📚 Parametric Differentiation | 参数微分
Parametric equations introduce a third variable, usually t, to describe curves in a way that often simplifies complex relationships between x and y. Mastering parametric differentiation is essential for A-Level Edexcel candidates, as it unlocks tangent, normal, and stationary point problems that appear regularly in exam papers. The technique relies on the chain rule and careful algebraic manipulation.
参数方程引入第三个变量（通常用 t）来描述曲线，这种方法往往能简化 x 与 y 之间的复杂关系。掌握参数微分对 Edexcel A-Level 考生至关重要，因为切线、法线和驻点问题是试卷中的常客。该技巧依赖链式法则和细致的代数运算。
1. Introduction to Parametric Equations | 参数方程简介
g(t), where t is a parameter. As t varies, the corresponding points (x, y) trace the curve. This representation is powerful for describing motion along a path and for curves that are not functions in the usual sense, such as circles and cycloids.
在标准笛卡尔表示中，曲线表示为 y = f(x)。而参数方程则定义 x = f(t) 和 y = g(t)，其中 t 为参数。当 t 变化时，相应的点 (x, y) 描绘出曲线。这种表示法对于描述沿路径的运动以及普通意义上非函数的曲线（如圆和摆线）非常有用。
2. The Derivative dy/dx for Parametric Curves | 参数曲线的导数 dy/dx
dt) ÷ (dx/dt), provided dx/dt ≠ 0. This expression gives the gradient of the tangent to the curve in terms of the parameter t.
对参数方程求导的关键是链式法则。由于 y 是 t 的函数，x 也是 t 的函数，因此可以写作 dy/dx = (dy/dt) ÷ (dx/dt)，前提是 dx/dt ≠ 0。该表达式给出了用参数 t 表示的曲线切线斜率。
dy/dx = (dy/dt) / (dx/dt)
3. Step-by-Step Procedure | 逐步求解步骤
To differentiate parametrically, first find dx/dt and dy/dt separately. Then form the ratio dy/dt over dx/dt. Finally, simplify the resulting expression. If a numerical gradient is required, substitute the given t-value into dy/dx. Always check that the denominator dx/dt is not zero at the point of interest, as this corresponds to a vertical tangent or a cusp.
进行参数微分时，首先分别求出 dx/dt 和 dy/dt。然后构成 dy/dt 与 dx/dt 的比值。最后将所得表达式化简。如果需要数值梯度，将给定的 t 值代入 dy/dx。务必检查分母 dx/dt 在所求点处不为零，因为这对应垂直切线或尖点。
- Step 1: Compute dx/dt and dy/dt.
- Step 2: Form dy/dx = (dy/dt) / (dx/dt).
- Step 3: Simplify algebraically.
- Step 4: Evaluate at a specific t, if needed.
- 第一步：计算 dx/dt 和 dy/dt。
- 第二步：构建 dy/dx = (dy/dt) / (dx/dt)。
- 第三步：进行代数化简。
- 第四步：如果需要，代入具体 t 值求值。
4. Example 1: Basic Polynomial Parametric Curve | 示例1：基本多项式参数曲线
used directly to find the gradient at any t. For instance, at t = 1, dy/dx = (2 - 4) / (3 + 2) = -2/5.
已知 x = t³ + 2t 和 y = t² - 4t + 1，用 t 表示 dy/dx。首先，dx/dt = 3t² + 2；dy/dt = 2t - 4。因此，dy/dx = (2t - 4) / (3t² + 2)。该表达式可直接用于求任意 t 处的梯度。例如，当 t = 1 时，dy/dx = (2 - 4) / (3 + 2) = -2/5。
5. Example 2: Trigonometric Parametric Equations | 示例2：三角参数方程
= (3 cos θ) / (-4 sin θ) = -3/4 cot θ. Notice that the derivative involves a trigonometric identity and is undefined when sin θ = 0, i.e., at points where x = ±4 and y = 0.
