Derivative as a Function of a Parameter | 导数作为参数的函数

📚 Derivative as a Function of a Parameter | 导数作为参数的函数

Parametric differentiation allows us to find the gradient dy/dx when a curve is expressed through a third variable, usually t or θ. Instead of eliminating the parameter, we treat the derivative itself as a function of that parameter.

参数方程求导可以帮助我们在曲线由第三个变量（通常是 t 或 θ）表示时求出梯度 dy/dx。我们不必消去参数，而是把导数本身看作该参数的函数。

This topic is central to AS Mathematics because many curves, especially circles, ellipses and loops, are much easier to describe parametrically than in the form y = f(x).

这个主题在 AS 数学中非常重要，因为许多曲线，尤其是圆、椭圆和有环的曲线，用参数形式描述比写成 y = f(x) 容易得多。

1. Parametric Equations | 参数方程

When a curve is described by two equations x = f(t) and y = g(t), the variable t is called a parameter. Instead of eliminating t to find y in terms of x, we can differentiate directly with respect to t.

当曲线由 x = f(t) 和 y = g(t) 两个方程描述时，变量 t 称为参数。我们可以直接对 t 求导，而不必消去 t 把 y 写成 x 的函数。

This is useful when eliminating the parameter is messy, or when the graph has loops and more than one y-value for a given x.

当消去参数很麻烦，或者图像有环、同一个 x 对应多个 y 时，这种方法非常有用。

2. The Chain Rule Connection | 链式法则的联系

The derivative dy/dx is connected to the parameter by the chain rule: dy/dt = (dy/dx)(dx/dt).

根据链式法则，dy/dt = (dy/dx)(dx/dt)。

Rearranging this identity gives the working rule for parametric differentiation.

整理这个等式即可得到参数方程求导的计算法则。

3. The Core Formula | 核心公式

To differentiate parametrically, find dx/dt and dy/dt first, then divide. Always leave the answer in terms of the parameter t or θ, unless the question asks you to evaluate at a point.

参数求导时，先求 dx/dt 和 dy/dt，再相除。除非题目要求在某点求值，否则结果应保留为参数 t 或 θ 的函数。

dy/dx = (dy/dt) ÷ (dx/dt)

The formula is valid only when dx/dt is not zero.

该公式仅在 dx/dt 不等于零时有效。

Step 1: Differentiate x with respect to t to get dx/dt. Step 2: Differentiate y with respect to t to get dy/dt. Step 3: Divide dy/dt by dx/dt.

第 1 步：对 x 关于 t 求导，得到 dx/dt。第 2 步：对 y 关于 t 求导，得到 dy/dt。第 3 步：用 dy/dt 除以 dx/dt。

4. Worked Example: Polynomial Path | 例题：多项式路径

Given x = t² + 1 and y = 2t³ − 3t, find dy/dx in terms of t.

已知 x = t² + 1，y = 2t³ − 3t，求 dy/dx 关于 t 的表达式。

Differentiate: dx/dt = 2t and dy/dt = 6t² − 3.

求导得：dx/dt = 2t，dy/dt = 6t² − 3。

Therefore dy/dx = (6t² − 3) / (2t) = 3t − 3/(2t).

因此 dy/dx = (6t² − 3) / (2t) = 3t − 3/(2t)。

At t = 2, dy/dx = 6 − 3/4 = 21/4, which gives the gradient of the tangent at that parameter value.

当 t = 2 时，dy/dx = 6 − 3/4 = 21/4，这就是该参数值处切线的斜率。

5. Worked Example: Trigonometric Path | 例题：三角函数路径

For x = a cos θ and y = b sin θ, find dy/dx.

对于 x = a cos θ，y = b sin θ，求 dy/dx。

dx/dθ = −a sin θ and dy/dθ = b cos θ.

dx/dθ = −a sin θ，dy/dθ = b cos θ。

Thus dy/dx = (b cos θ) / (−a sin θ) = −(b/a) cot θ.

所以 dy/dx = (b cos θ) / (−a sin θ) = −(b/a) cot θ。

This shows the derivative is still a function of the parameter θ, not a number.

这表明导数仍然是参数 θ 的函数，而不是一个固定的数。

6. Tangent and Normal Lines | 切线与法线

To find a tangent, first evaluate x, y and dy/dx at the given parameter value, then use y − y₁ = m(x − x₁).

