方向导数
定理
- 若函数f(x,y,z)在点P(x,y,z)处可微，沿任意方向l的方向导数
- ∂ f ∂ l = ∂ f ∂ x c o s α + ∂ f ∂ y c o s β + ∂ f ∂ z c o s γ \frac{\partial f}{\partial l} = \frac{\partial f}{\partial x} cos \alpha + \frac{\partial f}{\partial y} cos \beta + \frac{\partial f}{\partial z} cos \gamma ∂l∂f=∂x∂fcosα+∂y∂fcosβ+∂z∂fcosγ
- 其中 α , β , γ \alpha, \beta, \gamma α,β,γ 为l的方向角
- 证明
- 由函数 f ( x , y , z ) f(x,y,z) f(x,y,z)在点P可微
- △ f = ∂ f ∂ x △ x + ∂ f ∂ y △ y + ∂ f ∂ z △ z + o ( ρ ) \triangle f = \frac{\partial f}{\partial x} \triangle x + \frac{\partial f}{\partial y} \triangle y + \frac{\partial f}{\partial z} \triangle z + o(\rho) △f=∂x∂f△x+∂y∂f△y+∂z∂f△z+o(ρ)
- = ρ ( ∂ f ∂ x c o s α + ∂ f ∂ y c o s β + ∂ f ∂ z c o s γ ) + o ( ρ ) = \rho(\frac{\partial f}{\partial x} cos \alpha + \frac{\partial f}{\partial y} cos \beta + \frac{\partial f}{\partial z} cos \gamma) + o(\rho) =ρ(∂x∂fcosα+∂y∂fcosβ+∂z∂fcosγ)+o(ρ)
- ∂ f ∂ l = lim ρ → 0 △ f ρ = ∂ f ∂ x c o s α + ∂ f ∂ y c o s β + ∂ f ∂ z c o s γ \frac{\partial f}{\partial l} = \lim_{\rho \to 0} \frac{\triangle f}{\rho} = \frac{\partial f}{\partial x} cos \alpha + \frac{\partial f}{\partial y} cos \beta + \frac{\partial f}{\partial z} cos \gamma ∂l∂f=limρ→0ρ△f=∂x∂fcosα+∂y∂fcosβ+∂z∂fcosγ
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- 对于二元函数f(x,y)在点P(x,y)处沿着方向l(方向角为 α , β \alpha, \beta α,β)的方向导数为
- ∂ f ∂ l = lim ρ → 0 f ( x + △ x , y + △ y ) − f ( x , y ) ρ = f x ′ ( x , y ) c o s α + f y ′ ( x , y ) c o s β \frac{\partial f}{\partial l} = \lim_{\rho \to 0} \frac{f(x+\triangle x, y + \triangle y) – f(x,y)}{\rho} = f_x'(x,y)cos \alpha + f_y'(x,y) cos \beta ∂l∂f=limρ→0ρf(x+△x,y+△y)−f(x,y)=fx′(x,y)cosα+fy′(x,y)cosβ
- ρ = ( △ x ) 2 + ( △ y ) 2 \rho = \sqrt{(\triangle x)^2 + (\triangle y)^2} ρ=(△x)2+(△y)2
- △ x = ρ c o s α \triangle x = \rho cos \alpha △x=ρcosα
- △ y = ρ c o s β \triangle y = \rho cos \beta △y=ρcosβ
- 特别地
- l与x轴同向( α = 0 , β = π 2 \alpha = 0, \beta = \frac{\pi}{2} α=0,β=2π)时，有 ∂ f ∂ l = ∂ f ∂ x \frac{\partial f}{\partial l} = \frac{\partial f}{\partial x} ∂l∂f=∂x∂f
- l与x轴反向( α = π , β = π 2 \alpha = \pi, \beta = \frac{\pi}{2} α=π,β=2π)时，有 ∂ f ∂ l = − ∂ f ∂ x \frac{\partial f}{\partial l} = -\frac{\partial f}{\partial x} ∂l∂f=−∂x∂f
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方向导数
- 方向导数(directional derivative): 有时不仅仅需要知道函数在坐标轴上的变化率(即偏导数)，还需要设法求得函数在其他特定方向上的变化率；
- 而方向导数就是函数在其他特定方向上的变化率。
- 如果函数 z = f ( x , y ) z=f(x,y) z=f(x,y)在点P(x,y)是可微分的，那么，函数在该点沿着任意方向L的方向导数都存在
- 且计算公式为： ∂ f ∂ l = ∂ f ∂ x c o s α + ∂ f ∂ y c o s β \frac{\partial f}{\partial l} = \frac{\partial f}{\partial x} cos \alpha + \frac{\partial f}{\partial y} cos \beta ∂l∂f=∂x∂fcosα+∂y∂fcosβ
备注：图片托管于github，请确保网络的可访问性
例1
- 求函数 u = x 2 y z u = x^2yz u=x2yz 在点P(1,1,1)沿向量 l ⃗ = ( 2 , − 1 , 3 ) \vec{l} = (2, -1, 3) l =(2,−1,3)的方向导数.
