002. `query2
This assignment consisted of the worksheets
Students (often including some of the very best students, so there's no shame in it if this applies to you) frequently tell the instructor that they don't know where to find the data for some of these problems. This is usually because they have missed the instruction to do the second of these worksheets, which would include the exercises at the end of the worksheet.
If you find that you are among these students, go ahead and complete the parts of this 'query' that are based on the work you have completed, and submit that part. Then before completing and submitting the rest, simply go back and complete the second worksheet.
Within the worksheet entitled
'Overview and Introduction: The Modeling Process applied
to Flow From a Cylinder' are data for the temperature model and a series of
instructions
for constructing and assessing your model.
At the end of the worksheet entitled 'Completion of the Introductory Flow Model'
are two
data sets, which correspond to the other problems in the Query.
Question: `qAssignment 2
For the temperature vs. clock time model, what were temperature and time for the first, third and fifth data
points (express as temp vs clock time ordered pairs)?
Your solution:
Confidence Assessment:
Given Solution:
** Continue to the next question **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qAccording to your graph what would be the temperatures at clock
times 7, 19 and 31?
Your solution:
Confidence Assessment:
Given Solution:
** Continue to the next question **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat three points did you use as a basis for your quadratic model
(express as ordered pairs)?
Your solution:
Confidence Assessment:
Given Solution:
** 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 nswers'.
STUDENT SOLUTION (this student probably used a version
different from the one you used; this solution is given here
for comparison of
the steps, you should not expect that the numbers given here
will be the same as the numbers you obtained when you solved the
problem.)
For my quadratic model, I used the three points
(10, 75)
(20, 60)
(60, 30). **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the first equation you got when you substituted into the
form of a quadratic?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: The equation that I got from the first
data point (10,75) was 100a + 10b +c = 75.**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the second equation you got when you substituted into the
form of a quadratic?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: The equation that I got from my second
data point was 400a + 20b + c = 60 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the third equation you got when you substituted into the
form of a quadratic?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: The equation that I got from my third
data point was 3600a + 60b + c = 30. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat 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?
Your solution:
Confidence Assessment:
Given Solution:
** 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. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qTo 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?
Your solution:
Confidence Assessment:
Given Solution:
** 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**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhich variable did you eliminate from these two equations, and what
was the value of the variable for which you solved these equations?
Your solution:
Confidence Assessment:
Given Solution:
** 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. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat 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?
Your solution:
Confidence Assessment:
Given Solution:
** 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. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the value of c obtained from substituting into one of the
original equations?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: By substituting both a and b into the
original equations, I found that c = 93 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is the resulting quadratic model?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I
obtained was
y = (.015) x^2 -
(1.95)x + 93. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat 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?
Your solution:
Confidence Assessment:
Given Solution:
** 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. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat was your average deviation?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qIs there a pattern to your deviations?
Your solution:
Confidence Assessment:
Given Solution:
** 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. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qHave 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?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the
process after studying these outlines and explanations. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qHave you memorized the steps of the modeling process, and are you
gonna remember them forever?
Convince me.
Your solution:
Confidence Assessment:
Given Solution:
** 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. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qQuery Completion of
Model first problem: Completion of
model from your data.Give your data in the form of depth vs. clock time ordered
pairs.
Your solution:
Confidence Assessment:
Given Solution:
** 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). **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat three points on your graph did you use as a basis for your
model?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: As the basis for my graph, I used
( 5.3, 63.7)
(15.9, 46)
(26.5, 32)**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the first of your three equations.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the
equation 28.09a + 5.3b + c = 63.7 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the second of your three equations.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the
equation 252.81a +15.9b + c = 46 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the third of your three equations.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation
702.25a + 26.5b + c = 32. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the first of the equations you got when you eliminated c.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the
third gave me 449.44a + 10.6b = -14. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the second of the equations you got when you eliminated c.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the
third gave me 674.16a + 21.2b = -31.7. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qExplain how you solved for one of the variables.
