A-Level Maths: Applications of Differentiation in Real-World Models | A-Level数学：微分在实际模型中的应用

📚 A-Level Maths: Applications of Differentiation in Real-World Models | A-Level数学：微分在实际模型中的应用

Differentiation is one of the most powerful tools in A-Level Mathematics. It allows us to measure how one quantity changes with respect to another, to locate maximum and minimum points, and to build mathematical models of practical situations. This article follows the Edexcel specification and explains how to apply differentiation to real-world models.

微分是A-Level数学中最强大的工具之一。它使我们能够度量一个量相对于另一个量的变化、找到最大值与最小值，并为实际情景建立数学模型。本文按照Edexcel考纲，讲解如何将微分应用于实际模型。

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1. The Derivative as a Rate of Change | 导数作为变化率

The derivative f'(x) measures the instantaneous rate of change of a function y = f(x) at a given point. It is defined as the limit of the gradient of a chord as the interval shrinks to zero:

导数 f'(x) 度量函数 y = f(x) 在某一点处的瞬时变化率。它定义为当区间趋于零时弦的斜率的极限：

f'(x) = lim(h → 0) [f(x + h) – f(x)] / h

In real-world models, this tells us how quickly a quantity changes when another quantity changes by a tiny amount. For example, if V(t) is the volume of water in a tank at time t, then dV/dt is the rate at which the volume changes, with units such as litres per minute or cubic metres per second.

在实际模型中，这告诉我们当一个量发生微小变化时，另一个量变化得有多快。例如，若 V(t) 是 t 时刻水箱中水的体积，则 dV/dt 就是体积变化的速率，单位可以是升/分钟或立方米/秒。

Always attach units to your answer. If V is measured in litres and t in minutes, then dV/dt is measured in litres per minute. A positive derivative means the quantity is increasing; a negative derivative means it is decreasing.

务必在答案中注明单位。若 V 以升为单位、t 以分钟为单位，则 dV/dt 以升/分钟为单位。导数大于零表示量在增加；导数小于零表示量在减少。

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2. Tangents and Normals | 切线与法线

If y = f(x) is differentiable at x = a, then the gradient of the tangent to the curve at that point is m = f'(a). The equation of the tangent is:

若 y = f(x) 在 x = a 处可导，则该点处曲线切线的斜率为 m = f'(a)。切线方程为：

y – f(a) = f'(a)(x – a)

The normal is perpendicular to the tangent. If f'(a) ≠ 0, its gradient is -1/f'(a), so the equation of the normal is:

法线与切线垂直。若 f'(a) ≠ 0，则法线的斜率为 -1/f'(a)，因此法线方程为：

y – f(a) = -1/f'(a) (x – a)

Worked example: Let f(x) = x² + 2x. At x = 1, f(1) = 3 and f'(x) = 2x + 2, so f'(1) = 4. The tangent is y – 3 = 4(x – 1), which simplifies to y = 4x – 1. The normal has gradient -1/4, so its equation is y – 3 = -1/4(x – 1), or y = -x/4 + 13/4.

例题：设 f(x) = x² + 2x。在 x = 1 处，f(1) = 3，且 f'(x) = 2x + 2，所以 f'(1) = 4。切线为 y – 3 = 4(x – 1)，化简得 y = 4x – 1。法线斜率为 -1/4，因此其方程为 y – 3 = -1/4(x – 1)，即 y = -x/4 + 13/4。

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3. Stationary Points and Their Classification | 驻点及其分类

A stationary point occurs where the derivative is zero, f'(x) = 0. At such a point, the tangent is horizontal. Stationary points are important in real-world models because they often represent maximum profit, minimum cost, or maximum volume.

驻点出现在导数为零的地方，即 f'(x) = 0。在这些点上，切线是水平的。驻点在实际模型中非常重要，因为它们常常对应最大利润、最小成本或最大体积。

The second derivative f”(x) is used to classify a stationary point:

二阶导数 f”(x) 用于判断驻点的类型：

• If f”(x) > 0, the point is a local minimum.

