The Carbon Cycle: A Mathematical Modelling Approach | 碳循环：数学建模方法

📚 The Carbon Cycle: A Mathematical Modelling Approach | 碳循环：数学建模方法

The carbon cycle describes the movement of carbon between the atmosphere, oceans, land and living organisms. At A-level, Edexcel Mathematics opens powerful ways to understand these flows through differential equations, exponential functions and logarithms. This article explores how to build and analyse simple mathematical models of the carbon cycle, linking key pure mathematics skills to a real global system.

碳循环描述碳在大气、海洋、陆地和生物体之间的迁移。在 A-level 阶段，Edexcel 数学通过微分方程、指数函数和对数提供了强有力的方法来理解这些通量。本文探讨如何建立和分析碳循环的简单数学模型，将核心纯数技能与真实的全球系统联系起来。

---

1. Carbon Reservoirs and Fluxes | 碳库与碳通量

In a mathematical model, the carbon cycle is split into reservoirs – the atmosphere, vegetation, soil, surface ocean and deep ocean – each storing a certain mass of carbon, usually measured in gigatonnes of carbon (GtC). Fluxes are the rates of carbon transfer between reservoirs, expressed in GtC per year.

在数学模型中，碳循环被划分为几个库——大气、植被、土壤、表层海洋和深层海洋——每个库储存一定质量的碳，通常以十亿吨碳（GtC）为单位。通量是碳在各库之间转移的速率，以 GtC/年 表示。

The table below gives approximate pre-industrial steady-state values; such numbers become parameters in our differential equations.

下表给出了工业革命前近似稳态值；这些数值将成为我们微分方程中的参数。

Reservoir	Mass (GtC)
Atmosphere (A)	597
Vegetation (V)	650
Surface ocean (S)	900
Deep ocean (D)	37,000

---

2. Modelling Carbon Exchange Between Two Reservoirs | 两个碳库间碳交换的建模

Consider the atmosphere and the surface ocean. Carbon moves from atmosphere to ocean by dissolution, and from ocean to atmosphere by outgassing. A simple assumption is that each flux is proportional to the mass of carbon in the source reservoir. Let A be the mass of carbon in the atmosphere and S that in the surface ocean. The net rate of change of atmospheric carbon can be written as dA/dt = –k₁A + k₂S, where k₁ and k₂ are rate constants with units yr⁻¹.

考虑大气和表层海洋。碳通过溶解从大气进入海洋，又通过释气从海洋返回大气。一个简单的假设是：每个通量与源库的碳质量成正比。设 A 为大气碳质量，S 为表层海洋碳质量。大气碳量的净变化率可以写作 dA/dt = –k₁A + k₂S，其中 k₁ 和 k₂ 是速率常数，单位为 yr⁻¹。

This is a linked first-order linear differential equation; Edexcel students meet such structures in modelling with differential equations.

这是一个耦合的一阶线性微分方程；Edexcel 学生在微分方程建模中会遇到这类结构。

---

3. Deriving the Differential Equation | 推导微分方程

Assuming the total carbon in the two reservoirs is constant (A + S = T), we can eliminate S. Then dA/dt = –k₁A + k₂(T – A) = k₂T – (k₁+ k₂)A. This is a standard first-order linear ODE of the form dA/dt = c – m A, where c = k₂T and m = k₁ + k₂.

假设两个库的总碳量恒定 (A + S = T)，我们可以消去 S。于是 dA/dt = –k₁A + k₂(T – A) = k₂T – (k₁+ k₂)A。这是一个标准的一阶线性常微分方程，形式为 dA/dt = c – m A，其中 c = k₂T，m = k₁ + k₂。

Recognising this pattern allows us to use the integrating factor method or to identify it as a separable equation, both core techniques in Edexcel Pure Mathematics.

识别出这一形式后，我们可以使用积分因子法或视为可分离方程，这两者都是 Edexcel 纯数的核心技巧。

---

4. Solving the Differential Equation – Exponential Change | 求解微分方程——指数变化

Separation of variables gives ∫ dA/(c – mA) = ∫ dt. Integrating leads to –(1/m) ln|c – mA| = t + constant, so c – mA = C e^(–mt). Hence A(t) = c/m + (A₀ – c/m) e^(–mt). This shows that atmospheric carbon approaches the equilibrium value A_eq = c/m = k₂T/(k₁+ k₂) exponentially, governed by e^(–mt).

