section-3-1.xml

<?xml version="1.0" encoding="UTF-8" ?>
<!-- <mathbook><book> -->

<section xml:id="Variation"   xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Variation</title><introduction>
<p>
    Two types of functions are widely used in modeling and are known by special names: <term>direct variation</term> and <term>inverse variation</term>.
</p></introduction>
<subsection><title>Direct Variation</title><idx>direct variation</idx>
<p>
    Two variables are <term>directly proportional</term><idx>directly proportional</idx> (or just <term>proportional</term><idx>proportional</idx>) if the ratios of their corresponding values are always equal. Consider the functions described in Tables <xref ref="table-gasoline" text="global"/> and <xref ref="table-population" text="global"/>. The first table shows the price of gasoline as a function of the number of gallons purchased.
</p>
<sidebyside>
<table xml:id="table-gasoline"><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><line>Gallons of</line><line>gasoline</line></cell>
            <cell><line>Total</line><line>price</line></cell>
            <cell><line>Price/</line><line>Gallons</line></cell>
        </row>
        <row>
            <cell><m>4</m></cell>
            <cell><m>\$9.60</m></cell>
            <cell><m>\dfrac{9.60}{4}=2.40</m></cell>
        </row>
        <row>
            <cell><m>6</m></cell>
            <cell><m>\$14.40</m></cell>
            <cell><m>\dfrac{14.40}{6}=2.40</m></cell>
        </row>
        <row>
            <cell><m>8</m></cell>
            <cell><m>\$19.20</m></cell>
            <cell><m>\dfrac{19.20}{8}=2.40</m></cell>
        </row>
        <row>
            <cell><m>12</m></cell>
            <cell><m>\$28.80</m></cell>
            <cell><m>\dfrac{28.80}{12}=2.40</m></cell>
        </row>
        <row>
            <cell><m>15</m></cell>
            <cell><m>\$36.00</m></cell>
            <cell><m>\dfrac{36.00}{15}=2.40</m></cell>
        </row>
    </tabular><caption></caption>
</table>

<table xml:id="table-population"><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell>Years</cell>
            <cell>Population</cell>
            <cell>People/Years</cell>
        </row>
        <row>
            <cell><m>10</m></cell>
            <cell><m>432</m></cell>
            <cell><m>\dfrac{432}{10}\approx 43</m></cell>
        </row>
        <row>
            <cell><m>20</m></cell>
            <cell><m>932</m></cell>
            <cell><m>\dfrac{932}{20}\approx 47</m></cell>
        </row>
        <row>
            <cell><m>30</m></cell>
            <cell><m>2013</m></cell>
            <cell><m>\dfrac{2013}{30}\approx 67</m></cell>
        </row>
        <row>
            <cell><m>40</m></cell>
            <cell><m>4345</m></cell>
            <cell><m>\dfrac{4345}{40}\approx 109</m></cell>
        </row>
        <row>
            <cell><m>50</m></cell>
            <cell><m>9380</m></cell>
            <cell><m>\dfrac{9380}{50}\approx 188</m></cell>
        </row>
        <row>
            <cell><m>60</m></cell>
            <cell><m>20,251</m></cell>
            <cell><m>\dfrac{20,251}{60}\approx 338</m></cell>
        </row>
    </tabular><caption></caption>
</table>
</sidebyside>

<p>
	The ratio <m>\dfrac{\text{total price}}{\text{number of gallons}}</m>, or price per gallon, is the same for each pair of values in <xref ref="table-gasoline" text="type-global"/>. This agrees with everyday experience: The price per gallon of gasoline is the same no matter how many gallons you buy. Thus, the total price of a gasoline purchase is directly proportional to the number of gallons purchased.
</p>

<p>
	The second table, <xref ref="table-population" text="type-global"/>, shows the population of a small town as a function of the town’s age. The ratio <m>\dfrac{\text{number of people}}{\text{number of years}}</m> gives the average rate of growth of the population in people per year. You can see that this ratio is not constant; in fact, it increases as time goes on. Thus, the population of the town is not proportional to its age.
</p>
<figure xml:id="fig-gasoline-and-population-graphs"><caption></caption><image source="images/fig-gasoline-and-population-graphs"  width="80%"><description>graphs of gas price and population growth</description></image></figure>

