📚 Parametric Equations: Differentiation and Tangent/Normal Lines | 参数方程：求导与切线法线
In Edexcel A-Level Mathematics, parametric equations appear frequently across Pure papers, especially in the context of differentiation, tangents, normals, and stationary points. Mastering the technique of differentiating parametric functions is essential not only for solving exam problems but also for understanding how curves are described by a third variable, typically t or θ. This article provides a complete revision guide, with step-by-step explanations, worked examples, and exam-ready tips, all presented in both English and Chinese to support bilingual learners.
在爱德思 A-Level 数学中，参数方程在纯数试卷中出现频率很高，尤其是在求导、切线、法线和驻点问题中。掌握参数函数求导的技巧不仅有助于解答考试题目，也有助于理解曲线如何由一个第三变量（通常是 t 或 θ）来描述。本文提供完整的复习指南，包含逐步讲解、例题和应考技巧，并以中英双语呈现，帮助双语学习者。
1. What Are Parametric Equations? | 什么是参数方程？
nking x and y, we have x = f(t) and y = g(t). As t varies, the point (x, y) traces out the curve. This representation is particularly useful for curves that fail the vertical line test, such as loops or circles.
参数方程将曲线上点的坐标表示为第三个变量（通常为 t 或 θ）的函数。与直接关联 x 和 y 的单一方程不同，我们使用 x = f(t) 和 y = g(t)。随着 t 的变化，点 (x, y) 描绘出曲线。这种表示法对于不满足垂直线测试的曲线（如环或圆）特别有用。
2. The Core Differentiation Rule | 核心求导法则
ula arises because dy/dx = (dy/dt) × (dt/dx) and dt/dx = 1 / (dx/dt). Always ensure dx/dt ≠ 0 at the point of interest; otherwise the gradient may be undefined (vertical tangent).
要对参数曲线求导，需使用链式法则写成 dy/dx = (dy/dt) ÷ (dx/dt)，或者更明确地写为 dy/dx = (dy/dt) / (dx/dt)。这个公式的来源是 dy/dx = (dy/dt) × (dt/dx) 且 dt/dx = 1 / (dx/dt)。务必确保在关注点处 dx/dt ≠ 0；否则斜率可能未定义（切线垂直）。
3. Worked Example: dy/dx from a Parametric Curve | 例题：从参数曲线求 dy/dx
Given x = t³ – 2t and y = t² + 1, find dy/dx in terms of t.
First, compute dx/dt = 3t² – 2, and dy/dt = 2t.
Then, dy/dx = (2t) / (3t² – 2). This expression allows us to find the gradient at any allowable t value.
给定 x = t³ – 2t 和 y = t² + 1，用 t 表示 dy/dx。
首先计算 dx/dt = 3t² – 2 以及 dy/dt = 2t。
然后 dy/dx = (2t) / (3t² – 2)。这个表达式可用来求任意允许的 t 值处的斜率。
4. Second Derivative in Parametric Form | 参数形式的二阶导数
dt ÷ (dx/dt). In other words, differentiate the expression for dy/dx with respect to t, then divide by dx/dt. This yields the rate of change of the gradient with respect to x.
二阶导数并不是简单地对 dy/dx 关于 t 求导。正确的方法是使用 d²y/dx² = d(dy/dx)/dt ÷ (dx/dt)。换言之，先对 dy/dx 表达式关于 t 求导，再除以 dx/dt。这样就得到了斜率关于 x 的变化率。
Worked continuation: using dy/dx = 2t / (3t² – 2). Let u = 2t and v = 3t² – 2. Then du/dt = 2, dv/dt = 6t. The derivative with respect to t is (v·2 – 2t·6t) / (3t² – 2)² = (6t² – 4 – 12t²) / (3t² – 2)² = (–6t² – 4) / (3t² – 2)². Finally, d²y/dx² = [ (–6t² – 4) / (3t² – 2)² ] ÷ (3t² – 2) = (–6t² – 4) / (3t² – 2)³.
接着前面的例题：dy/dx = 2t / (3t² – 2)。令 u = 2t，v = 3t² – 2。则 du/dt = 2，dv/dt = 6t。对 t 求导得 (v·2 – 2t·6t) / (3t² – 2)² = (6t² – 4 – 12t²) / (3t² – 2)² = (–6t² – 4) / (3t² – 2)²。最后 d²y/dx² = [ (–6t² – 4) / (3t² – 2)² ] ÷ (3t² – 2) = (–6t² – 4) / (3t² – 2)³。
5. Equation of a Tangent Line | 切线方程
ₜ₀. The point of tangency is (x(t₀), y(t₀)). The tangent equation can then be written as y – y(t₀) = m (x – x(t₀)). This standard form is expected in Edexcel mark schemes.
一旦知道在特定参数值 t = t₀ 时的 dy/dx，切线的斜率 m 即为 m = (dy/dx)ₜ₌ₜ₀。切点坐标为 (x(t₀), y(t₀))。然后切线方程可写为 y – y(t₀) = m (x – x(t₀))。Edexcel 的评分标准要求使用这种标准形式。
Example: for x = t³ – 2t, y = t² + 1, at t = 1 we have x = –1, y = 2, and dy/dx = 2(1)/(3(1)² – 2) = 2/1 = 2. Tangent: y – 2 = 2(x + 1), or y = 2x + 4.
