course Mth 163
y = a(t^2) + b(t) + cxxxx
a(5.2^2) + b(5.2) + c = 80.1
a(15.6^2) + b(15.6) + c = 62.2
a(31.2^2) + b(31.2) + c = 44.2
28.6a + 5.2b + c = 80.1
243.36a + 15.6b + c = 62.2
973.44a + 31.2b + c = 44.2
973.44a + 31.2b + c = 44.2
-243.36a - 15.6b - c = -62.2
730.08a + 15.6b = -18
973.44a + 31.2b + c = 44.2
-28.6a - 5.2b - c = -80.1
944.84a + 26b = -35.9
730a + 16b = -18
945a + 26b = -36
26(730a + 16b) = 26(-18)
-16(945a + 26b) = -16(-36)
18980a + 416b = -468
-15120a - 416b = 576
3860a = 108
I can't find the error withuot checking your arithmetic to complete precision, and I need to focus my time on substantial error. I can tell you that the correct solution for a is a = 0.0004161946933. Your solution to the system of two equations in a and b is correct. So you've got a minor arithmetic error in reducing the system of three equations to two; all your numbers and signs make sense so it's probably just a typo--something I can't easily spot. So I'll leave it to you to go back over the bookkeeping details if you so choose. No real problem, though.
a = .02797
730(.02797) + 16b = -18
20.4181 + 16b = -18
16b = -38
b = -2.375
945(.028) + 26(-2.375) = -36
26.46 - 61.75 = -36
26 - 62 = -36
28.6(.028) + 5.2(-2.38) + c = 80.1
.80 - 12.38 + c = 80.1
-11.58 + c = 80.1
c = 91.68
243.36a + 15.6b + c = 62.2
243.36(.028) + 15.6(-2.38) + c = 62.2
c= 92.51
Why do I get two different answers for c when I put the figures in place, but only one will be the right one to double check back?
It simply goes back to whatever minor error you made earlier. Minor errors do get magnified with every step.
243.36a + 15.6b + c = 62.2
243.36(.028) + 15.6(-2.38) + 92.5 = 62.2
6.81 - 37.13 + 91.68 = 62.2
a = .028, b = -2.38, c = 92.68
y = .028(t^2) - 2.38(t) + 92.68
y = .028(46^2) - 2.38(46) + 92.68
y = 59.248 - 109.48 + 92.68
y = depth = 42.49cm
14 = .028(t^2) - 2.38(t) + 92.68
.028(t^2) - 2.38(t) + 92.68 - 14 = 0
.028(t^2) - 2.38(t) + 78.68 = 0
t = [-(-2.38) +/- 'sqrt[(-2.38)^2 - 4(.028)(78.68)] / (2 * .028)
t = 2.38 +/- 'sqrt(5.6644 - 8.81216) / .056
t = 2.38 +/- 'sqrt(3.14776) / .056
t = 2.38 + 1.774 / .056 = 74.18
t = 2.38 - 1.774 / .056 = 10.82
t = 10.82 when depth = 14cm
Question 2
y = a(t^2) + b(t) + c
a(10^2) + b(10) + c = 1.790569
a(100) + b(10) + c = 1.790569
a(60^2) + b(60) + c = 2.936492
a(3600) + b(60) + c = 2.936492
a(90^2) + b(90) + c = 3.371708
a(8100) + b(90) + c = 3.371708
100a + 10b + c = 1.790569
3600a + 60b + c = 2.936492
8100a + 90b + c = 3.371708
8100a + 90b + c = 3.371708
-3600a - 60b - c = -2.936492
4500a + 30b = .435216
8100a + 90b + c = 3.371708
-100a - 10b - c = -1.790569
8000a + 80b = 1.581139
4500a + 30b = .435216
8000a + 80b = 1.581139
8(4500a + 30b) = 8(.435216)
-3(8000a + 80b) = -3(1.581139)
36000a + 240b = 3.481728
-24000a - 240b = -4.743417
12000a = - 1.261689
a = -0.00010514075
8000(-0.00010514075) + 80b = 1.581139
-.841126 + 80b = 1.581139
80b = 2.422265
b = 0.0302783125
4500(-0.00010514075) + 30(0.0302783125) = .435216
- .473133375 + .908349375 = .435216
100a + 10b + c = 1.790569
100(-0.00010514075) + 10(0.0302783125) + c = 1.790569
-0.010514075 + .302783125 + c = 1.790569
.29226905 + c = 1.790569
c = 1.49829995
3600a + 60b + c = 2.936492
3600(-0.00010514075) + 60(0.0302783125) + 1.49829995 = 2.936492
-0.3785067 + 1.81669875 + 1.49829995 = 2.936492
a = -0.00010514075, b = 0.0302783125, c = 1.49829995
y = a(t^2) + b(t) + c
3.0 = -0.00010514075(t^2) + 0.0302783125(t) + 1.49829995
-0.00010514075(t^2) + 0.0302783125(t) + 1.49829995 - 3.0 = 0
-0.00010514075(t^2) + 0.0302783125(t) -1.50170005 = 0
t = [-b +/- 'sqrt(b^2 - 4(ac)] / (2a)
t = [-(0.0302783125) +/- 'sqrt[(0.0302783125)^2 - 4(-0.00010514075)(-1.50170005)] / (2 * (-0.00010514075))
t = [-(0.0302783125) +/- 'sqrt[0.000916776208 - 0.000631559478] / (-0.0002102815)
t = -(0.0302783125) +/- 'sqrt(0.00028521673)/ (-0.0002102815)
t = -(0.0302783125) +/- 0.01688836078487/ (-0.0002102815)
