Partial Differentiation in KS3 Mathematics | KS3 数学：偏微分考点精讲

📚 Partial Differentiation in KS3 Mathematics | KS3 数学：偏微分考点精讲

Partial differentiation is a concept often seen in advanced maths, but the seeds of this idea can already be found in Key Stage 3 topics. Understanding how a quantity depends on more than one variable, and exploring what happens when just one of those variables changes, is a powerful way to build mathematical thinking. In this guide, we will gently introduce the idea of partial differentiation in a way that connects with graphs, formulas and real-life situations you already know.

偏微分是一个通常在高等数学中出现概念，但其实它的思想种子早就在 KS3 阶段的课题里埋下了。理解一个量如何依赖于多个变量，并探索当其中只有一个变量变化时会发生什么，能够很好地培养数学思维。在这篇精讲中，我们将以一种温和的方式，结合你已经熟悉的图像、公式和实际情境，初步介绍偏微分的思想。

1. Understanding the Basics of Change | 理解变化的基础

In Key Stage 3, you have already learned how one quantity can change with another. For example, the cost of apples depends on how many kilograms you buy. If the price is £2 per kg, the total cost C in pounds is given by C = 2n, where n is the number of kilograms. This is a function of one variable: C depends solely on n. The rate of change is constant, and you can calculate it easily.

在 KS3 阶段，你已经学习过一个量如何随另一个量变化。例如，苹果的总价取决于你买了多少千克。如果单价是每千克2英镑，那么总价 C（英镑）就可以用 C = 2n 来表示，其中 n 是千克数。这就是一个一元函数：C 只由 n 决定。变化率是恒定的，你可以轻松计算出来。

But real life is often more complicated. Many quantities depend on two or more things at the same time. The area of a sports hall depends on both its length and its width. The speed of a car depends on distance and time. To understand such situations, we need to think about how the outcome changes when we change just one of the inputs while keeping the others fixed. This is the heart of partial differentiation.

但真实生活往往更复杂。许多量同时依赖于两个或更多的因素。体育馆的面积既取决于它的长度，也取决于它的宽度。汽车的速度依赖于路程和时间。要理解这些情况，我们就需要考虑：当我们固定其他输入，只改变其中一个输入时，最终结果会如何变化。这正是偏微分的核心思想。

2. What is Partial Differentiation? | 什么是偏微分？

Partial differentiation is a method for finding the rate of change of a function that has more than one variable. Instead of looking at how the whole function changes when everything varies, we focus on one variable at a time and treat all other variables as if they were constants. This gives us a ‘partial’ rate of change. For now, we will not use complicated limits; we will build the idea using simple arithmetic and tables.

偏微分是一种求多元函数变化率的方法。我们不是看所有变量一起变化时整个函数怎么变，而是一次只关注一个变量，把其他变量都暂时当作常数来看待。这样我们就得到了一个“偏”的变化率。目前我们不会使用复杂的极限定义，而是通过简单的算术和表格来建立这个概念。

In KS3, you often use formulas with several letters. A classic example is the area A of a rectangle: A = l × w. Suppose l is the length and w is the width. If we keep w fixed and change l, the area will change at a rate proportional to w. The partial rate of change of A with respect to l is just w. Similarly, if we keep l fixed, the partial rate of change of A with respect to w is l.

在 KS3 中，你经常使用含有几个字母的公式。一个典型的例子是矩形的面积 A：A = l × w。设 l 是长度，w 是宽度。如果我们固定 w 不变，仅改变 l，面积就会以 w 为比例常数产生变化。A 关于 l 的偏变化率就是 w。同样，固定 l 时，A 关于 w 的偏变化率就是 l。

3. Real-Life Example: Area of a Rectangle | 实际例子：矩形的面积

Imagine a rectangular garden where the width is permanently fixed at 3 metres because of a wall. The length can be varied. The area function becomes A(l) = 3 × l. This is just a one-variable function, and the rate of change (the ordinary gradient) is 3. So, the partial derivative of A = l × w with respect to l when w is held constant at 3 is 3.

