📚 10D Related Rates of Change | 10D 相关变化率
In many real-world situations, two or more quantities change over time and their rates of change are connected by an equation. For example, as a balloon is inflated, both its volume and radius increase, and the rates at which they grow are mathematically linked. This topic, usually found in IB Mathematics Section 10D, shows how to use derivatives to find an unknown rate of change when another rate is known.
在许多现实情境中，两个或多个量随时间变化，它们的变化率通过某个方程相关联。例如，给气球充气时，气球的体积和半径都在增加，且它们的增长速率在数学上紧密相连。IB 数学第 10D 节的相关内容，就是研究如何利用导数，在已知一个变化率的情况下，求出另一个未知的变化率。
1. Understanding Related Rates | 理解相关变化率
A "related rate" problem asks you to find the rate at which one quantity changes by relating it to another quantity whose rate of change is known. The quantities involved are functions of time, and they are connected by an equation such as a geometric formula, a trigonometric relation, or a physical law.
"相关变化率"问题要求你通过将一个量与另一个已知变化率的量联系起来，找出某个量的变化速率。所涉及的量都是时间的函数，并由一个方程联系在一起，例如几何公式、三角关系或物理定律。
The key idea is that time t is the hidden variable. Even when the problem does not explicitly mention t, every changing quantity is treated as a function of t, so implicit differentiation with respect to t is the main technique.
关键思想是：时间 t 是隐藏变量。即使题目没有明确提到 t，每个变化中的量都应视为 t 的函数，因此对 t 进行隐函数求导是主要技巧。
(x) and both x and y depend on t, then differentiating y with respect to t gives dy/dt = f 'x) · dx/dt. This single equation is the foundation of almost every related rates question.
例如，若 y = f(x)，且 x 和 y 都依赖于 t，则对 t 求导可得 dy/dt = f '(x) · dx/dt。这个方程几乎是所有相关变化率问题的基础。
Related rates problems are therefore a direct application of the chain rule to situations involving time-dependent variables. They appear in kinematics, geometry, economics, and the natural sciences.
因此，相关变化率问题是链式法则在含时间变量情境中的直接应用。它们出现在运动学、几何、经济学和自然科学中。
2. The Chain Rule Connection | 链式法则的关联
s form is especially useful because it clearly shows which derivative links the variables.
莱布尼茨记号下的链式法则指出：若 y 依赖于 x，x 依赖于 t，则 dy/dt = dy/dx · dx/dt。这种形式特别有用，因为它清楚地显示了是哪个导数把变量联系起来。
erentiating both sides with respect to t does not mean replacing r by a constant; it means treating r as r(t).
在相关变化率中，我们常有一个联系两个或多个变量的方程，例如球体体积 V = (4/3)πr³。对两边关于 t 求导时，并不是把 r 当作常数，而是把 r 看作 r(t)。
ative 4πr² is the familiar surface area formula, which appears naturally as the linking factor between volume rate and radius rate.
对 V = (4/3)πr³ 应用链式法则可得 dV/dt = 4πr² · dr/dt。导数 4πr² 正是熟悉的表面积公式，它自然地作为体积变化率与半径变化率之间的连接因子出现。
This pattern repeats throughout the topic: differentiate the geometric or physical relationship with respect to time, then substitute the known rate and the known instant values to solve for the unknown rate.
这一模式贯穿整个主题：先对几何或物理关系关于时间求导，再代入已知变化率和已知瞬时值，解出未知变化率。
3. A General Method for Related Rates | 相关变化率的通用方法
Most related rates problems can be solved by following a consistent sequence of steps. The method below is the one expected in IB examinations and in most calculus courses.
大多数相关变化率问题都可以通过一套固定步骤解决。下面的方法也是 IB 考试和多数微积分课程所期望的解题流程。
- Step 1: Draw a diagram and label all quantities that change with time.
第一步：画出示意图，并标注所有随时间变化的量。 - Step 2: Define variables and write down what is given and what is required, including units.
