広島大学 先進理工系科学研究科 情報科学プログラム 2022年1月実施 専門科目I 問題2 Author samparker, 祭音Myyura
Description (1) 関数 z = z ( u , v ) z = z(u, v) z = z ( u , v ) ( u , v ) ∈ R 2 (u, v) \in \mathbb{R}^2 ( u , v ) ∈ R 2 C 2 C^2 C 2 ( x , y ) ∈ R 2 (x, y) \in \mathbb{R}^2 ( x , y ) ∈ R 2 ( u , v ) = ( x + y , x − y ) (u, v) = (x + y, x - y) ( u , v ) = ( x + y , x − y )
z x x = ∂ 2 z ∂ x 2 , z x y = ∂ 2 z ∂ x ∂ y , z y y = ∂ 2 z ∂ y 2 z_{xx} = \frac{\partial^2 z}{\partial x^2}, \quad z_{xy} = \frac{\partial^2 z}{\partial x \partial y}, \quad z_{yy} = \frac{\partial^2 z}{\partial y^2} z xx = ∂ x 2 ∂ 2 z , z x y = ∂ x ∂ y ∂ 2 z , z yy = ∂ y 2 ∂ 2 z を
z u u = ∂ 2 z ∂ u 2 , z u v = ∂ 2 z ∂ u ∂ v , z v v = ∂ 2 z ∂ v 2 z_{uu} = \frac{\partial^2 z}{\partial u^2}, \quad z_{uv} = \frac{\partial^2 z}{\partial u \partial v}, \quad z_{vv} = \frac{\partial^2 z}{\partial v^2} z uu = ∂ u 2 ∂ 2 z , z uv = ∂ u ∂ v ∂ 2 z , z vv = ∂ v 2 ∂ 2 z を用いて表せ。
(2) 点 O ( 0 , 0 ) O(0, 0) O ( 0 , 0 )
z = ( x + y ) ln { 2 ( x 2 + y 2 ) } z = (x + y) \ln\{2(x^2 + y^2)\} z = ( x + y ) ln { 2 ( x 2 + y 2 )} の全ての極値およびそのときの ( x , y ) (x, y) ( x , y )
(1) Suppose that the function z = z ( u , v ) z = z(u, v) z = z ( u , v ) C 2 C^2 C 2 ( u , v ) ∈ R 2 (u, v) \in \mathbb{R}^2 ( u , v ) ∈ R 2 ( u , v ) = ( x + y , x − y ) (u, v) = (x + y, x - y) ( u , v ) = ( x + y , x − y ) ( x , y ) ∈ R 2 (x, y) \in \mathbb{R}^2 ( x , y ) ∈ R 2
z x x = ∂ 2 z ∂ x 2 , z x y = ∂ 2 z ∂ x ∂ y , z y y = ∂ 2 z ∂ y 2 z_{xx} = \frac{\partial^2 z}{\partial x^2}, \quad z_{xy} = \frac{\partial^2 z}{\partial x \partial y}, \quad z_{yy} = \frac{\partial^2 z}{\partial y^2} z xx = ∂ x 2 ∂ 2 z , z x y = ∂ x ∂ y ∂ 2 z , z yy = ∂ y 2 ∂ 2 z in terms of
z u u = ∂ 2 z ∂ u 2 , z u v = ∂ 2 z ∂ u ∂ v , z v v = ∂ 2 z ∂ v 2 . z_{uu} = \frac{\partial^2 z}{\partial u^2}, \quad z_{uv} = \frac{\partial^2 z}{\partial u \partial v}, \quad z_{vv} = \frac{\partial^2 z}{\partial v^2}. z uu = ∂ u 2 ∂ 2 z , z uv = ∂ u ∂ v ∂ 2 z , z vv = ∂ v 2 ∂ 2 z . (2) Find all local extrema and their extremum points of the function
z = ( x + y ) ln { 2 ( x 2 + y 2 ) } z = (x + y) \ln\{2(x^2 + y^2)\} z = ( x + y ) ln { 2 ( x 2 + y 2 )} defined for ( x , y ) ≠ ( 0 , 0 ) (x,y) \neq (0,0) ( x , y ) = ( 0 , 0 )
题目描述
