A right cylindrical water tank with a diameter of 3 feet and a height of 6 feet is being drained. At what rate is the volume of the water in the tank changing when the water level of the tank is dropping at a rate of 4 inches per minute?
Anonymous:
___k off
Answers
Answered by
0
Answer:
Step-by-step explanation:
Decreasing @ 2.356 cft/min, nearly, at the instant when the descent of surface is @ 4"/min.
Explanation:
Volume at time t'
V
(
t
)
=
π
r
2
h
(
t
)
=
π
(
3
2
)
2
h
(
t
)
cubic feet, where h is the height in feet,
at time t'.
Differentiating with respect to time,
V'=9/4ph'=9/4pi(--4/12)=-3/4pi cubic feet//minute.
When h'=-4''/minute=-4/12=-1/3'/minute,
V
'
=
3/4pi=-2.356# cft/min, nearly.
Note that there is no need to use use of initial surface height 6'.
The nozzle draining speed is governed by Bernoulli'equation.
Similar questions