📚 A-Level Mathematics: Modelling and Solving Rate of Change Problems | A-Level 数学：变化率问题建模与求解
Rates of change appear throughout A-Level Mathematics, from pure calculus to mechanics and statistics. This article focuses on how to translate real-world situations into mathematical models using derivatives, and how to solve these models systematically.
变化率贯穿 A-Level 数学的始终，从纯数到力学与统计学都有涉及。本文聚焦于如何将现实情境转化为使用导数的数学模型，并系统性地求解这些模型。
1. The Derivative as a Rate of Change | 导数作为变化率
The derivative dy/dx measures how y changes as x changes. In modelling, x often represents time t, so dy/dt gives the instantaneous rate of change of y with respect to time. If dy/dt > 0, y is increasing; if dy/dt < 0, y is decreasing.
导数 dy/dx 衡量的是 y 随 x 变化的速率。在建模中，x 常代表时间 t，因此 dy/dt 表示 y 关于时间的瞬时变化率。若 dy/dt > 0，则 y 在增加；若 dy/dt < 0，则 y 在减少。
is a mathematical model of population growth. The sign of k determines whether the population increases (k > 0) or decays (k < 0).
例如，若人口 P 的增长速率与其规模成正比，我们写 dP/dt = kP。这个方程就是人口增长的数学模型。k 的正负决定人口是增长（k > 0）还是衰减（k < 0）。
2. Setting Up a Model from a Word Problem | 从文字题建立模型
To model a rate problem, follow four steps: identify the variables, translate the rate statement into derivative notation, find any geometric or physical relations between variables, and write down initial conditions.
建模变化率问题需四步：确定变量、将变化率语句转译为导数符号、找出变量间的几何或物理关系、写下初始条件。
Step 1: Define variables. Use clear letters, e.g. V for volume, r for radius, t for time.
第一步：定义变量。使用清晰字母，如 V 表示体积，r 表示半径，t 表示时间。
Step 2: Write the given rate. Phrases like 'increases at a rate of 5 cm³/s' become dV/dt = 5.
第二步：写出已知速率。“以 5 cm³/s 的速率增加”写作 dV/dt = 5。
Step 3: Relate variables. Use known formulas such as V = 4/3 πr³ for a sphere.
第三步：联系变量。使用已知公式，如球的体积 V = 4/3 πr³。
Step 4: Initial conditions. At t = 0, state the starting values, e.g. r(0) = 2.
第四步：初始条件。在 t = 0 时给出初始值，如 r(0) = 2。
3. The Chain Rule for Related Rates | 链式法则与相关变化率
and r depends on t, then dV/dt = dV/dr × dr/dt. This is the core of related-rate problems.
当两个量相互关联时，它们的变化率通过链式法则相连。若 V 依赖于 r，且 r 依赖于 t，则 dV/dt = dV/dr × dr/dt。这是相关变化率问题的核心。
dV/dt = dV/dr × dr/dt
at dr/dt = 2 cm/s when r = 3 cm, then dV/dt = 4π(3)² × 2 = 72π cm³/s. Always check units.
对于球体，dV/dr = 4πr²。若半径以 dr/dt = 2 cm/s 增加，且 r = 3 cm，则 dV/dt = 4π(3)² × 2 = 72π cm³/s。务必检查单位。
4. Implicit Differentiation in Geometric Models | 几何模型中的隐函数求导
and h change with time, differentiate both sides with respect to t using the product rule.
有些模型变量间的关系并未显式解出一个变量。例如，圆锥体积 V = 1/3 πr²h。若 r 和 h 都随时间变化，则需用乘积法则对 t 求导。
dV/dt = 1/3 π(2r dr/dt × h + r² dh/dt)
This formula is essential for problems where both radius and height vary. It shows that the total rate of volume change comes from two contributions: the changing radius and the changing height.
此公式对 r 和 h 同时变化的问题至关重要。它表明体积的总变化率来自两部分：半径变化与高度变化的贡献。
5. Exponential Growth and Decay Models | 指数增长与衰减模型
s gives the differential equation dN/dt = kN, where N is the quantity and k is a constant.
许多自然过程满足：量的变化率与量本身成正比。由此得到微分方程 dN/dt = kN，其中 N 是量，k 为常数。
The general solution is:
通解为：
N(t) = N₀ eᵏᵗ
f k > 0, this models exponential growth; if k < 0, it models exponential decay. Radioactive decay, cooling and population growth often use this model.
其中 N₀ 是 t = 0 时的初始值。若 k > 0，为指数增长；若 k < 0，为指数衰减。放射性衰变、冷却和人口增长常用此模型。
6. Differential Equations from Rate Statements | 由速率描述建立微分方程
In A-Level questions you may be given a sentence like ' ture T of an object is proportional to the difference between T and the room temperature R' This translates to dT/dt = -λ(T - R), with a minus sign because the difference decreases.
在 A-Level 试题中，可能遇到类似描述：“物体温度 T 的上升速率与 T 和室温 R 之差成正比”。这转化为 dT/dt = -λ(T - R)，负号表示差值在减小。
Another common statement: ' lume of water V in a leaking tank decreases at a rate proportional to the square root of V'. This becomes dV/dt = -k√V.
另一种常见描述：“漏水箱中水的体积 V 以与 √V 成正比的速率减少”。这写作 dV/dt = -k√V。
Always identify whether the rate is positive or negative, and what variable the rate is proportional to. This determines the form of the differential equation.