Your solution:
Confidence Assessment:
Given Solution:
** 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. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat values did you get for a and b?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat did you then get for c?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: c = 73.4 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is your function model?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is your depth prediction for the given clock time (give clock
time also)?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and
my depth prediction was 16.314 cm.**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat clock time corresponds to the given depth (give depth also)?
Your solution:
Confidence Assessment:
Given Solution:
** 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. **
STUDENT QUESTION
I have done what I could with the completion of flow model
page 7 directions when I hit the sqrt
button on calculator so sqrt -.652 I would get (0,.807465169527) I do not
remember ever doing a problem with 0,.8…. so I
hope I used the correct numbers to solve the rest of quad equation using quad
formula
INSTRUCTOR RESPONSE:
Short answer:
The square root of a negative isn't a real number, so there is no solution to the equation for your given depth. Your calculator indicated a complex-number solution.
Longer answer:
The square root of a negative number is an imaginary number;
the result you got is a point in the complex-number plane, on the imaginary
axis. You might not understand what that means, but the point is that there is
no real-number solution. You can't square a real number and get a negative, so
the square root of a negative isn't a real number.
What this means is that the equation has no real-number solution. There is no
clock time t for which the depth takes the y value you used in the equation.
In terms of the graph, note that the graph of the quadratic function is a
parabola, which opens upward. So there are y values that lie completely below
the parabola. If you try to solve the quadratic for one of these y values, you
won't get a solution.
This sort of thing can certainly happen with a mathematical model. When it does,
the answer is simply that there is no such solution.
Self-critique (if necessary):
Self-critique Rating:
Question:
`qCompletion of Model second problem: grade average Give your data in the form
of grade vs. clock time ordered pairs.
Your solution:
Confidence Assessment:
Given Solution:
** 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). **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat three points on your graph did you use as a basis for your
model?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED:
(20, 2.118034)
(50, 2.767767)
(100, 3.5)**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the first of your three equations.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the second of your three equations.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the third of your three equations.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the first of the equations you got when you eliminated c.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the second of the equations you got when you eliminated c.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: Subracting the first equation from the
third I go
9600a + 80b = 1.381966 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qExplain how you solved for one of the variables.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated the
variable b. In order to do this, I
multiplied the first new equation by 80 and the second new equation by -50. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat values did you get for a and b?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED:
a = -.0000876638
b = .01727 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat did you then get for c?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: c = 1.773. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is your function model?
Your solution:
Confidence Assessment:
Given Solution:
** y = -.0000876638 x^2 + (.01727)x + 1.773 **
STUDENT QUESTION
Hello! I am working on the Modeling
Project #1 still and I am having such issues with the data sets for the rest of
the
worksheet. I keep reading and I see that your doing grade average versus
percentage of assignments, but I am confused on what
it is asking or what method I am supposed to be using. I got the first question,
solving for a, b, and c and I am familiar
with the quadratic forumla, I am just missing something on how to start these
next two problems.
Could you give me a boost to what to do?
INSTRUCTOR RESPONSE
What that boils down to can be summarized by a table.
For example, consider the following
x y
2 20
5 50
12 130
From this table and the form y = a x^2 + b x + c you get the equations
20 = 4 a + 2 b + c
50 = 25 a + 5 b + c
130 = 144 a + 12 b + c
which you can solve by elimination, as you did with the first question.
Now you are given data for grade ave. vs. percent of review.
You could make a table of y vs. x, with y the grade average and x the percent of review.
You could replace the heading 'x' in the first column with the identifier 'percent of review' and the 'y' in the second column with 'grade ave', so your table would represent percent of review vs. grade average.
Your table would have several additional rows (one additional row for every 'data point').
You are instructed to choose three data points, and to base your model on those three points.
You could for example make a 'shortened table' with just the three points you choose, very similar to the table given above (but with different numbers).
To get a quadratic model you would again use the form y = a x^2 + b x + c to get three equations, one for each point.
Solving the equations for a, b and c and plugging those values back into the form y = a x^2 + b x + c gives you your model.
Let me know if this doesn't help.
