東京大学 情報理工学研究科 2019年8月実施 数学 第2問 Author etsurin , 祭音Myyura
Description x y xy x y p = ( p ( t ) , q ( t ) ) \boldsymbol{p} = (p(t), q(t)) p = ( p ( t ) , q ( t )) t ∈ [ a , b ] t \in [a, b] t ∈ [ a , b ] t = a ′ t = a' t = a ′ b ′ b' b ′ p \boldsymbol{p} p l a ′ , b ′ l_{a', b'} l a ′ , b ′
l a ′ , b ′ = ∫ a ′ b ′ ( d p d t ) 2 + ( d q d t ) 2 d t l_{a', b'} = \int_{a'}^{b'} \sqrt{ \left( \frac{dp}{dt} \right)^2 + \left( \frac{dq}{dt} \right)^2 } dt l a ′ , b ′ = ∫ a ′ b ′ ( d t d p ) 2 + ( d t d q ) 2 d t と定義され、p \boldsymbol{p} p l a , b l_{a, b} l a , b L L L p \boldsymbol{p} p d p d t = ( 0 , 0 ) \frac{d\boldsymbol{p}}{dt} = (0, 0) d t d p = ( 0 , 0 ) a a a t t t p \boldsymbol{p} p l a , t l_{a, t} l a , t s = s ( t ) s = s(t) s = s ( t ) p \boldsymbol{p} p s ∈ [ 0 , L ] s \in [0, L] s ∈ [ 0 , L ] s s s
(1) 以下の等式を示せ。
( d p d s ) 2 + ( d q d s ) 2 = 1 \sqrt{ \left( \frac{dp}{ds} \right)^2 + \left( \frac{dq}{ds} \right)^2 } = 1 ( d s d p ) 2 + ( d s d q ) 2 = 1 (2) θ = θ ( s ) \theta = \theta(s) θ = θ ( s ) s s s p \boldsymbol{p} p d p d s = ( d p d s , d q d s ) \frac{d\boldsymbol{p}}{ds} = \left( \frac{dp}{ds}, \frac{dq}{ds} \right) d s d p = ( d s d p , d s d q ) x x x
d p d s d 2 q d s 2 − d q d s d 2 p d s 2 = d θ d s \frac{dp}{ds} \frac{d^2q}{ds^2} - \frac{dq}{ds} \frac{d^2p}{ds^2} = \frac{d\theta}{ds} d s d p d s 2 d 2 q − d s d q d s 2 d 2 p = d s d θ 以下では、曲線 p \boldsymbol{p} p K K K p \boldsymbol{p} p K K K
(3) 任意の時刻 s s s d θ d s ≥ 0 \frac{d\theta}{ds} \geq 0 d s d θ ≥ 0
(4) K K K x = ( x , y ) \boldsymbol{x} = (x, y) x = ( x , y ) s ∈ [ 0 , L ] s \in [0, L] s ∈ [ 0 , L ] x \boldsymbol{x} x K K K r r r
x = p ( s ) + r u ( s ) \boldsymbol{x} = \boldsymbol{p}(s) + r\boldsymbol{u}(s) x = p ( s ) + r u ( s ) と一意に表すことができる。
ここで、u ( s ) \boldsymbol{u}(s) u ( s ) s s s p \boldsymbol{p} p K K K x = ( x , y ) \boldsymbol{x} = (x, y) x = ( x , y )
∣ det ( ∂ x ∂ s ∂ x ∂ r ∂ y ∂ s ∂ y ∂ r ) ∣ = 1 + r d θ d s \left \lvert \text{det} \begin{pmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial r} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial r} \end{pmatrix} \right \rvert = 1 + r \frac{d\theta}{ds} det ( ∂ s ∂ x ∂ s ∂ y ∂ r ∂ x ∂ r ∂ y ) = 1 + r d s d θ (5) 非負実数 D D D K D K_D K D K K K D D D K D K_D K D A D = ∬ K D d x d y A_D = \iint_{K_D} dxdy A D = ∬ K D d x d y K K K A A A p \boldsymbol{p} p L L L