第二步：定义变量，写出已知量和所求量，并标明单位。 - Step 3: Find an equation relating the variables, using geometry, trigonometry, or physics.
第三步：利用几何、三角或物理知识，找出联系这些变量的方程。 - Step 4: Differentiate both sides of the equation with respect to time t, using the chain rule.
第四步：利用链式法则，对方程两边关于时间 t 求导。 - Step 5: Substitute the known values at the given instant, then solve for the unknown rate.
第五步：代入给定瞬时的已知值，解出未知变化率。 - Step 6: Check the sign, unit, and reasonableness of the answer.
第六步：检查答案的符号、单位和合理性。
A common warning is that you should not substitute numerical values too early. If you replace a changing variable by a constant before differentiating, its derivative becomes zero and the answer will be wrong.
一个常见警告是：不要过早代入数值。如果在求导前就把一个变化中的量替换成常数，它的导数就会变成零，答案就会出错。
Only constants such as π, fixed heights, or fixed distances may be substituted before differentiation. Variables such as r, h, x, and y must remain variables until after the derivative is taken.
只有像 π、固定高度或固定距离这样的常数可以在求导前代入。像 r、h、x、y 这样的变量必须在求导完成后才能代入数值。
4. Key Geometric Formulas | 常用几何公式
Related rates questions often depend on standard geometric formulas. You should know these formulas well enough to write them immediately without looking them up.
相关变化率问题常常依赖标准几何公式。你应当熟练掌握这些公式，做到无需查阅就能立即写出。
| Shape | 图形 | Formula | 公式 | Derivative with respect to t | 对 t 的导数 |
|---|---|---|
| Circle area | 圆面积 | A = πr² | dA/dt = 2πr · dr/dt |
| Circumference | 圆周长 | C = 2πr | dC/dt = 2π · dr/dt |
| Sphere volume | 球体积 | V = (4/3)πr³ | dV/dt = 4πr² · dr/dt |
| Sphere surface area | 球表面积 | S = 4πr² | dS/dt = 8πr · dr/dt |
| Cylinder volume | 圆柱体积 | V = πr²h | dV/dt = π(2rh · dr/dt + r² · dh/dt) |
| Cone volume | 圆锥体积 | V = (1/3)πr²h | dV/dt = (1/3)π(2rh · dr/dt + r² · dh/dt) |
| Pythagoras | 毕达哥拉斯定理 | x² + y² = L² | 2x · dx/dt + 2y · dy/dt = 0 |
h = k is constant, substitute r = kh so that the volume depends on only one changing variable.
当圆锥或影子问题涉及相似三角形时，通常最好在求导前消去一个变量。例如，若 r/h = k 为常数，可代入 r = kh，使体积只依赖于一个变化变量。
This simplification reduces the number of unknown rates and makes the differentiation shorter and less error-prone. It is one of the most valuable techniques in the topic.
这种简化减少了未知变化率的数量，使求导更短、更不易出错。它是本主题中最有价值的技巧之一。
5. Worked Example 1: Inflating a Balloon | 例 1：气球充气
A spherical balloon is inflated so that its volume increases at a constant rate of 100 cm³/s. Find the rate at which the radius is increasing when the radius is 5 cm.
一个球形气球被充气，体积以 100 cm³/s 的恒定速率增加。求半径等于 5 cm 时半径的增长速率。
radius at time t. The geometric relation is V = (4/3)πr³. Differentiate with respect to t:
设 V 为体积，r 为时刻 t 的半径。几何关系为 V = (4/3)πr³。对 t 求导：
dV/dt = 4πr² · dr/dt
Substitute dV/dt = 100 and r = 5:
代入 dV/dt = 100 和 r = 5：
100 = 4π(5)² · dr/dt = 100π · dr/dt
.318 cm/s. The positive sign means the radius is increasing, which agrees with the physical situation.