考虑某曲线的参数方程：x = 4 cos θ，y = 3 sin θ。对 θ 求导：dx/dθ = -4 sin θ，dy/dθ = 3 cos θ。梯度 dy/dx = (3 cos θ) / (-4 sin θ) = -3/4 cot θ。注意到导数包含三角恒等式，且在 sin θ = 0 时无定义，即 x = ±4、y = 0 的点处。
6. Finding the Tangent and Normal | 求切线与法线
ted at t₀. The normal gradient is -1/m, provided m ≠ 0. Express the line in Cartesian form after substituting the coordinates.
一旦求得 dy/dx，就可以写出参数 t = t₀ 对应点处的切线方程 y - y(t₀) = m [x - x(t₀)]，其中 m 为 t₀ 处的 dy/dx 值。法线斜率为 -1/m，前提是 m ≠ 0。代入坐标后用笛卡尔形式表示直线方程。
, the gradient is 1 and the point is (1,2). The tangent is y - 2 = 1 (x - 1) or y = x + 1.
例如，若 x = t²，y = 2t 且 t₀ = 1，则 dx/dt = 2t，dy/dt = 2，故 dy/dx = 1/t。在 t = 1 处，梯度为 1，点为 (1,2)。切线为 y - 2 = 1 (x - 1)，即 y = x + 1。
7. Second Order Derivative d²y/dx² | 二阶导数 d²y/dx²
ntiating dy/dx with respect to t and then dividing by dx/dt: d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt). This formula captures the curvature in terms of the parameter.
二阶导数 d²y/dx² 衡量梯度的变化率，并决定曲线的凹性。对于参数曲线，我们通过先对 t 求导 dy/dx，再除以 dx/dt 得到：d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt)。该公式用参数表达了曲率。
d²y/dx² = [d(dy/dx)/dt] / (dx/dt)
t(dy/dx) = [2(3t²+2) - (2t-4)(6t)] / (3t²+2)². Simplify to get the expression, then divide by dx/dt = 3t²+2. The result is d²y/dx² = [-6t² + 24t + 4] / (3t²+2)³.
沿用前例 x = t³ + 2t，y = t² - 4t + 1，我们已得 dy/dx = (2t - 4)/(3t² + 2)。用商的求导法则对其关于 t 求导：d/dt(dy/dx) = [2(3t²+2) - (2t-4)(6t)] / (3t²+2)²。化简后，再除以 dx/dt = 3t²+2。最终结果为 d²y/dx² = [-6t² + 24t + 4] / (3t²+2)³。
8. Application: Stationary Points | 应用：驻点
find the parameter values, then substitute back into x(t) and y(t) to get the coordinates. Determine the nature of each stationary point using the second derivative test or by examining the sign change of dy/dx around the point.
驻点发生在 dy/dx = 0 时，即等同于 dy/dt = 0（前提是 dx/dt ≠ 0）。求解 dy/dt = 0 得到参数值，再代回 x(t) 和 y(t) 求坐标。通过二阶导数检验或考察 dy/dx 在点两侧的符号变化来确定每个驻点的性质。
stationary point is at (2, -1). To classify it, find d²y/dx² at t = -1. dy/dx = (2t+2)/(3t²-3). With careful simplification and differentiation, the second derivative test reveals whether it is a maximum, minimum, or inflection.
以 x = t³ - 3t，y = t² + 2t 为例。dx/dt = 3t² - 3，dy/dt = 2t + 2。令 dy/dt = 0 得 t = -1。在 t = -1 处，点为 x = (-1)³ - 3(-1) = 2，y = (-1)² + 2(-1) = -1。驻点为 (2, -1)。要判断其类型，可求 t = -1 时的 d²y/dx²。dy/dx = (2t+2)/(3t²-3)，经过仔细化简和微分，二阶导数检验能揭示它是极大值、极小值还是拐点。
9. Common Pitfalls and Tips | 常见错误与提示
A frequent mistake is forgetting to divide by dx/dt when finding d²y/dx². Students often differentiate dy/dx with respect to t and mistakenly believe they have the second derivative without the additional division step. Another error is missing the condition dx/dt ≠ 0, leading to incorrect tangent calculations. Always simplify dy/dx as much as possible before differentiating further.
