📚 11: Modelling with differentiation | 11：微分建模
Modelling with differentiation allows us to apply calculus to real-world scenarios, such as optimising profit, minimising surface area for a given volume, or analysing motion. In A-Level Mathematics, you will often be required to construct a function from given information, differentiate it, and then find maximum or minimum values or rates of change. This topic combines algebraic manipulation, geometric insight, and an understanding of the physical or economic context. Mastering these techniques is essential for tackling extended problems in the Edexcel exam.
微分建模使我们能够将微积分应用于现实场景，例如优化利润、在给定体积下最小化表面积，或分析运动。在 A-Level 数学中，你经常需要根据给定信息构造函数，对其求导，然后找到最大值、最小值或变化率。这个主题结合了代数运算、几何洞察以及对物理或经济背景的理解。掌握这些技巧对于解决 Edexcel 考试中的综合题目至关重要。
1. Introduction to Modelling with Differentiation | 微分建模简介
Modelling with differentiation is the process of using derivatives to represent and solve problems about changing quantities. It involves identifying a quantity to be maximised or minimised, expressing it as a function of one variable, and then finding stationary points. The derivative dy/dx tells us the instantaneous rate of change of y with respect to x, which is fundamental to optimisation and related rates problems.
微分建模是使用导数表示并解决变化量问题的过程。它涉及确定要最大化或最小化的量，将其表示为单一变量的函数，然后找到驻点。导数 dy/dx 告诉我们 y 关于 x 的瞬时变化率，这对于最优化和相关变化率问题至关重要。
In many exam questions, you will be given a physical or economic situation described in words. Your first job is to translate this description into a mathematical function, often called a model. The model will usually contain a constant or a parameter, and you must use given constraints to reduce the number of variables before differentiating.
在许多考题中，你会得到用文字描述的物理或经济情境。你的首要任务是将这个描述转化为一个数学函数，通常称为模型。模型通常包含常数或参数，你必须在求导之前利用给定的约束条件来减少变量的数量。
2. Understanding Stationary Points | 理解驻点
A stationary point occurs where the first derivative dy/dx = 0. At this point, the tangent to the curve is horizontal, indicating a potential maximum, minimum, or point of inflection. In modelling, we are most interested in maximum and minimum points that correspond to optimal conditions.
驻点出现在一阶导数 dy/dx = 0 处。在该点，曲线的切线是水平的，表示潜在的最大值、最小值或拐点。在建模中，我们最感兴趣的是对应于最优条件的最大值和最小值点。
ative test (evaluating d²y/dx² at the point). A positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum. If d²y/dx² = 0, further investigation is needed.
为了确定驻点的性质，我们可以使用一阶导数判定法（检查驻点两侧 dy/dx 的符号）或二阶导数判定法（计算驻点处的 d²y/dx²）。二阶导数为正表示局部最小值，为负表示局部最大值。如果 d²y/dx² = 0，则需要进行深入分析。
3. First Derivative Test for Nature | 一阶导数判定法
The first derivative test involves taking x-values slightly less than and slightly greater than the stationary point and calculating the sign of dy/dx. If dy/dx changes from positive to negative, the point is a local maximum. If it changes from negative to positive, the point is a local minimum. If the sign does not change, it is a point of inflection.
一阶导数判定法需要取略小于和略大于驻点的 x 值，计算 dy/dx 的符号。如果 dy/dx 从正变为负，该点是局部最大值。如果从负变为正，该点是局部最小值。如果符号不改变，则是拐点。
This method is particularly useful when the second derivative is difficult to compute or when the second derivative is zero. In modelling contexts, it gives a clear connection between the behaviour of the function on either side of the optimum, which can be checked against the real-world meaning.
当二阶导数难以计算或二阶导数为零时，这种方法特别有用。在建模情境中，它清晰地展示了函数在最优值两侧的变化行为，可与实际意义相对照。
4. Second Derivative Test | 二阶导数判定法
The second derivative test is often quicker: if f'(x) = 0 and f''(x) < 0, the stationary point is a local maximum; if f'' alue you have found corresponds to the required maximum or minimum, rather than a point of inflection.
二阶导数判定法通常更快：如果 f'(x) = 0 且 f''(x) < 0，该驻点是局部最大值；若 f''(x) > 0，则是局部最小值。该判定依赖于图像的曲率。在最优化问题中，你必须始终确认所找到的驻点值对应所需的最大值或最小值，而不是拐点。
When applying this test, be careful to state clearly whether a value is indeed a maximum or minimum. For example, in a profit maximisation problem, you would note that the negative second derivative confirms that you have found the maximum profit.
