course 173 This is today's SEND file and also the randomized problem for assignment 3 with the Send file at the end
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21:55:08 Note that there are four questions in this assignment. `q001. Sketch a graph similar to that you constructed for the stock values, this time for the depth of the water vs. clock time (depths 80, 40, 20 at clock times 10, 40, 90). Your first point, for example, will be (10, 80). Connect these points with straight lines and determine the slopes of the lines.
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RESPONSE --> The slopes are as follows: (10,80) to (40,40) ... m= -4/3 (40,40) to (90,20) ... m= -2/5
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21:55:40 `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change?
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RESPONSE --> ok i understand
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21:56:34 06-07-2006 21:56:34 The slopes and the rates of change are numerically equal. For example between the second and third points the rise of -20 represents the -20 cm change in depth and the run of 50 represents the 50 seconds required to make this change, so the slope represents the -20 cm / (50 sec) average rate of change over the second time interval. We therefore see that slope represents average rate of change.
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NOTES -------> slope represents average rate of change
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21:59:57 `q003. To what extent do you think your graph with three points and straight line segments between them accurately depicts the detailed behavior of the water over the 80-second period of observation? How do you think the actual behavior of the system differs from that of the graph? How do you think the graph of the actual behavior of the system would differ from that of the graph you made?
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RESPONSE --> When the depth is higher, the water will be leaving the container faster than when it is lower because the pressure will be higher. The more water that leaves the container, the slower it will leave. A true graph of this would have a constantly changing slope, going from a greater to lesser slope. There would be no straight lines on the true graph
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22:00:09 The straight line segments would indicate a constant rate of change of depth. It is fairly clear that as depth decreases, the rate of change of depth will decrease, so that the rate of change of depth will not be constant. The graph will therefore never be straight, but will be a curve which is decreasing at a decreasing rate.
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RESPONSE --> ok
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22:02:43 `q004. From the given information, do you think you can accurately infer the detailed behavior of the water depth over the 80-second period? Do you think you can infer the detailed behavior better than you could the values of the stocks? Why or why not?
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RESPONSE --> Yes you could accurately infer the behavior of the water in detail. With the information we have we can exract a line equation. Yes, stocks can never be figured out by an equation because there is too much fluctuation in the market and it holds no true patterns.
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22:03:05 It will turn out that three data points will be sufficient to infer the detailed behavior, provided the data are accurate. However you might or might not be aware of that at this point, so you could draw either conclusion. However it should be clear that the behavior of the water depth is much more predictable than the behavior of the stock market. We don't know on a given day whether the market will go up or down, but we do know that if we shoot a hole in the bottom of a full bucket the water level will decrease, and we expect that identical holes in identical buckets should result in the same depth vs. clock time behavior.
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RESPONSE --> ok
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If the function y = .015 t2 + -1.7 t + 93 represents depth y vs. clock time t, then what is the average rate of depth change between clock time t = 13.9 and clock time t = 27.8? What is the rate of depth change at the clock time halfway between t = 13.9 and t = 27.8? What function represents the rate r of depth change at clock time t? What is the clock time halfway between t = 13.9 and t = 27.8, and what is the rate of depth change at this instant? If the function r(t) = .193 t + -2.1 represents the rate at which depth is changing at clock time t, then how much depth change will there be between clock times t = 13.9 and t = 27.8? What function represents the depth? What would this function be if it was known that at clock time t = 0 the depth is 130 ? The average rate of depth change or slope between clock time t=13.9 and t=27.8 is: y(t)= .03t 1.7 therefore the average rate of depth change would be (.03(27.8) - 1.7)+(.03(13.9) - 1.7)/2 = -1.07 The function y(t)=.03t 1.7 represents the rate r of depth change at clock time t. The clock time halfway b/t t=13.9 and t=27.8 is .03(6.95) 1.7 = -1.49 The antiderivative of r(t) = .193 + -2.1 is y = .0965t + -2.1t + C which represents the depth equation If t=0 when the depth is 130 then the depth equation would be y= .0965t + -2.1t + 130 "