想象一个长方形花园，由于有一堵墙，它的宽度永久固定为 3 米，而长度可以变化。那么面积函数就成了 A(l) = 3 × l。这只是一个一元函数，变化率（普通梯度）就是 3。因此，在 w 被固定为 3 的情况下，A = l × w 关于 l 的偏导数就是 3。

Instead of fixing w at a number, we can keep it as a variable but treat it as a constant during the calculation. The general rule for the partial rate of change of A with respect to l is: take the derivative of l × w as if w is just a number. The result is w. We write this using a special curly ‘d’, called ‘partial d’, like this:

与其把 w 固定为一个具体数字，我们可以让它保持为变量，但在计算中把它当作常数处理。A 关于 l 的偏变化率的一般规则是：把 w 当作一个数字，对 l × w 进行求导。结果是 w。我们用一个特殊的弯体字母 ‘d’ 来表示这种偏导数，写作：

∂A/∂l = w

This symbol tells us we are only differentiating with respect to l, while everything else is treated as constant.

这个符号告诉我们，我们只是在对 l 进行微分，而其他一切都被当作常数。

4. How to Find a Partial Rate of Change | 如何求偏变化率

To find a partial rate of change, follow these steps: identify the function with two or more variables; choose the variable you are interested in; regard all other variables as fixed numbers; then find the ordinary rate of change as you would for a one-variable function. Let’s practise with a simple formula from physics: distance = speed × time, or d = v × t.

要找到偏变化率，请遵循以下步骤：识别含有两个或更多变量的函数；选择你感兴趣的那个变量；将所有其他变量视为固定的数；然后按照处理一元函数的方法求出普通变化率。我们用一个来自物理学的简单公式来练习：路程 = 速度 × 时间，即 d = v × t。

If we want to know how distance changes when speed changes, while time stays the same, we treat t as a constant. The function is d(v) = t × v, so the rate of change of d with respect to v is t. That is: ∂d/∂v = t. If we want to know how distance changes when time changes, with speed constant, we get ∂d/∂t = v.

如果我们想知道：时间不变时，路程如何随速度变化，我们就把 t 当作常数。函数是 d(v) = t × v，因此 d 关于 v 的变化率是 t。即：∂d/∂v = t。如果我们想知道：速度不变时，路程如何随时间变化，我们得到 ∂d/∂t = v。

These partial rates make intuitive sense: doubling the speed while keeping time the same doubles the distance. The constant multiplier is the fixed time value.

这些偏变化率符合直觉：保持时间不变，速度翻倍，路程也翻倍。那个常数乘子就是被固定的时间值。

5. Treating Other Variables as Constants | 将其他变量视为常数

The key skill in partial differentiation is mastering the ability to ‘freeze’ variables. When you see an expression like volume V = l × w × h (for a cuboid), you might be asked: how quickly does the volume increase if we only increase the height? You temporarily pretend l and w are just numbers. Then V = (l × w) × h, and the partial rate ∂V/∂h = l × w.

偏微分的关键技能就是掌握“冻结”变量的能力。当你看到表达式如长方体体积 V = l × w × h 时，可能会被问到：如果我们只增加高度，体积增加的速度有多快？你暂时把 l 和 w 当成普通的数。那么 V = (l × w) × h，因此偏变化率 ∂V/∂h = l × w。

This technique can be practised with any formula you meet in KS3: pressure = force ÷ area, density = mass ÷ volume, or even the area of a triangle A = ½ × b × h. In each case, ask ‘what if one quantity varies and the rest are fixed?’ and you are already doing the thinking behind partial differentiation.

这个技巧可以用 KS3 中遇到的任何公式来练习：压强 = 压力 ÷ 面积，密度 = 质量 ÷ 体积，甚至三角形面积 A = ½ × b × h。在每种情况下，问自己“如果其中一个量变化而其他保持不变会怎样？”，你就已经在进行偏微分背后的思考了。

Here is a table to help you practise ‘freezing’ variables:

下面这张表格可以帮助你练习“冻结”变量：

Formula \| 公式	Variable of interest \| 关注变量	Constants \| 常数	Partial rate \| 偏变化率
A = l × w	l	w	w
V = l × w × h	w	l, h	l × h
A = ½ × b × h	b	h	½ × h
d = v × t	t	v	v

6. Notation and Simple Calculations | 符号与简单计算

In KS3, you are used to seeing the slope of a straight line as the coefficient of x in y = mx + c. When we move to functions of two variables like z = 2x + 3y, we can still talk about slopes, but now there are two directions. If we keep y constant, the slope in the x-direction is the coefficient of x, which is 2. We write this as ∂z/∂x = 2.