识别速率是正还是负，以及它与哪个变量成正比，这决定了微分方程的形式。
7. Solving Separable Differential Equations | 求解可分离变量微分方程
Most A-Level rate models use separable differential equations. To solve, rearrange so that each variable appears on one side with its differential, then integrate both sides.
A-Level 中大多数变化率模型使用可分离变量微分方程。求解时，将变量整理到各自一侧并配上对应的微分，然后两边积分。
Example: solve dV/dt = -k√V. Separate:
例如：解 dV/dt = -k√V。分离变量：
∫ V⁻¹ᐟ² dV = ∫ -k dt
Integrate to obtain 2√V tion to find C. Always include the constant of integration and solve for the explicit form when required.
积分得 2√V = -kt + C。然后用初始条件确定 C。别忘了积分常数，需要时解出显式形式。
8. Using Initial Conditions to Find Particular Solutions | 用初始条件求特解
allows us to determine the constant C. For the example above, if V = 16 when t = 0, then 2√16 = C, so C = 8. The particular solution is 2√V = 8 - kt.
初始条件如 V(0) = 16 可以确定常数 C。对于上例，若 t = 0 时 V = 16，则 2√16 = C，所以 C = 8。特解为 2√V = 8 - kt。
Particular solutions are essential for predicting the state of the system at a future time. They also help determine when a quantity reaches a certain value, such as the time when a tank becomes empty.
特解对预测系统未来状态至关重要，也可用于确定量达到某值的时间，例如水箱何时排空。
9. Example: Filling a Conical Tank | 例题：圆锥水箱注水
A conical tank has height 12 m and top radius 4 m. Water enters at a constant rate of 2 m³/min. Find the rate at which the water level rises when the depth is 6 m.
一个圆锥水箱高 12 m，顶部半径 4 m。水以 2 m³/min 的恒定速率注入。求水深为 6 m 时水面上升的速率。
/h = 4/12 = 1/3, so r = h/3. The volume of water is V = 1/3 πr²h = 1/3 π(h/3)² h = πh³/27.
根据相似三角形，水深 h 时水面半径 r 满足 r/h = 4/12 = 1/3，所以 r = h/3。水的体积 V = 1/3 πr²h = 1/3 π(h/3)² h = πh³/27。
dV/dt = 2 and h = 6, we have 2 = π(36)/9 × dh/dt = 4π × dh/dt. Hence dh/dt = 1/(2π) m/min.
对 t 求导：dV/dt = πh²/9 × dh/dt。已知 dV/dt = 2 且 h = 6，得 2 = π(36)/9 × dh/dt = 4π × dh/dt。因此 dh/dt = 1/(2π) m/min。
10. Example: Newton's Law of Cooling | 例题：牛顿冷却定律
= -0.05(T - 20), with T in °C and t in minutes. Find the time when the drink reaches 40°C.
一杯热饮初始温度为 90°C，房间温度为 20°C。冷却规律为 dT/dt = -0.05(T - 20)，T 的单位为 °C，t 的单位为分钟。求饮品降到 40°C 所需时间。
Separate variables: ∫ 1/(T - 20) dT = ∫ -0.05 dt. Integrating gives ln|T - 20| = -0.05t + C. Since T > 20, we can drop the absolute value.
分离变量：∫ 1/(T - 20) dT = ∫ -0.05 dt。积分得 ln|T - 20| = -0.05t + C。由于 T > 20，可去掉绝对值。
Using T(0) = 90, we find C = ln 70. Thus ln(T - 20) = -0.05t + ln 70, so T - 20 = 70e-0.05t. Set T = 40: 20 = 70e-0.05t, so e-0.05t = 2/7, giving t = -20 ln(2/7) minutes, approximately 25.1 minutes.
由 T(0) = 90，得 C = ln 70。因此 ln(T - 20) = -0.05t + ln 70，即 T - 20 = 70e-0.05t。令 T = 40：20 = 70e-0.05t，所以 e-0.05t = 2/7，解得 t = -20 ln(2/7) 分钟，约为 25.1 分钟。
11. Common Pitfalls and Exam Tips | 常见错误与考试技巧
Pitfall 1: Forgetting the chain rule — always differentiate with respect to t, not with respect to another variable.
错误一：忘记链式法则——始终对 t 求导，而不是对另一个变量求导。
Pitfall 2: Confusing positive and negative rates. Read whether the quantity is increasing or decreasing.
错误二：混淆正负速率。仔细分辨量是在增加还是减少。
Pitfall 3: Mixing up initial conditions with the value at a later time. Label time clearly.
错误三：混淆初始条件与之后时刻的值。清晰标记时间。
Pitfall 4: Dropping the constant of integration when solving differential equations. Always include + C before applying the initial condition.
错误四：解微分方程时忘掉积分常数。在代入初始条件前始终写出 + C。
Tip: Write down every step including units. In exams, correct units often earn method marks, and showing your model-building process helps clarify your reasoning.
技巧：写出每一步并标注单位。在考试中，正确的单位常能获得方法分，展示建模过程有助于理清思路。
12. Summary | 总结
Solving rate-of-change problems requires three skills: translating words into derivative equations, applying the chain rule or implicit differentiation correctly, and solving separable differential equations with initial conditions. Practice these steps on a variety of contexts, from geometry to cooling.
解决变化率问题需要三种技能：将文字转化为导数方程、正确运用链式法则或隐函数求导、以及用初始条件求解可分离变量微分方程。在不同情境中练习这些步骤，从几何问题到冷却问题。
Modelling → Differentiation → Solve → Interpret
Mastering this process will allow you to handle any A-Level rate-of-change question with confidence.
掌握这个过程，你就能自信应对任何 A-Level 变化率问题。
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