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<!DOCTYPE html>
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<!-- Mirrored from www.mathsisfun.com/activity/walk-in-desert.html by HTTrack Website Copier/3.x [XR&CO'2014], Sat, 29 Oct 2022 00:58:18 GMT -->
<head>
<meta http-equiv="content-type" content="text/html; charset=UTF-8">
<!-- #BeginEditable "doctitle" -->
<title>Activity: A Walk in the Desert</title>
<meta name="Description" content="Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.">
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<!-- #BeginEditable "Body" -->
<h1 class="center">Activity: A Walk in the Desert</h1>
<p style="float:right; margin: 10px;"><img src="images/walk-3.jpg" alt="walk in desert plane" height="310" width="150"></p>
<h2>Crash!</h2>
<p>Jade has crash-landed in the desert.</p>
<p>There is a village somewhere nearby, direction unknown.</p>
<p>So Jade comes up with a cunning plan:</p>
<ul>
<div class="bigul">
<li>Fill up a water
bottle from the plane, and take a compass,</li>
<li>Then walk 1 km north, change direction and
walk 2 km east, then 3 km south, 4 km west, 5 km north, 6 km east,
and so on, like this:</li>
</div>
</ul>
<p class="center"><img src="images/walk-1.gif" alt="walk pattern" height="248" width="271"></p>
<p align="left">This way Jade will find the village no matter what direction it is in, and can (hopefully) find the way back to the plane for fresh water and shade when needed.</p>
<p align="left">But Jade needs to know, at the end of each stage:</p>
<ol>
<li>The total distance walked</li>
<li>How far (in a
straight line) back to the plane</li>
</ol>
<h2>OK, Let's Do the Calcs ...</h2>
<p>After one stage of the journey, Jade has reached point A:</p>
<ul>
<li>Jade has walked 1 km
altogether.</li>
<li>And is 1 km (in a straight
line) from the plane.</li>
</ul>
<p>After two stages, Jade has reached point B:</p>
<ul>
<li>Jade has walked 3 km
altogether.</li>
<li>To answer the second question, we make a right-angled triangle OAB:</li>
</ul>
<p class="center"><img src="images/walk2.gif" alt="pythagoras triangle 1,2,root5" height="100" width="134"></p>
<p>We can calculate the length of OB using <a href="../pythagoras.html">Pythagoras' Theorem</a>, as follows:</p>
<div class="so">OB<sup>2</sup> = OA<sup>2</sup> + AB<sup>2</sup></div>
<div class="so">OB<sup>2</sup> = 1<sup>2</sup> + 2<sup>2</sup></div>
<div class="so">OB<sup>2</sup> = 1 + 4</div>
<div class="so">OB<sup>2</sup> = 5</div>
<div class="so">OB = √5<br>
</div>
<p>So the answer in this case is:</p>
<p class="center larger">At Point B, the distance back to the plane (in a straight line) is <b>√5 km</b></p>
<p>&nbsp;</p>
<p>After three stages, Jade is at point C:</p>
<div class="indent50px">
<p class="larger">Hey, it's your turn Now!</p>
<p class="larger">Fill in all the other values ... if it gets hard, read below for some help</p>
</div>
<table align="center" cellspacing="2" cellpadding="2" border="1">
<tbody>
<tr>
<th align="center">Point</th>
<th align="center">Distance<br>
walked
altogether</th>
<th align="center">Distance (in a<br>
straight
line) from O</th>
</tr>
<tr>
<td align="center" height="30">O</td>
<td align="center" height="25">0</td>
<td align="center" height="25">0</td>
</tr>
<tr>
<td align="center" height="30">A</td>
<td align="center" height="25">1</td>
<td align="center" height="25">1<br>
</td>
</tr>
<tr>
<td align="center" height="30">B</td>
<td align="center" height="25">3</td>
<td align="center" height="25">√5</td>
</tr>
<tr>
<td align="center" height="30">C</td>
<td align="center" height="25"> 6</td>
<td align="center" height="25">&nbsp;</td>
</tr>
<tr>
<td align="center" height="30">D</td>
<td align="center" height="25">&nbsp;</td>
<td align="center" height="25">&nbsp;</td>
</tr>
<tr>
<td align="center" height="30">E</td>
<td align="center" height="25">&nbsp;</td>
<td align="center" height="25">&nbsp;</td>
</tr>
<tr>
<td align="center" height="30">F</td>
<td align="center" height="25">&nbsp;</td>
<td align="center" height="25">&nbsp;</td>
</tr>
<tr>
<td align="center" height="30">G</td>
<td align="center" height="25">&nbsp;</td>
<td align="center" height="25">&nbsp;</td>
