📚 Tackling a Parametric Equations Challenge from Edexcel Mathematics (PDF Joiner 4-130) | 从Edexcel数学参数方程难题解析（PDF合并文件4-130页）
Many A-Level Edexcel revision packs compiled from past papers contain hidden gems on pages like the one labelled "PDF Joiner (4)-130". This article reconstructs and solves a parametric equations problem that mirrors the style and difficulty of such exam questions. We will work through differentiation, tangents, normals, stationary points and area integration, giving you a step-by-step model answer that covers key pure mathematics techniques.
许多由历年真题汇编而成的A-Level Edexcel复习资料中，常常隐藏着像“PDF Joiner (4)-130”这类页面上的好题。本文重构并详细解析了一道参数方程综合题，紧密贴合此类考试题目的风格与难度。我们将逐步完成求导、切线、法线、驻点以及面积积分，为你展示一个涵盖核心纯数技巧的完整范例。
1. Introducing the Challenge on Page 130 | 介绍第130页的挑战
d the equation of the normal to C at the point where t = 2. (c) Determine the coordinates of the points on C where the tangent is parallel to the x-axis. (d) The region R is bounded by C, the x-axis, and the lines x = 1 and x = 5. Calculate the total area of R, giving your answer in exact form.
我们要解决的题目是：一条曲线C的参数方程为 x = t² + 1, y = t³ − 3t，其中 t 是实数参数。(a) 证明 dy/dx = (3t² − 3) / (2t)。(b) 求曲线C在 t = 2 对应点处的法线方程。(c) 确定C上切线与x轴平行的所有点的坐标。(d) 区域R由曲线C、x轴以及直线 x = 1 和 x = 5 围成。计算R的总面积，结果保留精确值。
This type of question frequently appears in Edexcel Pure Mathematics 3 or the equivalent components. It tests your ability to manipulate parametric derivatives, link geometry to calculus, and handle signed areas. Let's unpack it systematically.
这类题目经常出现在Edexcel 纯数3或相应模块中，它考验你对参数导数的处理、几何与微积分的结合以及带符号面积的处理能力。下面我们逐步拆解。
2. Core Technique: Differentiating Parametric Equations | 核心技术：参数方程求导
≠ 0. This is simply an application of the chain rule. You must always express the derivative in terms of the parameter t unless the question asks you to substitute back to x and y.
对于由 x = f(t), y = g(t) 给出的曲线，梯度 dy/dx 可通过 dy/dx = (dy/dt) / (dx/dt) 求得，前提是 dx/dt ≠ 0。这只是链式法则的应用。除非题目要求将结果换回 x 和 y，否则你应当始终用参数 t 来表示导数。
In our problem, we compute the two elementary derivatives directly:
dx/dt = 2t, dy/dt = 3t² − 3
Therefore, the gradient function in terms of t is:
dy/dx = (3t² − 3) / (2t)
这为解答第一部分提供了直接的证明。
3. Working Out Part (a) – Gradient Expression | 解答 (a) – 梯度表达式
t). It is good practice to note that this expression is undefined when t = 0, which corresponds to a point where the tangent is vertical. Showing this awareness can impress examiners.
第(a)小题只要求你写出导数公式并化简。应写为：dx/dt = 2t，dy/dt = 3t² − 3，因此 dy/dx = (3t² − 3) / (2t)。良好的习惯是注明当 t = 0 时此式无定义，这对应于曲线的一条垂直切线，展现出你对题目更深的理解会给考官留下好印象。
4. Solving Part (b) – Tangent and Normal at t = 2 | 解答 (b) – t = 2处的切线与法线
e parametric equations: x = 2² + 1 = 5, y = 2³ − 3(2) = 8 − 6 = 2. So the point is (5, 2).
首先，求出对应点坐标。将 t = 2 代入参数方程：x = 2² + 1 = 5，y = 2³ − 3×2 = 8 − 6 = 2。因此点的坐标为 (5, 2)。
3) / (2×2) = (12 − 3) / 4 = 9/4. The gradient of the normal is the negative reciprocal: mnormal = −4/9.
