Space curves and torsion Consider smooth closed space curve (parametrised by its arc length) in 3 dimensional space.Does there exist atleast one point at which both the curvature & torsion attain extremum values?
 A: No.  Consider a curve on the unit $2$-sphere, parametrized by arc length $ds$ with geodesic curvature $\rho(s)$.  One easily computes that, as a curve in $3$-space, one has
$$
\kappa(s)^2  =  1 + \rho(s)^2
\qquad\text{and}\qquad
\tau(s) = \frac{\rho'(s)}{1+\rho(s)^2}.
$$
Thus, if what you were asking were true, then every closed curve on the $2$-sphere would have a point where both $\rho$ and $\rho'$ had critical points.  However, considering simple examples shows that this is not so.  
You can convince yourself that this isn't true in the plane by considering an ellipse, where $\rho$ has critical points only where $\rho'$ vanishes (and, at those $4$ points, i.e., the vertices, $\rho''$ is not zero).  Now consider the analog of the ellipse on the $2$-sphere, i.e., the curve cut out by $x^2+y^2+z^2=1$ and $x^2/a^2+y^2/b^2-z^2 = 0$ where $a>0$ and $b>0$ are very small.  This will have geodesic curvature on the $2$-sphere closely approximating that of the corresponding ellipse in the plane and will furnish an explicit counterexample.
A: Imagine a 2D curve defined as follows: $x(t) := \kappa(s), y(t) := \tau(s)$;
if it were true that a space curve always contains a point where curvature and torsion are simultaneously extremal, then the associated 2D curve (as defined above), would always contain a corner of its bounding box.  
Even if infinite curvatures and torsions are allowed, counterexamples can easily be constructed: take a space curve, whose associated 2D curve as defined previously, is an archimedian spiral.  
