7.5: Direct and Inverse Variation

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We start with the deﬁnition of the phrase “is proportional to.”

Proportional

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.

Example $\PageIndex{1}$

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 the $y$ 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 $k$.

\[\begin{array}{rl}{y=k x} & \color {Red} {y \text { is proportional to } x} \\ {12=k(5)} & \color {Red} {\text { Substitute } 12 \text { for } y, 5 \text { for } x} \\ {\dfrac{12}{5}=k} & \color {Red} {\text { Divide both sides by } 5}\end{array} \nonumber \]

Next, substitute the constant of proportionality $12/5$ for $k$ in $y = kx$, then substitute $10$ for $x$ to determine $y$ when $x = 10$.

\[\begin{array}{ll}{y=\dfrac{12}{5} x} & \color {Red} {\text { Substitute } 12 / 5 \text { for } k} \\ {y=\dfrac{12}{5}(10)} &\color {Red} {\text { Substitute } 10 \text { for } x} \\ {y=24} & \color {Red} {\text { Cancel and simplify. }}\end{array} \nonumber \]

Exercise $\PageIndex{1}$

Given that $y$ is proportional to $x$ and that $y = 21$ when $x = 9$, determine the value of $y$ when $x = 27$.

Answer: $63$

Example $\PageIndex{2}$

A ball is dropped from a balloon ﬂoating above the surface of the earth. 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 the $s$ 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.

\[\begin{array}{rl}{s=k t^{2}} & \color {Red} {s \text { is proportional to the square of } t} \\ {144=k(3)^{2}} & \color {Red} {\text { Substitute } 144 \text { for } s, 3 \text { for } t} \\ {144=9 k} & \color {Red} {\text { Simplify: } 3^{2}=9} \\ {16=k} & \color {Red} {\text { Divide both sides by } 9}\end{array} \nonumber \]

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.

\[\begin{array}{ll}{s=16 t^{2}} & \color {Red} {\text { Substitute } 16 \text { for } k} \\ {s=16(9)^{2}} & \color {Red} {\text { Substitute } 9 \text { for } t} \\ {s=1296} & \color {Red} {\text { Simplify }}\end{array} \nonumber \]

Thus, the ball falls $1,296$ feet during the ﬁrst $9$ seconds.

Exercise $\PageIndex{2}$

A ball is dropped from the edge of a cliﬀ on a certain planet. 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 $50$ feet during the ﬁrst $5$ seconds, how far does the ball fall in $8$ seconds?

Answer: $128$ feet

Example $\PageIndex{3}$

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:

\[y=k W \quad \color {Red} y \text { is proportional to } W \nonumber \]

Substitute $3$ for $y$, $0.5$ for $W$, then solve fork.

\[\begin{array}{rlrl}{3} & {=k(0.5)} & {} & \color {Red} {\text { Substitute } 3 \text { for } y, 0.5 \text { for } W} \\ {\dfrac{3}{0.5}} & {=k} & {} & \color {Red} {\text { Divide both sides by } 0.5} \\ {k} & {=6} & {} & \color {Red} {\text { Simplify. }}\end{array} \nonumber \]

Substitute $6$ for $k$ in $y = kW$ to produce:

\[y=6 W \quad \color {Red} \text { Substitute } 6 \text { for } k \text { in } y=k W \nonumber \]

To determine the distance the spring will stretch when $0.75$ pounds are hung on the spring, substitute $0.75$ for $W$.

\[\begin{array}{ll}{y=6(0.75)} & \color {Red} {\text { Substitute } 0.75 \text { for } W} \\ {y=4.5} & \color {Red} {\text { Simplify. }}\end{array} \nonumber \]

Thus, the spring will stretch $4.5$ inches.

Exercise $\PageIndex{3}$

If a $0.75$ pound weight stretches a spring $5$ inches, how far will a $1.2$ pound weight stretch the spring?

Answer: $8$ inches

Inversely Proportional

In Examples $\PageIndex{1}$, $\PageIndex{2}$, and $\PageIndex{3}$, 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.

