[半径为 $r$ 的球体体积由 $\frac{4}{3}\pi r^3$ 给出，半径为 $r$ 的球体表面积由 $4\pi r^2$ 给出。]

1. Differentiate the given volume and surface area equations with respect to the radius.

$V = \frac{4}{3}\pi r^3 \implies \frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2$
$A = 4\pi r^2 \implies \frac{\mathrm{d}A}{\mathrm{d}r} = 8\pi r$

2. Use the chain rule to relate the rate of change of volume to the rate of change of radius.

By the chain rule:
$\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t}$
Substitute the known values ($\frac{\mathrm{d}V}{\mathrm{d}t} = 5$):
$5 = 4\pi r^2 \frac{\mathrm{d}r}{\mathrm{d}t}$
$\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{5}{4\pi r^2}$

3. Use the chain rule to find the rate of change of surface area and determine $k$.

Apply the chain rule for surface area:
$\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{d}A}{\mathrm{d}r} \times \frac{\mathrm{d}r}{\mathrm{d}t}$
Substitute $\frac{\mathrm{d}A}{\mathrm{d}r}$ and $\frac{\mathrm{d}r}{\mathrm{d}t}$:
$\frac{\mathrm{d}A}{\mathrm{d}t} = 8\pi r \left( \frac{5}{4\pi r^2} \right)$
$\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{40\pi r}{4\pi r^2} = \frac{10}{r}$
Comparing this result with $\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{k}{r}$, we obtain $k = 10$.

1. Determine the value of $k$ from part (a) to find the explicit formula for $\frac{dA}{dt}$.

Given $\frac{dV}{dt} = 5$.
Using $V = \frac{4}{3}\pi r^3$, $\frac{dV}{dr} = 4\pi r^2$.
Using $A = 4\pi r^2$, $\frac{dA}{dr} = 8\pi r$.
By the chain rule:
$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = \frac{dA}{dr} \times \frac{\frac{dV}{dt}}{\frac{dV}{dr}}$
$\frac{dA}{dt} = 8\pi r \left( \frac{5}{4\pi r^2} \right) = \frac{10}{r}$

2. Find the exact radius $r$ when the surface area is $20\text{ cm}^2$.

$A = 4\pi r^2 = 20$
$r^2 = \frac{5}{\pi}$
$r = \sqrt{\frac{5}{\pi}}$ (since $r > 0$)

3. Calculate the exact rate of change of surface area.

Substitute $r = \sqrt{\frac{5}{\pi}}$ into $\frac{dA}{dt}$:
$\frac{dA}{dt} = \frac{10}{\sqrt{\frac{5}{\pi}}}$
$\frac{dA}{dt} = 10\sqrt{\frac{\pi}{5}} = 2\sqrt{5\pi}$