目录

5.1多元函数的极限、连续、偏导数与全微分（概念）例2试求下列二次极限：（1）lim⁡y→0x→0x2+y2∣x∣+∣y∣;\lim\limits_{y\to0 \atop x\to0}\cfrac{x^2+y^2}{|x|+|y|};x→0y→0​lim​∣x∣+∣y∣x2+y2​;（4）lim⁡y→0x→02xy2sin⁡xx2+y4.\lim\limits_{y\to0 \atop x\to0}\cfrac{2xy^2\sin x}{x^2+y^4}.x→0y→0​lim​x2+y42xy2sinx​.例11设fx′(x,y)f'_x(x,y)fx′​(x,y)存在，fy′(x,y)f'_y(x,y)fy′​(x,y)在点(x0,y0)(x_0,y_0)(x0​,y0​)处连续，证明：f(x,y)f(x,y)f(x,y)在点(x0,y0)(x_0,y_0)(x0​,y0​)处可微。5.2多元函数的微分法例26设函数u=f(x,y)u=f(x,y)u=f(x,y)具有二阶连续偏导数，且满足4∂2u∂x2+12∂2u∂x∂y+5∂2u∂y2=04\cfrac{\partial^2u}{\partial x^2}+12\cfrac{\partial^2u}{\partial x\partial y}+5\cfrac{\partial^2u}{\partial y^2}=04∂x2∂2u​+12∂x∂y∂2u​+5∂y2∂2u​=0。确定a,ba,ba,b的值，使等式ξ=x+ay,η=x+by\xi=x+ay,\eta=x+byξ=x+ay,η=x+by在变换下简化为∂2u∂ξ∂η\cfrac{\partial^2u}{\partial\xi\partial\eta}∂ξ∂η∂2u​。例28设(r,θ)(r,\theta)(r,θ)为极坐标，u=u(r,θ)u=u(r,\theta)u=u(r,θ)当r>0r>0r>0时具有二阶连续偏导数，并满足∂u∂θ≡0\cfrac{\partial u}{\partial\theta}\equiv0∂θ∂u​≡0，且∂2u∂x2+∂2u∂y2=0\cfrac{\partial^2u}{\partial x^2}+\cfrac{\partial^2u}{\partial y^2}=0∂x2∂2u​+∂y2∂2u​=0，求u(r,θ)u(r,\theta)u(r,θ)。例29若对任意t>0t>0t>0，有f(tx,ty)=tnf(x,y)f(tx,ty)=t^nf(x,y)f(tx,ty)=tnf(x,y)，则称函数f(x,y)f(x,y)f(x,y)是nnn次齐次函数。试证：若f(x,y)f(x,y)f(x,y)可微，则f(x,y)f(x,y)f(x,y)是nnn次齐次函数的充要条件是x∂f∂x+y∂f∂y=nf(x,y)x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}=nf(x,y)x∂x∂f​+y∂y∂f​=nf(x,y)。例36设f(x,y)f(x,y)f(x,y)有二阶连续偏导数，且fy′≠0f'_y\ne0fy′​​=0，证明：对任给的常数CCC，f(x,y)=Cf(x,y)=Cf(x,y)=C为一条直线的充要条件是f2′2f11′′+2f1′f2′f12′′+f1′2f22′′=0f'^2_2f''_{11}+2f'_1f'_2f''_{12}+f'^2_1f''_{22}=0f2′2​f11′′​+2f1′​f2′​f12′′​+f1′2​f22′′​=0。5.3极值与最值例49求中心在坐标原点的椭圆x2+4xy+5y2=1x^2+4xy+5y^2=1x2+4xy+5y2=1的长半轴与短半轴。5.4方向导数与梯度 多元微分在几何上的应用 泰勒定理例53函数z=x2+y2z=\sqrt{x^2+y^2}z=x2+y2​在点(0,0)(0,0)(0,0)处（）

(A)(A)(A)不连续；

(B)(B)(B)偏导数存在；

(C)(C)(C)沿任一方向方向导数存在；

(D)(D)(D)可微。例57设f(x,y)f(x,y)f(x,y)是R2\bold{R}^2R2（全平面）上的一个可微函数，且lim⁡ρ→+∞(x∂f∂x+y∂f∂y)=α>0\lim\limits_{\rho\to+\infty}\left(x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}\right)=\alpha>0ρ→+∞lim​(x∂x∂f​+y∂y∂f​)=α>0，其中ρ=x2+y2\rho=\sqrt{x^2+y^2}ρ=x2+y2​，α\alphaα为常数。试证明f(x,y)f(x,y)f(x,y)在R2\bold{R}^2R2上有最小值。写在最后