</tr>
<tr>
<td align="center" height="30">H</td>
<td align="center" height="25">&nbsp;</td>
<td align="center" height="25">&nbsp;</td>
</tr>
<tr>
<td align="center" height="30">I</td>
<td align="center" height="25">&nbsp;</td>
<td align="center" height="25">&nbsp;</td>
</tr>
<tr>
<td align="center" height="30">J</td>
<td align="center" height="25">&nbsp;</td>
<td align="center" height="25">&nbsp;</td>
</tr>
</tbody>
</table>
&nbsp;
<h2>How To Make It Easier</h2>
<h3>Distance walked
altogether</h3>
<p>At the end of each stage, the total distance is the sum of the series 1 + 2 + 3
+ 4 + 5 + 6 + ...</p>
<div class="dotpoint">
<p>So just add the new distance each time.</p>
</div>
<div class="dotpoint">
<p><b>OR</b> you can calculate each value using:</p>
<p class="center"><span class="large">n(n + 1)/2</span><br>
where <b>n</b> is the number of stages.</p>
<p>Like this:</p>
<table align="center" cellspacing="2" cellpadding="2" border="1">
<tbody>
<tr>
<td style="text-align: center;">Number
of<br>
stages (n)</td>
<td style="text-align: center;">Total
distance walked<br>
=&nbsp;n(n + 1)/2</td>
</tr>
<tr>
<td style="text-align: center;">1</td>
<td style="text-align: center;">1 × 2 / 2 = <b>1</b></td>
</tr>
<tr>
<td style="text-align: center;">2</td>
<td style="text-align: center;">2 × 3
/ 2 =&nbsp;<b>3</b></td>
</tr>
<tr>
<td style="text-align: center;">3</td>
<td style="text-align: center;">3 × 4
/ 2 =&nbsp;<b>6</b></td>
</tr>
<tr>
<td style="text-align: center;">4</td>
<td style="text-align: center;">etc ...<br>
</td>
</tr>
</tbody>
</table>
<p>This works because it is the "<a href="../algebra/triangular-numbers.html">Triangular Number Sequence</a>":</p>
<div class="simple"></div>
<center>
<p><img src="../numbers/images/triangular-number-dots.svg" alt="triangular numbers" style="max-width:100%" height="146" width="395"></p>
</center>
</div>
<p>Try doing the calculations both ways, for fun.</p>
<p>&nbsp;</p>
<h3>Distance (in a
straight
line) from O</h3>
<p>To work out the distance back to the plane, we can map out the journey on a <a href="../data/cartesian-coordinates.html">Coordinate Grid</a>, like this:</p>
<p class="center"><img src="images/walk-2.gif" alt="walk pattern on grid" height="273" width="303"></p>
<p>Now it is just a matter of finding the <a href="../algebra/distance-2-points.html">distance
between two points</a></p>
<p>The distance between the points (x<sub>A</sub>,y<sub>A</sub>)
and (x<sub>B</sub>,y<sub>B</sub>) is given by the formula:</p>
<p class="center large">
c = <span style="font-size:120%; ">√</span><span class="overline">(x<sub>A</sub> − x<sub>B</sub>)<sup>2</sup> + (y<sub>A</sub> − y<sub>B</sub>)<sup>2</sup></span>
</p>
<p>and one of those points is always the origin, which is at (0,0), so when <span class="larger">x<sub>B</sub></span> and <span class="larger">y<sub>B</sub></span> are zero we get:</p>
<div class="center large">c = <span style="font-size:120%;">√</span><span class="overline">x<sup>2</sup> + y<sup>2</sup></span></div>
<!-- c = SQR(x^2 + y^2) -->
<div class="example">
<p>Example, for the point <b>E (−2, 3)</b>, x = −2 and y = 3, and so:</p>
<div class="center large">c = <span style="font-size:120%;">√</span><span class="overline">(−2<sup>2</sup>) + 3<sup>2</sup></span> = <span style="font-size:120%;">√</span><span class="overline">4+9</span> = <span style="font-size:120%;">√</span><span class="overline">13</span></div>
<!-- c = SQR((-2)^2 + 3^2) = SQR(4+9) = SQR(13) -->
</div>
<p>Hopefully that will help make your job easier.</p>
<p style="float:right; margin: 0 0 5px 10px;"><img src="../measure/images/compass-bearing.svg" alt="compass bearing" style="float:left; margin: 10px;" height="220" width="219"></p>
<h2>Direction?</h2>
<p>There is one more thing Jade must know: to get back to the plane, what compass bearing to use?</p>
<p>This will be covered in <a href="walk-in-desert-2.html">Activity: A Walk in the
Desert 2</a></p>
<div style="clear:both"></div>
<p>&nbsp;</p>
<div class="related">
<a href="../pythagoras.html">Pythagoras' Theorem</a>
<a href="index.html">Activity Index</a>
</div>
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