section-4-1.xml

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

<section xml:id="Exponential-Growth-and-Decay"   xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Exponential Growth and Decay</title>


<subsection><title>Exponential Growth</title>
<p>
    The functions in <xref ref="investigation-population-growth" text="type-global"/> describe <term>exponential growth</term>. During each time interval of a fixed length, the population is multiplied by a certain constant amount. In Part A, the bacteria population grows by a factor of <m>3</m> every day.
</p>
<sidebyside width="60%"><image source="images/fig-table-bacteria-population"><description>table of bacteria population</description></image></sidebyside>
<p>
    For this reason, we say that <m>3</m> is the <term>growth factor</term><idx>growth factor</idx> for the function. Functions that describe exponential growth can be expressed in a standard form.
</p>
<assemblage><title>Exponential Growth</title>
<p>
    The function
    <me>P(t) = P_0 b^t</me>
    models exponential growth.
</p>
<p>
    <m>P_0 = P(0)</m> is the <term>initial value</term><idx>initial value</idx> of <m>P</m>;
</p>
<p>
    <m>b</m> is the <term>growth factor</term><idx>growth factor</idx>.
</p>
</assemblage>

<p>
    For the bacteria population, we have
    <me>P(t) = 100 \cdot 3^t</me>
    so <m>P_0 = 100</m> and <m>b = 3</m>.
</p>

<example xml:id="example-bacteria">
<p>
    A colony of bacteria starts with <m>300</m> organisms and doubles every week.
    <ol label="a">
        <li><p>Write a formula for the population of the bacteria colony after <m>t</m> weeks.</p></li>
        <li><p>How many bacteria will there be after <m>8</m> weeks? After <m>5</m> days?</p></li>
    </ol>
</p>
<solution>
    <ol label="a">
        <li><p>The initial value of the population was <m>P_0 = 300</m>, and its weekly growth factor is <m>b = 2</m>. Thus, a formula for the population after <m>t</m> weeks is
        <me>P(t) = 300 \cdot 2^t</me>
        </p></li>
        <li><p>After <m>8</m> weeks, the population will be
        <me>P(8) = 300 \cdot 2^8 = 76,800 \text{ bacteria}</me>
        Because <m>5</m> days is <m>\frac{5}{7}</m> of a week, after <m>5</m> days the population will be
        <me>P\left(\frac{5}{7}\right)= 300 \cdot 2^{5/7} = 492.2</me>
        We cannot have a fraction of a bacterium, so we round to the nearest whole number, <m>492</m>.</p></li>
    </ol>
</solution>
</example>

<warning><statement>
<p>
    In <xref ref="example-bacteria" text="type-global"/>a, note that
    <me>300 \cdot 2^8 \ne 600^8</me>
    According to the order of operations, we compute the power <m>2^8</m> first, then multiply by <m>300</m>.
</p></statement>
</warning>

<exercise xml:id="exercise-fruit-flies"><statement>
<p>
    A population of <m>24</m> fruit flies triples every month. 
    <ol label="a">
        <li><p>
            Write a formula for the population of fruit flies after <m>t</m> weeks.
        </p></li>
        <li><p>
            How many fruit flies will there be after <m>6</m> months? After <m>3</m> weeks? (Assume that a month equals <m>4</m> weeks.)
        </p></li>
    </ol>
</p>
</statement>
<answer><p>
    <ol cols="2" label="a">
        <li><p><m>P(t) = 24\cdot 3^t</m></p></li>
        <li><p><m>17,496</m>; <m>55</m></p></li>
    </ol>
</p></answer>
</exercise>

</subsection>


<subsection><title>Growth Factors</title>
<p>
    In Part B of <xref ref="investigation-population-growth" text="type-global"/>, the rabbit population grew by a factor of <m>2</m> every <m>3</m> months.
</p>
<sidebyside width="60%"><image source="images/fig-table-rabbit-population"><description>table of rabbit population</description></image></sidebyside>
<p>
    To write the growth formula for this population, we divide the value of <m>t</m> by <m>3</m> to find the number of doubling periods.
    <me>P(t) = 60 \cdot 2^{t/3}</me>
    To see the growth factor for the function, we use the third law of exponents to write <m>2^{t/3}</m> in another form. Recall that to raise a power to a power, we multiply exponents, so
    <me>\left(2^{1/3}\right)^t = 2^{t (1/3)} = 2^{t/3}</me>
    The growth law for the rabbit population is thus
    <me>P(t) = 60 \cdot \left(2^{1/3}\right)^t</me>
    The initial value of the function is <m>P_0 = 60</m>, and the growth factor is <m>b = 2^{1/3}</m>, or approximately <m>1.26</m>. The rabbit population grows by a factor of about <m>1.26</m> every month.
</p>
<p>
    If the units are the same, a population with a larger growth factor grows faster than one with a smaller growth factor.
</p>   

