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    <name>
      <text>12.2.4.2 maximum temperature</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<p class="HELM_exercise">Exercise</p><p>Write <code>omega</code> for \(\omega\), <code>pi</code> for \(\pi\) and <code>atan</code> for \(\tan^{-1}\).</p><p>A perturbation in the temperature of a stream leaving a chemical reactor follows a decaying sinusoidal variation, according to \[ T(t)=5\text{exp}(-at)\sin(\omega t)\]where \(a\) and \(\omega\) are positive constants.</p><br>


<ol class="HELM_parts">
    <li>Sketch the variation of temperature with time.</li>
    <li>By examining the behaviour of \(\dfrac{\mathrm{d}T}{\mathrm{d}t}\), show that the maximum temperatures occur at times of \(\left(\tan^{-1}\left(\dfrac{\omega}{a}\right)+2\pi n\right)/\omega\).<br><br> \(\quad \dfrac{\mathrm{d}T}{\mathrm{d}t} = \) [[input:ans1]] [[validation:ans1]] [[feedback:prt1]]<br><br>        \(\quad\)At stationary points, \(\quad\omega t = \) [[input:ans2]] [[validation:ans2]] \(+\) [[input:ans3]] [[validation:ans3]] [[feedback:prt2]]<br><br> \(\quad\)Maximum temperatures are located at \(t\) such that \(\dfrac{\mathrm{d}^2T}{\mathrm{d}t^2}\) [[input:ans4]]
        [[validation:ans4]] which is true [[input:ans5]] [[validation:ans5]] [[feedback:prt3]]<br><br> \(\quad\)Hence, maximum temperatures occur at times \(t=\left(\tan^{-1}\left(\dfrac{\omega}{a}\right)+2\pi n\right)/\omega\), where \(2n=\) [[input:ans6]] [[validation:ans6]]
        is an integer. [[feedback:prt4]]</li>
</ol>]]></text>
    </questiontext>
    <generalfeedback format="html">
      <text><![CDATA[<p><strong>Worked solution</strong></p><br>

<ol class="HELM_parts">
    <li>The sketch:<br><br>
      <figure class="HELM_figure"><img src="@@PLUGINFILE@@/12-2-diagram-decay.svg" alt="Oscillatory decay of temperature with time." width="201" height="151" class="img-responsive atto_image_button_text-bottom"><br></figure></li>
    <li>\(\dfrac{\mathrm{d}T}{\mathrm{d}t}=0\) implies that \({@spexp1@}\), so \(\tan(\omega t) &gt; 0\) and \[\omega t = {@rhs(spexp2)@} + k\pi,\quad k\mbox{ an integer }\]Examination of \(\dfrac{\mathrm{d}^2T}{\mathrm{d}t^2}\) reveals that only even values
        of \(k\) give \(\dfrac{\mathrm{d}^2T}{\mathrm{d}t^2}&lt;0\) for a maximum. So setting \(k=2n\) gives the required answer.
    </li>
</ol>]]></text>
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      <text>2020120600</text>
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      <text><![CDATA[declare(a, constant);
declare(omega, constant);
assume(a>0, omega>0);
triginverses: all;

T: 5*exp(-a*t)*sin(omega*t);
dT: diff(T,t);
spts: solve(dT)[1];
spexp1: trigreduce(trigsimp(spts/cos(omega*t)));
spexp2: atan(spexp1);
spexp3: rhs(spexp2)+k*pi;

ta1: dT;
ta2: rhs(spexp2);
ta3: [[0,false],[k*pi/2,false],[k*pi,true],[2*k*pi,false]];
ta4: [[1,false,"is positive"],[2,true,"is negative"],[3,false, "is zero"]];
ta5: [[0,false,"for all k"],[1,false,"only for k=0"],[2,false,"for odd k"],[3,true,"for even k"]];
ta6: k;]]></text>
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      <text><![CDATA[<span style="font-size: 1.5em; color:green;"><i class="fa fa-check"></i></span> Correct answer, well done.]]></text>
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