题目
Problem
3. The curve C C C O O O
r = 1 + cos θ , 0 ⩽ θ ⩽ π 2 . \begin{align*}
r=&\,1+\cos\theta,\\[4mm]
&\,0\leqslant\theta\leqslant\frac{\pi}{2}.
\end{align*} r = 1 + cos θ , 0 ⩽ θ ⩽ 2 π . At the point A A A C C C C C C
(a) Find the polar coordinates of A A A
(4)
(b) Find the finite area enclosed by the initial line, the line O A OA O A C C C a π + b 3 a\pi + b\sqrt{3} aπ + b 3 a a a b b b
(6)
题目中文翻译
3. 曲线 C C C O O O
r = 1 + cos θ , 0 ⩽ θ ⩽ π 2 . \begin{align*}
r=&\,1+\cos\theta,\\[4mm]
&\,0\leqslant\theta\leqslant\frac{\pi}{2}.
\end{align*} r = 1 + cos θ , 0 ⩽ θ ⩽ 2 π . 曲线 C C C A A A
(a) 求 A A A
(4)
(b) 求由始线、直线 O A OA O A C C C a π + b 3 a\pi+b\sqrt{3} aπ + b 3 a a a b b b
(6)
解答
(a)
解法一
思路
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Tangent parallel to the initial line 表示切线水平。极坐标里可以先写
y = r sin θ , \begin{align*}
y=r\sin\theta,
\end{align*} y = r sin θ , 水平切线对应 d y d θ = 0 \dfrac{\mathrm{d}y}{\mathrm{d}\theta}=0 d θ d y = 0
答题过程
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Since
y = r sin θ , \begin{align*}
y=r\sin\theta,
\end{align*} y = r sin θ , we have
y = ( 1 + cos θ ) sin θ = sin θ + sin θ cos θ = sin θ + 1 2 sin 2 θ . \begin{align*}
y
=&\,(1+\cos\theta)\sin\theta\\[4mm]
=&\,\sin\theta+\sin\theta\cos\theta\\[4mm]
=&\,\sin\theta+\frac12\sin2\theta.
\end{align*} y = = = ( 1 + cos θ ) sin θ sin θ + sin θ cos θ sin θ + 2 1 sin 2 θ . Differentiate with respect to θ \theta θ
d y d θ = cos θ + cos 2 θ . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}\theta}
=&\,\cos\theta+\cos2\theta.
\end{align*} d θ d y = cos θ + cos 2 θ . For a tangent parallel to the initial line,
d y d θ = 0. \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}\theta}=0.
\end{align*} d θ d y = 0. Thus
cos θ + cos 2 θ = 0 cos θ + 2 cos 2 θ − 1 = 0. \begin{align*}
\cos\theta+\cos2\theta=&\,0\\[4mm]
\cos\theta+2\cos^2\theta-1=&\,0.
\end{align*} cos θ + cos 2 θ = cos θ + 2 cos 2 θ − 1 = 0 0. Let c = cos θ c=\cos\theta c = cos θ
2 c 2 + c − 1 = 0 ( 2 c − 1 ) ( c + 1 ) = 0. \begin{align*}
2c^2+c-1=&\,0\\[4mm]
(2c-1)(c+1)=&\,0.
\end{align*} 2 c 2 + c − 1 = ( 2 c − 1 ) ( c + 1 ) = 0 0. Since 0 ⩽ θ ⩽ π 2 0\leqslant\theta\leqslant\dfrac{\pi}{2} 0 ⩽ θ ⩽ 2 π cos θ ⩾ 0 \cos\theta\geqslant0 cos θ ⩾ 0 cos θ = − 1 \cos\theta=-1 cos θ = − 1
cos θ = 1 2 , θ = π 3 . \begin{align*}
\cos\theta=\frac12,
\qquad
\theta=\frac{\pi}{3}.
\end{align*} cos θ = 2 1 , θ = 3 π . At this point,
r = 1 + cos π 3 = 3 2 . \begin{align*}
r=1+\cos\frac{\pi}{3}=\frac32.
\end{align*} r = 1 + cos 3 π = 2 3 . Therefore,
A = ( 3 2 , π 3 ) . \begin{align*}
A=\big(\frac32,\frac{\pi}{3}\big).
\end{align*} A = ( 2 3 , 3 π ) .
(b)
解法一
思路
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区域由 initial line、O A OA O A 0 0 0 π 3 \dfrac{\pi}{3} 3 π
1 2 ∫ r 2 d θ . \begin{align*}
\frac12\int r^2\,\mathrm{d}\theta.
