📚 Parametric Differentiation: Edexcel A-Level Pure Maths | 参数方程求导：爱德思 A-Level 纯数学
Parametric differentiation is a central Pure Mathematics topic in the Edexcel A-Level specification. It gives you a reliable method for finding gradients, tangents, normals and stationary points when a curve is defined by x = f(t) and y = g(t). This article covers the key formulas, common exam traps, and worked applications.
参数方程求导是爱德思 A-Level 纯数学大纲中的核心内容。当曲线由 x = f(t) 和 y = g(t) 定义时，它可以让你可靠地求出斜率、切线、法线和驻点。本文涵盖关键公式、常见考试陷阱以及典型例题应用。
1. Understanding Parametric Equations | 理解参数方程
Parametric equations express the coordinates x and y separately in terms of a third variable, usually t. The variable t often represents time, but it can also be an angle or any other parameter.
参数方程用第三个变量（通常是 t）分别表示坐标 x 和 y。变量 t 通常代表时间，但也可以是角度或其他任意参数。
Each value of t produces one point (x, y). As t changes, the point moves along the curve, so the parametric form is especially useful for describing loops, circles and motion paths that cannot be written as a single function y = f(x).
每个 t 值都会产生一个点 (x, y)。当 t 变化时，点沿曲线运动，因此参数形式特别适用于描述无法写成 y = f(x) 单一函数的回路、圆和运动轨迹。
x = f(t), y = g(t)
2. From Parametric to Cartesian Form | 从参数形式到笛卡尔形式
To see the shape of a curve, it often helps to eliminate the parameter t. Solve one equation for t, then substitute into the other equation.
为了看清曲线形状，通常需要消去参数 t。先从一个方程中解出 t，再代入另一个方程。
For example, if x = t² and y = 2t, then t = y ÷ 2, so x = (y ÷ 2)². The Cartesian equation is therefore 4x = y², which represents a sideways parabola.
例如，若 x = t² 且 y = 2t，则 t = y ÷ 2，所以 x = (y ÷ 2)²。因此笛卡尔方程为 4x = y²，表示一条开口向右的抛物线。
Always check for domain or range restrictions. If t ≥ 0, then x ≥ 0, so the Cartesian equation only describes the right-hand branch.
始终要检查定义域或值域限制。若 t ≥ 0，则 x ≥ 0，因此笛卡尔方程只描述右侧分支。
| Parametric form | Eliminate t | Cartesian form |
|---|---|---|
| x = t², y = 2t | t = y ÷ 2 | 4x = y² |
| x = 2t + 1, y = t³ | t = (x − 1) ÷ 2 | 8y = (x − 1)³ |
| x = a cos t, y = a sin t | cos² t + sin² t = 1 | x² + y² = a² |
3. The Key Formula for dy/dx | 求 dy/dx 的核心公式
The chain rule connects the rate of change of y with respect to t and x with respect to t. Provided dx/dt ≠ 0, the gradient of a parametric curve is given by:
链式法则将 y 对 t 的变化率与 x 对 t 的变化率联系起来。只要 dx/dt ≠ 0，参数曲线的斜率就可以由下式给出：
dy/dx = (dy/dt) ÷ (dx/dt) = ẏ ÷ ẋ
In words, differentiate y with respect to t, differentiate x with respect to t, then divide the first result by the second.
也就是说，先对 y 关于 t 求导，再对 x 关于 t 求导，然后用第一个结果除以第二个结果。
Geometrically, dx/dt is the horizontal velocity component and dy/dt is the vertical velocity component. Their ratio dy/dx is the slope of the tangent to the curve.
从几何意义上看，dx/dt 是水平速度分量，dy/dt 是竖直速度分量。它们的比值 dy/dx 就是曲线切线的斜率。
4. Step-by-Step Differentiation Strategy | 分步求导策略
When Edexcel asks you to find dy/dx from parametric equations, use a clear four-step method to avoid sign and division errors.
当爱德思考试要求你根据参数方程求 dy/dx 时，使用清晰的四步方法可以避免符号和除法错误。
Step 1: Differentiate x with respect to t to obtain dx/dt.
第 1 步：对 x 关于 t 求导，得到 dx/dt。
Step 2: Differentiate y with respect to t to obtain dy/dt.
第 2 步：对 y 关于 t 求导，得到 dy/dt。
Step 3: Write dy/dx = (dy/dt) ÷ (dx/dt).
第 3 步：写出 dy/dx = (dy/dt) ÷ (dx/dt)。
Step 4: Simplify fully and state any values of t for which the gradient is undefined, usually when dx/dt = 0.
第 4 步：完全化简，并说明哪些 t 值使斜率无定义，通常是在 dx/dt = 0 时。
5. Second Derivative d²y/dx² | 二阶导数 d²y/dx²
The second derivative measures how the gradient itself is changing. For parametric curves it is not simply d²y/dt² divided by d²x/dt².
二阶导数衡量的是斜率本身如何变化。对于参数曲线，它并不是简单地将 d²y/dt² 除以 d²x/dt²。
The correct formula comes from differentiating dy/dx with respect to x using the chain rule:
正确的公式来自用链式法则对 dy/dx 关于 x 求导：
d²y/dx² = d/dx(dy/dx) = [d/dt(dy/dx)] ÷ (dx/dt)
So first find dy/dx in terms of t. Then differentiate that expression with respect to t, and finally divide by dx/dt.
