📚 Exercise 21D: Related Rates in Differential Calculus | 练习21D：微分学中的相关变化率
Exercise 21D focuses on one of the most practical and examinable applications of differential calculus: related rates. In these problems, two or more quantities vary with time, and a relationship between their time derivatives allows us to find an unknown rate of change when other rates are known. This topic deepens your understanding of the chain rule and implicit differentiation, linking abstract calculus to real-world situations such as changing areas, volumes, distances, and motion.
练习21D聚焦于微分学中最具实用性和考试频次的应用之一：相关变化率。在这类问题中，两个或多个量随时间变化，它们时间导数之间的关系使我们能够在已知其他变化率的情况下求出未知的变化率。本主题加深你对链式法则和隐函数求导的理解，将抽象的微积分与变化的面积、体积、距离和运动等现实情境联系起来。
1. What Are Related Rates? | 什么是相关变化率？
Related rates problems involve variables that are functions of a common parameter, typically time t. Instead of finding a static equation, we are interested in how the rate of change of one variable influences the rate of change of another. The key mathematical tool is the chain rule: if y = f(x) and both x and y depend on time, then dy/dt = (dy/dx)·(dx/dt). This simple formula lies at the heart of every related rates question.
相关变化率问题涉及到的变量都是同一个参数（通常是时间 t）的函数。我们关心的不是静态方程，而是一个变量的变化率如何影响另一个变量的变化率。核心的数学工具是链式法则：若 y = f(x) 且 x 和 y 都依赖于时间，则 dy/dt = (dy/dx)·(dx/dt)。这个简单的公式是所有相关变化率问题的核心。
2. The Chain Rule Connection | 链式法则的联系
If a quantity Q is expressed in terms of a geometric measurement that itself changes over time, we differentiate with respect to t using the chain rule. For example, the area of a circle A = πr² becomes dA/dt = 2πr·(dr/dt). Here, dr/dt is the rate at which the radius changes, and dA/dt is the related rate we may need to calculate. Always identify the ‘link’ variable that connects the known and unknown rates.
如果一个量 Q 由一个几何测量值表达，而这个测量值本身又随时间变化，我们就需要对 t 求导并使用链式法则。例如，圆的面积 A = πr² 求导得 dA/dt = 2πr·(dr/dt)。其中 dr/dt 是半径的变化率，dA/dt 是我们可能需要计算的相关变化率。一定要找出连接已知和未知变化率的“中间”变量。
3. General Procedure for Solving Related Rates Problems | 求解相关变化率问题的一般步骤
A structured approach turns a seemingly complex word problem into a straightforward derivative calculation:
1. Read the problem carefully and identify the given rates and the required unknown rate.
2. Draw a clear diagram and label the variables.
3. Write an equation relating the variables (often a geometric formula).
4. Differentiate both sides with respect to time t, applying the chain rule.
5. Substitute the known values, including specific instant values of variables and rates.
6. Solve for the unknown rate.
7. Interpret the sign: negative usually means decreasing, positive means increasing.
一个结构化的方法能将看似复杂的文字题转化为简单的导数计算：
1. 仔细读题，确定已知变化率和要求解的变化率。
2. 画出清晰的示意图，标注各变量。
3. 写出变量之间的关系方程（通常是几何公式）。
4. 对方程两边关于时间 t 求导，运用链式法则。
5. 代入已知值，包括变量的具体瞬时值和变化率。
6. 解出未知变化率。
7. 注意符号的含义：负值通常表示减少，正值表示增加。
4. Identifying Variables and Their Rates | 确定变量及其变化率
Before any calculation, write down what you are given in mathematical notation. For instance, “the radius r is increasing at a constant rate of 3 cm/s” translates to dr/dt = 3 cm/s. “Find the rate at which the area is increasing when r = 5 cm” means calculate dA/dt at the instant r = 5. Clear notation such as dV/dt, dh/dt, dx/dt prevents confusion and sets up the problem correctly.
在计算之前，用数学符号写出已知条件。例如，“半径 r 以 3 cm/s 的恒定速率增加”应写为 dr/dt = 3 cm/s。“求当 r = 5 cm 时面积的增长率”意味着在 r = 5 这一时刻计算 dA/dt。清晰的符号表示，如 dV/dt、dh/dt、dx/dt，能避免混淆并正确设置题目。
5. Drawing a Diagram | 绘制示意图
For geometric situations – a filling conical tank, a sliding ladder, an expanding ripple – a well-labeled diagram is invaluable. Mark constants such as height, length, or angles. Indicate variables like water depth h, radius r, or distance x. Show rates with arrows and labels (e.g. dr/dt ↑). The diagram helps you spot the relationship and ensures you substitute the correct value at the required instant.
对于几何情境——正在注水的锥形水箱、滑动的梯子、扩大的波纹——标注清晰的示意图极为重要。标出高、长、角度等常量，以及水深 h、半径 r 或距离 x 等变量。用箭头和标注标示变化率（如 dr/dt ↑）。示意图可以帮助你发现变量之间的关系，并确保你在要求的时刻代入正确的数值。
6. Forming the Relationship Equation | 建立关系方程
Use known formulas: Pythagoras’ theorem for a ladder against a wall (x² + y² = L²), area of a circle (A = πr²), volume of a cone (V = ⅓πr²h), or similar triangles. If the problem involves more than two variables, eliminate one using a secondary relation (e.g. similar triangles give r/h = R/H). The final equation should connect only the variables whose rates are given or asked, ready for differentiation.
