math178 hw7
TRANSCRIPT
Kaya OtaMath 1784/12/2015
Mathematical Modeling HW 7 Problem 4.1 #10
Data Count Length x in Weight y oz1 12.5 172 12.625 16.53 14.125 234 14.5 26.55 17.25 416 17.75 49
12 13 14 15 16 17 18 190
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Because the graph Length vs. Weight shows liner-like, we apply the method of fitting a straight line.So, we want to fine the slope and the intercept from the equation from ch3.
The slope a=m∑xy−∑x∑ ym∑x2−(∑x )2
=5.856
The intercept b=∑x2∑ y−∑xy∑ xm∑x2−(∑x )2
=−57.78
Then we have the model y=ax+b=5.856 x−57.78 Length x (in) Weight y = ax + b (oz)
12.5 15.4139530112.625 16.1459192114.125 24.9295136
14.5 27.125412217.25 43.22866859
17.75 46.15653339
12 13 14 15 16 17 18 190
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Then the model fits to the data points
Problem 4.3 #3X Y 1st divided diff 2nd divided diff 3rd divided diff0 7 8 5 01 15 18 5 02 33 28 5 03 61 38 5 04 99 48 5 05 147 58 5 06 205 68 0 07 273 0 0 0From the divided difference table, we can know it shows quadratic polynomials Problem 4.3 #6 x y 1st divided diff
46 40 3.33333333349 50 2.551 55 852 63 4.554 72 -156 70 757 77 -458 73 1759 90 360 93 361 96 -8
62 88 1163 99 1164 110 1.566 113 767 120 768 127 3.33333333371 137 -572 132 0
Negative numbers in the columns of the 1st divided diff makes invalid to model with the lower-polynomials. The graph is the original data of x vs. y
40 45 50 55 60 65 70 750
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Problem 4.3 #7x y 1st divided diff
17 19 319 25 720 32 9.522 51 623 57 725 71 11.6666666731 141 -1832 123 6433 187 1.66666666736 192 1337 205 4738 252 -439 248 23
41 294 0Because the 1st divided diff contains negative value, it tells the invalid of identifying the lower-polynomials.
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However, the graph x vs. y looks like the lower-order polynomials, so we could apply the n-th divided difference manually.
Problem 4.4 #1(b)