一个常见错误是在求 d²y/dx² 时忘记除以 dx/dt。学生常对 dy/dx 关于 t 求导后，以为直接得到了二阶导数，而忽略了额外的除法步骤。另一个错误是遗漏条件 dx/dt ≠ 0，导致切线计算错误。在进行进一步求导之前，务必将 dy/dx 尽可能化简。
- Always check the domain of t for which dx/dt ≠ 0.
- Simplify dy/dx using algebraic or trigonometric identities.
- When evaluating d²y/dx², use the parametric formula systematically.
- Draw a diagram if possible to verify the tangent or stationary point.
- 务必检查 dx/dt ≠ 0 的 t 的取值范围。
- 利用代数或三角恒等式化简 dy/dx。
- 在计算 d²y/dx² 时，系统地使用参数化公式。
- 如有可能，画图验证切线或驻点的位置。
10. Practice Problem with Solution | 练习题与解答
e where the tangent is horizontal. Also, determine the equation of the tangent at t = π/4.
题目：某曲线由参数方程 x = 2 cos t + sin 2t，y = 2 sin t + cos 2t 定义，t ∈ [0,2π]。求曲线上切线为水平的点。并求在 t = π/4 处的切线方程。
2 sin 2t = 0, i.e., cos t = sin 2t = 2 sin t cos t. This gives cos t (1 - 2 sin t) = 0, so cos t = 0 or sin t = 1/2. Obtain t-values and check dx/dt ≠ 0. Then find corresponding coordinates. For the tangent at t = π/4, evaluate dy/dx = (dy/dt)/(dx/dt). At t = π/4, sin t = √2/2, cos t = √2/2, sin 2t = 1, cos 2t = 0. dy/dt = 2(√2/2) - 2(1) = √2 - 2; dx/dt = -2(√2/2) + 2(0) = -√2. Hence dy/dx = (√2 - 2)/(-√2) = -1 + √2. The point is x = 2(√2/2) + 1 = √2 + 1, y = 2(√2/2) + 0 = √2. The tangent line is y - √2 = (-1 + √2)(x - √2 - 1).
解答要点：首先计算 dx/dt = -2 sin t + 2 cos 2t，dy/dt = 2 cos t - 2 sin 2t。切线水平时 dy/dt = 0 且 dx/dt ≠ 0。解方程 2 cos t - 2 sin 2t = 0，即 cos t = sin 2t = 2 sin t cos t。得 cos t (1 - 2 sin t) = 0，故 cos t = 0 或 sin t = 1/2。求出 t 值并验证 dx/dt ≠ 0，然后得到对应坐标。对于 t = π/4 处的切线，求 dy/dx = (dy/dt)/(dx/dt)。在 t = π/4 时，sin t = √2/2，cos t = √2/2，sin 2t = 1，cos 2t = 0。dy/dt = 2(√2/2) - 2(1) = √2 - 2；dx/dt = -2(√2/2) + 2(0) = -√2。因此 dy/dx = (√2 - 2)/(-√2) = -1 + √2。点为 x = 2(√2/2) + 1 = √2 + 1，y = 2(√2/2) + 0 = √2。切线方程为 y - √2 = (-1 + √2)(x - √2 - 1)。
11. Summary and Exam Advice | 总结与考试建议
Parametric differentiation is a structured process: differentiate x and y with respect to t, form the ratio, and simplify. The second derivative follows a similar deductive route. In Edexcel exams, questions often integrate parametric gradients with coordinate geometry, requiring careful substitution and simplification. Practise with a mix of polynomial, trigonometric, and exponential parametric functions to build fluency. Always check the validity of the parameter domain and remember that the diagram in the question might provide insight into expected results.
参数微分是一个结构化过程：对 x 和 y 分别关于 t 求导，构成比值，然后化简。二阶导数遵循类似的演绎路径。在 Edexcel 考试中，题目常将参数梯度与坐标几何结合，要求仔细代入和化简。通过混合练习多项式、三角函数和指数型参数方程来提升熟练度。务必检查参数定义域的有效性，并记住题目中的图形可能为预期结果提供直观线索。
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