<example xml:id="example-compare-growth-factors">
<p>
    A lab technician compares the growth of <m>2</m> species of bacteria. She starts <m>2</m> colonies of <m>50</m> bacteria each. Species A doubles in population every <m>2</m> days, and species B triples every <m>3</m> days. Find the growth factor for each species.
</p>
<solution><p>
    A function describing the growth of species A is
    <me>P(t) = 50 \cdot 2^{t/2} = 50 \cdot \left(\alert{2^{1/2}}\right)^t</me>
    so the growth factor for species A is <m>\alert{2^{1/2}}</m>, or approximately <m>1.41</m>. For species B,
    <me>P(t) = 50 \cdot 3^{t/3} = 50 \cdot \left(\alert{3^{1/3}}\right)^t</me>
     so the growth factor for species B is <m>\alert{3^{1/3}}</m>, or approximately <m>1.44</m>. Species B grows faster than species A.
</p></solution>
</example>

<exercise xml:id="exercise-internet-providers"><statement><p>
    In 1999, analysts expected the number of Internet service providers to double in five years.
    <ol label="a">
        <li><p>What was the annual growth factor for the number of Internet service providers?</p></li>
        <li><p>If there were <m>5078</m> Internet service providers in April 1999, estimate the number of providers in April 2000 and in April 2001.</p></li>
        <li><p>Write a formula for <m>I (t)</m>, the number of Internet service providers <m>t</m> years after 1999.
        </p></li>
    </ol>
    Source: LA Times, Sept. 6, 1999
</p></statement>
<answer><p><ol cols="3" label="a">
    <li><p><m>2^{1/5}</m></p></li>
    <li><p><m>5833</m> and <m>6700</m></p></li>
    <li><p><m>I (t) = 5078\cdot 2^{t/5}</m></p></li>
</ol></p></answer>
</exercise>

</subsection>


<subsection><title>Percent Increase</title>
<p>
    Exponential growth occurs in other circumstances, too. For example, if the interest on a savings account is compounded annually, the amount of money in the account grows exponentially.
</p>
<p>
    Consider a principal of <dollar/>100 invested at 5<percent/> interest compounded annually. At the end of <m>1</m> year, the amount is
    \begin{align*}
    \text{Amount} \amp = \text{Principal} + \text{Interest} \\
    A \amp = P + Pr\\
    \amp = 100 + 100(0.05) = 105
    \end{align*}
    It will be more useful to write the formula for the amount after <m>1</m> year in factored form.
    \begin{align*}
    A \amp = P + Pr \amp\amp \text{Factor out P.} \\
    \amp = P(1 + r)
    \end{align*}
    With this version of the formula, the calculation for the amount at the end of <m>1</m> year looks like this:
    \begin{align*}
    A \amp = P(1 + r ) \\
    \amp = 100(1 + 0.05) \\
    \amp = 100(1.05) = \alert{105}
    \end{align*}
</p>
<p>
    The amount, <dollar/>105, becomes the new principal for the second year. To find the amount at the end of the second year, we apply the formula again, with <m>P = 105</m>.
    \begin{align*}
    A \amp = P(1 + r ) \\
    \amp = 105(1 + 0.05) \\
    \amp = \alert{105}(1.05) = 110.25
    \end{align*}
    Observe that to find the amount at the end of each year, we multiply the principal by a factor of <m>1 + r = 1.05</m>. Thus, we can express the amount at the end of the second year as
    \begin{align*}
    A \amp = [100(1.05)](1.05) \\
    \amp = 100(1.05)^2
    \end{align*}
    and at the end of the third year as
    \begin{align*}
    A \amp = \left[100(1.05)^2\right](1.05) \\
    \amp = 100(1.05)^3
    \end{align*}
</p>
<sidebyside><paragraphs width="70%">
<p>
    At the end of each year, we multiply the old balance by another factor of <m>1.05</m> to get the new amount. We organize our results into <xref ref="table-percent-increase" text="type-global"/>, where <m>A(t)</m> represents the amount of money in the account after <m>t</m> years. For this example, a formula for the amount after <m>t</m> years is
    <me>A(t) = 100(1.05)^t</me>
</p></paragraphs>
<table xml:id="table-percent-increase" valign="top"><caption></caption><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
    <row bottom="minor">
        <cell><m>t</m></cell>
        <cell><m>P(1 + r)^t</m></cell>
        <cell><m>A(t)</m></cell>
    </row>
    <row>
        <cell><m>0</m></cell>
        <cell><m>100</m></cell>
        <cell><m>100</m></cell>
    </row>
    <row>
        <cell><m>1</m></cell>
        <cell><m>100(1.05)</m></cell>
        <cell><m>105</m></cell>
    </row>
    <row>
        <cell><m>2</m></cell>
        <cell><m>100(1.05)^2</m></cell>
        <cell><m>110.25</m></cell>
    </row>
    <row>
        <cell><m>3</m></cell>
        <cell><m>100(1.05)^3</m></cell>
        <cell><m>115.76</m></cell>
    </row>
</tabular></table>
</sidebyside>

