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CIE 9709 2024 November Paper 32 Q10

A Level / CIE / P3

CIE 9709 2024 Nov Paper 32 Paper · Question 10

题目

Problem

A balloon in the shape of a sphere has volume VV and radius rr. Air is pumped into the balloon at a constant rate of 40π40\pi starting when time t=0t=0 and r=0r=0. At the same time, air begins to flow out of the balloon at a rate of 0.8πr0.8\pi r. The balloon remains a sphere at all times.

(a) Show that rr and tt satisfy the differential equation

drdt=50−r5r2.\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{50-r}{5r^2}.
(3)

(b) Find the quotient and remainder when 5r25r^2 is divided by 50−r50-r.

(3)

(c) Solve the differential equation in part (a), obtaining an expression for tt in terms of rr.

(6)

(d) Find the value of tt when the radius of the balloon is 1212.

(1)
题目中文翻译

一个球形气球的体积为 VV,半径为 rr。从时刻 t=0t=0 且 r=0r=0 开始,以 40π40\pi 的恒定速率向气球内充气。同时,空气开始以 0.8πr0.8\pi r 的速率从气球流出。气球始终保持球形。

(a) 证明 rr 和 tt 满足微分方程

drdt=50−r5r2.\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{50-r}{5r^2}.
(3)

(b) 求 5r25r^2 除以 50−r50-r 所得的商和余数。

(3)

(c) 解 (a) 问中的微分方程,求出用 rr 表示 tt 的表达式。

(6)

(d) 求气球半径为 1212 时的 tt 值。

(1)
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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净流入是流入减流出。用球体体积对半径的导数连接两种变化率。

答题过程

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The net rate of volume increase is

dVdt=40π−0.8πr.\frac{\mathrm{d}V}{\mathrm{d}t}=40\pi-0.8\pi r.

Since V=43πr3V=\frac43\pi r^3,

dVdr=4πr2.\frac{\mathrm{d}V}{\mathrm{d}r}=4\pi r^2.

For r>0r>0, the chain rule gives

drdt=40π−0.8πr4πr2=50−r5r2.\frac{\mathrm{d}r}{\mathrm{d}t} =\frac{40\pi-0.8\pi r}{4\pi r^2} =\boxed{\frac{50-r}{5r^2}}.

(b)

解法一

思路

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商是一次式,先比较二次项和一次项系数,再求常数余式。

答题过程

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Write

5r2=(50−r)(Ar+B)+C.5r^2=(50-r)(Ar+B)+C.

Comparing coefficients gives

−A= 5,50A−B= 0,50B+C= 0.\begin{align*} -A=&\,5,\\[4mm] 50A-B=&\,0,\\[4mm] 50B+C=&\,0. \end{align*}

Thus A=−5A=-5, B=−250B=-250 and C=12500C=12500. Therefore

5r2=(50−r)(−5r−250)+12500.5r^2=(50-r)(-5r-250)+12500.

The quotient is −5r−250\boxed{-5r-250} and the remainder is 12500\boxed{12500}.

(c)

解法一

思路

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分离变量后使用(b)的除法结果积分。初始半径为零时用所得时间函数确定常数。

答题过程

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From (a),

dtdr= 5r250−r= −5r−250+1250050−r.\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}r} =&\,\frac{5r^2}{50-r}\\[4mm] =&\,-5r-250+\frac{12500}{50-r}. \end{align*}

Using (b), the last term integrates to −12500ln⁡∣50−r∣-12500\ln|50-r| because d(50−r)/dr=−1\mathrm{d}(50-r)/\mathrm{d}r=-1. Integrating gives

t=−52r2−250r−12500ln⁡∣50−r∣+C.t=-\frac52r^2-250r-12500\ln|50-r|+C.

Using t=0t=0, r=0r=0 gives C=12500ln⁡50C=12500\ln50.

The physically relevant solution starts at r=0r=0 and has 0⩽r<500\leqslant r<50, so

t= −52r2−250r +12500ln⁡(5050−r).\begin{align*} t=&\,-\frac52r^2-250r\\[4mm] &\,\hspace{2pt}+12500\ln\bigg(\frac{50}{50-r}\bigg). \end{align*}

The relation extends continuously to r=0r=0, where t=0t=0.

(d)

解法一

思路

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把半径十二代入(c)的时间表达式。

答题过程

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Using (c),

t= −52(12)2−250(12) +12500ln⁡5038= −3360+12500ln⁡2519= 70.4605….\begin{align*} t=&\,-\frac52(12)^2-250(12)\\[4mm] &\,\hspace{2pt}+12500\ln\frac{50}{38}\\[4mm] =&\,-3360+12500\ln\frac{25}{19}\\ =&\,70.4605\ldots. \end{align*}

Therefore t=70.5\boxed{t=70.5}.