Modelling Water Security with A-Level Mathematics | 利用A-Level数学建模水安全

📚 Modelling Water Security with A-Level Mathematics | 利用A-Level数学建模水安全

Water security is a critical global challenge, encompassing the sustainable availability and quality of freshwater resources. As populations grow and climate patterns shift, the management of water resources demands precise quantitative analysis. A-Level Mathematics provides a powerful toolkit for modelling water systems, from predicting demand and assessing drought risks to optimising storage and testing water quality. This article explores how key topics in the Edexcel A-Level Mathematics syllabus—including algebra, calculus, sequences, probability distributions, and hypothesis testing—can be applied to real-world water security problems. Through practical examples, we will see how mathematical models aid decision-making in the face of uncertainty.

水安全是全球性的严峻挑战，涉及淡水资源的可持续供应与质量。随着人口增长和气候模式的变化，水资源管理需要精确的定量分析。A-Level数学提供了一套强大的工具，可为水系统建模，从预测需求和评估干旱风险到优化储存和测试水质。本文探讨了Edexcel A-Level数学大纲中的关键主题——包括代数、微积分、数列、概率分布和假设检验——如何应用于现实世界的水安全问题。通过实际示例，我们将看到数学模型如何在不确定性中辅助决策。

1. Water Supply and Demand Modelling Using Linear Equations | 使用线性方程对水供应和需求建模

A simple model for water supply and demand can be constructed using linear equations. Suppose the annual water supply from a reservoir is constant, described by S(t) = a, while domestic water demand is projected to grow linearly, D(t) = b + ct, where t is time in years. The point of intersection of these lines indicates when demand will exceed supply. Solving S(t) = D(t) yields the critical year t = (a − b)/c. For instance, if supply is 500 million cubic metres, current demand is 400, and demand increases by 10 million annually, the deficit occurs after 10 years. This linear approach allows policymakers to visualise the time frame for investing in new infrastructure.

一个简单的水供应与需求模型可用线性方程构建。假设水库的年供水量为常数，用S(t)=a表示，而生活用水需求预计呈线性增长，D(t)=b + ct，其中t为时间（年）。这两条直线的交点表示需求何时超过供应。解方程S(t)=D(t)得出关键年份t = (a − b)/c。例如，若供应量为5亿立方米，当前需求为4亿，且需求每年增加1千万，那么10年后会出现短缺。这种线性方法使决策者能够直观看到投资新基础设施的时间框架。

2. Geometric Sequences and Reservoir Depletion | 等比数列与水库枯竭

When a reservoir experiences a net outflow without replenishment, the water volume decreases by a constant percentage each month, forming a geometric sequence. If the initial volume is V₀ and it loses 5% each month, the volume after n months is Vₙ = V₀ × (0.95)ⁿ. This model helps to predict how many months remain before the reservoir reaches a critical level. For example, if V₀ = 2000 mega litres and the critical level is 500 ML, solve 2000 × (0.95)ⁿ < 500. Taking logs, n > log(500/2000) / log(0.95) ≈ 27.3, so after 28 months the water becomes critically low.

当水库没有补充而经历净流出时，每月水量按固定百分比减少，形成等比数列。若初始水量为V₀，每月损失5%，则n个月后的水量为Vₙ = V₀ × (0.95)ⁿ。该模型可预测水库达到警戒水位前还有多少个月。例如，V₀=2000兆升，警戒水位500兆升，解2000×(0.95)ⁿ < 500。取对数得 n > log(500/2000) / log(0.95) ≈ 27.3，因此28个月后水量将降至警戒线以下。

The sum of a geometric series can also estimate cumulative water losses over a period. If the monthly losses in megalitres follow a geometric progression 100, 95, 90.25,…, the total loss over 12 months is S₁₂ = 100(1 − 0.95¹²)/(1 − 0.95) ≈ 1103.5 ML. Such calculations are vital for water audits.

