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Q: Integration/Summation problem ( No Answer,   2 Comments )
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
There is no answer at this time.

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:

1. Make two graphs. In the first one, plot the following points

t & v \\
0 & 12 \\
15 & 12 \\
15 & 11 \\
30 & 11 \\
45 & 10 \\
60 & 10 \\
60 & 8 \\
75 & 8 \\
75 & 7 \\
90 & 7 \\

In the second, plot the following points.

t & v \\
0 & 11 \\
15 & 11 \\
15 & 10 \\
30 & 10 \\
45 & 10 \\
45 & 8 \\
60 & 8 \\
60 & 7 \\
75 & 7 \\
75 & 0 \\
90 & 0 \\

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.
Subject: Re: Integration/Summation problem
From: sgolaev-ga on 26 Aug 2006 23:25 PDT
I checked the calculations and agree with erictai4-ga.

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