what happened here?

A

allegansveritatem

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Problem:sandopile prob1.PNGsandopile prob1.PNG
12926
Here is book solution:
12927

Here is what I did:
12928

My problem is: Where did r go in cube root of 3v over pi? I mean how can V = 243 pi ft^3? What about r here?
 
You seem to be forgetting that V in your formula is the volume, and confusing it with the rate of change of the volume. Those are not the same!

The volume is a quantity that is increasing at a constant rate of 243 pi, starting at 0. This is like "distance = rate x time"; the volume at time t is the rate, 243 pi, times the time, t. The rate is given in ft^3/min; multiplying that by t (in minutes) gives a result in ft^3.

As for where r went: you solved for r, so it's on the other side of the equal sign!
 
Here's my dilemma: In the solution given in the book--reproduced in my OP--the author uses the formula for the volume of a cone to get a term to substitute for r so that he can then "express r in terms of t" ie., so he can write a the function r(t). No? So...I am laboring under the notion that V, as expressed in the formula V=1/3 pi r^2h has r^2, as it were, baked into it so that if you write an expression that is to stand for V, it must contain an r^2. That is the crux of the matter of my OP. But you are saying that V in this case has been defined differently than it is defined in the volume formula. But isn't V= 243 pi ft^3 the same as 243 pi r^3? I mean....well, I still don't see quite how the solutions manual, starting from V=1/3 pi r^2h comes up with substituting 243 pi for V.....unless the author is up to his usual tricks and has elided a step or two or three and is here going back to the wording of the problem and dropping the ft^3/m. Is it just that the problem has not been stated in a manner that is so unequivocal that the even a person (like me) of modest intellectual capacity can't fail to see what is being asked for? And then, again, am I actually wrong in supposing that ft^3/min is a direct reference to r^2h (which in this case is r^3).
 
No, [MATH]ft^3[/MATH] is a unit; [MATH]r^3[/MATH] is a variable (measured in that unit). They are not interchangeable.

Every author has to leave out some steps at some point, to save paper, especially in solutions to exercises. This one is typically terse. I'll try filling in the gaps with words:

The formula for the volume of a cone is [MATH]V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^3[/MATH] (since [MATH]h = r[/MATH]).​
Solving this for r, we have [MATH]r^3 = \frac{3V}{\pi}[/MATH], so [MATH]r = \sqrt[3]{\frac{3V}{\pi}}[/MATH].​
This allows us to calculate [MATH]r[/MATH] for a given volume.​
Now, we are told that [MATH]V[/MATH] starts at 0 ft3 and increases at a rate of [MATH]243 \pi[/MATH] ft3/min, so [MATH]V = 243\pi t[/MATH] ft3. (This is the equation of a straight line through the origin with slope [MATH]243 \pi[/MATH].)​
Replacing [MATH]V[/MATH] with this expression in the expression for [MATH]r[/MATH], we have [MATH]r = \sqrt[3]{\frac{3\cdot243\pi t}{\pi}}[/MATH]. This simplifies to [MATH]r = \sqrt[3]{729\pi}[/MATH].​

There are two different expressions used here for V: the formula that expresses it as a function of r (based on the geometry), and the formula that expresses it as a function of t (based on the constant rate of increase as sand is added to the pile). The goal here is to combine these two given facts to express r as a function of t.

I would drop all the units as soon as the appropriate unit for each variable has been established; so the fourth line could be written as

Now, we are told that V starts at 0 and increases at a rate of [MATH]243 \pi[/MATH], so [MATH]V = 243\pi t[/MATH].​
 
Dr.Peterson said:
No, [MATH]ft^3[/MATH] is a unit; [MATH]r^3[/MATH] is a variable (measured in that unit). They are not interchangeable.

Every author has to leave out some steps at some point, to save paper, especially in solutions to exercises. This one is typically terse. I'll try filling in the gaps with words:

The formula for the volume of a cone is [MATH]V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^3[/MATH] (since [MATH]h = r[/MATH]).​
Solving this for r, we have [MATH]r^3 = \frac{3V}{\pi}[/MATH], so [MATH]r = \sqrt[3]{\frac{3V}{\pi}}[/MATH].​
This allows us to calculate [MATH]r[/MATH] for a given volume.​
Now, we are told that [MATH]V[/MATH] starts at 0 ft3 and increases at a rate of [MATH]243 \pi[/MATH] ft3/min, so [MATH]V = 243\pi t[/MATH] ft3. (This is the equation of a straight line through the origin with slope [MATH]243 \pi[/MATH].)​
Replacing [MATH]V[/MATH] with this expression in the expression for [MATH]r[/MATH], we have [MATH]r = \sqrt[3]{\frac{3\cdot243\pi t}{\pi}}[/MATH]. This simplifies to [MATH]r = \sqrt[3]{729\pi}[/MATH].​

There are two different expressions used here for V: the formula that expresses it as a function of r (based on the geometry), and the formula that expresses it as a function of t (based on the constant rate of increase as sand is added to the pile). The goal here is to combine these two given facts to express r as a function of t.

