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  <question type="description">
    <name>
      <text>12.5.1.1 Differentiation of Vectors</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>Differentiation of vectors</h3>

<p>Consider the figure below.</p>

<figure class="HELM_figure"><img src="@@PLUGINFILE@@/Figure%201.svg" alt="See the text below." width="330" height="189" class="img-responsive atto_image_button_text-bottom"><br></figure>


<p>If \({\underline r}\) represents the <strong>position vector</strong> of an object which is moving along a curve \(C\), then the position vector will be dependent upon the time, \(t\). We write \({\underline r}={\underline r}(t)\) to show the dependence
    upon time. Suppose that the object is at the point \(P\), with position vector \(\underline{r}(t)\) at time \(t\) and at the point \(Q\), with position vector \({\underline r}(t+\delta t)\), at the later time \(t+\delta t\), as shown in the figure
    below.
</p>


   <figure class="HELM_figure"><img src="@@PLUGINFILE@@/Figure_12_5_2.svg" alt="See the text above and below." width="322" height="181" class="img-responsive atto_image_button_text-bottom"></figure>


<p>Then \(\overrightarrow{PQ}\) represents the <strong>displacement vector</strong> of the object during the interval of time \(\delta t.\) The length of the displacement vector represents the distance travelled, and its direction gives the direction of
    motion. The average velocity during the time from \(t\) to \(t+\delta t\) is defined as the displacement vector divided by the time interval \(\delta t,\) that is,</p>

<p>\[\mathrm{average} \; \mathrm{velocity} \; =\dfrac{\overrightarrow{PQ}}{\delta t}=\dfrac{{\underline r} (t+\delta t)-{\underline r}(t)}{\delta t}\]</p>

<p>If we now take the limit as the interval of time \(\delta t\) tends to zero then the expression on the right hand side is the <strong>derivative</strong> of \(\underline r\) with respect to \(t\). Not surprisingly we refer to this derivative as <strong>the instantaneous velocity</strong>,
    \({\underline v}\). By its very construction we see that the velocity vector is always tangential to the curve as the object moves along it. We have:</p>

<p> \[ \underline{v} = \lim_{\delta t \to 0} \frac{{\underline r}(t+\delta t)-{\underline r}(t)}{\delta t}=\frac{\mathrm{d}{\underline r}}{\mathrm{d}t}\]</p>

<p>Now, since the \(x\) and \(y\) coordinates of the object depend upon time, we can write the position vector \({\underline r}\) in Cartesian coordinates as:</p>

<p> \(\qquad {\underline r}(t)=x(t){\underline i}+y(t){\underline j}\)</p>

<p>Therefore, </p>
<p>\(\qquad {\underline r}(t+\delta t)=x(t+\delta t){\underline i}+y(t+\delta t){\underline j}\)</p>

<p>so that,</p>
<p>\[\begin{aligned} {\underline v}(t)&amp;=\lim_{\delta t\to 0}\dfrac{x(t+\delta t){\underline i}+y(t+\delta t){\underline j}-x(t){\underline i}-y(t){\underline j}}{\delta t}\\ &amp;= \lim_{\delta t\to 0}\left\{\dfrac{x(t+\delta t)-x(t)}{\delta t}{\underline
    i}+\dfrac{y(t+\delta t)-y(t)}{\delta t}{\underline j}\right\}\\ &amp;=\dfrac{\mathrm{d}x}{\mathrm{d}t}{\underline i}+\dfrac{\mathrm{d}y}{\mathrm{d}t}{\underline j}\end{aligned}\]</p>
<p>This is often abbreviated to \({\underline v}=\dot{\underline r}=\dot x{\underline i}+\dot y{\underline j}\), using notation for derivatives with respect to time. So we see that the velocity vector is the derivative of the position vector with respect
    to time. This result generalizes in an obvious way to three dimensions as summarized in the following key point.</p>

<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p>Given \(\qquad {\underline r}(t)=x(t){\underline i}+y(t){\underline j}+z(t){\underline k}\)</p>
    <p> then the velocity vector is \[\qquad {\underline v} =\dot{\underline r}(t)=\dot x(t){\underline i}+\dot y(t){\underline j}+\dot z(t){\underline k}\] The magnitude of the velocity vector gives the speed of the object. </p>
    <p>We can define the acceleration vector in a similar way, as the rate of change (i.e. the derivative) of the velocity with respect to the time: \[\qquad {\underline a}=\dfrac{\mathrm{d}{\underline v}}{\mathrm{d}t}=\dfrac{\mathrm{d}^2{\underline r}}{\mathrm{d}t^2}=\ddot {\underline r}=\ddot
        x{\underline i}+\ddot y{\underline j}+\ddot z{\underline k}\] </p>
</div>

<hr>
<h4 class="HELM_example">Example</h4>
<p>If \({\underline w}=3t^2{\underline i}+\cos 2t{\underline j},\) find</p>
<ol class="HELM_parts_inline">
    <li>\({\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}}\)</li>
    <li>\({\left|\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\right|}\)</li>
    <li>\({\dfrac{\mathrm{d}^2{\underline w}}{\mathrm{d}t^2}}\)</li>
</ol>

