Sustainability: Exponential Growth, Decay and Resource Models | 可持续性：指数增长、衰减与资源模型

📚 Sustainability: Exponential Growth, Decay and Resource Models | 可持续性：指数增长、衰减与资源模型

Sustainability is not a stand-alone Edexcel module, but the mathematics behind it - exponential growth, exponential decay, logarithmic graphs and first-order differential equations - runs through the whole Pure and Applied specification. This article collects those techniques into one place and shows exactly how examiners wrap them inside environmental contexts.

可持续性并不是 Edexcel 的一个独立模块，但它背后的数学——指数增长、指数衰减、对数图像和一阶微分方程——贯穿了整个纯数学与应用数学的考试规范。本文将把这些技巧集中整理，并展示考官如何把它们包装进环境类情境中。

Every sustainability question is, at heart, a standard question about a quantity that changes at a rate proportional to its current size. Once you recognise that structure, the context stops being a distraction.

每一道可持续性题目本质上都是一道标准题：某个量的变化率与它当前的数值成正比。一旦你识别出这个结构，题目背景就不再是干扰。

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1. Why Sustainability Appears in A-Level Maths | 为什么可持续性出现在 A-Level 数学中

Sustainability is not a separate Edexcel topic, but it links many parts of the specification: exponential growth, logarithmic graphs, differential equations and rates of change. Models of population, pollution, radioactive decay and resource use all fit this theme, and the same four or five mathematical skills keep reappearing under different disguises.

可持续性并不是 Edexcel 的一个独立专题，但它连接了规范中的许多部分：指数增长、对数图像、微分方程和变化率。人口、污染、放射性衰变和资源使用等模型都属于这一主题，而同样那四五个数学技巧总是换着外衣反复出现。

to their current value, which leads directly to exponential models of the form N = N₀eᵏᵗ.

关键思想是：许多环境量的变化率与它当前的数值成正比，这直接导向形如 N = N₀eᵏᵗ 的指数模型。

Before you start, note where each skill sits in the Pearson Edexcel A Level Mathematics specification:

开始之前，先看清每项技能在 Pearson Edexcel A Level 数学规范中的位置：

• Exponential models and graphs - Pure Mathematics 1 and 2.
• 对数与指数模型和图像 —— 纯数学 1 与 2。
• Laws of logarithms and linearising data - Pure Mathematics 1.
• 对数运算法则与数据线性化 —— 纯数学 1。
• Solving equations with ln and e - Pure Mathematics 2.
• 用 ln 和 e 解方程 —— 纯数学 2。
• First-order differential equations and modelling - Pure Mathematics 2.
• 一阶微分方程与建模 —— 纯数学 2。
• Connected rates of change - Pure Mathematics 2 and Mechanics.
• 相关变化率 —— 纯数学 2 与力学。

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2. Exponential Growth: The Core Model | 指数增长：核心模型

e a is the growth factor; in continuous form, N = N₀eᵏᵗ where k is the continuous growth rate per unit time.

指数增长模型表示：在时刻 t 的数量等于初始量 N₀ 在每个相等时间步上乘以同一个因子。离散形式为 N = N₀aᵗ，其中 a 是增长因子；连续形式为 N = N₀eᵏᵗ，其中 k 是每单位时间的连续增长率。

If a population increases by r% r small percentages these two numbers are almost identical, which is why examiners often let you work with either form.

如果人口每年增长 r%，则 a = 1 + r/100，对应的连续增长率为 k = ln a。当 r 很小时这两个数几乎相等，这也是考官允许你用任意一种形式作答的原因。

for N to equal 2N₀, so 2 = eᵏᵗ and therefore t = ln 2 ÷ k. This single result answers most "how long until..." questions in sustainability contexts.

倍增时间是指 N 变为 2N₀ 所需的时间，因此 2 = eᵏᵗ，从而 t = ln 2 ÷ k。这一个结果就能回答大部分可持续性情境中的"多久之后……"问题。

Worked outline: a town has 240 000 residents in 2020 and grows by 1.8% ice how close 0.018 and 0.01784 are - small-rate approximations are usually accepted, but exact working is safer.

示例思路：某城镇 2020 年有 24 万居民，每年增长 1.8%。则 P = 240 000 × 1.018ᵗ，且 k = ln 1.018 ≈ 0.01784，因此倍增时间为 ln 2 ÷ 0.01784 ≈ 38.9 年。注意 0.018 与 0.01784 非常接近——小增长率近似通常可以被接受，但精确运算更稳妥。

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3. Exponential Decay and Radioactive Half-Life | 指数衰减与放射性半衰期

N₀e⁻ᵏᵗ with k > 0. The half-life T is the time for N to fall to half of N₀, giving ½ = e⁻ᵏᵀ and hence k = ln 2 ÷ T.

