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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 variables by the operations of multiplication and division. There are three different kinds of variation problems: 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.

distance=rate×time\begin{align*}\text{distance}= \text{rate} \times \text{time}\end{align*}

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

Direct Variation

The general equation for a direct variation is

y=kx.\begin{align*}y = kx.\end{align*}

k\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 y\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 k\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, as one quantitiy increases, the other one decreases and vice-versa.

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

time=distancerate\begin{align*}\text{time}= \frac{\text{distance}}{\text{rate}}\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 t=90s\begin{align*}t = \frac{90}{s}\end{align*}.

Inverse Variation

The general equation for inverse variation is

y=kx\begin{align*}y=\frac{k}{x}\end{align*}

where k\begin{align*}k\end{align*} is called the constant of proportionality.

In this chapter, we will investigate how the graph of these relationships behave.

Another type 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:

V=πr2h\begin{align*} V=\pi r^2 \cdot h\end{align*}

In this formula, 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 π\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

F=Gm1m2d2\begin{align*} F=G\frac {m_1m_2}{d^2}\end{align*}

In this example the constant of proportionality, G\begin{align*}G\end{align*}, is called the gravitational constant and its value is given by G=6.673×1011Nm2/kg2\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=\left( \frac{k}{x} \right )\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 will 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 and 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 the value of 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 = \left ( \frac{k}{x} \right )\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. After the proportionality constant is known, we can 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 \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 the general relationship tells us} & & y & =\frac {k}{x}\\ \text{Plug in the values} \ y = 10 \ \text{and} \ x = 5. & & 10 & =\frac {k}{5}\\ \text{Solve for} \ k \ \text{by multiplying both sides of the equation by} \ 5. & & k&=50\\ \text{Now we put} \ k \ \text{back into the general equation.}\\ \text{The inverse relationship is given by} & & y & =\frac {50}{x}\\ \text{When} \ x = 2 & & y& =\frac {50}{2} \ \text{or} \ y=25\end{align*}

Answer \begin{align*}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 the general equation is} & & p& =\frac {k}{q^2}\\ \text{Plug in the values} \ p = 64 \ \text{and} \ q = 3. & & 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. & & k& =576\\ \text{The inverse relationship is given by} & & p&=\frac {576}{q^2}\\ \text{When} \ q = 5 & & p&=\frac {576}{25} \ \text{or} \ p=23.04\end{align*}

Answer \begin{align*}p=23.04\end{align*}.

## Solve Real-World Problems Using Inverse Variation Equations

Many formulas in physics are described by variations. In this section we will 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} & & f&=\frac {k}{\lambda}\\ \text{Plug in the values} \ \lambda = 34 \ \text{and} \ f = 10. & & 10&=\frac {k}{34}\\ \text{Multiply both sides by} \ 34. & & k&=340 \\ \text{Thus, the relationship is given by} & & f&=\frac {340}{\lambda} \\ \text{Plug in} \ \lambda = 120 \ meters. & & f&=\frac {340}{120}\Rightarrow f=2.83\end{align*}

Answer \begin{align*} f=2.83\ 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=\left ( \frac{Kq_1q_2}{d^2}\right )\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 proportionality constant. The charges do not change and are, thus, constants and can then be combined with the other constant \begin{align*}k\end{align*} to form a new constant \begin{align*}K\end{align*}. The equation is rewritten as \begin{align*}F=\left ( \frac{K}{d^2} \right )\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} \ meters\end{align*}, what is \begin{align*}F\end{align*} when \begin{align*}d = 2.0 \times 10^{-10} \ meters\end{align*}?

Solution

\begin{align*}\text{The inverse variation relationship is} & & f&=\frac {k}{d^2}\\ \text{Plug in the values} \ F = 740 \ \text{and} \ d = 5.3\times10^{-11}. & & 740&=\frac {k}{\left ( {5.3\times 10^{-11}} \right )^2}\\ \text{Multiply both sides by} \ (5.3\times10^{-11})^2. & & K&=740 \left ( {5.3\times 10^{-11}} \right )^2\\ \text{The electrostatic force is given by} & & F&=\frac {2.08\times 10^{-18}}{d^2}\\ \text{When} \ d = 2.0 \times 10^{-10} & & F&=\frac {2.08\times 10^{-18}}{\left ( 2.0\times 10^{-10} \right )^2}\\ \text{Enter} \ \frac{2.08*10^{(-18)}}{\left (2.0*10^{(-10)} \right )^2} \ \text{into a calculator}. & & F&=52\end{align*}

Answer \begin{align*}F=52 \ Newtons\end{align*}

Note: In the last example, you can also compute \begin{align*}F=\frac {2.08\times 10^{-18}}{\left ( 2.0\times 10^{-10} \right )^2}\end{align*} by hand.

\begin{align*}F & = \frac {2.08 \times 10^{-18}}{\left ( 2.0 \times 10^{-10} \right )^2} \\ & = \frac {2.08 \times 10^{-18}}{4.0 \times 10^{-20}}\\ & = \frac {2.08 \times 10^{20}}{4.0 \times 10^{18}}\\ & = \frac {2.08}{4.0} \left( 10^{2} \right) \\ & = 0.52(100) \\ & =52 \end{align*}

This illustrates the usefulness of scientific notation.

## 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*}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*}.
10. 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*}.
11. The intensity of light is inversely proportional to the square of the distance between the light source and the object being illuminated. 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?
12. 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.
13. The volume of a gas varies directly to its temperature and inversely to its pressure. At 273 degrees Kelvin and pressure of 2 atmospheres, the volume of the gas is 24 Liters. Find the volume of the gas when the temperature is 220 kelvin and the pressure is 1.2 atmospheres.
14. The volume of a square pyramid varies jointly as the height and the square of the length of the base. A cone 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*}. Find the volume of a square pyramid that has a height of 9 inches and whose base has a side length of 5 inches.

1. \begin{align*} W=\frac {243}{8}\end{align*}
2. \begin{align*} y=\frac {4}{3}\end{align*}
3. \begin{align*}b = 256\end{align*}
4. \begin{align*} w=\frac {1}{4}\end{align*}
5. \begin{align*} x=\frac {8}{7}\end{align*}
6. \begin{align*} a=\frac {2}{3}\end{align*}
7. \begin{align*}I = 5.6 \ lux\end{align*}
8. \begin{align*}R = 10 \ ohms\end{align*}
9. \begin{align*}V = 32.2 \ L\end{align*}
10. \begin{align*}V = 75 \ in^3\end{align*}

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