在 KS3，你已经习惯于将直线的斜率视为 y = mx + c 中 x 的系数。当我们转向像 z = 2x + 3y 这样的二元函数时，仍然可以谈论斜率，但现在有两个方向。如果固定 y，沿 x 方向的斜率就是 x 的系数，即 2。我们写作 ∂z/∂x = 2。

If we keep x constant, the slope in the y-direction is 3, so ∂z/∂y = 3. Notice how easy it is: just look at the coefficient of the variable you are focusing on, and treat the other term as part of the constant. Even when the formula is not a simple sum, the idea is the same. For example, for F = x × y, keeping y fixed gives ∂F/∂x = y, and keeping x fixed gives ∂F/∂y = x.

如果固定 x，沿 y 方向的斜率就是 3，因此 ∂z/∂y = 3。注意这有多么简单：只需看你所关注的变量的系数，把其他项看成常数的一部分即可。即使公式不是一个简单的和，想法也是一样的。比如，对于 F = x × y，固定 y 得到 ∂F/∂x = y，固定 x 得到 ∂F/∂y = x。

We can summarise this in a short calculation guide:

我们可以将其总结成简短的计算指南：

Write the function, e.g. T = a × b + c. | 写出函数，例如 T = a × b + c。
To find ∂T/∂a, treat b and c as constants. | 要找到 ∂T/∂a，把 b 和 c 当作常数。
The derivative of a × (constant) with respect to a is just that constant. | a × (常数) 对 a 的导数就是那个常数。
Any term without the variable a becomes zero when differentiating. | 求导时，任何不含变量 a 的项变为零。
So ∂T/∂a = b. | 因此 ∂T/∂a = b。

7. Exploring Functions of Two Variables | 探索二元函数

Let’s go a step further and create a function that describes the cost of a school trip. Suppose each pupil pays a basic fee plus an amount for each activity. If the number of pupils is p and the number of activities chosen is a, the total income I in pounds might be given by I = 20p + 5a + 50, where 50 is a fixed grant. Notice how this function depends on two variables.

我们更进一步，创建一个描述学校旅行费用的函数。假设每个学生支付基本费用再加上每项活动的费用。如果学生人数是 p，选择的活动数量是 a，那么总收入 I（英镑）可以表示为 I = 20p + 5a + 50，其中 50 是一笔固定补贴。请注意这个函数如何依赖于两个变量。

We can find partial rates: ∂I/∂p = 20. This means that for each extra pupil, with activities held fixed, the income rises by £20. Similarly, ∂I/∂a = 5, so each additional activity chosen (with pupil numbers fixed) adds £5 to the total. The constant 50 disappears because its rate of change is zero no matter which variable we consider.

我们可以求出偏变化率：∂I/∂p = 20。这意味着，在活动数量保持固定的情况下，每增加一名学生，收入就增加 20 英镑。类似地，∂I/∂a = 5，因此在学生人数固定时，每多选一项活动，总额就增加 5 英镑。常数 50 消失了，因为无论考虑哪个变量，它的变化率都是零。

This kind of analysis helps planners understand where extra resources have the most effect. Partial differentiation allows you to isolate the effect of each factor.

这种分析有助于规划者了解额外资源在哪里能发挥最大作用。偏微分让你能够分离出每个因素的影响。

8. Graphical Interpretation for KS3 Students | KS3 学生的图形解释

You are familiar with drawing graphs of y = mx + c on the coordinate plane. A function with two inputs requires three dimensions to draw, which is challenging, but we can still think about it. Imagine a hilly landscape. The height of the land depends on the east–west coordinate and the north–south coordinate. If you walk only east, the slope under your feet is the partial derivative of height with respect to east.

