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Section 3.7 Variation

Before Starting this Section, Review

1 Equation of a line (Section 2.3 , page 201)
2 Solving linear equations (Section 1.1 , page 83)

Objectives

1 Solve direct variation problems.
2 Solve inverse variation problems.
3 Solve joint and combined variation problems.

Newton and the Apple

According to legend, Isaac Newton was sitting under an apple tree, and when an apple fell on his head, he suddenly thought of the Law of Universal Gravitation. As in many such legends, the story may not be true in detail, but it contains elements of the truth.

Before we tell you a more realistic version of this story, let’s recall some terminology associated with the motion of objects.

Recall that speed = \frac{distance}{time} .

$Recall that speed = \frac{distance}{time} .$

The velocity of an object is the speed of an object in a specific direction.
Acceleration: If the velocity of an object is changing, we say that the object is accelerating. Acceleration is the rate of change of velocity.
$Force = Mass \times Acceleration$ $Force = Mass \times Acceleration$

What really happened with the apple? Perhaps the correct version of the story is that upon observing an apple falling from a tree, Newton concluded that the apple accelerated because the velocity of the apple changed from when it began hanging on the tree (zero velocity) and then moved to the ground. He decided that a force must act on the apple to cause this acceleration. He called this force “gravity” and the associated acceleration $(Force = Mass \times Acceleration)$ $(Force = Mass \times Acceleration)$ the “acceleration due to gravity.” He conjectured further that if the force of gravity reaches the top of the apple tree, it might reach even farther; might it reach all the way to the Moon? In that case, to keep the Moon moving in a circular orbit, rather than wandering into outer space, Earth must exert a gravitational force on the Moon. Newton realized that the force that brought the apple to the ground and the force that holds the Moon in orbit were the same. He concluded that any two objects in the universe exert gravitational attraction on each other, whereupon he gave us the Law of Universal Gravitation. See Example 6.

Some scientists believe that it was Kepler’s Third Law of Planetary Motion (see Exercise 53), not an apple, that led Newton to his Law of Universal Gravitation.

Sir Isaac Newton 1642–1727

Isaac Newton was a mathematician and physicist, one of the foremost scientific intellects of all time. He entered Cambridge University in 1661 and was elected a Fellow of Trinity College in 1667 and Lucasian Professor of Mathematics in 1669. He remained at the university, lecturing most years, until 1696.

As a firm opponent of the attempt by King James II to make the universities into Catholic institutions, Newton was elected Member of Parliament for the University of Cambridge to the Convention Parliament of 1689, and he sat again in 1701–1702.

Meanwhile, in 1696, he moved to London as warden of the Royal Mint. He became master of the Mint in 1699, an office he retained until his death. He was elected a Fellow of the Royal Society of London in 1671, and in 1703, he became president, being annually reelected for the rest of his life. One of his major works, Opticks, appeared the next year; he was knighted in Cambridge in 1705.

Direct Variation

1 Solve direct variation problems.

Two types of relationships between quantities (variables) occur so frequently in mathematics that they are given special names: direct variation and inverse variation.

Direct variation describes any relationship between two quantities in which any increase (or decrease) in one causes a proportional increase (or decrease) in the other. For example, the sales tax in Hillsborough County, Florida, in 2012 was 7%, so that the sales tax on a $1000 purchase was $70. If we double the purchase to $2000, the sales tax also doubles, to $140. If we reduce the purchase by half to $500, the sales tax is also reduced by half, to $35. We say that the sales tax varies directly as the purchase price. The equation

Sales tax = (0.07) (Purchase price)

$Sales tax = (0.07) (Purchase price)$

expresses the fact that the sales tax was a constant (0.07) multiple of the purchase price.

Side Note

The statements

“y varies directly as x,”
“y varies as x,”
“y is directly proportional to x,”
“y is proportional to x,”
and “ $y = k x, k \neq 0$ $y = k x, k \neq 0$ ”

have the same meaning.

Procedure in Action

Example 1 Solving the Variation Problems

OBJECTIVE	EXAMPLE
Solve variation problems.	Suppose `y` varies as `x` and $y = 20$ $y = 20$ when $x = \frac{4}{3} .$ $x = \frac{4}{3} .$ Find `y` when $x = 8 .$ $x = 8 .$
Step 1 Write the equation with the constant of variation, `k`.	$y = k x y varies as x .$ $y = k x y varies as x .$
Step 2 Substitute the given values of the variables into the equation in Step 1 to find the value of the constant `k`.	$\begin{array}{rcll} 20 & = & k (\frac{4}{3}) & Replace y with 20 and x with \frac{4}{3} . \\ 20 (\frac{3}{4}) & = & k (\frac{4}{3}) (\frac{3}{4}) & Multiply both sides by \frac{3}{4} . \\ 15 & = & k & Simplify to solve for k . \end{array}$ $\begin{array}{rcll} 20 & = & k (\frac{4}{3}) & Replace y with 20 and x with \frac{4}{3} . \\ 20 (\frac{3}{4}) & = & k (\frac{4}{3}) (\frac{3}{4}) & Multiply both sides by \frac{3}{4} . \\ 15 & = & k & Simplify to solve for k . \end{array}$
Step 3 Rewrite the equation in Step 1 with the value of `k` from Step 2.	$y = 15 x Replace k with 15.$ $y = 15 x Replace k with 15.$
Step 4 Use the equation from Step 3 to answer the question posed in the problem.	$\begin{array}{l} y = 15 (8) & Replace x with 8. \\ y = 120 & Simplify . \end{array}$ $\begin{array}{l} y = 15 (8) & Replace x with 8. \\ y = 120 & Simplify . \end{array}$ So $y = 120$ $y = 120$ when $x = 8 .$ $x = 8 .$

