Exercise 23B: Applications of Differentiation – Tangent, Normal & Optimization | 练习23B: 微分的应用——切线、法线与最优化

📚 Exercise 23B: Applications of Differentiation – Tangent, Normal & Optimization | 练习23B: 微分的应用——切线、法线与最优化

Exercise 23B in the IB Mathematics course takes you deeper into the practical power of differentiation. This exercise focuses on using derivatives to determine the equations of tangents and normals to curves, and to solve a broad range of optimization problems. These skills are essential for both the Analysis and Approaches and Applications and Interpretation syllabuses, and they appear frequently in both paper-based exams and internal assessments. Mastering the methods in this exercise will strengthen your ability to connect algebraic differentiation with geometric interpretation and real-world modelling.

IB 数学课程的练习 23B 将带你深入探索微分运算的实际应用。该练习重点在于利用导数求曲线的切线与法线方程，并解决各类最优化问题。这些技能对于分析与方法和应用与解释两类大纲都至关重要，在书面考试和内部评估中频繁出现。掌握本练习中的方法，将提升你将代数求导与几何意义及现实建模相联系的能力。

1. Core Concepts Behind Exercise 23B | 练习23B的核心概念

Before solving the problems, you must recall that the derivative f'(x) or dy/dx represents the slope of the tangent line to the graph of a function at any given point. A tangent touches the curve at exactly one point, sharing the instantaneous rate of change. The normal line, by contrast, is perpendicular to the tangent and has a slope that is the negative reciprocal of the tangent's slope at the point of contact.

在解题之前，你必须牢记导数 f'(x) 或 dy/dx 表示函数图形在任意给定点处切线的斜率。切线恰好与曲线交于一点，共享该点的瞬时变化率。相反，法线垂直于切线，其斜率为切线斜率在切点处的负倒数。

• Tangent slope m_t = f'(a) at x = a.
• Normal slope m_n = −1 / f'(a), provided f'(a) ≠ 0.

• 切线斜率 m_t = f'(a)，在 x = a 处。
• 法线斜率 m_n = −1 / f'(a)，假设 f'(a) ≠ 0。

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2. Quick Review of Essential Differentiation Rules | 基本求导法则速览

rule, and the chain rule for composite functions. For any term like axⁿ, the derivative is a·n·xⁿ⁻¹ where a is a constant coefficient.

练习23B假设你已经熟练掌握了幂函数、多项式以及基本三角函数的求导。你需要反复应用的关键法则包括幂法则：d/dx (xⁿ) = n xⁿ⁻¹、和法则，以及用于复合函数的链式法则。对于形如 axⁿ 的项，其导数为 a·n·xⁿ⁻¹，其中 a 是常数系数。

• If y = x³ − 4x² + 7, then dy/dx = 3x² − 8x.
• If y = sin(2x), then dy/dx = 2 cos(2x) using the chain rule.

• 若 y = x³ − 4x² + 7，则 dy/dx = 3x² − 8x。
• 若 y = sin(2x)，利用链式法则得 dy/dx = 2 cos(2x)。

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3. Finding the Equation of a Tangent Line | 求切线方程

oint where x = a, first compute the y-coordinate y₁ = f(a). Then evaluate the derivative f'(a) to obtain the slope m_t. Substitute these into the point-slope formula: y − y₁ = m_t(x − a). Finally, rearrange into gradient-intercept form y = mx + c if required.

要求出 x = a 处的切线方程，首先计算纵坐标 y₁ = f(a)。然后计算导数 f'(a) 以获得斜率 m_t。将其代入点斜式公式：y − y₁ = m_t(x − a)。最后，根据要求整理成斜截式 y = mx + c。

Example: f(x) = x³ − 3x + 2 at x = 1

例：f(x) = x³ − 3x + 2 在 x = 1 处

• f(1) = 1 − 3 + 2 = 0, so point is (1, 0).
• f'(x) = 3x² − 3 → f'(1) = 0.
• Tangent: y − 0 = 0(x − 1) → y = 0 (horizontal line).

• f(1) = 1 − 3 + 2 = 0，所以点是 (1, 0)。
• f'(x) = 3x² − 3 → f'(1) = 0。
• 切线：y − 0 = 0(x − 1) → y = 0（水平线）。

If the derivative is zero, the tangent is horizontal. If f'(a) is undefined (e.g., at a cusp), the tangent is vertical and must be treated separately.