示例：对于 x = t³ – 2t，y = t² + 1，在 t = 1 时，x = –1，y = 2，dy/dx = 2(1)/(3(1)² – 2) = 2。切线：y – 2 = 2(x + 1)，即 y = 2x + 4。
6. Equation of a Normal Line | 法线方程
ne is perpendicular to the tangent at the point of contact. Its gradient mₙ satisfies mₙ = – dy/dt ≠ 0), the normal is horizontal. Use the same point to write the normal equation as y – y₁ = mₙ (x – x₁).
法线在切点处与切线垂直。其斜率 mₙ 满足 mₙ = –1 / m，前提是 m ≠ 0。如果 m = 0，切线水平，法线垂直。如果切线垂直（dx/dt = 0 而 dy/dt ≠ 0），则法线水平。使用同一个点写出法线方程：y – y₁ = mₙ (x – x₁)。
From the previous example, m = 2 so mₙ = –1/2. Normal: y – 2 = –1/2 (x + 1) → y = –1/2 x + 3/2.
由前例 m = 2 得 mₙ = –1/2。法线：y – 2 = –1/2 (x + 1) → y = –1/2 x + 3/2。
7. Stationary Points on Parametric Curves | 参数曲线的驻点
classify a stationary point, you can either evaluate the second derivative at that t value, or analyse the sign of dy/dx on either side of the parameter. Since the curve is described by t, always check that dx/dt is not simultaneously zero; otherwise, the point might be a cusp or a point of self-intersection.
驻点出现在 dy/dx = 0 处，即 dy/dt = 0 而 dx/dt ≠ 0。要对驻点进行分类，可以计算该 t 值处的二阶导数，或者分析参数两侧 dy/dx 的符号。由于曲线由 t 描述，务必检查 dx/dt 并不同时为零；否则该点可能是尖点或自交点。
Using the same curve, dy/dt = 2t = 0 ⇒ t = 0. At t = 0, dx/dt = –2 ≠ 0, so there is a stationary point at (0, 1). Second derivative at t = 0: d²y/dx² = (–4)/(–8) = 1/2 > 0, indicating a local minimum.
使用同一曲线，dy/dt = 2t = 0 ⇒ t = 0。在 t = 0 处，dx/dt = –2 ≠ 0，所以存在驻点 (0, 1)。t = 0 处的二阶导数：d²y/dx² = (–4)/(–8) = 1/2 > 0，表明是一个局部极小点。
8. Converting between Parametric and Cartesian Forms | 参数形式与笛卡尔形式的转换
t. Common techniques include expressing t in terms of x from x = f(t) and substituting into y = g(t), or using trigonometric identities when the parameter is θ. For instance, if x = a cos θ and y = b sin θ, we can use cos²θ + sin²θ = 1 to obtain x²/a² + y²/b² = 1.
考试题目有时会要求将参数方程转换为笛卡尔方程。方法涉及消去参数 t。常用技巧包括从 x = f(t) 中解出 t 关于 x 的表达式并代入 y = g(t)，或者当参数为 θ 时使用三角恒等式。例如，若 x = a cos θ 且 y = b sin θ，可利用 cos²θ + sin²θ = 1 得到 x²/a² + y²/b² = 1。
For the previous polynomial example, from x = t³ – ulation is required. Edexcel questions typically provide a hint if elimination is complex.
对于前面的多项式例子，由 x = t³ – 2t 解出 t 并不简单，但有时需要代入或巧妙的变换。如果消参过程复杂，Edexcel 题目通常会给出提示。
9. Common Pitfalls and How to Avoid Them | 常见错误与如何避免
- Forgetting the chain rule structure: ite dy/dx = dy/dt × dx/dt; the correct form is division. Remember dy/dx = (dy/dt)/(dx/dt).
- 错误记忆链式法则结构： 切勿写成 dy/dx = dy/dt × dx/dt；正确形式是除法。记住 dy/dx = (dy/dt)/(dx/dt)。
- Miscomputing the second derivative: It is not the derivative of dy/dx with respect to t alone. You must divide by dx/dt again.
- 错误计算二阶导数： 它不是 dy/dx 仅对 t 的导数，必须再次除以 dx/dt。
- Ignoring domain of t: Check for values of t that make dx/dt = 0; these correspond to vertical tangents or cusps.
- 忽略 t 的定义域： 检查使 dx/dt = 0 的 t 值；这些点对应垂直切线或尖点。
- Tangent/normal with vertical/horizontal gradients: al : Apply special cases carefully; for vertical tangent (dx/dt = 0, dy/dt ≠ 0), equation is x = constant.
