inverse variation problem solving

Inverse Variation: Practice Problems

If $$ y $$ varies inversely as $$ x $$ and $$ y = 14 $$ when $$ x = 20 $$ . What is the value of $$ y $$ when $$ x $$ is 5?

Write down the variation equation.

$$ y = \frac k x $$

Use $$ \red{y = 14} $$ and $$ \blue{x = 20} $$ to find the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{14} & = \frac k {\blue{20}}\\[6pt] \red{14}(\blue{20}) & = \frac k {\blue{20}}\cdot \blue {20}\\[6pt] 280 & = \frac k {\cancelred{20}}\cdot \cancelred{20}\\[6pt] k & = 280 \end{align*} $$

The updated variation equation is $$ y = \frac{280} x $$

Determine the value of $$ y $$ when $$ \blue{x = 5} $$ .

$$ \begin{align*} y & = \frac{280}{\blue x}\\[6pt] y & = \frac{280}{\blue 5} = 56 \end{align*} $$

$$ y = 56 $$ when $$ x = 5 $$ .

Suppose $$ s $$ varies indirectly as $$ t $$, and $$ s(1.5)=25 $$ . Determine the value of $$ s(2.25) $$ .

$$ s = \frac k t $$

Use $$ \red{s = 25} $$ and $$ \blue{t = 1.5} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{25} & = \frac k {\blue{1.5}}\\[6pt] \red{25}(\blue{1.5}) & = \frac k {\blue{1.5}}\cdot \blue{1.5}\\[6pt] 27.5 & = \frac k {\cancelred{1.5}}\cdot \cancelred{1.5}\\[6pt] k & = 37.5 \end{align*} $$

The updated variation equation is $$ s = \frac{37.5} t $$

Determine the value of $$ s $$ when $$ \blue{t = 2.25} $$ .

$$ \begin{align*} s & = \frac{37.5}{\blue t}\\[6pt] s & = \frac{37.5}{\blue{2.25}} = \frac{3750}{225} = \frac{50} 3 = 16\,\frac 2 3 \approx 16.667 \end{align*} $$

$$ s = \frac{50} 3 $$ when $$ t = 2.25 $$ .

Suppose $$ y $$ varies inversely as $$ x $$, and $$ y = 0.025 $$ when $$ x = 2 $$ . What is the value of $$ x $$ when $$ y = 0.4 $$ ?

Use $$ \red{y = 0.025} $$ and $$ \blue{x = 2} $$ to find the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{0.025} & = \frac k {\blue 2}\\[6pt] \red{(0.025)}\blue 2 & = \frac k {\blue 2} \cdot \blue 2\\[6pt] 0.05 & = \frac k {\cancelred 2} \cdot \cancelred 2\\[6pt] k & = 0.05 = \frac 1 {20} \end{align*} $$

The updated variation equation is $$ y = \frac{0.05} x = \frac{1/20} x = \frac 1 {20x} $$

Determine the value of $$ x $$ when $$ \red{y = 0.4} $$ .

$$ \begin{align*} \red{0.4} & = \frac 1 {20x}\\[6pt] \red{0.4} \cdot (20x) & = 1\\[6pt] 8x & = 1\\[6pt] x & = \frac 1 8 = 0.125 \end{align*} $$

$$ x = 0.125 $$ when $$ y = 0.4 $$ .

Suppose $$ I $$ is inversely proportional to $$ R $$ and when $$ R = 200 $$ , $$ I = 35 $$ . Determine the value of $$ R $$ when $$ I = 100 $$ .

$$ I = \frac k R $$

Use $$ \red{I = 35} $$ and $$ \blue{R = 200} $$ to find the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{35} & = \frac k {\blue{200}}\\[6pt] \red{35}(\blue{200}) & = \frac k {\blue{200}}\cdot \blue{200}\\[6pt] 7000 & = \frac k {\cancelred{200}}\cdot \cancelred{200}\\[6pt] k & = 7000 \end{align*} $$

The updated variation equation is $$ I = \frac{7000} R $$ .

Determine the value of $$ R $$ when $$ \red{I = 100} $$ .

$$ \begin{align*} \red{100} & = \frac{7000} R\\[6pt] 100R & = 7000\\[6pt] R & = \frac{7000}{100} = 70 \end{align*} $$

$$ R = 70 $$ when $$ I = 100 $$ .

Suppose $$ y $$ varies inversely as the square-root of $$ x $$. If $$ y = 8 $$ when $$ x = 49 $$ , determine the value of $$ y $$ when $$ x = 121 $$

$$ y = \frac k {\sqrt x} $$

Use $$ \red{y = 8} $$ and $$ \blue{x = 49} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red 8 & = \frac k {\sqrt{\blue{49}}}\\[6pt] \red 8 & = \frac k 7\\[6pt] \red 8 (7) & = \frac k 7\cdot 7\\[6pt] 56 & = \frac k {\cancelred 7}\cdot \cancelred 7\\[6pt] k & = 56 \end{align*} $$

The updated variation equation is $$ y = \frac{56}{\sqrt x} $$

Determine the value of $$ y $$ when $$ \blue{x = 100} $$ .

$$ \begin{align*} y & = \frac{56}{\sqrt{\blue{100}}}\\[6pt] & = \frac{56}{10}\\[6pt] & = 5.6 \end{align*} $$

$$ y = 5.6 $$ when $$ x = 100 $$ .

Suppose $$ p $$ varies inversely as the cube of $$ q $$, and $$ p = 2 $$ when $$ q = 4 $$ . Determine the value of $$ q $$ when $$ p = 8 $$ .

$$ p = \frac k {q^3} $$

Use $$ \red{p = 2} $$ and $$ \blue{q = 4} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red 2 & = \frac k {\blue 4^3}\\[6pt] \red 2 & = \frac k {64}\\[6pt] \red 2 (64) & = \frac k {\cancelred{64}} \cdot \cancelred{64}\\[6pt] k & = 128 \end{align*} $$

The updated variation equation is $$ p = \frac{128}{q^3} $$ .

Determine the value of $$ q $$ when $$ \red{p = 8} $$ .

$$ \begin{align*} \red 8 & = \frac{128}{q^3}\\[6pt] 8q^3 & = 128\\[6pt] q^3 & = \frac{128} 8\\[6pt] & = 16\\[6pt] q & = \sqrt[3]{16} = 2\sqrt[3] 2 \end{align*} $$

$$ q = 2\sqrt[3] 2 $$ when $$ p = 8 $$ .

Word Problems

Speed and travel time are inversely related (the faster you go, the shorter the travel time). Suppose the daily trip to school normally takes 20 minutes when your speed is 30 miles per hour. How long should the trip take if you drive 45 miles per hour instead (and don't get pulled over for speeding)?

Define variables to describe travel time and speed, and state the equation relating them.

Let's use $$ t $$ for time (in minutes).

Let's use $$ s $$ for speed (in miles per hour).

The equation we need then is $$ s = \frac k t $$ .

Use $$ \red{s = 30} $$ and $$ \blue{t = 20} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{30} & = \frac k {\blue{20}}\\[6pt] \red{30}(\blue{20}) & = \frac k {\blue{20}}\cdot\blue{20}\\[6pt] 600 & = \frac k {\cancelred{20}}\cdot\cancelred{20}\\[6pt] k & = 600 \end{align*} $$

The updated variation equation is $$ s = \frac{600} t $$ .

Use the equation to determine the travel time for the trip when speed is 45 miles per hour.

$$ \begin{align*} \red{45} & = \frac{600} t\\[6pt] \red{45}t & = 600\\[6pt] t & = \frac{600}{\red{45}}\\[6pt] & = \frac{40} 3\\[6pt] & = 13\frac 1 3 \end{align*} $$

The trip will take $$ 13\frac 1 3 $$ minutes (or 13 minutes and 20 seconds) if you drive at 45 miles per hour.

The Ideal Gas Law from chemistry can be written as $$ P = \frac k V $$ , where $$ P $$ is the pressure (in pascals) and $$ V $$ is the volume of the container (in cubic meters).

Suppose that when a particular gas is stored in a 0.25 cubic meter container, and exerts 40 pascals of pressure. How much pressure will the gas exert if it were transferred to a container that only holds a volume of $$ 0.1 $$ cubic meters.

Use $$ \red{P = 40} $$ and $$ \blue{V = 0.25} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red{P} & = \frac k {\blue{V}}\\[6pt] \red{40} & = \frac k {\blue{0.25}}\\[6pt] \red{40}(\blue{0.25}) & = \frac k {\blue{0.25}}\cdot\blue{0.25}\\[6pt] 10 & = \frac k {\cancelred{0.25}}\cdot\cancelred{0.25}\\[6pt] k & = 10 \end{align*} $$

So for this particular gas, $$ P = \frac{10} V $$

Use the equation to determine the pressure when $$ \blue{V = 0.1} $$ .

$$ P = \frac{10}{\blue{0.1}} = 100 $$

The gas exerts a pressure of 100 pascals in a 0.1 cubic meter container.

The time it takes a block of ice to melt varies indirectly with temperature. If a block of ice that is 1/8 cubic meters in volume requires 6 hours to melt when it is $$72 ^\circ F $$ , at what temperature will the block melt in only 2 hours?

Define variables to represent temperature and time and write down the inverse variation equation.

Let $$ t = $$ time (in hours)

Let $$ T = $$ temperature (in degrees Fahrenheit)

The inverse variation equation will be

$$ t = \frac k T $$

Use $$ \red{t = 6} $$ and $$ \blue{T = 72} $$ to find the value of $$ k $$, then write down the updated version of the inverse variation equation.

$$ \begin{align*} \red t & = \frac k {\blue T}\\[6pt] \red 6 & = \frac k {\blue{72}}\\[6pt] \red 6(\blue{72}) & = \frac k {\blue{72}}\cdot\blue{72}\\[6pt] 432 & = \frac k {\cancelred{72}}\cdot\cancelred{72}\\[6pt] k & = 432 \end{align*} $$

The inverse variation equation is now $$ t = \frac{432} T $$ .

Determine the temperature required to have the block of ice melt in only $$ \blue{t = 2} $$ hours.

$$ \begin{align*} \blue 2 & = \frac{432} T\\[6pt] \blue 2T & = 432\\[6pt] T & = \frac{432}{\blue 2}\\[6pt] & = 216 \end{align*} $$

The 1/8 cubic meter block of ice will melt in only 2 hours if the temperature is $$ 216 ^\circ F $$ .

Suppose 4 adults can build a shed in 16 hours. How much quicker would it be to have 5 adults building the shed?

Confirm this is an inverse variation problem.

We note that (in an ideal world) having more adults working on a particular project will mean the task takes less time.

Since the time decreases as the number of workers increase, we can model this as an inverse variation.

Define variables to represent the quantities involved, and write down the inverse variation equation.

Let $$ w = $$ the number of workers building the shed.

Let $$ t = $$ the number of hours it takes to complete the shed.

The inverse variation equation is

$$ t = \frac k w $$

Use $$ \red{t = 16} $$ and $$ \blue{w = 4} $$ to determine the value of $$ k $$. Then write down the updated variation equation.

$$ \begin{align*} \red t & = \frac k {\red w}\\[6pt] \red{16} & = \frac k {\blue 4}\\[6pt] \red{16}(\blue 4) & = \frac k {\blue 4}\cdot \blue 4\\[6pt] 64 & = \frac k {\cancelred 4}\cdot \cancelred 4\\[6pt] k & = 64 \end{align*} $$

The updated variation equation is $$ t = \frac{64} w $$ .