  若 f”(x) > 0，则该点是局部极小值点。

• If f”(x) < 0, the point is a local maximum.

  若 f”(x) < 0，则该点是局部极大值点。

• If f”(x) = 0, the test is inconclusive; use a sign table for f'(x) instead.

  若 f”(x) = 0，则该判据失效；此时应使用 f'(x) 的符号表来判断。

Example: For y = x³ – 3x, dy/dx = 3x² – 3 = 3(x – 1)(x + 1). Setting dy/dx = 0 gives x = 1 or x = -1. d²y/dx² = 6x. At x = 1, d²y/dx² = 6 > 0, so (1, -2) is a local minimum. At x = -1, d²y/dx² = -6 < 0, so (-1, 2) is a local maximum.

例题：对于 y = x³ – 3x，dy/dx = 3x² – 3 = 3(x – 1)(x + 1)。令 dy/dx = 0，得 x = 1 或 x = -1。d²y/dx² = 6x。在 x = 1 处，d²y/dx² = 6 > 0，所以 (1, -2) 是局部极小值点；在 x = -1 处，d²y/dx² = -6 < 0，所以 (-1, 2) 是局部极大值点。

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4. Optimisation Problems | 最优化问题

Optimisation problems ask us to find the maximum or minimum value of a quantity subject to certain constraints. Differentiation provides a systematic method for solving them.

最优化问题要求我们在一定约束条件下求某个量的最大值或最小值。微分提供了系统的求解方法。

• Step 1: Define the variables and draw a diagram where appropriate.

  第一步：定义变量，并在适当时画出图形。

• Step 2: Write down a formula for the quantity to be optimised.

  第二步：写出要优化的量的公式。

• Step 3: Use the given constraint to write the formula in terms of a single variable.

  第三步：利用所给约束，将公式写成只含一个变量的形式。

• Step 4: Differentiate and solve dQ/dx = 0.

  第四步：求导并解 dQ/dx = 0。

• Step 5: Use the second derivative or a sign table to confirm it is a maximum or minimum.

  第五步：利用二阶导数或符号表确认它是极大值还是极小值。

• Step 6: State the answer with units and check it makes sense in the context.

  第六步：注明单位给出答案，并检验其在情境中是否合理。

Example (fencing): A rectangular enclosure has a fixed area of 800 m². Let the width be x metres and the length be 800/x metres. The perimeter is P = 2x + 1600/x. Differentiating gives dP/dx = 2 – 1600/x². Setting dP/dx = 0 gives x² = 800, so x = 20√2 ≈ 28.3 m. The length is also 20√2 m, so the minimum perimeter is P = 80√2 ≈ 113.1 m. Since d²P/dx² = 3200/x³ > 0, this is a minimum.

例题（围栏问题)：一个矩形围场的面积固定为 800 m²。设宽为 x 米，长为 800/x 米。周长 P = 2x + 1600/x。求导得 dP/dx = 2 – 1600/x²。令 dP/dx = 0，得 x² = 800，因此 x = 20√2 ≈ 28.3 m。长也为 20√2 m，所以最小周长为 P = 80√2 ≈ 113.1 m。由于 d²P/dx² = 3200/x³ > 0，故为最小值。

Example (open box): A square sheet of card is 20 cm by 20 cm. Equal squares of side x cm are cut from each corner and the sides are folded up to make an open box. The volume is V = x(20 – 2x)², with 0 < x < 10. Expanding gives V = 4x³ - 80x² + 400x, so dV/dx = 12x² - 160x + 400. Setting dV/dx = 0 and solving gives x = 10/3 or x = 10. The value x = 10 makes the box degenerate, so the maximum occurs at x = 10/3 cm, giving a maximum volume of V = 10/3 × (40/3)² = 16000/27 ≈ 592.6 cm³.

例题（开口盒子）：一张边长为20 cm的正方形纸板，从每个角剪去边长为 x cm 的小正方形并折起，得到一个

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