📚 The Environment | 环境
Mathematics provides powerful tools for understanding and preserving the natural world. From modelling the growth of species to analysing climate data, A-Level topics such as exponential functions, differential equations, correlation and hypothesis testing find direct applications in environmental science. This article explores how mathematical concepts from the Edexcel syllabus are used to study and predict environmental phenomena.
数学为理解和保护自然界提供了强大的工具。从模拟物种的增长到分析气候数据，A-Level 专题如指数函数、微分方程、相关性和假设检验在环境科学中有着直接的应用。本文探讨 Edexcel 考纲中的数学概念如何被用来研究和预测环境现象。
1. Exponential Growth and Decay | 指数增长与衰减
Exponential functions describe quantities that change at a rate proportional to their current value. In environmental contexts, this applies to the unchecked growth of a bacteria population in a lake or the decay of a pesticide in soil. The general form is P(t) = P₀ eᵏᵗ, where P₀ is the initial amount, k is the growth constant (k > 0 for growth, k < 0 for decay), and t is time. The base e ≈ 2.718 ensures that the rate of change dP/dt equals kP.
指数函数描述以与其当前值成比例的速率变化的量。在环境背景中，这适用于湖泊中细菌种群的无限制增长或农药在土壤中的衰减。其一般形式为 P(t) = P₀ eᵏᵗ，其中 P₀ 是初始量，k 是增长常数（k > 0 表示增长，k < 0 表示衰减），t 是时间。底数 e ≈ 2.718 保证了变化率 dP/dt 等于 kP。
Suppose a population of algae in a pond doubles every 3 days. To find the growth constant k, set P(3) = 2P₀ = P₀ e³ᵏ, giving e³ᵏ = 2. Taking natural logarithms yields 3k = ln 2, so k = (ln 2)/3 ≈ 0.231 day⁻¹. To predict when the population reaches 10,000 from an initial 500, solve 10000 = 500 e⁰˙²³¹ᵗ, leading to t = (ln 20)/0.231 ≈ 13.0 days.
假设一个池塘中的藻类数量每 3 天翻倍。为求增长常数 k，设 P(3) = 2P₀ = P₀ e³ᵏ，得 e³ᵏ = 2。取自然对数得 3k = ln 2，故 k = (ln 2)/3 ≈ 0.231 天⁻¹。若预测该种群从初始 500 达到 10000 所需的时间，解 10000 = 500 e⁰˙²³¹ᵗ，得 t = (ln 20)/0.231 ≈ 13.0 天。
2. Radioactive Decay and Half-Life | 放射性衰变与半衰期
Radioactive substances decay exponentially with a constant λ > 0: N(t) = N₀ e^(−λt). The half-life T½ is the time taken for half the nuclei to decay, found by setting N(T½) = N₀/2. This gives e^(−λ T½) = ½, so T½ = (ln 2)/λ. Environmental examples include carbon-14 dating of organic remains or monitoring the decay of radioactive waste.
放射性物质以正常数 λ > 0 指数衰减：N(t) = N₀ e^(−λt)。半衰期 T½ 是半数原子核衰变所需的时间，由 N(T½) = N₀/2 求得。可得 e^(−λ T½) = ½，因此 T½ = (ln 2)/λ。环境实例包括有机物遗骸的碳-14 定年法，或监测放射性废物的衰减。
If a contaminated soil sample initially contains 80 mg of a pollutant with a decay constant λ = 0.05 year⁻¹, the amount after 10 years is N(10) = 80 e^(−0.05×10) = 80 e^(−0.5) ≈ 48.5 mg. The half-life is T½ = (ln 2)/0.05 ≈ 13.9 years, indicating the time required for the pollutant to halve under natural breakdown.
若一份受污染土壤样品最初含有 80 mg 的某污染物，其衰变常数 λ = 0.05 年⁻¹，则 10 年后的量为 N(10) = 80 e^(−0.05×10) = 80 e^(−0.5) ≈ 48.5 mg。半衰期为 T½ = (ln 2)/0.05 ≈ 13.9 年，表明该污染物在自然分解下减半所需的时间。
3. Logistic Population Growth | 逻辑斯蒂人口增长
In reality, populations cannot grow indefinitely due to limited resources. The logistic model introduces a carrying capacity K, the maximum sustainable population. The differential equation is dP/dt = r P (1 − P/K), where r is the intrinsic growth rate. When P is small, the term (1 − P/K) ≈ 1, and growth is nearly exponential. As P approaches K, the growth rate slows to zero.
实际上，种群因资源有限而不能无限增长。逻辑斯蒂模型引入了容纳量 K，即可持续的最大种群数量。其微分方程为 dP/dt = r P (1 − P/K)，其中 r 为内禀增长率。当 P 很小时，(1 − P/K) ≈ 1，增长近乎指数形式。随着 P 趋近 K，增长率减慢至零。
The solution to the logistic equation is P(t) = K / [1 + ((K − P₀)/P₀) e^(−rt)]. This S-shaped (sigmoid) curve is used to model fish populations in a lake where food supply limits growth. Environmental managers use the model to set fishing quotas, ensuring the population remains near K/2 where growth is fastest.
逻辑斯蒂方程的解为 P(t) = K / [1 + ((K − P₀)/P₀) e^(−rt)]。这条 S 形（sigmoid）曲线用于模拟食物供给限制增长的湖泊鱼类种群。环境管理者利用该模型设定捕捞配额，保证种群数量保持在增长最快的 K/2 附近。
4. Differential Equations for Pollutant Concentration | 污染物浓度的微分方程
A common environmental problem involves a lake with a constant volume V receiving polluted water at a rate Q and draining at the same rate, maintaining constant volume. If the inflow concentration is Cᵢₙ and the lake is well mixed, the rate of change of pollutant mass y is dy/dt = Q Cᵢₙ − (Q/V) y. This first-order linear differential equation can be solved with an integrating factor to find the concentration over time.
