A little problem with inverted water reservoirs

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Air conditionner

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Hi!
It's been a few hours I'm blocking on a math problem. I've tried asking parents, tutors on Gauthmath, but I'm still blocked.
The problem is written in a calculus book. The answer, and the solving steps are there, but there's a little thing I can't seem to understand.
Could you please help me?
Inverted water reservoir problem.PNGIt took me a good 30 mins with my mom to find out the reservoir was inverted. I didn't know they existed. And then I had to convert the unitsXD. Anyway, I managed to find the derivative.
My problem is: I don't understand why they take the mean level, from 14 to 10 feet. Why couldn't they just use the formula for the volume at 14 feet, and then substract the volume at 10 feet? And finally, divide by 24 hours?
Maybe there's 2 ways of doing it. Anyway, I would really like to understand the ``calculus`` way (which is why I'm reading the book).
I think I'm just stuck on what the mean level represents. And the dV over dh...

Thank you in advance,
Air conditionner
 
Air conditionner said:
Hi!
It's been a few hours I'm blocking on a math problem. I've tried asking parents, tutors on Gauthmath, but I'm still blocked.
The problem is written in a calculus book. The answer, and the solving steps are there, but there's a little thing I can't seem to understand.
Could you please help me?
View attachment 28020It took me a good 30 mins with my mom to find out the reservoir was inverted. I didn't know they existed. And then I had to convert the unitsXD. Anyway, I managed to find the derivative.
My problem is: I don't understand why they take the mean level, from 14 to 10 feet. Why couldn't they just use the formula for the volume at 14 feet, and then substract the volume at 10 feet? And finally, divide by 24 hours?
Maybe there's 2 ways of doing it. Anyway, I would really like to understand the ``calculus`` way (which is why I'm reading the book).
I think I'm just stuck on what the mean level represents. And the dV over dh...

Thank you in advance,
Air conditionner
Click to expand...
I wouldn't say the reservoir is inverted, just that its shape is an "inverted" frustum, in the sense that the bottom is smaller than the top.

I would do exactly what you suggest: divide the total volume removed by the time, (642,058 2/3 - 441,333 1/3)/24 = 8363 5/9 ft^3/h = 52,272 2/9 gal/h using their conversion factor). They seem to be trying to do it the hard way. In fact, I'm not sure whether they do would work in general. They certainly should say something to justify it.

What topic does this appear under? And what book is this from? (I just searched for the problem and found it is in Calculus Made Easy, published in 1910. I don't think I'd recommend it.)

Since nothing is said about how fast the water is actually removed (e.g. at a constant rate, or such that the level falls at a constant rate, or irregularly), the question seems to be really about the average rate, not the actual rate at any given moment.

By the way, the rate of change of volume is equal to the surface area of the water, (200 + 2h)^2; do you see why?
 
Thank you very much for your answer!
Regarding the book, yes, from Thompson, but I use the "new" edition (1998).
Also, why wouldn't you recommend it?
I have a study plan; I need to learn calculus (for physics). I intended to read this, then Calculus for the Practical man (because, apparently it's also good for beginners), then Khan academy, and finally Thomas' calculus. Would you not recommend any of these books?

Secundly, I think I finally understood why they took the average of 10 and 14 feet! It's such a weird concept.

Finally, "By the way, the rate of change of volume is equal to the surface area of the water, (200 + 2h)^2; do you see why?" I'm not sure I understood your question. I know that it did took me a lot of time to figure out why it was 200"" because I didn't know that such reservoirs were physically possible.#)

Thank you,
Air conditioner
 
Air conditionner said:
Secundly, I think I finally understood why they took the average of 10 and 14 feet! It's such a weird concept.
Click to expand...
Can you explain it to us? It seems to make some sense informally, but we need a theorem to justify it. The fact that this book makes such a leap without justification is the reason I don't recommend it!

Air conditionner said:
Finally, "By the way, the rate of change of volume is equal to the surface area of the water, (200 + 2h)^2; do you see why?" I'm not sure I understood your question.
Click to expand...
If the level goes down a very small amount (which is the basic idea behind differentiation), the water removed is essentially a rectangular prism with volume equal to the area times the small change in height; the derivative is this divided by that change in height, which is just the area. If I wrote a book with this problem, that is something I would absolutely point out, as it helps in better understanding the derivative.

Air conditionner said:
I know that it did took me a lot of time to figure out why it was 200"" because I didn't know that such reservoirs were physically possible.#)
Click to expand...
The length and width both start at 200. As the water rises by h units, it increases by h units at both ends, so that we add 2h to 200 to get the new length and width. A picture would have helped. Once again, this example as presented doesn't belong in a book that is supposed to "make calculus easy".
 
How did you put the orange things? I couldn't find out...
"Can you explain it to us? It seems to make some sense informally, but we need a theorem to justify it. The fact that this book makes such a leap without justification is the reason I don't recommend it!"
I'm sorry, I have no theorem. I'm not sure if I understood why correctly, but I'll just do that next time.
 