t = -80.3433861 or 80.2828295
y = a(t^2) + b(t) + c
4.0 = -0.00010514075(t^2) + 0.0302783125(t) + 1.49829995
-0.00010514075(t^2) + 0.0302783125(t) + 1.49829995 - 4.0 = 0
-0.00010514075(t^2) + 0.0302783125(t) -2.50170005 = 0
t = [-b +/- 'sqrt(b^2 - 4(ac)] / (2a)
t = [-(0.0302783125) +/- 'sqrt[(0.0302783125)^2 - 4(-0.00010514075)(-2.50170005)] / (2 * (-0.00010514075))
t = [-(0.0302783125) +/- 'sqrt[0.000916776208 - 0.00105212248] / (-0.0002102815)
t = [-(0.0302783125) +/- 'sqrt[-0.000135346272] / (-0.0002102815)
***** 'sqrt[-0.000135346272] = i ********
80% of class that reviewed would get a 3.0 using my model
Even with 100% review no one would get a 4.0
I feel like that right now, I understand how to do it all, but it is taking me hours to work through each problem. just can't wrap my head around the numbers, transposing them all the time does not help.
y = a(t^2) + b(t) + c
y = -0.00010514075(80^2) + 0.0302783125(80) + 1.49829995
y = -0.00010514075(6400) + 0.0302783125(80) + 1.49829995
y = -0.6729008 + 2.422265 + 1.49829995
y = 3.24766415
80% of review by students would net a 3.25 grade
My model hates some of the data
Curve would follow real students
Data set 2
y = a(t^2) + b(t) + c
(1, 935.1395) (5, 43.06238) (10, 9.484465)
a(t^2) + b(t) + c = y
a(1^2) + b(1) + c = 935.1395
a + b + c = 935.1395
a(t^2) + b(t) + c = y
a(5^2) + b(5) + c = 43.06238
25a + 5b + c = 43.06238
a(t^2) + b(t) + c = y
a(10^2) + b(10) + c = 9.484465
100a + 10b + c = 9.484465
75a + 5b = -33.577915
99a + 9b = -925.655035
9(75a + 5b) = 9(-33.577915)
675a + 45b = -302.201235
-5(99a + 9b) = -5(-925.655035)
-495a - 45b = 4,628.27517
675a + 45b = -302.201235
-495a - 45b = 4,628.27517
180a = 4,326.07394
a = 24.0337441
99(24.0337441) + 9b = -925.655035
2379.34067 + 9b = -925.655035
9b = -925.655035 - 2379.34067 = -3 304.99571
b = - 367.221746
75(24.0337441) + 5(- 367.221746) = -33.577915
24.0337441 - 367.221746 + c = 935.1395
-343.188002 + c = 935.1395
c = 935.1395 + 343.188002
c = 1278.3275
100a + 10b + c = 9.484465
2403.37441 - 3672.21746 + 1278.3275 = 9.484465
9.48445 = 9.484465
I'm off by 0.00001 Does this matter for this problem? Would it be counted wrong on a test?
a = 24.0337441, b = - 367.221746, c = 1278.3275
y = a(t^2) + b(t) + c
t = [-b +/- 'sqrt(b^2 - 4(ac)] / (2a)
a(t^2) + b(t) + c = y
24.0337441(1.6^2) - 367.221746(1.6) + 1278.3275 = y
61.5263849 - 587.554794 + 1278.3275 = y
y = 752.299091
I don't know if this is right, it should be closer to 264.4411 since we are using 1.6 which is more than half way to 2? On my graph it looks to be about
5??.???
Reading would occur between 6 and 3 AU from sun
I think my model does not fit the data very well.
Note that you aren't asked to submit the work in this format. The open query will ask you to the questions that need to be answered. Ordinarily, unless you have specific questions, the format imposed by the open query is preferable for you. After another week or two you won't be able to look at the work as it will be posted from this document, and immediately understand what question is being answered by which series of equations. I could, since I'm very familiar with these problems it would be a lot easier for me, but even for me it would be easier to keep track of which equation answers which problem if it was don't in the 'open query' format.
Nevertheless, if you haven't submitted the Open Query already, you don't need to do so now. You clearly understand the process.
There are at least some minor arithmetic errors in your work, but there are a lot of steps and this is inevitable.
You are using your models in the appropriate manner (e.g., solving quadratic equations to find the time at which a given event occurs, something that escapes a majority of the students who come into this course), and other that inevitable arithmetic errors your work is just about perfect.