course Mth 163 zťaϨDŽӓ岍assignment #009 009. `query 9 Precalculus I 02-17-2008 IЩBwJd assignment #009 009. `query 9 Precalculus I 02-17-2008
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22:33:50 Symbolic calculation of slope, preliminary exercise What was the function, between which two points were you to calculate the average slope and how did you get this slope?
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RESPONSE --> The slope equaled .5 I got this number by using the coordinates (-2,-2.6) and (7,1.9) and used the formula rise divided by run. confidence assessment: 3
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22:35:16 problem 2 symbolic expression for slope, fn depth(t). What is the expression for the slope between the two specified t values?
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RESPONSE --> (f(30)-f(10)/20 confidence assessment: 3
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22:36:56 What is the rise between the two specified t values?
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RESPONSE --> (f(10) - f(3)) confidence assessment: 1
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22:37:25 ** The rise is the change in depth. The two depths are depth(10) and depth(30). The change in depth is final depth - initial depth, which gives us the expression depth(30)-depth(10) **
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RESPONSE --> Didn't really understand what you were aksing on that one, but I understand now. self critique assessment: 3
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22:37:47 What is the run between the two specified t values?
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RESPONSE --> 30-10 which equals 20 confidence assessment: 3
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22:39:26 What therefore is the slope and what does it mean?
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RESPONSE --> The slope is (f(30) - f(10)) / 20 which means that what ever the depth at f(30) is subtracted by the depth at f(10) and then that number is divided by 20. confidence assessment: 1
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22:39:39 ** rise = depth(30)-(depth(10) indicates change in depth. run = 30 - 10 = 20 = change in clock time. Slope = rise / run = (depth(30) - depth(10) ) / 20, which is the average rate at which depth changes with respect to clock time between t = 10 and t = 30. **
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RESPONSE --> self critique assessment: 3
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22:40:25 problem 5 graph points corresponding to load1 and load2 What are the coordinates of the requested graph points?
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RESPONSE --> (L1, f(L1)), (L2, f(L2)) confidence assessment: 3
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22:41:52 What is your expression for the average slope of the graph between load1 and load2?
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RESPONSE --> (f(L2)-f(L1)) / (L2-L1) confidence assessment: 3
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22:43:26 ** the name of the function is depth(t). We need the slope between t = t1 and t = t2. The depths are depth(t1) and depth(t2). Thus rise is depth(t2) - depth(t1) and run is t2 - t1. Slope is [ depth(t2) - depth(t1) ] / (t2 - t1). **
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RESPONSE --> Slope equals [ depth(t2) - depth(t1) ] / (t2 - t1) self critique assessment: 3
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22:53:24 problem 8 average rate from formula f(t) = 40 (2^(-.3 t) ) + 25 intervals of partition (10,20,30,40) What average rate do you get from the formula? Show your steps.
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RESPONSE --> t=10 = y = 40(2^(-.3(10)) + 25 y = 40(1/8) + 25 y = 30 t=20 = y = 40(2^(-.3(20)) + 25 y = 40(1/64) + 25 y = 25.625 t=30 = y = 40(2^-.3(30)) + 25 y = 40(1/512) + 25 y = 25.078125 t=40 = y = 40(2^-.3(40)) + 25 y = 40(1/4096) + 25 y = 25.009765625 The average rate of the three intervals was -.1663411458 confidence assessment: 3
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22:54:38 ** ave rate = change in depth / change in t. For the three intervals we get (f(20)-f(10))/(20-10) = (25.625 - 30 )/(20 - 10) = -4.375 / 10 = -.4375 (f(30)-f(20))/(30-20) = (25.07813 - 25.625)/(30 - 20) = -.5469 / 10 = -.05469. (f(40)-f(30))/(40-30) = (25.00977 - 25.07813)/(40 - 30) = -.0684 / 10 = -.00684. **
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RESPONSE --> I calculated the exact same numbers, but instead of putting the average rate of change for each data observation and their intervals I took the average of the three. self critique assessment: 3
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22:57:36 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> After using the DERIVE program for problem 9 it said to compare the answer to the one I got for number 8 and they were way off. I am not sure if I used the program incorrectly, but after calculating the problem on the DERIVE program I go the number -.00000000373 Was I supposed to get this answer or do you think I authored the wrond expressions? self critique assessment: 3
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