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Direct Variation

Identify and solve y=kx form equations

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Direct Variation

According to Newton's second law, the net force (F) of an object is equal to its mass (m) times its acceleration (a), where F is measured in Newtons, m is measured in kilograms, and a is measured in meters/sec/sec. If an object with an acceleration of 8 meters/sec/sec has a force of 24 Newtons, what is the object's force when its acceleration is 12 meters/sec/sec?


We say that a set of data is related directly if the independent and dependent variables both grow large or small together. For example, the equation of the line y=2x would represent a direct variation relationship. As x gets bigger, so would y . In fact, direct variation equation is y=kx, k \neq 0, which looks just like the equation of a line without a y -intercept. We call k the constant of variation and y is said to vary directly with x . k can also be written k=\frac{y}{x} .

Example A

The variables x and y vary directly, and y = 10 when x = 2 . Write an equation that relates x and y and find y when x = 9 .

Solution: Using the direct variation equation, we can substitute in x and y and solve for k .

y &= kx \\10 &= k(2) \\5 &= k

Therefore, the equation is y=5x . To find y when x is 9, we have y=5 \cdot 9=45 .

Example B

Determine if the set of data varies directly. If so, find the direct variation equation.

x 4 8 16 20
y 1 2 4 5

Solution: Looking at the set of data, the x values increase. For the data to vary directly, the y values would also have to increase, and they do. To find the equation, use the first point and find k .

y &= kx\\1 &= k(4)\\\frac{1}{4} &= k

So, the equation for the first point is y=\frac{1}{4}x . Plug each point into the equation to make sure it works.

2= \frac{1}{4}(8) \qquad 4= \frac{1}{4}(16) \qquad 5= \frac{1}{4}(20)

Example C

The number of calories, C , a person burns working out varies directly with length of time it was done, t (in minutes). A 150 pound person can burn 207 calories swimming laps for 30 minutes. Write a variation model for C as a function of t . Then, determine how long it will take that person to burn 520 calories.

Solution: Plug in what you know to the direct variation model and solve for k .

C&=kt \\207&=k(30) && \text{The model for a} \ 150\text{-pound person is} \ C=6.9 t. \\6.9&=k

To find how long it will take to this person to burn 520 calories, solve for t .

520&=6.9 t && \text{It will take} \ 75.4 \ \text{minutes to burn} \ 520 \ \text{calories}.\\75.4&=t

Intro Problem Revisit If we write Newton's second law as an equation, we get F = ma . We can now see that this is an example of a direct variation equation, where y = F , m = k , and a = x . Using the direct variation equation, we can substitute in F and a and solve for m .

F &= ma \\24 &= m(8) \\3 &= m

So the mass of the object is 3 kg but we're looking for its force when its acceleration is 12 meters/sec/sec. Hence, we use the formula again.

F &= ma \\F &= (3)(12) \\F &= 36

Therefore, the object's force is 36 Newtons.

Guided Practice

1. x and y vary directly. When x=-8, y=-6 . Find the equation and determine x when y=12 .

2. Determine if the set below varies directly.

x 1 2 3 4 5
y 2 4 8 16 20

3. Taylor’s income varies directly with the number of hours he works. If he worked 60 hours last week and made $900, how much does he make per hour? Set up a direct variation equation.


1. First, solve for k .

k=\frac{y}{x}=\frac{-6}{-8}=\frac{3}{4} \rightarrow y=\frac{3}{4}x

Now, substitute in 12 for y and solve for x .

12&=\frac{3}{4}x \\\frac{4}{3} \cdot 12&=x \\16&=x \\

2. At first glance, it looks like both values increase together. Let’s check to see if k is the same for each set of points.

k=\frac{y}{x}=\frac{2}{1}=\frac{4}{2} \ne \frac{8}{3}

At this point, we can stop because the point (3, 8) does not have the same ratio as the first two points. Therefore, this set of data does not vary directly.

3. We want to find Taylor’s hourly wage, which is the constant of variation.

k=\frac{900}{60}=15 , he makes $15/hour. The equation would be y=15x .


Direct Variation
When the dependent variable grows large or small as the independent variable does.
Constant of Variation
The rate of which the dependent variable grows, k , in y=kx .


For problems 1-4, use the given x and y values to write a direct variation equation and find y given that x =12 .

  1. x=3,y=15
  2. x=9,y=-3
  3. x=\frac{1}{2},y=\frac{1}{3}
  4. x=-8,y=\frac{4}{3}

For problems 5-8, use the given x and y values to write a direct variation equation and find x given that y=2 .

  1. x=5,y=4
  2. x=18,y=3
  3. x=7,y=-28
  4. x=\frac{2}{3},y=\frac{5}{6}

Determine if the following data sets vary directly.

  1. .
x 12 16 5 20
y 3 4 1 5
  1. .
x 2 10 5 6
y 14 70 35 42
  1. .
x 2 8 18 34
y 3 12 27 51

Solve the following word problems using a direct variation equation.

  1. Based on her weight and pace, Kate burns 586 calories when she runs 5 miles. How many calories will she burn if she runs only 3 miles? How many miles (to the nearest mile) does she need to run each week if she wants to burn one pound (3500 calories) of body fat each week?
  2. One a road trip, Mark and Bill cover 450 miles in 8 hours, including stops. If they maintain the same pace, how far (to the nearest mile) will they be from their starting point after 15 hours of driving?
  3. About three hours into a fundraising car wash, the Mathletes Club earned $240 washing 48 cars. How much was charged for each carwash? How many more cars will they have to wash to reach their goal of earning $400?
  4. Dorothy earned $900 last week for working 36 hours. What is her hourly wage? If she works full time (40 hours) in a week how much will she make?


Constant of Variation

Constant of Variation

The rate of which the dependent variable k grows, in a direct variation of the form y=kx.
Direct Variation

Direct Variation

When the dependent variable grows large or small as the independent variable does.

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