Opening the book…
- Model direct and inverse variation with appropriate equations
- Analyze a real-world scenario using direct or inverse variation
We start with a definition of the phrase “is proportional to.”
We say that \(y\) is proportional to \(x\) if and only if
\[y = kx \nonumber \]
where \(k\) is a constant called the constant of proportionality. The phrase “\(y\) varies directly as \(x\)” is an equivalent way of saying “\(y\) is proportional to \(x\).”
Here are a few examples that translate the phrase “is proportional to.”
- Given that \(d\) is proportional to \(t\), we write \(d = kt\), where \(k\) is a constant.
- Given that \(y\) is proportional to the cube of \(x\), we write \(y = kx^3\), where \(k\) is a constant.
- Given that \(s\) is proportional to the square of \(t\), we write \(s = kt^2\), where \(k\) is a constant.
We are not restricted to always using the letter \(k\) for our constant of proportionality.
Given that \(y\) is proportional to \(x\) and the fact that \(y = 12\) when \(x = 5\), determine the constant of proportionality, then determine the value of \(y\) when \(x = 10\).
Solution
Given the fact that \(y\) is proportional to \(x\), we know immediately that \[y = kx \nonumber \]where \(k\) is the proportionality constant. Because we are given that \(y = 12\) when \(x = 5\), we can substitute \(12\) for \(y\) and \(5\) for \(x\) to determine \(k\).
\[\begin{array}{rl}{y=k x} & \color {Blue} {y \text { is proportional to } x} \\ {12=k(5)} & \color {Blue} {\text { Substitute } 12 \text { for } y, 5 \text { for } x} \\ {\dfrac{12}{5}=k} & \color {Blue} {\text { Divide both sides by } 5}\end{array} \nonumber \]
Next, substitute the constant of proportionality \(12/5\) for \(k\) in \(y = kx\), then substitute \(10\) for \(x\) to determine \(y\) when \(x = 10\).
\[\begin{array}{ll}{y=\dfrac{12}{5} x} & \color {Blue} {\text { Substitute } 12 / 5 \text { for } k} \\ {y=\dfrac{12}{5}(10)} &\color {Blue} {\text { Substitute } 10 \text { for } x} \\ {y=24} & \color {Blue} {\text { Cancel and simplify. }}\end{array} \nonumber \]
A ball is dropped from a balloon floating above the Earth's surface. The distance \(s\) the ball falls is proportional to the square of the time \(t\) that has passed since the ball’s release. If the ball falls \(144\) feet during the first \(3\) seconds, how far does the ball fall in \(9\) seconds?
Solution
Given the fact that \(s\) is proportional to the square of \(t\), we know immediately that
\[s=k t^{2} \nonumber \]
where \(k\) is the proportionality constant. Because we are given that the ball falls \(144\) feet during the first \(3\) seconds, we can substitute \(144\) for \(s\) and \(3\) for \(t\) to determine the constant of proportionality.
\[\begin{array}{rl}{s=k t^{2}} & \color {Blue} {s \text { is proportional to the square of } t} \\ {144=k(3)^{2}} & \color {Blue} {\text { Substitute } 144 \text { for } s, 3 \text { for } t} \\ {144=9 k} & \color {Blue} {\text { Simplify: } 3^{2}=9} \\ {16=k} & \color {Blue} {\text { Divide both sides by } 9}\end{array} \nonumber \]
Next, substitute the constant of proportionality \(16\) for \(k\) in \(s = kt^2\), and then substitute \(9\) for \(t\) to determine the distance fallen when \(t = 9\) seconds.
\[\begin{array}{ll}{s=16 t^{2}} & \color {Blue} {\text { Substitute } 16 \text { for } k} \\ {s=16(9)^{2}} & \color {Blue} {\text { Substitute } 9 \text { for } t} \\ {s=1296} & \color {Blue} {\text { Simplify }}\end{array} \nonumber \]
Thus, the ball falls \(1,296\) feet during the first \(9\) seconds.
