Function of Several Variavles

Let

be a subset of ${\mathbb{R}}^2$ . Then for each point $(x,y)$

, there is a rule $f$

such that there exists a unique $z$

corresponding to $(x,y)$

. We call this rule $f$

a function of two variables and denote by $z = f(x,y)$

NOTE Let $D$ be a domian. $x$ and $y$ are called independent variable and $z$ is called dependent variable. A function $z = f(x,y)$ is called a function of two variables of $x,y$ . The domain of $f$ is the set of variables of $x,y$ for which $f(x,y)$ is also a real.

Domain
The domain of is given by $D(f) = \{(x,y) \in {\mathbb{R}}^{2} : f(x,y) \in {\mathbb{R}}\}$ .

Range
The range of is given by $R(f) = \{z \in {\mathbb{R}} : z = f(x,y), (x,y) \in D(f) \}$ .

Example 4..1 Find the domain of the following function

$\displaystyle z = f(x,y) = 1-(x^2 + y^2)$

Note that if $x$

is a real number, then $x^2$

is also a real number. Similarly, if $y$

is a real number, then $y^2$

is a real number. Furthermore, sum, difference, product, and quotient of non-zero real numbers are real numbers, Thus the domain of $z = f(x,y)$

-plane.

SOLUTION The domain of $f(x,y)$ is the set of real numbers $(x,y)$ such that $1 - (x^2 + y^2)$ is a real number. Thus

$\displaystyle D(f)$	$\displaystyle =$	$\displaystyle \{(x,y) \in {\mathbb{R}}^2 : 1 - (x^2 + y^2) \in {\mathbb{R}} \}$
	$\displaystyle =$	$\displaystyle \{(x,y) \in {\mathbb{R}}^2 \} = {\mathbb{R}}^2\ensuremath{ \blacksquare}$

Exercise 4..1 Find the domain of the following function.

$\displaystyle z = f(x,y) = \sqrt{1-(x^2 + y^2)}$

SOLUTION Note that $\sqrt{1-(x^2 + y^2)}$ is real if and only if $1 - (x^2 + y^2) \geq 0$ . Thus

$\displaystyle D(f)$	$\displaystyle =$	$\displaystyle \{(x,y) \in {\mathbb{R}}^2 : \sqrt{1 - (x^2 + y^2)} \in {\mathbb{R}} \}$
	$\displaystyle =$	$\displaystyle \{(x,y) \in {\mathbb{R}}^2 : x^2 + y^2 \leq 1 \}$

$\blacksquare$

For the function $z = f(x,y)$ , the set of points $(x,y,z)$ such that $\{(x,y,z): z = f(x,y)\}$ is called graph. Thus the graph of a function of two variables is a surface.

NOTE A surface is a collection of points. But it is not easy to draw a surface by plotting points. Then to draw the surface of a function of $z = f(x,y)$ , we use the following techniques. If we look at the surface of a function from the direction of $x$ -axis, then we can only see the curve on $yz$ -plane. If we look at the surface of a function from the direction of $y$ -axis, then we can only see the curve on $xz$ -plane. From these observation, we can draw the surface of a function by drawing the curve of a function $z = f(0,y)$ on the $yz$ -plane and the curve of a function $z = f(x,0)$ on $xz$ -plane. Finally, we let $z = c$ and draw the curve on the plane parallel to the $xy$ -plane.

Graph of Function
A graph of a function or two variables is given by $G(f) = \{(x,y,z) : z = f(x,y), (x,y) \in D(f) \}$ .

A line of intersection of a plane $z = c$ and $z = f(x,y)$ is called contour or level curve. NOTE Let $(x,y)$ be a position of object on the $xy$ -plane and be the atomospheric pressure at the point. Then $z = c$ is the plane whose atomosphric pressure is $c$ . Thus $f(x,y) = c$ is the level curve of the atomospheric pressure $c$ .

Example 4..2 Sketch the graph of the following function.

$\displaystyle z = f(x,y) = 1-(x^2 + y^2)$

**Figure 4.1:** Example4-2
$% latex2html id marker 24496 \includegraphics[width=4.0cm]{SOFTFIG-4/reidai4-2.eps}$

SOLUTION Let $x = 0$ to get $z = 1 - y^2$ . Then we have a parabola on $yz$ -plane. We next let $y = 0$ to get $z = 1 - x^2$ . Then we have a parabola on $xz$ -plane. Finally let $z = c$ to get $x^2 + y^2 = 1 - c$ . Thus we have a circle with the radius $\sqrt{1-c}$ on $z = c$ $\blacksquare$

Exercise 4..2 Sketch the graph of the following functions.
1. $\displaystyle{z = f(x,y) = \sqrt{1-(x^2 + y^2)}}$ 2. $\displaystyle{z = f(x,y) = x^2 - y^2}$ .

**Figure 4.2:** Exercise4-2-1
$\includegraphics[width=3.5cm]{SOFTFIG-4/kyumen.eps}$

**Figure 4.3:** Exercise4-2-2
$% latex2html id marker 24535 \includegraphics[width=3.5cm]{SOFTFIG-4/enshu4-2-2_gr1.eps}$

SOLUTION 1. We can sketch the graph of $z = f(x,y) = \sqrt{1 - (x^2 + y^2)}$ by squaring both sides to get $z^2 = 1 - (x^2 + y^2)$ . Then this is a sphere with the radius 1. $x^2 + y^2 + z^2 = 1$ . Since $z \geq 0$ , we have upper semisphere. $\blacksquare$
2. Let $x = 0$ . Then $z = f(0,y) = -y^2$ . Thus we have a concave up parabola on $yz$ -plane. Let $y = 0$ . Then $z = f(x,0) = - x^2$ . Thus we have concave down parabola on $xz$ -plane. Finally let $z = c$ . Then we have hyperbola $x^2 - y^2 = c$ . To sketch this, imagine the saddle on a horse back $\blacksquare$

We classify the surface represented by the following quadratic equation.

