Direct and Inverse Variation Problems

Direct and Inverse Variation Problems: Definition & Examples

In this lesson, you'll learn how to approach questions about direct and inverse variation with a simple explanation of what the terms mean and how to apply them to problems.

Variation

Equations with direct and inverse variation sound a little intimidating, but really, they're just two different ways of talking about how one number changes relative to another number.

In direct variation, as one number increases, so does the other. This is also called direct proportion: they're the same thing. An example of this is relationship between age and height. As the age in years of a child increases, the height will also increase.

In the abstract, we can express direct variation by using the equation y = kx.

x and y are the two quantities - in our example, they'd be the age and the height of the child. k is called the constant of proportionality: it tells you specifically how much bigger y will get for every increase in x. For example, maybe y = 2x: this means that for every increase in x, y will increase by double that amount.

You can see that the bigger the number you plug in for x, the bigger the resulting value of y will be.

In inverse variation, it's exactly the opposite: as one number increases, the other decreases. This is also called inverse proportion. An example would be the relationship between time spent goofing off in class and your grade on the midterm. The more you goof off, the lower your score on the test.

If we wanted to give this one an equation, we would say:

y = k/x, where x and y are the two quantities, and k is still the constant of proportionality, telling how much one varies when the other changes.

You can see that in this equation, you divide a constant number by x to get the value of y. So the bigger the value of x, the smaller the value of y will be. That's inverse variation: as one goes up, the other goes down.

Examle 1

This might seem really complicated and confusing, but just remember the two formulas: y = kx for direct variation, and y = k/x for inverse variation. As you practice with example problems, you'll learn how to apply them to specific problems.

A very simple example is a problem like this one:

The population of a certain species of bacteria varies directly with the temperature. When the temperature is 35 degrees Celsius, there are 7 million bacteria. How many millions of bacteria are there when the temperature is 38 degrees Celsius?

First of all, we can see that we'll be using the direct variation equation, y = kx.

Now let's plug in what we have from the problem:

The problem gives us two values: temperature and number of bacteria. We'll plug in the temperature for x and the number of bacteria for y. This gives us 7 = k(35).

Now all we have to do is divide to find the value of k for this particular problem: it turns out to be 0.2.

The next step is to use that value to find out how many millions of bacteria there are at 38 degrees Celsius. So, we use the equation again.

This time, we'll plug in x and k, since we're looking for y. We find that y = (0.2)(38). Do the multiplication, and we learn that y, or the value of the population in millions is 7.6. So the answer to this question would be 7.6 million bacteria.

That wasn't so painful, right? It's just about using the equations properly. Now let's try one that's a little harder.

Example 2

The square of a varies inversely with the sum of three and b. If k represents the constant of proportionality, then in terms of b and k, what is the value of a?

This one is a tough nut to crack because the problem gives you a value for a squared, but asks you about the value of a. It looks pretty impenetrable at first, but don't panic. Pull out your equations, and start plugging things in.

First of all, we know we're going to be using the inverse variation here. So we have our equation ready at hand: y = k/x.

Now let's go slowly through the problem and plug things in. First of all, we don't have just plain old a, we have asquared. But that's not so terrible: the variation is defined in terms of a squared, so we'll just put in a squared for y. We could also plug it in for x; it doesn't matter, as long as you're consistent. In this problem, we're trying to isolate a, so it makes more sense to put it all by itself on one side of the equation; this will make the math easier later on.

Next, we have 'the sum of 3 and b, or in math terms, b + 3 instead of plain old b. Again, not a problem: just plug in 'b + 3' for the value of x on the bottom of the fraction.

That's most of the work done already: now we have an equation relating a and b. Now we just need to solve for a, so we'll take the square root of both sides. Voila! a equals the square root of k over b plus 3! It's challenging, but it's not impossible.

Lesson Summary

In this lesson, you learned how to tackle direct and inverse variation problems by using the equations for each.

For direct variation, use the equation y = kx, where k is the constant of proportionality.

For inverse variation, use the equation y = k/x, again, with k as the constant of proportionality.

Remember that these problems might use the word 'proportion' instead of 'variation,' but it means the same thing.

Sometimes you'll have to work out proportional relationships that get a little bit complicated, like plugging in xsquared instead of plain x, but with these two equations under your belt, you should be able to handle them.