Tony and Paul are hanging weights on a spring in the physics lab. Each time a weight is hung, they measure the distance the spring stretches. They discover that the distance \(y\) that the spring stretches is proportional to the weight hung on the spring (Hooke’s Law). If a \(0.5\) pound weight stretches the spring \(3\) inches, how far will a \(0.75\) pound weight stretch the spring?
Solution
Let \(W\) represent the weight hung on the spring. Let \(y\) represent the distance the spring stretches. We’re told that the distance y the spring stretches is proportional to the amount of weight \(W\) hung on the spring. Hence, we can write:
\[y=k W \quad \color {Blue} y \text { is proportional to } W \nonumber \]
Substitute \(3\) for \(y\), \(0.5\) for \(W\), then solve for \(k\).
\[\begin{array}{rlrl}{3} & {=k(0.5)} & {} & \color {Blue} {\text { Substitute } 3 \text { for } y, 0.5 \text { for } W} \\ {\dfrac{3}{0.5}} & {=k} & {} & \color {Blue} {\text { Divide both sides by } 0.5} \\ {k} & {=6} & {} & \color {Blue} {\text { Simplify. }}\end{array} \nonumber \]
Substitute \(6\) for \(k\) in \(y = kW\) to produce:
\[y=6 W \quad \color {Blue} \text { Substitute } 6 \text { for } k \text { in } y=k W \nonumber \]
To determine the distance the spring will stretch when \(0.75\) pounds are hung on the spring, substitute \(0.75\) for \(W\).
\[\begin{array}{ll}{y=6(0.75)} & \color {Blue} {\text { Substitute } 0.75 \text { for } W} \\ {y=4.5} & \color {Blue} {\text { Simplify. }}\end{array} \nonumber \]
Thus, the spring will stretch \(4.5\) inches.
Inversely Proportional
In the previous examples, where one quantity was proportional to a second quantity, you may have noticed that when one quantity increased, the second quantity also increased. Vice versa, when one quantity decreased, the second quantity also decreased.
However, not all real-world situations follow this pattern. There are times when, as one quantity increases, the related quantity decreases. For example, consider the situation where you increase the number of workers on a job and note that the time to finish the job decreases. This is an example of a quantity being inversely proportional to a second quantity.
We say the \(y\) is inversely proportional to \(x\) if and only if\[y=\dfrac{k}{x} \nonumber \]where \(k\) is a constant called the constant of proportionality. The phrase “\(y\) varies inversely as \(x\)” is an equivalent way of saying “\(y\) in inversely proportional to \(x\).”
Here are a few examples that translate the phrase “is inversely proportional to.”
- Given that \(d\) is inversely proportional to \(t\), we write \(d = \dfrac{k}{t}\), where \(k\) is a constant.
- Given that \(y\) is inversely proportional to the cube of \(x\), we write \(y = \dfrac{k}{x^3}\), where \(k\) is a constant.
- Given that \(s\) is inversely proportional to the square of \(t\), we write \(s = \dfrac{k}{t^2}\), where \(k\) is a constant.
We are not restricted to always using the letter \(k\) for our constant of proportionality.
Given that \(y\) is inversely proportional to \(x\) and the fact that \(y = 4\) when \(x = 2\), determine the constant of proportionality, then determine the value of \(y\) when \(x = 4\).
Solution
Given the fact the \(y\) is inversely proportional to \(x\), we know immediately that\[y=\dfrac{k}{x} \nonumber \]where \(k\) is the proportionality constant. Because we are given that \(y = 4\) when \(x = 2\), we can substitute \(4\) for \(y\) and \(2\) for \(x\) to determine \(k\).
\[\begin{align*} y &= \dfrac{k}{x} \quad \color {Blue} y \text { is inversely proportional to } x.\\ 4 &= \dfrac{k}{2} \quad \color {Blue} \text {Substitute }4 \text { for } y, 2 \text { for }x.\\ 8 &= k \quad \color {Blue} \text {Multiply both sides by } 2. \end{align*} \nonumber\]
Substitute \(8\) for \(k\) in \(y = k/x\), then substitute \(4\) for \(x\) to determine \(y\) when \(x = 4\).
\[\begin{align*} y &= \dfrac{8}{x} \quad \color {Blue} \text {Substitute } 8 \text { for } k.\\ y &= \dfrac{8}{4} \quad \color {Blue} \text {Substitute }4 \text { for } x.\\ y &= 2 \quad \color {Blue} \text {Reduce.} \end{align*} \nonumber\]
Note that as \(x\) increased from \(2\) to \(4\), \(y\) decreased from \(4\) to \(2\).
