Parametric Integration (pdfjoiner_(4)-132) | 参数积分（pdfjoiner_(4)-132）

📚 Parametric Integration (pdfjoiner_(4)-132) | 参数积分（pdfjoiner_(4)-132）

This article works through a classic A‑ iating a pair of parametric functions, finding the gradient of the tangent, and then move on to setting up and evaluating a definite integral to compute the area enclosed by the curve and the x‑ y step is explained to reinforce the key techniques required by the Edexcel specification.

本文详细讲解一道典型 A‑Level 数学参数方程题目，其题型与文件 pdfjoiner_(4)-132 中的问题高度相似。我们将对参数函数求导、计算切线斜率，然后建立定积分并求出曲线与 x 轴围成的面积。每一步都紧扣 Edexcel 考试大纲的核心技巧，帮助巩固答题思路。

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1. The parametric equations and the problem | 参数方程与问题陈述

The curve C is given by the parametric equations x = t² + 2t, y = 2t³ − 3t² at the point where t = 1. Part (c) asks for the area of the region bounded by C and the x‑axis between the points where the curve crosses the axis.

曲线 C 由参数方程 x = t² + 2t，y = 2t³ − 3t² 给出，t 为实数。典型考题的 (a) 部分要求用 t 表示 dy/dx；(b) 部分要求写出 t = 1 处的切线方程；(c) 部分要求计算曲线与 x 轴相交的两点之间、由 C 与 x 轴围成的区域面积。

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2. Finding dy/dx using parametric differentiation | 利用参数微分求 dy/dx

dy/dx is given by (dy/dt) ÷ (dx/dt). First compute dx/dt = 2t + 2. Next compute dy/dt = 6t² − 6t. Therefore dy/dx = (6t² (t + 1) = 3t(t − 1)/(t + 1). The simplified gradient function is valid wherever dx/dt ≠ 0.

参数方程中，导数 dy/dx 由 (dy/dt) ÷ (dx/dt) 给出。先计算 dx/dt = 2t + 2。再计算 dy/dt = 6t² − 6t。因此 dy/dx = (6t² − 6t) / (2t + 2)。该表达式可分解因式化简：6t(t − 1) / 2(t + 1) = 3t(t − 1)/(t + 1)。简化后的斜率函数在 dx/dt ≠ 0 时有效。

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3. Gradient and tangent at t = 1 | t = 1 处的斜率与切线

t of the tangent is zero, meaning a horizontal tangent. The coordinates at t = 1 are x = 1² + 2×1 = 3, y = 2×1³ − 3×1² = −1. Hence the tangent line is simply y = −1.

将 t = 1 代入 dy/dx：3×1×(1−1)/(1+1) = 0。因此切线斜率为零，即水平切线。t = 1 时的坐标为 x = 1² + 2×1 = 3，y = 2×1³ − 3×1² = −1。故切线方程为 y = −1。

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4. Points where the curve meets the x‑axis | 曲线与 x 轴的交点

The curve crosses the x‑axis when y = 0. Solve 2t³ − 3t² = 0 → t²(2t − 3) = 0 giving t = 0 and t = 3/2. The corresponding x‑coordinates are found from x = t² + 2t. For t = 0, x = 0. For t = 3/2, x = (3/2)² + 2×(3/2) = 9/4 + 3 = 21/4. The region of interest lies between x = 0 and x = 21/4.

曲线与 x 轴相交时 y = 0。解 2t³ − 3t² = 0 → t²(2t − 3) = 0，得 t = 0 和 t = 3/2。对应的 x 坐标由 x = t² + 2t 求得。t = 0 时 x = 0；t = 3/2 时 x = (3/2)² + 2×(3/2) = 9/4 + 3 = 21/4。我们关心的区域在 x = 0 与 x = 21/4 之间。

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5. Setting up the area integral | 建立面积积分

When a curve is given parametrically, the area between the curve and the x‑ ∫ y (dx/dt) dt, with the integration limits transformed accordingly. For our curve, y = 2t³ − 3t² and dx/dt = 2t + 2. The limits are t = 0 (lower) and t = 3/2 (upper). Thus the area A = ∫₀^3/2 (2t³ − 3t²)(2t + 2) dt.

当曲线由参数方程给出时，曲线与 x 轴之间从 x = a 到 x = b 的面积为 ∫ y dx，但需要将变量换为 t：面积 = ∫ y (dx/dt) dt，同时转换积分限。对于本题，y = 2t³ − 3t²，dx/dt = 2t + 2。积分限为 t = 0（下限）到 t = 3/2（上限）。因此面积 A = ∫₀^3/2 (2t³ − 3t²)(2t + 2) dt。

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6. Expanding the integrand | 展开被积函数

Multiply the brackets: (2t³)(2t + 2) = 4t⁴ + 4t³. Next (−3t²)(2t + 2) = −6t³ − 6t². Combine like terms: 4t⁴ + (4t³ − 6t³) − 6t² = 4t⁴ − 2t³ − 6t². So the integral simplifies to A = ∫₀^3/2 (4t⁴ − 2t³ − 6t²) dt.

