空气正以每秒 12 cm$^3$ 的恒定速率被注入一个球形气球中。
求当气球半径为 4 cm 时，气球半径增加的速率。 [半径为 $r$ 的球体体积为 $\frac{4}{3}\pi r^3$] 当气球半径膨胀超过 5 cm 时，人们发现，在注入空气的同时，部分空气也开始从气球中泄漏。 已知当半径为 7 cm 时，气球半径正以每秒 0.015 cm 的速率增加，求空气从气球中泄漏的速率。

1. Differentiate the volume of the sphere with respect to its radius

Given $V = \frac{4}{3}\pi r^3$,
$\frac{dV}{dr} = 4\pi r^2$

2. Set up the chain rule for connected rates of change

By the chain rule,
$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$

3. Substitute the given values to find the rate of change of the radius

We are given $\frac{dV}{dt} = 12$ and $r = 4$.
$12 = 4\pi (4)^2 \times \frac{dr}{dt}$
$12 = 64\pi \times \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{12}{64\pi} = \frac{3}{16\pi} \approx 0.0597 \text{ cm/s}$

1. Find the derivative of the volume with respect to the radius

$V = \frac{4}{3}\pi r^3$
$\frac{dV}{dr} = 4\pi r^2$

2. Calculate the net rate of change of volume when $r = 7$

Using the chain rule: $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$
When $r = 7$, $\frac{dV}{dr} = 4\pi (7)^2 = 196\pi$
Given $\frac{dr}{dt} = 0.015$
Net $\frac{dV}{dt} = 196\pi \times 0.015 = 2.94\pi \text{ cm}^3/\text{s}$

3. Determine the rate at which air is leaking out

Net rate of volume change = Rate pumped in - Rate leaking out
$2.94\pi = 12 - \text{Rate leaking out}$
$\text{Rate leaking out} = 12 - 2.94\pi \approx 2.7637 \text{ cm}^3/\text{s}$