# Reparametrizing curves to make them smooth

I am trying to understand what it means for a curve to be smooth. Intuitively, it would seem that a curve with a sharp bend is not smooth, for example: $$t \mapsto \left[\begin{array}{c}t \\ |t| \end{array}\right] ~~ t \in [-1,1]$$ But, this can be reparametrized to be smooth: $$t \mapsto \left[\begin{array}{c}t^3 \\ -t^3 \end{array}\right]~~t \in[-1,0];~~~ t \mapsto \left[\begin{array}{c}t^3 \\ t^3 \end{array}\right]~~t \in[0,1]$$

Is this true in general or is this an instance of confusing the image of a curve with the curve.

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## 2 Answers

Given a curve $\gamma(t):\mathbb{R}\mapsto\mathbb{R}^n$, we wish to measure its differential structure separate from its parameterization. To do this we usually parameterize the curve by arclength before taking derivatives. Thus, the usual differential operator is $\dfrac{1}{\left|\dot{\gamma}(t)\right|}\dfrac{\mathrm{d}}{\mathrm{d}t}$. The simplest application of this operator is to the curve itself, yielding the unit tangent given by $\dfrac{\dot{\gamma}(t)}{\left|\dot{\gamma}(t)\right|}$. If the parameterization of the curve, at some point, has $\dot{\gamma}(t)=0$, you can see that even if $\gamma(t)$ is smooth, $\dfrac{\dot{\gamma}(t)}{\left|\dot{\gamma}(t)\right|}$ might swing around the unit sphere quite erratically for $t$ near $0$. So although your parameterization is a smooth function, since $\dot{\gamma}(0)=0$, the curve, that is the image of $\gamma(t)$, might not be smooth at $t=0$.

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One should not mix up the attributes ‘‘being smooth’’ with ‘‘possesses a tangent.’’ You just gave an example of a smooth function which doesn't posses a tangent line. –  Michael Hoppe Sep 22 '13 at 14:35
@MichaelHoppe: would this be a better comment on Student's question than on my answer? –  robjohn Sep 22 '13 at 16:09
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For a curve in two or three dimensional to have "no sharp bends" it has to be regular, not just smooth. That is, its derivative must not vanish.

In order to get an intuitive feeling for what it means to be a smooth curve, think of it as if you were in a car driving along the curve at the speed of the derivative. Would your car ride be "smooth"?

In your first example, the tangent vector is either $(1,-1)$ or $(1,1)$ which means it always has the same non-zero length, even at the sharp turn, which wouldn't be fun to drive through (you would feel an instant acceleration).

In the second example, the derivative is $(3t^2, -3t^2)$ for negative $t$ and $(3t^2,3t^2)$ for positive $t$, which means that your speed goes to zero as you approach the origin, so you'll come to a stop before doing the "sharp turn" and the ride will still be smooth.

As for your question about reparametrizations, I'm pretty sure that if the curve is piecewise smooth you can always reparametrize it to be smooth (but not regular). Otherwise, I don't know.

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Its not clear to me why a reparametrizion of this curve cannot exist such that the derivative at $t=0$ will not be non-zero vector and yet be smooth or at least $C^1$ –  Student Oct 15 '11 at 20:33
@Student: because for $t<0$, the limit as $t\to0^-$ of the unit tangent is $(1,-1)$, where as, for $t>0$, the limit as $t\to0^+$ is $(1,1)$. –  robjohn Oct 15 '11 at 21:12
@robjohn: So even though I might find a parametrization to get the tangent vector to vary continously with t near $0$, the unit tangent will always be some scalar times $(1,-1)$ for $t<0$ and $(1,1)$ for $t>0$. –  Student Oct 15 '11 at 21:55
@Student: yes, the unit vectors will always be $\frac{1}{\sqrt{2}}(1,-1)$ and $\frac{1}{\sqrt{2}}(1,1)$ no matter what parameterization you use (or maybe their opposites if you change the orientation of the parameterization). (I forgot the $\frac{1}{\sqrt{2}}$ in my previous comment.) –  robjohn Oct 15 '11 at 21:58
@robjohn: Thank you for the help! –  Student Oct 15 '11 at 22:02
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