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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 the amount of strawberries, you would pay two-fifths of$12.50 for your strawberries, or $5.00. Similarly, if you bought 10 pounds of strawberries (twice the amount) you would pay twice$12.50, 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 would pass 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*}.

Solution

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*} to get 7.5=k(2.5)\begin{align*}7.5 = k(2.5)\end{align*}. Then divide both sides by 2.5 to get k=7.52.5=3\begin{align*} k= \frac{7.5}{2.5} = 3\end{align*}. The constant of proportionality, k\begin{align*}k\end{align*}, is 3.

We can graph the relationship quickly, using the intercept (0, 0) and the point (2.5, 7.5). The graph is shown below. It is a straight line with slope 3.

The graph of a direct variation always passes through the origin, and always 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 8 inches, calculate how much water is in the tank when the depth is 20 inches.

Solution

This is a good example of a direct variation, but for this problem we’ll have 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’ll use an equation of the form y=kx\begin{align*}y = k \cdot x\end{align*}, but in place of y\begin{align*}y\end{align*} we’ll use V\begin{align*}V\end{align*} and in place of x\begin{align*}x\end{align*} we’ll use d\begin{align*}d\end{align*}:

V=kd\begin{align*}V = k \cdot d\end{align*}

We know that when the depth is 8 inches the volume is 15 gallons, so to solve for k\begin{align*}k\end{align*}, we plug in 15 for V\begin{align*}V\end{align*} and 8 for d\begin{align*}d\end{align*} to get 15=k(8)\begin{align*}15 = k(8)\end{align*}. Dividing by 8 gives us k=158=1.875\begin{align*}k = \frac{15}{8} = 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*} again, but this time we can plug in our new d\begin{align*}d\end{align*} and the value we found for k\begin{align*}k\end{align*}:

VV=1.875×20=37.5\begin{align*}V & = 1.875 \times 20\\ V & = 37.5\end{align*}

At a depth of 20 inches, the volume of water in the tank is 37.5 gallons.

Example 3

The graph shown below is 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: a) the number of pounds you could buy for$600

