Initial Flow Model

course Mth 163

I'm still planning to do the calculations on the remaining 2 data sets but wanted to make sure I was not making any major mistakes before starting them. I typed in my solutions beneath the query file.

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үY`{Ե Precalculus I 06-01-2006

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06:35:42 Query Introduction to General Themes; Examples (no summary needed) What were some of the things in this introduction that you found interesting or surprising?

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RESPONSE --> I really didn't know exactly what precalculus covered so most of what I read was informational and new for me.

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06:37:44 Query Introductory Flow Experiment (no summary needed) Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate? Support your conclusion.

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RESPONSE --> The depth changes at a slower and slower rate. This was evident by the increasing rate of time in seconds between points.

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06:38:03 ** If you time the water at equal time intervals you should find that the depth changes by less and less with each new interval. If you timed the depth at equal intervals of depth you should find that each interval takes longer than the one before it. } Either way you would conclude that water flows from the hole at a decreasing rate. The reason is that as the water depth decreases the pressure forcing the water out of the hole decreases. **

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RESPONSE --> OK

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06:40:59 What does the graph of depth vs. clock time look like? Is it increasing or decreasing? Does the rate of increase or decrease speed up or slow down? Does your graph intercept the y axis? Does it intercept the x axis? How would you describe its overall shape?

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RESPONSE --> It is decreasing and the rate gradually decreases. The graph, if extended with enough data points, touches the x axis and y axis. It is in the shape of a parabola.

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06:41:43 ** The graph will start on the positive y axis and will decrease at a decreasing rate. The shape of the graph is the left-hand side of a parabola that opens upward. The right-hand half of the parabola does not correspond to the flow. The lef-hand half of the parabola, which corresponds to the flow, gets less and less steep with increasing clock time, matching the fact that that the rate of decrease is slowing. At the instant the containers empties, the water will be at the level of the hole. If depth is measured relative to the hole, then at the instant depth will reach zero. The corresponding graph point will lie on the t axis and will correspond to the vertex of the parabola. **

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RESPONSE --> OK

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lSLD~ۚr assignment #011 үY`{Ե Precalculus I 06-01-2006 Qwzh assignment #001 үY`{Ե Precalculus I 06-01-2006 The following pertains to the initial depth vs. time flow model exercise. I used version 5 of the randomized problems (did experiment but found it awkward and don't know how reliable it would have been). The clock times and corresponding depths for this version were (2.8, 82.1),(5.6, 76.8), (8.4, 72.7), (11.2, 69.8), (14, 66.8), and (16.8, 63.7). I used the following three points (5.6, 76.8), (11.2, 69.8), and (16.8, 63.7) to formulate the initial three equations: 31.36a+5.6b+c=76.8 125.44a+11.2b+c=69.8 282.24a+16.8b+c=63.7 Using the elimination method to eliminate the variable c, I subtracted equation 1 from 3 and 2 from 3 to give the following 2 equations: 250.88a+11.2b=-13.1 156.8a+5.6b=-6.1 I then eliminated the variable b by multiplying the second equation by -2 and adding the 2 equations to get: -62.81a=-.09, and a (.00143). Solving for b, I inserted a (.00143) into the equation 250.88(.00143)+11.2b=-13.1 to get b=1.201. Solving for c, I plugged a and b into the first of the original equations: 31.36(.00143)+5.6(1.201)+c=76.8 to get 70.03 (c). I used the depth vs. clock time model [y(depth)=at^2+bt+c] to predict depth at clock time of 46 seconds: y=(.00143)46^2+1.201(46)+70.03 and arrived at 128.302 which didn't make sense looking at the other numbers. I predicted a clock time of 47 seconds at a depth of 14 cm using the quadratic formula as follows: t=+/-b(1.201)sqrt[(1.201)^2-4(.00143)(70.03)/.02(.00143) with the possible solutions of 753.67 and 47 seconds. So the predicted clock time seemed likely correct but the depth calculation did not.

I can't spot the discrepancy--your steps look good. Your error seems to occur after you got the solution for a. I don't think you solved 250.88(.00143)+11.2b=-13.1 correctly, and I believe b is about -1.49.

All your procedures appear not only correct, but very well thought out and very well understood. Even with a minor arithmetic error, this is very good-looking work and bodes well for your success in the course.