<p>
	The graphs of these two functions are shown in <xref ref="fig-gasoline-and-population-graphs" text="type-global"/>. We see that the price, <m>P</m>, of a fill-up is a linear function of the number of gallons, <m>g</m>, purchased. This should not be surprising if we write an equation relating the variables <m>g</m> and <m>P</m>. Because the ratio of their values is constant, we can write
	<me>\frac{P}{g}= k</me>
	where <m>k</m> is a constant. In this example, the constant <m>k</m> is <m>2.40</m>, the price of gasoline per gallon. Solving for <m>P</m> in terms of <m>g</m>, we have
	<me>P = kg = 2.40g</me>
	which we recognize as the equation of a line through the origin.
</p>

<p>
	In general, we make the following definition.
</p>

<assemblage><title>Direct Variation</title>
<idx>direct variation</idx>
<p>
	<m>y</m> varies directly with <m>x</m> if
	<me>y = kx</me>
	where <m>k</m> is a positive constant called the <term>constant of variation</term><idx>constant of variation</idx>.
</p></assemblage>

<note><p>
	From the preceding discussion, we see that <em>vary directly</em> means exactly the same thing as <em>are directly proportional</em>. The two phrases are interchangeable.
</p></note>

<example xml:id="example-circumference-and-simple-interest">
	<ol label="a">
		<li>The circumference, <m>C</m>, of a circle varies directly with its radius, <m>r</m>, because 
		<me>C = 2\pi r </me>
		The constant of variation is <m>2\pi</m>, or about 6.28.</li>
		<li>The amount of interest, <m>I</m>, earned in one year on an account paying 7<percent/> simple interest, varies directly with the principal, <m>P</m>, invested, because 
		<me>I = 0.07P</me>
		</li>
	</ol>
</example>

<p>
	Direct variation defines a linear function of the form
	<me>y = f (x) = kx</me>
	The positive constant <m>k</m> in the equation <m>y = kx</m> is just the slope of the graph, so it tells us how rapidly the graph increases. Compared to the standard form for a linear function, <m>y = b + mx</m>, the constant term, <m>b</m>, is zero, so the graph of a direct variation passes through the origin.
</p>

<exercise xml:id="exercise-three-lines"><statement>
	Which of the graphs in <xref ref="fig-three-lines" text="type-global"/> could represent direct variation? Explain why.
	<figure xml:id="fig-three-lines"><caption></caption><image source="images/fig-three-lines"  width="80%"><description>graphs of three lines</description></image></figure>
</statement>
<answer><p>(b): The graph is a straight line through the origin.</p></answer>
</exercise>

</subsection>


<subsection><title>The Scaling Property of Direct Variation</title>
<p>
	The fact that the constant term is zero in a direct variation is significant: If we double the value of <m>x</m>, then the value of <m>y</m> will double also. In fact, increasing <m>x</m> by any factor causes <m>y</m> to increase by the same factor. For example, in <xref ref="table-gasoline" text="type-global"/> doubling the number of gallons of gas purchased, say, from <m>4</m> gallons to <m>8</m> gallons or from <m>6</m> gallons to <m>12</m> gallons, causes the total price to double also. </p>

    <p>Or, consider investing <m>\$800</m> for one year at <m>7</m><percent/> simple interest, as in <xref ref="example-circumference-and-simple-interest" text="type-global"/>b. The interest earned is
	<me>I = 0.07(800) = \$56</me>
	If we increase the investment by a factor of <m>1.6</m> to <m>1.6 (800)</m>, or <m>\$1280</m>, the interest will be 
	<me>I = 0.07(1280) = \$89.60</me>
	You can check that multiplying the original interest of <m>\$56</m> by a factor of <m>1.6</m> does give the same figure for the new interest, <m>\$89.60</m>.
</p>

