方向导数与梯度
1 方向导数(二元为例)
Σ : z = f ( x , y ) , ( x , y ) ∈ D , M 0 ( x 0 , y 0 ) ∈ D Σ:z=f(x,y),(x,y)∈D,M_0(x_0,y_0)∈D Σ:z=f(x,y),(x,y)∈D,M0(x0,y0)∈D
在 x o y xoy xoy面内从 M 0 M_0 M0作射线 l l l ,取 M ( x 0 + △ x , y + △ y ) ∈ l M(x_0+△x,y+△y)∈l M(x0+△x,y+△y)∈l
△ z = f ( x 0 + △ x , y 0 + △ y ) − f ( x 0 , y 0 ) △z=f(x_0+△x,y_0+△y)-f(x_0,y_0) △z=f(x0+△x,y0+△y)−f(x0,y0)
ρ = ∣ M 0 M ∣ = ( △ x ) 2 + ( △ y ) 2 ρ=|M_0M|=\sqrt[]{(△x)^2+(△y)^2} ρ=∣M0M∣=(△x)2+(△y)2
如果 l i m ρ → 0 △ z ρ \underset{ρ→0}{lim}\frac{△z}{ρ} ρ→0limρ△z存在,称此极限为 z = f ( x , y ) z=f(x,y) z=f(x,y)在 M 0 M_0 M0处沿射线 l l l的方向导数,记作 ∂ z ∂ l ∣ M 0 \frac{∂z}{∂l}|_{M_0} ∂l∂z∣M0,
即 ∂ z ∂ l ∣ M 0 ≜ l i m ρ → 0 △ z ρ \frac{∂z}{∂l}|_{M_0}\triangleq\underset{ρ→0}{lim}\frac{△z}{ρ} ∂l∂z∣M0≜ρ→0limρ△z
计算
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z=f(x,y),M0(x0,y0),射线
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l的方向余弦
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f ( x , y ) f(x,y) f(x,y)连续可偏导
△ z ρ = f ( x 0 + △ x , y 0 + △ y ) − f ( x 0 , y 0 + △ y ) ρ + f ( x 0 , y 0 + △ y ) − f ( x 0 , y 0 ) ρ \frac{△z}{ρ}=\frac{f(x_0+△x,y_0+△y)-f(x_0,y_0+△y)}{ρ}+\frac{f(x_0,y_0+△y)-f(x_0,y_0)}{ρ} ρ△z=ρf(x0+△x,y0+△y)−f(x0,y0+△y)+ρf(x0,y0+△y)−f(x0,y0)
= f x ( ξ , y 0 + △ y ) △ x ρ + f ( x 0 , y 0 + △ y ) − f ( x 0 , y 0 ) △ y ⋅ △ y ρ f_x(ξ,y_0+△y)\frac{△x}{ρ}+\frac{f(x_0,y_0+△y)-f(x_0,y_0)}{△y}·\frac{△y}{ρ} fx(ξ,y0+△y)ρ△x+△yf(x0,y0+△y)−f(x0,y0)⋅ρ△y (ξ介于 x 0 x_0 x0与 x 0 + △ x x_0+△x x0+△x之间)
∵ f x ( x , y ) , f y ( x , y ) f_x(x,y),f_y(x,y) fx(x,y),fy(x,y)连续
∴ ∂ z ∂ l ∣ M 0 = l i m ρ → 0 f x ( ξ , y 0 + △ y ) △ x ρ + l i m ρ → 0 f ( x 0 , y 0 + △ y ) − f ( x 0 , y 0 ) △ y ⋅ △ y ρ \frac{∂z}{∂l}|_{M_0}= \underset{ρ→0}{lim}f_x(ξ,y_0+△y)\frac{△x}{ρ}+\underset{ρ→0}{lim}\frac{f(x_0,y_0+△y)-f(x_0,y_0)}{△y}·\frac{△y}{ρ} ∂l∂z∣M0=ρ→0limfx(ξ,y0+△y)ρ△x+ρ→0lim△yf(x0,y0+△y)−f(x0,y0)⋅ρ△y
= f x ( x 0 , y 0 ) c o s α + f y ( x 0 , y 0 ) c o s β f_x(x_0,y_0)cosα+f_y(x_0,y_0)cosβ fx(x0,y0)cosα+fy(x0,y0)cosβ
即 ∂ z ∂ l ∣ M 0 = ∂ z ∂ x ∣ M 0 c o s α + ∂ z ∂ y ∣ M 0 c o s β \frac{∂z}{∂l}|_{M_0}=\frac{∂z}{∂x}|_{M_0}cosα+\frac{∂z}{∂y}|_{M_0}cosβ ∂l∂z∣M0=∂x∂z∣M0cosα+∂y∂z∣M0cosβ
2 梯度(三元为例)
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