Real Function of Two Variables/Examples/Root of 1 minus y^2 over Root of 1 minus x^2
Jump to navigation
Jump to search
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
- 1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $2 \text C$: Function of Two Independent Variables: Example $2.62: 1$