📚 Applications of Related Rates | 相关变化率的应用
Related rates problems are a cornerstone of IB Mathematics HL and Analysis & Approaches HL courses. They require you to connect the rates at which two or more quantities change, using implicit differentiation and the chain rule. This article will guide you through the essential techniques, worked examples, and common pitfalls to help you master this topic.
相关变化率问题是IB数学高级水平（HL）与分析与方法高级水平课程的核心内容之一。它们要求你利用隐函数求导和链式法则，将两个或多个量变化的速率联系起来。本文将引导你掌握关键技巧、典型例题以及常见易错点，帮助你完全攻克这一主题。
1. What Are Related Rates? | 什么是相关变化率？
A related rates problem gives you the rate of change of one quantity and asks you to find the rate of change of another quantity that is linked to the first by a geometric or physical relationship. The key idea is to write an equation connecting the quantities, then differentiate both sides with respect to time t.
相关变化率问题会给出某个量的变化率，并要求你求出另一个量的变化率，而这两个量之间由几何或物理关系相联系。核心思路是先写出连接这些量的方程，然后对时间 t 求导。
For example, if a balloon is being inflated, the radius r and the volume V are related by V = 43π r3. Differentiating with respect to t gives dVdt = 4π r2 drdt.
例如，如果一个气球正在被充气，半径 r 和体积 V 满足关系 V = 43π r3。对时间 t 求导得到 dVdt = 4π r2 drdt。
2. The General Procedure | 一般解题步骤
Solving a related rates problem systematically reduces errors. Follow these five steps:
系统化地解决相关变化率问题可以减少错误。请遵循以下五个步骤：
- Step 1: Write down all given information and identify the unknown rate you need to find.
- Step 2: Draw a diagram if possible and label variables. Use variables whose values may change with time.
- Step 3: Write an equation relating the variables. Do not substitute known values until after differentiating.
- Step 4: Differentiate implicitly with respect to time t, using the chain rule.
- Step 5: Substitute the given values at the specific instant and solve for the unknown rate.
- 第一步：写下所有已知信息，并确定需要求的未知变化率。
- 第二步：若可能，画出图形并标注变量。使用可能随时间变化的变量符号。
- 第三步：写出连接各变量的方程。在求导之前不要代入已知数值。
- 第四步：利用链式法则对时间 t 隐式求导。
- 第五步：代入给定时刻的具体数值，解出未知变化率。
A common mistake is substituting constants too early. For instance, if a radius is always 5 cm, then r is constant and drdt=0, but in a related rates problem r usually changes with time.
一个常见错误是过早地代入常数。例如，如果半径始终是5厘米，那么 r 是常数且 drdt=0，但在相关变化率问题中 r 通常随时间改变。
3. Example 1: Expanding Circle | 例1：扩张的圆
A circular oil slick is expanding so that its radius is increasing at a constant rate of 2 cm/s. How fast is the area increasing when the radius is 10 cm?
一个圆形油污带正在扩张，其半径以恒定的速率2厘米/秒增加。当半径为10厘米时，面积增加的速率是多少？
Let r be the radius and A the area. The relationship is A = π r2. Differentiating both sides with respect to t:
设 r 为半径，A 为面积。它们的关系是 A = π r2。对两边关于 t 求导：
dAdt = 2π r drdt
Substitute r = 10 cm and drdt = 2 cm/s:
代入 r = 10 厘米和 drdt = 2 厘米/秒：
dAdt = 2π (10)(2) = 40π cm2/s
The area is increasing at 40π ≈ 125.66 cm²/s.
面积正以 40π ≈ 125.66 平方厘米/秒的速率增加。
4. Example 2: Ladder Sliding Down a Wall | 例2：梯子沿墙滑下
A 5-metre ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at a rate of 0.5 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 3 metres from the wall?
一把5米长的梯子靠在垂直墙上。梯子底部以0.5米/秒的速率远离墙脚滑动。当底部距离墙面3米时，梯子顶端沿墙下滑的速率是多少？
Let x be the distance from the wall to the bottom, and y the height of the top above the ground. By Pythagoras:
设 x 为墙脚到底部的距离，y 为梯子顶端离地面的高度。由勾股定理：
x2 + y2 = 25
Differentiate implicitly with respect to t:
对 t 隐式求导：
2x dxdt + 2y dydt = 0
When x = 3, we have y = √25 - 9 = 4. Given dxdt = 0.5 m/s, solve for dydt:
当 x = 3 时，y = √25 - 9 = 4。已知 dxdt = 0.5 米/秒，解出 dydt：
2(3)(0.5) + 2(4)dydt = 0 ⟹ dydt = -38 m/s
The negative sign means the top is moving downward at 0.375 m/s.