- ∂ u ∂ l = ∂ u ∂ x c o s α + ∂ u ∂ y c o s β + ∂ u ∂ z c o s γ \frac{\partial u}{\partial l} = \frac{\partial u}{\partial x} cos \alpha + \frac{\partial u}{\partial y} cos \beta + \frac{\partial u}{\partial z} cos \gamma ∂l∂u=∂x∂ucosα+∂y∂ucosβ+∂z∂ucosγ
- 解
- 向量 l ⃗ \vec{l} l 的方向余弦为: c o s α = 2 14 , cos β = − 1 14 , c o s γ = 3 14 cos \alpha = \frac{2}{\sqrt{14}}, \cos \beta = \frac{-1}{\sqrt{14}}, cos \gamma = \frac{3}{\sqrt{14}} cosα=14 2,cosβ=14 −1,cosγ=14 3
- ∂ u ∂ l ∣ P = ( 2 x y z ∗ 2 14 ) − x 2 z ∗ 1 14 + x 2 y ∗ 3 14 ∣ ( 1 , 1 , 1 ) = 6 14 \left. \frac{\partial u}{\partial l} \right|_P = \left. (2xyz * \frac{2}{\sqrt{14}}) – x^2z * \frac{1}{\sqrt{14}} + x^2y * \frac{3}{\sqrt{14}} \right|_{(1,1,1)} = \frac{6}{\sqrt{14}} ∂l∂u∣∣P=(2xyz∗14 2)−x2z∗14 1+x2y∗14 3∣∣∣(1,1,1)=14 6
例2
- 求函数 z = x e 2 y z=xe^{2y} z=xe2y在点P(1,0)处沿从点P(1,0)到点Q(2, -1)的方向的方向导数
- 解
- 方向l即向量 P Q = ( 1 , − 1 ) PQ = (1, -1) PQ=(1,−1)的方向，与l同方向的单位向量 e l = ( 1 2 , − 1 2 ) . = ( c o s α , c o s β ) e_l = (\frac{1}{\sqrt{2}}, – \frac{1}{\sqrt{2}}). = (cos \alpha, cos \beta) el=(2 1,−2 1).=(cosα,cosβ)
- 因函数可微，且 ∂ z ∂ x ∣ ( 1 , 0 ) = e 2 y ∣ ( 1 , 0 ) = 1 , ∂ z ∂ y ∣ ( 1 , 0 ) = 2 x e 2 y ∣ ( 1 , 0 ) = 2 \left. \frac{\partial z}{\partial x} \right|_{(1,0)} = \left. e^{2y} \right|_{(1,0)} = 1, \left. \frac{\partial z}{\partial y} \right|_{(1,0)} = \left. 2xe^{2y} \right|_{(1,0)} = 2 ∂x∂z∣∣(1,0)=e2y∣∣(1,0)=1,∂y∂z∣∣∣(1,0)=2xe2y∣∣(1,0)=2
- 所以，所求方向导数为： ∂ z ∂ l ∣ ( 1 , 0 ) = 1 ∗ 1 2 + 2 ∗ ( − 1 2 ) = − 2 2 \left. \frac{\partial z}{\partial l} \right|_{(1,0)} = 1 * \frac{1}{\sqrt{2}} + 2 * (- \frac{1}{\sqrt{2}}) = – \frac{\sqrt{2}}{2} ∂l∂z∣∣(1,0)=1∗2 1+2∗(−2 1)=−22
例3
- 求 f ( x , y , z ) = x y + y z + z x f(x,y,z) = xy + yz + zx f(x,y,z)=xy+yz+zx 在点(1,1,2)沿方向l的方向导数，其中l的方向角分别为：60°, 45°, 60°
- 解：
- 与l同方向的单位向量 e l = ( c o s 60 ° , c o s 45 ° , c o s 60 ° ) = ( 1 2 , 2 2 , 1 2 ) e_l = (cos 60°, cos 45°, cos 60°) = (\frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{1}{2}) el=(cos60°,cos45°,cos60°)=(21,22 ,21)
- 因函数可微，且
