📚 Edexcel Maths 5.7: Modelling Respiration with Exponential Functions | Edexcel 数学 5.7：用指数函数建模呼吸作用
In Edexcel Mathematics, Topic 5.7 focuses on exponential growth and decay, often examined through real-world contexts such as population growth, radioactive decay, drug clearance, and respiration. This article uses respiration as a modelling scenario to help you master the key techniques of writing, solving, and interpreting first-order rate equations, and to show exactly how the mark scheme rewards each step.
在 Edexcel 数学中，5.7 主题重点考察指数增长与指数衰减，常以现实情境出现，例如种群增长、放射性衰变、药物代谢和呼吸作用。本文以呼吸作用为建模背景，帮助你掌握一阶速率方程的建立、求解与解释等关键技巧，并说明评分标准在每一步上如何给分。
1. What Is Edexcel Maths 5.7? | 什么是 Edexcel 数学 5.7？
Section 5.7 of the Edexcel A Level Mathematics specification deals with exponential growth and decay. You are expected to model situations in which the rate of change of a quantity is proportional to its current value. The governing differential equation is dy/dx = ky, whose general solution is y = Aekx. When k > 0 the model represents growth; when k < 0 it represents decay.
在 Edexcel A Level 数学大纲中，5.7 节涉及指数增长与指数衰减。你需要对变化速率与当前值成正比的情形建立模型。其控制微分方程为 dy/dx = ky，通解为 y = Aekx。当 k > 0 时表示增长；当 k < 0 时表示衰减。
Exam questions often embed this topic in applied contexts. You may be given a table of experimental data and asked to decide whether an exponential model is appropriate, to find unknown constants, or to comment on the limitations of the model.
考试题常将本主题嵌入应用背景。你可能会得到一组实验数据表，并被要求判断指数模型是否合适、求出未知常数，或评价该模型的局限性。
The assessment objectives matter here. Writing down the correct differential equation, solving it by separating variables, and interpreting the constants in the context of the question are three separate skills, and each one carries marks.
这里要特别注意考核目标。写出正确的微分方程、用分离变量法求解、以及结合题目情境解释常数的含义，是三项独立的技能，每一项都单独计分。
2. Why Respiration Is Used in Exam Questions | 为什么考试题会以呼吸作用为背景
Respiration is a familiar biological process in which cells consume oxygen and release carbon dioxide. In simplified modelling, the rate of oxygen uptake can be treated as proportional to the remaining oxygen concentration. This gives a natural exponential decay context that links mathematics to real laboratory measurements.
呼吸作用是人们熟悉的生物过程，细胞消耗氧气并释放二氧化碳。在简化建模中，氧气的吸收速率可视为与剩余氧气浓度成正比。这提供了将数学与现实实验室测量联系起来的自然指数衰减情境。
You will not be expected to know detailed biochemistry. The question will normally state the modelling assumption explicitly, for example 'the rate of oxygen consumption is proportional to the concentration of oxygen remaining in the flask'. Your first job is to turn that sentence into an equation.
你不需要掌握详细的生物化学知识。题目通常会明确给出建模假设，例如 '氧气消耗速率与瓶中剩余氧气浓度成正比'。你的第一项任务就是把这句话翻译成方程。
The same context also supports the growth branch of 5.7. If a culture of yeast or bacteria is respiring, the cell count may grow exponentially as N = N₀ekt while the oxygen concentration falls as C = C₀e−kt. Examiners like this pairing because it tests whether you can handle a positive and a negative constant in the same scenario.
同一情境也能支撑 5.7 的增长分支。如果酵母或细菌培养液正在呼吸，细胞数量可能按 N = N₀ekt 指数增长，而氧气浓度按 C = C₀e−kt 下降。考官喜欢这种配对，因为它能检验你能否在同一情境中处理正、负两种常数。
3. Building the Model: From Words to a Differential Equation | 建立模型：从文字到微分方程
The phrase 'rate of change' always signals a derivative. If the rate of change of y is proportional to y, then we write dy/dt = ky, where k is the constant of proportionality. The sign of k tells you whether the quantity is increasing or decreasing.
'变化速率' 这个说法永远意味着导数。如果 y 的变化速率与 y 成正比，我们就写 dy/dt = ky，其中 k 是比例常数。k 的正负告诉你该量是在增加还是减少。
For oxygen uptake in a closed container, the concentration C is decreasing, so the derivative must be negative. The standard form is shown below.
