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 ( (x_1, y_1) ) on a curve ( y = f(x) ), the tangent line has slope ( m = f'(x_1) ). Its equation is:

对于曲线 ( y = f(x) ) 上的点 ( (x_1, y_1) )，切线斜率为 ( m = f'(x_1) )，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'(x_1) neq 0 ), the normal slope is:

因此，当切线斜率 ( m_t = f'(x_1) neq 0 ) 时，法线斜率为：

m_n = -1 / f'(x₁)

If ( f'(x_1) = 0 ), the tangent is horizontal, so the normal is vertical (x = x₁). If the tangent is vertical, the normal is horizontal (y = y₁).

若 ( f'(x_1) = 0 )，切线水平，则法线竖直，方程为 ( x = x_1 )；若切线竖直，则法线水平，方程为 ( y = y_1 )。

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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 ( (x_1, y_1) ) and the slope ( m ) in the point-slope form.

  第三步：利用点 ( (x_1, y_1) ) 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'(x_1) ).

  第一步：先求切线斜率 ( m_t = f'(x_1) )。

• Step 2: Compute ( m_n = -1/m_t ) if ( m_t neq 0 ).

  第二步：若 ( m_t neq 0 )，计算 ( m_n = -1/m_t )。

• Step 3: Write ( y – y_1 = m_n (x – x_1) ).

  第三步：写出 ( y – y_1 = m_n (x – x_1) )。

• 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) = x^2 + 3x – 2 ). Find the tangent and normal at ( x = 1 ).

设 ( f(x) = x^2 + 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 = -frac{1}{5}(x – 1) ), or ( y = -frac{1}{5}x + frac{11}{5} ).

法线斜率 ( m_n = -1/5 )。法线方程：( y – 2 = -frac{1}{5}(x – 1) )，即 ( y = -frac{1}{5}x + frac{11}{5} )。

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 = pi/3 ).

考虑 ( y = sin x ) 在 ( x = pi/3 ) 处的切线与法线。

We have ( f(pi/3) = sin(pi/3) = sqrt{3}/2 ). The derivative is ( f'(x) = cos x ), so ( m_t = cos(pi/3) = 1/2 ).

因为 ( f(pi/3) = sin(pi/3) = sqrt{3}/2 )。导数 ( f'(x) = cos x )，所以 ( m_t = cos(pi/3) = 1/2 )。

Tangent: ( y – sqrt{3}/2 = (1/2)(x – pi/3) ).

切线：( y – sqrt{3}/2 = (1/2)(x – pi/3) )。

Normal slope: ( m_n = -2 ). Normal equation: ( y – sqrt{3}/2 = -2(x – pi/3) ).

法线斜率：( m_n = -2 )。法线方程：( y – sqrt{3}/2 = -2(x – pi/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 ( t_0 ), compute ( x(t_0) ), ( y(t_0) ), and the slope, then use the point-slope form as usual.

在特定参数值 ( t_0 ) 处，先计算 ( x(t_0) )、( y(t_0) ) 和斜率，然后照常使用点斜式。

For example, if ( x = t^2 ), ( y = t^3 ), then ( dy/dx = (3t^2)/(2t) = (3/2)t ) for ( t neq 0 ). At ( t = 2 ), ( x = 4, y = 8 ), slope ( = 3 ). Tangent: ( y – 8 = 3(x – 4) ).

例如，若 ( x = t^2 ), ( y = t^3 )，则 ( dy/dx = (3t^2)/(2t) = (3/2)t )（( t neq 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 ( x^2 + y^2 = 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 ( x^2 + y^2 = 25 )，对等式两边关于 ( x ) 求导，将 ( y ) 视为 ( x ) 的函数。

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

求导得 ( 2x + 2y frac{dy}{dx} = 0 )，所以 ( frac{dy}{dx} = -x/y )。在点 ( (3, 4) ) 处，斜率为 ( -3/4 )。

Tangent: ( y – 4 = -frac{3}{4}(x – 3) ). Normal slope: ( 4/3 ). Normal: ( y – 4 = frac{4}{3}(x – 3) ).

切线：( y – 4 = -frac{3}{4}(x – 3) )。法线斜率为 ( 4/3 )。法线：( y – 4 = frac{4}{3}(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 = e^{2x} ) at ( x = 0 ).	Derivative is ( 2e^{2x} ); 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 ( x^3 + y^3 = 9 ), find tangent at ( (1, 2) ).	Use implicit differentiation: ( dy/dx = -x^2/y^2 ).
4. Parametric curve ( x = t^2 – 1, y = t^2 + 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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