allegansveritatem
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- Jan 10, 2018
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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.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].
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.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.
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 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).