Curve Slope and Tangent Lines in A-Level Maths | A-Level 数学：曲线斜率与切线

📚 Curve Slope and Tangent Lines in A-Level Maths | A-Level 数学：曲线斜率与切线

The gradient of a curve at a point is a measure of how steep it is at that exact location. Unlike a straight line, a curve does not have a single slope; instead, its slope changes continuously along its length. In A-Level mathematics, we quantify this changing slope using differentiation, and we use the derivative to find the equation of the tangent line at any given point.

曲线在某一点的斜率是衡量该点陡峭程度的量。与直线不同，曲线没有单一的斜率；相反，其斜率沿曲线连续变化。在 A-Level 数学中，我们通过微分来量化这种变化的斜率，并利用导数求曲线在任意给定点的切线方程。

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1. Understanding the Gradient of a Curve | 理解曲线的斜率

For a straight line y = mx + c, the gradient m is constant. For a curve, however, the gradient changes from point to point. To find the gradient at a specific point, we consider the limit of the gradient of a chord connecting two points on the curve as the second point approaches the first.

对于直线 y = mx + c，斜率 m 是常数。但对曲线而言，斜率逐点变化。要找到某一点的斜率，我们需要考察连接曲线上两点的割线斜率，并令第二个点无限趋近于第一个点时该割线斜率的极限值。

m = lim_{h→0} [f(a + h) − f(a)] / h

This limiting value, if it exists, is called the derivative of f at x = a, written as f'(a) or dy/dx evaluated at x = a. Geometrically, it gives the slope of the tangent line to the curve y = f(x) at the point (a, f(a)).

这个极限值如果存在，就称为函数 f 在 x=a 处的导数，记作 f'(a) 或在 x=a 处计算的 dy/dx。几何上，它给出了曲线 y = f(x) 在点 (a, f(a)) 处切线的斜率。

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2. Notation for Derivatives | 导数的记法

There are several common notations for derivatives in Edexcel A-Level mathematics. The most frequently used are Leibniz notation dy/dx and Lagrange notation f'(x). In mechanics contexts, you will also see Newton’s dot notation ẋ and ẍ.

在 Edexcel A-Level 数学中，导数的记法有几种常见形式。最常用的是莱布尼茨记法 dy/dx 和拉格朗日记法 f'(x)。在力学问题中，你还会看到牛顿的点记法 ẋ 和 ẍ。

• f'(x) is read as ‘f prime of x’ and represents the derivative of f with respect to x.

• f'(x) 读作”f 撇 x”，表示 f 关于 x 的导数。

• dy/dx emphasises the ratio of a small change in y to a small change in x.

• dy/dx 强调 y 的微小变化与 x 的微小变化之比。

• f'(a) means the derivative evaluated at the specific point x = a.

• f'(a) 表示导数在具体点 x = a 处的取值。

It is essential to be comfortable switching between these notations, as exam questions may use any of them.

熟练地在这些记法之间切换非常重要，因为考试题目可能使用其中任何一种形式。

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3. Differentiating Powers of x | 对 x 的幂函数求导

The fundamental rule for differentiating polynomials is the power rule. If y = xⁿ where n is a constant, then dy/dx = n xⁿ⁻¹. This rule applies for all real values of n, including negative and fractional powers, provided the expression is defined at the point of interest.

多项式求导的基本法则是幂法则。如果 y = xⁿ，其中 n 是常数，那么 dy/dx = n xⁿ⁻¹。该法则适用于所有实数 n，包括负指数和分数指数，前提是表达式在所考虑的点处有定义。

If y = xⁿ, then dy/dx = n xⁿ⁻¹

When differentiating a sum of terms, differentiate each term separately. Constants have derivative zero since a constant function has zero gradient.

