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FP2 2021 Oct Q8

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 8

题目

Problem

8.

The curve CC shown in Figure 1 has polar equation

r= 1+sin⁡θ, −π2<θ⩽π2.\begin{align*} r=&\,1+\sin\theta,\\[4mm] &\,-\frac{\pi}{2}<\theta\leqslant\frac{\pi}{2}. \end{align*}

The point PP lies on CC such that the tangent to CC at PP is perpendicular to the initial line.

(a) Use calculus to determine the polar coordinates of PP.

(5)

The tangent to CC at the point QQ where θ=π2\theta = \frac{\pi}{2} is parallel to the initial line.

The tangent to CC at QQ meets the tangent to CC at PP at the point SS, as shown in Figure 1.

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

(b) Use algebraic integration to show that the area of RR is

132(a3+bπ)\begin{align*} \frac{1}{32} (a\sqrt{3} + b\pi) \end{align*}

where aa and bb are integers to be determined.

(6)
题目中文翻译

8. 图 1 所示曲线 CC 的极坐标方程为

r= 1+sin⁡θ, −π2<θ⩽π2.\begin{align*} r=&\,1+\sin\theta,\\[4mm] &\,-\frac{\pi}{2}<\theta\leqslant\frac{\pi}{2}. \end{align*}

点 PP 在 CC 上，且 CC 在 PP 处的切线垂直于始边。

(a) 用微积分求 PP 的极坐标。

(5)

曲线 CC 在 θ=π2\theta=\frac{\pi}{2} 对应的点 QQ 处的切线平行于始边。

如图 1 所示，CC 在 QQ 处的切线与 CC 在 PP 处的切线交于点 SS。

图 1 中阴影所示的有限区域 RR 由线段 QSQS、SPSP 及曲线 CC 围成。

(b) 用代数积分证明，RR 的面积为

132(a3+bπ),\begin{align*} \frac{1}{32}(a\sqrt{3}+b\pi), \end{align*}

其中 aa 和 bb 是待求的整数。

(6)

解答

(a)

解法一

思路

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切线 perpendicular to the initial line，就是切线竖直。极坐标中

x=rcos⁡θ.\begin{align*} x=r\cos\theta. \end{align*}

竖直切线对应 dxdθ=0\dfrac{\mathrm{d}x}{\mathrm{d}\theta}=0。这里 r=1+sin⁡θr=1+\sin\theta，所以先对 x=(1+sin⁡θ)cos⁡θx=(1+\sin\theta)\cos\theta 求导。

答题过程

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For the curve

r=1+sin⁡θ,\begin{align*} r=1+\sin\theta, \end{align*}

we have

x= rcos⁡θ= (1+sin⁡θ)cos⁡θ.\begin{align*} x=&\,r\cos\theta\\[4mm] =&\,(1+\sin\theta)\cos\theta. \end{align*}

Differentiate:

 dxdθ= cos⁡θcos⁡θ −(1+sin⁡θ)sin⁡θ= cos⁡2θ−sin⁡θ −sin⁡2θ= cos⁡2θ−sin⁡θ.\begin{align*} &\,\frac{\mathrm{d}x}{\mathrm{d}\theta}\\[4mm] =&\,\cos\theta\cos\theta\\[4mm] &\,\hspace{2pt}-(1+\sin\theta)\sin\theta\\[4mm] =&\,\cos^2\theta-\sin\theta\\[4mm] &\,\hspace{2pt}-\sin^2\theta\\[4mm] =&\,\cos2\theta-\sin\theta. \end{align*}

For a tangent perpendicular to the initial line,

dxdθ=0.\begin{align*} \frac{\mathrm{d}x}{\mathrm{d}\theta}=0. \end{align*}

So

cos⁡2θ−sin⁡θ−sin⁡2θ=0.\begin{align*} \cos^2\theta-\sin\theta-\sin^2\theta=0. \end{align*}

Use cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta:

 1−sin⁡2θ−sin⁡θ −sin⁡2θ= 0, 1−sin⁡θ−2sin⁡2θ= 0.\begin{align*} &\,1-\sin^2\theta-\sin\theta\\[4mm] &\,\hspace{2pt}-\sin^2\theta\\[4mm] =&\,0,\\[4mm] &\,1-\sin\theta-2\sin^2\theta\\[4mm] =&\,0. \end{align*}

Therefore

2sin⁡2θ+sin⁡θ−1=0.\begin{align*} 2\sin^2\theta+\sin\theta-1=0. \end{align*}

Factorise:

