# 7.7: Variation

- Page ID
- 22445

Learning Objectives

- Solve applications involving direct variation.
- Solve applications involving inverse variation.
- Solve applications involving joint variation.

## Direct Variation

Consider a freight train moving at a constant speed of 30 miles per hour. The equation that expresses the distance traveled at that speed in terms of time is given by

\(D=30t\)

After \(1\) hour the train has traveled \(30\) miles, after \(2\) hours the train has traveled \(60\) miles, and so on. We can construct a chart and graph this relation.

Time, \(t\), in hours | Distance \(D=30t\) |
---|---|

0 | 0 |

1 | 30 |

2 | 60 |

3 | 90 |

4 | 120 |

*Table 7.7.1*

In this example, we can see that the distance varies over time as the product of the constant rate, \(30\) miles per hour, and the variable, \(t\). This relationship is described as direct variation and \(30\) is called the **variation constant**. In addition, if we divide both sides of \(D=30t\) by \(t\) we have

\(\frac{D}{t}=30\)

In this form, it is reasonable to say that \(D\) is proportional to \(t\), where \(30\) is the constant of proportionality. In general, we have

Key Words | Translation |
---|---|

"\(y\) varies directly as \(x\)" |
\[y=kx\] |

"\(y\) is directly proportional to \(x\)" | |

"\(y\) is proportional to \(x\)" |

*Table 7.7.2*

Here \(k\) is nonzero and is called the constant of variation or the constant of proportionality.

Example \(\PageIndex{1}\)

The circumference of a circle is directly proportional to its diameter, and the constant of proportionality is \(π\). If the circumference is measured to be \(20\) inches, then what is the radius of the circle?

**Solution**:

Let \(C\) represent the circumference of the circle.

Let \(d\) represent the diameter of a circle.

Use the fact that “the circumference is directly proportional to the diameter” to write an equation that relates the two variables.

\(C=kd\)

We are given that “the constant of proportionality is \(π\),” or \(k=π\). Therefore, we write

\(C=πd\)

Now use this formula to find \(d\) when the circumference is \(20\) inches.

\(\begin{aligned} 20&=\pi d \\ \frac{20}{\color{Cerulean}{\pi}}&=\frac{\pi d}{\color{Cerulean}{\pi}}\\ \frac{20}{\pi}&=d \end{aligned}\)

The radius of the circle, \(r\), is one-half of its diameter.

\(\begin{aligned} r&=\frac{d}{2} \\ &=\frac{\color{OliveGreen}{\frac{20}{\pi}}}{2} \\ &=\frac{20}{\pi}\cdot \frac{1}{2} \\ &= \frac{10}{\pi} \end{aligned}\)

**Answer**:

The radius is \(\frac{10}{π}\) inches, or approximately \(3.18\) inches

Typically, we will not be given the constant of variation. Instead, we will be given information from which it can be determined.

Example \(\PageIndex{2}\)

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.

**Solution**:

Let \(y\) represent the weight on Earth.

Let \(x\) represent the weight on the Moon.

We are given that the “weight on earth varies directly to the weight on the moon.”

\(y=kx\)

To find the constant of variation \(k\), use the given information. A \(180\)-pound man on earth weighs \(30\) pounds on the moon, or \(y=180\) when \(x=30\).

\(180 = k \cdot 30\)

Solve for \(k\).

\(\begin{aligned} \frac{180}{30} &=k \\ 6&=k \end{aligned}\)

Next, set up a formula that models the given information.

\(y=6x\)

This implies that a person’s weight on earth is \(6\) times her weight on the moon. To answer the question, use the woman’s weight on earth, \(y=120\) pounds, and solve for \(x\).

\(\begin{aligned} 120&=6x \\ \frac{120}{6} &=x \\ 20&=x \end{aligned}\)

**Answer**:

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

## Inverse Variation

Next, consider the relationship between time and rate,

\(r=\frac{D}{t}\)

If we wish to travel a fixed distance, then we can determine the average speed required to travel that distance in a given amount of time. For example, if we wish to drive 240 miles in 4 hours, we can determine the required average speed as follows:

\(r=\frac{240}{4}=6\)

The average speed required to drive 240 miles in 4 hours is 60 miles per hour. If we wish to drive the 240 miles in 5 hours, then determine the required speed using a similar equation:

\(r=\frac{240}{5}=48\)

In this case, we would only have to average 48 miles per hour. We can make a chart and view this relationship on a graph.

