Thread: Question 12
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Old December 29th, 2006, 04:03 PM
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Hey Cap'n.

No one bit on this problem, so I will give my farthings worth.

I think I have seen a problem like this before, a while back.

I think the width would be proportional to how far it is from the tap.

If I remember right the formula is R(y)=ky^{\frac{-1}{4}}

Where R is the radius at some point y down the stream and y is the distance down the stream form the tap.

We could probably stack an infinite number of itty-bitty disks of thickness h up the stream with volume {\pi}(R(y))^{2}h

Because of gravity and all that, the rate this disk is passing y is:

F={\pi}(R(y))^{2}h\sqrt{19.6y}

Where F is the rate the water is leaving the tap.

Now, solve for R(y).


Now, maybe we could integrate. Maybe construct a DE. That's all I have for now.

This does look like an interesting little problem, though.