4.1 Functions of several variables  (Page 4/12)

First set $x=-\frac{\pi }{4}$ in the equation $z=\text{sin}x\text{cos}y\text{:}$

This describes a cosine graph in the plane $x=-\frac{\pi }{4}.$ The other values of $z$ appear in the following table.

In a similar fashion, we can substitute the $y\text{-values}$ in the equation $f(x,y)$ to obtain the traces in the $yz\text{-plane,}$ as listed in the following table.

The three traces in the $xz\text{-plane}$ are cosine functions; the three traces in the $yz\text{-plane}$ are sine functions. These curves appear in the intersections of the surface with the planes $x=-\frac{\pi }{4},x=0,x=\frac{\pi }{4}$ and $y=-\frac{\pi }{4},y=0,y=\frac{\pi }{4}$ as shown in the following figure.

1. For the function $f\left(x,y,z\right)=\frac{3x-4y+2z}{\sqrt{9-x^2-y^2-z^2}}$ to be defined (and be a real value), two conditions must hold:
   1. The denominator cannot be zero.
   2. The radicand cannot be negative.
   Combining these conditions leads to the inequality
   
   $9-x^2-y^2-z^2>0.$
   
   Moving the variables to the other side and reversing the inequality gives the domain as
   
   $\text{domain}\left(f\right)=\left\{\left(x,y,z\right)\in {ℝ}^3|x^2+y^2+z^2<9\right\},$
   
   which describes a ball of radius $3$ centered at the origin. ( Note : The surface of the ball is not included in this domain.)
2. For the function $g\left(x,y,t\right)=\frac{\sqrt{2t-4}}{x^2-y^2}$ to be defined (and be a real value), two conditions must hold:
   1. The radicand cannot be negative.
   2. The denominator cannot be zero.
   Since the radicand cannot be negative, this implies $2t-4\ge 0,$ and therefore that $t\ge 2.$ Since the denominator cannot be zero, $x^2-y^2\ne 0,$ or $x^2\ne y^2,$ Which can be rewritten as $y=\text{±}x,$ which are the equations of two lines passing through the origin. Therefore, the domain of $g$ is
   
   $\text{domain}\left(g\right)=\{\left(x,y,t\right)|y\ne \text{±}x,t\ge 2\}.$