Day 5: Understanding Direct Variation


1) Learn what direction variation means

2) To learn how to calculate the constant of variation, write a direct variation equation, and use this to solve for values of x or y

3) Apply this knowledge in writing a direct variation model

1) What two things vary directly when two variables, x and y for instance, vary according to some constant known as the constant of variation? In an algebraic form, y = kx is a model of direct variation where x and y vary directly according to the constant of variation, k. In other words, say that k was 4, then y would be four times what x would be. If x was 2 and k was 4, then y will be 8.

y = kx

y = 2x

y = 2(4)

y = 8

2) If you were given x and y and were told that they were in direct variation of each other and you had the job of finding the constant of variation, then we would just have to invoke a little variable manipulation. Let's say that x = 4 and y = 18. How would we solve for k? Well, using the model given above, let's first solve the equation for k.

y = kx

Now divide both sides of the equation by x and we have

y/x = k

Now we can substitute in our values for x and y given in the problem

18/4 = k

9/2 = k

So, the constant of variation, in this case, is 9/2 or 4.5.

Now, using this, how could we find out what y would be when x = 12? We would just use the constant of variation just found and substitute in the value of x.

y = (9/2)x

y = (9/2)(12)

y = 54

To find out what x would be when y = 3/2 we would do something similar to above.

y = (9/2)x

(y)/(9/2) = x

(3/2)/(9/2) = x

1/3 = x

3) How can we use this to model an application? What if we had the following problem: It is known that the maximum distance a rabbit can hop varies directly with the rabbit's weight. From the data below, find the constant of variation and write a direct variation equation.











Maximum Hopping distance









To find the constant of variation use the general equation and solve for k. In general you should try most of the different combinations of data from above, but I will only do one here.

Hop. Dist. = k(Rabb. wt.)

(Hop. Dist.)/(Rabb. wt.) = k

From the first data set, I get the following:

6/3 = k

2 = k

It turns out that the constant of variation, in this case, is 2.

For my direct variation equation, I would have the following:

Hop. Dist. = 2(Rabb. wt.)

Now using this equation find out

a) What would be the maximum hopping distance for a 7 pound rabbit?

b) How much would a rabbit weigh if it could hop a maximum of 8.5 feet?

Solving part a...

We use the equation from above:

Hop. Dist. = 2(Rabb. wt.)

Hop. Dist. = (2)(7)

Hop. Dist. = 14

So, this 7 pound rabbit could hop a maximum of 14 feet.

Solving part b...

Again, we would use the equation from above:

Hop. Dist. = 2(Rabb. wt.)

First we must solve the general equation for rabbit's weight and then solve for the rabbit's weight.

(Hop. Dist.)/2 = Rabb. wt.

8.5/2 = Rabb. wt.

4.25 = Rabb. wt.

Therefore, if a rabbit that can hop a maximum of 8.5 feet, then it must weigh 4.25 pounds.

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