第6章 6.18

(1)

f(x,y)=log⁡(x2+y2)f_{(x,y)} = \log(x^2+y^2)

fx=2xx2+y2fy=2yx2+y2fxx=2(x2+y2)−4x2(x2+y2)2fyy=2(x2+y2)−4y2(x2+y2)2f_x = \frac{2x}{x^2+y^2} \qquad f_y = \frac{2y}{x^2+y^2} \qquad f_{xx} = \frac{2(x^2+y^2)-4x^2}{(x^2+y^2)^2} \qquad f_{yy} = \frac{2(x^2+y^2)-4y^2}{(x^2+y^2)^2}

Δf=−2x2+2y2+2x2−2y2(x2+y2)2=0\Delta f = \frac{-2x^2+2y^2+2x^2-2y^2}{(x^2+y^2)^2} = 0

(2)

f(x,y)=tan⁡−1yxf_{(x,y)} = \tan^{-1}\dfrac{y}{x}

fx=−yx2+y2fy=xx2+y2fxx=2xy(x2+y2)2fyy=−2xy(x2+y2)2f_x = \frac{-y}{x^2+y^2} \qquad f_y = \frac{x}{x^2+y^2} \qquad f_{xx} = \frac{2xy}{(x^2+y^2)^2} \qquad f_{yy} = \frac{-2xy}{(x^2+y^2)^2}

Δf=2xy−2xy(x2+y2)2=0\Delta f = \frac{2xy-2xy}{(x^2+y^2)^2} = 0

(3)

f(x,y)=eax(cos⁡by+sin⁡by)f_{(x,y)} = e^{ax}(\cos by + \sin by)

fx=aeax(cos⁡by+sin⁡by)fy=beax(−sin⁡by+cos⁡by)f_x = ae^{ax}(\cos by+\sin by) \qquad f_y = be^{ax}(-\sin by+\cos by)

fxx=a2eax(cos⁡by+sin⁡by)fyy=−b2eax(cos⁡by+sin⁡by)f_{xx} = a^2 e^{ax}(\cos by+\sin by) \qquad f_{yy} = -b^2 e^{ax}(\cos by+\sin by)

Δf=(a2−b2)eax(cos⁡by+sin⁡by)\Delta f = (a^2-b^2)e^{ax}(\cos by+\sin by)

(4)

f(x,y,z)=1x2+y2+z2f_{(x,y,z)} = \dfrac{1}{\sqrt{x^2+y^2+z^2}}

fx=−x(x2+y2+z2)3fy=−y(x2+y2+z2)3fz=−z(x2+y2+z2)3f_x = -\frac{x}{\sqrt{(x^2+y^2+z^2)^3}} \qquad f_y = -\frac{y}{\sqrt{(x^2+y^2+z^2)^3}} \qquad f_z = -\frac{z}{\sqrt{(x^2+y^2+z^2)^3}}

fxx=−(x2+y2+z2)3−x⋅32⋅2xx2+y2+z2(x2+y2+z2)3=−x2+y2+z2 (−2x2+y2+z2)(x2+y2+z2)3f_{xx} = -\frac{\sqrt{(x^2+y^2+z^2)^3} - x\cdot\dfrac{3}{2}\cdot 2x\sqrt{x^2+y^2+z^2}}{(x^2+y^2+z^2)^3} = -\frac{\sqrt{x^2+y^2+z^2}\,(-2x^2+y^2+z^2)}{(x^2+y^2+z^2)^3}

同様に考えて

fyy=−x2+y2+z2 (x2−2y2+z2)(x2+y2+z2)3f_{yy} = -\frac{\sqrt{x^2+y^2+z^2}\,(x^2-2y^2+z^2)}{(x^2+y^2+z^2)^3}

fzz=−x2+y2+z2 (x2+y2−2z2)(x2+y2+z2)3f_{zz} = -\frac{\sqrt{x^2+y^2+z^2}\,(x^2+y^2-2z^2)}{(x^2+y^2+z^2)^3}

Δf=x2+y2+z2 {(−2x2+y2+z2)+(x2−2y2+z2)+(x2+y2−2z2)}(x2+y2+z2)3=0\Delta f = \frac{\sqrt{x^2+y^2+z^2}\,\{(-2x^2+y^2+z^2)+(x^2-2y^2+z^2)+(x^2+y^2-2z^2)\}}{(x^2+y^2+z^2)^3} = 0

解説: r31bn1z