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8.7 方向导数和梯度
2025-12-03
高等数学
数学

§\S8.7 方向导数和梯度#

一、方向导数#

  • 定义1：设 ll 为 xOyxOy 平面上以 (x0,y0)(x_0,y_0) 为起点的一条射线，其中 el⃗=(cos⁡α,cos⁡β)\vec {e_l}=(\cos \alpha,\cos \beta) 是与 ll 同方向的单位向量，则射线 ll 的参数方程为 {x=x0+ρcos⁡αy=y0+ρcos⁡β(ρ≥0)\begin{cases} x=x_0+\rho \cos \alpha\\ y=y_0+\rho \cos \beta \end{cases} (\rho \ge 0)
  • 定义2：z=f(x,y)z=f(x,y) 在 U(P0)U(P_0) 有定义，P(x0+ρcos⁡α,y0+ρcos⁡β)P(x_0+\rho \cos\alpha,y_0+\rho \cos\beta) 为射线 ll 上一点，且 P∈U(P0)P\in U(P_0)。若 f(x0+ρcos⁡α,y0+ρcos⁡β)−f(x0,y0)f(x_0+\rho \cos\alpha,y_0+\rho \cos\beta)-f(x_0,y_0) 与 ∣PP0∣=ρ|PP_0|=\rho 的比值在 ρ→0+\rho\to 0^+ 下极限存在，则称此极限为 f(x,y)f(x,y) 在 P0P_0 沿射线 ll 的方向导数存在，记作 ∂f∂l∣(x0,y0)\dfrac{\partial f}{\partial l}|_{(x_0,y_0)} 或 ∂f∂n⃗∣(x0,y0)\dfrac{\partial f}{\partial \vec{n}}|_{(x_0,y_0)}
  • 方向导数与偏导数的关系 ∂f∂l∣(x0,y0)=lim⁡ρ→0+f(x0+ρcos⁡α,y0+ρcos⁡β)−f(x0,y0)ρ∂f∂x−=lim⁡ρ→0+f(x0−ρ,y0)−f(x0,y0)ρ=−ρ=Δxlim⁡ρ→0+f(x0+Δx,y0)−f(x0,y0)−Δx=−∂f∂x\begin{align*} \dfrac{\partial f}{\partial l}|_{(x_0,y_0)}&=\lim_{\rho\to0^+}\dfrac{f(x_0+\rho\cos\alpha,y_0+\rho\cos\beta)-f(x_0,y_0)}{\rho} \\ \dfrac{\partial f}{\partial x^-}&=\lim_{\rho\to0^+}\dfrac{f(x_0-\rho,y_0)-f(x_0,y_0)}{\rho}\\&\overset{-\rho=\Delta x}{=}\lim_{\rho\to0^+}\dfrac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{-\Delta x}\\ &=-\dfrac{\partial f}{\partial x} \end{align*}
    • ∂f∂y+=∂f∂y\dfrac{\partial f}{\partial y^+}=\dfrac{\partial f}{\partial y}，∂f∂y−=−∂f∂y\dfrac{\partial f}{\partial y^-}=-\dfrac{\partial f}{\partial y}
    • 若偏导存在，则 ff 沿 x,y,zx,y,z 的方向导数存在
  • 定理：可微→方向导数存在

    • 证明：可微 → f(x0+Δx,y0+Δy)−f(x0,y0)=fx(x0,y0)Δx+fy(x0,y0)Δy+o((Δx)2+(Δy)2)f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0)=f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y+o(\sqrt{(\Delta x)^2+(\Delta y)^2})，取 Δx=ρcos⁡α\Delta x=\rho\cos\alpha，Δy=ρcos⁡β\Delta y=\rho\cos\beta：

      $$
      \begin{align*}
      \dfrac{\partial f}{\partial l}|_{(x_0,y_0)}&=\lim_{\rho \to 0^+}\dfrac{f(x_0+\rho\cos\alpha,y_0+\rho\cos\beta)-f(x_0,y_0)}{\rho}\\
      &=\lim_{\rho \to 0^+}\dfrac{[f_x(x_0,y_0)\rho\cos\alpha+f_y(x_0,y_0)\rho\cos\beta]+o(\rho)}{\rho}\\
      &=f_x(x_0,y_0)\cos\alpha+f_y(x_0,y_0)\cos\beta
      \end{align*}
      $$