衰减使用同一套工具，只是指数为负：N = N₀e⁻ᵏᵗ，其中 k > 0。半衰期 T 是 N 降到 N₀ 一半所需的时间，由 ½ = e⁻ᵏᵀ 可得 k = ln 2 ÷ T。

Equivalent discrete form: N = N₀(½)^{t ÷ T}, which is often quicker when the question gives a half-life directly.

等价的离散形式：N = N₀(½)^{t ÷ T}，当题目直接给出半衰期时，这种写法往往更快。

ife of about 5730 years, so k = ln 2 ÷ 5730 ≈ 1.21 × 10⁻⁴ per year. If a sample retains 40% ginal carbon-14, then 0.4 = e⁻ᵏᵗ and t = ln(1 ÷ 0.4) ÷ k ≈ 0.9163 ÷ 0.000121 ≈ 7580 years.

碳-14 测年是经典的考试情境。碳-14 的半衰期约为 5730 年，因此 k = ln 2 ÷ 5730 ≈ 1.21 × 10⁻⁴ 每年。如果某样本还保留原始碳-14 的 40%，则 0.4 = e⁻ᵏᵗ，t = ln(1 ÷ 0.4) ÷ k ≈ 0.9163 ÷ 0.000121 ≈ 7580 年。

Other decay contexts worth recognising: a drug leaving the bloodstream, radioactive waste from a power station, and the breakdown of a plastic pollutant in the ocean. In every case the mathematics is identical and only the numbers change.

其他值得识别的衰减情境：药物从血液中代谢、核电站的放射性废物、以及海洋中塑料污染物的降解。在任何一种情况下，数学完全相同，只是数字不同。

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4. Logarithms: Turning Curves into Straight Lines | 对数：把曲线变成直线

aking logarithms linearises it, which is why log graphs are so heavily examined. If y = abˣ b, so a graph of ln y against x is a straight line with gradient ln b and intercept ln a.

实验数据很少直接呈现指数形态。取对数可以把它线性化，这正是对数图像被大量考查的原因。若 y = abˣ，则 ln y = ln a + x ln b，因此以 ln y 对 x 作图是一条直线，斜率为 ln b，截距为 ln a。

h of ln y against ln x has gradient n and intercept ln a. This power-law form is used for species-area relationships in ecology and for many resource-availability models.

若 y = axⁿ，则 ln y = ln a + n ln x，因此以 ln y 对 ln x 作图，斜率为 n，截距为 ln a。这种幂律形式常用于生态学中的物种—面积关系以及许多资源可得性模型。

y means base 10, while ln always means base e. Converting is easy: log₁₀ x = ln x ÷ ln 10.

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务必看清题目要求以什么为底。在 A-Level 数学中，未标明底数的 log 通常指以 10 为底，而 ln 始终指以 e 为底。换算很简单：log₁₀ x = ln x ÷ ln 10。

Typical exam instruction: "Plot ln N against t and use your line of best fit to estimate the decay constant." The gradient is -k, and the intercept is ln N₀. Interpret the gradient with its unit, and remember that a negative gradient in a decay model means a positive decay constant.

典型考法："以 ln N 对 t 作图，并用最佳拟合直线估计衰减常数。"斜率就是 -k，截距是 ln N₀。作答时要写出斜率的单位，并记住：衰减模型中斜率为负意味着衰减常数为正。

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5. Solving Equations with ln and e | 用 ln 和 e 解方程

questions reduce to one of three algebraic moves. First, isolate the exponential: from 3e^0.2t = 45 we get e^0.2t = 15, then 0.2t = ln 15 ≈ 2.708, so t ≈ 13.54.

大多数可持续性题目最终都归结为三种代数操作。第一，把指数项孤立出来：由 3e^0.2t = 45 得 e^0.2t = 15，于是 0.2t = ln 15 ≈ 2.708，t ≈ 13.54。

Second, substitute to create a quadratic. For e²ˣ - 5eˣ + 6 = 0, let y = eˣ = 2 or y = 3, giving x = ln 2 ≈ 0.693 or x = ln 3 ≈ 1.099. Both roots are valid because eˣ is always positive.

第二，换元构造二次方程。对于 e²ˣ - 5eˣ + 6 = 0，令 y = eˣ。则 y² - 5y + 6 = 0，故 y = 2 或 y = 3，得 x = ln 2 ≈ 0.693 或 x = ln 3 ≈ 1.099。两个根都有效，因为 eˣ 恒为正。

, ln(N ÷ N₀) = -kt rearranges to N = N₀e⁻ᵏᵗ, which is exactly the decay model in disguise.

