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# 12.1: Inverse Variation Models

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Distinguish direct and inverse variation.
• Graph inverse variation equations.
• Write inverse variation equations.
• Solve real-world problems using inverse variation equations.

## Introduction

Many variables in real-world problems are related to each other by variations. A variation is an equation that relates a variable to one or more other variables by the operations of multiplication and division. There are three different kinds of variation: direct variation, inverse variation and joint variation.

## Distinguish Direct and Inverse Variation

In direct variation relationships, the related variables will either increase together or decrease together at a steady rate. For instance, consider a person walking at three miles per hour. As time increases, the distance covered by the person walking also increases, at the rate of three miles each hour. The distance and time are related to each other by a direct variation:

\begin{align*}distance = speed \times time\end{align*}

Since the speed is a constant 3 miles per hour, we can write: \begin{align*}d = 3t\end{align*}.

The general equation for a direct variation is \begin{align*}y = kx\end{align*}, where \begin{align*}k\end{align*} is called the constant of proportionality.

You can see from the equation that a direct variation is a linear equation with a \begin{align*}y-\end{align*}intercept of zero. The graph of a direct variation relationship is a straight line passing through the origin whose slope is \begin{align*}k\end{align*}, the constant of proportionality.

A second type of variation is inverse variation. When two quantities are related to each other inversely, one quantity increases as the other one decreases, and vice versa.

For instance, if we look at the formula \begin{align*}distance = speed \times time\end{align*} again and solve for time, we obtain:

\begin{align*}time = \frac{distance}{speed}\end{align*}

If we keep the distance constant, we see that as the speed of an object increases, then the time it takes to cover that distance decreases. Consider a car traveling a distance of 90 miles, then the formula relating time and speed is: \begin{align*}t = \frac{90}{s}\end{align*}.

The general equation for inverse variation is \begin{align*}y= \frac{k}{x}\end{align*}, where \begin{align*}k\end{align*} is the constant of proportionality.

In this chapter, we’ll investigate how the graphs of these relationships behave.

Another type of variation is a joint variation. In this type of relationship, one variable may vary as a product of two or more variables.

For example, the volume of a cylinder is given by:

\begin{align*}V = \pi R^2 \cdot h\end{align*}

In this example the volume varies directly as the product of the square of the radius of the base and the height of the cylinder. The constant of proportionality here is the number \begin{align*}\pi\end{align*}.

In many application problems, the relationship between the variables is a combination of variations. For instance Newton’s Law of Gravitation states that the force of attraction between two spherical bodies varies jointly as the masses of the objects and inversely as the square of the distance between them:

\begin{align*}F = G \frac{m_1 m_2}{d^2}\end{align*}

In this example the constant of proportionality is called the gravitational constant, and its value is given by \begin{align*}G = 6.673 \times 10^{-11} \ N \cdot m^2 / kg^2\end{align*}.

## Graph Inverse Variation Equations

We saw that the general equation for inverse variation is given by the formula \begin{align*}y = \frac{k}{x}\end{align*}, where \begin{align*}k\end{align*} is a constant of proportionality. We will now show how the graphs of such relationships behave. We start by making a table of values. In most applications, \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are positive, so in our table we’ll choose only positive values of \begin{align*}x\end{align*}.

Example 1

Graph an inverse variation relationship with the proportionality constant \begin{align*}k = 1\end{align*}.

Solution

\begin{align*}x\end{align*} \begin{align*}y =\frac {1}{x}\end{align*}
0 \begin{align*}y=\frac {1}{0} = \text{undefined}\end{align*}
\begin{align*}\frac {1}{4}\end{align*} \begin{align*}y =\frac {1}{\frac{1}{4}}=4\end{align*}
\begin{align*}\frac {1}{2}\end{align*} \begin{align*}y=\frac {1}{\frac{1}{2}}=2\end{align*}
\begin{align*}\frac {3}{4}\end{align*} \begin{align*}y =\frac {1}{\frac{3}{4}}=1.33\end{align*}
1 \begin{align*}y =\frac {1}{1}=1\end{align*}
\begin{align*}\frac {3}{2}\end{align*} \begin{align*}y=\frac {1}{\frac{3}{2}}=0.67\end{align*}
2 \begin{align*}y =\frac {1}{2}=0.5\end{align*}
3 \begin{align*}y=\frac {1}{3}=0.33\end{align*}
4 \begin{align*}y =\frac {1}{4}=0.25\end{align*}
5 \begin{align*}y =\frac {1}{5}=0.2\end{align*}
10 \begin{align*}y=\frac {1}{10}=0.1\end{align*}