<p>
    In general, for an initial investment of <m>P</m> dollars at an interest rate <m>r</m> compounded annually, we have the following formula for the amount accumulated after <m>t</m> years.
</p>
<assemblage><title>Compound Interest</title><idx>compound interest</idx>
<p>
    The <term>amount A(t)</term> accumulated (principal plus interest) in an account bearing interest compounded annually is
    <me>A(t) = P(1 + r)^t</me>
    where 
    \begin{align*}
    \amp P \amp\amp \text{is the principal invested,} \\
    \amp r \amp\amp \text{is the interest rate,} \\
    \amp t \amp\amp \text{is the time period, in years.}
    \end{align*}
</p>
</assemblage>

<p>
    This function describes exponential growth with an initial value of <m>P</m> and a growth factor of <m>b = 1 + r</m>. The interest rate <m>r</m>, which indicates the <term>percent increase</term><idx>percent increase</idx> in the account each year, corresponds to a <term>growth factor</term><idx>growth factor</idx> of <m>1 + r</m>. The notion of percent increase is often used to describe the growth factor for quantities that grow exponentially.
</p>

<example xml:id="example-inflation">
<p>
    During a period of rapid inflation, prices rose by <m>12</m><percent/> over <m>6</m> months. At the beginning of the inflationary period, a pound of butter cost <dollar/><m>2</m>.
    <ol label="a">
        <li><p>Make a table of values showing the rise in the cost of butter over the next <m>2</m> years.</p></li>
        <li><p>Write a function that gives the price of a pound of butter <m>t</m> years after inflation began.</p></li>
        <li><p>How much did a pound of butter cost after <m>3</m> years? After <m>15</m> months?</p></li>
        <li><p>Graph the function you found in part (b).</p></li>
    </ol>
</p>
<solution>
    <ol label="a">
        <li><p>The percent increase in the price of butter is <m>12</m><percent/> every <m>6</m> months. Therefore, the growth factor for the price of butter is <m>1 + 0.12 = 1.12</m> every half-year. If <m>P(t)</m> represents the price of butter after <m>t</m> years, then <m>P(0) = 2</m>, and every half-year we multiply the price by <m>1.12</m>, as shown in <xref ref="table-inflation" text="type-global"/>.
        <table xml:id="table-inflation"><image source="images\fig-table-inflation"  width="70%"><description>table for inflation</description></image><caption></caption></table>
        </p></li>
        <li><p>Look closely at the second column of <xref ref="table-inflation" text="type-global"/>. After <m>t</m> years of inflation, the original price of <dollar/><m>2</m> has been multiplied by <m>2t</m> factors of <m>1.12</m>. Thus,
        <me>P = 2(1.12)^{2t}</me>
        </p></li>
        <li><p>To find the price of butter at any time after inflation began, we evaluate the function at the appropriate value of <m>t</m>.
        \begin{align*}
        P(\alert{3}) \amp = 2(1.12)^{2(\alert{3})} \\
        \amp = 2(1.12)^6 \approx 3.95
        \end{align*}
        After <m>3</m> years, the price was <dollar/><m>3.95</m>. Fifteen months is <m>1.25</m> years, so we evaluate <m>P(1.25)</m>.
        \begin{align*}
        P(\alert{1.25}) \amp = 2(1.12)^{2(\alert{1.25})} \\
        \amp = 2(1.12)^{2.5} \approx 2.66 
        \end{align*}
        After <m>15</m> months, the price of butter was <dollar/><m>2.66</m>.
        </p></li>
        <li><p>Evaluate the function
            <me>P(t) = 2(1.12)^{2t}</me>
            for several values, as shown in <xref ref="table-inflation2" text="type-global"/>. We plot the points and connect them with a smooth curve to obtain the graph shown in <xref ref="fig-inflation" text="type-global"/>.</p>
            <p>
            <sidebyside widths="45% 35%">
            <table valign="center" xml:id="table-inflation2"><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
                <row bottom="minor">
                    <cell><m>t</m></cell>
                    <cell><m>P(t)</m></cell>
                </row>
                <row>
                    <cell><m>0</m></cell>
                    <cell><m>2.00</m></cell>
                </row>
                <row>
                    <cell><m>1</m></cell>
                    <cell><m>2.51</m></cell>
                </row>
                <row>
                    <cell><m>2</m></cell>
                    <cell><m>3.15</m></cell>
                </row>
                <row>
                    <cell><m>3</m></cell>
                    <cell><m>3.95</m></cell>
                </row>
                <row>
                    <cell><m>4</m></cell>
                    <cell><m>4.95</m></cell>
                </row>
            </tabular><caption></caption></table>
            <figure xml:id="fig-inflation"><caption></caption><image source="images/fig-inflation"><description>graph of inflation</description></image></figure>
            </sidebyside>
        </p></li>
    </ol>
</solution>
</example>

