📚 Differentiation of Parametric Equations | 参数方程求导
ribe both x and y in terms of a third variable, typically t or θ, called a parameter. To analyse the gradient, find tangents or determine stationary points for such curves, we need a technique known as parametric differentiation. This article covers the Edexcel A-Level method for finding dy/dx and d²y/dx² from parametric equations, including key formulae, worked examples and common pitfalls.
在 A-Level 数学中，许多有趣的曲线（如圆、椭圆和摆线）很难直接写成 y = f(x) 的形式。我们通常使用第三个变量 t 或 θ（称为参数）来同时描述 x 和 y。要分析这类曲线的斜率、求切线或确定驻点，就需要用到参数求导法。本文涵盖 Edexcel A-Level 大纲中由参数方程求一阶导数 dy/dx 和二阶导数 d²y/dx² 的方法，包括核心公式、例题和常见错误。
1. What Are Parametric Equations? | 什么是参数方程？
² − 1, y = t³ + 2t. As t varies, the point (x, y) traces out a curve. Parametric forms are extremely useful for modelling motion, where t represents time, and for describing curves that fail the vertical line test.
参数曲线通过一个独立参数（通常用 t）来表示各点的 x 坐标和 y 坐标。例如 x = t² − 1, y = t³ + 2t。当 t 变化时，点 (x, y) 会描绘出一条曲线。参数形式在建模运动（t 表示时间）以及描述不满足垂直线检验的曲线时非常有用。
In Edexcel A-Level, you will encounter parametric equations involving polynomials, trigonometric functions, exponentials and sometimes rational functions. The parameter may be denoted t, θ or another letter, but the differentiation technique remains the same.
在 Edexcel A-Level 考试中，你会遇到含多项式、三角函数、指数函数甚至有理函数的参数方程。参数可能用 t、θ 或其他字母表示，但求导方法不变。
2. Why Standard Differentiation Fails | 为何常规求导不再适用
directly apply rules like the chain, product or quotient rule. However, in a parametric representation, y is not explicitly a function of x; both x and y are functions of t. To find the gradient of the curve, we must relate the small changes in y and x via the parameter. This introduces the classic parametric differentiation formula.
当曲线以 y = f(x) 给出时，我们可以直接使用链式法则、乘法法则或除法法则。但在参数表示中，y 并不是 x 的显函数；x 和 y 都是 t 的函数。要找到曲线的斜率，我们必须通过参数把 y 和 x 的微小变化关联起来，这就引出了经典的参数求导公式。
ly, and then use the chain rule in the form dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0.
本质上，我们先分别计算 dy/dt 和 dx/dt，然后利用链式法则的形式 dy/dx = (dy/dt) / (dx/dt)，要求 dx/dt ≠ 0。
3. The Fundamental Formula for dy/dx | 一阶导数 dy/dx 的基本公式
are differentiable functions of t, then the derivative of y with respect to x is given by:
若 x = x(t) 和 y = y(t) 都是 t 的可导函数，则 y 关于 x 的导数由下式给出：
dy/dx = (dy/dt) ÷ (dx/dt)
This result can be remembered as “derivative of y with respect to t over derivative of x with respect to t” ing gives the required expression. This formula works as long as dx/dt is not zero at the point of interest.
该结果可记作“y 对 t 的导数除以 x 对 t 的导数”。它直接来自链式法则：dy/dt = dy/dx · dx/dt，移项即可得到所需表达式。只要在所讨论的点 dx/dt 不为零，该公式就成立。
4. Step-by-Step Method for Parametric Differentiation | 参数求导的逐步方法
Follow these steps when tackling any parametric differentiation problem in Edexcel A-Level exams:
解答 Edexcel A-Level 考试中的参数求导题时，可按以下步骤进行：
- Step 1: Differentiate x with respect to t to find dx/dt.
- 步骤 1：对 x 关于 t 求导，得到 dx/dt。
- Step 2: Differentiate y with respect to t to find dy/dt.
