course MTH 174 Hello. Hello.
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13:59:36 `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 --> Since in the four month period from March to July the stocks grew $300, and in the five month period from July to December the stocks only grew $200, the March-July period had a faster growing rate. confidence assessment: 3
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13:59:59 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 --> self critique assessment:
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14:04:49 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> There are four months between mid-March and mid-July and the stock grew from $5000 to $5300. Therefore ($5300-$5000)/4 months = $75/month. There are five months between mid-July and mid-December and the stock grew from $5300 to $5500. Therefore {$5300-$5500)/5 months = $40/month. confidence assessment: 3
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14:05:40 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 --> The question did not ask again which rate was greater, so I did not include that in my answer since I had already answered it in the first question. self critique assessment: 3
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14:11:22 `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 rate of the water depth changing between t = 10 sec and t = 40 sec is (40cm-80cm)/(40sec-10sec) = -1.3cm/sec. The rate of the water depth changing between t = 40 sec and t = 90 sec is (20cm-40cm)/(90sec-40sec) = -0.40cm/sec. The depth of the water is changing more quickly between the t = 10 sec and t = 40 sec time period. confidence assessment: 3
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14:12:27 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 --> I don't guess I got my significant figures right on this problem, but the concept, and basically the answer, was understood. self critique assessment: 3
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14:16:19 `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 --> The questions are very similar in that they both dealt with an amount of time and a change in a quantity. In both problems, I had to find which time period had the quicker average rate. The mathematical reasoning is exactly the same in both of the problems. I had to find how much time had passed between each interval and how much the quantity had changed between each interval. Then, in both problems, I divided the change in quantity by the change in time. confidence assessment: 3
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14:16:46 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: 3
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