Exercise 21H: Applications of Derivatives – Related Rates and Optimization | 练习 21H：导数应用——相关变化率与最优化

📚 Exercise 21H: Applications of Derivatives – Related Rates and Optimization | 练习 21H：导数应用——相关变化率与最优化

Exercise 21H in the IB Mathematics: Analysis and Approaches HL course (commonly from the widely used Haese Mathematics textbook) provides a focused practice set on applying derivatives to model and solve real-world problems. This exercise spans two major applications: related rates and optimization. Mastering these concepts is essential for success in the IB analysis paper, as questions often require blending geometric insight with calculus techniques to connect changing quantities or to find maximum and minimum values under given constraints. The following guide breaks down the key ideas, problem-solving frameworks, worked examples and common pitfalls, helping you tackle Exercise 21H with confidence.

IB 数学分析与方法的 HL 课程中，练习 21H（常见于 Haese Mathematics 教材）集中训练使用导数对实际问题进行建模与求解。这项练习涵盖两大应用主题：相关变化率和最优化。掌握这些概念是 IB 分析卷取得高分的关键，因为考题经常要求将几何直观与微积分技巧结合起来，建立变化量之间的联系，或在给定约束下求最大值与最小值。下面的指南将分解核心思想、解题框架、典型例题和常见误区，助你自信攻克练习 21H。

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1. Overview of Exercise 21H | 练习 21H 概述

Exercise 21H is typically the last set of problems in Chapter 21, which focuses on applications of differential calculus. It brings together differentiation rules, implicit differentiation and the interpretation of derivatives as rates of change. The problems are not purely algebraic; they involve geometry, physics and economics settings, requiring you to translate a scenario into a mathematical relationship before differentiating. The exercise often includes related rates tasks (such as water flowing into a tank or a ladder sliding down a wall) and optimization tasks (like finding maximum volume, minimum surface area or minimum cost).

练习 21H 通常是第 21 章应用微分学部分的最后一组习题。它综合了微分法则、隐函数微分以及将导数解释为变化率的思想。题目并非纯代数运算，而是涉及几何、物理和经济背景，要求你将具体情境转化为数学关系，然后再求导。练习中常出现相关变化率问题（如水流入容器、梯子滑离墙壁）和最优化问题（如求最大体积、最小表面积或最小成本）。

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2. Prerequisite Knowledge: Differentiation Rules | 预备知识：微分法则

Before tackling related rates and optimization, you must be absolutely confident with basic differentiation. The power rule (d/dx xⁿ = n xⁿ⁻¹), sum rule, product rule and quotient rule are the foundations. In addition, the chain rule in its simplest form is non-negotiable: if y = f(g(x)), then dy/dx = f'(g(x))·g'(x). In related rates, most variables are functions of time t, so every derivative is taken with respect to t, often requiring the chain rule to link rates like dy/dt and dx/dt when y and x themselves are linked by an equation.

在处理相关变化率和最优化之前，你必须对基本的微分法则了如指掌。幂法则、和差法则、乘积法则和商法则是根基。此外，链式法则的掌握是硬性要求：若 y = f(g(x))，则 dy/dx = f'(g(x))·g'(x)。在相关变化率问题中，大多数变量都是时间 t 的函数，因此每个导数都是对 t 求导，经常需要借助链式法则将 dy/dt 和 dx/dt 等速率联系起来，而此时 y 与 x 之间由某个方程关联。

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3. The Chain Rule and Implicit Differentiation | 链式法则与隐函数微分

The chain rule is the engine of related rates. Consider a circle with radius r that expands over time. Its area A = πr². To find how fast the area changes (dA/dt) when the radius changes at dr/dt, we do not have A written directly as a function of t. Instead we use the chain rule: dA/dt = (dA/dr)·(dr/dt) = 2πr·dr/dt. This pattern – multiplying the derivative with respect to an intermediate variable by the rate of that variable – appears in every related rates problem. Implicit differentiation generalizes this: when an equation links x and y, and both are functions of t, differentiate both sides with respect to t, treating x and y as ‘inner functions’ and adding dy/dt or dx/dt accordingly.

