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 Subject: Integration/Summation problem Category: Science > Math Asked by: willymi-ga List Price: $30.00 Posted: 13 Aug 2006 16:50 PDT Expires: 12 Sep 2006 16:50 PDT Question ID: 755653  I know that this is a summation/integral problem and I have no problem finding the upper and lower estimates usually, but im just not sure how to extend them beyond what the graph has given me. Also, im not sure how to make the rectangles under the curve small enough in order to get the estimated distance to within 0.1 given the chart info: (Marathon Tarining) t= 0, 15, 30, 45, 60, 75, 90 S= 12, 11, 10, 10, 8, 7, 0 t=time (seconds) s=speed (mph) 1. Assuming that the speed is never increasing, give upper and lower estimates for the distance ran during the first half hour. 2. Give estimates for the distance ran for an hour and a half. 3. How often would speed measurments have to be taken in order to find the upper and lower estimates to within 0.1 of the actual distance ran? Clarification of Question by willymi-ga on 14 Aug 2006 09:02 PDT I meant to say that t=time (minutes)........not seconds  Answer  There is no answer at this time.  Comments  Subject: Re: Integration/Summation problem From: erictai4-ga on 15 Aug 2006 20:57 PDT  Note, for anything enclosed in dollar signs ($), use the following URL to render: http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php 1. Make two graphs. In the first one, plot the following points $\begin{tabular}{c|c} t & v \\ \hline 0 & 12 \\ 15 & 12 \\ 15 & 11 \\ 30 & 11 \\ 45 & 10 \\ 60 & 10 \\ 60 & 8 \\ 75 & 8 \\ 75 & 7 \\ 90 & 7 \\ \end{tabular}$ In the second, plot the following points. $\begin{tabular}{c|c} t & v \\ \hline 0 & 11 \\ 15 & 11 \\ 15 & 10 \\ 30 & 10 \\ 45 & 10 \\ 45 & 8 \\ 60 & 8 \\ 60 & 7 \\ 75 & 7 \\ 75 & 0 \\ 90 & 0 \\ \end{tabular}$ Connect the dots in the obvious way and you should get two stepped graphs. $\frac{area\ of\ each\ graph}{60\ minutes}$ is the distance travelled. The first graph represents the upper bound while the latter is the lower bound. You will find that the distance is $\frac{1}{4} \sum_{i=1}^6 v_i$ for the upper bound and $\frac{1}{4} \sum_{i=2}^7 v_i$ for the lower bound, where $v_i=\{12,11,10,10,8,7,0\}$. 2. If you did 1 correctly, you'll find the upper bound is 14.5 and the lower bound is 11.5. A good estimate one might say is the average, which assumes all the accelerations between data points were constant. The average is merely $\frac{11.5+14.5}{2}=13$. 3. This is a really ambiguous question. If one does not make the same assumption denoted in 1, then it is impossible. If one does make that assumption, but does not insist that the given datapoints be part of the speed log, then it is still impossible. So I will make the assumption that we make the same assumption about never increasing velocity and that the given datapoints are still valid. In this case, we need to make a speed vs. time graph with a total drop of 12 mph. Thus, we must have $12\left(\frac{x}{60}\right)=0.1$, where $x$ is the number of minutes between logging of datapoints. In our case, $x=0.5$. Since I'm not actually getting paid for this answer, I won't give any more explanation.
 I checked the calculations and agree with erictai4-ga.