应用此判定时，注意要明确陈述该值确实是最大值还是最小值。例如，在利润最大化问题中，你可以指出负的二阶导数确认已找到最大利润。
5. Optimisation: Maximising and Minimising Functions | 最优化：求函数的最大最小值
Optimisation involves finding the value of a variable that makes a quantity as large or as small as possible. The general steps are: (1) write an expression for the quantity to be optimised in terms of one variable; (2) differentiate that expression; (3) set the derivative equal to zero and solve for the variable; (4) verify the nature of the stationary point; and (5) answer the question in context.
最优化涉及找到使某个量尽可能大或尽可能小的变量值。一般步骤为：(1) 用单一变量写出待优化量的表达式；(2) 对该表达式求导；(3) 令导数为零，求解该变量；(4) 验证驻点的性质；(5) 在上下文情境中回答问题。
Many Edexcel problems require you to show that a quantity can be expressed in a particular form, for instance A = 2x(20 - x), before finding the maximum. Often you will be given a fixed perimeter, area or cost, which acts as a constraint and allows you to eliminate one variable.
许多 Edexcel 问题要求你先证明某个量可以表示为特定形式，如 A = 2x(20 - x)，然后再求最大值。通常会给定一个固定的周长、面积或成本，作为约束条件，借此可以消去一个变量。
6. Modelling Profit, Revenue and Cost | 利润、收入和成本建模
of units x produced and sold, profit P(x) = R(x) - C(x). The maximum profit occurs when P'(x) = 0 and P'' nalysis shows that profit is maximised when marginal revenue equals marginal cost, i.e., R'(x) = C'(x).
在商业和经济学中，微分被用于最大化利润或最小化成本。如果收入 R(x) 和成本 C(x) 是生产和销售数量 x 的函数，那么利润 P(x) = R(x) - C(x)。当 P'(x) = 0 且 P''(x) < 0 时利润最大。或者，边际分析表明，当边际收入等于边际成本，即 R'(x) = C'(x) 时，利润达到最大。
profit function before differentiating. Always check that your solution makes sense within the domain, for example x must be non-negative and often an integer or within a certain range.
考题可能会给出需求函数或总成本函数。你需要推导出收入函数（价格 × 数量），然后是利润函数，再进行求导。务必检查解在定义域内是否有意义，例如 x 必须非负，且通常为整数或在某个范围内。
7. Geometry Problems: Area and Volume Optimisation | 几何问题：面积与体积优化
Classic geometric modelling tasks include designing a box of maximum volume from a flat sheet by cutting squares from the corners, or minimising the surface area of a cylinder for a fixed volume. The key is to write the quantity to be optimised as a function of one variable using the constraint.
经典的几何建模题目包括从一块平板上切去四个角制作体积最大的盒子，或者在体积固定的情况下最小化圆柱体的表面积。关键在于利用约束条件将待优化的量表示为单一变量的函数。
u differentiate with respect to x, set dV/dx = 0, and solve. You must then test which root gives the maximum volume and check if the value of x is feasible (e.g., 0 < x < 10).
对于从 30 cm × 20 cm 的薄板四角各切去边长为 x cm 的正方形所制成的矩形盒子，体积 V = x(30 - 2x)(20 - 2x)。展开后，对 x 求导，令 dV/dx = 0 并求解。然后必须检验哪个根给出最大体积，并检查 x 是否可行（例如 0 < x < 10）。
8. Related Rates of Change | 相关变化率
Related rates problems involve two or more variables that change with respect to time t. The chain rule, dy/dt = (dy/dx) × (dx/dt) ume dV/dt and the rate of change of radius dr/dt are related by dV/dt = (dV/dr) × (dr/dt).
相关变化率问题涉及随时间 t 变化的两个或多个变量。链式法则 dy/dt = (dy/dx) × (dx/dt) 使我们能够将变化率联系起来。例如，如果给一个球形气球充气，体积的变化率 dV/dt 与半径的变化率 dr/dt 通过 dV/dt = (dV/dr) × (dr/dt) 相关联。
To solve such problems, first identify the known rate and the quantity whose rate you need to find. Write an equation linking the static variables (e.g., volume and radius), differentiate both sides with respect to t, substitute the known values, and solve for the unknown rate. Pay close attention to units and whether a rate is increasing or decreasing.
