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
It is not necessary to…
In reply to I believe the answer should by Anonymous (not verified)
It is not necessary to indicate a negative to the rate of change because the problem already said that it is decreasing.
However, the negative is necessary if the problem asks for "How fast is the surface area is changing". For this statement, it is important to determine if the rate of change is decreasing or increasing or perhaps it is an integral part of the problem. The negative sign is necessary for such a case, or an absolute value with word decreasing or increasing will also do.