Modeling using Variation

Direct Variation

A General Note: Direct Variation

If x and y are related by an equation of the form

$y=k{x}^{n}$

then we say that the relationship is direct variation and y varies directly with the nth power of x. In direct variation relationships, there is a nonzero constant ratio $k={\frac {y}{{x}^{n}}}$, where k is called the constant of variation, which help defines the relationship between the variables.

How To: Given a description of a direct variation problem, solve for an unknown.

1. Identify the input, x, and the output, y.
2. Determine the constant of variation. You may need to divide y by the specified power of x to determine the constant of variation.
3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.

Example 1: Solving a Direct Variation Problem

The quantity y varies directly with the cube of x. If y = 25 when x = 2, find y when x is 6.

Solution

The general formula for direct variation with a cube is $y=k{x}^{3}$. The constant can be found by dividing y by the cube of x.

${\begin{cases}k={\frac {y}{{x}^{3}}}\\={\frac {25}{{2}^{3}}}\\={\frac {25}{8}}\end{cases}}$

Now use the constant to write an equation that represents this relationship.

$y={\frac {25}{8}}{x}^{3}$

Substitute x = 6 and solve for y.

${\begin{cases}y={\frac {25}{8}}{\left(6\right)}^{3}\\{\text{ }}=675\end{cases}}$

Analysis of the Solution

The graph of this equation is a simple cubic, as shown below.

<img src="https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/1227/2015/04/03010805/CNX_Precalc_Figure_03_09_0022.jpg" alt="Graph of y=25/8(x^3) with the labeled points (2, 25) and (6, 675)." width="487" height="367" data-media-type="image/jpg" />

Figure 2

Inverse variation

A General Note: Inverse Variation

If x and y are related by an equation of the form

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

where k is a nonzero constant, then we say that y varies inversely with the nth power of x. In inversely proportional relationships, or inverse variations, there is a constant multiple $k={x}^{n}y$.

Example 2: Writing a Formula for an Inversely Proportional Relationship

A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.

Solution

Recall that multiplying speed by time gives distance. If we let t represent the drive time in hours, and v represent the velocity (speed or rate) at which the tourist drives, then vt = distance. Because the distance is fixed at 100 miles, vt = 100. Solving this relationship for the time gives us our function.

${\begin{cases}t\left(v\right)={\frac {100}{v}}\\{\text{ }}=100{v}^{-1}\end{cases}}$

We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction.

How To: Given a description of an indirect variation problem, solve for an unknown.

1. Identify the input, x, and the output, y.
2. Determine the constant of variation. You may need to multiply y by the specified power of x to determine the constant of variation.
3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.

Example 3: Solving an Inverse Variation Problem

A quantity y varies inversely with the cube of x. If y = 25 when x = 2, find y when x is 6.

Solution

The general formula for inverse variation with a cube is $y={\frac {k}{{x}^{3}}}$. The constant can be found by multiplying y by the cube of x.

${\begin{cases}k={x}^{3}y\\{\text{ }}={2}^{3}\cdot 25\\{\text{ }}=200\end{cases}}$

Now we use the constant to write an equation that represents this relationship.

${\begin{cases}y={\frac {k}{{x}^{3}}},k=200\\y={\frac {200}{{x}^{3}}}\end{cases}}$

Substitute x = 6 and solve for y.

${\begin{cases}y={\frac {200}{{6}^{3}}}\\{\text{ }}={\frac {25}{27}}\end{cases}}$

Analysis of the Solution

The graph of this equation is a rational function.

<img src="https://s3-us-west-2.amazonaws.com/courses-images-archive-read-only/wp-content/uploads/sites/1227/2015/04/03010806/CNX_Precalc_Figure_03_09_0042.jpg" alt="Graph of y=25/(x^3) with the labeled points (2, 25) and (6, 25/27)." width="488" height="292" data-media-type="image/jpg" />

Figure 4

Joint variation

Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c, cost, varies jointly with the number of students, n, and the distance, d.

A General Note: Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables.

For instance, if x varies directly with both y and z, we have x = kyz. If x varies directly with y and inversely with z, we have $x={\frac {ky}{z}}$. Notice that we only use one constant in a joint variation equation.

Example 4: Solving Problems Involving Joint Variation

A quantity x varies directly with the square of y and inversely with the cube root of z. If x = 6 when y = 2 and z = 8, find x when y = 1 and z = 27.

Solution

Begin by writing an equation to show the relationship between the variables.

$x={\frac {k{y}^{2}}{\sqrt[{3}]{z}}}$

Substitute x = 6, y = 2, and z = 8 to find the value of the constant k.

${\begin{cases}6={\frac {k{2}^{2}}{\sqrt[{3}]{8}}}\\6={\frac {4k}{2}}\\3=k\end{cases}}$

Now we can substitute the value of the constant into the equation for the relationship.

$x={\frac {3{y}^{2}}{\sqrt[{3}]{z}}}$

To find x when y = 1 and z = 27, we will substitute values for y and z into our equation.

${\begin{cases}x={\frac {3{\left(1\right)}^{2}}{\sqrt[{3}]{27}}}\\{\text{ }}=1\end{cases}}$

Try It 3

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.

<a href="https://courses.lumenlearning.com/precalcone/chapter/solutions-18/" target="_blank" rel="noopener">Solution</a>

 

<section id="fs-id1165137898092" class="key-equations" data-depth="1">

Key Equations

<tbody> </tbody>

Direct variation	$y=k{x}^{n},k{\text{ is a nonzero constant}}$.
Inverse variation	$y={\frac {k}{{x}^{n}}},k{\text{ is a nonzero constant}}$.

</section> <section id="fs-id1165137419773" class="key-concepts" data-depth="1">

Key Concepts

• A relationship where one quantity is a constant multiplied by another quantity is called direct variation.
• Two variables that are directly proportional to one another will have a constant ratio.
• A relationship where one quantity is a constant divided by another quantity is called inverse variation.
• Two variables that are inversely proportional to one another will have a constant multiple.
• In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation.

Glossary

constant of variation

the non-zero value k that helps define the relationship between variables in direct or inverse variation

direct variation

the relationship between two variables that are a constant multiple of each other; as one quantity increases, so does the other

inverse variation

the relationship between two variables in which the product of the variables is a constant

inversely proportional

a relationship where one quantity is a constant divided by the other quantity; as one quantity increases, the other decreases

joint variation

a relationship where a variable varies directly or inversely with multiple variables

varies directly

a relationship where one quantity is a constant multiplied by the other quantity

varies inversely

a relationship where one quantity is a constant divided by the other quantity

Resources

• Modeling Using Variation, Lumen Learning
• Variation Word Problems, Open Text Intermediate Algebra
• Intro to Direct & Inverse Variation, Khan Academy
• Direct Inverse and Joint Variation Word Problems, The Organic Chemistry Tutor