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CIE 9709 2022 November Paper 32 Q7

A Level / CIE / P3

CIE 9709 2022 Nov Paper 32 Paper · Question 7

题目

Problem

The variables xx and θ\theta satisfy the differential equation

xsin⁡2θdxdθ=tan⁡2θ−2cot⁡θ,x\sin^2\theta\frac{\mathrm{d}x}{\mathrm{d}\theta}=\tan^2\theta-2\cot\theta,

for 0<θ<12π0<\theta<\frac12\pi and x>0x>0. It is given that x=2x=2 when θ=14π\theta=\frac14\pi.

(a) Show that

ddθ(cot⁡2θ)=−2cot⁡θsin⁡2θ.\frac{\mathrm{d}}{\mathrm{d}\theta}(\cot^2\theta)=-\frac{2\cot\theta}{\sin^2\theta}.

(You may assume without proof that the derivative of cot⁡θ\cot\theta with respect to θ\theta is −cosec⁡2θ-\operatorname{cosec}^2\theta.)

(1)

(b) Solve the differential equation and find the value of xx when θ=16π\theta=\frac16\pi.

(7)
题目中文翻译

变量 xx 和 θ\theta 满足微分方程

xsin⁡2θdxdθ=tan⁡2θ−2cot⁡θ,x\sin^2\theta\frac{\mathrm{d}x}{\mathrm{d}\theta}=\tan^2\theta-2\cot\theta,

其中 0<θ<12π0<\theta<\frac12\pi 且 x>0x>0。已知当 θ=14π\theta=\frac14\pi 时，x=2x=2。

(a) 证明

ddθ(cot⁡2θ)=−2cot⁡θsin⁡2θ.\frac{\mathrm{d}}{\mathrm{d}\theta}(\cot^2\theta)=-\frac{2\cot\theta}{\sin^2\theta}.

（可直接使用 cot⁡θ\cot\theta 关于 θ\theta 的导数为 −cosec⁡2θ-\operatorname{cosec}^2\theta，无须证明。）

(1)

(b) 解此微分方程，并求 θ=16π\theta=\frac16\pi 时 xx 的值。

(7)
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

(a)

解法一

思路

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利用已给的余切导数，对外层平方使用链式法则。

答题过程

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By the chain rule,

ddθ(cot⁡2θ)= 2cot⁡θ(−cosec⁡2θ)= −2cot⁡θsin⁡2θ.\begin{align*} \frac{\mathrm{d}}{\mathrm{d}\theta}(\cot^2\theta) =&\,2\cot\theta(-\operatorname{cosec}^2\theta)\\[4mm] =&\,\boxed{-\frac{2\cot\theta}{\sin^2\theta}}. \end{align*}

(b)

解法一

思路

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分离变量后，tan⁡2θ/sin⁡2θ\tan^2\theta/\sin^2\theta 化为 sec⁡2θ\sec^2\theta；另一项直接承接(a)。最后因 x>0x>0 选正平方根。

答题过程

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Separating variables,

x dx=(tan⁡2θsin⁡2θ−2cot⁡θsin⁡2θ) dθ.x\,\mathrm{d}x =\bigg(\frac{\tan^2\theta}{\sin^2\theta} -\frac{2\cot\theta}{\sin^2\theta}\bigg)\,\mathrm{d}\theta.

Since tan⁡2θ/sin⁡2θ=sec⁡2θ\tan^2\theta/\sin^2\theta=\sec^2\theta, and using part (a), integration gives

12x2=tan⁡θ+cot⁡2θ+C.\frac12x^2=\tan\theta+\cot^2\theta+C.

At x=2x=2, θ=π/4\theta=\pi/4,

2= 1+1+C,C= 0.\begin{align*} 2=&\,1+1+C,\\[4mm] C=&\,0. \end{align*}

Thus

x=2tan⁡θ+2cot⁡2θ,\boxed{x=\sqrt{2\tan\theta+2\cot^2\theta}},

where the positive root is required. When θ=π/6\theta=\pi/6,

x=23+6=2.6748….x=\sqrt{\frac2{\sqrt3}+6} =2.6748\ldots.

Therefore x=2.67\boxed{x=2.67} to 3 significant figures.