📚 Exercise 10F: Applications of Differentiation | 练习10F：导数应用
Mastering the applications of differentiation is a crucial step in IB Mathematics. This article explores the key concepts covered in a typical Exercise 10F, from finding equations of tangents and normals to solving optimisation and kinematics problems. We will break down each topic with clear explanations, worked examples, and practical tips to help you tackle any question from this exercise confidently.
掌握导数的应用是IB数学中至关重要的一步。本文深入探讨典型练习10F所涵盖的核心概念，内容从求解切线与法线方程到处理最优化及运动学问题。我们将通过清晰的解释、例题和实用技巧逐一分解每个主题，帮助你自信地应对本练习中的任何题目。
1. Finding Equations of Tangents | 求切线方程
he curve y = f(x) at a given point, you need the gradient at that point, which is simply f'(x) evaluated at the x-coordinate. The equation then follows the point–slope form y – y₁ = m(x – e point lies exactly on the curve by substituting into the original function if necessary.
要确定曲线 y = f(x) 在某一点处的切线方程，你需要该点处的斜率，即在该x坐标处求得的 f'(x)。然后使用点斜式 y – y₁ = m(x – x₁) 得到方程。如有必要，始终确保该点确实在曲线上，可代入原函数进行验证。
For example, to find the tangent to y = x³ – 3x at x = 1, first compute f(1) = –2, and f'(x) = 3x² – 3, so f'(1) = 0. The tangent line is y – (–2) = 0(x – 1), giving y = –2, a horizontal line.
例如，求曲线 y = x³ – 3x 在 x = 1 处的切线：先计算 f(1) = –2，且 f'(x) = 3x² – 3，因此 f'(1) = 0。切线方程为 y – (–2) = 0(x – 1)，得到 y = –2，即一条水平线。
Common errors include forgetting to substitute the x-value into the derivative or misapplying the coordinate signs. A quick sketch can often prevent sign mistakes.
常见错误包括忘记将x值代入导数计算，或写错坐标的符号。快速画一个草图通常可以避免符号错误。
2. Equations of Normals | 求法线方程
The normal line is perpendicular to the tangent at the point of contact. If the gradient of the tangent is mt, then the gradient of the normal is mn = –1/mt, provided mt the tangent is horizontal, the normal is vertical and its equation is simply x = constant.
法线在接触点处与切线垂直。如果切线的斜率为 mt，则法线的斜率为 mn = –1/mt（前提是 mt ≠ 0）。当切线为水平线时，法线为竖直线，其方程就是 x = 常数。
In the previous example where the tangent was y = –2 (mt = 0), the normal at (1, – = 1. Always check for vertical or horizontal normals separately to avoid division by zero.
在前一个例子中切线为 y = –2 (mt = 0)，点 (1, –2) 处的法线则是竖直线 x = 1。务必单独检查垂直或水平法线的情况，以避免除以零的错误。
Practise converting between gradient forms efficiently: if mt = 2, the normal's gradient is –1/2; if mt = –3, the normal's gradient is 1/3. This simple reciprocal relationship is fundamental.
练习在斜率形式之间高效转换：若 mt = 2，法线斜率为 –1/2；若 mt = –3，法线斜率为 1/3。这种简单的倒数关系是基础。
3. Stationary Points – Local Maxima and Minima | 驻点——局部极大值与极小值
Stationary points occur where the first derivative f' ied as local maxima, local minima, or horizontal points of inflection. A sign diagram of f' e most reliable method to determine their nature: a sign change from positive to negative indicates a local maximum, while negative to positive indicates a local minimum.
驻点出现在一阶导数 f'(x) = 0 处。它们被分类为局部极大值、局部极小值或水平拐点。在驻点附近对 f'(x) 使用符号表是判断其性质最可靠的方法：符号由正变负表示局部极大值，由负变正表示局部极小值。
For example, if f'(x) = x² – 4, setting f'(x) = 0 gives x = –2 and x = 2. Testing intervals: for x < –2, f'(x) > 0; between –2 and 2, f'(x) < 0; for x > 2, f'(x) > 0. Hence, x = –2 is a local maximum and x = 2 is a local minimum.
例如，若 f'(x) = x² – 4，令 f'(x) = 0 得 x = –2 和 x = 2。检验区间：当 x < –2 时 f'(x) > 0；在 –2 和 2 之间 f'(x) < 0；当 x > 2 时 f'(x) > 0。因此，x = –2 为局部极大值点，x = 2 为局部极小值点。
A common mistake is to rely solely on the second derivative without checking cases where it equals zero. Always construct a sign diagram if the second derivative test is inconclusive.
一个常见错误是仅依赖二阶导数判断，而没有检验二阶导数等于零的情况。如果二阶导数检验无法确定，务必画出符号表。
4. Points of Inflection and Concavity | 拐点与凹凸性
nflection occurs where the concavity of the graph changes. This typically corresponds to f''(x) = 0, but you must check that the sign of f''(x) actually changes on either side of the point. Not every root of f''(x) is an inflection point.
