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# 4.6: Direct Variation Models

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Identify direct variation.
• Graph direct variation equations.
• Solve real-world problems using direct variation models.

Suppose you see someone buy five pounds of strawberries at the grocery store. The clerk weighs the strawberries and charges $12.50 for them. Now suppose you wanted two pounds of strawberries for yourself. How much would you expect to pay for them? ## Identify Direct Variation The preceding problem is an example of a direct variation. We would expect that the strawberries are priced on a “per pound” basis, and that if you buy two-fifths of the amount of strawberries, you would pay two-fifths of$12.50 for your strawberries.

25×$12.50=$5.00\begin{align*}\frac{2}{5}\times \ 12.50 = \ 5.00\end{align*}

Similarly, if you bought 10 pounds of strawberries (twice the amount) you would pay 2×12.50\begin{align*}2 \times \ 12.50\end{align*} and if you did not buy any strawberries you would pay nothing. If variable y\begin{align*}y\end{align*} varies directly with variable x\begin{align*}x\end{align*}, then we write the relationship as: y=kx\begin{align*}y=k\cdot x\end{align*} k\begin{align*}k\end{align*} is called the constant of proportionality. If we were to graph this function you can see that it passes through the origin, because y=0\begin{align*}y = 0\end{align*}, when x=0\begin{align*}x = 0\end{align*} whatever the value of k\begin{align*}k\end{align*}. So we know that a direct variation, when graphed, has a single intercept at (0, 0). Example 1 If y\begin{align*}y\end{align*} varies directly with x\begin{align*}x\end{align*} according to the relationship y=kx\begin{align*}y = k\cdot x\end{align*}, and y=7.5\begin{align*}y = 7.5\end{align*} when x=2.5\begin{align*}x = 2.5\end{align*}, determine the constant of proportionality, k\begin{align*}k\end{align*}. We can solve for the constant of proportionality using substitution. Substitute x=2.5\begin{align*}x = 2.5\end{align*} and y=7.5\begin{align*}y = 7.5\end{align*} into the equation y=kx\begin{align*}y = k\cdot x\end{align*} 7.57.52.5=k(2.5)=k=3Divide both sides by 2.5.\begin{align*} 7.5 & = k(2.5) && \text{Divide both sides by 2.5.} \\ \frac{7.5} {2.5} & = k = 3 \end{align*} Solution The constant of proportionality, k=3\begin{align*}k = 3\end{align*}. We can graph the relationship quickly, using the intercept (0, 0) and the point (2.5, 7.5). The graph is shown right. It is a straight line with a slope = 3. The graph of a direct variation has a slope that is equal to the constant of proportionality, k\begin{align*}k\end{align*}. Example 2 The volume of water in a fish-tank, V\begin{align*}V\end{align*}, varies directly with depth, d\begin{align*}d\end{align*}. If there are 15 gallons in the tank when the depth is eight inches, calculate how much water is in the tank when the depth is 20 inches. This is a good example of a direct variation, but for this problem we will need to determine the equation of the variation ourselves. Since the volume, V\begin{align*}V\end{align*}, depends on depth, d\begin{align*}d\end{align*}, we will use the previous equation to create new one that is better suited to the content of the new problem. yV=kx=kdIn place of y we will use V and in place of x we will use d.\begin{align*}y & = k\cdot x && \text{In place of}\ y\ \text{we will use}\ V\ \text{and in place of}\ x\ \text{we will use} \ d.\\ V & = k\cdot d\end{align*} We know that when the depth is 8 inches, the volume is 15 gallons. Now we can substitute those values into our equation. Substitute V=15\begin{align*}V = 15\end{align*} and x=8\begin{align*}x = 8\end{align*}: V15158=kd=k(8)=k=1.875Divide both sides by 8.\begin{align*}V & = k\cdot d \\ 15 & = k(8) && \text{Divide both sides by} \ 8. \\ \frac{15} {8} & = k = 1.875 \end{align*} Now to find the volume of water at the final depth we use V=kd\begin{align*}V=k\cdot d\end{align*} and substitute for our new d\begin{align*}d\end{align*}. VVV=kd=1.875×20=37.5\begin{align*}V & = k\cdot d \\ V & = 1.875 \times 20 \\ V & = 37.5\end{align*} Solution At a depth of 20 inches, the volume of water in the tank is 37.5 gallons. Example 3 The graph shown to the right shows a conversion chart used to convert U.S. dollars (US) to British pounds (GB£) in a bank on a particular day. Use the chart to determine the following.

