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IAL 2021 Oct Q8

A Level / Edexcel / FP2

IAL 2021 Oct Paper · Question 8

题目

Problem

Figure 1

The curve CC shown in Figure 1 has polar equation

r=1+sin⁡θ−π2<θ⩽π2\begin{align*} r = 1 + \sin \theta \qquad -\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)

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=(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} =&\,\cos\theta\cos\theta -(1+\sin\theta)\sin\theta\\[2mm] =&\,\cos^2\theta-\sin\theta-\sin^2\theta\\[2mm] =&\,\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-\sin^2\theta =&\,0,\\[2mm] 1-\sin\theta-2\sin^2\theta =&\,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)(\sin\theta+1)=0. \end{align*}

Hence

sin⁡θ=12orsin⁡θ=−1.\begin{align*} \sin\theta=\frac12 \quad\text{or}\quad \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*} \left(\frac32,\frac{\pi}{6}\right). \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。

要求的区域 RR 可以看成：

area of trapezium OQSP−area of polar sector OQP.\begin{align*} \text{area of trapezium }OQSP -\text{area of polar sector }OQP. \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=\left(\frac32,\frac{\pi}{6}\right) \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=\left(\frac{3\sqrt3}{4},\frac34\right). \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=\left(\frac{3\sqrt3}{4},2\right). \end{align*}

The area of trapezium OQSPOQSP is

12(2+(2−34))(334)= 12(134)(334)= 39332.\begin{align*} \frac12 \left(2+\left(2-\frac34\right)\right) \left(\frac{3\sqrt3}{4}\right) =&\,\frac12\left(\frac{13}{4}\right) \left(\frac{3\sqrt3}{4}\right)\\[2mm] =&\,\frac{39\sqrt3}{32}. \end{align*}

Now find the area of polar sector OQPOQP:

12∫π6π2r2 dθ= 12∫π6π2(1+sin⁡θ)2 dθ.\begin{align*} \frac12\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}r^2\,\mathrm{d}\theta =&\,\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 =&\,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 =&\,1+2\sin\theta+\frac12-\frac12\cos2\theta\\[2mm] =&\,\frac32+2\sin\theta-\frac12\cos2\theta. \end{align*}

So

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

Apply the limits:

sector area= 12[32θ−2cos⁡θ−14sin⁡2θ]π6π2.\begin{align*} \text{sector area} =&\,\frac12 \left[ \frac32\theta-2\cos\theta-\frac14\sin2\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}. \end{align*}

At θ=π2\theta=\dfrac{\pi}{2},

32θ−2cos⁡θ−14sin⁡2θ= 3π4.\begin{align*} \frac32\theta-2\cos\theta-\frac14\sin2\theta =&\,\frac{3\pi}{4}. \end{align*}

At θ=π6\theta=\dfrac{\pi}{6},

32θ−2cos⁡θ−14sin⁡2θ= π4−2(32)−14(32)= π4−3−38= π4−938.\begin{align*} \frac32\theta-2\cos\theta-\frac14\sin2\theta =&\,\frac{\pi}{4} -2\left(\frac{\sqrt3}{2}\right) -\frac14\left(\frac{\sqrt3}{2}\right)\\[2mm] =&\,\frac{\pi}{4}-\sqrt3-\frac{\sqrt3}{8}\\[2mm] =&\,\frac{\pi}{4}-\frac{9\sqrt3}{8}. \end{align*}

Therefore

sector area= 12[3π4−(π4−938)]= 12(π2+938)= π4+9316.\begin{align*} \text{sector area} =&\,\frac12 \left[ \frac{3\pi}{4} -\left(\frac{\pi}{4}-\frac{9\sqrt3}{8}\right) \right]\\[2mm] =&\,\frac12 \left(\frac{\pi}{2}+\frac{9\sqrt3}{8}\right)\\[2mm] =&\,\frac{\pi}{4}+\frac{9\sqrt3}{16}. \end{align*}

Hence the area of RR is

39332−(π4+9316)= 39332−8π32−18332= 213−8π32.\begin{align*} \frac{39\sqrt3}{32} -\left(\frac{\pi}{4}+\frac{9\sqrt3}{16}\right) =&\,\frac{39\sqrt3}{32} -\frac{8\pi}{32} -\frac{18\sqrt3}{32}\\[2mm] =&\,\frac{21\sqrt3-8\pi}{32}. \end{align*}

Thus

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