Direct and Inverse Variation

Variation describes how two or more variables relate to one another. When one variable changes, variation tells us exactly how the other variables will respond.

Direct Variation

In direct variation, two variables change in the same direction. As one increases, the other increases proportionally.

The equation for direct variation is: $y = kx$

Here, $k$ is the constant of variation (where $k \neq 0$). You can read this as "$y$ varies directly with $x$" or "$y$ is directly proportional to $x$."

Inverse Variation

In inverse variation, two variables change in opposite directions. As one increases, the other decreases.

The equation for inverse variation is: $y = \frac{k}{x}$

This is read as "$y$ varies inversely with $x$" or "$y$ is inversely proportional to $x$."

Joint and Combined Variation

Sometimes, a variable depends on more than one other variable.

• Joint Variation: A quantity varies directly with the product of two or more variables. Formula: $y = kxz$
• Combined Variation: A mix of direct and inverse variation in the same relationship. Formula example: $y = \frac{kx}{z}$ (Here, $y$ varies directly with $x$ and inversely with $z$).

How to Solve Variation Problems

Most variation problems can be solved using a simple four-step process:

1. Write the equation using $k$ based on the type of variation.
2. Substitute the first set of given values to solve for $k$.
3. Rewrite the equation using the specific value of $k$ you just found.
4. Solve for the final missing variable using the new equation.

Example Problems

Example 1: Basic Inverse Variation If $y$ varies inversely with $x$, and $y = 6$ when $x = 4$, find $y$ when $x = 8$.

1. Write the equation: $y = \frac{k}{x}$
2. Find $k$: $6 = \frac{k}{4} \implies k = 24$
3. Rewrite the equation: $y = \frac{24}{x}$
4. Solve for the new $y$: $y = \frac{24}{8} = 3$

Example 2: Inverse Variation with a Square The force of gravity ($F$) varies inversely with the square of the distance ($d$). If $F = 100$ when $d = 2$, find $F$ when $d = 5$.

1. Write the equation: $F = \frac{k}{d^2}$
2. Find $k$: $100 = \frac{k}{2^2} \implies 100 = \frac{k}{4} \implies k = 400$
3. Rewrite the equation: $F = \frac{400}{d^2}$
4. Solve for the new $F$: $F = \frac{400}{5^2} = \frac{400}{25} = 16$