对于密闭容器中的氧气消耗，浓度 C 在下降，因此导数必须为负。标准形式如下。
dC/dt = −kC (k > 0), C = C₀e−kt
Here C easured in units of reciprocal time, such as min⁻¹ or s⁻¹. Note that the minus sign is part of the model, not an accident: it encodes the fact that oxygen is being consumed.
其中 C₀ 是 t = 0 时的浓度，k 的单位是时间的倒数，例如 min⁻¹ 或 s⁻¹。注意这个负号是模型的一部分，并非笔误：它体现的是氧气被消耗这一事实。
A very common exam instruction is 'write down a differential equation for this situation'. One mark is usually available simply for the correct proportionality statement with the correct sign.
考试中非常常见的一道指令是 '写出该情形的微分方程'。通常只要写出带正确符号的比例关系，就能拿到一分。
4. Solving by Separating the Variables | 用分离变量法求解
Edexcel expects you to derive the solution rather than quote it. Starting from dC/dt = −kC, separate the variables and integrate both sides.
Edexcel 要求你推导出解，而不是直接引用。从 dC/dt = −kC 出发，分离变量并对两边积分。
∫ (1/C) dC = ∫ −k dt ⟹ ln C = −kt + c
Since concentrations are positive, the modulus signs can be dropped. Exponentiating both sides gives C = ece−kt, and writing A for the arbitrary constant ec produces the familiar result.
由于浓度为正，可以去掉绝对值符号。两边取指数得 C = ece−kt，把任意常数 ec 记作 A，就得到熟悉的结果。
C = Ae−kt, where A = C₀
Substituting t = 0 confirms that A is the initial concentration. This is the fastest way to find A in any question: it is always the value of the quantity at the start of the model.
代入 t = 0 可确认 A 就是初始浓度。这是在任何题目中求 A 的最快方法：它永远是模型开始时刻该量的取值。
5. Interpreting the Constants A and k | 解释常数 A 与 k 的含义
Examiners frequently ask you to 'state the meaning of the constant A' or 'interpret the value of k'. A is the initial value of the quantity, so for oxygen it is the concentration at the moment measurements began. k is the rate constant; it controls how quickly the exponential decays.
考官经常要求你 '说明常数 A 的含义' 或 '解释 k 的取值'。A 是该量的初始值，对氧气而言就是开始测量那一刻的浓度。k 是速率常数，它决定指数衰减的快慢。
The numerical value of k also has a percentage interpretation. After one unit of time, the fraction remaining is e−k. If k = 0.0297 min⁻¹, then e−0.0297 ≈ 0.9707, so about 97.1% of the oxygen remains after one minute, meaning roughly 2.9% is consumed in the first minute.
k 的数值还有百分比解释。经过一个时间单位后，剩余比例为 e−k。若 k = 0.0297 min⁻¹，则 e−0.0297 ≈ 0.9707，即一分钟后约 97.1% 的氧气仍然存在，也就是第一分钟内大约消耗 2.9%。
- Context sentence (EN): 'k is the constant rate of oxygen consumption per unit concentration, measured in min⁻¹.'
- 情境句（中）：'k 是单位浓度下氧气消耗的速率常数，单位为 min⁻¹。'
- Context sentence (EN): 'A is the initial oxygen concentration in the flask, measured in mg dm⁻³.'
- 情境句（中）：'A 是瓶内氧气的初始浓度，单位为 mg dm⁻³。'
Marks are lost when students write a mathematically correct but context-free answer such as 'A is the y-intercept'. Always name the physical quantity and give its unit.
如果学生写出数学上正确但没有情境的答案，比如 'A 是 y 轴截距'，就会失分。一定要说出具体的物理量并给出单位。
6. Half-Life and Doubling Time | 半衰期与倍增时间
The half-life is the time taken for the quantity to fall to half of its starting value. Setting C = C₀/2 in the model and solving gives a result that is independent of the starting amount.
半衰期是量降到初始值一半所需的时间。在模型中令 C = C₀/2 并求解，得到的结果与初始量无关。
t½ = ln 2 / k ≈ 0.693 / k
For a growth model N = N₀ekt with k > 0, the same algebra gives the doubling time t k. Notice that the two formulas are identical in form; only the physical meaning changes.
