题目
Problem
The curve C C C O O O
r = 1 + cos θ , 0 ⩽ θ ⩽ π 2 \begin{align*}
r = 1 + \cos\theta, \qquad 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)
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=\left(\frac32,\frac{\pi}{3}\right).
\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
=&\,1+2\cos\theta+\cos^2\theta\\[4mm]
=&\,1+2\cos\theta+\frac12(1+\cos2\theta)\\[4mm]
=&\,\frac32+2\cos\theta+\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 . Therefore,
Area = 1 2 ∫ 0 π 3 ( 3 2 + 2 cos θ + 1 2 cos 2 θ ) d θ = 1 2 [ 3 2 θ + 2 sin θ + 1 4 sin 2 θ ] 0 π 3 . \begin{align*}
\text{Area}
=&\,\frac12
\int_0^{\frac{\pi}{3}}
\left(
\frac32+2\cos\theta+\frac12\cos2\theta
\right)\,\mathrm{d}\theta\\[4mm]
=&\,\frac12
\left[
\frac32\theta+2\sin\theta+\frac14\sin2\theta
\right]_0^{\frac{\pi}{3}}.
\end{align*} Area = = 2 1 ∫ 0 3 π ( 2 3 + 2 cos θ + 2 1 cos 2 θ ) d θ 2 1 [ 2 3 θ + 2 sin θ + 4 1 sin 2 θ ] 0 3 π . 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}
=&\,\frac12
\left(
\frac{\pi}{2}
+2\cdot\frac{\sqrt3}{2}
+\frac14\cdot\frac{\sqrt3}{2}
\right)\\[4mm]
=&\,\frac12
\left(
\frac{\pi}{2}+\sqrt3+\frac{\sqrt3}{8}
\right)\\[4mm]
=&\,\frac{\pi}{4}+\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 . 