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is your percent-of-review prediction for the given range of
grades (give grade range also)?
Your solution:
Confidence Assessment:
Given Solution:
** The precise solution depends on the model desired average.
For example if the model is y = -.00028 x^2 + .06 x + .5
(your model will probably be different from this) and the grade average desired is 3.3 we
would find the percent of review x corresponding to grade average y = 3.3 then
we have
3.3 = -.00028 x^2 + .06 x + .5.
This equation is easily solved using the quadratic formula,
remembering to put the equation into the required form a x^2 + b x + c = 0.
We get two solutions, x = 69 and x = 146. Our solutions are therefore 69% grade
review, which is realistically within the 0 - 100% range, and 146%, which we
might reject as being outside the range of possibility.
To get a range you would solve two equations, on each for the
percent of review for the lower and higher ends of the range.
In many models the attempt to solve for a 4.0 average results
in an expression which includes the square root of a negative number; this
indicates that there is no real solution and that a 4.0 is not possible. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat grade average corresponds to the given percent of review (give
grade average also)?
Your solution:
Confidence Assessment:
Given Solution:
** Here you plug in your percent of review for the variable.
For example if your model is y = -.00028 x^2 + .06 x + .5 and the percent of
review is 75, you plug in 75 for x and evaluate the result.
The result gives you the grade average corresponding to the percent of review
according to the model. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qHow well does your model fit the data (support your answer)?
Your solution:
Confidence Assessment:
Given Solution:
** You should have evaluated your function at each given
percent of review-i.e., at 0, 10, 20, 30, . 100 to get the predicted grade
average for each. Comparing your
results with the given grade averages shows whether your model fits the data. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qillumination vs. distance
Give your data in the form of illumination vs. distance
ordered pairs.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION:
(1, 935.1395)
(2, 264..4411)
(3, 105.1209)
(4, 61.01488)
(5, 43.06238)
(6, 25.91537)
(7, 19.92772)
(8, 16.27232)
(9, 11.28082)
(10, 9.484465)**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat three points on your graph did you use as a basis for your
model?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED:
(2, 264.4411)
(4, 61.01488)
(8, 16.27232) **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the first of your three equations.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the second of your three equations.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the third of your three equations.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the first of the equations you got when you eliminated c.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qGive the second of the equations you got when you eliminated c.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qExplain how you solved for one of the variables.
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: I solved for a by eliminating the
variable b. I multiplied the first
new equation by 4 and the second new equation by -6 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat values did you get for a and b?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat did you then get for c?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: c = 588.5691**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is your function model?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691
**
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat is your illumination prediction for the given distance (give
distance also)?
Your solution:
Confidence Assessment:
Given Solution:
** STUDENT SOLUTION CONTINUED: The given distance was 1.6 Earth distances from the sun.
My illumination prediction was 319.61 w/m^2, obtained by evaluating my function
model for x = 1.6. **
Self-critique (if necessary):
Self-critique Rating:
Question: `qWhat distances correspond to the given
illumination range (give illumination range also)?
Your solution:
Confidence Assessment:
Given Solution:
** The precise solution depends on the model and the range of
averages.
For example if the model is y =9.4 r^2 - 139 r + 500 and the
illumination range is 25 to 100 we would find the distance r corresponding to
illumination y = 25, then the distance r corresponding to illumination y = 100,
by solving the equations
25=9.4 r^2 - 139 r + 500
and
100 =9.4 r^2 -
139 r + 500
Both of these equations are easily solved using the quadratic
formula, remembering to put both into the required form a r^2 + b r + c = 0.
Both give two solutions, only one solution of each having and correspondence at
all with the data.
The solutions which correspond to the data are
r = 3.9 when y = 100 and r = 5.4 when y = 25.
So when the distance x has range 3.9 - 5.4 the illumination
range is 25 to 100.
Note that a quadratic model does not fit this data well.
Sometimes data is quadratic in nature, sometimes it is not.
We will see as the course goes on how some situations are accurately modeled by
quadratic functions, while others are more accurately modeled by exponential or
power functions. **