Sand falls from a conveyor belt at a rate of 14 m cubed divided by min onto the top of a conical pile. the height of the pile is always three-eighths of the base diameter. how fast are the height and the radius changing when the pile is 5 m high?

7.16 cm/min or 0.0716 min.

Explanation:

From the equation the parameters given are; rate= 14m^-3/min, height of the pile,h= 3/8, rate when the pile is 5m high= ??.

d= diameter, r= radius. From the definition of diameter, diameter, d=2× radius, r(that is, 2r)

h= 3/8×d = 3/8(2r)= 3/4 × r

r= 4/3 × h

V= 1/3×πr^2×h (1).

V= 1/3×π×[4/3h]^2×h

V= 16/27×π×h^3

Differentiate dV with respect to time, t;

dV/dt= 3(16)/27 × π ×h^2 ×dh/dt.

dV/dt = 16/9×π×h^2 × dh/dt

Then we differentiate V implicitly with respect to time,t

14=16/9×π[5]^2×dh/dt

dh/dt = 10× 9/400π

dh/dt= 45/200π (m/min)

dh/dt= 0.0716 m/min.

Conversion to cm/min

= dh/dt= 4500/200π (cm/min)

dh/dt= 1125/50π

dh/dt= 7.16 cm/min.

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