<example xml:id="example-two-linear-functions">
<ol label="a">
	<li>Tuition at Woodrow University is <dollar/>400 plus <dollar/>30 per unit. Is the tuition proportional to the number of units you take?</li>
	<li>Imogen makes a 15<percent/> commission on her sales of environmentally friendly products marketed by her co-op. Do her earnings vary directly with her sales?</li>
</ol>
<solution>
<ol label="a">
	<li>Let <m>u</m> represent the number of units you take, and let <m>T(u)</m> represent your tuition. Then
	<me>T(u) = 400 + 30u</me>
	Thus, <m>T(u)</m> is a linear function of <m>u</m>, but the <m>T</m>-intercept is <m>400</m>, not <m>0</m>. Your tuition is <em>not</em> proportional to the number of units you take, so this is not an example of direct variation. You can check that doubling the number of units does not double the tuition. For example,
	<me>T (6) = 400 + 30(6) = 580</me>
	and
	<me>T (12) = 400 + 30(12) = 760</me>
	Tuition for <m>12</m> units is not double the tuition for <m>6</m> units. The graph of <m>T (u)</m> in <xref ref="fig-GC-graphs-of-two-linear-functions" text="type-global"/>a does not pass through the origin.
	<figure xml:id="fig-GC-graphs-of-two-linear-functions"><caption></caption><image source="images/fig-GC-graphs-of-two-linear-functions"  width="100%"><description>calculator graphs of two linear functions</description></image></figure>
	</li>
	<li>
		Let <m>S</m> represent Imogen’s sales, and let <m>C(S)</m> represent her commission. Then 
		<me>C(S) = 0.15S</me>
		Thus, <m>C(S)</m> is a linear function of <m>S</m> with a <m>C</m>-intercept of zero, so Imogen's earnings do vary directly with her sales. This is an example of direct variation. (See <xref ref="fig-GC-graphs-of-two-linear-functions" text="type-global"/>b.)
	</li>
</ol>
</solution>
</example>

<exercise xml:id="exercise-two-tables"><statement>
Which table could represent direct variation? Explain why. (Hint: What happens to
<m>y</m> if you multiply <m>x</m> by a constant?)
<ol label="a">
	<li><p>
		<sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>x</m></cell>
            <cell><m>1</m></cell>
            <cell><m>2</m></cell>
            <cell><m>3</m></cell>
            <cell><m>6</m></cell>
            <cell><m>8</m></cell>
            <cell><m>9</m></cell>
        </row>
        <row>
            <cell><m>y</m></cell>
            <cell><m>2.5</m></cell>
            <cell><m>5</m></cell>
            <cell><m>7.5</m></cell>
            <cell><m>15</m></cell>
            <cell><m>20</m></cell>
            <cell><m>22.5</m></cell>
        </row>
    </tabular></sidebyside></p>
	</li>


	<li><p>
		<sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>x</m></cell>
            <cell><m>2</m></cell>
            <cell><m>3</m></cell>
            <cell><m>4</m></cell>
            <cell><m>6</m></cell>
            <cell><m>8</m></cell>
            <cell><m>9</m></cell>
        </row>
        <row>
            <cell><m>y</m></cell>
            <cell><m>2</m></cell>
            <cell><m>3.5</m></cell>
            <cell><m>5</m></cell>
            <cell><m>7</m></cell>
            <cell><m>8.5</m></cell>
            <cell><m>10</m></cell>
        </row>
    </tabular></sidebyside></p>
	</li>

</ol>
</statement>
<answer><p>(a): If we multiply <m>x</m> by <m>c</m>, <m>y</m> is also multiplied by <m>c</m>.</p></answer>
</exercise>

</subsection>



<subsection><title>Finding a Formula for Direct Variation</title>
<p>
	If we know any one pair of values for the variables in a direct variation, we can find the constant of variation. We can then use the constant to write a formula for one of the variables as a function of the other.
</p>

<example xml:id="example-falling-into-grand-canyon">
<p>
	If an object is dropped from a great height, its speed, <m>v</m>, varies directly with the time, <m>t</m>, the object has been falling. A rock dropped off the rim of the Grand Canyon is falling at a speed of <m>39.2</m> meters per second when it passes a lizard on a ledge <m>4</m> seconds later.
</p>
<ol label="a">
	<li>Express <m>v</m> as a function of <m>t</m>.</li>
	<li>What is the speed of the rock after it has fallen for 6 seconds?</li>
	<li>Sketch a graph of <m>v(t)</m> versus <m>t</m>.</li>
</ol>

