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      <text>14.4.1 Introduction</text>
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      <text><![CDATA[<h4>The length of a curve</h4>
<p>To find the length of a curve in the \(xy\) plane we first divide the curve into a large number of pieces. We measure (or, at least, approximate) the length of each piece and then by an obvious summation process obtain an estimate for the length of the
    curve. Theoretically, we allow the number of pieces to increase without bound, implying that the length of each piece will tend to zero. In this limit the summation process becomes an integration process.</p>
<p style="text-align: center;"><img src="@@PLUGINFILE@@/14-4-fig-11-CurveLength.svg" alt="See text below" width="500" height="220" class="img-responsive atto_image_button_text-bottom"><br></p>
<p>The Figure above shows the portion of the curve \(y(x)\) between \(x = a\) and \(x = b\). A small piece of this curve has been selected and can be considered as the hypotenuse of a triangle with base \(\delta x\) and height \(\delta y\). (Here \(\delta
    x\) and \(\delta y\) are intended to be ‘small’ so that the <b>curved segment</b> can be regarded as a <b>straight segment</b>.) </p>
<p>Using Pythagoras’ theorem, the length of the hypotenuse is: \[\sqrt{\delta x^2 + \delta y^2}= \sqrt{1+\left(\frac{\delta y}{\delta x}\right)^2} \: \delta x \]</p>
<p>By summing all such contributions between \(x=a\) and \(x=b\), and letting \(\delta x \to 0\) we obtain an expression for the total length of the curve: \[ \lim_{\delta x \to 0} \sum_{x=a}^{x=b} \sqrt{1+\left(\frac{\delta y}{\delta x}\right)^2} \: \delta
    x \]</p>
<p>But we already know how to write such an expression in terms of an integral. We obtain the following result:
</p>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p>Given a curve with equation \(y = f(x)\), then the length of the curve between the points where \(x = a\) and \(x = b\) is given by the formula: \[\int_a^b \sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)^2}\ \mathrm{d}x\]</p>
</div>
<p>Because of the complicated form of the integrand, and in particular the square root, integrals of this type are often difficult to calculate. In practice, approximate numerical methods rather than exact methods are normally needed to perform the integration.
    We shall first illustrate the application of the formula in the Key Point above by a problem which could be calculated in a much simpler way, before looking at some harder problems.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Find the length of the curve \(y = 3x + 2\) between \(x = 1\) and \(x = 5\)</p>
<h4 class="HELM_solution">Solution</h4>
<p>In this Example, the curve is in fact a straight line segment, and its length could be obtained using Pythagoras’ theorem without the need for integration.</p>
<p>Notice from the formula in the Key Point that it is necessary to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\), which in this case is \(3\).</p>
<p>Applying the formula we find \[\begin{aligned}\text{length of curve} &amp; \int_1^5 \sqrt{1+(3)^2} \mathrm{d}x \\ &amp; = \int_1^5 \sqrt{10} \mathrm{d}x \\ &amp; = \left[\sqrt{10}x\right]^5_1 \\ &amp; = (5-1)\sqrt{10} = 4 \sqrt{10} = 12.65 \; \text{to}
    \: 2 \text{ d.p} \end{aligned} \]</p>
<p>Thus the length of the curve \(y = 3x + 2\) between the points where \(x = 1\) and \(x = 5\) is \(12.65\) units.</p>]]></text>
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