$\displaystyle Ax^{2} + By^{2} + Cz^{2} + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0.$

By suitable transformation, non degenerated equation can be clssified into 9 surfaces.

Ellipsoid $\displaystyle{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1}$
hyperboloid of one sheet $\displaystyle{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1}$
hyperboloid of two sheets $\displaystyle{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = -1}$
quadratic cone $\displaystyle{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = z^{2}}$
elliptic paraboloid $\displaystyle{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = z}$
hyperbolic paraboloid $\displaystyle{\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = z}$
parabolic cylinder $\displaystyle{x^{2} = 4cy}$
elliptic cylinder $\displaystyle{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1}$
hyperboli cylinder $\displaystyle{\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1}$

Degenerated Quadratic
Degenerate quadratic equation is a equation without any solution or only one solution. For example, $1 + x^{2} + y^{2} + z^{2} = 0$ or $x^{2} + y^{2} + z^{2} = 0$ .

Example 4..3 Classify the following surfaces.
1. $\displaystyle{x^{2} + 4y^{2} - 16z^{2} = 0}$
2. $\displaystyle{x^{2} + 4y^{2} + 16z^{2}- 12 = 0}$

SOLUTION 1. For $x = 0$ , we have $4y^2 = 16z^2$ . Then $\frac{y^2}{4} = z^2$ , $z = \pm \frac{y}{2}$ lines. Also for $y = 0$ , we have $z = \pm \frac{x}{4}$ lines. Thus, it is a quadratic cone $\blacksquare$

**Figure 4.4:** Example4-3-1
$\includegraphics[width=3.5cm]{SOFTFIG-4/cone_gr3.eps}$

2. For $x = 0$ , we have $4y^{2} + 16z^{2}- 12 = 0$ . Then $\frac{y^2}{3} + \frac{z^2}{\frac{3}{4}} = 1$ , ellipse. Also for $y = 0$ , we have $x^2 + 16z^2 = 12$ . Thus $\frac{x^2}{12} + \frac{y^2}{\frac{3}{4}} = 1$ ellipse. Thus $\blacksquare$ , it is a ellipsoid

**Figure 4.5:** Example4-3-2
$\includegraphics[width=3.5cm]{SOFTFIG-4/ellipsoid_gr3.eps}$

Exercise 4..3 Classify the following surfaces.
1. $\displaystyle{x - 4y^{2} = 0}$ 2. $\displaystyle{x^{2} - 4y^{2} - 2z = 0}$

SOLUTION Since $x - 4y^2 = 0$ , the value of $z$ can be arbitrary. Thus we have a parabola for all $z$ . Thus it is a parabolic cylinder $\blacksquare$

**Figure 4.6:** Exercise4-3-1
$% latex2html id marker 24653 \includegraphics[width=3.5cm]{SOFTFIG-4/enshu4-3-1_gr3.eps}$

2. Since $x^2 - 4y^2 = 2z$ , for $x = 0$ , we have $z = -2y^2$ a parabola. for $y = 0$ , we have $z = \frac{x^2}{2}$ a parabola. Also for $z = c$ , we have $x^2 - 4y^2 = c$ a hyperbola. Thus it is a hyperbolic paraboloid $\blacksquare$

Exercise A

1.

Find the domain and the region of the following functions．

(a) $\displaystyle{f(x,y) = \sqrt{xy}}$ (b) $\displaystyle{f(x,y) = \frac{1}{x+y}}$ (c) $\displaystyle{f(x,y) = \frac{1}{x^{2} + y^{2}}}$

(d) $\displaystyle{f(x,y) = \frac{x^{2}}{x^{2} + y^{2}}}$ (e) $\displaystyle{f(x,y) = \log(1-xy)}$ (f) $\displaystyle{f(x,y,z) = \frac{z}{x^{2} - y^{2}}}$

2.

Classify the following surfaces．

(a) $\displaystyle{x^{2} + 4y^{2} - 16z^{2} = 0}$ (b) $\displaystyle{x^{2} + 4y^{2} + 16z^{2}- 12 = 0}$ (c) $\displaystyle{x - 4y^{2} = 0}$

(d) $\displaystyle{x^{2} - 4y^{2} - 2z = 0}$ (e) $\displaystyle{2x^{2} + 4y^{2} - 1 = 0}$ (f) $\displaystyle{x^{2} + 4y^{2} - 4z = 0}$

(g) $\displaystyle{2x^{2} - 4y^{2} - 6 = 0}$ (h) $\displaystyle{x^{2} + y^{2} - 2z^{2} -10 = 0}$ (i) $\displaystyle{x^{2} + y^{2} - 2z^{2} +10 = 0}$

Exercise B

1.: Find the domain of the region of the following functions and draw the domain．
(a) $\displaystyle{f(x,y) = x^2 - y^2}$ (b) $\displaystyle{f(x,y) = \frac{x^2}{x^2 + y^2}}$ (c) $\displaystyle{f(x,y) = \log{(1 - xy)}}$