你已经熟悉在坐标平面上画出 y = mx + c 的图像。一个有两个输入的函数需要三维空间才能画出来，这有点挑战性，但我们仍然可以思考。想象一片丘陵地形。地面的高度依赖于东西向坐标和南北向坐标。如果你只向东走，脚下感受到的坡度就是高度关于东向坐标的偏导数。

In KS3 we can visualise this with a contour map. Contour lines join points of equal height. When you move perpendicular to the contour lines in an easterly direction, the spacing of the lines tells you how steep the terrain is in that specific direction. That is exactly what a partial derivative measures – the steepness in one coordinate direction while the other coordinate is unchanged.

在 KS3，我们可以用等高线地图来想象这一点。等高线连接高度相等的点。当你沿着正东方向垂直于等高线移动时，线条的疏密就告诉你在这个特定方向上的坡度有多大。这正是偏导数所度量的——在一个坐标方向上、保持另一个坐标不变时的陡峭程度。

Consider a flat roof tilted only in the north direction. If you walk east across it, you stay at the same height, so the partial derivative in the east direction is zero. If you walk north, you go up or down a slope, giving a non-zero partial derivative.

考虑一个仅向北倾斜的平屋顶。如果你向东走，高度不变，因此东方向的偏导数为零。如果你向北走，就会走上或走下一个斜坡，得到一个非零偏导数。

9. Linking to KS3 Topics: Proportions | 与 KS3 课题的联系：比例

In KS3, you spend a lot of time learning about direct proportion. If y is directly proportional to x, we write y = kx. But what if y is proportional to two different quantities? For instance, the amount of paint needed to cover a wall is directly proportional to both the height and the width. We can write P = k × h × w, where k is a constant. Here, P is a function of two variables, and partial differentiation helps unpack the relationship.

在 KS3，你花了很多时间学习正比例关系。如果 y 与 x 成正比例，我们写作 y = kx。但如果 y 同时与两个不同的量成正比例呢？例如，粉刷一面墙所需的油漆量既与高度成正比，也与宽度成正比。我们可以写作 P = k × h × w，其中 k 是常数。这里 P 是一个二元函数，而偏微分可以帮助我们解开这个关系。

If we keep width w fixed and double the height, the paint amount doubles. The partial rate ∂P/∂h = k × w. This value itself is proportional to w, which makes sense: for a wider wall, the effect of changing height is greater because there is more area to cover per extra metre of height.

如果我们固定宽度 w 不变，将高度加倍，油漆量就加倍。偏变化率 ∂P/∂h = k × w。这个值本身与 w 成正比例，这很合理：墙越宽，改变高度的影响就越大，因为高度每增加一米，需要覆盖的面积就更多。

Such thinking ties directly into KS3 work on algebra and formulae. You are learning to substitute numbers, rearrange equations and understand how changing one quantity affects another. Partial differentiation simply formalises the question: ‘How does changing just this one thing affect the answer, assuming the other things don’t move?’

这种思考直接联系到 KS3 的代数与公式学习。你正在学习代入数字、重新排列方程，并理解改变一个量如何影响另一个量。偏微分只不过是将这个问题正式化：“假设其他东西不动，只改变这一个东西，会怎样影响答案？”

10. Practice Questions and Examples | 练习题与示例

Let’s practice with some KS3-friendly questions. Remember, the aim is to build familiarity with the idea, not to master university-level calculus.

让我们用一些适合 KS3 的问题来练习。请记住，目的是熟悉这一思想，而不是掌握大学水平的微积分。

Example 1: The perimeter P of a rectangle with length l and width w is given by P = 2l + 2w. (a) Find the partial rate of change of P with respect to l. (b) Find ∂P/∂w.

示例1：一个长方形的周长 P 由长度 l 和宽度 w 表示为 P = 2l + 2w。(a) 求 P 关于 l 的偏变化率。 (b) 求 ∂P/∂w。

Solution: (a) Treat w as a constant. Then P(l) = 2l + (constant 2w). The derivative of 2l with respect to l is 2, and the derivative of the constant is 0. So ∂P/∂l = 2. (b) Treat l as constant. ∂P/∂w = 2.