对多项之和求导时，逐项分别求导。常数的导数为零，因为常数函数的斜率为零。

• y = x⁵ ⇒ dy/dx = 5x⁴

• y = 3x² ⇒ dy/dx = 6x

• y = 7 ⇒ dy/dx = 0

• y = 1/x² = x⁻² ⇒ dy/dx = −2x⁻³ = −2/x³

• y = √x = x^(1/2) ⇒ dy/dx = (1/2)x^(−1/2) = 1/(2√x)

Before differentiating, rewrite expressions in index form to apply the power rule easily.

在求导之前，先将表达式改写为指数形式，以便轻松应用幂法则。

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4. The Equation of a Tangent | 切线的方程

Once we know the gradient of the curve at a point and the coordinates of that point, we can write down the equation of the tangent line. The tangent is a straight line that touches the curve at exactly one point and has the same gradient as the curve at that point.

一旦我们知道了曲线在某一点的斜率以及该点的坐标，就可以写出切线的方程。切线是一条恰好接触曲线于一点的直线，并且在该点处与曲线具有相同的斜率。

y − y₁ = m(x − x₁)

Here, m is the derivative f'(x₁), and (x₁, y₁) is the point of contact on the curve. This form of a line equation is called the point-slope form.

这里，m 是导数 f'(x₁)，(x₁, y₁) 是曲线上的切点。这种直线方程形式称为点斜式。

To obtain the tangent equation, follow these steps:

求切线方程时，按以下步骤进行：

• Differentiate y = f(x) to find f'(x).

• 对 y = f(x) 求导，得到 f'(x)。

• Substitute the x-coordinate of the point into f'(x) to find the gradient m.

• 将点的 x 坐标代入 f'(x)，求出斜率 m。

• Use y − y₁ = m(x − x₁) to write the equation of the tangent.

• 利用 y − y₁ = m(x − x₁) 写出切线方程。

• Simplify to the required form, such as y = mx + c or ax + by + c = 0.

• 将方程化简为所需形式，例如 y = mx + c 或 ax + by + c = 0。

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5. Worked Example 1 | 示例 1

Find the equation of the tangent to the curve y = x² + 3x − 2 at the point where x = 2.

求曲线 y = x² + 3x − 2 在 x = 2 处的切线方程。

Solution | 解答:

First find the y-coordinate: y = 2² + 3(2) − 2 = 4 + 6 − 2 = 8. So the point is (2, 8).

首先求 y 坐标：y = 2² + 3(2) − 2 = 4 + 6 − 2 = 8。因此点为 (2, 8)。

Differentiate: dy/dx = 2x + 3. At x = 2, the gradient is m = 2(2) + 3 = 7.

求导：dy/dx = 2x + 3。在 x = 2 处，斜率 m = 2(2) + 3 = 7。

Using y − y₁ = m(x − x₁): y − 8 = 7(x − 2).

利用 y − y₁ = m(x − x₁)：y − 8 = 7(x − 2)。

Expanding: y − 8 = 7x − 14, so y = 7x − 6.

展开：y − 8 = 7x − 14，因此 y = 7x − 6。

The tangent is y = 7x − 6 | 切线方程为 y = 7x − 6

Notice the tangent has the same x-coefficient slope as the derivative at the point, and the constant term adjusts so that the line passes exactly through the point of contact.

注意切线的 x 系数斜率与该点处的导数相同，常数项则进行调整，使得直线恰好通过切点。

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6. Worked Example 2 | 示例 2

Find the equation of the tangent to y = √x at the point where x = 9.

求曲线 y = √x 在 x = 9 处切线的方程。

Solution | 解答:

The y-coordinate is y = √9 = 3, so the point is (9, 3).

y 坐标为 y = √9 = 3，因此点为 (9, 3)。

Rewrite y = x^(1/2), so dy/dx = (1/2)x^(−1/2). At x = 9, m = (1/2)(9)^(−1/2) = (1/2)(1/3) = 1/6.

将 y 改写为 y = x^(1/2)，因此 dy/dx = (1/2)x^(−1/2)。在 x = 9 处，m = (1/2)(9)^(−1/2) = (1/2)(1/3) = 1/6。

Tangent equation: y − 3 = (1/6)(x − 9).

