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A spherical balloon is being inflated. At a certain instant, the rate of increase of its volume is 16 times the rate of increase of its radius. What is the radius of the balloon at that instant?

1. 1/sqrt(pi)
2. 2/sqrt(pi)
3. 2/pi
4. 4/(3*sqrt(pi))

Correct answer: 2/sqrt(pi)

Solution

V = (4/3)*pi*r³. dV/dt = 4*pi*r² * dr/dt. Given dV/dt = 16 * dr/dt. So 4*pi*r² * dr/dt = 16 * dr/dt. Dividing by dr/dt (non-zero): 4*pi*r² = 16 => r² = 4/pi => r = 2/sqrt(pi).

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