5.1多元函数的极限、连续、偏导数与全微分（概念）

例2试求下列二次极限：

（1）lim⁡y→0x→0x2+y2∣x∣+∣y∣;\lim\limits_{y\to0 \atop x\to0}\cfrac{x^2+y^2}{|x|+|y|};x→0y→0​lim​∣x∣+∣y∣x2+y2​;

解由于0⩽x2+y2∣x∣+∣y∣=x2∣x∣+∣y∣+y2∣x∣+∣y∣⩽x2∣x∣+y2∣y∣=∣x∣+∣y∣0\leqslant\cfrac{x^2+y^2}{|x|+|y|}=\cfrac{x^2}{|x|+|y|}+\cfrac{y^2}{|x|+|y|}\leqslant\cfrac{x^2}{|x|}+\cfrac{y^2}{|y|}=|x|+|y|0⩽∣x∣+∣y∣x2+y2​=∣x∣+∣y∣x2​+∣x∣+∣y∣y2​⩽∣x∣x2​+∣y∣y2​=∣x∣+∣y∣，而lim⁡y→0x→0(∣x∣+∣y∣)=0\lim\limits_{y\to0 \atop x\to0}(|x|+|y|)=0x→0y→0​lim​(∣x∣+∣y∣)=0，由夹挤定理知lim⁡y→0x→0x2+y2∣x∣+∣y∣=0\lim\limits_{y\to0 \atop x\to0}\cfrac{x^2+y^2}{|x|+|y|}=0x→0y→0​lim​∣x∣+∣y∣x2+y2​=0。（这道题主要利用了夹挤定理求解）

（4）lim⁡y→0x→02xy2sin⁡xx2+y4.\lim\limits_{y\to0 \atop x\to0}\cfrac{2xy^2\sin x}{x^2+y^4}.x→0y→0​lim​x2+y42xy2sinx​.

解由于∣2xy2sin⁡xx2+y4∣⩽1(2ab⩽a2+b2)\left|\cfrac{2xy^2\sin x}{x^2+y^4}\right|\leqslant1(2ab\leqslant a^2+b^2)∣∣∣∣∣​x2+y42xy2sinx​∣∣∣∣∣​⩽1(2ab⩽a2+b2)，即为有界量，而lim⁡x→0sin⁡x=0\lim\limits_{x\to0}\sin x=0x→0lim​sinx=0，即为无穷小量，则原式=0=0=0。（这道题主要利用了无穷小量代换求解）

例11设fx′(x,y)f'_x(x,y)fx′​(x,y)存在，fy′(x,y)f'_y(x,y)fy′​(x,y)在点(x0,y0)(x_0,y_0)(x0​,y0​)处连续，证明：f(x,y)f(x,y)f(x,y)在点(x0,y0)(x_0,y_0)(x0​,y0​)处可微。

证

Δz=f(x0+Δx,y0+Δy)−f(x0,y0)=f(x0+Δx,y0+Δy)−f(x0+Δx,y0)+f(x0+Δx,y0)−f(x0,y0).\begin{aligned} \Delta z&=f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0)\\ &=f(x_0+\Delta x,y_0+\Delta y)-f(x_0+\Delta x,y_0)+f(x_0+\Delta x,y_0)-f(x_0,y_0). \end{aligned} Δz​=f(x0​+Δx,y0​+Δy)−f(x0​,y0​)=f(x0​+Δx,y0​+Δy)−f(x0​+Δx,y0​)+f(x0​+Δx,y0​)−f(x0​,y0​).​

由一元拉格朗日中值定理知f(x0+Δx,y0+Δy)−f(x0+Δx,y0)=fy′(x0+Δx,y0+θ2Δy)Δyf(x_0+\Delta x,y_0+\Delta y)-f(x_0+\Delta x,y_0)=f'_y(x_0+\Delta x,y_0+\theta_2\Delta y)\Delta yf(x0​+Δx,y0​+Δy)−f(x0​+Δx,y0​)=fy′​(x0​+Δx,y0​+θ2​Δy)Δy。由fy′(x,y)f'_y(x,y)fy′​(x,y)在(x0,y0)(x_0,y_0)(x0​,y0​)处连续，则lim⁡Δx→0Δy→0fy′(x0+Δx,y0+θ2Δy)=fy′(x0,y0)\lim\limits_{\Delta x\to0 \atop \Delta y\to0}f'_y(x_0+\Delta x,y_0+\theta_2\Delta y)=f'_y(x_0,y_0)Δy→0Δx→0​lim​fy′​(x0​+Δx,y0​+θ2​Δy)=fy′​(x0​,y0​)，从而有f(x0+Δx,y0+Δy)−f(x0+Δx,y0)=fy′(x0,y0)Δy+ϵ2Δyf(x_0+\Delta x,y_0+\Delta y)-f(x_0+\Delta x,y_0)=f'_y(x_0,y_0)\Delta y+\epsilon_2\Delta yf(x0​+Δx,y0​+Δy)−f(x0​+Δx,y0​)=fy′​(x0​,y0​)Δy+ϵ2​Δy，其中ϵ2\epsilon_2ϵ2​为Δx→0,Δy→0\Delta x\to0,\Delta y\to0Δx→0,Δy→0时的无穷小量。

