A-Level Mathematics: Applying Differentiation to Modelling | A-Level数学：微分在建模中的应用

📚 A-Level Mathematics: Applying Differentiation to Modelling | A-Level数学：微分在建模中的应用

Differentiation is one of the most powerful tools in A-Level Mathematics because it turns a static description of a system into a dynamic one. In modelling, a derivative measures how one quantity changes as another quantity changes, allowing us to predict behaviour, find optimal values and interpret real-world data. This article connects the formal rules of differentiation to practical modelling situations that appear in the examination.

微分是A-Level数学中最强大的工具之一，因为它能将一个系统的静态描述转化为动态描述。在建模中，导数度量一个量随另一个量变化的快慢，使我们能够预测行为、求最优值并解读现实数据。本文把微分的运算法则与考试中常见的实际建模情境联系起来。

1. Rates of Change | 变化率

A derivative is a rate of change. In a mathematical model, if a quantity y depends on a quantity x, then dy/dx measures the sensitivity of y to small changes in x. For instance, if V(t) is the volume of water in a tank after t seconds, then dV/dt is the rate at which the volume changes. When a model links several variables, the chain rule allows you to convert one rate into another. This single idea reappears in every topic in this article.

导数就是变化率。在数学模型中，若量y依赖于量x，则dy/dx度量y对x微小变化的敏感程度。例如，若V(t)表示t秒后水箱中的水量，那么dV/dt就是水量变化的速率。当一个模型连接多个变量时，链式法则允许你把一个变化率换算成另一个。这一核心思想将贯穿本文的所有主题。

dy/dx = lim(δx→0) δy/δx

For a composite model such as V = πr²h, the chain rule gives dV/dt = (∂V-like reasoning) dV/dr × dr/dt, which converts a known rate into an unknown one. In examination questions, always state which variable you are differentiating with respect to before writing the derivative.

对于形如V = πr²h的复合模型，链式法则给出dV/dt = dV/dr × dr/dt，从而把已知变化率转化为未知变化率。在考试题中，写出导数之前务必先说明你是对哪个变量求导。

2. Optimisation: Finding Maximum and Minimum Values | 优化：求最大值与最小值

Optimisation is the search for extreme values of a model. The key idea is to find where a quantity stops increasing and starts decreasing, or vice versa. Since the derivative gives the gradient of the curve, a stationary point occurs where f'(x) = 0. The real challenge in modelling is to create a single-variable function from a real situation, then apply a reliable procedure.

优化就是寻找模型的极值。其核心思想是找到某个量停止增加而开始减少（或相反）的位置。由于导数给出曲线的斜率，驻点出现在f'(x) = 0处。建模的真正挑战在于从实际问题构造出单变量函数，然后执行一套可靠的步骤。

• Identify the quantity Q to be maximised or minimised. 确定需要最大化或最小化的量Q。
• Use the constraints of the problem to write Q as a function of one variable. 利用题目中的约束条件把Q写成单变量函数。
• Differentiate to obtain Q'(x) and solve Q'(x) = 0. 求导得到Q'(x)并解方程Q'(x) = 0。
• Verify the nature of the stationary point and state the answer with units. 验证驻点的性质，并带上单位给出答案。

f'(x) = 0, f”(x) < 0 → local maximum; f”(x) > 0 → local minimum

A classic example is finding the largest rectangular area that can be enclosed with a fixed perimeter. If the perimeter is 100

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