链式法则是相关变化率的驱动力。设想一个半径 r 随时间扩大的圆，其面积 A = πr²。求面积变化率 dA/dt 时，并没有直接将 A 表示为 t 的函数，而是用链式法则：dA/dt = (dA/dr)·(dr/dt) = 2πr·dr/dt。这种“用中间变量的导数乘以该变量的变化率”的模式出现在每一个相关变化率题目中。隐函数微分将这一思想推广：当一个方程将 x 和 y 联系在一起，而两者均为 t 的函数时，两端对 t 求导，将 x 和 y 视作“内层函数”并相应地加上 dy/dt 或 dx/dt。

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4. Related Rates: The Central Idea | 相关变化率：核心思想

Related rates problems ask you to find the rate at which one quantity is changing by relating it to other quantities whose rates of change are known. All the quantities are functions of time, even if time does not appear explicitly in the given equation. Your job is to build an equation connecting the relevant variables – often using geometry (Pythagoras, similar triangles, volume formulas) – and then differentiate that equation with respect to time t. Always substitute known constant values after differentiating, never before, because you need the general rate relationship first.

相关变化率问题要求你通过一个量的已知变化率，求出另一个量的变化率。所有量都是时间的函数，即便时间并未显式出现在给定的方程中。你的任务是构建一个联系相关变量的方程——常借助几何（勾股定理、相似三角形、体积公式）——然后将该方程对时间 t 求导。务必在求导后再代入已知的常数值，切勿求导前代入，因为你首先需要得到一般的变化率关系式。

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5. Related Rates Problem-Solving Framework | 相关变化率解题框架

A disciplined approach prevents sign errors and overlooked constants. Follow these steps:
1. Draw and label a diagram. Identify all variables and the given rate (e.g., dV/dt = 5 cm³/s) and the required rate (e.g., dh/dt).
2. Write an equation relating the variables without the rates. Use geometry or given constraints.
3. Differentiate implicitly with respect to t. Apply the chain rule to every term that involves a variable.
4. Substitute known rates and the instantaneous values of variables (like r = 3 cm when h = 4 cm) – but only after differentiation.
5. Solve for the unknown rate and include units.

一套严谨的解题流程能防止符号错误和遗漏常数。请按以下步骤操作：
1. 画出示意图并标注变量，识别已知变化率（如 dV/dt = 5 cm³/s）和待求变化率（如 dh/dt）。
2. 写出纯变量关系式（不含变化率）。利用几何或给定的约束条件。
3. 对方程关于 t 进行隐函数求导。对每个含变量的项都使用链式法则。
4. 将已知变化率以及变量的瞬时值（如 h = 4 cm 时 r = 3 cm）代入——必须在求导之后。
5. 解出未知变化率并注明单位。

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6. Worked Example: The Sliding Ladder | 典型例题：滑落的梯子

A 5-meter ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 0.6 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 3 m from the wall?
Step 1: Let x be the distance from the wall to the bottom of the ladder, and y the height of the top on the wall. Given dx/dt = 0.6 m/s (positive because x increases). Need dy/dt when x = 3 m.
Step 2: By Pythagoras: x² + y² = 5² → x² + y² = 25.
Step 3: Differentiate with respect to t: 2x·dx/dt + 2y·dy/dt = 0 → x·dx/dt + y·dy/dt = 0.
Step 4: When x = 3, y = √(25 – 9) = 4 m. Substitute: 3·(0.6) + 4·dy/dt = 0 → 1.8 + 4·dy/dt = 0 → dy/dt = –0.45 m/s.
Interpretation: The top slides down at 0.45 m/s (negative sign indicates downward motion).

一架长 5 米的梯子斜靠在竖直墙壁上。梯子底部以 0.6 m/s 的恒定速度滑离墙壁。当梯子底部离墙 3 米时，梯子顶部沿墙壁下滑的速度是多少？
步骤1： 设 x 为墙壁到梯子底部的距离，y 为梯子顶部在墙上的高度。已知 dx/dt = 0.6 m/s（正数因 x 增加）。求当 x = 3 m 时的 dy/dt。
步骤2： 根据勾股定理：x² + y² = 5² → x² + y² = 25。
步骤3： 两边对 t 求导：2x·dx/dt + 2y·dy/dt = 0 → x·dx/dt + y·dy/dt = 0。
步骤4： 当 x = 3 时，y = √(25 – 9) = 4 m。代入：3·(0.6) + 4·dy/dt = 0 → 1.8 + 4·dy/dt = 0 → dy/dt = –0.45 m/s。
解读： 梯子顶部以 0.45 m/s 下滑（负号表示向下运动）。

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7. Worked Example: Filling a Conical Tank | 典型例题：圆锥形容器注水

A water tank in the shape of an inverted cone has height 10 m and base radius 4 m. Water is being pumped in at a rate of 2 m³/min. At what rate is the water level rising when the water is 5 m deep?
Solution: Volume of a cone: V = (1/3)πr²h. Because the cone’s dimensions are fixed, r and h are proportional by similar triangles: r/h = 4/10 → r = (2/5)h. Substitute: V = (1/3)π[(2/5)h]²·h = (4π/75)h³.
Differentiate w.r.t t: dV/dt = (4π/75)·3h²·dh/dt = (4π/25)h²·dh/dt.
Given dV/dt = 2 m³/min and h = 5 m: 2 = (4π/25)·25·dh/dt = 4π·dh/dt → dh/dt = 2/(4π) = 1/(2π) m/min ≈ 0.159 m/min. The water level rises at about 0.159 m/min when 5 m deep.

