Related Rates | 相关变化率的计算技巧

📚 Related Rates | 相关变化率的计算技巧

Related rates problems are among the most common applications of differentiation in IB Mathematics. They ask how fast one quantity changes when another quantity changes at a known rate. Mastering these problems requires a clear chain of reasoning, careful notation, and a reliable step-by-step method.

相关变化率问题是IB数学中微分应用最常见的考点之一。这类问题要求我们在已知某个量变化速率的情况下，求出另一个量的变化速率。掌握这类问题需要清晰的推理链条、规范的记号以及一套可靠的解题步骤。

1. Understanding the Concept | 理解相关变化率的概念

A related rates problem involves two or more quantities that are connected by an equation. If all quantities depend on time \(t\), then differentiating the equation implicitly with respect to \(t\) gives a relationship between their rates of change.

相关变化率问题涉及两个或多个由方程联系在一起的量。如果所有量都随时间 \(t\) 变化，那么对方程关于 \(t\) 隐式求导，就能得到各个变化率之间的关系。

For example, suppose a circle’s radius \(r\) is increasing at a constant rate. The area \(A = \pi r^2\) then changes at a rate given by:

例如，假设一个圆的半径 \(r\) 以恒定速率增加，那么面积 \(A = \pi r^2\) 的变化率由下式给出：

\(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)

Notice that the chain rule connects the two rates. The key is to identify which rates are known and which one you need to find.

注意，链式法则将两个变化率联系起来。关键是要判断哪些变化率已知、哪一个需要求解。

2. The General Strategy | 通用解题策略

All related rates problems follow a similar structure. Memorising this structure will save you time in an exam.

所有相关变化率问题都遵循类似的结构。在考试中记住这个结构可以节省时间。

Step 1: Draw a diagram if possible, and label all quantities that change with time.
Step 2: Write down the known rates and the unknown rate using derivative notation.
Step 3: Find an equation that relates the quantities themselves (not their rates).
Step 4: Differentiate both sides of that equation implicitly with respect to \(t\).
Step 5: Substitute the known values at the specific instant, then solve for the unknown rate.

第1步：尽可能画图，并标出所有随时间变化的量。
第2步：用导数记号写出现有变化率和待求变化率。
第3步：找出来联系这些量本身的等式（而不是它们的导数）。
第4步：对等式两边关于 \(t\) 隐式求导。
第5步：代入特定时刻的已知值，然后解出未知变化率。

Be careful: you cannot substitute numerical values before differentiating, because the rates are valid only at a specific moment.

注意：在求导之前不能代入数值，因为变化率只在特定时刻成立。

3. Worked Example 1: Expanding Square | 例1：扩张的正方形

A square’s side length \(x\) is increasing at a rate of 3 cm/s. How fast is its area increasing when the side length is 5 cm?

一个正方形的边长 \(x\) 以 3 cm/s 的速度增加。当边长为 5 cm 时，它的面积增加得多快？

Let \(A = x^2\). Differentiate with respect to time:

设 \(A = x^2\)。关于时间求导：

\(\frac{dA}{dt} = 2x \frac{dx}{dt}\)

Substitute \(x = 5\) and \(\frac{dx}{dt} = 3\):

代入 \(x = 5\) 和 \(\frac{dx}{dt} = 3\)：

\(\frac{dA}{dt} = 2(5)(3) = 30 \text{ cm}^2/\text{s}\)

The area is increasing at 30 cm²/s at that instant. Notice that the rate itself depends on the current side length, so it changes over time.

在该时刻，面积以 30 cm²/s 的速度增加。注意，这个变化率本身依赖于当前的边长，因此它会随时间改变。

4. Worked Example 2: Sliding Ladder | 例2：滑动的梯子

A ladder of length 10 m rests against a vertical wall. The foot of the ladder slides away from the wall at a constant speed of 1 m/s. How fast is the top of the ladder sliding down the wall when the foot is 6 m from the wall?

一把 10 m 长的梯子靠在竖直墙壁上。梯脚以 1 m/s 的恒定速度远离墙壁滑动。当梯脚距墙 6 m 时，梯顶沿墙下滑的速度是多少？

Let \(x\) be the distance from the wall to the foot, and let \(y\) be the height of the top above the ground. Pythagoras gives:

设 \(x\) 为墙壁到梯脚的距离，\(y\) 为梯顶离地面的高度。勾股定理给出：

\(x^2 + y^2 = 10^2 = 100\)

Differentiate implicitly with respect to \(t\):

关于 \(t\) 隐式求导：

\(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0\)

Divide by 2:

除以 2：

\(x \frac{dx}{dt} + y \frac{dy}{dt} = 0\)

At the instant \(x = 6\), use Pythagoras to find \(y = \sqrt{100 – 36} = 8\). Substitute \(\frac{dx}{dt} = 1\):

在 \(x = 6\) 的瞬间，用勾股定理求出 \(y = \sqrt{100 – 36} = 8\)。代入 \(\frac{dx}{dt} = 1\)：

\(6(1) + 8 \frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{6}{8} = -0.75\)

The negative sign means the top is moving downward at 0.75 m/s.

