高専の数学解説

第7章 7.6

(1)

y2=4axy^2 = 4ax

F=4ax−y2F = 4ax - y^2 とおくと

Fx=4aFy=−2yF_x = 4a \qquad F_y = -2y

dydx=−4a−2y=2ay\frac{dy}{dx} = -\frac{4a}{-2y} = \frac{2a}{y}

d2ydx2=(2ay)x=(2a4ax)x=−8a22(4ax)3=−4a2y3\frac{d^2y}{dx^2} = \left(\frac{2a}{y}\right)_x = \left(\frac{2a}{\sqrt{4ax}}\right)_x = \frac{-8a^2}{2\sqrt{(4ax)^3}} = \frac{-4a^2}{y^3}

(2)

x2+y2−a2=0x^2 + y^2 - a^2 = 0

F=x2+y2−a2F = x^2 + y^2 - a^2 とおくと

Fx=2xFy=2yF_x = 2x \qquad F_y = 2y

dydx=−2x2y=−xy\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}

d2ydx2=(−xy)x=(−xa2−x2)x=−a2−x2−x⋅−xa2−x2(a2−x2)2=−a2−x2+x2(a2−x2)3=−a2y3\frac{d^2y}{dx^2} = \left(-\frac{x}{y}\right)_x = \left(-\frac{x}{\sqrt{a^2-x^2}}\right)_x = -\frac{\sqrt{a^2-x^2} - x \cdot \dfrac{-x}{\sqrt{a^2-x^2}}}{(\sqrt{a^2-x^2})^2} = -\frac{a^2-x^2+x^2}{(\sqrt{a^2-x^2})^3} = -\frac{a^2}{y^3}

(3)

x2−4xy+5y2=2x^2 - 4xy + 5y^2 = 2

F=x2−4xy+5y2−2F = x^2 - 4xy + 5y^2 - 2 とおくと

Fx=2x−4yFy=−4x+10yFxx=2Fxy=−4Fyy=10F_x = 2x-4y \qquad F_y = -4x+10y \qquad F_{xx} = 2 \qquad F_{xy} = -4 \qquad F_{yy} = 10

dydx=−(2x−4y)−4x+10y=x−2y2x−5y\frac{dy}{dx} = \frac{-(2x-4y)}{-4x+10y} = \frac{x-2y}{2x-5y}

d2ydx2=−FxxFy 2−2FxyFxFy+FyyFx 2Fy 3=2(−4x+10y)2+8(2x−4y)(−4x+10y)+10(2x−4y)2(4x−10y)3\frac{d^2y}{dx^2} = -\frac{F_{xx}F_y^{\,2} - 2F_{xy}F_xF_y + F_{yy}F_x^{\,2}}{F_y^{\,3}} = \frac{2(-4x+10y)^2 + 8(2x-4y)(-4x+10y) + 10(2x-4y)^2}{(4x-10y)^3}

=2(2x−5y)3= \frac{2}{(2x-5y)^3}

(4)

x+y=a\sqrt{x}+\sqrt{y}=\sqrt{a}

F=x+y−aF=\sqrt{x}+\sqrt{y}-\sqrt{a}

Fx=12xFy=12yFxx=−14x3Fxy=0Fyy=−14y3F_x=\frac{1}{2\sqrt{x}} \qquad F_y=\frac{1}{2\sqrt{y}} \qquad F_{xx}=-\frac{1}{4\sqrt{x^3}} \qquad F_{xy}=0 \qquad F_{yy}=-\frac{1}{4\sqrt{y^3}}

dydx=−yx\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}

d2ydx2=(−yx)x=(x−ax)x=a2x3\frac{d^2y}{dx^2}=\left(-\frac{\sqrt{y}}{\sqrt{x}}\right)_x=\left(\frac{\sqrt{x}-\sqrt{a}}{\sqrt{x}}\right)_x=\frac{\sqrt{a}}{2\sqrt{x^3}}

解説: r31bn1z