Determine the length of time required for $$\blue{w = 5}$$ people to build the shed.

$$ t = \frac{64}{\blue 5} = 12\frac 4 5 = 12.8 $$

Determine the savings in time when 5 adults work on the shed versus only 4.

Note that the question didn't ask us, "How long does it take..." , but rather "How much quicker.." . This means we are interested in the difference between having 4 and 5 workers

$$ \mbox{Time saved} = (\mbox{time for 4 workers}) - (\mbox{time for 5 workers}) = 16 - 12.8 = 3.2 $$

Having 5 adults on the project instead of 4 will result in the shed be completed 3.2 hours more quickly (or 3 hours, 12 minutes more quickly).

The gravitational force (in newtons) between two objects is inversely proportional to square of the distance (in meters) between the centers of the objects.

Suppose the gravitational force between two particular objects is $$ 1.95\times 10^4 $$ newtons when they are separated by $$ 8.94\times 10^{12} $$ meters. If the gravitational force increases to $$ 4.1\times 10^6 $$ newtons, what has happened to the distance between the objects?

Define variables to represent the two quantities under investigation. Then write down the variation equation.

Let $$ F_g $$ represent the force due to gravity, measured in newtons.

Let $$ r $$ represent the distance between the centers of the objects, measured in meters.

The variation equation is $$ F_g = \frac k {r^2} $$ .

Use $$ \red{F_g = 1.95\times 10^4} $$ and $$ \blue{r = 8.94\times 10^{12}} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red{F_g} & = \frac k {\blue r^2}\\[6pt] \red{1.95\times 10^4} & = \frac k {(\blue{8.94\times 10^{12}})^2}\\[6pt] \red{1.95\times 10^4} & = \frac k {\blue{79.9236\times 10^{24}}}\\[6pt] \red{1.95\times 10^4}(\blue{79.9236\times 10^{24}}) & = \frac k {\blue{79.9236\times 10^{24}}}\cdot\blue{79.9236\times 10^{24}}\\[6pt] 1.5585102 \times 10^{30} & = \frac k {\cancelred{79.9236\times 10^{24}}}\cdot\cancelred{79.9236\times 10^{24}}\\[6pt] k & = 1.5585102 \times 10^{30}\\[6pt] & \approx 1.56 \times 10^{30} \end{align*} $$

The updated variation equation is $$ \displaystyle F_g = \frac{1.56\times 10^{30}}{r^2} $$ .

Determine the distance between the two objects if the force of gravity is $$ \red{F_g = 4.1\times 10^6} $$ newtons.

$$ \begin{align*} \red{4.1\times 10^6} & = \frac{1.56\times 10^{30}}{r^2}\\[6pt] (\red{4.1\times 10^6})r^2 & = 1.56\times 10^{30}\\[6pt] r^2 & = \frac{1.56\times 10^{30}}{\red{4.1\times 10^6}}\\[6pt] r^2 & = \frac{1.56}{4.1}\times \frac{10^{30}}{10^6}\\[6pt] & = 0.3805\times 10^{24}\\[6pt] r & = \pm\sqrt{0.3805\times 10^{24}}\\[6pt] & = 0.6169\times 10^{12}\\[6pt] & = 6.169 \times 10^{12} \end{align*} $$

The distance decreases to $$ 6.169\times 10^{12} $$ meters.

In the study of electricity, Ohm's Law says the electrical current (measured in amps) across a conductor is inversely proportional to electrical resistance (measured in ohms).

Suppose a particular circuit has a variable resistor that is currently set to 360 ohm, and this results in 1/3 amps of current. If the resistance were decreased to 80 ohms, what would the resulting change in current be?

Define variables to represent the quantities we are investigating, and then write down the appropriate variation equation.

Let $$ I = $$ electrical current, measured in amps.

Let $$ R = $$ electrical resistance, measured in volts.

The equation of variation is $$ I = \frac k R $$ .

Note: This is Ohm's Law, and the constant of proportionality represents the voltage on the circuit.

Use $$ \red{I = \frac 1 3} $$ and $$ \blue{R = 360} $$ to determine the value of $$ k $$. Then rewrite the variation equation.

$$ \begin{align*} \red I & = \frac k {\blue R}\\[6pt] \red{\frac 1 3} & = \frac k {\blue{360}}\\[6pt] \red{\frac 1 3}(\blue{360}) & = \frac k {\blue{360}}\cdot \blue{360}\\[6pt] 120 & = \frac k {\cancelred{360}}\cdot \cancelred{360}\\[6pt] k & = 120 \end{align*} $$

The variation equation is now $$ I = \frac{120} R $$ .

Use the equation to determine the current running through the circuit when the resistance drops to only $$ \blue{R = 80} $$ ohms.

$$ I = \frac{120}{\blue{80}} = \frac 3 2 $$

When the resistance decreases to 80 ohms, the current increases to 1.5 amps.

In economics, the basic Law of Demand tells us that as the price for a particular good (or service) increases, the demand for that good (or service) will decrease. This is an inverse variation relationship.

Suppose a new app is released for cell phones, and at a price of $$ \$4.99 $$ there are 3.2 million downloads each month. Later, the company that produced the app puts it on sale for $$ \$0.99 $$

(a) What will the accompanying change in the number of downloads per month be?

(b) If the wanted to phase out the app (for a newer, better one in development), and wanted to decrease the number of downloads per month to only $$ 500{,}000 $$ , what price should they charge for the app?

Define variables for the quantities we are investigating, then write down the appropriate variation equation.

Let $$ p = $$ the price, in dollars, for which the app is being sold.

Let $$ d = $$ the number of downloads per month, measured in millions of downloads per month.

So, $$ d = 1 $$ represents 1 million downloads per month.

The variation equation is $$ d = \frac k p $$ .

Use $$ \red{d = 3.2} $$ and $$ \blue{4.99} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red d & = \frac k {\blue p}\\[6pt] \red{3.2} & = \frac k {\blue{4.99}}\\[6pt] \red{3.2}(\blue{4.99}) & = \frac k {\blue{4.99}}\cdot \blue{4.99}\\[6pt] 15.968 & = \frac k {\cancelred{4.99}}\cdot \cancelred{4.99}\\[6pt] k & = 15.968 \end{align*} $$

The variation equation becomes $$ d = \frac{15.968} p $$ .

Use the variation equation to determine the number of downloads per month the company can expect from putting the app on sale for $$ \blue{p = \$0.99} $$ .

$$ d = \frac{15.968}{\blue{0.99}} = \frac{1596.8}{99} \approx 16.13 $$

The downloads should increase to approximately 16.13 million per month.

Use the variation equation to determine the price which will cause the number of downloads per month to decrease to only 5 hundred thousand ( $$ \red{d = 1/2} $$ ).

$$ \begin{align*} \red{\frac 1 2} & = \frac{15.968} p\\[6pt] \frac p 2 & = 15.968\\[6pt] p & = 31.936 \end{align*} $$

The price would need to be about $$ \$31.94 $$ to drive down the demand to only half a million.

(a) Demand will increase to about 16.13 million downloads per month.

(b) Increasing the price to about $$ \$31.94 $$ will drive down demand to $$ 500{,}000 $$ downloads per month.

When working with electrical circuits, it turns out that the electrical resistance is inversely proportional to the square of the current.

Suppose a particular circuit has a current measured at $$ 0.05 $$ amps with a resistance of 90 ohms.

(a) What will happen to the resistance if the current drops to only 0.01 amps?

(b) Suppose someone swaps out the 90-ohm resistor for one that is rated at only 60 ohms. What happens to the current?

Define variables for the quantities, then write down the variation equation.

Let $$ R = $$ the electrical resistance measured in ohms.

Let $$ I = $$ the electrical current measured in amps.

The variation equation is $$ R = \frac k {I^2} $$

Note: In a previous question about electrical circuits we mentioned the constant of proportionality was in fact the voltage on the circuit. This time, the constant of proportionality represents the electrical power (in watts).

Use $$ \red{R = 90} $$ and $$ \blue{I = 0.05} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red R & = \frac k {\blue I^2}\\[6pt] \red{90} & = \frac k {(\blue{0.05})^2}\\[6pt] \red{90}\cdot(\blue{0.0025}) & = \frac k {\blue{0.0025}}\cdot (\blue{0.0025})\\[6pt] 0.225 & = \frac k {\cancelred{0.0025}}\cdot (\cancelred{0.0025})\\[6pt] k & = 0.225 \end{align*} $$

The variation equation is now $$ \displaystyle R = \frac{0.225}{I^2} = \frac 9 {40I^2} $$

Determine what happens to the current if the resistance is changed to $$ \red{R = 60} $$ ohms.

$$ \begin{align*} \red{60} & = \frac 9 {40I^2}\\[6pt] \red{60}I^2 & = \frac 9 {40}\\[6pt] I^2 & = \frac 9 {2400}\\[6pt] I & = \sqrt{\frac 9 {2400}}\\[6pt] & = \frac 3 {20\sqrt 6}\\[6pt] & \approx 0.0612 \end{align*} $$

The current increases to $$ \frac 3 {20\sqrt 6} $$ or approximately 0.0612 amps.

Suppose an object is falling toward the earth, and it's speed is inversely proportional to the square root of the distance from the earth's surface.

Also, suppose the object's speed is 60 meters/second when it is 900 meters above the earth's surface. How close will the object be when it's velocity reaches 100 meters/second?

Define variables for the quantities we are interested in, then write down the variation equation.

$$ v = $$ velocity, in meters/second

$$ d = $$ distance from the earth's surface, in meters.

The variation equation is $$ \displaystyle v = \frac k {\sqrt d} $$ .

Use $$ \red{v = 60} $$ and $$ \blue{d =900} $$ to determine the value of $$ k $$, then write down the updated variation equation.

$$ \begin{align*} \red v & = \frac k {\sqrt{\blue d}}\\[6pt] \red{60} & = \frac k {\sqrt{\blue{900}}}\\[6pt] \red{60}(\blue{30}) & = \frac k {\blue{30}}\cdot\blue{30}\\[6pt] 1800 & = \frac k {\cancelred{30}}\cdot\cancelred{30}\\[6pt] k & = 1800 \end{align*} $$

The variation equation is now $$ v = \frac{1800}{\sqrt d} $$

Use the equation to determine the distance of the object from the earth's surface when it's velocity reaches $$ \red{v = 100} $$ meters per second.

$$ \begin{align*} \red{100} & = \frac{1800}{\sqrt d}\\[6pt] \sqrt d & = \frac{1800}{\red{100}} = 18\\[6pt] d & = 18^2 = 324 \end{align*} $$

The object is only 324 meters above the earth's surface when it's velocity reaches 100 meters per second.

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Word Problems: Inverse Variation

While direct variation describes a linear relationship between two variables, inverse variation describes another kind of relationship. For two quantities with inverse variation, as one quantity increases, the other quantity decreases.

For example, when you travel to a particular location, as your speed increases, the time it takes to arrive at that location decreases. When you decrease your speed, the time it takes to arrive at that location increases. So, the quantities are inversely proportional.

An inverse variation can be expressed by the equation x y = k or y = k x .

That is, y varies inversely as x if there is some nonzero constant k such that, x y = k or y = k x where x ≠ 0 and y ≠ 0 .

Some word problems require the use of inverse variation. Here are the ways to solve inverse variation word problems.

• Understand the problem.
• Write the formula.
• Identify the known values and substitute in the formula.
• Solve for the unknown.

