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CIE 9709 2022 March Paper 32 Q4

A Level / CIE / P3

CIE 9709 2022 March Paper 32 Paper · Question 4

题目

Problem

The parametric equations of a curve are

x=1−cos⁡θ,y=cos⁡θ−14cos⁡2θ.x=1-\cos\theta,\qquad y=\cos\theta-\tfrac14\cos2\theta.

Show that dydx=−2sin⁡2(12θ)\dfrac{\mathrm{d}y}{\mathrm{d}x}=-2\sin^2\big(\tfrac12\theta\big).

(5)
题目中文翻译

一条曲线的参数方程为

x=1−cos⁡θ,y=cos⁡θ−14cos⁡2θ.x=1-\cos\theta,\qquad y=\cos\theta-\tfrac14\cos2\theta.

证明 dydx=−2sin⁡2(12θ)\dfrac{\mathrm{d}y}{\mathrm{d}x}=-2\sin^2\big(\tfrac12\theta\big)。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

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分别对参数求导，取导数之比后使用正弦二倍角与余弦半角公式。

答题过程

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Differentiating,

dxdθ= sin⁡θ,dydθ= −sin⁡θ+12sin⁡2θ.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}\theta}=&\,\sin\theta,\\[4mm] \frac{\mathrm{d}y}{\mathrm{d}\theta}=&\,-\sin\theta+\frac12\sin2\theta. \end{align*}

At regular parameter values,

dydx= −sin⁡θ+12sin⁡2θsin⁡θ= −sin⁡θ+sin⁡θcos⁡θsin⁡θ= cos⁡θ−1= −2sin⁡2θ2.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,\frac{-\sin\theta+\frac12\sin2\theta}{\sin\theta}\\[4mm] =&\,\frac{-\sin\theta+\sin\theta\cos\theta}{\sin\theta}\\[4mm] =&\,\cos\theta-1\\[4mm] =&\,\boxed{-2\sin^2\frac\theta2}. \end{align*}

At parameters where sin⁡θ=0\sin\theta=0, the same tangent-gradient value follows from the Cartesian equation y=3/4−x2/2y=3/4-x^2/2.

解法二

思路

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也可消去参数得到笛卡尔方程，再直接求导。这同时说明参数导数为零时如何理解切线斜率。

答题过程

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From x=1−cos⁡θx=1-\cos\theta, we have cos⁡θ=1−x\cos\theta=1-x. Using cos⁡2θ=2cos⁡2θ−1\cos2\theta=2\cos^2\theta-1,

y= cos⁡θ−14(2cos⁡2θ−1)= (1−x)−12(1−x)2+14= 34−12x2.\begin{align*} y =&\,\cos\theta-\frac14(2\cos^2\theta-1)\\[4mm] =&\,(1-x)-\frac12(1-x)^2+\frac14\\[4mm] =&\,\frac34-\frac12x^2. \end{align*}

Therefore

dydx=−x=cos⁡θ−1=−2sin⁡2θ2.\frac{\mathrm{d}y}{\mathrm{d}x} =-x=\cos\theta-1 =\boxed{-2\sin^2\frac\theta2}.