📚 Gradients, Tangents and Normals | 梯度、切线与法线
In A-Level Edexcel Pure Mathematics, the derivative dy/dx gives the gradient of a curve at any point. This idea lets us write equations of tangent and normal lines, a core skill in differentiation.
在 A-Level Edexcel 纯数学中，导数 dy/dx 给出曲线在任意一点的梯度。利用这一概念，我们可以写出切线和法线方程，这是微分部分的核心技能。
This revision guide covers the rules, the line equations, worked examples for explicit, parametric and implicit curves, and common exam mistakes.
本复习指南涵盖微分法则、直线方程、显式、参数与隐函数曲线的例题，以及常见考试错误。
1. The Derivative as a Gradient Function | 导数作为梯度函数
For a curve with equation y = f(x), the derivative dy/dx is a function that gives the gradient of the curve at any point where the function is differentiable.
对于方程为 y = f(x) 的曲线，导数 dy/dx 是一个函数，它给出曲线上任意可导点的梯度。
At a fixed point x = a, the gradient of the tangent to the curve is m = f'(a), which means dy/dx evaluated at x = a.
在固定点 x = a 处，曲线切线的梯度为 m = f'(a)，也就是 dy/dx 在 x = a 时的值。
The derivative is therefore often called the gradient function. Without it, finding the gradient of a curved graph is not possible using simple linear methods.
因此，导数常被称为梯度函数。没有它，仅用简单线性方法就无法求出曲线图形的梯度。
2. Differentiation Rules You Need | 必备微分法则
The most important rules for A-Level Edexcel differentiation are:
A-Level Edexcel 微分中最常用的法则如下：
| Function y | Derivative dy/dx |
|---|---|
| y = xⁿ | dy/dx = nxⁿ⁻¹ |
| y = kxⁿ | dy/dx = knxⁿ⁻¹ |
| y = f(x) + g(x) | dy/dx = f'(x) + g'(x) |
| y = c, where c is constant | dy/dx = 0 |
| y = (ax + b)ⁿ | dy/dx = an(ax + b)ⁿ⁻¹ |
You should be able to apply these rules fluently before attempting tangent and normal problems.
在开始解决切线和法线问题之前，你应该能够熟练应用这些法则。
3. Finding the Gradient at a Given Point | 求给定点的梯度
To find the gradient of a curve at a specific point, first differentiate the function, then substitute the given x-coordinate into the derivative.
要求曲线在某一点的梯度，首先对函数求导，然后将给定的 x 坐标代入导数。
Example: find the gradient of y = 3x² − 2x + 1 at x = 2.
例子：求 y = 3x² − 2x + 1 在 x = 2 处的梯度。
dy/dx = 6x − 2
Now substitute x = 2:
现在代入 x = 2：
m = 6(2) − 2 = 10
The gradient of the curve at x = 2 is 10, so the tangent at that point also has gradient 10.
曲线在 x = 2 处的梯度为 10，因此该点的切线梯度也为 10。
4. Equation of a Tangent | 切线方程
At point P(x₁, y₁) on y = f(x), if the tangent gradient is m_t, then the tangent is the straight line through P with gradient m_t.
在 y = f(x) 上的点 P(x₁, y₁) 处，如果切线梯度为 m_t，那么切线就是过 P 且梯度为 m_t 的直线。
y − y₁ = m_t(x − x₁)
Always substitute x₁ into the derivative to find m_t, and use the original equation to find y₁.
一定要把 x₁ 代入导数求 m_t，并用原函数求 y₁。
You may be asked for the final answer in the form y = mx + c or ax + by + c = 0, so rearrange carefully.
题目可能要求最终答案写成 y = mx + c 或 ax + by + c = 0 的形式，因此要仔细整理。
5. Equation of a Normal | 法线方程
The normal at P is perpendicular to the tangent at P. For perpendicular lines, the product of their gradients is −1.
法线在 P 点与切线垂直。对于互相垂直的直线，其梯度乘积为 −1。
m_normal = −1 / m_tangent
This formula works provided m_tangent is not equal to 0. Then use the same point and the new gradient:
该公式在 m_tangent 不等于 0 时适用。然后使用同一个点和新的梯度：
y − y₁ = m_normal(x − x₁)
Special cases: if the tangent is horizontal with gradient 0, the normal is vertical and has equation x = x₁. If the tangent is vertical, the normal is horizontal and has equation y = y₁.