The volume V of a gas varies inversely as the pressure P on it. If the volume is 240     cm 3 under pressure of 30     kg / cm 2 , what pressure has to be applied to have a volume of 160     cm 3 ?

The volume V varies inversely as the pressure P means when the volume increases, the pressure decreases and when the volume decreases, the pressure increases.

Now write the formula for inverse variation.

Substitute 240 for V 30 for P in the formula and find the constant

( 240 ) ( 30 ) = k

Now write an equation and solve for the unknown.

We have to find the pressure when the volume is 160     cm 3 .

( 160 ) ( P ) = 7200 .

Solve for P .

P = 7200 160           = 45

Therefore, pressure 45     kg / cm 2 be applied to have a volume of 160     cm 3 .

The length of a violin string varies inversely as the frequency of its vibrations. A violin string 14 inches long vibrates at a frequency of 450 cycles per second. Find the frequency of a 12 -inch violin string.

The length( l ) varies inversely as the frequency( f ), when the length increases, the frequency decreases and when the length decreases, the frequency increases.

Substitute 450 for f 14 for l in the formula and find the constant.

( 450 ) ( 14 ) = k

We have to find the frequency of 12 -inch violin string.

( 12 ) ( f ) = 6300 .

Solve for f .

f = 6300 12             = 525

Therefore, 12 -inch violin string vibrates at a frequency of 525 cycles per second.

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8.9 Use Direct and Inverse Variation

Learning objectives.

By the end of this section, you will be able to:

• Solve direct variation problems
• Solve inverse variation problems

Be Prepared 8.9

Before you get started, take this readiness quiz.

If you miss a problem, go back to the section listed and review the material.

• Find the multiplicative inverse of −8 −8 . If you missed this problem, review Example 1.126 .
• Solve for n n : 45 = 20 n 45 = 20 n . If you missed this problem, review Example 2.13 .
• Evaluate 5 x 2 5 x 2 when x = 10 x = 10 . If you missed this problem, review Example 1.20 .

When two quantities are related by a proportion, we say they are proportional to each other. Another way to express this relation is to talk about the variation of the two quantities. We will discuss direct variation and inverse variation in this section.

Solve Direct Variation Problems

Lindsay gets paid $15 per hour at her job. If we let s be her salary and h be the number of hours she has worked, we could model this situation with the equation

Lindsay’s salary is the product of a constant, 15, and the number of hours she works. We say that Lindsay’s salary varies directly with the number of hours she works. Two variables vary directly if one is the product of a constant and the other.

Direct Variation

For any two variables x and y , y varies directly with x if

The constant k is called the constant of variation.

In applications using direct variation, generally we will know values of one pair of the variables and will be asked to find the equation that relates x and y . Then we can use that equation to find values of y for other values of x .

Example 8.86

How to solve direct variation problems.

If y varies directly with x and y = 20 y = 20 when x = 8 x = 8 , find the equation that relates x and y .

Try It 8.171

If y y varies directly as x x and y = 3 , when x = 10 . y = 3 , when x = 10 . find the equation that relates x and y .

Try It 8.172

If y y varies directly as x x and y = 12 when x = 4 y = 12 when x = 4 find the equation that relates x and y .

We’ll list the steps below.

Solve direct variation problems.

• Step 1. Write the formula for direct variation.
• Step 2. Substitute the given values for the variables.
• Step 3. Solve for the constant of variation.
• Step 4. Write the equation that relates x and y.

Now we’ll solve a few applications of direct variation.

Example 8.87

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?

Try It 8.173

The number of calories, c , burned varies directly with the amount of time, t, spent exercising. Arnold burned 312 calories in 65 minutes exercising.

• ⓐ Write the equation that relates c and t .
• ⓑ How many calories would he burn if he exercises for 90 minutes?

Try It 8.174

The distance a moving body travels, d , varies directly with time, t , it moves. A train travels 100 miles in 2 hours

ⓐ Write the equation that relates d and t . ⓑ How many miles would it travel in 5 hours?

In the previous example, the variables c and m were named in the problem. Usually that is not the case. We will have to name the variables in the next example as part of the solution, just like we do in most applied problems.

Example 8.88

The number of gallons of gas Eunice’s car uses varies directly with the number of miles she drives. Last week she drove 469.8 miles and used 14.5 gallons of gas.

• ⓐ Write the equation that relates the number of gallons of gas used to the number of miles driven.
• ⓑ How many gallons of gas would Eunice’s car use if she drove 1000 miles?

ⓑ Find g when m = 1000 . Write the equation that relates g and m . g = 0.031 m Substitute the given value for m . g = 0.031 ( 1000 ) Simplify. g = 31 Eunice’s car would use 31 gallons of gas if she drove it 1,000 miles. Find g when m = 1000 . Write the equation that relates g and m . g = 0.031 m Substitute the given value for m . g = 0.031 ( 1000 ) Simplify. g = 31 Eunice’s car would use 31 gallons of gas if she drove it 1,000 miles.

Notice that in this example, the units on the constant of variation are gallons/mile. In everyday life, we usually talk about miles/gallon.

Try It 8.175

The distance that Brad travels varies directly with the time spent traveling. Brad travelled 660 miles in 12 hours,

• ⓐ Write the equation that relates the number of miles travelled to the time.
• ⓑ How many miles could Brad travel in 4 hours?

Try It 8.176

The weight of a liquid varies directly as its volume. A liquid that weighs 24 pounds has a volume of 4 gallons.

• ⓐ Write the equation that relates the weight to the volume.
• ⓑ If a liquid has volume 13 gallons, what is its weight?

In some situations, one variable varies directly with the square of the other variable. When that happens, the equation of direct variation is y = k x 2 y = k x 2 . We solve these applications just as we did the previous ones, by substituting the given values into the equation to solve for k .

Example 8.89

The maximum load a beam will support varies directly with the square of the diagonal of the beam’s cross-section. A beam with diagonal 4” will support a maximum load of 75 pounds.

• ⓐ Write the equation that relates the maximum load to the cross-section.
• ⓑ What is the maximum load that can be supported by a beam with diagonal 8”?

ⓑ Find L when c = 8 . Write the equation that relates L and c . L = 4.6875 c 2 Substitute the given value for c . L = 4.6875 ( 8 ) 2 Simplify. L = 300 A beam with diagonal 8” could support a maximum load of 300 pounds. Find L when c = 8 . Write the equation that relates L and c . L = 4.6875 c 2 Substitute the given value for c . L = 4.6875 ( 8 ) 2 Simplify. L = 300 A beam with diagonal 8” could support a maximum load of 300 pounds.

Try It 8.177

The distance an object falls is directly proportional to the square of the time it falls. A ball falls 144 feet in 3 seconds.

• ⓐ Write the equation that relates the distance to the time.
• ⓑ How far will an object fall in 4 seconds?

Try It 8.178

The area of a circle varies directly as the square of the radius. A circular pizza with a radius of 6 inches has an area of 113.04 square inches.

• ⓐ Write the equation that relates the area to the radius.
• ⓑ What is the area of a pizza with a radius of 9 inches?

Solve Inverse Variation Problems

Many applications involve two variable that vary inversely . As one variable increases, the other decreases. The equation that relates them is y = k x y = k x .

Inverse Variation

For any two variables x and y , y varies inversely with x if

The word ‘inverse’ in inverse variation refers to the multiplicative inverse. The multiplicative inverse of x is 1 x 1 x .

We solve inverse variation problems in the same way we solved direct variation problems. Only the general form of the equation has changed. We will copy the procedure box here and just change ‘direct’ to ‘inverse’.

Solve inverse variation problems.

• Step 1. Write the formula for inverse variation.

Example 8.90

If y varies inversely with x x and y = 20 y = 20 when x = 8 x = 8 , find the equation that relates x and y .

Try It 8.179

If p p varies inversely with q q and p = 30 p = 30 when q = 12 q = 12 find the equation that relates p p and q . q .

Try It 8.180

If y y varies inversely with x x and y = 8 y = 8 when x = 2 x = 2 find the equation that relates x x and y y .

Example 8.91

The fuel consumption (mpg) of a car varies inversely with its weight. A car that weighs 3100 pounds gets 26 mpg on the highway.

• ⓐ Write the equation of variation.
• ⓑ What would be the fuel consumption of a car that weighs 4030 pounds?

ⓑ Find f when w = 4030 . Write the equation that relates f and w . f = 80,600 w Substitute the given value for w . f = 80,600 4030 Simplify. f = 20 A car that weighs 4030 pounds would have fuel consumption of 20 mpg. Find f when w = 4030 . Write the equation that relates f and w . f = 80,600 w Substitute the given value for w . f = 80,600 4030 Simplify. f = 20 A car that weighs 4030 pounds would have fuel consumption of 20 mpg.

Try It 8.181

A car’s value varies inversely with its age. Elena bought a two-year-old car for $20,000.

ⓐ Write the equation of variation. ⓑ What will be the value of Elena’s car when it is 5 years old?

Try It 8.182

The time required to empty a pool varies inversely as the rate of pumping. It took Lucy 2.5 hours to empty her pool using a pump that was rated at 400 gpm (gallons per minute).

• ⓑ How long will it take her to empty the pool using a pump rated at 500 gpm?

Example 8.92

The frequency of a guitar string varies inversely with its length. A 26” long string has a frequency of 440 vibrations per second.

• ⓑ How many vibrations per second will there be if the string’s length is reduced to 20” by putting a finger on a fret?

ⓑ Find f when L = 20 . Write the equation that relates f and L . f = 11,440 L Substitute the given value for L . f = 11,440 20 Simplify. f = 572 A 20” guitar string has frequency 572 vibrations per second. Find f when L = 20 . Write the equation that relates f and L . f = 11,440 L Substitute the given value for L . f = 11,440 20 Simplify. f = 572 A 20” guitar string has frequency 572 vibrations per second.

Try It 8.183

The number of hours it takes for ice to melt varies inversely with the air temperature. Suppose a block of ice melts in 2 hours when the temperature is 65 degrees.

• ⓑ How many hours would it take for the same block of ice to melt if the temperature was 78 degrees?

Try It 8.184

The force needed to break a board varies inversely with its length. Richard uses 24 pounds of pressure to break a 2-foot long board.

• ⓑ How many pounds of pressure is needed to break a 5-foot long board?

Section 8.9 Exercises

Practice makes perfect.

In the following exercises, solve.

If y y varies directly as x x and y = 14 , when x = 3 y = 14 , when x = 3 , find the equation that relates x and y x and y .

If p p varies directly as q q and p = 5 , when q = 2 p = 5 , when q = 2 , find the equation that relates p and q p and q .

If v v varies directly as w w and v = 24 , when w = 8 v = 24 , when w = 8 , find the equation that relates v and w . v and w .

If a a varies directly as b b and a = 16 , when b = 4 a = 16 , when b = 4 , find the equation that relates a and b . a and b .

If p p varies directly as q q and p = 9.6 , when q = 3 p = 9.6 , when q = 3 , find the equation that relates p and q . p and q .

If y y varies directly as x x and y = 12.4 , when x = 4 , y = 12.4 , when x = 4 , find the equation that relates x and y x and y

If a a varies directly as b b and a = 6 , when b = 1 3 a = 6 , when b = 1 3 , find the equation that relates a and b . a and b .

If v v varies directly as w w and v = 8 , when w = 1 2 v = 8 , when w = 1 2 , find the equation that relates v and w . v and w .