    • ∂f∂l∣(x0,y0)=fx(x0,y0)cos⁡α+fy(x0,y0)cos⁡β\dfrac{\partial f}{\partial l}|_{(x_0,y_0)}=f_x(x_0,y_0)\cos \alpha+f_y(x_0,y_0)\cos \beta

    • ∂f∂l∣(x0,y0,z0)=fx(x0,y0,z0)cos⁡α+fy(x0,y0,z0)cos⁡β+fz(x0,y0,z0)cos⁡γ\dfrac{\partial f}{\partial l}|_{(x_0,y_0,z_0)}=f_x(x_0,y_0,z_0)\cos \alpha+f_y(x_0,y_0,z_0)\cos \beta+f_z(x_0,y_0,z_0)\cos \gamma（el⃗=(cos⁡α,cos⁡β,cos⁡γ)=1∣l⃗∣l⃗\vec{e_l}=(\cos\alpha,\cos\beta,\cos\gamma)=\dfrac{1}{|\vec{l}|}\vec{l}）

例题#

  1. 讨论 z=x2+y2z=\sqrt{x^2+y^2} 在 (0,0)(0,0) 的方向导数情况

    解：el⃗=(cos⁡α,cos⁡β)\vec{e_l}=(\cos\alpha,\cos\beta)

    ∂f∂l∣(0,0)=lim⁡ρ→0+f(0+ρcos⁡α,0+ρcos⁡β)−f(0,0)ρ=lim⁡ρ→0+ρ−0ρ=1∂f∂x+∣(0,0)=∂f∂x−∣(0,0)=1\begin{align*} \dfrac{\partial f}{\partial l}|_{(0,0)}&=\lim_{\rho \to 0^+}\dfrac{f(0+\rho\cos\alpha,0+\rho\cos\beta)-f(0,0)}{\rho}\\ &=\lim_{\rho \to 0^+}\dfrac{\rho-0}{\rho}=1\\ \dfrac{\partial f}{\partial x^+}|_{(0,0)}=\dfrac{\partial f}{\partial x^-}|_{(0,0)}=1 \end{align*}

  2. f={xyx2+y2sin⁡1x2+y2x2+y2≠00x2+y2=0f=\begin{cases}\dfrac{xy}{\sqrt{x^2+y^2}}\sin\dfrac{1}{x^2+y^2}\quad x^2+y^2\ne0\\0\quad x^2+y^2=0\end{cases}，讨论偏导性、方向导数情况

    解：fx(0,0)=fy(0,0)=0f_x(0,0)=f_y(0,0)=0

    ∂f∂l∣(0,0)=lim⁡ρ→0+ρcos⁡αρcos⁡βρsin⁡1ρ2ρ=lim⁡ρ→0+cos⁡αcos⁡βsin⁡1ρ2\begin{align*} \dfrac{\partial f}{\partial l}|_{(0,0)}&=\lim_{\rho \to 0^+}\dfrac{\frac{\rho\cos\alpha \rho\cos\beta}{\rho}\sin\frac{1}{\rho^2}}{\rho}\\ &=\lim_{\rho \to 0^+}\cos\alpha\cos\beta\sin\dfrac{1}{\rho^2} \end{align*}

    当 cos⁡α=0\cos\alpha=0 或 cos⁡β=0\cos\beta=0 时，极限存在

    当 cos⁡α≠0\cos\alpha\ne 0 且 cos⁡β≠0\cos\beta \ne 0 时，极限不存在，方向导数也不存在

    ∂f∂x+=∂f∂x−=∂f∂y+=∂f∂y−=0\dfrac{\partial f}{\partial x^+}=\dfrac{\partial f}{\partial x^-}=\dfrac{\partial f}{\partial y^+}=\dfrac{\partial f}{\partial y^-}=0