因此 dr/dt = 100/(100π) = 1/π ≈ 0.318 cm/s。正号表示半径在增加，这与实际情境一致。
Notice that the answer depends on the current radius. As the balloon gets larger, the same volume rate produces a smaller radius rate because the surface area is larger.
注意答案取决于当前半径。气球越大，同样的体积速率产生的半径增长速率越小，因为表面积更大。
6. Worked Example 2: Expanding Circle | 例 2：圆面积扩张
The radius of a circular oil slick is increasing at 0.5 cm/s. Find the rate at which the area is increasing when the radius is 10 cm.
一个圆形油污的半径以 0.5 cm/s 增加。求半径等于 10 cm 时面积的增长速率。
rentiating with respect to t gives dA/dt = 2πr · dr/dt. Substitute r = 10 and dr/dt = 0.5:
设 A = πr²。对 t 求导得 dA/dt = 2πr · dr/dt。代入 r = 10 和 dr/dt = 0.5：
dA/dt = 2π(10)(0.5) = 10π ≈ 31.4 cm²/s
The area is increasing at approximately 31.4 cm²/s. The unit cm²/s comes from multiplying cm by cm/s.
面积以约 31.4 cm²/s 增加。单位 cm²/s 来自 cm 与 cm/s 相乘。
If the question instead gives the area rate and asks for the radius rate, rearrange the same equation to make dr/dt the subject. The structure of the problem is unchanged.
如果题目改为已知面积变化率而求半径变化率，只需把同一方程变形，使 dr/dt 成为主元。问题结构并没有改变。
7. Worked Example 3: Sliding Ladder | 例 3：梯子下滑
A 5 m ladder leans against a vertical wall. The bottom of the ladder is 3 m from the wall and is sliding away at 0.5 m/s. Find the speed at which the top of the ladder is sliding down the wall.
一架 5 m 长的梯子靠在竖直墙上。梯子底端距墙 3 m，并以 0.5 m/s 向外滑动。求梯子顶端沿墙下滑的速度。
the top of the ladder. By Pythagoras, x² + y² = 5² = 25. Differentiate with respect to t:
设 x 为墙到梯子底端的距离，y 为梯子顶端的高度。由毕达哥拉斯定理，x² + y² = 5² = 25。对 t 求导：
2x · dx/dt + 2y · dy/dt = 0
When x = 3, y = √25 − 9 = √16 = 4. Substitute x = 3, y = 4, and dx/dt = 0.5:
当 x = 3 时，y = √25 − 9 = √16 = 4。代入 x = 3、y = 4、dx/dt = 0.5：
2(3)(0.5) + 2(4) · dy/dt = 0
. The negative sign means the height y is decreasing, so the top slides down at 0.375 m/s.
由此得 3 + 8 · dy/dt = 0，所以 dy/dt = −3/8 = −0.375 m/s。负号表示高度 y 在减小，因此顶端以 0.375 m/s 下滑。
Notice that the speed of the top is not constant. As the ladder approaches the ground, the top moves faster for the same bottom speed.
注意梯子顶端的速度不是恒定的。当梯子接近地面时，在相同的底端速度下，顶端会移动得更快。
8. Worked Example 4: Conical Tank | 例 4：圆锥水槽
Water is poured into a conical tank at 3 m³/min. The tank has radius 2 m at the top and height 4 m, with the vertex pointing down. Find the rate at which the water level is rising when the water is 1 m deep.
水以 3 m³/min 注入一个圆锥形水槽。水槽顶部半径为 2 m，高为 4 m，顶点朝下。求水深 1 m 时水面上升的速率。
/h = 2/4 = 1/2, so r = h/2. The volume of water is V = (1/3)πr²h = (1/3)π(h/2)²h = πh³/12.
