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IAL 2023 Jan Q8

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 8

题目

Problem

Figure 1

The curve CC shown in Figure 1 has polar equation

r= 1−sin⁡θ,0⩽θ<π2\begin{align*} r =\,& 1 - \sin \theta, \qquad 0 \leqslant \theta < \frac{\pi}{2} \end{align*}

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

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

(4)

The finite region RR, shown shaded in Figure 1, is bounded by

  • the line with equation θ=π2\theta = \frac{\pi}{2}
  • the tangent to CC at PP
  • part of the curve CC
  • the initial line

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

132(aπ+b3+c)\begin{align*} \frac{1}{32}(a\pi + b\sqrt{3} + c) \end{align*}

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

(6)

解答

(a)

解法一

思路

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切线平行于 initial line 表示水平切线。极坐标中先写

y=rsin⁡θ.\begin{align*} y=r\sin\theta. \end{align*}

然后令 dydθ=0\dfrac{\mathrm{d}y}{\mathrm{d}\theta}=0。

答题过程

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Since

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

we have

y= rsin⁡θ= (1−sin⁡θ)sin⁡θ= sin⁡θ−sin⁡2θ.\begin{align*} y=&\,r\sin\theta\\[2mm] =&\,(1-\sin\theta)\sin\theta\\[2mm] =&\,\sin\theta-\sin^2\theta. \end{align*}

Differentiate:

dydθ= cos⁡θ−2sin⁡θcos⁡θ= cos⁡θ(1−2sin⁡θ).\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta} =&\,\cos\theta-2\sin\theta\cos\theta\\[2mm] =&\,\cos\theta(1-2\sin\theta). \end{align*}

At PP, the tangent is parallel to the initial line, so

dydθ=0.\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}\theta}=0. \end{align*}

Since 0⩽θ<π20\leqslant\theta<\dfrac{\pi}{2}, cos⁡θ≠0\cos\theta\neq0 at PP. Hence

1−2sin⁡θ= 0sin⁡θ= 12θ= π6.\begin{align*} 1-2\sin\theta=&\,0\\[2mm] \sin\theta=&\,\frac{1}{2}\\[2mm] \theta=&\,\frac{\pi}{6}. \end{align*}

Then

r= 1−sin⁡π6= 1−12= 12.\begin{align*} r=&\,1-\sin\frac{\pi}{6}\\[2mm] =&\,1-\frac{1}{2}\\[2mm] =&\,\frac{1}{2}. \end{align*}

Therefore

P=(12,π6).\begin{align*} \boxed{P=\left(\frac{1}{2},\frac{\pi}{6}\right)}. \end{align*}

(b)

解法一

思路

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区域 RR 可以看成两部分：

从 θ=0\theta=0 到 θ=π6\theta=\frac{\pi}{6} 的极坐标面积，加上点 PP 到 yy 轴形成的小三角形面积。

点 PP 的直角坐标是

(34,14).\begin{align*} \left(\frac{\sqrt3}{4},\frac14\right). \end{align*}

答题过程

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From part (a),

P=(12,π6).\begin{align*} P=\left(\frac{1}{2},\frac{\pi}{6}\right). \end{align*}

The Cartesian coordinates of PP are

xP= 12cos⁡π6=34,yP= 12sin⁡π6=14.\begin{align*} x_P=&\,\frac{1}{2}\cos\frac{\pi}{6} =\frac{\sqrt3}{4},\\[2mm] y_P=&\,\frac{1}{2}\sin\frac{\pi}{6} =\frac{1}{4}. \end{align*}

The triangular part has area

12xPyP= 12⋅34⋅14= 332.\begin{align*} \frac{1}{2}x_Py_P =&\,\frac{1}{2}\cdot\frac{\sqrt3}{4}\cdot\frac{1}{4}\\[2mm] =&\,\frac{\sqrt3}{32}. \end{align*}

Now calculate the polar area from θ=0\theta=0 to θ=π6\theta=\frac{\pi}{6}:

12∫0π6(1−sin⁡θ)2 dθ= 12∫0π6(1−2sin⁡θ+sin⁡2θ) dθ.\begin{align*} \frac{1}{2}\int_0^{\frac{\pi}{6}}(1-\sin\theta)^2\,\mathrm{d}\theta =&\,\frac{1}{2}\int_0^{\frac{\pi}{6}} (1-2\sin\theta+\sin^2\theta)\,\mathrm{d}\theta. \end{align*}

Using

sin⁡2θ=1−cos⁡2θ2,\begin{align*} \sin^2\theta=\frac{1-\cos2\theta}{2}, \end{align*}

the polar area is

12∫0π6(32−2sin⁡θ−12cos⁡2θ) dθ= 12[3θ2+2cos⁡θ−14sin⁡2θ]0π6= 12(π4+3−38−2)= π8+7316−1.\begin{align*} \frac{1}{2}\int_0^{\frac{\pi}{6}} \left(\frac{3}{2}-2\sin\theta-\frac{1}{2}\cos2\theta\right) \,\mathrm{d}\theta =&\,\frac{1}{2} \left[ \frac{3\theta}{2}+2\cos\theta-\frac{1}{4}\sin2\theta \right]_0^{\frac{\pi}{6}}\\[2mm] =&\,\frac{1}{2} \left(\frac{\pi}{4}+\sqrt3-\frac{\sqrt3}{8}-2\right)\\[2mm] =&\,\frac{\pi}{8}+\frac{7\sqrt3}{16}-1. \end{align*}

Therefore the area of RR is

332+(π8+7316−1)= 4π32+332+14332−3232= 132(4π+153−32).\begin{align*} \frac{\sqrt3}{32} +\left(\frac{\pi}{8}+\frac{7\sqrt3}{16}-1\right) =&\,\frac{4\pi}{32}+\frac{\sqrt3}{32} +\frac{14\sqrt3}{32}-\frac{32}{32}\\[2mm] =&\,\frac{1}{32}(4\pi+15\sqrt3-32). \end{align*}

Hence

a=4,b=15,c=−32.\begin{align*} \boxed{a=4,\qquad b=15,\qquad c=-32}. \end{align*}