📚 Exercise 8D.2: Tangents and Normals | 练习8D.2：切线与法线
In IB Mathematics, Exercise 8D.2 typically focuses on applying differentiation to find the equations of tangents and normals to a curve at a given point. This is a fundamental skill that bridges the gap between algebra and calculus, and it appears frequently in both Analysis & Approaches and Applications & Interpretation courses. Mastering this exercise deepens your understanding of the geometric meaning of the derivative.
在IB数学中，练习8D.2通常集中于运用微分求曲线在某一点处的切线和法线方程。这是一项连接代数与微积分的基本技能，频繁出现在分析与方法和应用与解释课程中。掌握这个练习能加深你对导数几何意义的理解。
1. The Derivative as Slope | 导数即斜率
The derivative of a function f' int where the function is differentiable. If we need the slope at a specific point P(a, f(a)), we simply evaluate f'(a). This instantaneous rate of change is the gradient of the tangent.
函数的导数 f'(x) 或 dy/dx 给出曲线 y = f(x) 在任意可微点处的切线斜率。若需在某点 P(a, f(a)) 处的斜率，只需计算 f'(a)。这一瞬时变化率就是切线的梯度。
2. Equation of a Tangent | 切线方程
Once the slope mT = f' t line passing through (a, f(a)). Using the point-slope form, the equation is y - f(a) = mT xams, you are usually asked to leave the answer in the form y = mx + c or ax + by + d = 0.
一旦求得斜率 mT = f'(a)，切线就是一条经过 (a, f(a)) 的直线。利用点斜式，其方程为 y - f(a) = mT (x - a)。在IB考试中，通常要求将答案写成 y = mx + c 或 ax + by + d = 0 的形式。
3. Equation of a Normal | 法线方程
The normal line is perpendicular to the tangent at the point of contact. Therefore its slope mN is the negative reciprocal of mT, provided mT ≠ 0: mN = -1/mT. The normal also passes through (a, f(a)), so its equation is y - f(a) = mN (x - a). Special cases: if the tangent is horizontal (mT rmal is vertical (x = a); if the tangent is vertical, the normal is horizontal (y = f(a)).
法线是在切点处与切线垂直的直线。因此其斜率 mN 是 mT 的负倒数（假设 mT ≠ 0）：mN = -1/mT。法线同样经过 (a, f(a))，故其方程为 y - f(a) = mN (x - a)。特殊情况：若切线水平 (mT = 0)，法线垂直 (x = a)；若切线垂直，法线水平 (y = f(a))。
4. Step-by-Step Strategy | 解题步骤
Step 1: If only the x-coordinate is given, find the corresponding y-coordinate by substituting into the original function f(x).
Step 2: Differentiate f(x) to obtain f'(x).
Step 3: Evaluate f'(a) to get the tangent gradient mT.
Step 4: Write the tangent equation using point-slope form and simplify.
Step 5: Find the normal gradient mN = -1/mT and write the normal equation, simplifying to the required form.
第一步：若只给出x坐标，代入原函数 f(x) 求出对应的y坐标。
第二步：对 f(x) 求导得到 f'(x)。
第三步：计算 f'(a) 得切线斜率 mT。
第四步：用点斜式写出切线方程并化简。
第五步：求法线斜率 mN = -1/mT，写出法线方程并化为所要求的形式。
5. Worked Example 1 – Polynomial Function | 示例1 – 多项式函数
equations of the tangent and normal to the curve y = x³ - 3x + 2 at the point where x = 1.
求曲线 y = x³ - 3x + 2 在 x = 1 处的切线和法线方程。
Step 1: y-coordinate: f(1) = 1³ - 3(1) + 2 = 0. Point is (1, 0).
Step 2: f'(x) = 3x² - 3.
Step 3: mT = f'(1) = 3(1)² - 3 = 0.
Step 4: Tangent equation: y - 0 = 0(x - 1) → y = 0. This is a horizontal line.
Step 5: Since mT = 0, the normal is vertical: x = 1.
第一步：y坐标：f(1) = 1³ - 3(1) + 2 = 0。点为 (1, 0)。
第二步：f'(x) = 3x² - 3。
第三步：mT = f'(1) = 3(1)² - 3 = 0。
第四步：切线方程：y - 0 = 0(x - 1) → y = 0。这是水平线。
第五步：由于 mT = 0，法线为垂直线：x = 1。
6. Worked Example 2 – Rational Function | 示例2 – 有理函数
Consider f(x) = (2x + 1)/(x - 3). Find the tangent and normal at the point where x = 4.
考虑 f(x) = (2x + 1)/(x - 3)。求在 x = 4 处的切线和法线。
Step 1: f(4) = (9)/(1) = 9. Point (4, 9).
Step 2: Use quotient rule: f'(x) = [2(x - 3) - (2x + 1)(1)] / (x - 3)² = (2x - 6 - 2x - 1)/(x - 3)² = -7/(x - 3)².
Step 3: mT = f'(4) = -7/(1)² = -7.
Step 4: Tangent: y - 9 = -7(x - 4) → y = -7x + 28 + 9 → y = -7x + 37.
Step 5: mN = 1/7. Normal: y - 9 = (1/7)(x - 4) → y = (1/7)x - 4/7 + 63/7 → y = (1/7)x + 59/7.
