Problem Walkthrough: pdfjoiner_(4)-136 – Parametric Equations and Applications | 题目精讲：pdfjoiner_(4)-136 – 参数方程及其应用

📚 Problem Walkthrough: pdfjoiner_(4)-136 – Parametric Equations and Applications | 题目精讲：pdfjoiner_(4)-136 – 参数方程及其应用

sts core Pure Mathematics skills involving parametric equations, differentiation, tangents, stationary points and area integration. Every step is explained in detail so that you can understand the reasoning and apply it confidently in your exam.

本文针对 Edexcel A-Level 数学题目 pdfjoiner_(4)-136 进行完整解析。该题涵盖参数方程、求导、切线、驻点以及参数曲线下的面积等重要考点。我们将逐步展开讲解，帮助你彻底掌握每一个解题环节，为考试做好充分准备。

---

1. Problem Statement | 题目陈述

A curve C ametric equations x = 2 cos θ, y = 3 sin θ, where 0 ≤ θ ≤ 2π. The following tasks are set:
(a) Find dy/dx in terms of θ.
(b) Determine the equation of the tangent to C at the point where θ = π/4.
(c) Find the coordinates of the points on C at which the tangent is parallel to the x‑axis.
(d) Show that the exact area of the region enclosed by C is 6π.

曲线 C 由参数方程 x = 2 cos θ, y = 3 sin θ 定义，其中 0 ≤ θ ≤ 2π。题目要求如下：
(a) 求 dy/dx，用 θ 表示。
(b) 求曲线在 θ = π/4 处的切线方程。
(c) 求曲线上切线平行于 x 轴的所有点的坐标。
(d) 证明曲线所围区域的精确面积为 6π。

---

2. Understanding Parametric Equations | 理解参数方程

θ, y = 3 sin θ describe an ellipse centred at the origin. By eliminating θ (using the identity cos²θ + sin²θ = 1) we obtain the Cartesian equation x²/4 + y²/9 = 1, which has a horizontal semi‑axis of length 2 and a vertical semi‑axis of length 3.

参数方程用一个第三方变量（称为参数）来表示 x 和 y。本题中参数为 θ。方程 x = 2 cos θ, y = 3 sin θ 描述的是一个中心在原点的椭圆。若利用恒等式 cos²θ + sin²θ = 1 消去 θ，便可得到直角坐标方程 x²/4 + y²/9 = 1，其水平半轴长为 2，垂直半轴长为 3。

---

3. Differentiating Parametric Equations | 参数方程求导

hain rule: dy/dx = (dy/dθ) ÷ (dx/dθ). First differentiate each function with respect to θ: dx/dθ = −2 sin θ dy/dθ = 3 cos θ Thus

dy/dx = 3 cos θ / (−2 sin θ) = − (3/2) cot θ

the tangent at any point on C for which sin θ ≠ 0. When sin θ = 0 the tangent is vertical.

要从参数方程求导数 dy/dx，我们利用链式法则：dy/dx = (dy/dθ) ÷ (dx/dθ)。首先分别对 θ 求导： dx/dθ = −2 sin θ dy/dθ = 3 cos θ 因此

dy/dx = 3 cos θ / (−2 sin θ) = − (3/2) cot θ

该表达式给出了曲线上任意一点（sin θ ≠ 0）切线的斜率。当 sin θ = 0 时，切线为铅垂线。

---

4. Finding the Tangent Line | 求切线方程

we need the tangent at θ = π/4. First calculate the coordinates of the point of tangency: x = 2 cos(π/4) = 2 × (√2/2) = √2 y = 3 sin(π/4) = 3 × (√2/2) = (3√2)/2 The gradient at this point is m = −(3/2) cot(π/4) = −(3/2) × 1 = −3/2. Using the point‑slope form y − y₁ = m(x − x₁), the equation of the tangent is y − (3√2)/2 = −(3/2)(x − √2) Multiply both sides by 2 and simplify: 2y − 3√2 = −3(x − √2) 2y − 3√2 = −3x + 3√2 Bring all terms to one side: 3x + 2y − 6√2 = 0.

