Example 7.3.3: Gradients, Tangents and Normals | 例7.3.3：梯度、切线与法线

📚 Example 7.3.3: Gradients, Tangents and Normals | 例7.3.3：梯度、切线与法线

In this worked example, we explore how to apply differentiation to find the equation of a tangent and a normal to a curve at a given point. This is a fundamental skill in AQA A-Level Mathematics and appears frequently in both pure mathematics and applied contexts.

在本例题中，我们探讨如何应用微分求曲线在某一点处的切线与法线方程。这是AQA A-Level数学中的基本技能，在纯数学和应用场景中都频繁出现。

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1. The Problem Statement | 题目陈述

Consider the curve given by the equation y = x³ − 4x² + 5x − 2. Find the equation of the tangent and the equation of the normal to the curve at the point where x = 1.

考虑由方程 y = x³ − 4x² + 5x − 2 给出的曲线。求该曲线在 x = 1 处的切线方程和法线方程。

This example tests your ability to differentiate polynomial functions, evaluate the derivative at a specific point, and then use the point-slope form of a straight line to construct the required equations.

本例题考查你区分多项式函数、在特定点处求导数值，以及利用点斜式直线方程构建所需方程的能力。

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2. Step 1: Find the Coordinates of the Point | 第一步：求该点坐标

First, we must determine the y-coordinate of the point on the curve when x = 1. Substitute x = 1 into the equation of the curve:

首先，我们必须确定曲线上 x = 1 时该点的 y 坐标。将 x = 1 代入曲线方程：

y = (1)³ − 4(1)² + 5(1) − 2 = 1 − 4 + 5 − 2 = 0

Therefore, the point of interest is P(1, 0). This is the point through which both the tangent and the normal will pass.

因此，所求点为 P(1, 0)。切线和法线都将经过该点。

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3. Step 2: Differentiate the Curve | 第二步：对曲线求导

To find the gradient of the tangent at any point on the curve, we differentiate y with respect to x using the power rule:

为了求曲线上任意一点处切线的梯度，我们使用幂法则对 y 关于 x 求导：

dy/dx = 3x² − 8x + 5

This derivative function gives the gradient of the tangent at any x-coordinate. We now evaluate it at x = 1.

该导函数给出了任意 x 坐标处切线的梯度。现在我们计算 x = 1 时的值。

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4. Step 3: Evaluate the Gradient at x = 1 | 第三步：求 x = 1 处的梯度

Substitute x = 1 into the derivative:

将 x = 1 代入导数：

dy/dx = 3(1)² − 8(1) + 5 = 3 − 8 + 5 = 0

The gradient of the tangent at P(1, 0) is m = 0. This means the tangent is a horizontal line at this point. We will use this result in the next step to write the equation of the tangent.

在 P(1, 0) 处切线的梯度为 m = 0。这意味着该点处的切线是一条水平线。我们将在下一步用这个结果写出切线方程。

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5. Step 4: Equation of the Tangent | 第四步：切线方程

The equation of a straight line with gradient m passing through a point (x₁, y₁) is given by the point-slope formula:

经过点 (x₁, y₁) 且梯度为 m 的直线方程由点斜式公式给出：

y − y₁ = m(x − x₁)

Substituting m = 0, x₁ = 1 and y₁ = 0, we obtain:

代入 m = 0，x₁ = 1，y₁ = 0，我们得到：

y − 0 = 0(x − 1) → y = 0

Thus the equation of the tangent at P(1, 0) is simply y = 0, which is the x-axis itself.

因此，在 P(1, 0) 处的切线方程就是 y = 0，即 x 轴本身。

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6. Step 5: Gradient of the Normal | 第五步：法线的梯度

The normal to a curve at a point is perpendicular to the tangent at that point. If the gradient of the tangent is m, the gradient of the normal is −1/m, provided m ≠ 0.

曲线在某一点的法线垂直于该点处的切线。如果切线的梯度为 m，则法线的梯度为 −1/m，前提是 m ≠ 0。

However, in this case m = 0. This is a special case: the tangent is horizontal, so the normal is vertical. A vertical line has an undefined gradient and its equation is of the form x = constant.

然而，在本例中 m = 0。这是一个特殊情况：切线是水平的，因此法线是垂直的。垂直线的梯度无定义，其方程为 x = 常数 的形式。

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7. Step 6: Equation of the Normal | 第六步：法线方程

Since the normal is vertical and passes through the point P(1, 0), all points on the normal have the same x-coordinate, namely x = 1.

由于法线是垂直的且经过点 P(1, 0)，法线上所有点的 x 坐标都相同，即 x = 1。

x = 1

Therefore, the equation of the normal is x = 1, which is a vertical line passing through P(1, 0).

因此，法线方程为 x = 1，即经过 P(1, 0) 的垂直线。

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8. Verification and Geometric Interpretation | 验证与几何解释

Let us verify our results by considering the geometry of the situation. The tangent y = 0 is horizontal, and the normal x = 1 is vertical. These two lines intersect at the point (1, 0), which is exactly the point of contact on the curve, confirming the correctness of our calculations.