第一步：f(4) = (9)/(1) = 9。点 (4, 9)。
第二步：用商法则求导：f'(x) = [2(x - 3) - (2x + 1)(1)] / (x - 3)² = (2x - 6 - 2x - 1)/(x - 3)² = -7/(x - 3)²。
第三步：mT = f'(4) = -7/(1)² = -7。
第四步：切线：y - 9 = -7(x - 4) → y = -7x + 37。
第五步：mN = 1/7。法线：y - 9 = (1/7)(x - 4) → y = (1/7)x + 59/7。
7. Worked Example 3 – Using the Chain Rule | 示例3 – 应用链式法则
Find the normal to the curve y = √2x + 5 at the point where the curve cuts the x-axis.
求曲线 y = √2x + 5 在它与x轴交点处的法线。
The curve cuts the x-axis when y = 0: √2x + 5 = 0 → 2x + 5 = 0 → x = -5/2. Point is (-2.5, 0).
Differentiate: y = (2x + 5)1/2. f'(x) = (1/2)(2x + 5)-1/2 × 2 = 1 / √2x + 5.
At x = -5/2, the denominator becomes 0, so f' s undefined. The tangent is vertical; therefore the normal is horizontal. Equation: y = 0.
曲线与x轴相交时 y = 0：√2x + 5 = 0 → 2x + 5 = 0 → x = -5/2。点为 (-2.5, 0)。
求导：y = (2x + 5)1/2。f'(x) = (1/2)(2x + 5)-1/2 × 2 = 1 / √2x + 5。
在 x = -5/2 处分母为零，f' 无定义。切线垂直，故法线为水平。方程：y = 0。
8. Common Mistakes and How to Avoid Them | 常见错误及规避方法
Many students forget to find the y-coordinate before differentiating. Always compute f(a) first. Another error is misapplying the negative reciprocal when finding the normal gradient: remember mN = -1/mT, not simply -mT al equation is expressed in the form requested by the IB mark scheme (usually y = mx + c).
许多学生忘记在求导前先求出y坐标。务必先计算 f(a)。另一个错误是求法线斜率时错误使用负倒数：记住 mN = -1/mT，而非只是 -mT。此外，化简分式时（尤其是有理函数和根式函数）要仔细检查代数运算。最后，确保最终方程符合IB评分标准所要求的形式（通常为 y = mx + c）。
9. IB Exam-Style Practice Questions | IB考试风格练习题
Try these to consolidate your understanding of Exercise 8D.2:
尝试以下题目以巩固对练习8D.2的理解：
- on of the tangent to y = x² ln x at x = 1. (Hint: use product rule and remember ln 1 = 0.)
- 求 y = x² ln x 在 x = 1 处的切线方程。（提示：使用积法则，记住 ln 1 = 0。）
- For the curve y = e2x cos x, find the normal at the point where x = 0. Give the answer in exact form.
- 对于曲线 y = e2x cos x，求在 x = 0 处的法线方程。答案用精确形式表示。
- + ax + b at (1, 3) has a gradient of 5. Find a and b, then determine the normal equation.
- 曲线 y = x³ + ax + b 在 (1, 3) 处的切线斜率为5。求 a 和 b，然后确定法线方程。
10. Checking Your Answers with Technology | 用技术验证答案
In IB examinations, calculator use is often permitted (depending on the paper). You can verify your tangent and normal equations by graphing the function and the lines. The tangent should just touch the curve at the point, while the normal should appear perpendicular. Use the derivative function on your GDC to confirm the slope, and check that the line passes through the given point using the trace feature. This is a valuable self-checking strategy.
在IB考试中，通常允许使用计算器（视具体试卷而定）。你可以通过绘制函数与直线图像来验证切线和法线方程。切线应刚好在点处接触曲线，而法线应与之垂直。利用GDC上的导数函数确认斜率，并通过追踪功能检查直线是否经过给定点。这是一种有价值的自检策略。
11. The Connection to Increasing/Decreasing Functions | 与函数增减性的联系
The sign of the tangent slope indicates whether the function is increasing or decreasing at that point. If f'(a) > 0, the function is increasing, and the tangent slopes upward. If f'(a) < 0, the function is decreasing. This contextual understanding is often tested alongside tangent/normal problems in IB, especially in kinematics and optimisation questions.
切线斜率的符号表示函数在该点是递增还是递减。若 f'(a) > 0，函数递增，切线向上倾斜；若 f'(a) < 0，函数递减。这种语境理解常在IB的切线/法线问题中同时考查，特别是在运动学和优化问题中。
12. Summary and Revision Tips | 总结与复习提示
ent. From there, the normal follows as perpendicular. Practise a variety of function types—polynomials, exponentials, logarithms, trigonometric, and combinations— s and negative reciprocal conversions handy during revision. With consistent practice, you'll be able to solve any tangent/normal problem efficiently and accurately in your IB exam.
练习8D.2是你将微分熟练转化为几何方程的机会。始终牢记核心关系：导数 = 切线斜率。由此出发，法线作为垂直线自然得出。练习各种函数类型——多项式、指数、对数、三角函数及其组合——以建立信心。复习时身边常备微分法则和负倒数转换的公式表。通过持续练习，你将能在IB考试中高效、准确地解决任何切线/法线问题。
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