Question

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 4 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 22 feet high?

Answer 1

`h^(\prime)=(4)/(1089\pi)ft\text{ /min}`

STEP - BY - STEP EXPLANATION

What to find?

dh/dt

Given that;

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 4 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude.

Since we have that the rate is 4 cubic per minute, then dv/dt = 4 (since v is the volume of the cone at time t.)

The formula for the volume of a cone is

`V=(1)/(3)\pi r^2h---------------(1)`

Where r is the radius and h is the height.

We have that, the diameter of the cone is approximately three times the altitude.

That is;

diameter = 3h

But, d = 2r

⇒2r = 3h

⇒ r = 3h/2

Now, we substitute r=3h/2 into equation (1).

`V=(1)/(3)\pi((3h)/(2))^2h`
`=(1)/(3)\pi((9h^2)/(4))h`
`V=(3)/(4)\pi h^3`

Now, differentiate the above with respect to t.

`(dv)/(dt)=(3)/(4)\pi3h^2(dh)/(dt)`

Simplify .

`(dv)/(dt)=(9)/(4)\pi h^2(dh)/(dt)`

Make dh/dt subject of formula.

`(dh)/(dt)=(dv)/(dt)*(4)/(9\pi h^2)----------(2)`

Recall that, dv/dt = 4 and h=22

Substituting the values into equation (2), we have;

`(dh)/(dt)=4*(4)/(9\pi(22)^2)`
`=\frac{16}{9\pi\text{ (484)}}`
`=\frac{4}{9\pi\text{ (121)}}`
`=(4)/(1089\pi)`

Therefore, h' = dh/dt = 4/1089π ft/min.