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FP2 2024 Jan Q5

A Level / Edexcel / FP2

IAL 2024 Jan Paper · Question 5

题目

Problem

5.

Figure 1 shows a sketch of the curve with polar equation

r= 10cos⁡θ+tan⁡θ0⩽θ<π2\begin{align*} r =&\, 10\cos \theta + \tan \theta \qquad 0 \leqslant \theta < \frac{\pi}{2} \end{align*}

The point PP lies on the curve where θ=π3\theta = \frac{\pi}{3}.

The region RR , shown shaded in Figure 1, is bounded by the initial line, the curve and the line OPOP , where OO is the pole.

Use algebraic integration to show that the exact area of RR is

112(aπ+b3+c)\begin{align*} \frac{1}{12}(a\pi + b\sqrt{3} + c) \end{align*}

where aa , bb and cc are integers to be determined.

(9)
题目中文翻译

图 1 为曲线的示意图，其极坐标方程为

r= 10cos⁡θ+tan⁡θ,0⩽θ<π2. \begin{align*} r=&\,10\cos\theta+\tan\theta,\qquad0\leqslant\theta<\frac{\pi}{2}. \end{align*}

点 PP 在曲线上，且 θ=π3\theta=\frac{\pi}{3}。

图 1 中阴影所示区域 RR 由极轴、曲线及线段 OPOP 围成，其中 OO 为极点。

用代数积分证明，RR 面积的精确值为

112(aπ+b3+c), \begin{align*} \frac1{12}(a\pi+b\sqrt3+c), \end{align*}

其中 aa、bb、cc 为待求的整数。

(9)

解答

解法一

思路

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区域由 θ=0\theta=0、θ=π3\theta=\frac{\pi}{3} 和曲线围成，所以直接用极坐标面积公式 12∫r2 dθ\frac12\int r^2\,\mathrm{d}\theta。关键是把 r2r^2 展开后改写成容易积分的形式：cos⁡2θ=12(1+cos⁡2θ)\cos^2\theta=\frac12(1+\cos2\theta)，tan⁡2θ=sec⁡2θ−1\tan^2\theta=\sec^2\theta-1。

答题过程

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The area is

Area= 12∫0π3r2 dθ\begin{align*} \text{Area} =&\, \frac12\int_0^{\frac{\pi}{3}} r^2\,\mathrm{d}\theta \end{align*}

First expand r2r^2:

r2= (10cos⁡θ+tan⁡θ)2= 100cos⁡2θ+20cos⁡θtan⁡θ+tan⁡2θ= 100cos⁡2θ+20sin⁡θ+tan⁡2θ\begin{align*} r^2 =&\, (10\cos\theta+\tan\theta)^2\\[4mm] =&\, 100\cos^2\theta +20\cos\theta\tan\theta +\tan^2\theta\\[4mm] =&\, 100\cos^2\theta +20\sin\theta +\tan^2\theta \end{align*}

Use

cos⁡2θ= 12(1+cos⁡2θ)tan⁡2θ= sec⁡2θ−1\begin{align*} \cos^2\theta=&\,\frac12(1+\cos2\theta)\\[2mm] \tan^2\theta=&\,\sec^2\theta-1 \end{align*}

Therefore

Area= 12∫0π3(100cos⁡2θ+20sin⁡θ+tan⁡2θ) dθ= 12∫0π3(50(1+cos⁡2θ)+20sin⁡θ+sec⁡2θ−1) dθ= 12∫0π3(49+50cos⁡2θ+20sin⁡θ+sec⁡2θ) dθ\begin{align*} \text{Area} =&\, \frac12\int_0^{\frac{\pi}{3}} \bigg( 100\cos^2\theta+20\sin\theta+\tan^2\theta \bigg)\,\mathrm{d}\theta\\[4mm] =&\, \frac12\int_0^{\frac{\pi}{3}} \bigg( 50(1+\cos2\theta)+20\sin\theta+\sec^2\theta-1 \bigg)\,\mathrm{d}\theta\\[4mm] =&\, \frac12\int_0^{\frac{\pi}{3}} \bigg( 49+50\cos2\theta+20\sin\theta+\sec^2\theta \bigg)\,\mathrm{d}\theta \end{align*}

Integrate:

Area= 12[49θ+25sin⁡2θ−20cos⁡θ+tan⁡θ]0π3\begin{align*} \text{Area} =&\, \frac12 \bigg[ 49\theta+25\sin2\theta-20\cos\theta+\tan\theta \bigg]_0^{\frac{\pi}{3}} \end{align*}

Let

F(θ)= 49θ+25sin⁡2θ−20cos⁡θ+tan⁡θ. \begin{align*} F(\theta)=&\,49\theta+25\sin2\theta-20\cos\theta+\tan\theta. \end{align*}

The values at the limits are

F(π3)= 49π3+25sin⁡2π3−20cos⁡π3+tan⁡π3= 49π3+2532−10+3= 49π3+2732−10,F(0)= 0+0−20+0=−20. \begin{align*} F\bigg(\frac{\pi}{3}\bigg) =&\,\frac{49\pi}{3}+25\sin\frac{2\pi}{3} -20\cos\frac{\pi}{3}+\tan\frac{\pi}{3}\\[2mm] =&\,\frac{49\pi}{3}+\frac{25\sqrt3}{2}-10+\sqrt3\\[2mm] =&\,\frac{49\pi}{3}+\frac{27\sqrt3}{2}-10,\\[2mm] F(0)=&\,0+0-20+0=-20. \end{align*}

Therefore

Area⁡= 12(F(π3)−F(0))= 12(49π3+2732−10+20)= 49π6+2734+5= 98π+813+6012. \begin{align*} \operatorname{Area} =&\,\frac12\bigg(F\bigg(\frac{\pi}{3}\bigg)-F(0)\bigg)\\[2mm] =&\,\frac12\bigg(\frac{49\pi}{3} +\frac{27\sqrt3}{2}-10+20\bigg)\\[2mm] =&\,\frac{49\pi}{6}+\frac{27\sqrt3}{4}+5\\[2mm] =&\,\frac{98\pi+81\sqrt3+60}{12}. \end{align*}

This is in the required form

112(aπ+b3+c)\begin{align*} \frac{1}{12}(a\pi+b\sqrt3+c) \end{align*}

where

a=98,b=81,c=60\begin{align*} \boxed{a=98,\qquad b=81,\qquad c=60} \end{align*}