📚 Exercise 24F: Related Rates | 练习24F：相关变化率
In IB Mathematics, Exercise 24F typically focuses on related rates – a powerful application of implicit differentiation where we connect the rates of change of two or more quantities that vary with time. This chapter is a classic part of calculus that shows up frequently in examinations, requiring you to translate a real‑world scenario into mathematical equations and then differentiate with respect to time to find an unknown rate.
在IB数学中，练习24F通常聚焦于相关变化率——这是隐函数微分的重要应用，我们将随时间变化的两者或多者量的变化率联系起来。这一章是微积分的经典内容，常出现在考试中，要求你将现实场景转化为数学方程，然后对时间求导以找出未知变化率。
1. What Are Related Rates? | 什么是相关变化率？
r/dt). This equation links the rate of change of the area, dA/dt, with the rate of change of the radius, dr/dt. If we know one rate, we can solve for the other at a particular instant.
相关变化率问题涉及两个或更多随时间 t 变化的量。例如，一个正在膨胀的圆的半径 r(t) 和面积 A(t) 通过公式 A = πr² 相关联。运用链式法则对时间 t 求导，得到 dA/dt = 2πr (dr/dt)。这个方程将面积变化率 dA/dt 与半径变化率 dr/dt 联系起来。如果我们知道其中一个速率，就可以求出另一个在某一时刻的值。
2. The Fundamental Strategy | 基本解题策略
Every related rates problem can be tackled using a structured approach. First, identify the quantities that change with time and assign variables. Second, write an equation that relates those variables (geometry, trigonometry, or physics). Third, differentiate the entire equation with respect to time t, remembering to use the chain rule for every variable that depends on time. Fourth, substitute the known values and rates, then solve for the unknown rate. Finally, check that the sign and units make sense.
每个相关变化率问题都可以通过结构化的方法解决。首先，找出随时间变化的量，并设定变量。其次，写出这些变量之间的关系式（几何、三角或物理关系）。第三，对整个方程关于时间 t 求导，对于每个依赖于时间的变量务必使用链式法则。第四，代入已知数值和变化率，求解未知变化率。最后，检查符号和单位是否合理。
3. Implicit Differentiation with Time | 对时间的隐函数求导
the derivative is 2x (dx/dt) + 2y (dy/dt) = 0. Every variable that depends on time gets a “d( )/dt” factor. This is the essence of finding related rates.
当你对像 x² + y² = 25 这样的关系式关于 t 求导时，你不会简单地得到 2x + 2y = 0。你必须使用链式法则：d(x²)/dt = 2x (dx/dt)，d(y²)/dt = 2y (dy/dt)。因此导数为 2x (dx/dt) + 2y (dy/dt) = 0。每个依赖于时间的变量都会带上一个“d( )/dt”因子。这正是求相关变化率的精髓。
4. Geometric Relationships – Circles and Spheres | 几何关系——圆与球
V/dt = 4πr² (dr/dt). You are often given dV/dt and the radius at that moment, then asked for dr/dt.
经典例子涉及圆的半径扩张。如果一块石头掉入池塘，涟漪形成同心圆。面积 A = πr²，周长 C = 2πr。求导得 dA/dt = 2πr (dr/dt)，dC/dt = 2π (dr/dt)。如果向球形气球中注水，体积 V = (4/3)πr³，所以 dV/dt = 4πr² (dr/dt)。题目通常会给出 dV/dt 和该时刻的半径，然后让你求 dr/dt。
5. The Sliding Ladder Problem | 滑梯问题
Consider a 5‑ iven dx/dt = 0.2 m/s and need dy/dt when x = 3, we first solve for y using Pythagoras: y = √25 − 9 8(dy/dt) = 0 → dy/dt = −0.15 m/s. The negative sign indicates the top is sliding downward.
设想一架长5米的梯子斜靠在墙上，梯子底部以0.2米/秒的速度向外滑动。梯子底部到墙的距离 x 与顶部的高度 y 之间的关系为 x² + y² = 5² = 25。对 t 求导：2x (dx/dt) + 2y (dy/dt) = 0。如果已知 dx/dt = 0.2 m/s，需要在 x = 3 时求 dy/dt，我们先用勾股定理求 y：y = √25 − 9 = 4。代入得：2(3)(0.2) + 2(4)(dy/dt) = 0 → 1.2 + 8(dy/dt) = 0 → dy/dt = −0.15 米/秒。负号表示梯子顶部正在下滑。
6. Conical Tank and Filling Rates | 锥形水箱与注水速率
r is V = (1/3)πr²h = (1/3)π(h/2)² h = (π/12)h³. Differentiate with respect to t: dV/dt = (π/12) · 3h² (dh/dt) = (π/4)h² (dh/dt). We know dV/dt = 2 m³/min and h = 3, so 2 = (π/4)(9)(dh/dt) ⇒ dh/dt = 8/(9π) ≈ 0.283 m/min.
