📚 4 Population Change | 人口变化
Population change is a core application of differential equations in A-Level Mathematics. It shows how the size of a population grows or declines over time, and how we can model that behaviour using rates of change.
人口变化是 A-Level 数学中微分方程的核心应用之一。它展示了人口规模如何随时间增长或减少，以及我们如何用变化率来刻画这种变化。
In this revision guide we build the exponential model from first principles, solve it step by step, compare it with discrete percentage models, and then extend it to the logistic model with a carrying capacity. Every stage is linked to the kind of question that appears in Edexcel, AQA, OCR and Cambridge papers.
在本篇复习指南中，我们将从基本原理出发建立指数模型，逐步求解，并与离散的百分比模型作对比，最后扩展到带环境容纳量的逻辑斯蒂（Logistic）模型。每一个环节都对应 Edexcel、AQA、OCR 与剑桥试卷中常见的题型。
1. What is a population model? | 什么是人口模型？
A population model describes how the number of individuals P in a population changes over time t. In A-Level Mathematics, we usually model population change with differential equations because the rate of change dP/dt often depends on the current population size.
人口模型描述一个种群中个体数量 P 如何随时间 t 变化。在 A-Level 数学中，我们通常用微分方程来建立人口变化模型，因为变化率 dP/dt 往往取决于当前的人口规模。
Continuous population models assume that P can be treated as a differentiable function of t, even though real populations are discrete. This approximation works well when the population is large, and it is the standard assumption behind every A-Level question of this type.
连续人口模型假设 P 可以看作 t 的可微函数，尽管真实种群是离散的。当种群规模很大时，这种近似效果很好，也是这类 A-Level 题目的标准假设。
Every population question contains three ingredients: an initial value, a rule for the rate of change, and a quantity you are asked to find. Identify all three before you start differentiating or integrating, and the algebra becomes routine.
每一道人口题都包含三个要素：初始值、变化率规则，以及题目要求你求出的量。在开始求导或积分之前先把这三者找出来，后面的代数运算就会变得程式化。
The notation below is used consistently throughout this article and in most textbooks.
下表符号在本篇及多数教材中通用。
| Symbol 符号 | Meaning 含义 |
| P | Population size at time t ／ t 时刻的人口规模 |
| P₀ | Initial population at t = 0 ／ 初始人口 |
| t | Time in years, months, hours ／ 时间单位（年、月、小时） |
| k | Net growth rate per unit time ／ 单位时间净增长率 |
| dP/dt | Instantaneous rate of change ／ 瞬时变化率 |
| M | Carrying capacity ／ 环境容纳量 |
2. The exponential growth equation | 指数增长方程
isfies the differential equation dP/dt = kP, where k is the net growth rate per unit time.
如果每个个体以恒定速率贡献增长，人口满足微分方程 dP/dt = kP，其中 k 是单位时间净增长率。
dP/dt = kP
This is a first-order separable differential equation. The term kP means the growth rate is proportional to the population size, so larger populations grow faster in absolute terms, even though every individual grows at the same relative rate.
这是一个一阶可分离变量微分方程。kP 这一项表示增长速率与人口规模成正比，因此人口越多，绝对增长越快，尽管每个个体的相对增长速率相同。
at an instantaneous rate of 200 individuals per year, because dP/dt = 0.02 × 10,000 = 200.
例如，如果 k = 0.02 / 年，则 10,000 人口的瞬时增长速率为每年 200 人，因为 dP/dt = 0.02 × 10,000 = 200。
The word "net" matters. k is not the birth rate on its own; it is the birth rate minus the death rate per individual per unit time. A negative k therefore describes a declining population, not an error in the model.
"净"字很关键。k 不是单指出生率，而是单位时间内每个个体的出生率减去死亡率。因此 k 为负数描述的是人口下降，而不是模型出错。
The equation also tells you something qualitative before you solve it: the gradient of the graph of P against t is always proportional to the current height of the graph, so the curve gets steeper as P increases.
这个方程在你求解之前就给出了定性信息：P 对 t 的图像的斜率始终与图像当前高度成正比，因此 P 越大，曲线越陡。
3. Solving dP/dt = kP | 求解 dP/dt = kP
Separating variables gives (1/P) dP = k dt, and integrating both sides yields ln|P| = kt + C.
分离变量得到 (1/P) dP = k dt，两边积分得到 ln|P| = kt + C。
∫ (1/P) dP = ∫ k dt → ln|P| = kt + C
ondition, gives P = Aeᵏᵗ. Substituting P(0) = P₀ gives A = P₀, so the general solution is:
由于真实人口恒为正，可以去掉绝对值符号。把任意常数写成 ln A，或直接代入初始条件，得到 P = Aeᵏᵗ。代入 P(0) = P₀ 得 A = P₀，因此通解为：
P(t) = P₀eᵏᵗ
atches the original differential equation exactly. This thirty-second check protects you from sign errors and from forgetting the initial condition.
