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If there is a cylinder with a base radius of 70cm, and water is being poured into it at 10 liters per minute, how fast is the water level rising?

I've gotten myself horribly confused as to how to do this; can anyone provide any insight?

Thanks!

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  • $\begingroup$ Where are you confused? $\endgroup$
    – GeoffDS
    Oct 18 '12 at 14:58
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Hints: You have a cylinder with height $h$ and radius of the base $r$ and volume $V$. Then $$ V = \pi r^2h = \pi (7dm)^2h. $$ (Using this the volume will be in liters since $1$ liter is $(10$cm$)^3$ and $10$cm$=1$dm (decimeter). The height is now measured in decimeters).

You know that $$ \frac{dV}{dt} = 10 $$ and you want to find $$\frac{dh}{dt}.$$ What you can do is to use the colume equation above and use implicit differentiation to relate $\frac{dV}{dt}$ and $\frac{dh}{dt}$.

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$10$ liter = $10000$ cubic centimeters. Area of cylinder's base is $70 \times 70 \times \pi$.

So, the height of the cylinder increases by $\frac{10000}{70 \times 70 \times \pi}$ per minute.

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I think this question and the answers highlight two important things.

  1. @Legendre 's answer shows that there is a simple, non-calculus solution. I think it is useful to recognize the times when you do not need calculus to solve the problem, though you can use the calculus solution if you wish (and obviously you would use the calculus solution in a calculus class).

  2. When using the calculus solution as @Thomas does above, it is important to recognize that the radius of the cylinder is not changing and therefore, when differentiated, $r$ is treated as a constant. (Volume is a function of both height and radius. Because the radius is not changing, you hold it constant and effectively take the partial derivative.)

Both methods produce the same result (which is a nice way to check you answer).

By the way, @Thomas refers to the "colume equation" above. Obviously he meant "volume equation." (I tried to edit it, but the edit was rejected. ¯_(ツ)_/¯ 🤷)

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