(i) The number of pounds you could buy for 600. (ii) The number of dollars it would cost to buy £200. (iii) The exchange rate in pounds per dollar. (iv) Is the function continuous or discrete? Solution In order to solve (i) and (ii) we could simply read off the graph: it looks as if at x=600\begin{align*}x=600\end{align*} the graph is about one fifth of the way between £350 and £400. So600 would buy £360. Similarly, the line y=200\begin{align*}y =200\end{align*} would appear to intersect the graph about a third of the way between $300 and$400. We would probably round this to $330. So it would cost approximately$330 to buy £200.

To solve for the exchange rate we should note that as this is a direct variation, because the graph is a straight line passing through the origin. The slope of the line gives the constant of proportionality (in this case the exchange rate) and it is equal to the ratio of the y\begin{align*}y-\end{align*}value to x\begin{align*}x-\end{align*}value. Looking closely at the graph, it is clear that there is one lattice point that the line passes through (500, 300). This will give us the most accurate estimate for the slope (exchange rate).

yrate=kxk=yx=300 pounds500 dollars=0.60 pounds per dollar\begin{align*} y & = k \cdot x \Rightarrow k = \frac{y} {x} \\ \text{rate} & = \frac{300\ pounds} {500\ dollars} = 0.60\ pounds \ per \ dollar\end{align*}

## Graph Direct Variation Equations

We know that all direct variation graphs pass through the origin, and also that the slope of the line is equal to the constant of proportionality, k\begin{align*}k\end{align*}. Graphing is a simple matter of using the point-slope or point-point methods discussed earlier in this chapter.

Example 4

Plot the following direct relations on the same graph.

a. y=3x\begin{align*}y = 3x\end{align*}

b. y=2x\begin{align*}y = -2x\end{align*}

c. y=0.2x\begin{align*}y= -0.2x\end{align*}

d. y=29x\begin{align*}y =\frac{2}{9}x\end{align*}

Solution

a. The line passes through (0, 0). All these functions will pass through this point. It is plotted in red. This function has a slope of 3. When we move across by one unit, the function increases by three units.

b. The line has a slope of -2. When we move across the graph by one unit the function falls by two units.

c. The line has a slope of -0.2. As a fraction this is equal to 15\begin{align*}-\frac{1}{5}\end{align*}. When we move across by five units, the function falls by one unit.

d. The line passes through (0, 0)and has a slope of 29\begin{align*}\frac{2}{9}\end{align*}. When we move across the graph by 9 units, the function increases by two units.

## Solve Real-World Problems Using Direct Variation Models

Direct variations are seen everywhere in everyday life. Any time that we have one quantity that doubles when another related quantity doubles, we say that they follow a direct variation.

Newton's Second Law

In 1687, Sir Isaac Newton published the famous Principea Mathematica. It contained, among other things, his Second Law of Motion. This law is often written as:

F=ma\begin{align*}F=m\cdot a\end{align*}

A force of F\begin{align*}F\end{align*} (Newtons) applied to a mass of m\begin{align*}m\end{align*} (kilograms) results in acceleration of a\begin{align*}a\end{align*} (meters per second2)\begin{align*}\text{(meters per second}^2)\end{align*}.

Example 5

If a 175 Newton force causes a heavily loaded shopping cart to accelerate down the aisle with an acceleration of 2.5 m/s2\begin{align*}2.5\ m/s^2\end{align*}, calculate

(i) The mass of the shopping cart.

(ii) The force needed to accelerate the same cart at \begin{align*}6 \ m/s^2\end{align*}.

Solution

(i) This question is basically asking us to solve for the constant of proportionality. Let us compare the two formulas.

\begin{align*}y & = k\cdot x && \text{The direct variation equation}\\ F & = m\cdot a && \text{Newton's Second law}\end{align*}

We see that the two equations have the same form; \begin{align*}y\end{align*} is analogous to force and \begin{align*}x\end{align*} analogous to acceleration.

We can solve for \begin{align*}m\end{align*} (the mass) by substituting our given values for force and acceleration:

Substitute \begin{align*}F=175, a=2.5\end{align*}

\begin{align*}175 & = m(2.5) && \text{Divide both sides by}\ 2.5.\\ 70 & = m\end{align*}

The mass of the shopping cart is 70 kg.

(ii) Once we have solved for the mass we simply substitute that value, plus our required acceleration back into the formula \begin{align*}F=m\cdot a\end{align*} and solve for \begin{align*}F\end{align*}:

Substitute \begin{align*}m=70\end{align*}, \begin{align*}a = 6\end{align*}

\begin{align*}F = 70 \times 6 = 420\end{align*}

The force needed to accelerate the cart at \begin{align*}6 \ m/s^2\end{align*} is 420 Newtons.

## Ohm's Law

The electrical current, \begin{align*}I\end{align*} (amps), passing through an electronic component varies directly with the applied voltage, \begin{align*}V\end{align*} (volts), according to the relationship:

\begin{align*}V = I\cdot R && \text{where} \ R \ \text{is the resistance (measured in Ohms)}\end{align*}

The resistance is considered to be a constant for all values of \begin{align*}V\end{align*} and \begin{align*}I\end{align*}.

