📚 Environment and Population Modelling | 环境与人口建模
This article connects the Edexcel A-Level Mathematics specification to real-world problems in environment and population. You will see how exponential functions, differential equations, statistical sampling and hypothesis testing can be applied to model population growth, environmental capacity and ecological data. The focus is on exam-ready methods and notation.
本文将 Edexcel A-Level 数学考纲与环境、人口领域的实际问题相结合。你将看到指数函数、微分方程、统计抽样和假设检验如何用于人口增长、环境容量和生态数据建模。重点是考试可直接使用的方法与符号。
1. Exponential Growth in Populations | 人口指数增长
Many biological populations grow approximately exponentially when resources are unlimited. If P is the population size at time t, the basic model is P = P₀e^(kt), where P₀ is the initial population and k is the continuous growth rate.
当资源不受限制时，许多生物种群近似呈指数增长。如果 P 是时间 t 时的种群数量，基本模型为 P = P₀e^(kt)，其中 P₀ 是初始种群，k 是连续增长率。
A positive k means growth; a negative k means decay, which can model pollutant decay or a declining species. In Edexcel questions you often have to find k from two known population values.
k 为正表示增长；k 为负表示衰减，可用于污染物降解或物种数量下降。Edexcel 考题常要求你从两个已知种群值求出 k。
P = P0ekt
k = (ln P₂ – ln P₁) / (t₂ – t₁)
2. The Differential Equation dP/dt = kP | 微分方程 dP/dt = kP
Exponential growth also arises as the solution of the differential equation dP/dt = kP. This equation states that the rate of change of population is proportional to the current population.
指数增长也可由微分方程 dP/dt = kP 解得。该方程表示种群变化率与当前种群数量成正比。
To solve it, separate variables: ∫(1/P) dP = ∫ k dt, giving ln P = kt + C. Exponentiating both sides recovers P = Ae^(kt), where A = e^C = P₀.
求解时先分离变量：∫(1/P) dP = ∫ k dt，得 ln P = kt + C。两边取指数还原为 P = Ae^(kt)，其中 A = e^C = P₀。
dP/dt = kP
∫ (1/P) dP = ∫ k dt ⇒ ln P = kt + C
3. Logarithmic Transformation and Linear Fitting | 对数变换与线性拟合
Taking natural logarithms gives ln P = ln P₀ + kt. This is a straight line in t with gradient k and intercept ln P₀.
取自然对数得 ln P = ln P₀ + kt。这是关于 t 的直线，斜率为 k，截距为 ln P₀。
In exam questions, a table of population data may be given. Plotting ln P against t and finding the least squares regression line allows you to estimate k and P₀.
考题可能给出种群数据表。绘制 ln P 对 t 的散点图并用最小二乘回归线可估计 k 和 P₀。
ln P = ln P₀ + kt
| t (years) | P | ln P |
| 0 | 200 | 5.30 |
| 1 | 245 | 5.50 |
| 2 | 300 | 5.70 |
4. Logistic Growth and Carrying Capacity | Logistic 增长与环境容量
Real environments have limited resources, so growth slows near a maximum carrying capacity L. The logistic model is dP/dt = kP(1 – P/L), where L is the carrying capacity.
真实环境资源有限，因此增长在最大环境容量 L 附近放缓。Logistic 模型为 dP/dt = kP(1 – P/L)，其中 L 为环境容量。
The solution is P = L / (1 + Ae^(-kt)), where A = (L – P₀)/P₀. When P is much smaller than L the growth is nearly exponential; when P approaches L, the rate tends to zero.
其解为 P = L / (1 + Ae^(-kt))，其中 A = (L – P₀)/P₀。当 P 远小于 L 时接近指数增长；当 P 接近 L 时增长率趋于零。
dP/dt = kP(1 – P/L)
P = L / (1 + Ae-kt)
5. Differentiation to Find Growth Rate | 微分求增长率
Given a population function P(t), the instantaneous growth rate is P'(t). For P = 1000e^(0.05t), differentiating gives P’ = 50e^(0.05t).
给定种群函数 P(t)，瞬时增长率为 P'(t)。对于 P = 1000e^(0.05t)，求导得 P’ = 50e^(0.05t)。
You may also need the relative growth rate P’/P. In the exponential model this is constant and equal to k. In the logistic model it decreases linearly as P increases.
你可能还需要相对增长率 P’/P。在指数模型中它恒为 k。在 Logistic 模型中它随 P 增大而线性减小。
P'(t) = dP/dt
Relative growth rate = P’/P
6. Numerical Methods: Euler’s Method | 数值方法：欧拉法
When a differential equation cannot be solved exactly, Edexcel expects you to use numerical methods. For dP/dt = f(t, P), Euler’s method uses P_new = P_old + h f(t_old, P_old).
当微分方程无法精确求解时，Edexcel 要求使用数值方法。对于 dP/dt = f(t, P)，欧拉法公式为 P_new = P_old + h f(t_old, P_old)。
Choose a small step length h. Repeatedly applying the formula produces approximate population values at successive times. Smaller h improves accuracy but increases calculation.
选择较小步长 h。反复应用该公式可得出各时刻的近似种群值。步长越小精度越高，但计算量增大。
Pn+1 = Pn + h f(tn, Pn)
- Step length h affects accuracy: halving h roughly halves the local error.
- Always show full substitutions in exam working.
- Euler’s method gives an approximate solution, not an exact one.
7. Sampling Environmental Populations | 环境种群抽样
In statistics questions, populations may be animals, plants or pollution sites. A simple random sample of size n gives every member an equal chance of selection.
在统计题中，总体可以是动物、植物或污染点。容量为 n 的简单随机样本使每个个体被抽中的概率相等。
For large or mobile populations, systematic sampling or stratified sampling may be more practical. Be prepared to comment on bias and sampling errors.
对于大规模或移动种群，系统抽样或分层抽样可能更实用。要准备对偏差和抽样误差进行评述。
- Simple random sampling: unbiased, but may miss rare subpopulations.
- Systematic sampling: easier in the field, but can introduce periodicity bias.
- Stratified sampling: ensures subgroups are represented, improves precision.
8. Capture-Recapture Estimation | 标记重捕法估计
Capture-recapture uses two samples to estimate unknown population size N. If M animals are marked and released, then n are recaptured and m are marked, the Lincoln-Petersen estimate is N ≈ Mn/m.
标记重捕法用两个样本来估计未知总体 N。若标记并释放 M 只动物，再捕获 n 只，其中 m 只有标记，则 Lincoln-Petersen 估计为 N ≈ Mn/m。
This method assumes the population is closed, marks are not lost, and marked animals mix randomly. Edexcel questions often ask for the estimate and assumptions.
该方法假设总体封闭、标记不丢失、标记动物随机混合。Edexcel 考题常要求计算估计值并说明假设。
N ≈ M × n ÷ m
- Closed population: no births, deaths, immigration or emigration.
- Marks must be visible and permanent.
- Marked and unmarked individuals must mix completely.
9. Hypothesis Testing for Population Parameters | 种群参数的假设检验
If sample data suggest a change in mean population, use a one-sample t-test: H₀: μ = μ₀ vs H₁: μ ≠ μ₀ or a one-sided alternative. The test statistic is t = (x̄ – μ₀) / (s/√n).
如果样本数据表明平均种群发生变化，使用单样本 t 检验：H₀: μ = μ₀ 对 H₁: μ ≠ μ₀ 或单侧。检验统计量为 t = (x̄ – μ₀) / (s/√n)。
Alternatively, for proportions, use a binomial test or a normal
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