A couple of clarifications about curvature, T(t), N(t)

Question

Suppose $f: \mathbb{R} \rightarrow \mathbb{R^3}$ is for example 3 times differentiable, not necessarily smooth function (smooth meaning $f'(t)$ exists and $f'(t) \ne \overrightarrow{0}$ for each $t$).

Here $f$ is not a path-length parametrization.

Suppose also that

$T(t)$ - the unit tangent at $t$
$N(t)$ - the principal normal at $t$
$\kappa(t)$ - the curvature at $t$

Is it correct that:

1. when $f'(a) = \overrightarrow{0}$
   then $N(a)$ and $\kappa(a)$ are not defined?

2. when $f'(a) \ne \overrightarrow{0}$
   then $N(a)$ and $\kappa(a)$ are defined?

3. the image of $f$ (which is the curve $C$ in $\mathbb{R^3}$) has "a corner" at the point $t = a$, if and only if $f'(a) = 0$? Note: What do I mean by "a corner"? Well, I am not sure of the exact definition but e.g. the image of the function $g(t) = (t^2, t^3, 0)$ has a corner at $t = 0$.

So are the above statements true or not?

I am asking because of this formula.

I got confused by chapter 2.8 of this book
https://www.amazon.com/Vector-Calculus-Dover-Books-Mathematics/dp/0486466205/

Somehow the authors first introduce curvature for path-length parametrizations and this curvature is related to the 2nd derivative. But when they move on to any parametrization the theory somehow becomes fuzzy to me.

And then that formula above (for the curvature) comes up in the exercises for an arbitrary (not necessarily path-length) parametrization, and I just got confused by this. I mean $f'$ is in the denominator so... for this formula to be true it must be non-zero. OK, but what do we do when $f'(t)=\overrightarrow{0}$ ?! Can we have there (at that $t$) a principal normal or a curvature, and if so, how could we calculate them?

Side note: here is another quote from the same book (page 50).

This leads to another question.

4. Why is it so that by requiring smoothness we eliminate the possibility of "corners" or "cusps"? This is just stated without a proof or an intuitive explanation. Not very clear to me.

Answer 1

It happens that I'm also studying using this book, so you may take it with a grain of salt.

By definition, if $f:\mathbb{R}\to\mathbb{R}^n$ is an smooth (has non-zero first derivative everywhere) $C_n$ (n-differentiable) parametrization, then we use a generalized formula for Frénet-Serret (via the following theorem)

Let $f:D\subseteq\mathbb{R}\to\mathbb{R}^n$ be a smooth parametrization of the curve $C=f(D)$

By the existance of arc-length parametrizations theorem, there exists a function $h=f\circ \lambda^{-1}:\mathbb{R}\to\mathbb{R}^n$, where $\lambda'(t)=||f'(t)||$

If $||h''(s)||=\kappa_h(s)>0$, then the function $h$ posesses its Frenét-Serrét apparatus

Then, we denote $v(t)=||\lambda'(t)||$, and:

1. $f'(t)=v(t)T_h(\lambda(t))$
2. $f''(t)=v'(t)T_h(\lambda(t))+v^{2}(t)\kappa_h(\lambda(t))N_{h}(\lambda(t))$
3. If $\kappa_{h}(s)=0$, then the second term vanishes

If $n=3$, then we can talk about the Binormal and use cross products.

When confused, refer back to this theorem. Plus:

1. Given that Frénet-Serret is defined for smooth parametrizations (path-length or simply smooth) $f'(t)$ can be zero, with a caveat (refer to the bottom of the comment), since $h'(s)=1$ is never zero. So we cannot talk about some Frénet-Serret formulation in that point.

2. The book is kind of careful when talking about when its talking about if, for example, $\kappa$ is either the path-length $\kappa$ or the general $\kappa$. In the general case, you can try using this theorem to derive everything in terms of $f$

3. The corner refers to the parametrization which has a zero derivative; or a derivative that does not exists at a point (ie $f(t)=(t,|t|)$) the graph of the absolute value function.

Finally, by requiring smoothness, we eliminate the possibility of this happening, having either zero-derivative or non-existance of the derivative. If there exists a finite number of corners, then we can have piecewise smoothness, yet that does not guarantee the fact that the curvature characterizes the function at every point. There is an exercise in the book that refers to this, if I remember.

I'm not sure if this answers everything, but I hope it clears some things up.

Answer 2

There's a lot of salient questions here, but I think that addressing the following might be satisfactory.

Suppose that we are given a 3-times differentiable function $f:\Bbb R \to \Bbb R^3$. Let $s:\Bbb R \to \Bbb R$ be the path-length function of $f$, so that $f \circ s^{-1}$ is parameterized by path-length.

We define a cusp in the path to be a value $a \in \Bbb R$ such that $f \circ s^{-1}$ has derivative zero at $s(a)$.