b) the number of dollars it would cost to buy £200

c) the exchange rate in pounds per dollar

Solution

We can read the answers to a) and b) right 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. So 600 would buy £360. Similarly, the line y=200\begin{align*}y = 200\end{align*} appears to intersect the graph about a third of the way between300 and $400. We can 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 - 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 at any point. Looking closely at the graph, we can see that the line passes through one convenient lattice point: (500, 300). This will give us the most accurate value for the slope and so the exchange rate. y=kx=yxAnd so rate=300 pounds500 dollors=0.60 pounds per dollar\begin{align*}y= k \cdot x \Rightarrow = \frac{y}{x} \qquad \text{And so rate} =\frac {300 \ \text{pounds}}{500 \ \text{dollors}} = 0.60 \ \text{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 variations 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), as will all these functions. 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 \begin{align*}\frac{2}{9}\end{align*}. When we move across the graph by 9 units, the function increases by two units. For more examples of how to plot and identify direct variation functions, see the video at http://neaportal.k12.ar.us/index.php/2010/06/slope-and-direct-variation/. ## Solve Real-World Problems Using Direct Variation Models Direct variations are seen everywhere in everyday life. Any time one quantity increases at the same rate another quantity increases (for example, doubling when it doubles and tripling when it triples), we say that they follow a direct variation. Newton's Second Law In 1687 Sir Isaac Newton published the famous Principia Mathematica. It contained, among other things, his second law of motion. This law is often written as \begin{align*}F = m \cdot a\end{align*}, where a force of \begin{align*}F\end{align*} Newtons applied to a mass of \begin{align*}m\end{align*} kilograms results in acceleration of \begin{align*}a\end{align*} meters per second\begin{align*}^2\end{align*}. Notice that if the mass stays constant, then this formula is basically the same as the direct variation equation, just with different variables—and \begin{align*}m\end{align*} is the constant of proportionality. Example 5 If a 175 Newton force causes a shopping cart to accelerate down the aisle with an acceleration of \begin{align*}2.5 \ m/s^2\end{align*}, calculate: a) The mass of the shopping cart. b) The force needed to accelerate the same cart at \begin{align*}6 \ m/s^2\end{align*}. Solution a) We can solve for \begin{align*}m\end{align*} (the mass) by plugging in our given values for force and acceleration. \begin{align*}F = m \cdot a\end{align*} becomes \begin{align*}175 = m(2.5)\end{align*}, and then we divide both sides by 2.5 to get \begin{align*}70 = m\end{align*}. So the mass of the shopping cart is 70 kg. b) 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*}. We get \begin{align*}F = 70 \times 6 = 420\end{align*}. So 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\end{align*}, where \begin{align*}R\end{align*} is the resistance (measured in Ohms). The resistance is considered to be a constant for all values of \begin{align*}V\end{align*} and \begin{align*}I\end{align*}, so once again, this formula is a version of the direct variation formula, with \begin{align*}R\end{align*} as the constant of proportionality. 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 Ohm’s law? b) What would the current be at 6 volts? Solution Ohm’s law is a simple direct proportionality law, with the resistance \begin{align*}R\end{align*} as our constant of proportionality. To know if this 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*}? We can determine this in two different ways. Graph It: If we plot our two points on a graph and join them with a line, does the line pass through (0, 0)? Voltage is the independent variable and current is the dependent variable, so normally we would graph \begin{align*}V\end{align*} on the horizontal axis and \begin{align*}I\end{align*} on the vertical axis. However, if we swap the variables around just this once, we’ll get a graph whose slope conveniently happens to be equal to the resistance, \begin{align*}R\end{align*}. So we’ll treat \begin{align*}I\end{align*} as the independent variable, and our two points will be (1.3, 2.6) and (6, 12). Plotting those points and joining them gives the following graph: The graph does appear to pass through the origin, so yes, the component obeys Ohm’s law. Solve for \begin{align*}R\end{align*}: If this component does obey Ohm’s law, the constant of proportionality \begin{align*}(R)\end{align*} should be the same when we plug in the second set of values as when we plug in the first set. Let’s see if it is. (We can quickly find the value of \begin{align*}R\end{align*} in each case; since \begin{align*}V = I \cdot R\end{align*}, that means \begin{align*} R = \frac{V}{I}\end{align*}.) \begin{align*}\text{Case 1:} \ R & = \frac{V}{I} = \frac{2.6}{1.3} = 2 \ \text{Ohms}\\ \text{Case 2:} \ R & = \frac{V}{I} = \frac{12}{6} = 2 \ \text{Ohms}\end{align*} The values for \begin{align*}R\end{align*} agree! This means that we are indeed looking at a direct variation. The component obeys Ohm’s law. b) Now 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*}. We found that \begin{align*}R = 2\end{align*}, so we plug in 2 for \begin{align*}R\end{align*} and 6 for \begin{align*}V\end{align*} to get \begin{align*}6 = I(2)\end{align*}, and divide both sides by 2 to get \begin{align*}3 = I\end{align*}. So 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. 1. In the first 10 minutes, he spends3.50 playing games. If his allowance for the day is $20, how long can he keep playing games before his money is gone? 2. He spends the next 15 minutes playing Alien Invaders. In the first two minutes, he shoots 130 aliens. If he keeps going at this rate, how many aliens will he shoot in fifteen minutes? 3. The high score on this machine is 120000 points. If each alien is worth 100 points, will Dasan beat the high score? What if he keeps playing for five more minutes? 3. The current standard for low-flow showerheads is 2.5 gallons per minute. 1. How long would it take to fill a 30-gallon bathtub using such a showerhead to supply the water? 2. If the bathtub drain were not plugged all the way, so that every minute 0.5 gallons ran out as 2.5 gallons ran in, how long would it take to fill the tub? 3. After the tub was full and the showerhead was turned off, how long would it take the tub to empty through the partly unplugged drain? 4. If the drain were immediately unplugged all the way when the showerhead was turned off, so that it drained at a rate of 1.5 gallons per minute, how long would it take to empty? 4. Amin is using a hose to fill his new swimming pool for the first time. He starts the hose at 10 PM and leaves it running all night. 1. 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? 2. At 10 AM he measures again and realizes his earlier calculations were wrong. The pool is still only three quarters full. When will it actually be full? 3. After filling the pool, he needs to chlorinate it to a level of 2.0 ppm (parts per million). He adds two gallons of chlorine solution and finds that the chlorine level is now 0.7 ppm. How many more gallons does he need to add? 4. If the chlorine level in the pool decreases by 0.05 ppm per day, how much solution will he need to add each week? 5. Land in Wisconsin is for sale to property investors. A 232-acre lot is listed for sale for$200,500.
1. Assuming the same price per acre, how much would a 60-acre lot sell for?
2. Again assuming the same price, what size lot could you purchase for 100,000? 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, or 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. 7. Angela’s cell phone is completely out of power when she puts it on the charger at 3 PM. An hour later, it is 30% charged. When will it be completely charged? 8. It costs100 to rent a recreation hall for three hours and \$150 to rent it for five hours.
1. Is this a direct variation?
2. Based on the cost to rent the hall for three hours, what would it cost to rent it for six hours, assuming it is a direct variation?
3. Based on the cost to rent the hall for five hours, what would it cost to rent it for six hours, assuming it is a direct variation?
4. Plot the costs given for three and five hours and graph the line through those points. Based on that graph, what would you expect the cost to be for a six-hour rental?

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