Time \(t\) in hours | Speed \(r=\frac{240}{t}\) |
---|---|

2 | 120 |

3 | 80 |

4 | 60 |

5 | 48 |

*Table 7.7.3*

This is an example of an inverse relationship. We say that \(r\) is inversely proportional to the time \(t\), where \(240\) is the constant of proportionality. In general, we have

Key Words | Translation |
---|---|

"\(y\) varies inversely as \(x\)" | \[y=\frac{k}{x}\] |

"\(y\) is inversely proportional to \(x\)" |

*Table 7.7.4*

Again, \(k\) is nonzero and is called the **constant of variation** or the **constant of proportionality**.

Example \(\PageIndex{3}\)

If \(y\) varies inversely as \(x\) and \(y=5\) when \(x=2\), then find the constant of proportionality and an equation that relates the two variables.

**Solution**:

If we let \(k\) represent the constant of proportionality, then the statement “\(y\) varies inversely as \(x\)” can be written as follows:

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

Use the given information, \(y=5\) when \(x=2\), to find \(k\).

\(5=\frac{k}{2}\)

Solve for \(k\).

\(\begin{aligned} \color{Cerulean}{2}\color{black}{\cdot 5}&=\color{Cerulean}{2}\color{black}{\cdot\frac{k}{2}\\10&=k} \end{aligned}\)

Therefore, the formula that models the problem is

\(y=\frac{10}{x}\)

**Answer**:

The constant of proportionality is \(10\), and the equation is \(y=\frac{10}{x}\).

Example \(\PageIndex{4}\)

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), then how much will it weigh at \(1,000\) miles above earth’s surface?

**Solution**:

Let \(w\) represent weight of the object.

Let \(d\) represent the object's distance from the center of Earth.

Since "\(w\) varies inversely as the square of \(d\)," we can write

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

Use the given information to find \(k\). An object weighs \(100\) pounds on the surface of earth, approximately \(4,000\) miles from the center. In other words, \(w = 100\) when \(d = 4,000\):

\(100=\frac{k}{(4000)^{2}}\)

Solve for \(k\):

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

Therefore, we can model the problem with the following formula:

\(w=\frac{1.6\times 10^{9}}{d^{2}}\)

To use the formula to find the weight, we need the distance from the center of earth. Since the object is \(1,000\) miles above the surface, find the distance from the center of earth by adding \(4,000\) miles:

\(d=4,000+1,000=5,000\) miles

To answer the question, use the formula with \(d = 5,000\).

\(\begin{aligned} y &= \frac{1.6\times 10^{9}}{(\color{OliveGreen}{5,000}\color{black}{)^{2}}} \\ &=\frac{1.6\times 10^{9}}{25,000,000}\\ &=\frac{1.6\times 10^{9}}{2.5\times 10^{7}}\\ &=0.64\times 10^{2}\\ &=64 \end{aligned}\)

**Answer**:

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. In general, we have

Vocabulary | Translation |
---|---|

"\(y\) varies jointly as \(x\) and \(z\)" | \[y=kyz\] |

"\(y\) is jointly proportional to \(x\) and \(z\)" |

*Table 7.7.5*

Here \(k\) is nonzero and is called the **constant of variation** or the **constant of proportionality**.

Example \(\PageIndex{5}\)

The area of an ellipse varies jointly as \(a\), half of the ellipse’s major axis, and \(b\), half of the ellipse’s minor axis. If the area of an ellipse is \(300π \text{cm}^{2}\), where \(a=10\) cm and \(b=30\) cm, then what is the constant of proportionality? Give a formula for the area of an ellipse.

**Solution**:

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

\(A=kab\)

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

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

Therefore, the formula for the area of an ellipse is

\(A=\pi ab\)

**Answer**:

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

Exercise \(\PageIndex{1}\)

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

**Answer**-
\(\frac{3}{2}\)

## Key Takeaways

- The setup of variation problems usually requires multiple steps. First, identify the key words to set up an equation and then use the given information to find the constant of variation \(k\). After determining the constant of variation, write a formula that models the problem. Once a formula is found, use it to answer the question.

Exercise \(\PageIndex{2}\) Variation Problems

Translate the following sentences into a mathematical formula.