因此先求出 dy/dx 关于 t 的表达式。然后对该表达式关于 t 求导，最后再除以 dx/dt。
Example: if x = t² and y = t³, then dx/dt = 2t and dy/dt = 3t², so dy/dx = 3t ÷ 2. Differentiating with respect to t gives d/dt(3t ÷ 2) = 3 ÷ 2. Therefore d²y/dx² = (3 ÷ 2) ÷ (2t) = 3 ÷ (4t).
例如：若 x = t²，y = t³，则 dx/dt = 2t，dy/dt = 3t²，所以 dy/dx = 3t ÷ 2。对该式关于 t 求导得 d/dt(3t ÷ 2) = 3 ÷ 2。因此 d²y/dx² = (3 ÷ 2) ÷ (2t) = 3 ÷ (4t)。
6. Tangents and Normals | 切线与法线
To find the tangent at a point where t = t₀, first calculate the gradient m = dy/dx at t = t₀. Then find the coordinates x₁ = x(t₀) and y₁ = y(t₀).
要在 t = t₀ 处求切线，首先计算该点的斜率 m = dy/dx。然后求出坐标 x₁ = x(t₀) 和 y₁ = y(t₀)。
Substitute these into the straight-line equation y − y₁ = m(x − x₁).
将这些值代入直线方程 y − y₁ = m(x − x₁)。
The normal is perpendicular to the tangent, so its gradient is −1 ÷ m when m ≠ 0. If the tangent is horizontal, the normal is vertical.
法线垂直于切线，因此当 m ≠ 0 时，法线斜率为 −1 ÷ m。如果切线是水平的，则法线是竖直的。
When dy/dx = 0, the tangent is horizontal. When dx/dt = 0 but dy/dt ≠ 0, the tangent is vertical, and the gradient is undefined.
当 dy/dx = 0 时，切线为水平。当 dx/dt = 0 但 dy/dt ≠ 0 时，切线为竖直，斜率无定义。
7. Stationary Points and Turning Points | 驻点与极值点
A stationary point occurs when the gradient dy/dx equals zero. Since dy/dx = (dy/dt) ÷ (dx/dt), this usually happens when dy/dt = 0 and dx/dt ≠ 0.
驻点出现在斜率 dy/dx 等于零时。由于 dy/dx = (dy/dt) ÷ (dx/dt)，这通常发生在 dy/dt = 0 且 dx/dt ≠ 0 时。
Solve dy/dt = 0 to find the relevant t values, then substitute back into x(t) and y(t) to get the coordinates.
解方程 dy/dt = 0 求出相应的 t 值，然后代回 x(t) 和 y(t) 得到坐标。
To determine the nature of a stationary point, use the second derivative test: if d²y/dx² > 0, the point is a local minimum; if d²y/dx² < 0, it is a local maximum.
要判断驻点的性质，可使用二阶导数检验：若 d²y/dx² > 0，该点为局部最小值；若 d²y/dx² < 0，该点为局部最大值。
If both dx/dt and dy/dt are zero at the same t value, the curve may have a cusp or a more complicated behaviour, and the standard test may fail.
如果 dx/dt 和 dy/dt 在同一 t 值处都为零，曲线可能存在尖点或更复杂的行为，标准检验可能失效。
8. Sketching Parametric Curves | 参数曲线作图
To sketch a parametric curve, create a table of values for t, x and y. Plot the points and indicate the direction of increasing t with an arrow.
要绘制参数曲线，可以列出 t、x 和 y 的取值表。描出各点，并用箭头标出 t 增大的方向。
Use the Cartesian equation, if available, to identify the overall shape, such as a parabola, ellipse or hyperbola.
如果可以得到笛卡尔方程，就用它来识别整体形状，例如抛物线、椭圆或双曲线。
Find axis intercepts by setting x = 0 and y = 0 separately, then solving for t. Also mark any points where the tangent is horizontal or vertical.
通过分别令 x = 0 和 y = 0 并解出 t 来求坐标轴截距。同时标出切线为水平或竖直的点。
9. Modelling with Parametric Equations | 参数方程建模
In mechanics and applied mathematics, parametric equations often describe projectile motion. For a particle projected with initial speed u at angle θ to the horizontal:
在力学和应用数学中，参数方程通常描述抛体运动。对于以初速度 u 与水平面成 θ 角抛出的质点：
x = u cos θ t, y = u sin θ t − ½gt²
Here t is time, g is the acceleration due to gravity, and the curve is the trajectory of the projectile.
这里 t 是时间，g 是重力加速度，曲线就是抛体的轨迹。
Parametric differentiation gives dy/dx, which represents the slope of the trajectory at a given time. Edexcel exam questions sometimes combine this with velocity and acceleration vectors.
参数求导给出 dy/dx，它表示给定时刻轨迹的斜率。爱德思考试题有时会将其与速度矢量和加速度矢量结合起来考查。
10. Worked Example: Tangent and Second Derivative | 例题精讲：切线与二阶导数
Let x = 3t² + 1 and y = 2t³ − t. Find dy/dx as a function of t, then find the equation of the tangent at t = 1.
设 x = 3t² + 1，y = 2t³ − t。求 dy/dx 关于 t 的函数，并求出 t = 1 处的
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