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FP2 2020 Oct Q7

A Level / Edexcel / FP2

IAL 2020 Oct Paper · Question 7

题目

Problem

7.

The curve CC, shown in Figure 1, has polar equation

 r=2a(1+cos⁡θ) 0⩽θ⩽π\begin{align*} &\,r=2a(1+\cos\theta)\\[3mm] &\,0\leqslant\theta\leqslant\pi \end{align*}

where aa is a positive constant.

The tangent to CC at the point AA is parallel to the initial line.

(a) Determine the polar coordinates of AA.

(6)

The point BB on the curve has polar coordinates (a(2+3),π6)\left(a(2+\sqrt3),\dfrac{\pi}{6}\right).

The finite region RR, shown shaded in Figure 1, is bounded by the curve CC and the line ABAB.

(b) Use calculus to determine the exact area of the shaded region RR.

Give your answer in the form

a24(dπ−e+f3)\begin{align*} \frac{a^2}{4}(d\pi-e+f\sqrt3) \end{align*}

where dd, ee and ff are integers.

(7)
题目中文翻译

图 1 中的曲线 CC 的极坐标方程为

 r=2a(1+cos⁡θ) 0⩽θ⩽π,\begin{align*} &\,r=2a(1+\cos\theta)\\[3mm] &\,0\leqslant\theta\leqslant\pi, \end{align*}

其中 aa 为正常数。

曲线 CC 在点 AA 处的切线平行于始线。

(a) 求 AA 的极坐标。

(6)

曲线上点 BB 的极坐标为 (a(2+3),π6)\big(a(2+\sqrt3),\frac{\pi}{6}\big)。

图 1 中的有限阴影区域 RR 由曲线 CC 与直线 ABAB 围成。

(b) 用微积分求阴影区域 RR 的精确面积。

将答案写成

a24(dπ−e+f3)\begin{align*} \frac{a^2}{4}(d\pi-e+f\sqrt3) \end{align*}

的形式，其中 dd、ee、ff 为整数。

(7)

解答

(a)

解法一

思路

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切线平行于初始线，说明切线水平。极坐标中可以写 y=rsin⁡θy=r\sin\theta，水平切线对应 dydθ=0\dfrac{\mathrm{d}y}{\mathrm{d}\theta}=0。

答题过程

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Since

y= rsin⁡θ= 2a(1+cos⁡θ)sin⁡θ= 2asin⁡θ+2asin⁡θcos⁡θ= 2asin⁡θ+asin⁡2θ\begin{align*} y =&\,r\sin\theta\\[4mm] =&\,2a(1+\cos\theta)\sin\theta\\[4mm] =&\,2a\sin\theta+2a\sin\theta\cos\theta\\[4mm] =&\,2a\sin\theta+a\sin2\theta \end{align*}

Differentiate:

dydθ= 2acos⁡θ+2acos⁡2θ\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} =&\,2a\cos\theta+2a\cos2\theta \end{align*}

For a horizontal tangent,

2acos⁡θ+2acos⁡2θ= 0cos⁡θ+cos⁡2θ= 0cos⁡θ+(2cos⁡2θ−1)= 02cos⁡2θ+cos⁡θ−1= 0(2cos⁡θ−1)(cos⁡θ+1)= 0\begin{align*} 2a\cos\theta+2a\cos2\theta =&\,0\\[4mm] \cos\theta+\cos2\theta =&\,0\\[4mm] \cos\theta+(2\cos^2\theta-1) =&\,0\\[4mm] 2\cos^2\theta+\cos\theta-1 =&\,0\\[4mm] (2\cos\theta-1)(\cos\theta+1) =&\,0 \end{align*}

In the required part of the curve,

cos⁡θ=12⇒θ=π3\begin{align*} \cos\theta=\frac12 \quad \Rightarrow \quad \theta=\frac{\pi}{3} \end{align*}

Then

r= 2a(1+cos⁡π3)= 2a(1+12)= 3a\begin{align*} r =&\,2a\big(1+\cos\frac{\pi}{3}\big)\\[4mm] =&\,2a\big(1+\frac12\big)\\[4mm] =&\,3a \end{align*}