分离变量得 ∫ dA/(c – mA) = ∫ dt。积分得到 –(1/m) ln|c – mA| = t + 常数，从而 c – mA = C e^(–mt)。因此 A(t) = c/m + (A₀ – c/m) e^(–mt)。这表明大气碳量以指数形式趋近平衡值 A_eq = c/m = k₂T/(k₁+ k₂)，由 e^(–mt) 决定。

The time constant 1/m represents the adjustment timescale; a larger m means a quicker return to equilibrium. Edexcel students interpret m as the rate constant in exponential decay towards steady state.

时间常数 1/m 表示调整的时间尺度；m 越大，返回平衡越快。Edexcel 学生将其解读为趋向稳态的指数衰减速率常数。

---

5. Radiocarbon Dating and Exponential Decay | 放射性碳定年与指数衰减

Carbon-14 (¹⁴C) is continually formed in the atmosphere and incorporated into living tissue. When an organism dies, ¹⁴C decays with a half-life of about 5730 years. This process is modelled by dN/dt = –λN, where N is the number of ¹⁴C atoms and λ is the decay constant. The solution is N = N₀ e^(–λt).

碳-14 (¹⁴C) 在大气中不断形成并被生物体吸收。生物体死亡后，¹⁴C 发生衰变，半衰期约为 5730 年。这一过程可用 dN/dt = –λN 建模，其中 N 为 ¹⁴C 原子数目，λ 为衰变常数。解为 N = N₀ e^(–λt)。

Linking λ to the half-life t₁/₂ uses the exponential equation: N₀/2 = N₀ e^(–λ t₁/₂) ⇒ e^(λ t₁/₂) = 2 ⇒ λ = (ln 2)/t₁/₂. This connection is a standard exam application of exponentials and natural logarithms.

将 λ 与半衰期 t₁/₂ 关联需运用指数方程：N₀/2 = N₀ e^(–λ t₁/₂) ⇒ e^(λ t₁/₂) = 2 ⇒ λ = (ln 2)/t₁/₂。这一联系是指数和对数的标准考试应用题。

---

6. Determining the Age of a Sample – Using Logarithms | 确定样本年龄——使用对数

Given the remaining fraction of ¹⁴C, N/N₀, the age t is found by rearranging: t = (1/λ) ln(N₀/N). For example, if N/N₀ = 0.35, then t = (t₁/₂ / ln 2) × ln(1/0.35). With t₁/₂ = 5730 years, t ≈ (5730/0.6931)×1.0498 ≈ 8670 years.

给定 ¹⁴C 的剩余比例 N/N₀，年龄 t 可通过移项求得：t = (1/λ) ln(N₀/N)。例如，若 N/N₀ = 0.35，则 t = (t₁/₂ / ln 2) × ln(1/0.35)。代入 t₁/₂ = 5730 年，t ≈ (5730/0.6931)×1.0498 ≈ 8670 年。

Edexcel questions often require log manipulation and careful unit handling, exactly as demonstrated here.

Edexcel 考试题常要求对数运算和仔细的单位处理，如本例所示。

---

7. Steady States and Equilibrium Analysis | 稳态与平衡分析

In the two-reservoir model, equilibrium occurs when dA/dt = 0, giving A_eq = (k₂/(k₁+k₂)) T. If human emissions add a constant flux F (GtC/yr), the differential equation becomes dA/dt = F + k₂T – (k₁+k₂)A. The new equilibrium is A_eq’ = (F + k₂T)/(k₁+k₂), showing a raised atmospheric carbon level.

在两库模型中，当 dA/dt = 0 时达到平衡，得到 A_eq = (k₂/(k₁+k₂)) T。如果人类排放增加一个恒定的通量 F (GtC/年)，微分方程变为 dA/dt = F + k₂T – (k₁+k₂)A。新的平衡点为 A_eq’ = (F + k₂T)/(k₁+k₂)，显示大气碳水平升高。

Setting dA/dt = 0 is a simple algebraic task; it demonstrates how external forcing shifts the steady state, a concept easily visualised with graphs of A against t.

令 dA/dt = 0 是一个简单的代数任务；它展示了外部强迫如何移动稳态，通过绘制 A 随时间 t 变化的图像可以直观呈现。

---

8. Sensitivity to Parameters – Rate Constants | 参数敏感性——速率常数

The rate constants k₁ and k₂ determine how quickly the system responds. If k₁ is large (rapid dissolution), atmospheric carbon is drawn down faster. A sensitivity analysis varies k₁ while holding k₂ constant; students can compute A_eq for different values and observe the hyperbolic relationship A_eq ∝ k₂/(k₁+k₂).