I would drop all the units as soon as the appropriate unit for each variable has been established; so the fourth line could be written as

Now, we are told that V starts at 0 and increases at a rate of [MATH]243 \pi[/MATH], so [MATH]V = 243\pi t[/MATH].​
Click to expand...
Good. I will read over your post again today and then go back at this and see how it applies. I knew that this is a composition of functions problem but could not quite get my head around the nuts and bolts--to compose metaphors--of what the solutions author was doing. I think I see more clearly now. Thanks very much. I will work this out later and post results.
 
So, I went through this problem again and eventually I think I came out right. The following is the result of the enterprise--at one point I came to a false conclusion but quickly set things right (I hope). Please ignore the expression in the box in the middle, ie 9/2pi ft^3/min. This is extraneous to the business at hand.
12950
 
You got the correct answer, [MATH]r = 9\sqrt[3]{t}[/MATH], and were not messed up by my typo where I put pi in place of t. Good.

I hope the writing in the lower right is unrelated to this question, because it is. You can't talk about how much the radius increases "every 3914 seconds" or "per minute", because it is increasing at a varying rate. That can be handled with calculus, but not until then. All they asked for is the function, which is a cube root, not a linear function that would have a constant rate.
 
Dr.Peterson said:
You got the correct answer, [MATH]r = 9\sqrt[3]{t}[/MATH], and were not messed up by my typo where I put pi in place of t. Good.

I hope the writing in the lower right is unrelated to this question, because it is. You can't talk about how much the radius increases "every 3914 seconds" or "per minute", because it is increasing at a varying rate. That can be handled with calculus, but not until then. All they asked for is the function, which is a cube root, not a linear function that would have a constant rate.
Click to expand...
Well, I thought that the function, 9 times cuberoot of t meant that you take the time (I took 60 seconds because I was looking for how much radius changed in one minute (although right now I don't know exactly what I thought that would prove) and get the cube root of that and thus find how much the radius grows in one minute...I really don't have a clear idea of the situation. A cube root certainly isn't going to produce anything that would look like steady growth...what would happen is that at first there would be a dramatic increase that would gradually, as the minutes went by, start to flatten out. Right, I see it. Well, if I live long enough I will learn how to deal with varying rates. Thanks for pointing it out.
 
The important thing for the moment is that you just stop when you have answered the question that was asked, which is just the formula
[MATH]r = 9\sqrt[3]{t}[/MATH].

You should also be aware, though, that if t is defined as time in minutes, you don't replace it with 60! (I still don't see where your numbers came from, but we don't need to pursue that.)

And, as you now know, when a function is not linear, you can't talk (yet) about a single rate of increase. You could talk about an average rate of some time interval, or about an instantaneous rate at a moment (using calculus).
 
Dr.Peterson said:
The important thing for the moment is that you just stop when you have answered the question that was asked, which is just the formula
[MATH]r = 9\sqrt[3]{t}[/MATH].

You should also be aware, though, that if t is defined as time in minutes, you don't replace it with 60! (I still don't see where your numbers came from, but we don't need to pursue that.)

And, as you now know, when a function is not linear, you can't talk (yet) about a single rate of increase. You could talk about an average rate of some time interval, or about an instantaneous rate at a moment (using calculus).
Click to expand...
The numbers I got at the end of my solution were derived by taking the cube root of t which in this case defined as one minute. Now, let anyone try to take the cube root of one and see where that gets them. So I used 60 seconds as my radicand. Thus, I got 3.914 sec. It was an exercise in futility, I admit, but not without a certain method in its madness.
 
The cube root of 1 is 1, since 1 cubed is 1!

If a variable is defined as a number of minutes, then setting it to 60 means 60 minutes, not 60 seconds. You can't just decide to change its meaning.

But, yes, all this is moot, as far as this problem is concerned.
 
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