Query 2b

course Mth 271

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{Á¦÷¤Å÷HÈï}yä Õ´ïàÌ—ÕÕ

Applied Calculus I

06-07-2006

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11:29:27

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 -->

(99,0)

(83,10)

(63,20)

(49,30)

(37,40)

(27,50)

(16,60)

8,70)

(5,80)

(1,90)

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assignment #002

{Á¦÷¤Å÷HÈï}yä Õ´ïàÌ—ÕÕ

Applied Calculus I

06-07-2006

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11:48:30

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 -->

(63.7, 5.3)

(54.8, 10.6)

(46, 15.9)

(37.7, 21.2)

(32, 26.5)

(26.6, 31.8)

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11:49:54

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

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RESPONSE -->

(54.8, 10.6)

(37.7, 21.2)

(26.6, 31.8)

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11:50:34

** 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 -->

I used spread apart points to get a better reading like you have done here.

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11:51:18

Give the first of your three equations.

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RESPONSE -->

54.8= a (10.6^2)+ b(10.6)+ c

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11:52:20

** 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 -->

It seems that we have put them in the same order in the equations.

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11:53:13

Give the second of your three equations.

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RESPONSE -->

37.7= a (449.44)+ b(21.2)+ c

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11:53:25

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

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RESPONSE -->

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11:54:05

Give the third of your three equations.

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RESPONSE -->

26.6= 1011.24a+ 31.8 b+ c

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11:54:19

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

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RESPONSE -->

I understand

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11:55:30

Give the first of the equations you got when you eliminated c.

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RESPONSE -->

I subtracted the first equation from the second and I got 17.1= -337.08a+ -10.6b

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11:56:06

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

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RESPONSE -->

I did this to find my second equation I think that you can do this

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11:57:00

Give the second of the equations you got when you eliminated c.

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RESPONSE -->

I subtracted the second equation from the third and got 11.1= -561.8a+ -10.6 b

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11:57:46

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

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RESPONSE -->

I am not sure that I have done this right now....I think that I may have subtracted the wrong equations from each other.

It doesn't make any difference which two equations you use, as long as you don't use the same two equations on the second step as on the first.

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11:58:55

Explain how you solved for one of the variables.

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RESPONSE -->

I subtracted my first new equation from my second new equation and I got 6= 224.72a ....then I divided and got that a= .027

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12:00:04

** 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 -->

I understand what I did but you expalined it alot better. I too multiplied to get the b's to cancel out or go away then I solved for a.

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12:00:29

What values did you get for a and b?

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RESPONSE -->

a= .027.....b= -2.47

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12:01:27

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

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RESPONSE -->

I am not sure if got the correct answers or not. I do not know if ours need to be the same or not

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12:01:38

What did you then get for c?

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RESPONSE -->

c= 77.95

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12:02:17

** STUDENT SOLUTION CONTINUED: c = 73.4 **

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RESPONSE -->

Our values are pretty close my c value is just 4 away from yours

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12:02:52

What is your function model?

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RESPONSE -->

y= .027t^2+ -2.47t+ 77.95

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12:03:23

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

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RESPONSE -->

I used t instead of x but I used the same ideal as you

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12:03:48

What is your depth prediction for the given clock time (give clock time also)?

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RESPONSE -->

21.462

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12:04:29

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

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RESPONSE -->

I pluged in 46 for my t value and got 21.462

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12:04:46

What clock time corresponds to the given depth (give depth also)?

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RESPONSE -->

21.462

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12:05:54

** 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 -->

I used y= -b +- sqroot( b^2-4ac) / 2(a)

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12:09:22

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 -->

(1, 0)

(1.79, 10)

(2.118, 20)

(2.369, 30)

(2.58, 40)

(2.768, 50)

(2.936, 60)

(3.092, 70)

(3.236, 80)

(3.372, 90)

(3.5, 100)

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12:09:45

** 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 -->

Thats what I got too

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12:10:50

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

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RESPONSE -->

I used (2.118, 20) ....(2.7678, 50)....(3.237, 80)

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12:11:15

** STUDENT SOLUTION CONTINUED:

(20, 2.118034)

(50, 2.767767)

(100, 3.5)**

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RESPONSE -->

I choose points that were spread apart like you did

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12:11:49

Give the first of your three equations.

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RESPONSE -->

2.12= a (400)+ b (20) +c

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12:12:11

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

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RESPONSE -->

I rounded but we have basicly the same thing

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12:12:35

Give the second of your three equations.

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RESPONSE -->

2.77= a 2500+ b 50 +c

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12:12:46

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

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RESPONSE -->

once again same idea

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12:13:12

Give the third of your three equations.

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RESPONSE -->

3.24= a 6400 + b 80 +c

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12:13:19

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

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RESPONSE -->

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12:13:57

Give the first of the equations you got when you eliminated c.

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RESPONSE -->

-.65=a-2100+ -30b

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12:14:09

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

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RESPONSE -->

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12:14:48

Give the second of the equations you got when you eliminated c.

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RESPONSE -->

-.47= -3900a+ -30b

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12:15:17

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

9600a + 80b = 1.381966 **

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RESPONSE -->

I subtracted the 2 from the 3 and got mine

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12:17:24

Explain how you solved for one of the variables.

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RESPONSE -->

I muliplied my first new equation threw by a negative then I subtracted my first new equation form the second new equation then I solved for a

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12:17:34

** 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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12:18:08

What values did you get for a and b?

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RESPONSE -->

a= -1e-4 and b= .029

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12:18:50

** STUDENT SOLUTION CONTINUED:

a = -.0000876638

b = .01727 **

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RESPONSE -->

I didnt get these answers but I did get a negative and they are preety close to each other

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12:19:06

What did you then get for c?

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RESPONSE -->

c= 1.58

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12:19:32

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

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RESPONSE -->

We are only a few tenths off from each other

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12:21:11

What is your function model?

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RESPONSE -->

y= -1e-4 t^2 + .029 b+ 1.58

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12:21:27

** STUDENT ANSWER: y = (0) x^2 + (.01727)x + 1.773 **

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RESPONSE -->

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This looks good. You're not showing quite all the detail I would need to be sure, but I think you're doing the right things thoughout. Let me know if you have questions.