# Is it possible to visit all points on a differentiable function by “rolling”?

I recall from a discussion thread some week ago we talked about different ways to pedagogically explain differentiability.

So I came up with this idea that if there for each point there exists a circular "wheel" of some radius $0 < \epsilon_R < \infty$ ( that there exists such an $\epsilon_R$ ), so that circle's center is "above" the function, the wheel lays tangent to the curve and "rolls" along the curve without crossing it (while possibly having to change radius once in a while).

Would this be possible for all differentiable functions?

Own work:

A step discontinous function (which obviously is not differentiable) does not qualify as no matter how small we make $\epsilon_R$, we won't reach every point on each side of the step discontinuity ( and also the circle would need "jumping" which is not allowed ).

• I'm not sure what exactly do you mean by "rolling along curve without crossing it". But what about $f(x) = |x|^{1+\varepsilon}$ at point $x=0$ (with $0 < \varepsilon < 1$)? It is differentiable everywhere, but it grows faster than $x^2$ for near-zero $x$. – stgatilov Dec 17 '15 at 18:44
• Yes this is an interesting example you bring up. I have to think about it for a bit. – mathreadler Dec 17 '15 at 19:48