6.8 解答

6.8

1.

(a) $f(x,y) = 2x^2 + 5xy - 3y^2 - 1 = 0$とおき，$f(x,y)$の全微分を求めると，

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

これより，

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

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

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

(b)

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

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

これより，

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

次に， $\frac{d^2 y}{dx^2}$を求める．ここでは，

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

を用いる．

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

より

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

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

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

(c)

$f(x,y) = x^2 - y^2 - xy= 0$とおき，$f(x,y)$の全微分を求めると，

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

これより，

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

次に， $\frac{d^2 y}{dx^2}$を求める．ここでは，

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

を用いる．

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

より

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

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

(d)

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

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

$$= \frac{x}{x^2 + y^2} + \frac{y}{x^2 + y^2} = \frac{x+y}{x^2 + y^2}$$

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

$$= \frac{y}{x^2 + y^2} - \frac{x}{x^2 + y^2} = \frac{y - x}{x^2 + y^2}$$

より

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

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

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

$$= \frac{(x - y + x + y)(x-y) - (x-y)(x-y-x-y) }{(x - y)^3}$$

$$= \frac{2x^2 - 2xy + 2xy + 2y^2}{(x-y)^3}$$

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

2.

(a) $f(x,y,z) = x^2 + y^2 + z^2 - 4 = 0, g(x,y) = x^2 + y^2 - 4x = 0$とおき全微分を求めると，

$$df = f_{x}dx + f_{y}dy + f_{z}dz = 2xdx + 2ydy + 2zdz = 0$$ (10.5)

$$dg = g_{x}dx + g_{y}dy = (2x -4)dx + 2y dy = 0$$ (10.6)

式(10.6)より，

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

また，式(10.5)と式(10.6)より$dy$の項を消去すると

$$4dx + 2zdz = 0$$

よって

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

(b) $f(x,y,z) = xyz -1 = 0, g(x,y,z) = xy + yz + zx - 1 = 0$とおき全微分を求めると，

$$df = f_{x}dx + f_{y}dy + f_{z}dz = yzdx + xzdy + xydz = 0$$ (10.7)

$$dg = g_{x}dx + g_{y}dy + g_{z}dz = (y + z)dx + (x + z) dy + (y + x)dz = 0$$ (10.8)

ここで，式(10.7)と式(10.8)より$dz$の項を消去すると

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

よって

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

また，式(10.7)と式(10.8)より$dy$の項を消去すると

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

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

よって

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

3.

$2x^2 + 5y^2 = 12$より $\frac{dy}{dx}$を求めると，

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

これより，接線の傾きは

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

よって，点 $(1,\sqrt{2})$を通る接線の方程式は

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

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

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

4.

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

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

よって

$$\nabla f(1,1,\frac{\pi}{2}) = (-\frac{1}{2}, \frac{1}{2}, -1)$$

ここで，$\nabla f$は曲面 $f(x,y,z) = 0$に直交するので，接平面$\Gamma$上に任意の点$(x,y,z)$を取ると，ベクトル $(x - 1, y - 1, z - \frac{\pi}{2})$と $\nabla f(1,1,\frac{\pi}{2})$は直交する．よって，接平面の方程式は

$$(x - 1, y - 1, z - \frac{\pi}{2}) \cdot (-\frac{1}{2}, \frac{1}{2}, -1) = 0$$

又は，

$$-\frac{1}{2}(x-1) + \frac{1}{2}(y-1) - (z - \frac{\pi}{2}) = 0$$

法線は$\nabla f$と同方向にあるので，法線上に任意の点$(x,y,z)$を取ると，

$$(x - 1, y - 1, z - \frac{\pi}{2}) = t(-\frac{1}{2}, \frac{1}{2}, -1)$$

又は，

$$t = \frac{x-1}{-\frac{1}{2}} = \frac{y-1}{\frac{1}{2}} = \frac{z - \frac{\pi}{2}}{-1}$$

5.

