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      <text>28.2.1.1 The gradient of a scalar field</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>The gradient of a scalar field</h3>

<p>Consider the height \(\phi\) above sea level at various points on a hill. Some contours for such a hill are shown in the figure below.</p>

<p><figure class="HELM_figure">
<img src="@@PLUGINFILE@@/28.2%20fig%2014.svg" alt="Contour map of a hill" width="388" height="171" class="img-fluid atto_image_button_text-bottom">
</figure></p>

<p>We are interested in how \(\phi\) changes from one point to another. Starting from \(A\) and making a displacement \(\underline{d}\) the change in height (\(\phi\) ) depends on the direction of the displacement. The magnitude of each \(\underline{d}\) is the same.</p>

<p>
<table class="HELM_table">
  <thead>
    <tr>
      <th>Displacement</th>
      <th>Change in \(\phi\)</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>\(AB\)</td>
      <td>\[40 - 30 = 10\]</td>
    </tr>
    <tr>
      <td>\(AC\)</td>
      <td>\[40 - 30 = 10\]</td>
    </tr>
    <tr>
      <td>\(AD\)</td>
      <td>\[30 - 30 = 0\]</td>
    </tr>
    <tr>
      <td>\(AE\)</td>
      <td>\[20 - 30 = -10\]</td>
    </tr>
  </tbody>
</table>
</p>

<p>The change in \(\phi\) clearly depends on the direction of the displacement. For the paths shown \(\phi\) increases most rapidly along \(AB\), does not increase at all along \(AD\) (as \(A\) and \(D\) are both on the same contour and so are both at the same height) and decreases along \(AE\). </p>
<p>The direction in which \(\phi\) changes fastest is along the line of greatest slope which is orthogonal (i.e. perpendicular) to the contours. Hence, at each point of a scalar field we can define a vector field giving the magnitude and direction of the greatest rate of change of \(\phi\) locally.</p>
<p>A vector field, called the gradient, written grad \(\phi\), can be associated with a scalar field \(\phi\) so that at every point the direction of the vector field is orthogonal to the scalar field contour. This vector field is the direction of the maximum rate of change of \(\phi\).</p>
<p>For a second example consider a metal plate heated at one corner and cooled by an ice bag at the opposite corner. All edges and surfaces are insulated. After a while a steady state situation exists in which the temperature \(\phi\) at any point remains the same. Some temperature contours are shown in the figure below.</p>

<p><img src="@@PLUGINFILE@@/28.2%20fig%2015.svg" alt="Temperature contours and heat flow lines for a metal plate" width="541" height="165" class="img-fluid atto_image_button_text-bottom"></p>

<p>The direction of the heat flow is along the flow lines which are orthogonal to the contours (see the dashed lines in part (b) of the above figure); this heat flow is proportional to the vector field grad \(\phi\).</p>

<p><strong>Definition</strong><br> The gradient of the scalar field \(\phi = f(x,y,z)\) is \(\text{grad}\ \phi = {\underline \nabla} \phi = \dfrac{\partial \phi} {\partial x}\underline{i} + \dfrac{\partial \phi} {\partial y}\underline{j} + \dfrac{\partial \phi} {\partial z}\underline{k}\)</p>

<p>Often, instead of grad \(\phi\), the notation \({\underline \nabla} \phi\) is used. (\(\underline{\nabla}\) is a vector differential operator called ‘del’ or ‘nabla’ defined by \(\displaystyle \frac{\partial } {\partial x}\underline{i} + \frac{\partial } {\partial y}\underline{j} + \frac{\partial } {\partial z}\underline{k}\). As a vector differential operator, it retains the characteristics of a vector while also carrying out differentiation.)<br></p>
<p>The vector grad \(\phi\) gives the magnitude and direction of the greatest rate of change of \(\phi\) at any point, and is always orthogonal to the contours of \(\phi\). For example, in Figure 14, grad \(\phi\) points in the direction of \(AB\) while the contour line is parallel to \(AD\) i.e. perpendicular to \(AB\). Similarly, in part (b) of the above figure, the various intersections of the contours with the lines representing grad \(\phi\) occur at right-angles.<br></p>
<p>For the hill considered earlier the direction and magnitude of grad \(\phi\) are shown at various points in the figure below. Note that the magnitude of grad \(\phi\) is greatest (as indicated by the length of the arrow) when the hill is at its steepest (as indicated by the closeness of the contours).</p>