The amount of money Sally earns, P , varies directly with the number, n , of necklaces she sells. When Sally sells 15 necklaces she earns $150.

• ⓐ Write the equation that relates P and n .
• ⓑ How much money would she earn if she sold 4 necklaces?

The price, P , that Eric pays for gas varies directly with the number of gallons, g , he buys. It costs him $50 to buy 20 gallons of gas.

• ⓐ Write the equation that relates P and g .
• ⓑ How much would 33 gallons cost Eric?

Terri needs to make some pies for a fundraiser. The number of apples, a , varies directly with number of pies, p . It takes nine apples to make two pies.

• ⓐ Write the equation that relates a and p .
• ⓑ How many apples would Terri need for six pies?

Joseph 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.

• ⓐ Write the equation that relates d and v .
• ⓑ How far would he travel before stopping for lunch at a rate of 65 mph?

The price of gas that Jesse purchased varies directly to how many gallons he purchased. He purchased 10 gallons of gas for $39.80.

• ⓐ Write the equation that relates the price to the number of gallons.
• ⓑ How much will it cost Jesse for 15 gallons of gas?

The distance that Sarah travels varies directly to how long she drives. She travels 440 miles in 8 hours.

• ⓐ Write the equation that relates the distance to the number of hours.
• ⓑ How far can Sally travel in 6 hours?

The mass of a liquid varies directly with its volume. A liquid with mass 16 kilograms has a volume of 2 liters.

• ⓐ Write the equation that relates the mass to the volume.
• ⓑ What is the volume of this liquid if its mass is 128 kilograms?

The length that a spring stretches varies directly with a weight placed at the end of the spring. When Sarah placed a 10 pound watermelon on a hanging scale, the spring stretched 5 inches.

• ⓐ Write the equation that relates the length of the spring to the weight.
• ⓑ What weight of watermelon would stretch the spring 6 inches?

The distance an object falls varies directly to the square of the time it falls. A ball falls 45 feet in 3 seconds.

• ⓑ How far will the ball fall in 7 seconds?

The maximum load a beam will support varies directly with the square of the diagonal of the beam’s cross-section. A beam with diagonal 6 inch will support a maximum load of 108 pounds.

• ⓐ Write the equation that relates the load to the diagonal of the cross-section.
• ⓑ What load will a beam with a 10 inch diagonal support?
• ⓑ What is the area of a personal pizza with a radius 4 inches?

The distance an object falls varies directly to the square of the time it falls. A ball falls 72 feet in 3 seconds,

• ⓑ How far will the ball have fallen in 8 seconds?

If y y varies inversely with x x and y = 5 y = 5 when x = 4 x = 4 find the equation that relates x x and y . y .

If p p varies inversely with q q and p = 2 p = 2 when q = 1 q = 1 find the equation that relates p p and q . q .

If v v varies inversely with w w and v = 6 v = 6 when w = 1 2 w = 1 2 find the equation that relates v v and w . w .

If a a varies inversely with b b and a = 12 a = 12 when b = 1 3 b = 1 3 find the equation that relates a a and b . b .

Write an inverse variation equation to solve the following problems.

The fuel consumption (mpg) of a car varies inversely with its weight. A Toyota Corolla weighs 2800 pounds and gets 33 mpg on the highway.

• ⓐ Write the equation that relates the mpg to the car’s weight.
• ⓑ What would the fuel consumption be for a Toyota Sequoia that weighs 5500 pounds?

A car’s value varies inversely with its age. Jackie bought a 10 year old car for $2,400.

• ⓐ Write the equation that relates the car’s value to its age.
• ⓑ What will be the value of Jackie’s car when it is 15 years old ?

The time required to empty a tank varies inversely as the rate of pumping. It took Janet 5 hours to pump her flooded basement using a pump that was rated at 200 gpm (gallons per minute),

• ⓐ Write the equation that relates the number of hours to the pump rate.
• ⓑ How long would it take Janet to pump her basement if she used a pump rated at 400 gpm?

The volume of a gas in a container varies inversely as pressure on the gas. A container of helium has a volume of 370 cubic inches under a pressure of 15 psi.

• ⓐ Write the equation that relates the volume to the pressure.
• ⓑ What would be the volume of this gas if the pressure was increased to 20 psi?

On a string instrument, the length of a string varies inversely as the frequency of its vibrations. An 11-inch string on a violin has a frequency of 400 cycles per second.

• ⓐ Write the equation that relates the string length to its frequency.
• ⓑ What is the frequency of a 10-inch string?

Paul, a dentist, determined that the number of cavities that develops in his patient’s mouth each year varies inversely to the number of minutes spent brushing each night. His patient, Lori, had 4 cavities when brushing her teeth 30 seconds (0.5 minutes) each night.

• ⓐ Write the equation that relates the number of cavities to the time spent brushing.
• ⓑ How many cavities would Paul expect Lori to have if she had brushed her teeth for 2 minutes each night?

The number of tickets for a sports fundraiser varies inversely to the price of each ticket. Brianna can buy 25 tickets at $5each.

• ⓐ Write the equation that relates the number of tickets to the price of each ticket.
• ⓑ How many tickets could Brianna buy if the price of each ticket was $2.50?

Boyle’s Law states that if the temperature of a gas stays constant, then the pressure varies inversely to the volume of the gas. Braydon, a scuba diver, has a tank that holds 6 liters of air under a pressure of 220 psi.

• ⓐ Write the equation that relates pressure to volume.
• ⓑ If the pressure increases to 330 psi, how much air can Braydon’s tank hold?

Mixed Practice

If y y varies directly as x x and y = 5 , when x = 3 . y = 5 , when x = 3 . , find the equation that relates x and y x and y .

If v v varies directly as w w and v = 21 , when w = 8 . v = 21 , when w = 8 . find the equation that relates v and w . v and w .

If p p varies inversely with q q and p = 5 p = 5 when q = 6 q = 6 , find the equation that relates p p and q . q .

If y y varies inversely with x x and y = 11 y = 11 when x = 3 x = 3 find the equation that relates x x and y . y .

If p p varies directly as q q and p = 10 , when q = 2 . p = 10 , when q = 2 . find the equation that relates p and q p and q .

If v v varies inversely with w w and v = 18 v = 18 when w = 1 3 w = 1 3 find the equation that relates v v and w . w .

The force needed to break a board varies inversely with its length. If Tom uses 20 pounds of pressure to break a 1.5-foot long board, how many pounds of pressure would he need to use to break a 6 foot long board?

The number of hours it takes for ice to melt varies inversely with the air temperature. A block of ice melts in 2.5 hours when the temperature is 54 degrees. How long would it take for the same block of ice to melt if the temperature was 45 degrees?

The length a spring stretches varies directly with a weight placed at the end of the spring. When Meredith placed a 6-pound cantaloupe on a hanging scale, the spring stretched 2 inches. How far would the spring stretch if the cantaloupe weighed 9 pounds?

The amount that June gets paid varies directly the number of hours she works. When she worked 15 hours, she got paid $111. How much will she be paid for working 18 hours?

The fuel consumption (mpg) of a car varies inversely with its weight. A Ford Focus weighs 3000 pounds and gets 28.7 mpg on the highway. What would the fuel consumption be for a Ford Expedition that weighs 5,500 pounds? Round to the nearest tenth.

The volume of a gas in a container varies inversely as the pressure on the gas. If a container of argon has a volume of 336 cubic inches under a pressure of 2,500 psi, what will be its volume if the pressure is decreased to 2,000 psi?

The distance an object falls varies directly to the square of the time it falls. If an object falls 52.8 feet in 4 seconds, how far will it fall in 9 seconds?

The area of the face of a Ferris wheel varies directly with the square of its radius. If the area of one face of a Ferris wheel with diameter 150 feet is 70,650 square feet, what is the area of one face of a Ferris wheel with diameter of 16 feet?

Everyday Math

Ride Service It costs $35 for a ride from the city center to the airport, 14 miles away.

• ⓐ Write the equation that relates the cost, c , with the number of miles, m .
• ⓑ What would it cost to travel 22 miles with this service?

Road Trip The number of hours it takes Jack to drive from Boston to Bangor is inversely proportional to his average driving speed. When he drives at an average speed of 40 miles per hour, it takes him 6 hours for the trip.

• ⓐ Write the equation that relates the number of hours, h , with the speed, s.
• ⓑ How long would the trip take if his average speed was 75 miles per hour?

Writing Exercises

In your own words, explain the difference between direct variation and inverse variation.

Make up an example from your life experience of inverse variation.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well-prepared for the next chapter? Why or why not?

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Inverse Variation Word Problems

Related Pages: Direct Variation Proportions Proportion Word Problems Joint And Combined Variation More Algebra Lessons

In these lessons, we will learn about inverse variation and how to solve applications that involve inverse variation.

The following diagrams show the differences between Direct Variation and Indirect Variation. Scroll down the page for examples and solutions.

What Is Inverse Variation?

There are many situations in our daily lives that involve inverse variation (indirect variation).

For example, the number of days required to build a bridge varies inversely to the number of workers. As the number of workers increases, the number of days required to build would decrease and as the number of workers decreases, the number of days increases.

Another example would be that speed varies inversely to time. As speed increases, the time taken to cover the same distance decreases and as speed decreases, the time taken increases.

In general, when two variables x and y are such that xy = k where k is a non-zero constant, we say that y varies inversely with x.

Example: Suppose that y varies inversely as x and that y = 8 when x = 3. a) Form an equation connecting x and y. b) Calculate the value of y when x = 10.

a) Substitute x = 3 and y = 8 into the equation to obtain k 3 × 8 = k ⇒ k = 24 The equation is xy = 24

a) Find the equation connecting x and y. b) Find the value of y when x = 3.

How to define inverse variation and how to solve inverse variation problems?

• y varies inversely as x. y = 4 when x = 2. Determine the inverse variation equation. Then determine y when x = 16.
• The time, t, required to empty a tank varies inversely as the rate, r, of pumping. If a pump can empty a tank in 2.5 hours at a rate of 400 gallons per minute, how long will it take to empty a tank at 500 gallons per minute?
• The force, F, needed to break a board varies inversely with the length, L, of the board. If it takes 24 pounds of pressure to break a board 2 feet long, how many pounds of pressure would it take to break a board that is 5 feet long?
• y varies inversely as the square root of x. y = 6 when x = 16. Determine the inverse variation equation. Then determine y when x = 4.

How to solve a basic inverse variation problem?

Example: y varies inversely as x. y = 3 when x = 10. Determine the inverse variation equation. Then determine y when x = 6.

How to solve an inverse variation problem with a change of variables?

Example: Given m varies inversely as t, and m = 9 when t = 6, find the variation constant and the inverse variation equation. Then determine m when t = 27.

How to solve a inverse variation problem when k is a fraction?

Example: y varies inversely as x. y = 1/2 when x = 2/3. Find the variation constant and the inverse variation equation. Then determine y when x = 2/15.

What is the difference between direct and inverse variation?

In general, if two quantities vary directly, if one goes up so does the other. If one goes down so does the other. The following statements are equivalent

• y varies directly as x.
• y is directly proportional to x.
• y = kx for some constant k.

In general, if two quantities vary indirectly, if one goes up and the other goes down. The following statements are equivalent

• y varies indirectly as x.
• y is inversely proportional to x.
• y = k/x for some constant k.

How to tell if two variables vary inversely or directly?

Recognizing Direct and Inverse Variation

How to solve direct variation and indirect/inverse variation word problems?