使用已知公式：梯子靠墙用勾股定理（x² + y² = L²），圆面积（A = πr²），圆锥体积（V = ⅓πr²h），或相似三角形。如果问题涉及两个以上的变量，用辅助关系消去一个（例如相似三角形给出 r/h = R/H）。最终的方程应只保留那些给出了变化率或需要求解变化率的变量，为求导做好准备。
7. Implicit Differentiation with Respect to Time | 关于时间进行隐函数求导
Differentiate the entire equation with respect to t, treating every variable as an implicit function of time. For x² + y² = L², we get 2x·(dx/dt) + 2y·(dy/dt) = 0. For V = πr²h, the product rule and chain rule give dV/dt = π(2r·(dr/dt)·h + r²·(dh/dt)). Pay careful attention to derivatives of products and compound functions – missing a factor is a common mistake.
对整个方程关于 t 求导，把每个变量都看作时间的隐函数。对 x² + y² = L² 求导得到 2x·(dx/dt) + 2y·(dy/dt) = 0。对 V = πr²h 使用乘积法则和链式法则可得 dV/dt = π(2r·(dr/dt)·h + r²·(dh/dt))。要特别留意乘积和复合函数的求导——遗漏因子是常见的错误。
8. Substituting Known Values | 代入已知值
After differentiation, plug in the instantaneous values of variables and the known rates. This is done at a specific moment, not as a general function. For example, if dr/dt = 0.5 m/min and at the instant r = 4 m, we substitute r = 4 and dr/dt = 0.5 into the expression for dA/dt. Be careful: never substitute values before differentiating, as that would treat variables as constants and break the chain rule.
求导之后，代入变量的瞬时值和已知变化率。这一步在某一个特定瞬间进行，而不是得到一个通用函数。例如，若 dr/dt = 0.5 m/min，且在 r = 4 m 的瞬间，我们将 r = 4 和 dr/dt = 0.5 代入 dA/dt 的表达式中。注意：绝对不要在求导之前代入数值，否则就会把变量当成常量，破坏链式法则。
9. Solving for the Unknown Rate | 求解未知变化率
Rearrange the substituted equation to isolate the required rate. The result should include correct units, e.g. m²/min, cm³/s. The arithmetic is usually straightforward, but keep fractions and square roots exact where possible. If the answer is negative, state clearly whether the quantity is decreasing. A concluding sentence like “The area is increasing at 20π cm²/s” is expected.
整理代入数值后的方程，将待求的变化率单独解出来。结果应包含正确的单位，如 m²/min、cm³/s。计算通常很简单，但在可能的情况下应保留分数和根号的精确值。如果答案是负数，应清楚说明该量正在减少。最终需要一个总结性的陈述，如“面积正以 20π cm²/s 的速度增加”。
10. Checking Units and Sign | 检查单位和符号
Units must be consistent. If lengths are in metres and time in seconds, area rates will be in m²/s. Convert all measurements to a single system before starting. The sign of a rate tells direction: a positive dx/dt means x is increasing; a negative dh/dt means height is decreasing. Interpreting the sign correctly is crucial in context-based problems, such as a leaking tank (negative dV/dt) or a rising balloon (positive dh/dt).
单位必须统一。如果长度用米、时间用秒，那么面积变化率就是 m²/s。开始解题前，将所有测量值转换为同一单位系统。变化率的符号代表方向：正的 dx/dt 表示 x 在增大；负的 dh/dt 表示高度在降低。在具体情境中正确解释符号至关重要，例如水箱泄漏时 dV/dt 为负，气球上升时 dh/dt 为正。
11. Worked Example: Expanding Circular Ripple | 典型例题：扩散的圆形涟漪
Problem: A stone dropped into a pond creates a circular ripple whose radius increases at a constant rate of 4 cm/s. Find the rate at which the area of the ripple is increasing at the instant when the radius is 15 cm.
Solution: Let A = πr². Differentiate: dA/dt = 2πr (dr/dt). Given dr/dt = 4 cm/s, at r = 15 cm substitute: dA/dt = 2π·15·4 = 120π cm²/s. The area is increasing at 120π cm²/s.
题目：一块石头掉入池塘，产生一个圆形涟漪，其半径以 4 cm/s 的恒定速率增加。求在半径为 15 cm 的时刻，涟漪面积的增长速率。
解：设 A = πr²。求导得 dA/dt = 2πr (dr/dt)。已知 dr/dt = 4 cm/s，当 r = 15 cm 时代入：dA/dt = 2π·15·4 = 120π cm²/s。因此面积正以 120π cm²/s 的速率增加。
12. Worked Example: Sliding Ladder | 典型例题：滑动的梯子
Problem: A 5-metre ladder rests against a vertical wall. The bottom of the ladder slides away from the wall at 1 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 3 metres from the wall?
Solution: Let x be the distance from the wall to the bottom, y the height of the top. By Pythagoras: x² + y² = 25. Differentiate: 2x·(dx/dt) + 2y·(dy/dt) = 0 ⇒ x·(dx/dt) + y·(dy/dt) = 0. Given dx/dt = 1, when x = 3 we find y = √(25–9) = 4. Substitute: 3·1 + 4·(dy/dt) = 0 ⇒ dy/dt = –3/4 m/s. The negative sign indicates the top is sliding down at 0.75 m/s.
题目：一架长 5 米的梯子靠在竖直墙上。梯子底部以 1 m/s 的速度向外滑动。当梯子底部距离墙 3 米时，梯子顶部沿墙下滑的速率是多少？
解：设 x 为底端到墙的距离，y 为顶端的高度。由勾股定理得 x² + y² = 25。求导：2x·(dx/dt) + 2y·(dy/dt) = 0 ⇒ x·(dx/dt) + y·(dy/dt) = 0。已知 dx/dt = 1，当 x = 3 时，y = √(25–9) = 4。代入得 3·1 + 4·(dy/dt) = 0 ⇒ dy/dt = –3/4 m/s。负号表示顶端正以 0.75 m/s 的速度下滑。
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