Math, asked by BMsingpho9526, 11 months ago

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 finusajid
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.

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