What do level curves signify?

Question

Suppose I have a function $z=f(x,y)$, say like $z=\sqrt{x^2+y^2}$. By fixing some value for $z$ and varying all possible $x$ and $y$, we would get a level curve of $z=f(x,y)$. By changing values for $z$, one can get different level curves. For $z=\sqrt{x^2+y^2}$, the level curves would be concentric circles. My question is, What do level curves signify? What all conclusions can be made about the function $z=f(x,y)$ just from the level curves?

Answer 1

In addition to the applied examples from the physical sciences that appear in @drak's comment, here are some mathematical applications of level curves.

You can infer all sorts of data from level curves, depending on your function. The spacing between level curves is a good way to estimate gradients: level curves that are close together represent areas of steeper descent/ascent.

If the function is a bivariate probability distribution, level curves can give you an estimate of variance.

If the function is a classification boundary in a data-mining application, level curves can define the classification boundary between inclusion and exclusion.

Level curves can show you boundaries of constant flux in some types of flow problems. Level curves can show you areas where temperature, stress, or concentrations are within some interval.

Finally, level curves are useful if your function is sufficiently complicated that it is difficult to visualize a 3-D rendering of the surface that it makes.

I am community wiki'ing this answer, because certainly other folks might have more to add.

Answer 2

Regarding your questions in the comments i will elaborate on the two facts you asked me about.

Let $L_c(f) = \{x\in \text{dom}f | f(x) \leq c \}$ be the $c$ level set of a convex function $f$. Then $L_c(f)$ is convex for all $c$.

Proof: For $x,y \in L_c(f)$ and $\lambda \in [0,1]$ we have from the convexity of $f$: $$f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y) \leq \lambda c + (1-\lambda) c = c.$$ So the convex combination is in $L_c(f)$.

Let $g:\Bbb{R}\rightarrow \Bbb{R}^n$ be a parametrisation of $L_c(f)$ around $x_0$ ($f(x_0)=c$) and $g(t_0) = x_0$. Then $\nabla f(x_0) \cdot \dot{g}(t_0)=0 $. ($\nabla f$ is orthogonal to tangent vectors of $L_c(f)$)

Proof: By looking at $f(g(t))$ you find (since $g$ is a parametrization of $L_c(f)$) we have $f(g(t))=c$ and therefore $\frac{d}{dt}f(g(t))=0$. From the chain rule we have $\nabla f(g(t)) \cdot \dot{g}(t)=0$. In particular this follows for the point $x_0$: $\nabla f(x_0) \cdot \dot{g}(t_0)=0$.