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FP2 2023 Jan Q8

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 8

题目

Problem

8.

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)
题目中文翻译

图 1 所示曲线 CC 的极坐标方程为

r= 1−sin⁡θ,0⩽θ<π2. \begin{align*} r=&\,1-\sin\theta,\qquad0\leqslant\theta<\frac{\pi}{2}. \end{align*}

点 PP 在曲线 CC 上，且曲线在 PP 处的切线平行于极轴。

(a) 用微积分方法求 PP 的极坐标。

(4)

图 1 中阴影所示的有限区域 RR 由以下部分围成：

  • 方程为 θ=π2\theta=\frac{\pi}{2} 的直线；
  • 曲线 CC 在 PP 处的切线；
  • 曲线 CC 的一部分；
  • 极轴。

(b) 用代数积分证明，RR 的面积为

132(aπ+b3+c), \begin{align*} \frac1{32}(a\pi+b\sqrt3+c), \end{align*}

其中 aa、bb、cc 为待求的整数。

(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=\bigg(\frac{1}{2},\frac{\pi}{6}\bigg)}. \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*} \bigg(\frac{\sqrt3}{4},\frac14\bigg). \end{align*}

答题过程

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From part (a),

P=(12,π6).\begin{align*} P=\bigg(\frac{1}{2},\frac{\pi}{6}\bigg). \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}:

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

S= 12[3θ2+2cos⁡θ−14sin⁡2θ]0π6= 12(π4+3−38−2)= π8+7316−1.\begin{align*} S=&\,\frac{1}{2} \bigg[ \frac{3\theta}{2}+2\cos\theta-\frac{1}{4}\sin2\theta \bigg]_0^{\frac{\pi}{6}}\\[2mm] =&\,\frac{1}{2} \bigg(\frac{\pi}{4}+\sqrt3-\frac{\sqrt3}{8}-2\bigg)\\[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} +\bigg(\frac{\pi}{8}+\frac{7\sqrt3}{16}-1\bigg) =&\,\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*}

解法二

思路

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也可以先求曲线在 0⩽θ<π/20\leqslant\theta<\pi/2 内的整个极坐标面积，再加上曲线与水平切线之间的补区域。补区域等于顶点为 OO、PP 和切线与竖轴交点的小三角形，减去 π/6\pi/6 到 π/2\pi/2 的极坐标面积。

答题过程

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From part (a), the Cartesian coordinates of PP are

xP= 12cos⁡π6=34,yP= 12sin⁡π6=14.\begin{align*} x_P=&\,\frac12\cos\frac{\pi}{6}=\frac{\sqrt3}{4},& y_P=&\,\frac12\sin\frac{\pi}{6}=\frac14. \end{align*}

The triangle formed by OO, PP and the intersection of the tangent with the vertical axis has area

T= 12xPyP=332.\begin{align*} T=&\,\frac12 x_Py_P =\frac{\sqrt3}{32}. \end{align*}

For the polar area, use

(1−sin⁡θ)2= 1−2sin⁡θ+1−cos⁡2θ2= 32−2sin⁡θ−12cos⁡2θ.\begin{align*} (1-\sin\theta)^2 =&\,1-2\sin\theta+\frac{1-\cos2\theta}{2}\\[2mm] =&\,\frac32-2\sin\theta-\frac12\cos2\theta. \end{align*}

An antiderivative is

F(θ)= 3θ2+2cos⁡θ−14sin⁡2θ.\begin{align*} F(\theta)=&\,\frac{3\theta}{2}+2\cos\theta-\frac14\sin2\theta. \end{align*}

The whole polar area is

S= 12∫0π/2(1−sin⁡θ)2 dθ= 12(F(π/2)−F(0))= 12(3π4−2)= 3π8−1.\begin{align*} S=&\,\frac12\int_0^{\pi/2}(1-\sin\theta)^2\,\mathrm{d}\theta\\[2mm] =&\,\frac12\big(F(\pi/2)-F(0)\big)\\[2mm] =&\,\frac12\bigg(\frac{3\pi}{4}-2\bigg)\\[2mm] =&\,\frac{3\pi}{8}-1. \end{align*}

The polar area between θ=π/6\theta=\pi/6 and θ=π/2\theta=\pi/2 is

U= 12∫π/6π/2(1−sin⁡θ)2 dθ= 12(F(π/2)−F(π/6))= 12(3π4−π4−3+38)= π4−7316.\begin{align*} U=&\,\frac12\int_{\pi/6}^{\pi/2}(1-\sin\theta)^2\,\mathrm{d}\theta\\[2mm] =&\,\frac12\big(F(\pi/2)-F(\pi/6)\big)\\[2mm] =&\,\frac12\bigg(\frac{3\pi}{4} -\frac{\pi}{4}-\sqrt3+\frac{\sqrt3}{8}\bigg)\\[2mm] =&\,\frac{\pi}{4}-\frac{7\sqrt3}{16}. \end{align*}

Thus the area above the curve and below the horizontal tangent is

T−U= 332−(π4−7316)= 15332−π4.\begin{align*} T-U=&\,\frac{\sqrt3}{32} -\bigg(\frac{\pi}{4}-\frac{7\sqrt3}{16}\bigg)\\[2mm] =&\,\frac{15\sqrt3}{32}-\frac{\pi}{4}. \end{align*}

Therefore

Area⁡(R)= S+(T−U)= 3π8−1+15332−π4= 132(4π+153−32).\begin{align*} \operatorname{Area}(R)=&\,S+(T-U)\\[2mm] =&\,\frac{3\pi}{8}-1+\frac{15\sqrt3}{32}-\frac{\pi}{4}\\[2mm] =&\,\frac{1}{32}(4\pi+15\sqrt3-32). \end{align*}

Hence a=4,b=15,c=−32\boxed{a=4,\quad b=15,\quad c=-32}.