Real Function of Two Variables/Examples/Root of x^2 + y^2 - 25
Examples of Real Functions of Two Variables
Let $z$ denote the function defined as:
- $z = \sqrt {x^2 + y^2 - 25}$
The domain of $z$ is:
- $\Dom z = C$
where $C$ consists of the set of points outside and on the circumference of the circle of radius $5$ whose center is at $\tuple {0, 0}$ in the Cartesian plane.
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:
- $\forall \tuple {x, y} \in S: \map z {x, y} = \sqrt {x^2 + y^2 - 25}$
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:
\(\ds x^2 + y^2 - 25\) | \(\ge\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2 + y^2\) | \(\ge\) | \(\ds 25\) |
From Equation of Circle center Origin, $x^2 + y^2 = 25$ is the equation for the circle of radius $5$ whose center is at $\tuple {0, 0}$ in the Cartesian plane.
Points inside this circle correspond are such that $x^2 + y^2 < 25$.
Hence the domain of $z$ is the set of points consisting of the exterior of that circle and the points on its circumference.
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 $2.62: 3$