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Modeling Using Variation

Solve direct variation problems.

Modeling Using Variation

Solve a formula for a specific variable (IA 2.3.1).
Solve direct variation problems (IA 7.5.5).

It is often helpful to solve a formula for a specific variable. If you need to put a formula in a spreadsheet, it is not unusual to have to solve it for a specific variable first. We isolate that variable on one side of the equals sign and all other variables and constants are on the other side of the equal sign.

Example

Try it.

Solve the formula $V=\frac{1}{3}\pi {r}^{2}h$ for h.

Solution

Write the formula.
Remove the fraction on the right.
Simplify.
Divide both sides by $\pi {r}^{2}.$

We could now use this formula to find the height of a right circular cone when we know the volume and the radius of the base, by using the formula $h=\frac{3V}{\pi {r}^{2}}.$

Solve the given formula for the indicated variable.

Try it.

$A=\frac{1}{2}bh$ for b

Try it.

$F=\frac{9}{5}C+32$ for C

Try it.

$C=\frac{5}{9}(F-32)$ for F

Condensed — the full section is in OpenStax College Algebra 2e.

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, $e,$ 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. See .

$s$ , sales price	$e=0.16s$	Interpretation
$4,600	$e=0.16(4,600)=736$	A sale of a $4,600 vehicle results in $736 earnings.
$9,200	$e=0.16(9,200)=1,472$	A sale of a $9,200 vehicle results in $1472 earnings.
$18,400	$e=0.16(18,400)=2,944$	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.

represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula $y=k{x}^{n}$ 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 $n=1.$ We saw functions like this one when we discussed power functions.

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Inverse Variation Problems

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 , we observe that, as the depth increases, the water temperature decreases.

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

Example

Try it.

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 $vt=\text{distance}\text{.}$ Because the distance is fixed at 100 miles, $vt=100$ so $t=100/v.$ Because time is a function of velocity, we can write $t(v).$

\[\begin{array}{lll}t(v) & = & \frac{100}{v} \\ & = & 100{v}^{-1}\end{array}\]

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. We say that time varies inversely with velocity.

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Problems Involving 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, $d.$

Example

Try it.

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 $k.$

\[\begin{array}{lll}6 & = & \frac{k{2}^{2}}{\sqrt[3]{8}} \\ 6 & = & \frac{4k}{2} \\ 3 & = & k\end{array}\]

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.

\[\begin{array}{lll}x & = & \frac{3{(1)}^{2}}{\sqrt[3]{27}} \\ & = & 1\end{array}\]

Key Concepts

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

Chapter Test

Give the degree and leading coefficient of the following polynomial function.

Determine the end behavior of the polynomial function.

Write the quadratic function in standard form. Determine the vertex and axes intercepts and graph the function.

Given information about the graph of a quadratic function, find its equation.

Solve the following application problem.

Find all zeros of the following polynomial functions, noting multiplicities.

Based on the graph, determine the zeros of the function and multiplicities.

Condensed — the full section is in OpenStax College Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve the formula $V=\frac{1}{3}\pi {r}^{2}h$ for h.

Zbulo përgjigjen

Write the formula.
Remove the fraction on the right.
Simplify.
Divide both sides by $\pi {r}^{2}.$
We could now use this formula to find the height of a right circular cone when we know the volume and the radius of the base, by using the formula $h=\frac{3V}{\pi {r}^{2}}.$
$A=\frac{1}{2}bh$ for b
$F=\frac{9}{5}C+32$ for C
$C=\frac{5}{9}(F-32)$ for F

When Raoul runs on the treadmill at the gym, the number of calories, c, he burns varies directly with the number of minutes, m, he uses the treadmill. He burned 315 calories when he used the treadmill for 18 minutes.

ⓐ Write the equation that relates c and m. ⓑ How many calories would he burn if he ran on the treadmill for 25 minutes?

Zbulo përgjigjen

ⓐ

	The number of calories, c, varies directly with the number of minutes, m, on the treadmill, and $c=315$ when $m=18$ .
Write the formula for direct variation.
We will use c in place of y and m in place of x.
Substitute the given values for the variables.
Solve for the constant of variation.
Write the equation that relates c and m.
Substitute in the constant of variation.