设 z = z ( u , v ) z=z(u,v) z = z ( u , v ) ( u , v ) ∈ R 2 (u,v)\in\mathbb R^2 ( u , v ) ∈ R 2 C 2 C^2 C 2 ( x , y ) ∈ R 2 (x,y)\in\mathbb R^2 ( x , y ) ∈ R 2
( u , v ) = ( x + y , x − y ) . (u,v)=(x+y,x-y). ( u , v ) = ( x + y , x − y ) . 用
z u u = ∂ 2 z ∂ u 2 , z u v = ∂ 2 z ∂ u ∂ v , z v v = ∂ 2 z ∂ v 2 z_{uu}=\frac{\partial^2z}{\partial u^2},\qquad
z_{uv}=\frac{\partial^2z}{\partial u\partial v},\qquad
z_{vv}=\frac{\partial^2z}{\partial v^2} z uu = ∂ u 2 ∂ 2 z , z uv = ∂ u ∂ v ∂ 2 z , z vv = ∂ v 2 ∂ 2 z 表示
z x x = ∂ 2 z ∂ x 2 , z x y = ∂ 2 z ∂ x ∂ y , z y y = ∂ 2 z ∂ y 2 . z_{xx}=\frac{\partial^2z}{\partial x^2},\qquad
z_{xy}=\frac{\partial^2z}{\partial x\partial y},\qquad
z_{yy}=\frac{\partial^2z}{\partial y^2}. z xx = ∂ x 2 ∂ 2 z , z x y = ∂ x ∂ y ∂ 2 z , z yy = ∂ y 2 ∂ 2 z .
函数
z = ( x + y ) ln { 2 ( x 2 + y 2 ) } z=(x+y)\ln\{2(x^2+y^2)\} z = ( x + y ) ln { 2 ( x 2 + y 2 )} 定义在除原点 O ( 0 , 0 ) O(0,0) O ( 0 , 0 ) ( x , y ) (x,y) ( x , y )
Kai (1)
∂ z ∂ x = ∂ z ∂ u ∂ u ∂ x + ∂ z ∂ v ∂ v ∂ x = ∂ z ∂ u + ∂ z ∂ v \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} = \frac{\partial z}{\partial u} + \frac{\partial z}{\partial v} ∂ x ∂ z = ∂ u ∂ z ∂ x ∂ u + ∂ v ∂ z ∂ x ∂ v = ∂ u ∂ z + ∂ v ∂ z ∂ z ∂ y = ∂ z ∂ u ∂ u ∂ y + ∂ z ∂ v ∂ v ∂ y = ∂ z ∂ u − ∂ z ∂ v \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial z}{\partial u} - \frac{\partial z}{\partial v} ∂ y ∂ z = ∂ u ∂ z ∂ y ∂ u + ∂ v ∂ z ∂ y ∂ v = ∂ u ∂ z − ∂ v ∂ z z x x = ∂ 2 z ∂ x 2 = ∂ 2 z ∂ u 2 + ∂ 2 z ∂ u ∂ v + ∂ 2 z ∂ v ∂ u + ∂ 2 z ∂ v 2 = ∂ 2 z ∂ u 2 + 2 ∂ 2 z ∂ u ∂ v + ∂ 2 z ∂ v 2 z_{xx} = \frac{\partial^2 z}{\partial x^2} = \frac{\partial^2 z}{\partial u^2} + \frac{\partial^2 z}{\partial u \partial v} + \frac{\partial^2 z}{\partial v \partial u} + \frac{\partial^2 z}{\partial v^2} = \frac{\partial^2 z}{\partial u^2} + 2\frac{\partial^2 z}{\partial u \partial v} + \frac{\partial^2 z}{\partial v^2} z xx = ∂ x 2 ∂ 2 z = ∂ u 2 ∂ 2 z + ∂ u ∂ v ∂ 2 z + ∂ v ∂ u ∂ 2 z + ∂ v 2 ∂ 2 z = ∂ u 2 ∂ 2 z + 2 ∂ u ∂ v ∂ 2 z + ∂ v 2 ∂ 2 z Similarly,
z y y = ∂ 2 z ∂ u 2 − 2 ∂ 2 z ∂ u ∂ v + ∂ 2 z ∂ v 2 z_{yy} = \frac{\partial^2 z}{\partial u^2} - 2\frac{\partial^2 z}{\partial u \partial v} + \frac{\partial^2 z}{\partial v^2} z yy = ∂ u 2 ∂ 2 z − 2 ∂ u ∂ v ∂ 2 z + ∂ v 2 ∂ 2 z z x y = ∂ 2 z ∂ u 2 − ∂ 2 z ∂ v 2 z_{xy} = \frac{\partial^2 z}{\partial u^2} - \frac{\partial^2 z}{\partial v^2} z x y = ∂ u 2 ∂ 2 z − ∂ v 2 ∂ 2 z (2)
