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<title>Exponential Growth and Decay</title>
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<!-- #BeginEditable "Body" -->
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<h1 class="center">Exponential Growth and Decay</h1>
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<h3 align="center">Exponential growth can be amazing!</h3>
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<p>The idea: something always grows
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in relation to its <b>current </b>value, such as always doubling.</p>
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<div class="example">
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<h3>Example: If a population of rabbits doubles every month, we would have 2, then 4, then 8, 16, 32, 64, 128, 256, etc!</h3>
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</div>
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<h2>Amazing Tree</h2>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/tree.jpg" alt="tree" height="240" width="161"></p>
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<p> </p>
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<p>Let us say we have this special tree.</p>
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<p>It grows <a href="../exponent.html">exponentially</a> , following this formula:</p>
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<div class="center80">
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<p class="center large">Height (in mm) = e<sup>x</sup></p>
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</div>
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<p class="center"><i><b>e</b> is <a href="../numbers/e-eulers-number.html">Euler's number</a>, about 2.718</i></p>
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<div style="clear:both"></div>
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<p> </p>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/ex.gif" alt="e^x graph" height="394" width="140"></p>
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<ul>
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<li>At 1 year old it is: <span class="large">e<sup>1 </sup> = 2.7 mm</span> high ... really tiny!</li>
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<li>At 5 years it is: <span class="large">e<sup>5 </sup> = 148 mm</span> high ... as high as a cup</li>
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<li>At 10 years: <span class="large">e<sup>10 </sup> = 22 m</span> high ... as tall as a building</li>
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<li>At 15 years: <span class="large">e<sup>15 </sup> = 3.3 km</span> high ... 10 times the height of the Eiffel Tower</li>
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<li>At 20 years: <span class="large">e<sup>20 </sup> = 485 km</span> high ... up into space!</li>
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</ul>
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<p> </p>
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<p class="center">No tree could ever grow that tall.<br>
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So when people say "it grows exponentially" ... just think what that means.</p>
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<div style="clear:both"></div>
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<h2>Growth and Decay</h2>
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<p>But sometimes things <b>can</b> grow (or the opposite: decay) exponentially, <b><i>at least for a while</i></b>.</p>
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<p>So we have a generally useful formula:</p>
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<div class="def">
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<p class="center large">y(t) = a × e<sup>kt</sup></p>
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<p class="center">Where <b>y(t)</b> = value at time "t"<br>
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<b>a</b> = value at the start<br>
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<b>k</b> = rate of growth (when >0) or decay (when <0)<br>
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<b>t</b> = time</p>
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</div>
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<p> </p>
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<div class="example">
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<h3>Example: 2 months ago you had 3 mice, you now have 18.</h3>
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<div class="example2">
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<table style="border: 0;">
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<tbody>
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<tr>
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<td><img src="images/mice.jpg" alt="Mice" height="135" width="166"></td>
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<td>
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<p>Assuming the growth continues like that</p>
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<ul>
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<li>What is the "k" value?</li>
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<li>How many mice 2 Months from now?</li>
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<li>How many mice 1 Year from now?</li>
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</ul></td>
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</tr>
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</tbody></table>
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</div>
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<p> </p>
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<p>Start with the formula:</p>
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<p class="center larger">y(t) = a × e<sup>kt</sup></p>
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<p>We know <b>a=3</b> mice, <b>t=2</b> months, and right now <b>y(2)=18</b> mice:</p>
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<p class="center"><span class="larger">18 = 3 × e<sup>2k</sup></span></p>
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<p>Now some algebra to solve for <b>k</b>:</p>
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<div class="tbl">
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<div class="row"><span class="left">Divide both sides by 3:</span><span class="right">6 = e<sup>2k</sup></span></div>
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<div class="row"><span class="left">Take the natural logarithm of both sides:</span><span class="right">ln(6) = ln(e<sup>2k</sup>)</span></div>
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<div class="row"><span class="left"><b>ln(e<sup>x</sup>)=x</b>, so:</span><span class="right">ln(6) = 2k</span></div>
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<div class="row"><span class="left">Swap sides:</span><span class="right">2k = ln(6)</span></div>
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<div class="row"><span class="left">Divide by 2:</span><span class="right"><b>k = ln(6)/2</b></span></div>
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</div>
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<p><i>Notes:</i></p>
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<ul>
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<li><i>The step where we used <b>ln(e<sup>x</sup>)=x</b> is explained at <a href="exponents-logarithms.html">Exponents and Logarithms</a></i>.</li>
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<li><i>we could calculate <b>k ≈ 0.896</b>, but it is best to keep it as <b>k = ln(6)/2</b> until we do our final calculations.</i></li>
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</ul>
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<p> </p>
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<p>We can now put <b>k = ln(6)/2</b> into our formula from before:</p>
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<p class="center larger">y(t) = 3 e<sup>(ln(6)/2)t<b></b></sup></p>
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<p>Now let's calculate the population in 2 more months (at <b>t=4</b> months):</p>
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<p class="center larger">y(<b>4</b>) = 3 e<sup>(ln(6)/2)×<b>4</b></sup> = 108</p>
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<p>And in 1 year from now (<b>t=14</b> months):</p>
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<p class="center"><span class="larger">y(<b>14</b>) = 3 e<sup>(ln(6)/2)×<b>14</b></sup> = 839,808</span></p>
