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Implicit Function Theorem
Given $u = f(x)$ for $f: {\bf R}^n \rightarrow {\bf R}$, then for each $c \in {\bf R}$ the level set relative to $c$ of $f$ is
In other point these are points with the same altitude.
Is it always a curve?
Could be a point - if this is a local maximum or minimum.
Example. Let
$$z = x^2 + y^2,$$
paraboloid. The $0$-level curve for $c = 0$ is a point $( 0, 0 )$.
What would be a level set that is not a curve, but that isn't a maximum or a minimum?
Answer: A saddle point.
Ex. Let
$$z = x^2 - y^2,$$
the $0$-level set is
What do these two examples have in common? They both are critical points, i.e.
$$f'( a, b ) = 0,$$
so that the tangent plane is horizontal.
Theorem (Implicit Function Theorem). Let
$$f: {\bf R}^2 \rightarrow {\bf R}, f(a) = c, a \in {\bf R}^2, c \in {\bf R}.$$
Suppose $f$ is differentiable on an open disk $D$ around $a$, and that $f'$ is continuous on $D$. Suppose further $f′(a) \neq 0$. Then the intersection of the $c$-level set with $D$ is a curve (i.e., it can be parametrized).
The name, Implicit Function Theorem, makes sense as an implicitly ($f(a) = c$) represented function can be represented explicitly.
Ex.
- Implicit: $x^2 + y^2 = 1,$
- Explicit:
$$\begin{array}{} y = g(x) &= \pm ( 1 - x^2 )^{\frac{1}{2}} \\ &= + ( 1 - x^2 )^{\frac{1}{2}} {\rm \hspace{3pt} if \hspace{3pt}} y > 0 \\ &= - ( 1 - x^2 )^{\frac{1}{2}} {\rm \hspace{3pt} if \hspace{3pt}} y < 0. \end{array}$$
How does the theorem apply here?
The theorem is easy to modify to get an analogue for parametric surfaces: If
$$f: {\bf R}^3 \rightarrow {\bf R}, f(a) = c, a \in {\bf R}^3, c \in {\bf R}$$
and $f$ is differentiable on an open ball $B$ around $a$, and $f'$ is continuous on $B$, and $f'(a) \neq 0$, then the intersection of the $c$-level set with $B$ is a surface.
Now, we have the theorem for curves in dimension $1$ and for surfaces in dimension $2$. The $n$ dim analogue is obvious.
