Water drips from a conical tank with height 16 feet and diameter 8 feet into a cylindrical tank which has a base of area $900\pi$ sq.ft.. The depth $h$ of the water in the conical tank is changing at a rate of $h-16$ feet per minute.

Two questions:

At what rate is the volume of water in the conical tank changing when $h = 3 $ feet?

If $y$ is the depth in feet, of the water in the cylindrical tank, at what rate is $y$ changing when $h = 3$ feet.

So for the first question, what I did was this:

$V = (1/3)16\pi h$

$dV/dT = 16/4*\pi*(dH/dT)$

Where $dH/dT = 3-16$ so

$dV/dT = -217.8 ~ \text{cu.ft. / min}$

Is this correct?

And how do I go through with the second question?

Excuse my typing because I don't know how to format it in the math language coding stuff.

  • $\begingroup$ Here is a short tutorial on typing math here. $\endgroup$ – Ross Millikan Jul 2 '14 at 22:31
  • $\begingroup$ If $V$ is the volume in the conical tank, your first equation is not correct. Presumably the point of the cone is down, so the area of the base depends on the depth in the tank. $V$ should be proportional to $h^3$ For the second question, you take the rate water is flowing into the bottom tank (which is $-dV/dT$) divide by the area of the cylindrical tank $\endgroup$ – Ross Millikan Jul 2 '14 at 22:34

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