Modeling Using Variation

Section 4.3 Modeling Using Variation

Objectives

Solve direct variation problems
Solve inverse variation problems

A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.

Subsection 4.3.1 Solving Direct Variation Problems

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula $e=0.16s$ tells us her earnings, $e\text{,}$ come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.

Table 4.3.1.

Sales Price	$e = 0.16 s$	Interpretation
$4,600	$e = 0.16\cdot 4600 = 736$	A sale of a $4,600 vehicle results in $736 earnings.
$9,200	$e = 0.16 \cdot 9200= 1472$	A sale of a $9,200 vehicle results in $1472 earnings.
$18,400	$e = 0.16 \cdot 18400$	A sale of a $18,400 vehicle results in $2944 earnings.

Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation. Each variable in this type of relationship varies directly with the other.

The figure below represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula $y=kx^n$ is used for direct variation. The value $k$ is a nonzero constant greater than zero and is called the constant of variation. In this case, $k = 0.16$ and $n = 1\text{.}$

Definition 4.3.2. Direct Variation.

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

\begin{equation*} y = kx^n \end{equation*}

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

Note 4.3.3. How To Solve Direct Variation Problems.

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

Identify the input, $x\text{,}$ and the output, $y\text{.}$
Determine the constant of variation. You may need to divide $y$ by the specified power of $x$ to determine the constant of variation.
Use the constant of variation to write an equation for the relationship.
Substitute known values into the equation to find the unknown.

Example 4.3.4.

The quantity $y$ varies directly with $x\text{.}$ If $y=22$ when $x=4\text{,}$ find $y$ when $x$ is 6.

Solution.

The general formula for direct variation is

\begin{equation*} y = kx \end{equation*}

First plug-in $y = 22$ and $x = 4$ and solve for $k\text{.}$

\begin{equation*} 22 = k\cdot 4 \end{equation*}

\begin{equation*} k = \frac{22}{4} \end{equation*}

\begin{equation*} k = \frac{11}{2} \end{equation*}

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

\begin{equation*} y = \frac{11}{2}x \end{equation*}

Substitute $x = 6$ and solve for $y\text{.}$

\begin{equation*} y = \frac{11}{2}\cdot 6 \end{equation*}

\begin{equation*} y = \frac{66}{2} \end{equation*}

\begin{equation*} y = 33 \end{equation*}

Example 4.3.5.

The quantity $p$ varies directly with the square of $q\text{.}$ If $p=20$ when $q=3\text{,}$ find $p$ when $q$ is 6.

Solution.

The general formula for direct variation is

\begin{equation*} y = kx^2\text{,} \end{equation*}

or in this case:

\begin{equation*} p = kq^2 \end{equation*}

First plug-in $p = 20$ and $q = 3$ and solve for $k\text{.}$

\begin{equation*} 20 = k\cdot 3^2 \end{equation*}

\begin{equation*} 20 = k \cdot 9 \end{equation*}

\begin{equation*} k = \frac{20}{9} \end{equation*}

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

\begin{equation*} p = \frac{20}{9}q^2 \end{equation*}

Substitute $q = 6$ and solve for $p\text{.}$

\begin{equation*} p = \frac{20}{9}\cdot 6^2 \end{equation*}

\begin{equation*} p = \frac{20}{9}\cdot 36 \end{equation*}

\begin{equation*} p = 80 \end{equation*}

Subsection 4.3.2 Solving Inverse Variation Prolblems

Water temperature in an ocean varies inversely to the water’s depth. The formula $T=\frac{14,000}{d}$ gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.

If we create the table below, we observe that, as the depth increases, the water temperature decreases.

Table 4.3.6.

Depth	$T = \frac{14,000}{d}$	Interpretation
500 ft	$T = \frac{14,000}{500}=28$	At a depth of 500 ft, the water temperature is $28^{\circ} F$
1000 ft	$T = \frac{14,000}{1000}=14$	At a depth of 1,000 ft, the water temperature is $14^{\circ} F$
2000 ft	$T = \frac{14,000}{2000}=7$	At a depth of 2,000 ft, the water temperature is $7^{\circ} F$

We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations.

For our example, the figure depicts the inverse variation. We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula $y = \frac{k}{x}$ for inverse variation in this case uses $k=14,000\text{.}$

Definition 4.3.7. Inverse Variation.

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

\begin{equation*} y = \frac{k}{x^n} \end{equation*}

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

Example 4.3.8.

The quantity $y$ varies inversely with the square $x\text{.}$ If $y=10$ when $x=2\text{,}$ find $y$ when $x$ is 4.

Solution.

The general formula for this inverse variation is

\begin{equation*} y = \frac{k}{x^2} \end{equation*}

First plug-in $y = 10$ and $x = 2$ and solve for $k\text{.}$

\begin{equation*} 10 = \frac{k}{2^2} \end{equation*}

\begin{equation*} 10 = \frac{k}{4} \end{equation*}

\begin{equation*} 4\cdot 10 = \frac{k}{4} \cdot 4 \end{equation*}

\begin{equation*} k = 40 \end{equation*}

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

\begin{equation*} y = \frac{40}{x^2} \end{equation*}

Substitute $x = 4$ and solve for $y\text{.}$

\begin{equation*} y = \frac{40}{4^2} \end{equation*}

\begin{equation*} y = \frac{40}{16} \end{equation*}

\begin{equation*} y = \frac{5}{2} \end{equation*}

Example 4.3.9.

The volume of a gas held at constant temperature varies inversely as the pressure of the gas. If the volume of a gas is 1200 cubic centimeters when the pressure is 200 millimeters of mercury, what is the volume when the pressure is 300 millimeters of mercury?

Solution.

We can expression the fact that "volume varies inversely as pressure" with the equation,

\begin{equation*} V = \frac{k}{P} \end{equation*}

Use the $V = 1200$ when $P = 200$ to solve for $k\text{:}$

\begin{equation*} 1200 = \frac{k}{200} \end{equation*}

\begin{equation*} 200 \cdot 1200 = \frac{k}{200} \cdot 200 \end{equation*}

\begin{equation*} k = 240,000 \end{equation*}

We can write the relationship,

\begin{equation*} V = \frac{240,000}{P} \end{equation*}

and plug in $P = 300$ to find the volume when pressure is 300 millimeters of mercury.

\begin{equation*} V = \frac{240,000}{300} \end{equation*}

\begin{equation*} V = 800 \end{equation*}

The volume is 800 cubic centimeters.

Depth	\(T = \frac{14,000}{d}\)	Interpretation
500 ft	\(T = \frac{14,000}{500}=28\)	At a depth of 500 ft, the water temperature is \(28^{\circ} F\)
1000 ft	\(T = \frac{14,000}{1000}=14\)	At a depth of 1,000 ft, the water temperature is \(14^{\circ} F\)
2000 ft	\(T = \frac{14,000}{2000}=7\)	At a depth of 2,000 ft, the water temperature is \(7^{\circ} F\)