Example: On a string instrument, the length of a string varies inversely as the frequency of its vibrations. An 11-inch string has a frequency of 400 cycles per second. Find the constant of proportionality and the frequency of a 10-inch string.

Inverse Variation Equations and Ordered Pairs

This video looks at inverse variation: identifying inverse variations from ordered pairs, writing inverse variation equations, graphing inverse variations, and finding missing values.

Example: Let x 1 = 4, y 1 = 12 and x 2 = 3. Let y vary inversely as x. Find y 2 .

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

Inverse Variation Lesson

• Demonstrate an understanding of direct variation
• Learn how to solve an inverse variation problem
• Learn how to solve an inverse variation as a power problem
• Learn how to solve an inverse variation word problem

How to Solve an Inverse Variation Problem

Solving an inverse variation problem.

• Write the variation equation: y = k/x or k = xy
• Substitute in for the given values and find the value of k
• Rewrite the variation equation: y = k/x with the known value of k
• Substitute the remaining values and find the unknown

Inverse Variation as a Power

Inverse variation word problems, skills check:.

Solve each inverse variation problem.

y varies inversely with x and y = 3 when x = 8. Find y when x = 6.

Please choose the best answer.

Jacobs Medical Devices produces medical sponges. The cost of producing a medical sponge varies inversely with the number produced. When 10,000 medical sponges are produced, the cost is $2 per medical sponge. What is the cost per medical sponge for Jacobs Medical Devices to produce 25,000 medical sponges?

The current in a simple electrical circuit varies inversely with the resistance. If the current is 20 amps when the resistance is 5 ohms, find the current when the resistance is 8 ohms.

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Chapter 2: Linear Equations

2.7 Variation Word Problems

Direct variation problems.

There are many mathematical relations that occur in life. For instance, a flat commission salaried salesperson earns a percentage of their sales, where the more they sell equates to the wage they earn. An example of this would be an employee whose wage is 5% of the sales they make. This is a direct or a linear variation, which, in an equation, would look like:

[latex]\text{Wage }(x)=5\%\text{ Commission }(k)\text{ of Sales Completed }(y)[/latex]

[latex]x=ky[/latex]

A historical example of direct variation can be found in the changing measurement of pi, which has been symbolized using the Greek letter π since the mid 18th century. Variations of historical π calculations are Babylonian [latex]\left(\dfrac{25}{8}\right),[/latex] Egyptian [latex]\left(\dfrac{16}{9}\right)^2,[/latex] and Indian [latex]\left(\dfrac{339}{108}\text{ and }10^{\frac{1}{2}}\right).[/latex] In the 5th century, Chinese mathematician Zu Chongzhi calculated the value of π to seven decimal places (3.1415926), representing the most accurate value of π for over 1000 years.

Pi is found by taking any circle and dividing the circumference of the circle by the diameter, which will always give the same value: 3.14159265358979323846264338327950288419716… (42 decimal places). Using an infinite-series exact equation has allowed computers to calculate π to 10 13 decimals.

[latex]\begin{array}{c} \text{Circumference }(c)=\pi \text{ times the diameter }(d) \\ \\ \text{or} \\ \\ c=\pi d \end{array}[/latex]

All direct variation relationships are verbalized in written problems as a direct variation or as directly proportional and take the form of straight line relationships. Examples of direct variation or directly proportional equations are:

• [latex]x[/latex] varies directly as [latex]y[/latex]
• [latex]x[/latex] varies as [latex]y[/latex]
• [latex]x[/latex] varies directly proportional to [latex]y[/latex]
• [latex]x[/latex] is proportional to [latex]y[/latex]
• [latex]x[/latex] varies directly as the square of [latex]y[/latex]
• [latex]x[/latex] varies as [latex]y[/latex] squared
• [latex]x[/latex] is proportional to the square of [latex]y[/latex]
• [latex]x[/latex] varies directly as the cube of [latex]y[/latex]
• [latex]x[/latex] varies as [latex]y[/latex] cubed
• [latex]x[/latex] is proportional to the cube of [latex]y[/latex]
• [latex]x[/latex] varies directly as the square root of [latex]y[/latex]
• [latex]x[/latex] varies as the root of [latex]y[/latex]
• [latex]x[/latex] is proportional to the square root of [latex]y[/latex]

Example 2.7.1

Find the variation equation described as follows:

The surface area of a square surface [latex](A)[/latex] is directly proportional to the square of either side [latex](x).[/latex]

[latex]\begin{array}{c} \text{Area }(A) =\text{ constant }(k)\text{ times side}^2\text{ } (x^2) \\ \\ \text{or} \\ \\ A=kx^2 \end{array}[/latex]

Example 2.7.2

When looking at two buildings at the same time, the length of the buildings’ shadows [latex](s)[/latex] varies directly as their height [latex](h).[/latex] If a 5-story building has a 20 m long shadow, how many stories high would a building that has a 32 m long shadow be?

The equation that describes this variation is:

[latex]h=kx[/latex]

Breaking the data up into the first and second parts gives:

[latex]\begin{array}{ll} \begin{array}{rrl} \\ &&\textbf{1st Data} \\ s&=&20\text{ m} \\ h&=&5\text{ stories} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ h&=&kx \\ 5\text{ stories}&=&k\text{ (20 m)} \\ k&=&5\text{ stories/20 m}\\ k&=&0.25\text{ story/m} \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ s&=&\text{32 m} \\ h&=&\text{find 2nd} \\ k&=&0.25\text{ story/m} \\ \\ &&\text{Find }h\text{:} \\ h&=&kx \\ h&=&(0.25\text{ story/m})(32\text{ m}) \\ h&=&8\text{ stories} \end{array} \end{array}[/latex]

Inverse Variation Problems

Inverse variation problems are reciprocal relationships. In these types of problems, the product of two or more variables is equal to a constant. An example of this comes from the relationship of the pressure [latex](P)[/latex] and the volume [latex](V)[/latex] of a gas, called Boyle’s Law (1662). This law is written as:

[latex]\begin{array}{c} \text{Pressure }(P)\text{ times Volume }(V)=\text{ constant} \\ \\ \text{ or } \\ \\ PV=k \end{array}[/latex]

Written as an inverse variation problem, it can be said that the pressure of an ideal gas varies as the inverse of the volume or varies inversely as the volume. Expressed this way, the equation can be written as:

[latex]P=\dfrac{k}{V}[/latex]

Another example is the historically famous inverse square laws. Examples of this are the force of gravity [latex](F_{\text{g}}),[/latex] electrostatic force [latex](F_{\text{el}}),[/latex] and the intensity of light [latex](I).[/latex] In all of these measures of force and light intensity, as you move away from the source, the intensity or strength decreases as the square of the distance.

In equation form, these look like:

[latex]F_{\text{g}}=\dfrac{k}{d^2}\hspace{0.25in} F_{\text{el}}=\dfrac{k}{d^2}\hspace{0.25in} I=\dfrac{k}{d^2}[/latex]

These equations would be verbalized as:

• The force of gravity [latex](F_{\text{g}})[/latex] varies inversely as the square of the distance.
• Electrostatic force [latex](F_{\text{el}})[/latex] varies inversely as the square of the distance.
• The intensity of a light source [latex](I)[/latex] varies inversely as the square of the distance.

All inverse variation relationship are verbalized in written problems as inverse variations or as inversely proportional. Examples of inverse variation or inversely proportional equations are:

• [latex]x[/latex] varies inversely as [latex]y[/latex]
• [latex]x[/latex] varies as the inverse of [latex]y[/latex]
• [latex]x[/latex] varies inversely proportional to [latex]y[/latex]
• [latex]x[/latex] is inversely proportional to [latex]y[/latex]
• [latex]x[/latex] varies inversely as the square of [latex]y[/latex]
• [latex]x[/latex] varies inversely as [latex]y[/latex] squared
• [latex]x[/latex] is inversely proportional to the square of [latex]y[/latex]
• [latex]x[/latex] varies inversely as the cube of [latex]y[/latex]
• [latex]x[/latex] varies inversely as [latex]y[/latex] cubed
• [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex]
• [latex]x[/latex] varies inversely as the square root of [latex]y[/latex]
• [latex]x[/latex] varies as the inverse root of [latex]y[/latex]
• [latex]x[/latex] is inversely proportional to the square root of [latex]y[/latex]

Example 2.7.3

The force experienced by a magnetic field [latex](F_{\text{b}})[/latex] is inversely proportional to the square of the distance from the source [latex](d_{\text{s}}).[/latex]

[latex]F_{\text{b}} = \dfrac{k}{{d_{\text{s}}}^2}[/latex]

Example 2.7.4

The time [latex](t)[/latex] it takes to travel from North Vancouver to Hope varies inversely as the speed [latex](v)[/latex] at which one travels. If it takes 1.5 hours to travel this distance at an average speed of 120 km/h, find the constant [latex]k[/latex] and the amount of time it would take to drive back if you were only able to travel at 60 km/h due to an engine problem.

[latex]t=\dfrac{k}{v}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} &&\textbf{1st Data} \\ v&=&120\text{ km/h} \\ t&=&1.5\text{ h} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ k&=&tv \\ k&=&(1.5\text{ h})(120\text{ km/h}) \\ k&=&180\text{ km} \end{array} & \hspace{0.5in} \begin{array}{rrl} \\ \\ \\ &&\textbf{2nd Data} \\ v&=&60\text{ km/h} \\ t&=&\text{find 2nd} \\ k&=&180\text{ km} \\ \\ &&\text{Find }t\text{:} \\ t&=&\dfrac{k}{v} \\ \\ t&=&\dfrac{180\text{ km}}{60\text{ km/h}} \\ \\ t&=&3\text{ h} \end{array} \end{array}[/latex]

Joint or Combined Variation Problems

In real life, variation problems are not restricted to single variables. Instead, functions are generally a combination of multiple factors. For instance, the physics equation quantifying the gravitational force of attraction between two bodies is:

[latex]F_{\text{g}}=\dfrac{Gm_1m_2}{d^2}[/latex]

• [latex]F_{\text{g}}[/latex] stands for the gravitational force of attraction
• [latex]G[/latex] is Newton’s constant, which would be represented by [latex]k[/latex] in a standard variation problem
• [latex]m_1[/latex] and [latex]m_2[/latex] are the masses of the two bodies
• [latex]d^2[/latex] is the distance between the centres of both bodies

To write this out as a variation problem, first state that the force of gravitational attraction [latex](F_{\text{g}})[/latex] between two bodies is directly proportional to the product of the two masses [latex](m_1, m_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating the two masses. From this information, the necessary equation can be derived. All joint variation relationships are verbalized in written problems as a combination of direct and inverse variation relationships, and care must be taken to correctly identify which variables are related in what relationship.

Example 2.7.5

The force of electrical attraction [latex](F_{\text{el}})[/latex] between two statically charged bodies is directly proportional to the product of the charges on each of the two objects [latex](q_1, q_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating these two charged bodies.

[latex]F_{\text{el}}=\dfrac{kq_1q_2}{d^2}[/latex]

Solving these combined or joint variation problems is the same as solving simpler variation problems.

First, decide what equation the variation represents. Second, break up the data into the first data given—which is used to find [latex]k[/latex]—and then the second data, which is used to solve the problem given. Consider the following joint variation problem.

