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CIE 9709 2022 November Paper 13 Q4

A Level / CIE / P1

CIE 9709 2022 Nov Paper 13 Paper · Question 4

题目

Problem

A large industrial water tank is such that, when the depth of the water in the tank is xx metres, the volume V m3V\text{ m}^3 of water in the tank is given by V=243−13(9−x)3V=243-\dfrac{1}{3}(9-x)^3. Water is being pumped into the tank at a constant rate of 3.6 m33.6\text{ m}^3 per hour.

Find the rate of increase of the depth of the water when the depth is 4 m4\text{ m}, giving your answer in cm per minute.

(5)
题目中文翻译

一个大型工业水箱中,当水深为 xx 米时,水的体积 V m3V\text{ m}^3 满足 V=243−13(9−x)3V=243-\dfrac{1}{3}(9-x)^3。水以每小时 3.6 m33.6\text{ m}^3 的恒定速率泵入水箱。

求水深为 4 m4\text{ m} 时水深增加的速率,答案以厘米每分钟表示。

(5)
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

解法一

思路

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用链式法则把注水速率换算成水深变化率,最后把米每小时换成厘米每分钟。

答题过程

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Differentiating gives dVdx=(9−x)2\frac{\mathrm{d}V}{\mathrm{d}x}=(9-x)^2. At x=4x=4, the chain rule gives

dxdt=3.6(9−4)2=0.144 m h−1.\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{3.6}{(9-4)^2}=0.144\,\mathrm{m}\,\mathrm{h}^{-1}.

Multiplying by 100/60100/60 gives 0.24 cm min−10.24\,\mathrm{cm}\,\mathrm{min}^{-1}.