一个倒圆锥形水箱高 10 m，底面半径 4 m。水以 2 m³/min 的速率注入。当水深为 5 m 时，水位上升的速率是多少？
解答： 圆锥体积：V = (1/3)πr²h。由于水箱形状固定，由相似三角形得 r 与 h 成比例：r/h = 4/10 → r = (2/5)h。代入：V = (1/3)π[(2/5)h]²·h = (4π/75)h³。
对 t 求导：dV/dt = (4π/75)·3h²·dh/dt = (4π/25)h²·dh/dt。
已知 dV/dt = 2 m³/min，h = 5 m：2 = (4π/25)·25·dh/dt = 4π·dh/dt → dh/dt = 2/(4π) = 1/(2π) m/min ≈ 0.159 m/min。水位在 5 m 深时约以 0.159 m/min 上升。

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8. Optimization: Maximizing and Minimizing | 最优化：求最大与最小值

Optimization uses derivatives to find the maximum or minimum value of a function under given constraints. In IB problems, you often need to design a shape (box, cylinder, container) that maximizes volume while minimizing surface area or cost. The process requires expressing the quantity to be optimized as a function of a single variable, then finding the critical points where the first derivative is zero or undefined. The second derivative or a sign chart then confirms whether the critical point is a maximum or minimum. Always check endpoints if the domain is closed.

最优化就是运用导数求函数在给定约束下的最大值或最小值。在 IB 问题中，你常需要设计一个形状（盒子、圆柱、容器），在最大化体积的同时最小化表面积或成本。解题过程要求将待优化量表达为单一变量的函数，然后求出一阶导数为零或不存在的临界点。再用二阶导数或符号表确认该临界点是最大值还是最小值。如果定义域是闭区间，务必检查端点值。

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9. Optimization Problem-Solving Steps | 最优化问题解题步骤

1. Understand the problem: Identify what is to be maximized or minimized and the given constraints.
2. Draw and introduce variables: Assign letters to all varying quantities.
3. Write the objective function: Express the quantity to optimize in terms of one variable, using the constraint to eliminate other variables.
4. Determine the domain: Find the realistic interval for the independent variable (e.g., length must be positive).
5. Find the derivative: Compute f'(x) and solve f'(x) = 0.
6. Verify maximum or minimum: Use the second derivative test (f”(x) < 0 ⇒ local maximum; f''(x) > 0 ⇒ local minimum) or the first derivative sign change.
7. Answer in context: Include units and check that the value makes practical sense.

1. 理解问题： 确定要最大化或最小化的量以及已知约束。
2. 画图并引入变量： 为所有变化的量分配字母。
3. 写出目标函数： 利用约束消去其他变量，将待优化量表示为一个变量的函数。
4. 确定定义域： 找到自变量在实际情况中的取值范围（如长度必须为正）。
5. 求导： 计算 f'(x) 并解方程 f'(x) = 0。
6. 验证最大值或最小值： 使用二阶导数检验法（f”(x) < 0 ⇒ 局部极大；f''(x) > 0 ⇒ 局部极小）或一阶导数符号变化法。
7. 结合语境作答： 标注单位，检查数值是否合乎实际。

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10. Worked Example: Maximizing Volume of a Box | 典型例题：最大化盒子的体积

An open box is to be made from a rectangular sheet of cardboard measuring 30 cm by 20 cm by cutting equal squares from each corner and folding up the sides. Find the side length x of the squares that maximizes the volume.
Solution: After cutting squares of side x, the base dimensions become (30 – 2x) by (20 – 2x), and the height is x. Volume V = x(30 – 2x)(20 – 2x) = 4x³ – 100x² + 600x. The domain is 0 < x < 10 (since 20 – 2x > 0 → x < 10).
Differentiate: V'(x) = 12x² – 200x + 600. Set V'(x) = 0 and divide by 4: 3x² – 50x + 150 = 0. Using the quadratic formula: x = [50 ± √(2500 – 1800)] / 6 = [50 ± √700] / 6 = [50 ± 10√7] / 6. Approximate values: (50 + 26.46)/6 ≈ 12.74 (reject, outside domain) and (50 – 26.46)/6 ≈ 3.92 cm. Check V”(x) = 24x – 200; at x ≈ 3.92, V” is negative, confirming a maximum. The optimal square side is about 3.92 cm.