Example 2.7.6

[latex]y[/latex] varies jointly with [latex]m[/latex] and [latex]n[/latex] and inversely with the square of [latex]d[/latex]. If [latex]y = 12[/latex] when [latex]m = 3[/latex], [latex]n = 8[/latex], and [latex]d = 2,[/latex] find the constant [latex]k[/latex], then use [latex]k[/latex] to find [latex]y[/latex] when [latex]m=-3[/latex], [latex]n = 18[/latex], and [latex]d = 3[/latex].

[latex]y=\dfrac{kmn}{d^2}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} \\ \\ \\ && \textbf{1st Data} \\ y&=&12 \\ m&=&3 \\ n&=&8 \\ d&=&2 \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ 12&=&\dfrac{k(3)(8)}{(2)^2} \\ \\ k&=&\dfrac{12(2)^2}{(3)(8)} \\ \\ k&=& 2 \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ y&=&\text{find 2nd} \\ m&=&-3 \\ n&=&18 \\ d&=&3 \\ k&=&2 \\ \\ &&\text{Find }y\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ y&=&\dfrac{(2)(-3)(18)}{(3)^2} \\ \\ y&=&12 \end{array} \end{array}[/latex]

For questions 1 to 12, write the formula defining the variation, including the constant of variation [latex](k).[/latex]

• [latex]x[/latex] is jointly proportional to [latex]y[/latex] and [latex]z[/latex]
• [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex]
• [latex]x[/latex] is jointly proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
• [latex]x[/latex] is inversely proportional to [latex]y[/latex] to the sixth power
• [latex]x[/latex] is jointly proportional with the cube of [latex]y[/latex] and inversely to the square root of [latex]z[/latex]
• [latex]x[/latex] is inversely proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
• [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex] and is inversely proportional to the cube of [latex]p[/latex]
• [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex] and square of [latex]z[/latex]

For questions 13 to 22, find the formula defining the variation and the constant of variation [latex](k).[/latex]

• If [latex]A[/latex] varies directly as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=15[/latex] and [latex]B=5.[/latex]
• If [latex]P[/latex] is jointly proportional to [latex]Q[/latex] and [latex]R,[/latex] find [latex]k[/latex] when [latex]P=12, Q=8[/latex] and [latex]R=3.[/latex]
• If [latex]A[/latex] varies inversely as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=7[/latex] and [latex]B=4.[/latex]
• If [latex]A[/latex] varies directly as the square of [latex]B,[/latex] find [latex]k[/latex] when [latex]A=6[/latex] and [latex]B=3.[/latex]
• If [latex]C[/latex] varies jointly as [latex]A[/latex] and [latex]B,[/latex] find [latex]k[/latex] when [latex]C=24, A=3,[/latex] and [latex]B=2.[/latex]
• If [latex]Y[/latex] is inversely proportional to the cube of [latex]X,[/latex] find [latex]k[/latex] when [latex]Y=54[/latex] and [latex]X=3.[/latex]
• If [latex]X[/latex] is directly proportional to [latex]Y,[/latex] find [latex]k[/latex] when [latex]X=12[/latex] and [latex]Y=8.[/latex]
• If [latex]A[/latex] is jointly proportional with the square of [latex]B[/latex] and the square root of [latex]C,[/latex] find [latex]k[/latex] when [latex]A=25, B=5[/latex] and [latex]C=9.[/latex]
• If [latex]y[/latex] varies jointly with [latex]m[/latex] and the square of [latex]n[/latex] and inversely with [latex]d,[/latex] find [latex]k[/latex] when [latex]y=10, m=4, n=5,[/latex] and [latex]d=6.[/latex]
• If [latex]P[/latex] varies directly as [latex]T[/latex] and inversely as [latex]V,[/latex] find [latex]k[/latex] when [latex]P=10, T=250,[/latex] and [latex]V=400.[/latex]

For questions 23 to 37, solve each variation word problem.

• The electrical current [latex]I[/latex] (in amperes, A) varies directly as the voltage [latex](V)[/latex] in a simple circuit. If the current is 5 A when the source voltage is 15 V, what is the current when the source voltage is 25 V?
• The current [latex]I[/latex] in an electrical conductor varies inversely as the resistance [latex]R[/latex] (in ohms, Ω) of the conductor. If the current is 12 A when the resistance is 240 Ω, what is the current when the resistance is 540 Ω?
• Hooke’s law states that the distance [latex](d_s)[/latex] that a spring is stretched supporting a suspended object varies directly as the mass of the object [latex](m).[/latex] If the distance stretched is 18 cm when the suspended mass is 3 kg, what is the distance when the suspended mass is 5 kg?
• The volume [latex](V)[/latex] of an ideal gas at a constant temperature varies inversely as the pressure [latex](P)[/latex] exerted on it. If the volume of a gas is 200 cm 3 under a pressure of 32 kg/cm 2 , what will be its volume under a pressure of 40 kg/cm 2 ?
• The number of aluminum cans [latex](c)[/latex] used each year varies directly as the number of people [latex](p)[/latex] using the cans. If 250 people use 60,000 cans in one year, how many cans are used each year in a city that has a population of 1,000,000?
• The time [latex](t)[/latex] required to do a masonry job varies inversely as the number of bricklayers [latex](b).[/latex] If it takes 5 hours for 7 bricklayers to build a park wall, how much time should it take 10 bricklayers to complete the same job?
• The wavelength of a radio signal (λ) varies inversely as its frequency [latex](f).[/latex] A wave with a frequency of 1200 kilohertz has a length of 250 metres. What is the wavelength of a radio signal having a frequency of 60 kilohertz?
• The number of kilograms of water [latex](w)[/latex] in a human body is proportional to the mass of the body [latex](m).[/latex] If a 96 kg person contains 64 kg of water, how many kilograms of water are in a 60 kg person?
• The time [latex](t)[/latex] required to drive a fixed distance [latex](d)[/latex] varies inversely as the speed [latex](v).[/latex] If it takes 5 hours at a speed of 80 km/h to drive a fixed distance, what speed is required to do the same trip in 4.2 hours?
• The volume [latex](V)[/latex] of a cone varies jointly as its height [latex](h)[/latex] and the square of its radius [latex](r).[/latex] If a cone with a height of 8 centimetres and a radius of 2 centimetres has a volume of 33.5 cm 3 , what is the volume of a cone with a height of 6 centimetres and a radius of 4 centimetres?
• The centripetal force [latex](F_{\text{c}})[/latex] acting on an object varies as the square of the speed [latex](v)[/latex] and inversely to the radius [latex](r)[/latex] of its path. If the centripetal force is 100 N when the object is travelling at 10 m/s in a path or radius of 0.5 m, what is the centripetal force when the object’s speed increases to 25 m/s and the path is now 1.0 m?
• The maximum load [latex](L_{\text{max}})[/latex] that a cylindrical column with a circular cross section can hold varies directly as the fourth power of the diameter [latex](d)[/latex] and inversely as the square of the height [latex](h).[/latex] If an 8.0 m column that is 2.0 m in diameter will support 64 tonnes, how many tonnes can be supported by a column 12.0 m high and 3.0 m in diameter?
• The volume [latex](V)[/latex] of gas varies directly as the temperature [latex](T)[/latex] and inversely as the pressure [latex](P).[/latex] If the volume is 225 cc when the temperature is 300 K and the pressure is 100 N/cm 2 , what is the volume when the temperature drops to 270 K and the pressure is 150 N/cm 2 ?
• The electrical resistance [latex](R)[/latex] of a wire varies directly as its length [latex](l)[/latex] and inversely as the square of its diameter [latex](d).[/latex] A wire with a length of 5.0 m and a diameter of 0.25 cm has a resistance of 20 Ω. Find the electrical resistance in a 10.0 m long wire having twice the diameter.
• The volume of wood in a tree [latex](V)[/latex] varies directly as the height [latex](h)[/latex] and the diameter [latex](d).[/latex] If the volume of a tree is 377 m 3 when the height is 30 m and the diameter is 2.0 m, what is the height of a tree having a volume of 225 m 3 and a diameter of 1.75 m?

Answer Key 2.7

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Module 10: Rational and Radical Functions

Inverse and joint variation, learning outcomes.

• Solve an Inverse variation problem.
• Write a formula for an inversely proportional relationship.

Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula [latex]T=\frac{14,000}{d}[/latex] 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 a table we observe that, as the depth increases, the water temperature decreases.

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 graph 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 [latex]y=\dfrac{k}{x}[/latex] for inverse variation in this case uses [latex]k=14,000[/latex].

A General Note: Inverse Variation

If [latex]x[/latex] and [latex]y[/latex] are related by an equation of the form

[latex]y=\dfrac{k}{{x}^{n}}[/latex]

where [latex]k[/latex] is a nonzero constant, then we say that [latex]y[/latex]  varies inversely with the [latex]n[/latex]th power of [latex]x[/latex]. In inversely proportional relationships, or inverse variations , there is a constant multiple [latex]k={x}^{n}y[/latex].

isolating the constant of variation

To isolate the constant of variation in an inverse variation, use the properties of equality to solve the equation for [latex]k[/latex].

Isolate [latex]k[/latex] using algebra.

[latex]yx^n=k[/latex]

Example: Writing a Formula for an Inversely Proportional Relationship

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.

Recall that multiplying speed by time gives distance. If we let [latex]t[/latex] represent the drive time in hours, and [latex]v[/latex] represent the velocity (speed or rate) at which the tourist drives, then [latex]vt=[/latex] distance. Because the distance is fixed at 100 miles, [latex]vt=100[/latex]. Solving this relationship for the time gives us our function.

[latex]\begin{align}t\left(v\right)&=\dfrac{100}{v} \\[1mm] &=100{v}^{-1} \end{align}[/latex]

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.

How To: Given a description of an inverse variation problem, solve for an unknown.

• Identify the input, [latex]x[/latex], and the output, [latex]y[/latex].
• Determine the constant of variation. You may need to multiply [latex]y[/latex] by the specified power of [latex]x[/latex] 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: Solving an Inverse Variation Problem

A quantity [latex]y[/latex] varies inversely with the cube of [latex]x[/latex]. If [latex]y=25[/latex] when [latex]x=2[/latex], find [latex]y[/latex] when [latex]x[/latex] is 6.

The general formula for inverse variation with a cube is [latex]y=\dfrac{k}{{x}^{3}}[/latex]. The constant can be found by multiplying [latex]y[/latex] by the cube of [latex]x[/latex].

[latex]\begin{align}k&={x}^{3}y \\[1mm] &={2}^{3}\cdot 25 \\[1mm] &=200 \end{align}[/latex]

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

[latex]\begin{align}y&=\dfrac{k}{{x}^{3}},\hspace{2mm}k=200 \\[1mm] y&=\dfrac{200}{{x}^{3}} \end{align}[/latex]

Substitute [latex]x=6[/latex] and solve for [latex]y[/latex].

[latex]\begin{align}y&=\dfrac{200}{{6}^{3}} \\[1mm] &=\dfrac{25}{27} \end{align}[/latex]

Analysis of the Solution

The graph of this equation is a rational function.

A quantity [latex]y[/latex] varies inversely with the square of [latex]x[/latex]. If [latex]y=8[/latex] when [latex]x=3[/latex], find [latex]y[/latex] when [latex]x[/latex] is 4.

[latex]\dfrac{9}{2}[/latex]

https://ohm.lumenlearning.com/multiembedq.php?id=91393&theme=oea&iframe_resize_id=mom1

The following video presents a short lesson on inverse variation and includes more worked examples.

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 [latex]c[/latex], cost, varies jointly with the number of students, [latex]n[/latex], and the distance, [latex]d[/latex].

