The height of the water decreases at a rate of 0.012 ft/sec if a cylindrical tank of a radius of 2.5 feet is being drained of water at a rate of 0.25 ft³/sec.
In geometry, it is defined as the three-dimensional shape having two circular shapes at a distance called the height of the cylinder.
We know the volume of the cylinder is given by:
[tex]\rm V = \pi r^2 h[/tex]
It is given that:
A cylindrical tank of a radius of 2.5 feet is being drained of water at a rate of 0.25 ft³/sec
As we know,
The volume of the cylinder is given by:
V = πr²h
Let V is the water draining rate
V = 0.25 ft³/sec
The radius r = 2.5 feet
0.25 ft³/sec = π(2.5 feet)²(h)
h = [0.25/(2.5²π)] [ft³/(sec×feet²]
h = 0.012 ft/sec
Thus, the height of the water decreases at a rate of 0.012 ft/sec if a cylindrical tank of a radius of 2.5 feet is being drained of water at a rate of 0.25 ft³/sec.
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