6.8 解答

6.8

(a) とおき，の全微分を求めると，

$\displaystyle df = f_{x}dx + f_{y}dy = (4x + 5y)dx + (5x - 6y)dy = 0$

これより，

$\displaystyle \frac{dy}{dx} = -\frac{4x + 5y}{5x - 6y}.$

$\displaystyle \frac{d^2 y}{dx^2}$	$\displaystyle =$	$\displaystyle -\frac{(4 + 5\frac{dy}{dx})(5x - 6y) - (4x + 5y)(5 - 6\frac{dy}{dx})}{(5x - 6y)^2}$
	$\displaystyle =$	$\displaystyle -\frac{4(5x-6y)^2 - 10(4x+5y)(5x-6y) - 6(4x+5y)^2}{(5x - 6y)^3}$

(b)

$f(x,y) = y - e^{x+y} = 0$ とおき，の全微分を求めると，

$\displaystyle df = f_{x}dx + f_{y}dy = -e^{x+y}dx + (1 - e^{x+y})dy = 0$

これより，

$\displaystyle \frac{dy}{dx} = -\frac{e^{x+y}}{1 - e^{x+y}}.$

次に， $\frac{d^2 y}{dx^2}$ を求める．ここでは，

$\displaystyle \frac{d^2 y}{dx^2} = - \frac{f_{xx}f_{y}^2 - 2f_{xy}f_{x}f_{y} + f_{xx}f_{y}^2}{f_{y}^{3}}$

を用いる．

$\displaystyle f_{xx} = -e^{x+y}, f_{xy} = -e^{x+y}, f_{yy} = -e^{x+y}$

より

$\displaystyle \frac{d^2 y}{dx^2}$	$\displaystyle =$	$\displaystyle -\frac{-e^{x+y}(1-e^{x+y})^2 + 2e^{x+y}(-e^{x+y})(1 - e^{x+y}) - e^{x+y}e^{2(x+y)}}{(1 - e^{x+y})^3}$
	$\displaystyle =$	$\displaystyle -\frac{-e^{x+y} + 2e^{2(x+y)} - e^{3(x+y)} - 2e^{2(x+y)} + 2e^{3(x+y)} - e^{3(x+y)}}{(1 - e^{x+y})^3}$
	$\displaystyle =$	$\displaystyle \frac{e^{x+y}}{(1 - e^{x+y})^{3}}$

(c)

とおき，の全微分を求めると，

$\displaystyle df = f_{x}dx + f_{y}dy = (2x-y)dx + (-2y-x)dy = 0$

これより，

$\displaystyle \frac{dy}{dx} = \frac{2x - y}{2y+x}.$

次に， $\frac{d^2 y}{dx^2}$ を求める．ここでは，

$\displaystyle \frac{d^2 y}{dx^2} = - \frac{f_{xx}f_{y}^2 - 2f_{xy}f_{x}f_{y} + f_{xx}f_{y}^2}{f_{y}^{3}}$

を用いる．

$\displaystyle f_{xx} = 2, f_{xy} = -1, f_{yy} = -2$

より

$\displaystyle \frac{d^2 y}{dx^2}$	$\displaystyle =$	$\displaystyle -\frac{2(-2y-x)^2 + 2(2x-y)(-2y-x) -2(2x-y)^2}{(-2y-x)^3}$
	$\displaystyle =$	$\displaystyle \frac{2(x+2y)^2 -2(2x-y)(x+2y) -2(2x-y)^2}{(x+2y)^3}$

(d)

$f(x,y) = \log{\sqrt{x^2 + y^2}} - \tan^{-1}{\frac{y}{x}} = 0$ とおき，の全微分を求める.

$\displaystyle f_{x}$	$\displaystyle =$	$\displaystyle \frac{1}{2}(\frac{2x}{x^2 + y^2}) - \frac{-\frac{y}{x^2}}{1 + (\frac{y}{x})^2}$
	$\displaystyle =$	$\displaystyle \frac{x}{x^2 + y^2} + \frac{y}{x^2 + y^2} = \frac{x+y}{x^2 + y^2}$
$\displaystyle f_{y}$	$\displaystyle =$	$\displaystyle \frac{1}{2}(\frac{2y}{x^2 + y^2} - \frac{\frac{1}{x}}{1 + (\frac{y}{x})^2}$
	$\displaystyle =$	$\displaystyle \frac{y}{x^2 + y^2} - \frac{x}{x^2 + y^2} = \frac{y - x}{x^2 + y^2}$