解答：(a) 把 w 当作常数。那么 P(l) = 2l + (常数2w)。2l 对 l 的导数是 2，常数的导数是 0。所以 ∂P/∂l = 2。(b) 把 l 当作常数。∂P/∂w = 2。

Example 2: The volume V of a cuboid is V = l × w × h. Find ∂V/∂l when l = 5, w = 3 and h = 2. What does this number mean?

示例2：一个长方体的体积 V = l × w × h。当 l = 5，w = 3，h = 2 时，求 ∂V/∂l。这个数字表示什么意思？

Solution: First find the general partial derivative: ∂V/∂l = w × h. Substitute the fixed values: w = 3, h = 2, so ∂V/∂l = 3 × 2 = 6. This means that at the moment when the dimensions are 5 by 3 by 2, increasing the length by 1 unit while holding width and height fixed adds 6 cubic units to the volume.

解答：先求一般的偏导数：∂V/∂l = w × h。代入固定值：w = 3，h = 2，因此 ∂V/∂l = 3 × 2 = 6。这意味着，当尺寸为 5×3×2 时，将长度增加 1 个单位并保持宽度和高度不变，体积就会增加 6 个立方单位。

11. Avoiding Common Misunderstandings | 避免常见误解

One mistake is to think that partial differentiation is completely different from ordinary rate of change. In truth, it is just ordinary differentiation applied while pretending other variables are numbers. If you can differentiate 5x or 3t, you can find partial derivatives of expressions like 5xy or 3t²s.

一个错误是认为偏微分与普通变化率完全不同。事实上，它就是把普通微分应用在“假装其他变量是数字”的情境中。只要你会对 5x 或 3t 求导，你就能找到像 5xy 或 3t²s 这样的表达式的偏导数。

Another misunderstanding is confusing the symbol ∂ with the letter d. They are related, but ∂ reminds us that the function depends on more than one variable. It is a useful reminder to keep track of which variable we are allowing to change.

另一个误解是把符号 ∂ 和字母 d 搞混。它们是相关的，但 ∂ 提醒我们这个函数依赖于不只一个变量。这有助于我们留意正在让哪个变量发生变化。

Finally, remember that partial derivatives are not fixed numbers in general; they are expressions involving the other variables. For example, ∂(x²y)/∂x = 2xy, which still contains y. This is because if y changes, the steepness in the x-direction also changes. Thinking about this prepares you for more advanced study.

最后，请记住偏导数通常不是固定的数字，而是包含其他变量的表达式。例如，∂(x²y)/∂x = 2xy，它仍然包含 y。这是因为如果 y 改变了，沿 x 方向的斜率也会改变。思考这一点能为更深入的学习做好准备。

12. Summary and Key Takeaways | 总结与核心要点

Partial differentiation is a tool for exploring how multivariable functions change when only one input varies. At KS3, you do not need to calculate formal limits, but you can grasp the core idea through familiar formulas. Always identify the variable you are differentiating with respect to, treat the others as constants, and apply the rules you already know for one-variable functions.

偏微分是探索多元函数在只有一个输入变化时如何变化的工具。在 KS3 阶段，你不需要计算正式的极限定义，但可以通过熟悉的公式掌握其核心思想。始终要明确你对哪个变量进行微分，把其他变量当作常数，然后应用你已知的一元函数求导规则。

This skill strengthens your algebraic fluency and prepares you for GCSE and A-level topics involving rates of change, optimisation and multi-step modelling. More importantly, it helps you think systematically about the world, where outcomes rarely depend on just one factor.

这项技能能增强你的代数流畅度，并为 GCSE 和 A-level 中涉及变化率、优化以及多步骤建模的课题做好准备。更重要的是，它能帮助你系统性地思考这个世界，因为世界上的结果几乎从不只依赖于单一因素。

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Partial Differentiation in KS3 Mathematics | KS3 数学：偏微分 考点精讲