Tangent and Normal Equations | 切线与法线方程

📚 Tangent and Normal Equations | 切线与法线方程

In differential calculus, the tangent line to a curve at a given point represents the instantaneous direction of the curve, while the normal line is perpendicular to the tangent at that point. This article covers how to derive both equations, a core topic for IB Mathematics.

在微分学中，曲线在某一点的切线表示曲线在该点的瞬时方向，而法线则是过该点与切线垂直的直线。本文系统讲解如何求切线与法线方程，这是 IB 数学的核心考点。

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1. Derivative as Slope | 导数作为斜率

The derivative f'(x) describes the rate of change of f(x) at a given value of x . Geometrically, f'(a) is the slope of the tangent line to the curve y = f(x) at the point (a, f(a)) .

导数 f'(x) 表示函数 f(x) 在给定 x 值处的变化率。从几何上看， f'(a) 就是曲线 y = f(x) 在点 (a, f(a)) 处切线的斜率。

If the derivative does not exist at a point, the tangent line may still have a limit (vertical tangent) or not exist at all. In IB, you are usually asked to find equations where the derivative is well defined.

如果某一点的导数不存在，则切线可能是竖直切线（极限存在）或根本不存在。在 IB 考试中，通常要求求导数存在点的切线方程。

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2. General Equation of a Tangent Line | 切线方程的一般形式

For a point (x₁, y₁) on a curve y = f(x) , the tangent line has slope m = f'(x₁) . Its equation is:

对于曲线 y = f(x) 上的点 (x₁, y₁) ，切线斜率为 m = f'(x₁) ，其方程为：

y - y₁ = f'(x₁)(x - x₁)

This is the point-slope form of a straight line. It can also be rearranged into slope-intercept or general form depending on the question.

这是直线方程的点斜式，也可以根据题目要求改写为斜截式或一般式。

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3. General Equation of a Normal Line | 法线方程的一般形式

The normal line at a point is perpendicular to the tangent line. Therefore its slope m_n satisfies the relation:

法线是过该点且与切线垂直的直线，因此其斜率 m_n 满足：

m_t × m_n = -1

Thus, for a tangent slope m_t = f'(x₁) ≠ 0 , the normal slope is:

因此，当切线斜率 m_t = f'(x₁) ≠ 0 时，法线斜率为：

m_n = -1 / f'(x₁)

If f'(x₁) = 0 ormal is vertical (x = x₁). If the tangent is vertical, the normal is horizontal (y = y₁).

若 f'(x₁) = 0 ，切线水平，则法线竖直，方程为 x = x₁ ；若切线竖直，则法线水平，方程为 y = y₁ 。

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4. Steps to Find a Tangent Equation | 求切线方程的步骤

Follow these steps when asked to find the tangent line at a given point:

求给定点处的切线方程可遵循以下步骤：

• Step 1: Differentiate y = f(x) to obtain f'(x) .

  第一步：对 y = f(x) 求导，得到 f'(x) 。

• Step 2: Substitute the x-coordinate of the given point into f'(x) to find the slope m .

  第二步：将给定点的横坐标代入 f'(x) ，求出斜率 m 。

• Step 3: Use the point (x₁, y₁) and the slope m in the point-slope form.

  第三步：利用点 (x₁, y₁) 和斜率 m 写出点斜式方程。

• Step 4: Simplify if required (e.g. into y = mx + c ).

  第四步：按要求化简（例如化为 y = mx + c 的形式）。

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5. Steps to Find a Normal Equation | 求法线方程的步骤

The normal line shares the same point but has a different slope. The steps are:

法线过同一点，但斜率不同，具体步骤如下：

• Step 1: Find the tangent slope m_t = f'(x₁) .

  第一步：先求切线斜率 m_t = f'(x₁) 。

• Step 2: Compute m_n = -1/m_t if m_t ≠ 0 .

  第二步：若 m_t ≠ 0 ，计算 m_n = -1/m_t 。

• Step 3: Write y - y₁ = m_n (x - x₁) .

  第三步：写出 y - y₁ = m_n (x - x₁) 。

• Step 4: Handle special cases: horizontal tangent gives vertical normal, etc.

  第四步：处理特殊情况：切线水平则法线竖直，等等。

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6. Example: Polynomial Curve | 例子：多项式曲线

Let f(x) = x² + 3x - 2 . Find the tangent and normal at x = 1 .

设 f(x) = x² + 3x - 2 ，求该曲线在 x = 1 处的切线与法线方程。

First find f(1) = 1 + 3 - 2 = 2 . So the point is (1, 2) .

先求 f(1) = 1 + 3 - 2 = 2 ，因此点为 (1, 2) 。

Differentiate: f'(x) = 2x + 3 . At x = 1 , m_t = 2(1) + 3 = 5 .

求导得 f'(x) = 2x + 3 。在 x = 1 处， m_t = 2(1) + 3 = 5 。

Tangent equation: y - 2 = 5(x - 1) , i.e. y = 5x - 3 .

切线方程： y - 2 = 5(x - 1) ，即 y = 5x - 3 。

Normal slope: m_n = -1/5 . Normal equation: y - 2 = -15(x - 1) , or y = -15x + 115 .

