$\exists$ a continuous differentiable map from $[0,1]$ to $U$ such that $\gamma(0)=p,\gamma(1)=q$ Let $U=\{(x,y)\in\Bbb{R}^2~|~1<x^2+y^2<4\}$. Let $p,q\in U$. Show that there is a continuous map $\gamma:[0,1]\to U$ such that $\gamma(0)=p,\gamma(1)=q$ and such that $\gamma$ is differentiable over $(0,1)$.
The existence of a continuous map is trivial because the given set $U$ is path connected. But how do I find a map that is also differentiable.
 A: You can explicitly construct it here: $p=(r\cos(\alpha),r\sin(\alpha))$ and $q=(s\cos(\beta),s\sin(\beta))$ and thus by linear interpolation for both the radius and angle:
$$\gamma(t)=((1-t)r+ts)(\cos((1-t)\alpha+t\beta),\sin((1-t)\alpha+t\beta))$$
which is obviously differentiable and lies within $U$, since $1<((1-t)r+ts)<2$ for $t\in[0,1]$.
A: Presumably, you showed path-connectedness using polylines piecewise linear curves). 
For open $U\subseteq \Bbb R^n$, you can always obtain an approximating smooth path from a polyline $\gamma\colon[0,1]\to U$.
First, extend $\gamma$ symmetrically beyond $0$ and $1$ to $[-1,2]$.
$$\hat\gamma(t):=\begin{cases}\gamma(t)&0\le t\le 1\\
2\gamma(1)-\gamma(t)&1\le t\le 2\\
2\gamma(0)-\gamma(t)&-1\le t\le 0\end{cases} $$
Then pick $h>0$ and define $\tilde\gamma\colon[0,1]\to\Bbb R^n$,
$$\tilde\gamma(t)=\frac1{2h}\int_{-h}^h\hat\gamma(t+u)\,\mathrm du=\frac1{2h}\int_{t-h}^{t+h}\hat\gamma(u)\,\mathrm du. $$
We automatically have $\tilde\gamma(0)=\gamma(0)$,  $\tilde\gamma(1)=\gamma(1)$, $\tilde\gamma '(t)=\frac1{2h}(\hat\gamma(t+h)-\hat\gamma(t-h))$, i.e., $\tilde\gamma$ is one level smoother that $\gamma$. Remains to show that we can achieve that $\tilde \gamma$ is in $U$ by making $h$ small enough. Use compactness of the path and openness of $U$