切线方程：y − 3 = (1/6)(x − 9)。

Multiplying through by 6: 6y − 18 = x − 9, so x − 6y + 9 = 0.

两边同乘 6：6y − 18 = x − 9，因此 x − 6y + 9 = 0。

The tangent is x − 6y + 9 = 0 | 切线方程为 x − 6y + 9 = 0

Surds and fractional indices are common in these problems, so keep your algebraic manipulation precise.

根式和分数指数在这些问题中很常见，因此请保持代数运算的精确性。

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7. The Normal Line | 法线

The normal to a curve at a point is the straight line perpendicular to the tangent at that point. While the tangent shares the gradient of the curve, the normal intersects the curve at the same point but at right angles to the tangent.

曲线在某一点的法线是过该点且垂直于切线的直线。切线与曲线在该点具有相同的斜率，而法线则以直角与切线相交于同一点。

If the tangent has gradient m, the normal has gradient −1/m, provided m ≠ 0. The product of the gradients of two perpendicular lines is −1.

如果切线的斜率为 m，则法线的斜率为 −1/m，前提是 m ≠ 0。两条互相垂直的直线的斜率乘积为 −1。

• If m = 0, the tangent is horizontal and the normal is vertical with equation x = x₁.

• 如果 m = 0，切线水平，法线为竖直直线，方程为 x = x₁。

• If m is undefined (vertical tangent), the normal is horizontal with equation y = y₁.

• 如果 m 不存在（竖直切线），法线为水平直线，方程为 y = y₁。

To find the normal equation, replace m with −1/m in the point-slope form.

要求法线方程，只需在点斜式中将 m 替换为 −1/m。

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8. Worked Example 3 | 示例 3

Find the equation of the normal to the curve y = x³ − 2x + 1 at the point where x = 1.

求曲线 y = x³ − 2x + 1 在 x = 1 处法线的方程。

Solution | 解答:

The y-coordinate: y = 1³ − 2(1) + 1 = 0. The point is (1, 0).

y 坐标：y = 1³ − 2(1) + 1 = 0。点为 (1, 0)。

Differentiate: dy/dx = 3x² − 2. At x = 1, the tangent gradient is m = 3(1)² − 2 = 1.

求导：dy/dx = 3x² − 2。在 x = 1 处，切线斜率 m = 3(1)² − 2 = 1。

The normal gradient is −1/m = −1.

法线斜率为 −1/m = −1。

Equation: y − 0 = −1(x − 1), which simplifies to y = −x + 1, or x + y − 1 = 0.

方程：y − 0 = −1(x − 1)，化简得 y = −x + 1，即 x + y − 1 = 0。

The normal is x + y − 1 = 0 | 法线方程为 x + y − 1 = 0

Always check whether the question asks for the tangent or the normal, as mixing them up is a common trap.

始终确认题目要求的是切线还是法线，因为将两者混淆是常见的陷阱。

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9. Points Where the Gradient Has a Specific Value | 斜率为特定值的点

Exam questions often ask you to find the coordinates of the point on a curve where the gradient equals a given number. To solve this, differentiate the curve, set f'(x) equal to the given gradient, and solve for x. Then substitute x back into the original equation to find y.

考题经常要求你求出曲线上斜率为某一给定值的点的坐标。解决方法是：先对曲线求导，令 f'(x) 等于给定的斜率，再解出 x。然后将 x 代回原方程求出 y。

Set f'(x) = m and solve for x | 令 f'(x) = m，解出 x

For example, find the points on y = x² − 4x + 5 where the gradient is 2. Differentiating gives 2x − 4 = 2, so 2x = 6 and x = 3. The y-coordinate is 3² − 4(3) + 5 = 2, so the point is (3, 2).

例如，求曲线 y = x² − 4x + 5 上斜率为 2 的点。求导得 2x − 4 = 2，因此 2x = 6，x = 3。y 坐标为 3² − 4(3) + 5 = 2，所以点为 (3, 2)。

A quadratic equation may yield two solutions, meaning there are two points on the curve with the required gradient.