要解决此类问题，首先要确定已知的变化率以及需要求变化率的量。写出将静态变量联系起来的方程（例如体积和半径），对 t 两边求导，代入已知数值，解出未知的变化率。务必注意单位以及变化率是增加还是减小。
9. Kinematics: Displacement, Velocity and Acceleration | 运动学：位移、速度和加速度
In Kinematics, differentiation links displacement s(t), velocity v(t), and acceleration a(t). Velocity is the first derivative of displacement: v = ds/dt. Acceleration is the derivative of velocity, or the second derivative of displacement: a = dv/dt = d²s/dt². These relationships allow you to find maximum displacement, maximum velocity, or when a particle changes direction.
在运动学中，微分将位移 s(t)、速度 v(t) 和加速度 a(t) 联系起来。速度是位移的一阶导数：v = ds/dt。加速度是速度的导数，或是位移的二阶导数：a = dv/dt = d²s/dt²。这些关系使你能够找到最大位移、最大速度，或粒子何时改变方向。
A typical problem provides s as a function of t, such as s = 2t³ - 9t² + 12t. By differentiating, you can find the times when the particle is at rest (v = 0) and then use the second derivative to assess whether acceleration is positive or negative at that instant. The sign of velocity and acceleration indicates direction and speeding up or slowing down.
一个典型问题会给出 s 关于 t 的函数，例如 s = 2t³ - 9t² + 12t。通过求导，你可以找到粒子静止的时刻（v = 0），然后利用二阶导数判断该时刻的加速度是正还是负。速度和加速度的符号指示方向以及是在加速还是减速。
10. Dealing with Constraints | 处理约束条件
Constraints are additional conditions that limit the values variables can take. In optimisation, a constraint is used to express one variable in terms of another, reducing the function to a single variable. Examples include a fixed length of fencing, a given surface area, or a total cost budget.
约束条件是限制变量可取值范围的附加条件。在最优化中，约束用于将一变量用另一变量表示，从而将函数简化为单变量。例子包括固定长度的围栏、给定的表面积或总成本预算。
Always consider the domain implied by the practical context. For instance, if a length x must be positive and less than a certain value, you should state the domain explicitly. Check endpoints as well as stationary points: the optimum could occur at a boundary, especially if the derivative never equals zero within the domain.
始终要考虑实际情境所隐含的定义域。例如，如果长度 x 必须为正且小于某个特定值，你应该明确写出定义域。既要检查驻点，也要检查端点：最优值可能出现在边界处，尤其是如果导数在定义域内从未等于零。
11. Practical Examples and Exam Techniques | 实例与考试技巧
Work through a variety of past paper questions to become familiar with the language of modelling. Common phrasing includes "show that the volume V can be written as ...", "find the maximum rate of change of ...", and "explain why this value is a maximum". Always answer the final part in context, with correct units.
通过练习各种历年真题来熟悉建模的语言。常见的措辞包括 “证明体积 V 可写为……”、“求……的最大变化率” 以及 “解释为何此值为最大值”。始终结合语境回答最后一部分，并带上正确单位。
Sketching a rough graph can help visualise the problem. When dealing with practical maxima and minima, confirm that your mathematical answer is physically possible. Avoid losing marks by not simplifying expressions before differentiating, and double-check any algebraic expansions before use.
画一个简图有助于可视化问题。处理实际的最大值和最小值时，要确认数学答案在物理上是否可行。求导前务必化简表达式，并在使用前仔细检查代数展开，以免失分。
12. Summary and Key Takeaways | 总结与要点
Modelling with differentiation integrates many core calculus skills. The essential steps are: read the problem carefully, identify the quantity to optimise, express it as a function of a single variable using any constraints, differentiate, find stationary points, and confirm their nature. Always interpret your answer in the original context, and state the maximum or minimum value clearly.
微分建模整合了许多核心微积分技能。基本步骤为：仔细阅读题目，确定要优化的量，利用约束条件将其表示为单一变量的函数，求导，找到驻点并确认其性质。始终在原问题的语境下解释答案，并清楚地陈述最大值或最小值。
Key topics to revise include stationary points, first and second derivative tests, related rates of change, and kinematic applications. Understanding the link between algebraic expressions and real-world quantities is the key to success. Practice constructing functions from geometric and economic scenarios, and remember that modelling questions often carry high marks, so a logical, well-reasoned presentation is vital.
需要复习的关键主题包括驻点、一阶和二阶导数判定法、相关变化率以及运动学应用。理解代数表达式与现实量之间的联系是成功的关键。练习从几何和经济情境中构造函数，并记住建模题目通常分值很高，因此逻辑清晰、推理充分的展示至关重要。
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