Parametric Equations | 参数方程

📚 Parametric Equations | 参数方程

Parametric equations are equations where the coordinates x and y are both expressed as functions of a third variable, usually t, known as the parameter. In the Edexcel GCSE Mathematics (Higher Tier) curriculum, parametric equations appear within the topic of differentiation and coordinate geometry, enabling students to describe curves that are not easily expressed as simple y = f(x) functions. Understanding parametric equations enhances problem-solving skills in calculus and kinematics, and is fundamental for tackling more advanced pure mathematics.

参数方程是指 x 和 y 坐标都表示成第三个变量（通常为 t，称为参数）的函数的方程。在 Edexcel GCSE 数学（高等级）大纲中，参数方程出现在微分和坐标几何的内容中，可以描述那些难以用简单的 y = f(x) 函数表达的曲线。理解参数方程能够提高在微积分和运动学中的问题解决能力，也是处理更高阶纯数学内容的基础。

1. Introduction to Parametric Equations | 参数方程简介

A parametric equation defines a curve using a separate variable t. For example, x = t², y = 2t. As t varies, the point (x, y) traces out a curve. Unlike Cartesian equations of the form y = f(x), parametric forms can represent loops, self-intersections and other complex paths without breaking the function rule. This makes them powerful tools in both pure mathematics and applied contexts such as projectile motion.

参数方程用一个独立变量 t 定义曲线。例如 x = t², y = 2t。随着 t 变化，点 (x, y) 描绘出一条曲线。与形如 y = f(x) 的笛卡儿方程不同，参数形式可以表示环形、自相交以及其他复杂路径，而无需打破函数的定义。这使得它们在纯数学和抛体运动等应用情境中成为强大的工具。

In Edexcel GCSE Higher, you are expected to be able to differentiate parametric equations and find equations of tangents and normals. You will also need to convert between parametric and Cartesian forms and apply these skills to describe the gradient and behaviour of curves.

在 Edexcel GCSE 高等级中，你需要能够对参数方程求导并求出切线和法线的方程。你还需要能够在参数形式和笛卡儿形式之间进行转换，并应用这些技能来描述曲线的斜率和性态。

2. Eliminating the Parameter to Find the Cartesian Equation | 消去参数得到笛卡儿方程

Often it is useful to convert parametric equations into a single Cartesian equation linking x and y directly. To eliminate the parameter t, rearrange one of the parametric equations to express t in terms of x (or y) and then substitute this expression into the other equation. If the parametric equations involve trigonometric functions, use identities such as cos²θ + sin²θ = 1 to eliminate t.

通常把参数方程转化为直接关联 x 与 y 的单一笛卡儿方程很有用。要消去参数 t，可将其中一个参数方程变形，用 x（或 y）表示 t，然后将此表达式代入另一个方程。如果参数方程包含三角函数，可利用恒等式 cos²θ + sin²θ = 1 来消去 t。

For instance, given x = 2t + 1, y = 4t², from x = 2t + 1 we obtain t = (x – 1)/2. Substituting into y yields y = 4[(x – 1)/2]² = (x – 1)². Always note the domain restrictions implied by the original parametric definitions; for example, if t ≥ 0, then x ≥ 1 and y ≥ 0, so the Cartesian equation (x – 1)² = y is only valid for that restricted range.

例如，给定 x = 2t + 1, y = 4t²，由 x = 2t + 1 得到 t = (x – 1)/2。代入 y 得到 y = 4[(x – 1)/2]² = (x – 1)²。务必注意原始参数定义所隐含的定义域限制；例如，若 t ≥ 0，则 x ≥ 1 且 y ≥ 0，因此笛卡儿方程 (x – 1)² = y 仅在该受限范围内有效。

3. Differentiation of Parametric Equations | 参数方程的求导

To find the gradient dy/dx when x and y are given in terms of t, use the chain rule. Since y is a function of t and t is a function of x, we have dy/dx = dy/dt × dt/dx. Because dt/dx = 1 / (dx/dt), the formula becomes:

当 x 和 y 用 t 表示时，使用链式法则求梯度 dy/dx。由于 y 是 t 的函数，t 是 x 的函数，我们有 dy/dx = dy/dt × dt/dx。由于 dt/dx = 1 / (dx/dt)，公式成为：

dy/dx = (dy/dt) / (dx/dt)

This method is essential for calculating the slope of a tangent at any point on a parametric curve. Remember to differentiate both x and y with respect to t separately before dividing. The result will be expressed in terms of the parameter t, which you can then evaluate at a specific t-value to find a numerical gradient.

这一方法对于计算参数曲线上任一点切线的斜率至关重要。记得先分别对 t 求导 x 和 y，然后再相除。结果将用参数 t 表示，之后你可以代入特定的 t 值以求得数值梯度。

4. Worked Example of Parametric Differentiation | 参数求导的示例

Consider the parametric equations x = t³ – t, y = t² + 1. Find dy/dx in terms of t, and then evaluate it at t = 2.

考虑参数方程 x = t³ – t, y = t² + 1。求出 dy/dx 关于 t 的表达式，并计算 t = 2 时的值。

Step 1: Differentiate x with respect to t: dx/dt = 3t² – 1.