\end{align*} 2 1 ∫ r 2 d θ . 关键是把 cos 2 θ \cos^2\theta cos 2 θ 1 2 ( 1 + cos 2 θ ) \dfrac12(1+\cos2\theta) 2 1 ( 1 + cos 2 θ )
答题过程
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The required area is
1 2 ∫ 0 π 3 ( 1 + cos θ ) 2 d θ . \begin{align*}
\frac12\int_0^{\frac{\pi}{3}}(1+\cos\theta)^2
\,\mathrm{d}\theta.
\end{align*} 2 1 ∫ 0 3 π ( 1 + cos θ ) 2 d θ . Now
( 1 + cos θ ) 2 = 1 + 2 cos θ + cos 2 θ = 1 + 2 cos θ + 1 2 ( 1 + cos 2 θ ) = 3 2 + 2 cos θ + 1 2 cos 2 θ . \begin{align*}
&\,(1+\cos\theta)^2\\[4mm]
=&\,1+2\cos\theta+\cos^2\theta\\[4mm]
=&\,1+2\cos\theta\\[4mm]
&\,\hspace{2pt}+\frac12(1+\cos2\theta)\\[4mm]
=&\,\frac32+2\cos\theta\\[4mm]
&\,\hspace{2pt}+\frac12\cos2\theta.
\end{align*} = = = ( 1 + cos θ ) 2 1 + 2 cos θ + cos 2 θ 1 + 2 cos θ + 2 1 ( 1 + cos 2 θ ) 2 3 + 2 cos θ + 2 1 cos 2 θ . An antiderivative of ( 1 + cos θ ) 2 (1+\cos\theta)^2 ( 1 + cos θ ) 2
G ( θ ) = 3 2 θ + 2 sin θ + 1 4 sin 2 θ . \begin{align*}
G(\theta)=&\,\frac32\theta+2\sin\theta\\[4mm]
&\,\hspace{2pt}+\frac14\sin2\theta.
\end{align*} G ( θ ) = 2 3 θ + 2 sin θ + 4 1 sin 2 θ . Hence
Area = 1 2 [ G ( π 3 ) − G ( 0 ) ] . \begin{align*}
&\,\text{Area}\\[4mm]
=&\,\frac12\bigg[G\bigg(\frac\pi3\bigg)-G(0)\bigg].
\end{align*} = Area 2 1 [ G ( 3 π ) − G ( 0 ) ] . Substitute the limits:
Area = 1 2 ( π 2 + 2 ⋅ 3 2 + 1 4 ⋅ 3 2 ) = 1 2 ( π 2 + 3 + 3 8 ) = π 4 + 9 3 16 . \begin{align*}
&\,\text{Area}\\[4mm]
=&\,\frac12\bigg(\frac\pi2+2\cdot\frac{\sqrt3}{2}\\[4mm]
&\,\hspace{2pt}+\frac14\cdot\frac{\sqrt3}{2}\bigg)\\[4mm]
=&\,\frac12\bigg(\frac\pi2+\sqrt3+\frac{\sqrt3}{8}\bigg)\\[4mm]
=&\,\frac\pi4+\frac{9\sqrt3}{16}.
\end{align*} = = = Area 2 1 ( 2 π + 2 ⋅ 2 3 + 4 1 ⋅ 2 3 ) 2 1 ( 2 π + 3 + 8 3 ) 4 π + 16 9 3 . Thus
a = 1 4 , b = 9 16 . \begin{align*}
a=\frac14,\qquad b=\frac9{16}.
\end{align*} a = 4 1 , b = 16 9 .
解法二
思路
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不使用倍角降幂，也可以通过分部积分求出 cos 2 θ \cos^2\theta cos 2 θ sin 2 θ \sin^2\theta sin 2 θ 1 − cos 2 θ 1-\cos^2\theta 1 − cos 2 θ
答题过程
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From part (a), the line O A OA O A π 3 \dfrac\pi3 3 π
1 2 ∫ 0 π / 3 ( 1 + cos θ ) 2 d θ . \frac12\int_0^{\pi/3}(1+\cos\theta)^2\,\mathrm{d}\theta. 2 1 ∫ 0 π /3 ( 1 + cos θ ) 2 d θ . To integrate cos 2 θ \cos^2\theta cos 2 θ
u = cos θ , d v = cos θ d θ , d u = − sin θ d θ , v = sin θ . \begin{align*}
u=&\,\cos\theta,\\[4mm]
\mathrm{d}v=&\,\cos\theta\,\mathrm{d}\theta,\\[4mm]
\mathrm{d}u=&\,-\sin\theta\,\mathrm{d}\theta,\\[4mm]
v=&\,\sin\theta.
