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4.6: Direct Variation

Difficulty Level: At Grade Created by: CK-12
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At the local farmer’s market, you saw someone purchase 5 pounds of strawberries and pay $12.50. You want to buy strawberries too, but you want only 2 pounds. How much would you expect to pay?

This situation is an example of a direct variation. You 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, or $5.00. Similarly, if you bought 10 pounds of strawberries (twice the amount), you would pay $25.00 (twice $12.50), and if you did not buy any strawberries you would pay nothing.

Direct Variation can be expressed as the equation \begin{align*}y=(k)x\end{align*}, where \begin{align*}k\end{align*} is called the constant of variation.

Direct variation occurs when:

  • The fraction \begin{align*}\frac{rise}{run}\end{align*} or \begin{align*}\frac{change \ in \ y}{change \ in \ x}\end{align*} is always the same, and
  • The ordered pair (0, 0) is a solution to the situation.

Example: If \begin{align*}y\end{align*} varies directly with \begin{align*}x\end{align*} according to the relationship \begin{align*}y=k \cdot x\end{align*}, and \begin{align*}y=7.5\end{align*} when \begin{align*}x=2.5\end{align*}, determine the constant of proportionality, \begin{align*}k\end{align*}.

Solution: We can solve for the constant of proportionality using substitution.

Substitute \begin{align*}x=2.5\end{align*} and \begin{align*}y=7.5\end{align*} into the equation \begin{align*}y=k \cdot x\end{align*}.

\begin{align*}7.5 & = k(2.5) && \text{Divide both sides by} \ 2.5.\\ \frac{7.5}{2.5}& =k=3\end{align*}

The constant of variation (or the constant of proportionality) is 3.

You can use this information to graph this direct variation situation. Remember that all direct variation situations cross the origin. You can plot the ordered pair (0, 0) and use the constant of variation as your slope.

Example: Explain why each of the following equations are not examples of direct variation.

\begin{align*}y& =\frac{2}{x}\\ y& =5x-1\\ 2x+y& =6\end{align*}

Solution: In equation 1, the variable is in the denominator of the fraction, violating the definition.

In equation 2, there is a \begin{align*}y-\end{align*}intercept of –1, violating the definition.

In equation 3, there is also a \begin{align*}y-\end{align*}intercept, violating the definition.

Translating a Sentence into a Direct Variation Equation

Direct variation equations use the same phrase to give the reader a clue. The phrase is either “directly proportional” or “varies directly.”

Example: The area of a square varies directly as the square of its side.

Solution: The first variable you encounter is “area.” Think of this as your \begin{align*}y\end{align*}. Think the phrase “varies directly” means \begin{align*}= (k)\times\end{align*}. The second variable is “square of its side.” Call this letter \begin{align*}s\end{align*}.

Now translate into an equation: \begin{align*}y=(k)\times s^2\end{align*}.

You’ve written your first direct variation equation.

Example 2: The distance you travel is directly proportional to the time you have been traveling. Write this situation as a direct variation equation.

Solution: The first variable is distance; call it \begin{align*}d\end{align*}. The second variable is time you have been traveling, call it \begin{align*}t\end{align*}. Apply the direct variation definition:

\begin{align*}d=(k) \times t\end{align*}

Solving Real-World Situations Using Direct Variation

Direct variation has numerous real-world examples. You have already seen three examples: the area of a square is directly proportional to its side length; the distance you travel varies directly as the time you have been driving; and the total cost is directly proportional to the number of pounds of strawberries you purchase.

Newton's Second Law

In 1687, Sir Isaac Newton published the famous Principea Mathematica. It contained his Second Law of Motion. This law is often written as: \begin{align*}F=m \cdot a\end{align*}, where \begin{align*}F=\end{align*} the amount of force applied to an object with mass \begin{align*}(m)\end{align*} and \begin{align*}a=\end{align*} acceleration.

Acceleration is given in the units meters/second\begin{align*}^2\end{align*} and force is given in units of Newtons.

Example: If a 175 Newton force causes a heavily loaded shopping cart to accelerate down the aisle with an acceleration of \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. The variable \begin{align*}y\end{align*} is equal to force and \begin{align*}x\end{align*} is equal to acceleration.

\begin{align*}175=m \cdot 2.5\end{align*}

Now solve for \begin{align*}m\end{align*}, the constant of variation.

\begin{align*}\frac{175}{2.5} & =\frac{m \cdot 2.5}{2.5}\\ m & = 70\end{align*}

(ii) Now you know the constant of variation is 70. In this formula, 70 represents the mass. To find the force needed to move the cart at an accelerated rate of 6 meters/second, substitute 6 for \begin{align*}a\end{align*} and evaluate the equation.

When \begin{align*}a=6\end{align*}, \begin{align*}F=70 \cdot 6=420\end{align*}.

The force needed to accelerate the cart is 420 Newtons.

