東京大学 新領域創成科学研究科 人間環境学専攻 2016年8月実施 微分積分 Author 思齐塾 , 祭音Myyura
Description Oを原点とした x y xy x y P P P ( r , θ ) (r, \theta) ( r , θ ) θ \theta θ x x x O P ‾ \overline{OP} OP 0 ≤ θ < 2 π 0 \leq \theta < 2\pi 0 ≤ θ < 2 π r r r O P OP OP r = e θ r = e^{\theta} r = e θ θ \theta θ 0 0 0 C C C
(1) 軌跡 C C C ( θ = 0 ) (\theta = 0) ( θ = 0 ) θ \theta θ
(2) 点 P P P x x x y y y s s s x ( s ) x(s) x ( s ) y ( s ) y(s) y ( s ) p ( s ) = O P ‾ = ( x ( s ) , y ( s ) ) p(s) = \overline{OP} = (x(s), y(s)) p ( s ) = OP = ( x ( s ) , y ( s )) ∣ d p ( s ) d s ∣ = 1 \left| \frac{dp(s)}{ds} \right| = 1 d s d p ( s ) = 1
(3) 軌跡 C C C O P ‾ \overline{OP} OP θ \theta θ
(4) 軌跡 C C C P P P θ \theta θ Q Q Q s s s O O O Q Q Q q ( s ) q(s) q ( s ) q ( s ) q(s) q ( s ) R R R
(5) 軌跡 C C C P P P ( x a , y a ) (x_a, y_a) ( x a , y a ) ( x a , y a ) (x_a, y_a) ( x a , y a ) C ′ C' C ′ ( r ′ , θ ′ ) (r', \theta') ( r ′ , θ ′ ) r ′ r' r ′ θ ′ \theta' θ ′ θ \theta θ
図1：弧長表示と曲率半径の概念図。
题目描述
在以 O O O x y xy x y ( r , θ ) (r,\theta) ( r , θ ) P P P θ \theta θ x x x O P ‾ \overline{OP} OP 0 ≤ θ < 2 π 0\leq\theta<2\pi 0 ≤ θ < 2 π r = O P r=OP r = OP r = e θ r=e^\theta r = e θ θ \theta θ 0 0 0 P P P C C C
将轨迹 C C C θ = 0 \theta=0 θ = 0 P P P θ \theta θ
以第 1 问所得弧长 s s s P P P x ( s ) , y ( s ) x(s),y(s) x ( s ) , y ( s )
p ( s ) = O P ‾ = ( x ( s ) , y ( s ) ) , p(s)=\overline{OP}=(x(s),y(s)), p ( s ) = OP = ( x ( s ) , y ( s )) , 并证明
∣ d p ( s ) d s ∣ = 1. \left|\frac{dp(s)}{ds}\right|=1. d s d p ( s ) = 1.
证明轨迹 C C C P P P O P ‾ \overline{OP} OP θ \theta θ
将轨迹 C C C P P P θ \theta θ Q Q Q s s s Q Q Q O O O q ( s ) q(s) q ( s ) q ( s ) q(s) q ( s ) R R R
设轨迹 C C C P P P ( x a , y a ) (x_a,y_a) ( x a , y a ) C ′ C' C ′ ( r ′ , θ ′ ) (r',\theta') ( r ′ , θ ′ ) r ′ r' r ′ θ ′ \theta' θ ′ θ \theta θ
Kai (1) 軌跡 C C C ( θ = 0 ) (\theta = 0) ( θ = 0 ) s s s r = e θ r = e^{\theta} r = e θ
d r d θ = e θ \frac{dr}{d\theta} = e^{\theta} d θ d r = e θ したがって，
d s d θ = r 2 + ( d r d θ ) 2 = ( e θ ) 2 + ( e θ ) 2 = 2 e θ \frac{ds}{d\theta} = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} = \sqrt{(e^{\theta})^2 + (e^{\theta})^2} = \sqrt{2}e^{\theta} d θ d s = r 2 + ( d θ d r ) 2 = ( e θ ) 2 + ( e θ ) 2 = 2 e θ θ \theta θ 0 0 0 θ \theta θ s s s
s = ∫ 0 θ 2 e t d t = 2 [ e t ] 0 θ = 2 ( e θ − 1 ) s = \int_0^{\theta} \sqrt{2}e^t dt = \sqrt{2} [e^t]_0^{\theta} = \sqrt{2}(e^{\theta} - 1) s = ∫ 0 θ 2 e t d t = 2 [ e t ] 0 θ = 2 ( e θ − 1 ) (2) r = e θ r = e^{\theta} r = e θ x ( θ ) = r cos θ = e θ cos θ x(\theta) = r \cos \theta = e^{\theta} \cos \theta x ( θ ) = r cos θ = e θ cos θ y ( θ ) = r sin θ = e θ sin θ y(\theta) = r \sin \theta = e^{\theta} \sin \theta y ( θ ) = r sin θ = e θ sin θ θ = log ( s 2 + 1 ) \theta = \log \left(\frac{s}{\sqrt{2}} + 1\right) θ = log ( 2 s + 1 )
