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So, I'm solving the following problem that I found here, and it has the solution in the page, so I'm more interested in fixing my way of understanding the problem.

Water is being poured into a cylindrical can that is 20 inches in height and has a radius of 8 inches. The water is being poured at a rate of 3 cubic inches per second. How fast is the height of the water in the can changing when the height is 8 feet deep?

I have the following data:

  • Change in volume $ \frac{dV}{dt} = 3 $ cubic inches/s
  • h = 20 inches
  • r = 8 inches
  • I want to find $ \frac{dh}{dt} $

Given the formula for the volume of a cylinder: $ V = \pi r^2 h $.

And this has been my thought process: As we are pouring water, and therefore changing the volume of the cylinder of water, the height of the cylinder of water is changing and that's what I'm looking for.

In the cylinder formula, $ \pi $ and $ r $ are constants so the derivative is:

$ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} $

Therefore:

$ \frac{\frac{dV}{dt}}{\pi r^2} = \frac{dh}{dt} $

Replacing the values that we know:

$$ \frac{3}{\pi 8^2} = \frac{dh}{dt} \\ 0.015 = \frac{dh}{dt} $$

So my answer seems to be the rate of change in height of the water and is constant. But based on the answer in the page. It doesn't seem to be right as I don't have a "height" variable to plug in the 8 feet deep and in the solution they link the radius with the height which doesn't make sense to me as the radius doesn't change. What is wrong in my train of thought?

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  • $\begingroup$ As your height changes, you get cones of different sizes. Do they have all have the same radius, too? $\endgroup$ Aug 23, 2020 at 22:20
  • $\begingroup$ What do you mean with cones? The problem only uses cylinders $\endgroup$
    – Jon
    Aug 23, 2020 at 22:25
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    $\begingroup$ So you're saying that the height is changing at a constant rate. Doesn't that sound reasonable? There's a constant volume of water flowing into a cylindrical can. The change in volume is proportional to the change in height. $\endgroup$
    – saulspatz
    Aug 23, 2020 at 22:32
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    $\begingroup$ It may be that your solution is right and that this was a "trick" question to check if you would realise that the depth is unused. (The level grows at the same speed, no matter how much water is already in.) Or, it may be a typo (they thought of come but said cylinder). Or it may be a typo of the other kind (they had a cone problem, and wanted to simplify it, but forgot to delete the additional condition). We will never know. In the circumstances, you've done the best you possibly could. $\endgroup$
    – user700480
    Aug 23, 2020 at 22:34
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    $\begingroup$ Having visited the Web site, indeed they say "cylinder" and even have a picture of a cylinder, yet their calculations are for a cone! This is just not right. $\endgroup$
    – user700480
    Aug 23, 2020 at 22:40

1 Answer 1

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Your calculation is correct, and your method of getting it is also correct. The 'answer' on the website is odd, confusing, and totally wrong.

Actually - even the question on the website is wrong, because they ask for rate of change when the water depth is "8 feet deep" (presumably they mean 8 inches deep, as 8 ft would be 76 inches above the top of the 20 inch deep cylinder). I would try another web-site.

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