ⓑ

	$\$ Find c when m = 25.
Write the equation that relates c and m. $\$	$\$
Substitute the given value for m.	$\$
Simplify.	$\$
	$\$ Raoul would burn 437.5 calories if he used the treadmill for 25 minutes.

The number of calories $c$, burned varies directly with the amount of time $t$, spent exercising. Arnold burned 312 calories in 65 minutes exercising.
1. ⓐ Write the equation that relates c and t.
2. ⓑ How many calories would he burn if he exercises for 90 minutes?
The distance a moving body travels, $d$, varies directly with the time, $t$, that it moves. A train travels 100 miles in 2 hours.
1. ⓐ Write the equation that relates d and t.
2. ⓑ How many miles would it travel in 5 hours?
Leland is traveling on a road trip. The distance, $d$, he travels before stopping for lunch varies directly with the speed, $v$, he travels. He can travel 120 miles at a speed of 60 mph.
1. ⓐ Write the equation that relates d and v.
2. ⓑ How far would he travel before stopping for lunch at a rate of 65 mph?
The quantity $y$ varies directly with the cube of $x.$ If $y=25$ when $x=2,$ find $y$ when $x$ is 6.

Zbulo përgjigjen

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 $x.$
\[\begin{array}{lll}k & = & \frac{y}{{x}^{3}} \\ & = & \frac{25}{{2}^{3}} \\ & = & \frac{25}{8}\end{array}\]
Now use the constant to write an equation that represents this relationship.
\[y=\frac{25}{8}{x}^{3}\]
Substitute $x=6$ and solve for $y.$
\[\begin{array}{lll}y & = & \frac{25}{8}{(6)}^{3} \\ & = & 675\end{array}\]
The quantity $y$ varies directly with the square of $x.$ If $y=24$ when $x=3,$ find $y$ when $x$ is 4.

Zbulo përgjigjen

$\frac{128}{3}$
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.

Zbulo përgjigjen

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 $vt=\text{distance}\text{.}$ Because the distance is fixed at 100 miles, $vt=100$ so $t=100/v.$ Because time is a function of velocity, we can write $t(v).$
\[\begin{array}{lll}t(v) & = & \frac{100}{v} \\ & = & 100{v}^{-1}\end{array}\]
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. We say that time varies inversely with velocity.
A quantity $y$ varies inversely with the cube of $x.$ If $y=25$ when $x=2,$ find $y$ when $x$ is 6.

Zbulo përgjigjen

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 $x.$
\[\begin{array}{lll}k & = & {x}^{3}y \\ & = & {2}^{3}⋅25 \\ & = & 200\end{array}\]
Now we use the constant to write an equation that represents this relationship.
\[\begin{array}{lll}y & = & \frac{k}{{x}^{3}},\ k=200 \\ y & = & \frac{200}{{x}^{3}}\end{array}\]
Substitute $x=6$ and solve for $y.$
\[\begin{array}{lll}y & = & \frac{200}{{6}^{3}} \\ & = & \frac{25}{27}\end{array}\]
A quantity $y$ varies inversely with the square of $x.$ If $y=8$ when $x=3,$ find $y$ when $x$ is 4.

Zbulo përgjigjen

$\frac{9}{2}$
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 $z=27.$

Zbulo përgjigjen

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 $k.$
\[\begin{array}{lll}6 & = & \frac{k{2}^{2}}{\sqrt[3]{8}} \\ 6 & = & \frac{4k}{2} \\ 3 & = & k\end{array}\]
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.
\[\begin{array}{lll}x & = & \frac{3{(1)}^{2}}{\sqrt[3]{27}} \\ & = & 1\end{array}\]
A quantity $x$ 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 $z=25.$

Zbulo përgjigjen

$x=20$
What is true of the appearance of graphs that reflect a direct variation between two variables?

Zbulo përgjigjen

The graph will have the appearance of a power function.
If two variables vary inversely, what will an equation representing their relationship look like?
Is there a limit to the number of variables that can vary jointly? Explain.

