Rates in Context | 情境中的变化率

📚 Rates in Context | 情境中的变化率

In Edexcel A-Level Mathematics, rates of change appear throughout pure and applied units, especially in differentiation and mechanics. Students are often asked to interpret a derivative as a real-world rate, such as speed, growth, or flow, and to link several changing quantities using the chain rule. This article explains the key ideas and common modelling contexts you need to master.

在爱德思 A-Level 数学中，变化率主题贯穿纯数与力学单元，尤其是在微分部分。考生经常需要把导数解释为现实中的速率，例如速度、增长或流量，并利用链式法则把多个变化量联系起来。本文讲解你必须掌握的核心概念和常见建模情境。

1. Average and Instantaneous Rates of Change | 平均变化率与瞬时变化率

The average rate of change of a function y = f(x) over an interval [a, b] is the gradient of the chord joining the two points, given by (f(b) – f(a))/(b – a). The instantaneous rate of change at x = a is the gradient of the tangent, defined as the limit of this quotient as b approaches a. This is exactly the derivative f'(a).

函数 y = f(x) 在区间 [a, b] 上的平均变化率是连接两点的弦的斜率，即 (f(b) – f(a))/(b – a)。在 x = a 处的瞬时变化率是切线的斜率，定义为 b 趋近于 a 时上述商的极限。这正是导数 f'(a)。

dy/dx = lim (h→0) [f(x+h) – f(x)] / h

2. The Derivative as a Rate in Context | 情境中的导数即变化率

In applied problems, the derivative dy/dx means the rate at which y changes with respect to x. If x represents time t, then dy/dt is the rate of change of y over time. For example, if s(t) is displacement, ds/dt = v(t) is velocity, and dv/dt = d²s/dt² = a(t) is acceleration. Understanding the units of the derivative is essential: the unit of dy/dx is the unit of y divided by the unit of x.

在应用题中，导数 dy/dx 表示 y 关于 x 的变化率。如果 x 代表时间 t，那么 dy/dt 就是 y 随时间的变化率。例如，若 s(t) 是位移，则 ds/dt = v(t) 是速度，而 dv/dt = d²s/dt² = a(t) 是加速度。理解导数的单位至关重要：dy/dx 的单位是 y 的单位除以 x 的单位。

3. The Chain Rule | 链式法则

The chain rule allows us to differentiate a composite function. If y = f(u) and u = g(x), then dy/dx = dy/du × du/dx. In rate-of-change problems, the chain rule is often used to connect the rate of change of a volume, radius, and height. For instance, if V = (4/3)πr³, then dV/dt = dV/dr × dr/dt = 4πr² × dr/dt. This is the core of connected rates of change.

链式法则用于求复合函数的导数。若 y = f(u) 且 u = g(x)，则 dy/dx = dy/du × du/dx。在变化率问题中，链式法则常用来联系体积、半径和高度等量的变化率。例如，若 V = (4/3)πr³，则 dV/dt = dV/dr × dr/dt = 4πr² × dr/dt。这是相关变化率问题的核心。

4. Connected Rates of Change | 相关变化率

In a typical Edexcel problem, you are given one rate, such as dV/dt, and asked to find another rate, such as dr/dt or dh/dt. You must first write a geometric or physical relationship between the variables, differentiate it with respect to time, and then substitute known values. Always check which variables are constant and which are changing, and express all quantities in consistent units.

在典型的爱德思题目中，通常会给出一个速率，如 dV/dt，要求求另一个速率，如 dr/dt 或 dh/dt。你必须先写出变量之间的几何或物理关系，将其对时间求导，然后代入已知值。务必检查哪些量是常量，哪些量在变化，并使用一致的单位表示所有量。

• Identify the given rate and the required rate.
• Find an equation linking the variables.
• Differentiate both sides with respect to time t.
• Substitute known values and solve for the unknown rate.

• 确定已知速率和待求速率。
• 建立变量之间的关系式。
• 两边对时间 t 求导。
• 代入已知值并求解未知速率。

5. Worked Example: Expanding Circular Oil Slick | 例题：圆形油膜扩张

Consider a circular oil slick whose radius r increases at a constant rate of 0.5 m/min. The area is A = πr². Differentiating with respect to time gives dA/dt = 2πr × dr/dt. When r = 10 m, dA/dt = 2π × 10 × 0.5 = 10π m²/min. This shows that the rate of increase of the area depends on the current radius, even if the radius grows at a constant rate.

考虑一个圆形油膜，其半径 r 以 0.5 m/min 的恒定速率增大。面积公式为 A = πr²。对时间求导得 dA/dt = 2πr × dr/dt。当 r = 10 m 时，dA/dt = 2π × 10 × 0.5 = 10π m²/min。这表明面积增大的速率依赖于当前半径，即使半径以恒定速率增长。

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