A D = A + L D + π D 2 A_D = A + LD + \pi D^2 A D = A + L D + π D 2 と表せることを示せ。
题目描述
考虑平面光滑曲线
p ( t ) = ( p ( t ) , q ( t ) ) \boldsymbol p(t)=(p(t),q(t)) p ( t ) = ( p ( t ) , q ( t )) t ∈ [ a , b ] t\in[a,b] t ∈ [ a , b ] d p / d t ≠ ( 0 , 0 ) \mathrm d\boldsymbol p/\mathrm dt\ne(0,0) d p / d t = ( 0 , 0 ) a ′ a' a ′ b ′ b' b ′
l a ′ , b ′ = ∫ a ′ b ′ ( d p d t ) 2 + ( d q d t ) 2 d t , l_{a',b'}
=\int_{a'}^{b'}
\sqrt{\left(\frac{\mathrm dp}{\mathrm dt}\right)^2+
\left(\frac{\mathrm dq}{\mathrm dt}\right)^2}\,\mathrm dt, l a ′ , b ′ = ∫ a ′ b ′ ( d t d p ) 2 + ( d t d q ) 2 d t , 全长记为 L = l a , b L=l_{a,b} L = l a , b s = l a , t s=l_{a,t} s = l a , t s ∈ [ 0 , L ] s\in[0,L] s ∈ [ 0 , L ]
（1）证明弧长参数下
( d p d s ) 2 + ( d q d s ) 2 = 1. \sqrt{\left(\frac{\mathrm dp}{\mathrm ds}\right)^2+
\left(\frac{\mathrm dq}{\mathrm ds}\right)^2}=1. ( d s d p ) 2 + ( d s d q ) 2 = 1. （2）令 θ ( s ) \theta(s) θ ( s ) d p / d s \mathrm d\boldsymbol p/\mathrm ds d p / d s x x x
d p d s d 2 q d s 2 − d q d s d 2 p d s 2 = d θ d s . \frac{\mathrm dp}{\mathrm ds}\frac{\mathrm d^2q}{\mathrm ds^2}
-\frac{\mathrm dq}{\mathrm ds}\frac{\mathrm d^2p}{\mathrm ds^2}
=\frac{\mathrm d\theta}{\mathrm ds}. d s d p d s 2 d 2 q − d s d q d s 2 d 2 p = d s d θ . 以下假定曲线是凸集 K K K
（3）说明为何任意 s s s d θ / d s ≥ 0 \mathrm d\theta/\mathrm ds\ge0 d θ / d s ≥ 0
（4）任一 K K K
x = p ( s ) + r u ( s ) , \boldsymbol{x}
=\boldsymbol p(s)+r\boldsymbol u(s), x = p ( s ) + r u ( s ) , 其中 r r r K K K u ( s ) \boldsymbol u(s) u ( s )
∣ det ( ∂ x / ∂ s ∂ x / ∂ r ∂ y / ∂ s ∂ y / ∂ r ) ∣ = 1 + r d θ d s . \left|
\det\begin{pmatrix}
\partial x/\partial s&\partial x/\partial r\\
\partial y/\partial s&\partial y/\partial r
\end{pmatrix}
\right|
=1+r\frac{\mathrm d\theta}{\mathrm ds}. det ( ∂ x / ∂ s ∂ y / ∂ s ∂ x / ∂ r ∂ y / ∂ r ) = 1 + r d s d θ . （5）对 D ≥ 0 D\ge0 D ≥ 0 K D K_D K D K K K D D D K K K A A A
area ( K D ) = A + L D + π D 2 . \operatorname{area}(K_D)=A+LD+\pi D^2. area ( K D ) = A + L D + π D 2 . Kai (1)
s ( t ) = ∫ a t ( d p d ξ ) 2 + ( d q d ξ ) 2 d ξ s(t) = \int_a^t \sqrt{\left( \frac{dp}{d\xi} \right)^2 + \left( \frac{dq}{d\xi} \right)^2} d\xi s ( t ) = ∫ a t ( d ξ d p ) 2 + ( d ξ d q ) 2 d ξ d s d t = ( d p d t ) 2 + ( d q d t ) 2 \frac{ds}{dt} = \sqrt{ \left( \frac{dp}{dt} \right)^2 + \left( \frac{dq}{dt} \right)^2 } d t d s = ( d t d p ) 2 + ( d t d q ) 2 Since d s / d t > 0 ds/dt>0 d s / d t > 0 d p / d s = ( d p / d t ) / ( d s / d t ) dp/ds=(dp/dt)/(ds/dt) d p / d s = ( d p / d t ) / ( d s / d t ) d q / d s = ( d q / d t ) / ( d s / d t ) dq/ds=(dq/dt)/(ds/dt) d q / d s = ( d q / d t ) / ( d s / d t )