Practice Problem 1

Suppose y varies directly as x. If y is 6 when x is 30, find y when $x = 120 .$ $x = 120 .$

Example 2 Direct Variation in Electrical Circuits

The current in a circuit connected to a 220-volt battery is 50 amperes. If the current is directly proportional to the voltage of the attached battery, what voltage battery is needed to produce a current of 75 amperes?

Solution

Let $I = current$ $I = current$ in amperes and $V = voltage$ $V = voltage$ in volts of the battery.

Step 1 $I = k V I varies directly as V .$ $I = k V I varies directly as V .$
Step 2 $\begin{array}{rcll} 50 & = & k (220) & Substitute I = 50, V = 220. \\ \frac{50}{220} & = & k & Solve for k . \\ \frac{5}{22} & = & k & Simplify . \end{array}$ $\begin{array}{rcll} 50 & = & k (220) & Substitute I = 50, V = 220. \\ \frac{50}{220} & = & k & Solve for k . \\ \frac{5}{22} & = & k & Simplify . \end{array}$
Step 3 $I = \frac{5}{22} V Replace k with \frac{5}{22} in I = k V .$ $I = \frac{5}{22} V Replace k with \frac{5}{22} in I = k V .$
Step 4 $\begin{array}{rcll} 75 & = & \frac{5}{22} V & Substitute I = 75 in Step 3 . \\ \frac{22}{5} \cdot 75 & = & V & Solve for V . \\ 330 & = & V & Simplify . \end{array}$ $\begin{array}{rcll} 75 & = & \frac{5}{22} V & Substitute I = 75 in Step 3 . \\ \frac{22}{5} \cdot 75 & = & V & Solve for V . \\ 330 & = & V & Simplify . \end{array}$

A battery of 330 volts is needed to produce 75 amperes of current.

Practice Problem 2

In Example 2 , if the current in the circuit is 60 amperes, what voltage battery is needed to produce a current of 75 amperes?

We can generalize the concept of direct variation to variation with powers.

Note that when $n = 1,$ $n = 1,$ y varies directly as x. In the formula for the area of a circle, $A = π r^{2},$ $A = π r^{2},$ A varies directly as the square (or second power) of the radius r, with $π$ $π$ as the constant of variation.

Example 3 Solving a Problem Involving Direct Variation with Powers

Suppose you had forgotten the formula for the volume of a sphere but were told that the volume V of a sphere varies directly as the cube of its radius r. In addition, you are given that $V = 972 π$ $V = 972 π$ when $r = 9 .$ $r = 9 .$ Find V when $r = 6 .$ $r = 6 .$

Solution

Step 1 $V = k r^{3} V varies directly as r^{3} .$ $V = k r^{3} V varies directly as r^{3} .$
Step 2 $\begin{array}{rcll} 972 π & = & k {(9)}^{3} & Replace V with 972 π and r with 9. \\ 972 π & = & k (729) & 9^{3} = 729 \\ k & = & \frac{972 π}{729} & Solve for k . \\ = & \frac{4}{3} π & Simplify . \end{array}$ $\begin{array}{rcll} 972 π & = & k {(9)}^{3} & Replace V with 972 π and r with 9. \\ 972 π & = & k (729) & 9^{3} = 729 \\ k & = & \frac{972 π}{729} & Solve for k . \\ = & \frac{4}{3} π & Simplify . \end{array}$
Step 3 $V = \frac{4}{3} π r^{3} Substitute k = \frac{4}{3} π in Step 1.$ $V = \frac{4}{3} π r^{3} Substitute k = \frac{4}{3} π in Step 1.$
Step 4 $\begin{array}{rcll} V & = & \frac{4}{3} π {(6)}^{3} & Replace r with 6. \\ = & 288 π cubic units \end{array}$ $\begin{array}{rcll} V & = & \frac{4}{3} π {(6)}^{3} & Replace r with 6. \\ = & 288 π cubic units \end{array}$

Practice Problem 3

If y varies directly as the square of x and $y = 48$ $y = 48$ when $x = 2,$ $x = 2,$ find y when $x = 5 .$ $x = 5 .$

Inverse Variation

2 Solve inverse variation problems.

Suppose a plane takes four hours to fly from Atlanta to Denver at an average speed of 300 miles per hour. If we increase the average speed to 600 miles per hour (twice the previous speed), the flight time decreases to two hours (half the previous time). For the Atlanta–Denver trip, the time of flight varies inversely as the speed of the plane.

\begin{array}{rcl} time & = & \frac{distance}{speed} \\ = & \frac{k}{speed}, where k = distance between Atlanta and Denver . \end{array}

$\begin{array}{rcl} time & = & \frac{distance}{speed} \\ = & \frac{k}{speed}, where k = distance between Atlanta and Denver . \end{array}$

We see that the distance is the constant of variation when time and speed are the variables.

