The height of the pile increasing at the rate of 0.0039 feet/minute when the base radius is 6 feet.
Rate of sand dropping dv/dt = 12 cubic feet/minute
dv/dt = 12 cubic feet/ min
A conical pile whose altitude is 2/3 of the radius of the base,
h= (2/3) r
or r = (3/2)h
Volume of a cone = (1/3) π [tex]r^{2} h[/tex]
v = (1/3) π × [tex](3h/2)^{2}[/tex] × h
v = (3/4) π × [tex]h^{3}[/tex]
Differentiating both sides w.r.t. t ,
dv/dt =(9/4) π × h^2 dh/dt
dh/dt = ( 4 / (9(π × h^2)) ) dv/dt
dh/dt at h= 6 feet
dh/dt = {4/(9(π × 36))}(10)
dh/dt = {1/(9 π × 9)}
dh/dt = 1/{254.34}
dh/dt = 0.0039 feet/minute
Thus, we can conclude that the height of the pile increasing at the rate of 0.0039 feet/minute.
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