- f x ′ ( 1 , 1 , 2 ) = ( y + z ) ∣ ( 1 , 1 , 2 ) = 3 f_x'(1,1,2) = (y + z)|_{(1,1,2)} = 3 fx′(1,1,2)=(y+z)∣(1,1,2)=3
- f y ′ ( 1 , 1 , 2 ) = ( x + z ) ∣ ( 1 , 1 , 2 ) = 3 f_y'(1,1,2) = (x + z)|_{(1,1,2)} = 3 fy′(1,1,2)=(x+z)∣(1,1,2)=3
- f z ′ ( 1 , 1 , 2 ) = ( y + x ) ∣ ( 1 , 1 , 2 ) = 2 f_z'(1,1,2) = (y + x)|_{(1,1,2)} = 2 fz′(1,1,2)=(y+x)∣(1,1,2)=2
- 所以 ∂ f ∂ l ∣ ( 1 , 1 , 2 ) = 3 ∗ 1 2 + 3 ∗ 2 2 + 2 ∗ 1 2 = 1 2 ( 5 + 3 2 ) \frac{\partial f}{\partial l} |_{(1,1,2)} = 3*\frac{1}{2} + 3*\frac{\sqrt{2}}{2} + 2*\frac{1}{2} = \frac{1}{2}(5 + 3\sqrt{2}) ∂l∂f∣(1,1,2)=3∗21+3∗22 +2∗21=21(5+32 )
梯度
1 ） 概念
- 在空间的每一个点都可以确定无限多个方向，因此，一个多元函数在某个点也必然有无限多个方向导数.
- 在这无限多个方向导数中，最大的一个(它直接反映了函数在这个点的变化率的数量级)等于多少? 它是沿什么方向达到的?
- 描述这个最大方向导数及其所沿方向的矢量，就是我们所讨论的梯度.
- 梯度是场论里的一个基本概念.所谓”场”, 它表示空间区域上某种物理量的一种分布
- 从数学上看，这种分布常常表示为 Ω \Omega Ω 上的一种数值函数或向量函数
- 能表示为数值函数u=u(x,y,z)的场，称为数量场，如温度场、密度场等
2 ) 方向导数公式
- ∂ f ∂ l = ∂ f ∂ x c o s α + ∂ f ∂ y c o s β + ∂ f ∂ z c o s γ \frac{\partial f}{\partial l} = \frac{\partial f}{\partial x} cos \alpha + \frac{\partial f}{\partial y} cos \beta + \frac{\partial f}{\partial z} cos \gamma ∂l∂f=∂x∂fcosα+∂y∂fcosβ+∂z∂fcosγ
- 令向量 G ⃗ = ( ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ z ) \vec{G} = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) G =(∂x∂f,∂y∂f,∂z∂f)
- l ° ⃗ = ( c o s α , c o s β , c o s γ ) \vec{l°} = (cos \alpha, cos \beta, cos \gamma) l° =(cosα,cosβ,cosγ)
- ∂ f ∂ l = G ⃗ ⋅ l ° ⃗ = ∣ G ⃗ ∣ c o s ( G ⃗ , l ° ⃗ ) ( ∣ l ° ⃗ ∣ = 1 ) \frac{\partial f}{\partial l} = \vec{G}·\vec{l°} = |\vec{G}|cos(\vec{G}, \vec{l°}) \ \ \ (|\vec{l°}| = 1) ∂l∂f=G ⋅l° =∣G ∣cos(G ,l° ) (∣l° ∣=1)
- 当 l ° ⃗ \vec{l°} l° 与 G ⃗ \vec{G} G 方向一致时，方向导数取最大值： m a x ( ∂ f ∂ l ) = ∣ G ⃗ ∣ max(\frac{\partial f}{\partial l}) = |\vec{G}| max(∂l∂f)=∣G ∣
- 可见： G ⃗ \vec{G} G
- 方向：f 变化率最大的方向
- 模：f 的最大变化率之值
3 ) 梯度定义
- 向量 G ⃗ \vec{G} G ：称为函数 f ( P ) f(P) f(P)在点P处的梯度(gradient), 记做：grad f
- 即 g r a d f = ( ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ z ) = ∂ f ∂ x i ⃗ + ∂ f ∂ y j ⃗ + ∂ f ∂ z k ⃗ grad \ f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z}\vec{k} grad f=(∂x∂f,∂y∂f,∂z∂f)=∂x∂fi +∂y∂fj +∂z∂fk
