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      <text>11.6.1.1 Parametric differentiation</text>
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      <text><![CDATA[<h4>Parametric differentiation</h4>
<p></p>
In this subsection we consider the parametric approach to describing a curve:
\[ x = h(t), \hspace{0.6cm} y = g(t) \hspace{2cm}t_0 \leq t \leq t_1\]
The functions \(x = h(t)\) and \(y = g(t)\) are known as the <b>parametric equations</b>.
<p></p>
The interval \(t_0 \leq t \leq t_1\) is known as the <b>parametric range</b>
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As various values of \(t\) are chosen within the parameter range the corresponding values of \(x\), \(y\) are
calculated from the parametric equations. When these points are plotted on an \(xy\) plane they trace
out a curve. The Cartesian equation of this curve is obtained by eliminating the parameter \(t\) from
the parametric equations. For example, consider the curve:
\[ x = 2 \cos t \hspace{0.6cm} y = 2 \sin t  \hspace{2cm} 0 \leq t \leq 2\pi .\]
We can eliminate the \(t\) variable in an obvious way - square each parametric equation and then add:
\[ x^2 + y^2 = 4 \cos^2 t + 4 \sin^2 t = 4 \hspace{1cm} \therefore \hspace{1cm}  x^2 + y^2 = 4\]
which we recognise as the standard equation of a <b>circle</b> with centre at \((0, 0)\) with radius \(2\).
In a similar fashion the parametric equations
\[ x = 2t \hspace{0.6cm} y = 4t^2 \hspace{2cm}   - \infty &lt; t &lt; \infty \]
describes a <b>parabola</b>. This follows since, eliminating the parameter \(t\):
\[ t = \dfrac{x}{2} \hspace{0.6cm} \therefore \hspace{0.6cm} y = 4 \left( \dfrac{x^2}{4} \right)  \hspace{0.6cm} \text{so } y = x^2 \]
which we recognise as the standard equation of a parabola.
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The question we wish to address in this Section is `how do we obtain the derivative \( \dfrac{dy}{dx} \) if a curve is given in parametric form?' To answer this we note the key result in this area:

<div class="HELM_keypoint">
<h4>Key Point</h4>
<b>Parametric Differentiation</b>
<p></p>
If \(x = h(t)\) and \(y = g(t)\) then
\[ \dfrac{dy}{dx} = \left.\dfrac{dy}{dt} \middle/\dfrac{dx}{dt}\right.\]
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<p></p>
We note that this result allows the determination of \( \dfrac{dy}{dx}\) without the need to find \(y\) as an explicit function of \(x\).]]></text>
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