Related Rates of Change | 相关变化率

📚 Related Rates of Change | 相关变化率

In IB Mathematics, related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. The core idea is to differentiate an equation with respect to time using the chain rule, turning a static relationship into a dynamic connection between rates. Mastering this topic requires a clear strategy: identify the variables, write an equation, differentiate implicitly, substitute known values, and solve for the unknown rate.

在IB数学中，相关变化率问题涉及通过将一个量与已知变化率的其他量联系起来，来求该量的变化速率。核心思想是利用链式法则对方程关于时间求导，从而将静态关系转化为速率之间的动态联系。掌握这一主题需要清晰的策略：识别变量，写出方程，进行隐式求导，代入已知数值，最后解出未知速率。

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1. Understanding Related Rates | 理解相关变化率

A related rates problem typically asks: ‘How fast is A changing when B is changing at a given rate?’ Here, both A and B are quantities that depend on time t. They are linked by a geometric or physical equation. By differentiating both sides with respect to t, we obtain an equation that involves dA/dt and dB/dt. The challenge is to correctly set up the relationship and then differentiate.

相关变化率问题通常会问：“当B以给定的速率变化时，A的变化有多快？”这里A和B都是依赖于时间t的量，它们通过几何或物理方程联系在一起。对等式两边关于t求导，我们得到一个涉及dA/dt和dB/dt的方程。难点在于正确建立关系式并进行求导。

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2. The Chain Rule as the Engine | 链式法则作为核心引擎

The chain rule is the mathematical engine behind every related rates solution. If a quantity Y depends on X, and X depends on time t, then dY/dt = (dY/dX) · (dX/dt). This allows us to express a rate we cannot measure directly in terms of a rate we do know. In IB problems, you might see this in the form dV/dt = (dV/dr)(dr/dt) for a balloon’s volume and radius.

链式法则是每个相关变化率解答背后的数学引擎。如果量Y依赖于X，而X又依赖于时间t，那么dY/dt = (dY/dX) · (dX/dt)。这使我们能够用已知的速率来表示无法直接测量的速率。在IB问题中，你可能会看到气球体积和半径的形式dV/dt = (dV/dr)(dr/dt)。

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3. General Problem-Solving Strategy | 通用解题策略

Use this five-step method to tackle any related rates question:

使用以下五步法解决任何相关变化率问题：

Step 1: Draw a diagram and assign symbols. Identify all variables that depend on time, and note which rates are given and which rate is to be found.

步骤1：画图并分配符号。识别所有依赖于时间的变量，并标出哪些速率已知、哪个速率需要求。

Step 2: Write an equation linking the variables. This could be the Pythagorean theorem, a volume formula, a similar-triangle proportion, or any geometric relationship.

步骤2：写出联系变量的方程。可以是勾股定理、体积公式、相似三角形比例或任何几何关系。

Step 3: Differentiate both sides with respect to time t. Use implicit differentiation and the chain rule. Remember that every variable except t itself must be differentiated as a function of t.

步骤3：对等式两边关于时间t求导。使用隐函数求导和链式法则。记住，除t本身外，每个变量都必须当作t的函数来求导。

Step 4: Substitute the known values and rates. Be careful to plug in the values at the specific moment described in the problem, not general expressions.

步骤4：代入已知数值和速率。注意代入题目描述的特定时刻的数值，而不是一般表达式。

Step 5: Solve for the unknown rate and include correct units. Check if the sign (positive/negative) indicates increasing or decreasing.

步骤5：解出未知速率并标明正确单位。检查符号（正/负）表示增加还是减少。

Step	Action
1	Identify variables and rates
2	Write a static equation
3	Differentiate with respect to t
4	Substitute known values
5	Solve for the desired rate

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4. Example 1: Inflating a Balloon | 示例1：膨胀的气球

Air is being pumped into a spherical balloon at a rate of 100 cm³/s. Find the rate at which the radius increases when the radius is 10 cm.

空气以100 cm³/s的速率注入一个球形气球。求当半径为10 cm时半径增大的速率。

We know dV/dt = 100 cm³/s and r = 10 cm. The volume of a sphere is V = (4/3)πr³. Differentiating with respect to t gives dV/dt = 4πr² (dr/dt).

