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IAL 2022 June Q3

A Level / Edexcel / P4

IAL 2022 June Paper · Question 3

题目

Problem

A tablet is dissolving in water.

The tablet is modelled as a cylinder, shown in Figure 1.

At tt seconds after the tablet is dropped into the water, the radius of the tablet is xx mm and the length of the tablet is 3x3x mm.

The cross-sectional area of the tablet is decreasing at a constant rate of 0.50.5 mm2^2 s−1^{-1}

(a) Find

dxdt\frac{dx}{dt}

when x=7x=7

(4)

(b) Find, according to the model, the rate of decrease of the volume of the tablet when x=4x=4

(4)
题目中文翻译

一片药片正在水中溶解。

药片被建模为一个圆柱体，如图 1 所示。

药片投入水中 tt 秒后，药片半径为 xx mm，长度为 3x3x mm。

药片横截面积正以恒定速率 0.50.5 mm2^2 s−1^{-1} 减小。

(a) 当 x=7x=7 时，求

dxdt\frac{dx}{dt}

。

(b) 当 x=4x=4 时，根据该模型求药片体积的减小速率。

解答

(a)

解法一

思路

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横截面是半径为 xx 的圆，所以面积为 A=πx2A=\pi x^2。把已知的面积减小速率写成 dAdt=−0.5\frac{\mathrm{d}A}{\mathrm{d}t}=-0.5，再利用链式法则求 dxdt\frac{\mathrm{d}x}{\mathrm{d}t}。

答题过程

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The cross-sectional area is

A=πx2.A=\pi x^2.

Since the area is decreasing at 0.5 mm2 s−10.5\text{ mm}^2\text{ s}^{-1},

dAdt=−0.5.\frac{\mathrm{d}A}{\mathrm{d}t}=-0.5.

Also,

dAdx=2πx.\frac{\mathrm{d}A}{\mathrm{d}x}=2\pi x.

Using the chain rule,

dxdt= dA/dtdA/dx= −0.52πx= −14πx.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} =&\, \frac{\mathrm{d}A/\mathrm{d}t} {\mathrm{d}A/\mathrm{d}x}\\ =&\,\frac{-0.5}{2\pi x}\\ =&\,-\frac{1}{4\pi x}. \end{align*}

When x=7x=7,

dxdt= −128π= −0.011368…\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} =&\,-\frac{1}{28\pi}\\ =&\,-0.011368\ldots \end{align*}

Therefore,

dxdt=−0.0114 mm s−1\boxed{ \frac{\mathrm{d}x}{\mathrm{d}t} =-0.0114\mathrm{\ mm\,s^{-1}} }

to three significant figures.

(b)

解法一

思路

展开

圆柱体长度为 3x3x，所以先把体积写成只含 xx 的函数。再将 dVdx\frac{\mathrm{d}V}{\mathrm{d}x} 与 (a) 中得到的通式 dxdt=−14πx\frac{\mathrm{d}x}{\mathrm{d}t}=-\frac1{4\pi x} 相乘；最后把负的体积变化率转换为正的“减小速率”。

答题过程

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The volume of the cylindrical tablet is

V= πx2(3x)= 3πx3.\begin{align*} V =&\,\pi x^2(3x)\\ =&\,3\pi x^3. \end{align*}

Hence

dVdx=9πx2.\frac{\mathrm{d}V}{\mathrm{d}x}=9\pi x^2.

Using the general expression for dxdt\frac{\mathrm{d}x}{\mathrm{d}t} from part (a),

dVdt= dVdxdxdt= (9πx2)(−14πx)= −94x.\begin{align*} \frac{\mathrm{d}V}{\mathrm{d}t} =&\, \frac{\mathrm{d}V}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}t}\\ =&\,(9\pi x^2) \bigg(-\frac{1}{4\pi x}\bigg)\\ =&\,-\frac94x. \end{align*}

When x=4x=4,

dVdt=−9 mm3 s−1.\frac{\mathrm{d}V}{\mathrm{d}t} =-9\mathrm{\ mm^3\,s^{-1}}.

Therefore, the rate of decrease of the volume is

9 mm3 s−1.\boxed{9\mathrm{\ mm^3\,s^{-1}}}.