Is it possible to detect whether $f$ has a cusp at $a$ without calculating $s(t)$ or its inverse? If $f$ does not have a cusp at $a$ but $f'(a) = 0$, can the curvature and principal normal of the path be computed?

First of all, let's take stock of what can go "wrong" here.

It is possible that $f$ is 3-times differentiable, but the curve has a cusp. For instance, $(t^2,t^3)$ describes a curve with a cusp.

It is possible that the curvature and principal normal fail to exist even where no cusp is present and $f$ is 3-times differentiable. For instance, consider the function $$ f(t) = (t^3,|t|^3t^3). $$ The path is differentiable. We can see that the reparameterization $h(t) = (t,|t|t)$ is such that $h'(t)$ is never zero, which is enough for us to conclude that no cusp exists. However, we can see that the tangent normal vector for this reparameterization $$ \mathbf T(t) = \frac{h'(t)}{\|h'(t)\|} = \frac{(t,|t|t)}{\sqrt{t^2 + t^4}} = \frac 1{\sqrt{1 + t^2}}\left(\frac{t}{|t|}, t\right) $$ fails to be differentiable at $t = 0$.

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Now, some questions remain: if the curve does not have a cusp, how can we find its tangent vector? And if the curve does has a well-defined principal normal, how can this normal be computed?

A partial answer is that if the underlying curve is sufficiently smooth, then these can be computed using limits.

Let $s$ denote the path-length function associated with $f$ and let $g = f \circ s^{-1}$. Note that $f = g \circ s$, and that $g$ is parameterized by path-length. The tangent vector at $t = a$ is equal to $g'(s(a))$. If $g$ is sufficiently smooth (e.g., $g$ is twice differentiable), then we can guarantee that $g'$ is continuous so that $$ g'(s(a)) = \lim_{t \to a}g'(s(t)) = \lim_{t \to a}\frac{f'(t)}{\|f'(t)\|}. $$ Similarly, the principal normal at $t = a$ is equal to the unit vector in the direction of $g''(s(a))$ and the curvature is equal to $\|g''(s(a))\|$. If $g$ is sufficiently smooth (e.g. $g$ is 3-times differentiable), then we can guarantee that $g''$ is continuous, so that $$ g''(s(a)) = \lim_{t \to a}g''(s(t)) = \lim_{t \to a}\frac{\mathbf T'(t)}{s(t)}. $$ Once this limit is obtained, both the principal normal and curvature can be obtained.

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With all that said, I think that the spirit of this question is really about whether the formal $$ \kappa(a) = \frac{\|f'(a) \times f''(a)\|}{\|f'(a)\|^3} $$ can be extended to the case where $f'(a) = 0$. To some extent, the answer is yes. Again, assuming that $g$ is sufficiently smooth, the answer must be obtainable via a limit, $$ \kappa(a) = \lim_{t\to a}\kappa(t) = \lim_{t \to a}\frac{\|f'(t) \times f''(t)\|}{\|f'(t)\|^3}. $$ I think this becomes a bit more interesting with more specifics: if $a$ is a point where the curvature exists, $f'(a) = 0$, and $f''(a) \neq 0$, then L'Hôpital allows us to rewrite the limit as follows: \begin{align} [\kappa(a)]^2 &= \lim_{t \to a}\frac{\|f'(t) \times f''(t)\|^2}{\|f'(t)\|^6} \\ & = \lim_{t \to a}\frac{\frac d{dt}\|f'(t) \times f''(t)\|^2}{\frac d{dt}\|f'(t)\|^6}. \end{align} Note that $$ \frac d{dt}\|f'(t) \times f''(t)\|^2 = \\ 2(f'(t) \times f'''(t) + f''(t) \times f''(t)) \cdot (f'(t) \times f''(t)) = \\ 2 (f'(t) \times f'''(t)) \cdot (f'(t) \times f''(t)) = \\ 2([f'(t) \cdot f'(t)][f'''(t) \cdot f''(t)] - [f'(t) \cdot f''(t)][f'''(t) \cdot f'(t)]). $$ and $$ \frac d{dt}\|f'(t)\|^6 = 6 \|f'(t)\|^4 (f'(t) \cdot f''(t)) $$ With that, we can rewrite \begin{align} [\kappa(a)]^2 &= \lim_{t \to a}\frac{\frac d{dt}\|f'(t) \times f''(t)\|^2}{\frac d{dt}\|f'(t)\|^6} \\ & = \lim_{t \to a} \frac{2([f'(t) \cdot f'(t)][f'''(t) \cdot f''(t)] - [f'(t) \cdot f''(t)][f'''(t) \cdot f'(t)])}{6 \|f'(t)\|^4 (f'(t) \cdot f''(t))} \\ & = \frac 13 \lim_{t \to a}\left[\frac{f'''(t) \cdot f''(t)}{\|f'(t)\|^2(f'(t) \cdot f''(t))} - \frac{f'(t) \cdot f'''(t)}{\|f'(t)\|^4}\right]. \end{align}