Direct Variation - Free Math Help

Direct Variation

In This Lesson:

• What Is Direct Variation?
• The Formula: y = kx
• Finding the Constant
• Graphing Direct Variation
• Real-World Examples
• Connection to Proportions
• Practice Problems

What Is Direct Variation?

When two variables are related in such a way that their ratio always stays the same, we say they vary directly. In simpler terms, when one quantity doubles, the other doubles. When one triples, the other triples. They move together in perfect sync.

Think about buying apples at a grocery store. If one apple costs $0.50, then two apples cost $1.00, three apples cost $1.50, and so on. The total cost varies directly with the number of apples. The ratio of cost to apples is always the same: $0.50 per apple.

Here's another way to think about it: if A is always twice as much as B, then A and B vary directly. If your shadow is always three times as long as your height, shadow length and height vary directly.

The key characteristic is that the ratio between the two quantities never changes. This constant ratio is what makes direct variation so predictable and useful.

The Formula: y = kx

Every direct variation relationship can be written as:

$$y = kx$$

where \(k\) is called the constant of variation (or constant of proportionality). This constant tells you exactly how the two variables relate to each other.

Let's break down what this means. If \(y = 3x\), then \(k = 3\), and \(y\) is always three times as large as \(x\). Double \(x\), and \(y\) doubles too. The constant \(k\) is the multiplier that connects them.

Notice that this is actually a special type of linear equation. The general form of a line is \(y = mx + b\), but in direct variation, \(b = 0\). There's no y-intercept except at the origin. The line always passes through the point \((0, 0)\).

You might also see direct variation written as:

$$\frac{y}{x} = k$$

This is the same formula, just rearranged. It emphasizes that the ratio of \(y\) to \(x\) is constant.

Finding the Constant

The most common type of problem gives you one pair of values and asks you to find \(k\) or to predict other values.

Example 1: If \(y\) varies directly with \(x\), and \(y = 12\) when \(x = 4\), find the constant of variation.

Use the formula \(y = kx\) and substitute: $$12 = k \cdot 4$$ $$k = 3$$

The constant of variation is 3. This means \(y\) is always three times \(x\), so the equation is \(y = 3x\).

Example 2: If \(y\) varies directly with \(x\), and \(y = 20\) when \(x = 5\), what is \(y\) when \(x = 8\)?

First, find \(k\): $$20 = k \cdot 5$$ $$k = 4$$

Now use \(y = 4x\) with \(x = 8\): $$y = 4 \cdot 8 = 32$$

When \(x = 8\), \(y = 32\).

Example 3: The distance a car travels varies directly with the time it's been driving. If the car travels 120 miles in 2 hours, how far will it travel in 5 hours?

Let \(d\) = distance and \(t\) = time. We have \(d = kt\).

Find \(k\): $$120 = k \cdot 2$$ $$k = 60$$

This makes sense: the car is traveling 60 miles per hour. Now find the distance at \(t = 5\): $$d = 60 \cdot 5 = 300$$

The car will travel 300 miles in 5 hours.

Graphing Direct Variation

One of the defining features of direct variation is what its graph looks like. Since the equation is \(y = kx\), you're graphing a line that passes through the origin with slope \(k\).

The graph always goes through \((0, 0)\) because when \(x = 0\), \(y\) must also equal 0. Think about it: if you buy zero apples, you pay zero dollars. If you drive for zero hours, you travel zero miles.

The steepness of the line depends on \(k\). A larger value of \(k\) means a steeper line because \(y\) increases faster as \(x\) increases. If \(k = 5\), the line rises faster than if \(k = 2\).

If \(k\) is negative, the line slopes downward from left to right. This represents situations where as one variable increases, the other decreases at a constant rate.

Example 4: Graph \(y = 2x\).

When \(x = 0\), \(y = 0\). When \(x = 1\), \(y = 2\). When \(x = 2\), \(y = 4\). When \(x = 3\), \(y = 6\).

Plot these points and draw a straight line through them. The line passes through the origin with a slope of 2.

Real-World Examples

Direct variation shows up in countless real-world situations where two quantities change together at a constant rate.

Hourly Wages

If you earn $15 per hour, your total pay varies directly with the hours worked. The equation is \(P = 15h\), where \(P\) is pay and \(h\) is hours. Work 3 hours, earn $45. Work 8 hours, earn $120. The constant \(k = 15\) is your hourly rate.

Example 5: Sarah earns $18 per hour. Write the direct variation equation and find how much she earns working 6.5 hours.

The equation is \(P = 18h\).

For 6.5 hours: $$P = 18 \cdot 6.5 = 117$$

Sarah earns $117.

Buying in Bulk

When you buy items at a constant price per unit, the total cost varies directly with the quantity.

Example 6: Oranges cost $0.75 each. Write the equation and find the cost of 12 oranges.

Let \(C\) = total cost and \(n\) = number of oranges. $$C = 0.75n$$

For 12 oranges: $$C = 0.75 \cdot 12 = 9$$

Twelve oranges cost $9.00.

Perimeter of Regular Shapes

The perimeter of a square varies directly with the length of one side. Since a square has four equal sides, \(P = 4s\), where \(P\) is perimeter and \(s\) is the side length. The constant is 4.