设 h 为水深，r 为水面半径。由相似三角形，r/h = 2/4 = 1/2，所以 r = h/2。水的体积为 V = (1/3)πr²h = (1/3)π(h/2)²h = πh³/12。
Differentiate with respect to t:
对 t 求导：
dV/dt = (πh²/4) · dh/dt
Substitute dV/dt = 3 and h = 1:
代入 dV/dt = 3 和 h = 1：
3 = (π(1)²/4) · dh/dt = (π/4) · dh/dt
m/min. The water level rises quickly at first because the tank is narrow near the vertex.
因此 dh/dt = 12/π ≈ 3.82 m/min。起初水面上升很快，因为靠近顶点处水槽很窄。
If the tank were draining instead of filling, dV/dt would be negative, and dh/dt would also be negative, indicating a falling water level.
如果水槽是在排水而不是注水，dV/dt 将为负，dh/dt 也为负，表示水位下降。
9. Similar Triangles and Pythagoras | 相似三角形与毕达哥拉斯定理
Two geometric tools appear again and again in related rates: similar triangles and Pythagoras' theorem. Recognizing which one to use is often the key to setting up the equation correctly.
相关变化率中反复出现两个几何工具：相似三角形与毕达哥拉斯定理。识别该用哪一个，往往是正确建立方程的关键。
Similar triangles are used when a cone, a shadow, or a projected image has a constant ratio between two lengths. For example, in a conical tank, r/h is constant, so one variable can be eliminated before differentiation.
当圆锥、影子或投影图像的两个长度之间存在固定比例时，就使用相似三角形。例如在圆锥水槽中，r/h 为常数，因此可以在求导前消去一个变量。
Pythagoras' he sliding ladder problem. Differentiating x² + y² = L² gives 2x · dx/dt + 2y · dy/dt = 0.
当两个长度和一条固定斜边构成直角三角形时，就使用毕达哥拉斯定理，如梯子下滑问题。对 x² + y² = L² 求导得 2x · dx/dt + 2y · dy/dt = 0。
les give s/(x + s) = 1.8/6, so s = 3x/7. Then ds/dt = (3/7)(1.5) ≈ 0.643 m/s, and the tip of the shadow moves at (10/7)(1.5) ≈ 2.14 m/s.
考虑一个身高 1.8 m 的人以 1.5 m/s 远离一盏 6 m 高的路灯。若 x 为人到灯的距离，s 为影子长度，由相似三角形得 s/(x + s) = 1.8/6，所以 s = 3x/7。于是 ds/dt = (3/7)(1.5) ≈ 0.643 m/s，影子尖端以 (10/7)(1.5) ≈ 2.14 m/s 移动。
In both cases, the equation is differentiated only after all geometric relationships have been used to express the target quantity in terms of a single changing variable.
在这两种情况下，都要先利用所有几何关系，把目标量表示成单一变化变量的函数，然后再求导。
10. Units, Signs and Sanity Checks | 单位、符号与合理性检验
Every related rates answer should include a unit. The unit of a rate is the unit of the quantity divided by time, such as cm/s, cm²/s, m³/min, or m/s.
每个相关变化率的答案都应带有单位。变化率的单位是量的单位除以时间，例如 cm/s、cm²/s、m³/min 或 m/s。
The sign of the derivative tells you the direction of change. A positive rate means the quantity is increasing; a negative rate means it is decreasing. In the ladder problem, dy/dt is negative because the height is falling.
导数的符号告诉你变化方向。正的变化率表示量在增加；负的变化率表示量在减少。在梯子问题中，dy/dt 为负，因为高度在下降。
A quick sanity check can catch many errors. If the radius of a balloon is increasing, then dV/dt and dr/dt must have the same sign. If the water level in a draining tank is falling, then dV/dt and dh/dt must both be negative.
快速合理性检验可以抓住许多错误。如果气球半径在增加，则 dV/dt 与 dr/dt 必须同号。如果排水水槽的水位在下降，则 dV/dt 与 dh/dt 都必须为负。
when r = 5 cm is plausible; a rate of 100 cm/s would be far too large for a volume rate of
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