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.

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )