題目
Problem
Suppose that f ( x , y ) f(x,y) f ( x , y ) is a differentiable function of ( x , y ) (x,y) ( x , y ) for ( x , y ) ∈ R 2 (x,y)\in\mathbb{R}^2 ( x , y ) ∈ R 2 , and f x x f_{xx} f xx , f x y f_{xy} f x y , and f y y f_{yy} f y y are continuous on { ( x , y ) : − 2 < x < 2 and − 1 < y < 1 } \{(x,y):-2<x<2\text{ and }-1<y<1\} {( x , y ) : − 2 < x < 2 and − 1 < y < 1 } . Suppose that
f x ( 1 , 0 ) = 0 = f y ( 1 , 0 ) , f x x ( 1 , 0 ) = 4 , f x y ( 1 , 0 ) = f y x ( 1 , 0 ) = 3 , f y y ( 1 , 0 ) = 2. f_x(1,0)=0=f_y(1,0),\qquad f_{xx}(1,0)=4,\qquad f_{xy}(1,0)=f_{yx}(1,0)=3,\qquad f_{yy}(1,0)=2. f x ( 1 , 0 ) = 0 = f y ( 1 , 0 ) , f xx ( 1 , 0 ) = 4 , f x y ( 1 , 0 ) = f y x ( 1 , 0 ) = 3 , f y y ( 1 , 0 ) = 2.
(a) (10 pts) Let g ( t ) = f ( e 3 t , sin ( t ) ) g(t)=f(e^{3t},\sin(t)) g ( t ) = f ( e 3 t , sin ( t )) for t ∈ R t\in\mathbb{R} t ∈ R . Does g g g have a relative minimum at 0 0 0 ? Justify your answer.
(b) (10 pts) Does f f f have a relative minimum at ( 1 , 0 ) (1,0) ( 1 , 0 ) ? Justify your answer.
解答
解法一
思路
展開
(a) 先以多變數鏈鎖規則求 g ′ ( 0 ) g'(0) g ′ ( 0 ) 、g ′ ′ ( 0 ) g''(0) g ′′ ( 0 ) ;一變數二階導數判別可判斷 g g g 的局部極小。
(b) 直接檢查 f f f 的 Hessian 判別式;若行列式為負,該點是鞍點。
答題過程
展開
(a) 設 x ( t ) = e 3 t x(t)=e^{3t} x ( t ) = e 3 t 、y ( t ) = sin t y(t)=\sin t y ( t ) = sin t 。在 t = 0 t=0 t = 0 ,
x = 1 , y = 0 , x ′ = 3 , y ′ = 1 , x=1,\quad y=0,\quad x'=3,\quad y'=1, x = 1 , y = 0 , x ′ = 3 , y ′ = 1 ,
且已知 f x ( 1 , 0 ) = f y ( 1 , 0 ) = 0 f_x(1,0)=f_y(1,0)=0 f x ( 1 , 0 ) = f y ( 1 , 0 ) = 0 ,故 g ′ ( 0 ) = 0 g'(0)=0 g ′ ( 0 ) = 0 。再由鏈鎖規則,所有含 f x , f y f_x,f_y f x , f y 的項在 t = 0 t=0 t = 0 消失:
g ′ ′ ( 0 ) = f x x ( 1 , 0 ) ( x ′ ) 2 + 2 f x y ( 1 , 0 ) x ′ y ′ + f y y ( 1 , 0 ) ( y ′ ) 2 = 4 ⋅ 9 + 2 ⋅ 3 ⋅ 3 ⋅ 1 + 2 ⋅ 1 2 = 56 > 0. \begin{align*}
g''(0)
=&\,f_{xx}(1,0)(x')^2+2f_{xy}(1,0)x'y'+f_{yy}(1,0)(y')^2\\[4mm]
=&\,4\cdot9+2\cdot3\cdot3\cdot1+2\cdot1^2
=56>0.
\end{align*} g ′′ ( 0 ) = = f xx ( 1 , 0 ) ( x ′ ) 2 + 2 f x y ( 1 , 0 ) x ′ y ′ + f y y ( 1 , 0 ) ( y ′ ) 2 4 ⋅ 9 + 2 ⋅ 3 ⋅ 3 ⋅ 1 + 2 ⋅ 1 2 = 56 > 0.
因此 g 在 t = 0 有相對極小值 \boxed{g\text{ 在 }t=0\text{ 有相對極小值}} g 在 t = 0 有相對極小值 。
(b) f f f 在 ( 1 , 0 ) (1,0) ( 1 , 0 ) 的 Hessian 判別式是
∣ 4 3 3 2 ∣ = 8 − 9 = − 1 < 0. \begin{vmatrix}4&3\\3&2\end{vmatrix}=8-9=-1<0. 4 3 3 2 = 8 − 9 = − 1 < 0.
故 ( 1 , 0 ) (1,0) ( 1 , 0 ) 是鞍點,f 在 ( 1 , 0 ) 沒有相對極小值 \boxed{f\text{ 在 }(1,0)\text{ 沒有相對極小值}} f 在 ( 1 , 0 ) 沒有相對極小值 。