步骤1：对 t 求导 x：dx/dt = 3t² – 1。

Step 2: Differentiate y with respect to t: dy/dt = 2t.

步骤2：对 t 求导 y：dy/dt = 2t。

Step 3: Apply the formula: dy/dx = 2t / (3t² – 1).

步骤3：应用公式：dy/dx = 2t / (3t² – 1)。

At t = 2, dy/dx = (2×2) / (3×4 – 1) = 4 / (12 – 1) = 4/11. This is the gradient of the tangent at the point where t = 2.

当 t = 2 时，dy/dx = (2×2) / (3×4 – 1) = 4 / (12 – 1) = 4/11。这就是 t = 2 对应点处切线的斜率。

5. Finding the Equation of a Tangent | 求切线方程

To find the tangent line to a parametric curve at a point corresponding to a specific value of t, follow these steps: First, compute the gradient m = dy/dx using the derivative formula and evaluate it at the given t. Then find the coordinates of the point by substituting t into the expressions for x and y. Finally, use the point-slope form: y – y₁ = m(x – x₁).

要找到参数曲线上对应特定 t 值的点的切线，请遵循以下步骤：首先，使用导数公式计算斜率 m = dy/dx 并在给定 t 值处求值。然后，将 t 代入 x 和 y 的表达式求出该点的坐标。最后，使用点斜式：y – y₁ = m(x – x₁)。

Example: Find the equation of the tangent to the curve x = t², y = t³ at the point where t = 2.

示例：求曲线 x = t², y = t³ 在 t = 2 处的切线方程。

dy/dx = dy/dt ÷ dx/dt = 3t² / 2t = (3/2)t. So m = (3/2) × 2 = 3. The point is (2², 2³) = (4, 8). Tangent: y – 8 = 3(x – 4) → y = 3x – 4.

dy/dx = dy/dt ÷ dx/dt = 3t² / 2t = (3/2)t。所以 m = (3/2) × 2 = 3。点是 (2², 2³) = (4, 8)。切线：y – 8 = 3(x – 4) → y = 3x – 4。

6. Finding the Equation of a Normal | 求法线方程

The normal to a curve at a point is perpendicular to the tangent. Its gradient is the negative reciprocal of the tangent’s gradient: m_normal = -1 / m_tangent, provided m_tangent ≠ 0. Once you have the normal gradient, use the same point (x₁, y₁) and the point-slope form to write the equation.

曲线上一点的法线垂直于切线。其斜率是切线斜率的负倒数：m_normal = -1 / m_tangent（设 m_tangent ≠ 0）。一旦得到法线斜率，即可使用同一点 (x₁, y₁) 和点斜式写出方程。

Using the previous example where the tangent gradient at (4, 8) was 3, the normal gradient is -1/3. The equation of the normal is y – 8 = (-1/3)(x – 4), which simplifies to 3y + x = 28 or y = -x/3 + 28/3.

沿用前面的例子，在 (4, 8) 处切线斜率为 3，则法线斜率为 -1/3。法线方程为 y – 8 = (-1/3)(x – 4)，化简为 3y + x = 28 或 y = -x/3 + 28/3。

7. Stationary Points on Parametric Curves | 参数曲线上的驻点

A stationary point occurs where the gradient dy/dx equals zero. In parametric form, this condition is equivalent to dy/dt = 0, provided that dx/dt ≠ 0. Solve dy/dt = 0 to find the values of t that give stationary points. Always check that dx/dt is not simultaneously zero; if both derivatives are zero, the point might be a cusp or require further investigation.

驻点出现在梯度 dy/dx 等于零的位置。在参数形式下，此条件等价于 dy/dt = 0，前提是 dx/dt ≠ 0。求解 dy/dt = 0 以得到产生驻点的 t 值。务必检查此时 dx/dt 是否不同时为零；如果两个导数都为零，该点可能是尖点或需要进一步研究。

For example, for x = t² – 4t, y = t³ – 3t, dy/dt = 3t² – 3 = 3(t² – 1). Set dy/dt = 0 → t = ±1. Check dx/dt = 2t – 4: at t = 1, dx/dt = -2 (nonzero); at t = -1, dx/dt = -6 (nonzero). So both t = 1 and t = -1 give stationary points. Substitute t to find coordinates and then determine nature using the second derivative or a sign table.

例如，对于 x = t² – 4t, y = t³ – 3t，dy/dt = 3t² – 3 = 3(t² – 1)。令 dy/dt = 0 → t = ±1。检验 dx/dt = 2t – 4：在 t = 1 时 dx/dt = -2（非零）；在 t = -1 时 dx/dt = -6（非零）。因此 t = 1 和 t = -1 都给出驻点。代入 t 求出坐标，然后使用二阶导数或符号表判断点的性质。

8. Second Derivatives of Parametric Functions | 参数函数的二阶导数

The second derivative d²y/dx² measures the rate of change of the gradient and helps determine concavity and the nature of stationary points. For parametric equations, d²y/dx² is not simply the second derivative of y with respect to t divided by the second derivative of x. The correct formula is derived by differentiating dy/dx with respect to t and then dividing by dx

Published by TutorHao | GCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Parametric Equations | 参数方程