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4.6: Direct Variation Models

Created by: CK-12

Learning Objectives

At the end of this lesson, students will be able to:

  • Identify direct variation.
  • Graph direct variation equations.
  • Solve real-world problems using direct variation models.

Vocabulary

Terms introduced in this lesson:

direct variation
constant of proportionality
directly proportional

Teaching Strategies and Tips

Use Examples 1-3 to motivate direct variation.

  • Direct variation is an expression of a simple linear relationship with y-intercept = 0. Since the line passes through origin, (0,0) is a point on it.
  • The slope of the line is the constant of proportionality and is the only parameter in the problem. Therefore, it takes only one more point to determine the line (to determine the constant of proportionality).
  • Emphasize the switch from m (slope) to k (constant of proportionality) is only notational.

Determine the constant of proportionality by substituting the given information (look for one point) into y = kx. The result is the equation of variation; use it to answer the question in the problem.

By solving applied problems involving direct variation, teachers demonstrate the usefulness of these models.

Additional Examples:

The cost of raisins varies directly with the number of kilograms of raisins purchased. If the cost is \$34.25 when the number of kilograms purchased is 7.5, calculate the amount of raisins that can be purchased for \$10.50.

Solution: Let C denote the cost of the raisins and N the number of kilograms of raisins purchased. The direct variation equation is C = k \cdot N. Substitute C = 34.25, N = 7.5.

34.25 & = k \cdot 7.5\\4.6 & \approx k

With the constant of proportionality known, calculate N when C = 10.50.

C & =  4.6 \cdot N\\10.5 & = 4.6 \cdot N\\2.2 & = N

So the number of kilograms of raisins that can be purchased for \$10.50 is 2.2\;\mathrm{kg}.

A ball was thrown from the roof of a high building. The velocity v of the falling ball is directly proportional to the time t of the fall. After 4\;\mathrm{seconds}, the velocity of the ball is 85\;\mathrm{meters} per second. What will the velocity be after 6\;\mathrm{seconds}, assuming it hasn’t hit the ground yet?

Hint: Solve for k in the direct variation equation v = k \cdot t first.

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Feb 22, 2012

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Apr 29, 2014
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CK.MAT.ENG.TE.1.Algebra-I.4.6

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