Sand is being dumped from a conveyor belt and forms a conical pile. Assuming that the height of this cone is always exactly 3 times the size of the radius of its base, and that thesand is added at the rate of 10 m^3/min, how fast is the height increasing when the pile is15 m high?

dh/dt =  0.4 m/min

Step-by-step explanation:

The volume of the cone is:

V(c) = (1/3)*r² *h              if always  h = 3r    then    r = h/3

The volume of the cone as a function of h will be:

V(h)  =  (1/3)* (h/3)²*h

V(h) =  (1/27)*h³

The increasing rate of the volume is equal to the rate of sand added the:

D(V)/dt   = (1/27)*3*h²*dh/dt

D(v) / dt  =  10 m³/min

h =  15 m      and   dh/dt is the rate of increasing of the height

By substitution

10 m³/min  = ( 1/9)* 225 * dh/dt  (m²)

dh/dt =  90 / 225   m/min

dh/dt =  0.4 m/min

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