📚 Exercise 11B: Tangents, Normals, and Rates of Change | 练习11B：切线、法线与变化率
The derivative is much more than an abstract limit; it is the key to understanding how quantities change and how curves behave at specific points. In this exercise set, we apply differentiation to find tangents, normals, and rates of change, building essential skills for IB Mathematics Analysis & Approaches and Applications & Interpretation. Mastery of these topics allows you to connect algebra with geometry and motion, paving the way for optimisation, kinematics, and more advanced calculus concepts.
导数远不只是抽象的极限；它是理解量如何变化以及曲线在特定点如何表现的关键。在本练习中，我们将运用微分来求切线、法线和变化率，培养IB数学分析与方法和应用与解释所需的核心技能。掌握这些专题能够将代数与几何、运动联系起来，为最优化、运动学和更高级的微积分概念奠定基础。
1. Understanding the Derivative as a Slope | 理解导数的斜率意义
The derivative of a function f at a point x = a, denoted f' te of change of the function. If you zoom in very close to the point on a smooth curve, the tangent line almost coincides with the curve itself.
函数 f 在点 x = a 处的导数，记作 f'(a) 或在 a 处求值的 dy/dx，给出了曲线 y = f(x) 在该点处切线的斜率。从几何上看，它表示函数在该点的瞬时变化率。如果你在光滑的曲线上无限放大某一点，切线几乎与曲线重合。
t at x = a, first differentiate the function, then substitute a into the derivative: m = f' o, indicating whether the function is increasing, decreasing, or stationary at that point.
要计算在 x = a 处切线的斜率 m，首先对函数求导，然后将 a 代入导数：m = f'(a)。斜率可以是正值、负值或零，分别表示函数在该点处递增、递减或驻定。
Understanding the derivative as a slope is the cornerstone for solving problems involving tangent lines, normal lines, and the interpretation of motion graphs. Without this visual connection, differentiation becomes a purely mechanical process.
将导数理解为斜率是解决切线、法线问题以及解释运动图像的基石。如果没有这种直观联系，微分就变成了纯粹的机械操作。
2. Finding the Equation of a Tangent Line | 求切线方程
Once the slope m = f' ) are known, the equation of the tangent line can be written using the point-slope form: y – y₁ = m (x – e-intercept form y = mx + c or the general form ax + by + c = 0, depending on the question's requirement.
一旦知道了斜率 m = f'(a) 和切点 (a, f(a))，就可以利用点斜式写出切线方程：y – y₁ = m (x – x₁)。根据题目要求，常将其改写为斜截式 y = mx + c 或一般式 ax + by + c = 0。
For example, for the function f(x) = x² + 3x, the derivative is f'(x) = 2x + 3. At x = 1, the slope is f'(1) = 5 and the point is (1, 4). The tangent equation is y – 4 = 5(x – 1), which simplifies to y = 5x – 1.
例如，对于函数 f(x) = x² + 3x，其导数为 f'(x) = 2x + 3。在 x = 1 处，斜率为 f'(1) = 5，点为 (1, 4)。切线方程为 y – 4 = 5(x – 1)，化简得 y = 5x – 1。
Always check your answer by verifying that the point satisfies the tangent equation and that the slope matches the derivative. A quick sketch can also help catch sign errors or arithmetic mistakes.
始终通过验证该点满足切线方程且斜率与导数一致来检查答案。快速绘制草图也有助于发现符号错误或计算失误。
3. Finding the Equation of a Normal Line | 求法线方程
the tangent at that point. If the tangent slope is m, the normal slope mₙ is given by mₙ = – normal is vertical (equation x = a). If the tangent is vertical, the normal is horizontal.
曲线在给定点处的法线是与该点切线垂直的直线。如果切线斜率为 m，则法线斜率 mₙ 为 mₙ = –1/m，前提是 m ≠ 0。如果 m = 0，则切线为水平线，法线为竖直线（方程为 x = a）。如果切线为竖直线，法线则为水平线。
ing the same example f(x) = x² + 3x at x = 1, the tangent slope was 5. The normal slope is –1/5. The normal passes through (1,4), giving the equation y – 4 = –1/5 (x – 1), or y = – x + 21/5. Many students forget to take the negative reciprocal, so double-check this step.