<img src="@@PLUGINFILE@@/28.2%20fig%2016.svg" alt="Grad phi and the steepest ascent direction for a hill" width="487" height="223" class="img-fluid atto_image_button_text-bottom"></p>
<p><br></p>

<div class="HELM_keypoint">
<h4>Key Point</h4>
    <p>\(\phi\) is a scalar field but grad \(\phi\) is a vector field.</p>
</div><br>

<hr>
<h4 class="HELM_example">Example</h4>
<p> Find grad \(\phi\) for </p>
<ol class="HELM_parts">
<li>\(\phi = x^2 - 3y\)</li>
<li> (b) \(\phi = x y^2 z^3\)</li>
</ol>

<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
<li>\(\underline{\nabla} \phi = \dfrac{\partial}{\partial x}(x^2-3y)\underline{i} + \dfrac{\partial}{\partial y} (x^2 - 3y)\underline{j} + \dfrac{\partial}{\partial z} (x^2 - 3y)\underline{k} = 2x \underline{i} + (-3)\underline{j} +0\underline{k} = 2x\underline{i} - 3\underline{j}.\)</li>
<li>\( \underline{\nabla} \phi = \dfrac{\partial}{\partial x}(xy^2z^3)\underline{i} + \dfrac{\partial}{\partial y} (xy^2z^3)\underline{j} + \dfrac{\partial}{\partial z} (xy^2z^3)\underline{k} = y^2z^3 \underline{i} + 2xyz^3\underline{j} +3xy^2z^2\underline{k}.\)</li>
</ol>

<p>&nbsp;</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>For \(f = x^2 + y^2\) find grad \(f\) at the point \(A(1,2)\). Show that the direction of grad \(f\) is orthogonal to the contour at this point.</p>

<h4 class="HELM_solution">Solution</h4>
<p>\[\underline{\nabla} f = \dfrac{\partial f}{\partial x} \underline{i} + \dfrac{\partial f}{\partial y} \underline{j} + \dfrac{\partial f}{\partial z} \underline{k} = 2x\underline{i} + 2y \underline{j} + 0\underline{k} = 2x \underline{i} + 2y\underline{j}\]
and at \(A (1,2)\) this equals \(2 \times 1\underline{i} + 2 \times 2\underline{j} = 2\underline{i} + 4\underline{j}.\)</p>
<p>Since \(f=x^2+y^2\) then the contours are defined by \(x^2+y^2=\) constant, so the contours are circles centred at the origin. The vector grad \(f\) at \(A(1,2)\) points directly away from the origin and hence grad \(f\) and the contour are orthogonal; see the figure below. Note that \(\underline{r} (A)=\underline{i} +2\underline{j} =\dfrac{1}{2} \text{grad }f\).</p>
<p><strong>Figure 17:</strong></p>


<p>The change in a function \(\phi\) in a given direction (specified as a unit vector \(\underline{a}\)) is determined from the scalar product \((\text{grad} \; \phi) \cdot \underline{a}\). This scalar quantity is called the directional derivative.</p>
<p>Note:</p>
<ul>
<li>\(\underline{a}\) along a contour implies \(\underline{a}\) is perpendicular to \(\text{grad} \; \phi\) which implies \(\underline{a} \cdot \text{grad} \; \phi=0\).</li>
<li>\(\underline{a}\) perpendicular to a contour implies \(\underline{a} \cdot\text{grad} \; \phi\) is a maximum.</li>
</ul>]]></text>
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