如果导数为零，切线是水平的。如果 f'(a) 无定义（例如在尖点处），切线为垂直线，需单独处理。

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4. Determining the Equation of a Normal Line | 确定法线方程

The normal line is perpendicular to the tangent. Its slope m_n = −1 / m_t, provided m_t ≠ 0. Once you have the point of contact (a, f(a)) and the normal slope, apply the same point-slope formula to write the equation. If the tangent is horizontal (m_t = 0), the normal is vertical with equation x = a.

法线垂直于切线。其斜率 m_n = −1 / m_t，前提是 m_t ≠ 0。一旦得到切点 (a, f(a)) 和法线斜率，应用同样的点斜式写出方程。如果切线是水平的 (m_t = 0)，则法线为垂直线，方程为 x = a。

Example: Same f(x) = x³ − 3x + 2 but at x = 2

例：同一 f(x) = x³ − 3x + 2 但在 x = 2 处

• f(2) = 8 − 6 + 2 = 4, point (2, 4).
• f'(2) = 3(4) − 3 = 9, so m_t = 9.
• Normal slope m_n = −1/9.
• Normal equation: y − 4 = (−1/9)(x − 2) → y = (−1/9)x + 2/9 + 4 → y = (−1/9)x + 38/9.

• f(2) = 8 − 6 + 2 = 4，点为 (2, 4)。
• f'(2) = 3(4) − 3 = 9，故 m_t = 9。
• 法线斜率 m_n = −1/9。
• 法线方程：y − 4 = (−1/9)(x − 2) → y = (−1/9)x + 38/9。

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5. Introduction to Optimization Using Derivatives | 利用导数进行最优化的简介

d maximum or minimum values of a quantity. The derivative-based method relies on the fact that at a local extremum, f' ing the critical points, you must determine whether each yields a maximum or minimum using the second derivative test or a sign chart of the first derivative.

最优化问题要求你求出某个量的最大值或最小值。基于导数的方法依赖于这样一个事实：在局部极值处，f'(x) = 0（临界点）。找到临界点后，你必须通过二阶导数检验或一阶导数符号表来判断每个点对应的是最大值还是最小值。

• Find the expression to be maximized/minimized, say A(x).
• Compute A'(x) and set it to zero: A'(x) = 0.
• Solve for x to get critical values.
• Verify maxima/minima: if A''(x) < 0, it's a maximum; if A''(x) > 0, it's a minimum.

• 找到需要最大化或最小化的表达式，比如 A(x)。
• 计算 A'(x) 并令其等于零：A'(x) = 0。
• 解出 x 得到临界值。
• 验证极大/极小：若 A''(x) < 0，则为极大值；若 A''(x) > 0，则为极小值。

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6. Classic Optimization: Maximizing Area Given Perimeter | 经典最优化：给定周长求最大面积

A typical Exercise 23B problem might involve fencing a rectangular field with a fixed length of fencing wire. Suppose you have 200 m of fencing to enclose a rectangular plot against a barn (one side uses the barn, so no fence needed). You need to express area A in terms of one variable, differentiate, and find the dimensions that maximize area.

练习23B中一个典型问题可能涉及用固定长度的围栏围矩形场地。假设你有 200 米围栏，想靠着一间谷仓围出一块矩形场地（沿谷仓的一侧无需围栏）。你需要将面积 A 表示成单变量的函数，求导，并找出使面积最大化的尺寸。

Step-by-step

步骤分解

• Let width perpendicular to barn be w meters, length parallel be L meters.
• Constraint: 2w + L = 200 (only three sides fenced).
• Area A = w × L = w(200 − 2w) = 200w − 2w².
• A'(w) = 200 − 4w. Set A'(w) = 0 → w = 50.
• A''(w) = −4 < 0, so maximum when w = 50.
• Length L = 200 − 2·50 = 100 m. Maximum area = 50 × 100 = 5000 m².

• 设垂直于谷仓的宽为 w 米，平行于谷仓的长为 L 米。
• 约束条件：2w + L = 200（仅三边围栏）。
• 面积 A = w × L = w(200 − 2w) = 200w − 2w²。
• A'(w) = 200 − 4w。令 A'(w) = 0 → w = 50。
• A''(w) = −4 < 0，故 w = 50 时取得最大值。
• 长 L = 200 − 2·50 = 100 米。最大面积为 50 × 100 = 5000 平方米。

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7. Box and Cylinder Optimization | 盒体和圆柱体的最优化

IB exam questions often extend to manufacturing boxes from flat sheets or designing optimal cans. A common task: an open-top box is made by cutting squares from the corners of a rectangular sheet and folding up the sides. You must express the volume V in terms of the square's side length x, differentiate V(x), and find x that maximizes volume while respecting domain restrictions (x must be positive and less than half the shorter sheet dimension).