A General Note: Joint Variation

Joint variation occurs when a variable varies directly or inversely with multiple variables.

For instance, if [latex]x[/latex] varies directly with both [latex]y[/latex] and [latex]z[/latex], we have [latex]x=kyz[/latex]. If [latex]x[/latex] varies directly with [latex]y[/latex] and inversely with [latex]z[/latex], we have [latex]x=\dfrac{ky}{z}[/latex]. Notice that we only use one constant in a joint variation equation.

To isolate the constant of variation in a joint variation, use the properties of equality to solve the equation for [latex]k[/latex].

[latex]x=kyz[/latex]

[latex]\dfrac{x}{yz}=k[/latex]

Example: Solving Problems Involving Joint Variation

A quantity [latex]x[/latex] varies directly with the square of [latex]y[/latex] and inversely with the cube root of [latex]z[/latex]. If [latex]x=6[/latex] when [latex]y=2[/latex] and [latex]z=8[/latex], find [latex]x[/latex] when [latex]y=1[/latex] and [latex]z=27[/latex].

Begin by writing an equation to show the relationship between the variables.

[latex]x=\dfrac{k{y}^{2}}{\sqrt[3]{z}}[/latex]

Substitute [latex]x=6[/latex], [latex]y=2[/latex], and [latex]z=8[/latex] to find the value of the constant [latex]k[/latex].

[latex]\begin{align}6&=\dfrac{k{2}^{2}}{\sqrt[3]{8}} \\[1mm] 6&=\dfrac{4k}{2} \\[1mm] 3&=k \end{align}[/latex]

Now we can substitute the value of the constant into the equation for the relationship.

[latex]x=\dfrac{3{y}^{2}}{\sqrt[3]{z}}[/latex]

To find [latex]x[/latex] when [latex]y=1[/latex] and [latex]z=27[/latex], we will substitute values for [latex]y[/latex] and [latex]z[/latex] into our equation.

[latex]\begin{align}x&=\dfrac{3{\left(1\right)}^{2}}{\sqrt[3]{27}} \\[1mm] &=1 \end{align}[/latex]

[latex]x[/latex] varies directly with the square of [latex]y[/latex] and inversely with [latex]z[/latex]. If [latex]x=40[/latex] when [latex]y=4[/latex] and [latex]z=2[/latex], find [latex]x[/latex] when [latex]y=10[/latex] and [latex]z=25[/latex].

[latex]x=20[/latex]

https://ohm.lumenlearning.com/multiembedq.php?id=91394&theme=oea&iframe_resize_id=mom1

The following video provides another worked example of a joint variation problem.

• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• Question ID 91393,91394. Authored by : Jenck,Michael (for Lumen Learning). License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
• College Algebra. Authored by : Abramson, Jay et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
• Inverse Variation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/awp2vxqd-l4 . License : CC BY: Attribution
• Joint Variation: Determine the Variation Constant (Volume of a Cone). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Provided by : Joint Variation: Determine the Variation Constant (Volume of a Cone). Located at : https://youtu.be/JREPATMScbM . License : CC BY: Attribution

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Inverse Variation

Inverse variation is a type of proportionality where one quantity decreases while the other increases or vice versa. This implies that the magnitude or the absolute value of one quantity decreases if the other quantity increases such that their product will always remain the same. This product is also known as the constant of proportionality.

Inverse variation represents an inverse relationship between two quantities. There are several real-life applications of inverse variation. For example, if the speed of a car increases, the time taken to reach the destination decreases. In this article, we will elaborate on inverse variation, its formula, graph, and various examples.

What is Inverse Variation?

Inverse variation establishes a proportionality between two quantities that follow an inverse relationship. There are two types of proportionalities. These are direct variation and inverse variation. Two quantities are said to be directly proportional to each other if an increase or decrease in one quantity leads to a corresponding increase or decrease in the other quantity. This is in contrast to an inverse variation where one quantity increases with a decrease in the other.

Inverse Variation Definition

Two non-zero quantities are said to be in an inverse variation if their product yields a constant term ( constant of proportionality ). In other words, two quantities follow inverse variation if one quantity is directly proportional to the reciprocal of the other quantity. This means that an increase in one quantity leads to a decrease in the other while a decrease in one quantity leads to an increase in the other.

Inverse Variation Example

Suppose x and y are in inverse variation. It is given that x = 10 and the constant of proportionality equals 50. Then the value of y will be 50 / 10 = 5.

Inverse Variation Formula

The symbol "\(\propto\)" is used to indicate proportionality. If two quantities x and y follow an inverse variation then they are represented as follows:

x \(\propto\) \(\frac{1}{y}\) or y \(\propto\) \(\frac{1}{x}\)

To convert this expression into an equation , a constant or coefficient of proportionality needs to be introduced. Thus, the formula for inverse variation is given as follows:

x = \(\frac{k}{y}\) or y = \(\frac{k}{x}\)

Here, k is the constant of proportionality. Also, x \(\neq\) 0 and y \(\neq\) 0

Product Rule for Inverse Variation

Suppose the two solutions of inverse variation are (\(x_{1}\), \(y_{1}\)) and (\(x_{2}\), \(y_{2}\)).

This can also be expressed as \(x_{1}\) \(y_{1}\) = k and \(x_{2}\) \(y_{2}\) = k.

Using these two equations,

\(x_{1}\) \(y_{1}\) = \(x_{2}\) \(y_{2}\) or \(\frac{x_{1}}{x_{2}}\) = \(\frac{y_{2}}{y_{1}}\)

This is the product rule for inverse variation.

Let us understand this inverse variation formula with the help of an example.

Example: Suppose x and y are in an inverse proportion such that, when x = 15, then y = 4. Find the value of y when x = 20.

Given: \(x_{1}\) = 15, \(y_{1}\) = 4, \(x_{2}\) = 20, \(y_{2}\) = ?

Using inverse variation formula,

\(x_{1}\)\(y_{1}\) = \(x_{2}\) \(y_{2}\)

⇒ 15 × 4 = 20 × \(y_{2}\)

⇒ 60 = 20 × \(y_{2}\)

⇒ \(y_{2}\) = 3

Answer : Thus, the value of y is 3, when x is 20.

Inverse Variation Graph

The graph of an inverse variation is a rectangular hyperbola . If there are two quantities x and y that are in inverse variation then their product will be equal to a constant k. As neither x nor y can be equal to zero thus, the graph never crosses the x-axis or y-axis . The graph of an inverse variation with the function y = k / x is given below:

Inverse Variation Table

An inverse variation table is used to examine how one quantity changes with a change in another quantity that follow an inverse proportionality. Suppose two quantities x and y are in an inverse variation given by y = 18 / x. Then the inverse variation table can be given as follows:

Related Articles:

• Inverse proportion
• Ratio and Proportion
• Percent Proportion

Important Notes on Inverse Variation

• If one quantity decreases while another increases and vice versa, then both quantities are said to follow an inverse variation.
• An inverse variation is represented as x \(\propto\) \(\frac{1}{y}\) or x = \(\frac{k}{y}\).
• Both quantities that follow an inverse variation should not be equal to 0 and their product is equal to a constant (constant of proportionality).
• Two quantities following an inverse variation will result in a rectangular hyperbola in the cartesian coordinate plane .

Examples on Inverse Variation

• Example 1: Find the constant of proportionality if x = 12 and y = 4 follow an inverse variation. Solution: As x and y are in inverse variation thus, xy = k (12)(4) = 48 = k Answer: Constant of proportionality k = 48
• Example 2: If 12 workers complete a job in 5 hours, how much time will 10 workers take to complete the same job? Solution: Let x be the number of workers and y be the time taken to complete the job. The time taken to finish the work will increase if the number of workers decreases. Thus, this is a case of inverse variation. Using the product rule, \(\frac{x_{1}}{x_{2}}\) = \(\frac{y_{2}}{y_{1}}\) \(\frac{12}{10}\) = \(\frac{y}{5}\) y = 6 hours Answer: 10 workers will take 6 hours to complete the job.
• Example 3: If the constant of proportionality equals 7 and y = 14, what will be the value of x when x and y are in inverse variation? Solution: As x and y are in inverse variation thus, xy = k x = k / y x = 7 / 14 = 0.5 Answer: x = 1 / 2 or 0.5

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Practice Questions on Inverse Variation

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FAQs on Inverse Variation

What does inverse variation mean.

Inverse variation refers to the relationship between two quantities wherein one increases while the other correspondingly decreases or vice versa. Both are non-zero quantities and their product is the constant of proportionality.

What is the Formula for Inverse Variation?

The formula for inverse variation is xy = k. The product rule of inverse variation is given by \(\frac{x_{1}}{x_{2}}\) = \(\frac{y_{2}}{y_{1}}\).

What is an Example of Inverse Variation?

Suppose a certain number of pipes are used to fill a tank. If the number of pipes is increased then the time taken to fill the tank will reduce. Thus, the number of pipes and the time taken to fill the tank are an example of inverse variation.

How to Graph Inverse Variation?

The graph of an inverse variation is a rectangular hyperbola . In the given inverse variation function of the form y = k / x, substitute values of x to get the corresponding values of y. The test points so obtained can be plotted on a cartesian plane to get the graph of an inverse variation.

How to Find the Constant of Inverse Variation?

By multiplying the two quantities in inverse variation, the constant of proportionality can be obtained. This is expressed as xy = k.

Is xy = 12 an Inverse Variation?

Yes, xy = 12 is an inverse variation. This can be given as x = 12 / y. If the constant 12 is removed then x \(\propto\) \(\frac{1}{y}\). This implies that when x increases y decreases and vice versa.

What is the Difference Between Direct and Inverse Variation?

When one quantity increases with an increase in another quantity it is known as direct variation. This is expressed as y \(\propto\) x or y / x = k. In an inverse variation, one quantity increases while the other decreases. This is represented as x \(\propto\) \(\frac{1}{y}\) or xy = k.

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1.8: Variation - Constructing and Solving Equations

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Solving Problems involving Direct, Inverse, and Joint variation

Certain relationships occur so frequently in applied situations that they are given special names. Variation equations show how one quantity changes in relation to other quantities. The relationship between the quantities can be described as direct , inverse , or joint variation .

Direct Variation

Many real-world problems encountered in the sciences involve two types of functional relationships. The first type of functional relationship can be explored using the fact that the distance \(s\) in feet an object falls from rest, without regard to air resistance, can be approximated using the following formula:

\(s=16t^{2}\)

Here \(t\) represents the time in seconds the object has been falling. For example, after \(2\) seconds the object will have fallen \(s = 16 ( 2 ) ^ { 2 } = 16 \cdot 4 = 64\) feet.

In this example, we can see that the distance varies over time as the product of a constant \(16\) and the square of the time \(t\). This relationship is described as direct variation  and \(16\) is called the constant of variation or the  constant of proportionality . 

Definition: Direct Variation (\(y=kx\))

Direct variation is a relationship where quantities behave in a like manner. If one increases, so does the other. If one decreases, so does the other.

For two quantities \(x\) and \(y\), this relationship is described as "\(y\) varies directly as \(x\)" or "\(y\) is directly proportional to \(x\)" .

The equation that describes this relationship is \(y=kx\) , where \(k\) is a non-zero constant called the constant of variation or the proportionality constant .

• Translate the given English statement containing the words varies  or proportional , into a model equation.
• Substitute a given set of values into the equation and solve for \(k\), the constant of variation.
• Rewrite the equation obtained in step 1 as a formula with a value for \(k\) found in step 2 defined. Make note of the units used for each variable in the formula.
• Use the equation from step 3, and another set of values (with one value missing) to solve for the unknown quantity.

