📚 Worked Example 4.4.2: The Chain Rule and Related Rates of Change | 示例 4.4.2：链式法则与相关变化率
In AQA A Level Mathematics, Example 4.4.2 is a classic worked problem on connected rates of change. It tests your ability to apply the chain rule when one quantity changes with respect to time. The example uses a spherical balloon to show how a known volume flow rate can be used to find the rate at which the radius grows. Understanding this example fully will prepare you for similar questions on differentiating functions that are not written explicitly in terms of the independent variable.
在 AQA A Level 数学中，示例 4.4.2 是有关相关变化率的经典例题。它考查你应用链式法则处理一个量随时间变化的能力。该示例以一个球形气球为背景，展示如何利用已知的体积变化率求出半径增大的速率。彻底理解这个示例将帮助你应对同类题目，即对不是显式以自变量表示的函数进行求导。
1. What Is Example 4.4.2? | 示例 4.4.2 简介
Example 4.4.2 asks you to find the rate at which the radius of a spherical balloon increases when air is pumped into it at a constant rate. The problem gives the rate of change of volume with respect to time, dV/dt, and asks for the rate of change of radius with respect to time, dr/dt. You are also given the instant at which the radius equals 5 cm.
示例 4.4.2 要求你求出当空气以恒定速率充入球形气球时，气球半径增大的速率。题目给出体积随时间的变化率 dV/dt，并要求求出半径随时间的变化率 dr/dt。此外，题目还给出了半径恰为 5 厘米的瞬间。
The full statement of the example can be written as follows: A spherical balloon is inflated at a rate of 20 cm³ s⁻¹. Find the rate at which the radius increases when the radius is 5 cm.
该示例的完整表述如下：一个球形气球以 20 cm³ s⁻¹ 的速率被充气。求当半径为 5 厘米时，半径增大的速率。
This problem is a standard application of the chain rule in AQA Pure Mathematics. The chain rule links the derivative of the volume function with respect to time to the derivative of the radius function with respect to time. You will often see similar problems involving cylinders, cones, or other geometric shapes.
这是 AQA 纯数学中链式法则的标准应用。链式法则将体积函数对时间的导数与半径函数对时间的导数联系起来。你会经常遇到涉及圆柱、圆锥或其他几何形状的类似问题。
2. Learning Objectives | 学习目标
ld also be able to differentiate a volume formula with respect to the radius, substitute known values, and interpret the result in the context of the problem.
通过学习示例 4.4.2，你应该能够识别相关变化率，并应用链式法则的形式 dV/dt = dV/dr × dr/dt。你还应该能够对体积公式关于半径求导，代入已知值，并在题目情境下解释结果。
The specific skills tested include using standard differentiation rules, rearranging equations to isolate the required rate, and keeping units consistent throughout the calculation. AQA examiners also expect clear notation and an explicit final statement with correct units.
该题考查的具体技能包括使用标准求导法则、重新整理方程以分离出所需的变化率，以及在整个计算过程中保持单位一致。AQA 考官还期望看到清晰的符号表示和带有正确单位的最终结论。
3. Key Concept: The Chain Rule | 核心概念：链式法则
ates of change problems, the same principle is used with time as the independent variable.
链式法则是求复合函数导数的基本工具。如果变量 y 依赖于 u，而 u 依赖于 x，那么 y 关于 x 的变化率由 dy/dx = dy/du × du/dx 给出。在相关变化率问题中，同样的原理以时间作为自变量来使用。
In Example 4.4.2, the volume V depends on the radius r, and the radius r depends on time t. Therefore, the chain rule connects the known rate dV/dt to the unknown rate dr/dt by introducing dV/dr. This link is written as:
在示例 4.4.2 中，体积 V 依赖于半径 r，而半径 r 依赖于时间 t。因此，链式法则通过引入 dV/dr，将已知变化率 dV/dt 与未知变化率 dr/dt 连接起来。这个联系可以写成：
dV/dt = dV/dr × dr/dt
Once you find dV/dr from the volume formula, you can substitute the known values and solve for dr/dt. This is the central strategy for all related rates questions in AQA A Level Mathematics.
一旦你从体积公式中求出 dV/dr，就可以代入已知值并解出 dr/dt。这是 AQA A Level 数学中所有相关变化率问题的核心策略。
4. Setting Up the Worked Example | 建立示例问题
The given information in Example 4.4.2 is:
示例 4.4.2 中给出的信息如下：
dV/dt = 20 cm³ s⁻¹ and r = 5 cm
The quantity to find is dr/dt when r = 5 cm. The sphere has a known volume formula:
需要求出的量是当 r = 5 cm 时的 dr/dt。球体有已知的体积公式：
V = (4/3)πr³
This formula is given in the AQA mathematical formulae booklet, but you should be able to recall it from memory. The next step is to differentiate V with respect to r, because the chain rule requires dV/dr.
