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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.

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

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

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.

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.

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

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

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

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$ 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

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$ 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) $\displaystyle{f(x,y) = \log{(1 - xy)}}$



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Next: Limit of Functions of Up: Partial Differentiation Previous: Partial Differentiation   Contents   Index