6.1 Answer

6.1

1.

(a)

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

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

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)

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

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

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)

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

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

$$= \{(x,y) : xy < 1\}$$

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.