Line with a knot in $\mathbb R^3$ isotopic to standard embedding $\mathbb R\subset \mathbb R^3$?

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!

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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.
OK, that's what I would have called a smooth isotopy. So maybe one has to play with $e^{-\frac1t}$ or the like here? – Hagen von Eitzen Oct 18 '12 at 6:01