又由于fx′(x,y)f'_x(x,y)fx′​(x,y)存在，则lim⁡Δx→0f(x0+Δx,y0)−f(x0,y0)Δx=fx′(x0,y0)\lim\limits_{\Delta x\to0}\cfrac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}=f'_x(x_0,y_0)Δx→0lim​Δxf(x0​+Δx,y0​)−f(x0​,y0​)​=fx′​(x0​,y0​)，从而有f(x0+Δx,y0)−f(x0,y0)Δx=fx′(x0,y0)+ϵ1\cfrac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}=f'_x(x_0,y_0)+\epsilon_1Δxf(x0​+Δx,y0​)−f(x0​,y0​)​=fx′​(x0​,y0​)+ϵ1​，其中ϵ1\epsilon_1ϵ1​为Δx→0,Δy→0\Delta x\to0,\Delta y\to0Δx→0,Δy→0时的无穷小量。则f(x0+Δx,y0)−f(x0,y0)=fx′(x0,y0)Δx+ϵ1Δxf(x_0+\Delta x,y_0)-f(x_0,y_0)=f'_x(x_0,y_0)\Delta x+\epsilon_1\Delta xf(x0​+Δx,y0​)−f(x0​,y0​)=fx′​(x0​,y0​)Δx+ϵ1​Δx。

由于∣ϵ1Δx+ϵ2Δy(Δx)2+(Δy)2∣⩽∣ϵ1∣∣Δx∣+∣ϵ2∣∣Δy∣(Δx)2+(Δy)2⩽∣ϵ1∣+∣ϵ2∣→0(Δx→0,Δy→0)\left|\cfrac{\epsilon_1\Delta x+\epsilon_2\Delta y}{\sqrt{(\Delta x)^2+(\Delta y)^2}}\right|\leqslant\cfrac{|\epsilon_1||\Delta x|+|\epsilon_2||\Delta y|}{\sqrt{(\Delta x)^2+(\Delta y)^2}}\leqslant|\epsilon_1|+|\epsilon_2|\to0(\Delta x\to0,\Delta y\to0)∣∣∣∣∣​(Δx)2+(Δy)2​ϵ1​Δx+ϵ2​Δy​∣∣∣∣∣​⩽(Δx)2+(Δy)2​∣ϵ1​∣∣Δx∣+∣ϵ2​∣∣Δy∣​⩽∣ϵ1​∣+∣ϵ2​∣→0(Δx→0,Δy→0)，则当Δx→0,Δy→0\Delta x\to0,\Delta y\to0Δx→0,Δy→0时，ϵ1Δx+ϵ2Δy=ο(ρ)\epsilon_1\Delta x+\epsilon_2\Delta y=\omicron(\rho)ϵ1​Δx+ϵ2​Δy=ο(ρ)，故f(x,y)f(x,y)f(x,y)在(x0,y0)(x_0,y_0)(x0​,y0​)处可微。（这道题主要利用了多元函数极限定义求解）

5.2多元函数的微分法

例26设函数u=f(x,y)u=f(x,y)u=f(x,y)具有二阶连续偏导数，且满足4∂2u∂x2+12∂2u∂x∂y+5∂2u∂y2=04\cfrac{\partial^2u}{\partial x^2}+12\cfrac{\partial^2u}{\partial x\partial y}+5\cfrac{\partial^2u}{\partial y^2}=04∂x2∂2u​+12∂x∂y∂2u​+5∂y2∂2u​=0。确定a,ba,ba,b的值，使等式ξ=x+ay,η=x+by\xi=x+ay,\eta=x+byξ=x+ay,η=x+by在变换下简化为∂2u∂ξ∂η\cfrac{\partial^2u}{\partial\xi\partial\eta}∂ξ∂η∂2u​。