一个倒锥形水箱高8米，底部半径4米。水以2立方米/分钟的恒定速率注入。求水深为3米时，水位 h 的上升速率。利用相似三角形，深度 h 处的水面半径 r 满足 r/h = 4/8 = 1/2，所以 r = h/2。水的体积 V = (1/3)πr²h = (1/3)π(h/2)² h = (π/12)h³。对 t 求导：dV/dt = (π/12) · 3h² (dh/dt) = (π/4)h² (dh/dt)。已知 dV/dt = 2 m³/min，h = 3，故 2 = (π/4)(9)(dh/dt) ⇒ dh/dt = 8/(9π) ≈ 0.283 米/分钟。
7. Shadows and Similar Triangles | 影子与相似三角形
he pole, and s the length of his shadow. Using similar triangles: (x + s)/6 = s/1.8. Cross‑ (3/7)(1.5) = 4.5/7 ≈ 0.643 m/s. The tip moves at dx/dt + ds/dt = 1.5 + 0.643 = 2.143 m/s.
一个身高1.8米的人以1.5米/秒的速度离开一盏6米高的路灯。我们想知道他的影子尖端移动的速率。设 x 为行人与灯杆的距离，s 为影子长度。利用相似三角形：(x + s)/6 = s/1.8。交叉相乘得 1.8(x + s) = 6s → 1.8x + 1.8s = 6s → 1.8x = 4.2s → s = (1.8/4.2)x = (3/7)x。求导：ds/dt = (3/7) dx/dt。已知 dx/dt = 1.5 m/s，ds/dt = (3/7)(1.5) = 4.5/7 ≈ 0.643 m/s。影子尖端移动速率为 dx/dt + ds/dt = 1.5 + 0.643 = 2.143 m/s。
8. Rates Involving Angles and Trigonometry | 涉及角度与三角函数的变化率
, when h = 2, tan θ = 1, so θ = π/4, sec²θ = 2. If dh/dt = 0.5 km/s, then 2 (dθ/dt) = (1/2)(0.5) = 0.25 → dθ/dt = 0.125 rad/s.
假设一枚火箭垂直发射，由2公里外的雷达站追踪。若火箭的高度为 h(t)，仰角为 θ(t)，则 tan θ = h/2。求导得 sec²θ (dθ/dt) = (1/2)(dh/dt)。如果我们知道速度 dh/dt 和瞬时角度，就可以求出 dθ/dt。例如，当 h = 2 时，tan θ = 1，因此 θ = π/4，sec²θ = 2。如果 dh/dt = 0.5 公里/秒，那么 2 (dθ/dt) = (1/2)(0.5) = 0.25 → dθ/dt = 0.125 弧度/秒。
9. Connecting Multiple Rates – The Chain Rule Pattern | 连接多个变化率——链式法则模式
dt). Each link is a derivative that you know or can calculate from geometric formulas. This modular approach simplifies complex multi‑step problems.
有时你需要连接三个或更多变化率。例如，圆的面积依赖于半径，半径又通过热膨胀依赖于时间。你可能会得到 dT/dt（温度变化率）和 dT/dr，然后让你求 dA/dt。链式法则自然延伸：dA/dt = (dA/dr)(dr/dT)(dT/dt)。每个链接都是你知道的、或可通过几何公式计算的导数。这种模块化方法简化了复杂的多步问题。
10. Dealing with Constants and Instantaneous Values | 处理常量和瞬时值
A very common mistake is to substitute the given numerical values of variables (like r or h) before differentiating. Always differentiate the symbolic equation first, then substitute the specific numerical values at the given instant. If you substitute too early, you are treating a varying quantity as constant, and its derivative would vanish, leading to an incorrect result. Only values that are truly constant throughout the motion (like the length of a ladder) may be inserted before differentiation.
一个很常见的错误是在求导之前将给定变量的数值（如 r 或 h）代入。一定要先对符号方程求导，然后在给定时刻代入具体数值。如果你过早代入，就是将变化中的量当作常量处理，其导数就会消失，导致错误结果。只有在整个运动过程中真正恒定的值（如梯子的长度）才可以在求导前插入。
11. Units and Dimensional Analysis | 单位与量纲分析
s if lengths are in meters. This check prevents scaling errors and reinforces the physical meaning of your answer.
在相关变化率中要始终留意单位。如果半径以厘米计、时间以秒计，dr/dt 便为厘米/秒。如果体积为立方米、时间为分钟，dV/dt 则为立方米/分钟。求解后，核实计算出的变化率单位是否符合预期。例如，对 V = lwh 求导，若长度单位是米，则速率单位为立方米/秒。这种检验可防止比例错误，并加强答案的物理意义。
12. Tips for IB Exam Success | IB 考试成功技巧
Read the problem carefully to identify the “given” rate and the “wanted” nstraints (Pythagoras, similarity, volume formulas). After solving, check the sign: a positive rate means increase, negative means decrease. Practice a wide variety of scenarios: spheres, cones, ladders, shadows, angles, and even distance between moving cars.
仔细阅读题目，识别“已知”变化率和“待求”变化率。画出草图，标注随时间变化的变量。用正确的符号写出已知变化率（如 dy/dt = 5）。寻找几何约束（勾股定理、相似性、体积公式）。求解后检查符号：正号表示增加，负号表示减少。练习多种情景：球体、锥体、梯子、影子、角度，甚至两车之间的距离问题。
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