IB 考试题目常延伸到用平板制作盒子或设计最优罐体。常见任务：从矩形纸板四角切去正方形，再折起边缘制成开口盒。你需要用正方形边长 x 表示体积 V，对 V(x) 求导，并求出使体积最大化的 x，同时注意定义域限制（x 必须为正且小于纸板较短边的一半）。

Example: Sheet 30 cm by 20 cm, cut squares of side x

例：纸板 30 厘米 × 20 厘米，切去边长为 x 的正方形

• Volume V(x) = x(30 − 2x)(20 − 2x) = 4x³ − 100x² + 600x.
• V'(x) = 12x² − 200x + 600.
• Solve 12x² − 200x + 600 = 0 → divide by 4: 3x² − 50x + 150 = 0.
• Use quadratic formula: x = [50 ± √2500 − 1800] / 6 = [50 ± √700] / 6 = [50 ± 10√7] / 6.
• Approximately x ≈ 3.92 cm (valid) or x ≈ 12.75 cm (reject because 2x = 25.5 > 20).
• Test V''(3.92) < 0, confirming a maximum, giving max volume ≈ 1056 cm³.

• 体积 V(x) = x(30 − 2x)(20 − 2x) = 4x³ − 100x² + 600x。
• V'(x) = 12x² − 200x + 600。
• 解 12x² − 200x + 600 = 0 → 除以 4：3x² − 50x + 150 = 0。
• 使用二次公式：x = [50 ± √2500 − 1800] / 6 = [50 ± √700] / 6 = [50 ± 10√7] / 6。
• 近似为 x ≈ 3.92 cm（有效）或 x ≈ 12.75 cm（舍去，因为 2x = 25.5 > 20）。
• 检验 V''(3.92) < 0，确认是极大值，最大体积约为 1056 cm³。

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8. Optimization with Trigonometric Functions | 三角函数的优化

within the interval 0 ≤ θ ≤ π/2 or as specified. Always check endpoints if the domain is closed.

练习23B中有些问题可能涉及三角函数，例如求给定边长的三角形最大面积，或半圆的内接矩形。你需要将要优化的量表示为角度 θ 的函数，利用链式法则和三角函数的导数求导，并在区间 0 ≤ θ ≤ π/2 或指定范围内求解。若定义域为闭区间，务必检查端点。

le: Area of a symmetric triangle with equal sides r, with included angle θ: A = ½ r² sin θ

例：等腰三角形两腰长为 r，夹角为 θ，面积 A = ½ r² sin θ

• For fixed r, maximize A with respect to θ: A'(θ) = ½ r² cos θ.
• Set A'(θ) = 0 → cos θ = 0 → θ = π/2 (or 90°).
• Second derivative: A''(θ) = −½ r² sin θ. At θ = π/2, A'' = −½ r² < 0 → maximum.
• Thus, maximum area occurs when the angle is 90°, making the triangle right-angled.

• 固定 r，关于 θ 最大化 A：A'(θ) = ½ r² cos θ。
• 令 A'(θ) = 0 → cos θ = 0 → θ = π/2（即 90°）。
• 二阶导数：A''(θ) = −½ r² sin θ。在 θ = π/2 处，A'' = −½ r² < 0 → 极大值。
• 因此，当夹角为 90° 时面积最大，此时三角形为直角三角形。

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9. Common Mistakes and How to Avoid Them | 常见错误及如何避免

Many students lose marks on tangent/normal and optimization problems not because they cannot differentiate, but due to algebraic slip-ups or misinterpretation. One classic error is confusing the tangent and normal slopes—always pause to confirm m_n = −1/m_t. Another frequent mistake is forgetting to check the second derivative or the nature of critical points, leading to incorrectly claiming a minimum as a maximum.

许多学生在切线/法线和最优化问题上丢分，并非因为不会求导，而是由于代数失误或理解偏差。一个经典错误是混淆切线与法线的斜率——务必停下确认 m_n = −1/m_t。另一个常见错误是忘记检验二阶导数或临界点的性质，从而错误地将极小值当作极大值。

• Mistake: Using the derivative value itself as the equation of the line. The derivative gives only the slope, not the full equation.
• Tip: Always write the point-slope form first, then simplify.
• Mistake: Forgetting the domain of the practical problem (e.g., x can't be negative or exceed physical limits).
• Tip: State the feasible interval before differentiating and check critical points against it.

