# On the implicit function theorem and the gradient.

I was following some MIT notes and came across this proof

I had a doubt about the existence of $r(t) = \langle x(t), y(t), z(t) \rangle$ a parametrization of a curve on the level surface.

Then I stumbled across this question that dealt with the same issue that seems to be resolved by the implicit function theorem, but I have not yet understood fully.

Does the implicit function theorem allow us to always find a parametrization of a curve passing through a point on the level surface of any given function? Or is it guaranteeing that a tangent to that curve exists?

I have read the statement of the implicit function theorem but I do not see where a parametrization comes into play.

• That title is awful. – Pedro Tamaroff Sep 5 '15 at 16:41
• Sorry :(, I will try an edit. – Monolite Sep 5 '15 at 16:46
• I mean the title in the MIT notes. – Pedro Tamaroff Sep 5 '15 at 22:47

One consequence of the implicit function theorem is that if $d f (p) \not = 0$, then near $p$ the zero set is locally a graph of a function of $R^2$. Then we can get this curves by applying that function to curves in $R^2$, which are easy to define.
Another way to think about this: $df(p) \not = 0$, means that there is a tangent space at $p$ , and then we can realize the a small piece of the level surface around $p$ as a "graph" of a function $F$ with domain the tangent plane at $p$ and range some real numbers representing the orthogonal distance from the tangent plane. (Try visualizing this for a sphere for instance, or on for the graph of $y = x^3$.)
From this point of view, there are clearly curves that have the property you want - just pick lines in the tangent space and look at their image under $F$. Every sufficiently small curve in the surface will arise this way - we have identified a piece of the tangent space with a piece of the surface using the map $F$.
• Thanks for your answer! what is intended by the zero set is locally a graph of a function of $R^2$? And when you say "this curve" what curve are you talking about? – Monolite Sep 5 '15 at 16:51