一个高为 30 cm 的液体容器可以建模为一个圆台 A，它放置在圆柱体 B 的下方，如图 1 所示。 图 2 展示了圆台 A 作为其一部分的圆锥示意图。圆台顶部的直径 QRS 为 12 cm，底部的直径 VUT 为 8 cm。圆台的高度 RU 为 9 cm。R 和 U 分别是 QS 和 VT 的中点。 将液体以每秒 $q$ cm^3 的恒定速率倒入容器中。完全填满 A 需要 50 秒，另外还需要 30 秒才能完全填满 B。

1. Find the height of the smaller cone below the frustum using similar triangles.

Let the height of the smaller cone be $x$ cm.

The radii of the bottom and top of the frustum are
$\frac{8}{2}=4\text{ cm},\qquad \frac{12}{2}=6\text{ cm}.$

By similarity,
$\frac{x}{x+9}=\frac{4}{6}.$
$6x=4x+36$
$2x=36$
$x=18.$

2. Calculate the volume of the large cone and the smaller cone.

Height of the large cone
$=18+9=27\text{ cm}.$

Volume of large cone:
$\frac{1}{3}\pi(6)^2(27)=324\pi.$

Volume of smaller cone:
$\frac{1}{3}\pi(4)^2(18)=96\pi.$

3. Subtract the smaller cone from the larger cone to get the frustum volume.

Volume of frustum A
$=324\pi-96\pi$
$=228\pi$
$=716.283\ldots$

So the volume is approximately
$716\text{ cm}^3.$

1. Mark the key points where each section is filled.

The frustum $A$ has height $9\text{ cm}$ and is filled in $50\text{ s}$.

So the graph passes through
$(0,0) \quad \text{and} \quad (50,9).$

The whole container is $30\text{ cm}$ tall and cylinder $B$ takes another $30\text{ s}$ to fill.

So the graph also passes through
$(80,30).$

2. Decide the shape of the graph in the frustum and cylinder sections.

In the frustum, the diameter increases from $8\text{ cm}$ to $12\text{ cm}$ as the liquid rises, so the cross-sectional area increases. Since liquid is poured in at a constant rate, the height rises more slowly as time increases.

Therefore, from $(0,0)$ to $(50,9)$, draw an increasing curve that becomes less steep.

In the cylinder, the cross-sectional area is constant, so the height rises at a constant rate.

Therefore, from $(50,9)$ to $(80,30)$, draw a straight line.