Query Assn 1

course MTH 173

Under the asingment it said not to do the randomized problems unless specifically instructed to do so, which I wasn't so I only used the Query program on the Modeling Project #1. If you would like me to do the rest of the query please let me know and I will do one of the 9 random problem sets and do the entire query. Sorry for the Confusion. Thanks, Coleman Hamilton

assignment #001001. `query 1

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

Calculus I

09-07-2007

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18:01:29

What were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?

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{ ( 95, 0 ), ( 20 , 60 ), ( 40 , 41) }

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18:11:56

According to your graph what would be the temperatures at clock times 7, 19 and 31?

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At 7 sec. the temp would be around 83 degrees.

At 19 sec. the temp would be around 61 degrees.

At 31 sec. the temp would be around 48 degrees.

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18:14:30

What three points did you use as a basis for your quadratic model (express as ordered pairs)?

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I used the points:

{ (20, 60), (40, 41), (60, 30) }

Had I used the first, last and middle points my equations would have had a better resembalance to the graph.

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18:17:28

** A good choice of points `spreads' the points out rather than using three adjacent points. For example choosing the t = 10, 20, 30 points would not be a good idea here since the resulting model will fit those points perfectly but by the time we get to t = 60 the fit will probably not be good. Using for example t = 10, 30 and 60 would spread the three points out more and the solution would be more likely to fit the data. The solution to this problem by a former student will be outlined in the remaining `answers'.

STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps)

For my quadratic model, I used the three points

(10, 75)

(20, 60)

(60, 30). **

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If I had spread my points to the extremes and the middle I would have had better results. All results using my quadratic model are within +- .5 that are within the points I used to make the model. However the results for points outside of points used for the model are off a good bit.

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18:20:59

What is the first equation you got when you substituted into the form of a quadratic?

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My quadratic model was

Y = (.01) t^2 + (-1.55) t + (87)

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18:24:40

** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**

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I mis understood and entered my Quadratic Model instead. My first equation for point (20, 60) was:

60 = A (20)^2 + B (20) + C

60 = A (400) + B (20) + C

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18:26:05

What is the second equation you got when you substituted into the form of a quadratic?

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The second equation I got was:

41 = A (40)^2 + B (40) + C

41 = A (1600) + B (40) + C

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18:29:13

What is the third equation you got when you substituted into the form of a quadratic?

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The equation I got from my third data point was:

30 = A (60)^2 + B (60) + C

30 = A (3600) + B (60) + C

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18:29:18

** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **

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18:32:36

What multiple of which equation did you first add to what multiple of which other equation to eliminate c, and what is the first equation you got when you eliminated c?

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I subtracted (60 = A (400) + B (20) + C) from

(41 = A (1600) + B (40) + C) and got:

19 = A (-1200) + B (-20)

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18:32:53

** STUDENT SOLUTION CONTINUED: First, I subtracted the second equation from the third equation in order to eliminate c.

By doing this, I obtained my first new equation

3200a + 40b = -30. **

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18:34:38

To get the second equation what multiple of which equation did you add to what multiple of which other quation, and what is the resulting equation?

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I subtracted (30 = A (3600) + B (60) + C) from

(41 = A (1600) + B (40) + C) and got:

11 = A (-2000) + B (-20)

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18:34:57

Which variable did you eliminate from these two equations, and what was its value?

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18:39:08

** STUDENT SOLUTION CONTINUED: In order to solve for a and b, I decided to eliminate b because of its smaller value. In order to do this, I multiplied the first new equation by -5

-5 ( 3200a + 40b = -30)

and multiplied the second new equation by 4

4 ( 3500a + 50b = -45)

making the values of -200 b and 200 b cancel one another out. The resulting equation is -2000 a = -310. **

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In order to solve for A and B I decided to eliminate B because of there similarity:

I subtracted (11 = A (-2000) + B (-20)) from

(19 = A (-1200) + B (-20) making the B's cancell out and got:

8 = 800 A

.01 = A

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18:43:28

What equation did you get when you substituted this value into one of the 2-variable equations, and what did you get for the other variable?

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Then I subbed in A in the equation

(19 = A (-1200) + B (-20)) and got:

19 = (.01)(-1200) + B (-20)

19 = (-12) + B (-20)

31 = B - 20

-1.55 = B

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18:43:34

** STUDENT SOLUTION CONTINUED: After eliminating b, I solved a to equal .015

a = .015

I then substituted this value into the equation

3200 (.015) + 40b = -30

and solved to find that b = -1.95. **

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18:47:50

What is the value of c obtained from substituting into one of the original equations?

