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.

Consider a set of points $P=\{p_1,p_2,\ldots,p_n\}$ in the plane. Define $x_1$ (resp. $x_2$) to be the minimal (resp. maximal) $x$ coordinate of the points in $P$.

Now, let $c(P)$ be any curve, with differentiable curvature, that passes through all points in $P$, and is restricted to the domain $[x_1, x_2]$. Note that the restriction ensures that $c(P)$ does not extrapolate the $p_i$'s. Also, define $\kappa(c(P))$ to be the maximal curvature of $c(P)$ in its domain.

Can the $|\kappa(c(P))|$ be arbitrarily small, or does it have a lower bound? Does this change if $c(P)$ has continuous, but not necessarily differentiable, curvature in its domain?

Plausible answer (Updated after Rahul's comment:) Suppose not all $p_i$'s lie in a straight line. It seems to me that $\kappa$ is bounded below by the radius of the minimal circle that encloses all points in $P$.I don't know enough differential geometry to prove or refute this. Note that differentiable curvature is key; otherwise we could draw straight lines segments between the points, where the curvature is undefined as we switch from one segment to another . Side note this question is a variant of my older question.

share|improve this question
    
What do you mean by "minimal circle"? If you mean minimal in terms of area, then that's not true when (for example) the points are all collinear; $\kappa$ is zero but the smallest circle that encloses the points has nonzero curvature. –  Rahul Apr 7 '12 at 1:38
    
Also, do you want to restrict the curve to be monotonic in $x$? If not, then the problem is a little less well-behaved: consider the points $(x_i,y_i)$ where $x_i = -10, -9, \ldots, 9, 10$, and $y_i$ is $1$ at $x_i = 0$ and $0$ elsewhere. You might expect the curve to have a curvature of at least $1$ to go through $(-1,0)$, $(0,1)$, and $(1,0)$, but instead it could go straight through $(-1,0)$ and $(0,1)$, loop back through a really wide arc, and then go straight through $(0,1)$ and $(1,0)$. –  Rahul Apr 7 '12 at 1:46
    
No, I don't want this constraint, for loops are fine –  Ganesh Apr 7 '12 at 1:47
    
Your edit doesn't help. Instead of lying on a straight line, they could just lie on a very gently curving arc of a really large circle. –  Rahul Apr 7 '12 at 1:47
    
Well, instead of addressing many super-special cases, I'll just say that my intuition could be wrong –  Ganesh Apr 7 '12 at 1:48
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.