 (2sin⁡θ−1) ×(sin⁡θ+1)= 0.\begin{align*} &\,(2\sin\theta-1)\\[4mm] &\,\hspace{2pt}\times(\sin\theta+1)\\[4mm] =&\,0. \end{align*}

Hence

sin⁡θ= 12or sin⁡θ= −1.\begin{align*} \sin\theta=&\,\frac12\\[4mm] \text{or }\sin\theta=&\,-1. \end{align*}

Since

−π2<θ⩽π2,\begin{align*} -\frac{\pi}{2}<\theta\leqslant\frac{\pi}{2}, \end{align*}

the value sin⁡θ=−1\sin\theta=-1 would give θ=−π2\theta=-\dfrac{\pi}{2}, which is not in the domain. Therefore

θ=π6.\begin{align*} \theta=\frac{\pi}{6}. \end{align*}

Then

r=1+sin⁡π6=32.\begin{align*} r=1+\sin\frac{\pi}{6}=\frac32. \end{align*}

So the polar coordinates of PP are

(32,π6).\begin{align*} \bigg(\frac32,\frac{\pi}{6}\bigg). \end{align*}

(b)

解法一

思路

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先把 PP 和 QQ 的 Cartesian coordinates 找出来。QQ 在 θ=π2\theta=\dfrac{\pi}{2}，所以 Q=(0,2)Q=(0,2)。PP 的竖直切线与 QQ 的水平切线相交于 SS。

记梯形 OQSPOQSP 的面积为 ATA_T，极坐标扇形 OQPOQP 的面积为 ASA_S，所求区域 RR 的面积为 ARA_R，则

AR= AT−AS.\begin{align*} A_R=&\,A_T-A_S. \end{align*}

其中极坐标扇形面积用

12∫r2 dθ.\begin{align*} \frac12\int r^2\,\mathrm{d}\theta. \end{align*}

答题过程

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From part (a),

P=(32,π6)\begin{align*} P=\bigg(\frac32,\frac{\pi}{6}\bigg) \end{align*}

in polar coordinates. In Cartesian coordinates,

xP= 32cos⁡π6=334,\begin{align*} x_P =&\,\frac32\cos\frac{\pi}{6} =\frac{3\sqrt3}{4}, \end{align*}

and

yP= 32sin⁡π6=34.\begin{align*} y_P =&\,\frac32\sin\frac{\pi}{6} =\frac34. \end{align*}

So

P=(334,34).\begin{align*} P=\bigg(\frac{3\sqrt3}{4},\frac34\bigg). \end{align*}

At QQ, θ=π2\theta=\dfrac{\pi}{2}, so

r=1+sin⁡π2=2.\begin{align*} r=1+\sin\frac{\pi}{2}=2. \end{align*}

Hence

Q=(0,2).\begin{align*} Q=(0,2). \end{align*}

The tangent at QQ is horizontal and the tangent at PP is vertical, so

S=(334,2).\begin{align*} S=\bigg(\frac{3\sqrt3}{4},2\bigg). \end{align*}

Let ATA_T be the area of trapezium OQSPOQSP. Then

AT= 12[2+(2−34)] ×334= 12(134) ×334= 39332.\begin{align*} A_T=&\,\frac12\bigg[2+\bigg(2-\frac34\bigg)\bigg]\\[4mm] &\,\hspace{2pt}\times\frac{3\sqrt3}{4}\\[4mm] =&\,\frac12\bigg(\frac{13}{4}\bigg)\\[4mm] &\,\hspace{2pt}\times\frac{3\sqrt3}{4}\\[4mm] =&\,\frac{39\sqrt3}{32}. \end{align*}

Let ASA_S be the area of polar sector OQPOQP. Then

AS= 12∫π6π2r2 dθ= 12∫π6π2(1+sin⁡θ)2 dθ.\begin{align*} A_S=&\,\frac12\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}r^2\,\mathrm{d}\theta\\[4mm] =&\,\frac12\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(1+\sin\theta)^2\,\mathrm{d}\theta. \end{align*}

Expand the integrand:

 (1+sin⁡θ)2= 1+2sin⁡θ+sin⁡2θ.\begin{align*} &\,(1+\sin\theta)^2\\[4mm] =&\,1+2\sin\theta+\sin^2\theta. \end{align*}

Use

sin⁡2θ=12−12cos⁡2θ.\begin{align*} \sin^2\theta=\frac12-\frac12\cos2\theta. \end{align*}

