Answer: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.
The height is changing at a rate of 0.1 m/min whine the radius is changing at a rate of 0.133 m/min.
Given that h = 5 m, dV/dt = 14 m³/min,
h = (3/8)d = (3/8)*2r
h = 3r/4; r = 4h/3
a)
[tex]Volume\ of\ cone:\\\\V=\frac{1}{3} \pi r^2h\\\\V=\frac{1}{3} \pi(\frac{4}{3}h )^2h\\\\V=\frac{16}{27} \pi h^3\\\\\frac{dV}{dt}=\frac{16}{9} \pi h^2\frac{dh}{dt} \\\\14=\frac{16}{9} \pi (5)^2 \frac{dh}{dt}\\\\\frac{dh}{dt}=0.1\ m/min[/tex]
b)
[tex]h=\frac{3}{8}d \\\\h=\frac{3}{8}(2r)\\\\h=\frac{3}{4}r\\\\\frac{dh}{dt}=\frac{3}{4} \frac{dr}{dt} \\\\\frac{dr}{dt}=\frac{4}{3} \frac{dh}{dt} \\\\\frac{dr}{dt}=\frac{4}{3} (0.1)\\\\\frac{dr}{dt}=0.133\ m/min \\[/tex]
Hence the height is changing at a rate of 0.1 m/min whine the radius is changing at a rate of 0.133 m/min.
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