📚 Exercise 12C: Implicit Differentiation and Related Rates | 习题12C：隐函数求导与相关变化率
are not written as y = f(x) and to tackle real-world problems where multiple quantities change with respect to time. This article walks you through the core concepts, step‑by‑step methods, and carefully chosen examples that mirror the style of IB exam questions.
在 IB 数学分析与方法（HL）课程中，第 12 章通常会整合高级求导技巧。习题 12C 聚焦于两个紧密关联的主题：隐函数求导与相关变化率。掌握这些技能后，你不仅可以对未写成 y = f(x) 形式的方程求导，还能解决实际中多个量随时间变化的动态问题。本文带你梳理核心概念、分步方法，并配合精心挑选的例题，题型贴近 IB 考试风格。
1. Understanding Explicit and Implicit Functions | 理解显函数与隐函数
s still a function of x, but it is not isolated on one side. Many curves we study in calculus—circles, ellipses, cardioids— equation without solving for y, which can be messy or even impossible in elementary terms.
显函数直接将因变量用自变量表示，例如 y = 3x² + 2x。隐函数则将 x 和 y 混合在同一方程中，例如 x² + y² = 25 或 x³ + y³ = 6xy。虽然 y 仍然是 x 的函数，但它没有被单独解出。微积分中研究的许多曲线——圆、椭圆、心形线——都用隐函数描述最为自然。隐函数求导使我们能直接从方程求出 dy/dx，无需解出 y，而后者常常十分繁琐，甚至无法用初等形式表达。
2. Why Implicit Differentiation? | 为什么需要隐函数求导？
If you tried to differentiate x² + y² = 25 by first rewriting it as y = ±√25 − x², you would have to handle two separate branches and then piece together results. Implicit differentiation avoids splitting the curve. More importantly, in related rates problems, we often work with equations that link several variables, none of which is explicitly given as a function of time. Implicit differentiation with respect to t allows us to relate the rates of change directly, making it an indispensable technique for IB HL exam questions involving expanding circles, moving shadows, or filling tanks.
如果你试图先把 x² + y² = 25 改写为 y = ±√25 − x² 再求导，就得分别处理两支，再把结果拼接起来。隐函数求导无需拆分曲线。更重要的是，在相关变化率问题中，我们面对的方程往往联系多个变量，却没有哪一个直接表示成时间的函数。对时间 t 进行隐函数求导可以直接建立变化率之间的关系，因此它是解决 IB HL 考试中膨胀的圆、移动的影子、注水水箱等问题的必备方法。
3. The Key Idea: Chain Rule Everywhere | 核心思路：处处使用链式法则
rms that contain only x are differentiated normally. The result is an equation that includes dy/dx, which we then solve algebraically. The same logic extends to differentiation with respect to time t: for any variable that changes with time, we multiply its derivative by d(variable)/dt.
当我们对隐函数方程逐项关于 x 求导时，需要把 y 看作 x 的函数，对所有含 y 的项使用链式法则。例如 d/dx (y²) = 2y · dy/dx，类似地 d/dx (sin y) = cos y · dy/dx。只含 x 的项正常求导。结果是一个含有 dy/dx 的方程，再通过代数方法解出 dy/dx。同样的逻辑适用于对时间 t 求导：只要某个变量随时间变化，就乘上该变量对 t 的导数。
4. Step-by-Step Procedure | 隐函数求导步骤
To perform implicit differentiation for an equation F(x, y) = 0:
1. Differentiate both sides of the equation with respect to x.
2. Whenever you differentiate a term involving y, multiply by dy/dx (chain rule).
3. Collect all terms containing dy/dx on one side of the equation.
4. Factor out dy/dx.
5. Solve for dy/dx by dividing through.
Always check if the original equation can be used to simplify the final expression—this is a mark of an elegant solution in IB exams.
对形如 F(x, y) = 0 的方程进行隐函数求导时：
1. 方程两边同时对 x 求导。
2. 每遇到含 y 的项，求导后乘以 dy/dx（链式法则）。
3. 将所有含 dy/dx 的项移到方程的一边。
4. 提取公因子 dy/dx。
5. 通过除法解出 dy/dx。
务必检查原方程能否用于化简最终表达式——这常常是 IB 考试中得分的关键细节。
5. Worked Example 1: The Circle | 示例1：圆的切线
Consider the circle x² + y² = 25. Differentiating both sides with respect to x gives:
考虑圆 x² + y² = 25。两边同时对 x 求导：
2x + 2y · dy/dx = 0
Solving for dy/dx:
解出 dy/dx：
dy/dx = −x/y, y ≠ 0
This tells us that at the point (3, 4) on the circle, the slope of the tangent is −3/4. The result matches the geometric fact that the radius from the origin to (3,4) has slope 4/3, and the tangent is perpendicular. Notice how implicit differentiation gives the gradient in terms of both x and y, which is typical for curves defined implicitly.
