Function of Several Variavles

Let $D$ be a subset of ${\mathbb{R}}^2$. Then for each point $(x,y)$ in $D$, 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 $z = f(x,y)$is given by $D(f) = \{(x,y) \in {\mathbb{R}}^{2} : f(x,y) \in {\mathbb{R}}\}$.

Range

The range of $z = f(x,y)$ 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

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

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 $z = f(x,y)$ is $xy$-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

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

$$= \{(x,y) \in {\mathbb{R}}^2 \} = {\mathbb{R}}^2\ensuremath{ \blacksquare}$$

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

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

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

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

$$= \{(x,y) \in {\mathbb{R}}^2 : x^2 + y^2 \leq 1 \}$$

$\blacksquare$

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 $z = f(x,y)$ 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.

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

Figure 4.1: Example4-2

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

Figure 4.2: Exercise4-2-1

Figure 4.3: Exercise4-2-2

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.

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

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

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

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. $${x^{2} + 4y^{2} - 16z^{2} = 0}$$
2. $${x^{2} + 4y^{2} + 16z^{2}- 12 = 0}$$

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

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

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

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

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

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

2.

Classify the following surfaces．

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

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

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

Exercise B

1.

Find the domain of the region of the following functions and draw the domain．

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