- The distance, \(D\), an automobile can travel is directly proportional to the time, \(t\), that it travels at a constant speed.
- The extension of a hanging spring, \(d\), is directly proportional to the weight, \(w\), attached to it.
- An automobile’s breaking distance, \(d\), is directly proportional to the square of the automobile’s speed, \(v\).
- The volume, \(V\), of a sphere varies directly as the cube of its radius, \(r\).
- The volume, \(V\), of a given mass of gas is inversely proportional to the pressure, \(p\), exerted on it.
- The intensity, \(I\), of light from a light source is inversely proportional to the square of the distance, \(d\), from the source.
- Every particle of matter in the universe attracts every other particle with a force, \(F\), that is directly proportional to the product of the masses, \(m_{1}\) and \(m_{2}\), of the particles and inversely proportional to the square of the distance, \(d\), between them.
- Simple interest, \(I\), is jointly proportional to the annual interest rate, \(r\), and the time, \(t\), in years a fixed amount of money is invested.
- The period, \(T\), of a pendulum is directly proportional to the square root of its length, \(L\).
- The time, \(t\), it takes an object to fall is directly proportional to the square root of the distance, \(d\), it falls.

**Answer**-
1. \(D=kt\)

3. \(d=kv^{2}\)

5. \(V=\frac{k}{p}\)

7. \(F=\frac{km_{1}⋅m_{2}}{d^{2}}\)

9. \(T=k\sqrt{L}\)

Exercise \(\PageIndex{3}\) Variation Problems

Construct a mathematical model given the following.

- \(y\) varies directly as \(x\), and \(y = 30\) when \(x = 6\).
- \(y\) varies directly as \(x\), and \(y = 52\) when \(x = 4\).
- \(y\) is directly proportional to \(x\), and \(y = 12\) when \(x = 3\).
- \(y\) is directly proportional to \(x\), and \(y = 120\) when \(x = 20\).
- \(y\) varies directly as \(x\), and \(y = 14\) when \(x = 10\).
- \(y\) varies directly as \(x\), and \(y = 2\) when \(x = 8\).
- \(y\) varies inversely as \(x\), and \(y = 5\) when \(x = 7\).
- \(y\) varies inversely as \(x\), and \(y = 12\) when \(x = 2\).
- \(y\) is inversely proportional to \(x\), and \(y = 3\) when \(x = 9\).
- \(y\) is inversely proportional to \(x\), and \(y = 21\) when \(x = 3\).
- \(y\) varies inversely as \(x\), and \(y = 2\) when \(x = \frac{1}{8}\).
- \(y\) varies inversely as \(x\), and \(y = \frac{3}{2}\) when \(x = \frac{1}{9}\).
- \(y\) varies jointly as \(x\) and \(z\), where \(y = 8\) when \(x = 4\) and \(z = \frac{1}{2}\).
- \(y\) varies jointly as \(x\) and \(z\), where \(y = 24\) when \(x =\frac{1}{3}\) and \(z = 9\).
- \(y\) is jointly proportional to \(x\) and \(z\), where \(y = 2\) when \(x = 1\) and \(z = 3\).
- \(y\) is jointly proportional to \(x\) and \(z\), where \(y = 15\) when \(x = 3\) and \(z = 7\).
- \(y\) varies jointly as \(x\) and \(z\), where \(y = \frac{2}{3}\) when \(x = \frac{1}{2}\) and \(z = 12\).
- \(y\) varies jointly as \(x\) and \(z\), where \(y = 5\) when \(x = \frac{3}{2}\) and \(z = \frac{2}{9}\).
- \(y\) varies directly as the square of \(x\), where \(y = 45\) when \(x = 3\).
- \(y\) varies directly as the square of \(x\), where \(y = 3\) when \(x = \frac{1}{2}\).
- \(y\) is inversely proportional to the square of \(x\), where \(y = 27\) when \(x = \frac{1}{3}\).
- \(y\) is inversely proportional to the square of \(x\), where \(y = 9\) when \(x = \frac{2}{3}\).
- \(y\) varies jointly as \(x\) and the square of \(z\), where \(y = 54\) when \(x = 2\) and \(z = 3\).
- \(y\) varies jointly as \(x\) and the square of \(z\), where \(y = 6\) when \(x = \frac{1}{4}\) and \(z = \frac{2}{3}\).
- \(y\) varies jointly as \(x\) and \(z\) and inversely as the square of \(w\), where \(y = 30\) when \(x = 8, z = 3\), and \(w = 2\)
- \(y\) varies jointly as \(x\) and \(z\) and inversely as the square of \(w\), where \(y = 5\) when \(x= 1, z = 3\), and \(w = \frac{1}{2}\).
- \(y\) varies directly as the square root of \(x\) and inversely as \(z\), where \(y = 12\) when \(x= 9\) and \(z = 5\).
- \(y\) varies directly as the square root of \(x\) and inversely as the square of \(z\), where \(y = 15\) when \(x = 25\) and \(z = 2\).
- \(y\) varies directly as the square of \(x\) and inversely as \(z\) and the square of \(w\), where \(y = 14\) when \(x = 4, w = 2\), and \(z = 2\).
- \(y\) varies directly as the square root of \(x\) and inversely as \(z\) and the square of \(w\), where \(y = 27\) when \(x = 9, w = \frac{1}{2}\), and \(z = 4\).

