題目
Problem
5. Let u ( x , y ) u(x, y) u ( x , y ) ∂ u ∂ x ( 4 , 1 ) = 1 \frac{\partial u}{\partial x}(4, 1) = 1 ∂ x ∂ u ( 4 , 1 ) = 1 ∂ u ∂ y ( 4 , 1 ) = 2 \frac{\partial u}{\partial y}(4, 1) = 2 ∂ y ∂ u ( 4 , 1 ) = 2 x = s t x = st x = s t y = s t y = \frac{s}{t} y = t s h ( s , t ) = u ( x ( s , t ) , y ( s , t ) ) h(s, t) = u(x(s, t), y(s, t)) h ( s , t ) = u ( x ( s , t ) , y ( s , t )) ( 2 , 2 ) (2, 2) ( 2 , 2 ) u \mathbf{u} u s t st s t h h h u \mathbf{u} u
解答
思路
展開
函數 h ( s , t ) h(s,t) h ( s , t ) ∇ h ( s , t ) \nabla h(s,t) ∇ h ( s , t )
當 ( s , t ) = ( 2 , 2 ) (s, t) = (2, 2) ( s , t ) = ( 2 , 2 ) x = s t = 4 , y = s t = 1 x = st = 4, \quad y = \frac{s}{t} = 1 x = s t = 4 , y = t s = 1 ( 4 , 1 ) (4,1) ( 4 , 1 )
利用偏微鏈鎖律計算 h s h_s h s h t h_t h t
h s = u x x s + u y y s = u x ⋅ t + u y ⋅ 1 t h_s = u_x x_s + u_y y_s = u_x \cdot t + u_y \cdot \frac{1}{t} h s = u x x s + u y y s = u x ⋅ t + u y ⋅ t 1 h t = u x x t + u y y t = u x ⋅ s − u y ⋅ s t 2 h_t = u_x x_t + u_y y_t = u_x \cdot s - u_y \cdot \frac{s}{t^2} h t = u x x t + u y y t = u x ⋅ s − u y ⋅ t 2 s
將 ( s , t ) = ( 2 , 2 ) (s,t) = (2,2) ( s , t ) = ( 2 , 2 ) ∇ h ( 2 , 2 ) \nabla h(2,2) ∇ h ( 2 , 2 )
答題過程
展開
當 ( s , t ) = ( 2 , 2 ) (s, t) = (2, 2) ( s , t ) = ( 2 , 2 )
x = 2 ( 2 ) = 4 , y = 2 2 = 1 x = 2(2) = 4, \quad y = \frac{2}{2} = 1 x = 2 ( 2 ) = 4 , y = 2 2 = 1 根據鏈鎖律計算 h s h_s h s
∂ h ∂ s = ∂ u ∂ x ∂ x ∂ s + ∂ u ∂ y ∂ y ∂ s \frac{\partial h}{\partial s} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s} ∂ s ∂ h = ∂ x ∂ u ∂ s ∂ x + ∂ y ∂ u ∂ s ∂ y 其中 ∂ x ∂ s = t \frac{\partial x}{\partial s} = t ∂ s ∂ x = t ∂ y ∂ s = 1 t \frac{\partial y}{\partial s} = \frac{1}{t} ∂ s ∂ y = t 1
∂ h ∂ s ∣ ( 2 , 2 ) = 1 ( 2 ) + 2 ( 1 2 ) = 3 \left. \frac{\partial h}{\partial s} \right|_{(2,2)} = 1(2) + 2\left(\frac{1}{2}\right) = 3 ∂ s ∂ h ( 2 , 2 ) = 1 ( 2 ) + 2 ( 2 1 ) = 3 根據鏈鎖律計算 h t h_t h t
∂ h ∂ t = ∂ u ∂ x ∂ x ∂ t + ∂ u ∂ y ∂ y ∂ t \frac{\partial h}{\partial t} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial t} ∂ t ∂ h = ∂ x ∂ u ∂ t ∂ x + ∂ y ∂ u ∂ t ∂ y 其中 ∂ x ∂ t = s \frac{\partial x}{\partial t} = s ∂ t ∂ x = s ∂ y ∂ t = − s t 2 \frac{\partial y}{\partial t} = -\frac{s}{t^2} ∂ t ∂ y = − t 2 s
∂ h ∂ t ∣ ( 2 , 2 ) = 1 ( 2 ) + 2 ( − 2 4 ) = 2 − 1 = 1 \left. \frac{\partial h}{\partial t} \right|_{(2,2)} = 1(2) + 2\left( -\frac{2}{4} \right) = 2 - 1 = 1 ∂ t ∂ h ( 2 , 2 ) = 1 ( 2 ) + 2 ( − 4 2 ) = 2 − 1 = 1 因此，在 ( 2 , 2 ) (2,2) ( 2 , 2 )
∇ h ( 2 , 2 ) = ⟨ 3 , 1 ⟩ \nabla h(2, 2) = \langle 3, 1 \rangle ∇ h ( 2 , 2 ) = ⟨ 3 , 1 ⟩ 函數在梯度方向增長最快，將其單位化即為所求之方向向量 u \mathbf{u} u
u = ∇ h ( 2 , 2 ) ∣ ∇ h ( 2 , 2 ) ∣ = ⟨ 3 , 1 ⟩ 3 2 + 1 2 = 1 10 ⟨ 3 , 1 ⟩ \mathbf{u} = \frac{\nabla h(2, 2)}{|\nabla h(2, 2)|} = \frac{\langle 3, 1 \rangle}{\sqrt{3^2 + 1^2}} = \frac{1}{\sqrt{10}} \langle 3, 1 \rangle u = ∣∇ h ( 2 , 2 ) ∣ ∇ h ( 2 , 2 ) = 3 2 + 1 2 ⟨ 3 , 1 ⟩ = 10 1 ⟨ 3 , 1 ⟩ 結論：
單位向量為 1 10 ⟨ 3 , 1 ⟩ \displaystyle \frac{1}{\sqrt{10}} \langle 3, 1 \rangle 10 1 ⟨ 3 , 1 ⟩
