course Mth 173 This work is incomplete until I get some further clarification on the modelling project and the program requested. ϵ{ʱĩ]Ďassignment #001
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13:14:49 `qNote 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 to July is four months; during this time, the investments grew a total of $300 or average rate of growth $75 per month ($300/4 months). July to December is five months; the growth was $200 for an average rate of growth of $40 per month ($200/5 months.) Therefore, the March to July period saw the faster rate of growth. confidence assessment: 3
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13:15:13 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 self critique assessment: 3
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13:16:02 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> $75/month vs $40/month confidence assessment: 3
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13:16:19 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) = $40 / month. Thus the rate was greater during the first period than during the second.
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RESPONSE --> ok self critique assessment: 3
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13:25:44 `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec?
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RESPONSE --> Between t=10 and t=40, the depth was decreasing at an average rate of (4/3) cm/sec ; between t=40 and t=90, the depth was decreasing at an average rate of (2/5) cm/sec. Therefore, the former time period saw the average depth change quicker. confidence assessment: 3
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13:26:06 Between clock times t = 10 sec and t = 40 sec the depth changes by -40 cm, from 80 cm to 40 cm, in 30 seconds. The average rate is therefore -40 cm / (30 sec) = -1.33 cm/s. Between t = 40 sec and t = 90 sec the change is -20 cm and the time interval is 50 sec so the average rate of change is -20 cm/ 50 sec = -.4 cm/s, approx. The depth is changing more quickly during the first time interval.
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RESPONSE --> ok self critique assessment: 3
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13:28:28 `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases?
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RESPONSE --> Both of the questions involved average rates of change where the difference in units ($'s in the first problem, depth in the second problem) was divided by the elapsed time to come up with a rate of change. confidence assessment: 3
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13:28:48 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 --> self critique assessment: 2
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