- 垂直/水平斜率时的切线法线： 小心处理特殊情况；对于垂直切线（dx/dt = 0, dy/dt ≠ 0），方程为 x = 常数。
10. Exam-Style Tips for Edexcel Papers | Edexcel 试卷的应考技巧
In Edexcel Pure Mathematics papers, parametric differentiation questions often carry 6– tional expression before substituting t = value; this reduces arithmetic errors. When asked to find a tangent or normal, always quote the point explicitly. If verifying a given Cartesian equation, show clear steps of eliminating t, even if the algebra is messy.
在 Edexcel 纯数试卷中，参数求导题通常占 6–10 分。正确求导、代入参数以及写出未化简的方程都能得分。在代入 t 值之前，保持梯度为有理表达式；这可以减少计算错误。当要求求切线或法线时，务必明确写出切点。如果需要验证给定的笛卡尔方程，应清晰展示消去 t 的步骤，即使代数过程繁琐。
Below is a summary table for the key formulas:
以下是关键公式总结表：
| Concept | 概念 | Formula | 公式 |
| First derivative | 一阶导数 | dy/dx = (dy/dt) / (dx/dt) |
| Second derivative | 二阶导数 | d²y/dx² = [d(dy/dx)/dt] / (dx/dt) |
| Tangent gradient | 切线斜率 | m = dy/dx evaluated at t₀ | 代入 t₀ 计算 |
| Normal gradient | 法线斜率 | mₙ = –1 / m (if m ≠ 0) | 若 m ≠ 0 |
11. Practice Problem with Solution | 练习题与解答
A curve is defined by x = 2t² + 1, y = 4t – e t = 1. (c) Determine the coordinates of any stationary points and identify their nature.
一条曲线由 x = 2t² + 1，y = 4t – t³ 定义。(a) 用 t 表示 dy/dx。(b) 求 t = 1 处的切线方程。(c) 求任何驻点的坐标并判断其性质。
Solution:
(a) dx/dt = 4t, dy/dt = 4 – 3t². Thus dy/dx = (4 – 3t²) / (4t).
(b) At t = 1, x = 3, y = 3, dy/dx = (4 – 3)/4 = 1/4. Tangent: y – 3 = 1/4 (x – 3) → y = 1/4 x + 9/4.
(c) Stationary points when dy/dt = 0 ⇒ 4 – 3t² = 0 ⇒ t² = 4/3 ⇒ t = ±2/√3. At t = 2/√3, x = 2(4/3)+1 = 8/3+1=11/3, y = 4(2/√3) – (8/(3√3)) = (8/√3 – 8/(3√3)) = (24-8)/(3√3)=16/(3√3). Use second derivative: dy/dx = (4 – 3t²)/(4t). Differentiate with respect to t: numerator derivative = –6t, denominator derivative = 4. Using quotient rule, d/dt (dy/dx) = [4t(–6t) – (4 – 3t²)(4)]/(16t²) = [–24t² – 16 + 12t²]/(16t²) = [–12t² – 16]/(16t²) = –(12t²+16)/(16t²). Then d²y/dx² = this divided by dx/dt = 4t, giving –(12t²+16)/(64t³). At t = 2/√3, t³ positive → d²y/dx² negative, so local maximum. At t = –2/√3, by symmetry minimum nature can be checked similarly.
解答：
(a) dx/dt = 4t，dy/dt = 4 – 3t²。因此 dy/dx = (4 – 3t²) / (4t)。
(b) 在 t = 1 处，x = 3，y = 3，dy/dx = (4 – 3)/4 = 1/4。切线：y – 3 = 1/4 (x – 3) → y = 1/4 x + 9/4。
(c) 驻点满足 dy/dt = 0 ⇒ 4 – 3t² = 0 ⇒ t² = 4/3 ⇒ t = ±2/√3。t = 2/√3 时，x = 2(4/3)+1 = 11/3，y = 4(2/√3) – 8/(3√3) = 16/(3√3)。使用二阶导数：dy/dx = (4 – 3t²)/(4t)。对 t 求导：分子导数 = –6t，分母导数 = 4。用商法则得 d/dt (dy/dx) = [–24t² – 16 + 12t²]/(16t²) = –(12t²+16)/(16t²)。然后 d²y/dx² = 该式除以 dx/dt = 4t，得 –(12t²+16)/(64t³)。在 t = 2/√3 处，t³ 为正，二阶导数为负，所以是局部极大点。在 t = –2/√3 处，由对称性可知为极小点。
12. Summary and Final Revision Notes | 总结与复习要点
he division structure: dy/dx = (dy/dt) / (dx/dt). The second derivative requires an extra division by dx/dt. When dealing with tangents and normals, compute the gradient at the given parameter value and use the point-slope form. Finally, check for vertical tangents by examining where dx/dt = 0. With consistent practice, these problems become some of the most predictable and high-scoring on your Edexcel exam.
掌握参数求导归结为熟练运用链式法则和细致的代数处理。始终记住除法的结构：dy/dx = (dy/dt) / (dx/dt)。二阶导数需要额外除以 dx/dt。处理切线和法线时，计算给定参数值处的斜率并使用点斜式。最后，通过检查 dx/dt = 0 来找出垂直切线。通过持续练习，这类题目将成为 Edexcel 考试中最为可预测且得高分的部分。
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