\end{align*} u = d v = d u = v = cos θ , cos θ d θ , − sin θ d θ , sin θ . Let J = ∫ cos 2 θ d θ J=\displaystyle\int\cos^2\theta\,\mathrm{d}\theta J = ∫ cos 2 θ d θ
J = sin θ cos θ + ∫ sin 2 θ d θ = sin θ cos θ + ∫ ( 1 − cos 2 θ ) d θ = sin θ cos θ + θ − J , \begin{align*}
J=&\,\sin\theta\cos\theta\\[4mm]
&\,\hspace{2pt}+\int\sin^2\theta\,\mathrm{d}\theta\\[4mm]
=&\,\sin\theta\cos\theta\\[4mm]
&\,\hspace{2pt}+\int(1-\cos^2\theta)\,\mathrm{d}\theta\\[4mm]
=&\,\sin\theta\cos\theta+\theta-J,
\end{align*} J = = = sin θ cos θ + ∫ sin 2 θ d θ sin θ cos θ + ∫ ( 1 − cos 2 θ ) d θ sin θ cos θ + θ − J , up to an additive constant. Hence
2 J = sin θ cos θ + θ + C , J = 1 2 ( sin θ cos θ + θ ) + C 1 . \begin{align*}
&\,2J\\[4mm]
=&\,\sin\theta\cos\theta+\theta+C,\\[4mm]
&\,J\\[4mm]
=&\,\frac12(\sin\theta\cos\theta+\theta)\\[4mm]
&\,\hspace{2pt}+C_1.
\end{align*} = = 2 J sin θ cos θ + θ + C , J 2 1 ( sin θ cos θ + θ ) + C 1 . Expanding the area integrand,
( 1 + cos θ ) 2 = 1 + 2 cos θ + cos 2 θ . \begin{align*}
&\,(1+\cos\theta)^2\\[4mm]
=&\,1+2\cos\theta+\cos^2\theta.
\end{align*} = ( 1 + cos θ ) 2 1 + 2 cos θ + cos 2 θ . An antiderivative is
H ( θ ) = θ + 2 sin θ + 1 2 ( sin θ cos θ + θ ) = 3 2 θ + 2 sin θ + 1 2 sin θ cos θ . \begin{align*}
&\,H(\theta)\\[4mm]
=&\,\theta+2\sin\theta\\[4mm]
&\,\hspace{2pt}+\frac12(\sin\theta\cos\theta+\theta)\\[4mm]
=&\,\frac32\theta+2\sin\theta\\[4mm]
&\,\hspace{2pt}+\frac12\sin\theta\cos\theta.
\end{align*} = = H ( θ ) θ + 2 sin θ + 2 1 ( sin θ cos θ + θ ) 2 3 θ + 2 sin θ + 2 1 sin θ cos θ . Therefore,
Area = 1 2 [ H ( π 3 ) − H ( 0 ) ] . \begin{align*}
&\,\text{Area}\\[4mm]
=&\,\frac12\bigg[H\bigg(\frac\pi3\bigg)-H(0)\bigg].
\end{align*} = Area 2 1 [ H ( 3 π ) − H ( 0 ) ] . Since H ( 0 ) = 0 H(0)=0 H ( 0 ) = 0
Area = 1 2 [ π 2 + 2 ⋅ 3 2 + 1 2 ⋅ 3 2 ⋅ 1 2 ] = 1 2 ( π 2 + 9 3 8 ) = π 4 + 9 3 16 . \begin{align*}
&\,\text{Area}\\[4mm]
=&\,\frac12\bigg[\frac\pi2+2\cdot\frac{\sqrt3}{2}\\[4mm]
&\,\hspace{2pt}+\frac12\cdot\frac{\sqrt3}{2}\cdot\frac12\bigg]\\[4mm]
=&\,\frac12\bigg(\frac\pi2+\frac{9\sqrt3}{8}\bigg)\\[4mm]
=&\,\frac\pi4+\frac{9\sqrt3}{16}.
\end{align*} = = = Area 2 1 [ 2 π + 2 ⋅ 2 3 + 2 1 ⋅ 2 3 ⋅ 2 1 ] 2 1 ( 2 π + 8 9 3 ) 4 π + 16 9 3 . Thus
a = 1 4 , b = 9 16 . a=\frac14,\qquad b=\frac9{16}. a = 4 1 , b = 16 9 . 