第三，利用对数法则合并对数：ln a + ln b = ln(ab)，ln a - ln b = ln(a ÷ b)，n ln a = ln(aⁿ)。例如 ln(N ÷ N₀) = -kt 可整理为 N = N₀e⁻ᵏᵗ，这正是伪装后的衰减模型。

Watch the domain: ln x is defined only for x > 0, so any solution that requires taking the logarithm of a negative number must be rejected. This is the single most common lost mark in this topic.

注意定义域：ln x 仅在 x > 0 时有定义，因此任何需要取负数对数的解都必须舍去。这是本专题中最常见的失分点。

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6. Differential Equations: dN ÷ dt = kN | 微分方程：dN ÷ dt = kN

The phrase "rate of change proportional to the current amount" y. Separating variables gives ∫ (1 ÷ N) dN = ∫ k dt, so ln N = kt + c and hence N = N₀eᵏᵗ.

"变化率与当前数量成正比"这句话可以直接翻译成微分方程 dN ÷ dt = kN，其中 k 是比例常数。分离变量得 ∫ (1 ÷ N) dN = ∫ k dt，故 ln N = kt + c，从而 N = N₀eᵏᵗ。

found from the initial condition. If dP ÷ dt = 0.05P and P = 200 when t = 0, then P = 200e^0.05t, and P reaches 400 when t = ln 2 ÷ 0.05 ≈ 13.9 time units.

常数 c 由初始条件确定。若 dP ÷ dt = 0.05P 且 t = 0 时 P = 200，则 P = 200e^0.05t，当 t = ln 2 ÷ 0.05 ≈ 13.9 个时间单位时 P 达到 400。

You must be able to run the argument in reverse as well. Given N = 500e^0.3t, differentiating gives dN ÷ dt = 0.3 × 500e^0.3t = 0.3N, so the rate is proportional to N with constant 0.3 per unit time. This "show that" style is extremely common.

你也必须能反向论证。给定 N = 500e^0.3t，求导得 dN ÷ dt = 0.3 × 500e^0.3t = 0.3N，因此变化率与 N 成正比，比例常数为每单位时间 0.3。这种"证明"型题目极为常见。

Exam tip: always write the general solution before substituting the initial condition, and keep the "+c" until after you integrate. Marks are awarded for the method line even when the arithmetic goes wrong.

考试提示：先写出通解再代入初始条件，并在积分完成前一直保留 "+c"。即使算术出错，方法行依然可以得分。

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7. The Logistic Model and Limiting Resources | 逻辑斯蒂模型与资源限制

= kP(1 - P ÷ L), where L is the carrying capacity - the maximum population the environment can sustain.

纯粹的指数增长不可能永远持续，因为栖息地的资源是有限的。逻辑斯蒂模型把恒定增长替换为 dP ÷ dt = kP(1 - P ÷ L)，其中 L 是环境容纳量——即环境能够维持的最大种群数量。

When P is small, the factor (1 - P ÷ L) is close to 1 and growth is almost exponential. As P approaches L, the factor approaches zero and growth slows to nothing. The population curve is a sigmoid: slow start, rapid middle, plateau at L.

当 P 很小时，因子 (1 - P ÷ L) 接近 1，增长几乎是指数式的。当 P 趋近 L 时，该因子趋近于零，增长趋于停止。种群曲线呈 S 形：开始缓慢，中段迅速，最终在 L 处趋于平缓。

ᵏᵗ), where A depends on the initial population. You can verify it by differentiation, and examiners often ask you to show that a given expression satisfies the differential equation.

该方程的解形如 P = L ÷ (1 + Ae⁻ᵏᵗ)，其中 A 由初始种群决定。你可以通过求导验证它，考官也常要求你证明某个给定表达式满足该微分方程。

L - P) is largest, which happens at P = L ÷ 2. This midpoint is a favourite exam question: "Find the population at which the rate of growth is greatest."

增长最快发生在乘积 P(L - P) 最大时，即 P = L ÷ 2。这个中点位置是考官偏爱的考点："求增长速率最大时的种群数量。"

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8. Modified Models: Harvesting, Depletion and Pollution | 修正模型：采伐、消耗与污染

equilibrium population P* = h ÷ k, the level at which births exactly replace the harvest.

真实的资源模型会加上一个移除项。以恒定捕捞率 h 捕捞的渔场满足 dP ÷ dt = kP - h。令 dP ÷ dt = 0 得平衡种群 P* = h ÷ k，即出生量恰好补偿捕捞量的水平。

If P₀ is below the equilibrium the population grows towards it; if P₀ is above it the population falls towards it. If h is set larger than kL, the equilibrium exceeds the carrying capacity and the resource collapses - a direct mathematical statement of over-exploitation.

其解为 P = h ÷ k + (P₀ - h ÷ k)eᵏᵗ。若 P₀ 低于平衡值，

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