题目
Problem
A balloon in the shape of a sphere has volume V and radius r. Air is pumped into the balloon at a constant rate of 40π starting when time t=0 and r=0. At the same time, air begins to flow out of the balloon at a rate of 0.8πr. The balloon remains a sphere at all times.
(a) Show that r and t satisfy the differential equation
dtdr=5r250−r.
(3)
(b) Find the quotient and remainder when 5r2 is divided by 50−r.
(3)
(c) Solve the differential equation in part (a), obtaining an expression for t in terms of r.
(6)
(d) Find the value of t when the radius of the balloon is 12.
(1)
题目中文翻译
一个球形气球的体积为 V,半径为 r。从时刻 t=0 且 r=0 开始,以 40π 的恒定速率向气球内充气。同时,空气开始以 0.8πr 的速率从气球流出。气球始终保持球形。
(a) 证明 r 和 t 满足微分方程
dtdr=5r250−r.
(3)
(b) 求 5r2 除以 50−r 所得的商和余数。
(3)
(c) 解 (a) 问中的微分方程,求出用 r 表示 t 的表达式。
(6)
(d) 求气球半径为 12 时的 t 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
dtdV=40π−0.8πr.
Since V=34πr3,
drdV=4πr2.
For r>0, the chain rule gives
dtdr=4πr240π−0.8πr=5r250−r.
(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.
Comparing coefficients gives
−A=50A−B=50B+C=5,0,0.
Thus A=−5, B=−250 and C=12500. Therefore
5r2=(50−r)(−5r−250)+12500.
The quotient is −5r−250 and the remainder is 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),
drdt==50−r5r2−5r−250+50−r12500.
Using (b), the last term integrates to −12500ln∣50−r∣ because d(50−r)/dr=−1. Integrating gives
t=−25r2−250r−12500ln∣50−r∣+C.
Using t=0, r=0 gives C=12500ln50.
The physically relevant solution starts at r=0 and has 0⩽r<50, so
t=−25r2−250r+12500ln(50−r50).
The relation extends continuously to r=0, where t=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===−25(12)2−250(12)+12500ln3850−3360+12500ln192570.4605….
Therefore t=70.5.