  3. f(x,y,z)=x2cos⁡y+e−yln⁡(x+z)f(x,y,z)=x^2\cos y+e^{-y}\ln(x+z)，求其在 (−1,0,2)(-1,0,2) 处沿该点到 (1,2,1)(1,2,1) 方向的方向导数

    解：fx∣(−1,0,2)=2×(−1)×cos⁡0+e−01−1+2=−1f_x|_{(-1,0,2)}=2\times(-1)\times\cos 0+e^{-0}\dfrac{1}{-1+2}=-1，fy∣(−1,0,2)=0f_y|_{(-1,0,2)}=0，fz∣(−1,0,2)=1f_z|_{(-1,0,2)}=1

    MM0⃗=(2,2,−1)\vec{MM_0}=(2,2,-1)

    el⃗=(23,23,−13)\vec{e_l}=(\dfrac{2}{3},\dfrac{2}{3},-\dfrac{1}{3})

    ∂f∂l∣(−1,0,2)=−1×23+0×23+1×(−13)=−1\dfrac{\partial f}{\partial l}|_{(-1,0,2)}=-1\times\dfrac{2}{3}+0\times\dfrac{2}{3}+1\times(-\dfrac{1}{3})=-1

  4. n⃗\vec {n} 是 2x2+3y2+z2=62x^2+3y^2+z^2=6 在 P0(1,1,1)P_0(1,1,1) 处指向外侧的法向量，求 u=6x2+8y2zu=\dfrac{\sqrt{6x^2+8y^2}}{z} 在 P0P_0 处沿 n⃗\vec{n} 的方向导数

    解：ux(1,1,1)=614u_x(1,1,1)=\dfrac{6}{\sqrt{14}}，uy(1,1,1)=814u_y(1,1,1)=\dfrac{8}{\sqrt{14}}，uz(1,1,1)=−14u_z(1,1,1)=-\sqrt{14}

    12n⃗=(2,3,1)\dfrac{1}{2}\vec{n}=(2,3,1)，en⃗=(214,814,114)\vec{e_n}=(\dfrac{2}{\sqrt{14}},\dfrac{8}{\sqrt{14}},\dfrac{1}{\sqrt{14}})

    ∂f∂n⃗∣(1,1,1)=614×214+814×314−14×114=117\dfrac{\partial f}{\partial \vec{n}}|_{(1,1,1)}=\dfrac{6}{\sqrt{14}}\times\dfrac{2}{\sqrt{14}}+\dfrac{8}{\sqrt{14}}\times\dfrac{3}{\sqrt{14}}-\sqrt{14}\times\dfrac{1}{\sqrt{14}}=\dfrac{11}{7}

  5. f(x,y)={xy2x2+y4x2+y2≠00x2+y2=0f(x,y)=\begin{cases}\dfrac{xy^2}{x^2+y^4} \quad x^2+y^2\ne0\\0\quad x^2+y^2=0\end{cases}，求 ff 在 (0,0)(0,0) 处沿任意方向的方向导数

    解：

    ∂f∂l∣(0,0)=lim⁡ρ→0+ρcos⁡αρ2cos⁡βρ2cos⁡2α+ρ4cos⁡4β−0ρ=lim⁡ρ→0+cos⁡αcos⁡βcos⁡2α+ρ2cos⁡4β\begin{align*} \dfrac{\partial f}{\partial l}|_{(0,0)}&=\lim_{\rho \to 0^+}\dfrac{\frac{\rho\cos\alpha\rho^2\cos\beta}{\rho^2\cos^2\alpha+\rho^4\cos^4\beta}-0}{\rho}\\ &=\lim_{\rho \to 0^+}\dfrac{\cos\alpha\cos\beta}{\cos^2\alpha+\rho^2\cos^4\beta} \end{align*}
    • 当 cos⁡α≠0\cos\alpha\ne0 时，原式 =cos⁡2βcos⁡α=1−cos⁡2αcos⁡α=sin⁡2αcos⁡α=\dfrac{\cos^2\beta}{\cos\alpha}=\dfrac{1-\cos^2\alpha}{\cos\alpha}=\dfrac{\sin^2\alpha}{\cos\alpha}
    • 当 cos⁡α=0\cos\alpha=0 时，sin⁡2α=cos⁡2β=1\sin^2\alpha=\cos^2\beta=1，原式 =0=0