x ( s ) = ( s 2 + 1 ) cos ( log ( s 2 + 1 ) ) , y ( s ) = ( s 2 + 1 ) sin ( log ( s 2 + 1 ) ) x(s) = \left(\frac{s}{\sqrt{2}} + 1\right) \cos \left(\log \left(\frac{s}{\sqrt{2}} + 1\right) \right), \quad y(s) = \left(\frac{s}{\sqrt{2}} + 1\right) \sin \left(\log \left(\frac{s}{\sqrt{2}} + 1\right) \right) x ( s ) = ( 2 s + 1 ) cos ( log ( 2 s + 1 ) ) , y ( s ) = ( 2 s + 1 ) sin ( log ( 2 s + 1 ) ) よって，
p ( s ) = ( x ( s ) , y ( s ) ) = ( ( s 2 + 1 ) cos ( log ( s 2 + 1 ) ) , ( s 2 + 1 ) sin ( log ( s 2 + 1 ) ) ) p(s) = (x(s), y(s)) = \left( \left(\frac{s}{\sqrt{2}} + 1\right) \cos \left(\log \left(\frac{s}{\sqrt{2}} + 1\right) \right), \left(\frac{s}{\sqrt{2}} + 1\right) \sin \left(\log \left(\frac{s}{\sqrt{2}} + 1\right) \right) \right) p ( s ) = ( x ( s ) , y ( s )) = ( ( 2 s + 1 ) cos ( log ( 2 s + 1 ) ) , ( 2 s + 1 ) sin ( log ( 2 s + 1 ) ) ) d x d s = 1 2 cos ( log ( s 2 + 1 ) ) − 1 2 sin ( log ( s 2 + 1 ) ) \frac{dx}{ds} = \frac{1}{\sqrt{2}} \cos \left(\log \left(\frac{s}{\sqrt{2}} + 1\right) \right) - \frac{1}{\sqrt{2}} \sin \left(\log \left(\frac{s}{\sqrt{2}} + 1\right) \right) d s d x = 2 1 cos ( log ( 2 s + 1 ) ) − 2 1 sin ( log ( 2 s + 1 ) ) d y d s = 1 2 sin ( log ( s 2 + 1 ) ) + 1 2 cos ( log ( s 2 + 1 ) ) \frac{dy}{ds} = \frac{1}{\sqrt{2}} \sin \left(\log \left(\frac{s}{\sqrt{2}} + 1\right) \right) + \frac{1}{\sqrt{2}} \cos \left(\log \left(\frac{s}{\sqrt{2}} + 1\right) \right) d s d y = 2 1 sin ( log ( 2 s + 1 ) ) + 2 1 cos ( log ( 2 s + 1 ) ) よって，
∣ d p ( s ) d s ∣ 2 = 1 2 { ( cos θ − sin θ ) 2 + ( sin θ + cos θ ) 2 } = 1 \left| \frac{dp(s)}{ds} \right|^2
=\frac12\{(\cos\theta-\sin\theta)^2
+(\sin\theta+\cos\theta)^2\}
=1 d s d p ( s ) 2 = 2 1 {( cos θ − sin θ ) 2 + ( sin θ + cos θ ) 2 } = 1 ∣ d p ( s ) d s ∣ = 1 \left| \frac{dp(s)}{ds} \right| = 1 d s d p ( s ) = 1
(3) 軌跡 C C C t t t
t = ( d x d θ , d y d θ ) = ( e θ cos θ − e θ sin θ , e θ sin θ + e θ cos θ ) = e θ ( cos θ − sin θ , sin θ + cos θ ) t = \left( \frac{dx}{d\theta}, \frac{dy}{d\theta} \right) = (e^{\theta} \cos \theta - e^{\theta} \sin \theta, e^{\theta} \sin \theta + e^{\theta} \cos \theta) = e^{\theta} (\cos \theta - \sin \theta, \sin \theta + \cos \theta) t = ( d θ d x , d θ d y ) = ( e θ cos θ − e θ sin θ , e θ sin θ + e θ cos θ ) = e θ ( cos θ − sin θ , sin θ + cos θ ) O P ‾ = ( e θ cos θ , e θ sin θ ) \overline{OP} = (e^{\theta} \cos \theta, e^{\theta} \sin \theta) OP = ( e θ cos θ , e θ sin θ ) C C C O P ‾ \overline{OP} OP α \alpha α
