I have come across the following exercise in Kosinski's 'Differential Manifolds':
Exercise: Consider an imbedding $\mathbb R\to \mathbb R^3$ where the image is "the line with a knot":
Show that this imbedding is isotopic to the standard imbedding $\mathbb R\subset \mathbb R^3$.
But I cannot see what such an isotopy could possibly do to unknot this knot? The only thing I can think of is pulling the knot at both ends to tie it completely tight at the origin. But then I would necessarily cause the second derivative to explode, wouldn't I? What else is there to do? Or am I possibly wrong in thinking that the pulling gives me trouble with higher derivatives?
Thanks for your help!
Isotopy requires in my understanding only homeomorphisms, not diffeomorphisms, hence your concern about exploding derivatives can be put aside.
Thus asume that the knot embedding $f\colon\mathbb R\to\mathbb R^3$ is identical to the standard embedding for $x\notin(-1,1)$. Letting $$F(t,x)=\begin{cases}(x,0,0)&\text{if }|x|\ge t\\tf(\frac xt)&\text{if }|x|<t\end{cases}$$ this gives an isotopy with $F(1,x)=f(x)$, $F(0,x)=(x,0,0)$, as one can check. (And this is exactly what your idea was).
For a smooth $F$, as seems to be required by Kosinsky, one can just push the knot out to infinity, i.e. (again assumin $f(x)=(x,0,0)$ for $x\notin(-1,1)$) let $$F(t,x)=\begin{cases}(x,0,0)&\text{if }t=0\\f(x+\frac1t-1)+1-\frac1t&\text{if }0<t\le 1\end{cases}$$ This is obviously smooth at $(t,x)$ with $t\ne0$ and is just $(t,x)\mapsto(x,0,0)$ in a neighbourhood of points at the $t=0$ boundary.