Parametric Differentiation for Edexcel A-Level Mathematics | Edexcel A-Level 数学：参数微分

📚 Parametric Differentiation for Edexcel A-Level Mathematics | Edexcel A-Level 数学：参数微分

nstead, Edexcel Pure Year 2 introduces parametric equations, where x and y are each written in terms of a third variable, usually t. Parametric differentiation allows us to find the gradient of such a curve without first eliminating t.

在 A-Level 数学中，许多曲线很难直接写成 y = f(x) 的形式。因此 Edexcel Pure Year 2 引入了参数方程：x 与 y 分别用第三个变量 t 表示。参数微分让我们无需先消去 t，就能求出这类曲线的斜率。

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1. Cartesian and Parametric Equations | 笛卡尔方程与参数方程

a parameter, such as t, to define x and y separately. A simple example is x = t², y = 2t.

笛卡尔方程直接描述 x 与 y 之间的关系，例如 y = x² − 4x 或 x² + y² = 25。参数方程则利用参数 t 分别定义 x 与 y，例如 x = t²，y = 2t。

are much easier to study in parametric form, especially when their motion depends on time.

同一条曲线通常可以用两种形式表示。从 x = t²，y = 2t 中消去 t，可得 t = y/2，因此 x = (y/2)² = y²/4，即 y² = 4x。有些曲线在参数形式下更容易研究，特别是在运动随时间变化的问题中。

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2. The Key Rule for Parametric Differentiation | 参数微分的核心法则

/dt = dy/dx × dx/dt. Rearranging gives the central formula for parametric differentiation:

设 x = f(t)，y = g(t)。根据链式法则，dy/dt = dy/dx × dx/dt。整理后得到参数微分的核心公式：

dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0

may have a vertical tangent, and the gradient cannot be found using this method directly.

当 dx/dt ≠ 0 时，该公式成立。如果 dx/dt = 0，曲线可能具有竖直切线，此时不能直接用这个方法求斜率。

Expression	Meaning	含义
dx/dt	Rate of change of x with respect to t	x 关于 t 的变化率
dy/dt	Rate of change of y with respect to t	y 关于 t 的变化率
dy/dx	Gradient of the curve	曲线的斜率

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3. Worked Example 1: Polynomial Parametric Curve | 示例 1：多项式参数曲线

Find dy/dx for the curve x = t² + 1, y = t³ − t, giving your answer in terms of t.

求曲线 x = t² + 1，y = t³ − t 的 dy/dx，并将答案用 t 表示。

First differentiate each equation with respect to t:

首先分别对 t 求导：

dx/dt = 2t, dy/dt = 3t² − 1

Now apply the parametric differentiation formula:

接下来代入参数微分公式：

dy/dx = (3t² − 1) / (2t)

s us that the tangent at t = 2 rises by 11 units for every 4 units of horizontal movement.

当 t = 2 时，斜率为 (3 × 4 − 1) / (2 × 2) = 11/4。这意味着在 t = 2 处，切线每水平移动 4 个单位，竖直上升 11 个单位。

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4. Worked Example 2: Trigonometric Parametric Curve | 示例 2：三角参数曲线

Consider the circle defined by x = 2cos t, y = 2sin t. Differentiate to find dy/dx.

考虑圆 x = 2cos t，y = 2sin t。求 dy/dx。

We have dx/dt = −2sin t and dy/dt = 2cos t. Therefore:

我们有 dx/dt = −2sin t，dy/dt = 2cos t。因此：

dy/dx = 2cos t / (−2sin t) = −cos t / sin t = −cot t

At t = π/4, the point on the circle is (√2, √2 1. This matches the fact that the radius to that point has slope 1, and the tangent is perpendicular to it.

当 t = π/4 时，圆上的点为 (√2, √2)。该点的斜率是 −cot(π/4) = −1。这正好符合“半径斜率为 1，切线与半径垂直”的性质。

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5. Equation of a Tangent | 切线方程

ormula y − y₀ = m(x − x₀), where (x₀, y₀) is the point on the curve and m is the gradient.

一旦求出某个参数值处的 dy/dx，就可以用直线公式 y − y₀ = m(x − x₀) 求出切线方程，其中 (x₀, y₀) 是曲线上的点，m 是斜率。

For the circle x = 2cos t, y = 2sin t at t = π/4, the point is (√2, √2) and m = −1. The tangent equation is:

对于圆 x = 2cos t，y = 2sin t，当 t = π/4 时，点为 (√2, √2)，m = −1。切线方程为：

y − √2 = −(x − √2)

Simplifying gives x + y = 2√2. This is a neat way to test whether your differentiated expression is correct: the tangent line should touch the curve at exactly one point locally.

化简得到 x + y = 2√2。这是一个检验微分结果是否正确的有效方法：切线应在局部与曲线恰好相切于一点。

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6. Equation of a Normal | 法线方程

The normal to a curve is perpendicular to the tangent. If the tangent gradient is m, then the normal gradient is −1/m, provided m ≠ 0.

法线与切线互相垂直。如果切线斜率为 m，则法线斜率为 −1/m，前提是 m ≠ 0。

4, the tangent gradient is −1, so the normal gradient is 1. The normal passes through the point (√2, √2):

继续使用上面圆的例子，在 t = π/4 处，切线斜率为 −1，因此法线斜率为 1。法线经过点 (√2, √2)：

y − √2 = 1(x − √2)

circle, the normal always passes through the centre, so the line y = x is exactly correct.

化简得 y = x。对于圆来说，法线一定经过圆心，因此 y = x 是正确的。

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7. Stationary Points | 驻点

dy/dx = (dy/dt)/(dx/dt), this means dy/dt = 0, provided dx/dt ≠ 0 at the same value of t.

参数曲线上的驻点出现在 dy/dx = 0 时。由于 dy/dx = (dy/dt)/(dx/dt)，所以当 dx/dt ≠ 0 时，驻点对应于 dy/dt = 0。

dt = 0 gives t = 1. At this value, dx/dt = 2, so the curve has a genuine stationary point.

考虑 x = t²，y = t² − 2t。则 dx/dt = 2t，dy/dt = 2t − 2。令 dy/dt = 0，得 t = 1。此时 dx/dt = 2，所以曲线在该处确实是一个驻点。

nd y = 1² − 2 = −1, so the point is (1, −1). To classify it, we use the second derivative.

该驻点坐标为 x = 1² = 1，y = 1² − 2 = −1，即点 (1, −1)。要判断它是极大值还是极小值，需要使用二阶导数。

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8. The Second Derivative | 二阶导数

The second derivative is obtained by differentiating dy/dx with respect to t and then dividing by dx/dt:

二阶导数的求法是：先对 dy/dx 关于 t 求导，再除以 dx/dt：

d²y/dx² = [d/dt (dy/dx)] / (dx/dt)

This is not equal to (d²y/dt²)/(d²x/dt²). This is a common error, and examiners design questions to catch students who use the wrong version.

注意：d²y/dx² 不等于 (d²y/dt²)/(d²x/dt²)。这是一个常见的错误，出题人经常设计陷阱题来考查学生是否真正理解这个公式。

t² − 2t, we have dy/dx = (t − 1)/t. Then d/dt(dy/dx) = 1/t². Dividing by dx/dt = 2t gives:

对于曲线 x = t²，y = t² − 2t，有 dy/dx = (t − 1)/t。因此 d/dt(dy/d

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