course MTH272
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17:37:58 `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-July at 6% 5300-5000/5000= .06 5500-5300/5300= .037 confidence assessment: 3
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17:42:32 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 $200). A greater increase in a shorter period implies a greater rate of change. So the rate was greater during the first period.
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RESPONSE --> I thought of the question from more of an economic standpoint, such as total growth, rather than growth per month, but it is a simple concept. self critique assessment: 3
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17:46:12 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> During the first period, they experienced a growth of 6%(75$ per month), while the second period of 5 months experienced a 3.7% growth(400$ per month). confidence assessment: 3
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17:47:23 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 --> I'm still looking at overall growth for some reason, but I included the same information in the given answer in my own response self critique assessment: 3
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17:51:02 `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 --> first time period=(80-40)/(10-40)= -1.3cm/s second time period= (40-20)/(40-90)= -.4cm/s The water is leaking faster during the first time period. confidence assessment: 3
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17:51:23 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 --> self critique assessment: 3
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17:55:24 `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 both asking for an average rate of change. The mathematical procedure are similar in both because they require us to use our order of operations to subtract then divide. confidence assessment: 3
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17:58:36 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 --> This is kind of bad for a second semester applied calculus student, but I never thought of 'dQ/'dt as the change in quantity over the change in time. self critique assessment: 3