特殊情况：如果切线水平且梯度为 0，则法线为竖直直线，方程为 x = x₁。如果切线竖直，则法线为水平直线，方程为 y = y₁。
6. Worked Example: Tangent and Normal to a Quadratic | 实例：二次曲线的切线与法线
Worked example: find the tangent and normal to y = x² + 3x − 2 at the point where x = 1.
例题：求 y = x² + 3x − 2 在 x = 1 处的切线和法线方程。
Step 1: find the y-coordinate.
第一步：求 y 坐标。
y = (1)² + 3(1) − 2 = 2
Step 2: differentiate.
第二步：求导。
dy/dx = 2x + 3
Step 3: substitute x = 1 to get the tangent gradient.
第三步：代入 x = 1 求切线梯度。
m_t = 2(1) + 3 = 5
Tangent equation:
切线方程：
y − 2 = 5(x − 1) → y = 5x − 3
Normal gradient:
法线梯度：
m_n = −1 / 5
Normal equation:
法线方程：
y − 2 = (−1/5)(x − 1) → x + 5y − 11 = 0
The tangent is y = 5x − 3 and the normal is x + 5y − 11 = 0.
切线为 y = 5x − 3，法线为 x + 5y − 11 = 0。
7. Using Parallel or Perpendicular Conditions | 利用平行或垂直条件
Exam questions often give a condition instead of a point. For example, a tangent may be parallel to a known line, so its gradient must equal that line’s gradient.
考试题目经常给出条件而不是点。例如，切线可能与已知直线平行，因此其梯度必须等于该直线的梯度。
Example: find the point on y = x² − 4x + 5 where the tangent is parallel to y = 2x + 3.
例子：求 y = x² − 4x + 5 上切线平行于 y = 2x + 3 的点。
Parallel lines have equal gradients, so set dy/dx = 2.
平行直线梯度相等，因此令 dy/dx = 2。
dy/dx = 2x − 4 = 2 → x = 3
Substitute x = 3 into the original curve:
将 x = 3 代入原曲线：
y = (3)² − 4(3) + 5 = 2
The required point is (3, 2). The tangent there is y − 2 = 2(x − 3), or y = 2x − 4. The normal is y − 2 = (−1/2)(x − 3), or x + 2y − 7 = 0.
所求点为 (3, 2)。该点切线为 y − 2 = 2(x − 3)，即 y = 2x − 4。法线为 y − 2 = (−1/2)(x − 3)，即 x + 2y − 7 = 0。
8. Tangents and Normals for Parametric Curves | 参数曲线的切线与法线
For a parametric curve x = x(t), y = y(t), Edexcel requires the use of:
对于参数曲线 x = x(t)，y = y(t)，Edexcel 要求使用：
dy/dx = (dy/dt) / (dx/dt)
Example: x = t², y = t³ at t = 2.
例子：x = t²，y = t³，在 t = 2 处。
First find dx/dt and dy/dt:
先求 dx/dt 和 dy/dt：
dx/dt = 2t = 4, dy/dt = 3t² = 12
So the tangent gradient is:
因此切线梯度为：
m = 12 / 4 = 3
The point at t = 2 is (4, 8).
t = 2 时的点为 (4, 8)。
Tangent:
切线：
y − 8 = 3(x − 4) → y = 3x − 4
Normal gradient is −1/3, so the normal is:
法线梯度为 −1/3，因此法线为：
y − 8 = (−1/3)(x − 4) → x + 3y − 28 = 0
9. Implicit Differentiation and Tangent/Normal | 隐函数微分与切线与法线
For implicit curves, differentiate every term with respect to x, applying the chain rule to y terms, then solve for dy/dx.
对于隐函数曲线，对每一项关于 x 求导，对含有 y 的项应用链式法则，然后解出 dy/dx。
Example: x² + y² = 25 at the point (3, 4).
例子：x² + y² = 25 在点 (3, 4) 处。
2x + 2y(dy/dx) = 0 → dy/dx = −x / y
Substitute x = 3, y = 4:
代入 x = 3，y = 4：
m_t = −3 / 4
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