#$&*
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.
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Explain the difference between a situation in which you would do each of the following; if there is no such situation for a given item tell why:
• average two values of a function y(t) and divide by the corresponding difference in t values
@& You get the approximate average value of the function, divided by the time interval over which the values are averaged. This calculation is not related to a rate of change and usually doesn't tell you anything important.*@
• subtract two values of a function y(t) and divide by the corresponding difference in t values
@& When you subtract the two values you are calculating a change in the value of the function, a change in y.
The difference in the t values is the change in t.
So you are calculating change in y / change in t.
This is the average rate of change of y with respect to t.
In the limit as the interval shrinks to zero, this is the derivative of y with respect to t.*@
• subtract two values of a function y(t) and multiply by the corresponding difference in t values
@& This would be change in y * change in t. This is unrelated to any rate of change and has no generally significant interpretation.*@
• average two values of a function y(t) and multiply by the corresponding difference in t values.
@& This corresponds to the average value of y(t) multiplied by `dt.
If y(t) is a rate of change of some other quantity q with respect to t, this is then the approximate average rate of change of q with respect to t * change in t, which is equal to the change in q.*@
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I had to answer this question on test 1 and I dont think I answered them correctly. I am curious to know what the correct answers are.
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What are the correct answers to this question?
@& (also inserted above)*@
@& average two values of a function y(t) and divide by the corresponding difference in t values
@& You get the approximate average value of the function, divided by the time interval over which the values are averaged. This calculation is not related to a rate of change and usually doesn't tell you anything important.*@
• subtract two values of a function y(t) and divide by the corresponding difference in t values
@& When you subtract the two values you are calculating a change in the value of the function, a change in y.
The difference in the t values is the change in t.
So you are calculating change in y / change in t.
This is the average rate of change of y with respect to t.
In the limit as the interval shrinks to zero, this is the derivative of y with respect to t.*@
• subtract two values of a function y(t) and multiply by the corresponding difference in t values
@& This would be change in y * change in t. This is unrelated to any rate of change and has no generally significant interpretation.*@
• average two values of a function y(t) and multiply by the corresponding difference in t values.
@& This corresponds to the average value of y(t) multiplied by `dt.
If y(t) is a rate of change of some other quantity q with respect to t, this is then the approximate average rate of change of q with respect to t * change in t, which is equal to the change in q.*@
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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.
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The dollar values of a certain antique at t=0, 2, 4 and 8 years after coming on the market are 45.51, 56.85, 65.18 and 76 dollars. Estimate as accurately as possible the average value of the antique over the 8 year period. Do NOT just average the four numbers--this does not give an acceptable answer
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What would be the best way to approach this party? Would I take the given info and try to make it into an equation then go from there?
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@& The average value of a function f(x) defined on an interval a <= x <= b is equal to the integral of the function on that interval, divided by the time interval.
The graph of this information can be taken to form three trapezoids on a graph of value vs. clock time.
The integral can be estimated by adding up the areas of these trapezoids, and dividing by the total interval.
Trapezoid areas are very roughly 104, 122 and 284. The total area is roughly 510. Each area represents average value * time interval, in dollars * years. So total area represents average value * total time interval.
Thus average value is 510 dollars * years / (8 years) = 64 dollars, approximately.
This process represents a weighted average. Each average value is weighted by the length of time during which this average applies. About $52 for the first two years, about $61 for the next two, about $71 for the next four.*@