course Mth 163 assignment #017017. `query 17
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19:10:19 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 --> y vs. t t y 0 0 1 2 2 8 3 18 sqrt(y) vs. t t sqrt(y) 0 0 1 1.4 2 2.8 3 4.2 confidence assessment: 2
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19:10:33 ** 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 --> self critique assessment: 3
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19:11:10 It the first difference of the `sqrt(y) sequence constant and nonzero?
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RESPONSE --> yes confidence assessment: 3
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19:11:35 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 --> should have shown the actual numbers confidence assessment: 2
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19:14:43 Give your values of m and b for the linear function that models your table.
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RESPONSE --> (0,0) (1,1.4) (2,2.8) (3,4.2) and slope is sqrt(y)=1.4t confidence assessment: 3
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19:14:49 ** 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 --> self critique assessment: 3
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19:15:18 Does the square of this linear functiongive you back the original function?
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RESPONSE --> yes confidence assessment: 3
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19:15:36 04-23-2009 19:15:36 ** 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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19:16:19 problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem.
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RESPONSE --> did not do this problem confidence assessment: 0
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19:16:30 04-23-2009 19:16:30 ** 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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19:18:42 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 --> it is not very close to the ideal function confidence assessment: 1
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19:18:49 04-23-2009 19:18:49 ** 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.**
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19:26:43 problem 9. Assuming exponential follow the entire 7-step procedure for given data set Give your x and y data. Show you solution. Be sure to give the average deviation of your function from the given data?
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RESPONSE --> for the points (0,.42) (1,.29) (2,.21) (3,.15) (4,.10) (5,.07) the table is t log(y) 0 -.375 1 -.538 2 -.678 3 -.824 4 -1 5 -1.15 and we get log(y)= -0.55x - 0.374 the we get y=.42 *.70 ^ t and the average deviation is .00078 confidence assessment: 2
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19:27:13 For (t, y) data set (0,.42), (1,.29),(2,.21),(3,.15),(4,.10),(5,.07) we get log(y) vs. t: t log(y) 0 -.375 1 -.538 2 -.678 3 -.824 4 -1 5 -1.15 A best fit to this data gives log(y) = - 0.155? - 0.374. Solving we get 10^log(y) = 10^(- 0.155? - 0.374) or y = 10^-.374 * (10^-.155)^t or y = .42 * .70^t. The columns below give t, y as in the original table, y calculated as y = .42 * .70^t and the difference between the predicted and original values of y: 0 0.42 0.42 0 1 0.29 0.294 -0.004 2 0.21 0.2058 0.0042 3 0.15 0.14406 0.00594 4 0.1 0.100842 -0.000842 5 0.07 0.0705894 -0.0005894 The deviations in the last column have an average value of -.00078. This indicates that the model is very good. **
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RESPONSE --> self critique assessment: 3
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19:27:50 problem 11. determine whether the log(y) vs. t or the log(y) vs. log(t) transformation works. Complete the problem and give the average discrepancy between the first function and your data.
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RESPONSE --> confidence assessment: 0
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19:27:58 ** The first table gives us x y log(x) log(y) 0.5 0.7 -0.30103 -0.1549 1 0.97 0 -0.01323 1.5 1.21 0.176091 0.082785 2 1.43 0.30103 0.155336 2.5 1.56 0.39794 0.193125 log(y) vs. x is not linear. log(y) vs. log(x) is linear with equation log(y) = 0.5074 log(x) - 0.0056. Applying the inverse transformation we get 10^log(y) =10^( 0.5074 log(x) - 0.0056) which we simplify to obtain y = 0.987?^0.507. The second table gives us x y log(x) log(y) 2 2.3 0.30103 0.361728 4 5 0.60206 0.69897 6 11.5 0.778151 1.060698 8 25 0.90309 1.39794 log(y) vs. x is linear, log(y) vs. log(x) is not. From the linear graph we get log(y) = 0.1735x + 0.0122, which we solve for y: 10^log(y) = 10^(0.1735x + 0.0122) or y = 10^.0122 * 10^(0.1735?) = 1.0285 * 1.491^x. **
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RESPONSE --> self critique assessment: accidentally skipped didn't mean to hit button twice
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19:31:28 Inverse Functions and Logarithms, Problem 7. Construct table for the squaring function f(x) = x^2, using x values between 0 and 2 with a step of .5. Reverse the columns of this table to form a partial table for the inverse function.