法线斜率 m_n = -1/5 。法线方程： y - 2 = -15(x - 1) ，即 y = -15x + 115 。

Tangent: y = 5x - 3 , Normal: y = -(1/5)x + 11/5

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7. Example: Trigonometric Function | 例子：三角函数

Consider y = sin x at the point where x = π/3 .

考虑 y = sin x 在 x = π/3 处的切线与法线。

We have f(π/3) = sin(π/3) = √3/2 . The derivative is f'(x) = cos x , so m_t = cos(π/3) = 1/2 .

因为 f(π/3) = sin(π/3) = √3/2 。导数 f'(x) = cos x ，所以 m_t = cos(π/3) = 1/2 。

Tangent: y - √3/2 = (1/2)(x - π/3) .

切线： y - √3/2 = (1/2)(x - π/3) 。

Normal slope: m_n = -2 . Normal equation: y - √3/2 = -2(x - π/3) .

法线斜率： m_n = -2 。法线方程： y - √3/2 = -2(x - π/3) 。

Remember to use radian measure for trigonometric derivatives in IB.

注意：IB 中三角函数的导数必须使用弧度制。

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8. Parametric Equations | 参数方程情形

When a curve is given in parametric form x = f(t), y = g(t) , the slope of the tangent is found using the chain rule:

当曲线由参数方程 x = f(t), y = g(t) 给出时，切线斜率通过链式法则求得：

dy/dx = (dy/dt) / (dx/dt) , provided dx/dt ≠ 0

At a specific parameter value t₀ , compute x(t₀) , y(t₀) , and the slope, then use the point-slope form as usual.

在特定参数值 t₀ 处，先计算 x(t₀) 、 y(t₀) 和斜率，然后照常使用点斜式。

For example, if x = t² , y = t³ , then dy/dx = (3t²)/(2t) = (3/2)t for t ≠ 0 . At t = 2 , x = 4, y = 8 , slope = 3 . Tangent: y - 8 = 3(x - 4) .

例如，若 x = t² , y = t³ ，则 dy/dx = (3t²)/(2t) = (3/2)t （ t ≠ 0 ）。在 t = 2 时， x = 4, y = 8 ，斜率为 3 。切线： y - 8 = 3(x - 4) 。

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9. Implicit Differentiation | 隐函数微分

For curves defined implicitly, such as x² + y² = 25 , differentiate both sides with respect to x , treating y as a function of x .

对于隐式定义的曲线，如 x² + y² = 25 ，对等式两边关于 x 求导，将 y 视为 x 的函数。

Differentiating gives 2x + 2y dydx = 0 , so dydx = -x/y . At a point such as (3, 4) , the slope is -3/4 .

求导得 2x + 2y dydx = 0 ，所以 dydx = -x/y 。在点 (3, 4) 处，斜率为 -3/4 。

Tangent: y - 4 = -34(x - 3) . Normal slope: 4/3 . Normal: y - 4 = 43(x - 3) .

切线： y - 4 = -34(x - 3) 。法线斜率为 4/3 。法线： y - 4 = 43(x - 3) 。

Be careful when y = 0 : the derivative may be undefined, often indicating a vertical tangent.

注意当 y = 0 时导数可能无定义，这通常意味着存在竖直切线。

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10. Common Mistakes and Pitfalls | 常见错误与陷阱

Students often make errors in the following areas:

学生在以下方面容易出错：

• Using the original function value instead of the derivative value as the slope.

  误将函数值当作斜率，而没有使用导数值。

• Forgetting to apply the chain rule when differentiating composite functions.

  对复合函数求导时忘记使用链式法则。

• Assuming m_n = -m_t instead of m_n = -1/m_t .

  错误地认为 m_n = -m_t ，而不是 m_n = -1/m_t 。

• Ignoring special cases where slopes are zero or infinite.

  忽略斜率为零或无穷大的特殊情况。

• Not checking that the given point actually lies on the curve.

  未检查给定点是否真的在曲线上。

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11. Workbook-Style Questions | 练习式问题

Try these typical IB-style questions to test your understanding:

尝试以下典型 IB 风格题目，检验你的理解：

Question	Hint
1. Find tangent to y = e2x at x = 0 .	Derivative is 2e2x ; slope at 0 is 2.
2. Find normal to y = ln x at x = 1 .	Slope of tangent is 1; normal slope is -1.
3. For x3 + y3 = 9 , find tangent at (1, 2) .	Use implicit differentiation: dy/dx = -x2/y2 .
4. Parametric curve x = t2 - 1, y = t2 + t : find normal at t = 1 .	Compute dy/dx = (2t+1)/(2t) ; at t=1 , slope = 3/2.

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12. Summary and Final Tips | 总结与最终建议

The key formulas for tangent and normal lines are based on the derivative. Always write the point-slope form carefully and convert to the required form.

切线与法线方程的核心公式都基于导数。务必准确写出点斜式，并按题目要求转换为相应形式。

In IB exams, show clear steps: differentiate, substitute, then form the equation. This maximizes method marks even if arithmetic errors occur.

在 IB 考试中，应展示清晰的步骤：求导、代入、写出方程。这样即使出现计算错误，也能获得方法分。

Practice with polynomials, trigonometric, exponential, logarithmic, parametric and implicit functions to be fully prepared.

建议练习多项式、三角函数、指数、对数、参数方程和隐函数等各类题型，以做好充分准备。

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