一块 30 cm × 20 cm 的矩形纸板，从每个角切去相等的小正方形后折起边缘，制成一个无盖盒子。求使体积最大的正方形边长 x。
解答： 切去边长为 x 的正方形后，底面尺寸为 (30 – 2x) 和 (20 – 2x)，高度为 x。体积 V = x(30 – 2x)(20 – 2x) = 4x³ – 100x² + 600x。定义域为 0 < x < 10（因为 20 – 2x > 0 → x < 10）。
求导：V'(x) = 12x² – 200x + 600。令 V'(x) = 0，除以 4 得：3x² – 50x + 150 = 0。用求根公式：x = [50 ± √(2500 – 1800)] / 6 = [50 ± √700] / 6 = [50 ± 10√7] / 6。近似值：(50 + 26.46)/6 ≈ 12.74（舍去，超出定义域），(50 – 26.46)/6 ≈ 3.92 cm。检验 V”(x) = 24x – 200；在 x ≈ 3.92 处 V” 为负，确认是极大值。最优正方形边长约为 3.92 cm。

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11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Mistake 1: Differentiating before substituting the constraint. In related rates, you must differentiate the general equation first; substituting a specific value too early can wipe out a variable that actually has a rate. Always keep the variables until after you have d/dt.
Mistake 2: Sign errors with rates. Be clear about whether a quantity is increasing or decreasing. In the ladder example, dx/dt was positive (moving away) and dy/dt came out negative (sliding down). Draw a diagram and assign positive directions to avoid confusion.
Mistake 3: Forgetting to use the chain rule. When differentiating V = (4π/75)h³, some students write dV/dt = (4π/75)h³ instead of (4π/75)·3h²·dh/dt. The dh/dt factor is crucial.

错误1：在对约束代入之前求导。 在相关变化率中，必须首先对一般方程求导；过早代入特定数值可能会消去一个实际上具有变化率的变量。务必先对 t 求导，再代入数值。
错误2：变化率的符号错误。 要明确一个量是在增加还是在减少。梯子例题中，dx/dt 为正（向外滑动），dy/dt 结果为负（向下滑）。画出示意图并规定正方向可避免混淆。
错误3：忘记使用链式法则。 在对 V = (4π/75)h³ 求导时，有的学生写成 dV/dt = (4π/75)h³，而正确应为 (4π/75)·3h²·dh/dt。dh/dt 因子至关重要。

Mistake 4: Not verifying max vs min in optimization. Solving f'(x) = 0 gives critical points, but not all are maxima. Always use a sign test or second derivative.
Mistake 5: Ignoring the domain. In the box problem, x = 12.74 was mathematically valid but physically impossible. Always check boundary conditions.
Mistake 6: Unit inconsistency. Convert all measurements to the same unit system before differentiating, and report the rate with proper units (cm/s, m³/min).

错误4：最优解未验证极大还是极小。 解 f'(x) = 0 得到的是临界点，但并非所有都是最大值。务必用符号检验或二阶导数判别。
错误5：忽略定义域。 盒子问题中 x = 12.74 在数学上成立但实际不可能。一定要检查边界条件。
错误6：单位不一致。 求导前将所有测量单位统一，并在结果中注明正确单位（cm/s, m³/min）。

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12. Practice Tips for Success | 成功备考练习建议

To master Exercise 21H, work through a variety of problems beyond the textbook. Start with simple related rates (sphere volume, rectangle area) and then move to composite shapes (trough, conical pile). For optimization, practice problems where the constraint equation is not directly given – sometimes you need to use similar triangles or the Pythagorean theorem to eliminate a variable. Write out the steps in full sentences to reinforce the logic. Peer-discussion helps: explain your solution to a friend, as teaching exposes any hidden uncertainties. Finally, keep a formula sheet with common volume and area formulas, and a checklist of the seven-step frameworks. With systematic practice, Exercise 21H will become a source of confidence rather than dread.

要想掌握练习 21H，请完成教材之外的多样化练习。从简单的相关变化率入手（球体积、矩形面积），然后过渡到复合形状（槽形容器、锥形沙堆）。对于最优化，应多练习约束方程未直接给出的问题——有时你需要借助相似三角形或勾股定理消去变量。用完整句子一步步写出解答以强化逻辑。同伴讨论也非常有效：向朋友讲解你的解题过程，因为在教授他人时隐藏的不确定因素会暴露出来。最后，准备一张常用体积和面积公式表，以及七步解题法的核对清单。经过系统训练，练习 21H 必将成为你的信心来源，而非畏惧对象。

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