由此得到，在圆上点 (3, 4) 处，切线斜率为 −3/4。几何上，从圆心到 (3,4) 的半径斜率为 4/3，而切线与之垂直，结果吻合。注意隐函数求导给出的斜率同时依赖于 x 和 y，这是隐式曲线导数的常见特征。
6. Worked Example 2: A Folium of Descartes | 示例2：笛卡尔叶形线
escartes is given by x³ + y³ = 6xy. Differentiating term by term with respect to x yields:
笛卡尔叶形线的方程为 x³ + y³ = 6xy。逐项对 x 求导得到：
3x² + 3y² · dy/dx = 6y + 6x · dy/dx
Now collect the dy/dx terms on the left and the remaining terms on the right:
现在把含 dy/dx 的项移到左边，其余移右边：
3y² · dy/dx − 6x · dy/dx = 6y − 3x²
Factor dy/dx and solve:
提取 dy/dx 并求解：
dy/dx (3y² − 6x) = 6y − 3x²
dy/dx = (6y − 3x²) / (3y² − 6x) = (2y − x²) / (y² − 2x)
ituting into the original equation then yields the coordinates. The process illustrates how implicit differentiation provides access to curve properties that would be nearly impossible to obtain by solving for y explicitly.
该导数可用于寻找水平和铅直切线。例如求水平切线时令 dy/dx = 0，即 2y − x² = 0（且 y² ≠ 2x），再代入原方程解出坐标。整个过程表明，隐函数求导使我们能获取那些几乎无法通过显式解 y 得到的曲线性质。
7. Introduction to Related Rates | 相关变化率简介
Related rates problems involve two or more quantities that change with time and are linked by an equation. The goal is to find the rate of change of one quantity given the rates of change of the others. We use implicit differentiation with respect to time t, applying the chain rule to every variable that depends on t. The classic IB scenarios include a ladder sliding down a wall, a conical tank filling with water, or a spotlight casting a moving shadow. The underlying mathematics is essentially implicit differentiation where the independent variable is t instead of x.
相关变化率问题涉及两个或多个随时间变化的量，它们由一个方程联系。已知某些量的变化率，要求另一个量的变化率。解决方案是对时间 t 进行隐函数求导，对所有依赖于 t 的变量使用链式法则。IB 经典场景包括梯子靠墙滑下、圆锥水箱注水，或聚光灯投射移动的影子。其数学本质就是隐函数求导，只不过自变量换成了 t 而非 x。
8. Strategy for Solving Related Rates Problems | 相关变化率解题策略
A structured approach is essential:
1. Draw a diagram and label variables. Identify constants.
2. Write an equation that relates the variables (often from geometry or trigonometry).
3. Differentiate both sides with respect to time t, remembering to apply the chain rule.
4. Substitute the known values of variables and rates. Be careful: only substitute quantities that are constant or instantaneous snapshots—never substitute a variable that is still changing before differentiation.
5. Solve for the unknown rate. Include units and interpret the sign.
This framework turns what can look like an intimidating word problem into a clean sequence of algebraic steps.
一套清晰的解题流程必不可少：
1. 画图并标注变量，识别常数。
2. 写出联系各变量的方程（常来源于几何或三角）。
3. 两边同时对时间 t 求导，注意使用链式法则。
4. 代入已知的变量值和变化率。特别注意：在求导之前，只能代入常数；瞬时快照值必须在求导之后代入。
5. 解出未知速率，标明单位并解释正负号的意义。
这一框架能把看似复杂的应用題转化为井井有条的代数步骤。
9. Example 3: Expanding Circle Area | 示例3：膨胀的圆面积
Suppose a circular ripple spreads outward. The radius r increases at a constant rate of 2 cm/s. How fast is the area A increasing when the radius is 10 cm?
假设圆形涟漪向外扩散，半径 r 以 2 cm/s 的恒定速率增加。当半径为 10 cm 时，面积 A 的增加速率是多少？
ea of a circle is A = πr². Both A and r vary with time. Differentiating with respect to t:
圆面积公式 A = πr²。A 和 r 都随时间变化。对 t 求导：
dA/dt = 2πr · dr/dt
Substitute r = 10 cm and dr/dt = 2 cm/s:
代入 r = 10 cm 和 dr/dt = 2 cm/s：
dA/dt = 2π(10)(2) = 40π cm²/s
Thus, the area is increasing at approximately 125.7 cm²/s when the radius is 10 cm. This example highlights that the rate of change of area depends on the instantaneous radius, even though dr/dt is constant—a key insight for related rates.
