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CIE 9709 2023 November Paper 11 Q3

A Level / CIE / P1

CIE 9709 2023 Nov Paper 11 Paper · Question 3

题目

Problem

The diagram shows a cubical closed container made of a thin elastic material which is filled with water and frozen. During the freezing process the length, x cmx\text{ cm}, of each edge of the container increases at the constant rate of 0.01 cm0.01\text{ cm} per minute. The volume of the container at time tt minutes is V cm3V\text{ cm}^3.

Find the rate of increase of VV when x=20x=20.

(3)
题目中文翻译

图中所示的是一个由薄弹性材料制成的封闭正方体容器,容器装满水后进行冷冻。在结冰过程中,容器每条棱的长度 x cmx\text{ cm} 以每分钟 0.01 cm0.01\text{ cm} 的恒定速率增加。在 tt 分钟时,容器的体积为 V cm3V\text{ cm}^3。

求当 x=20x=20 时,VV 的增加速率。

(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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把体积写成边长的函数,再用链式法则把边长的增加速率换算为体积的增加速率。

答题过程

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Since V=x3V=x^3, the volume increases at the rate

dVdt=3x2dxdt.\frac{\mathrm{d}V}{\mathrm{d}t}=3x^2\frac{\mathrm{d}x}{\mathrm{d}t}.

At x=20x=20 this gives 3(20)2(0.01)=12 cm3 min−13(20)^2(0.01)=12\,\mathrm{cm}^3\,\mathrm{min}^{-1}.