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Modeling Using Variation

Solve direct variation problems.

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 prices	$e=0.16s$	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 $n=1.$

Condensed — the full section is in OpenStax Precalculus 2e.

Solving Inverse Variation Problems

Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, 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.
350 ft	$\frac{14,000}{350}=40$	At a depth of 350 ft, the water temperature is 40° F.
250 ft	$\frac{14,000}{250}=56$	At a depth of 250 ft, the water temperature is 56° 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, $vt=100.$ Solving this relationship for the time gives us our function.

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

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.

Condensed — the full section is in OpenStax Precalculus 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 $k.$

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

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.

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

Key Equations

Direct variation	\[y=k{x}^{n},k\text{ is a nonzero constant}.\]
Inverse variation	\[y=\frac{k}{{x}^{n}},k\text{ is a nonzero constant}.\]

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

Perform the indicated operation or solve the equation.

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.

Condensed — the full section is in OpenStax Precalculus 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.

The quantity $y$ varies directly with the cube of $x.$ If $y=25$ when $x=2,$ find $y$ when $x$ is 6.

Откройте ответ.

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

Откройте ответ.

$\frac{128}{3}$
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.

Откройте ответ.

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, $vt=100.$ Solving this relationship for the time gives us our function.
\[\begin{array}{l}t(v)=\frac{100}{v} \\ =100{v}^{-1}\end{array}\]
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.
A quantity $y$ varies inversely with the cube of $x.$ If $y=25$ when $x=2,$ find $y$ when $x$ is 6.

Откройте ответ.

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

Откройте ответ.

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

Откройте ответ.

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

Откройте ответ.

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

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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 jointly vary? Explain.

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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 $x=4,\ y=80\text{.\,}$

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

Откройте ответ.

$y=\frac{1}{1944}{x}^{3}$
$y$ 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 $x=1,\ y=6.$

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

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

Откройте ответ.

$y=\frac{81}{{x}^{4}}$
$y$ 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 $x=64,\ y=5.$

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$y=\frac{20}{\sqrt[3]{x}}$
$y$ 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 $y=100.$

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$y=10xzw$
$y$ 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 $y=100.$

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

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

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

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$y=256$
$y$ varies directly as the cube of $x$. When $x=3$, then $y=5$. Find $y$ when $x=4$.
$y$ varies directly as the square root of $x.$ When $x=16$, then $y=4$. Find $y\text{ when }x=36$.

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$y=6$
$y$ varies directly as the cube root of $x.$ When $x=125$, then $y=15$. Find $y$ when $x=1$, $000.$
$y$ varies inversely with $x.$ When $x=3$, then $y=2$. Find $y$ when $x=1$.

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$y=6$
$y$ varies inversely with the square of $x$. When $x=4$, then $y=3$. Find $y$ when $x=2$.
$y$ varies inversely with the cube of $x.$ When $x=3$, then $y=1$. Find $y$ when $x=1$.

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$y=27$
$y$ varies inversely with the square root of $x.$ When $x=64,$ then $y=12.$ Find $y$ when $x=36.$
$y$ varies inversely with the cube root of $x.$ When $x=27,$ then $y=5.$ Find $y$ when $x=125.$

Откройте ответ.

$y=3$

Symbols used here

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

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

|x|

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

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

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

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

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

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Modeling Using Variation

Solve direct variation problems.
Solve inverse variation problems.
Solve problems involving joint variation.
Identify the input,
Determine the constant of variation. You may need to divide
Use the constant of variation to write an equation for the relationship.
Substitute known values into the equation to find the unknown.
Identify the input,

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Попробуй сам.

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

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Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates

\(s\), sales prices	\(e=0.16s\)	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.
350 ft	\(\frac{14,000}{350}=40\)	At a depth of 350 ft, the water temperature is 40° F.
250 ft	\(\frac{14,000}{250}=56\)	At a depth of 250 ft, the water temperature is 56° F.