6.1 Answer

6.1

(a)

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

Let $x= 0 $ . Then $z = f(x,y) = -y^2$ . Thus, a parabola on $y-z$ plane． Let $y=0$ . Then $z = f(x,y) = x^2$ . Thus a parabola on $x-z$ plane． Let $z = c$ . Then $z = f(x,y) = x^2 - y^2 = c$ . Thus, a hyperbola on the plane $z = c$ .

(b)

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

Let $x= 0 $ . Then $z = f(x,y) = 0$ ． Let $y=0$ . Then $z = f(x,y) = 1$ . Thus a line on $x-z$ plane. Let $z = c$ . Then $z = f(x,y) = \frac{x^2}{x^2 + y^2} = c$ . Thus $x^2 = c(x^2 + y^2)$ ． $(1 - c)x^2 - y^2 = 0$ and $y = \pm \sqrt{(1-c)}x$

(c)

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

Let $z = c$ . Then $z = f(x,y) = \log(1 - xy) = c$ . Thus $1 - xy = e^{c}$ ． $y = \frac{1 - e^{c}}{x}$ ．Now changing the value of $c$ , draw a graph.