**Answer**-
1. \(y=5x\)

3. \(y=4x\)

5. \(y=\frac{7}{5}x\)

7. \(y=\frac{35}{x}\)

9. \(y=\frac{27}{x}\)

11. \(y=\frac{1}{4x}\)

13. \(y=4xz\)

15. \(y=\frac{2}{3}xz\)

17. \(y=\frac{1}{9}xz\)

19. \(y=5x^{2}\)

21. \(y=3x^{2}\)

23. \(y=3xz^{2}\)

25. \(y=\frac{5xz}{w^{2}}\)

27. \(y=\frac{20x}{\sqrt{z}}\)

29. \(y=\frac{7x^{2}}{w^{2}z}\)

Exercise \(\PageIndex{4}\) Variation Problems

Applications involving variation.

- Revenue in dollars is directly proportional to the number of branded sweat shirts sold. If the revenue earned from selling \(25\) sweat shirts is $\(318.75\), then determine the revenue if \(30\) sweat shirts are sold.
- The sales tax on the purchase of a new car varies directly as the price of the car. If an $\(18,000\) new car is purchased, then the sales tax is $\(1,350\). How much sales tax is charged if the new car is priced at $\(22,000\)?
- The price of a share of common stock in a company is directly proportional to the earnings per share (EPS) of the previous 12 months. If the price of a share of common stock in a company is $\(22.55\) and the EPS is published to be $\(1.10\), then determine the value of the stock if the EPS increases by $\(0.20\).
- The distance traveled on a road trip varies directly with the time spent on the road. If a \(126\)-mile trip can be made in \(3\) hours, then what distance can be traveled in \(4\) hours?
- The circumference of a circle is directly proportional to its radius. If the circumference of a circle with radius \(7\) centimeters is measured as \(14π\) centimeters, then find the constant of proportionality.
- The area of circle varies directly as the square of its radius. If the area of a circle with radius \(7\) centimeters is determined to be \(49π\) square centimeters, then find the constant of proportionality.
- The surface area of a sphere varies directly as the square of its radius. When the radius of a sphere measures \(2\) meters, the surface area measures \(16π\) square meters. Find the surface area of a sphere with radius \(3\) meters.
- The volume of a sphere varies directly as the cube of its radius. When the radius of a sphere measures \(3\) meters, the volume is \(36π\) cubic meters. Find the volume of a sphere with radius \(1\) meter.
- With a fixed height, the volume of a cone is directly proportional to the square of the radius at the base. When the radius at the base measures \(10\) centimeters, the volume is \(200\) cubic centimeters. Determine the volume of the cone if the radius of the base is halved.
- The distance, \(d\), an object in free fall drops varies directly with the square of the time, \(t\), that it has been falling. If an object in free fall drops \(36\) feet in \(1.5\) seconds, then how far will it have fallen in \(3\) seconds?

**Answer**-
1. $\(382.50 \)

3. $\(26.65 \)

5. \(2π\)

7. \(36π\) square meters

9. \(50\) cubic centimeters

Exercise \(\PageIndex{5}\) Variation Problems

Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.

- If a hanging spring is stretched \(5\) inches when a \(20\)-pound weight is attached to it, then determine its spring constant.
- If a hanging spring is stretched \(3\) centimeters when a \(2\)-kilogram weight is attached to it, then determine the spring constant.
- If a hanging spring is stretched \(3\) inches when a \(2\)-pound weight is attached, then how far will it stretch with a \(5\)-pound weight attached?
- If a hanging spring is stretched \(6\) centimeters when a \(4\)-kilogram weight is attached to it, then how far will it stretch with a \(2\)-kilogram weight attached?

**Answer**-
1. \(\frac{1}{4}\)

3. \(7.5\) inches

Exercise \(\PageIndex{6}\) Variation Problems

The breaking distance of an automobile is directly proportional to the square of its speed.

- If it takes \(36\) feet to stop a particular automobile moving at a speed of \(30\) miles per hour, then how much breaking distance is required if the speed is \(35\) miles per hour?
- After an accident, it was determined that it took a driver \(80\) feet to stop his car. In an experiment under similar conditions, it takes \(45\) feet to stop the car moving at a speed of \(30\) miles per hour. Estimate how fast the driver was moving before the accident.