展开括号：(2t³)(2t + 2) = 4t⁴ + 4t³。然后 (−3t²)(2t + 2) = −6t³ − 6t²。合并同类项：4t⁴ + (4t³ − 6t³) − 6t² = 4t⁴ − 2t³ − 6t²。因此积分简化为 A = ∫₀^3/2 (4t⁴ − 2t³ − 6t²) dt。

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7. Evaluating the definite integral | 计算定积分

Integrate term by term: ∫ 4t⁴ dt = (4/5)t⁵; ∫ −2t³ dt = −(2/4)t⁴ = −(1/2)t⁴; ∫ −6t² dt = −(6/3)t³ = −2t³. The antiderivative is F(t) = (4/5)t⁵ − (1/2)t⁴ − 2t³. Apply the limits t = 3/2 and t = 0. At t = 0, F(0) = 0.

逐项积分：∫ 4t⁴ dt = (4/5)t⁵；∫ −2t³ dt = −(2/4)t⁴ = −(1/2)t⁴；∫ −6t² dt = −(6/3)t³ = −2t³。原函数为 F(t) = (4/5)t⁵ − (1/2)t⁴ − 2t³。代入上下限 t = 3/2 和 t = 0。t = 0 时 F(0) = 0。

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8. Substituting the upper limit | 代入上限计算

Calculate F(3/2): t = 3/2 gives t³ = 27/8, t⁴ = 81/16, t⁵ ) = 243/40 after simplifying by 4: 972÷4=243, 160÷4=40. Next (1/2)×(81/16) = 81/32. And 2t³ 160; 81/32 = 405/160; 27/4 = 1080/160. Hence F(3/2) = (972 − 405 − 1080)/160 = (−513)/160.

计算 F(3/2)：t = 3/2 时 t³ = 27/8，t⁴ = 81/16，t⁵ = 243/32。于是 (4/5)×(243/32) = 972/160，约分后为 243/40。接着 (1/2)×(81/16) = 81/32。而 2t³ = 2×(27/8) = 54/8 = 27/4。因此 F(3/2) = 243/40 − 81/32 − 27/4。通分取公分母 160：243/40 = 972/160；81/32 = 405/160；27/4 = 1080/160。故 F(3/2) = (972 − 405 − 1080)/160 = (−513)/160。

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9. Interpreting the negative sign and final area | 负号的解释与最终面积

integral yields A = −513/160. The negative sign indicates that the curve lies below the x‑ quare units. This can be simplified to a mixed number: 3 and 33/160, but the exact fraction is usually preferred.

定积分结果为 A = −513/160。负号说明在该区间内曲线位于 x 轴下方，可通过取 t = 1 验证，此时 y = −1。实际面积为绝对值：面积 = 513/160 平方单位。该结果可化为带分数 3 又 33/160，但考试中通常保留精确分数。

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10. Checking the dx/dt sign and orientation | 检查 dx/dt 的符号与方向

itive, so x is strictly increasing, and the parametric integral correctly represents the area swept from left to right. No reversal of limits is needed, and the absolute area is simply the negative of the computed integral.

建议检查在此区间内 x 是否随 t 增大而增大。对于 t ∈ [0, 3/2]，dx/dt = 2t + 2 恒正，因此 x 严格递增，参数积分正确地表示了从左到右扫过的面积。积分限无需对调，实际面积只需对计算出的积分值取负。

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11. Alternative approach: Cartesian equation | 备选方法：直角坐标方程

For completeness, one could eliminate t to find a Cartesian relationship. From x = t² + 2t, solve for t: t² + 2t − x = 0, so t = −1 ± √1 + x. Since t ranges from 0 to 3/2, we take t = −1 + √1 + x. Substituting into y yields y = 2(√1+x − 1)³ − 3(√1+x − 1)². This is much messier, confirming why parametric integration is the efficient method for such questions.

为完整起见，可以消去 t 得到直角坐标关系。由 x = t² + 2t 解 t：t² + 2t − x = 0，故 t = −1 ± √1 + x。由于 t 的取值范围是 0 到 3/2，取 t = −1 + √1 + x。代入 y 得 y = 2(√1+x − 1)³ − 3(√1+x − 1)²。该表达式非常繁杂，从而证明在此类题目中参数积分是高效解法。

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12. Key takeaways for Edexcel A‑Level Maths | 对 Edexcel A‑Level 数学的启示

etric derivatives and parametric area integrals. Remember: dy/dx = (dy/dt) / (dx/dt) and ∫_x1^x2 y dx = ∫_t1^t2 y (dx/dt) dt. Always check the sign of dx/dt to ensure the limits are in the correct order. The absolute area is taken when the curve lies below the x‑axis. Mastery of algebraic simplification and fraction arithmetic is essential for accurate final answers.

类似 pdfjoiner_(4)-132 中的题目考查学生熟练运用参数导数和参数面积积分的能力。请牢记：dy/dx = (dy/dt) / (dx/dt)，∫_x1^x2 y dx = ∫_t1^t2 y (dx/dt) dt。务必检查 dx/dt 的符号以确保积分限顺序正确。当曲线位于 x 轴下方时取绝对面积。熟练掌握代数化简与分数运算对得出准确答案至关重要。

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