( d p d s ) 2 + ( d q d s ) 2 = 1 \sqrt{ \left( \frac{dp}{ds} \right)^2 + \left( \frac{dq}{ds} \right)^2 } = 1 ( d s d p ) 2 + ( d s d q ) 2 = 1 (2)
On intervals where d p / d s ≠ 0 dp/ds\ne0 d p / d s = 0 θ \theta θ
tan θ = d q d s d p d s θ = arctan ( d q d s d p d s ) \tan \theta = \frac{\frac{dq}{ds}}{\frac{dp}{ds}} \qquad \theta = \arctan \left( \frac{\frac{dq}{ds}}{\frac{dp}{ds}} \right) tan θ = d s d p d s d q θ = arctan ( d s d p d s d q ) d θ d s = d 2 q d s 2 d p d s − d 2 p d s 2 d q d s ( d p d s ) 2 1 1 + ( d q d s ) 2 ( d p d s ) 2 = d 2 q d s 2 d p d s − d 2 p d s 2 d q d s ( d p d s ) 2 ( d p d s ) 2 ( d p d s ) 2 + ( d q d s ) 2 = d 2 q d s 2 d p d s − d 2 p d s 2 d q d s \begin{aligned}
\frac{d\theta}{ds} &= \frac{\frac{d^2q}{ds^2} \frac{dp}{ds} - \frac{d^2p}{ds^2} \frac{dq}{ds}}{\left( \frac{dp}{ds} \right)^2} \frac{1}{1 + \frac{\left( \frac{dq}{ds} \right )^2}{ \left( \frac{dp}{ds} \right)^2}} \\
&=\frac{\frac{d^2q}{ds^2} \frac{dp}{ds} - \frac{d^2p}{ds^2} \frac{dq}{ds}}{\left( \frac{dp}{ds} \right)^2} \frac{\left( \frac{dp}{ds} \right)^2}{\left( \frac{dp}{ds} \right)^2 + \left( \frac{dq}{ds} \right)^2} \\
&= \frac{d^2q}{ds^2} \frac{dp}{ds} - \frac{d^2p}{ds^2} \frac{dq}{ds}
\end{aligned} d s d θ = ( d s d p ) 2 d s 2 d 2 q d s d p − d s 2 d 2 p d s d q 1 + ( d s d p ) 2 ( d s d q ) 2 1 = ( d s d p ) 2 d s 2 d 2 q d s d p − d s 2 d 2 p d s d q ( d s d p ) 2 + ( d s d q ) 2 ( d s d p ) 2 = d s 2 d 2 q d s d p − d s 2 d 2 p d s d q The identity extends to vertical tangents by continuity.
(3)
Fix s 0 s_0 s 0 p ( s 0 ) = ( 0 , 0 ) \boldsymbol p(s_0)=(0,0) p ( s 0 ) = ( 0 , 0 ) p ′ ( s 0 ) = ( 1 , 0 ) \boldsymbol p'(s_0)=(1,0) p ′ ( s 0 ) = ( 1 , 0 ) K K K y = h ( x ) y=h(x) y = h ( x ) h ( 0 ) = h ′ ( 0 ) = 0 h(0)=h'(0)=0 h ( 0 ) = h ′ ( 0 ) = 0 h ′ ′ ( 0 ) ≥ 0 h''(0)\geq0 h ′′ ( 0 ) ≥ 0 θ ′ ( s 0 ) = q ′ ′ ( s 0 ) = h ′ ′ ( 0 ) ≥ 0 \theta'(s_0)=q''(s_0)=h''(0)\geq0 θ ′ ( s 0 ) = q ′′ ( s 0 ) = h ′′ ( 0 ) ≥ 0
There are following 4 possibilities:
Case 1:
