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P4 2024 Oct Q5

A Level / Edexcel / P4

IAL 2024 Oct Paper · Question 5

题目

Problem

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

The height of the container is 3030 cm and the radius is 1212 cm, as shown in Figure 3.

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

When the height of water is hh cm, the surface of the water has radius rr cm and the volume of water is V cm3V\text{ cm}^3.

(a) Show that

V=4πh375V=\frac{4\pi h^3}{75}

[The volume VV of a right circular cone with vertical height hh and base radius rr is given by the formula V=13πr2hV=\frac13\pi r^2h]

(2)

Given that water flows into the container at a constant rate of 2π cm3 s−12\pi\text{ cm}^3\text{ s}^{-1},

(b) find, in cm s−1\text{cm s}^{-1}, the rate at which hh is changing, exactly 1.51.5 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 3030 cm，半径 1212 cm。

开始向容器中注水时，容器最初为空。

当水深为 hh cm 时，水面的半径为 rr cm，水的体积为 V cm3V\text{ cm}^3。

(a) 证明

V=4πh375V=\frac{4\pi h^3}{75}

[竖直高为 hh、底面半径为 rr 的直圆锥体积公式为 V=13πr2hV=\frac13\pi r^2h]

(2)

已知流入容器的水的体积速率恒为 2π cm3 s−12\pi\text{ cm}^3\text{ s}^{-1}，

(b) 求开始注水恰好 1.51.5 分钟后，hh 的变化速率，单位为 cm s−1\text{cm s}^{-1}。

(4)

解答

(a)

解法一

思路

展开

利用相似圆锥的半径与高度比例，将 rr 用 hh 表示，再代入圆锥体积公式。

答题过程

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The cone has total height 3030 cm and total radius 1212 cm.

By similarity,

rh=1230\frac{r}{h}=\frac{12}{30}

So

r=25hr=\frac{2}{5}h

The volume of water is the volume of a cone with height hh and radius rr:

V=13πr2hV=\frac13\pi r^2h

Substitute r=25hr=\dfrac25h:

V= 13π(25h)2h= 13π⋅425h3= 4πh375\begin{align*} V =&\,\frac13\pi\big(\frac25h\big)^2h \\ =&\,\frac13\pi\cdot\frac4{25}h^3 \\ =&\,\frac{4\pi h^3}{75} \end{align*}

Hence

V=4πh375\boxed{V=\frac{4\pi h^3}{75}}

(b)

解法一

思路

展开

先把分钟换成秒，由进水量求体积和水深，再用体积公式的导数与链式法则求水深变化率。

答题过程

展开

The time after 1.51.5 minutes is t=90t=90 seconds. Since the container was initially empty and dV/dt=2π\mathrm{d}V/\mathrm{d}t=2\pi,

V=2π(90)=180π.V=2\pi(90)=180\pi.

Using part (a),

180π=4πh375.180\pi=\frac{4\pi h^3}{75}.

Thus h3=3375h^3=3375 and h=15h=15 cm.

Differentiate the volume formula:

dVdh=12πh275.\frac{\mathrm{d}V}{\mathrm{d}h} =\frac{12\pi h^2}{75}.

The chain rule gives

dVdt=dVdhdhdt.\frac{\mathrm{d}V}{\mathrm{d}t} =\frac{\mathrm{d}V}{\mathrm{d}h} \frac{\mathrm{d}h}{\mathrm{d}t}.

At h=15h=15, dV/dh=36π\mathrm{d}V/\mathrm{d}h=36\pi. Therefore

2π=36πdhdt,2\pi=36\pi\frac{\mathrm{d}h}{\mathrm{d}t},

so the rate is

dhdt=118 cm s−1.\boxed{\frac{\mathrm{d}h}{\mathrm{d}t} =\frac1{18}\text{ cm s}^{-1}}.

解法二

思路

展开

把体积写成时间的函数，与 (a) 的体积公式联立，直接求出 h(t)h(t) 后求导。

答题过程

展开

Let tt be the time in seconds. The initial volume is zero, so

V=2πt.V=2\pi t.

Equate this with the result of part (a):

2πt=4πh375.2\pi t=\frac{4\pi h^3}{75}.

Hence

h3=75t2,h=(75t2)1/3.h^3=\frac{75t}{2}, \qquad h=\bigg(\frac{75t}{2}\bigg)^{1/3}.

Differentiate this expression:

dhdt=252(75t2)−2/3.\frac{\mathrm{d}h}{\mathrm{d}t} =\frac{25}{2}\bigg(\frac{75t}{2}\bigg)^{-2/3}.

At t=90t=90, 75t/2=3375=15375t/2=3375=15^3, giving

dhdt= 252×15−2= 118.\begin{align*} \frac{\mathrm{d}h}{\mathrm{d}t} =&\,\frac{25}{2}\times15^{-2}\\ =&\,\frac1{18}. \end{align*}

Therefore the required rate is

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

也可以先从链式法得到微分方程，再分离变量：

6h2dhdt=75.6h^2\frac{\mathrm{d}h}{\mathrm{d}t}=75.

积分得

2h3=75t+C.2h^3=75t+C.

由空容器的初始条件 t=0, h=0t=0,\ h=0 得 C=0C=0，于是回到同一关系 h3=75t/2h^3=75t/2，后续求导和代入步骤相同。