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Water from one tank is being drained into another tank at a rate of 3 m3/min. The first tank is an inverted circular cone with height 3 m and radius 2 m. The second tank is a circular cylindrical tank with height 4 m and radius 2 m. a) How fast is the height of the water changing in the cylindrical tank when the height of water in the conical tank is 1 m?

Rather than coming to an answer, does anyone mind giving the large talking points for how to go about solving this problem?

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Just curious, what does "3 m3/min" mean? –  J. W. Perry Jul 18 at 1:39
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3m^3/min as in meters cubed –  user1419623 Jul 18 at 1:47
    
Ah excellent. Thanks for that, much clearer. –  J. W. Perry Jul 18 at 1:53

2 Answers 2

The first thing to do is to imagine the situation. OK, cone draining at a certain rate into a cylinder. Is the cylinder lying on its side? Who knows, it doesn't say. Careless of them, but not atypical. In real work, one gets a lot of irrelevant information, and the relevant bits are all too often missing.

And we don't know how much water was in the cone to begin with. That makes a big difference. So let us assume the cylinder is circular base down, like my hot water tank.

We are asked about the rate of increase in water level in the cylinder when the water level in the cone is whatever.

Who cares about how high the water is in the cone? Not I, the cylinder is filling at constant rate. Now it's over, minor calculation.

Remark: The above, though formula-free, is mathematics. If you end up doing any consulting, you will find that the majority of problems do not involve high level mathematics. And yet the perspective of a professional mathematician may be precisely what is needed, even if in principle the technical level is very low.

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So I just need to find when it is the voluem associated with that height? –  user1419623 Jul 18 at 1:52
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Rate is $3$ cubic metre/min. Cross-section area of cylinder is $4\pi$. The rate of increase of height is obviously constant, to $m$ metres per minute where $4\pi m=3$.' Please note that more complicated problems may need more complicated solutions, including the machinery of "related rates." I was trying to your (refreshing) request for what I interpreted as ideas, not formulas. –  André Nicolas Jul 18 at 2:00

I would say that André has exposed the singularly most important talking point here. That is, the cylinder is filling at a constant rate, and so the dimensions of the source are entirely irrelevant. Therefore this problem reduces to a question of rate of change of height with respect to time on a cylinder of radius $r$ while it is being filled at a constant rate, and so the cone can leave our minds.

Furthering the issue:

The rate of change of volume $V$ with respect to time $t$, and we are already in appropriate SI units so this is nice, is given by

$$\frac{dV}{dt}=3.$$

We wish to find

$$\frac{dh}{dt}.$$

We relate $h$ to $V$ by way of the formula for the volume of the cylinder,

$$V=\pi r^2 h.$$

Now we treat $V$ and $h$ as being implicitly differentiable functions of $t$. Differentiating both sides of the equation for $V$ produces

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

I will stop there. I wanted to talk, but I did not want to spoil your puzzle before you had a chance to work on it yourself.

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