Gradients, Tangents and Normals | 梯度、切线与法线

📚 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

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

e 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。

ed 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)

tion 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。

---

广告 / Sponsored

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

s 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

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）