The first oddity one runs across in the study of the differential geometry of curves is the requirement that such a curve be regular. We shall describe the basic idea behind this little word and its significance.
What is the meaning of this condition, and why have we chosen it? Recall that the point of using parametrized curves is that we want to think physically, we will pretend that the curve is the tracing out of some motion in space. In fact, some people refer to the image $\gamma(a,b)$ as the trace of the curve. In this interpretation, $\gamma'(t) = \frac{d\gamma}{dt}(t)$ is the velocity of the motion at time $t$. This is a vector quantity that tells us about the instantaneous motion: "which direction are we going and how fast?"
The detail of "how fast" is measured by the norm $||\gamma'||$ of the vector $\gamma'$. The physical interpretation is speed. The condition that a parametrized curve be regular is the condition that the motion modeled is not allowed to stop.
Again, it is not hard to see that this curve has a "California Stop" at $t=0$, which is the origin. But this is not some artifact of the parametrization--NO!, we are stuck with this madness.
How does one see this? Well, suppose that a curve has a regular parametrization near some point $P$, then we can approximate the curve well using a first order Taylor expasion:
\[ \gamma(t_0 + h) = \gamma(t_0) + h \cdot \gamma'(t_0) + o(h^2). \]
Geometrically, we are replacing the curve by its tangent line. But our cusped curve has no tangent line at the bad point. So, there is no way out. In effect, this motion has to stop, because the motion along the curve for $t<0$ is tending to exactly the $(-1,0)$ direction, but for $t>0$, the motion has to be going very nearly in the $(1,0)$ direction. The idealized particle taking this trip has to stop so it can turn around and go the opposite direction.
What is the meaning of this condition, and why have we chosen it? Recall that the point of using parametrized curves is that we want to think physically, we will pretend that the curve is the tracing out of some motion in space. In fact, some people refer to the image $\gamma(a,b)$ as the trace of the curve. In this interpretation, $\gamma'(t) = \frac{d\gamma}{dt}(t)$ is the velocity of the motion at time $t$. This is a vector quantity that tells us about the instantaneous motion: "which direction are we going and how fast?"
The detail of "how fast" is measured by the norm $||\gamma'||$ of the vector $\gamma'$. The physical interpretation is speed. The condition that a parametrized curve be regular is the condition that the motion modeled is not allowed to stop.
An Example to Avoid
Consider the following simple parametrized plane curve:
\[ \gamma(t) = \begin{pmatrix} 7 + 3 t^3 \\ -2 + t^3 \end{pmatrix} = P + t^3 Q, \]
where $P = (7,-2)$ and $Q=(3,1)$ are elements in $\mathbb{R}^2$. This traces out a line in the plane,
but in such a way that at $t=0$ the motion comes to a stop. To see this, we evaluate directly:
\[ \dfrac{d\gamma}{dt} = 3t^2 Q = \begin{pmatrix} 9 t^2 \\ 3 t^2 \end{pmatrix}, \]
which gives us
\[ || \gamma'(0) || = 0. \]
To get a better sense of this, let's plot the speed as a function of the parameter.
So, how does this parametrized curve trace the line? We come in from the lower left, first very fast, then slowing to a stop at $P$. But the stop is instantaneous. (Once, I heard my mom call this a "California stop" at an intersection.) Immediately the motion begins again, first slowly, but accelerating away from the point $P$. That is all very fine and interesting, but we don't like it. All of that motion business has nothing to do with the geometry of the line. It is an artifact of the parametrization that tells us nothing about the shape of the curve traced.
Better Parametrizations
So, we want a better parametrization. What will count as "better"?
To avoid all the problems above we pull the standard mathematician's gambit: make a definition to eliminate it.
Definition: A parametrized curve $\gamma:(a,b) \to \mathbb{R}^3$ is called regular when for all $t \in (a,b)$ we have $||\gamma'(t)|| \neq 0$.
See? Easy. From now on we will just consider regular parametrized curves. No stopping allowed.
A Cautionary Tale
Well, that is all very good if we are sure that regularity is only an artificial problem imposed by our importing of physics. (Wouldn't that be nice if we could just use a Python-style library import? "from Newton import mechanics" That would have made physics class easier.) But it is not the case that regularity only fails for silly reasons. Sometimes you are just stuck with it.
Our next example is the ordinary cuspidal cubic:
\[ y^2 - x^3 = 0.\]
This is a plane curve, and it is not hard to find a parametrization in this case:
\[ \gamma(t) = \begin{pmatrix} t^2 \\ t^3 \end{pmatrix} . \]
Again, it is not hard to see that this curve has a "California Stop" at $t=0$, which is the origin. But this is not some artifact of the parametrization--NO!, we are stuck with this madness.
How does one see this? Well, suppose that a curve has a regular parametrization near some point $P$, then we can approximate the curve well using a first order Taylor expasion:
\[ \gamma(t_0 + h) = \gamma(t_0) + h \cdot \gamma'(t_0) + o(h^2). \]
Geometrically, we are replacing the curve by its tangent line. But our cusped curve has no tangent line at the bad point. So, there is no way out. In effect, this motion has to stop, because the motion along the curve for $t<0$ is tending to exactly the $(-1,0)$ direction, but for $t>0$, the motion has to be going very nearly in the $(1,0)$ direction. The idealized particle taking this trip has to stop so it can turn around and go the opposite direction.
So what do we do?
We again appeal to the mathematician's gambit. A point which is not regular will be called singular and we will have to deal with those in a different manner.
In this case, we make a division between the "healthy" regular points, and the "pathological" singular points. Our differential geometry tools will serve us well for those portions of our curves which are regular. At singular points, we will have to call for help.
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