二、梯度#

  • 定义3：z=f(x,y)z=f(x,y) 在 DD 内具有一阶连续偏导数，则 ∀P0∈D\forall P_0\in D 都有 (fx(x0,y0),fy(x0,y0))(f_x(x_0,y_0),f_y(x_0,y_0))，称该向量为 f(x,y)f(x,y) 在 P0P_0 处的梯度（▽f\bigtriangledown f） grad⁡f(x0,y0)=(fx(x0,y0),fy(x0,y0))=fx(x0,y0)i⃗+fy(x0,y0)j⃗\begin{align*} \operatorname{grad} f(x_0,y_0)&=(f_x(x_0,y_0),f_y(x_0,y_0))\\ &=f_x(x_0,y_0)\vec{i}+f_y(x_0,y_0)\vec{j} \end{align*}
    • 推广：u=f(x1,x2,⋯ ,xn)u=f(x_1,x_2,\cdots,x_n)，▽u=(fx1,fx2,⋯ ,fxn)∣P0\bigtriangledown u=(f_{x_1},f_{x_2},\cdots,f_{x_n})|_{P_0} \begin{align*} \dfrac{\partial f}{\partial l}|_{(x_0,y_0)}&=f_x(x_0,y_0)\cos\alpha+f_y(x_0,y_0)\cos\beta\\ &=(f_x,f_y)\cdot (\cos\alpha,\cos\beta)\\ &=\operatorname{grad} f(x_0,y_0) \cdot \vec{e_l}\\ &=|\operatorname{grad}f(x_0,y_0)|\cdot 1 \cdot \cos\theta, \theta\in[0,\pi] \end{align*}
    • 讨论：
      • θ=0\theta=0 时，方向导数最大值为 grad⁡f(x0,y0)\operatorname{grad}f(x_0,y_0)。函数沿 ▽f(x0,y0)\bigtriangledown f(x_0,y_0) 方向变化时，函数增加最快
      • θ=π\theta=\pi 时，方向导数最小值为 −∣grad⁡f(x0,y0)∣-|\operatorname{grad}f(x_0,y_0)|。函数沿 −▽f(x0,y0)-\bigtriangledown f(x_0,y_0) 变化时，函数减少最快
      • θ=π2\theta=\dfrac{\pi}{2} 时，方向导数为0。函数沿与 grad⁡f(x0,y0)\operatorname{grad}f(x_0,y_0) 垂直方向变化时，函数变化率为0
  • 定义4【等值线】：z=f(x,y)z=f(x,y) 被平面 z=cz=c 所截得的曲线 ll 的方程 {z=f(x,y)z=c\begin{cases}z=f(x,y)\\z=c\end{cases} LL 在 xOyxOy 平面上的投影曲线 L∗L* {f(x,y)=cz=0\begin{cases}f(x,y)=c\\z=0\end{cases}（柱面）。
    • 若 fxf_x、fyf_y 不同时为0，f(x,y)=cf(x,y)=c 上任一点 P0(x0,y0)P_0(x_0,y_0) 处的一个单位法向量为 n⃗=1fx2(x0,y0)+fy2(x0,y0)⋅(fx(x0,y0),fy(x0,y0))grad⁡f(x0,y0)k=fyfx∣(x0,y0)y−y0=fyfx∣(x0,y0)(x−x0)x−x0fx(x0,y0)=y−y0fy(x0,y0)∂f∂n⃗∣(x0,y0)=∣grad⁡f(x0,y0)∣grad⁡f(x0,y0)=∂f∂n⃗∣(x0,y0)⋅n⃗ \vec{n}=\dfrac{1}{\sqrt{f_x^2(x_0,y_0)+f_y^2(x_0,y_0)}}\cdot \dfrac{(f_x(x_0,y_0),f_y(x_0,y_0))}{\operatorname{grad}f(x_0,y_0)}\\ k=\dfrac{f_y}{f_x}|_{(x_0,y_0)}\\ y-y_0=\dfrac{f_y}{f_x}|_{(x_0,y_0)}(x-x_0)\\ \dfrac{x-x_0}{f_x(x_0,y_0)}=\dfrac{y-y_0}{f_y(x_0,y_0)}\\ \dfrac{\partial f}{\partial \vec{n}}|_{(x_0,y_0)}=|\operatorname{grad}f(x_0,y_0)|\\ \operatorname{grad}f(x_0,y_0)=\dfrac{\partial f}{\partial \vec{n}}|_{(x_0,y_0)}\cdot \vec{n}