第 (b) 小题要求 θ = π/4 处的切线。首先求出切点坐标： x = 2 cos(π/4) = √2 y = 3 sin(π/4) = (3√2)/2 该点处切线的斜率为 m = −(3/2) cot(π/4) = −3/2。 利用点斜式 y − y₁ = m(x − x₁)，切线方程为 y − (3√2)/2 = −(3/2)(x − √2) 两边同乘 2 并整理： 2y − 3√2 = −3x + 3√2 ⇒ 3x + 2y − 6√2 = 0。

---

广告 / Sponsored

5. Stationary Points and Tangents Parallel to Axes | 驻点与平行切线

The question asks for points where the tangent is parallel to the x‑ he interval 0 ≤ θ ≤ 2π, cos θ = 0 when θ = π/2 and θ = 3π/2. (At these values sin θ is non‑zero, so the cotangent is defined in the limiting sense.)

Substitute θ = π/2: x = 2 cos(π/2) = 0, y = 3 sin(π/2) = 3 → (0, 3) Substitute θ = 3π/2: x = 2 cos(3π/2) = 0, y = 3 sin(3π/2) = −3 → (0, −3) angents, they occur when dx/dθ = 0, i.e. sin θ = 0, which yields the points (2,0) and (−2,0).)

题目要求切线平行于 x 轴的点，即 dy/dx = 0。令 −(3/2) cot θ = 0，得 cot θ = 0，故 cos θ = 0。在 0 ≤ θ ≤ 2π 内，cos θ = 0 的解为 θ = π/2 和 θ = 3π/2（这些点处 sin θ 不为零，余切在极限意义下有意义）。

代入 θ = π/2：x = 0, y = 3 → (0, 3) 代入 θ = 3π/2：x = 0, y = −3 → (0, −3) 这两个点是椭圆的上下顶点，其切线均为水平线，斜率为零。（若同时考查竖直切线，则令 dx/dθ = 0，即 sin θ = 0，得 (2,0) 与 (−2,0)。）

---

6. Area Under Parametric Curves | 参数曲线下的面积

) we need the area enclosed by the ellipse. The area bound by a parametric curve and the x‑axis for a given parameter range is given by the integral A = ∫ y dx = ∫ y (dx/dθ) dθ areful attention must be paid to the sign of the integrand to ensure the area is positive. For symmetric shapes we can integrate over a suitable half‑ or quarter‑cycle and multiply.

第 (d) 小题要求椭圆所围区域的面积。由参数曲线与 x 轴围成的面积公式为： A = ∫ y dx = ∫ y (dx/dθ) dθ 其中积分限用参数 θ 表示。由于曲线是一条封闭环，总围成面积可通过参数从 0 到 2π 历经一个完整循环的积分求得，但需要注意被积函数的正负以确保面积值为正。对于对称图形，通常会选择半周期或四分之一周期积分再乘以适当的倍数。

---

7. Enclosed Area Calculation | 封闭区域面积计算

We have dx/dθ = −2 sin θ and y = 3 sin θ. Thus y dx/dθ = 3 sin θ × (−2 sin θ) = −6 sin²θ. ely because the same area is traced twice with opposite orientation (above and below the x‑ y are positive. In the first quadrant, θ runs from 0 to π/2. During this interval, y ≥ 0 and dx/dθ = −2 sin θ ≤ 0, so y dx is negative, but the geometric area is the absolute value of the integral. It is easiest to integrate the upper half of the ellipse (θ from 0 to π) and double.

Let us integrate for the upper half: y = 3 sin θ, dx = −2 sin θ dθ. The area above the x‑axis (y ≥ 0, θ from 0 to π) is A_upper = ∫ (θ=0 to θ=π) |y dx| = ∫₀π 3 sin θ · |−2 sin θ| dθ = ∫₀π 6 sin²θ dθ. Because sin θ ≥ 0 for θ ∈ lue simply removes the negative sign from dx/dθ. The total enclosed area is then A = 2 × A_upper.

因 dx/dθ = −2 sin θ 且 y = 3 sin θ，乘积 y dx/dθ = −6 sin²θ。若直接计算 ∫₀²π −6 sin²θ dθ 会因正负抵消而得到零。为此利用对称性：椭圆关于两坐标轴对称，总面积等于第一象限面积的四倍。第一象限内 θ 从 0 到 π/2，此时 y ≥ 0 而 dx/dθ ≤ 0，积分值为负，几何面积取绝对值。更为简便的方法是先计算上半椭圆（θ 从 0 到 π）的面积再乘以 2。

上半部分：y = 3 sin θ, dx = −2 sin θ dθ。由于在 [0,π] 内 sin θ ≥ 0，去掉 dx 的负号后，上半面积 A_upper = ∫

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）