已知dV/dt = 100 cm³/s，r = 10 cm。球体体积为V = (4/3)πr³。对t求导得dV/dt = 4πr² (dr/dt)。

dV/dt = 4πr² (dr/dt)

Substitute the known values: 100 = 4π(10)² (dr/dt) → 100 = 400π (dr/dt). Solve: dr/dt = 100/(400π) = 1/(4π) cm/s. The radius increases at approximately 0.0796 cm/s.

代入已知值：100 = 4π(10)² (dr/dt) → 100 = 400π (dr/dt)。解得dr/dt = 100/(400π) = 1/(4π) cm/s。半径以约0.0796 cm/s增大。

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5. Example 2: Sliding Ladder | 示例2：滑落的梯子

A 5-metre ladder rests against a vertical wall. Its base is pushed away from the wall at a constant speed of 0.2 m/s. How fast is the top of the ladder sliding down the wall when the base is 3 m from the wall?

一架5米长的梯子靠在竖直墙上。其底端以0.2 m/s的恒定速度远离墙壁。当底端距墙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 base, and y the height of the top. The ladder length is constant: x² + y² = 5² = 25. Differentiate with respect to t: 2x (dx/dt) + 2y (dy/dt) = 0. Simplify: x (dx/dt) + y (dy/dt) = 0.

设x为墙到底端的距离，y为顶端的高度。梯长不变：x² + y² = 5² = 25。对t求导：2x (dx/dt) + 2y (dy/dt) = 0。化简得x (dx/dt) + y (dy/dt) = 0。

Given dx/dt = 0.2 m/s (away from wall, positive), and at the moment x = 3 m, we find y = √(25 – 9) = 4 m. Substituting: 3(0.2) + 4(dy/dt) = 0 → 0.6 + 4(dy/dt) = 0 → dy/dt = -0.15 m/s. The negative sign indicates the top is sliding down.

已知dx/dt = 0.2 m/s（远离墙，为正），当x = 3 m时，y = √(25 – 9) = 4 m。代入：3(0.2) + 4(dy/dt) = 0 → 0.6 + 4(dy/dt) = 0 → dy/dt = -0.15 m/s。负号表示顶端正在下滑。

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6. Example 3: Conical Water Tank | 示例3：圆锥水箱注水

Water is poured into an inverted conical tank of height 6 m and base radius 3 m at a rate of 2 m³/min. Find how fast the water level is rising when the water is 4 m deep.

水以2 m³/min的速率注入一个高6米、底面半径3米的倒立圆锥水箱。求水深为4米时水位上升的速率。

Let h be the water depth and r the radius of the water surface. By similar triangles, r/h = 3/6 = 1/2, so r = h/2. The volume of water is V = (1/3)πr²h = (1/3)π(h/2)² h = (1/12)πh³.

设h为水深，r为水面半径。由相似三角形，r/h = 3/6 = 1/2，即r = h/2。水体积为V = (1/3)πr²h = (1/3)π(h/2)² h = (1/12)πh³。

Differentiate with respect to t: dV/dt = (1/12)π · 3h² (dh/dt) = (1/4)πh² (dh/dt). We know dV/dt = 2 m³/min and at the instant h = 4 m. Plug in: 2 = (1/4)π(4)² (dh/dt) = (1/4)π(16)(dh/dt) = 4π (dh/dt). Therefore, dh/dt = 2/(4π) = 1/(2π) m/min. The water level rises at about 0.159 m/min.

对t求导：dV/dt = (1/12)π · 3h² (dh/dt) = (1/4)πh² (dh/dt)。已知dV/dt = 2 m³/min，当h = 4 m时代入：2 = (1/4)π(4)² (dh/dt) = (1/4)π(16)(dh/dt) = 4π (dh/dt)。因此，dh/dt = 2/(4π) = 1/(2π) m/min。水位上升速率约为0.159 m/min。

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7. Example 4: Two People Walking Apart | 示例4：两人相背而行

Two people start at the same point. Person A walks east at 4 km/h, and person B walks north at 3 km/h. How fast is the distance between them increasing after 2 hours?

两人从同一点出发。A以4 km/h向东走，B以3 km/h向北走。2小时后，他们之间的距离以多快的速率增加？

Let x be the distance A has walked east, y be the distance B has walked north, and s be the straight-line distance between them. The relationship is s² = x² + y². Differentiate: 2s (ds/dt) = 2x (dx/dt) + 2y (dy/dt), so ds/dt = (x (dx/dt) + y (dy/dt)) / s.