例题#

  1. f(x,y,z)f(x,y,z) 在 (x0,y0,z0)(x_0,y_0,z_0) 处的梯度为 g⃗\vec{g}，l⃗=(0,2,2)\vec{l}=(0,2,2) 且 g⃗⋅l⃗=1\vec{g}\cdot \vec{l}=1，∂f∂l∣(x0,y0,z0)=g⃗⋅el⃗=1∣l⃗∣l⃗⋅g⃗=122\dfrac{\partial f}{\partial l}|_{(x_0,y_0,z_0)}=\vec{g}\cdot\vec{e_l}=\dfrac{1}{|\vec{l}|}\vec{l}\cdot\vec{g}=\dfrac{1}{2\sqrt{2}}

  2. u=x2yz+z3u=x^2yz+z^3，求 uu 在 M0(2,−1,1)M_0(2,-1,1) 处沿 l⃗=(2,−2,1)\vec{l}=(2,-2,1) 的方向导数并求 uu 在 M0M_0 处最大的方向导数及其方向

    解：grad⁡u∣M0=(2xyz,x2z,x2y+3z2)∣(2,−1,1)=(−4,4,−1)\operatorname{grad}u|_{M_0}=(2xyz,x^2z,x^2y+3z^2)|_{(2,-1,1)}=(-4,4,-1)

    ∂f∂l=1∣l⃗∣l⃗⋅grad⁡u∣M0=13(−8−8−1)=−173\dfrac{\partial f}{\partial l}=\dfrac{1}{|\vec{l}|}\vec{l}\cdot \operatorname{grad}u|_{M_0}=\dfrac{1}{3}(-8-8-1)=-\dfrac{17}{3}

    n⃗=k(−4,4,−1)\vec{n}=k(-4,4,-1)

    max⁡∂u∂n⃗∣M0=∣grad⁡u∣M0∣=33\max \dfrac{\partial u}{\partial \vec{n}}|_{M_0}=|\operatorname{grad}u|_{M_0}|=\sqrt{33}

  3. f(x,y,z)=axy2+byz+cx3z2f(x,y,z)=axy^2+byz+cx^3z^2 在 (1,2,−1)(1,2,-1) 处沿z轴正方向的方向导数有最大值64，求a、b、c

    解：k⃗=(0,0,1)\vec{k}=(0,0,1)

    grad⁡f∣(1,2,−1)=(0,0,d)(d>0)\operatorname{grad}f|_{(1,2,-1)}=(0,0,d)\quad(d\gt 0)

    令 d=64d=64，grad⁡f∣(1,2,−1)=(ay2+3cx2z2,2axy+bz,by+2cx3z)∣(1,2,−1)=(4a+3c,4a−b,2b−2c)\operatorname{grad}f|_{(1,2,-1)}=(ay^2+3cx^2z^2,2axy+bz,by+2cx^3z)|_{(1,2,-1)}=(4a+3c,4a-b,2b-2c)

    {4a+3c=04a−b=02b−2c=64→{a=6b=24c=−8\begin{cases}4a+3c=0\\4a-b=0\\2b-2c=64\end{cases}\rightarrow \begin{cases}a=6\\b=24\\c=-8\end{cases}

8.7 方向导数和梯度
https://gitee.com/jason_ren/advanced-math-note
作者
Jason Ren
发布于
2025-12-03
许可协议
CC BY-SA 4.0

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