<solution>
	<ol label="a">
		<li>Because <m>v</m> varies directly with <m>t</m>, there is a positive constant <m>k</m> for which <m>v = kt</m>.
		We substitute <m>v = 39.2</m> when <m>t = 4</m> and solve for <m>k</m> to find
		\begin{align*}
		39.2 \amp = k(4)\amp\amp \text{Divide both sides by 4.}\\
		k \amp = 9.8
		\end{align*}
		Thus, <m>v (t) = 9.8 t</m>.</li>
		<li>We evaluate the function found in part (a) for <m>t = 6</m>.
		<me>v(\alert{6}) = 9.8(\alert{6}) = 58.8</me>
		At <m>t = 6</m> seconds, the rock is falling at a speed of <m>58.8</m> meters per second.</li>
		<li>You can use your calculator to graph the function <m>v(t) = 9.8t</m>. The graph is shown in <xref ref="fig-GC-falling-into-grand-canyon" text="type-global"/>.
		<figure xml:id="fig-GC-falling-into-grand-canyon"><caption></caption><image source="images/fig-GC-falling-into-grand-canyon"  width="50%"><description>GC graph of speed of rock falling into grand canyon</description></image></figure>
		</li>
	</ol>
</solution>
</example>

<exercise xml:id="exercise-rice"><statement>
	The volume of a bag of rice, in cups, is directly proportional to the weight of the bag. A <m>2</m>-pound bag contains <m>3.5</m> cups of rice.
	<ol label="a">
		<li>
			Express the volume, <m>V</m>, of a bag of rice as a function of its weight, <m>w</m>.
		</li>
		<li>
			How many cups of rice are in a <m>15</m>-pound bag?
		</li>
	</ol>
</statement>
<answer><p><ol label="a" cols="2">
    <li><p><m>V = 1.75w</m></p></li>
    <li><p><m>26.25</m></p></li>
</ol></p></answer>
</exercise>

</subsection>


<subsection><title>Direct Variation with a Power of <m>x</m></title>
<p>
	We can generalize the notion of direct variation to include situations in which <m>y</m> is proportional to a power of <m>x</m>, instead of <m>x</m> itself.
</p>	

<assemblage><title>Direct Variation with a Power</title>
<idx>direct variation with a power</idx><p></p>
<p>
	<m>y</m> <term>varies directly</term><idx>varies directly</idx> with a power of <m>x</m> if
	<me>y = kx^n</me>
	where <m>k</m> and <m>n</m> are positive constants.
</p>
</assemblage>

<example xml:id="example-sphere-surface-area">
<p>
	The surface area of a sphere varies directly with the <em>square</em> of its radius. A balloon of radius <m>5</m> centimeters has surface area <m>100\pi</m> square centimeters, or about <m>314</m> square centimeters. Find a formula for the surface area of a sphere as a function of its radius.
</p>
<solution>
	If <m>S</m> stands for the surface area of a sphere of radius <m>r</m>, then
	<me>S = f (r ) = kr^2</me>
	To find the constant of variation, <m>k</m>, we substitute the values of <m>S</m> and <m>r</m>.
	\begin{align*}
	100\pi \amp = k(5)^2\\
	4\pi \amp= k
	\end{align*}
	Thus, <m>S = f (r ) = 4πr^2</m>.
</solution>
</example>

<exercise xml:id="exercise-sphere-volume"><statement><p>
	The volume of a sphere varies directly with the <em>cube</em> of its radius. A balloon of radius <m>5</m> centimeters has volume <m>\dfrac{500\pi}{3}</m> cubic centimeters, or about <m>524</m> cubic centimeters. Find a formula for the volume of a sphere as a function of its radius.</p>
</statement>
<answer><p><m>V=\dfrac{4}{3}\pi r^3</m></p></answer>
</exercise>

