modeling project 1

modeling project 1

You were moving along pretty well, then kind of faded out at the end (understandable at 2 a.m.). I think you understand the overall process.

However you have to be able to answer questions about your model. For example on the grade vs. study time problem, how do you figure out how much study time goes with which grade (requires that you solve a quadratic), etc.. There are questions of that nature on the test.

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21:00:52 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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RESPONSE --> um, assuming this is the potato experiment (95,0) (60,20) (41,40)

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21:00:57 ** Continue to the next question **

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RESPONSE --> ok

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22:48:43 According to your graph what would be the temperatures at clock times 7, 19 and 31?

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RESPONSE --> ok. im really pissed the f**k off now. after 1.5hrs, my 200$ graph calculator wont plot coordinates,i couldnt find anything on ur stuff about converting points to formula algebracally and cant fig out how to use that stupid derive crap t o do it either. so i did what i knew i would have to resort to all along and hand graphed it. 79,7 61.3,19 48.1,31

Hand graphing at this stage is a good idea anyway.

What kind of calculator do you have?

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22:48:53 ** Continue to the next question **

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RESPONSE --> #@%#$%#@4

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22:51:57 What three points did you use as a basis for your quadratic model (express as ordered pairs)?

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RESPONSE --> the potato temp ones. points (95,0) (60,20) (41,40)

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22:52:56 ** 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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RESPONSE --> ok

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00:04:41 What is the first equation you got when you substituted into the form of a quadratic?

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RESPONSE --> using the matrix method: y= -.00532x^2 -.01733x +49.6522

Be sure you know how to do this with the simultaneous equations, which is what I'll be looking for on the Major Quiz. You can, of course, double-check your results with matrices.

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00:09:52 ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**

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RESPONSE --> ok. i even used the calc on mine; mabye because i assumed that it was"express as temp vs clock time ordered pairs" as #1 say.. oh wait um i did the whole thing sorry.

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00:11:20 What is the second equation you got when you substituted into the form of a quadratic?

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RESPONSE --> using 60,20 ; 3600a+60b+c+20

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00:12:12 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **

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RESPONSE --> ok. yup. im backwards

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00:14:54 What is the third equation you got when you substituted into the form of a quadratic?

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RESPONSE --> it was 1681a+41b+c=40 but should be 1600a+40b+c=41

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00:15:36 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **

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RESPONSE --> ok... so that is adiff point set

Right. You pick your own points so the given solution won't match yours, but it's there to show the procedure.

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00:18:30 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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RESPONSE --> did it with a matrix, but would use -1 would be -3600a -60b -c =20

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00:19:35 ** 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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RESPONSE --> yeah.

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00:21:50 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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RESPONSE --> wish i did it the long way- sorry.

No problem here. As long as you know how to do it the long way.

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00:23:10 ** 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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RESPONSE --> will do it like that (by hand) next time.

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00:28:06 Which variable did you eliminate from these two equations, and what was its value?

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RESPONSE --> 50 x 40 = 2000b..... and -2000b

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00:29:44 ** 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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RESPONSE --> oops yeah, xtra zero on multiple.

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00:47:27 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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RESPONSE --> 3200(.155)+40b= -30

496+40b= -30

40b= -526

b= -13.15

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00:52:20 ** 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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RESPONSE --> uuuuuuuhhhhhhhhhh..... i dont know where i got the .155 from.

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01:06:10 What is the value of c obtained from substituting into one of the original equations?

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RESPONSE --> 100^2( .015)+10( -1.95) +c= 75 150-19.5+c =75 130.5+c=75 c= -55.5

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01:08:26 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **

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RESPONSE --> ok im tired. im pissed. dont know what the hell is going on with my #s imgoingtosleep.

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01:10:15 What is the resulting quadratic model?

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RESPONSE --> daaaammmmmit. ittl be wrong anyway not my math damn computers

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01:10:27 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was

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

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RESPONSE --> yeah

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01:19:36 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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RESPONSE --> god i dont want to know what mine would have said. yours would not be: y = (.015) x^2 - (1.95)x + 93 75 = (.015) 10^2 - (1.95)10 + 93 75= 1.5 - 19.5 + 93 75=75 of course.

Good realization.

i swear im getting dumber

Not so, but doing math late at night it can sometimes feel that way.

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01:20:06 ** 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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RESPONSE --> yeah

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01:28:38 What was your average deviation?

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RESPONSE --> y= -.00532x^2 -.01733x +49.6522 20= -.00532 (60) ^2 -.01733 (60) +49.6522 20= -19.152 -1.0398 +49.6522 20= 29.46122

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01:28:56 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **

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RESPONSE --> xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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01:30:02 Is there a pattern to your deviations?

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RESPONSE --> yeah. they're all way off.

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01:30:24 ** 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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RESPONSE --> ok

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01:37:04 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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RESPONSE --> kinda. all the #s are running together now. but ive had this all before. in parts. but i never connected them like this before. dont know where i messed up though. funny my grapph calc must have an xtra chromosome

Maybe that's why the things irritate me. Mainly, though, I don't like them because they're too limited and don't have keyboards.

