6.1 解答

6.1

1.

$ f(x,y)$の定義域とは関数$ f(x,y)$が実数の値をとる$ (x,y)$の範囲のことである.

2変数関数のグラフは正面図($ x = 0$),側面図($ y=0$),等位面($ z = c$)を用いて描く. (a)

$\displaystyle D(f)$ $\displaystyle =$ $\displaystyle \{(x,y) : f(x,y) \in {\cal R}\}$  
  $\displaystyle =$ $\displaystyle \{(x,y) : x^2 - y^2 \in {\cal R}\} = {\cal R}^2$  

$ x = 0$とおくと, $ z = f(x,y) = -y^2$より$ y-z$平面に放物線. $ y=0$とおくと, $ z = f(x,y) = x^2$より$ x-z$平面に放物線. $ z = c$とおくと, $ z = f(x,y) = x^2 - y^2 = c$より,等位面$ z = c$に双曲線となる.これを用いて図を描く.

(b)


$\displaystyle D(f)$ $\displaystyle =$ $\displaystyle \{(x,y) : f(x,y) \in {\cal R}\}$  
  $\displaystyle =$ $\displaystyle \{(x,y) : \frac{x^2}{x^2 + y^2} \in {\cal R}\} = {\cal R}^2 - (0,0)$  

$ x = 0$とおくと, $ z = f(x,y) = 0$$ y=0$とおくと, $ z = f(x,y) = 1$より$ x-z$平面に直線. $ z = c$とおくと, $ z = f(x,y) = \frac{x^2}{x^2 + y^2} = c$より, $ x^2 = c(x^2 + y^2)$ $ (1 - c)x^2 - y^2 = 0$より $ y = \pm \sqrt{(1-c)}x$

(c)


$\displaystyle D(f)$ $\displaystyle =$ $\displaystyle \{(x,y) : f(x,y) \in {\cal R}\}$  
  $\displaystyle =$ $\displaystyle \{(x,y) : \log(1 - xy) \in {\cal R}\}$  
  $\displaystyle =$ $\displaystyle \{(x,y) : xy < 1\}$  

$ z = c$とおくと, $ z = f(x,y) = \log(1 - xy) = c$より, $ 1 - xy = e^{c}$ $ y = \frac{1 - e^{c}}{x}$.ここで,$ c$の値を変化させながらグラフを描く