## Day 4: Direct Variation

#### by: BJ Jackson

Objectives:

1. to understand what a direct variation is and to know the definition of direct variation.

2. to be able to solve problems involving direct variations

Definition:

Direct Variaton: a relationship between two objects where when the first ojbect increases the second object increases. The relationship is given by the equation:

y = kx

where x and y are the related objects and k is called the constant of variation.

Discussion

In a direct varation, both objects either increase or decrease. It is a type of linear function. The constant of variation, k, is the slope of the function, but we don't usually think of it in terms of slope because we use the equation to solve problems. Constant variations are useful because if we know one point we can then find other points that would be on the graph. We find the other points based on k. This is done by substituting the values of x and y into the equation for a direct varition and then solving for k. Example 1 will describe show how to find k.

Example1: The point (2, 8) is a point on the graph of a direct variation. Find the constant of variation.

We start with the equation: y = kx. The point (2, 8) tells us that x = 2 and that y = 8. Next, substitute these values into the equation to get: 8 = 2k. Finally, solve the equation for k by dividing both sides by 2 and one gets: 4 = k. So, the constant of variation is 4.

Now that we can find the constant of variation, we can find points on the graph. This will be done by using our equation twice. The first time we will substitute values for x and y to find k. The second time, we will substitute for k and either x or y to find the missing x or y. Problems of this kind will look like examples 2 and 3.

Example2: Given a direct variation where y = 12 when x = 3, what is y when x = 10.

Step 1: Find k.

So, to find k, we use the fact that y = 12 when x = 3 and substitute those values into our equation. This substitution yields the equation: 12 = 3k. Now, divide by 3 and find that 4 = k.

Step 2: Find the missing variable.

In this case, the missing variable is y. So, we will use the equation: y = kx again, but this time we know that k = 4 and that x = 10. Substituting in for k and x gives us the equation: y = (4)(10). Finally, do the multiplication to find that y = 40.

Now, we should ask ourselves if this answer seems reasonable. Should y have gotten bigger? Remember that in a direct variation ,as x increases y increases or as x decreases y decreases. Since our first x = 3 and our second x = 10 which is an increase, we would expect our second y to increase. Therefore, the answer seems reasonable.

Example3: Given a direct variation where y = 15 when x = 3, find x when y = 5.

Step 1: Find k.

Like before, substitute for x and y to get: 15 = 3k. Divide by 3 to get 5 = k.

Step 2: Find the missing variable.

In this case, the missing variable is x. Again, use the equation y = kx, but now we substitute for y and k to get the equation: 5 = 5x. Solve the equation by dividing both sides by 5 and find that x = 1.

Again, ask yourself if this seems reasonable. Should the second x have been smaller than the first x? If yes, Why?

The answer is yes it should have been smaller because the second y is smaller than the first y. Therefore, the answer seems reasonable.

Now, you should understand the concept of a direct variation and be able to solve problems involving direct variations.

Note: Remember that these are linear functions with a positive slope. So, what could these look like graphically?

Could it look like this?

Yes, because this graph has a positive slope. So, as x increases; y increases.