Precal Query 2

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course MTH 163

002. `query2This assignment consisted of the worksheets

• Overview and Introduction: The Modeling Process applied to Flow From a Cylinder and

• Completion of the Introductory Flow Model.

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.

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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)?

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Your solution:

- (0, 95)

- (20, 60)

- (40, 41)

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Given Solution:

** Continue to the next question **

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Question: `qAccording to your graph what would be the temperatures at clock times 7, 19 and 31?

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Your solution:

- If this question pertains to estimating the temperatures and the respective times, then I would first use the equation relating from my graph: I would take the values of a, b, and c (which were triple-checked for accuracy) and the equation y = 0.015t^2 - 1.93t + 93. Then I would insert the t values, 7, 19, and 31, to get the respective solutions:

- when x =7, y = 80.225.

- When x = 19, y = 61.745,

- x = 31 y = 47.585

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Given Solution:

** Continue to the next question **

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Question: `qWhat three points did you use as a basis for your quadratic model (express as ordered pairs)?

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Your solution:

- I used (10, 75) , (20, 60) , and (60, 30).

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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). **

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Question: `qWhat is the first equation you got when you substituted into the form of a quadratic?

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Your solution: 100a + 10b +c = 75, from the point (10, 75)

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Given Solution:

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

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Question: `qWhat is the second equation you got when you substituted into the form of a quadratic?

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Your solution:

- From the point (20, 60) I came out with 400a + 20b + c = 60

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Given Solution:

** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **

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Question: `qWhat is the third equation you got when you substituted into the form of a quadratic?

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Your solution:

- From (60, 30) I got 3600a + 60b + c = 30

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Given Solution:

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

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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?

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Your solution:

- I used the first and second equation, no multiples, and eliminated ‘c’ to get my equation (4), being 300a + 10b = -15

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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. **

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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?

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Your solution:

- I subtracted the first and third equation, no multiples, and got 3500a + 50b = -45 as my fifth equation.

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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**

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Question: `qWhich variable did you eliminate from these two equations, and what was the value of the variable for which you solved these equations?

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Your solution:

- I eliminated b first, a = 0.015

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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. **

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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?

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Your solution:

- 300 (0.015) + 10b = -15.

- b = -1.95

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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. **

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Question: `qWhat is the value of c obtained from substituting into one of the original equations?

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Your solution:

- c = 93

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Given Solution:

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

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Question: `qWhat is the resulting quadratic model?

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Your solution:

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

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Given Solution:

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

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

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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?

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Your solution:

- Using 0, 10, and 20 I received deviations of -2, 0, and 0 respectively.

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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. **

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Question: `qWhat was your average deviation?

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Your solution: -3

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Given Solution:

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

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Question: `qIs there a pattern to your deviations?

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Your solution:

- None that I could surmise.

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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. **

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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?

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Your solution:

- I have used them, which I believe is more important to focus on; as for word-for-word memorizing, no, I have not. But as far as collecting my data on a graph and table, picking points, making systems of equations and solving them, and applying actual and theoretical data? I’m set on.

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Given Solution:

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

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Question: `qHave you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.

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Your solution:

- In the interest of sounding humble and open to learn, I haven’t memorized them. But as I said in the previous question, I can hold my own. From time to time I would attempt to keep it fresh, but alas, I cannot promise to remember them forever.

@&
Obviously I don't expect that sort of committement.

You're in good shape.
*@

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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. **

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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.

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Your solution:

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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). **

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Question: `qWhat three points on your graph did you use as a basis for your model?

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Your solution:

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Given Solution:

** 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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Question: `qGive the first of your three equations.

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Your solution:

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Given Solution:

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

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Question: `qGive the second of your three equations.

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Your solution:

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Given Solution:

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

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Question: `qGive the third of your three equations.

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Your solution:

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Given Solution:

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

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Question: `qGive the first of the equations you got when you eliminated c.

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Your solution:

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Given Solution:

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

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Question: `qGive the second of the equations you got when you eliminated c.

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Your solution:

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Given Solution:

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

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Question: `qExplain how you solved for one of the variables.

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Your solution:

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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. **

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Question: `qWhat values did you get for a and b?

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Your solution:

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Given Solution:

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

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Question: `qWhat did you then get for c?

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Your solution:

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Given Solution:

** STUDENT SOLUTION CONTINUED: c = 73.4 **

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Question: `qWhat is your function model?

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Your solution:

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Given Solution:

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

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Question: `qWhat is your depth prediction for the given clock time (give clock time also)?

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Your solution:

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Given Solution:

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

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Question: `qWhat clock time corresponds to the given depth (give depth also)?

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Your solution:

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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.

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

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Your solution:

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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). **

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Question: `qWhat three points on your graph did you use as a basis for your model?

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Your solution:

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Given Solution:

** STUDENT SOLUTION CONTINUED:

(20, 2.118034)

(50, 2.767767)

(100, 3.5)**

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Question: `qGive the first of your three equations.

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Your solution:

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Given Solution:

** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**

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Question: `qGive the second of your three equations.

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Your solution:

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Given Solution:

** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **

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Question: `qGive the third of your three equations.

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Your solution:

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Given Solution:

** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **

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Question: `qGive the first of the equations you got when you eliminated c.