Inversely proportional

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 $d = k/t$, where $k$ is a constant.
Given that $y$ is inversely proportional to the cube of $x$, we write $y = k/x^3$, where $k$ is a constant.
Given that $s$ is inversely proportional to the square of $t$, we write $s = k/t^2$, where $k$ is a constant.

We are not restricted to always using the letter $k$ for our constant of proportionality.

Example $\PageIndex{4}$

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 $k$.

\[\begin{align*} y &= \dfrac{k}{x} \quad \color {Red} y \text { is inversely proportional to } x.\\ 4 &= \dfrac{k}{2} \quad \color {Red} \text {Substitute }4 \text { for } y, 2 \text { for }x.\\ 8 &= k \quad \color {Red} \text {Multiply both sides by } 2. \end{align*} \nonumber\]

Substitute $8$ for $k$ in $y = k/x$, then substitute $4$ for $x$ to determine $y$ when $x = 4$.

\[\begin{align*} y &= \dfrac{8}{x} \quad \color {Red} \text {Substitute } 8 \text { for } k.\\ y &= \dfrac{8}{4} \quad \color {Red} \text {Substitute }4 \text { for } x.\\ y &= 2 \quad \color {Red} \text {Reduce.} \end{align*} \nonumber\]

Note that as $x$ increased from $2$ to $4$, $y$ decreased from $4$ to $2$.

Exercise $\PageIndex{4}$

Given that $y$ is inversely proportional to $x$ and that $y = 5$ when $x = 8$, determine the value of $y$ when $x = 10$.

Answer: $4$

Example $\PageIndex{5}$

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 $k$.

\[\begin{align*} I &= \dfrac{k}{d^2} \quad \color {Red} I \text { is inversely proportional to } d^2 .\\ 3 &= \dfrac{k}{5^2} \quad \color {Red} \text {Substitute }3 \text { for } I, 5 \text { for } d.\\ 3 &= \dfrac{k}{25} \quad \color {Red} \text {Simplify.}\\ 75 &= k \quad \color {Red} \text {Multiply both sides by } 25. \end{align*} \nonumber\]

Substitute $75$ for $k$ in $I = k/d^2$, then substitute $15$ for $d$ to determine $I$ when $d = 15$.

\[\begin{align*} I &= \dfrac{75}{d^2} \quad \color {Red} \text {Substitute }75 \text { for } k.\\ I &= \dfrac{75}{15^2} \quad \color {Red} \text {Substitute }15 \text { for } d.\\ I &= \dfrac{75}{225} \quad \color {Red} \text {Simplify.}\\ I &= \dfrac{1}{3} \quad \color {Red} \text {Reduce.} \end{align*} \nonumber\]

Thus, the intensity of the light $15$ feet from the light source is $1/3$ foot-candle.

Exercise $\PageIndex{5}$

If the light intensity $4$ feet from a light source is $2$ foot-candles, what is the intensity of the light $8$ feet from the light source?

Answer: $1/2$ foot-candle

Example $\PageIndex{6}$

Suppose that the price per person for a camping experience is inversely proportional to the number of people who sign up for the experience. 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 $k$.

\[\begin{align*} p &= \dfrac{k}{N} \quad \color {Red} p \text { is inversely proportional to }N.\\ 350 &= \dfrac{k}{10} \quad \color {Red} \text {Substitute }350 \text { for } p, 10 \text { for } N.\\ 3500 &= k \quad \color {Red} \text {Multiply both sides by } 10. \end{align*} \nonumber\]

Substitute $3500$ for $k$ in $p = k/N$, then substitute $50$ for $N$ to determine $p$ when $N = 50$.

\[\begin{align*} p &= \dfrac{3500}{N} \quad \color {Red} \text {Substitute }3500 \text { for } k.\\ p &= \dfrac{3500}{50} \quad \color {Red} \text {Substitute }50 \text { for } N.\\ p &= 70 \quad \color {Red} \text {Simplify.} \end{align*} \nonumber\]

Thus, the price per person is $$70$ if $50$ people sign up for the camping experience.

Exercise $\PageIndex{6}$

Suppose that the price per person for a tour is inversely proportional to the number of people who sign up for the tour. If $8$ people sign up, the price per person is $$70$. What will be the price per person if $20$ people sign up?

Answer: $$28$

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