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RESPONSE --> x f ^-1(x) 0 0 .25 .5 1 1 2.25 1.5 4 2 confidence assessment: 3
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19:32:17 ** The table is x f(x) 0 0 .5 .25 1 1 1.5 2.25 2 4. Reversing columns we get the following partial table for the inverse function: x f^-1(x) 0 0 .25 .5 1 1 2.25 1.5 4 2 **
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RESPONSE --> Shoot! I did not put the original table. self critique assessment: 1
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19:35:27 Describe your graph consisting of the smooth curves corresponding to both functions. How are the pairs of points positioned with respect to the y = x function?
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RESPONSE --> curve of original: increasing at increasing rate curve for inverse: increasing at decreasing rate they look exactly alike except one is flipped confidence assessment: 3
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19:35:37 04-23-2009 19:35:37 ** The curve of the original function is increasing at an increasing rate, the curve for the inverse function is increasing at a decreasing rate. The curves meet at (0, 0) and at (1, 1). The line connecting the pairs of points passes through the y = x line at a right angle, and the y = x line bisects each connecting line. So the two graphs are mirror images of one another with respect to the line y = x. **
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19:37:42 8. If we reversed the columns of the 'complete' table of the squaring function from 0 to 12, precisely what table would we get?
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RESPONSE --> y=sqrt (x) is the table we would get confidence assessment: 1
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19:37:52 ** Our reversed table would give us the table for the square root function y = sqrt(x). The y = x^2 and y = sqrt(x) functions are inverse functions for x >= 0. **
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RESPONSE --> self critique assessment: 3
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19:39:53 9. If we could construct the 'complete' table of the squaring function from 0 to infinity, listing all possible positive numbers in the x column, then why would we be certain that every possible positive number would appear exactly one time in the second column?
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RESPONSE --> because the second column are the squares and they are all unique to the numbers in the first column confidence assessment: 3
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19:40:07 04-23-2009 19:40:07 ** The second column consists of all the squares. In order for a number to appear in the second column the square root of that number would have to appear in the first. Since every possible number appears in the first column, then no matter what number we select it will appear in the second column. So every possible positive number appears in the second column. If a number appears twice in the second column then its square root would appear twice in the first column. But no number can appear more than once in the first column. **
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19:40:36 What number would appear in the second column next to the number 4.31 in the first column?
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RESPONSE --> 18.57 confidence assessment: 3
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19:40:56 ** The table is the squaring function so next to 4.31 in the first column, 4.31^2 = 18.5761 would appear in the second. **
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RESPONSE --> I rounded self critique assessment: 2
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19:41:46 What number would appear in the second column next to the number `sqrt(18) in the first column?
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RESPONSE --> 18? confidence assessment: 2
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19:41:53 ** The square of sqrt(18) is 18, so 18 would appear in the second column. **
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RESPONSE --> self critique assessment: 3
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19:42:24 What number would appear in the second column next to the number `pi in the first column?
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RESPONSE --> pi^2 confidence assessment: 3
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19:42:32 ** pi^2 would appear in the second column. **
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RESPONSE --> self critique assessment: 3
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19:43:30 What would we obtain if we reversed the columns of this table?
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RESPONSE --> sqrt function table confidence assessment: 2
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19:43:35 Our table would have the square of the second-column value in the first column, so the second column would be the square root of the first column. Our function would now be the square-root function.
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RESPONSE --> self critique assessment:
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19:43:41 04-23-2009 19:43:41 Our table would have the square of the second-column value in the first column, so the second column would be the square root of the first column. Our function would now be the square-root function.