Example 7: A square has a perimeter of 48 inches. What is the length of one side?

Use \(P = 4s\): $$48 = 4s$$ $$s = 12$$

Each side is 12 inches long.

Currency Conversion

Exchange rates create direct variation. If 1 US dollar equals 0.85 euros, then the number of euros you get varies directly with the dollars you exchange.

Example 8: If $1 USD = 0.85 EUR, how many euros do you get for $250?

Let \(E\) = euros and \(D\) = dollars. $$E = 0.85D$$

For $250: $$E = 0.85 \cdot 250 = 212.50$$

You get 212.50 euros.

Connection to Proportions

Direct variation is really just another way of talking about proportions. When we say \(y\) varies directly with \(x\), we're saying that the ratio \(\frac{y}{x}\) is constant.

This means if you know \(y_1\) and \(x_1\) go together, and you want to find \(y_2\) for a different \(x_2\), you can set up a proportion:

$$\frac{y_1}{x_1} = \frac{y_2}{x_2}$$

Example 9: If 6 pounds of flour cost $12, how much do 10 pounds cost?

Method 1 (Direct Variation): Find \(k\): \(12 = k \cdot 6\), so \(k = 2\). Use \(C = 2p\) where \(p\) = pounds: \(C = 2 \cdot 10 = 20\).

Method 2 (Proportion): $$\frac{12}{6} = \frac{x}{10}$$

Cross-multiply: \(12 \cdot 10 = 6x\), so \(120 = 6x\), thus \(x = 20\).

Both methods give $20. Direct variation and proportions are two sides of the same coin.

Practice Problems

Try these on your own, then check your answers.

1. If \(y\) varies directly with \(x\), and \(y = 15\) when \(x = 3\), find \(k\).

2. Using the relationship from problem 1, find \(y\) when \(x = 7\).

3. The cost of gasoline varies directly with the number of gallons purchased. If 8 gallons cost $28, how much do 12 gallons cost?

4. A recipe calls for ingredients in direct variation. If 2 cups of flour are needed for 12 cookies, how much flour is needed for 30 cookies?

5. Write the direct variation equation if \(y = 45\) when \(x = 9\).

6. Which of these equations represents direct variation? a) \(y = 3x + 2\) b) \(y = 5x\) c) \(y = x^2\) d) \(y = \frac{12}{x}\)

7. The distance a spring stretches varies directly with the weight attached. If a 5-pound weight stretches it 2 inches, how far will a 15-pound weight stretch it?

8. If \(y\) varies directly with \(x\) and the constant of variation is -4, what is \(y\) when \(x = 6\)?

Check Your Work

1. \(k = 5\) Use \(15 = k \cdot 3\), so \(k = 5\)

2. \(y = 35\) Use \(y = 5x\) with \(x = 7\): \(y = 5 \cdot 7 = 35\)

3. $42 Find \(k\): \(28 = k \cdot 8\), so \(k = 3.50\) (price per gallon) Cost for 12: \(C = 3.50 \cdot 12 = 42\)

4. 5 cups Find \(k\): \(2 = k \cdot 12\), so \(k = \frac{1}{6}\) For 30 cookies: \(f = \frac{1}{6} \cdot 30 = 5\)

5. \(y = 5x\) Find \(k\): \(45 = k \cdot 9\), so \(k = 5\) Equation: \(y = 5x\)

6. b) \(y = 5x\) Only this one has the form \(y = kx\) with no added constant

7. 6 inches Find \(k\): \(2 = k \cdot 5\), so \(k = 0.4\) For 15 pounds: \(d = 0.4 \cdot 15 = 6\)

8. \(y = -24\) Use \(y = -4x\) with \(x = 6\): \(y = -4 \cdot 6 = -24\)

Things to Remember

Direct variation always passes through the origin. If the graph doesn't go through \((0, 0)\), it's not direct variation. This is the quickest way to identify it.

The constant \(k\) is the slope. When you see \(y = kx\), that \(k\) tells you how steep the line is. It's also the rate of change between the variables.

Doubling one variable doubles the other. This is what "direct" means. They're directly connected. If \(x\) increases by any factor, \(y\) increases by the same factor.

Not all proportional relationships are direct variation. A relationship like \(y = 2x + 3\) is linear but not direct variation because it doesn't pass through the origin.

The units of \(k\) matter. If you're calculating dollars per hour, \(k\) has units of dollars/hour. If you're finding miles per gallon, \(k\) has units of miles/gallon. The constant isn't just a number; it represents a real-world rate.

Why This Matters

Direct variation is one of the simplest and most useful mathematical relationships. It models countless real-world situations where two quantities change together at a constant rate. Understanding it helps you:

• Calculate wages, costs, and prices
• Work with speeds, distances, and times
• Convert between units
• Understand rates and ratios
• Predict outcomes based on patterns

More importantly, it's your first step toward understanding more complex relationships. Once you grasp direct variation, you can move on to inverse variation, joint variation, and other types of relationships that appear in science, economics, and engineering.

The beauty of direct variation is its simplicity. One formula, \(y = kx\), describes so many different situations. Master this concept, and you have a powerful tool for solving real problems.