**Answer**-
1. \(49\) feet

Exercise \(\PageIndex{7}\) Variation Problems

Boyle’s law states that if the temperature remains constant, the volume, \(V\), of a given mass of gas is inversely proportional to the pressure, \(p\), exerted on it.

- A balloon is filled to a volume of \(216\) cubic inches on a diving boat under \(1\) atmosphere of pressure. If the balloon is taken underwater approximately \(33\) feet, where the pressure measures \(2\) atmospheres, then what is the volume of the balloon?
- If a balloon is filled to \(216\) cubic inches under a pressure of \(3\) atmospheres at a depth of \(66\) feet, then what would the volume be at the surface, where the pressure is \(1\) atmosphere?
- To balance a seesaw, the distance from the fulcrum that a person must sit is inversely proportional to his weight. If a \(72\)-pound boy is sitting \(3\) feet from the fulcrum, then how far from the fulcrum must a \(54\)- pound boy sit to balance the seesaw?
- The current, \(I\), in an electrical conductor is inversely proportional to its resistance, \(R\). If the current is \(\frac{1}{4}\) ampere when the resistance is \(100\) ohms, then what is the current when the resistance is \(150\) ohms?
- The number of men, represented by \(y\), needed to lay a cobblestone driveway is directly proportional to the area, \(A\), of the driveway and inversely proportional to the amount of time, \(t\), allowed to complete the job. Typically, \(3\) men can lay \(1,200\) square feet of cobblestone in \(4\) hours. How many men will be required to lay \(2,400\) square feet of cobblestone given \(6\) hours?
- The volume of a right circular cylinder varies jointly as the square of its radius and its height. A right circular cylinder with a \(3\)-centimeter radius and a height of \(4\) centimeters has a volume of \(36π\) cubic centimeters. Find a formula for the volume of a right circular cylinder in terms of its radius and height.
- The period, \(T\), of a pendulum is directly proportional to the square root of its length, \(L\). If the length of a pendulum is \(1\) meter, then the period is approximately \(2\) seconds. Approximate the period of a pendulum that is \(0.5\) meter in length.
- The time, \(t\), it takes an object to fall is directly proportional to the square root of the distance, \(d\), it falls. An object dropped from \(4\) feet will take \(\frac{1}{2}\) second to hit the ground. How long will it take an object dropped from \(16\) feet to hit the ground?

**Answer**-
1. \(108\) cubic inches

3. \(4\) feet

5. \(4\) men

7. \(1.4\) seconds

Exercise \(\PageIndex{8}\) Variation Problems

Newton’s universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force, \(F\), that is directly proportional to the product of the masses, \(m_{1}\) and \(m_{2}\), of the particles and inversely proportional to the square of the distance, \(d\), between them. The constant of proportionality is called the gravitational constant.

- If two objects with masses \(50\) kilograms and \(100\) kilograms are \(\frac{1}{2}\) meter apart, then they produce approximately \(1.34×10^{−6}\) newtons (N) of force. Calculate the gravitational constant.
- Use the gravitational constant from the previous exercise to write a formula that approximates the force, \(F\), in newtons between two masses \(m_{1}\) and \(m_{2}\), expressed in kilograms, given the distance \(d\) between them in meters.
- Calculate the force in newtons between earth and the moon, given that the mass of the moon is approximately \(7.3×10^{22}\) kilograms, the mass of earth is approximately \(6.0×10^{24}\) kilograms, and the distance between them is on average \(1.5×10^{11}\) meters.
- Calculate the force in newtons between earth and the sun, given that the mass of the sun is approximately \(2.0×10^{30}\) kilograms, the mass of earth is approximately \(6.0×10^{24}\) kilograms, and the distance between them is on average \(3.85×10^{8}\) meters.
- If \(y\) varies directly as the square of \(x\), then how does \(y\) change if \(x\) is doubled?
- If \(y\) varies inversely as square of \(t\), then how does \(y\) change if \(t\) is doubled?
- If \(y\) varies directly as the square of \(x\) and inversely as the square of \(t\), then how does \(y\) change if both \(x\) and \(t\) are doubled?

**Answer**-
1. \(6.7×10^{−11} \frac{N m^{2}}{kg^{2}}\)

3. \(1.98×10^{20}\) N

5. \(y\) changes by a factor of \(4\)

7. \(y\) remains unchanged