The rate at which the spherical balloon's radius is increasing is
3/16π ft/min when the volume is increasing at the rate of 3 ft³/min.
Given, a spherical balloon is inflated so that its volume is increasing at the rate of 3 cubic feet per minute.
we have to find that how fast is the radius of the balloon increasing when the radius is 2 feet.
as, we know the volume of the balloon be,
V = 4/3πr³
now we have to find the rate of change of radius,
On differentiating both the sides with respect to time, we get
V' = 4/3π(3r²)dr/dt
V' = 4πr²dr/dt
as, the rate of change of volume of the balloon be, 3 ft³/min
V' = 3
3 = 4πr²dr/dt
dr/dt = 3/4πr²
Now, on putting r = 2, we get
dr/dt = 3/4π(2)²
dr/dt = 3/16π
so, the radius of the balloon increase at the rate of 3/16π.
Hence, the radius of the balloon increase at the rate of 3/16π.
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