Example 4 Solving an Inverse Variation Problem

Suppose y varies inversely as x and $y = 35$ $y = 35$ when $x = 11 .$ $x = 11 .$ Find y if $x = 55 .$ $x = 55 .$

Solution

Step 1 $y = \frac{k}{x} y varies inversely as x .$ $y = \frac{k}{x} y varies inversely as x .$
Step 2 $\begin{array}{rcll} 35 & = & \frac{k}{11} & Substitute y = 35, x = 11. \\ 35 \cdot 11 & = & k & Solve for k . \\ 385 & = & k & Simplify . \end{array}$ $\begin{array}{rcll} 35 & = & \frac{k}{11} & Substitute y = 35, x = 11. \\ 35 \cdot 11 & = & k & Solve for k . \\ 385 & = & k & Simplify . \end{array}$
Step 3 $y = \frac{385}{x} Replace k with 385 in y = \frac{k}{x} .$ $y = \frac{385}{x} Replace k with 385 in y = \frac{k}{x} .$
Step 4 $\begin{array}{l} y = \frac{385}{55} & Replace x with 55. \\ y = 7 & Simplify . \end{array}$ $\begin{array}{l} y = \frac{385}{55} & Replace x with 55. \\ y = 7 & Simplify . \end{array}$

Practice Problem 4

A varies inversely as B, and $A = 12$ $A = 12$ when $B = 5 .$ $B = 5 .$ Find A when $B = 3 .$ $B = 3 .$

Example 5 Solving a Problem Involving Inverse Variation with Powers

The intensity of light varies inversely as the square of the distance from the light source. If Rita doubles her distance from a lamp, what happens to the intensity of light at her new location?

Solution

Let I be the intensity of light at a distance d from the light source. Because I is inversely proportional to the square of d, we have

I = \frac{k}{d^{2}} .

$I = \frac{k}{d^{2}} .$

If we replace d with 2d, the intensity $I_{1}$ $I_{1}$ at the new location is given by

\begin{array}{l} I_{1} = \frac{k}{{(2 d)}^{2}} & Replace d with 2 d in I = \frac{k}{d^{2}} . \\ I_{1} = \frac{k}{4 d^{2}} & {(2 d)}^{2} = 2^{2} d^{2} = 4 d^{2} \end{array}

$\begin{array}{l} I_{1} = \frac{k}{{(2 d)}^{2}} & Replace d with 2 d in I = \frac{k}{d^{2}} . \\ I_{1} = \frac{k}{4 d^{2}} & {(2 d)}^{2} = 2^{2} d^{2} = 4 d^{2} \end{array}$

Therefore,

I_{1} = \frac{I}{4} Replace \frac{k}{d^{2}} with I .

$I_{1} = \frac{I}{4} Replace \frac{k}{d^{2}} with I .$

This equation tells us that if Rita doubles her distance from the light source, the intensity of light at the new location will be one-fourth the intensity at the original location.

Practice Problem 5

If y varies inversely as the square root of x and $y = \frac{3}{4}$ $y = \frac{3}{4}$ when $x = 16,$ $x = 16,$ find x when $y = 2 .$ $y = 2 .$

Joint and Combined Variation

3 Solve joint and combined variation problems.

Sometimes different types of variations occur in a problem that has more than one independent variable.

For example, the volume V of a right circular cylinder with radius r and height h is given by

V = π r^{2} h .

$V = π r^{2} h .$

Using the vocabulary of variation, we say, “The volume V of a right circular cylinder varies jointly as its height h and the square of its radius r.” The constant of variation is $π .$ $π .$

We can combine joint variation and inverse variation. For example, the relationship

w = \frac{k x^{2} y^{3}}{\sqrt{z}}

$w = \frac{k x^{2} y^{3}}{\sqrt{z}}$

can be described as “w varies jointly as the square of x and cube of y and inversely as the square root of z.” The same relationship also can be stated as “w varies directly as $x^{2} y^{3}$ $x^{2} y^{3}$ and inversely as $\sqrt{z} .$ $\sqrt{z} .$ ”

Example 6 Newton’s Law of Universal Gravitation

Newton’s Law of Universal Gravitation says that every object in the universe attracts every other object with a force acting along the line of the centers of the two objects. Also, this attracting force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between the two objects.

Write the law symbolically.
Estimate the value of g (the acceleration due to gravity) near the surface of Earth. Use the following estimates: radius of Earth $R_{E} = 6.38 \times 10^{6}$ $R_{E} = 6.38 \times 10^{6}$ meters, and mass of Earth $M_{E} = 5.98 \times 10^{24}$ $M_{E} = 5.98 \times 10^{24}$ kilograms.

