course Mth 163 I had emailed you a couple of weeks ago regarding my Test 1 results. You said they were under lock down to email you back in a couple of days. I had emailed you back a few days later. Just wondering if you got my email. Also, I was wondering what the last day to take the final exam will be? assignment #017017. `query 17
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14:40:06 Linearizing Data and Curve Fitting Problem 1. table for y = 2 t^2 vs. t, for t = 0 to 3, linearize. Give your table and the table for sqrt(y) vs. t.
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RESPONSE --> My table for y verses t consisted of t y 0 0 1 2 2 8 3 18 Table for sqrt(y) vs. t. is t sqrt(y) 0 0 1 1.4 2 2.8 3 4.2 confidence assessment: 2
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14:40:22 ** The table for y vs. t is t y 0 0 1 2 2 8 3 18 The table for sqrtIy) vs t, with sqrt(y) give to 2 significant figures, is t sqrt(y) 0 0 1 1.4 2 2.8 3 4.2 **
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RESPONSE --> I see self critique assessment: 2
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14:40:43 It the first difference of the `sqrt(y) sequence constant and nonzero?
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RESPONSE --> I think it is confidence assessment: 2
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14:41:17 The sqrt(y) sequence is 0, 1.4, 2.8, 4.2. The first-difference sequence is 1.4, 1.4, 1.4, which is constant and nonzero.
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RESPONSE --> I sort of understand, but it is much harder to just complete the assignment and query without a qa first. confidence assessment: 2
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14:42:16 Give your values of m and b for the linear function that models your table.
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RESPONSE --> my points are (0,0), (1, 1.4), (2, 2.8), (3, 4.2). confidence assessment: 2
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14:42:27 ** The points (t, sqrt(y) ) are (0,0), (1, 1.4), (2, 2.8), (3, 4.2). These points are fit by a straight line thru the origin with slope 1.4, so the equation of the line is sqrty) = 1.4 t + 0, or just sqrt(y) = 1.4 t. **
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RESPONSE --> I didn't go far enough self critique assessment: 2
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14:43:03 Does the square of this linear functiongive you back the original function?
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RESPONSE --> I don't think they are exact confidence assessment: 2
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14:43:27 ** Squaring both sides of sqrt(y) = 1.4 t we get y = 1.96 t^2. The original function was y = 2 t^2. Our values for the sqrt(y) function were accurate to only 2 significant figures. To 2 significant figures 1.96 would round off to 2, so the two functions are identical to 2 significant figures. *&*&
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RESPONSE --> When I got my answer I didn't round to 2 significant figures. self critique assessment: 2
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14:44:29 problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem.
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RESPONSE --> First you make a table then do an analysis on the values to get .85 confidence assessment: 2
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14:44:57 ** A table for the function is t y = 7 ( 3^t) 0 7 1 21 2 63 3 189 The table for log(y) vs. t is t log(7 ( 3^t)) 0 0..85 1 1.32 2 1.80 3 2.28/ Sequence analysis on the log(7 * 3^t) values: sequence 0.85 1.32 1.80 2.28 1st diff .47 .48 .48 The first difference appears constant with value about .473. log(y) is a linear function of t with slope .473 and vertical intercept .85. We therefore have log(y) = .473 t + .85. Thus 10^(log y) = 10^(.473 t + .85) so that y = 10^(.473 t) * 10^(.85) or y = (10^.473)^t * (10^.85), which evaluating the power of 10 with calculator gives us y = 2.97^t * 7.08. To 2 significant figures this is the same as the original function y = 3 * 7^t. **
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RESPONSE --> I didn't take my formula out far enough. This stuff is really hard. Having trouble understanding. self critique assessment: 2
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14:46:06 problem 7. Hypothesized fit is `sqrt(y) = 2.27 x + .05. Compare your result to the 'ideal' y = 5 t^2 function.
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RESPONSE --> You set up your tables for your data and solve for 10^log(y) = 10^(- 0.155? - 0.374) or confidence assessment: 2
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** For the simulated data the y values are .14, 5.66, 23.2, 52, 82.2 and 135.5. The square roots of these values are 0.374; 2.38, 4.82; 7.21; 9.34, 11.64. Plotting these square roots vs. t = 0, 1, 2, 3, 4, 5 we obtain a nearly straight-line graph. The best-fit linear function to sqrt(y) vs. x gives us sqrt(y) = 2.27? + 0.27. Your function should be reasonably close to this but will probably not be identical. Squaring both sides we get y = 5.1529?^2 + 1.2258? + 0.0729. If the small term .0729 is neglected we get y = 5.15 t^2 + 1.23 t. Because of the 1.23 t term this isn't a particularly good approximation to y = 5 t^2.**"