The surface area of a sphere is given by the formula 4πr^2, where r is the radius of the sphere. This means that the surface area of the balloon is increasing at the rate of 8πr cm^2/sec.
The volume of a sphere is given by the formula 4/3πr^3, where r is the radius of the sphere. If the radius of the balloon is 6 cm, then the volume of the balloon is (4/3)π(6^3) = (4/3)π216 = 288π cm^3.
To find the rate at which the volume of the balloon is increasing, we need to take the derivative of the formula for the volume of a sphere with respect to time. This gives us (4/3)π(3r^2)(dr/dt).
Since the radius of the balloon is 6 cm and the surface area of the balloon is increasing at the rate of 8πr cm^2/sec, the rate at which the radius of the balloon is increasing is (8πr/4πr^2) = 2/r = 2/6 = 1/3 cm/sec.
Plugging this into the formula for the rate of change of the volume of the sphere, we get
Therefore, the volume of the balloon is increasing at a rate of 144π cm^3/sec when the radius of the balloon is 6 cm.
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Complete Question:
The surface area of a spherical balloon is increasing at the rate of 2 cm 2/sec. At what rate is the volume of the balloon is increasing when the radius of the balloon is 6 cm?