Solution

Let $m_{1}$ $m_{1}$ and $m_{2}$ $m_{2}$ be the masses of the two objects and r be the distance between their centers. See Figure 3.36. Let F denote the gravitational force between the objects.

Figure 3.36

Gravitational attraction

Then according to Newton’s Law of Universal Gravitation,

$F = G \cdot \frac{m_{1} m_{2}}{r^{2}} .$ $F = G \cdot \frac{m_{1} m_{2}}{r^{2}} .$

The constant of proportionality G is called the universal gravitational constant. It is “universal” because it is thought to be the same at all places and all times. If the masses $m_{1}$ $m_{1}$ and $m_{2}$ $m_{2}$ are measured in kilograms, r is measured in meters, and the force F is measured in newtons, then the value of G is approximately $6.67 \times 10^{- 11}$ $6.67 \times 10^{- 11}$ meters cubed per kilogram per second squared. This value was experimentally verified by Henry Cavendish in 1795.
Estimating the value of g (acceleration due to gravity) near the surface of Earth. We use Newton’s laws: $Force = Mass \times Acceleration = m \cdot g$ $Force = Mass \times Acceleration = m \cdot g$

$Force = G \cdot \frac{m_{1} m_{2}}{r^{2}}$ $Force = G \cdot \frac{m_{1} m_{2}}{r^{2}}$

For an object of mass m near the surface of Earth (see Figure 3.37):

Figure 3.37

Finding g on Earth

$\begin{array}{rcll} Force & = & G \cdot \frac{m_{1} m_{2}}{r^{2}} & \begin{array}{l} Law of Universal \\ Gravitation \end{array} \\ m \cdot g & = & G \cdot \frac{m M_{E}}{R_{E}^{2}} & \begin{array}{l} Force = m \cdot g; replace m_{1} \\ with m, m_{2} with M_{E}, \\ and r with R_{E} . \end{array} \\ g & = & G \cdot \frac{M_{E}}{R_{E}^{2}} & Divide both sides by m . \\ = & \frac{(6.67 \times 10^{- 11} m^{3} / kg / {sec}^{2}) \cdot (5.98 \times 10^{24} kg)}{{(6.38 \times 10^{6} m)}^{2}} & \begin{array}{l} Substitute appropriate \\ values for G, M_{E}, and R_{E} . \end{array} \\ \approx & 9.8 {m/sec}^{2} & Use a calculator . \end{array}$ $\begin{array}{rcll} Force & = & G \cdot \frac{m_{1} m_{2}}{r^{2}} & \begin{array}{l} Law of Universal \\ Gravitation \end{array} \\ m \cdot g & = & G \cdot \frac{m M_{E}}{R_{E}^{2}} & \begin{array}{l} Force = m \cdot g; replace m_{1} \\ with m, m_{2} with M_{E}, \\ and r with R_{E} . \end{array} \\ g & = & G \cdot \frac{M_{E}}{R_{E}^{2}} & Divide both sides by m . \\ = & \frac{(6.67 \times 10^{- 11} m^{3} / kg / {sec}^{2}) \cdot (5.98 \times 10^{24} kg)}{{(6.38 \times 10^{6} m)}^{2}} & \begin{array}{l} Substitute appropriate \\ values for G, M_{E}, and R_{E} . \end{array} \\ \approx & 9.8 {m/sec}^{2} & Use a calculator . \end{array}$

Practice Problem 6

The mass of Mars is about $6.42 \times 10^{23}$ $6.42 \times 10^{23}$ kilograms, and its radius is about 3397 kilometers. What is the acceleration due to gravity near the surface of Mars?

We summarize the steps involved in solving most variation problems.

If a variation problem uses a more complicated relationship than those listed in Step 1, simply write the appropriate equation involving k and the variables and then proceed with Steps 2–4.

Section 3.7 Exercises

Concepts and Vocabulary

y varies directly as x if .
y varies inversely as x if .
y varies directly as the nth power of x if .
z varies jointly as x and y if .
True or False. In the equation $P = 2 T,$ $P = 2 T,$ P varies directly as T.
True or False. In the equation $A = 3 r^{2},$ $A = 3 r^{2},$ A varies directly as the third power of r.
True or False. In the equation $D = 2 t + t^{2},$ $D = 2 t + t^{2},$ D varies directly as the second power of t.
True or False. In the equation $Q = \frac{π ω^{3}}{\sqrt{u}},$ $Q = \frac{π ω^{3}}{\sqrt{u}},$ Q Varies jointly as the third power of $ω$ $ω$ and inversely as the square root of u.

