题目

Problem

A colony of bees is being studied. The number of bees in the colony is modelled by the equation

P = 200 - \frac{160e^{0.6t}}{15 + e^{0.8t}} \qquad t \in \mathbb{R}, \qquad t \ge 0

where $P$ is the number of bees, measured in thousands, $t$ years after the study started. A sketch of the graph of $P$ against $t$ is shown in Figure 3

(a) Calculate the number of bees in the colony at the start of the study.

(2)

(b) Find $\dfrac{\mathrm{d}P}{\mathrm{d}t}$

(3)

The population of bees initially decreases, reaching a minimum value after $T$ years, as shown in Figure 3.

Solutions relying entirely on calculator technology are not acceptable.

(4)

题目中文翻译

正在研究一个蜜蜂群落。群落中的蜜蜂数量由下式建模：

P = 200 - \frac{160e^{0.6t}}{15 + e^{0.8t}} \qquad t \in \mathbb{R}, \qquad t \ge 0

其中 $P$ 以千只计， $t$ 为研究开始后经过的年数。图 3 给出了 $P$ 随 $t$ 变化的图像草图。

(a) 计算研究开始时群落中的蜜蜂数量。

(b) 求 $\dfrac{\mathrm{d}P}{\mathrm{d}t}$ 。

蜜蜂数量开始时下降，并如图 3 所示，在 $T$ 年后达到最小值。

（不接受完全依赖计算器技术的解法。）

解答

(a)

解法一

思路

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研究开始时 $t=0$ 。把 $t=0$ 代入模型，并用 $e^0=1$ 求出 $P$ ；由于 $P$ 的单位是千只，最后要换算成实际只数。

答题过程

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At the start of the study, $t=0$ . Therefore,

\begin{align*} P =&\,200-\frac{160e^0}{15+e^0}\\ =&\,200-\frac{160}{16}\\ =&\,190. \end{align*}

Since $P$ is measured in thousands, the number of bees is

190\times1000=\boxed{190\,000}.

(b)

解法一

思路

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对分式使用商法则，并在过程中分别求 $160e^{0.6t}$ 与 $15+e^{0.8t}$ 的导数。最后合并指数项，可得到便于 (c) 使用的形式。

答题过程

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Let

u=160e^{0.6t} \quad\text{and}\quad v=15+e^{0.8t}.

Then

u'=96e^{0.6t} \quad\text{and}\quad v'=0.8e^{0.8t}.

Using the quotient rule,

\begin{align*} \frac{\mathrm{d}P}{\mathrm{d}t} =&\,-\frac{vu'-uv'}{v^2}\\ =&\,-\frac{(15+e^{0.8t})96e^{0.6t}} {(15+e^{0.8t})^2}\\ &\,+\frac{160e^{0.6t}\cdot0.8e^{0.8t}} {(15+e^{0.8t})^2}\\ =&\,\frac{32e^{1.4t}-1440e^{0.6t}} {(15+e^{0.8t})^2}. \end{align*}

Hence

\boxed{\frac{\mathrm{d}P}{\mathrm{d}t} =\frac{32e^{1.4t}-1440e^{0.6t}} {(15+e^{0.8t})^2}}.

解法二

思路

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也可以把分式写成乘积，再使用乘积法则与链式法则。这条路线避开了直接套用商法则，化简后得到同一个导数。

答题过程

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Rewrite the model as

P=200-160e^{0.6t}(15+e^{0.8t})^{-1}.

Using the product rule and the chain rule,

\begin{align*} \frac{\mathrm{d}P}{\mathrm{d}t} =&\,-96e^{0.6t}(15+e^{0.8t})^{-1}\\ &\,+128e^{1.4t}(15+e^{0.8t})^{-2}. \end{align*}

Writing both terms over a common denominator,

\begin{align*} \frac{\mathrm{d}P}{\mathrm{d}t} =&\,\frac{-96e^{0.6t}(15+e^{0.8t})} {(15+e^{0.8t})^2}\\ &\,+\frac{128e^{1.4t}} {(15+e^{0.8t})^2}\\ =&\,\frac{32e^{1.4t}-1440e^{0.6t}} {(15+e^{0.8t})^2}. \end{align*}

Therefore,

\boxed{\frac{\mathrm{d}P}{\mathrm{d}t} =\frac{32e^{1.4t}-1440e^{0.6t}} {(15+e^{0.8t})^2}}.

(c)

解法一

思路

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由题意， $t=T$ 时数量达到最小值，因此令 (b) 中的导数等于 $0$ 。分母恒为正，所以只需令分子为 $0$ ；整理指数方程后取自然对数即可求出 $T$ 。

答题过程

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At the minimum point,

\frac{\mathrm{d}P}{\mathrm{d}t}=0.

Since $(15+e^{0.8t})^2>0$ , the numerator must be zero:

\begin{align*} 32e^{1.4T}-1440e^{0.6T}=&\,0\\ 32e^{1.4T}=&\,1440e^{0.6T}\\ e^{0.8T}=&\,45. \end{align*}

Taking natural logarithms,

\begin{align*} 0.8T=&\,\ln45\\ T=&\,\frac{\ln45}{0.8}\\ =&\,4.758\ldots \end{align*}

Therefore, to 2 decimal places,

\boxed{T=4.76\text{ years}}.

P3 2025 Oct A 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

解法二

思路

答题过程

(c)

解法一

思路

答题过程