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      <text>5.1.1.11 Examples</text>
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<h4 class="HELM_example">Example</h4>

<p>A local authority has flood control plans in which the emergency and rescue services are alerted when the river level rises to critical values. A linear model is used to estimate the variation of height with time. After a period of continuous heavy rain
    the level one day was \(1.5 \, \textrm{m}\) at \(8\) a.m. and \(1.8 \, \textrm{m}\) at \(2\) p.m.</p>
<ol class="HELM_parts">
    <li>Use a linear model to write down an equation for estimating the level of the river at different times of the day:</li>
    <li>Suggest a suitable range of values of time for which the model could be used:</li>
    <li>What time does the model predict that the level of the river will reach \( 2 \, \textrm{m}\)?</li>
</ol>

<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
    <li>
        <p>If the level of water is represented by \(L \, \textrm{m}\) and time by \(t\) hours after \(8\) a.m. then a linear model for the level as a function of time may be written</p>
        <p>\( \quad L = at + b\)</p>
        <p>where \(a\) and \(b\) are constants to be found from the other information in the problem. Specifically, it is stated that \(L = 1.5\) when \(t = 0\) and \(L = 1.8\) when \(t = 6.\) The first statement implies that</p>
        <p>\(\quad 1.5 = 0 + b \quad \textrm{or} \quad b = 1.5\)</p>
        <p>The second statement implies that</p>
        <p>\(\quad 1.8 = 6a + b\)</p>
        <p>or, after substituting for \(b,\)</p>
        <p>\(\quad 1.8 = 6a + 1.5 \quad \textrm{or} \quad 0.3 = 6a \quad \textrm{or} \quad a = 0.05\)</p>
        <p>So the equation for estimating the level of the river at different times is</p>
        <p>\(\quad L = 0.05t + 1.5\)</p>
    </li>

    <li>
        <p>The model is valid between \(8\) a.m. and \(2\) p.m. and, subsequently, only as long as the river level rises steadily.</p>
    </li>

    <li>
        <p>The model will predict a level of \(2 \, \textrm{m}\) at time \(t\) given by</p>
        <p>\( \quad 2 = 0.05t + 1.5 \quad \textrm{or} \quad t =\dfrac{0.5}{0.05}= 10\)</p>
        <p>i.e. \(10\) hours after \(8\) a.m. which is \(6\) p.m.</p>
    </li>
</ol>

<hr>
<h4 class="HELM_example">Example</h4>
<p>During one winter, the roads in a rural area were completely free from snow when it started snowing at midnight. It snowed heavily all night and day. By \(10\) a.m. it was \(19 \, \textrm{cm}\) deep.</p>
<p>To save money the local authorities wait until the snow is \(30 \, \textrm{cm}\) deep before ploughing the snow away from the roads. Forecast when ploughing should start, stating any assumptions you have made.</p>

<h4 class="HELM_solution">Solution</h4>
<p>If the depth of snow is represented by \(D \, \textrm{cm}\) and time by \(t\) hours after midnight then a linear modelfor the depth as a function of time may be written:</p>
<p>\(\quad D = at + b\)</p>
<p>where \(a\) and \(b\) are constants to be found from the other information in the problem or from assumptions. As there was no snow at midnight</p>
<p>\(quad 0 = 0 + b \quad \textrm{or} \quad \(b = 0\)</p>
<p>It is stated that \(D = 19\) when \(t = 10\), i.e.</p>
<p>\(\quad 19 = 10a \quad \textrm{or} \quad a = 1.9\)</p>
<p>So the equation for estimating the depth of snow at different times is</p>
<p>\( \quad D = 1.9t\)</p>
<p>The model will predict a level of \(30 \, \textrm{cm}\) at a time \(t\) is given by</p>
<p>\( \quad 30 = 1.9t \quad \textrm{or} \quad t =\dfrac{30}{1.9}= 15.789\)</p>
<p>i.e. \(15.789\) hours after midnight which is a little after \(3.47\) p.m.</p>
<p>Assuming that the snow build up is steady e.g. no drifting or change in precipitation, this suggests that ploughing should start about \(3.45\) p.m.</p>]]></text>
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