<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
    <li>
        <p>If <span class="math inline">\({\underline w}=3t^2{\underline i}+\cos 2t{\underline j},\)</span> then differentiation with respect to <span class="math inline">\(t\)</span> yields: <span class="math inline">\({\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}}=6t{\underline i}-2\sin 2t{\underline j}\)</span></p>
    </li>
    <li>
        <p><span class="math inline">\({\left|\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\right|}=\sqrt{(6t)^2+(-2\sin
2t)^2}=\sqrt{36t^2+4\sin^2 2t}\)</span></p>
    </li>
    <li>
        <p><span class="math inline">\({\dfrac{\mathrm{d}^2{\underline w}}{\mathrm{d}t^2}=6{\underline i}-4\cos 2t{\underline j}}\)</span></p>
    </li>
</ol>
<hr>

<p>It is possible to differentiate more complicated expressions involving vectors provided certain rules are adhered to as summarized in the following key point.</p>

<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p>If <span class="math inline">\({\underline w}\)</span> and <span class="math inline">\({\underline z}\)</span> are vectors and <span class="math inline">\(c\)</span> is a scalar, all these being functions of time <span class="math inline">\(t\)</span>,
        then: <span class="math display">\[\begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}+{\underline z})&amp;=\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}\\
\dfrac{\mathrm{d}}{\mathrm{d}t}(c{\underline w})&amp;=c\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}+\dfrac{\mathrm{d}c}{\mathrm{d}t}{\underline w}\\
\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\cdot {\underline z})&amp;={\underline w}\cdot \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\cdot {\underline z}\\ \dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\times {\underline z})&amp;={\underline w}\times \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\times {\underline z}\end{aligned}\]</span></p>
</div>


<hr>
<h4 class="HELM_example">Example</h4>
<p>If <span class="math inline">\({\underline w}=3t{\underline i}-t^2{\underline j}\)</span> and <span class="math inline">\({\underline z}=2t^2{\underline i}+3{\underline j}\)</span>, verify the result</p>
<p><span class="math inline">\({\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\cdot {\underline z})={\underline w}\cdot \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\cdot {\underline z}}\)</span></p>

<h4 class="HELM_solution">Solution</h4>
<p><span class="math inline">\({\underline w}\cdot {\underline z}=(3t{\underline i}-t^2{\underline j})\cdot (2t^2{\underline i}+3{\underline j})=6t^3-3t^2.\)</span></p>

<p>Therefore <span class="math inline">\(\qquad \dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\cdot {\underline z})=18t^2-6t \qquad (1)\)</span></p>
<p>Also <span class="math inline">\(\qquad \dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}=3{\underline i}-2t{\underline j}\qquad\text{and}\qquad \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}=4t{\underline i}\)</span></p>

<p><span class="math display">\(\begin{aligned}
\text{so}\qquad{\underline w}\cdot \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+{\underline z}\cdot \dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}&amp;=(3t{\underline i}-t^2{\underline j})\cdot (4t{\underline i})+(2t^2{\underline i}+3{\underline j})\cdot
(3{\underline i}-2t{\underline j})\\ &amp;=12t^2+6t^2-6t\\ &amp;=18t^2-6t \qquad\qquad (2)\end{aligned}\)</span></p>

<p>We have verified <span class="math inline">\({\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\cdot {\underline z})={\underline w}\cdot \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\cdot {\underline z}}\)</span> since (1) is the same as (2).</p>
<hr>

<h4 class="HELM_example">Example</h4>
<p>If <span class="math inline">\({\underline w}=3t{\underline i}-t^2{\underline j}\)</span> and <span class="math inline">\({\underline z}=2t^2{\underline i}+3{\underline j}\)</span>, verify the result</p>
<p> <span class="math inline">\({\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\times {\underline z})={\underline w}\times \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}+\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\times {\underline z}}\)</span></p>

<h4 class="HELM_solution">Solution</h4>
<p><span class="math inline">\({\underline w}\times {\underline z}=\left|\begin{array}{ccc}{\underline
i}&amp;{\underline j}&amp;{\underline k}\\ 3t&amp;-t^2&amp;0\\ 2t^2&amp;3&amp;0
\end{array}\right|
=(9t+2t^4){\underline k}\)</span> \(\qquad \text{implying} \qquad\) <span class="math inline">\({\dfrac{\mathrm{d}}{\mathrm{d}t}({\underline w}\times {\underline z})=(9+8t^3){\underline k}} \qquad (1)\)</span></p>

<p><span class="math display">\(\begin{aligned}
{\underline w}\times \dfrac{\mathrm{d}{\underline z}}{\mathrm{d}t}&amp;=&amp;\left|\begin{array}{ccc}{\underline i}&amp;{\underline
j}&amp;{\underline k}\\ 3t&amp;-t^2&amp;0\\
4t&amp;0&amp;0
\end{array}\right|=4t^3{\underline k}\qquad (2)\end{aligned}\)</span></p>

<p><span class="math display">\(\begin{aligned}
\dfrac{\mathrm{d}{\underline w}}{\mathrm{d}t}\times {\underline z}&amp;=&amp;\left|\begin{array}{ccc}{\underline i}&amp;{\underline
j}&amp;{\underline k}\\ 3&amp;-2t&amp;0\\ 2t^2&amp;3&amp;0
\end{array}\right|=(9+4t^3){\underline k}\qquad (3)\end{aligned}\)</span></p>

<p>We can see that (1) is the same as (2) <span class="math inline">\(+\)</span> (3) as required.</p>]]></text>
<file name="Figure 1.svg" path="/" encoding="base64">[encoded data, 11040 characters]</file>
<file name="Figure 2.svg" path="/" encoding="base64">[encoded data, 14200 characters]</file>
<file name="Figure_12_5_2.svg" path="/" encoding="base64">[encoded data, 32228 characters]</file>
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