- 步骤 2：对 y 关于 t 求导，得到 dy/dt。
- Step 3: Write dy/dx = (dy/dt) / (dx/dt).
- 步骤 3：写出 dy/dx = (dy/dt) / (dx/dt)。
- Step 4: Simplify the expression, possibly cancelling common factors of t.
- 步骤 4：化简表达式，尽可能约去 t 的公因子。
- Step 5: (If required) Evaluate dy/dx at a specific value of t or a given point.
- 步骤 5：（如果需要）代入特定的 t 值或给定点计算 dy/dx。
Always state the condition dx/dt ≠ 0 when presenting your final derivative, as this is where the formula is valid.
在给出最终导数时，务必注明条件 dx/dt ≠ 0，因为这是公式成立的前提。
5. Example 1 – Basic Polynomial Parametric Curve | 例题 1：基础多项式参数曲线
Given the parametric equations: x = t² − 2t, y = t³ − 3t. Find dy/dx in terms of t.
给定参数方程：x = t² − 2t, y = t³ − 3t。求用 t 表示的 dy/dx。
ce the cancellation of (t − 1), which highlights why we exclude the point where dx/dt = 0.
首先，计算 dx/dt = 2t − 2 = 2(t − 1)。接着，dy/dt = 3t² − 3 = 3(t² − 1) = 3(t − 1)(t + 1)。于是 dy/dx = [3(t − 1)(t + 1)] / [2(t − 1)] = (3/2)(t + 1)，其中 t ≠ 1。注意 (t − 1) 被约去，这正好说明了为什么要排除 dx/dt = 0 的点。
This simple example illustrates that the parametric derivative can often be simplified into a neat linear or rational expression, making further calculations (like finding tangents) straightforward.
这个简单例子说明，参数导数通常可以化简为一个整齐的线性或有理表达式，从而使后续计算（如求切线）变得简单。
6. Example 2 – Trigonometric Parametric Equations | 例题 2：含三角函数的参数方程
= a cos³ t, y = a sin³ t, where 0 ≤ t ≤ π/2. Find the gradient of the curve in terms of t.
某曲线由 x = a cos³ t, y = a sin³ t 定义，其中 0 ≤ t ≤ π/2。求用 t 表示的曲线斜率。
cos t) / (−3a cos² t sin t) = − (sin t) / (cos t) = −tan t, after cancelling common factors (3a, cos t, sin t). The derivative exists for t where cos t ≠ 0 and sin t ≠ 0 (i.e., t non-zero and non-right angle in this interval).
对 x 求导：dx/dt = a · 3 cos² t · (−sin t) = −3a cos² t sin t。对 y 求导：dy/dt = a · 3 sin² t · (cos t) = 3a sin² t cos t。因此，dy/dx = (3a sin² t cos t) / (−3a cos² t sin t) = − (sin t) / (cos t) = −tan t，约去了公因子 3a、cos t、sin t。导数在 cos t ≠ 0 且 sin t ≠ 0（即该区间内 t 不为 0 且不为直角）时存在。
Parametric curves involving trigonometric functions often yield elegant derivatives like −tan t, which immediately tells us how the slope behaves along the curve.
涉及三角函数的参数曲线常常得出像 −tan t 这样简洁的导数，直接告诉我们曲线上斜率的变化规律。
7. Finding the Equation of a Tangent | 求切线方程
valuated at t = t₀. The coordinates of the point are (x(t₀), y(t₀)). The tangent line is then y − y(t₀) = m (x − x(t₀)).
一旦求出 dy/dx，就可以确定对应参数值 t₀ 的点处的切线方程。斜率 m 为 t = t₀ 时的 dy/dx 值。该点坐标为 (x(t₀), y(t₀))。切线方程即为 y − y(t₀) = m (x − x(t₀))。
o dy/dx = 2/(2t) = 1/t. At t = 1, gradient m = 1. Point: (1, 2). Tangent: y − 2 = 1(x − 1) ⇒ y = x + 1.
