Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
The level curves give you a good idea of where the extrema of your function are... –  J. M. Jul 23 '12 at 14:13
    
Another name for that is the contour line , contour plot or Equipotential Curve: A curve in two dimensions on which the value of a function f(x,y) is a constant. Other synonymous terms are isarithm, isopleth, and contour line. A plot of several equipotential curves is called a contour plot. –  draks ... Jul 23 '12 at 14:42

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
-) In convex analysis there is a link between the convexity of a (smooth?) function and the convexity of its level sets. Further more there are links between the (sub-)gradients and the (sub-)level sets. For exaple the gradient of $f$ at $x$ is always perpendicular to the level set. -) In the evolution of surfaces (for example the boundaries of a fire) so-called Level-Set-Methods are used to model the surface as level set of the solution of a PDE. en.wikipedia.org/wiki/Level_set_method Pleas also recognize algebraic geometry as the science of $0$-levelsets (;-) ) and isolines on maps. –  vanguard2k Jul 23 '12 at 15:55
    
Thanks Ed Gorcenski. Can you elaborate mathematically especially, the first two points? –  Kumara Jul 24 '12 at 6:12
    
@vanguard2k: What is the link between the convexity of a (smooth?) function and the convexity of its level sets? Can you elaborate on this? When you say gradient of at is always perpendicular to the level set, do you mean $\nabla f(x)\cdot (x,y,k)=0$ if $(x,y,k)$ is a point on the level curve $k=f(x,y)$? –  Kumara Jul 24 '12 at 6:23

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 $x_0$: $\nabla f(x_0) \cdot \dot{g}(t_0)=0$.

share|improve this answer
    
Thank you vanguard2k. –  Kumara Jul 26 '12 at 4:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.