解

∂u∂x=∂u∂ξ+∂u∂η,∂2u∂x2=∂2u∂ξ2+2∂2u∂ξ∂η+∂2u∂η2,∂u∂y=a∂u∂ξ+b∂u∂η,∂2u∂y2=a2∂2u∂ξ2+2ab∂2u∂ξ∂η+b2∂2u∂η2,∂2u∂x∂y=a2∂2u∂ξ2+(a+b)∂2u∂ξ∂η+b∂2u∂η2,\cfrac{\partial u}{\partial x}=\cfrac{\partial u}{\partial\xi}+\cfrac{\partial u}{\partial\eta},\cfrac{\partial^2u}{\partial x^2}=\cfrac{\partial^2u}{\partial\xi^2}+2\cfrac{\partial^2u}{\partial\xi\partial\eta}+\cfrac{\partial^2u}{\partial\eta^2},\\ \cfrac{\partial u}{\partial y}=a\cfrac{\partial u}{\partial\xi}+b\cfrac{\partial u}{\partial\eta},\cfrac{\partial^2u}{\partial y^2}=a^2\cfrac{\partial^2u}{\partial\xi^2}+2ab\cfrac{\partial^2u}{\partial\xi\partial\eta}+b^2\cfrac{\partial^2u}{\partial\eta^2},\\ \cfrac{\partial^2u}{\partial x\partial y}=a^2\cfrac{\partial^2u}{\partial\xi^2}+(a+b)\cfrac{\partial^2u}{\partial\xi\partial\eta}+b\cfrac{\partial^2u}{\partial\eta^2}, ∂x∂u​=∂ξ∂u​+∂η∂u​,∂x2∂2u​=∂ξ2∂2u​+2∂ξ∂η∂2u​+∂η2∂2u​,∂y∂u​=a∂ξ∂u​+b∂η∂u​,∂y2∂2u​=a2∂ξ2∂2u​+2ab∂ξ∂η∂2u​+b2∂η2∂2u​,∂x∂y∂2u​=a2∂ξ2∂2u​+(a+b)∂ξ∂η∂2u​+b∂η2∂2u​,

将以上三个二阶偏导数代入等式4∂2u∂x2+12∂2u∂x∂y+5∂2u∂y2=04\cfrac{\partial^2u}{\partial x^2}+12\cfrac{\partial^2u}{\partial x\partial y}+5\cfrac{\partial^2u}{\partial y^2}=04∂x2∂2u​+12∂x∂y∂2u​+5∂y2∂2u​=0得(5a2+12a+4)∂2u∂ξ2+[10ab+12(a+b)+8]∂2u∂x∂y+(5b2+12b+4)∂2u∂y2=0(5a^2+12a+4)\cfrac{\partial^2u}{\partial\xi^2}+[10ab+12(a+b)+8]\cfrac{\partial^2u}{\partial x\partial y}+(5b^2+12b+4)\cfrac{\partial^2u}{\partial y^2}=0(5a2+12a+4)∂ξ2∂2u​+[10ab+12(a+b)+8]∂x∂y∂2u​+(5b2+12b+4)∂y2∂2u​=0。

由题设知{5a2+12a+4=0,5b2+12b+4=0,\begin{cases}5a^2+12a+4=0,\\5b^2+12b+4=0,\end{cases}{5a2+12a+4=0,5b2+12b+4=0,​但10ab+12(a+b)+8≠010ab+12(a+b)+8\ne010ab+12(a+b)+8​=0，解得{a=−2,b=−25,\begin{cases}a=-2,\\b=-\cfrac{2}{5},\end{cases}⎩⎨⎧​a=−2,b=−52​,​或{a=−25,b=−2.\begin{cases}a=-\cfrac{2}{5},\\b=-2.\end{cases}⎩⎨⎧​a=−52​,b=−2.​（这道题主要利用了多元复合函数求导求解）

例28设(r,θ)(r,\theta)(r,θ)为极坐标，u=u(r,θ)u=u(r,\theta)u=u(r,θ)当r>0r>0r>0时具有二阶连续偏导数，并满足∂u∂θ≡0\cfrac{\partial u}{\partial\theta}\equiv0∂θ∂u​≡0，且∂2u∂x2+∂2u∂y2=0\cfrac{\partial^2u}{\partial x^2}+\cfrac{\partial^2u}{\partial y^2}=0∂x2∂2u​+∂y2∂2u​=0，求u(r,θ)u(r,\theta)u(r,θ)。

解由∂u∂θ≡0\cfrac{\partial u}{\partial\theta}\equiv0∂θ∂u​≡0知，uuu仅为rrr的函数，令u=φ(r)u=\varphi(r)u=φ(r)，其中r=x2+y2,r>0r=\sqrt{x^2+y^2},r>0r=x2+y2​,r>0。则

