course MTH 173 ÀÝ_¼¢ÑÀ‡–ÙÈà€ðܹþ§Óٞϓassignment #001
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11:06:25 `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 --> Y axis= Stock Growth in 100's T axis= Months If stocks rose from $5000 to $5300 that is a rise of 3 units on the graph. Also this happened over a 4 month time period therfore the slope of March-July is 3/4. Stocks rose from $5300 to $5500 that is a rise of 2 units. This happened in a 5 month period so we are left with 2/5. (3 /4) > (2/5) therefore it has a greater slope which means more money in less time so it is growing faster in March-July. confidence assessment: 2
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11:07:49 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 --> This is about what I wrote, I gave myself a 2 because I wasn't sure if you wanted us to go about finding the answer that way. self critique assessment: 2
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11:15:17 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> Since the slope of the March-July period is 3/4 and the slope of the July-December period is 2/5 we can average the two together and get the average slope for the two time periods. Avg rise/ Avg Run ((3+2)/2)/((4+5)/2)= (5/2)/(9/2)= 2.5/4.5= 5/9= Avg. Slope. This means in the average nine month period the stock will rise $500. confidence assessment: 2
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13:35:05 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 gave the average for the entire 9 month period, using the graph and slope method. I didn't use the actual figures I used the units that they represented on the graph.
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13:51:42 `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 --> To find the Average time for the first one first we see the total drop in cm. 80cm. - 40cm. = 40cm. Then we have to have the ellapsed time in seconds, 40s. - 10s. = 30s. So we have a 40cm. drop in 30s. for the first period. Do the same for the next period. 40cm. - 20cm. = 20cm. 90s. - 40s. = 50s. This gives us a 20cm. drop in 50s. When comparing the fractions must have a common denomenator, we can cancell out the zeros and then find a common denomenator. 40/30 compared to 20/50 zeros cancell out 4/3 compared to 2/5 Common denomonator is 15 20/15 compared to 6/15 20/15 > 6/15 So the average drop is higher for the first period. confidence assessment: 3
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13:55:45 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 went about the fractions in a different way, is it okay to use common denominator between the two or would you rather have fractions always over 1. Also my figures where not negative where the water dropped, it is like the difference between speed and velocity, but the math still works out, I will try to keep it always in the more velocity like format. self critique assessment: 2
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13:58:46 `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 --> In both questions you are asking for the average of a change within two different periods. For both questions you must find the difference between the two points on both the t and y axis, then use what is similiar to the slope formula of rise / run, so y / t. confidence assessment: 3
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14:01:27 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 close to what I said but I didn't use the 'd (which is delta or difference I assume) and I used the axis y and t instead of Q and t. self critique assessment: 2
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