仍以 f(x) = x² + 3x 在 x = 1 处为例，切线斜率为 5。法线斜率为 –1/5。法线经过点 (1,4)，方程为 y – 4 = –1/5 (x – 1)，即 y = –1/5 x + 21/5。许多学生会忘记取负倒数，所以务必仔细检查这一步。
When the question asks for the equation of the normal, it is often after a tangent problem. Be careful with fractions and signs to avoid losing marks on otherwise straightforward questions.
当题目要求求法线方程时，通常紧跟切线问题之后。要小心分数的处理与符号，以免在这些本可轻松得分的题目上失分。
4. Points of Horizontal Tangency | 水平切点（驻点）
A horizontal tangent occurs where the derivative equals zero: f' he x-coordinates of points where the tangent is parallel to the x-axis. These points are crucial in curve sketching and optimisation because they often correspond to local maxima, minima, or horizontal points of inflection.
水平切线出现在导数等于零的地方：f'(x) = 0。解这个方程可得出切线平行于 x 轴的点的 x 坐标。这些点在曲线作图和最优化中至关重要，因为它们通常对应局部极大值、极小值或水平拐点。
For f(x) = 2x³ – 9x² + 12x, compute f'(x) = 6x² – 18x + 12. Setting f'(x) = 0 gives 6(x² – 3x + 2) = 0 ⇒ ary points (1, 5) and (2, 4). To classify them, use the second derivative or a sign chart.
对于 f(x) = 2x³ – 9x² + 12x，求导得 f'(x) = 6x² – 18x + 12。令 f'(x) = 0，得 6(x² – 3x + 2) = 0 ⇒ x = 1 或 x = 2。代回原函数得驻点 (1, 5) 和 (2, 4)。可利用二阶导数或符号表对它们进行分类。
Remember that f' urning point, but not sufficient; always confirm the nature of the stationary point. For IB exam success, show substitution steps clearly and label the points with their coordinates.
记住，f'(x) = 0 是转折点的必要条件，但不是充分条件；务必确认驻点的性质。为了在IB考试中成功，清晰展示代入步骤并标出点的坐标。
5. Angle of Intersection between Curves | 两曲线相交角度
nt. If the slopes of the tangents are m₁ and m₂, the angle is given by the formula tan θ = | (m₁ – m₂) / (1 + m₁ m₂) |, where θ ∈ [0, π/2). This is derived from the tangent difference identity.
当两条曲线在某点相交时，交角 θ 定义为在该点处两切线之间的锐角。如果切线斜率分别为 m₁ 和 m₂，交角由公式 tan θ = | (m₁ – m₂) / (1 + m₁ m₂) | 给出，其中 θ ∈ [0, π/2)。该公式由正切的差角恒等式推导而来。
= x³ + x², m₂ = 3x² + 2x, at x=0 gives m₂ = 0. The angle is 0 because both are horizontal – a poor example. Let' ose y = x² and y = x. They intersect at (0,0) and (1,1). At (1,1), m₁ = 2, m₂ = 1. tan θ = |(2-1)/(1+2·1)| = 1/3, so θ = arctan(1/3). This shows how the formula works.
例如，考虑曲线 y = x² 和 y = x³ + x²。它们相交于 (0,0)。求导：对于 y = x²，m₁ = 2x，在 x=0 时 m₁ = 0；对于 y = x³ + x²，m₂ = 3x² + 2x，在 x=0 时 m₂ = 0。交角为 0，因为两者都是水平的——这个例子不好。我们选择 y = x² 和 y = x。它们相交于 (0,0) 和 (1,1)。在 (1,1) 处，m₁ = 2，m₂ = 1。tan θ = |(2-1)/(1+2·1)| = 1/3，因此 θ = arctan(1/3)。这演示了公式的用法。
le is π/2 (90°), meaning the tangents are perpendicular. This is a special case often tested in multiple-choice questions.
