Real Function of Two Variables/Examples/Root of 1 minus y^2 over Root of 1 minus x^2

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Examples of Real Functions of Two Variables

Let $z$ denote the function defined as:

$z = \dfrac {\sqrt {1 - y^2} } {\sqrt {1 - x^2} }$

The domain of $z$ is:

$\Dom z = \openint {-1} 1 \times \closedint {-1} 1$


Proof

The domain of $z$ is given implicitly and conventionally.

What is meant is:

$z: S \to \R$ is the function defined on the largest possible subset $S$ of $\R^2$ such that:
$\forall \tuple {x, y} \in S: \map z {x, y} = \dfrac {\sqrt {1 - y^2} } {\sqrt {1 - x^2} }$


From Domain of Real Square Root Function, in order for the real square root function to be defined, its argument must be non-negative.

Hence for $z$ to be defined, it is necessary for:

\(\ds 1 - y^2\) \(\ge\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds y^2\) \(\le\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \size y\) \(\le\) \(\ds 1\)

Similarly:

$\size x \le 1$

But because $\sqrt {1 - x^2}$ is the denominator of the right hand side, it cannot be zero.


So it follows that for $z$ to be defined, it must be the case that:

$-1 \le y \le 1$

and:

$-1 < x < 1$

Hence the result.

$\blacksquare$


Sources