<p>
    In <xref ref="example-inflation" text="type-global"/>, we can rewrite the formula for <m>P(t)</m> as follows:
    \begin{align*}
    P(t)\amp = 2(1.12)^{2t} \\
    \amp = 2\left[(1.12)^2\right]^t = 2(1.2544)^t
    \end{align*}
    Thus, the annual growth factor for the price of butter is <m>1.2544</m>, and the annual percent growth rate is <m>25.44</m><percent/>.
</p>

<exercise xml:id="exercise-inflation"><statement>
    In <m>1998</m>, the average annual cost of attending a public college was <m>\$10,069</m>, and costs were climbing by <m>6</m><percent/> per year.
    <ol label="a">
        <li><p>Write a formula for <m>C(t)</m>, the cost of one year of college <m>t</m> years after <m>1998</m>.</p></li>
        <li><p>Complete the table and sketch a graph of <m>C(t)</m>.</p><p>
        <sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>t</m></cell>
            <cell><m>0</m></cell>
            <cell><m>5</m></cell>
            <cell><m>10</m></cell>
            <cell><m>15</m></cell>
            <cell><m>20</m></cell>
            <cell><m>25</m></cell>
        </row>
        <row>
            <cell><m>C(t)</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
        </row>
    </tabular></sidebyside>

        </p></li>
        <li><p>If the percent growth rate remained steady, how much did a year of college cost in <m>2005</m>?</p></li>
        <li><p>If the percent growth rate continues to remain steady, how much will a year of college cost in <m>2020</m>?
        </p></li>
    </ol>
</statement>
<answer><p><ol label="a">
    <li><p><m>C(t) = 10,069\cdot 1.06^t</m></p></li>
    <li><p>
        <tabular left="minor" right="minor" top="major" bottom="minor" halign="right"><col halign="left"/>
            <row>
                <cell><m>t</m></cell>
                <cell><m>0</m></cell>
                <cell><m>5</m></cell>
                <cell><m>10</m></cell>
                <cell><m>15</m></cell>
                <cell><m>20</m></cell>
                <cell><m>25</m></cell>
            </row>
            <row>
                <cell><m>C(t)</m></cell>
                <cell><m>10,069</m></cell>
                <cell><m>13,475</m></cell>
                <cell><m>18,032</m></cell>
                <cell><m>24,131</m></cell>
                <cell><m>32,293</m></cell>
                <cell><m>43,215</m></cell>
            </row>
        </tabular>
        <image source="images/fig-in-ex-ans-4-1-3"  width="40%">
          <description>
            exponential growth
          </description>
        </image>
    </p></li>
    <li><p><m>$15,140</m> per year</p></li>
    <li><p><m>$36,284</m></p></li>
</ol></p></answer>
</exercise>

</subsection>

<subsection><title>Exponential Decay</title>
<p>
    In the preceding examples, exponential growth was modeled by increasing functions of the form
    <me>P(t) = P_0 b^t</me>
    where <m>b \gt 1</m>. The function <m>P(t) = P_0b^t</m> is a <em>decreasing</em> function if <m>0 \lt b \lt 1</m>. In this case, we say that the function describes <term>exponential decay</term>, and the constant <m>b</m> is called the <term>decay factor</term><idx>decay factor</idx>. In <xref ref="investigation-exponential-decay" text="type-global"/>, we consider two examples of exponential decay.
</p>


