The rate in the height of water changing for the given cylinder is equal to 0.318m/s.
As given in the question,
Change in volume represent the decrease in rate of change of water height in the cylinder in time t
dV/dt = 4m³/sec.
Decrease in volume represent the change in the height of the water
Let 'h' be the height
h = 10m
radius 'r' = 2m
V = πr²h
Rate of change of leaking water only change the height radius remain constant.
⇒V = (3.14)(2²) h
Differentiate both the side with respect to 't' to get the rate of change
dV/dt = (3.14)(4)dh/dt
⇒4 =(3.14)(4)dh/dt
⇒dh/dt = 4/ (3.14)(4)
⇒dh/dt = 0.318 m/s
Therefore, the rate of change of height is equal to 0.318m/s.
The complete question is:
A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate is the height of the water changing when the height is 3m?
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