题目

Problem

A scientist is studying two different populations of bacteria.

The number of bacteria $N$ in the first population is modelled by the equation

N=Ae^{kt}\qquad t\ge 0

where $A$ and $k$ are positive constants and $t$ is the time in hours from the start of the study.

Given that

• there were $2500$ bacteria in this population at the start of the study

• there were $10\,000$ bacteria 8 hours later

(a) find the exact value of $A$ and the value of $k$ to 4 significant figures.

(3)

The number of bacteria $N$ in the second population is modelled by the equation

N=60\,000e^{-0.6t}\qquad t\ge 0

where $t$ is the time in hours from the start of the study.

(b) Find the rate of decrease of bacteria in this population exactly 5 hours from the start of the study. Give your answer to 3 significant figures.

(2)

When $t=T$ , the number of bacteria in the two different populations was the same.

Solutions relying entirely on calculator technology are not acceptable.

(3)

题目中文翻译

一位科学家正在研究两种不同的细菌群体。

第一种细菌群体的数量 $N$ 由方程

N=Ae^{kt}\qquad t\ge 0

建模，其中 $A$ 和 $k$ 是正常数， $t$ 是从研究开始起经过的小时数。

已知

• 研究开始时该群体有 $2500$ 个细菌

• 8 小时后该群体有 $10\,000$ 个细菌

(a) 求 $A$ 的精确值，并求 $k$ 的值（保留 4 位有效数字）。

第二种细菌群体的数量 $N$ 由方程

N=60\,000e^{-0.6t}\qquad t\ge 0

建模，其中 $t$ 是从研究开始起经过的小时数。

(b) 求该群体在研究开始后恰好 5 小时的细菌减少速率，答案保留 3 位有效数字。

当 $t=T$ 时，两种细菌群体的数量相同。

不接受完全依赖计算器技术的解法。

解答

(a)

解法一

思路

展开

先把 $t=0$ 代入模型，研究开始时的数量会直接给出 $A$ 。再把 8 小时后的数量代入，得到关于 $k$ 的指数方程；用自然对数解出 $k$ 。

答题过程

展开

The first population is modelled by

N=Ae^{kt}.

At the start of the study, $t=0$ and $N=2500$ , so

2500=Ae^0=A.

Therefore

\boxed{A=2500}.

After 8 hours, $N=10000$ , so

10000=2500e^{8k}.

Divide by $2500$ :

4=e^{8k}.

Taking natural logarithms,

\ln4=8k.

Hence

k=\frac18\ln4=0.173286\cdots.

So, to 4 significant figures,

\boxed{k=0.1733}.

(b)

解法一

思路

展开

减少速率来自 $\frac{\mathrm{d}N}{\mathrm{d}t}$ 的负值。先对第二个模型求导，再代入 $t=5$ ；因为题目问 “rate of decrease”，最后写正的减少速率。

答题过程

展开

For the second population,

N=60000e^{-0.6t}.

Differentiate with respect to $t$ :

\frac{\mathrm{d}N}{\mathrm{d}t} =60000(-0.6)e^{-0.6t} =-36000e^{-0.6t}.

When $t=5$ ,

\frac{\mathrm{d}N}{\mathrm{d}t} =-36000e^{-3} =-1792.3\cdots.

So the rate of decrease is

1792.3\cdots.

To 3 significant figures, this is

\boxed{1790\text{ bacteria per hour}}.

(c)

解法一

思路

展开

两种群体数量相等时，把两个模型直接相等。利用 (a) 中的 $k=\frac18\ln4$ ，整理成 $e^{\text{常数}\cdot T}=24$ ，再取自然对数求 $T$ 。

答题过程

展开

From part (a), the first population is

N=2500e^{kt}, \qquad k=\frac18\ln4.

When $t=T$ , the two populations are equal, so

2500e^{kT}=60000e^{-0.6T}.

Divide by $2500$ :

e^{kT}=24e^{-0.6T}.

Multiply by $e^{0.6T}$ :

e^{(k+0.6)T}=24.

Taking natural logarithms,

(k+0.6)T=\ln24.

Therefore

T=\frac{\ln24}{k+0.6}.

Substitute $k=\dfrac18\ln4$ :

T=\frac{\ln24}{0.6+\frac18\ln4} =4.109\cdots.

So, to 3 significant figures,

\boxed{T=4.11}.

IAL 2023 May Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程