题目
Problem
A container in the shape of a cuboid has a square base of side x and a height of (10−x). It is given that x varies with time, t, where t>0. The container decreases in volume at a rate which is inversely proportional to t.
When t=101, x=21 and the rate of decrease of x is 3720.
(a) Show that x and t satisfy the differential equation
dtdx=−2t(20x−3x2)1.
(5)
(b) Solve the differential equation, obtaining an expression for t in terms of x.
(6)
题目中文翻译
一个长方体容器的底面为边长为 x 的正方形，高为 (10−x)。已知 x 随时间 t 变化，其中 t>0。容器体积减小的速率与 t 成反比。
当 t=101 时，x=21，且 x 减小的速率为 3720。
(a) 证明 x 和 t 满足微分方程
dtdx=−2t(20x−3x2)1.
(5)
(b) 解此微分方程，得到用 x 表示 t 的表达式。
(6)
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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体积随时间减少，先把比例关系写成带负号的导数。用链式法则连接体积变化率与边长变化率，再用初始数据求比例常数。
答题过程
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The volume is
V=dxdV=x2(10−x)=10x2−x3,20x−3x2.
Since the volume decreases at a rate inversely proportional to t,
dtdV=−tk,k>0.
The chain rule gives
(20x−3x2)dtdx=−tk.
At t=1/10, x=1/2 and dx/dt=−20/37, so
(10−43)(−3720)=−10k.
Thus −5=−10k, giving k=1/2. Consequently,
dtdx=−2t(20x−3x2)1.
(b)
解法一
思路
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分离变量并积分。先用初始数据求积分常数，再指数化，最后把 t 单独写出。
答题过程
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Separating variables in the equation from part (a),
2(20x−3x2)dx=−t1dt.
Since t>0, integration gives
20x2−2x3=−lnt+C.
Using t=1/10 and x=1/2,
5−41=C=ln10+C,419−ln10.
Therefore
lnt=2x3−20x2+419−ln10.
Exponentiating,
t=101e2x3−20x2+19/4.
