题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Two populations of insects are being studied over the same period of time.

The number of insects $N_A$ in population A, $t$ years after the start of the study, is modelled by the equation

N_A=8000+900e^{0.2t}

According to the model,

(a) find the rate of growth in the number of insects in population A exactly 5 years after the start of the study.

(2)

The number of insects $N_B$ in population B, $t$ years after the start of the study, is modelled by the equation

N_B=8000+Pe^{kt}

where $P$ and $k$ are positive constants.

Given that

there are 10 000 insects in population B at the start of the study
there are 11 570 insects in population B exactly 4 years after the start of the study
there are the same number of insects in population A and population B after $T$ years

find, according to the models,

(b) the exact value of $P$ and the value of $k$ to 3 decimal places,

(4)

(3)

题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

现同时研究两个昆虫种群。

研究开始后 $t$ 年时，A 种群中的昆虫数量 $N_A$ 由下式建模：

N_A=8000+900e^{0.2t}

根据该模型，

(a) 求研究开始后恰好 5 年时，A 种群数量的增长率。

B 种群中的昆虫数量 $N_B$ 由下式建模：

N_B=8000+Pe^{kt}

其中 $P,k$ 为正常数。

已知：

研究开始时，B 种群有 10 000 只昆虫；
研究开始 4 年后，B 种群有 11 570 只昆虫；
研究开始后 $T$ 年时，A 与 B 两个种群的昆虫数量相同。

根据模型，求

(b) $P$ 的精确值以及 $k$ 的值（保留 3 位小数）；

解答

(a)

解法一

思路

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增长率就是 $\frac{\mathrm{d}N_A}{\mathrm{d}t}$ 。先对

N_A=8000+900e^{0.2t}

关于 $t$ 求导，再代入 $t=5$ 。

答题过程

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Differentiate $N_A$ with respect to $t$ .

\begin{align*} N_A=&\,8000+900e^{0.2t} \\[2mm] \frac{\mathrm{d}N_A}{\mathrm{d}t} =&\,900(0.2)e^{0.2t} \\[2mm] =&\,180e^{0.2t} \end{align*}

At $t=5$ ,

\begin{align*} \frac{\mathrm{d}N_A}{\mathrm{d}t} =&\,180e^{0.2(5)} \\[2mm] =&\,180e \end{align*}

Therefore the rate of growth after exactly 5 years is

\boxed{180e}

This is approximately $489$ insects per year.

(b)

解法一

思路

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先用 $t=0$ 的资料求 $P$ 。因为 $e^0=1$ ，所以

10000=8000+P

再用 $t=4$ 和 $N_B=11570$ 求 $k$ 。

答题过程

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For population B,

N_B=8000+Pe^{kt}

At the start of the study, $t=0$ and $N_B=10000$ . Hence

\begin{align*} 10000=&\,8000+Pe^{0} \\[2mm] 10000=&\,8000+P \\[2mm] P=&\,2000 \end{align*}

\boxed{P=2000}

After exactly 4 years, $N_B=11570$ . Therefore

\begin{align*} 11570=&\,8000+2000e^{4k} \\[2mm] 3570=&\,2000e^{4k} \\[2mm] e^{4k}=&\,\frac{3570}{2000} \end{align*}

Taking natural logarithms,

\begin{align*} 4k=&\,\ln\left(\frac{3570}{2000}\right) \\[2mm] k=&\,\frac{1}{4}\ln\left(\frac{3570}{2000}\right) \end{align*}

Thus

\boxed{k=0.145\text{ to 3 d.p.}}

(c)

解法一

思路

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当两个种群数量相等时，令 $N_A=N_B$ 。两边都有 $8000$ ，所以可以先消去 $8000$ ，再用指数律和对数求 $T$ 。

这里要使用 (b) 中的模型参数 $P=2000$ 和 $k\approx0.145$ 。

答题过程

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When the two populations are equal,

N_A=N_B

Using the two models,

8000+900e^{0.2T}=8000+2000e^{0.145T}

Cancel $8000$ from both sides.

900e^{0.2T}=2000e^{0.145T}

Divide by $900e^{0.145T}$ .

\begin{align*} \frac{e^{0.2T}}{e^{0.145T}} =&\,\frac{2000}{900} \\[2mm] e^{(0.2-0.145)T} =&\,\frac{2000}{900} \end{align*}

Taking natural logarithms,

\begin{align*} (0.2-0.145)T =&\,\ln\left(\frac{2000}{900}\right) \\[2mm] T =&\,\frac{\ln\left(\frac{2000}{900}\right)}{0.2-0.145} \end{align*}

Using the unrounded value of $k$ from part (b) gives

T=14.480\ldots

Therefore

\boxed{T=14.5\text{ years to 1 d.p.}}

IAL 2026 Jan Q10

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