The volume of gasoline in the cylindrical tank is increasing at 23.56 ft.³/sec when the depth of the gasoline in the tank is 6 feet. Computed using differentiation.
Since the tank is cylindrical in shape, its volume can be written as:
V = πr²d,
where V is its volume, r is the radius, and d is the depth.
The radius is constant, given r = 5ft.
Thus the volume can be shown as:
V = π(5)²d,
or, V = 25πd.
Differentiating this with respect to time, we get:
δV/δt = 25πδd/δt ... (i),
where δV/δt, represents the rate of change of volume with respect to time, and δd/δt represents the rate of change of depth with respect to time.
Now, we are given that when the depth increases at 0.3 ft./sec when the depth of the gasoline is 6 feet.
Thus, we can take δd/δt = 0.3 ft./sec, in (i) to get:
δV/δt = 25πδd/δt = 25π(0.3) ft.³/sec = 23.56 ft.³/sec.
Thus, the volume of gasoline in the cylindrical tank is increasing at 23.56 ft.³/sec when the depth of the gasoline in the tank is 6 feet. Computed using differentiation.
The question written correctly is:
"Gasoline is pouring into a vertical cylindrical tank of radius 5 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.3 ft./sec. How fast is the volume of gasoline changing at that instant?"
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