<p>
	In any example of direct variation, as the input variable increases through positive values, the output variable increases also. Thus, a direct variation is an increasing function, as we can see when we consider the graphs of some typical direct variations in <xref ref="fig-vary-with-power" text="type-global"/>
</p>
<figure xml:id="fig-vary-with-power"><caption></caption><image source="images/fig-vary-with-power"  width="80%"><description>graphs of three variation with power</description></image></figure>

<warning><statement>
<p>
	The graph of a direct variation always passes through the origin, so when the input is zero, the output is also zero. Thus, the functions <m>y = 3x + 2</m> and <m>y = 0.4x^2 - 2.3</m>, for example, are not direct variation, even though they are increasing functions for positive <m>x</m>.
</p></statement>
</warning>

<p>
	Even without an equation, we can check whether a table of data describes direct variation or merely an increasing function. If <m>y</m> varies directly with <m>x^n</m> , then <m>y = kx^n</m> , or, equivalently,
	<m>\dfrac{y}{x^n} = k</m>.
</p>

<assemblage><title>Test for Direct Variation</title>
<p>
	If the ratio <m>\dfrac{y}{x^n}</m> is constant, then <m>y</m> varies directly with <m>x^n</m>.
</p></assemblage>

<example xml:id="example-test-variation-with-square">
	<p>
		Delbert collects the following data and would like to know if <m>y</m> varies directly with the square of <m>x</m>. What should he calculate?
	</p><p>
	<sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>x</m></cell>
            <cell><m>2</m></cell>
            <cell><m>5</m></cell>
            <cell><m>8</m></cell>
            <cell><m>10</m></cell>
            <cell><m>12</m></cell>
        </row>
        <row>
            <cell><m>y</m></cell>
            <cell><m>6</m></cell>
            <cell><m>16.5</m></cell>
            <cell><m>36</m></cell>
            <cell><m>54</m></cell>
            <cell><m>76</m></cell>
        </row>
    </tabular></sidebyside></p>
<solution>
	<p>If <m>y</m> varies directly with <m>x^2</m>, then <m>y = kx^2</m>, or <m>\dfrac{y}{x^2}= k</m>. Delbert should calculate the ratio <m>\dfrac{y}{x^2}</m> for each data point.</p><p>
	<sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>x</m></cell>
            <cell><m>2</m></cell>
            <cell><m>5</m></cell>
            <cell><m>8</m></cell>
            <cell><m>10</m></cell>
            <cell><m>12</m></cell>
        </row>
        <row>
            <cell><m>y</m></cell>
            <cell><m>6</m></cell>
            <cell><m>16.5</m></cell>
            <cell><m>36</m></cell>
            <cell><m>54</m></cell>
            <cell><m>76</m></cell>
        </row>
        <row>
            <cell><m>\dfrac{y}{x^2}</m></cell>
            <cell><m>1.5</m></cell>
            <cell><m>0.66</m></cell>
            <cell><m>0.56</m></cell>
            <cell><m>0.54</m></cell>
            <cell><m>0.53</m></cell>
        </row>
    </tabular></sidebyside></p>
	Because the ratio <m>\dfrac{y}{x^2}</m> is not constant, <m>y</m> does not vary directly with <m>x^2</m>.
</solution>
</example>

<exercise xml:id="exercise-vary-with-cube"><statement><p>
	Does <m>B</m> vary directly with the cube of <m>r</m>? Explain your decision.</p><p>
	<sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>r</m></cell>
            <cell><m>0.1</m></cell>
            <cell><m>0.3</m></cell>
            <cell><m>0.5</m></cell>
            <cell><m>0.8</m></cell>
            <cell><m>1.2</m></cell>
        </row>
        <row>
            <cell><m>B</m></cell>
            <cell><m>0.072</m></cell>
            <cell><m>1.944</m></cell>
            <cell><m>9.0</m></cell>
            <cell><m>16.864</m></cell>
            <cell><m>124.416</m></cell>
        </row>
     </tabular></sidebyside>
</p></statement>
<answer><p>Yes, <m>\dfrac{B}{r^3}</m> is constant.</p></answer>
</exercise>