让我们通过考虑几何情形来验证我们的结果。切线 y = 0 是水平的，法线 x = 1 是垂直的。这两条线相交于点 (1, 0)，这正是曲线上的接触点，证实了我们计算的正确性。

Note that when the tangent is horizontal, the curve has a stationary point at that location. We can check whether it is a local maximum, local minimum, or point of inflection by considering the second derivative, but this is not required for the current example.

注意当切线为水平时，曲线在该处有一个驻点。我们可以通过二阶导数判断它是局部最大值、局部最小值还是拐点，但这不是当前例题所要求的。

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9. Alternative Example: A Non-Horizontal Tangent | 另一例：非水平切线

To further consolidate understanding, consider the same curve y = x³ − 4x² + 5x − 2 at the point where x = 2.

为了进一步巩固理解，考虑同一条曲线 y = x³ − 4x² + 5x − 2 在 x = 2 处的点。

When x = 2:

当 x = 2 时：

y = 8 − 16 + 10 − 2 = 0

So the point is Q(2, 0). The gradient at x = 2 is:

因此该点为 Q(2, 0)。在 x = 2 处的梯度为：

dy/dx = 3(2)² − 8(2) + 5 = 12 − 16 + 5 = 1

The tangent through Q(2, 0) with gradient m = 1 is:

经过 Q(2, 0) 且梯度为 m = 1 的切线为：

y − 0 = 1(x − 2) → y = x − 2

The gradient of the normal is −1/1 = −1, so the normal through Q(2, 0) is:

法线的梯度为 −1/1 = −1，因此经过 Q(2, 0) 的法线为：

y − 0 = −1(x − 2) → y = −x + 2

These two lines are clearly perpendicular since the product of their gradients is 1 × (−1) = −1.

显然这两条直线是垂直的，因为它们的梯度乘积为 1 × (−1) = −1。

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10. Common Pitfalls and Exam Tips | 常见错误与考试提示

Students often make several mistakes when solving this type of problem. Being aware of these pitfalls will help you avoid them in the exam.

学生在解答此类问题时经常犯一些错误。了解这些陷阱将帮助你在考试中避免它们。

• Forgetting to find the y-coordinate of the point first by substituting x into the original equation.

  忘记先将 x 代入原方程求得该点的 y 坐标。

• Differentiating incorrectly, especially applying the power rule incorrectly to constant terms, which have derivative 0.

  求导错误，特别是对常数项错误地应用幂法则，常数项的导数为 0。

• When m = 0, incorrectly using the formula −1/m for the normal, ignoring the fact that this formula is undefined for m = 0. The normal is vertical instead.

  当 m = 0 时，错误地用 −1/m 求法线梯度，忽略了当 m = 0 时该公式无定义。此时法线是垂直的。

• Writing the final equation in an incorrect form. Always check that your equation passes through the given point by substituting the coordinates.

  最终方程写成错误的形式。始终通过代入坐标来检查方程是否经过给定点。

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11. Summary of Key Methods | 关键方法总结

To summarise, the procedure for finding the tangent and normal to a curve y = f(x) at a given point x = a is as follows:

总结如下，求曲线 y = f(x) 在给定点 x = a 处的切线和法线的步骤如下：

Step | 步骤	Action | 操作
1	Compute the y-coordinate by substituting x = a into y = f(x). | 将 x = a 代入 y = f(x) 计算 y 坐标。
2	Find dy/dx = f'(x). | 求 dy/dx = f'(x)。
3	Evaluate the gradient m = f'(a). | 计算梯度 m = f'(a)。
4	If m ≠ 0, gradient of normal = −1/m; if m = 0, normal is vertical. | 若 m ≠ 0，法线梯度为 −1/m；若 m = 0，法线为垂直。
5	Write both lines using y − y₁ = m(x − x₁), where (x₁, y₁) is the point of contact. | 使用 y − y₁ = m(x − x₁) 写出两条直线方程，其中 (x₁, y₁) 为接触点。

Mastering this procedure is essential for success in the differentiation section of AQA A-Level Mathematics. Practice with a variety of curves to become fluent.

熟练掌握此步骤对于在AQA A-Level数学微分部分取得成功至关重要。通过练习不同类型的曲线来达到熟练。

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12. Further Practice | 拓展练习

To test your understanding, try the following problem independently before checking the answers.

为测试你的理解，请先独立尝试以下问题，然后再检查答案。

For the curve y = x³ − 6x² + 9x + 1, find the equations of the tangent and normal at the point where x = 2. Verify that your tangent line passes through (2, 3) and has gradient 3.

对于曲线 y = x³ − 6x² + 9x + 1，求在 x = 2 处的切线和法线方程。验证你的切线经过点 (2, 3) 且梯度为 3。

Solution: dy/dx = 3x² − 12x + 9. At x = 2, dy/dx = 12 − 24 + 9 = −3. The tangent is y − 3 = −3(x − 2), i.e. y = −3x + 9. The normal has gradient 1/3, giving y − 3 = (1/3)(x − 2), i.e. y = (1/3)x + 7/3.

解答：dy/dx = 3x² − 12x + 9。在 x = 2 处，dy/dx = 12 − 24 + 9 = −3。切线为 y − 3 = −3(x − 2)，即 y = −3x + 9。法线梯度为 1/3，得到 y − 3 = (1/3)(x − 2)，即 y = (1/3)x + 7/3。

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