Day 5: Understanding Direct Variation

Objectives:

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.

 RabbitWeight (lbs) 3 4.5 2 5 3.5 4 6 2.5 Maximum Hopping distance (ft) 6 9 4 10 7 8 12 5

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.

Continue to next day

Return to previous day

Return to Instructional Unit overview