<investigation xml:id="investigation-exponential-decay"><title>Exponential Decay</title><statement>
<ol label="A">
    <li><p>
        A small coal-mining town has been losing population since 1940, when 5000 people lived there. At each census thereafter (taken at 10-year intervals), the population declined to approximately 0.90 of its earlier figure.</p><p>
        <sidebyside widths="45% 45%"><p>
        <ol label="1"> 
            <li><p>
                Fill in the table showing the population <m>P(t)</m> of the town <m>t</m> years after 1940.</p></li>
            <li><p>
                Plot the data points and connect them with a smooth curve. 
            </p></li>
            <li><p>
                Write a function that gives the population of the town at any time <m>t</m> in years after 1940.
                <hint>Express the values you calculated in part (1) using powers of <m>0.90</m>. Do you see a connection between the value of <m>t</m> and the exponent on <m>0.90</m>?</hint>
            </p></li>
            <li><p>
                Graph your function from part (3) using a calculator. (Use the table to choose an appropriate domain and range.) The graph should resemble your hand-drawn graph from part (2).
            </p></li>
            <li><p>
                Evaluate your function to find the population of the town in 1995. What was the population in 2000?
            </p></li>
        </ol></p>    
                
            <p><tabular halign="center" right="minor" left="minor" bottom="minor">
                <col top="minor" />
                <col top="minor" />
                <col />
                <col halign="left"/>
                        
                <row bottom="minor">
                    <cell><m>t</m></cell>
                    <cell><m>P(t)</m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                </row>
                <row> 
                    <cell><m>0</m></cell>
                    <cell><m>5000</m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(0)=5000</m></cell>
                </row>
                <row> 
                    <cell><m>10</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(10)=5000\cdot 0.90=</m></cell>
                </row>
                <row> 
                    <cell><m>20</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(20)=[5000\cdot 0.90]\cdot 0.90=</m></cell>
                </row>
                <row> 
                    <cell><m>30</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(3)=</m></cell>
                </row>
                <row> 
                    <cell><m>40</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(4)=</m></cell>
                </row>
                <row> 
                    <cell><m>50</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(5)=</m></cell>
                </row>
            </tabular> 
            
            <image source="images/fig-grid-for-population-decay">
              <description>
                grid for graph of population decay
              </description>
            </image></p>
        </sidebyside>
    </p></li>

    <li><p>
        A plastic window coating <m>1</m> millimeter thick decreases the light coming through a window by <m>25</m><percent/>. This means that <m>75</m><percent/> of the original amount of light comes through <m>1</m> millimeter of the coating. Each additional millimeter of coating reduces the light by another <m>25</m><percent/>.</p>
        <p></p>
        <sidebyside widths="48% 48%"><p>
        <ol>
            <li><p>
                Fill in the table showing the percent of the light, <m>P(x)</m>, that shines through <m>x</m> millimeters of the window coating.
                </p></li>
            <li><p>
                Plot the data points and connect them with a smooth curve. 
            </p></li>
            <li><p>
                Write a function that gives the percent of the light that shines through <m>x</m> millimeters of the coating. 
                <hint> Express the values you calculated in part (1) using powers of <m>0.75</m>. Do you see a connection between the value of <m>x</m> and the exponent on <m>0.75</m>?</hint>
            </p></li>
            <li><p>
                Graph your function from part (3) using a calculator. (Use your table of values to choose an appropriate domain and range.) The graph should resemble your hand-drawn graph from part (2).
            </p></li>
        </ol></p>
        
        <p><tabular halign="center" right="minor" left="minor" bottom="minor">
                <col top="minor" />
                <col top="minor" />
                <col />
                <col halign="left"/>
                        
                <row bottom="minor">
                    <cell><m>x</m></cell>
                    <cell><m>P(x)</m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                </row>
                <row> 
                    <cell><m>0</m></cell>
                    <cell><m>100</m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(0)=100</m></cell>
                </row>
                <row> 
                    <cell><m>1</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(1)=100\cdot 0.75=</m></cell>
                </row>
                <row> 
                    <cell><m>2</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(2)=[100\cdot 0.75]\cdot 0.75=</m></cell>
                </row>
                <row> 
                    <cell><m>3</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(3)=</m></cell>
                </row>
                <row> 
                    <cell><m>4</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(4)=</m></cell>
                </row>
                <row> 
                    <cell><m>5</m></cell>
                    <cell><m></m></cell>
                    <cell top="none" right="none" bottom="none"></cell>
                    <cell top="none" right="none" bottom="none"><m>P(5)=</m></cell>
                </row>
            </tabular>
            <p></p>
            <image source="images/fig-grid-for-light-decay">
              <description>
                grid for graph of light decay
              </description>
            </image></p>
        </sidebyside> 
    </li>
</ol></statement>
</investigation>