对于曲线 x = t², y = 2t，求 t = 1 处的切线。dx/dt = 2t, dy/dt = 2，所以 dy/dx = 2/(2t) = 1/t。t = 1 时，斜率 m = 1。点为 (1, 2)。切线为 y − 2 = 1(x − 1) ⇒ y = x + 1。
in exam questions, often combined with finding the normal line as well (gradient = −1/m).
这一步在考题中十分常见，通常还会结合求法线（斜率为 −1/m）。
8. The Second Derivative d²y/dx² | 二阶导数 d²y/dx²
To determine concavity or locate points of inflection, we need the second derivative. With parametric equations, d²y/dx² is NOT simply (d²y/dt²) / (d²x/dt²). The correct formula, derived from the chain rule, is:
要确定凹凸性或找拐点，就需要二阶导数。在参数方程中，d²y/dx² 不能简单地由 (d²y/dt²) / (d²x/dt²) 得到。正确的公式由链式法则导出，为：
d²y/dx² = (d/dt [dy/dx]) ÷ (dx/dt)
dx]) / (dx/dt). This means we first obtain dy/dx as a function of t, differentiate it with respect to t, and then divide by dx/dt. A common mistake is to forget the division by dx/dt.
也就是说，我们首先求出用 t 表示的 dy/dx，并将其对 t 求导，再除以 dx/dt。常见的错误是忘记除以 dx/dt。
It is also possible to express d²y/dx² using the quotient rule on dx/dt and dy/dt, but the stepwise approach is safer and recommended by Edexcel mark schemes.
也可以对 dx/dt 和 dy/dt 使用商法则来推导 d²y/dx²，但逐步计算的方法更安全，也是 Edexcel 评分标准中推荐的做法。
9. Example 3 – Second Derivative Calculation | 例题 3：二阶导数计算
For the curve x = t² + 1, y = t³ − 3t, find d²y/dx² in terms of t.
对于曲线 x = t² + 1, y = t³ − 3t，求用 t 表示的 d²y/dx²。
u/dt = d/dt [ (3t² − 3)/(2t) ] = [ (6t)(2t) − (3t² − 3)(2) ] / (4t²) = [12t² − 6t² + 6] / (4t²) = (6t² + 6) / (4t²) = (6(t² + 1))/(4t²) = (3(t² + 1))/(2t²). Then d²y/dx² = (du/dt) / (dx/dt) = [3(t² + 1)/(2t²)] / (2t) = 3(t² + 1)/(4t³).
首先，dx/dt = 2t，dy/dt = 3t² − 3。于是 dy/dx = (3t² − 3) / (2t) = (3(t² − 1))/(2t)。接下来，令 u = dy/dx = (3t² − 3)/(2t)。对 t 求导： (4t²) = [12t² − 6t² + 6] / (4t²) = (6t² + 6)/(4t²) = (6(t² + 1))/(4t²) = (3(t² + 1))/(2t²)。然后 d²y/dx² = (du/dt) / (dx/dt) = [3(t² + 1)/(2t²)] / (2t) = 3(t² + 1)/(4t³)。
This method can be applied to any parametric system, but careful algebraic simplification is key. For trigonometric cases, we often differentiate dy/dx using standard trig derivatives.
该方法适用于任何参数系统，但关键是细心的代数化简。对于含三角函数的情况，我们通常用标准三角导数对 dy/dx 求导。
10. Points of Horizontal and Vertical Tangency | 水平与铅垂切线的点
0. A vertical tangent occurs where dx/dt = 0 but dy/dt ≠ 0. If both derivatives vanish simultaneously, we may have a cusp or a point of self-intersection requiring further limit analysis.
参数曲线通过考察 dx/dt 和 dy/dt，能清晰识别特殊点。水平切线出现在 dy/dt = 0 且 dx/dt ≠ 0 处。铅垂切线则出现在 dx/dt = 0 且 dy/dt ≠ 0 处。若两个导数同时为零，可能出现尖点或自交点，需要进一步进行极限分析。
t = 1, dy/dt = 2 ≠ 0, so vertical tangent at (−2,1). At t = −1, vertical tangent at (2,1).