- 同样可定义二元函数f(x,y)在点P(x,y)处的梯度 g r a d f = ∂ f ∂ x i ⃗ + ∂ f ∂ y j ⃗ = ( ∂ f ∂ x , ∂ f ∂ y ) grad \ f = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) grad f=∂x∂fi +∂y∂fj =(∂x∂f,∂y∂f)
- 说明：函数的方向导数为梯度在该方向上的投影
- ∇ = ( ∂ ∂ x , ∂ ∂ y ) \nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) ∇=(∂x∂,∂y∂), 引用记号，称为奈布拉(Nebla)算符，或称为向量微分算子或哈密顿(W.R.Hamilton)算子
- 则梯度可记为： g r a d f = ( ∂ f ∂ x , ∂ f ∂ y ) ∇ f grad \ f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \nabla f grad f=(∂x∂f,∂y∂f)∇f
- 函数f沿梯度grad f方向，增加最快(上升)
- 函数f沿负梯度 -grad f方向，减小最快(下降)
- g r a d f ( x 0 , y 0 ) = f x ′ ( x 0 , y 0 ) i + f y ′ ( x 0 , y 0 ) j ) grad \ f(x_0, y_0) = f_x'(x_0, y_0)i + f_y'(x_0, y_0)j) grad f(x0,y0)=fx′(x0,y0)i+fy′(x0,y0)j)
- 或 ∇ f ( x 0 , y 0 ) = f x ′ ( x 0 , y 0 ) i + f y ′ ( x 0 , y 0 ) j = f x ′ ( x 0 , y 0 ) , f y ′ ( x 0 , y 0 ) \nabla f(x_0, y_0) = f_x'(x_0, y_0)i + f_y'(x_0, y_0) j = {f_x'(x_0, y_0), f_y'(x_0, y_0)} ∇f(x0,y0)=fx′(x0,y0)i+fy′(x0,y0)j=fx′(x0,y0),fy′(x0,y0)
- g r a d f = ( ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ z ) = ∂ f ∂ x i ⃗ + ∂ f ∂ y j ⃗ + ∂ f ∂ z k ⃗ grad \ f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z}\vec{k} grad f=(∂x∂f,∂y∂f,∂z∂f)=∂x∂fi +∂y∂fj +∂z∂fk
- 或 ∇ f ( x 0 , y 0 , z 0 ) = { f x ′ ( x 0 , y 0 , z 0 ) , f y ′ ( x 0 , y 0 , z 0 ) , f z ′ ( x 0 , y 0 , z 0 ) } = f x ′ ( x 0 , y 0 , z 0 ) i + f y ′ ( x 0 , y 0 , z 0 ) j + f z ′ ( x 0 , y 0 , z 0 ) k \nabla f(x_0, y_0, z_0) = \{f_x'(x_0, y_0, z_0), f_y'(x_0, y_0, z_0), f_z'(x_0, y_0, z_0)\} = f_x'(x_0, y_0, z_0)i + f_y'(x_0, y_0, z_0)j + f_z'(x_0, y_0, z_0)k ∇f(x0,y0,z0)={ fx′(x0,y0,z0),fy′(x0,y0,z0),fz′(x0,y0,z0)}=fx′(x0,y0,z0)i+fy′(x0,y0,z0)j+fz′(x0,y0,z0)k
说明
- 以三元函数为例，设 u = f ( x , y , z ) u=f(x,y,z) u=f(x,y,z)在点P(x,y,z)处可微分，则函数在该点的梯度为 g r a d f = ∇ f = ∂ f ∂ x i ⃗ + ∂ f ∂ y j ⃗ + ∂ f ∂ z k ⃗ = ( ∂ f ∂ x , ∂ f ∂ y , ∂ f ∂ z ) = ( ∂ ( f ) ∂ ( x , y , z ) ) grad \ f = \nabla f = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z}\vec{k} = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) = (\frac{\partial (f)}{\partial(x,y,z)}) grad f=∇f=∂x∂fi +∂y∂fj +∂z∂fk =(∂x∂f,∂y∂f,∂z∂f)=(∂(x,y,z)∂(f))