∂u∂x=φ′(r)xx2+y2=φ′(x)xr,∂2u∂x2=φ′′(x)x2r2+φ′(r)r−x2rr2=φ′′(r)x2r2+φ′(r)(1r−x2r3).\cfrac{\partial u}{\partial x}=\varphi'(r)\cfrac{x}{\sqrt{x^2+y^2}}=\varphi'(x)\cfrac{x}{r},\\ \cfrac{\partial^2u}{\partial x^2}=\varphi''(x)\cfrac{x^2}{r^2}+\varphi'(r)\cfrac{r-\cfrac{x^2}{r}}{r^2}=\varphi''(r)\cfrac{x^2}{r^2}+\varphi'(r)\left(\cfrac{1}{r}-\cfrac{x^2}{r^3}\right). ∂x∂u​=φ′(r)x2+y2​x​=φ′(x)rx​,∂x2∂2u​=φ′′(x)r2x2​+φ′(r)r2r−rx2​​=φ′′(r)r2x2​+φ′(r)(r1​−r3x2​).

由对称性可得∂2u∂y2=φ′′(x)y2r2+φ′(r)(1r−y2r3)\cfrac{\partial^2u}{\partial y^2}=\varphi''(x)\cfrac{y^2}{r^2}+\varphi'(r)\left(\cfrac{1}{r}-\cfrac{y^2}{r^3}\right)∂y2∂2u​=φ′′(x)r2y2​+φ′(r)(r1​−r3y2​)，则∂2u∂x2+∂2u∂y2=φ′′(x)+φ′(x)1r\cfrac{\partial^2u}{\partial x^2}+\cfrac{\partial^2u}{\partial y^2}=\varphi''(x)+\varphi'(x)\cfrac{1}{r}∂x2∂2u​+∂y2∂2u​=φ′′(x)+φ′(x)r1​，从而得φ′′(x)+φ′(x)1r=0\varphi''(x)+\varphi'(x)\cfrac{1}{r}=0φ′′(x)+φ′(x)r1​=0。

这是一个不显含φ(r)\varphi(r)φ(r)的可降阶方程，令φ′(x)=p\varphi'(x)=pφ′(x)=p，则φ′′(r)=dpdr\varphi''(r)=\cfrac{\mathrm{d}p}{\mathrm{d}r}φ′′(r)=drdp​。代入上式得dpdr+pr=0⇒p=C1r\cfrac{\mathrm{d}p}{\mathrm{d}r}+\cfrac{p}{r}=0\Rightarrow p=\cfrac{C_1}{r}drdp​+rp​=0⇒p=rC1​​，即φ′(r)=C1r\varphi'(r)=\cfrac{C_1}{r}φ′(r)=rC1​​，则φ(r)=C1ln⁡r+C2\varphi(r)=C_1\ln r+C_2φ(r)=C1​lnr+C2​，故u=C1ln⁡r+C2u=C_1\ln r+C_2u=C1​lnr+C2​。（这道题主要利用了微分方程求解）

例29若对任意t>0t>0t>0，有f(tx,ty)=tnf(x,y)f(tx,ty)=t^nf(x,y)f(tx,ty)=tnf(x,y)，则称函数f(x,y)f(x,y)f(x,y)是nnn次齐次函数。试证：若f(x,y)f(x,y)f(x,y)可微，则f(x,y)f(x,y)f(x,y)是nnn次齐次函数的充要条件是x∂f∂x+y∂f∂y=nf(x,y)x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}=nf(x,y)x∂x∂f​+y∂y∂f​=nf(x,y)。

证必要性：由于f(x,y)f(x,y)f(x,y)为nnn次齐次方程，则对任意t>0t>0t>0，有f(tx,ty)=tnf(x,y)f(tx,ty)=t^nf(x,y)f(tx,ty)=tnf(x,y)。该式两端对ttt求导得xf1′(tx,ty)+yf2′(tx,ty)=ntn−1f(x,y)xf'_1(tx,ty)+yf'_2(tx,ty)=nt^{n-1}f(x,y)xf1′​(tx,ty)+yf2′​(tx,ty)=ntn−1f(x,y)。令t=1t=1t=1得xf1′(tx,ty)+yf2′(tx,ty)=nf(x,y)xf'_1(tx,ty)+yf'_2(tx,ty)=nf(x,y)xf1′​(tx,ty)+yf2′​(tx,ty)=nf(x,y)，即有x∂f∂x+y∂f∂y=nf(x,y)x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}=nf(x,y)x∂x∂f​+y∂y∂f​=nf(x,y)。