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Your solution:

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Given Solution:

** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **

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Question: `qGive the second of the equations you got when you eliminated c.

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Your solution:

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Given Solution:

** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go

9600a + 80b = 1.381966 **

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Question: `qExplain how you solved for one of the variables.

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Your solution:

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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. **

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Question: `qWhat values did you get for a and b?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED:

a = -.0000876638

b = .01727 **

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: c = 1.773. **

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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.

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Question: `qWhat is your percent-of-review prediction for the given range of grades (give grade range also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. **

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Question: `qWhat grade average corresponds to the given percent of review (give grade average also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. **

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Question: `qHow well does your model fit the data (support your answer)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. **

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Question: `qillumination vs. distance

Give your data in the form of illumination vs. distance ordered pairs.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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)**

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Question: `qWhat three points on your graph did you use as a basis for your model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED:

(2, 264.4411)

(4, 61.01488)

(8, 16.27232) **

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Question: `qGive the first of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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Given Solution:

** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**

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Question: `qGive the second of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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Given Solution:

** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**

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Question: `qGive the third of your three equations.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**

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Question: `qGive the first of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**

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Question: `qGive the second of the equations you got when you eliminated c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**

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Question: `qExplain how you solved for one of the variables.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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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 **

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Question: `qWhat values did you get for a and b?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **

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Question: `qWhat did you then get for c?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: c = 588.5691**

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Question: `qWhat is your function model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **

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Question: `qWhat is your illumination prediction for the given distance (give distance also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. **

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Question: `qWhat distances correspond to the given illumination range (give illumination range also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. **

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Question: `qWhat distances correspond to the given illumination range (give illumination range also)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** The precise solution depends on the model and the range of averages.

25=9.4 r^2 - 139 r + 500

and

100 =9.4 r^2 - 139 r + 500

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.

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#*&!

@&
You appear to understand the process.

Be sure you also know how to solve to find a value of t for a given value of y, and know when to do so. (that solution will generally require the quadratic formula)
*@

Precal Query 2

#$&*

course MTH 163

002. `query2This assignment consisted of the worksheets

• Overview and Introduction: The Modeling Process applied to Flow From a Cylinder and

• Completion of the Introductory Flow Model.

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.

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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)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- (0, 95)

- (20, 60)

- (40, 41)

confidence rating #$&*:

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Given Solution:

** Continue to the next question **

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Question: `qAccording to your graph what would be the temperatures at clock times 7, 19 and 31?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- when x =7, y = 80.225.

- When x = 19, y = 61.745,

- x = 31 y = 47.585

confidence rating #$&*:

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Given Solution:

** Continue to the next question **

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Question: `qWhat three points did you use as a basis for your quadratic model (express as ordered pairs)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- I used (10, 75) , (20, 60) , and (60, 30).

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

For my quadratic model, I used the three points

(10, 75)

(20, 60)

(60, 30). **

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Question: `qWhat is the first equation you got when you substituted into the form of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 100a + 10b +c = 75, from the point (10, 75)

confidence rating #$&*:

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Given Solution:

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

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Question: `qWhat is the second equation you got when you substituted into the form of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- From the point (20, 60) I came out with 400a + 20b + c = 60

confidence rating #$&*:

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Given Solution:

** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **

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Question: `qWhat is the third equation you got when you substituted into the form of a quadratic?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- From (60, 30) I got 3600a + 60b + c = 30

confidence rating #$&*:

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Given Solution:

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

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- I used the first and second equation, no multiples, and eliminated ‘c’ to get my equation (4), being 300a + 10b = -15

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. **

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- I subtracted the first and third equation, no multiples, and got 3500a + 50b = -45 as my fifth equation.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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**

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Question: `qWhich variable did you eliminate from these two equations, and what was the value of the variable for which you solved these equations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- I eliminated b first, a = 0.015

confidence rating #$&*:

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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. **

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- 300 (0.015) + 10b = -15.

- b = -1.95

confidence rating #$&*:

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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. **

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Question: `qWhat is the value of c obtained from substituting into one of the original equations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- c = 93

confidence rating #$&*:

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Given Solution:

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

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Question: `qWhat is the resulting quadratic model?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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

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

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- Using 0, 10, and 20 I received deviations of -2, 0, and 0 respectively.

confidence rating #$&*:

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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. **

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Question: `qWhat was your average deviation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: -3

confidence rating #$&*:

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Given Solution:

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

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Question: `qIs there a pattern to your deviations?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

- None that I could surmise.

confidence rating #$&*:

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Given Solution:

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

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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Given Solution:

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

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Question: `qHave you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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Given Solution:

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

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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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). **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qWhat three points on your graph did you use as a basis for your model?

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Your solution:

confidence rating #$&*:

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Given Solution:

** 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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Self-critique (if necessary):

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Self-critique Rating:

Precal Query 2

@& Obviously I don't expect that sort of committement. You're in good shape. *@

@& You appear to understand the process. Be sure you also know how to solve to find a value of t for a given value of y, and know when to do so. (that solution will generally require the quadratic formula) *@

Precal Query 2

@&
Obviously I don't expect that sort of committement.

You're in good shape.
*@

@&
You appear to understand the process.

Be sure you also know how to solve to find a value of t for a given value of y, and know when to do so. (that solution will generally require the quadratic formula)
*@