速率常数 k₁ 和 k₂ 决定了系统的响应速度。若 k₁ 较大（溶解快），大气碳被吸收得更快。敏感性分析可固定 k₂，改变 k₁；学生可以计算不同 k₁ 值下的 A_eq，并观察到 A_eq 与 k₂/(k₁+k₂) 的双曲线关系。

This links to graph sketching and rational functions, where A_eq levels off as k₁ increases. It is a beautiful crossover between pure mathematics and environmental science.

这联系到图像绘制和有理函数，随着 k₁ 增大，A_eq 趋于平缓。这是纯数学与环境科学之间美妙的交融。

---

9. Multi-reservoir Carbon Cycle Models | 多库碳循环模型

The real carbon cycle involves more reservoirs. A three-reservoir atmosphere–vegetation–soil model can be described by a system of linear differential equations. Using matrix notation, dx/dt = M x + f, where x is a vector of carbon masses, M contains rate constants, and f represents inputs. Edexcel Further Mathematics students can apply eigenvalues and eigenvectors to analyse long-term behaviour.

真实的碳循环涉及多个库。一个大气–植被–土壤三库模型可用线性微分方程组描述。采用矩阵记法，dx/dt = M x + f，其中 x 是碳质量向量，M 包含速率常数，f 代表输入。Edexcel 进阶数学学生可运用特征值和特征向量分析长期行为。

Even at single mathematics A-level, a simpler approach with sequential compartment models using repeated exponential expressions is possible, reinforcing skills in manipulating e^(–kt) terms.

即使只学习单科数学 A-level，也可以用重复的指数表达式建立序贯分室模型，从而强化处理 e^(–kt) 项的技能。

---

10. Applying A-Level Maths Skills to Environmental Data | 将A-Level数学技能用于环境数据

Edexcel exam-style questions increasingly embed mathematical modelling in real-world contexts. A carbon cycle problem may give a table of atmospheric CO₂ concentration over time and ask students to estimate the net flux using numerical differentiation or to fit an exponential model using log-linear graphs.

Edexcel 考试风格的题目越来越多地将数学建模嵌入真实情境。一道碳循环题可能给出大气 CO₂ 浓度随时间变化的数据表，要求学生使用数值微分估算净通量，或通过半对数图拟合指数模型。

Converting exponential data to linear form by taking logarithms (ln C vs t) and using least-squares regression mirrors the statistics content in A-level Mathematics, unifying exponential growth, calculus and data analysis.

通过对数变换将指数数据转化为线性形式（ln C 与 t 作图），并使用最小二乘回归，这与 A-level 数学中的统计内容相呼应，将指数增长、微积分和数据分析融为一体。

---

11. Common Mistakes and Key Exam Tips | 常见错误与关键考试技巧

When solving dA/dt = c – mA, students often forget to divide by m after integration, or misplace the constant of integration. Always write the solution in the form A = A_eq + (A₀ – A_eq)e^(–mt) and check that when t=0, A=A₀, and as t→∞, A→A_eq. Also, ensure units are consistent – GtC, years and yr⁻¹ must match.

在求解 dA/dt = c – mA 时，学生常忘记积分后除以 m，或弄错积分常数。始终将解写成 A = A_eq + (A₀ – A_eq)e^(–mt) 的形式，并验证 t=0 时 A=A₀，以及 t→∞ 时 A→A_eq。此外，要确保单位一致——GtC、年和 yr⁻¹ 必须匹配。

Logarithmic steps in carbon dating require careful handling: ln(N₀/N) cannot be negative if N/N₀ < 1 because ln of a number less than 1 is negative, which makes t positive after including the negative sign from λ. Remember λ = ln2/t₁/₂, so always write the time formula as t = (t₁/₂ / ln2) · ln(N₀/N).

碳定年中的对数步骤需要小心处理：若 N/N₀ < 1，ln(N₀/N) 不能为负，因为小于 1 的数的对数为负，考虑 λ 前的负号后 t 为正。记住 λ = ln2/t₁/₂，所以时间公式始终写作 t = (t₁/₂ / ln2) · ln(N₀/N)。

---

12. Summary – The Carbon Cycle as a Mathematical Playground | 总结——碳循环作为数学演练场

The carbon cycle offers a rich, exam-friendly context for practising first-order differential equations, exponential growth and decay, logarithmic transformation, equilibrium analysis and parameter sensitivity. By mastering the models shown here, students sharpen the skills tested in Edexcel Pure Mathematics and begin to see how mathematics illuminates urgent planetary questions.

碳循环为练习一阶微分方程、指数增长与衰减、对数变换、平衡分析和参数敏感性提供了一个丰富且适合考试的情境。通过掌握这里展示的模型，学生们可以磨炼 Edexcel 纯数考查的技能，并开始领会数学如何照亮紧迫的全球问题。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）