如果分母 1 + m₁ m₂ = 0，则交角为 π/2（90°），表示两切线垂直。这是选择题中常考的特殊情况。
6. Rate of Change in Real-Life Contexts | 实际情境中的变化率
The derivative represents the instantaneous rate of change of one quantity with respect to another. In physical contexts, this could be velocity (rate of change of displacement with respect to time), acceleration (rate of change of velocity), or growth rate of a population. Interpreting the derivative in context is a major skill in IB exams.
导数表示一个量相对于另一个量的瞬时变化率。在物理情境中，这可以是速度（位移随时间的变化率）、加速度（速度的变化率）或人口增长率。在特定情境下解释导数是IB考试中的一项重要技能。
dly the volume increases as the radius grows. When asked for the rate of change at a specific radius, simply substitute the value.
如果一个球形气球体积 V 由 V = (4/3)πr³ 给出，则体积关于半径的变化率为 dV/dr = 4πr²。这告诉我们随着半径增大，体积如何迅速增加。若要求特定半径下的变化率，直接代入值即可。
Always include units when interpreting rates of change. For instance, if r is in cm, dV/dr is in cm² per cm, effectively representing the surface area – a nice geometric interpretation. Connecting calculus to geometry deepens understanding.
在解释变化率时务必包含单位。例如，如果 r 以厘米为单位，dV/dr 的单位是平方厘米每厘米，实际上代表表面积——这是一个很巧妙的几何解释。将微积分与几何联系起来能加深理解。
7. Motion along a Straight Line | 直线运动
In kinematics, the position of a particle moving along a line is given by a displacement function s(t). Its velocity v(t) is the first derivative s'(t), and acceleration a(t) is the second derivative s''(t) or v'(t). The sign of velocity indicates direction, while speed is the magnitude |v(t)|.
在运动学中，沿直线运动的粒子的位置由位移函数 s(t) 给出。其速度 v(t) 是一阶导数 s'(t)，加速度 a(t) 是二阶导数 s''(t) 或 v'(t)。速度的符号表示方向，而速率是速度的绝对值 |v(t)|。
For example, if s(t) = t³ – 6t² + 9t, then v(t) = 3t² – 12t + 9, and a(t) = 6t – 12. To find when the particle is at rest, solve v(t) = 0: 3(t² – on when the sign of v(t) changes. Analysing these times gives complete motion description.
例如，若 s(t) = t³ – 6t² + 9t，则 v(t) = 3t² – 12t + 9，a(t) = 6t – 12。求粒子何时静止，解 v(t) = 0：3(t² – 4t + 3) = 0 → t = 1 或 t = 3。当 v(t) 改变符号时，粒子改变运动方向。分析这些时刻可得完整的运动描述。
Questions often ask for the total distance travelled in a time interval. You must identify when the particle reverses direction and compute the sum of absolute displacements over subintervals, not just the net displacement. This is a common pitfall.
题目常要求在一段时间内所经过的总路程。必须找出粒子何时反向运动，并计算各子区间内位移绝对值之和，而不是只求净位移。这是一个常见的易错点。
8. Applying the Chain Rule to Related Rates | 用链式法则解决相关变化率问题
hain rule links the derivatives: if y = f(x) and x = g(t), then dy/dt = (dy/dx) · (dx/dt).
相关变化率问题涉及两个或多个随时间变化的量，且它们之间的关系已知。通过运用链式法则对方程关于 t 求导，我们可以求出未知的变化率。链式法则将导数联系起来：若 y = f(x) 且 x = g(t)，则 dy/dt = (dy/dx) · (dx/dt)。
the top. Since x² + y² = 25, differentiate with respect to t: 2x (dx/dt) + 2y (dy/dt) = 0 ⇒ dy/dt = – ). If the base moves away at 0.5 m/s when x = 3 m, find the speed of the top sliding down.