设x为A向东走的距离，y为B向北走的距离，s为两人间的直线距离。关系式为s² = x² + y²。求导：2s (ds/dt) = 2x (dx/dt) + 2y (dy/dt)，即ds/dt = (x (dx/dt) + y (dy/dt)) / s。

After 2 hours, x = 4 × 2 = 8 km, y = 3 × 2 = 6 km. Then s = √(8² + 6²) = 10 km. dx/dt = 4 km/h and dy/dt = 3 km/h (both constant). Substitute: ds/dt = (8×4 + 6×3)/10 = (32+18)/10 = 50/10 = 5 km/h. The distance increases at 5 km/h.

2小时后，x = 4 × 2 = 8 km，y = 3 × 2 = 6 km。于是s = √(8² + 6²) = 10 km。dx/dt = 4 km/h，dy/dt = 3 km/h（均为常数）。代入：ds/dt = (8×4 + 6×3)/10 = (32+18)/10 = 50/10 = 5 km/h。距离以5 km/h增加。

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8. Dealing with Implicit Differentiation and Multiple Variables | 处理隐式求导与多个变量

When an equation contains three or more time-dependent variables, you must decide which ones remain constant. A common IB twist involves a relationship like the cosine rule or similar-triangle constraints. Always express one variable in terms of another (e.g., r in terms of h) before differentiating, if possible. This reduces the number of unknown rates and simplifies the implicit differentiation step.

当方程包含三个或更多依赖于时间的变量时，你必须判断哪些量保持不变。IB常见的变体涉及余弦定理或相似三角形约束。如果可能，在求导之前总是将一个变量用另一个变量表示（例如用h表示r）。这能减少未知速率的数量，并简化隐式求导步骤。

For example, in a conical sand pile problem where the height is always three times the radius, you can write V = (1/3)πr²(3r) = πr³. Then differentiate directly: dV/dt = 3πr² (dr/dt). Always look for such simplifications.

例如，在一个圆锥沙堆问题中，如果高度始终是半径的三倍，你可以写成V = (1/3)πr²(3r) = πr³。然后直接求导：dV/dt = 3πr² (dr/dt)。要时刻寻找这样的简化机会。

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9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Mistake 1: Plugging in numbers before differentiating. Always differentiate first to get the general relationship among rates. Substituting constant values too early can destroy the derivative because you lose variable dependencies.

错误1：在求导之前代入数值。始终要先求导，得到速率之间的通用关系。过早代入常数值会破坏导数，因为你失去了变量间的依赖关系。

Mistake 2: Forgetting to use the chain rule on every variable. When differentiating a term like y² with respect to t, you must write 2y (dy/dt), not just 2y. Similarly, x³ becomes 3x² (dx/dt).

错误2：忘记对每个变量使用链式法则。当对t求导y²这样的项时，必须写出2y (dy/dt)，而不只是2y。同样，x³变为3x² (dx/dt)。

Mistake 3: Ignoring sign conventions. A negative rate means the quantity is decreasing. Always interpret the sign in the context of the problem—descending ladder, shrinking volume, or receding distance.

错误3：忽略符号规范。负速率意味着该量在减少。始终结合问题背景解释符号——梯子下降、体积缩小或距离拉远。

Mistake 4: Using inconsistent or wrong units. Check that all rates and quantities share consistent units (e.g., all in metres and seconds) before solving. Convert if necessary.

错误4：单位不一致或错误。在求解前检查所有速率和量是否使用一致的单位（例如，全部使用米和秒）。如有必要需进行换算。

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10. Summary and IB Exam Tips | 总结与IB考试提示

Related rates problems test your ability to connect geometry, algebra, and calculus under a time-dependent framework. In the IB exam, you can expect a structured question that guides you through the steps, but increasingly HL papers expect you to set up the equation independently. Always present your work clearly, label every derivative with its meaning (e.g., dV/dt in cm³/s), and check your final answer for reasonableness.

相关变化率问题考查你在时间依赖的框架下连接几何、代数和微积分的能力。在IB考试中，你可能会遇到引导你逐步求解的结构化问题，但越来越高阶的HL试卷期望你独立建立方程。始终清晰展示解题过程，给每个导数标注其含义（例如dV/dt的单位为cm³/s），并检查最终答案的合理性。

Remember: The chain rule is your best tool. With methodical practice, identifying the right equation and differentiating with respect to t will become second nature.

记住：链式法则是你最好的工具。通过有条理的练习，识别正确的方程并对t求导将成为第二天性。

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