拐点出现在图形凹凸性发生变化的位置。这通常对应 f''(x) = 0，但你必须检查 f''(x) 在该点两侧的符号是否确实发生改变。并非每个 f''(x) 的根都是拐点。
There are two types: stationary points of inflection (where f'(x) = 0) and non-stationary points of inflection (where f'(x) ≠ 0). For example, y = x³ has a stationary inflection at the origin because f'(0) = 0 and concavity changes.
存在两种类型：驻点拐点（f'(x) = 0）和非驻点拐点（f'(x) ≠ 0）。例如，y = x³ 在原点有一个驻点拐点，因为 f'(0) = 0 且凹凸性改变。
Use a concavity table: if f''(x) > 0 the graph is concave up (∪); if f''(x) < 0 it is concave down (∩). An inflection point marks the transition between these shapes.
使用凹凸性表格：若 f''(x) > 0，图形凹向上（∪）；若 f''(x) < 0，凹向下（∩）。拐点标志着这些形状之间的过渡。
5. Increasing and Decreasing Functions | 递增与递减函数
A function is increasing on an interval if f'(x) ≥ 0 for all x in that interval, and strictly increasing if f'(x) > 0. Similarly, decreasing corresponds to f'(x) ≤ 0. These intervals are identified by solving f'(x) > 0 or f'(x) < 0 and using sign charts.
如果在某区间内对所有 x 都有 f'(x) ≥ 0，则该函数在该区间上递增；若 f'(x) > 0，则为严格递增。类似地，递减对应 f'(x) ≤ 0。通过求解 f'(x) > 0 或 f'(x) < 0 并使用符号表来确定这些区间。
For the cubic f(x) = 2x³ – 9x² + 12x, its derivative f'(x) = 6x² – 18x + 12 = 6(x – 1)(x – up a sign diagram shows f is increasing for x < 1 and x > 2, and decreasing for 1 < x < 2.
对于三次函数 f(x) = 2x³ – 9x² + 12x，其导数 f'(x) = 6x² – 18x + 12 = 6(x – 1)(x – 2)。建立符号表后可看出函数在 x < 1 和 x > 2 时递增，在 1 < x < 2 时递减。
Always express intervals using proper notation, such as (–∞, 1) ∪ (2, ∞) for increasing. Clearly state the endpoints where the derivative is zero.
始终使用正确的区间表示法，例如递增区间为 (–∞, 1) ∪ (2, ∞)。清楚地说明导数等于零的端点。
6. Solving Optimisation Problems | 求解最优化问题
Optimisation involves finding the maximum or minimum value of a quantity subject to given constraints. The steps are: define the variable to be optimised, express it in terms of one independent variable using the constraint equation, differentiate, set the derivative equal to zero to locate critical points, and verify the nature using a sign test or second derivative.
最优化涉及在给定约束条件下求某个量的最大值或最小值。步骤为：定义要优化的变量，利用约束条件将其表示成单一自变量的函数，求导，令导数等于零以找出临界点，再利用符号检验或二阶导数验证其性质。
sions of a rectangle with perimeter P that maximise the area. Let width = x, length = (P/2 – x). Area A = x(P/2 – x) = (P/2)x – x². Then A'(x) = P/2 – 2x = 0 gives x = P/4, which yields a square.
一个经典问题：求周长为 P 的矩形中面积最大的尺寸。设宽为 x，则长为 (P/2 – x)。面积 A = x(P/2 – x) = (P/2)x – x²。则 A'(x) = P/2 – 2x = 0，得 x = P/4，此时为正方形。
Always check the domain: dimensions cannot be negative, and practical constraints may restrict the possible x-values. Don't forget units in word problems.
始终检查定义域：尺寸不能为负，且实际约束可能限制可能的 x 值。在应用题中不要忘记单位。
7. Kinematics: Displacement, Velocity, and Acceleration | 运动学：位移、速度和加速度
motion, if displacement s is given as a function of time t, then velocity v(t) = ds/dt = s'(t) and acceleration a(t) = dv/dt = v'(t) = s'' y to acceleration, and integrating to go backwards, form the core of kinematics questions.
在直线运动中，如果位移 s 作为时间 t 的函数给出，那么速度 v(t) = ds/dt = s'(t)，加速度 a(t) = dv/dt = v'(t) = s''(t)。通过求导可以从位移到速度再到加速度，而通过积分则可反向推导，这构成了运动学问题的核心。
e the absolute value of velocity or sum the absolute displacements between turning points.
若要找出质点何时静止，令 v(t) = 0 并求解 t。若要计算从 t = t₁ 到 t = t₂ 走过的总路程，必须对速度的绝对值积分，或者将各转折点之间位移的绝对值相加。
Example: s(t) = t³ – 6t² + 9t. Then v(t) = 3t² – 12t + 9 = 3(t – 1)(t – tantaneously at rest at t = 1 and t = 3. Analyse sign of v to determine direction changes.
示例：s(t) = t³ – 6t² + 9t。则 v(t) = 3t² – 12t + 9 = 3(t – 1)(t – 3)。质点在 t = 1 和 t = 3 时瞬间静止。分析 v 的符号以确定方向变化。
8. Related Rates | 相关变化率
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 strategy is to write an equation linking the variables, differentiate both sides with respect to time t using implicit differentiation, and substitute known values.
