题目

Problem

Figure 3 shows a container in the shape of a hollow, inverted, right circular cone.

The height of the container is $30$ cm and the radius is $12$ cm, as shown in Figure 3.

The container is initially empty when water starts flowing into it.

When the height of water is $h$ cm, the surface of the water has radius $r$ cm and the volume of water is $V\text{ cm}^3$ .

(a) Show that

V=\frac{4\pi h^3}{75}

\left[\text{The volume }V\text{ of a right circular cone with vertical height }h\text{ and base radius }r\text{ is given by the formula }V=\frac{1}{3}\pi r^2h\right]

(2)

Given that water flows into the container at a constant rate of $2\pi\text{ cm}^3\text{ s}^{-1}$ ,

(b) find, in $\text{cm s}^{-1}$ , the rate at which $h$ is changing, exactly $1.5$ minutes after water starts flowing into the container.

(4)

题目中文翻译

图 3 所示容器是一个空心倒置直圆锥。

如图 3 所示，容器高 $30$ cm，半径 $12$ cm。

开始向容器中注水时，容器最初为空。

当水深为 $h$ cm 时，水面的半径为 $r$ cm，水的体积为 $V\text{ cm}^3$ 。

(a) 证明

V=\frac{4\pi h^3}{75}

\left[\text{竖直高为 }h\text{、底面半径为 }r\text{ 的直圆锥体积公式为 }V=\frac{1}{3}\pi r^2h\right]

已知流入容器的水的体积速率恒为 $2\pi\text{ cm}^3\text{ s}^{-1}$ ，

(b) 求开始注水恰好 $1.5$ 分钟后， $h$ 的变化速率，单位为 $\text{cm s}^{-1}$ 。

解答

The cone has total height $30$ cm and total radius $12$ cm.

By similarity,

\frac{r}{h}=\frac{12}{30}

r=\frac{2}{5}h

The volume of water is the volume of a cone with height $h$ and radius $r$ :

V=\frac13\pi r^2h

Substitute $r=\dfrac25h$ :

\begin{aligned} V &=\frac13\pi\left(\frac25h\right)^2h \\ &=\frac13\pi\cdot\frac4{25}h^3 \\ &=\frac{4\pi h^3}{75} \end{aligned}

Hence

\boxed{V=\frac{4\pi h^3}{75}}

The water flows in at a constant rate

\frac{dV}{dt}=2\pi

From part (a),

V=\frac{4\pi h^3}{75}

Differentiate with respect to $h$ :

\frac{dV}{dh}=\frac{12\pi h^2}{75}

After $1.5$ minutes, the time is

1.5\times60=90\text{ seconds}

So the volume of water is

V=2\pi(90)=180\pi

Use

V=\frac{4\pi h^3}{75}

Then

180\pi=\frac{4\pi h^3}{75}

Cancel $\pi$ :

180=\frac{4h^3}{75}

Thus

h^3=3375

h=15

Using the chain rule,

\frac{dV}{dt} =\frac{dV}{dh}\cdot\frac{dh}{dt}

At $h=15$ ,

2\pi =\frac{12\pi(15)^2}{75}\frac{dh}{dt}

Therefore

\frac{dh}{dt} =\frac{2\pi\cdot75}{12\pi(15)^2}

\frac{dh}{dt} =\frac{150}{2700} =\frac1{18}

Hence the rate at which $h$ is changing is

\boxed{\frac1{18}\text{ cm s}^{-1}}