section-2-1.xml

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<section xml:id="nonlinear-models"   xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Nonlinear Models</title><introduction>
<p>
    In Chapter 1, we considered models described by linear functions. In this chapter, we begin our study of nonlinear models.
</p>
</introduction>
<subsection><title>Solving Nonlinear Equations</title> 
<p>When studying nonlinear models, we will need to solve nonlinear equations. For example, in <xref ref="investigation-Perimeter-and-Area" text="type-hybrid"/> we used a graph to solve the quadratic equation <me>18x - x^2 = 80</me></p>
<p>Here is another example. <xref ref="fig-parabola" text="type-global" /> shows a table and a graph for the function <m>y = 2x^2 - 5</m>. </p><p>
<sidebyside widths="38% 50%" valigns="top middle" margins="6%"><figure xml:id="fig-parabola"><caption></caption><image source="images/fig-parabola"><description>parabola</description></image></figure>
    <tabular top="major" halign="center" right="minor" left="minor" bottom="minor">
        <row bottom="minor">
            <cell><m>x</m></cell>
            <cell><m>-3</m></cell>
            <cell><m>-2</m></cell>
            <cell><m>-1</m></cell>
            <cell><m>0</m></cell>
            <cell><m>1</m></cell>
            <cell><m>2</m></cell>
            <cell><m>3</m></cell>
        </row>
        <row>
            <cell><m>y</m></cell>
            <cell><m>13</m></cell>
            <cell><m>3</m></cell>
            <cell><m>-3</m></cell>
            <cell><m>-5</m></cell>
            <cell><m>-3</m></cell>
            <cell><m>3</m></cell>
            <cell><m>13</m></cell>
        </row>
    </tabular>
</sidebyside></p>
<p>  You can see that there are two points on the graph for each <m>y</m>-value greater than <m>-5</m>.</p> 
<p>For example, the two points with <m>y</m>-coordinate <m>7</m> are shown. To solve the equation 
    <me>2x^2 - 5 = 7</me> 
    we need only find the <m>x</m>-coordinates of these points. From the graph, the solutions appear to be about <m>2.5</m> and <m>-2.5</m>.</p>
<p>
    How can we solve this equation algebraically? The opposite operation for squaring a number is taking a square root. So we can undo the operation of squaring by extracting square roots. We first solve for <m>x^2</m> to get
    \begin{align}
    2x^2 \amp= 12 \\
    x^2 \amp= 6\\
    \end{align}
    and then take square roots to find
    <me>x = \pm\sqrt{6}</me>
</p>
<warning><title></title>
<p>
    Don't forget that every positive number has two square roots. The symbol <m>\pm</m> (read “plus or minus”) is a shorthand notation used to indicate both square roots of <m>6</m>.</p>
</warning>
 <p>The exact solutions are thus <m>\sqrt{6}</m> and <m>-\sqrt{6}</m>. We can also find decimal approximations for the solutions using a calculator. Rounded to two decimal places, the approximate solutions are <m>2.45</m> and <m>-2.45</m>.
</p>
<p>
    In general, we can solve equations of the form 
    <m>ax^2 + c = 0</m>
    by isolating <m>x^2</m> on one side of the equation and then taking the square root of each side. This method for solving equations is called <term>extraction of roots</term><idx>extraction of roots</idx>.
</p>

<assemblage><title>Extraction of Roots</title>
<p>
    To solve the equation
    <me>ax^2 + c = 0</me>
    <ol>
        <li>Isolate <m>x^2</m>.</li>
        <li>Take square roots of both sides. There are two solutions.</li>
    </ol>
</p>
</assemblage>

<example xml:id="example-cat-falling">
    <p>
        If a cat falls off a tree branch 20 feet above the ground, its height <m>t</m> seconds later is given by <m>h = 20 - 16t^2</m>.
    </p>
    <ol label="a">
        <li>What is the height of the cat <m>0.5</m> second later?</li>
        <li>How long does the cat have to get in position to land on its feet before it reaches the ground?</li>
    </ol>
<solution><p><ol label="a">
    <li><p>In this question, we are given a value of <m>t</m> and asked to find the corresponding value of <m>h</m>. To do this, we evaluate the formula for <m>t = 0.5</m>. We substitute <m>\alert{0.5}</m> for <m>t</m> into the formula and simplify.
    \begin{align*}
    h \amp= 20 - 16(\alert{0.5})^2 \amp\amp \text{Compute the power.}\\
    \amp = 20 - 16 (0.25) \amp\amp \text{Multiply; then subtract.}\\
    \amp = 20 - 4 = 16
    \end{align*}
    The cat is <m>16</m> feet above the ground after <m>0.5</m> second.</p></li>
    <li><p>We would like to find the value of <m>t</m> when the height, <m>h</m>, is known. We substitute <m>h = \alert{0}</m> into the equation to obtain
    <me>\alert{0} = 20 - 16t^2</me>
    To solve this equation, we use extraction of roots. First we isolate <m>t^2</m> on one side of the equation.
    \begin{align*}
    16t^2 \amp= 20 \amp\amp \text{Divide by 16.}\\
    t^2 \amp= \frac{20}{16}=1.25
    \end{align*}
    Then we take the square root of both sides of the equation to find
    <me>t = \pm \sqrt{1.25} \approx \pm1.118</me></p></li>
</ol>
</p>

