#$&*
Mth 271
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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Mr. Smith,
I've been looking over some of the problems and I have a few questions:
###In problems involving sandpiles, when talking about grains of sand, are we talking about volume or surface??? When we have kilograms or sand vs. diameter, it is pretty clear to me that the proportionality is: y = ax^3.
The proportionality in case of surface vs. diameter is y = ax^2. Am I correct???
@& Absolutely.
Mass occupies volume, which is proportional to the cube of a given linear dimension.
The grains are visible only on the surface, and the area of the surface is proportional to the square of the linear dimension.*@
###So I am not sure what number of grains of sand should be: volume or surface????
###In the same type of problems, do we use the derivative of y = ax^3, which if I am correct is y = 3ax^2 to find the rate??? It makes sense to me but I am not sure...
@& You are again correct.*@
So again, in this type of problems so we use y = 3x^3 to find the rate???
@& If y = x^3, then the rate of change of y with respect to x is 3 x^2.
If y = 3 x^3, then the rate would be 3 * (3 x^2) = 9 x^2.*@
###Also, what is better was to find average rate of depth change between two clock times:
1.) to find y values (depths) for each value of x (time) and then use slope formula to find the average rate of change.
2.) to use the derivative y = ax + b, find values for the two given values of time, and then average them out?????
@& If you are given the depth function you would do the former. The result would be close to, but not necessarily the same as if you did the second procedure.
If the depth function is quadratic, both procedures would give you the same result; if it's not there would usually be slight difference and only the first procedure, which follows the definition of an average rate, would be accurate.
If you want to find the rate at the midpoint of the interval then you would evaluate the derivative function at the midpoint.
Now, if you were given the values of the rate-of-depth-change function at the beginning and end of the interval, you would average them to get the approximate average rate, then multiply by the time interval to get the approximate change in depth.
If you have the rate function, and if you can find its antiderivative, then the change in depth is the change in the antiderivative function.
If you really understand all this, as it seems you are in a position to do, then you're very much ahead of the curve at this point.
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Thank you so much for your help,
Maja
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@& Great questions, which I'm more than glad to answer.*@