Implicit Function/Examples/x^2 + y^2 - r^2 = 0

Example of Implicit Function

Consider the equation:

$(1): \quad x^2 + y^2 - r^2 = 0$

where:

$x, y \in \R$ are real variables

Then $(1)$ defines $y$ as an implicit function of $x$ on the closed interval $\closedint {-r} r$.

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 $x$.

$\blacksquare$

1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $2 \text D$: Implicit Function

for the special case of $r = 5$

for the special case of $r = 1$

for the special case of $r = 1$