Estimating Oil Production with Simpson's Rule

[shamieh]

The graph below gives the rate of production of oil in thousands of barrels per day by a large refinery. Use Simpson's rule to estimate the total amount of oil produced through the end of the fourth day.so I got

\(\displaystyle \frac{1}{3} [ 10 + 4(12) + 2(14) + 4(9) + 10.5)] \approx 44.1666666667\)

 

[MarkFL]

There appear to be 19 intervals, so I though I would extrapolate and say:

\(\displaystyle P(5.0)=P(4.5)\) so that we have 20 intervals, and then use 10 parabolic arcs rather than just 2 (unless you were instructed to do otherwise).

And be careful reading the graph...notice $P(4.75)\approx12.5$. :D

 

[shamieh]

There appear to be 19 intervals, so I though I would extrapolate and say:

\(\displaystyle P(5.0)=P(4.5)\) so that we have 20 intervals, and then use 10 parabolic arcs rather than just 2 (unless you were instructed to do otherwise).

And be careful reading the graph...notice $P(4.75)\approx12.5$. :D

19 intervals? But isn't the function something like \(\displaystyle \int^4_0 f(x)\) and it says to just go to the 4th day. So wouldn't i have n = 4 since i have 5 subscripts?

Then \(\displaystyle \Delta x =(4/4) / 3 = \frac{1}{3}\)
then \(\displaystyle \frac{1}{3}[f(0) + 4f(1) + 2f(2) + 4f(3) + f(4)]\) ?

Are you saying i need to split up into 19 pieces?

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Also what does P(5.0) =P(4.5) mean?

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I also don't understand why we are going past 4? It just says go to the end of the 4th day? Now we would be in the end of the 5th day from what you're saying wouldn't we?

Because from 0 to 1 that's the end of the first day isn't it? There can't be a 0th day? so then from 1 to 2 that's the end of the 2nd day etc etc? Or am I way wrong lol

 

[MarkFL]

It looks like each day is split up into 4 sub-intervals, however, I was reading the graph incorrectly. There is no extrapolation required and your ending value ($P(4)$) is correct. I was using $P(t)$ to represent the function being plotted (P for production). So, if I was working this problem, I would want 16 sub-intervals ($n$=16) or 8 parabolic arcs.

 

[shamieh]

So I can't do this right? Because wouldn't this be way easier then splitting it into 16 pieces? Plus n is even so why wouldn't this approach work?

\(\displaystyle \Delta x = b-a/n\)
\(\displaystyle \Delta x = 4 - 0/ 4 \)
\(\displaystyle \Delta x = 1\)

Cut into n pieces, so cut into 1 piece. (0 to 1, 1 to 2, etc up to 4.
Applying Simpsons

\(\displaystyle \frac{\Delta x}{3} = \frac{1}{3}[f(0) + 4f(1) + 2f(2) + 4f(3) + f(4)] \)

 

[shamieh]

So should I get:

\(\displaystyle \Delta x = \frac{4}{16} = \frac{1}{4}\)

then \(\displaystyle \frac{\Delta x}{3} = \frac{1}{12}\)

\(\displaystyle \frac{1}{12}[10 + 4(8.75) + 2(8.75) + 4(9.50) + 2(12) + 4(14.5) + 2(15.75) + 4(15.75) + 2(14) + 4(12.75) + 2(11) + 4(10) + 2(9) + 4(8.50) + 2(8.50) + 4(9.50) + 10.75]\)so
\(\displaystyle 1/12 * 535.75 \approx 44.6458333333\)

Is this correct?

 

[shamieh]

What in the world? If that is correct then wouldn't it be pointless to split them up into \(\displaystyle 16\) sub intervals when I just split them in \(\displaystyle 4\) pieces before and got the same answer essentially? I mean it only differed by \(\displaystyle 0.4791666666\).. Is that a big deal?

 

[MarkFL]

I guess it boils down to your personal preference. My own would be to use the better approximation. :D

 

[shamieh]

I guess it boils down to your personal preference. My own would be to use the better approximation. :D

So would you say that it would be wrong to approximate it my way? As in, incorrect. Or would you basically be doing the same thing? I mean because I guess an approximation is just that exactly right? An approximation?

 

[MarkFL]

No, what you have done is not wrong, it is just not as accurate. It just depends on whether you want to make the extra computations. To me the graph seems to indicate using 4 sub-intervals per day.