<sidebyside widths="30% 50%" margins="8%"><figure xml:id="fig-cat-falling"><caption></caption><image source="images/fig-cat-falling"><description>cat falling</description></image></figure>
<paragraphs><p>
    Only the positive solution makes sense here, so the cat has approximately 1.12 seconds to get into position for landing.</p>
    <p> A graph of the cat's height after <m>t</m> seconds is shown in <xref ref="fig-cat-falling" text="type-global"/>. The points corresponding to parts (a) and (b) are labeled.
</p> </paragraphs></sidebyside>
</solution></example>
<note><p>
    Note that in part (a) of <xref ref="example-cat-falling" text="type-global"/> we evaluated the expression <m>20 - 16t^2</m> to find a value for <m>h</m>, and in part (b) we solved the equation <m>0 = 20 - 16t^2</m> to find a value for <m>t</m>.
</p></note>

<exercise xml:id="exercise-extract-roots"><statement>
    <ol label="a">
        <li><p><sidebyside widths="60% 36%"><p>Solve by extracting roots
            <m>\dfrac{3x^2 - 8}{5}= 10.</m> </p><p><m>\text{First, isolate } x^2 \text{. Take the}</m><m>\text{square root of both sides.}</m></p></sidebyside></p></li>
        <li><p>Give exact answers; then give approximations rounded to two decimal places.</p></li>
    </ol>
</statement>
<answer><p><m>x=\pm\sqrt{\dfrac{58}{3}}\approx \pm 4.40 </m></p></answer>
</exercise>

</subsection>

<subsection><title>Solving Formulas</title>
<p>
    We can use extraction of roots to solve many formulas involving the square of the variable.
</p>

<example xml:id="example-volume-cone">
    <p>The formula <m>V = \dfrac{1}{3} πr^2h</m> gives the volume of a cone in terms of its height and radius. Solve the formula for <m>r</m> in terms of <m>V</m> and <m>h</m>.
    </p>

<solution><p>
    Because the variable we want is squared, we use extraction of roots. First, we multiply both sides by <m>3</m> to clear the fraction.
    <me>
        \begin{aligned}
        \alert{3}V \amp = \alert{3}(\frac{1}{3} πr^2h)\\
        3V \amp = πr^2h \amp\amp \text{Divide both sides by } πh.\\
        \frac{3V}{πh}\amp= r^2 \amp\amp \text{Take square roots.}\\
        \pm\sqrt{\dfrac{3V}{πh}}\amp= r
        \end{aligned}
    </me>
    Because the radius of a cone must be a positive number, we use only the positive square root: 
    <m>r = \sqrt{\dfrac{3V}{πh}}</m>.
</p></solution>
</example>

<exercise xml:id="exercise-area-circle"><statement>
    <p>Find a formula for the radius of a circle in terms of its area.</p>
</statement>
    <hint>
    <p>Start with the formula for the area of a circle: <m>A =</m> <fillin characters="8" /></p>
    <p>Solve for <m>r</m> in terms of <m>A</m>.</p></hint>
<answer><p><m>r=\sqrt{A/ \pi} </m></p></answer>
</exercise>

</subsection>

<subsection><title>More Extraction of Roots</title>
<p>
    Equations of the form
    <me>a( px + q)^2 + r = 0</me>
    can also be solved by extraction of roots after isolating the squared expression, <m>(px + q)^2</m>.
</p>

<example xml:id="example-extracting-roots">
<p>
    Solve the equation <m>3(x - 2)^2 = 48</m>.
</p>
<solution><p>
    First, we isolate the perfect square, <m>(x - 2)^2</m>.
    <me>
        \begin{aligned}
        3(x - 2)^2 \amp = 48 \amp\amp\text{Divide both sides by }3.\\
        (x - 2)^2 \amp = 16 \amp\amp\text{Take the square root of each side.}\\
        x - 2 \amp = \pm\sqrt{16} = \pm 4
        \end{aligned}
    </me>
    