负号表示梯顶正以 0.75 m/s 的速度向下移动。

5. Worked Example 3: Inflating Balloon | 例3：充气的气球

Air is pumped into a spherical balloon at a rate of 100 cm³/s. How fast is the radius increasing when the radius is 5 cm?

空气以 100 cm³/s 的速度充入球形气球。当半径为 5 cm 时，半径增长的速率是多少？

The volume of a sphere is \(V = \frac{4}{3}\pi r^3\). Differentiate with respect to \(t\):

球的体积为 \(V = \frac{4}{3}\pi r^3\)。关于 \(t\) 求导：

\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)

Substitute \(\frac{dV}{dt} = 100\) and \(r = 5\):

代入 \(\frac{dV}{dt} = 100\) 和 \(r = 5\)：

\(100 = 4\pi (25) \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi} \text{ cm/s}\)

The radius increases at approximately 0.318 cm/s at that instant.

在该时刻，半径以约 0.318 cm/s 的速度增加。

6. Worked Example 4: Conical Tank | 例4：圆锥形容器

Water flows into a conical tank at a rate of 20 m³/min. The tank has height 10 m and top radius 4 m. How fast is the water level rising when the water depth is 6 m?

水以 20 m³/min 的速度流入圆锥形容器。容器高 10 m，顶部半径 4 m。当水深为 6 m 时，水面上升的速度是多少？

Let \(r\) be the radius of the water surface and \(h\) the water depth. Similar triangles give:

设 \(r\) 为水面半径，\(h\) 为水深。相似三角形给出：

\(\frac{r}{h} = \frac{4}{10} = 0.4 \Rightarrow r = 0.4h\)

The volume of water in the cone is \(V = \frac{1}{3}\pi r^2 h\). Substituting \(r = 0.4h\):

锥形水的体积为 \(V = \frac{1}{3}\pi r^2 h\)。代入 \(r = 0.4h\)：

\(V = \frac{1}{3}\pi (0.4h)^2 h = \frac{1}{3}\pi (0.16)h^3 = \frac{0.16\pi}{3} h^3\)

Differentiate with respect to \(t\):

关于 \(t\) 求导：

\(\frac{dV}{dt} = 0.16\pi h^2 \frac{dh}{dt}\)

Substitute \(\frac{dV}{dt} = 20\) and \(h = 6\):

代入 \(\frac{dV}{dt} = 20\) 和 \(h = 6\)：

\(20 = 0.16\pi (36) \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{20}{5.76\pi} \approx 1.105 \text{ m/min}\)

The water level rises at about 1.105 m/min when the depth is 6 m.

当水深为 6 m 时，水面以约 1.105 m/min 的速度上升。

7. Choosing the Right Equation | 选择正确的等式

In many problems, more than one geometric formula can be used. The key is to choose an equation that contains the quantities whose rates are known or desired, and that eliminates any unwanted variables using given relationships.

在许多问题中，可供使用的几何公式不止一个。关键在于选择一个等式中包含已知变化率和待求变化率所对应的量，并利用已知关系消去不需要的变量。

For example, in the conical tank problem, expressing \(r\) in terms of \(h\) before differentiating simplifies the algebra. If you instead leave both \(r\) and \(h\), you would need \(\frac{dr}{dt}\), which is not directly given.

例如，在圆锥容器问题中，在求导之前先用 \(h\) 表示 \(r\) 可以简化代数运算。如果你同时保留 \(r\) 和 \(h\)，就需要知道 \(\frac{dr}{dt}\)，而这个量并没有直接给出。

Always check: does your final equation contain exactly one unknown derivative at the instant of interest? If not, you may need additional relationships.

始终检查：在所求时刻，你的最终等式中是否只含有一个未知导数？如果不是，你可能需要额外的关系式。

8. Common Mistakes | 常见错误

Several errors appear again and again in related rates problems. Avoid them and you will gain marks quickly.

有几个错误在相关变化率问题中反复出现。避免它们，你就能快速得分。

错误	正确做法
在求导前代入数值，导致求导结果为0。	先对一般表达式求导，再代入特定时刻的值。
忘记使用链式法则，漏掉 \(\frac{dx}{dt}\) 因子。	对每个随时间变化的量都要乘以其导数。
用错几何公式（如把球体积写成 \(\pi r^2\)）。	默写所有常用几何公式并反复练习。
忽略单位或符号（正负号）。	用文字说明增长/减少，并保持单位一致。

Tip: always write down units in the final answer. A missing unit costs you a mark in IB exams.

