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