</subsection>




<subsection><title>Decay Factors</title>
<p>
    Before <xref ref="example-inflation" text="type-global"/>, we noted that a percent increase of <m>r</m> (in decimal form) corresponds to a growth factor of <m>b = 1 + r</m>. A percent decrease of <m>r</m> corresponds to a decay factor of <m>b = 1 - r</m>. In Part B of <xref ref="investigation-exponential-decay" text="type-global"/>, each millimeter of plastic reduced the amount of light by <m>25</m><percent/>, so <m>r = 0.25</m>, and the decay factor for the function <m>P(x)</m> is
    \begin{align*}
    b \amp = 1 - r \\
    \amp= 1 - 0.25 = 0.75
    \end{align*}
</p>

<example xml:id="example-computing-prices">
<p>
    David Reed writes in Context magazine: "Computing prices have been falling exponentially—50<percent/> every 18 months—for the past 30 years and will probably stay on that curve for another couple of decades." An accounting firm invests <dollar/>50,000 in new computer equipment.
    <ol label="a">
        <li><p>
            Write a formula for the value of the equipment <m>t</m> years from now.
        </p></li>
        <li><p>
            By what percent does the equipment depreciate each year?
        </p></li>
        <li><p>
            What will the equipment be worth in <m>5</m> years?
        </p></li>
    </ol>
</p>
<solution>
<ol label="a">
    <li><p>
        The initial value of the equipment is <m>V_0 = 50,000</m>. Every <m>18</m> months, the value of the equipment is multiplied by
        <me>b = 1 - r = 1 - 0.50 = 0.50</me>
        However, because <m>18</m> months is <m>1.5</m> years, we must divide <m>t</m> by <m>1.5</m> in our formula, giving us
        <me>V(t) = 50,000(0.50)^{t/1.5}</me>
    </p></li>
    <li><p>
        After <m>1</m> year, we have
        <me>V(1) = 50,000(0.50)^{1/1.5} = 50,000(0.63)</me>
        The equipment is worth <m>63</m><percent/> of its original value, so it has depreciated by <m>1 - 0.63</m>, or <m>37</m><percent/>.
    </p></li>
    <li><p>
        After <m>5</m> years,
        <me>V(5) = 50,000(0.50)^{5/1.5} = 4960.628</me>
        To the nearest dollar, the equipment is worth <dollar/>4961.
    </p></li>
</ol>
</solution>
</example>

<exercise xml:id="exercise-butterflies"><statement>
The number of butterflies visiting a nature station is declining by 18<percent/> per year. In 1998, <m>3600</m> butterflies visited the nature station.
<ol label="a">
    <li><p>What is the decay factor in the annual butterfly count?</p></li>
    <li><p>Write a formula for <m>B(t)</m>, the number of butterflies <m>t</m> years after 1998.</p></li>
    <li><p>Complete the table and sketch a graph of <m>B(t)</m>.</p><p>
    <sidebyside><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>t</m></cell>
            <cell><m>0</m></cell>
            <cell><m>2</m></cell>
            <cell><m>4</m></cell>
            <cell><m>6</m></cell>
            <cell><m>8</m></cell>
            <cell><m>10</m></cell>
        </row>
        <row>
            <cell><m>B(t)</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
            <cell><m>\hphantom{0000}</m></cell>
        </row>
    </tabular></sidebyside>
    </p></li>
</ol>
</statement>
<answer><p><ol label="a">
    <li><p><m>0.82</m></p></li>
    <li><p><m>B(t) = 3600\cdot 0.82^t</m></p></li>
    <li><p>
    <tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>t</m></cell>
            <cell><m>0</m></cell>
            <cell><m>2</m></cell>
            <cell><m>4</m></cell>
            <cell><m>6</m></cell>
            <cell><m>8</m></cell>
            <cell><m>10</m></cell>
        </row>
        <row>
            <cell><m>B(t)</m></cell>
            <cell><m>3600</m></cell>
            <cell><m>2421</m></cell>
            <cell><m>1628</m></cell>
            <cell><m>1094</m></cell>
            <cell><m>736</m></cell>
            <cell><m>495</m></cell>
        </row>
    </tabular>
    <image source="images/fig-in-ex-ans-4-1-4"  width="40%">
      <description>
        exponential decay
      </description>
    </image>
    </p></li>
</ol></p></answer>
</exercise>

<p>
    We summarize our observations about exponential growth and decay functions as follows.
</p>

