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Mth 173
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
depth chng assumption mistake
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Question: `q007. How much depth change is there between t = 20 sec and the time at which depth stops changing?
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Your solution:
if I recall correctly the depth was originally 80...so 80 to zero would be...80cm change?
confidence rating #$&*:3
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Given Solution:
`aThe rate of depth change at t = 20 sec is - 4 cm/s; at t = 60 sec the rate is 0 cm/s. The average rate at which depth
changes during this 40-second interval is therefore the average of -4 cm/s and 0 cm/s, or -2 cm/s.
At an average rate of -2 cm/s for 40 s, the depth change will be -80 cm. Starting at 80 cm when t = 20 sec, we see that the
depth at t = 60 is therefore 80 cm - 80 cm = 0.
STUDENT QUESTION
I dont totally understand where the 2 cm/s comes from.
INSTRUCTOR RESPONSE
The two rates -4 cm/s and 0 cm/s, calculated from the given rate function, are applicable to the interval between t = 20 sec
and t = 60 sec.
The first is the rate at the beginning of the interval, and the second is the rate at the end of the interval.
Without additional information, our first conjecture would be that the average rate is the average of the initial and final rates.
For different situations this conjecture might be more or less valid; in this case since the rate function is linear, it turns out that it
is completely valid.
The average of the two rates -4 cm/s and 0 cm/s is -2 cm/s, and this is the rate we apply to our analysis of this interval.
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Self-critique (if necessary):
why must we do the rates if we know it dropped to 0, and we knew the current depth?
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Self-critique Rating:OK
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We don't know where the hole is relative to the bottom of the container.
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Besides, even if we know the flow ends when depth = 0, we don't know until we have verified that our model is consistent with this knowledge that it is so.
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I was hoping you could set a problem up where the hole is not at the bottom of the container so I could better see this, and not make such assumption. I understand what you are saying, but I feel like actually having the problem will help me a good deal.
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If you add 10 cm to the original function, the depth would stop changing at the same clock time, but the depth at this point would be 10 cm. This would correspond to a hole 10 cm higher than the point relative to which the depth is measured. If the hole happened to be 10 cm above the bottom of the container, then this would correspond to the situation I described.
The concept of depth can actually be generalized so that it can be measured relative to any chosen position. If it is generalized in this way, then it would be appropriate to refer to it as position rather than depth, meaning the vertical position of the water surface relative to a chosen point.
If the chosen point is changed, this changes the function by adding or subtracting the constant number in the quadratic function. This corresponds to raising or lowering the graph.
No matter what is done to that constant number, the shape of the graph is unchanged, and the clock time at which the depth stops changing is the same.
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