接下来，计算 t = 2 处的切线斜率。代入第(a)小题的结果，dy/dx = (3(2)² − 3) / (2×2) = (12 − 3) / 4 = 9/4。法线斜率是切线的负倒数：mnormal = −4/9。
y − 38 = 0. You can leave the equation in any equivalent form expected by the mark scheme.
现在利用点斜式：y − 2 = (−4/9)(x − 5)。两边同乘并整理，得到 9y − 18 = −4x + 20，即 4x + 9y − 38 = 0。你可以按评分标准要求的任何等价形式来书写方程。
5. Finding Points with Horizontal Tangents (Part c) | 寻找水平切线的点 (c)
o, provided the denominator is not zero. So we solve 3t² − 3 = 0, giving t² = 1 or t = ±1.
切线与x轴平行意味着梯度 dy/dx = 0。令梯度表达式的分子为零，同时分母不为零。于是解方程 3t² − 3 = 0，得 t² = 1，即 t = ±1。
2, −2) and (2, 2). Notice they share the same x-coordinate, forming a tiny vertical stack; the curve crosses itself there.
当 t = 1：x = 1² + 1 = 2，y = 1³ − 3(1) = −2。当 t = −1：x = (−1)² + 1 = 2，y = (−1)³ − 3(−1) = −1 + 3 = 2。因此所求点为 (2, −2) 和 (2, 2)。注意到它们的x坐标相同，形成一个竖直的“堆叠”，曲线在此处自交。
6. Setting Up the Area Integral (Part d) | 建立面积积分 (d)
tegrate y with respect to x across the interval, but we must take absolute values where the curve dips below the x-axis. It is usually simpler to convert the integral to the parameter t.
区域R由曲线C、x轴以及垂直线 x = 1 和 x = 5 围成。要求出总面积，需对 y 关于 x 在区间上积分，但在曲线位于x轴下方的部分必须取绝对值。通常把积分转换为以参数 t 表示会更简便。
When x = 5, t² = 4 gives t = 2. The curve crosses the x-axis when y = 0, i.e. t³ − 3t = 0 ⇒ t(t² − 3) = 0 ⇒ t = 0, t = √3 (since t = −√3 is not in our t-interval [0, 2]). So the relevant sign change occurs at t = √3.
由 x = t² + 1，当 x = 1 时有 t = 0（选取正向分支以连续描绘曲线）。当 x = 5，t² = 4 得 t = 2。曲线与x轴相交时 y = 0，即 t³ − 3t = 0 ⇒ t(t² − 3) = 0 ⇒ t = 0, t = √3（t = −√3 不在我们的区间 [0, 2] 内）。因此相关的符号变化出现在 t = √3。
The total area is given by: Area = ∫x=15 |y| dx = ∫t=0√3 (−y)(dx/dt) dt + ∫t=√32 y (dx/dt) dt. The first integral has a minus sign because y is negative there, and we need the positive area.
总面积为：Area = ∫x=15 |y| dx = ∫t=0√3 (−y)(dx/dt) dt + ∫t=√32 y (dx/dt) dt。第一个积分前有一个负号，因为在这段 y 为负，而我们想要正值面积。
7. Splitting the Integral at the x-axis Crossing | 在x轴交点处拆分积分
nd y = t³ − 3t. Thus y · dx/dt = (t³ − 3t)·2t = 2t⁴ − 6t². The two area contributions are:
A₁ = ∫0√3 −(2t⁴ − 6t²) dt = ∫0√3 (6t² − 2t⁴) dt
A₂ = ∫√32 (2t⁴ − 6t²) dt
我们有 dx/dt = 2t，y = t³ − 3t。因此 y · dx/dt = (t³ − 3t)·2t = 2t⁴ − 6t²。两部分面积贡献为上述两式。
s essential to set the integrals correctly. Always sketch the region mentally: from t=0 to √3 the curve lies below the x-axis, then from √3 to 2 it is above. Using a sign-reversed integral for the negative portion guarantees the area adds up correctly.