Solving Direct Variation Using Proportions

You can use the Cross Products Theorem of proportions to solve direct variation situations. Because the fraction \begin{align*}\frac{rise}{run}\end{align*} is constant in a direct variation situation, you can create a proportion.

Ohm’s Law states that the voltage \begin{align*}(V)\end{align*} is equal to the electrical current \begin{align*}(I)\end{align*} in amps times the resistance \begin{align*}(R)\end{align*} in ohms. Translating this to an equation, \begin{align*}V=I \cdot R\end{align*}.

Suppose an electronics device passed a current of 1.3 amps at a voltage of 2.6 volts. What was the current when the voltage was increased to 12 volts?

Ohm’s Law matches with the definition of direct variation, so you can write a proportion. \begin{align*}\frac{2.6 \ volts}{1.3 \ amps}=\frac{12 \ volts}{I \ amps}\end{align*}. Using the Cross Products Theorem, solve for \begin{align*}I\end{align*}. \begin{align*}I=6\end{align*}. Therefore, when the voltage was increased to 12 volts, the electronics device passed a current of 6 amps.

Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both. CK-12 Basic Algebra: Direct Variation Models (11:11)

  1. Describe direct variation.
  2. What is the formula for direct variation? What does \begin{align*}k\end{align*} represent?
  3. What two methods can be used to solve a direct variation situation?
  4. True or false? Every linear equation is a direct variation situation.

Translate the following direct variation situations into equations. Choose appropriate letters to represent the varying quantities.

  1. The amount of money you earn is directly proportional to the number of hours you work.
  2. The weight of an object on the Moon varies directly with its weight on Earth.
  3. The volume of a gas is directly proportional to its temperature in Kelvin.
  4. The number of people served varies directly with the amount of ground meat used to make burgers.
  5. The amount of a purchase varies directly with the number of pounds of peaches.

Explain why each equation is not an example of direct variation.

  1. \begin{align*}\frac{4}{x}=y\end{align*}
  2. \begin{align*}y=9\end{align*}
  3. \begin{align*}x=-3.5\end{align*}
  4. \begin{align*}y=\frac{1}{8} x+7\end{align*}
  5. \begin{align*}4x+3y=1\end{align*}

Graph the following direct variation equations.

  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*}
  5. Is \begin{align*}y=6x-2\end{align*} an example of direct variation? Explain your answer.

In 20 – 24, determine the constant of variation in each exercise.

  1. \begin{align*}y\end{align*} varies directly as \begin{align*}x\end{align*}; when \begin{align*}x=4,y=48\end{align*}
  2. \begin{align*}d\end{align*} varies directly as \begin{align*}t\end{align*}; when \begin{align*}t=7,d=329\end{align*}
  3. \begin{align*}l\end{align*} varies directly as \begin{align*}h\end{align*}; when \begin{align*}l=112\end{align*}, \begin{align*}h=-16\end{align*}
  4. \begin{align*}m\end{align*} is directly proportional to \begin{align*}h\end{align*}; when \begin{align*}m=461.50\end{align*}, \begin{align*}h=89.6\end{align*}
  5. \begin{align*}z\end{align*} is directly proportional to \begin{align*}r;\end{align*} when \begin{align*}r=412,z=51.5\end{align*}
  6. Determine the equation of the strawberry purchase in the opener of this lesson.
  7. 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?
  8. The current standard for low-flow showerheads 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.
  9. An electrical device passes a force of 288 volts at 32 amps. Using Ohm’s Law, determine:
    1. The constant of proportionality
    2. The force needed to pass 65 amps
  10. The diameter of a circle is directly proportional to its radius. If a circle with a 2-inch diameter has a circumference of approximately 6.28 inches, what is the circumference of a 15-inch circle?
  11. Amin is using a hose to fill his new swimming pool for the first time. He starts the hose at 10:00 P.M. and leaves it running all night. At 6:00 A.M. he measures the depth and calculates that the pool is four-sevenths full. At what time will his new pool be full?
  12. Land in Wisconsin is for sale to property investors. A 232-acre lot is listed for sale for $200,500. Assuming the same price per acre, how much would a 60-acre lot sell for?
  13. 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
  14. Determine the equations of graphs \begin{align*}a - d\end{align*} below.

Mixed Review

  1. Graph \begin{align*}3x+4y=48\end{align*} using its intercepts.
  2. Graph \begin{align*}y=\frac{2}{3} x-4\end{align*}.
  3. Solve for \begin{align*}u: 4(u+3)=3(3u-7)\end{align*}.
  4. Are these lines parallel? \begin{align*}y=\frac{1}{2} x-7\end{align*} and \begin{align*}2y=x+2\end{align*}
  5. In which quadrant is (–99, 100)?
  6. Find the slope between (2, 0) and (3, 7).
  7. Evaluate if \begin{align*}a=-3\end{align*} and \begin{align*}b=4\end{align*}: \begin{align*}\frac{1+4b}{2a-5b}\end{align*}.

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