Example \(\PageIndex{1}\): Direct Variation

An object’s weight on Earth varies directly to its weight on the Moon. If a man weighs \(180\) pounds on Earth, then he will weigh \(30\) pounds on the Moon. Set up an algebraic equation that expresses the weight on Earth in terms of the weight on the Moon and use it to determine the weight of a woman on the Moon if she weighs \(120\) pounds on Earth.

Step 1. Translate “the weight on Earth varies directly to the weight on the Moon.”    \(E = kM \) 

Step 2. Find \(k\) using "If a man weighs \(180\) pounds on Earth, then he will weigh \(30\) pounds on the Moon."    \(E=180\) pounds, \(M=30\) pounds

\(\begin{array} { cr } E = kM & \text{Model equation} \\180=k \cdot 30\\ \frac { 180 } { 30 } = k  \\ { 6 = k } \end{array}\)

Step 3. The formula is \(E = 6M \), where \(E\) is the weight on Earth in pounds and \(M\) is the weight on the moon in pounds.

Step 4. Answer the question: "determine the weight of a woman on the Moon if she weighs \(120\) pounds on Earth."    \(E=120\) pounds, find M

\(\begin{array} { cll } E = 6M & \text{Formula:} & \text{ \(E\) pounds on Earth}\\  && \text{ \(M\) pounds on the Moon}\\{ 120 = 6 M } \\ { \frac { 120 } { 6 } = M } \\ { 20 = M } \end{array}\)

The woman weighs \(20\) pounds on the Moon.

Indirect Variation

The second functional relationship can be explored using the model that relates the intensity of light \(I\) to the square of the distance from its source \(d\).

\(I = \frac { k } { d ^ { 2 } }\)

Here \(k\) represents some constant. A foot-candle is a measurement of the intensity of light. One foot-candle is defined to be equal to the amount of illumination produced by a standard candle measured one foot away. For example, a \(125\)-Watt fluorescent growing light is advertised to produce \(525\) foot-candles of illumination. This means that at a distance \(d=1\) foot, \(I=525\) foot-candles and we have:

\(\begin{array} { l } { 525 = \frac { k } { ( 1 ) ^ { 2 } } } \\ { 525 = k } \end{array}\)

Using \(k=525\) we can construct a formula which gives the light intensity produced by the bulb:

\(I = \frac { 525 } { d ^ { 2 } }\)

Here \(d\) represents the distance the growing light is from the plants. In the chart above, we can see that the amount of illumination fades quickly as the distance from the plants increases.

This type of relationship is described as an inverse variation . We say that I is inversely proportional  to the square of the distance \(d\), where \(525\) is the constant of proportionality.

Definition: Indirect Variation (\(y=\frac{k}{x}\))

Indirect variation is a relationship between quantities where if one increases, the other decreases.

For two quantities \(x\) and \(y\), this relationship is described as "\(y\) varies indirectly as \(x\)" or "\(y\) is inversely proportional to \(x\)" .

The equation that describes this relationship is \(y=\dfrac{k}{x}\) , where \(k\) is a non-zero constant called the constant of variation or the proportionality constant .

Example \(\PageIndex{2}\): Indirect Variation

The weight of an object varies inversely as the square of its distance from the center of Earth. If an object weighs \(100\) pounds on the surface of Earth (approximately \(4,000\) miles from the center), how much will it weigh at \(1,000\) miles above Earth’s surface?

Step 1. Translate  “\(w\) varies inversely as the square of \(d\)”    \(w = \frac { k } { d ^ { 2 } }\)

Step 2. Find \(k\) using "An object weighs \(100\) pounds on the surface of Earth, approximately \(4,000\) miles from the center".     \(w = 100\) when \(d = 4,000\)

\(\begin{aligned} \color{Cerulean}{( 4,000 ) ^ { 2 }}\color{black}{ \cdot} 100 & =\color{Cerulean}{ ( 4,000 ) ^ { 2 }}\color{black}{ \cdot} \frac { k } { ( 4,000 ) ^ { 2 } } \\ 1,600,000,000 &= k \\ 1.6 \times 10 ^ { 9 } &= k \end{aligned}\)

Step 3. The formula is \(w = \frac { 1.6 \times 10 ^ { 9 } } { d ^ { 2 } }\), where \(w\) is the weight of the object in pounds and \(d\) is the distance of the object from the center of the Earth in miles.

Step 4. Answer the question: "how much will it weigh at \(1,000\) miles above Earth’s surface?"     Since the object is \(1,000\) miles above the surface, the distance of the object from the center of Earth is  \(d = 4,000 + 1,000 = 5,000 \:\:\text{miles}\)

\(\begin{aligned} y & = \frac { 1.6 \times 10 ^ { 9 } } { ( \color{OliveGreen}{5,000}\color{black}{ )} ^ { 2 } } \\ & = \frac { 1.6 \times 10 ^ { 9 } } { 25,000,000 } \\ & = \frac { 1.6 \times 10 ^ { 9 } } { 2.5 \times 10 ^ { 9 } } \\ & = 0.64 \times 10 ^ { 2 } \\ & = 64 \end{aligned}\)

The object will weigh \(64\) pounds at a distance \(1,000\) miles above the surface of Earth.

Joint Variation

Lastly, we define relationships between multiple variables.

Definition: Joint Variation and Combined Variation

Joint variation is a relationship in which one quantity is proportional to the product of two or more quantities.

Combined variation exists when combinations of direct and/or inverse variation occurs

Example \(\PageIndex{3}\): Joint Variation

Step 1. If we let \(A\) represent the area of an ellipse, then we can use the statement “area varies jointly as \(a\) and \(b\)” to write

Step 2. To find the constant of variation \(k\), use the fact that the area is \(300π\) when \(a=10\) and \(b=30\).

\(\begin{array} { c } { 300 \pi = k ( \color{OliveGreen}{10}\color{black}{ )} (\color{OliveGreen}{ 30}\color{black}{ )} } \\ { 300 \pi = 300 k } \\ { \pi = k } \end{array}\)

Step 3. Therefore, the formula for the area of an ellipse is

\(A=πab\)

The constant of proportionality is \(π\) and the formula for the area of an ellipse is \(A=abπ\).

Given that \(y\) varies directly as the square of \(x\) and inversely with \(z\), where \(y=2\) when \(x=3\) and \(z=27\), find \(y\) when \(x=2\) and \(z=16\).

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Inverse Variation

In Maths, inverse variation is the relationships between variables that are represented in the form of y = k/x, where x and y are two variables and k is the constant value. It states if the value of one quantity increases, then the value of the other quantity decreases.

In our day-to-day life, we observe that the variation in values of some quantity depends upon the variation in values of some other quantity. Inverse variation means that a variable is inversely varying with respect to another variable. It represents the inverse relationship between two quantities. Hence, a variable is inversely proportional to another variable.

There are many real-life examples of inverse variation, that can be seen in our day to day life. For example:

• If the distance travelled by train at constant speed increases then the time taken by it increases too and vice versa
• If the number of people is added to a job, the time taken to accomplish the job decreases

Inverse Variation Definition

Unlike the direct variation, where one quantity varies directly as per changes in another quantity, in the case of inverse variation, the first quantity varies inversely as per another quantity. Hence, the first quantity is inversely proportional to the other.

Let us say, x and y are two variables, then as per inverse variation:

An example of inverse variation, y = 3/x, is represented graphically in the given figure.

Inverse Variation Formula

As per the inverse variation formula, if any variable x is inversely proportional to another variable y, then the variables x and y are represented by the formula:

xy = k or y=k/x

where k is any constant value.

Inverse Variation Equation

Sometimes, we observe that the variation in values of one quantity is just opposite to the variation in the values of another quantity. If the value of one quantity increases, the value of the other quantity decreases in the same proportion and vice versa. This is termed inverse variation and the two quantities are said to be inversely proportional to each other. Two quantities existing in inverse variation can be expressed as,

\(\begin{array}{l}x~∝~\frac{1}{y}\end{array} \)

\(\begin{array}{l}\Rightarrow~ xy~=~k\end{array} \)

Where x and y are the value of two quantities and k is a constant known as the constant of proportionality. If x 1 , y 1 are initial values and x 2 , y 2 are final values of quantities existing in inverse variation. They can be expressed as:

Relation Between Two Quantities in Inverse Variation

In an inverse variation, the relationship between two quantities or variables is defined by inverse proportion. If x and y are two quantities then, the inverse relationship between them is given by:

See the examples given below to find the relationship between two quantities in inverse variation.

Inverse Variation Table

The inverse variation table can be written by placing the values of one quantity and finding the other.

For example, y = 10/x

Let us put some values for x and find the value of y.

Related Articles

• Direct Proportion
• Direct And Inverse Proportion
• Ratios And Proportion
• Basic Proportionality Theorem

Inverse Variation Solved Examples

Q.1: If x varies inversely with y, and x = 10 and y = 4, then what is the value of constant of variation?

Solution: Given, x varies inversely with y.

Let the constant of variation be k.

But x = 10 and y = 4 (given)

k = 10 x 4 = 40

Hence, the constant of variation is 40.

Q.2: If x and y are in an inverse variation and k/3 = 2, and k/x = 3, then find the value of x when y = 2.

Solution: Given, k/3 = 2

then, k = 3 x 2 = 6

So, x = k/3 = 6/3 = 2

Hence, if y = 2, then ;

x = k/y = 6/2 = 3

Inverse Variation Word Problems

Example 1: 9 pipes are required to fill a tank in 4 hours. How long will it take if 12 pipes of the same type are used?

Solution: Let, the desired time to fill the tank be \(\begin{array}{l}x\end{array} \) minutes. We know that as the number of pipes increases, the time taken to fill the tank will decrease. Hence, this is a case of inverse variation. In other words, the number of pipes is inversely proportional to the time taken. Thus,

\(\begin{array}{l}\frac{x_1}{x_2}~=~\frac{y_2}{y_1}\end{array} \)

\(\begin{array}{l}\Rightarrow~\frac{9}{x}~=~\frac{12}{4}\end{array} \)

\(\begin{array}{l}x~=~3~hours\end{array} \)

Example 2: If 24 workers can build a house in 40 days, how many workers will be required to build the same house in 20 days?

Solution: Let the number of workers employed to build the wall in 20 days be \(\begin{array}{l}x\end{array} \) . We know that time taken to build the house is inversely proportional to the number of workers required. Thus,

\(\begin{array}{l}\Rightarrow~\frac{24}{x}~=~\frac{20}{40}\end{array} \)

\(\begin{array}{l}\Rightarrow~x~=~48~days\end{array} \)

Example 3: There are 100 students in a hostel. Food provision for them is for 20 days. After 8 days 20 new students entered the hostel. How long will this provision last now?

Solution: Suppose the provision lasts for x more days after the intrusion of 20 new students which would have lasted for 12 days if the number of students was 100. We know that the time till which food lasts is inversely proportional to the number of students in the hostel. Thus,

\(\begin{array}{l}\Rightarrow~\frac{100}{120}~=~\frac{x}{12}\end{array} \)

\(\begin{array}{l}\Rightarrow~x~=~10~days\end{array} \)

Total number of days for which provision lasts = \(\begin{array}{l}10~+~8\end{array} \) = \(\begin{array}{l}18~days\end{array} \)

Frequently Asked Questions on Inverse Variation

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