← 返回资讯屏轩教育 · 行业资讯

Worked Example 4.5.4: Connected Rates of Change | 例题 4.5.4：相关变化率

2026-08-21 20:52 · 本文部分内容由AI辅助创作

📚 Worked Example 4.5.4: Connected Rates of Change | 例题 4.5.4：相关变化率

In AQA A-Level Mathematics, connected rates of change questions test your ability to link two or more variables that both depend on time. Worked Example 4.5.4 is a classic application: finding how fast the radius of a sphere changes as its volume increases at a known constant rate.

在 AQA A-Level 数学中，相关变化率题目考查你能否将两个或多个都随时间变化的变量联系起来。例题 4.5.4 是一个经典应用：当球体的体积以已知恒定速率增加时，求其半径的变化速率。

These problems appear regularly in the differentiation section and require confident use of the chain rule, correct differentiation, and careful unit management. Mastering this example gives you a reliable template for many real-world rate problems.

这类题目经常出现在微分部分，要求你熟练运用链式法则、正确求导并谨慎处理单位。掌握这个例题可以为你解决许多现实中的变化率问题提供可靠模板。

1. The Exam Context | 考点背景

Connected rates of change form part of the AQA specification for both AS and A-Level Mathematics, under differentiation and its applications. They often appear in applied contexts such as expanding spheres, filling containers, moving ladders, or changing areas and volumes.

相关变化率是 AQA AS 和 A-Level 数学大纲中微分及其应用的一部分。它们经常出现在实际背景中，例如膨胀的球体、注水容器、移动的梯子，或变化的面积和体积。

The key skill is not simply memorising a formula, but understanding how a change in one variable causes a change in another through an intermediate variable. In Example 4.5.4, volume changes over time, and this causes the radius to change over time.

关键技能不是简单记住公式，而是理解一个变量的变化如何通过中间变量引起另一个变量的变化。在例题 4.5.4 中，体积随时间变化，这导致半径也随时间变化。

Examiners reward clear structure: identify the known rate, the required rate, the linking equation, and the chain rule statement before substituting numbers.

阅卷老师青睐清晰的结构：先确定已知速率、所求速率、连接方程和链式法则表达式，再代入数值。

2. The Worked Problem | 例题呈现

Worked Example 4.5.4 can be stated as follows: A spherical balloon is inflated so that its volume increases at a constant rate of 20 cm³ s⁻¹. Find the rate of increase of the radius at the instant when the radius is 8 cm. Give your answer in terms of π.

例题 4.5.4 可以表述如下：一个球形气球被充气，使其体积以 20 cm³ s⁻¹ 的恒定速率增加。求当半径为 8 cm 的瞬间，半径增加的速率。答案用 π 表示。

r/dt. The connecting relationship is the formula for the volume of a sphere, V = (4/3)πr³.

已知 dV/dt = 20 cm³ s⁻¹ 且 r = 8 cm。需要求 dr/dt。连接关系是球体体积公式 V = (4/3)πr³。

This problem is typical because it requires differentiating a power, applying the chain rule with respect to time, and rearranging to isolate the unknown rate.

这个问题很典型，因为它需要对幂函数求导、对时间应用链式法则，并通过变形分离出未知速率。

3. Core Principle: The Chain Rule for Rates | 核心原理：变化率的链式法则

of change of y with respect to time is given by the chain rule: dy/dt = (dy/dx) × (dx/dt).

如果量 y 依赖于 x，而 x 依赖于时间 t，那么 y 对时间的变化率由链式法则给出：dy/dt = (dy/dx) × (dx/dt)。

te dV/dt = (dV/dr) × (dr/dt). This links the known rate dV/dt with the unknown rate dr/dt.