二次方程可能给出两个解，这意味着曲线上有两个点具有所要求的斜率。

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10. Finding Points from a Tangent Condition | 由切线条件求点

Another common question type provides information about the tangent and asks you to find an unknown constant or the coordinates of the point. For instance, if a tangent is parallel to a given line, then the gradient of the tangent equals the gradient of that line.

另一种常见题型给出关于切线的信息，要求求出未知常数或点的坐标。例如，如果一条切线平行于给定直线，那么切线的斜率等于该直线的斜率。

Example | 示例: The tangent to y = ax² at x = 2 is parallel to y = 6x + 1. Find a.

示例：曲线 y = ax² 在 x = 2 处的切线平行于 y = 6x + 1。求 a。

Solution | 解答: dy/dx = 2ax. At x = 2, the gradient is 4a. Since parallel lines have equal gradients, 4a = 6, so a = 3/2.

解答：dy/dx = 2ax。在 x = 2 处，斜率为 4a。因为平行线斜率相等，所以 4a = 6，从而 a = 3/2。

This type of problem tests not only differentiation but also your understanding of the geometric meaning of the derivative.

这类问题不仅考查微分计算，还考查你对导数几何意义的理解。

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11. Second Derivatives and Concavity | 二阶导数与凹凸性

The second derivative, written as d²y/dx² or f”(x), is the derivative of the derivative. It measures the rate of change of the gradient and tells us about the concavity of the curve.

二阶导数，记作 d²y/dx² 或 f”(x)，是导数的导数。它衡量斜率的变化率，并告诉我们曲线的凹凸性。

• If f”(x) > 0 at a point, the curve is convex (curving upwards, like y = x²).

• 如果 f”(x) > 0，曲线在该点是凸的（向上弯曲，如 y = x²）。

• If f”(x) < 0 at a point, the curve is concave (curving downwards, like y = −x²).

• 如果 f”(x) < 0，曲线在该点是凹的（向下弯曲，如 y = −x²）。

• If f”(x) = 0, the point may be a point of inflection, where the curve changes concavity.

• 如果 f”(x) = 0，该点可能是拐点，即曲线改变凹凸性的点。

For a curve y = x³, we have dy/dx = 3x² and d²y/dx² = 6x. At x = 0, the second derivative is zero and the curve changes from concave to convex, confirming a point of inflection.

对于曲线 y = x³，有 dy/dx = 3x² 和 d²y/dx² = 6x。在 x = 0 处，二阶导数为零，曲线从凹变为凸，确认该点是拐点。

Although the second derivative is not strictly needed for finding tangent lines, it appears alongside tangent and normal questions in structured exam questions.

虽然求切线并不严格需要二阶导数，但在结构化的考试题目中，它常与切线和法线问题一起出现。

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12. Summary of Key Steps | 关键步骤总结

To find the tangent line at a point on a curve, always carry out the following procedure carefully.

要找到曲线上某一点的切线，务必认真执行以下步骤。

Step | 步骤	Action | 操作
1	Find the y-coordinate by substituting x into y = f(x). | 将 x 代入 y = f(x) 求 y 坐标。
2	Differentiate y = f(x) to obtain f'(x). | 对 y = f(x) 求导得到 f'(x)。
3	Substitute x into f'(x) to get the gradient m. | 将 x 代入 f'(x) 得到斜率 m。
4	Write y − y₁ = m(x − x₁) and simplify. | 写出 y − y₁ = m(x − x₁) 并化简。
5	For the normal, use gradient −1/m instead. | 对于法线，改用斜率 −1/m。

Practice differentiating polynomials, negative powers, and fractional powers until the rules are automatic. Then apply them confidently in tangent and normal problems, which are guaranteed to appear in Edexcel exams.

反复练习多项式、负指数和分数指数的求导，直到这些规则运用自如。然后自信地将其应用于切线和法线问题，这类题目在 Edexcel 考试中必然会出现。

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