一个常见的环境问题涉及体积 V 恒定的湖泊，以流量 Q 接收污水并以相同流量排出，保持体积不变。若流入浓度为 Cᵢₙ 且湖水混合均匀，则污染物质量 y 的变化率为 dy/dt = Q Cᵢₙ − (Q/V) y。这一阶线性微分方程可利用积分因子求解，得到浓度随时间的变化。
After a long period, the system reaches a steady state where dy/dt = 0, giving the equilibrium concentration Cₑ = Cᵢₙ. If clean water (Cᵢₙ = 0) starts flowing, the pollutant decays according to y = Y₀ e^(−(Q/V)t). The time needed to halve the pollutant is V ln 2 / Q, illustrating how flushing rate cleans the lake.
长时间后，系统达到稳态，dy/dt = 0，得到平衡浓度 Cₑ = Cᵢₙ。若洁净水（Cᵢₙ = 0）开始流入，污染物按 y = Y₀ e^(−(Q/V)t) 衰减。污染物减半所需时间为 V ln 2 / Q，这说明了冲刷速率对湖泊净化的作用。
5. Newton’s Law of Cooling in Environmental Contexts | 环境中的牛顿冷却定律
Newton’s law of cooling states that the rate of temperature change of an object is proportional to the difference between its own temperature T and the ambient temperature Tₐ: dT/dt = −k (T − Tₐ). This model applies to a warm river plume entering a colder ocean or the cooling of soil at night. The solution is T(t) = Tₐ + (T₀ − Tₐ) e^(−kt).
牛顿冷却定律指出，物体温度的变化率与其自身温度 T 和环境温度 Tₐ 之差成正比：dT/dt = −k (T − Tₐ)。该模型适用于暖流进入较冷海洋或夜间土壤降温的情形。其解为 T(t) = Tₐ + (T₀ − Tₐ) e^(−kt)。
For example, if industrial cooling water is discharged at 45 °C into a river at 15 °C, and k = 0.1 min⁻¹, the temperature after 10 minutes is T(10) = 15 + (45 − 15) e^(−0.1×10) = 15 + 30 e^(−1) ≈ 26.0 °C. This helps ecologists assess the thermal impact on aquatic life downstream.
例如，若工业冷却水以 45 °C 排入 15 °C 的河流，且 k = 0.1 min⁻¹，则 10 分钟后的温度为 T(10) = 15 + (45 − 15) e^(−0.1×10) = 15 + 30 e^(−1) ≈ 26.0 °C。这有助于生态学家评估对下游水生生物的热影响。
6. Correlation: CO₂ Emissions and Temperature Rise | 相关性：二氧化碳排放与气温上升
Bivariate data, such as atmospheric CO₂ concentration and global average temperature, often show a linear correlation. The Pearson product-moment correlation coefficient r measures the strength and direction of a linear relationship. It is calculated as r = Sₓᵧ / √(Sₓₓ Sᵧᵧ), where Sₓᵧ = Σ(xᵢ − x̄)(yᵢ − ȳ), Sₓₓ = Σ(xᵢ − x̄)², and Sᵧᵧ = Σ(yᵢ − ȳ)². A value of r close to +1 indicates a strong positive linear correlation.
双变量数据，如大气 CO₂ 浓度和全球平均温度，往往呈现线性相关。皮尔逊积矩相关系数 r 度量线性关系的强度和方向。其计算公式为 r = Sₓᵧ / √(Sₓₓ Sᵧᵧ)，其中 Sₓᵧ = Σ(xᵢ − x̄)(yᵢ − ȳ)，Sₓₓ = Σ(xᵢ − x̄)²，Sᵧᵧ = Σ(yᵢ − ȳ)²。r 值接近 +1 表示强正线性相关。
Using yearly data, if x represents CO₂ (ppm) and y represents temperature anomaly (°C), a sample produces r = 0.89. This suggests a strong association, but correlation does not imply causation. Environmental scientists must consider other variables, such as solar activity and ocean currents, before drawing conclusions about global warming.
使用年度数据，若 x 表示 CO₂ (ppm)，y 表示气温距平 (°C)，某样本得出 r = 0.89。这表明有很强的联系，但相关性并不意味着因果关系。环境科学家在得出全球变暖的结论前须考虑其他变量，如太阳活动和洋流。
7. Least Squares Regression for Prediction | 最小二乘回归预测
When a strong correlation exists, a regression line can be used for prediction. The least squares regression line of y on x has the form y = a + bx, where b = Sₓᵧ / Sₓₓ and a = ȳ − b x̄. This line minimises the sum of squared vertical distances from the data points. It enables estimation of the temperature anomaly for a given CO₂ level, assuming the relationship remains linear.
当存在较强相关性时，可用回归直线进行预测。y 对 x 的最小二乘回归直线形式为 y = a + bx，其中 b = Sₓᵧ / Sₓₓ，a = ȳ − b x̄。该直线使数据点垂直距离的平方和最小。它能够估算给定 CO₂ 水平下的气温距平，前提是这种关系保持线性。
For a data set with x̄ = 400 ppm, ȳ = 0.6 °C, Sₓₓ = 5000, Sₓᵧ = 3500, we find b = 3500/5000 = 0.7, and a = 0.6 − 0.7×400 = −279.4. The regression equation is y = −279.4 + 0.7x. If CO₂ reaches 450 ppm, the predicted anomaly is y = −279.4 + 0.7×450 = 35.6 °C? That seems
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