该公式在 AQA 数学公式手册中给出，但你应该能够凭记忆写出。下一步是对 V 关于 r 求导，因为链式法则需要 dV/dr。
5. Step 1: Identify the Variables and Rates | 第1步：识别变量与变化率
Before differentiating, write down the three quantities involved: V, r, and t. The volume V is a function of the radius r, and the radius r is a function of time t. The problem gives one derivative, dV/dt, and asks for another derivative, dr/dt.
在求导之前，先写下涉及的三个量：V、r 和 t。体积 V 是半径 r 的函数，而半径 r 是时间 t 的函数。题目给出了一个导数 dV/dt，并要求求出另一个导数 dr/dt。
A useful way to organise the information is to make a small table:
整理信息的一种有效方法是制作一个小表格：
| Quantity | Value | Status |
| dV/dt | 20 cm³ s⁻¹ | Given |
| r | 5 cm | Given |
| dr/dt | ? | To find |
This table helps you see that the missing link is dV/dr, which must be obtained by differentiating the volume formula. Keeping the notation clear from the beginning reduces errors later.
这个表格帮助你看到缺失的环节是 dV/dr，它必须通过对体积公式求导得到。从一开始就保持符号清晰可以减少后续错误。
6. Step 2: Differentiate the Volume Formula | 第2步：对体积公式求导
ect to r. The constant (4/3)π is treated as a multiplier, and the derivative of r³ is 3r². Therefore, the derivative is:
对 V = (4/3)πr³ 关于 r 求导。常数 (4/3)π 被视为乘数，r³ 的导数是 3r²。因此，导数为：
dV/dr = 4πr²
rface area of the sphere. At the instant when r = 5 cm, you can evaluate dV/dr as follows:
这个结果在几何上是有意义的：球体体积关于半径的变化率等于球体的表面积。在 r = 5 cm 的瞬间，你可以计算 dV/dr 如下：
dV/dr = 4π(5)² = 100π cm³ cm⁻¹
Note that the units of dV/dr are cm³ per cm, which simplifies to cm², matching the units of area. This consistency check is helpful in an exam.
注意 dV/dr 的单位是 cm³ 每 cm，简化后为 cm²，与面积单位一致。这个一致性检查在考试中很有帮助。
7. Step 3: Apply the Chain Rule and Substitute | 第3步：应用链式法则并代入
dV/dr × dr/dt. Substitute the known values dV/dt = 20 and dV/dr = 100π into the equation:
现在使用链式法则表达式 dV/dt = dV/dr × dr/dt。将已知值 dV/dt = 20 和 dV/dr = 100π 代入方程：
20 = 100π × dr/dt
To isolate dr/dt, divide both sides of the equation by 100π:
为了分离出 dr/dt，将方程两边同时除以 100π：
dr/dt = 20 ÷ (100π) = 1 ÷ (5π)
Simplifying this expression gives the exact answer. To three significant figures, the numerical value is approximately 0.0637 cm s⁻¹. The exact form 1/(5π) is preferred in AQA marks schemes unless a numerical answer is specifically requested.
化简该表达式得到精确答案。保留三位有效数字时，该数值约为 0.0637 cm s⁻¹。除非题目特别要求数值答案，否则 AQA 评分标准更倾向于精确形式 1/(5π)。
8. Step 4: Interpret the Result | 第4步：解释结果
The positive sign of dr/dt indicates that the radius is increasing. This is expected because air is being pumped into the balloon. The magnitude of 1/(5π) cm s⁻¹ means that at the exact moment when the radius is 5 cm, the radius grows at about 0.0637 cm every second.
dr/dt 的正号表明半径正在增大。这是意料之中的，因为空气正被充入气球。1/(5π) cm s⁻¹ 的大小意味着在半径恰好为 5 cm 的时刻，半径每秒增大约 0.0637 厘米。
This rate is not constant. As the balloon gets larger, the same volume flow rate produces a smaller increase in radius because the surface area 4πr² increases. Therefore, dr/dt decreases as r increases, even though dV/dt remains constant.
该速率并不是恒定的。随着气球变大，相同的体积流量产生的半径增量会变小，因为表面积 4πr² 在增大。因此，即使 dV/dt 保持恒定，dr/dt 也会随着 r 的增大而减小。
9. Common Mistakes to Avoid | 需要避免的常见错误
orrectly. Always write the chain rule statement dV/dt = dV/dr × dr/dt before substituting.
