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(A) Given the rate of increase in the volume of the sphere is $\frac{dV}{dt} = \pi 	ext{ cm}^3/	ext{s}$.We know that the volume of a sphere is $V =

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$2 \pi 	ext{ cm}^2/	ext{s}$

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(A) Given the rate of increase in the volume of the sphere is $\frac{dV}{dt} = \pi 	ext{ cm}^3/	ext{s}$.We know that the volume of a sphere is $V = \frac{4}{3} \pi r^3$.Differentiating with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Substituting $\frac{dV}{dt} = \pi$,we have $\pi = 4 \pi r^2 \frac{dr}{dt}$,which implies $\frac{dr}{dt} = \frac{1}{4r^2}$.The surface area of a sphere is $S = 4 \pi r^2$.Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.Substituting $\frac{dr}{dt} = \frac{1}{4r^2}$,we get $\frac{dS}{dt} = 8 \pi r \left( \frac{1}{4r^2} ight) = \frac{2 \pi}{r}$.When $r = 1 	ext{ cm}$,the rate of increase in surface area is $\frac{dS}{dt} = \frac{2 \pi}{1} = 2 \pi 	ext{ cm}^2/	ext{s}$.

$A$ sphere increases its volume at the rate of $\pi \text{ cm}^3/\text{s}$. The rate at which its surface area increases when the radius is $1 \text{ cm}$ is

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