- #1

1MileCrash

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## Homework Statement

Sand is pouring from a pipe at the rate of 16 cubic feet per second. If the falling sand forms a conical pile on the ground whose altitude is always 1/4 the diameter of the base, how fast is the altitude increasing when the pile is 4 feet high?

## Homework Equations

[itex]V = \frac{1}{3}\pi r^{2}h[/itex]

## The Attempt at a Solution

Firstly, since I am given that the altitude is 1/4 of the diameter, I rearrange the volume formula.

1/4(2)r = h

1/2r = h

r = 2h

So:

[itex]V = \frac{1}{3}\pi 2h^{2}h[/itex]

or

[itex]V = \frac{1}{3}\pi 2h^{3}[/itex]

Second, I'm given the rate of volume increase as 16, so [itex]\frac{dV}{dt}=16[/itex]

Differentiation of both sides of the volume equation gives me:

[itex]16 = 2\pi 2h^{2}\frac{dh}{dt}[/itex]

So it follows that

[itex]\frac{dh}{dt}= \frac{16}{2\pi h^{2}}[/itex]

Solving this for h = 4 gives me ~= .159155

Which is incorrect.

Can someone show me where I'm messing this up?

Thanks!