相关变化率问题是通过将某一量与已知变化率的其他量联系起来，从而求出该量的变化率。策略是写出连接各变量的方程，使用隐函数求导对时间 t 求导，然后代入已知值。
tarting with V = (4/3)πr³, differentiate: dV/dt = 4πr² (dr/dt). Then substitute and solve.
一个典型设定：一个球形气球被充气，已知 dV/dt。要求在某些半径值时求 dr/dt。从 V = (4/3)πr³ 开始，求导得 dV/dt = 4πr² (dr/dt)。然后代入求解。
Be consistent with units and pay attention to whether the rate is positive (increasing) or negative (decreasing). Always write down the “given rate” and “required rate” at the start.
单位要保持一致，并注意变化率是正（增加）还是负（减少）。开始时始终写下“已知速率”和“所求速率”。
9. Graph Sketching Using Derivatives | 利用导数绘制函数图像
Derivatives provide an analytical tool for sketching curves. Steps: find intercepts; determine stationary points and their nature; identify intervals of increase/decrease; locate inflection points and determine concavity; and examine asymptotic behaviour if any.
导数为绘制曲线提供了分析工具。步骤：确定截距；求出驻点及其性质；确定递增/递减区间；找出拐点并确定凹凸性；如有渐近行为，也加以分析。
With the function f(x) = x⁴ – 4x³, compute f'(x) = 4x³ – 12x² = 4x²(x – 3) and f''(x) = 12x² – 24x = 12x(x – = 3; inflection points at x = 0 and x = 2. Draw a combined sign chart to synthesise shape.
对于函数 f(x) = x⁴ – 4x³，计算得 f'(x) = 4x³ – 12x² = 4x²(x – 3)，f''(x) = 12x² – 24x = 12x(x – 2)。驻点在 x = 0 和 x = 3；拐点在 x = 0 和 x = 2。绘制一个综合符号表以归纳图像形状。
A sketch does not need exact scale but must clearly show all critical features: turning points, inflection, and correct end behaviour. Label coordinates of key points.
草图不需要精确比例，但必须清晰地展示所有关键特征：转折点、拐点以及正确的末端走势。标注关键点的坐标。
10. Common Pitfalls and Exam Tips | 常见错误与应试技巧
sign diagram of the first derivative; the second derivative test is a shortcut only when f''(x) ≠ 0. Also, remember that an inflection point requires a concavity change, not merely f''(x) = 0.
许多学生因未能正确检验驻点性质而丢分。务必使用一阶导数的符号表；二阶导数检验只是 f''(x) ≠ 0 时的一种捷径。还要记住，拐点需要凹凸性改变，而不仅仅是 f''(x) = 0。
In optimisation, always explicitly state the domain of your variable and verify that your solution lies within it. In kinematics, clearly distinguish between displacement and total distance travelled. For related rates, never substitute constant values before differentiation – always differentiate first, then plug in numbers.
在最优化题中，始终明确说明变量的定义域，并验证求得的解位于其中。在运动学中，清楚地区分位移与走过的总路程。对于相关变化率，绝对不要在求导之前代入常量值——总是先求导，再代入数值。
Manage your time well: typically, an Exercise 10F will test a mix of these applications. Practise identifying the type of problem quickly by looking for keywords (e.g., “rate,” “maximum,” “tangent”). Write down given information and required quantities before diving into calculations.
合理分配时间：通常，练习10F会综合考查上述多种应用。通过寻找关键词（如“变化率”“最大值”“切线”）快速识别问题类型。在投入计算之前，先写下已知信息和所求量。
Finally, always check your answers for reasonableness – a negative area, a negative maximum profit, or a speed exceeding physical limits indicates a mistake. Revising with past paper questions will build your intuition.
最后，始终检查答案的合理性——负的面积、负的最大利润或超出物理极限的速度都可能表明存在错误。使用历年真题进行复习能培养你的直觉。
11. Summary Table of Key Formulas | 关键公式汇总表
| Application | Key Idea | 数学表达 / 公式 |
|---|---|---|
| Tangent gradient | mt = f'(x₀) | y – y₀ = f'(x₀)(x – x₀) |
| Normal gradient | mn = –1/f'(x₀) (if f'(x₀) ≠ 0) | y – y₀ = –1/f'(x₀) (x – x₀) |
| Stationary points | Solve f'(x) = 0 | Sign diagram of f'(x) |
| Increasing interval | f'(x) ≥ 0 | Solve inequality |
| Inflection point | Concavity change, f''(x) = 0 (usually) | Sign diagram of f''(x) |
| Optimisation | Express quantity in one variable, differentiate | A'(x) = 0, verify max/min |
| Kinematics | v = ds/dt, a = dv/dt | v(t) = s'(t), a(t) = v'(t) |
| Related rates | Differentiate constraint equation w.r.t. t | e.g., dV/dt = 4πr² dr/dt |
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