d 2 q d s 2 ≥ 0 , d p d s ≥ 0 , d 2 p d s 2 ≤ 0 , d q d s ≥ 0 \frac{d^2q}{ds^2} \geq 0,\ \frac{dp}{ds} \geq 0,\ \frac{d^2p}{ds^2} \leq 0, \ \frac{dq}{ds} \geq 0 d s 2 d 2 q ≥ 0 , d s d p ≥ 0 , d s 2 d 2 p ≤ 0 , d s d q ≥ 0 d 2 q d s 2 d p d s ≥ 0 , d 2 p d s 2 d q d s ≤ 0 , d θ d s ≥ 0 \frac{d^2q}{ds^2} \frac{dp}{ds} \geq 0, \quad \frac{d^2p}{ds^2}\frac{dq}{ds} \leq 0, \quad \frac{d\theta}{ds} \geq 0 d s 2 d 2 q d s d p ≥ 0 , d s 2 d 2 p d s d q ≤ 0 , d s d θ ≥ 0 Case 2:
d 2 q d s 2 ≤ 0 , d p d s ≤ 0 , d 2 p d s 2 ≤ 0 , d q d s ≥ 0 \frac{d^2q}{ds^2} \leq 0,\ \frac{dp}{ds} \leq 0,\ \frac{d^2p}{ds^2} \leq 0, \ \frac{dq}{ds} \geq 0 d s 2 d 2 q ≤ 0 , d s d p ≤ 0 , d s 2 d 2 p ≤ 0 , d s d q ≥ 0 d 2 q d s 2 d p d s ≥ 0 , d 2 p d s 2 d q d s ≤ 0 , d θ d s ≥ 0 \frac{d^2q}{ds^2} \frac{dp}{ds} \geq 0, \quad \frac{d^2p}{ds^2}\frac{dq}{ds} \leq 0, \quad \frac{d\theta}{ds} \geq 0 d s 2 d 2 q d s d p ≥ 0 , d s 2 d 2 p d s d q ≤ 0 , d s d θ ≥ 0 Case 3:
d 2 q d s 2 ≤ 0 , d p d s ≤ 0 , d 2 p d s 2 ≥ 0 , d q d s ≤ 0 \frac{d^2q}{ds^2} \leq 0,\ \frac{dp}{ds} \leq 0,\ \frac{d^2p}{ds^2} \geq 0, \ \frac{dq}{ds} \leq 0 d s 2 d 2 q ≤ 0 , d s d p ≤ 0 , d s 2 d 2 p ≥ 0 , d s d q ≤ 0 d 2 q d s 2 d p d s ≥ 0 , d 2 p d s 2 d q d s ≤ 0 , d θ d s ≥ 0 \frac{d^2q}{ds^2} \frac{dp}{ds} \geq 0, \quad \frac{d^2p}{ds^2}\frac{dq}{ds} \leq 0, \quad \frac{d\theta}{ds} \geq 0 d s 2 d 2 q d s d p ≥ 0 , d s 2 d 2 p d s d q ≤ 0 , d s d θ ≥ 0 Case 4:
d 2 q d s 2 ≥ 0 , d p d s ≥ 0 , d 2 p d s 2 ≥ 0 , d q d s ≤ 0 \frac{d^2q}{ds^2} \geq 0,\ \frac{dp}{ds} \geq 0,\ \frac{d^2p}{ds^2} \geq 0, \ \frac{dq}{ds} \leq 0 d s 2 d 2 q ≥ 0 , d s d p ≥ 0 , d s 2 d 2 p ≥ 0 , d s d q ≤ 0 d 2 q d s 2 d p d s ≥ 0 , d 2 p d s 2 d q d s ≤ 0 , d θ d s ≥ 0 \frac{d^2q}{ds^2} \frac{dp}{ds} \geq 0, \quad \frac{d^2p}{ds^2}\frac{dq}{ds} \leq 0, \quad \frac{d\theta}{ds} \geq 0 d s 2 d 2 q d s d p ≥ 0 , d s 2 d 2 p d s d q ≤ 0 , d s d θ ≥ 0 Therefore, for arbitrary s s s d θ d s ≥ 0 \frac{d\theta}{ds} \geq 0 d s d θ ≥ 0
(4)
Let p ( s ) = ( p ( s ) , q ( s ) ) , u ( s ) = ( u ( s ) , v ( s ) ) \boldsymbol{p}(s) = (p(s), q(s)),\ \boldsymbol{u}(s) = (u(s), v(s)) p ( s ) = ( p ( s ) , q ( s )) , u ( s ) = ( u ( s ) , v ( s ))
∂ x ∂ s = p ′ ( s ) + r u ′ ( s ) ∂ y ∂ s = q ′ ( s ) + r v ′ ( s ) ∂ x ∂ r = u ( s ) ∂ y ∂ r = v ( s ) \begin{aligned}
\frac{\partial x}{\partial s} = p'(s) + ru'(s)& &\frac{\partial y}{\partial s} = q'(s) + rv'(s) \\
\frac{\partial x}{\partial r} = u(s)& &\frac{\partial y}{\partial r} = v(s)