Here is a graph showing these points connected with a smooth curve.

Both the table and the graph demonstrate the relationship between variables in an inverse variation. As one variable increases, the other variable decreases and vice versa.

Notice that when \begin{align*}x = 0\end{align*}, the value of \begin{align*}y\end{align*} is undefined. The graph shows that when the value of \begin{align*}x\end{align*} is very small, the value of \begin{align*}y\end{align*} is very big—so it approaches infinity as \begin{align*}x\end{align*} gets closer and closer to zero.

Similarly, as the value of \begin{align*}x\end{align*} gets very large, the value of \begin{align*}y\end{align*} gets smaller and smaller but never reaches zero. We will investigate this behavior in detail throughout this chapter.

## Write Inverse Variation Equations

As we saw, an inverse variation fulfills the equation \begin{align*}y = \frac{k}{x}\end{align*}. In general, we need to know the value of \begin{align*}y\end{align*} at a particular value of \begin{align*}x\end{align*} in order to find the proportionality constant. Once we know the proportionality constant, we can then find the value of \begin{align*}y\end{align*} for any given value of \begin{align*}x\end{align*}.

Example 2

If \begin{align*}y\end{align*} is inversely proportional to \begin{align*}x\end{align*}, and if \begin{align*}y = 10\end{align*} when \begin{align*}x = 5\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x = 2\end{align*}.

Solution

\begin{align*}\text{Since} \ y \ \text{is inversely proportional to} \ x, \text{then:} \qquad \qquad \qquad \qquad y =\frac {k}{x}\!\\ \\ \text{Plug in the values} \ y = 10 \ \text{and} \ x = 5: \qquad \qquad \ \qquad \qquad \qquad 10 =\frac {k}{5}\!\\ \\ \text{Solve for} \ k \ \text{by multiplying both sides of the equation by} \ 5: \ \ \ k =50\!\\ \\ \text{The inverse relationship is given by:} \qquad \qquad \qquad \qquad \ \qquad \ \ y =\frac {50}{x}\!\\ \\ \text{When} \ x = 2: \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \ \ \qquad y=\frac {50}{2} \ \text{or} \ y=25\end{align*}

Example 3

If \begin{align*}p\end{align*} is inversely proportional to the square of \begin{align*}q\end{align*}, and \begin{align*}p = 64\end{align*} when \begin{align*}q = 3\end{align*}, find \begin{align*}p\end{align*} when \begin{align*}q = 5\end{align*}.

Solution

\begin{align*}\text{Since} \ p \ \text{is inversely proportional to} \ q^2, \text{then:} \qquad \qquad \qquad \qquad \ \ p =\frac {k}{q^2}\!\\ \\ \text{Plug in the values} \ p = 64 \ \text{and} \ q = 3: \qquad \qquad \quad \qquad \qquad \qquad \ \ 64 =\frac {k}{3^2} \ \text{or} \ 64=\frac {k}{9}\!\\ \\ \text{Solve for} \ k \ \text{by multiplying both sides of the equation by} \ 9: \qquad \ k =576\!\\ \\ \text{The inverse relationship is given by:} \qquad \qquad \qquad \qquad \qquad \qquad \ p =\frac {576}{q^2}\!\\ \\ \text{When} \ q = 5: \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad p =\frac {576}{25} \ \text{or} \ y=23.04\end{align*}

To see some more variation problems worked out, including problems involving joint variation, watch the video at

.