务必通过求导检验解：dP/dt = kP₀eᵏᵗ = kP，与原始微分方程完全一致。这三十秒的检验可以帮你避免符号错误和遗漏初始条件。
If k > 0, the population grows without bound. If k < 0, the population decays towards zero. The sign of k alone determines whether the model represents growth or decline, so state it explicitly in your answer.
如果 k > 0，人口无限增长；如果 k < 0，人口衰减趋于零。仅凭 k 的符号就能判断模型代表增长还是衰退，因此在答案中要明确写出这一点。
4. Interpreting k: growth, decay and doubling time | 理解 k：增长、衰减与倍增时间
The constant k has units of "per unit time" — day. It is a relative rate, so k = 0.02 per year means that each individual contributes 2% of its own number per year at that instant.
常数 k 的单位是"每单位时间"——每年、每小时、每天。它是一个相对速率，因此 k = 0.02 / 年表示在那一瞬间，每个个体每年贡献自身数量的 2%。
the time needed for the population to double. Setting P(T) = 2P₀ in the solution gives eᵏᵀ = 2, so:
倍增时间 T 是人口翻倍所需的时间。在解中令 P(T) = 2P₀，得 eᵏᵀ = 2，因此：
T = ln 2 / k ≈ 0.693 / k (k > 0)
For decay, the same algebra gives the half-life t½ = ln 2 / |k| on and a large one take exactly the same time to double or halve under exponential growth.
对于衰减，同样的代数运算给出半衰期 t½ = ln 2 / |k|。注意倍增时间和半衰期都与 P₀ 无关：在指数增长下，小人口和大人口翻倍或减半所需的时间完全相同。
ves about 9.9 years. The reciprocal 1/k is the time constant of the model: after a time 1/k the population has changed by a factor of e.
快速检验一下：k = 0.02 / 年对应的倍增时间约为 34.7 年，而 k = 0.07 / 年对应的约为 9.9 年。倒数 1/k 是模型的时间常数：经过 1/k 的时间，人口变化了 e 倍。
f P(t) = P₀eᵏᵗ to get ln P = ln P₀ + kt, which is a straight line of gradient k against t.
考题常常要求你从数据中求出 k，而不是直接使用 k。对 P(t) = P₀eᵏᵗ 两边取对数得 ln P = ln P₀ + kt，这是一条关于 t 的直线，斜率为 k。
5. Worked example: exponential growth | 例题：指数增长
5 per hour. Find the population after 6 hours, the time needed to reach 10,000, and the doubling time.
某细菌培养液初始含有 500 个细菌，增长率为 k = 0.35 / 小时。求 6 小时后细菌数量、达到 10,000 所需时间，以及倍增时间。
k = 0.35 and t = 6. Then 0.35 × 6 = 2.1, so P(6) = 500e²·¹ ≈ 500 × 8.1662 ≈ 4083 bacteria.
第 (a) 问：代入 P(t) = P₀eᵏᵗ，其中 P₀ = 500，k = 0.35，t = 6。因为 0.35 × 6 = 2.1，所以 P(6) = 500e²·¹ ≈ 500 × 8.1662 ≈ 4083 个细菌。
Part (b): set P(t) = 10,000 and solve 10,000 = 500e0.35t. Dividing by 500 gives 20 = e0.35t, so 0.35t = ln 20 and t = 2.9957 / 0.35 ≈ 8.56 hours.
第 (b) 问：令 P(t) = 10,000，解 10,000 = 500e0.35t。两边除以 500 得 20 = e0.35t，所以 0.35t = ln 20，t = 2.9957 / 0.35 ≈ 8.56 小时。
how the answer to (c) is independent of both 500 and 10,000, which is exactly what the theory predicts.
第 (c) 问：倍增时间为 ln 2 / 0.35 ≈ 0.6931 / 0.35 ≈ 1.98 小时。注意 (c) 的答案与 500 和 10,000 都无关，这正是理论所预言的。
Marks are lost when students round the exponential function too early. Keep at least four significant figures during the calculation and only round at the final stage, then give the answer in context with units.
过早对指数函数取近似会丢分。计算过程中至少保留四位有效数字，只在最后一步取近似，并在答案中带上单位与实际情境。
6. Worked example: exponential decay and half-life | 例题：指数衰减与半衰期
= −0.04 per year. Find the population after 10 years and the half-life of the population.
某鱼群数量估计为 8000，并以 k = −0.04 / 年下降。求 10 年后的鱼群数量以及该种群的半衰期。
P(t) = P₀eᵏᵗ with P₀ = 8000 and k = −0.04: P(10) = 8000e⁻⁰·⁴ ≈ 8000 × 0.67032 ≈ 5363 fish.
代入 P(t) = P₀eᵏᵗ，P₀ = 8000，k = −0.04：P(10) = 8000e⁻⁰·⁴ ≈ 8000 × 0.67032 ≈ 5363 条鱼。
The half-life comes from ekt½ = ½, so kt½ = ln(1/2) = −ln 2 and t½ = ln 2 / 0.04 ≈ 17.3
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图