</subsection>


<subsection><title>Scaling</title><idx>scaling</idx>
<p>
	Recall that if <m>y</m> varies directly with <m>x</m>, then doubling <m>x</m> causes <m>y</m> to double also. But:
    <ul>
    <li> Is the area of a <m>16</m>-inch circular pizza double the area of an <m>8</m>-inch pizza?</li> 
    <li>If you double the dimensions of a model of a skyscraper, will its weight double also? </li> 
    </ul> </p>

    <p>You probably know that the answer to both of these questions is <em>No</em>. The area of a circle is proportional to the <em>square</em> of its radius, and the volume (and hence the weight) of an object is proportional to the <em>cube</em> of its linear dimension. Variation with a power of <m>x</m> produces a different scaling effect.
</p>

<example xml:id="example-scaling-weight">
<p>
	The Taipei 101 building is 1671 feet tall, and in 2006 it was the tallest skyscraper in the world. Show that doubling the dimensions of a model of the Taipei 101 building produces a model that weighs <m>8</m> times as much.
</p>
<solution>
	The Taipei 101 skyscraper is approximately box shaped, so its volume is given by the product of its linear dimensions, <m>V = lwh</m>. The weight of an object is proportional to its volume, so the weight, <m>W</m>, of the model is
	<me>W = klwh</me>
	where the constant <m>k</m> depends on the material of the model. If we double the length, width, and height of the model, then
	<me>
		\begin{aligned}
		W_\text{new} \amp = k(2l)(2w)(2h)\\
		\amp = 2^3(klwh) = 8W_\text{old}
		\end{aligned}
	</me>
	The weight of the new model is <m>2^3 = 8</m> times the weight of the original model.
</solution>
</example>

<exercise xml:id="exercise-double-pizza-diameter"><statement><p>
	Use the formula for the area of a circle to show that doubling the diameter of a pizza quadruples its area.
</p></statement>
<answer><p><m>A =\pi r^2</m>, so <m>A_\text{new} = \pi(2r )^2 = 4\pi r^2 = 4A_\text{old}</m></p></answer>
</exercise>

<p>
	In general, if <m>y</m> varies directly with a power of <m>x</m>, that is, if <m>y = kx^n</m>, then doubling the value of <m>x</m> causes <m>y</m> to increase by a factor of <m>2^n</m>. In fact, if we multiply <m>x</m> by any positive number <m>c</m>, then
	<me>
		\begin{aligned}
		y_\text{new} \amp =k(cx)^n\\
		\amp =c^n(kx^n) = c^n(y_\text{old})
		\end{aligned}
	</me>
	so the value of <m>y</m> is multiplied by <m>c^n</m>.</p> 
    <p>We will call <m>n</m> the <term>scaling exponent</term><idx>scaling exponent</idx>, and you will often see variation described in terms of scaling. For example, we might say that "the area of a circle scales as the square of its radius." (In many applications, the power <m>n</m> is called the <em>scale factor</em>, even though it is not a factor but an exponent.)
</p>

</subsection>



<subsection><title>Inverse Variation</title><idx>inverse variation</idx>
<p>
	How long does it take to travel a distance of <m>600</m> miles? The answer depends on your average speed. If you are on a bicycle trip, your average speed might be <m>15</m> miles per hour. In that case, your traveling time will be
	<me>T = \frac{D}{R}= \frac{600}{15}= 40 \text{ hours}</me>
	(Of course, you will have to add time for rest stops; the <m>40</m> hours are just your travel time.) 
</p>
<p>
	If you are driving your car, you might average <m>50</m> miles per hour. Your travel time is then
	<me>T = \frac{D}{R}= \frac{600}{50}= 12 \text{ hours}</me>
</p>
<p>
	If you take a commercial air flight, the plane’s speed might be <m>400</m> miles per hour, and the flight time would be
	<me>T = \frac{D}{R}= \frac{600}{400}= 1.5 \text{ hours}</me>
</p>
<p>
	You can see that for higher average speeds, the travel time is shorter. In other words, the time needed for a <m>600</m>-mile journey is a decreasing function of average speed. In fact, a formula for the function is
	<me>T = f (R) = \frac{600}{R}</me>
</p>