一个经典例题：一架长 5 米的梯子斜靠在墙上并向下滑动。设 x 为梯脚到墙的距离，y 为顶端的高度。由于 x² + y² = 25，关于 t 求导得：2x (dx/dt) + 2y (dy/dt) = 0 ⇒ dy/dt = –(x/y) (dx/dt)。若当 x = 3 米时，梯脚以 0.5 米/秒的速度向外移动，求顶端下滑的速度。
At x = 3, y = √25 – 9 = 4. So dy/dt = –(3/4) × 0.5 = – decreasing. Drawing a diagram and listing known rates at the snapshot moment is essential.
当 x = 3 时，y = √25 – 9 = 4。因此 dy/dt = –(3/4) × 0.5 = –0.375 米/秒。负号表示高度在减少。绘制示意图并列出该瞬间的已知变化率至关重要。
9. Optimisation Problems | 最优化问题
uantity to be optimised as a function of a single variable, finding stationary points by f'(x) = 0, and using the second derivative test or analysis of the sign of f'(x) to confirm whether the point is a maximum or minimum.
最优化利用导数求函数在给定区间内的最大值或最小值。步骤包括将待优化量表示成一个变量的函数，通过解 f'(x) = 0 求驻点，再利用二阶导数检验或分析 f'(x) 的符号来确认该点是最大值还是最小值。
Let the width perpendicular to the river be x, and length parallel be y. Then 2x + y = 200 ⇒ y = 200 – 2x. Area A = xy = x(200 – 2x) = 200x – 2x². Then A'(x) = 200 – 4x = 0 ⇒ x = 50. The area is maximised at x = 50, y = 100, giving Amax = 5000 m².
例题：一个矩形场地三面围栏，用200米围栏，第四面靠河。设垂直于河的宽度为 x，平行于河的长度为 y。则 2x + y = 200 ⇒ y = 200 – 2x。面积 A = xy = x(200 – 2x) = 200x – 2x²。A'(x) = 200 – 4x = 0 ⇒ x = 50。面积在 x = 50，y = 100 时最大，最大面积为 5000 平方米。
Always check the endpoints of the domain if it is restricted. Also, confirm that the value obtained indeed gives a maximum (second derivative negative or first derivative changes from positive to negative).
若定义域有界，务必检查端点处的值。同时验证所求值确实是最大值（二阶导数为负或一阶导数由正变负）。
10. Graphical Interpretation and Curve Sketching | 图形解释与曲线草图
Interpreting graphs of f, f', and f'' is a recurring theme in IB. The sign of f'(x) indicates where f is increasing or decreasing. The sign of f''(x) reveals concavity: f''(x) > 0 concave up, f'' concave down. Points of inflection occur where concavity changes, which may happen when f''(x) = 0 or is undefined.
理解 f、f' 和 f'' 的图像关系是IB中的常见主题。f'(x) 的符号表明 f 的递增或递减区间。f''(x) 的符号显示凹凸性：f''(x) > 0 为凹向上，f''(x) < 0 为凹向下。拐点出现在凹凸性改变的位置，可能发生在 f''(x) = 0 或无定义的地方。
Given the graph of f'(x), you can deduce the shape of f(x). Where f'(x) > 0, f is increasing; where f'(x) < 0, f is decreasing; and where f'(x) = 0, f has stationary points. If f' m positive to negative, f has a local maximum; negative to positive gives a local minimum.
已知 f'(x) 的图像，可以推断 f(x) 的形状。在 f'(x) > 0 的区域，f 递增；f'(x) < 0 时 f 递减；f'(x) = 0 处 f 有驻点。若 f'(x) 由正变负，f 有局部极大值；由负变正则对应局部极小值。
Practice sketching curves by first identifying intercepts, stationary points, asymptotes, and concavity. Using a table of signs for f' and f'' organises the analysis and leads to an accurate diagram, which is often required in the exam.