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<p>That's a lot of mice! I hope you will be feeding them properly.</p>
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</div>
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<h2>Exponential Decay</h2>
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<p>Some things "decay" (get smaller) exponentially.</p>
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<div class="example">
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<h3>Example: Atmospheric pressure (the pressure of air around you) decreases as you go higher.</h3>
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<p>It decreases about 12% for every 1000 m: an <b>exponential decay</b>.</p>
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<p>The pressure at sea level is about 1013 hPa (depending on weather).</p>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/mount-everest.jpg" alt="mount everest" height="214" width="243"></p>
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<p> </p>
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<ul>
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<li>Write the formula (with its "k" value),</li>
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<li>Find the pressure on the roof of the Empire State Building (381 m),</li>
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<li>and at the top of Mount Everest (8848 m)</li>
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</ul>
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<p> </p>
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<p>Start with the formula:</p>
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<p class="center larger">y(t) = a × e<sup>kt</sup></p>
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<p>We know</p>
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<ul>
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<li><b>a</b> (the pressure at sea level) = <b>1013</b> hPa</li>
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<li><b>t</b> is in meters (distance, not time, but the formula still works)</li>
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<li><b>y(1000)</b> is a 12% reduction on 1013 hPa = <b>891.44</b> hPa</li>
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</ul>
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<p>So:</p>
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<p class="center"><span class="larger">891.44 = 1013 e<sup>k×1000</sup></span></p>
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<p>Now some algebra to solve for <b>k</b>:</p>
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<div class="tbl">
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<div class="row"><span class="left">Divide both sides by 1013:</span><span class="right">0.88 = e<sup>1000k</sup></span></div>
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<div class="row"><span class="left">Take the natural logarithm of both sides:</span><span class="right">ln(0.88) = ln(e<sup>1000k</sup>)</span></div>
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<div class="row"><span class="left"><b>ln(e<sup>x</sup>)=x</b>, so:</span><span class="right">ln(0.88) = 1000k</span></div>
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<div class="row"><span class="left">Swap sides:</span><span class="right">1000k = ln(0.88)</span></div>
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<div class="row"><span class="left">Divide by 1000:</span><span class="right"><b>k = ln(0.88)/1000</b></span></div>
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</div>
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<p> </p>
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<p>Now we know "k" we can write<b></b>:</p>
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<p class="center"><span class="larger">y(t) = 1013 e<sup>(ln(0.88)/1000)×t</sup></span></p>
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<p> </p>
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<p>And finally we can calculate the pressure at <b>381 m</b>, and at <b>8848 m</b>:</p>
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<p> </p>
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<p class="center larger">y(<b>381</b>) = 1013 e<sup>(ln(0.88)/1000)×<b>381</b></sup> = 965 hPa</p>
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<p class="center"><span class="larger">y(<b>8848</b>) = 1013 e<sup>(ln(0.88)/1000)×<b>8848</b></sup> = 327 hPa</span></p>
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<p> </p>
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<p>(In fact pressures at Mount Everest are around 337 hPa ... good calculations!)</p>
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</div>
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<h2>Half Life</h2>
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<p>The "half life" is how long it takes for a value to halve with exponential decay.</p>
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<p>Commonly used with radioactive decay, but it has many other applications!</p>
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<div class="example">
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<h3>Example: The half-life of caffeine in your body is about 6 hours. If you had 1 cup of coffee 9 hours ago how much is left in your system?</h3>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/coffee.jpg" alt="cup of coffee" height="197" width="239"></p>
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<p>Start with the formula:</p>
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<p class="center larger">y(t) = a × e<sup>kt</sup></p>
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<p>We know:</p>
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<ul>
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<li><b>a</b> (the starting dose) = <b>1</b> cup of coffee!</li>
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<li><b>t</b> is in hours</li>
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<li>at <b>y(6)</b> we have a 50% reduction (because 6 is the half life)</li>
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</ul>
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<p>So:</p>
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<p class="center"><span class="larger">0.5 = 1 cup × e<sup><b>6</b>k</sup></span></p>
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<p>Now some algebra to solve for <b>k</b>:</p>
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<div class="tbl">
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<div class="row"><span class="left">Take the natural logarithm of both sides:</span><span class="right">ln(0.5) = ln(e<sup>6k</sup>)</span></div>
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<div class="row"><span class="left"><b>ln(e<sup>x</sup>)=x</b>, so:</span><span class="right">ln(0.5) = 6k</span></div>
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<div class="row"><span class="left">Swap sides:</span><span class="right">6k = ln(0.5)</span></div>
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<div class="row"><span class="left">Divide by 6:</span><span class="right"><b>k = ln(0.5)/6</b></span></div>
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</div>
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<p>Now we can write<b></b>:</p>
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<p class="center"><span class="larger">y(t) = 1 e<sup>(ln(0.5)/6)×t</sup></span></p>
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<p>In <b>6 hours:</b></p>
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<p class="center larger">y(<b>6</b>) = 1 e<sup>(ln(0.5)/6)×<b>6</b></sup> = 0.5</p>
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<p>Which is correct as 6 hours is the half life</p>
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<p>And in <b>9 hours:</b></p>
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<p class="center larger">y(<b>9</b>) = 1 e<sup>(ln(0.5)/6)×<b>9</b></sup> = 0.35</p>
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<p>After <b>9</b> hours the amount left in your system is <b>about 0.35</b> of the original amount. Have a nice sleep :)</p>
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</div>
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<p>Have a play with the <a href="../numbers/half-life-medicine-tool.html">Half Life of Medicine Tool</a> to get a good understanding of this.</p>
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<p> </p>
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