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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?


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Where are you confused? – Graphth Oct 18 '12 at 14:58

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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