Air is discharging from a large spherical ballon at the rate of 4 cubic meters per minute. Then the rate at which the surface area is shrinking when the radius of the balloon is 8 meters is

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$4 m^{2} \min$

$8 m^{2} \min$

$1 m^{2} \min$

$2 m^{2} \min$

answer is B.

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Detailed Solution

$\frac{dv}{dt} = 4 r = 8 V = \frac{4}{3} {πr}^{3} \Rightarrow \frac{dv}{dt} = 4 {πr}^{2} \frac{dr}{dt} \Rightarrow 4 = 4 π \times 64 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{64 π} S = 4 {πr}^{2} \Rightarrow \frac{ds}{dt} = 8 πr \frac{dr}{dt} = 8 π \times 8 \times \frac{1}{64 π} = 1$

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