cos α = t ⋅ O P ‾ ∣ t ∣ ∣ O P ‾ ∣ = e 2 θ ( cos 2 θ − sin θ cos θ + sin θ cos θ + sin 2 θ ) 2 e θ ⋅ e θ = e 2 θ 2 e 2 θ = 1 2 \cos \alpha = \frac{t \cdot \overline{OP}}{|t| |\overline{OP}|} = \frac{e^{2\theta}(\cos^2 \theta - \sin \theta \cos \theta + \sin \theta \cos \theta + \sin^2 \theta)}{\sqrt{2}e^{\theta} \cdot e^{\theta}} = \frac{e^{2\theta}}{\sqrt{2}e^{2\theta}} = \frac{1}{\sqrt{2}} cos α = ∣ t ∣∣ OP ∣ t ⋅ OP = 2 e θ ⋅ e θ e 2 θ ( cos 2 θ − sin θ cos θ + sin θ cos θ + sin 2 θ ) = 2 e 2 θ e 2 θ = 2 1 したがって， α = π 4 \alpha = \frac{\pi}{4} α = 4 π
(4) 曲率半径 R = d s d ϕ = ( x ˙ 2 + y ˙ 2 ) 3 / 2 ∣ x ˙ y ¨ − y ˙ x ¨ ∣ R = \frac{ds}{d\phi} = \frac{(\dot{x}^2 + \dot{y}^2)^{3/2}}{|\dot{x} \ddot{y} - \dot{y} \ddot{x}|} R = d ϕ d s = ∣ x ˙ y ¨ − y ˙ x ¨ ∣ ( x ˙ 2 + y ˙ 2 ) 3/2 x ˙ = d x d θ = e θ ( cos θ − sin θ ) , y ˙ = d y d θ = e θ ( sin θ + cos θ ) \dot{x} = \frac{dx}{d\theta} = e^{\theta}(\cos \theta - \sin \theta), \quad \dot{y} = \frac{dy}{d\theta} = e^{\theta}(\sin \theta + \cos \theta) x ˙ = d θ d x = e θ ( cos θ − sin θ ) , y ˙ = d θ d y = e θ ( sin θ + cos θ ) x ¨ = d 2 x d θ 2 = − 2 e θ sin θ , y ¨ = d 2 y d θ 2 = 2 e θ cos θ \ddot{x} = \frac{d^2 x}{d\theta^2} = -2e^{\theta} \sin \theta, \quad \ddot{y} = \frac{d^2 y}{d\theta^2} = 2e^{\theta} \cos \theta x ¨ = d θ 2 d 2 x = − 2 e θ sin θ , y ¨ = d θ 2 d 2 y = 2 e θ cos θ
x ˙ y ¨ − y ˙ x ¨ = 2 e 2 θ \dot{x}\ddot{y}-\dot{y}\ddot{x}
=2e^{2\theta} x ˙ y ¨ − y ˙ x ¨ = 2 e 2 θ であるから，
R = ( 2 e 2 θ ) 3 / 2 2 e 2 θ = 2 e θ . R
=\frac{(2e^{2\theta})^{3/2}}{2e^{2\theta}}
=\sqrt{2}e^\theta . R = 2 e 2 θ ( 2 e 2 θ ) 3/2 = 2 e θ . (5) 曲率は正なので，内接円の中心は左単位法線の方向にあり，
( x a , y a ) = ( x , y ) + R ( − d y d s , d x d s ) (x_a,y_a)
=(x,y)+R\left(-\frac{dy}{ds},\frac{dx}{ds}\right) ( x a , y a ) = ( x , y ) + R ( − d s d y , d s d x ) である．したがって，
x a = e θ cos θ − 2 e θ sin θ + cos θ 2 = − e θ sin θ , y a = e θ sin θ + 2 e θ cos θ − sin θ 2 = e θ cos θ . \begin{aligned}
x_a
&=e^\theta\cos\theta
-\sqrt{2}e^\theta\frac{\sin\theta+\cos\theta}{\sqrt{2}}
=-e^\theta\sin\theta,\\
y_a
&=e^\theta\sin\theta
+\sqrt{2}e^\theta\frac{\cos\theta-\sin\theta}{\sqrt{2}}
=e^\theta\cos\theta .
\end{aligned} x a y a = e θ cos θ − 2 e θ 2 sin θ + cos θ = − e θ sin θ , = e θ sin θ + 2 e θ 2 cos θ − sin θ = e θ cos θ . よって中心の軌跡 C ′ C' C ′
r ′ = e θ , θ ′ = θ + π 2 ( m o d 2 π ) . \boxed{r'=e^\theta,\qquad
\theta'=\theta+\frac{\pi}{2}\pmod{2\pi}} . r ′ = e θ , θ ′ = θ + 2 π ( mod 2 π ) . 