等比数列的求和也可估计一段时间内的累计水量损失。如果每月的损失量（兆升）遵循等比数列100, 95, 90.25,…，那么12个月的总损失量为S₁₂ = 100(1 − 0.95¹²)/(1 − 0.95) ≈ 1103.5兆升。这类计算对水资源审计至关重要。

3. Exponential Growth and Water Demand Forecasting | 指数增长与用水需求预测

Water demand often grows exponentially due to population and economic expansion. An exponential model D(t) = D₀eᵏᵗ can be used, where D₀ is initial demand and k is the continuous growth rate. If annual demand is 300 million m³ and grows at 2% per year continuously, after 15 years the demand is 300e^(0.02×15) = 300e^0.3 ≈ 404.96 million m³. Differentiating this function gives the instantaneous rate of change of demand, helping planners to anticipate infrastructure needs.

由于人口和经济扩张，用水需求通常呈指数增长。可使用指数模型 D(t) = D₀eᵏᵗ，其中D₀为初始需求，k为连续增长率。如果年需求量为3亿立方米，并以每年2%的速度连续增长，15年后需求为300e^(0.02×15)=300e^0.3 ≈ 4.0496亿立方米。对该函数求导可得出需求的瞬时变化率，帮助规划者预测基础设施需求。

The doubling time for exponential growth is a useful metric given by t_double = ln 2 / k. For a growth rate of 2%, t_double ≈ 0.693 / 0.02 = 34.65 years. This clearly indicates how urgently new water sources must be developed.

指数增长的翻倍时间是一个有用的指标，由t_double = ln 2 / k给出。对于2%的增长率，t_double ≈ 0.693/0.02 = 34.65年。这清楚地表明开发新水源的紧迫程度。

4. Probability Distributions for Rainfall Analysis | 降雨分析的概率分布

The occurrence of extreme rainfall events can be modelled using a Poisson distribution. If a region experiences an average of λ = 2 severe storms per year, the probability of exactly x storms in a year is P(X=x) = e⁻² × 2ˣ / x!. This helps in designing flood defences. The probability of at least one storm is 1 − P(X=0) = 1 − e⁻² ≈ 0.865.

极端降雨事件的发生可用泊松分布建模。如果一个区域平均每年发生λ=2次强风暴，则一年中恰好发生x次风暴的概率为P(X=x) = e⁻² × 2ˣ / x!。这有助于设计防洪工程。至少发生一次风暴的概率为1 − P(X=0) = 1 − e⁻² ≈ 0.865。

For a 5-year period, the expected number is λ_total = 10, so the probability of zero storms is e⁻¹⁰ ≈ 0.000045, indicating near certainty of extreme weather within that span. With climate change, λ may increase, requiring updated models.

对于5年时期，期望次数λ_total=10，因此无风暴的概率为e⁻¹⁰≈0.000045，表明在此期间几乎肯定会发生极端天气。随着气候变化，λ可能增加，需要更新模型。

5. The Normal Distribution and Drought Risk | 正态分布与干旱风险

Annual rainfall in a region often follows a normal distribution with mean μ and standard deviation σ. Drought is typically defined as rainfall below a certain threshold. If μ = 800 mm, σ = 150 mm, and drought occurs when rainfall < 500 mm, we calculate the z-score: z = (500 − 800)/150 = −2. The probability from standard normal tables is about 0.0228, meaning a 2.28% chance each year. This informs water rationing plans.

一个地区的年降雨量通常服从正态分布，均值为μ，标准差为σ。干旱通常定义为降雨量低于某个阈值。若μ=800 mm，σ=150 mm，当降雨量<500 mm时发生干旱，计算z值：z=(500 − 800)/150 = −2。查标准正态表可得概率约为0.0228，即每年有2.28%的几率。这为限水计划提供了依据。

Conversely, water resource managers might set a 95% reliability level for supply. They need to know the rainfall amount such that 95% of years exceed it: x = μ + zσ. For z = −1.645 (5th percentile), x = 800 − 1.645×150 ≈ 553.3 mm. Design capacity must meet demand even in dry years around this level.

相反，水资源管理者可能设定95%的供应可靠性水平。他们需要知道有95%的年份降雨量超过的数值：x = μ + zσ。对于z = −1.645（第5百分位），x = 800 − 1.

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）