提示：最终答案务必写单位。漏写单位在IB考试中会扣分。

9. Using Implicit Differentiation Correctly | 正确使用隐式求导

Implicit differentiation is the engine of related rates. Every variable that depends on time must be treated as a function \(x(t)\), \(y(t)\), etc. When differentiating \(x^2\), for example, you write \(2x \frac{dx}{dt}\).

隐式求导是相关变化率的引擎。每一个依赖时间的变量都必须看作函数 \(x(t)\)、\(y(t)\) 等。例如，对 \(x^2\) 求导时，要写成 \(2x \frac{dx}{dt}\)。

This applies to products and quotients too. If you have \(xy = 8\), then:

这同样适用于乘积和商。如果 \(xy = 8\)，那么：

\(x \frac{dy}{dt} + y \frac{dx}{dt} = 0\)

Practice this until it becomes automatic. Many students lose marks by forgetting the extra factor even when the rest of the method is correct.

多练习直到形成条件反射。许多学生即使其他步骤正确，也会因忘记额外因子而丢分。

10. Strategy for Unusual Shapes | 处理不规则图形的策略

Not every related rates problem involves a standard geometric shape. Sometimes you are given a relationship like \(A = 2x^3 + 5x\), where \(A\) is an area and \(x\) is a length. The method is the same: differentiate both sides with respect to \(t\).

并非所有相关变化率问题都涉及标准几何图形。有时题目会给出类似 \(A = 2x^3 + 5x\) 的关系，其中 \(A\) 是面积，\(x\) 是长度。方法是一样的：两边对 \(t\) 求导。

For such cases, write down the exact relationship first, then differentiate term by term. This reinforces the chain rule and avoids confusion.

对于这类情况，先完整写出关系式，然后逐项求导。这能强化链式法则并避免混淆。

If the problem involves angles, use trigonometric relationships. For example, if a ladder is sliding and you need the rate of change of the angle with the ground, write \(\sin\theta = \frac{y}{10}\) and differentiate implicitly.

如果问题涉及角度，使用三角关系。例如，如果梯子滑动且需要求与地面夹角的变化率，可以写出 \(\sin\theta = \frac{y}{10}\) 并隐式求导。

11. Exam-Style Practice Problem | 考试风格练习

A kite is flying horizontally at a constant height of 30 m. The string is being let out at a rate of 2 m/s. How fast is the kite moving horizontally when the string is 50 m long?

一只风筝在30 m的高度水平飞行。风筝线以 2 m/s 的速度放出。当线长为 50 m 时，风筝水平移动的速度是多少？

Let \(x\) be the horizontal distance from the person to the kite, and let \(s\) be the string length. Then:

设 \(x\) 为人到风筝的水平距离，\(s\) 为线长。那么：

\(x^2 + 30^2 = s^2\)

Differentiate: \(2x \frac{dx}{dt} = 2s \frac{ds}{dt} \Rightarrow x \frac{dx}{dt} = s \frac{ds}{dt}\).

求导：\(2x \frac{dx}{dt} = 2s \frac{ds}{dt} \Rightarrow x \frac{dx}{dt} = s \frac{ds}{dt}\)。

When \(s = 50\), find \(x = \sqrt{50^2 – 30^2} = \sqrt{2500 – 900} = \sqrt{1600} = 40\). Substitute:

当 \(s = 50\) 时，求出 \(x = \sqrt{50^2 – 30^2} = \sqrt{2500 – 900} = \sqrt{1600} = 40\)。代入：

\(40 \frac{dx}{dt} = 50(2) \Rightarrow \frac{dx}{dt} = \frac{100}{40} = 2.5 \text{ m/s}\)

Try this problem yourself before checking the solution. It is a typical IB-style question with a hidden triangle relationship.

先自己尝试解决这道题，再对照答案。这是一个典型的IB风格问题，隐藏了三角形关系。

12. Final Exam Tips | 最终考试建议

Before starting a related rates problem, pause and identify: what changes? what stays constant? Draw the situation. Write the master equation. Differentiate. Substitute. Interpret the answer with units and a sentence.

在开始相关变化率问题之前，停下来思考：什么在变化？什么保持不变？画出图示，写出主方程，求导，代入，最后用带单位的完整句子解释答案。

In the IB exam, related rates often appear in Paper 1 (no calculator) or Paper 2 (calculator allowed). In both cases, the method is identical. If the algebra becomes messy, do not panic — simplify using the given instant.

在IB考试中，相关变化率常出现在Paper 1（不可使用计算器）或Paper 2（允许使用计算器）中。两种情况方法完全相同。如果代数计算变得复杂，不要慌张——使用给定时刻的数值进行简化。

Finally, remember that practice is everything. The more problems you solve, the faster you will identify the correct equation and avoid traps. Related rates reward careful thinking, not memorised formulas.

最后，记住练习是最重要的。解决问题越多，你就能越快识别出正确等式并避开陷阱。相关变化率考验的是细致思考，而非死记公式。

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Related Rates | 相关变化率的计算技巧