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CIE 9709 2026 June Paper 15 Q2

A Level / CIE / P1

CIE 9709 2026 June Paper 15 Paper · Question 2

题目

Problem

A tank is being filled with water at a rate of 200 cm3 s−1200\text{ cm}^3\text{ s}^{-1}. The volume of water in the tank, V cm3V\text{ cm}^3, is given by

V=100πh+13πh3,V=100\pi h+\frac{1}{3}\pi h^3,

where hh is the height of water in the tank in centimetres.

Find the exact rate at which hh is increasing when h=30h=30.

(3)
题目中文翻译

一个水箱正以 200 cm3 s−1200\text{ cm}^3\text{ s}^{-1} 的速率注水。水箱内水的体积 V cm3V\text{ cm}^3 满足

V=100πh+13πh3,V=100\pi h+\frac{1}{3}\pi h^3,

其中 hh 是水箱内的水深,单位为厘米。

求当 h=30h=30 时,hh 增加的速率的精确值。

(3)
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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先求体积对水深的变化率,再用链式法则把注水速率换算成水深的增加速率。

答题过程

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Let tt be time in seconds. Differentiating with respect to hh gives dVdh=100π+πh2\frac{\mathrm{d}V}{\mathrm{d}h}=100\pi+\pi h^2. Therefore

200=dVdt=(100π+πh2)dhdt.200=\frac{\mathrm{d}V}{\mathrm{d}t} =(100\pi+\pi h^2)\frac{\mathrm{d}h}{\mathrm{d}t}.

At h=30h=30, the exact rate of increase is 2001000π=15π cm s−1\frac{200}{1000\pi}=\frac{1}{5\pi}\,\mathrm{cm}\,\mathrm{s}^{-1}.