<p>
	This function is an example of <term>inverse variation</term><idx>inverse variation</idx>. A table of values and a graph of the function are shown in <xref ref="fig-inverse-variation" text="type-global"/>.
</p>
<sidebyside widths="50% 35%">
<tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>R</m></cell>
            <cell><m>T</m></cell>
        </row>
        <row>
            <cell><m>10</m></cell>
            <cell><m>60</m></cell>
        </row>
        <row>
            <cell><m>15</m></cell>
            <cell><m>40</m></cell>
        </row>
        <row>
            <cell><m>20</m></cell>
            <cell><m>30</m></cell>
        </row>
        <row>
            <cell><m>50</m></cell>
            <cell><m>12</m></cell>
        </row>
        <row>
            <cell><m>200</m></cell>
            <cell><m>3</m></cell>
        </row>
        <row>
            <cell><m>400</m></cell>
            <cell><m>1.5</m></cell>
        </row>
    </tabular>
<figure xml:id="fig-inverse-variation"><caption></caption><image source="images/fig-inverse-variation"><description>graph of time vs average speed</description></image></figure>
</sidebyside>

<assemblage><title>Inverse Variation</title>
<p>
	<m>y</m> <term>varies inversely</term><idx>varies inversely</idx> with <m>x</m> if
	<me>y = \dfrac{k}{x}\text{, }x \ne 0</me>
	where <m>k</m> is a positive constant.
</p></assemblage>

<warning><statement>
	Inverse variation describes a decreasing function, but not every decreasing function represents inverse variation. People sometimes mistakenly use the phrase <em>varies inversely</em> to describe any decreasing function, but if <m>y</m> varies inversely with <m>x</m>, the variables must satisfy an equation of the form <m>y = \dfrac{k}{x}</m>, or <m>xy = k</m>. </statement></warning>

   <p> To decide whether two variables truly vary inversely, we can check whether their product is constant. For instance, in the preceding travel-time example, we see from the table that <m>RT = 600</m>.
</p><p>
<sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>R</m></cell>
            <cell><m>10</m></cell>
            <cell><m>15</m></cell>
            <cell><m>20</m></cell>
            <cell><m>50</m></cell>
            <cell><m>200</m></cell>
            <cell><m>400</m></cell>
        </row>
        <row>
            <cell><m>T</m></cell>
            <cell><m>60</m></cell>
            <cell><m>40</m></cell>
            <cell><m>30</m></cell>
            <cell><m>12</m></cell>
            <cell><m>3</m></cell>
            <cell><m>1.5</m></cell>
        </row>
        <row>
            <cell><m>RT</m></cell>
            <cell><m>600</m></cell>
            <cell><m>600</m></cell>
            <cell><m>600</m></cell>
            <cell><m>600</m></cell>
            <cell><m>600</m></cell>
            <cell><m>600</m></cell>
        </row>
    </tabular></sidebyside></p>

<p>
	We can also define inverse variation with a power of the variable.
</p>

<assemblage><title>Inverse Variation with a power</title>
<idx>inverse variation with a power</idx>
<p>
	<m>y</m> <term>varies inversely</term><idx>varies inversely</idx> with <m>x^n</m> if
	<me>y = \frac{k}{x^n}\text{, }x \ne 0</me>
	where <m>k</m> and <m>n</m> are positive constants.
</p></assemblage>

<p>
	We may also say that <m>y</m> is <term>inversely proportional</term><idx>inversely proportional</idx> to <m>x^n</m>.
</p>

<example xml:id="example-weight-vs-distance-from-earth">
	<p>
		The weight, <m>w</m>, of an object varies inversely with the square of its distance, <m>d</m>, from the center of the Earth. Thus,
		<me>w = \dfrac{k}{d^2}</me>
		If you double your distance from the center of the Earth, what happens to your weight? What if you triple the distance?
	</p>
<solution>
<p>
	Suppose you weigh <m>W</m> pounds at distance <m>D</m> from the center of the Earth. Then 	<m>W = \dfrac{k}{D^2}</m>. At distance <m>2D</m>, your weight will be
	<me>w = \frac{k}{(2D)^2}= \frac{k}{4D^2}= \frac{1}{4}\cdot\frac{k}{D^2}
		= \frac{1}{4}W</me>
	Your new weight will be <m>\dfrac{1}{4}</m> of your old weight. By a similar calculation, you can check that by tripling the distance, your weight will be reduced to <m>\dfrac{1}{9}</m> of its original value.
</p>
</solution>
</example>