より

$\displaystyle \frac{dy}{dx} = - \frac{f_{x}}{f_{y}} = - \frac{x+y}{y-x} = \frac{x+y}{x-y}$

次に， $\frac{d^2 y}{dx^2}$ を求める．ここでは，直接求める．

$\displaystyle \frac{d^2 y}{dx^2}$	$\displaystyle =$	$\displaystyle \frac{(1 + \frac{dy}{dx})(x-y) - (x+y)(1 - \frac{dy}{dx})}{(x-y)^2}$
	$\displaystyle =$	$\displaystyle \frac{(x - y + x + y)(x-y) - (x-y)(x-y-x-y) }{(x - y)^3}$
	$\displaystyle =$	$\displaystyle \frac{2x^2 - 2xy + 2xy + 2y^2}{(x-y)^3}$
	$\displaystyle =$	$\displaystyle \frac{2(x^2 + y^2)}{(x-y)^3}$

(a) とおき全微分を求めると，

$\displaystyle df$	$\displaystyle =$	$\displaystyle f_{x}dx + f_{y}dy + f_{z}dz = 2xdx + 2ydy + 2zdz = 0$	(7.5)
$\displaystyle dg$	$\displaystyle =$	$\displaystyle g_{x}dx + g_{y}dy = (2x -4)dx + 2y dy = 0$	(7.6)

式(7.6)より，

$\displaystyle \frac{dy}{dx} = \frac{4-2x}{2y} = \frac{2-x}{y}$

また，式(7.5)と式(7.6)より

の項を消去すると

$\displaystyle 4dx + 2zdz = 0$

よって

$\displaystyle \frac{dz}{dx} = -\frac{2}{z}.$

(b) とおき全微分を求めると，

$\displaystyle df$	$\displaystyle =$	$\displaystyle f_{x}dx + f_{y}dy + f_{z}dz = yzdx + xzdy + xydz = 0$	(7.7)
$\displaystyle dg$	$\displaystyle =$	$\displaystyle g_{x}dx + g_{y}dy + g_{z}dz = (y + z)dx + (x + z) dy + (y + x)dz = 0$	(7.8)

ここで，式(7.7)と式(7.8)より

の項を消去すると

$\displaystyle ((y+x)yz - xy(y+z))dx + ((y+x)xz - xy(x+z))dy = 0$

よって

$\displaystyle \frac{dy}{dx} = -\frac{y^2 z - xy^2}{x^2 z - x^2 y} = \frac{y^2 (x-z)}{x^2 (z-y)}.$

また，式(7.7)と式(7.8)より

の項を消去すると

$\displaystyle ((x+z)yz - xz(y+z))dx + ((x+z)xy - xz(y+x))dz = 0$

$\displaystyle z^2(y-x)dx + x^2(y-z)dz = 0$

よって

$\displaystyle \frac{dz}{dx} = \frac{z^2 (x-y)}{x^2 (y-z)}.$

より $\frac{dy}{dx}$ を求めると，

$\displaystyle \frac{dy}{dx} = -\frac{f_{x}}{f_{y}} = -\frac{4x}{10y} = -\frac{2x}{5y}$

これより，接線の傾きは

$\displaystyle \frac{dy}{dx}\mid_{(1,\sqrt{2})} = -\frac{2}{5\sqrt{2}} = -\frac{\sqrt{2}}{5}$

よって，点 $(1,\sqrt{2})$ を通る接線の方程式は

$\displaystyle y - \sqrt{2} = -\frac{\sqrt{2}}{5}(x - 1).$

また，法線は接線と垂直なので，その傾きは $-\frac{5\sqrt{2}}{2}$ ．よって点 $(1,\sqrt{2})$ を通る法線の方程式は

$\displaystyle y - \sqrt{2} = -\frac{5\sqrt{2}}{2}(x-1).$

$f(x,y,z) = \tan^{-1}{\frac{y}{x}} - z = 0$ とおくと

$\displaystyle \nabla f(x,y,z) = (\frac{\frac{-y}{x}}{1 + (\frac{y}{x})^2}, \fra... ...x}}{1 + (\frac{y}{x})^2}, -1) = (\frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}, -1)$