course 173
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13:04:14 Note that there are four questions in this assignment. `q001. If your stocks are worth $5000 in mid-March, $5300 in mid-July and $5500 in mid-December, during which period, March-July or July-December was your money growing faster?
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RESPONSE --> March-July
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13:04:39 The first period (4 months) was shorter than the second (5 months), and the value changed by more during the shorter first period (increase of $300) than during the longer second perios (increase of $200). A greater increase in a shorter period implies a greater rate of change. So the rate was greater during the first period.
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RESPONSE --> ok
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13:07:11 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> The average rate of change for the first period is $75/month and the second period is $40/month
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13:07:24 From mid-March thru mid-July is 4 months. A change of $300 in four months gives an average rate of change of $300 / (4 months) = $75/month. From mid-July through mid-December is 5 months, during which the value changes by $200, giving an average rate of $200 / (5 months) = $40 / month. Thus the rate was greater during the first period than during the second.
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RESPONSE --> ok
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13:09:00 `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec?
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RESPONSE --> The water level is change more quickly between t=10 sec and t=40 sec and slower between t=40 sec and t=90 sec.
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13:09:07 Between clock times t = 10 sec and t = 40 sec the depth changes by -40 cm, from 80 cm to 40 cm, in 30 seconds. The average rate is therefore -40 cm / (30 sec) = -1.33 cm/s. Between t = 40 sec and t = 90 sec the change is -20 cm and the time interval is 50 sec so the average rate of change is -20 cm/ 50 sec = -.4 cm/s, approx. The depth is changing more quickly during the first time interval.
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RESPONSE --> ok
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13:10:37 `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases?
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RESPONSE --> Both questions deal with average rate of change but use different units. Both questions are answered with (change in x)/(time)
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13:10:50 In each case we were given a quantity that changed with time. We were given the quantities at three different clock times and we were asked to compare the rates of change over the corresponding intervals. We did this by determining the changes in the quantities and the changes in the clock times, and for each interval dividing change in quantity by change in clock time. We could symbolize this process by representing the change in clock time by `dt and the change in the quantity by `dQ. The rate is thus rate = `dQ /`dt.
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RESPONSE --> ok
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wzǏ Student Name: assignment #002 002.
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13:15:31 Note that there are four questions in this assignment. `q001. Recall the stock value problem, where March, July and December values were $5000, $5300 and $5500. Construct a graph of stock value vs. number of month (e.g., 1 for Jan, 2 for Feb, etc.). You will have three points on your graph, one corresponding to the March value, one to the July value, and one to the December value. Stock value will be on the y axis and month number on the x axis. Your first point, for example, will be (3, 5000), corresponding to $5000 in March. Connect your three points with straight lines--i.e., connect the first point to the second and the second to the third. What is the slope of your line between the first and second point, and what is the slope of your line between the second in the third point? Recall that slope is rise / run.
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RESPONSE --> The slope between $5000 and $5300 is 75 and the slope between $5300 and $5500 is 40.
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13:15:44 The three points on the graph are (3, 5000), (7, 5300) and (12, 5500). The rise between the first point and the second is from 5000 to 5300, or 300, and the run is from 3 to 7, or 4, so the slope is 300 / 4 = 75. Note that the 300 represents $300 and the 4 represents 4 months, so the slope represents $300 / (4 months) = $75 / month, which is the average rate of change during the first time interval. The rise between the second point and the third is from 5300 to 5500, or 200, and the run from 7 to 12 is 5, so the slope is 200 / 5 = 40. This slope represents the $40/month average rate of change during the second time interval. Click on 'Next Picture' to see the graph.
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RESPONSE --> ok
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13:17:57 `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 --> The graph represents the average rate of change perfectly. Even though the ""rate of change"" may not be constant between these time periods, the ""average"" rate of change can be represented with a line that has a constant slope.
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13:18:09 We see from this example that the slope of a graph of value vs. clock time represents the rate at which value is changing with respect to clock time.
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RESPONSE --> ok
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13:18:52 `q003. To what extent do you think your graph, consisting of 3 points with straight line segments between them, accurately depicts the detailed behavior of the stocks over the 9-month period?
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RESPONSE --> It shows that they increased at a decreasing rate.
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13:19:00 Stocks can do just about anything from day to day-they can go up or down more in a single day than their net change in a month or even a year. So based on the values several months apart we can't say anything about what happens from day to day or even from month to month. We can only say that on the average, from one time to another, the stocks changed at a certain rate.
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RESPONSE --> ok
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13:20:19 `q004. From the given information, do you think you can accurately infer the detailed behavior of the stock values over the nine-month period?
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RESPONSE --> You can accurately infer detailed behavior because the graph represents an average, not exact values for specific times all throughout the 9 months.
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13:20:24 Not on a day-to-day basis, and not even on a month-to-month basis. All we can see from the given information is what might be an average trend.
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RESPONSE --> ok
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