Answer
a) $\frac{dV}{dr}$ is the instantaneous rate of change of the volume at a specific radius $r$.
$\frac{dV}{dt}$ is the instantaneous rate of change of the volume at a specific time $t$.
b) $\frac{dV}{dt}=4\pi r^{2}(t)\cdot \frac{dr}{dt}$
Work Step by Step
a) The derivative $\frac{dV}{dr}$ is the instantaneous rate of change of the volume at a specific radius $r$.
The derivative $\frac{dV}{dt}$ is the instantaneous rate of change of the volume at a specific time $t$.
$$\frac{dV}{dt}=4\pi r^{2}(t)\cdot \frac{dr}{dt}$$
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b) The volume of the ballon is:
$$V=\frac{4}{3}\cdot \pi \cdot r^{3}(t)$$
Using the chain rule it follows:
$$\frac{dV}{dt}=\frac{dV}{dr}\cdot \frac{dr}{dt}$$
$$V=\frac{4}{3}\cdot \pi \cdot r^{3}(t) \to \frac{dV}{dr}=\frac{4}{3}\cdot \pi \cdot 3r^{2}(t)$$
$$\frac{dV}{dt}=\frac{4}{3}\cdot \pi \cdot 3r^{2}(t)\cdot \frac{dr}{dt}$$
$$\frac{dV}{dt}=4\pi r^{2}(t)\cdot \frac{dr}{dt}$$
