course Mth 271 Self-critique Rating:*********************************************
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok Self-critique Rating: ok ********************************************* Question: `q002. What were the precise average rates of change during these two periods? ********************************************* Your solution: f(b) – f(a) = 5300-5000 = $75/ month for first period b-a 7-3 5500-5300 = $40/month for second period 12-7 Confidence Assessment: 3
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Given Solution: `aFrom 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating:ok ********************************************* Question: `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? ********************************************* Your solution: 40-80 = -1.33 20-40 = -.4 40-10 90 – 40 Therefore it is changing more quickly on the interval 10 < t < 40. Confidence Assessment: 3
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Given Solution: `aBetween 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating:ok ********************************************* Question: `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? ********************************************* Your solution: They both deal with the average rate of change over a given time period. One can use the formula for average rate of change to solve the question. Confidence Assessment: 2
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Given Solution: `aIn 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.