Edexcel Maths 6.5 Ecosystems: Modelling Population Dynamics | 爱德思数学 6.5 生态系统：种群动态建模

📚 Edexcel Maths 6.5 Ecosystems: Modelling Population Dynamics | 爱德思数学 6.5 生态系统：种群动态建模

In Edexcel Mathematics, the study of ecosystems offers a powerful context for applying rates of change, exponential functions, and differential equations. This unit focuses on how population sizes change over time and how mathematical models can describe predator-prey interactions, resource limits, and ecological stability.

在爱德思数学中，生态系统的研究为应用变化率、指数函数和微分方程提供了强有力的背景。本单元重点探讨种群数量如何随时间变化，以及数学模型如何描述捕食者与猎物之间的相互作用、资源限制和生态稳定性。

1. Ecosystems as Dynamic Systems | 生态系统作为动态系统

An ecosystem can be viewed as a set of interacting populations whose sizes change continuously. Typical variables include N for prey population, P for predator population, and t for time. The rate of change dN/dt is the key quantity that describes how quickly a population grows or shrinks.

生态系统可以看作一组相互作用的种群，其数量不断变化。典型变量包括猎物数量 N、捕食者数量 P 和时间 t。变化率 dN/dt 是描述种群增长或缩减速度的关键量。

In Edexcel modelling questions, you are often given a differential equation and asked to interpret its terms, solve it, or find equilibrium values. This connects directly to the 'rate of change' part of the A Level Mathematics specification.

在爱德思建模题中，通常会给出一个微分方程，要求解释各项、求解或找出平衡值。这与 A Level 数学大纲中“变化率”部分直接相关。

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2. Exponential Growth in Ecosystems | 生态系统中的指数增长

When resources are unlimited, a population may grow at a rate proportional to its current size. This gives the exponential model:

当资源无限时，种群可能以与其当前规模成正比的速率增长。这就得到了指数模型：

dN/dt = rN

Here r is the intrinsic growth rate. Solving this first-order differential equation by separation of variables gives:

这里 r 是内禀增长率。通过分离变量法求解这个一阶微分方程，得到：

N(t) = N₀eʳᵗ

where N₀ is the initial population. You should be able to substitute values, sketch the curve, and use logarithms to estimate r from data.

其中 N₀ 是初始种群数量。你应当能够代入数值、绘制曲线，并利用对数从数据中估计 r。

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3. Logistic Growth and Carrying Capacity | 逻辑斯蒂增长与容纳量

In practice, ecosystems have finite resources, so growth slows as the population approaches a carrying capacity K. The logistic differential equation is:

实际上，生态系统的资源有限，因此当种群接近容纳量 K 时，增长会减慢。逻辑斯蒂微分方程为：

dN/dt = rN(1 − N/K)

, so the population levels off. For small N, the model behaves like exponential growth because N/K is close to zero.

项 (1 − N/K) 随着 N 的增大而降低增长率。当 N = K 时，dN/dt = 0，种群数量趋于平稳。当 N 很小时，由于 N/K 接近于零，模型表现类似指数增长。

You may be asked to find K from a differential equation or explain why an ecosystem cannot sustain unlimited exponential growth.

你可能会被要求从微分方程中求出 K，或解释为什么生态系统无法维持无限的指数增长。

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4. Predator-Prey Models | 捕食者-猎物模型

A classic ecosystem model uses two coupled differential equations for prey N and predator P. The Lotka-Volterra equations are:

经典的生态系统模型使用两个耦合微分方程来描述猎物 N 和捕食者 P。Lotka-Volterra 方程为：

dN/dt = aN − bNP

dP/dt = cNP − dP

Here a is the prey birth rate, b is the predation rate, c is the predator growth rate per prey eaten, and d is the predator death rate. The product NP represents encounters between predators and prey.

这里 a 是猎物的出生率，b 是捕食率，c 是每吃掉一只猎物带来的捕食者增长率，d 是捕食者死亡率。乘积 NP 表示捕食者与猎物之间的相遇。

You do not need to solve these simultaneous equations analytically at A Level, but you should be able to interpret each term and analyse equilibrium points.

在 A Level 阶段你不需要解析求解这组联立方程，但应能够解释每一项并分析平衡点。

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5. Equilibrium Points in Ecosystem Models | 生态系统模型中的平衡点

s of change are zero. For the predator-prey system, setting dN/dt = 0 and dP/dt = 0 gives:

当所有变化率均为零时，系统达到平衡。对于捕食者-猎物系统，令 dN/dt = 0 和 dP/dt = 0，得到：

aN − bNP = 0 and cNP − dP = 0

Solving these equations yields two equilibria: (0,0) and (d/c, a/b). The non-zero equilibrium gives constant population sizes. In exam questions, you may be asked to verify these values or explain their ecological meaning.

求解这些方程可得到两个平衡点：(0,0) 和 (d/c, a/b)。非零平衡点表示种群数量保持恒定。在考试题中，你可能需要验证这些数值

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