练习画曲线图时，首先确定截距、驻点、渐近线和凹凸区间。使用 f' 和 f'' 的符号表来组织分析，从而绘制准确的示意图，这是在考试中常被要求的。
11. Common Mistakes and How to Avoid Them | 常见错误与避免方法
is the negative reciprocal, not just the negative, nor simply the reciprocal. Writing mₙ = –m instead of – will give a completely wrong equation. Always pause and ask: are these lines perpendicular?
一个常见错误是混淆切线斜率和法线斜率。法线斜率是负倒数，而不仅仅是负值或单纯的倒数。写成 mₙ = –m 而不是 –1/m，会得出完全错误的方程。始终停下来问自己：这两条线垂直吗？
ch is nonsensical. The correct sequence is: differentiate f(x) → substitute a → get slope.
另一个易错点是忘记先求导再代入 x 值。有些学生先将 x = a 代入原函数，然后再求导，这是不合理的。正确的顺序是：先对 f(x) 求导 → 代入 a → 得出斜率。
In kinematic problems, confusing displacement and total distance is frequent. Always analyse when velocity equals zero to break the interval into segments. In related rates, failing to differentiate implicitly with respect to t and omitting the chain rule leads to missing factors like dx/dt.
在运动学问题中，混淆位移和总路程的情况非常频繁。务必分析速度在何时为零，以将时间区间分成若干段。在相关变化率问题中，没有对 t 进行隐函数求导、遗漏链式法则就会丢失诸如 dx/dt 这样的因子。
Finally, not checking the domain in optimisation can lead to accepting an answer that is mathematically correct but physically impossible (e.g., a negative length). Always relate the answer back to the real-world context.
最后，优化问题中若不检查定义域，可能导致接受一个数学上正确但物理上不可能的答案（如负长度）。务必把答案放回实际情境中考量。
12. Exam-Style Questions Walkthrough | 考试题型精讲
A typical IB question might ask: The function f(x) = x³ – 3x² – d intervals where the function is increasing. This integrates several Exercise 11B skills.
一道典型的IB题目可能会这样问：给定函数 f(x) = x³ – 3x² – 9x + 5。求在 x = 2 处的切线和法线方程，求出驻点坐标及其性质，以及函数递增的区间。这融合了练习11B中的多项技能。
Step 1: f'(x) = 3x² – 6x – 9. At x = 2, f'(2) = 3(4) – 12 – 9 = –9, and point is (2, 8 – 12 – 18 + 5) = (2, –17). Tangent: y + 17 = –9(x – 2) → y = –9x + 1. Normal: slope = 1/9, equation y + 17 = (1/9)(x – 2) → y = (1/9)x – 155/9.
第1步：f'(x) = 3x² – 6x – 9。在 x = 2 处，f'(2) = 3(4) – 12 – 9 = –9，点为 (2, 8 – 12 – 18 + 5) = (2, –17)。切线：y + 17 = –9(x – 2) → y = –9x + 1。法线：斜率 = 1/9，方程 y + 17 = (1/9)(x – 2) → y = (1/9)x – 155/9。
Step 2: Stationary points: set f'(x) = 0 → 3(x² – 2x – 3) = 0 → x = –1, 3. Points: (–1, 10) and (3, –22). f''(x) = 6x – 6; f''(–1) = –12 < 0, local max; f''(3) = 12 > 0, local min.
第2步：驻点：令 f'(x) = 0 → 3(x² – 2x – 3) = 0 → x = –1, 3。对应点：(–1, 10) 和 (3, –22)。f''(x) = 6x – 6；f''(–1) = –12 < 0，局部极大值；f''(3) = 12 > 0，局部极小值。
Step 3: Increasing where f'(x) > 0 → 3(x + 1)(x – 3) > 0 → for x < – and ensures no steps are omitted. Clearly present each part, labeled, for maximum clarity.
第3步：递增区间为 f'(x) > 0 → 3(x + 1)(x – 3) > 0 → x < –1 或 x > 3。这种结构化的解答能获得方法分，并确保不遗漏任何步骤。清晰陈述每一部分，并加以标注，以求最大清晰度。
Review your work to see if the tangent
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