course
Melissa McelhenyWhen submitting your work electronically, show the details of your work and give a good verbal description of your graphs.
One very important goal of the course is to learn to communicate mathematical thinking and logical reasoning. If you can effectively communicate mathematics, you will be able to effectively communicate a wide range of important ideas, which is extremely valuable in your further education and in your career.
When writing out solutions, self-document. That is, write your solution so it can be read without reference by the reader to the problem statement. Use specific and descriptive statements like the following:
Using the depth vs. clock time data points (0, 13), (3, 12), (10,10), (25,8), (35, 6), (52, 3), (81, 1), we obtain a model as follows . . .
Using the depth vs. clock time data points (3, 12), (25, 8) and (52,3) we obtain the system of equations . . .
From the parameters a = -1.3, b = 12 and c = 15 we obtain the function . . .
Comparing the predicted depths at clock times t = 0, 3, 10, 25, 35, 52, 81 with the observed depths we see that . . .
Here are some data for the temperature of a hot potato vs. time:
Time (minutes) Temperature (Celsius)
0 108
17 108 Sum of Temperature (Celsius)
34 87.90224 Time (minutes) Total
51 79.75957 0 108
68 72.67098 17 108
85 66.5 34 87.90224
102 61.12785 51 79.75957
119 56.45112 68 72.67098
85 66.5
Graph these data below, using an appropriate scale: 102 61.12785
119 56.45112
Pick three representative points and circle them. (17 and 108), (85 and 66.5), (119 and 56.45112)
Write the equations that result from the assumption that the appropriate mathematical model is a quadratic function y = a t^2 + b t + c.
108 = a(17)^2 + b(17) + c
66.5= a(85)^2 + b(85) + c
56.45112=a(119)^2 + b(119) + c
Eliminate c from your equations to obtain two equations in a and b.
108= 289a + 17 b + c 108= 289a + 17 b + c 66.5 = 7225a +85b +c
66.5 = 7225a +85b +c -66.5 = -7225a -85b -c -56.45112 = -14161a - 119b- c
56.45112 = 14161a + 119b + c 41.5 = -6936a - 68b 10.04888 = -6936a -34b
Solve for a and b.
41.5 = -6936a - 68b 10.04888 = -6936a -34b 41.5 = -6936a - 68b
1411 = -235824a -2312b 41.5 = -6936(.003086) - 68b
-683.32384 = 471648a - 2312b 41.5 = -21.404496 - 68b
727.68616 = 235824a 62.904496 = -68b
a = .003086 b= -.92507
108 = 289(.003086) + 17 (-.92507) + c 108 = .891854 - 15.72619 + c
122.834336 = c
Write the resulting model for temperature vs. time.
y = .003086t^2 -.92507t + 122.834336
Make a table for this function:
Time (minutes) Model Function's Prediction of Temperature
0 122.8
17 107.9 y = .003086(17)^2 - .92507(17) +122.8
34 94.9 y = .003086(34)^2 - .92507(34) +122.8
51 83.6 y = .003086(51)^2 - .92507(51) +122.8
68 74.2 y = .003086(68)^2 - .92507(68) +122.8
85 66.5 y = .003086(85)^2 - .92507(85) +122.8
102 60.5 y = .003086(102)^2 - .92507(102) +122.8
119 56.4 y = .003086(119)^2 - .92507(119) +122.8
Sketch a smooth curve representing this function on your graph.
Expand your table to include the original temperatures and the deviations of the model function for each time:
Time (minutes) Temperature (Celsius) Prediction of Model Deviation of Observed Temperature from Model
0 108 122.8 -14.8
17 97.2557 107.9 -10.6443
34 87.90224 94.9 -6.99776
51 79.75957 83.6 -3.84043
68 72.67098 74.2 -1.52902
85 66.5 66.5 0
102 61.12785 60.5 0.62785
119 56.45112 56.4 0.05112
Find the average of the deviations.
-37.13254 divided by 8 = -4.6415675
1. If you have not already done so, obtain your own set of flow depth vs. time data as instructed in the Flow Experiment (either perform the experiment, as recommended, or E-mail the instructor for a set of data).
Complete the modeling process for your own flow depth vs. time data.
Use your model to predict depth when clock time is 46 seconds, and the clock time when the water depth first reaches 14 centimeters.
Comment on whether the model fits the data well or not.
2. Follow the complete modeling procedure for the two data sets below, using a quadratic model for each. Note that your results might not be as good as with the flow model. It is even possible that at least one of these data sets cannot be fit by a quadratic model.
Data Set 1
In a study of precalculus students, average grades were compared with the percent of classes in which the students took and reviewed class notes. The results were as follows:
Percent of Assignments Reviewed Grade Average Model
0 1.014738 1.03518 y= -.000166t^2 + .03946t + 1.03518
10 1.408518 1.41318
20 1.756831 1.75798 3 = -.000166t^2 + .03946t + 1.03518
30 2.064929 2.06958
40 2.337454 2.34798
50 2.578513 2.59318
60 2.79174 2.80518
70 2.980347 2.98398
80 3.147178 3.12958
90 3.294747 3.24198
100 3.425278 3.32118
Determine from your model the percent of classes reviewed to achieve grades of 3.0 and 4.0.
3 = -.000166t^2 + .03946t + 1.03518 4 = -.000166t^2 + .03946t + 1.03518
0 = -.000166t^2 + .03946t -1.96482 0 = -.000166t^2 + .03946t -2.96482
x=71 negative sign under the square root and not possible.
Determine also the projected grade for someone who reviews notes for 80% of the classes.
y= -.000166t^2 + .03946t + 1.03518
y= -.000166(80)^2 + .03946(80) + 1.03518
y=3.12958
Comment on how well the model fits the data. The model may fit or it may not. The model fits the data, not too much deviation
Comment on whether or not the actual curve would look like the one you obtained, for a real class of real students.
there is a good possibilty.
Data Set 2
The following data represent the illumination of a comet by a certain star, reasonably similar to our Sun, at various distances from the star:
Distance from Star (AU) Illumination of Comet (W/m^2)
1 1470 y=at^2 + bt + c
2 367.5
3 163.3333 y = a(1)^2 + b(1) + c y = a(5)^2 + b(5) + c y = a(10)^2 + b(10) + c
4 91.875 1470 = a + b + c 58.8 = 25a + 5b + c 14.7 = 100a + 10b + c
5 58.8
6 40.83333 -1411.2 =24a+ 4b
7 30 -44.1=75a+5b
8 22.96875 38.22 = a
9 18.14815 582.12 = b
10 14.7 849.66 = c
Obtain a model. y= 38.22t^2 - 582.12t + 2013.90
Determine from your model what illumination would be expected at 1.6 AU from the star.
y= 38.22t^2 - 582.12t + 2013.90
y= 38.22(1.6)^2 - 582.12(1.6) + 2013.90
y=1180.35
At what range of distances from the star would the illumination be comfortable for reading, if reading comfort occurs in the range from 25 to 100 Watts per square meter?
y= 38.22t^2 - 582.12t + 2013.90 y= 38.22t^2 - 582.12t + 2013.90
25= 38.22t^2 - 582.12t + 2013.90 100= 38.22t^2 - 582.12t + 2013.90
0=38.22t^2 - 582.12 + 1988.90 0=38.22t^2 - 582.12t + 1913.90
y = 10.0558 y = 10.429
Analyze how well your model fits the data and give your conclusion. The model might fit, and it might not. You determine whether it does or doesn't.
The model does not fit . The equations do not work with every number and as the range gets larger the AU should get
smaller, not larger.
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Your work here looks very good. Let me know if you have questions.