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FP3 2023 Jan Q6

A Level / Edexcel / FP3

IAL 2023 Jan Paper · Question 6

题目

Problem

6.

A curve has parametric equations

x=a(θ−sin⁡θ)y=a(1−cos⁡θ)x=a(\theta-\sin\theta) \qquad y=a(1-\cos\theta)

where aa is a positive constant.

(a) Show that

(dxdθ)2+(dydθ)2=ka2sin⁡2θ2\bigl(\frac{\mathrm{d}x}{\mathrm{d}\theta}\bigr)^2+\bigl(\frac{\mathrm{d}y}{\mathrm{d}\theta}\bigr)^2 =ka^2\sin^2\frac{\theta}{2}

where kk is a constant to be determined.

(4)

The part of the curve from θ=0\theta=0 to θ=2π\theta=2\pi is rotated through 2π2\pi radians about the xx-axis.

(b) Determine the area of the surface generated, giving your answer in terms of π\pi and aa.

[Solutions relying on calculator technology are not acceptable.]

(5)
题目中文翻译

6.

一条曲线的参数方程为

x=a(θ−sin⁡θ)y=a(1−cos⁡θ)x=a(\theta-\sin\theta) \qquad y=a(1-\cos\theta)

其中 aa 为正常数。

(a) 证明

(dxdθ)2+(dydθ)2=ka2sin⁡2θ2\bigl(\frac{\mathrm{d}x}{\mathrm{d}\theta}\bigr)^2+\bigl(\frac{\mathrm{d}y}{\mathrm{d}\theta}\bigr)^2 =ka^2\sin^2\frac{\theta}{2}

其中 kk 为待定常数。

(4)

从 θ=0\theta=0 到 θ=2π\theta=2\pi 的那段曲线绕 xx 轴旋转 2π2\pi 弧度。

(b) 求生成曲面的面积，答案用 π\pi 和 aa 表示。

[不接受依赖计算器技术的解答。]

(5)

解答

(a)

解法一

思路

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先分别对两个参数方程关于 θ\theta 求导，再平方相加。利用 sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 化成 1−cos⁡θ1-\cos\theta，最后用半角恒等式 1−cos⁡θ=2sin⁡2θ21-\cos\theta=2\sin^2\frac\theta2 得到指定形式。

答题过程

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Differentiating with respect to θ\theta,

dxdθ=a(1−cos⁡θ),dydθ=asin⁡θ.\frac{\mathrm{d}x}{\mathrm{d}\theta} =a(1-\cos\theta), \qquad \frac{\mathrm{d}y}{\mathrm{d}\theta} =a\sin\theta.

Therefore,

(dxdθ)2+(dydθ)2= a2(1−cos⁡θ)2+a2sin⁡2θ= a2(1−2cos⁡θ+cos⁡2θ+sin⁡2θ)= 2a2(1−cos⁡θ)= 4a2sin⁡2(θ2).\begin{align*} \biggl(\frac{\mathrm{d}x}{\mathrm{d}\theta}\biggr)^2 +\biggl(\frac{\mathrm{d}y}{\mathrm{d}\theta}\biggr)^2 =&\,a^2(1-\cos\theta)^2+a^2\sin^2\theta \\ =&\,a^2\bigl( 1-2\cos\theta+\cos^2\theta+\sin^2\theta \bigr) \\ =&\,2a^2(1-\cos\theta) \\ =&\,4a^2\sin^2\biggl(\frac\theta2\biggr). \end{align*}

Hence

k=4.\boxed{k=4}.

(b)

解法一

思路

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参数曲线绕 xx 轴旋转的曲面面积为 2π∫ydsdθ dθ2\pi\int y\frac{\mathrm{d}s}{\mathrm{d}\theta}\,\mathrm{d}\theta。承接 (a) 求出弧长因子时，必须根据给定参数范围确定绝对值的正负；再用半角恒等式把被积函数化成 sin⁡3θ2\sin^3\frac\theta2，以准确的代换完成积分。

答题过程

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From part (a),

dsdθ= (dxdθ)2+(dydθ)2= 2a∣sin⁡(θ2)∣.\begin{align*} \frac{\mathrm{d}s}{\mathrm{d}\theta} =&\,\sqrt{ \biggl(\frac{\mathrm{d}x}{\mathrm{d}\theta}\biggr)^2 +\biggl(\frac{\mathrm{d}y}{\mathrm{d}\theta}\biggr)^2 } \\ =&\,2a\bigg|\sin\biggl(\frac\theta2\biggr)\bigg|. \end{align*}

For 0≤θ≤2π0\le\theta\le2\pi, we have 0≤θ2≤π0\le\frac\theta2\le\pi, so sin⁡θ2≥0\sin\frac\theta2\ge0. Hence

dsdθ=2asin⁡(θ2).\frac{\mathrm{d}s}{\mathrm{d}\theta} =2a\sin\biggl(\frac\theta2\biggr).

Also,

y=a(1−cos⁡θ)=2asin⁡2(θ2).y=a(1-\cos\theta) =2a\sin^2\biggl(\frac\theta2\biggr).

Thus the surface area SS is

S= 2π∫02πydsdθ dθ= 8πa2∫02πsin⁡3(θ2) dθ.\begin{align*} S =&\,2\pi\int_0^{2\pi} y\frac{\mathrm{d}s}{\mathrm{d}\theta} \,\mathrm{d}\theta \\ =&\,8\pi a^2\int_0^{2\pi} \sin^3\biggl(\frac\theta2\biggr) \,\mathrm{d}\theta. \end{align*}

Let u=θ2u=\frac\theta2, so dθ=2 du\mathrm{d}\theta=2\,\mathrm{d}u. Then

∫02πsin⁡3(θ2) dθ= 2∫0πsin⁡3u du= 2∫0πsin⁡u(1−cos⁡2u) du= 2[−cos⁡u+13cos⁡3u]0π= 2[(1−13)−(−1+13)]= 83.\begin{align*} \int_0^{2\pi} \sin^3\biggl(\frac\theta2\biggr) \,\mathrm{d}\theta =&\,2\int_0^\pi\sin^3u\,\mathrm{d}u \\ =&\,2\int_0^\pi \sin u(1-\cos^2u)\,\mathrm{d}u \\ =&\,2\biggl[ -\cos u+\frac13\cos^3u \biggr]_0^\pi \\ =&\,2\biggl[\biggl(1-\frac13\biggr) -\biggl(-1+\frac13\biggr)\biggr] \\ =&\,\frac83. \end{align*}

Therefore,

S=643πa2.\boxed{S=\frac{64}{3}\pi a^2}.