Then

 (1+sin⁡θ)2= 1+2sin⁡θ+12 −12cos⁡2θ= 32+2sin⁡θ −12cos⁡2θ.\begin{align*} &\,(1+\sin\theta)^2\\[4mm] =&\,1+2\sin\theta+\frac12\\[4mm] &\,\hspace{2pt}-\frac12\cos2\theta\\[4mm] =&\,\frac32+2\sin\theta\\[4mm] &\,\hspace{2pt}-\frac12\cos2\theta. \end{align*}

So

 ∫(1+sin⁡θ)2 dθ= ∫32 dθ +2∫sin⁡θ dθ −12∫cos⁡2θ dθ= 32θ−2cos⁡θ −14sin⁡2θ+C.\begin{align*} &\,\int(1+\sin\theta)^2\,\mathrm{d}\theta\\[4mm] =&\,\int\frac32\,\mathrm{d}\theta\\[4mm] &\,\hspace{2pt}+2\int\sin\theta\,\mathrm{d}\theta\\[4mm] &\,\hspace{4pt}-\frac12\int\cos2\theta\,\mathrm{d}\theta\\[4mm] =&\,\frac32\theta-2\cos\theta\\[4mm] &\,\hspace{2pt}-\frac14\sin2\theta+C. \end{align*}

Define

F(θ)= 32θ−2cos⁡θ −14sin⁡2θ.\begin{align*} F(\theta)=&\,\frac32\theta-2\cos\theta\\[4mm] &\,\hspace{2pt}-\frac14\sin2\theta. \end{align*}

Applying the limits,

AS= 12[F(θ)]π6π2.\begin{align*} A_S=&\,\frac12\big[F(\theta)\big]_{\frac{\pi}{6}}^{\frac{\pi}{2}}. \end{align*}

At θ=π2\theta=\dfrac{\pi}{2},

F(π2)= 32(π2) −2cos⁡π2 −14sin⁡π= 3π4.\begin{align*} F\bigg(\frac{\pi}{2}\bigg) =&\,\frac32\bigg(\frac{\pi}{2}\bigg)\\[4mm] &\,\hspace{2pt}-2\cos\frac{\pi}{2}\\[4mm] &\,\hspace{4pt}-\frac14\sin\pi\\[4mm] =&\,\frac{3\pi}{4}. \end{align*}

At θ=π6\theta=\dfrac{\pi}{6},

F(π6)= π4−2(32) −14(32)= π4−3−38= π4−938.\begin{align*} F\bigg(\frac{\pi}{6}\bigg) =&\,\frac{\pi}{4}-2\bigg(\frac{\sqrt3}{2}\bigg)\\[4mm] &\,\hspace{2pt}-\frac14\bigg(\frac{\sqrt3}{2}\bigg)\\[4mm] =&\,\frac{\pi}{4}-\sqrt3-\frac{\sqrt3}{8}\\[4mm] =&\,\frac{\pi}{4}-\frac{9\sqrt3}{8}. \end{align*}

Therefore

AS= 12[3π4 −(π4−938)]= 12(π2+938)= π4+9316.\begin{align*} A_S=&\,\frac12\bigg[\frac{3\pi}{4}\\[4mm] &\,\hspace{2pt}-\bigg(\frac{\pi}{4}-\frac{9\sqrt3}{8}\bigg)\bigg]\\[4mm] =&\,\frac12\bigg(\frac{\pi}{2}+\frac{9\sqrt3}{8}\bigg)\\[4mm] =&\,\frac{\pi}{4}+\frac{9\sqrt3}{16}. \end{align*}

Hence the area ARA_R of RR is

AR= AT−AS= 39332 −(π4+9316)= 39332−8π32 −18332= 213−8π32.\begin{align*} A_R=&\,A_T-A_S\\[4mm] =&\,\frac{39\sqrt3}{32}\\[4mm] &\,\hspace{2pt}-\bigg(\frac{\pi}{4}+\frac{9\sqrt3}{16}\bigg)\\[4mm] =&\,\frac{39\sqrt3}{32}-\frac{8\pi}{32}\\[4mm] &\,\hspace{2pt}-\frac{18\sqrt3}{32}\\[4mm] =&\,\frac{21\sqrt3-8\pi}{32}. \end{align*}

Thus

a=21,b=−8.\begin{align*} a=21,\qquad b=-8. \end{align*}