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19:44:27 What number would appear in the second column next to the number 4.31 in the first column of this table?
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RESPONSE --> 2.07 confidence assessment: 3
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19:44:34 ** you would have sqrt(4.31) = 2.076 **
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RESPONSE --> self critique assessment: 3
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19:45:30 What number would appear in the second column next to the number `pi^2 in the first column of this table?
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RESPONSE --> I'm kinda confused I don't know confidence assessment: 0
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19:46:15 ** The number in the second column would be pi, since the first-column value is the square of the second-column value. **
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RESPONSE --> Ohhhhh! Ok now I see what is goin on. I think I had a major blonde moment. self critique assessment: 2
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19:47:53 What number would appear in the second column next to the number -3 in the first column of this table?
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RESPONSE --> this is definately a trick question.....I think? because negative 3 is not a square root confidence assessment: 1
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19:48:00 ** -3 would not appear in the first column of the reversed table of the squaring function, since it wouldn't appear in the second column of that table. **
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RESPONSE --> self critique assessment: 3
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19:48:50 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18
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RESPONSE --> x=log{base 2}(18) =log(18)/log(2) confidence assessment: 2
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19:49:06 ** b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as x = log{base 2}(18) = log(18) / log(2). **
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RESPONSE --> self critique assessment: 3
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19:50:03 2 ^ (4x) = 12
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RESPONSE --> 4x=log(base 2)(12) =log(12)/log(2) confidence assessment: 3
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19:50:10 ** b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as 4 x = log{base 2}(12) = log(12) / log(2). **
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RESPONSE --> self critique assessment: 3
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19:51:21 5 * 2^x = 52
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RESPONSE --> x=log(base 2)(52/5) =log(52/5)/log(2) confidence assessment: 2
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19:51:25 ** You get 2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2). **
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RESPONSE --> self critique assessment: 3
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19:53:14 2^(3x - 4) = 9.
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RESPONSE --> 3x=log(9)/log(6) confidence assessment: 3
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19:53:24 04-23-2009 19:53:24 ** You get 3x - 4 = log 9 / log 2 so that 3x = log 9 / log 2 + 4 and x = ( log 9 / log 2 + 4 ) / 3. **
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19:53:49 14. Solve each of the following equations: 2^(3x-5) + 4 = 0
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RESPONSE --> can not figure this one out confidence assessment: 0
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19:54:09 ** You get log(-4)/log(2)=3x - 5. However log(-4) is not a real number so there is no solution. Note that 2^(3x-5) cannot be negative so the equation is impossible. **
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RESPONSE --> I guess thats why :) self critique assessment: 2
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19:54:45 2^(1/x) - 3 = 0
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RESPONSE --> .64 confidence assessment: 2
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19:54:53 ** You get 2^(1/x) = 3 so that 1/x = log(3) / log(2) and x = log(2) / log(3) = .63 approx. **
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RESPONSE --> self critique assessment: 3
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19:55:23 2^x * 2^(1/x) = 15
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RESPONSE --> I could not figure this one out either confidence assessment: 0
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19:55:41 04-23-2009 19:55:41 ** 2^x * 2^(1/x) is the same as 2^(x + 1/x) so you get x + 1/x = log{base 2}(15). Multiplying both sides by x we get x^2 + 1 = log{base 2}(15). This is quadratic. We rearrange to get x^2 - log{base 2}(15) x + 1 = 0 then use quadratic formula with a=1, b=-log{base 2}(15) and c=4. Our solutions are x = 0.2753664762 OR x = 3.631524119. **
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19:56:18 (2^x)^4 = 5
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RESPONSE --> 1/4log(5)/log(2) confidence assessment: 2
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19:56:24 ** You take the 1/4 power of both sides to get 2^x = 5^(1/4) so that x = log(5^(1/4) ) / log(2) = 1/4 log(5) / log(2). **
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RESPONSE --> self critique assessment: 3
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