#$&*
course Mth 163
6/9 around 9:00 am.
The Modeling ProcessExercises:
Here are some data for the temperature of a hot potato vs. time:
Time (minutes) Temperature (Celsius)
0 95
10 75
20 60
30 49
40 41
50 35
60 30
70 26
Graph these data below, using an appropriate scale:
I did this on paper.
Pick three representative points and circle them.
I did this on paper. The three points I selected are (30, 49), (50, 35), and (70, 26).
Write the equations that result from the assumption that the appropriate mathematical model is a quadratic function y = a t^2 + b t + c.
900a+30b+c=49
2500a+50b+c=35
4900a+70b+c=26
Eliminate c from your equations to obtain two equations in a and b.
(4900a+70b+c=26) - (900a+30b+c=49) = 4000a+40b=-23
(4900a+70b+c=26) - (2500a+50b+c=35) = 2400a+20b=-9
Solve for a and b.
(2400a+20b=-9) *(-2) = (-4800a-40b=18)
(4000a+40b=-23) + (-4800a-40b=18) = -800a = -5
a= 0.00625
4000(0.00625)+40b=-23
b= -1.2
900(0.00625)+30(-1.2)+c=49
c= 79.375
Write the resulting model for temperature vs. time.
y= 0.00625t^2-1.2t+79.378
Make a table for this function:
Time (minutes) Model Function's Prediction of Temperature
0 79.375
10 68
20 57.875
30 49
40 41.375
50 35
60 29.875
70 26
Sketch a smooth curve representing this function on your graph.
I did this on paper.
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 95 79.375 -15.625
10 75 68 -7
20 60 57.875 -2.125
30 49 49 0
40 41 41.375 +0.375
50 35 35 0
60 30 29.875 -0.125
70 26 26 0
Find the average of the deviations.
The average of the deviations is -3.0625.
Comment on how well the function model fits the data. (Note: the model might or might not do a good job of fitting the data. Some types of data can be fit very well by quadratic functions, while some cannot).
Despite some deviations I think the model fits the data relatively well because it got 3 values exactly correct and 2 more values really close to the original.
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#$&*
course Mth 163
6/9 around 9:00 am.
The Modeling ProcessExercises:
Here are some data for the temperature of a hot potato vs. time:
Time (minutes) Temperature (Celsius)
0 95
10 75
20 60
30 49
40 41
50 35
60 30
70 26
Graph these data below, using an appropriate scale:
I did this on paper.
Pick three representative points and circle them.
I did this on paper. The three points I selected are (30, 49), (50, 35), and (70, 26).
Write the equations that result from the assumption that the appropriate mathematical model is a quadratic function y = a t^2 + b t + c.
900a+30b+c=49
2500a+50b+c=35
4900a+70b+c=26
Eliminate c from your equations to obtain two equations in a and b.
(4900a+70b+c=26) - (900a+30b+c=49) = 4000a+40b=-23
(4900a+70b+c=26) - (2500a+50b+c=35) = 2400a+20b=-9
Solve for a and b.
(2400a+20b=-9) *(-2) = (-4800a-40b=18)
(4000a+40b=-23) + (-4800a-40b=18) = -800a = -5
a= 0.00625
4000(0.00625)+40b=-23
b= -1.2
900(0.00625)+30(-1.2)+c=49
c= 79.375
Write the resulting model for temperature vs. time.
y= 0.00625t^2-1.2t+79.378
Make a table for this function:
Time (minutes) Model Function's Prediction of Temperature
0 79.375
10 68
20 57.875
30 49
40 41.375
50 35
60 29.875
70 26
Sketch a smooth curve representing this function on your graph.
I did this on paper.
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 95 79.375 -15.625
10 75 68 -7
20 60 57.875 -2.125
30 49 49 0
40 41 41.375 +0.375
50 35 35 0
60 30 29.875 -0.125
70 26 26 0
Find the average of the deviations.
The average of the deviations is -3.0625.
Comment on how well the function model fits the data. (Note: the model might or might not do a good job of fitting the data. Some types of data can be fit very well by quadratic functions, while some cannot).
Despite some deviations I think the model fits the data relatively well because it got 3 values exactly correct and 2 more values really close to the original.
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
#$&*
course Mth 163
6/9 around 9:00 am.
The Modeling ProcessExercises:
Here are some data for the temperature of a hot potato vs. time:
Time (minutes) Temperature (Celsius)
0 95
10 75
20 60
30 49
40 41
50 35
60 30
70 26
Graph these data below, using an appropriate scale:
I did this on paper.
Pick three representative points and circle them.
I did this on paper. The three points I selected are (30, 49), (50, 35), and (70, 26).
Write the equations that result from the assumption that the appropriate mathematical model is a quadratic function y = a t^2 + b t + c.
900a+30b+c=49
2500a+50b+c=35
4900a+70b+c=26
Eliminate c from your equations to obtain two equations in a and b.
(4900a+70b+c=26) - (900a+30b+c=49) = 4000a+40b=-23
(4900a+70b+c=26) - (2500a+50b+c=35) = 2400a+20b=-9
Solve for a and b.
(2400a+20b=-9) *(-2) = (-4800a-40b=18)
(4000a+40b=-23) + (-4800a-40b=18) = -800a = -5
a= 0.00625
4000(0.00625)+40b=-23
b= -1.2
900(0.00625)+30(-1.2)+c=49
c= 79.375
Write the resulting model for temperature vs. time.
y= 0.00625t^2-1.2t+79.378
Make a table for this function:
Time (minutes) Model Function's Prediction of Temperature
0 79.375
10 68
20 57.875
30 49
40 41.375
50 35
60 29.875
70 26
Sketch a smooth curve representing this function on your graph.
I did this on paper.
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 95 79.375 -15.625
10 75 68 -7
20 60 57.875 -2.125
30 49 49 0
40 41 41.375 +0.375
50 35 35 0
60 30 29.875 -0.125
70 26 26 0
Find the average of the deviations.
The average of the deviations is -3.0625.
Comment on how well the function model fits the data. (Note: the model might or might not do a good job of fitting the data. Some types of data can be fit very well by quadratic functions, while some cannot).
Despite some deviations I think the model fits the data relatively well because it got 3 values exactly correct and 2 more values really close to the original.
"
@& Very good.
Note that you are only asked to submit the qa and the query. It wasn't necessary to submit the document in this format.
Generally you will do the qa, work out the problems, then with the problems worked out in front of you you will complete the query.
Howver if you do wish to submit some or all of the problems in the format of this document, I'll be glad to review them. However that might be too time-consuming for you.*@