本题中，体积 V 依赖于半径 r，而半径 r 依赖于时间 t。因此我们可以写出 dV/dt = (dV/dr) × (dr/dt)。这将已知速率 dV/dt 与未知速率 dr/dt 联系起来。

Once we have this statement, the solution becomes a straightforward sequence: differentiate V with respect to r, substitute the known dV/dt, then solve for dr/dt.

一旦我们写出这个表达式，解题步骤就变得直接：先对 r 求导 V，代入已知的 dV/dt，然后解出 dr/dt。

Always keep the chain rule in its general symbolic form before substituting numbers. This helps you avoid sign errors and makes your working clear to an examiner.

在代入数值之前，始终保持链式法则的一般符号形式。这可以避免符号错误，并让阅卷老师清楚地看到你的解题过程。

广告 / Sponsored

🛒
TI-Nspire CX II CAS 图形计算器 A-Level数学
¥1298
📱 去京东购买
京东精选

4. Step-by-Step Solution | 分步解析

Step 1: Write down the formula for the volume of a sphere. This is the geometric link between V and r.

步骤 1：写出球体体积公式。这是 V 与 r 之间的几何联系。

V = (4/3)πr³

Step 2: Differentiate V with respect to r. Bring down the power 3 and reduce the power by 1.

步骤 2：对 r 求导 V。将幂指数 3 下移，并将幂指数减 1。

dV/dr = 4πr²

in rule to relate the rates. Substitute dV/dt = 20 cm³ s⁻¹ into dV/dt = (dV/dr) × (dr/dt).

步骤 3：使用链式法则联系相关速率。将 dV/dt = 20 cm³ s⁻¹ 代入 dV/dt = (dV/dr) × (dr/dt)。

20 = 4πr² × (dr/dt)

Step 4: Rearrange the equation to make dr/dt the subject. This means dividing both sides by 4πr².

步骤 4：变形方程，使 dr/dt 成为主项。这意味着两边同时除以 4πr²。

dr/dt = 20 ÷ (4πr²)

Substitute r = 8 cm into the expression. Remember to square the radius before multiplying.

步骤 5：将 r = 8 cm 代入表达式。记住先平方半径再相乘。

dr/dt = 20 ÷ (4π × 8²) = 20 ÷ (256π) = 5 / (64π) cm s⁻¹

Therefore the radius increases at a rate of 5 / (64π) cm s⁻¹. As a decimal this is approximately 0.0249 cm s⁻¹, but the exact answer in terms of π is preferred unless a decimal is requested.

因此半径以 5 / (64π) cm s⁻¹ 的速率增加。换算成小数约为 0.0249 cm s⁻¹，但除非题目要求小数，否则最好保留含 π 的精确答案。

5. Interpreting the Answer | 结果解读

The positive sign of dr/dt confirms that the radius is increasing over time, which makes physical sense because the balloon is being inflated.

dr/dt 的正号表明半径随时间增大，这在物理上是合理的，因为气球正在被充气。

grows by about 0.025 cm every second. This may seem small, but it is consistent with the geometry: a large balloon has a large surface area, so a fixed volume increase spreads thinly over the surface and produces a slower radial growth.

在 r = 8 cm 的瞬间，半径每秒约增长 0.025 cm。这个数值看似很小，但符合几何关系：大气球表面积很大，因此固定的体积增量会分散在很大的表面上，导致半径增长较慢。

ius would increase much faster because the surface area is smaller. This inverse relationship between radius and its rate of change is a useful check for plausibility.

如果同样的体积速率应用于更小的半径，例如 r = 2 cm，半径会增长得更快，因为表面积更小。半径与其变化率之间的这种反比关系是检验结果合理性的有效方法。

6. Common Mistakes | 常见错误

One frequent error is confusing dV/dt with dV/dr. The first is the rate of volume change over time, while the second is the derivative of volume with respect to radius only. Always label each derivative with its units.

一个常见错误是混淆 dV/dt 与 dV/dr。前者是体积对时间的变化率，而后者只是体积对半径的导数。始终为每个导数标注单位。