<exercise xml:id="exercise-rusty-bolt"><statement>
	The amount of force, <m>F</m>, (in pounds) needed to loosen a rusty bolt with a wrench is inversely proportional to the length, <m>l</m>, of the wrench. Thus,
	<me>F = \frac{k}{l}</me>
	If you increase the length of the wrench by 50<percent/> so that the new length is <m>1.5l</m>, what happens to the amount of force required to loosen the bolt?
</statement>
<answer><p><m>F_\text{new}=\dfrac{2}{3}F_\text{old} </m></p></answer>
</exercise>

<p>
	In <xref ref="example-weight-vs-distance-from-earth" text="type-global"/> and <xef ref="exercise-rusty-bolt" text="type-global"/> as the independent variable increases through positive values, the dependent variable decreases. An inverse variation is an example of a decreasing function. The graphs of some typical inverse variations are shown in <xref ref="fig-inverse-variation-with-power" text="type-global"/>.
</p>
<figure xml:id="fig-inverse-variation-with-power"><caption></caption><image source="images/fig-inverse-variation-with-power"  width="50%"><description>two graphs of inverse variation with power</description></image></figure>

</subsection>



<subsection><title>Finding a Formula for Inverse Variation</title>
<p>
	If we know that two variables vary inversely and we can find one pair of corresponding values for the variables, we can determine <m>k</m>, the constant of variation.
</p>	

<example xml:id="example-EMR-vs-distance">
<p>
	The intensity of electromagnetic radiation, such as light or radio waves, varies inversely with the square of the distance from its source. Radio station KPCC broadcasts a signal that is measured at <m>0.016</m> watt per square meter by a receiver <m>1</m> kilometer away.
	<ol label="a">
		<li>Write a formula that gives signal strength as a function of distance.</li>
		<li>If you live <m>5</m> kilometers from the station, what is the strength of the signal you will receive?</li>
	</ol>
</p>
<solution>
<ol label="a">
	<li>
		Let <m>I</m> stand for the intensity of the signal in watts per square meter, and <m>d</m> for the distance from the station in kilometers. Then <m>I = \dfrac{k}{d^2}</m>. To find the constant <m>k</m>, we substitute <m>0.016</m> for <m>I</m> and <m>1</m> for <m>d</m>. Solving for <m>k</m> gives us
		\begin{align*}
			0.016 \amp = \dfrac{k}{1^2} \\
			k \amp = 0.016 (1^2) = 0.016
		\end{align*}
		Thus, <m>I = \dfrac{0.016}{d^2}</m>.
	</li>
	<li>
		Now we can substitute <m>5</m> for <m>d</m> and solve for <m>I</m>.
		<me>I = \frac{0.016}{5^2}= 0.00064</me>
		At a distance of <m>5</m> kilometers from the station, the signal strength is <m>0.00064</m> watt per square meter.
	</li>
</ol>
</solution>
</example>

<exercise xml:id="exercise-share-gift"><statement><p>
	Delbert’s officemates want to buy a <dollar/>120 gold watch for a colleague who is retiring. The cost per person is inversely proportional to the number of people who contribute.
	<ol label="a">
		<li>Express the cost per person, <m>C</m>, as a function of the number of people, <m>p</m>, who contribute.</li>
		<li>Sketch the function on the domain <m>0 \le p \le 20</m>.</li>
	</ol>
</p></statement>
<answer><p><ol label="a">
    <li><p><m>C=\dfrac{120}{p} </m></p></li>
    <li><p><image source="images/fig-in-ex-ans-3-1-8"  width="40%"><description>reciprocal</description></image> </p></li>
</ol></p></answer>
</exercise>


</subsection>
<xi:include href="./summary-3-1.xml" /> <!-- summary  -->
<xi:include href="./section-3-1-exercises.xml" /> <!-- exercises  -->


</section> 
<!-- </book>  </mathbook> -->