(a) まず， $f(x,y) = 0, f_{x}(x,y) = 0$ を満たす $(x,y)$ を求めます．

$$f(x,y) = 8x^2 + 4xy + 5y^2 - 36 = 0, f_{x}(x,y) = 16x + 4y = 0$$

より $y = -4x$. これを $f(x,y) = 0$に代入すると $8x^2 - 16x^2 + 80x^2 - 36 = 36(2x^2 - 1) = 0$. よって， $(x,y) = (\frac{1}{\sqrt{2}},-2\sqrt{2}),(-\frac{1}{\sqrt{2}},2\sqrt{2})$．次に

$$f_{xx} = 16, f_{y} = 4x + 10y$$

より $${\frac{d^{2}y}{dx^{2}}}$$ を計算すると

$$\frac{d^{2}y}{dx^2}\mid_{(\frac{1}{\sqrt{2}},-2\sqrt{2})} = -\fra... ...\sqrt{2})}{f_{y}(\frac{1}{\sqrt{2}},-2\sqrt{2})} = -\frac{16}{-18\sqrt{2}} > 0$$

$$\frac{d^{2}y}{dx^2}\mid_{(-\frac{1}{\sqrt{2}},2\sqrt{2})} = -\fra... ...2\sqrt{2})}{f_{y}(-\frac{1}{\sqrt{2}},2\sqrt{2})} = -\frac{16}{18\sqrt{2}} < 0$$

したがって， $x = \frac{1}{\sqrt{2}}$ のときの $y = -2\sqrt{2}$ は極小値， $x = -\frac{1}{\sqrt{2}}$ のときの $y = 2\sqrt{2}$ は極大値.

(b) まず， $f(x,y) = 0, f_{x}(x,y) = 0$ を満たす $(x,y)$ を求めます．

$$f(x,y) = x^2 y + x + y = 0, f_{x}(x,y) = 2xy + 1= 0$$

より $y = -\frac{1}{2x}$．これを $f(x,y) = 0$に代入すると $x^2(-\frac{1}{2x}) + x - \frac{1}{2x} = \frac{-x^2 + 2x^2 - 1}{2x} = \frac{x^2 -1}{2x} = 0$. よって， $(x,y) = (1, -\frac{1}{2}),(-1, \frac{1}{2})$．次に

$$f_{xx} = 2y, f_{y} = x^2 + 1$$

より $${\frac{d^{2}y}{dx^{2}}}$$ を計算すると

$$\frac{d^{2}y}{dx^2}\mid_{(1, -\frac{1}{2})} = -\frac{f_{xx}(1, -\frac{1}{2})}{f_{y}(1, -\frac{1}{2})} = -\frac{-1}{2} > 0$$

$$\frac{d^{2}y}{dx^2}\mid_{(-1, \frac{1}{2})} = -\frac{f_{xx}(-1, \frac{1}{2})}{f_{y}(-1, \frac{1}{2})} = -\frac{1}{2} < 0$$

したがって， $x = 1$ のときの $y = -\frac{1}{2}$ は極小値， $x = -1$ のときの $y = \frac{1}{2}$ は極大値.

(c) まず， $f(x,y) = 0, f_{x}(x,y) = 0$ を満たす $(x,y)$ を求めます．

$$f(x,y) = x^3 + y^3 - 6xy = 0, f_{x}(x,y) = 3x^2 -6y= 0$$

より $y = \frac{x^2}{2}$．これを $f(x,y) = 0$に代入すると $x^3 + \frac{x^6}{8} - 3x^3 = \frac{x^3(x^3 - 16)}{8} = 0$. よって， $(x,y) = (0, 0),(2^{4/3}, 2^{5/3})$．次に

$$f_{xx} = 6x, f_{y} = 3x^2 -6x$$

より $(0,0)$で$f_{y} = 0$となるので，$(0,0)$では極値はとらない． $(2^{4/3}, 2^{5/3})$で $${\frac{d^{2}y}{dx^{2}}}$$ を計算すると

$$\frac{d^{2}y}{dx^2}\mid_{(2^{4/3}, 2^{5/3})} = -\frac{f_{xx}(2^{4... .../3}, 2^{5/3})} = -\frac{6 \cdot 2^{4/3}}{6 \cdot 2^{2/3} - 6\cdot 2^{5/3}} > 0$$

したがって， $x = 2^{4/3}$ のときの $y = 2^{5/3}$ は極小値