Building Skills

In Exercises 9–24, use the four-step procedure to solve for the variable requested.

x varies directly as y, and $x = 15$ $x = 15$ when $y = 30 .$ $y = 30 .$ Find x if $y = 28 .$ $y = 28 .$
y varies directly as x, and $y = 3$ $y = 3$ when $x = 2 .$ $x = 2 .$ Find y when $x = 7 .$ $x = 7 .$
s varies directly as the square of t, and $s = 64$ $s = 64$ when $t = 2 .$ $t = 2 .$ Find s when $t = 5 .$ $t = 5 .$
y varies directly as the cube of x, and $y = 270$ $y = 270$ when $x = 3 .$ $x = 3 .$ Find x when $y = 80 .$ $y = 80 .$
r varies inversely as u, and $r = 3$ $r = 3$ when $u = 11 .$ $u = 11 .$ Find r if $u = \frac{1}{3} .$ $u = \frac{1}{3} .$
y varies inversely as z, and $y = 24$ $y = 24$ when $z = \frac{1}{6} .$ $z = \frac{1}{6} .$ Find y if $z = 1 .$ $z = 1 .$
B varies inversely with the cube of A. If $B = 1$ $B = 1$ when $A = 2,$ $A = 2,$ find B when $A = 4 .$ $A = 4 .$
y varies inversely with the cube root of x, and $y = 10$ $y = 10$ when $x = 2 .$ $x = 2 .$ Find x if $y = 40 .$ $y = 40 .$
z varies jointly as x and y, and $z = 42$ $z = 42$ when $x = 2$ $x = 2$ and $y = 3 .$ $y = 3 .$ Find y if $z = 56$ $z = 56$ and $x = 2 .$ $x = 2 .$
m varies directly as q and inversely as p, and $m = \frac{1}{2}$ $m = \frac{1}{2}$ when $p = 26$ $p = 26$ and $q = 13 .$ $q = 13 .$ Find m when $p = 14$ $p = 14$ and $q = 7 .$ $q = 7 .$
z varies directly as the square of x, and $z = 32$ $z = 32$ when $x = 4 .$ $x = 4 .$ Find z if $x = 5 .$ $x = 5 .$
u varies inversely as the cube of t, and $u = 9$ $u = 9$ when $t = 2 .$ $t = 2 .$ Find u if $t = 6 .$ $t = 6 .$
P varies jointly as T and the square of Q, and $P = 36$ $P = 36$ when $T = 17$ $T = 17$ and $Q = 6 .$ $Q = 6 .$ Find P when $T = 4$ $T = 4$ and $Q = 9 .$ $Q = 9 .$
a varies jointly as b and the square root of c, and $a = 9$ $a = 9$ when $b = 13$ $b = 13$ and $c = 81 .$ $c = 81 .$ Find a when $b = 5$ $b = 5$ and $c = 9 .$ $c = 9 .$
z varies directly as the square root of x and inversely as the square of y, and $z = 24$ $z = 24$ when $x = 16$ $x = 16$ and $y = 3 .$ $y = 3 .$ Find x when $z = 27$ $z = 27$ and $y = 2 .$ $y = 2 .$
z varies jointly as u and the cube of v and inversely as the square of $w;$ $w;$ $z = 9$ $z = 9$ when $u = 4, v = 3,$ $u = 4, v = 3,$ and $w = 2 .$ $w = 2 .$ Find w when $u = 27, v = 2,$ $u = 27, v = 2,$ and $z = 8 .$ $z = 8 .$

In Exercises 25–28, solve for the variable requested without determining the constant of variation, k. Use the fact that if $x_{1} = k y_{1}$ $x_{1} = k y_{1}$ and $x_{2} = k y_{2},$ $x_{2} = k y_{2},$ then $\frac{x_{1}}{y_{1}} = k = \frac{x_{2}}{y_{2}}$ $\frac{x_{1}}{y_{1}} = k = \frac{x_{2}}{y_{2}}$ so that $\frac{x_{1}}{y_{1}} = \frac{x_{2}}{y_{2}} .$ $\frac{x_{1}}{y_{1}} = \frac{x_{2}}{y_{2}} .$

If y is proportional to x and if $y = 12$ $y = 12$ when $x = 16,$ $x = 16,$ find y when $x = 8 .$ $x = 8 .$
If z varies directly as w and if $z = 17$ $z = 17$ when $w = 22,$ $w = 22,$ find z when $w = 110 .$ $w = 110 .$
If y is directly proportional to x and if $y = 100$ $y = 100$ for the value $x_{0}$ $x_{0}$ of x, find y when $x_{0}$ $x_{0}$ is doubled—that is, when $x = 2 x_{0} .$ $x = 2 x_{0} .$
If x varies directly as the square root of y and if $x = 2$ $x = 2$ when $y = 9,$ $y = 9,$ find y when $x = 3 .$ $x = 3 .$