充分性：令F(t)=f(tx,ty)(t>0)F(t)=f(tx,ty)(t>0)F(t)=f(tx,ty)(t>0)，则dFdt=x∂f∂x+y∂f∂y\cfrac{\mathrm{d}F}{\mathrm{d}t}=x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}dtdF​=x∂x∂f​+y∂y∂f​，两边乘以ttt得tdFdt=tx∂f∂x+ty∂f∂y=nf(tx,ty)=nF(t)t\cfrac{\mathrm{d}F}{\mathrm{d}t}=tx\cfrac{\partial f}{\partial x}+ty\cfrac{\partial f}{\partial y}=nf(tx,ty)=nF(t)tdtdF​=tx∂x∂f​+ty∂y∂f​=nf(tx,ty)=nF(t)。于是dFF=ntdt\cfrac{\mathrm{d}F}{F}=\cfrac{n}{t}\mathrm{d}tFdF​=tn​dt，解得F(t)=CtnF(t)=Ct^nF(t)=Ctn。令t=1t=1t=1得，F(1)=CF(1)=CF(1)=C，而由F(t)=f(tx,ty)F(t)=f(tx,ty)F(t)=f(tx,ty)知F(1,1)=f(x,y)F(1,1)=f(x,y)F(1,1)=f(x,y)，则C=f(x,y)C=f(x,y)C=f(x,y)。于是F(t)=tnf(x,y)F(t)=t^nf(x,y)F(t)=tnf(x,y)，即f(tx,ty)=tnf(x,y)f(tx,ty)=t^nf(x,y)f(tx,ty)=tnf(x,y)。（这道题主要利用了多元函数求导求解）

例36设f(x,y)f(x,y)f(x,y)有二阶连续偏导数，且fy′≠0f'_y\ne0fy′​​=0，证明：对任给的常数CCC，f(x,y)=Cf(x,y)=Cf(x,y)=C为一条直线的充要条件是f2′2f11′′+2f1′f2′f12′′+f1′2f22′′=0f'^2_2f''_{11}+2f'_1f'_2f''_{12}+f'^2_1f''_{22}=0f2′2​f11′′​+2f1′​f2′​f12′′​+f1′2​f22′′​=0。

证设f(x,y)=Cf(x,y)=Cf(x,y)=C确定的隐函数为y=y(x)y=y(x)y=y(x)，等式f(x,y)=Cf(x,y)=Cf(x,y)=C两端对xxx求导得f1′+f2′dydx=0⇒dydx=−f1′f2′f'_1+f'_2\cfrac{\mathrm{d}y}{\mathrm{d}x}=0\Rightarrow\cfrac{\mathrm{d}y}{\mathrm{d}x}=-\cfrac{f'_1}{f_2'}f1′​+f2′​dxdy​=0⇒dxdy​=−f2′​f1′​​。从而有

d2ydx2=−ddx(f1′f2′)=−(f11′′+f12′′dydx)f2′−(f21′′+f22′′dydx)f1′f2′2=−f2′2f11′′+2f1′f2′f12′′+f1′2f22′′f2′3\begin{aligned} \cfrac{\mathrm{d}^2y}{\mathrm{d}x^2}&=-\cfrac{\mathrm{d}}{\mathrm{d}x}\left(\cfrac{f'_1}{f_2'}\right)\\ &=-\cfrac{\left(f''_{11}+f''_{12}\cfrac{\mathrm{d}y}{\mathrm{d}x}\right)f'_2-\left(f''_{21}+f''_{22}\cfrac{\mathrm{d}y}{\mathrm{d}x}\right)f'_1}{f'^2_2}\\ &=-\cfrac{f'^2_2f''_{11}+2f'_1f'_2f''_{12}+f'^2_1f''_{22}}{f'^3_2} \end{aligned} dx2d2y​​=−dxd​(f2′​f1′​​)=−f2′2​(f11′′​+f12′′​dxdy​)f2′​−(f21′′​+f22′′​dxdy​)f1′​​=−f2′3​f2′2​f11′′​+2f1′​f2′​f12′′​+f1′2​f22′′​​​

必要性：若f(x,y)=Cf(x,y)=Cf(x,y)=C是一条直线，则由f(x,y)=Cf(x,y)=Cf(x,y)=C所确定的函数y=y(x)y=y(x)y=y(x)应为线性函数（即y=ax+by=ax+by=ax+b），则d2ydx2=0\cfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=0dx2d2y​=0，从而有f2′2f11′′+2f1′f2′f12′′+f1′2f22′′=0f'^2_2f''_{11}+2f'_1f'_2f''_{12}+f'^2_1f''_{22}=0f2′2​f11′′​+2f1′​f2′​f12′′​+f1′2​f22′′​=0。