Example 6

A certain electronics component was found to pass a current of 1.3 amps at a voltage of 2.6 volts. When the voltage was increased to 12.0 volts the current was found to be 6.0 amps.

a) Does the component obey Ohms law?

b) What would the current be at 6 volts?

Solution

a) Ohm's law is a simple direct proportionality law. Since the resistance \begin{align*}R\end{align*} is constant, it acts as our constant of proportionality. In order to know if the component obeys Ohm's law we need to know if it follows a direct proportionality rule. In other words is \begin{align*}V\end{align*} directly proportional to \begin{align*}I\end{align*}?

### Method One - Graph It

If we plot our two points on a graph and join them with a line, does the line pass through (0, 0)?

Point 1 \begin{align*}V = 2.6, I = 1.3\end{align*} our point is (1.3, 2.6)*

Point 2 \begin{align*}V = 12.0, I = 6.0\end{align*} our point is (6, 12)

Plotting the points and joining them gives the following graph.

The graph does appear to pass through the origin, so...

Yes, the component obeys Ohms law.

### Method Two - Solve for R

We can quickly determine the value of \begin{align*}R\end{align*} in each case. It is the ratio of the voltage to the resistance.

\begin{align*}\text{Case} \ 1 \ R & = \frac{V} {I} = \frac{2.6} {1.3} = 2 \ Ohms\\ \text{Case} \ 2 \ R & = \frac{V} {I} = \frac{12} {6} = 2 \ Ohms\end{align*}

The values for \begin{align*}R\end{align*} agree! This means that the line that joins point 1 to the origin is the same as the line that joins point 2 to the origin. The component obeys Ohms law.

b) To find the current at 6 volts, simply substitute the values for \begin{align*}V\end{align*} and \begin{align*}R\end{align*} into \begin{align*}V=I\cdot R\end{align*}

Substitute \begin{align*}V=6, R=2\end{align*}

• In physics, it is customary to plot voltage on the horizontal axis as this is most often the independent variable. In that situation, the slope gives the conductance, \begin{align*}\sigma \end{align*}. However, by plotting the current on the horizontal axis, the slope is equal to the resistance, \begin{align*}R\end{align*}.

\begin{align*}6 & = I(2) && \text{Divide both sides by}\ 2.\\ 3 & = I \end{align*}

Solution

The current through the component at a voltage of 6 volts is 3 amps.

## Lesson Summary

• If a variable \begin{align*}y\end{align*} varies directly with variable \begin{align*}x\end{align*}, then we write the relationship as

\begin{align*} y = k \cdot x\end{align*}

Where \begin{align*}k\end{align*} is a constant called the constant of proportionality.

• Direct variation is very common in many areas of science.

## Review Questions

1. Plot the following direct variations on the same graph.
1. \begin{align*}y=\frac{4}{3}x \end{align*}
2. \begin{align*}y=-\frac{2}{3}x \end{align*}
3. \begin{align*}y= -\frac{1}{6}x \end{align*}
4. \begin{align*}y = 1.75x\end{align*}
2. Dasan’s mom takes him to the video arcade for his birthday. In the first 10 minutes, he spends $3.50 playing games. If his allowance for the day is$20.00, how long can he keep playing games before his money is gone?
3. The current standard for low-flow showerheads heads is 2.5 gallons per minute. Calculate how long it would take to fill a 30 gallon bathtub using such a showerhead to supply the water.
4. Amen is using a hose to fill his new swimming pool for the first time. He starts the hose at 10 P.M. and leaves it running all night. At 6 AM he measures the depth and calculates that the pool is four sevenths full. At what time will his new pool be full?
5. Land in Wisconsin is for sale to property investors. A 232 acre lot is listed for sale for 200500. Assuming the same price per acre, how much would a 60 acre lot sell for? 6. The force \begin{align*}(F)\end{align*} needed to stretch a spring by a distance \begin{align*}x\end{align*} is given by the equation \begin{align*}F = k\cdot x\end{align*}, where \begin{align*}k\end{align*} is the spring constant (measured in Newtons per centimeter, (N/cm). If a 12 Newton force stretches a certain spring by 10 cm, calculate: 1. The spring constant, \begin{align*}k\end{align*} 2. The force needed to stretch the spring by 7 cm. 3. The distance the spring would stretch with a 23 Newton force. ## Review Answers 1. 57 minutes 8 seconds 2. 12 minutes 3. 12:00 Midday 4.51, 853
1. \begin{align*}k = 1.2 \ N/cm\end{align*}
2. 8.4 Newtons
3. 19.17 cm

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