Applying the Concepts

Hubble constant. The American astronomer Edwin Powell Hubble (1889–1953) is renowned for having determined that there are other galaxies in the universe beyond the Milky Way. In 1929, he stated that the galaxies observed at a particular time recede from each other at a speed that is directly proportional to the distance between them. The constant of proportionality is denoted by the letter H. Write Hubble’s statement in the form of an equation.
Malthusian doctrine. The British scientist Thomas Robert Malthus (1766–1834) was a pioneer of population science and economics. In 1798, he stated that populations grow faster than the means that can sustain them. One of his arguments was that the rate of change, R, of a given population is directly proportional to the size P of the population. Write the equation that describes Malthus’s argument.
Converting units of length. Use the fact that $1 foot \approx 30.5 centimeters .$ $1 foot \approx 30.5 centimeters .$
1. Write an equation that expresses the fact that a length measured in centimeters is directly proportional to the length measured in feet.
2. Convert the following measurements into centimeters.
  1. 8 feet
  2. 5 feet 4 inches
3. Convert the following measurements into feet.
  1. 57 centimeters
  2. 1 meter 24 centimeters
Converting weight. Use the fact that $1 kilogram \approx 2.20 pounds .$ $1 kilogram \approx 2.20 pounds .$
1. Write an equation that expresses the fact that a weight measured in pounds is directly proportional to the weight measured in kilograms.
2. Convert the following measurements to pounds.
  1. 125 grams
  2. 4 kilograms
  3. 2.4 kilograms
3. Convert the following measurements to kilograms.
  1. 27 pounds
  2. 160 pounds
Protein from soybeans. The quantity P of protein obtained from dried soybeans is directly proportional to the quantity Q of dried soybeans used. If 20 grams of dried soybeans produces 7 grams of protein, how much protein can be produced from 100 grams of soybeans?
Wages. A person’s weekly wages W are directly proportional to the number of hours h worked per week. If Samantha earned $600 for a 40-hour week, how much would she earn if she worked only 25 hours in a particular week? What does the constant of proportionality mean here?
Physics. The distance d that an object falls varies directly as the square of the time t during which it is falling. If an object falls 64 feet in 2 seconds, how long will it take the object to fall 9 feet?
Physics. Hooke’s Law states that the force F required to stretch a spring by x units is directly proportional to x. If a force of 10 pounds stretches a spring by 4 inches, find the force required to stretch a spring by 6 inches.
Chemistry. Boyle’s Law states that at a constant temperature, the pressure P of a compressed gas is inversely proportional to its volume V. If the pressure is 20 pounds per square inch when the volume of the gas is 300 cubic inches, what is the pressure when the gas is compressed to 100 cubic inches?
Chemistry. In the Kelvin temperature scale, the lowest possible temperature (called absolute zero) is 0 K, where K denotes degrees Kelvin. The relationship between Kelvin temperature $(T_{K})$ $(T_{K})$ and Celsius temperature $(T_{C})$ $(T_{C})$ is given by $T_{K} = T_{C} + 273 .$ $T_{K} = T_{C} + 273 .$

The pressure P exerted by a gas varies directly as its temperature $T_{K}$ $T_{K}$ and inversely as its volume V. Assume that at a temperature of 260 K, a gas occupies 13 cubic inches at a pressure of 36 pounds per square inch.
1. Find the volume of the gas when the temperature is 300 K and the pressure is 40 pounds per square inch.
2. Find the pressure when the temperature is 280 K and the volume is 39 cubic inches.
Weight. The weight of an object varies inversely as the square of the object’s distance from the center of Earth. The radius of Earth is 3960 miles.
1. If an astronaut weighs 120 pounds on the surface of Earth, how much does she weigh 6000 miles above the surface of Earth?
2. If a miner weighs 200 pounds on the surface of Earth, how much does he weigh 10 miles below the surface of Earth?
Making a profit. Suppose you wanted to make a profit by buying gold by weight at one altitude and selling at another altitude for the same price per unit weight. Should you buy or sell at the higher altitude? (Use the information from Exercise 39.)

In Exercises 41 and 42, use Newton’s Law of Universal Gravitation. (See Example 6.)

Gravity on the moon. The mass of the moon is about $7.4 \times 10^{22}$ $7.4 \times 10^{22}$ kilograms, and its radius is about 1740 kilometers. How much is the acceleration due to gravity on the surface of the moon?
Gravity on the sun. The mass of the sun is about $2 \times 10^{30}$ $2 \times 10^{30}$ kilograms, and its radius is 696,000 kilometers. How much is the acceleration due to gravity on the surface of the sun?
Illumination. The intensity I of illumination from a light source is inversely proportional to the square of the distance d from the source. Suppose the intensity is 320 candela at a distance of 10 feet from a light source.
1. What is the intensity at 5 feet from the source?
2. How far away from the source will the intensity be 400 candela?
Speed and skid marks. Police estimate that the speed s of a car in miles per hour varies directly as the square root of d, where d in feet is the length of the skid marks left by a car traveling on a dry concrete pavement. A car traveling 48 miles per hour leaves skid marks of 96 feet.
1. Write an equation relating s and d.
2. Use the equation in part (a) to estimate the speed of a car whose skid marks stretched (i) 60 feet, (ii) 150 feet, and (iii) 200 feet.
3. Suppose you are driving 70 miles per hour and slam on your brakes. How long will your skid marks be?
Simple pendulum. Periodic motion is motion that repeats itself over successive equal intervals of time. The time required for one complete repetition of the motion is called the period. The period of a simple pendulum varies directly as the square root of its length. What is the effect on the length if the period is doubled?