充分性：若f2′2f11′′+2f1′f2′f12′′+f1′2f22′′=0f'^2_2f''_{11}+2f'_1f'_2f''_{12}+f'^2_1f''_{22}=0f2′2​f11′′​+2f1′​f2′​f12′′​+f1′2​f22′′​=0，则d2ydx2=0\cfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=0dx2d2y​=0。从而有y=ax+by=ax+by=ax+b，即f(x,y)=Cf(x,y)=Cf(x,y)=C所确定的隐函数y=y(x)y=y(x)y=y(x)为线性函数。故f(x,y)=Cf(x,y)=Cf(x,y)=C表示直线。（这道题主要利用了直线函数特征求解）

5.3极值与最值

例49求中心在坐标原点的椭圆x2+4xy+5y2=1x^2+4xy+5y^2=1x2+4xy+5y2=1的长半轴与短半轴。

解椭圆x2+4xy+5y2=1x^2+4xy+5y^2=1x2+4xy+5y2=1上点(x,y)(x,y)(x,y)到原点(0,0)(0,0)(0,0)距离平方d2=f(x,y)=x2+y2d^2=f(x,y)=x^2+y^2d2=f(x,y)=x2+y2。问题归结为求f(x,y)=x2+y2f(x,y)=x^2+y^2f(x,y)=x2+y2在条件x2+4xy+5y2=1x^2+4xy+5y^2=1x2+4xy+5y2=1下的最大值和最小值。

令F(x,y,λ)=x2+y2+λ(x2+4xy+5y2−1)F(x,y,\lambda)=x^2+y^2+\lambda(x^2+4xy+5y^2-1)F(x,y,λ)=x2+y2+λ(x2+4xy+5y2−1)，则

{Fx′=2x+λ(2x−4y)=0,(1)Fy′=2y+λ(−4x+10y)=0,(2)Fλ′=x2+4xy+5y2−1,(3)\begin{cases} F_x'=2x+\lambda(2x-4y)=0,&\qquad(1)\\ F_y'=2y+\lambda(-4x+10y)=0,&\qquad(2)\\ F_\lambda'=x^2+4xy+5y^2-1,&\qquad(3) \end{cases} ⎩⎪⎨⎪⎧​Fx′​=2x+λ(2x−4y)=0,Fy′​=2y+λ(−4x+10y)=0,Fλ′​=x2+4xy+5y2−1,​(1)(2)(3)​

(1)(1)(1)式乘x2\cfrac{x}{2}2x​加(2)(2)(2)式乘y2\cfrac{y}{2}2y​得x2+y2+λ(x2+4xy+5y2)=0x^2+y^2+\lambda(x^2+4xy+5y^2)=0x2+y2+λ(x2+4xy+5y2)=0。则x2+y2+λ=0x^2+y^2+\lambda=0x2+y2+λ=0，即x2+y2=−λx^2+y^2=-\lambdax2+y2=−λ。

由(1)(1)(1)式和(2)(2)(2)式知{(1+λ)x−2λy=0,−2λ+(1+5λ)y=0,\begin{cases}(1+\lambda)x-2\lambda y=0,\\-2\lambda+(1+5\lambda)y=0,\end{cases}{(1+λ)x−2λy=0,−2λ+(1+5λ)y=0,​这是一个关于x,yx,yx,y的二元线性齐次方程组，由题意知它有非零解，则∣1+λ−2λ−2λ1+5λ∣=0\begin{vmatrix}1+\lambda&-2\lambda\\-2\lambda&1+5\lambda\end{vmatrix}=0∣∣∣∣​1+λ−2λ​−2λ1+5λ​∣∣∣∣​=0，即λ2+6λ+1=0\lambda^2+6\lambda+1=0λ2+6λ+1=0，得λ=−3±22\lambda=-3\pm2\sqrt{2}λ=−3±22​。

故长半轴a=3+22=1+2a=\sqrt{3+2\sqrt{2}}=1+\sqrt{2}a=3+22​​=1+2​，短半轴a=3−22=1−2a=\sqrt{3-2\sqrt{2}}=1-\sqrt{2}a=3−22​​=1−2​。（这道题主要利用了椭圆几何特点求解）

5.4方向导数与梯度 多元微分在几何上的应用 泰勒定理

例53函数z=x2+y2z=\sqrt{x^2+y^2}z=x2+y2​在点(0,0)(0,0)(0,0)处（）

(A)(A)(A)不连续；

(B)(B)(B)偏导数存在；

(C)(C)(C)沿任一方向方向导数存在；

(D)(D)(D)可微。

解由于lim⁡x→0y→0x2+y2=0=z(0,0)\lim\limits_{x\to0 \atop y\to0}\sqrt{x^2+y^2}=0=z(0,0)y→0x→0​lim​x2+y2​=0=z(0,0)，则在(0,0)(0,0)(0,0)连续，选项(A)(A)(A)不正确。