负号表示顶端以0.375米/秒的速率向下移动。
5. Example 3: Filling a Conical Tank | 例3：圆锥形容器注水
A conical tank has height 12 m and base radius 5 m. Water is poured in at a rate of 2 m³/min. How fast is the water level rising when the depth is 6 m?
一个圆锥形容器高12米，底面半径5米。水以2立方米/分钟的速率注入。当水深为6米时，水面上升的速率是多少？
The water forms a smaller cone similar to the tank. Let r be the radius of the water surface and h the depth. Since triangles are similar, rh = 512, so r = 512h.
水形成一个与容器相似的较小圆锥。设 r 为水面半径，h 为水深。由于三角形相似，rh = 512，所以 r = 512h。
The volume of water is V = 13π r2 h = 13π(512h)2 h = 25π432h3.
水的体积为 V = 13π r2 h = 13π(512h)2 h = 25π432h3。
Differentiate with respect to t:
对 t 求导：
dVdt = 75π432h2 dhdt = 25π144h2 dhdt
Substitute dVdt = 2 m³/min and h = 6 m:
代入 dVdt = 2 立方米/分钟和 h = 6 米：
2 = 25π144(36)dhdt ⟹ dhdt = 825π ≈ 0.1019 m/min
The water level rises at approximately 0.102 m/min.
水面以大约0.102米/分钟的速率上升。
6. Example 4: Shadow Problem | 例4：影子问题
A person 1.8 m tall walks away from a streetlight that is 6 m high at a speed of 1.2 m/s. How fast is the tip of the shadow moving?
一个身高1.8米的人以1.2米/秒的速度远离一个高6米的街灯行走。影子顶端移动的速率是多少？
Let x be the person's distance from the pole, and s the length of the shadow. By similar triangles, 1.8s = 6x+s.
设 x 为人离灯杆的距离，s 为影子的长度。由相似三角形，1.8s = 6x+s。
Cross-multiply: 1.8(x+s) = 6s ⟹ 1.8x = 4.2s ⟹ s = 37x.
交叉相乘：1.8(x+s) = 6s ⟹ 1.8x = 4.2s ⟹ s = 37x。
The distance from the pole to the shadow tip is x + s = x + 37x = 107x. Differentiating:
从灯杆到影子顶端的距离为 x + s = x + 37x = 107x。求导：
ddt(x+s) = 107dxdt = 107(1.2) = 127 ≈ 1.714 m/s
The shadow tip moves at about 1.71 m/s, which is faster than the person because the shadow also grows.
影子顶端以约1.71米/秒的速度移动，比人快，因为影子同时也在变长。
7. More Advanced Example: Two Moving Objects | 例7：进阶：两个运动物体
Two ships, A and B, leave a port at the same time. Ship A travels north at 20 km/h, and ship B travels east at 15 km/h. How fast is the distance between them increasing after 2 hours?
两艘船A和B同时离开港口。A船以20公里/小时向北行驶，B船以15公里/小时向东行驶。2小时后，两船之间的距离增加的速率是多少？
Let x be the eastward distance of ship B and y the northward distance of ship A. The distance D between them satisfies D2 = x2 + y2.
设 x 为B船向东的距离，y 为A船向北的距离。两船之间的距离 D 满足 D2 = x2 + y2。
Differentiate: 2DdDdt = 2xdxdt + 2ydydt.
求导：2DdDdt = 2xdxdt + 2ydydt。
At t=2 h, x = 30 km and y = 40 km, so D = 50 km. Then:
在 t=2 小时时，x = 30 公里，y = 40 公里，因此 D = 50 公里。于是：
50dDdt = 30(15) + 40(20) ⟹ dDdt = 450+80050 = 25 km/h
The distance is increasing at 25 km/h.
两船距离正以25公里/小时的速率增加。
8. Common Mistakes and How to Avoid Them | 常见错误与规避方法
Students frequently lose marks in related rates questions due to avoidable errors. Below are the most common pitfalls:
学生在相关变化率题目中经常因为可避免的错误而失分。以下是最常见的陷阱：
| Mistake | 错误 | Correction | 纠正 |
| Substituting known values before differentiating | Differentiate first, then substitute. |
| Forgetting the chain rule (missing dxdt) | Always multiply by the derivative of the inner variable. |
| Ignoring units in the final answer | Write units such as m/s, cm²/s, km/h. |
| Using the wrong geometric formula | Check whether the shape is a cone, cylinder, sphere, etc. |
| 错误 | 纠正 |
| 求导前代入已知数值 | 先求导，再代入。 |
| 忘记链式法则（漏掉 dxdt） | 始终乘以内层变量的导数。 |
| 最终答案漏写单位 | 写出单位，如米/秒、平方厘米/秒、公里/小时。 |
| 使用错误的几何公式 | 检查形状是圆锥、圆柱、球体等。 |
Another key point: when a quantity is constant, its derivative is zero. If the problem states "the radius is always 5", then drdt=0, but in most related rates problems the radius changes with time.
