#$&* course Mth 271 9/12 10:45 001. Depth vs. Clock Time and Rate of Depth Change*********************************************
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Given Solution: 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 periods (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: ********************************************* Question: `q002. What were the precise average rates of change during these two periods? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: $300 grew from March to mid-July, 4 months. $300/ 4 months = $75 a month. During the second period $200 grew from mid-July to mid-December, 5 months. $200/ 5 months = $40 a month. This shows the average rate of change was greater in the first four months than the last five. confidence rating #$&*: 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) = $40 / month. Thus the rate was greater during the first period than during the second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I did not understand this question and I did have to look ahead to see how to get the answer. I do believe looking ahead helped me understand how to do the problem. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: I do not fully understand this question, I did have to look ahead to see how to get the answer, I will also need to start using my DVDs more. ********************************************* 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: Both questions involved a change in quantity according to time. And were questions on rate of change. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIn each case we were given a quantity that changed with time. We were given the quantities at three different clock times and we were asked to compare the rates of change over the corresponding intervals. We did this by determining the changes in the quantities and the changes in the clock times, and for each interval dividing change in quantity by change in clock time. We could symbolize this process by representing the change in clock time by `dt and the change in the quantity by `dQ. The rate is thus rate = `dQ /`dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Both questions involved a change in quantity according to time. And were questions on rate of change. confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!