<assemblage><title>Exponential Growth and Decay</title><idx>exponential growth and decay</idx>
<p>The function
<me>
    P(t) = P_0 b^t
</me>
models exponential growth and decay.</p>
<p><m>P_0 =P(0)</m> is the <term>initial value</term><idx>initial value</idx> of <m>P</m>;</p>
<p><m>b</m> is the <term>growth</term> <idx>growth factor</idx> or <term>decay factor</term><idx>decay factor</idx>.
<ol>
    <li><p>
        If <m>b \gt 1</m>, then <m>P(t)</m> is increasing, and <m>b = 1 + r</m>, where <m>r</m> represents percent increase.
    </p></li>
    <li><p>
        If <m>0 \lt b \lt 1</m>, then <m>P(t)</m> is decreasing, and <m>b = 1 - r</m>, where <m>r</m> represents percent decrease.
    </p></li>
</ol>
</p></assemblage>

</subsection>



<subsection><title>Comparing Linear Growth and Exponential Growth</title>
<p>
    It may be helpful to compare linear growth <idx>linear growth</idx> and <idx>exponential growth</idx> exponential growth. Consider the two functions
    <me>L(t) = 5 + 2t ~\text{ and } ~ E(t) = 5 \cdot 2^t ~~~ (t \ge 0)</me>
    whose graphs are shown in <xref ref="fig-linear-vs-exponential" text="type-global"/>.
</p>
<sidebyside widths="25% 25% 25%">
<table valign="top" xml:id="table-linear"><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>t</m></cell>
            <cell><m>L(t)</m></cell>
        </row>
        <row>
            <cell><m>0</m></cell>
            <cell><m>5</m></cell>
        </row>
        <row>
            <cell><m>1</m></cell>
            <cell><m>7</m></cell>
        </row>
        <row>
            <cell><m>2</m></cell>
            <cell><m>9</m></cell>
        </row>
        <row>
            <cell><m>3</m></cell>
            <cell><m>11</m></cell>
        </row>
        <row>
            <cell><m>4</m></cell>
            <cell><m>13</m></cell>
        </row>
    </tabular><caption><m>m=2</m></caption></table>

<table valign="top" xml:id="table-exponential"><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>t</m></cell>
            <cell><m>E(t)</m></cell>
        </row>
        <row>
            <cell><m>0</m></cell>
            <cell><m>5</m></cell>
        </row>
        <row>
            <cell><m>1</m></cell>
            <cell><m>10</m></cell>
        </row>
        <row>
            <cell><m>2</m></cell>
            <cell><m>20</m></cell>
        </row>
        <row>
            <cell><m>3</m></cell>
            <cell><m>40</m></cell>
        </row>
        <row>
            <cell><m>4</m></cell>
            <cell><m>80</m></cell>
        </row>
    </tabular><caption>Growth factor <m>b=2</m></caption></table>

<figure valign="top" xml:id="fig-linear-vs-exponential"><caption></caption><image source="images/fig-linear-vs-exponential"><description>graph of linear and exponential</description></image></figure>
</sidebyside>

<p>
    <m>L</m> is a linear function with initial value <m>5</m> and slope <m>2</m>; <m>E</m> is an exponential function with initial value <m>5</m> and growth factor <m>2</m>. In a way, the growth factor of an exponential function is analogous to the slope of a linear function: Each measures how quickly the function is increasing (or decreasing).</p>
    <p> However, for each unit increase in <m>t</m>, <m>2</m> units are <em>added</em> to the value of <m>L(t)</m>, whereas the value of <m>E(t)</m> is <em>multiplied</em> by <m>2</m>. An exponential function with growth factor <m>2</m> eventually grows much more rapidly than a linear function with slope <m>2</m>, as you can see by comparing the graphs in <xref ref="fig-linear-vs-exponential" text="type-global"/> or the function values in  <xref ref="table-linear">Tables</xref> and <xref ref="table-exponential" text="global"/>.
</p>