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I took A and B subbed them in the equation

(41 = A (1600) + B (40) + C) and got:

41 = (.01) (1600) + (-1.55) (40) + C

41 = (16) + (-62) + C

41 = (-46) + C

87 = C

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18:48:03

** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **

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18:48:56

What is the resulting quadratic model?

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My quadratic model is:

Y = (.01) T^2 + (-1.55) T + (87)

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18:49:07

** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was

y = (.015) x^2 - (1.95)x + 93. **

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18:51:39

What did your quadratic model give you for the first, second and third clock times on your table, and what were your deviations for these clock times?

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For my first 3 times I got:

Time Temp Pred. Deviation

0 95 87 -11

10 75 72.5 -2.5

20 60 60 0

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18:54:19

** STUDENT SOLUTION CONTINUED: This model y = (.015) x^2 - (1.95)x + 93 evaluated for clock times 0, 10 and 20 gave me these numbers:

First prediction: 93

Deviation: 2

Then, since I used the next two ordered pairs to make the model, I got back

}the exact numbers with no deviation. So. the next two were

Fourth prediction: 48

Deviation: 1

Fifth prediction: 39

Deviation: 2. **

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I used the 3rd point to make my equation so the deviation was zero. The 4th point was:

Time Temp Pred. Deviation

30 49 49.5 .5

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19:01:03

What was your average deviation?

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My deviations were {-11, -2.5, 0, .5, 0, .5, 0, -1.5} I added them together and got -11 then I divided by 8 and got for my avg. Deviation:

Avg. Dev. = -1.375

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19:01:08

** STUDENT SOLUTION CONTINUED: My average deviation was .6 **

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19:05:19

Is there a pattern to your deviations?

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Yes, the points that come before (from right to left on graph) the points I used to make model are off by a negative deviation. Where as the points after are off by a positive deviation. Where as every other point in the middle of the data set I used so thier deviation is zero and the others are off by .5

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19:05:34

** STUDENT SOLUTION CONTINUED: There was no obvious pattern to my deviations.

INSTRUCTOR NOTE: Common patterns include deviations that start positive, go negative in the middle then end up positive again at the end, and deviations that do the opposite, going from negative to positive to negative. **

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19:06:02

Have you studied the steps in the modeling process as presented in Overview, the Flow Model, Summaries of the Modeling Process, and do you completely understand the process?

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Yes I understand the model

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19:06:11

** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **

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19:09:45

Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.

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I worked for a few hours working with the equations and making sure every thing was right. While I'm sure that I can do it mathematically I am still working on memorizing the actual words and the outline BUT, soon I will have them Memorized too.

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19:10:58

** STUDENT SOLUTION CONTINUED: Yes, sir, I have memorized the steps of the modeling process at this point. I also printed out an outline of the steps in order to refresh my memory often, so that I will remember them forever!!!

INSTRUCTOR COMMENT: OK, I'm convinced. **

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I did print them off and staple them in my notebook, I'm thinking about tatooing them onto the back (of my notebook) so I can better remember them, also the writing down reinforces the learning process.

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19:14:29

Query Completion of Model first problem: Completion of model from your data.Give your data in the form of depth vs. clock time ordered pairs.

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The ordered pairs that I got when I pluged the time into my model are:

{ (0, 87), (10, 72.5), (20, 60), (30, 49.5), (40, 41), (50, 34.5), (60, 30), (70, 27.5)}

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19:20:05

** STUDENT SOLUTION: Here are my data which are from the simulated data provided on the website under randomized problems.

(5.3, 63.7)

(10.6. 54.8)

(15.9, 46)

(21.2, 37.7)

(26.5, 32)

(31.8, 26.6). **

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I entered the data set that I Got when I pluged the time into the model for the Potatoe temperature.

The data for the temperature vs. time for the potatoe are are

Time Temp.

0 95

10 75

20 60

30 49

40 41

50 35

60 30

70 26

Where the data from the flow model are under randomized problems.

depth Cm Time sec

82.1 2.8

76.8 5.6

72.7 8.4

69.8 11.2

66.8 14

63.7 16.8

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19:22:59

What three points on your graph did you use as a basis for your model?