- 梯度是函数 u = f ( x , y , z ) u=f(x,y,z) u=f(x,y,z)在点P处取得的最大方向导数的方向，最大方向导数为： ∣ g r a d f ∣ = ( ∂ f ∂ x ) 2 + ( ∂ f ∂ y ) 2 + ( ∂ f ∂ z ) 2 |grad \ f| = \sqrt{(\frac{\partial f}{\partial x})^2 + (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2} ∣grad f∣=(∂x∂f)2+(∂y∂f)2+(∂z∂f)2
- 函数 u = f ( x , y , z ) u=f(x,y,z) u=f(x,y,z)在点P处沿方向 l ⃗ \vec{l} l 的方向导数： ∂ f ∂ l ⃗ = g r a d f ⋅ l ° ⃗ = ∇ f ⋅ l ° ⃗ \frac{\partial f}{\partial \vec{l}} = grad \ f·\vec{l°} = \nabla f · \vec{l °} ∂l ∂f=grad f⋅l° =∇f⋅l°
例1
- 求 g r a d 1 x 2 + y 2 grad \ \frac{1}{\sqrt{x^2 + y^2}} grad x2+y2 1
- 解：
- 这里 f ( x , y ) = 1 x 2 + y 2 f(x,y) = \frac{1}{x^2 + y^2} f(x,y)=x2+y21
- 因 ∂ f ∂ x = − 2 x ( x 2 + y 2 ) 2 , ∂ f ∂ y = − 2 y ( x 2 + y 2 ) 2 \frac{\partial f}{\partial x} = – \frac{2x}{(x^2 + y^2)^2}, \frac{\partial f}{\partial y} = – \frac{2y}{(x^2 + y^2)^2} ∂x∂f=−(x2+y2)22x,∂y∂f=−(x2+y2)22y
- 所以， g r a d 1 x 2 + y 2 = − 2 x ( x 2 + y 2 ) 2 i ⃗ − 2 y ( x 2 + y 2 ) 2 j ⃗ grad \ \frac{1}{\sqrt{x^2 + y^2}} = – \frac{2x}{(x^2 + y^2)^2} \vec{i} – \frac{2y}{(x^2 + y^2)^2} \vec{j} grad x2+y2 1=−(x2+y2)22xi −(x2+y2)22yj
例2
- 设 f ( x , y , z ) = x 3 − x y 2 − z f(x,y,z) = x^3 – xy^2 – z f(x,y,z)=x3−xy2−z, p ( 1 , 1 , 0 ) p(1,1,0) p(1,1,0).
- 问f(x,y,z)在p处沿什么方向变化最快，在这方向的变化率是多少?
- 解
- ∇ f = f x ′ i + f y ′ j + f z ′ k = ( 3 x 2 − y 2 ) i − 2 x y j − k \nabla f = f_x’i + f_y’j + f_z’k = (3x^2 – y^2)i – 2xyj – k ∇f=fx′i+fy′j+fz′k=(3x2−y2)i−2xyj−k
- ∇ f ( 1 , 1 , 0 ) = 2 i − 2 j − k \nabla f(1,1,0) = 2i – 2j – k ∇f(1,1,0)=2i−2j−k
- 沿 ∇ f ( 1 , 1 , 0 ) \nabla f(1,1,0) ∇f(1,1,0) 方向，增加最快(上升)
- 沿 − ∇ f ( 1 , 1 , 0 ) – \nabla f(1,1,0) −∇f(1,1,0) 方向，增加最快(下降)
- m a x { ∂ f ∂ l ∣ p } = ∣ g r a d f ∣ = ∣ ∇ f ( 1 , 1 , 0 ) ∣ = 3 max\{\frac{\partial f}{\partial l} |_p\} = |grad \ f| = |\nabla f(1,1,0)| = 3 max{ ∂l∂f∣p}=∣grad f∣=∣∇f(1,1,0)∣=3
- m i n { ∂ f ∂ l ∣ p } = − ∣ g r a d f ∣ = − ∣ ∇ f ( 1 , 1 , 0 ) ∣ = − 3 min\{\frac{\partial f}{\partial l} |_p\} = -|grad \ f| = -|\nabla f(1,1,0)| = -3 min{ ∂l∂f∣p}=−∣grad f∣=−∣∇f(1,1,0)∣=−3