求切线时，先在给定参数值处求出 x、y 和 dy/dx，再用 y − y₁ = m(x − x₁)。

Example: x = 3 cos t, y = 4 sin t. At t = π/4, x = 3/√2 and y = 4/√2 = 2√2.

例如：x = 3 cos t，y = 4 sin t。当 t = π/4 时，x = 3/√2，y = 4/√2 = 2√2。

dx/dt = −3 sin t = −3/√2, dy/dt = 4 cos t = 4/√2, so dy/dx = −4/3.

dx/dt = −3 sin t = −3/√2，dy/dt = 4 cos t = 4/√2，所以 dy/dx = −4/3。

Tangent: y − 2√2 = −(4/3)(x − 3/√2).

切线为：y − 2√2 = −(4/3)(x − 3/√2)。

The normal line has gradient 3/4 at the same point.

同一点处法线的斜率为 3/4。

7. Second Derivative | 二阶导数

If the first derivative is a function of t, the second derivative is not simply d²y/dt² divided by d²x/dt².

如果一阶导数是 t 的函数，二阶导数不能简单地把 d²y/dt² 除以 d²x/dt²。

d²y/dx² = d/dt(dy/dx) ÷ (dx/dt)

Example: x = t², y = t³. dx/dt = 2t, dy/dt = 3t², dy/dx = 3t/2.

例如：x = t²，y = t³。dx/dt = 2t，dy/dt = 3t²，dy/dx = 3t/2。

Then d²y/dx² = (d/dt(3t/2)) / (2t) = (3/2)/(2t) = 3/(4t).

因此 d²y/dx² = (d/dt(3t/2)) / (2t) = (3/2)/(2t) = 3/(4t)。

8. Horizontal and Vertical Tangents | 水平与竖直切线

The tangent is horizontal when dy/dt = 0 and dx/dt ≠ 0. The tangent is vertical when dx/dt = 0 and dy/dt ≠ 0.

当 dy/dt = 0 且 dx/dt ≠ 0 时，切线水平；当 dx/dt = 0 且 dy/dt ≠ 0 时，切线竖直。

Example: x = t², y = t³ − 3t. Differentiate to get dx/dt = 2t and dy/dt = 3t² − 3.

例如：x = t²，y = t³ − 3t。求导可得 dx/dt = 2t，dy/dt = 3t² − 3。

For horizontal tangents, dy/dt = 0 gives t = 1 or t = −1. Since dx/dt is not zero at these values, horizontal tangents occur at (1, −2) and (1, 2).

水平切线：dy/dt = 0 得 t = 1 或 t = −1。此时 dx/dt 不为零，因此水平切线出现在 (1, −2) 和 (1, 2)。

For a vertical tangent, dx/dt = 0 gives t = 0. Since dy/dt = −3 ≠ 0, the vertical tangent is at (0, 0).

竖直切线：dx/dt = 0 得 t = 0。因为 dy/dt = −3 ≠ 0，竖直切线在 (0, 0)。

9. Common Mistakes | 常见错误

One common mistake is to write dy/dx = dy/dt only, or to multiply by dx/dt incorrectly.

常见错误之一是只写 dy/dx = dy/dt，或者错误地乘以 dx/dt。

Another mistake is using d²y/dx² = (d²y/dt²)/(d²x/dt²). This is wrong; the correct formula uses the derivative of dy/dx with respect to t divided by dx/dt.

另一个错误是用 d²y/dx² = (d²y/dt²)/(d²x/dt²)。这是错误的；正确公式应先对 dy/dx 关于 t 求导，再除以 dx/dt。

Also remember that x = f(t) and y = g(t) are evaluated at the same parameter value for any point on the curve.

还要记住，曲线上任意点的 x = f(t) 和 y = g(t) 必须使用同一个参数值。

10. Exam Strategy Summary | 考试策略总结

Keep a clear structure: state the formula, find dx/dt and dy/dt, then substitute. Always check whether you need the tangent, normal, or second derivative before doing extra work.

保持清晰的解题结构：写出公式，求出 dx/dt 和 dy/dt，再代入。始终先判断题目要求的是切线、法线还是二阶导数，避免多余计算。

Quantity	Formula
First derivative | 一阶导数	dy/dx = (dy/dt) ÷ (dx/dt)
Second derivative | 二阶导数	d²y/dx² = d/dt(dy/dx) ÷ (dx/dt)

If a question says “find the gradient at point P”, identify the parameter value first, then substitute into dy/dx.

如果题目说“求点 P 处的梯度”，应首先确定参数值，再代入 dy/dx。

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