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CIE 9709 2024 November Paper 31 Q10

A Level / CIE / P3

CIE 9709 2024 Nov Paper 31 Paper · Question 10

题目

Problem

A large cylindrical tank is used to store water. The base of the tank is a circle of radius 44 metres. At time tt minutes, the depth of the water in the tank is hh metres. There is a tap at the bottom of the tank. When the tap is open, water flows out of the tank at a rate proportional to the square root of the volume of water in the tank.

(a) Show that dhdt=−λh\frac{\mathrm{d}h}{\mathrm{d}t}=-\lambda\sqrt h, where λ\lambda is a positive constant.

(4)

(b) At time t=0t=0 the tap is opened. It is given that h=4h=4 when t=0t=0 and that h=2.25h=2.25 when t=20t=20.

Solve the differential equation to obtain an expression for tt in terms of hh, and hence find the time taken to empty the tank.

(6)
题目中文翻译

一个大型圆柱形水箱用来储水。水箱的底面是半径为 44 米的圆。在时刻 tt 分钟，水箱内的水深为 hh 米。水箱底部有一个水龙头。当水龙头打开时，水从水箱流出的速率与水箱内水的体积的平方根成正比。

(a) 证明 dhdt=−λh\frac{\mathrm{d}h}{\mathrm{d}t}=-\lambda\sqrt h，其中 λ\lambda 为正常数。

(4)

(b) 在时刻 t=0t=0 打开水龙头。已知当 t=0t=0 时，h=4h=4；当 t=20t=20 时，h=2.25h=2.25。

解此微分方程，求出用 hh 表示 tt 的表达式，并由此求排空水箱所需的时间。

(6)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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先把体积写成底面积乘水深，再用流出为负的体积变化率和链式法则，把比例常数合并。

答题过程

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The volume of water is

V=π(42)h=16πh.V=\pi(4^2)h=16\pi h.

Since water flows out at a rate proportional to V\sqrt V, write

dVdt=−kV,k>0.\frac{\mathrm{d}V}{\mathrm{d}t}=-k\sqrt V,\qquad k>0.

Also,

dVdt=16πdhdt.\frac{\mathrm{d}V}{\mathrm{d}t} =16\pi\frac{\mathrm{d}h}{\mathrm{d}t}.

Hence

dhdt= −k16πh16π= −k4πh.\begin{align*} \frac{\mathrm{d}h}{\mathrm{d}t} =&\,-\frac{k\sqrt{16\pi h}}{16\pi}\\[4mm] =&\,-\frac{k}{4\sqrt\pi}\sqrt h. \end{align*}

Setting λ=k/(4π)>0\lambda=k/(4\sqrt\pi)>0 gives

dhdt=−λh.\boxed{\frac{\mathrm{d}h}{\mathrm{d}t}=-\lambda\sqrt h}.

(b)

解法一

思路

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分离变量后积分，再用两个时间条件确定常数。求空槽时间时取水深到零的极限。

答题过程

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While h>0h>0, separating variables gives

h−1/2 dh=−λ dt.h^{-1/2}\,\mathrm{d}h=-\lambda\,\mathrm{d}t.

Integrating,

2h=−λt+C.2\sqrt h=-\lambda t+C.

Using h=4h=4 at t=0t=0 gives C=4C=4. Using h=2.25h=2.25 at t=20t=20,

2(1.5)=−20λ+4⟹λ=120.2(1.5)=-20\lambda+4 \quad\Longrightarrow\quad \lambda=\frac1{20}.

Thus

2h=4−t20,2\sqrt h=4-\frac t{20},

or

t=80−40h.\boxed{t=80-40\sqrt h}.

As the tank empties, h→0h\to0, giving 80 minutes\boxed{80\text{ minutes}}.