course MTH 173 6/17 2 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: $75 avg. growth per month for the first, $40 avg. growth per month in the second period (see calculations above in first question). Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Between clock times 10 and 40 is a difference of 30 seconds, in which the water dropped 40 cm, which means the avg. rate of change is 40cm/30s = 1.3cm/s. Between clock times 40 and 90 is a diff. of 50 seconds, with a drop 20 cm in that period. 20cm/50s = .40cm/s avg. rate of change. Therefore, the depth of water changed much more quickly on avg. in the time between 10 seconds and 40 seconds. 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: They both involved rate of change vs. time, only with different units ($ vs. cm). The reasoning is the same because the total displacement divided by the total time gives the avg rate of change per unit of time. Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.