The intensity \(I\) of light is inversely proportional to the square of the distance \(d\) from the light source. If the light intensity \(5\) feet from the light source is \(3\) foot-candles, what is the intensity of the light \(15\) feet from the light source?
Solution
Given the fact that the intensity \(I\) of the light is inversely proportional to the square of the distance d from the light source, we know immediately that\[I = \dfrac{k}{d^2} \nonumber \]where \(k\) is the proportionality constant. Because we are given that the intensity is \(I = 3\) foot-candles at \(d = 5\) feet from the light source, we can substitute \(3\) for \(I\) and \(5\) for \(d\) to determine \(k\).
\[\begin{align*} I &= \dfrac{k}{d^2} \quad \color {Blue} I \text { is inversely proportional to } d^2 .\\ 3 &= \dfrac{k}{5^2} \quad \color {Blue} \text {Substitute }3 \text { for } I, 5 \text { for } d.\\ 3 &= \dfrac{k}{25} \quad \color {Blue} \text {Simplify.}\\ 75 &= k \quad \color {Blue} \text {Multiply both sides by } 25. \end{align*} \nonumber\]
Substitute \(75\) for \(k\) in \(I = k/d^2\), then substitute \(15\) for \(d\) to determine \(I\) when \(d = 15\).
\[\begin{align*} I &= \dfrac{75}{d^2} \quad \color {Blue} \text {Substitute }75 \text { for } k.\\ I &= \dfrac{75}{15^2} \quad \color {Blue} \text {Substitute }15 \text { for } d.\\ I &= \dfrac{75}{225} \quad \color {Blue} \text {Simplify.}\\ I &= \dfrac{1}{3} \quad \color {Blue} \text {Reduce.} \end{align*} \nonumber\]
Thus, the intensity of the light \(15\) feet from the light source is \(1/3\) foot-candle.
Suppose the price per person for a camping experience is inversely proportional to the number of people who sign up for it. If \(10\) people sign up, the price per person is \($350\). What will be the price per person if \(50\) people sign up?
Solution
Let \(p\) represent the price per person and let \(N\) be the number of people who sign up for the camping experience. Because we are told that the price per person is inversely proportional to the number of people who sign up for the camping experience, we can write:
\[p = \dfrac{k}{N} \nonumber \]
where \(k\) is the proportionality constant. Because we are given that the price per person is \($350\) when \(10\) people sign up, we can substitute \(350\) for \(p\) and \(10\) for \(N\) to determine \(k\).
\[\begin{align*} p &= \dfrac{k}{N} \quad \color {Blue} p \text { is inversely proportional to }N.\\ 350 &= \dfrac{k}{10} \quad \color {Blue} \text {Substitute }350 \text { for } p, 10 \text { for } N.\\ 3500 &= k \quad \color {Blue} \text {Multiply both sides by } 10. \end{align*} \nonumber\]
Substitute \(3500\) for \(k\) in \(p = k/N\), then substitute \(50\) for \(N\) to determine \(p\) when \(N = 50\).
\[\begin{align*} p &= \dfrac{3500}{N} \quad \color {Blue} \text {Substitute }3500 \text { for } k.\\ p &= \dfrac{3500}{50} \quad \color {Blue} \text {Substitute }50 \text { for } N.\\ p &= 70 \quad \color {Blue} \text {Simplify.} \end{align*} \nonumber\]
Thus, the price per person is \($70\) if \(50\) people sign up for the camping experience.
This page titled 3.1: Direct and Inverse Variation was last modified on Fri, 29 Aug 2025 03:30:36 GMT and is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Math Department via source content that was edited to the style and standards of the LibreTexts platform.
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