Direct & Inverse Variation: Videos & Practice Problems

Direct variation describes a relationship between two variables where one quantity increases or decreases in direct proportion to the other. This means if one variable goes up, the other also goes up; if one goes down, the other decreases as well. Mathematically, this relationship is expressed as y = kx, where y and x are the variables, and k is the constant of variation. This constant k represents the rate at which y changes with respect to x, making the equation a linear function that models direct proportionality.

When solving direct variation problems, the key step is to determine the constant k. For example, if it is given that y = 10 when x = 2, substituting these values into the equation y = kx allows you to solve for k by dividing both sides by x, yielding k = \(\frac{10}{2}\) = 5. Once k is found, the specific direct variation equation becomes y = 5x. This equation can then be used to find y for any value of x. For instance, when x = 6, substituting into the equation gives y = 5 \(\times\) 6 = 30.

Understanding direct variation is essential because it models many real-world scenarios where two quantities change proportionally. Recognizing phrases like "y varies directly as x" or "y is directly proportional to x" signals that direct variation applies. This concept extends the idea of linear equations by emphasizing the constant ratio between variables, which is fundamental in fields such as physics, economics, and biology.

In problems involving direct variation, the relationship between two variables can be expressed as y = kx, where k is the constant of variation. When applying this to real-world scenarios, such as pricing, it is essential to carefully define the variables to match the context. For example, if a bakery sells eggs at \$3 per dozen, the price (p) varies directly with the quantity (q) measured in dozens. This can be modeled as p = kq.

To find the constant of variation k, use the given information: the price is \$3 for one dozen eggs. Since the quantity is measured in dozens, set q = 1 for one dozen. Substituting these values gives 3 = k × 1, so k = 3. Thus, the variation model becomes
\( p = 3q \).

Using this model, you can calculate the cost for any number of dozens. For instance, to find the price of five dozen eggs, substitute q = 5 into the equation:
\( p = 3 \times 5 = 15 \).
Therefore, five dozen eggs cost \$15.

Similarly, if a customer spends \$39 on eggs, you can determine the quantity purchased by setting p = 39 and solving for q:
\( 39 = 3q \implies q = \frac{39}{3} = 13 \).
This means the customer bought 13 dozen eggs.

Understanding how to set up and manipulate direct variation equations is crucial for solving practical problems involving proportional relationships. Always pay close attention to the units and quantities involved to maintain consistency throughout the problem. This approach ensures accurate modeling and problem-solving in real-world contexts such as pricing, rates, and other directly proportional scenarios.

Inverse variation describes a relationship between two quantities where one increases as the other decreases, which is the opposite of direct variation. In direct variation, the equation relating two variables y and x is expressed as y = kx, where k is the constant of variation. In contrast, inverse variation is represented by the equation y = \(\frac{k}{x}\), indicating that y varies inversely as x. Here, k remains the constant of variation, but instead of multiplying x, it is divided by x. It is important to note that x cannot be zero in inverse variation because division by zero is undefined.

To find the constant k in an inverse variation problem, you substitute known values of x and y into the equation and solve for k. For example, if y = 8 when x = 4, substituting these values gives 8 = \(\frac{k}{4}\). Multiplying both sides by 4 yields k = 32. Thus, the equation relating x and y is y = \(\frac{32}{x}\). This equation can then be used to find y for any other value of x. For instance, when x = 2, substituting into the equation gives y = \(\frac{32}{2} = 16\).

Inverse variation is also commonly described using terms like "y varies inversely as x" or "y is inversely proportional to x," both indicating the same mathematical relationship. Understanding the distinction between direct and inverse variation is crucial: in direct variation, both variables increase or decrease together, while in inverse variation, one variable increases as the other decreases. Mastery of these concepts allows for solving a variety of problems involving proportional relationships, especially when determining the constant of variation and applying it to find unknown values.

In problems involving the inverse variation between volume and pressure of a gas, the volume varies inversely as the pressure applied. This means that as pressure increases, volume decreases proportionally, and vice versa. The general form of the equation representing this inverse relationship is given by:

\[ V = \frac{k}{P} \]

where V is the volume, P is the pressure, and k is the constant of variation. To determine the constant k, you can use known values of volume and pressure from the problem. For example, if a helium gas has a volume of 370 cubic inches at a pressure of 15 pounds per square inch (PSI), substitute these values into the equation:

\[ 370 = \frac{k}{15} \]

Multiplying both sides by 15 to isolate k gives:

\[ k = 370 \times 15 = 5550 \]

With the constant found, the equation relating volume and pressure becomes:

\[ V = \frac{5550}{P} \]

This formula allows you to calculate the volume for any given pressure. For instance, if the pressure increases to 20 PSI, substitute P = 20 into the equation:

\[ V = \frac{5550}{20} = 277.5 \text{ cubic inches} \]

This volume can also be expressed as the fraction:

\[ V = \frac{555}{2} \text{ cubic inches} \]

Understanding inverse variation is crucial in real-world applications such as gas laws, where pressure and volume are inversely proportional under constant temperature conditions. This relationship is foundational in physics and chemistry, particularly in understanding how gases behave under different pressures.