A答案 Q3 — 北里大学 2019
(1) 点 A , P A,P A , P
B = ( cos θ , sin θ ) B=(\cos\theta,\sin\theta) B = ( cos θ , sin θ ) A = ( a , 0 ) A=(a,0) A = ( a , 0 ) 0 ≦ a ≦ 2 0\leq a\leq2 0 ≦ a ≦ 2 A B = O B = 1 AB=OB=1 A B = O B = 1 ( a − cos θ ) 2 + sin 2 θ = 1 より a 2 − 2 a cos θ = 0. (a-\cos\theta)^2+\sin^2\theta=1
\quad\text{より}\quad
a^2-2a\cos\theta=0. ( a − cos θ ) 2 + sin 2 θ = 1 より a 2 − 2 a cos θ = 0. a > 0 a>0 a > 0 a = 2 cos θ a=2\cos\theta a = 2 cos θ y B ≧ 0 y_B\geq0 y B ≧ 0 a > 0 a>0 a > 0 0 ≦ θ < π 2 \displaystyle 0\leq\theta<\frac\pi2 0 ≦ θ < 2 π a = 0 a=0 a = 0 θ = π 2 \displaystyle \theta=\frac\pi2 θ = 2 π B = P = ( 0 , 1 ) B=P=(0,1) B = P = ( 0 , 1 ) 0 ≦ θ ≦ π 2 \displaystyle 0\leq\theta\leq\frac\pi2 0 ≦ θ ≦ 2 π A = ( 2 cos θ , 0 ) . \boxed{A=(2\cos\theta,0)}. A = ( 2 cos θ , 0 ) .
A B = 1 AB=1 A B = 1 B P = O A / 2 = a / 2 = cos θ BP=OA/2=a/2=\cos\theta B P = O A /2 = a /2 = cos θ P P P B B B A A A cos θ \cos\theta cos θ P = B + cos θ ( A − B ) . P=B+\cos\theta(A-B). P = B + cos θ ( A − B ) . P = ( cos θ + cos 2 θ , ( 1 − cos θ ) sin θ ) . \boxed{P=(\cos\theta+\cos^2\theta,\ (1-\cos\theta)\sin\theta)}. P = ( cos θ + cos 2 θ , ( 1 − cos θ ) sin θ ) . θ = π 2 \displaystyle \theta=\frac\pi2 θ = 2 π P = ( 0 , 1 ) P=(0,1) P = ( 0 , 1 )
(2) 曲線の長さ
x ( θ ) = cos θ + cos 2 θ , y ( θ ) = ( 1 − cos θ ) sin θ . x(\theta)=\cos\theta+\cos^2\theta,\qquad
y(\theta)=(1-\cos\theta)\sin\theta. x ( θ ) = cos θ + cos 2 θ , y ( θ ) = ( 1 − cos θ ) sin θ . x ′ ( θ ) = − sin θ ( 1 + 2 cos θ ) , y ′ ( θ ) = ( 1 − cos θ ) ( 1 + 2 cos θ ) . x'(\theta)=-\sin\theta(1+2\cos\theta),\qquad
y'(\theta)=(1-\cos\theta)(1+2\cos\theta). x ′ ( θ ) = − sin θ ( 1 + 2 cos θ ) , y ′ ( θ ) = ( 1 − cos θ ) ( 1 + 2 cos θ ) . 0 ≦ θ ≦ π 2 \displaystyle 0\leq\theta\leq\frac\pi2 0 ≦ θ ≦ 2 π ( x ′ ) 2 + ( y ′ ) 2 = ( 1 + 2 cos θ ) sin 2 θ + ( 1 − cos θ ) 2 = 2 ( 1 + 2 cos θ ) sin θ 2 . \displaystyle
\begin{aligned}
\sqrt{(x')^2+(y')^2}
&=(1+2\cos\theta)\sqrt{\sin^2\theta+(1-\cos\theta)^2}\\
&=2(1+2\cos\theta)\sin\frac\theta2.
\end{aligned} ( x ′ ) 2 + ( y ′ ) 2 = ( 1 + 2 cos θ ) sin 2 θ + ( 1 − cos θ ) 2 = 2 ( 1 + 2 cos θ ) sin 2 θ . u = θ / 2 u=\theta/2 u = θ /2 L = 4 ∫ 0 π / 4 sin u ( 1 + 2 cos 2 u ) d u = 4 ∫ 0 π / 4 ( 3 sin u − 4 sin 3 u ) d u = 4 [ cos u − 4 3 cos 3 u ] 0 π / 4 = 4 + 2 2 3 . \displaystyle
\begin{aligned}
L
&=4\int_0^{\pi/4}\sin u(1+2\cos2u)\,du\\
&=4\int_0^{\pi/4}(3\sin u-4\sin^3u)\,du\\
&=4\left[\cos u-\frac43\cos^3u\right]_0^{\pi/4}
=\boxed{\frac{4+2\sqrt2}{3}}.
