Let the radius of the conical pile be r and height be h, then,

`2r = 3h`

`h = 2/3r` this gives, `(dh)/(dt) = 2/3(dr)/(dt)`

Assuming that the conical pile retains its shape through out whole time and also its tip does the same. So the volume increase is on...

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Let the radius of the conical pile be r and height be h, then,

`2r = 3h`

`h = 2/3r` this gives, `(dh)/(dt) = 2/3(dr)/(dt)`

Assuming that the conical pile retains its shape through out whole time and also its tip does the same. So the volume increase is on the conical side of the pile. Also I assume taht sand is evenly distributed around the cone at everypoint on the conical surface.

Volume of the cone is = V

`V = 1/3pir^2h`

now differentiating wrt t,

`(dV)/(dt) = pi/3 (d(r^2h))/(dt)`

`(dV)/(dt) = pi/3(2rh(dr)/(dt)+r^2(dh)/(dt))`

since `(dh)/(dt) = 2/3(dr)/(dt)` ,

`(dV)/(dt) = pi/3(2r*2/3r*(dr)/(dt)+r^2*2/3(dr)/(dt))`

`(dV)/(dt) = (2pi)/9(2r^2(dr)/(dt)+r^2(dr)/(dt))`

`(dV)/(dt) = (2pi)/9*3r^2(dr)/(dt)`

`(dV)/(dt) = (2pir^2)/3(dr)/(dt)`

Now at h =22 ft, `r = 3/2h`

r = 33 ft.

`(dV)/(dt) = 12 ft^3min^-1`

12 = (2*pi*33^2)/3*(dr)/(dt)

`(dr)/(dt) = 5.261 * 10^(-3) ftmin^(-1)`

or we can convert this to per hour.

`(dr)/(dt) = 0.3157 ` ft per hour

Therefore,

`(dh)/(dt) = 2/3*0.3157` ft per hour

`(dh)/(dt) = 0.2104` ft per hour.

The rate of height increase is 0.2104 ft per hour.