Does smoothness depend on parameterization?

Question

Consider an algebraic curve:

$$ \vec{r(t)} = t \vec{i} + t^2 \vec{j}$$

It has a tangent vector:

$$ \frac{d \vec{r}}{dt} = \vec{i} + 2 t \vec{j} \tag{1}$$

Now, consider parameterizing the first equation by sub $ t \to t^3$, then:

$$ \vec{r(t^3)} = t^3 \vec{i} + t^6 \vec{j}$$

Now the derivative of this is also (w/ chain rule considerations):

$$ \frac{dr}{dt}|_{t^3} = \vec{i} + 2t \vec{j} \tag{2}$$

This leads me to a weird conclusion, if I put t=2 then the tangent vector corresponding to that equation in (1) is $ \vec{i} + 2t \vec{j}$. However in the second parameterization, for t=2, he points on the curve is where $t=8$,* at that point it is also having the same tangent vector!! How can different points on the curve have the same tangent vector?**

Secondly, consider calling:

$$ \vec{q(t)} = \vec{r(t^3) } = t^3 \vec{i} + t^6 \vec{j}$$

Then, q is not smooth! because the $ \frac{dq}{dt}|_0 = 0$, how is it that when I reparamterizaed the curve using another function , it became not smooth? Aren't they both representing the same geometric object?

Answer 1

The behavior of a parametric curve depends not only on the set of points that the parametric curve passes through (the "geometric curve"), but also the chosen parameterization of that curve. Properties of the geometric curve can be defined in terms of the set of points only. But if a property requires the parameter to define, it is a property of the parametrization, not of the geometric curve.

It is possible to define geometric properties in terms of parameters. But such properties will not change if you change the parametrization. If a change of parametrization also changes the property, then that property is not geometric. It is a property of the parametrization.

The tangent vector to a curve is defined in terms of the parameter: $$\left.\dfrac{d\vec r}{dt}\right|_{t_0} := \lim_{t\to t_0} \dfrac{\vec r(t) - \vec r(t_0)}{t-t_0}$$

And a change of parametrization changes the tangent vector. If $t = t(u)$ and $\vec r(t(u)) = \vec q(u)$, then

$$\left.\dfrac{d\vec q}{du}\right|_{u_0} = \left.\dfrac{d\vec r}{dt}\right|_{t(u_0)}\left.\dfrac{dt}{du}\right|_{u_0}$$

Thus if $\frac{dt}{du} \ne 1$ (which is easily accomplished), you get a different vector for one parametrization than for the other.

What you can define without respect to the parametrization is the tangent line. Note in the equation above that while the magnitude of the tangent vector changes, the direction does not - except for possibly flipping to point in the anti-parallel direction. But both parallel and anti-parallel directions point along the same line. Thus the tangent line at the point is unchanged by the reparametrization. (Yes, you can reparametrize to make a $0$ tangent vector, or to make it undifferentiable at the point, but those just fail to properly define a tangent line, they don't change it.) The tangent vector is a parametric property, but the tangent line is a geometric property.

"Smoothness" is the existence of the tangent line, not the tangent vector, so it is also a geometric property.

Also, when you change the parametrization, you are changing which points are assigned to which values of the parameter. So to compare the values of two different parametrizations when the parameter is $2$ is nonsensical - at least as far as geometrical properties are concerned. These are at two diffent points on the geometric curve. They might have some properties in common, but that would only be by coincidence.

Finally, you are mistaken in thinking your $\vec r(t)$ and $\vec r(t^3)$ parametric curves have the same tangent vector at the two points. Though you claimed to have accounted for the chain rule, you did not do so correctly.