Therefore

A=(3a,π3)\begin{align*} A=\big(3a,\frac{\pi}{3}\big) \end{align*}

(b)

解法一

思路

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阴影面积等于曲线从 θ=π6\theta=\frac{\pi}{6} 到 θ=π3\theta=\frac{\pi}{3} 扫过的极坐标面积，减去三角形 OABOAB 的面积。先用倍角公式处理 cos⁡2θ\cos^2\theta，再代入上下限。

答题过程

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Let II be the polar area swept out between π/6\pi/6 and π/3\pi/3, and let f(θ)=(1+cos⁡θ)2f(\theta)=(1+\cos\theta)^2. Then

I= 12∫π/6π/3r2 dθ= 2a2∫π/6π/3f(θ) dθ.\begin{align*} I=&\,\frac12\int_{\pi/6}^{\pi/3}r^2\,\mathrm{d}\theta\\[4mm] =&\,2a^2\int_{\pi/6}^{\pi/3}f(\theta)\,\mathrm{d}\theta. \end{align*}

Expand the integrand and use the double-angle formula:

 (1+cos⁡θ)2= 1+2cos⁡θ+cos⁡2θ= 32+2cos⁡θ +12cos⁡2θ.\begin{align*} &\,(1+\cos\theta)^2\\[2mm] =&\,1+2\cos\theta+\cos^2\theta\\[2mm] =&\,\frac32+2\cos\theta\\[2mm] &\,+\frac12\cos2\theta. \end{align*}

An antiderivative is

J(θ)= 32θ+2sin⁡θ +14sin⁡2θ.\begin{align*} J(\theta)=&\,\frac32\theta+2\sin\theta\\[2mm] &\,+\frac14\sin2\theta. \end{align*}

At the limits,

J(π3)= π2+3+38,J(π6)= π4+1+38.\begin{align*} J\big(\frac{\pi}{3}\big)=&\,\frac{\pi}{2}+\sqrt3+\frac{\sqrt3}{8},\\[4mm] J\big(\frac{\pi}{6}\big)=&\,\frac{\pi}{4}+1+\frac{\sqrt3}{8}. \end{align*}

Therefore

I= 2a2[J(π3)−J(π6)]= 2a2(π4+3−1).\begin{align*} I=&\,2a^2\Big[J\big(\frac{\pi}{3}\big)-J\big(\frac{\pi}{6}\big)\Big]\\[4mm] =&\,2a^2\big(\frac{\pi}{4}+\sqrt3-1\big). \end{align*}

Let TT be the area of △OAB\triangle OAB. Since OA=3aOA=3a, OB=a(2+3)OB=a(2+\sqrt3) and ∠AOB=π/6\angle AOB=\pi/6,

T= 12 OA OBsin⁡π6= 12(3a) a(2+3) 12= 3a24(2+3).\begin{align*} T=&\,\frac12\,OA\,OB\sin\frac{\pi}{6}\\[4mm] =&\,\frac12(3a)\,a(2+\sqrt3)\,\frac12\\[4mm] =&\,\frac{3a^2}{4}(2+\sqrt3). \end{align*}

The shaded area is

R= I−T= 2a2(π4+3−1) −3a24(2+3)= a2π2+2a23 −2a2−3a22 −3a234= a24(2π−14+53).\begin{align*} R=&\,I-T\\[2mm] =&\,2a^2\big(\frac{\pi}{4}+\sqrt3-1\big)\\[2mm] &\,-\frac{3a^2}{4}(2+\sqrt3)\\[4mm] =&\,\frac{a^2\pi}{2}+2a^2\sqrt3\\[2mm] &\,-2a^2-\frac{3a^2}{2}\\[2mm] &\,\hspace{2pt}-\frac{3a^2\sqrt3}{4}\\[4mm] =&\,\frac{a^2}{4}(2\pi-14+5\sqrt3). \end{align*}

Thus d=2d=2, e=14e=14 and f=5f=5.

