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lgloftin · Oct 28, 2011, 07:59 PM

Water is pouring into a conical cistern at the rate of 8 cubic feet per minute. If the height of the cistern is 12 feet anf the radius of its circular opening is 6 feet, how fast is the water level rising when the water is 4 feet deep?

jcaron2 · Oct 28, 2011, 10:16 PM

At some depth x, we know the volume of a water in the cistern is

$V=\frac13 \pi r^2 x$

We also know in this case that the radius r is half the depth. Thus we have

$V=\frac{1}{12} \pi x^3$

or, rewriting in terms of x,

$x=\sqrt[3]{\frac{12V}{\pi}}$

Now if we want to find the rate at which x is growing, $\frac{dx}{dt}$ , we can find it as follows:

$\frac{dx}{dt}=\frac{dx}{dV}\cdot \frac{dV}{dt}$

We know $\frac{dV}{dt}$ is 8 cubic feet per minute, so we just need to calculate $\frac{dx}{dV}$ using the formula for x above.

I'll let you do the derivative calculation. Once you've computed the derivative, that will be the formula to calculate the rate at which the depth changes as a function of the volume of water in the cistern. Since the problem is asking for that rate of change when the depth is 4 feet, you just have to calculate the water volume at that depth and plug that into your derivative formula.

Let me know if you get stuck. Otherwise, reply with your answer if you want one of us to check it for you.