    This gives us two equations for <m>x</m>, 
    <me>
        \begin{aligned}
        x - 2 \amp =4 \amp\amp \text{or }\amp x - 2 \amp =-4\amp\amp\text {Solve each equation.}\\
        x \amp=6 \amp\amp \text{or }\amp x \amp =-2
        \end{aligned}
    </me>
        The solutions are <m>6</m> and <m>-2</m>.
</p></solution>
</example>

<p>Here is a general strategy for solving equations by extraction of roots.</p>

<assemblage><title>Extraction of Roots</title><line></line>
<p>
    To solve the equation
    <me>a(px+q)^2 +r=0</me>
<ol>
    <li>Isolate the squared expression, <m>(px + q)^2</m>.</li>
    <li>Take the square root of each side of the equation. Remember that a positive number has two square roots.</li>
    <li>Solve each equation. There are two solutions.</li>
</ol>
</p>
</assemblage>

<exercise xml:id="exercise-extract-roots2"><statement>
    Solve <m>2(5x + 3)^2 = 38</m> by extracting roots.
    <ol label="a">
        <li>Give your answers as exact values.</li>
        <li>Find approximations for the solutions to two decimal places.</li>
    </ol>
</statement>
<answer><p>
    <ol label="a" cols="2">
        <li><p><m>x=\dfrac{-3\pm \sqrt{19}}{5} </m></p></li>
        <li><p><m>x\approx -1.47</m> or <m>x\approx 0.27</m></p></li>
    </ol>
</p></answer>
</exercise>
</subsection>

<subsection><title>Compound Interest and Inflation</title>
<p>
    Many savings institutions offer accounts on which the interest is <em>compounded annually</em>. At the end of each year, the interest earned is added to the principal, and the interest for the next year is computed on this larger sum of money.
</p>

<assemblage><title>Compound Interest</title>
<p>
    If interest is compounded annually for <m>n</m> years, the amount, <m>A</m>, of money in an account is given by
    <me>A = P(1 + r)^n</me>
    where <m>P</m> is the principal and <m>r</m> is the interest rate, expressed as a decimal fraction.
</p>
</assemblage>

<example xml:id="example-compound-interest">
<p>
    Carmella invests <dollar/>3000 in an account that pays an interest rate, <m>r</m>, compounded annually.
    <ol label="a">
        <li>Write an expression for the amount of money in Carmella’s account after two years.</li>
        <li>What interest rate would be necessary for Carmella's account to grow to <dollar/>3500 in two years?</li>
    </ol>
</p>
<solution><p>
    <ol label="a">
        <li>We use the formula above with <m>P = 3000</m> and <m>n = 2</m>. Carmella’s account balance will be
            <me>A = 3000(1 + r)^2</me></li>
        <li>We substitute <m>3500</m> for <m>A</m> in the equation.
        <me>\alert{3500} = 3000(1 + r)^2</me>
        We can solve this equation in <m>r</m> by extraction of roots. First, we isolate the perfect square.
        \begin{align}
        3500 \amp = 3000(1 + r )^2 \amp\amp \text{Divide both sides by 3000.} \\
        1.1\overline{6} \amp = (1 + r )^2 \amp\amp \text{Take the square root of both sides.} \\
        \pm 1.0801 \amp\approx 1 + r \amp\amp \text{Subtract 1 from both sides.}\\
        r \amp\approx 0.0801   \text{  or  } r \approx -2.0801
        \end{align}
        Because the interest rate must be a positive number, we discard the negative solution. Carmella needs an account with interest rate <m>r \approx 0.0801</m>, or just over 8<percent/>, to achieve an account balance of <dollar/>3500 in two years.</li>
</ol>
</p></solution>
</example>

<p>
    The formula for compound interest also applies to the effects of inflation. For instance, if there is a steady inflation rate of 4<percent/> per year, in two years an item that now costs <dollar/>100 will cost
    \begin{align}
    A \amp= P(1 + r)^2\\
    \amp = 100(1 + 0.04)2 = \$108.16
    \end{align}
</p>

