Volume of the sphere:

$V = \frac{4}{3}\pi r^3$

$\dfrac{dV}{dt} = 4\pi r^2 \, \dfrac{dr}{dt}$

$2 = 4\pi r^2 \, \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \dfrac{1}{2\pi r^2}$

When *r* = 8 cm

$\dfrac{dr}{dt} = \dfrac{1}{2\pi (8^2)}$

$\dfrac{dr}{dt} = \dfrac{1}{128\pi}$

Surface area of the sphere:

$A = 4\pi r^2$

$\dfrac{dA}{dt} = 8\pi r \, \dfrac{dr}{dt}$

When *r* = 8 cm

$\dfrac{dA}{dt} = 8\pi(8)\left(\dfrac{1}{128\pi}\right)$

$\dfrac{dA}{dt} = \frac{1}{2} ~ \text{cm}^2\text{/min}$ ← *answer*

For alternate solution, go to node 18010

## Comments

## I believe the answer should

I believe the answer should be negative since the surface area is decreasing