因此当半径为 10 cm 时，面积以约 125.7 cm²/s 的速率增大。这个例子揭示，即使 dr/dt 恒定，面积的变化率仍依赖于当前的半径值，这是相关变化率问题中的一个重要洞察。
10. Example 4: Ladder Problem | 示例4：梯子问题
A 5‑metre ladder leans against a vertical wall. The bottom of the ladder is pulled away from the wall at 1 m/s. How fast is the top sliding down when the bottom is 3 m from the wall?
一个 5 米长的梯子斜靠在竖直墙壁上。梯子底端以 1 m/s 的速度被拉离墙壁。当底端离墙 3 米时，顶端下滑的速率是多少？
Let x be the distance from the wall to the bottom, and y the height of the top. By Pythagoras:
设 x 为底端到墙的距离，y 为顶端的高度。由勾股定理：
x² + y² = 5² = 25
Differentiate with respect to t:
对 t 求导：
2x · dx/dt + 2y · dy/dt = 0
Simplify and solve for dy/dt:
化简并解出 dy/dt：
dy/dt = − (x/y) · dx/dt
When x = 3, the corresponding y = √25 − 9 = 4. Substitute dx/dt = 1 m/s:
当 x = 3 时，y = √25 − 9 = 4。代入 dx/dt = 1 m/s：
dy/dt = − (3/4) · 1 = −0.75 m/s
The negative sign means the top is descending. So the ladder top slides down at 0.75 m/s at that instant. This classic problem demonstrates the power of implicit differentiation in linking rates.
负号表示顶端在下降。因此该瞬间梯子顶端以 0.75 m/s 的速度下滑。这一经典问题充分展示了隐函数求导在联系变化率方面的威力。
11. Example 5: Conical Water Tank | 示例5：圆锥水箱
Water is poured into a conical tank (vertex down) of height 6 m and base radius 3 m at a rate of 2 m³/min. How fast is the water level rising when the water depth is 4 m?
水以 2 m³/min 的速率注入一个顶点朝下、高 6 m、底面半径 3 m 的圆锥水箱。当水深为 4 m 时，水面上升速率是多少？
y similar triangles: r/h = 3/6 = 1/2, so r = h/2. Substituting gives V in terms of h only:
圆锥内水的体积 V = (1/3)πr²h。其中 r 和 h 都随时间变化，但由相似三角形得 r/h = 3/6 = 1/2，故 r = h/2。代入得 V 仅用 h 表示：
V = (1/3)π (h/2)² h = (1/3)π (h²/4) h = (π/12) h³
Differentiate with respect to t:
对 t 求导：
dV/dt = (π/4) h² · dh/dt
Now plug in dV/dt = 2 m³/min and h = 4 m:
代入 dV/dt = 2 m³/min 和 h = 4 m：
2 = (π/4) (16) · dh/dt ⇒ 2 = 4π · dh/dt
dh/dt = 2 / (4π) = 1/(2π) m/min
9 m/min. This example reinforces the need to reduce the number of variables before differentiation and to use geometric constraints wisely.
水面以 1/(2π) ≈ 0.159 m/min 的速率上升。该例题强调了在求导前减少变量个数、巧妙利用几何约束的重要性。
12. Common Mistakes and Final Tips | 常见错误与总结
Students often forget to multiply by dy/dx (or d(variable)/dt) when differentiating a term involving a dependent variable. Another common error is substituting given instantaneous values before differentiating—this kills the derivative because the variable becomes a constant. Practise writing down the entire differentiated equation first, then plug in numbers. Also, watch your signs: negative rates indicate decrease or downward motion. For IB success, memorise the step‑by‑step process for both implicit differentiation and related rates, and always answer with correct units and a brief interpretation. Master these exercises, and Exercise 12C will become a source of confident marks rather than anxiety.
学生常常在求导涉及因变量的项时忘记乘以 dy/dx（或 d(变量)/dt）。另一个常见错误是在求导之前就把给定的瞬时值代入——这会使变量变为常数，导致导数为零。正确的做法是先写出完整的求导方程，再代入数值。还要注意正负号：负号表示减少或向下运动。想在 IB 考试中稳操胜券，就要牢记隐函数求导和相关变化率的分步流程，并始终给出正确的单位和简要解释。掌握这些习题后，练习 12C 将不再是焦虑的来源，而会成为自信拿分的利器。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图