Variation

There are various equations that describe the relationship between two variables where one quantity changes proportionally to another. We refer to these relationships as variations.

Direct Variation

If $x$ and $y$ are related by an equation of the form $y = kx^n$ then we state that the relationship is direct variation which means $y$ varies directly with, or is proportional to, the $n$th power of $x$. In these relationships, $k$ is called the constant of variation and it is a non-zero constant ratio where $k = \frac{y}{x^n}$. This constant is what defines the relationship between the two variables.

Given a description of a direction variation problem, we can solve for an unknown by...

1. Identifying the input and output variables.
2. Determining the constant of variation which is either given or can be found using the ratio $k = \frac{y}{x^n}$.
3. Use the constant of variation to write an equation for the relationship.
4. Solve for the unknown.

$\underline{Example}$

The quantity $y$ varies directly with the square of $x$. If $y = 24$ when $x = 3$, find $y$ when $x$ is $4$.

The input is defined by $x$, the output is defined as $y$, and the power of the function is $2$. We still need to find the constant of variation using the ratio:

$k = \frac{y}{x^n} = \frac{24}{3^2} = \frac{24}{9} = \frac{8}{3}$

With all the information, we can write the equation for the relationship as $y = \frac{8}{3}x^2$. We can use this equation to solve for $y$ when $x = 4$ by substituting the value into the equation:

$y = \frac{8}{3}(4)^2 = \frac{8}{3}(16) = \frac{128}{3}$

Therefore, when $x = 4$, $y = \frac{128}{3}$.

Inverse Variation

If $x$ and $y$ are related by an equation of the form $y = \frac{k}{x^n}$ then we state that the relationship is inverse variation which means $y$ varies inversely proportional with the $n$th power of $x$. The constant of variation, $k$, is a non-zero constant ratio where $k = x^ny$.

We can solve for an unknown in an inverse variation problem the same way we do for direct variation problems. The only difference is that the ratio we use to find the constant of variation is $k = x^ny$.

$\underline{Example}$

A quantity $y$ varies inversely with the square of $x$. If $y = 8$ where $x = 3$, find $y$ when $x$ is $4$.

The input is defined by $x$, the output is defined as $y$, and the power of the function is $2$. We still need to find the constant of variation using the ratio:

$k = x^ny = 3^2(8) = 9(8) = 72$

With all the information, we can write the equation for the relationship as $y = \frac{72}{x^2}$. We can use this equation to solve for $y$ when $x = 4$ by substituting the value into the equation:

$y = \frac{72}{4^2} = \frac{72}{16} = \frac{9}{2}$

Therefore, when $x = 4$, $y = \frac{9}{2}$.

Joint Variation

The final type of variation is joint variation which occurs when a variable varies directly or inversely with multiple variables. For example, if $x$ varies directly with both $y$ and $z$, then $x = kyz$. This can get more general like if $x$ varies directly with $y$ and inversely with $z$, we have $x = \frac{ky}{z}$.

note

We only use one constant of variation in a joint variation equation.

$\underline{Example}$

A quantity $x$ varies directly with the square of $y$ and inversely with $z$. If $x = 40$ when $y = 4$ and $z = 2$, find $x$ when $y = 10$ and $z = 25$.

The value of $x$ is dependent on both $y$ and $z$. The relationship between $x$ and $y$ is a direct variation which means $x = ky^2$ and the relationship between $x$ and $z$ is an inverse variation which means $x = \frac{k}{z}$. Combining these two relationships, we get $x = \frac{ky^2}{z}$ which only has one constant of variation because it is a joint variation.

We can use the equation $x = \frac{ky^2}{z}$ to solve for $k$ by substituting the values of $x$, $y$, and $z$:

$40 = \frac{k(4)^2}{2} \to 40 = \frac{16k}{2} \to 40 = 8k \to k = 5$

The relationship between $x$, $y$, and $z$ is $x = \frac{5y^2}{z}$. We can use this equation to solve for $x$ when $y = 10$ and $z = 25$ by substituting the values into the equation:

$x = \frac{5y^2}{z} = \frac{5(10)^2}{25} = \frac{5(100)}{25} = \frac{500}{25} = 20$

Therefore, when $y = 10$ and $z = 25$, $x = 20$.