#### Answer

(a) $\frac{dV}{dr}$ represents the rate of change of the volume of the balloon according to its radius.
$\frac{dV}{dt}$ represents the rate of change of the volume of the balloon according to time.
(b) $$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$

#### Work Step by Step

(a) $\frac{dV}{dr}$ represents the rate of change of the volume of the balloon according to its radius. That means, as the balloon reaches radius $r_0$, the volume of the balloon is expanding at a rate of $\frac{dV}{dr_0}(volume/radius)$.
$\frac{dV}{dt}$ represents the rate of change of the volume of the balloon according to time. That means, at time $t_0$, the volume of the balloon is expanding at a rate of $\frac{dV}{dt_0}(volume/time)$.
(b) The weather balloon is spherical. The volume of the balloon with radius $r$ would be $$V=\frac{4}{3}\pi r^3$$
The derivative of $V(t)$ then would be $$\frac{dV}{dt}=\frac{d(\frac{4}{3}\pi r^3)}{dt}$$ $$\frac{dV}{dt}=\frac{4}{3}\pi\frac{d(r^3)}{dt}$$
Here we apply the Chain Rule, $$\frac{dV}{dt}=\frac{4\pi}{3}\frac{d(r^3)}{dr}\frac{dr}{dt}$$ $$\frac{dV}{dt}=\frac{4\pi}{3}\times3r^2\frac{dr}{dt}$$ $$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$