Let
x = r cos θ , y = r sin θ x = r \cos \theta, y = r \sin \theta x = r cos θ , y = r sin θ Then we have
z = r ( cos θ + sin θ ) ln ( 2 r 2 ) z = r(\cos \theta + \sin \theta) \ln(2r^2) z = r ( cos θ + sin θ ) ln ( 2 r 2 ) ∂ z ∂ r = ( sin θ + cos θ ) ( ln ( 2 r 2 ) + 2 ) \frac{\partial z}{\partial r} = (\sin \theta + \cos \theta) \left(\ln (2r^2) + 2\right) ∂ r ∂ z = ( sin θ + cos θ ) ( ln ( 2 r 2 ) + 2 ) ∂ z ∂ θ = r ( cos θ − sin θ ) ln ( 2 r 2 ) \frac{\partial z}{\partial \theta} = r(\cos \theta - \sin \theta) \ln(2r^2) ∂ θ ∂ z = r ( cos θ − sin θ ) ln ( 2 r 2 ) Solve the following equations
{ ∂ z ∂ r = ( sin θ + cos θ ) ( ln ( 2 r 2 ) + 2 ) = 0 ∂ z ∂ θ = r ( cos θ − sin θ ) ln ( 2 r 2 ) = 0 \begin{cases}
\frac{\partial z}{\partial r} = (\sin \theta + \cos \theta) \left(\ln (2r^2) + 2\right) = 0 \\
\frac{\partial z}{\partial \theta} = r(\cos \theta - \sin \theta) \ln(2r^2) = 0
\end{cases} { ∂ r ∂ z = ( sin θ + cos θ ) ( ln ( 2 r 2 ) + 2 ) = 0 ∂ θ ∂ z = r ( cos θ − sin θ ) ln ( 2 r 2 ) = 0 Since r > 0 r>0 r > 0
( r , θ ) = ( 1 2 e , π 4 ) , ( 1 2 e , 5 π 4 ) , ( 1 2 , 3 π 4 ) , ( 1 2 , 7 π 4 ) ( m o d 2 π ) . \left(r,\theta\right)=
\left(\frac{1}{\sqrt2e},\frac{\pi}{4}\right),
\left(\frac{1}{\sqrt2e},\frac{5\pi}{4}\right),
\left(\frac{1}{\sqrt2},\frac{3\pi}{4}\right),
\left(\frac{1}{\sqrt2},\frac{7\pi}{4}\right)
\pmod{2\pi}. ( r , θ ) = ( 2 e 1 , 4 π ) , ( 2 e 1 , 4 5 π ) , ( 2 1 , 4 3 π ) , ( 2 1 , 4 7 π ) ( mod 2 π ) . At the first two points, z r θ = 0 z_{r\theta}=0 z r θ = 0 ( z r r , z θ θ ) (z_{rr},z_{\theta\theta}) ( z rr , z θθ ) ( 4 e , 2 / e ) (4e,2/e) ( 4 e , 2/ e ) ( − 4 e , − 2 / e ) (-4e,-2/e) ( − 4 e , − 2/ e ) − 8 < 0 -8<0 − 8 < 0
Therefore all local extrema are
z min = − 2 e at ( x , y ) = ( 1 2 e , 1 2 e ) , \boxed{z_{\min}=-\frac2e\quad\text{at}\quad
(x,y)=\left(\frac1{2e},\frac1{2e}\right)}, z m i n = − e 2 at ( x , y ) = ( 2 e 1 , 2 e 1 ) , z max = 2 e at ( x , y ) = ( − 1 2 e , − 1 2 e ) . \boxed{z_{\max}=\frac2e\quad\text{at}\quad
(x,y)=\left(-\frac1{2e},-\frac1{2e}\right)}. z m a x = e 2 at ( x , y ) = ( − 2 e 1 , − 2 e 1 ) .