## Solve Real-World Problems Using Inverse Variation Equations

Many formulas in physics are described by variations. In this section we’ll investigate some problems that are described by inverse variations.

Example 4

The frequency, \begin{align*}f\end{align*}, of sound varies inversely with wavelength, \begin{align*}\lambda\end{align*}. A sound signal that has a wavelength of 34 meters has a frequency of 10 hertz. What frequency does a sound signal of 120 meters have?

Solution

\begin{align*}\text{The inverse variation relationship is:} \qquad \qquad \ f =\frac {k}{\lambda}\!\\ \\ \text{Plug in the values:} \ \lambda = 34 \ \text{and} \ f = 10: \qquad \quad 10=\frac {k}{34}\!\\ \\ \text{Multiply both sides by} \ 34: \qquad \quad \qquad \qquad \quad \ k=340\!\\ \\ \text{Thus, the relationship is given by:} \qquad \quad \qquad \ f=\frac {340}{\lambda}\!\\ \\ \text{Plug in} \ \lambda = 120 \ \text{meters:} \qquad \qquad \qquad \qquad \qquad f=\frac {340}{120} \Rightarrow f=2.83 \ \text{Hertz}\end{align*}

Example 5

Electrostatic force is the force of attraction or repulsion between two charges. The electrostatic force is given by the formula \begin{align*}F = \frac{Kq_1 q_2}{d^2}\end{align*}, where \begin{align*}q_1\end{align*} and \begin{align*}q_2\end{align*} are the charges of the charged particles, \begin{align*}d\end{align*} is the distance between the charges and \begin{align*}k\end{align*} is a proportionality constant. The charges do not change, so they too are constants; that means we can combine them with the other constant \begin{align*}k\end{align*} to form a new constant \begin{align*}K\end{align*}, so we can rewrite the equation as \begin{align*}F = \frac{K}{d^2}\end{align*}.

If the electrostatic force is \begin{align*}F = 740\end{align*} Newtons when the distance between charges is \begin{align*}5.3 \times 10^{-11}\end{align*} meters, what is \begin{align*}F\end{align*} when \begin{align*}d = 2.0 \times 10^{-10}\end{align*} meters?

Solution

\begin{align*}\text{The inverse variation relationship is:} \qquad \qquad \qquad \qquad F=\frac {K}{d^2}\!\\ \\ \text{Plug in the values} \ F = 740 \ \text{and} \ d = 5.3\times10^{-11}: \qquad \quad 740=\frac {K}{\left ( {5.3\ \times \ 10^{-11}} \right )^2}\!\\ \\ \text{Multiply both sides by} \ (5.3\times10^{-11})^2: \qquad \qquad \qquad \quad \ K=740 \left ( {5.3\times 10^{-11}} \right )^2\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad K = 2.08\times 10^{-18}\!\\ \\ \text{The electrostatic force is given by:} \qquad \qquad \qquad \quad \qquad F=\frac {2.08\ \times \ 10^{-18}}{d^2}\!\\ \\ \text{When} \ d = 2.0 \times 10^{-10}: \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ F=\frac {2.08\ \times \ 10^{-18}}{\left ( 2.0 \ \times \ 10^{-10} \right )^2}\!\\ \\ \text{Use scientific notation to simplify:} \qquad \qquad \qquad \qquad \quad F=52 \ \text{Newtons}\end{align*}

## Review Questions

Graph the following inverse variation relationships.

