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# 12.1: Inverse Variation Models

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What if you were paid \$500 per week regardless of the number of hours you worked? The more hours you worked in a week (increasing quantity), the less your hourly rate (decreasing quantity) would be. How could you write and solve a function to model this situation? After completing Concept, you'll be able to write inverse variation equations and solve inverse variation applications like this one.

### Watch This

Watch this video to see some more variation problems worked out, including problems involving joint variation.

### Guidance

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:

$distance = speed \times time$

Since the speed is a constant 3 miles per hour, we can write: $d = 3t$ .

The general equation for a direct variation is $y = kx$ , where $k$ is called the constant of proportionality.

You can see from the equation that a direct variation is a linear equation with a $y-$ intercept of zero. The graph of a direct variation relationship is a straight line passing through the origin whose slope is $k$ , 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 $distance = speed \times time$ again and solve for time, we obtain:

$time = \frac{distance}{speed}$

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: $t = \frac{90}{s}$ .

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

$V = \pi R^2 \cdot h$

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 $\pi$ .

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:

$F = G \frac{m_1 m_2}{d^2}$

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

Graph Inverse Variation Equations

We saw that the general equation for inverse variation is given by the formula $y = \frac{k}{x}$ , where $k$ 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, $x$ and $y$ are positive, so in our table we’ll choose only positive values of $x$ .

#### Example A

Graph an inverse variation relationship with the proportionality constant $k = 1$ .

Solution

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

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 $x = 0$ , the value of $y$ is undefined. The graph shows that when the value of $x$ is very small, the value of $y$ is very big—so it approaches infinity as $x$ gets closer and closer to zero.

Similarly, as the value of $x$ gets very large, the value of $y$ 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 $y = \frac{k}{x}$ . In general, we need to know the value of $y$ at a particular value of $x$ in order to find the proportionality constant. Once we know the proportionality constant, we can then find the value of $y$ for any given value of $x$ .

#### Example B

If $y$ is inversely proportional to $x$ , and if $y = 10$ when $x = 5$ , find $y$ when $x = 2$ .

Solution

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

Compare Graphs of Inverse Variation Equations

Inverse variation problems are the simplest example of rational functions. We saw that an inverse variation has the general equation: $y= \frac{k}{x}$ . In most real-world problems, $x$ and $y$ take only positive values. Below, we will show graphs of three inverse variation functions.

#### Example C

On the same coordinate grid, graph inverse variation relationships with the proportionality constants $k = 1, k = 2,$ and $k = \frac{1}{2}$ .

Solution

We’ll skip the table of values for this problem, and just show the graphs of the three functions on the same coordinate axes. Notice that for larger constants of proportionality, the curve decreases at a slower rate than for smaller constants of proportionality. This makes sense because the value of $y$ is related directly to the proportionality constants, so we should expect larger values of $y$ for larger values of $k$ .

Watch this video for help with the Examples above.

### Vocabulary

• The general equation for a direct variation is $y = kx$ , where $k$ is called the constant of proportionality.
• The general equation for inverse variation is $y= \frac{k}{x}$ , where $k$ is the constant of proportionality .

### Guided Practice

If $p$ is inversely proportional to the square of $q$ , and $p = 64$ when $q = 3$ , find $p$ when $q = 5$ .

Solution

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

### Practice

For 1-4, graph the following inverse variation relationships.

1. $y= \frac{3}{x}$
2. $y= \frac{10}{x}$
3. $y= \frac{1}{4x}$
4. $y= \frac{5}{6x}$
5. If $z$ is inversely proportional to $w$ and $z = 81$ when $w = 9$ , find $w$ when $z = 24$ .
6. If $y$ is inversely proportional to $x$ and $y = 2$ when $x = 8$ , find $y$ when $x = 12$ .
7. If $a$ is inversely proportional to the square root of $b$ , and $a = 32$ when $b = 9$ , find $b$ when $a = 6$ .
8. If $w$ is inversely proportional to the square of $u$ and $w = 4$ when $u = 2$ , find $w$ when $u = 8$ .
9. If $a$ is proportional to both $b$ and $c$ and $a = 7$ when $b = 2$ and $c = 6$ , find $a$ when $b = 4$ and $c = 3$ .
10. If $x$ is proportional to $y$ and inversely proportional to $z$ , and $x = 2$ when $y = 10$ and $z = 25$ , find $x$ when $y = 8$ and $z = 35$ .
11. If $a$ varies directly with $b$ and inversely with the square of $c$ , and $a = 10$ when $b = 5$ and $c = 2$ , find the value of $a$ when $b = 3$ and $c = 6$ .
12. If $x$ varies directly with $y$ and $z$ varies inversely with $x$ , and $z = 3$ when $y = 5$ , find $z$ when $y = 10$ .

Oct 01, 2012

Mar 05, 2014