解法二

思路

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也可以不先把 cos⁡2θ\cos^2\theta 换成倍角形式，而是用分部积分求它的原函数。将产生的 sin⁡2θ\sin^2\theta 写成 1−cos⁡2θ1-\cos^2\theta，便能把同一个积分移到左边求出。最后仍减去三角形 OABOAB。

答题过程

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Let

J= ∫cos⁡2θ dθ.\begin{align*} J=&\,\int\cos^2\theta\,\mathrm{d}\theta. \end{align*}

Integrating by parts with u=cos⁡θu=\cos\theta and dv=cos⁡θ dθ\mathrm{d}v=\cos\theta\,\mathrm{d}\theta gives

J= sin⁡θcos⁡θ +∫sin⁡2θ dθ= sin⁡θcos⁡θ +∫(1−cos⁡2θ) dθ= sin⁡θcos⁡θ+θ−J.\begin{align*} J=&\,\sin\theta\cos\theta\\[2mm] &\,+\int\sin^2\theta\,\mathrm{d}\theta\\[4mm] =&\,\sin\theta\cos\theta\\[2mm] &\,+\int(1-\cos^2\theta)\,\mathrm{d}\theta\\[4mm] =&\,\sin\theta\cos\theta+\theta-J. \end{align*}

Hence, up to an additive constant,

2J= θ+sin⁡θcos⁡θ,J= θ2+12sin⁡θcos⁡θ.\begin{align*} 2J=&\,\theta+\sin\theta\cos\theta,\\[4mm] J=&\,\frac{\theta}{2}+\frac12\sin\theta\cos\theta. \end{align*}

At both θ=π/6\theta=\pi/6 and θ=π/3\theta=\pi/3, sin⁡θcos⁡θ=3/4\sin\theta\cos\theta=\sqrt3/4. Therefore

K= ∫π/6π/3cos⁡2θ dθ= 12(π3−π6)= π12.\begin{align*} K=&\,\int_{\pi/6}^{\pi/3}\cos^2\theta\,\mathrm{d}\theta\\[4mm] =&\,\frac12\big(\frac{\pi}{3}-\frac{\pi}{6}\big)\\[4mm] =&\,\frac{\pi}{12}. \end{align*}

Also,

L= ∫π/6π/3(1+2cos⁡θ) dθ= [θ+2sin⁡θ]π/6π/3= π6+3−1.\begin{align*} L=&\,\int_{\pi/6}^{\pi/3}(1+2\cos\theta)\,\mathrm{d}\theta\\[4mm] =&\,\big[\theta+2\sin\theta\big]_{\pi/6}^{\pi/3}\\[4mm] =&\,\frac{\pi}{6}+\sqrt3-1. \end{align*}

Expanding (1+cos⁡θ)2(1+\cos\theta)^2, the polar area is

I= 2a2(L+K)= 2a2(π4+3−1).\begin{align*} I=&\,2a^2(L+K)\\[4mm] =&\,2a^2\big(\frac{\pi}{4}+\sqrt3-1\big). \end{align*}

The triangle has area

T= 12 OA OBsin⁡π6= 12(3a) a(2+3) 12= 3a24(2+3).\begin{align*} T=&\,\frac12\,OA\,OB\sin\frac{\pi}{6}\\[4mm] =&\,\frac12(3a)\,a(2+\sqrt3)\,\frac12\\[4mm] =&\,\frac{3a^2}{4}(2+\sqrt3). \end{align*}

Thus the shaded area is

R= I−T= 2a2(π4+3−1) −3a24(2+3)= a24(2π−14+53).\begin{align*} R=&\,I-T\\[2mm] =&\,2a^2\big(\frac{\pi}{4}+\sqrt3-1\big)\\[2mm] &\,-\frac{3a^2}{4}(2+\sqrt3)\\[4mm] =&\,\frac{a^2}{4}(2\pi-14+5\sqrt3). \end{align*}

Thus d=2d=2, e=14e=14 and f=5f=5.