Zbulo përgjigjen

No. Multiple variables may jointly vary.
$y$ varies directly as $x$ and when $x=6,y=12.$
$y$ varies directly as the square of $x$ and when $x=4,\ y=80\text{.\,}$

Zbulo përgjigjen

$y=5{x}^{2}$
$y$ varies directly as the square root of $x$ and when $x=36,\ y=24.$
$y$ varies directly as the cube of $x$ and when $x=36,\ y=24.$

Zbulo përgjigjen

$y=\frac{1}{1944}{x}^{3}$
$y$ varies directly as the cube root of $x$ and when $x=27,\ y=15.$
$y$ varies directly as the fourth power of $x$ and when $x=1,\ y=6.$

Zbulo përgjigjen

$y=6{x}^{4}$
$y$ varies inversely as $x$ and when $x=4,\ y=2.$
$y$ varies inversely as the square of $x$ and when $x=3,\ y=2.$

Zbulo përgjigjen

$y=\frac{18}{{x}^{2}}$
$y$ varies inversely as the cube of $x$ and when $x=2,\ y=5.$
$y$ varies inversely as the fourth power of $x$ and when $x=3,\ y=1.$

Zbulo përgjigjen

$y=\frac{81}{{x}^{4}}$
$y$ varies inversely as the square root of $x$ and when $x=25,\ y=3.$
$y$ varies inversely as the cube root of $x$ and when $x=64,\ y=5.$

Zbulo përgjigjen

$y=\frac{20}{\sqrt[3]{x}}$
$y$ varies jointly with $x$ and $z$ and when $x=2$ and $z=3,\ y=36.$
$y$ varies jointly as $x,z,$ and $w$ and when $x=1,\ z=2,\ w=5,$ then $y=100.$

Zbulo përgjigjen

$y=10xzw$
$y$ varies jointly as the square of $x$ and the square of $z$ and when $x=3$ and $z=4,$ then $y=72.$
$y$ varies jointly as $x$ and the square root of $z$ and when $x=2$ and $z=25,$ then $y=100.$

Zbulo përgjigjen

$y=10x\sqrt{z}$
$y$ varies jointly as the square of $x$ the cube of $z$ and the square root of $W.$ When $x=1,z=2,$ and $w=36,$ then $y=48.$
$y$ varies jointly as $x$ and $z$ and inversely as $w$. When $x=3,\ z=5$, and $w=6$, then $y=10.$

Zbulo përgjigjen

$y=4\frac{xz}{w}$
$y$ varies jointly as the square of $x$ and the square root of $z$ and inversely as the cube of $w\text{.\,}$ When $x=3,z=4,$ and $w=3,$ then $y=6.$
$y$ varies jointly as $x$ and $z$ and inversely as the square root of $w$ and the square of $t\text{.}$ When $x=3,z=1,w=25,$ and $t=2,$ then $y=6.$

Zbulo përgjigjen

$y=40\frac{xz}{\sqrt{w}{t}^{2}}$
$y$ varies directly as $x.$ When $x=3,$ then $y=12.$ Find $y$ wneh $x=20.$
$y$ varies directly as the square of $x.$ When $x=2,$ then $y=16.$ Find $y$ when $x=8.$

Zbulo përgjigjen

$y=256$

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Modeling Using Variation

Solve direct variation problems.
Solve inverse variation problems.
Solve problems involving joint variation.
Solve a formula for a specific variable (IA 2.3.1).
Solve direct variation problems (IA 7.5.5).
Write the formula for direct variation.
Substitute the given values for the variables.
Solve for the constant of variation.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Provo timen.

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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	\(\\) Find c when m = 25.
Write the equation that relates c and m. \(\\)	\(\\)
Substitute the given value for m.	\(\\)
Simplify.	\(\\)
	\(\\) Raoul would burn 437.5 calories if he used the treadmill for 25 minutes.

\(s\) , sales price	\(e=0.16s\)	Interpretation
$4,600	\(e=0.16(4,600)=736\)	A sale of a $4,600 vehicle results in $736 earnings.
$9,200	\(e=0.16(9,200)=1,472\)	A sale of a $9,200 vehicle results in $1472 earnings.
$18,400	\(e=0.16(18,400)=2,944\)	A sale of a $18,400 vehicle results in $2944 earnings.

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