对于 k > 0 的增长模型 N = N₀ekt，同样的代数运算给出倍增时间 t₂ = ln 2 / k。注意两个公式形式完全相同，只是物理含义不同。
Half-life questions are often worth two or three marks: one for the correct equation, one for the correct rearrangement, and one for the answer with units. Never leave the answer as a bare number.
半衰期问题通常值两到三分：一分给正确的方程，一分给正确的变形，一分给带单位的答案。切勿只留一个光秃秃的数字。
7. Testing Whether an Exponential Model Fits | 检验指数模型是否合适
Many exam questions supply experimental data and ask whether an exponential model is appropriate. The standard technique is to transform the data. Taking natural logarithms of C = C₀e−kt gives a linear relationship.
许多考题会给出实验数据，并问指数模型是否合适。标准做法是对数据进行变换。对 C = C₀e−kt 取自然对数，得到一个线性关系。
ln C = ln C₀ − kt
This is of the form y = mx + c, so plotting ln C against t should produce a straight line with gradient −k and intercept ln C₀. If the plotted points lie close to a straight line, the exponential model is supported by the data.
这正是 y = mx + c 的形式，因此以 ln C 对 t 作图应得到一条直线，斜率为 −k，截距为 ln C₀。若数据点接近直线，则说明数据支持指数模型。
the gradient becomes −k / ln 10 ≈ −0.434k, so state clearly which logarithm you are using.
你也可以使用以 10 为底的对数。图像仍是直线，但斜率变为 −k / ln 10 ≈ −0.434k，因此必须明确说明你使用的是哪种对数。
- Plot ln C on the vertical axis and t on the horizontal axis. | 纵轴取 ln C，横轴取 t。
- Gradient = −k, so k = −gradient. | 斜率 = −k，故 k = −斜率。
- Intercept = ln C₀, so C₀ = eintercept. | 截距 = ln C₀，故 C₀ = e截距。
- Never read k directly off the original curved graph. | 不要直接从原始曲线图上读取 k。
8. Worked Example 1: Oxygen Uptake in a Sealed Flask | 例题一：密闭瓶中的氧气消耗
A sealed flask contains oxygen at a concentration of 8.0 mg dm⁻³. After 12 minutes the concentration has fallen to 5.6 mg dm⁻³. Assuming the rate of consumption is proportional to the remaining concentration, find k and the half-life.
一个密闭瓶内氧气浓度为 8.0 mg dm⁻³。12 分钟后浓度降至 5.6 mg dm⁻³。假设消耗速率与剩余浓度成正比，求 k 和半衰期。
Step 1: the model is C = 8.0e−kt. Step 2: substitute the data point.
第一步：模型为 C = 8.0e−kt。第二步：代入已知数据点。
5.6 = 8.0e−12k ⟹ e−12k = 0.7 ⟹ −12k = ln 0.7
97 min⁻¹ to three significant figures. The half-life is then ln 2 / 0.0297 ≈ 23.3 minutes.
由于 ln 0.7 = −0.35667，得到 k = 0.0297 min⁻¹（三位有效数字）。半衰期则为 ln 2 / 0.0297 ≈ 23.3 分钟。
To predict when the concentration reaches 2.0 mg dm⁻³, solve e−kt = 0.25, so kt . Always check that your answer is longer than the time already elapsed, which it is here.
若要求浓度何时降到 2.0 mg dm⁻³，解 e−kt = 0.25，得 kt = ln 4 = 1.3863，t ≈ 46.7 分钟。务必检查答案是否大于已流逝的时间，这里确实如此。
9. Worked Example 2: Fitting a Model to a Data Table | 例题二：用数据表拟合模型
The table below shows measured oxygen concentrations over 20 minutes. We test the exponential model by plotting ln C against t.
下表给出 20 分钟内测得的氧气浓度。我们通过作 ln C 对 t 的图来检验指数模型。
| t / min | 0 | 5 | 10 | 15 | 20 |
| C / (mg dm⁻³) | 10.0 | 8.2 | 6.7 | 5.5 | style="
广告 / Sponsored 培生爱德思考试教材 统计数学3 Edexcel International A Level Mathematics S… 数学 备考教辅 · 正版现货 · 与本文内容相关 查看详情 / View on Taobao参考价 ¥156.0更多咨询请联系16621398022（同微信） 备考资讯每日更新，关注不错过 考纲变化 · 科目分析 · 学习方法  在微信里长按二维码识别关注 或截图后用「扫一扫 → 相册」选择本图 公众号：tutorhao |