<example xml:id="example-compare-linear-vs-exponential">
<p>
    A solar energy company sold <dollar/><m>80,000</m> worth of solar collectors last year, its first year of operation. This year its sales rose to <dollar/><m>88,000</m>, an increase of <m>10</m><percent/>. The marketing department must estimate its projected sales for the next <m>3</m> years.
    <ol label="a">
        <li><p>
            If the marketing department predicts that sales will grow linearly, what should it expect the sales total to be next year? Graph the projected sales figures over the next <m>3</m> years, assuming that sales will grow linearly.
        </p></li>
        <li><p>
            If the marketing department predicts that sales will grow exponentially, what should it expect the sales total to be next year? Graph the projected sales figures over the next <m>3</m> years, assuming that sales will grow exponentially.
        </p></li>
    </ol>
</p>
<solution>
    <ol label="a">
        <li><p>
            Let <m>L(t)</m> represent the company's total sales <m>t</m> years after starting business, where <m>t = 0</m> is the first year of operation. If sales grow linearly, then <m>L(t)</m> has the form <m>L(t) = mt + b</m>. Now <m>L(0) = 80,000</m>, so the intercept <m>b</m> is <m>80,000</m>. The slope <m>m</m> of the graph is
            <me>
                \frac{\Delta S}{\Delta t}= 
                \frac{8000 \text{ dollars}}{1\text{ year}}= 
                8000 \text{ dollars/year}
            </me>
            where <m>\Delta S = 8000</m> is the increase in sales during the first year. Thus, <m>L(t) = 8000t + 80,000</m>, and sales grow by adding <dollar/><m>8000</m> each year. The expected sales total for the next year is
            <me>
                L(2) = 8000(2) + 80,000 = 96,000
            </me>
            The values of <m>L(t)</m> for <m>t=0</m> to <m>t=4</m> are shown in the middle column of <xref ref="table-linear-exponential" text="type-global"/>. The linear graph of <m>L(t)</m> is shown in <xref ref="fig-linear-vs-exponential2" text="type-global"/>.
        </p></li>
        <li><p>
            Let <m>E(t)</m> represent the company's sales assuming that sales will grow exponentially. Then <m>E(t)</m> has the form <m>E(t) = E_0b^t</m> . The percent increase in sales over the first year was <m>r = 0.10</m>, so the growth rate is
            <me>b = 1 + r = 1.10</me>
            The initial value, <m>E_0</m>, is <m>80,000</m>. Thus, <m>E(t) = 80,000(1.10)^t</m>, and sales grow by being multiplied each year by <m>1.10</m>. The expected sales total for the next year is
            <me>
                E(2) = 80,000(1.10)^2= 96,800
            </me>
            The values of <m>E(t)</m> for <m>t=0</m> to <m>t=4</m> are shown in the last column of <xref ref="table-linear-exponential" text="type-global"/>. The exponential graph of <m>E(t)</m> is shown in <xref ref="fig-linear-vs-exponential2" text="type-global"/>.</p>
            <p>
            <sidebyside widths="45% 45%">
            <table xml:id="table-linear-exponential"><tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
                <row bottom="minor">
                    <cell><m>t</m></cell>
                    <cell><m>L(t)</m></cell>
                    <cell><m>E(t)</m></cell>
                </row>
                <row>
                    <cell><m>0</m></cell>
                    <cell><m>80,000</m></cell>
                    <cell><m>80,000</m></cell>
                </row>
                <row>
                    <cell><m>1</m></cell>
                    <cell><m>88,000</m></cell>
                    <cell><m>88,000</m></cell>
                </row>
                <row>
                    <cell><m>2</m></cell>
                    <cell><m>96,000</m></cell>
                    <cell><m>96,800</m></cell>
                </row>
                <row>
                    <cell><m>3</m></cell>
                    <cell><m>104,000</m></cell>
                    <cell><m>106,480</m></cell>
                </row>
                <row>
                    <cell><m>4</m></cell>
                    <cell><m>112,000</m></cell>
                    <cell><m>117,128</m></cell>
                </row>
            </tabular><caption></caption></table>
            <figure xml:id="fig-linear-vs-exponential2"><caption></caption><image source="images/fig-linear-vs-exponential2"><description>linear and exponential growth</description></image></figure>
            </sidebyside>
        </p></li>
    </ol>
</solution>
</example>

<exercise xml:id="exercise-depreciation"><statement>
A new car begins to depreciate in value as soon as you drive it off the lot. Some models depreciate linearly, and others depreciate exponentially. Suppose you buy a new car for <dollar/><m>20,000</m>, and <m>1</m> year later its value has decreased to <dollar/><m>17,000</m>.
<ol label="a">
    <li><p>
        If the value decreased linearly, what was its annual rate of decrease?
    </p></li>
    <li><p>
        If the value decreased exponentially, what was its annual decay factor? What was its annual percent depreciation?
    </p></li>
    <li><p>
        Calculate the value of your car when it is <m>5</m> years old under each assumption, linear or exponential depreciation.
    </p></li>
</ol>
</statement>
<answer><p><ol label="a">
    <li><p><m>\$3000</m> per year</p></li>
    <li><p><m>0.85</m>; <m>15\%</m></p></li>
    <li><p>Linear: <m>\$5000</m>; Exponential: <m>\$8874</m></p></li>
</ol></p></answer>
</exercise>
    


</subsection>

<xi:include href="./summary-4-1.xml" /> <!-- summary  -->
<xi:include href="./section-4-1-exercises.xml" /> <!-- exercises  -->
</section> 
<!-- </book>  </mathbook> -->