对于 x = t³ − 3t, y = t²，有 dx/dt = 3t² − 3，dy/dt = 2t。令 dy/dt = 0 得 t = 0（水平切线在 (−3,0)）。令 dx/dt = 0 得 t = ±1。t = 1 时 dy/dt = 2 ≠ 0，故铅垂切线在 (−2,1)；t = −1 时铅垂切线在 (2,1)。
Edexcel exam questions frequently ask for the coordinates of points where the tangent is parallel to the x-axis or y-axis, testing this exact concept.
Edexcel 试题经常会要求找出切线平行于 x 轴或 y 轴的点坐标，直接考查这一概念。
11. Common Mistakes and How to Avoid Them | 常见错误与避免方法
- Mistake 1: Using (dy/dt) / (dx/dt) but forgetting to check dx/dt ≠ 0, leading to invalid expressions at certain points.
- 错误 1：使用 (dy/dt) / (dx/dt) 时忘记检查 dx/dt ≠ 0，导致在特定点得到无效表达式。
- Mistake 2: For the second derivative, incorrectly writing (d²y/dt²) / (d²x/dt²) instead of using the proper chain rule formula.
- 错误 2：对于二阶导数，错误地写成 (d²y/dt²) / (d²x/dt²)，而不是使用正确的链式法则公式。
- Mistake 3: etric parametric functions; for instance, when x = a cos t, dx/dt = −a sin t, not a cos t.
- 错误 3：混淆三角参数函数的求导，例如 x = a cos t 时 dx/dt = −a sin t，而不是 a cos t。
- Mistake 4: Simplifying dy/dx too early before finding d²y/dx², which often makes differentiation messier. Keep dy/dx as a clear function of t before differentiating again.
- 错误 4：在求 d²y/dx² 之前过早化简 dy/dx，往往使后续求导变得更繁琐。再次求导前应将 dy/dx 保留为清晰的 t 函数。
- Mistake 5: Substituting the point coordinates directly into dy/dx before finding the parameter value that corresponds to that point.
- 错误 5：在找到点对应的参数值之前，直接把点坐标代入 dy/dx。
By systematically writing dx/dt and dy/dt and following the two-step process for the second derivative, many of these pitfalls can be avoided.
只要系统地写出 dx/dt 和 dy/dt，并遵循求二阶导数的两步法，多数错误都能避免。
12. Exam-Style Question and Solution | 考题风格例题与解答
The curve C has parametric equations x = 2t + 1, y = t² − 4t + 3.
曲线 C 的参数方程为 x = 2t + 1, y = t² − 4t + 3。
oint where t = 2. (c) Find the value of t for which the tangent is parallel to the x-axis.
(a) 求用 t 表示的 dy/dx，并化简答案。(b) 由此求 C 在 t = 2 处的切线方程。(c) 求使切线平行于 x 轴的 t 值。
Solution: At t = 2, x = 5, y = 4 − 8 + 3 = −1. Gradient m = 2 − 2 = 0. Tangent: y − (−1) = 0(x − 5) ⇒ y = −1 (horizontal line). (c) Tangent parallel to x-axis ⇒ dy/dx = 0 ⇒ t − 2 = 0 ⇒ t = 2.
解答：(a) dx/dt = 2, dy/dt = 2t − 4。故 dy/dx = (2t − 4)/2 = t − 2。(b) t = 2 时，x = 5, y = 4 − 8 + 3 = −1。斜率 m = 2 − 2 = 0。切线为 y − (−1) = 0(x − 5) ⇒ y = −1（水平线）。(c) 切线平行于 x 轴 ⇒ dy/dx = 0 ⇒ t − 2 = 0 ⇒ t = 2。
This question illustrates how parametric differentiation feeds directly into coordinate geometry, a frequent theme in Edexcel exams.
此题展示了参数求导如何直接与坐标几何衔接，这是 Edexcel 考试中的常见主题。
Published by TutorHao | Edexcel A-Level 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