced when w1d1 = w2d2.

Exercise 5: The fulcrum is placed in the middle of a 20-foot seesaw. Cholena, who weights 120 pounds, is seated 9 feet from the fulcrum. How far from teh fulcrum should Antonio sit if he weights 135 pounds?

Let w1 = 120, d1 = 9, and w2 = 135. Solve for d2.

w1d1 = w2d2

120 * 9 = 135 * d2

1080 = 135d2

d2 = 8

Antonio should sit 8 feet from the fulcrum.

Closing Activity: Check for understanding by using this as a quick review before class is over. It should take about the last five to ten minutes. I would use it for my students as their 'ticket out the door'. Click Here.

Homework: The homework to be assigned for tonight would be: 13 - 29 odd, 30, 31, 33 - 40

Alternative Homework: Enriched: 12 - 28 even, 30 - 40

Extra Practice: Students book page 766 Lesson 4-8

Extra Practice Worksheet: Click Here.

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