<exercise xml:id="exercixe-inflation"><statement>
    Two years ago, the average cost of dinner and a movie was <dollar/>24. This year the average cost is <dollar/>25.44. What was the rate of inflation over the past two years?
</statement>
<answer><p><m>r\approx 2.96\%</m></p></answer>
</exercise>
</subsection>
<subsection><title>Other Nonlinear Equations</title>
<p>
    Because squaring and taking square roots are opposite operations, we can solve the equation
    <me>\sqrt{x} = 8.2</me>
    by squaring both sides to get
    \begin{align*}
    (\sqrt{x})^2 \amp= 8.2^2 \\
    x \amp = 67.24
    \end{align*}
    Similarly, we can solve
    <me>x^3 = 258</me>
    by taking the cube root of both sides, because cubing and taking cube roots are opposite operations. Rounding to three places, we find 
    \begin{align}
    \sqrt[3]{x^3}\amp=258\\
    x \amp \approx 6.366
    \end{align}
    The notion of undoing operations can help us solve a variety of simple nonlinear equations. The operation of taking a reciprocal is its own opposite, so we solve the equation
    <me>\frac{1}{x} = 50</me>
    by taking the reciprocal of both sides to get
    <me>x = \frac{1}{50}= 0.02</me>
</p>

<example xml:id="example-nonlinear-equation">
    <p>
        Solve <m>\dfrac{3}{x - 2}= 4</m>.
    </p>
<solution><p>
    We begin by taking the reciprocal of both sides of the equation to get
    <me>\frac{x - 2}{3}= \frac{1}{4}</me>
    We continue to undo the operations in reverse order. First, we multiply both sides by <m>3</m>.
    \begin{align}
    x - 2 \amp = \frac{3}{4} \amp\amp \text{Add 2 to both sides.}\\
    x \amp = 2 + \frac{3}{4}= \frac{11}{4} \amp\amp \tfrac{2}{1}=\tfrac{8}{4}
    \text{, so } \tfrac{2}{1}+ \tfrac{3}{4}=\tfrac{8}{4}+\tfrac{3}{4}=\tfrac{11}{4}
    \end{align}
    The solution is <m>\dfrac{11}{4}</m>, or <m>2.75</m>.
</p></solution>
</example>  

<exercise xml:id="exercise-nonlinear-equation"><statement>
    Solve <m>2\sqrt{x + 4} = 6</m>.
</statement>
<answer><p><m>x=5</m></p></answer>
</exercise>

<technology><title><!--<image source="images/icon-GC.jpg"  width="8%"><description>Graphing Calculator</description></image>-->Using the Intersect Feature</title>
    <p>We can use the <em>intersect</em> feature on a graphing calculator to solve equations.</p>
</technology>

<example xml:id="example-GC-nonlinear-equation">
    <p>
        Use a graphing calculator to solve <m>\dfrac{3}{x -2}= 4.</m>
    </p>

<solution><p>
    We would like to find the points on the graph of <m>y = \dfrac{3}{x -2}</m> that have <m>y</m>-coordinate equal to <m>4</m>. We graph the two functions
    \begin{align}
    Y_1 \amp = 3/(X - 2)  \\
    Y_2 \amp = 4
    \end{align}

    in the window
    \begin{align}
    \text{Xmin} \amp = -9.4 \amp\amp \text{Xmax} = 9.4\\
    \text{Ymin} \amp = -10 \amp\amp \text{Ymax} = 10
    \end{align}
    The point where the two graphs intersect locates the solution of the equation. If we trace along the graph of <m>Y_1</m>, the closest we can get to the intersection point is <m>(2.8, 3.75)</m>, as shown in <xref ref="fig-GC-nonlinear" text="type-global" />a. We get a better approximation using the <em>intersect</em> feature.
    <figure xml:id="fig-GC-nonlinear"><caption></caption><image source="images/fig-GC-nonlinear"  width="80%"><description>GC displays</description></image></figure>

    Use the arrow keys to position the Trace bug as close to the intersection point as you can. Then press <c>2nd</c> <c>TRACE</c> to see the Calculate menu. Press <m>5</m> for intersect; then respond to each of the calculator's questions, <em>First curve?</em>, <em>Second curve?</em>, and <em>Guess?</em> by pressing <c>ENTER</c>. The calculator will then display the intersection point, <m>x = 2.75</m>, <m>y = 4</m>, as shown in <xref ref="fig-GC-nonlinear" text="type-global" />b. The solution of the original equation is <m>x = 2.75</m>.
</p></solution>
</example>

<exercise xml:id="exercise-GC-nonlinear"><statement>
    Use the intersect feature to solve the equation <m>2x^2 - 5 = 7</m>. Round your answers to three decimal places.
</statement>
<answer><p><m>x=\pm 2.449</m></p></answer>
</exercise>

</subsection>
<xi:include href="./summary-2-1.xml" /> <!-- summary  -->
<xi:include href="./section-2-1-exercises.xml" /> <!-- exercises  -->


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