一个常见错误是在不使用链式法则的情况下直接对 V 关于时间求导。这常常导致错误地混合使用 r 和 t 的表达式。在代入之前，始终要先写出链式法则关系式 dV/dt = dV/dr × dr/dt。
s an expression in r instead of substituting r = 5. This gives an incomplete final answer.
另一个常见错误是忘记在给定半径处计算 dV/dr。一些学生将 dV/dr 保留为 4πr²，然后试图将 dr/dt 解为 r 的表达式，而不是代入 r = 5。这会导致最终答案不完整。
Units are also frequently mishandled. Ensure that dV/dt has units of volume per time, dV/dr has units of area, and dr/dt has units of length per time. A final answer without units can lose a mark in AQA exams.
单位也经常被错误处理。确保 dV/dt 的单位是体积每时间，dV/dr 的单位是面积，dr/dt 的单位是长度每时间。在 AQA 考试中，最终答案不带单位可能会丢分。
Finally, avoid using the diameter instead of the radius. If the problem states a radius of 5 cm, do not substitute 10 cm into the formula. Read the question carefully and identify whether r or d is given.
最后，避免使用直径代替半径。如果题目说明半径为 5 cm，不要将 10 cm 代入公式。仔细读题，确定给出的是 r 还是 d。
10. Practice Variation: Cone-Shaped Container | 练习变式：圆锥形容器
To strengthen your understanding, consider a variation of Example 4.4.2. A cone-shaped container is being filled with water at a rate of 30 cm³ s⁻¹. The height of the cone is twice its radius. Find the rate at which the water level rises when the radius of the water surface is 6 cm.
为了加深理解，可以考虑示例 4.4.2 的一个变式。一个圆锥形容器以 30 cm³ s⁻¹ 的速率注水。圆锥的高度是其半径的两倍。求当水面半径为 6 cm 时，水位上升的速率。
= (2/3)πr³. Differentiate with respect to r to get dV/dr = 2πr². Then use the chain rule:
对于圆锥，体积为 V = (1/3)πr²h。由于 h = 2r，你可以将 V 仅用 r 表示为 V = (2/3)πr³。对 r 求导得到 dV/dr = 2πr²。然后使用链式法则：
dV/dt = dV/dr × dr/dt
π) cm s⁻¹. This type of question regularly appears in AQA A Level Mathematics exam papers.
代入 dV/dt = 30 和 r = 6，得到 dr/dt = 30 ÷ [2π(6)²] = 5 ÷ (12π) cm s⁻¹。这类题目经常出现在 AQA A Level 数学考试试卷中。
11. Exam Tips for AQA Related Rates | AQA 相关变化率考试技巧
In the AQA exam, related rates of change questions are typically worth between 4 and 6 marks. The marks are awarded for the correct chain rule statement, correct differentiation, correct substitution, and a final answer with units. Show all four steps clearly.
在 AQA 考试中，相关变化率题目通常占 4 到 6 分。分数分别给予正确的链式法则表达式、正确的求导、正确的代入以及带单位的最终答案。请清楚展示所有四个步骤。
can still earn you a method mark, even if the subsequent algebra is incomplete. Never leave a related rates question blank.
如果时间不够，写出链式法则方程 dV/dt = dV/dr × dr/dt 仍然可以为你赢得方法分，即使后续代数运算没有完成。千万不要在相关变化率题目上留白。
Always check that your final rate has the correct sign. A positive rate means the quantity is increasing, while a negative rate means it is decreasing. In physical problems, the sign must match the situation described in the question.
始终检查最终变化率的符号是否正确。正变化率表示该量正在增大，负变化率表示该量正在减小。在物理问题中，符号必须与题目描述的情境一致。
12. Summary and Key Points | 总结与要点
Example 4.4.2 demonstrates that related rates of change can be solved in four clear steps: identify the known and unknown rates, differentiate the geometric formula, apply the chain rule, and substitute the given values. The chain rule is the essential link between rates that are not directly connected.
示例 4.4.2 说明，相关变化率可以通过四个清晰的步骤解决：识别已知和未知变化率、对几何公式求导、应用链式法则以及代入给定值。链式法则是不直接相关的变化率之间的核心联系。
, which is approximately 0.0637 cm s⁻¹. Remember that this rate changes as the radius changes, even if the volume flow rate stays constant.
球形气球问题的最终答案是 dr/dt = 1/(5π) cm s⁻¹，约等于 0.0637 cm s⁻¹。请记住，即使体积流量保持恒定，该速率也会随着半径的变化而变化。
Mastering this example will give you confidence in tackling any AQA connected rates question, whether it involves spheres, cones, cylinders, or other shapes. Practise the steps until they become automatic.
掌握这个示例将使你有信心应对任何 AQA 相关变化率问题，无论涉及球体、圆锥、圆柱还是其他形状。反复练习这些步骤，直到它们变得自动化。
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