1. \begin{align*}y= \frac{3}{x}\end{align*}
2. \begin{align*}y= \frac{10}{x}\end{align*}
3. \begin{align*}y= \frac{1}{4x}\end{align*}
4. \begin{align*}y= \frac{5}{6x}\end{align*}
5. If \begin{align*}z\end{align*} is inversely proportional to \begin{align*}w\end{align*} and \begin{align*}z = 81\end{align*} when \begin{align*}w = 9\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}z = 24\end{align*}.
6. If \begin{align*}y\end{align*} is inversely proportional to \begin{align*}x\end{align*} and \begin{align*}y = 2\end{align*} when \begin{align*}x = 8\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x = 12\end{align*}.
7. If \begin{align*}a\end{align*} is inversely proportional to the square root of \begin{align*}b\end{align*}, and \begin{align*}a = 32\end{align*} when \begin{align*}b = 9\end{align*}, find \begin{align*}b\end{align*} when \begin{align*}a = 6\end{align*}.
8. If \begin{align*}w\end{align*} is inversely proportional to the square of \begin{align*}u\end{align*} and \begin{align*}w = 4\end{align*} when \begin{align*}u = 2\end{align*}, find \begin{align*}w\end{align*} when \begin{align*}u = 8\end{align*}.
9. If \begin{align*}a\end{align*} is proportional to both \begin{align*}b\end{align*} and \begin{align*}c\end{align*} and \begin{align*}a = 7\end{align*} when \begin{align*}b = 2\end{align*} and \begin{align*}c = 6\end{align*}, find \begin{align*}a\end{align*} when \begin{align*}b = 4\end{align*} and \begin{align*}c = 3\end{align*}.
10. If \begin{align*}x\end{align*} is proportional to \begin{align*}y\end{align*} and inversely proportional to \begin{align*}z\end{align*}, and \begin{align*}x = 2\end{align*} when \begin{align*}y = 10\end{align*} and \begin{align*}z = 25\end{align*}, find \begin{align*}x\end{align*} when \begin{align*}y = 8\end{align*} and \begin{align*}z = 35\end{align*}.
11. If \begin{align*}a\end{align*} varies directly with \begin{align*}b\end{align*} and inversely with the square of \begin{align*}c\end{align*}, and \begin{align*}a = 10\end{align*} when \begin{align*}b = 5\end{align*} and \begin{align*}c = 2\end{align*}, find the value of \begin{align*}a\end{align*} when \begin{align*}b = 3\end{align*} and \begin{align*}c = 6\end{align*}.
12. If \begin{align*}x\end{align*} varies directly with \begin{align*}y\end{align*} and \begin{align*}z\end{align*} varies inversely with \begin{align*}x\end{align*}, and \begin{align*}z = 3\end{align*} when \begin{align*}y = 5\end{align*}, find \begin{align*}z\end{align*} when \begin{align*}y = 10\end{align*}.
13. The intensity of light is inversely proportional to the square of the distance between the light source and the object being illuminated.
1. A light meter that is 10 meters from a light source registers 35 lux. What intensity would it register 25 meters from the light source?
2. A light meter that is registering 40 lux is moved twice as far away from the light source illuminating it. What intensity does it now register? (Hint: let \begin{align*}x\end{align*} be the original distance from the light source.)
3. The same light meter is moved twice as far away again (so it is now four times as far from the light source as it started out). What intensity does it register now?
14. Ohm’s Law states that current flowing in a wire is inversely proportional to the resistance of the wire. If the current is 2.5 Amperes when the resistance is 20 ohms, find the resistance when the current is 5 Amperes.
15. The volume of a gas varies directly with its temperature and inversely with its pressure. At 273 degrees Kelvin and pressure of 2 atmospheres, the volume of a certain gas is 24 liters.
1. Find the volume of the gas when the temperature is 220 Kelvin and the pressure is 1.2 atmospheres.
2. Find the temperature when the volume is 24 liters and the pressure is 3 atmospheres.
16. The volume of a square pyramid varies jointly with the height and the square of the side length of the base. A pyramid whose height is 4 inches and whose base has a side length of 3 inches has a volume of \begin{align*}12 \ in^3\end{align*}.
1. Find the volume of a square pyramid that has a height of 9 inches and whose base has a side length of 5 inches.
2. Find the height of a square pyramid that has a volume of \begin{align*}49 \ in^3\end{align*} and whose base has a side length of 7 inches.
3. A square pyramid has a volume of \begin{align*}72 \ in^3\end{align*} and its base has a side length equal to its height. Find the height of the pyramid.

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Date Created:
Feb 23, 2012