Direct Variation

Direct variation describes a relationship between two variables where one variable is always a constant multiple of the other. In symbols, we say that \(y\) varies directly with \(x\) if there is a nonzero number \(k\) such that

\(y = kx\)

The number \(k\) is called the constant of variation (sometimes called the constant of proportionality). As \(x\) increases, \(y\) increases at a steady rate, and if \(x\) doubles, \(y\) doubles too. This is why direct variation is also described as \(y\) being "directly proportional to" \(x\).

The direct variation formula \(y = kx\) is really just a special kind of linear function, one where the graph passes through the origin. Because \(y = kx\) has no added constant term, whenever \(x = 0\), \(y\) must also equal \(0\). If you have studied the general shape and behavior of functions, you will recognize direct variation as a specific case covered in the broader idea of how to write domain and range for a linear function.

To find the constant of variation, rearrange the formula:

\(k = \dfrac{y}{x}\)

As long as you know one matching pair of \(x\) and \(y\) values, you can solve for \(k\), and then use that value of \(k\) to predict any other pair.

Suppose \(y\) varies directly with \(x\), and \(y = 12\) when \(x = 4\). Find \(k\), then find \(y\) when \(x = 9\).

Step 1: Solve for \(k\) using \(k = \dfrac{y}{x}\).

\(k = \dfrac{12}{4} = 3\)

Step 2: Write the specific equation.

\(y = 3x\)

Step 3: Substitute \(x = 9\).

\(y = 3(9) = 27\)

Given a table of \(x\) and \(y\) values, you can test for direct variation by checking whether the ratio \(\dfrac{y}{x}\) is the same for every pair. If it is constant, the relationship is a direct variation and that constant ratio is \(k\). If the ratio changes from row to row, the relationship is not a direct variation.

For example, in the table below, dividing each \(y\) by its matching \(x\) always gives \(2\), so \(k = 2\) and the equation is \(y = 2x\).

Because \(y = kx\) has no constant term added on, its graph is always a straight line through the origin \((0,0)\). The value of \(k\) controls the steepness (slope) of the line. A positive \(k\) gives a line rising from left to right, while a negative \(k\) gives a line falling from left to right, which is the same as a reflection across x axis of the line with the opposite, positive \(k\).

Notice that this line represents a 1 to 1 function when \(k \ne 0\): every input \(x\) produces exactly one output \(y\), and every output comes from exactly one input.

Many real world situations involve direct variation, such as distance and time at a constant speed, or the cost of items sold at a fixed price per unit. To solve these problems, follow the same three steps used above: find \(k\) from the given information, write the equation \(y = kx\), then substitute to answer the question.

Example: The cost \(c\) of gasoline varies directly with the number of liters \(n\) purchased. If \(15\) liters cost \(\$18\), how much would \(25\) liters cost?

Step 1: Find \(k\).

\(k = \dfrac{c}{n} = \dfrac{18}{15} = 1.2\)

Step 2: Write the equation.

\(c = 1.2n\)

Step 3: Substitute \(n = 25\).

\(c = 1.2(25) = 30\)

So \(25\) liters of gasoline would cost \(\$30\).

It is easy to confuse direct variation with inverse variation, so it helps to keep the two definitions separate. In direct variation, \(y = kx\), and \(y\) increases as \(x\) increases. In inverse variation, \(y = \dfrac{k}{x}\), and \(y\) decreases as \(x\) increases. A quick way to tell them apart in a table is that direct variation has a constant ratio \(\dfrac{y}{x}\), while inverse variation has a constant product \(xy\).

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