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20:49:19 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 --> The stocks grew faster during the March-July period. In the 4 month interval between March and July, they gained $300. If we graphed these points, we would get a line with a slope of 300 / 4 or 75. In the 5 month interval between July and December, they gained $200. If we graphed these points, we would get a line with a slope of 200 / 5 or 40. The first line has the steeper slope, therefore, that is when the stock prices rose the fastest.
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20:49:29 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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20:50:31 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> For the first period, the average rate of change was 300 / 4 or 75. During the second period, the average rate of change was 200 / 5 or 40.
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20:50:39 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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20:57:13 `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 --> On the average, the depth of the water is changing less quickly between t = 40 sec and t = 90 sec. This is because the change in depth is less over a longer period of time than the first interval. For the first interval, the change is -40 / 30 or -4/3. For the second interval, the change is -20 / 50 or -2/5.
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20:57:32 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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20:59:26 `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 --> They are essentially the same question because you must compare a change in the quantity of something over a period of time. The mathematical reasoning is the same in both cases because you compare the difference in the quantity to the difference in time between two given points.
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20:59:35 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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