Air conditionner said:
How did you put the orange things? I couldn't find out...
"Can you explain it to us? It seems to make some sense informally, but we need a theorem to justify it. The fact that this book makes such a leap without justification is the reason I don't recommend it!"
I'm sorry, I have no theorem. I'm not sure if I understood why correctly, but I'll just do that next time.
Click to expand...
"Orange things"? You mean the quotes? Use the Reply button at the bottom of a message, or select text and use the Reply button that pops up there.

I asked you to explain how you concluded that "I think I finally understood why they took the average of 10 and 14 feet!" Tell me what your thinking is, so I can either support it, or help you see that your thinking is wrong (even though it seems to provide the right answer in this case). Both of these are helpful for learning, which is your goal. Not talking about your thinking is the way to accidentally "learn" wrong things.
 
So, I thought that if I have dV over dh, that gives me the variation of the Volume for each unit of depth (If I had to use this way (the "differential" , I would have calculated the variation of the volume at 14, 13, 12, 11, added them, and then divided by 24. ) But what I think they did, is that they just wanted the average variation per foot, between 14 and 10, so they just took the average number. And so, multiply the average variation from 10 to 14 feet must give the same result as adding the volume variation at each point. Is that correct?
That's what I think they did.

(If I had to use this way (the "differential" , I would have calculated the variation of the volume at 14, 13, 12, 11, added them, and then divided by 24. ) I did this, and it gave me sadly 52 739 gallons. That's far from the real answer (52 267 gallons). Does that mean that that way of thinking about the problem is incorrect?
 
Air conditionner said:
So, I thought that if I have dV over dh, that gives me the variation of the Volume for each unit of depth (If I had to use this way (the "differential" , I would have calculated the variation of the volume at 14, 13, 12, 11, added them, and then divided by 24. ) But what I think they did, is that they just wanted the average variation per foot, between 14 and 10, so they just took the average number. And so, multiply the average variation from 10 to 14 feet must give the same result as adding the volume variation at each point. Is that correct?
That's what I think they did.

(If I had to use this way (the "differential" , I would have calculated the variation of the volume at 14, 13, 12, 11, added them, and then divided by 24. ) I did this, and it gave me sadly 52 739 gallons. That's far from the real answer (52 267 gallons). Does that mean that that way of thinking about the problem is incorrect?
Click to expand...
The trouble is, although the average rate of change over some interval will equal the instantaneous rate of change (derivative) at some point in that interval (have you learned that theorem somewhere?), that point is not in general the exact midpoint of the interval. They haven't given any reason to believe that it is, in this case. And we've seen that it is close, but not exact.

You are now talking about "variations" over nonzero intervals, which are not "differentials". This is essentially the same issue.
 
Hello, Air conditioner :)

Sorry, but I think the solution given by the book is not quite correct. They get the correct result but the procedure is inadequate. If you already have the function [MATH]V(h)[/MATH], just doing

[MATH]\frac{(V(14)-V(10))*6.25}{24}[/MATH]
leads you to the correct solution. Differentiation is unnecessary.

The mean value theorem establishes the following statement: if you have a derivable function in an open interval (a; b) and continuous in the closed interval [a; b] then, exist c in (a; b) that

[MATH]\frac{f(b) - f(a)}{b-a}= f'(c)[/MATH]
If f is a quadratic function, c is always the midpoint between a and b, but it's not true otherwise. For example, if you have [MATH]f(x)=x^3[/MATH] in the interval [0; 3], then

[MATH]f'(c) = \frac{3^3-0^3}{3-0}[/MATH]
[MATH]3c^2 = 9[/MATH]
[MATH]c=\sqrt{3}[/MATH]
and c is between 0 and 3 but it´s not the midpoint.

I hope this late answer help you ;)
 
Hi, Eugenio,
The first way is what I would have done too.
So, if I understood well, their procedure is false because they used the average, and the average isn't the midpoint of the interval, unless we're talking about a quadratic function?
But I'm not sure to understand why this doesn't apply to the following:20210704_122704.jpgIsn't it a quadratic function?
Thank you,
Air conditioner
 
Air conditionner said:
Hi, Eugenio,
The first way is what I would have done too.
So, if I understood well, their procedure is false because they used the average, and the average isn't the midpoint of the interval, unless we're talking about a quadratic function?
But I'm not sure to understand why this doesn't apply to the following:View attachment 28095Isn't it a quadratic function?
Thank you,
Air conditioner
Click to expand...
The derivative is quadratic, but the function itself (V) is cubic!

In fact, I had considered mentioning this fact about quadratics, but chose not to because it doesn't help with the problem, and might just be confusing.
 
Dr.Peterson said:
The derivative is quadratic, but the function itself (V) is cubic!

In fact, I had considered mentioning this fact about quadratics, but chose not to because it doesn't help with the problem, and might just be confusing.
Click to expand...

Yes, you're right! So, the solution given by the book is incorrect! ?

The prosecution rests, Your Honor. ?
 
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