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<h1>🎯 Adventure 13 — Activity 2 (Solution)</h1>
<div class="subtitle">DiVA reasoning with D(t) given</div>
<div class="chart-top"><img alt="ATP DiVA chart with values (solution)" src="images/adventure_13_chart_1.png"/></div>
<div class="one-col">
<div class="callout">
<b>Solution:</b> Use derivatives to locate min/max and the inflection point, then sketch <b>D(t)</b>.
</div>
<h2>1) Derivatives</h2>
<p><b>V(t)=D′(t)=</b> −2t² + 80t − 600</p>
<p><b>A(t)=V′(t)=D″(t)=</b> −4t + 80</p>
<h2>2) Critical points for D(t)</h2>
<ul>
<li>Solve V(t)=0 → <b>t=10</b> and <b>t=30</b></li>
<li>D″(10)=40>0 → <b>minimum at t=10</b></li>
<li>D″(30)=−40<0 → <b>maximum at t=30</b></li>
<li>Inflection point: A(t)=0 → <b>t=20</b></li>
</ul>
<h2>3) Helpful values</h2>
<table class="gridtable">
<tr><th>t</th><th>D(t)</th><th>V(t)</th><th>A(t)</th></tr>
<tr><td>0</td><td>26000</td><td>-600</td><td>80</td></tr>
<tr><td>10</td><td>23333.33</td><td>0</td><td>40</td></tr>
<tr><td>20</td><td>24666.67</td><td>200</td><td>0</td></tr>
<tr><td>30</td><td>26000</td><td>0</td><td>-40</td></tr>
<tr><td>40</td><td>23333.33</td><td>-600</td><td>-80</td></tr>
</table>
<h2>4) Reasoning statements</h2>
<ul>
<li>When <b>A(t)>0</b>, V(t) is <b>increasing</b>.</li>
<li>When <b>V(t)>0</b>, D(t) is <b>increasing</b>.</li>
<li>V(t) reaches its <b>maximum at t=20</b> (because A(t)=0 and changes sign).</li>
</ul>
</div>
<h2>Given (simplified units)</h2>
<div class="callout">
<div><b>D(t)</b> = <span class="mono">−(2/3)t³ + 40t² − 600t + 26,000</span></div>
<div><b>V(t)=D′</b> = <span class="mono">−2t² + 80t − 600</span></div>
<div><b>A(t)=V′=D″</b> = <span class="mono">−4t + 80</span></div>
</div>
<h2>Table 1. Sign Table for Drawing D (ATP Availability)</h2>
<table class="tbl">
<tr>
<th>Interval (seconds)</th>
<th>0–10</th><th>10</th><th>10–20</th><th>20</th><th>20–30</th><th>30</th><th>30–40</th>
</tr>
<tr>
<th>D (ATP availability)</th>
<td>−</td><td><b>min</b></td><td>+</td><td><b>Inflection</b></td><td>+</td><td><b>max</b></td><td>−</td>
</tr>
<tr>
<th>V = D′</th>
<td>−</td><td>0</td><td>+</td><td>+</td><td>+</td><td>0</td><td>−</td>
</tr>
<tr>
<th>A = V′ = D″</th>
<td>+</td><td>+</td><td>+</td><td>0</td><td>−</td><td>−</td><td>−</td>
</tr>
</table>
<div class="callout">
▭(V(t)=−2t²+80t−600) ▭(A(t)=−4t+80)
</div>
<h2>Table 2. Values for V(t) and A(t)</h2>
<table class="tbl">
<tr><th>t</th><th>0</th><th>5</th><th>10</th><th>15</th><th>20</th><th>25</th><th>30</th><th>35</th><th>40</th></tr>
<tr><th>V(t)</th>
<td>-600</td><td>250</td><td>0</td><td>150</td><td>200</td><td>150</td><td>0</td><td>250</td><td>-600</td>
</tr>
<tr><th>A(t)</th>
<td>80</td><td>60</td><td>40</td><td>20</td><td>0</td><td>-20</td><td>-40</td><td>-60</td><td>-80</td>
</tr>
</table>
<div class="callout">
Use the values in Table 2 to put signs in Table 1.
<br/><br/>
<b>First derivative</b> decides if there is an extremum (max/min). <b>Second derivative</b> decides which kind.
<br/><br/>
At <b>t=10</b> we have a <b>minimum</b> since V(10)=0 and A(10)=40>0.<br/>
At <b>t=30</b> we have a <b>maximum</b> since V(30)=0 and A(30)=−40<0.<br/>
At <b>t=20</b> we have an <b>inflection point</b> since A(20)=0.
</div>
<h2>Table 3. Values for D(t)</h2>
<table class="tbl">
<tr><th>t</th><th>0</th><th>10</th><th>20</th><th>30</th><th>40</th></tr>
<tr><th>D(t)</th>
<td>26,000</td><td>23,333</td><td>24,666</td><td>26,000</td><td>23,333</td>
</tr>
</table>
<div class="callout">
▭(D(t)=−2/3 t³ + 40t² − 600t + 26,000)
<br/>
D(0)=26,000 D(10)=23,333 D(20)=24,666 D(30)=26,000 D(40)=23,333
</div>
<h2>Sketch notes</h2>
<ul>
<li>On the top graph of the DiVA: mark where <b>V(t)=0</b> (min/max).</li>
<li>Mark where <b>A(t)=0</b> (inflection).</li>
<li>Use Table 1 and Table 3 to draw a smooth, reasoned sketch.</li>
</ul>
<h2>The Parting Shot</h2>
<div class="callout">
You can sketch the function’s shape using only the critical points (zeros of V), the inflection point (zero of A),
and the signs. This is powerful because calculus lets you understand the whole curve without plotting lots of points.
</div>
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