course Mth 173
I can't separate your work in this document from the output of the program. If you designate your work with &&&& before and after and resubmit, I can critique it.
<-<-<-<- Q/A: ** 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. ** YOUR ANSWER ->->->-> I have made notes to memorize each step. <-<-<-<- Q/A: Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me. <-<-<-<- Q/A: ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. ** YOUR ANSWER ->->->-> I understand each step in the process. <-<-<-<- Q/A: 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? <-<-<-<- Q/A: ** 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. ** YOUR ANSWER ->->->-> There is no pattern to my deviations. <-<-<-<- Q/A: Is there a pattern to your deviations? <-<-<-<- Q/A: ** STUDENT SOLUTION CONTINUED: My average deviation was .6 ** YOUR ANSWER ->->->-> Average deviation = 7.3. <-<-<-<- Q/A: What was your average deviation? <-<-<-<- Q/A: ** 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. ** YOUR ANSWER ->->->-> By using the clock times of 0, 10, and 20, the equation y = (.003) x^2 - (2.58)x + 81.5, I came up with these predictions.
Estimate:69.6
Deviation:11.9
Estimate:55
Deviation:1
Estimate:41.4
Deviation:10.3 <-<-<-<- Q/A: 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? <-<-<-<- Q/A: ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was
y = (.015) x^2 - (1.95)x + 93. ** YOUR ANSWER ->->->-> y = (.003)x^2 + (-2.58)x + c = 81.5 <-<-<-<- Q/A: What is the resulting quadratic model? <-<-<-<- Q/A: ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 ** YOUR ANSWER ->->->-> By substituting a and b, we are able to solve for c, which equals 81.5. <-<-<-<- Q/A: What is the value of c obtained from substituting into one of the original equations? <-<-<-<- Q/A: ** 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. ** YOUR ANSWER ->->->-> I solved a to equal .003. Substituting this into my equation resulted in b= -2.58. <-<-<-<- Q/A: 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? <-<-<-<- Q/A: ** 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. ** YOUR ANSWER ->->->-> I eliminated the b variable. I multiplied the 1st new equation by -3 to get,
3600a - 60b = 84.
The 2nd equation was multiplied by 2 to get,
3000a + 60b = -86.
The final equation is 600a = 2. <-<-<-<- Q/A: Which variable did you eliminate from these two equations, and what was its value? <-<-<-<- Q/A: ** STUDENT SOLUTION CONTINUED: This time, I subtracted the first equation from the third equation in order to again eliminate c.
I obtained my second new equation:
3500a + 50b = -45** YOUR ANSWER ->->->-> To get th 2nd equation, I subtracted the first equation from the the third to get 1500a +30b = -43. <-<-<-<- Q/A: 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? <-<-<-<- Q/A: ** 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. ** YOUR ANSWER ->->->-> I subtracted the ist equation from the second. My 1st new equation was: 1200a + 20b = -28. <-<-<-<- Q/A: 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? <-<-<-<- Q/A: ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. ** YOUR ANSWER ->->->-> 1600a + 40b + c = 13 <-<-<-<- Q/A: What is the third equation you got when you substituted into the form of a quadratic? <-<-<-<- Q/A: ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 ** YOUR ANSWER ->->->-> 400a + 20b + c = 41
<-<-<-<- Q/A: What is the second equation you got when you substituted into the form of a quadratic? <-<-<-<- Q/A: ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.** YOUR ANSWER ->->->-> 100a + 10b + c = 56
<-<-<-<- Q/A: What is the first equation you got when you substituted into the form of a quadratic? <-<-<-<- Q/A: ** 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). ** YOUR ANSWER ->->->-> I used these three ordered pairs as a basis for my quadratic model:
(10,56)
(20,41)
(40,13) <-<-<-<- Q/A: What three points did you use as a basis for your quadratic model (express as ordered pairs)? <-<-<-<- Q/A: ** Continue to the next question ** YOUR ANSWER ->->->-> 7sec = 59.7
19sec = 42.8
31sec = 25.8 <-<-<-<- Q/A: According to your graph what would be the temperatures at clock times 7, 19 and 31? <-<-<-<- Q/A: ** Continue to the next question ** YOUR ANSWER ->->->-> (5.3, 63.7), (15.9, 46), (31.8, 26.6) <-<-<-<- Q/A: What were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?"
see my note at the beginning