\end{aligned} L = 4 ∫ 0 π /4 sin u ( 1 + 2 cos 2 u ) d u = 4 ∫ 0 π /4 ( 3 sin u − 4 sin 3 u ) d u = 4 [ cos u − 3 4 cos 3 u ] 0 π /4 = 3 4 + 2 2 .
(3) 囲まれた部分の面積
θ \theta θ π 2 \displaystyle \frac\pi2 2 π y y y ( 0 , 1 ) (0,1) ( 0 , 1 ) x x x ( 2 , 0 ) (2,0) ( 2 , 0 ) ∫ y d x \displaystyle \int y\,dx ∫ y d x S = ∫ 0 π / 2 y ( θ ) ( − x ′ ( θ ) ) d θ = ∫ 0 π / 2 sin 2 θ ( 1 − cos θ ) ( 1 + 2 cos θ ) d θ . \displaystyle
S=\int_0^{\pi/2}y(\theta)(-x'(\theta))\,d\theta
=\int_0^{\pi/2}\sin^2\theta(1-\cos\theta)(1+2\cos\theta)\,d\theta. S = ∫ 0 π /2 y ( θ ) ( − x ′ ( θ )) d θ = ∫ 0 π /2 sin 2 θ ( 1 − cos θ ) ( 1 + 2 cos θ ) d θ . sin 2 θ ( 1 − cos θ ) ( 1 + 2 cos θ ) = 1 + cos θ − 3 cos 2 θ − cos 3 θ + 2 cos 4 θ . \sin^2\theta(1-\cos\theta)(1+2\cos\theta)
=1+\cos\theta-3\cos^2\theta-\cos^3\theta+2\cos^4\theta. sin 2 θ ( 1 − cos θ ) ( 1 + 2 cos θ ) = 1 + cos θ − 3 cos 2 θ − cos 3 θ + 2 cos 4 θ . ∫ 0 π / 2 cos θ d θ = 1 , ∫ 0 π / 2 cos 2 θ d θ = π 4 , ∫ 0 π / 2 cos 3 θ d θ = 2 3 , ∫ 0 π / 2 cos 4 θ d θ = 3 π 16 \displaystyle
\int_0^{\pi/2}\cos\theta\,d\theta=1,\quad
\int_0^{\pi/2}\cos^2\theta\,d\theta=\frac\pi4,\quad
\int_0^{\pi/2}\cos^3\theta\,d\theta=\frac23,\quad
\int_0^{\pi/2}\cos^4\theta\,d\theta=\frac{3\pi}{16} ∫ 0 π /2 cos θ d θ = 1 , ∫ 0 π /2 cos 2 θ d θ = 4 π , ∫ 0 π /2 cos 3 θ d θ = 3 2 , ∫ 0 π /2 cos 4 θ d θ = 16 3 π S = π 2 + 1 − 3 π 4 − 2 3 + 3 π 8 = π 8 + 1 3 . \displaystyle
S=\frac\pi2+1-\frac{3\pi}{4}-\frac23+\frac{3\pi}{8}
=\boxed{\frac\pi8+\frac13}. S = 2 π + 1 − 4 3 π − 3 2 + 8 3 π = 8 π + 3 1 .
最終答案
(1) A = ( 2 cos θ , 0 ) A=(2\cos\theta,0) A = ( 2 cos θ , 0 ) P = ( cos θ + cos 2 θ , ( 1 − cos θ ) sin θ ) P=(\cos\theta+\cos^2\theta,(1-\cos\theta)\sin\theta) P = ( cos θ + cos 2 θ , ( 1 − cos θ ) sin θ ) 0 ≦ θ ≦ π 2 \displaystyle 0\leq\theta\leq\frac\pi2 0 ≦ θ ≦ 2 π 4 + 2 2 3 \displaystyle \frac{4+2\sqrt2}{3} 3 4 + 2 2 π 8 + 1 3 \displaystyle \frac\pi8+\frac13 8 π + 3 1
使用解法：三角関数による距離条件、線分の内分表示、媒介変数表示の曲線長、定積分による面積。 issues：なし。