由于∂z∂x∣(0,0)=ddx(z(x,0))∣x=0=ddx(∣x∣)∣x=0\cfrac{\partial z}{\partial x}\biggm\vert_{(0,0)}=\cfrac{\mathrm{d}}{\mathrm{d}x}{(z(x,0))}\biggm\vert_{x=0}=\cfrac{\mathrm{d}}{\mathrm{d}x}{(|x|)}\biggm\vert_{x=0}∂x∂z​∣∣∣∣​(0,0)​=dxd​(z(x,0))∣∣∣∣​x=0​=dxd​(∣x∣)∣∣∣∣​x=0​，而∣x∣|x|∣x∣在x=0x=0x=0不可导，则∂z∂x∣(0,0)\cfrac{\partial z}{\partial x}\biggm\vert_{(0,0)}∂x∂z​∣∣∣∣​(0,0)​不存在。

设lll为从原点引出的射线，其方向余弦为cos⁡α,cos⁡β\cos\alpha,\cos\betacosα,cosβ，则

∂l∂x∣(0,0)=lim⁡t→0+f(tcos⁡α,tcos⁡β)−f(0,0)t=lim⁡t→0+t2cos⁡2α+t2cos⁡2β−0t=1.\begin{aligned} \cfrac{\partial l}{\partial x}\biggm\vert_{(0,0)}&=\lim\limits_{t\to0^+}\cfrac{f(t\cos\alpha,t\cos\beta)-f(0,0)}{t}\\ &=\lim\limits_{t\to0^+}\cfrac{\sqrt{t^2\cos^2\alpha+t^2\cos^2\beta}-0}{t}=1. \end{aligned} ∂x∂l​∣∣∣∣​(0,0)​​=t→0+lim​tf(tcosα,tcosβ)−f(0,0)​=t→0+lim​tt2cos2α+t2cos2β​−0​=1.​

这说明函数z=x2+y2z=\sqrt{x^2+y^2}z=x2+y2​在点(0,0)(0,0)(0,0)沿任一方向的方向导数都存在且为1，故应选(C)(C)(C)。（这道题主要利用了方向导数的定义求解）

例57设f(x,y)f(x,y)f(x,y)是R2\bold{R}^2R2（全平面）上的一个可微函数，且lim⁡ρ→+∞(x∂f∂x+y∂f∂y)=α>0\lim\limits_{\rho\to+\infty}\left(x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}\right)=\alpha>0ρ→+∞lim​(x∂x∂f​+y∂y∂f​)=α>0，其中ρ=x2+y2\rho=\sqrt{x^2+y^2}ρ=x2+y2​，α\alphaα为常数。试证明f(x,y)f(x,y)f(x,y)在R2\bold{R}^2R2上有最小值。

证因为lim⁡ρ→+∞(x∂f∂x+y∂f∂y)=α>0\lim\limits_{\rho\to+\infty}\left(x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}\right)=\alpha>0ρ→+∞lim​(x∂x∂f​+y∂y∂f​)=α>0，由极限保号性知，存在R>0R>0R>0，当ρ⩾R\rho\geqslant Rρ⩾R时（即x2+y2⩾R2x^2+y^2\geqslant R^2x2+y2⩾R2时）x∂f∂x+y∂f∂y>0x\cfrac{\partial f}{\partial x}+y\cfrac{\partial f}{\partial y}>0x∂x∂f​+y∂y∂f​>0。从而有xρ∂f∂x+yρ∂f∂y>0\cfrac{x}{\rho}\cfrac{\partial f}{\partial x}+\cfrac{y}{\rho}\cfrac{\partial f}{\partial y}>0ρx​∂x∂f​+ρy​∂y∂f​>0。若r\bm{r}r用表示点(x,y)(x,y)(x,y)处极径方向的方向导数，由(1)(1)(1)式知∂f∂r>0\cfrac{\partial f}{\partial r}>0∂r∂f​>0，则f(x,y)f(x,y)f(x,y)在区域x2+y2⩾R2x^2+y^2\geqslant R^2x2+y2⩾R2上任一点沿极径方向为增函数，由f(x,y)f(x,y)f(x,y)可微性知，f(x,y)f(x,y)f(x,y)连续，则f(x,y)f(x,y)f(x,y)在有界闭域x2+y2⩾R2x^2+y^2\geqslant R^2x2+y2⩾R2上有最小值，由于f(x,y)f(x,y)f(x,y)在x2+y2⩾R2x^2+y^2\geqslant R^2x2+y2⩾R2上任一点沿极径方向为增函数，则f(x,y)f(x,y)f(x,y)在x2+y2⩾R2x^2+y^2\geqslant R^2x2+y2⩾R2上的最小值也是在全平面上的最小值。（这道题主要利用了极限的保号性求解）

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