Sand is falling off a conveyor at a rate of 10m^3/min. It forms a conical pile where the diameter of the base is always three times the height. At what rate is the height of the pile changing when the pile is 15 metres high?

Thanks

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- November 6th 2008, 07:11 PM #1

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## Response

Sand is falling off a conveyor at a rate of 10m^3/min. It forms a conical pile where the diameter of the base is always three times the height. At what rate is the height of the pile changing when the pile is 15 metres high?

Well, start by using the equation for the volume of a cone:

V = (Pi*r^2*h)/3

The diameter can be expressed as:

diam = 3h. But we need radius for this, and diam = 2r, So:

2r = 3h

r = (3h/2)

We are given dV/dt, which is the rate at which the sand falls off the conveyor belt: dV/dt = 10

And we need to find dh/dt when h = 15

So, start by substituting 3h/2 for r in the volume equation:

V = Pi*(9/4)*h^2*h/3

V = (3*Pi*h^3)/4

Now take the derivative with respect to time:

dV/dt = (3*Pi*3h^2 * dh/dt)/4 Substitute for dV/dt

10 = (9*Pi*h^2 *dh/dt) / 4 Substitute for h

10 = (9*Pi*225*dh/dt)/4

Solve for dh/dt:

40 = 2025* Pi* dh/dt

dh/dt = 40/(2025*Pi)

dh/dt = 8/(405*Pi)

and we should be done. hope this helps