1
$\begingroup$

Let $f:[0,2\pi]\rightarrow R^2$ be a smooth function such that $f([0,2\pi])$ is a smooth closed simple curve $C$. Suppose $(0,0)$ lies inside the the bounded open region enclosed by $C$ and $f(t)=(x(t),y(t))$. Is it true that $g(t)=x^2(t)+y^2(t)$ has at least 4 critical points in $[0,2\pi)$?

$\endgroup$

1 Answer 1

6
$\begingroup$

My original answer had a silly mistake in it.

Updated Answer: There exists a smooth closed simple curve, which encloses the origin, yet whose squared distance to the origin has only 2 critical points in $[0,2\pi)$.

Let $f(t)=(\frac{1}{2}+\cos{t}, \sin{t})$. This is simply the unit circle shifted to the right by $\frac{1}{2}$, so it is a smooth closed simple curve enclosing the origin. Then, $g(t) = \frac{5}{4} + \cos{t}$, which has critical points at integer multiples of $\pi$. So, $g(t)$ only has 2 critical points in $[0,2\pi)$. We can see this intuitively since the squared distance from $f(t)$ to the origin gets smaller as we go from the point $(\frac{3}{2}, 0)$ to the point $(-\frac{1}{2},0)$ and then increases until we go back to the point $(\frac{3}{2}, 0)$.

It's clear that there does not exist a smooth closed simple curve enclosing the origin where the squared distance to the origin has only one critical point.

$\endgroup$
3
  • $\begingroup$ Doesn't this function have infinite number of critical points? g'(t)=0 for all t in (0,1). $\endgroup$ Feb 18, 2015 at 6:51
  • $\begingroup$ My mistake. I was thinking about sign changes for some reason. $\endgroup$
    – Alec Payne
    Feb 18, 2015 at 7:13
  • 1
    $\begingroup$ Note that if the origin is chosen to be the center of mass of C or of the region enclosed by C, then g(t) has at least 4 critical points (see Domokos, Papadopoulos, Ruina, J. Elasticity 36 [1994], 59–66, link.springer.com/article/10.1007%2FBF00042491) $\endgroup$ Feb 18, 2015 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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