正确设定积分至关重要。在脑海中对区域做一个简单的草图：从 t=0 到 √3 曲线在x轴下方，随后从 √3 到 2 在x轴上方。对负值部分做符号反转的积分能确保面积正确累加。
8. Evaluating the Definite Integrals | 计算定积分
Find the antiderivative of 2t⁴ − 6t². Integrating term by term:
∫ (2t⁴ − 6t²) dt = (2/5)t⁵ − 2t³ + C
For A₁, we use the opposite sign integrand 6t² − 2t⁴, whose antiderivative is:
∫ (6t² − 2t⁴) dt = 2t³ − (2/5)t⁵ + C
求 2t⁴ − 6t² 的原函数。逐项积分：∫ (2t⁴ − 6t²) dt = (2/5)t⁵ − 2t³ + C。
对于 A₁，我们使用符号相反的积分式 6t² − 2t⁴，其原函数为：∫ (6t² − 2t⁴) dt = 2t³ − (2/5)t⁵ + C。
Now evaluate A₁ from 0 to √3:
A₁ = [2(√3)³ − (2/5)(√3)⁵] − [0]
计算 (√3)³ = 3√3, (√3)⁵ = (√3)²·(√3)³ = 3·3√3 = 9√3. Thus:
A₁ = 2·3√3 − (2/5)·9√3 = 6√3 − (18/5)√3 = (30/5 − 18/5)√3 = (12/5)√3
现在计算 A₁ 从 0 到 √3：A₁ = [2(√3)³ − (2/5)(√3)⁵] − [0]。计算 (√3)³ = 3√3，(√3)⁵ = 9√3。因此 A₁ = 2·3√3 − (2/5)·9√3 = 6√3 − (18/5)√3 = (30/5 − 18/5)√3 = (12/5)√3。
Now evaluate A₂ from √3 to 2 using the antiderivative of 2t⁴ − 6t²:
A₂ = [ (2/5)t⁵ − 2t³ ]√32
At t = 2: (2/5)(32) − 2(8) = 64/5 − 16 = 64/5 − 80/5 = −16/5. At t = √3: (2/5)(9√3) − 2(3√3) = (18/5)√3 − 6√3 = (18/5 − 30/5)√3 = (−12/5)√3.
Subtracting: A₂ = (−16/5) − (−12/5 √3) = −16/5 + (12/5)√3. Since A₂ should be positive (the area above the axis), we check: (12/5)√3 ≈ 4.156, minus 3.2, positive. Good.
现在计算 A₂ 从 √3 到 2，用 2t⁴ − 6t² 的原函数。t = 2 时：(2/5)(32) − 2(8) = 64/5 − 16 = 64/5 − 80/5 = −16/5。t = √3 时：(2/5)(9√3) − 2(3√3) = (18/5)√3 − 6√3 = (18/5 − 30/5)√3 = (−12/5)√3。相减：A₂ = (−16/5) − (−12/5 √3) = −16/5 + (12/5)√3。A₂ 应为正值（上方面积），验算：(12/5)√3 ≈ 4.156，减去3.2，结果为正，无误。
9. Final Exact Area and Verification | 最终精确面积与验证
Total area = A₁ + A₂ = (12/5)√3 + ( −16/5 + (12/5)√3 ) = (24/5)√3 − 16/5. Factorising 8/5 gives a neat exact answer:
Total Area = (8/5)(3√3 − 2) square units
总面积 = A₁ + A₂ = (12/5)√3 + ( −16/5 + (12/5)√3 ) = (24/5)√3 − 16/5。提取公因子 8/5 可写成整齐的精确答案：总面积 = (8/5)(3√3 − 2) 平方单位。
To verify, you can quickly approximate: √3 ≈ 1.732, 3√3 − 2 ≈ 3.196, times 8/5 = 1.6 gives ~5.11. A rough sketch confirms the magnitude is plausible.
快速近似验证：√3 ≈ 1.732，3√
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