[Note: The amplitude of a pendulum has no effect on its period. This is what makes pendulums such good timekeepers. Because they invariably lose energy due to friction, their amplitude decreases but their period remains constant.]
Biology. The volume V of a lung is directly proportional to its internal surface area A. A lung with volume 400 cubic centimeters from a certain species has an average internal surface area of 100 square centimeters. Find the volume of a lung of a member of this species if the lung’s internal surface area is 120 square centimeters.
Horsepower. The horsepower H of an automobile engine varies directly as the square of the piston radius R and the number N of pistons.
1. Write the given information in equation form.
2. What is the effect on the horsepower if the piston radius is doubled?
3. What is the effect on the horsepower if the number of pistons is doubled?
4. What is the effect on the horsepower if the radius of the pistons is cut in half and the number of pistons is doubled?
Safe load. The safe load that a rectangular beam can support varies jointly as the width and square of the depth of the beam and inversely as its length. A beam 4 inches wide, 6 inches deep, and 25 feet long can support a safe load of 576 pounds. Find the safe load for a beam that is of the same material but is 6 inches wide, 10 inches deep, and 20 feet long.

Beyond the Basics

Energy from a windmill. The energy E from a windmill varies jointly as the square of the length l of the blades and the cube of the wind velocity v.
1. Express the given information as an equation with k as the constant of variation.
2. If blades of length 10 feet and wind velocity 8 miles per hour generate 1920 watts of electric power, find k.
3. How much electric power would be generated if the blades were 8 feet long and the wind velocity was 25 miles per hour?
4. If the velocity of the wind doubles, what happens to E?
5. If the length of the blades doubles, what happens to E?
6. What happens to E if both the length of the blades and the wind velocity are doubled?
Intensity of light. The intensity of light, $I_{d},$ $I_{d},$ at a distance d from the source of light varies directly as the intensity I of the source and inversely as $d^{2} .$ $d^{2} .$ It is known that at a distance of 2 meters, a 100-watt bulb produces an intensity of approximately 2 watts per square meter.
1. Find the constant of variation, k.
2. Suppose a 200-watt bulb is located on a wall 2 meters above the floor. What is the intensity of light at a point A that is 3 meters from the wall?
3. What is the intensity of illumination at the point A in the figure if the bulb is raised by 1 meter?
Metabolic rate. Metabolism is the sum total of all physical and chemical changes that take place within an organism. According to the laws of thermodynamics, all of these changes will ultimately release heat, so the metabolic rate is a measure of heat production by an animal. Biologists have found that the normal resting metabolic rate of a mammal is directly proportional to the $\frac{3}{4}$ $\frac{3}{4}$ power of its body weight.

The resting metabolic rate of a person weighing 75 kilograms is 75 watts.
1. Find the constant of proportionality, k.
2. Estimate the resting metabolic rate of a brown bear weighing 450 kilograms.
3. What is the effect on metabolic rate if the body weight is multiplied by 4?
4. What is the approximate weight of an animal with a metabolic rate of 250 watts?
Comparing gravitational forces. The masses of the sun, Earth, and moon are $2 \times 10^{30}$ $2 \times 10^{30}$ kilograms, $6 \times 10^{24}$ $6 \times 10^{24}$ kilograms, and $7.4 \times 10^{22}$ $7.4 \times 10^{22}$ kilograms, respectively. The Earth–sun distance is about 400 times the Earth–moon distance. Use Newton’s Law of Universal Gravitation to compare the gravitational attraction between the sun and Earth with that between Earth and the moon.
Kepler’s Third Law. Suppose an object of mass $M_{1}$ $M_{1}$ orbits around an object of mass $M_{2} .$ $M_{2} .$ Let r be the average distance in meters between the centers of the two objects and let T be the orbital period (the time in seconds the object completes one orbit). Kepler’s Third Law states that $T^{2}$ $T^{2}$ is directly proportional to $r^{3}$ $r^{3}$ and inversely proportional to $M_{1} + M_{2} .$ $M_{1} + M_{2} .$ The constant of proportionality is $\frac{4 π^{2}}{G} .$ $\frac{4 π^{2}}{G} .$
1. Write the equation that expresses Kepler’s Third Law.
2. Earth orbits the sun once each year at a distance of about $1.5 \times 10^{8}$ $1.5 \times 10^{8}$ kilometers. Find the mass of the sun. Use these estimates:
  
  $\begin{array}{l} Mass of Earth + Mass of Sun \approx Mass of Sun, \\ G = 6.67 \times 10^{- 11}, 1 year = 3.15 \times 10^{7} seconds \end{array}$ $\begin{array}{l} Mass of Earth + Mass of Sun \approx Mass of Sun, \\ G = 6.67 \times 10^{- 11}, 1 year = 3.15 \times 10^{7} seconds \end{array}$
Use Kepler’s Third Law. The moon orbits Earth in about 27.3 days at an average distance of about 384,000 kilometers. Find the mass of Earth. [Hint: Convert the orbital period, defined in Exercise 53, to seconds; also use $Mass of Earth + Mass of Moon \approx Mass of Earth .$ $Mass of Earth + Mass of Moon \approx Mass of Earth .$ ]
Spreading a rumor. Let P be the population of a community. The rate R (per day) at which a rumor spreads in the community is jointly proportional to the number N of people who have already heard the rumor and the number $(P - N)$ $(P - N)$ of people who have not yet heard the rumor.

