题目
Problem
Bacteria are growing on the surface of a dish in a laboratory.
The area of the dish, A cm2, covered by the bacteria, t days after the bacteria start to grow, is modelled by the differential equation
dtdA=5t2A3/2t>0
Given that A=2.25 when t=3
(a) show that
A=(qt+rpt)2
where p, q and r are integers to be found.
(7)
According to the model, there is a limit to the area that will be covered by the bacteria.
(b) Find the value of this limit.
(2)
题目中文翻译
实验室培养皿表面正在生长细菌。
细菌开始生长后 t 天，细菌覆盖的培养皿面积 A cm2 由下列微分方程建模：
dtdA=5t2A3/2t>0
已知当 t=3 时 A=2.25。
(a) 证明
A=(qt+rpt)2
其中 p,q,r 为待求整数。
根据该模型，细菌最终覆盖的面积有一个极限。
(b) 求这个极限值。
解答
(a)
解法一
思路
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微分方程可分离变量。分别对 A 和 t 积分后，用 t=3、A=2.25 求积分常数，再把方程整理成 A 为 t 的显式函数，并与题目指定的形式比较。
答题过程
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Separating the variables gives
A−3/2dA=5t21dt.
Integrating both sides,
−2A−1/2=−5t1+c.
When t=3, A=2.25=49, so
−2(49)−1/2=−34=−151+c,−151+c.
Hence,
c=−1519.
Therefore,
−A2=A2==−5t1−1519,5t1+151915t3+19t.
It follows that
A=19t+330t,
and hence
A=(19t+330t)2.
Thus p=30, q=19 and r=3.
(b)
解法一
思路
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由 (a) 的结果，当 t→∞ 时，分子与分母同为一次式，分式的极限等于最高次项系数之比。再把该比值平方，即得覆盖面积的极限。
答题过程
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Using the result from part (a),
t→∞limA===t→∞lim(19t+330t)2(1930)2361900 cm2.
This is approximately 2.49 cm2.