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01:37:57 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **

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RESPONSE --> uuuhhh

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01:38:56 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.

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RESPONSE --> um, could i sssssslllllllllllllleeeeeeeeeeeeppppppp on that?

Good idea at this point.

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01:39:46 ** 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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RESPONSE --> wow. my pc is talking to itself or typing.

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01:44:44 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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RESPONSE --> 0,11 10,8.5 20,6.2 30,4.6 40,3.4 50,2.8 60,2 70'1.4 80,1 90.8 100,.6 110,.5 120,.4

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01:45:22 ** 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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RESPONSE --> oh, that was in sec, cm

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01:47:39 What three points on your graph did you use as a basis for your model?

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RESPONSE --> i didnt. it said temp in # 1, so i used the pot. ex.the corresp. points on it

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01:48:25 ** 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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RESPONSE --> WILL IT EVER END??????????????/

Eventually, but it does go on ...

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01:48:39 Give the first of your three equations.

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RESPONSE --> what?

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01:49:07 ** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **

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RESPONSE --> zzzzzzzzzzzzz

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01:50:35 Give the second of your three equations.

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RESPONSE --> i forgot

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01:50:44 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **

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RESPONSE --> dddddd

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01:51:34 Give the third of your three equations.

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RESPONSE --> its wrong

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01:51:43 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **

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RESPONSE --> a

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01:52:01 Give the first of the equations you got when you eliminated c.

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RESPONSE --> isnt that above

Yes, but for a different situation.

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01:52:06 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **

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RESPONSE --> a

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01:52:19 Give the second of the equations you got when you eliminated c.

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RESPONSE --> ditto

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01:53:05 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **

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RESPONSE --> god its 200am

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01:53:39 Explain how you solved for one of the variables.

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RESPONSE --> i will not have enough time to do it all

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01:53:59 ** 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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RESPONSE --> zzz

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01:54:45 What values did you get for a and b?

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RESPONSE --> mine wre above

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01:55:08 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **

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RESPONSE --> im still wrong

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01:55:17 What did you then get for c?

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RESPONSE --> d

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01:56:04 ** STUDENT SOLUTION CONTINUED: c = 73.4 **

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RESPONSE --> braindead

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01:56:40 What is your function model?

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RESPONSE --> whats urs

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01:56:47 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **

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RESPONSE --> ok

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01:57:31 What is your depth prediction for the given clock time (give clock time also)?

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RESPONSE --> is this progream looped?

Questions are similar, but they apply to different situations.

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01:58:09 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**

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RESPONSE --> time

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01:58:43 What clock time corresponds to the given depth (give depth also)?

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RESPONSE --> im lost

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01:59:30 ** 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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RESPONSE --> zzzzzz

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02:01:20 Completion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.

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RESPONSE --> ewfa

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02:01:40 ** 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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RESPONSE --> gfe

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02:02:05 What three points on your graph did you use as a basis for your model?

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RESPONSE --> i didhnt USE MINE

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02:02:21 ** STUDENT SOLUTION CONTINUED:

(20, 2.118034) (50, 2.767767) (100, 3.5)**

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RESPONSE --> bm,jh

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02:02:26 Give the first of your three equations.

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RESPONSE -->

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02:02:33 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**

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RESPONSE --> fsd

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02:02:35 Give the second of your three equations.

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RESPONSE --> f

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02:02:38 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **

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RESPONSE --> f

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02:02:40 Give the third of your three equations.

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RESPONSE --> f

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02:02:43 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **

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RESPONSE --> f

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02:02:45 Give the first of the equations you got when you eliminated c.

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RESPONSE --> efewf

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02:02:46 Give the first of the equations you got when you eliminated c.

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RESPONSE -->

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02:02:46 Give the first of the equations you got when you eliminated c.

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RESPONSE -->

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02:02:48 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **

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RESPONSE -->

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02:02:50 Give the second of the equations you got when you eliminated c.

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RESPONSE -->

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02:02:53 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go

9600a + 80b = 1.381966 **

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RESPONSE -->

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02:02:56 Explain how you solved for one of the variables.

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RESPONSE -->

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02:02:58 ** 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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RESPONSE -->

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02:03:00 What values did you get for a and b?

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RESPONSE -->

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02:03:01 ** STUDENT SOLUTION CONTINUED:

a = -.0000876638

b = .01727 **

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RESPONSE -->

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02:03:28 What did you then get for c?

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RESPONSE --> goddam thisisetertal

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02:03:32 ** STUDENT SOLUTION CONTINUED: c = 1.773. **

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RESPONSE -->

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02:03:35 What is your function model?

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RESPONSE -->

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02:03:37 y = (0) x^2 + (.01727)x + 1.773

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RESPONSE -->

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02:03:54 What is your percent-of-review prediction for the given range of grades (give grade range also)?

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RESPONSE -->

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