Does there exist a vector field tangent to a given curve? Let $\gamma : \mathbb{R} \to \mathbb{R}^2$ be an injective $C^2$ curve. Does there exist a $C^1$ vector field $X : \mathbb{R}^2 \to \mathbb{R}^2$ tangent to $\gamma$, ie. such that $\gamma'(t)= X(\gamma(t))$ for all $t \in \mathbb{R}$?
If such a vector field exists, it can be really useful to build counterexamples for ordinary differential equations: it would be sufficient to draw a (sufficiently regular) curve and to take such a $X$.
In fact, this problem can be viewed as an extension problem: Does $\left\{ \begin{array}{ccc} \gamma(\mathbb{R}) \subset \mathbb{R}^2 & \to & \mathbb{R}^2 \\ \gamma(t) & \mapsto & \gamma'(t) \end{array} \right.$ admit a $C^1$ extension on $\mathbb{R}^2$?
A partially answer is given by the following result:

Theorem: Let $M$ and $N$ be two subvarieties of $\mathbb{R}^m$ and $\mathbb{R}^n$ respectively, and $f : M \to N$ be a $C^k$ function. Then, there exists $U \subset \mathbb{R}^m$ an open set containing $M$ and a $C^k$ extension of $f$, $g : U \to \mathbb{R}^n$. Moreover, if $M$ is closed in $\mathbb{R}^n$, we can take $U= \mathbb{R}^n$.

So if $\gamma$ is $C^2$ and $\gamma'$ is an immersion, there exist an open set $U$ containing $\gamma(\mathbb{R})$ and a $C^1$ vector field $X : U \to \mathbb{R}^2$ tangent to $\gamma$. Moreover, if $\gamma([0,1])$ is closed, we can take $U= \mathbb{R}^2$.
Is it possible to improve the result? Is it possible to explicitely build $X$?
 A: You can almost always do this, as long as you do not object to the vector field needing to become $0$ at times.
One case where it becomes impossible is where $\gamma$ follows the $y$ axis along the segment with $-1 \leq y \leq 1,$ then $\gamma$ goes in a curved arc out to the positive $x$ axis, and returns along the curve $y = \sin \left( \frac{1}{x} \right)$ with $x > 0.$ The trouble then is that there are parts of the curve arbitrarily close to the $y$ axis where the tangent is pointing almost straight up, other parts of the curve arbitrarily close to the $y$ axis where the tangent is pointing almost straight down. So there is no continuous vector field extending $\gamma'$ in a neighborhood of that segment of the $y$ axis.
It can always be done if, along a short arc of $\gamma,$ a very narrow tubular neighborhood contains no other points of $\gamma.$ In that case you can extend $\gamma'$ orthogonally times a bump function, as Sammy Black points out. 
Note that, even in very pleasant situations, the extended vector field may be forced to become $0$ somewhere anyway. This is the case if $\gamma$ is the unit circle, travelling counterclockwise. Somewhere inside, any vector field that extends  $\gamma'$ becomes zero. Essentially Brouwer fixed point theorem: there is no continuous retraction of the unit disk onto the unit circle.   
