上图显示了一个带有封闭圆形底面的直圆锥。它的底面圆半径为 $4\text{ cm}$，高为 $12\text{ cm}$，最初装满了水。水正从圆锥的底面以 $2\pi\text{ cm}^3\text{ min}^{-1}$ 的速率泄漏。如果 $h$ 是在时间 $t$ 分钟时圆锥中水的深度，证明在时间 $t$ 分钟时圆锥中剩余水的体积 $V$ 由 $V = 64\pi - \frac{\pi}{27}(12-h)^3$ 给出。 由此求当水深为 $6\text{ cm}$ 时，水深的变化率。

1. Calculate the total volume of the cone

V_{\text{total}} = \frac{1}{3}\pi r^2 H = \frac{1}{3}\pi (4)^2 (12) = 64\pi

2. Find the dimensions and volume of the empty space at the top of the cone

\text{The empty space is a cone of height } (12-h).
\text{Let its base radius be } x.
\text{By similar triangles, } \frac{x}{4} = \frac{12-h}{12} \implies x = \frac{12-h}{3}.

V_{\text{empty}} = \frac{1}{3}\pi x^2 (12-h) = \frac{1}{3}\pi \left(\frac{12-h}{3}\right)^2 (12-h) = \frac{\pi}{27}(12-h)^3

3. Formulate the expression for the volume of water remaining

V = V_{\text{total}} - V_{\text{empty}}
V = 64\pi - \frac{\pi}{27}(12-h)^3 \text{ (shown)}

4. Differentiate the volume $V$ with respect to the depth $h$

\frac{\text{d}V}{\text{d}h} = -\frac{\pi}{27} \times 3(12-h)^2 \times (-1) = \frac{\pi}{9}(12-h)^2

5. Apply the chain rule to relate the rates of change

\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}h} \times \frac{\text{d}h}{\text{d}t}
\text{Since water is leaking, } \frac{\text{d}V}{\text{d}t} = -2\pi.
\text{Substitute } h = 6:
-2\pi = \frac{\pi}{9}(12-6)^2 \times \frac{\text{d}h}{\text{d}t}

6. Solve for the rate of change of depth

-2\pi = \frac{\pi}{9}(36) \times \frac{\text{d}h}{\text{d}t}
-2\pi = 4\pi \frac{\text{d}h}{\text{d}t}
\frac{\text{d}h}{\text{d}t} = -0.5