For basic solution, see

node 2360.

**Solution by Chain Rule**

$V = \frac{4}{3}\pi r^3$
$r = \left( \dfrac{3V}{4\pi} \right)^{1/3}$

when *r* = 12 ft

$V = \frac{4}{3}\pi (12^3)$

$V = 2304 \pi ~ \text{ft}^3$

$A = 4\pi r^2$

$A = 4\pi \left( \dfrac{3V}{4\pi} \right)^{2/3}$

$\dfrac{dA}{dt} = 4\pi \cdot \dfrac{d}{dt} \left[ \left( \dfrac{3V}{4\pi} \right)^{2/3} \right]_{V = 2304 \pi} \cdot \dfrac{dV}{dt}$

Using the derivative function of calculator

$\dfrac{dA}{dt} = 4\pi \cdot (0.0133) \cdot 2$

$\dfrac{dA}{dt} = 0.3333 ~ \text{ft}^2/\text{min}$ ← *answer*

**Solution by Volume Flow Rate of Hydraulics**

$V = \frac{4}{3}\pi r^3$
$V = \frac{1}{3}(4\pi r^2)r$

$V = \frac{1}{3}Ar$

$\dfrac{dV}{dt} = \dfrac{1}{3}\left( A \dfrac{dr}{dt} + r \dfrac{dA}{dt} \right)$

Note that *dr*/*dt* is a velocity, and from Fluid Mechanics, *Av* = *Q* :

$Q = \dfrac{1}{3}\left( Q + r \dfrac{dA}{dt} \right)$

When *r* = 12 ft

$2 = \dfrac{1}{3}\left( 2 + 12 \cdot \dfrac{dA}{dt} \right)$

$\dfrac{dA}{dt} = \dfrac{2(3) - 2}{12}$

$\dfrac{dA}{dt} = 0.3333 ~ \text{ft}^2\text{/min}$ ← *answer*