• 错误: 将导数值本身当作直线方程。导数只给出斜率，不是完整方程。
• 建议: 始终先写出点斜式，再进行化简。
• 错误: 忘记实际问题的定义域（例如 x 不能为负或超出物理限制）。
• 建议: 在求导前就明确可行区间，并将临界点与之对照检验。

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10. Connecting to IB Exam Paper Structure | 连接 IB 考试结构

Tangent and normal lines often appear in Paper 1 Section A as short-response questions testing fundamental calculus skills. Optimization problems, on the other hand, typically show up in Section B of Paper 1 or Paper 2 as extended, multi-part problems with a real-world context. These questions can blend algebraic manipulation, graphical interpretation, and verbal justification, so clear communication of your mathematical reasoning is crucial.

切线与法线常出现在试卷一的 A 部分，作为测试基础微积分技能的简答题。另一方面，最优化问题通常出现在试卷一的 B 部分或试卷二中，作为具有现实情境的拓展性多部分题目。这些问题可能融合了代数运算、图形解读和文字论证，因此清晰表达数学推理至关重要。

Mark allocation insight

评分分配洞察

• Setting up the objective function and constraint: 2–3 marks.
• Differentiation and solving: 2–3 marks.
• Confirming maximum or minimum: 1 mark.
• Final statement in context: 1 mark.

• 设立目标函数和约束条件：2–3 分。
• 求导并求解：2–3 分。
• 确认最大值或最小值：1 分。
• 结合实际情况的最终陈述：1 分。

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11. Additional Worked Example for Consolidation | 巩固附加例题

Let' of the tangent and normal at x = 2, and then determine the local minimum of this function.

让我们通过一个融合多项式和有理函数元素的题目来模拟练习23B的混合难度：f(x) = (x² + 1) / x，x > 0。求 x = 2 处的切线与法线方程，然后确定该函数的局部最小值。

Simplify first: f(x) = x + 1/x

先化简：f(x) = x + 1/x

• f'(x) = 1 − 1/x².
• At x = 2: f(2) = 2 + 1/2 = 2.5; f'(2) = 1 − 1/4 = 0.75.
• Tangent: y − 2.5 = 0.75(x − 2) → y = 0.75x + 1.
• 1/0.75 = −4/3. Normal: y − 2.5 = (−4/3)(x − 2) → y = (−4/3)x + 8/3 + 2.5 = (−4/3)x + 31/6.
• For optimum: f'(x) = 0 → 1 − 1/x² = 0 → x = 1 (positive).
• f''(x) = 2/x³, f''(1) = 2 > 0 → local minimum at (1, 2).

• f'(x) = 1 − 1/x²。
• 在 x = 2 处：f(2) = 2 + 1/2 = 2.5；f'(2) = 1 − 1/4 = 0.75。
• 切线：y − 2.5 = 0.75(x − 2) → y = 0.75x + 1。
• 法线斜率 = −1/0.75 = −4/3。法线：y − 2.5 = (−4/3)(x − 2) → y = (−4/3)x + 31/6。
• 求最优：f'(x) = 0 → 1 − 1/x² = 0 → x = 1（正值）。
• f''(x) = 2/x³，f''(1) = 2 > 0 → 在 (1, 2) 处取得局部极小值。

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12. Summary and Key Takeaways | 总结与关键要点

Exercise 23B equips you with a direct link between the abstract derivative and concrete geometry. Whenever you find the tangent or normal, you are essentially translating the slope into a linear model that locally mimics the curve. Optimization shows how calculus helps make the best decision—whether in engineering design, business profit analysis, or resource allocation. Consistent practice will make these procedures second nature.

练习23B 让你在抽象导数与具体几何之间建立了直接联系。每当你求出切线或法线时，本质上都是将斜率转化为局部模拟曲线的线性模型。最优化则展示了微积分如何帮助做出最优决策——无论是在工程设计、商业利润分析还是资源配置中。持续练习将使这些步骤成为你的第二天性。

• Key formulas: Tangent: y − y₁ = f'(a)(x − a); Normal: y − y₁ = −1/f'(a)(x − a).
• Optimization process: duce to one variable → derivative → critical points → second derivative test → conclusion.
• Marks depend on clear communication: always state the domain, explain why a value is max/min, and write a final answer in context.

• 关键公式：切线：y − y₁ = f'(a)(x − a)；法线：y − y₁ = −1/f'(a)(x − a)。
• 最优化流程：建模 → 化为单变量 → 求导 → 求临界点 → 二阶导数检验 → 得出结论。
• 评分依赖清晰表达：始终说明定义域，解释为何某值是极大或极小，并在情境中写出最终答案。

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