A rumor about the president of a college is spread in his college community of 10,000 people. Five days after the rumor started, 1000 people had heard it, and it was spreading at the rate of 45 additional people per day.
1. Write an equation relating R, P, and N.
2. Find the constant k of variation.
3. Find the rate at which the rumor was spreading when one-half the college community had heard it.
4. How many people had heard the rumor when it was spreading at the rate of 100 people per day?

Critical Thinking / Discussion / Writing

Electricity. The current I in an electrical circuit varies directly as the voltage V and inversely as the resistance R. If the resistance is increased by 20%, what percent increase must occur in the voltage to increase the current by 30%?
Precious stones. The value of a precious stone is proportional to the square of its weight.
1. Calculate the loss incurred by cutting a diamond worth $1000 into two pieces whose weights are in the ratio 2:3.
2. A precious stone worth $25,000 is accidentally dropped and broken into three pieces, the weights of which are in the ratio 5:9:11. Calculate the loss incurred due to breakage.
3. A diamond breaks into five pieces, the weights of which are in the ratio 1:2:3:4:5. If the resulting loss is $85,000, find the value of the original diamond. Also calculate the value of a diamond whose weight is twice that of the original diamond.
Bus service. The profit earned in running a bus service is jointly proportional to the distance and the number of passengers over a certain fixed number. The profit is $80 when 30 passengers are carried a distance of 40 km and is $180 when 35 passengers are carried 60 km. What is the minimum number of passengers that results in no loss?
Weight of a sphere. The weight of a sphere is directly proportional to the cube of its radius. A metal sphere has a hollow space about its center in the form of a concentric sphere, and its weight is $\frac{7}{8}$ $\frac{7}{8}$ times the weight of a solid sphere of the same radius and material. Find the ratio of the inner to the outer radius of the hollow sphere.
Train speed. A locomotive engine can go 24 miles per hour, and its speed is reduced by a quantity that varies directly as the square root of the number of cars it pulls. Pulling four cars, its speed is 20 miles per hour. Find the greatest number of cars the engine can pull.

Getting Ready for the Next Section

In Exercises 61–70, simplify each expression.

$5^{0}$ $5^{0}$
$2^{3}$ $2^{3}$
$3^{- 2}$ $3^{- 2}$
${(\frac{1}{2})}^{3}$ ${(\frac{1}{2})}^{3}$
${(\frac{1}{2})}^{- 4}$ ${(\frac{1}{2})}^{- 4}$
$2^{x - 1} \cdot 2^{3 - x}$ $2^{x - 1} \cdot 2^{3 - x}$
$5^{2 x - 3} \cdot 5^{3 - x}$ $5^{2 x - 3} \cdot 5^{3 - x}$
$\frac{2^{3 x - 2}}{2^{x - 5}}$ $\frac{2^{3 x - 2}}{2^{x - 5}}$
${(2^{x - 1})}^{x}$ ${(2^{x - 1})}^{x}$
$\sqrt[3]{\sqrt[4]{2}}$ $\sqrt[3]{\sqrt[4]{2}}$

In Exercises 71–76, solve each equation for the requested variable.

$y = m x + b$ $y = m x + b$ for m
$a x + b y = 3$ $a x + b y = 3$ for x
$A = B (1 + C)$ $A = B (1 + C)$ for C
$A = B {(1 + C)}^{3}$ $A = B {(1 + C)}^{3}$ for C
$A = B \cdot 10^{- n}$ $A = B \cdot 10^{- n}$ for B
$A = B \cdot 10^{m} + C$ $A = B \cdot 10^{m} + C$ for B

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`y` varies directly with `x`.	$y = k x$ $y = k x$
`y` varies with the `n`th power of `x`.	$y = k x^{n}$ $y = k x^{n}$
`y` varies inversely with `x`.	$y = \frac{k}{x}$ $y = \frac{k}{x}$
`y` varies inversely with the `n`th power of `x`.	$y = \frac{k}{x^{n}}$ $y = \frac{k}{x^{n}}$
`z` varies jointly with the `n`th power of `x` and the `m`th power of `y`.	$z = k x^{n} y^{m}$ $z = k x^{n} y^{m}$
`z` varies directly with the `n`th power of `x` and inversely with the `m`th power of `y`.	$z = \frac{k x^{n}}{y^{m}}$ $z = \frac{k x^{n}}{y^{m}}$

Table of Contents for Section 3.7 Variation

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Table of Contents for
Section 3.7 Variation