另一个关键点：当某个量是常量时，它的导数为零。如果题目说明“半径始终为5”，则 drdt=0，但在大多数相关变化率问题中，半径是随时间变化的。
9. Exam-Style Question: Balloon with Helium | 例9：考试风格题：氦气球
A spherical balloon is being filled with helium at a rate of 0.1 m³/s. Find the rate at which the surface area is increasing when the radius is 0.5 m.
一个球形气球正以0.1立方米/秒的速率充入氦气。求当半径为0.5米时，表面积增加的速率。
For a sphere, V = 43π r3 and A = 4π r2. We know dVdt = 0.1. First find drdt:
对于球体，V = 43π r3，A = 4π r2。已知 dVdt = 0.1。先求 drdt：
dVdt = 4π r2 drdt ⟹ 0.1 = 4π (0.5)2 drdt ⟹ drdt = 0.1π = 110π m/s
Then dAdt = 8π r drdt. At r=0.5:
然后 dAdt = 8π r drdt。在 r=0.5 时：
dAdt = 8π (0.5)(110π) = 0.4 m2/s
The surface area is increasing at 0.4 m²/s.
表面积正以0.4平方米/秒的速率增加。
10. Tips for IB Exam Success | IB考试成功技巧
To get full marks on related rates questions in IB exams, follow these recommendations:
要在IB考试的相关变化率题目中获得满分，请遵循以下建议：
- Use a clear notation: define every variable with its unit.
- Write the known rates as derivatives, e.g. dxdt=0.5 m/s.
- State the geometric relationship explicitly before differentiating.
- Do not skip the chain rule step; write it out.
- Check your final answer for reasonableness and include units.
- 使用清晰的记号：为每个变量定义单位。
- 将已知速率写成导数形式，例如 dxdt=0.5 米/秒。
- 在求导前明确写出几何关系。
- 不要跳过链式法则步骤；把它写出来。
- 检查最终答案的合理性并包含单位。
In IB mark schemes, you often earn method marks for correct differentiation even if your final answer is wrong. Therefore, always show the differentiated equation.
在IB评分标准中，即使最终答案错误，正确的求导过程也能获得方法分。因此，务必写出求导后的方程。
11. Practice Problems | 练习题目
Try these problems on your own before looking at the hints below.
请先自己尝试以下题目，再看下方的提示。
Problem 1: A cube's edge increases at 3 cm/s. How fast does the volume increase when the edge is 5 cm?
问题1：立方体的棱长以3厘米/秒增加。当棱长为5厘米时，体积增加的速率是多少？
Problem 2: A boat is pulled toward a dock by a rope through a ring at the dock level. The ring is 4 m above the water. If the rope is pulled in at 1 m/s, how fast is the boat approaching the dock when 5 m of rope is out?
问题2：一艘船被一根绳子通过岸上的环拉向码头。环高出水面4米。如果绳子以1米/秒的速度被收进，当绳长（系船端到环）为5米时，船接近码头的速率是多少？
Hints | 提示
Problem 1: V = x3. Answer: 225 cm3/s.
问题1：V = x3。答案：225 立方厘米/秒。
Problem 2: Let x be horizontal distance from dock to boat, ell the rope length. Then x2 + 42 = ell2. Differentiate and use x=3 when ell=5. Answer: 5/3 ≈ 1.67 m/s.
问题2：设 x 为船到码头的水平距离，ell 为绳长。则 x2 + 42 = ell2。求导并利用当 ell=5 时 x=3。答案：5/3 ≈ 1.67 米/秒。
12. Summary | 总结
Related rates problems are solved by linking variables through an equation, differentiating implicitly with respect to time, and substituting the specific values. The most important skill is to treat all variables as functions of time and always use the chain rule.
相关变化率问题的解法是：通过方程将变量联系起来，对时间隐式求导，然后代入特定数值。最重要的技能是将所有变量视为时间的函数，并始终使用链式法则。
Mastering this topic requires practice. Work through the examples above, then try additional problems from past IB papers. With a systematic approach, you can confidently tackle any related rates question.
掌握这一主题需要练习。请认真完成以上例题，再尝试IB历年真题中的相关题目。只要使用系统化的方法，你就能自信地解决任何相关变化率问题。
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