\end{aligned} ∂ s ∂ x = p ′ ( s ) + r u ′ ( s ) ∂ r ∂ x = u ( s ) ∂ s ∂ y = q ′ ( s ) + r v ′ ( s ) ∂ r ∂ y = v ( s ) Note that for time s s s p \boldsymbol{p} p ( d p d s , d q d s ) \left(\frac{dp}{ds}, \frac{dq}{ds} \right) ( d s d p , d s d q ) u ( s ) \boldsymbol{u}(s) u ( s ) ( d q d s , − d p d s ) \left( \frac{dq}{ds}, -\frac{dp}{ds} \right) ( d s d q , − d s d p )
∣ det ( ∂ x ∂ s ∂ x ∂ r ∂ y ∂ s ∂ y ∂ r ) ∣ = ∣ − d p d s ( d p d s + r d 2 q d s 2 ) − d q d s ( d q d s − r d 2 p d s 2 ) ∣ = ( ( d p d s ) 2 + ( d q d s ) 2 ) + r ( d 2 q d s 2 d p d s − d 2 p d s 2 d q d s ) = 1 + r d θ d s \begin{aligned}
\left \lvert \text{det} \begin{pmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial r} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial r} \end{pmatrix} \right \rvert &=
\left \lvert -\frac{dp}{ds} \left ( \frac{dp}{ds} + r \frac{d^2q}{ds^2} \right ) - \frac{dq}{ds} \left ( \frac{dq}{ds} - r \frac{d^2p}{ds^2} \right) \right \rvert \\
&= \left( \left( \frac{dp}{ds} \right)^2 + \left( \frac{dq}{ds} \right)^2 \right) + r \left( \frac{d^2q}{ds^2} \frac{dp}{ds} - \frac{d^2p}{ds^2} \frac{dq}{ds} \right) \\
&= 1 + r\frac{d\theta}{ds}
\end{aligned} det ( ∂ s ∂ x ∂ s ∂ y ∂ r ∂ x ∂ r ∂ y ) = − d s d p ( d s d p + r d s 2 d 2 q ) − d s d q ( d s d q − r d s 2 d 2 p ) = ( ( d s d p ) 2 + ( d s d q ) 2 ) + r ( d s 2 d 2 q d s d p − d s 2 d 2 p d s d q ) = 1 + r d s d θ (5)
Note that the matrix in (4) is a Jacobian matrix.
Consider the area S S S r ∈ [ 0 , D ] , s ∈ [ 0 , L ] r \in [0,D], \ s \in [0, L] r ∈ [ 0 , D ] , s ∈ [ 0 , L ] K K K K D K_D K D
S = ∬ K , K D d x d y = ∫ 0 L ∫ 0 D ( 1 + r d θ d s ) d r d s = ∫ 0 L ( D + D 2 2 d θ d s ) d s = ∫ 0 L D d s + ∫ 0 2 π D 2 2 d θ = D L + D 2 π \begin{aligned}
S &= \iint_{K, K_D} dxdy \\
&= \int_0^L \int_0^D \left( 1 + r\frac{d\theta}{ds} \right) drds \\
&= \int_0^L \left(D + \frac{D^2}{2} \frac{d\theta}{ds} \right) ds \\
&= \int_0^L D\ ds + \int_0^{2\pi} \frac{D^2}{2} d\theta \\
&=DL + D^2 \pi
\end{aligned} S = ∬ K , K D d x d y = ∫ 0 L ∫ 0 D ( 1 + r d s d θ ) d r d s = ∫ 0 L ( D + 2 D 2 d s d θ ) d s = ∫ 0 L D d s + ∫ 0 2 π 2 D 2 d θ = D L + D 2 π Here θ ( L ) − θ ( 0 ) = 2 π \theta(L)-\theta(0)=2\pi θ ( L ) − θ ( 0 ) = 2 π
Since the area inside K K K A A A
A D = A + S = A + D L + D 2 π A_D = A + S = A + DL + D^2 \pi A D = A + S = A + D L + D 2 π 