Implicit Function/Examples/x^2 + y^2 - r^2 = 0/Proof
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Example of Implicit Function
Consider the equation:
- $(1): \quad x^2 + y^2 - r^2 = 0$
where:
- $x, y \in \R$ are real variables
- $r \in \R_{>0}$ is a strictly positive real constant.
Then $(1)$ defines $y$ as an implicit function of $x$ on the closed interval $\closedint {-r} r$.
Proof
Solving for $y$, we obtain:
- $y = \pm \sqrt {r^2 - x^2}$
As it stands, $(1)$ does not define a real function, because:
- $\text{(a)}: \quad y$ is not defined for values of $x$ outside the range $-r \le x \le r$
- $\text{(b)}: \quad$ For all $x$ within the range $-r \le x \le r$, there are two possible values of $y$ that can be taken.
This can be corrected by:
- $\text{(a)}: \quad$ Specifying the domain to be the closed interval $\closedint {-r} r$ (or a subset thereof)
- $\text{(b)}: \quad$ Specifying which value of $y$ that is to be chosen, for example:
- $\forall x \in \closedint {-r} r: y = \sqrt {r^2 - x^2}$ (where $\sqrt {}$ in this context specifically means the positive square root)
- $\forall x \in \closedint {-r} r: y = -\sqrt {r^2 - x^2}$
- $\forall x \in \closedint {-r} r: y = \begin {cases} \sqrt {r^2 - x^2} & : x \le 0 \\ -\sqrt {r^2 - x^2} & : x > 0 \end {cases}$
Thus we have:
- $y = \map f {x, \map g x}$
where:
- $\map f {x, \map g x} = x^2 + \paren {\sqrt {r^2 - x^2} }^2 - r^2 = 0$
thereby demonstrating that $y$ is an implicit function of $x$.
$\blacksquare$
Sources
- 1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $2 \text D$: Implicit Function: Example $2.84$
- for the special case of $r = 5$