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I chose the first 3 points (a bad choice):

{(2.8, 82.1), (5.6, 76.8), (8.4, 72.7)}

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19:23:11

** STUDENT SOLUTION CONTINUED: As the basis for my graph, I used

( 5.3, 63.7)

(15.9, 46)

(26.5, 32)**

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19:24:30

Give the first of your three equations.

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My first equation is:

A (70.56) + B (2.8) + C = 82.1

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19:25:05

** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **

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That was for point (2.8, 82.1)

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19:26:53

Give the second of your three equations.

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My second equation is for point (5.6, 76.8)

A (31.36) + B (5.6) + C = 76.8

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19:27:04

** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **

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19:28:41

Give the third of your three equations.

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My third equation was for point (8.4, 72.7)

A (70.56) + B (8.4) + C = 72.7

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19:28:45

** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **

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19:30:29

Give the first of the equations you got when you eliminated c.

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when I eliminated C I subtracted equation 2 from equation 3 and got:

A (39.2) + B (2.8) = -4.1

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19:30:34

** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **

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19:32:15

Give the second of the equations you got when you eliminated c.

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I used equation 1 as a second equation because the second t, what you multiply B by was the same as the synthetic equation.

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19:32:58

** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **

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Doing it this different way may have thrown off my final answer.

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19:34:47

Explain how you solved for one of the variables.

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I subtracted the first equation from the synthetic equation that was produced by subtracting the third and second equation and got:

A (31.36) = -86.2

A = -2.74872449

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19:36:15

** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated b by multiplying the first equation by 21.2, which was the b value in the second equation. Then, I multiplied the seond equation by -10.6, which was the b value of the first equation, only I made it negative so they would cancel out. **

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Again by not making another synthetic equation with out C it may have thrown off my final answer.

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19:37:09

What values did you get for a and b?

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for A I got: -2.74872449

and B I got: 37.01785714

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19:37:18

** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **

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19:37:33

What did you then get for c?

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c = -44.3

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19:37:38

** STUDENT SOLUTION CONTINUED: c = 73.4 **

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19:39:16

What is your function model?

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My function model was:

Y = (-2.74872449) T^2 + (37.01785714) T + (-44.3)

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19:39:21

** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **

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19:44:31

What is your depth prediction for the given clock time (give clock time also)?

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My depth prediction for 14 sec. was 64.80000008 and the actual depth was 66.8

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19:44:53

** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**

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19:46:03

What clock time corresponds to the given depth (give depth also)?

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14s. corresponds to 66.8cm.

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19:58:15

** INSTRUCTOR COMMENT: The exercise should have specified a depth.

The specifics will depend on your model and the requested depth. For your model y = (.0165)x^2 + (-2)x + 73.4, if we wanted to find the clock time associated with depth 68 we would note that depth is y, so we would let y be 68 and solve the resulting equation:

68 = .01t^2 - 1.6t + 126

using the quadratic formula. There are two solutions, x = 55.5 and x = 104.5, approximately. **

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I didn't have a depth on my screen so I will use 68 like is in the screen review box.

Plugging into the equation I get.

68 = (-2.74872449) T^2 + (37.01785714) T + (-44.3)

112.3 = (-2.74872449) T^2 + (37.01785714) T

3.033 = (-2.74872449) T^2 + T

-1.104 = 2T^2

-0.5518 = t^2

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20:08:29

Completion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.

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I don't understand the question I don't have a grade average model. I will look forawrd and see if I understand then give an answer.

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20:09:58

** STUDENT SOLUTION: Grade vs. percent of assignments reviewed

(0, 1)

(10, 1.790569)

(20, 2.118034)

(30, 2.369306)

(40, 2.581139)

(50, 2.767767)

(60, 2.936492)

(70, 3.09165)

(80, 3.236068)

(90, 3.371708)

(100, 3.5). **

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This may have been on the Cd's that I don't have and they are not in Modeling Project #1

Part of the Modeling Project 1 was 'Completion of the Initial Flow Model', which included excercises at the end for the two data sets represented in these given solutions.

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20:10:22

What three points on your graph did you use as a basis for your model?

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See Past answere

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20:10:27

** STUDENT SOLUTION CONTINUED:

(20, 2.118034)

(50, 2.767767)

(100, 3.5)**

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