A smooth curve $C$ does not double back or reverse the direction of motion over the time interval $[a, b]$ since $(f')^2 + (g')^2 > 0$ throughout.

Question

In my introductory calculus textbook, I have just begun reading a section about finding the lengths of parametrically defined curves using integration. In the introduction to this section, my textbook makes the following assertion:

$C$ is a curve specified parametrically by the equation $x = f(t)$ and $y = g(t)$ for $a \le t \le b$. A smooth curve $C$ does not double back or reverse the direction of motion over the time interval $[a, b]$ since $(f')^2 + (g')^2 > 0$ throughout the interval. At a point where a curve does start to double back on itself, either the curve fails to be differentiable or both derivatives must simultaneously equal zero.

My understanding is that a smooth function is a function that has all order derivatives defined on its domain.

However, my textbook does not justify its assertion with any conceptual explanation or proof. I think this is a useful piece of information that would be enlightening and help me understand the concept much more effectively. As such, I would greatly appreciate it if people would be gracious enough to take the time to elaborate on this assertion and/or prove/show why this assertion is true.

For context, my calculus understanding is at a level of basic multivariable calculus.

Answer 1

The stated condition means that there is no $t \in [a,b]$ such that $f'(t) = g'(t) = 0$.

If $f'(t^*) = g'(t^*) = 0$, that means that $x$ and $y$, at $t^*$ can reach a (local) extremum, and one of them can go from decreasing to increasing, and that can cause a "pointy" behaviour.

Another way to look at it, is that $\sqrt{(f')^2 + (g')^2}$ is the norm of the "instantaneous" speed. If this reaches $0$, then the curve can go in any other direction at this point, and again that would lead to a non-differentiability of the curve.

Answer 2

A smooth function and a smooth curve are not the same. A function $t\mapsto f(t)\in{\mathbb R}^n$ is smooth if $f=(f_1,\ldots,f_n)$ is at least $C^1$ (i.e., continuously differentiable), and preferably $C^\infty$ (i.e., continuously differentiable to arbitrary order), on its domain of definition. This is equivalent with all coordinate functions $f_i$ $(1\leq i\leq n)$ of $f$ having this degree of smoothness.

A curve $\gamma\subset{\mathbb R}^n$, on the other hand, is not the same as a function, but is a geometrical entity with its own special properties. A curve $\gamma\subset{\mathbb R}^n$ is called smooth if there is a smooth parametrization $t\mapsto f(t)$ of $\gamma$ with $\|\dot f(t)\|>0$ for all $t$. Such a curve has a well defined unit tangent vector $T(t)={\dot f(t)\over\|\dot f(t)\|}$ at all of its points, and $t\mapsto T(t)\in S^{n-1}$ is an at least continuous (maybe even $C^\infty$) function. In order to compute this $T(t)$ we need $\|\dot f(t)\|>0$.

As an example consider the $C^\infty$-map $t\mapsto z(t):=(t^3,t^3)$. This map is not regular at $t=0$ since $\dot z(0)=0$. Nevertheless we are lucky: The image set is the full line $x=y$, which is clearly a "smooth curve", not "by general principles" applied to $z(\cdot)$, but by inspection. If we consider the map $t\mapsto z(t):=(t^3,t^2)$ instead then this map is again not regular at $t=0$. Inspection shows that we have a cusp there, and computation shows that $$\lim_{t\to 0-}{\dot z(t)\over\|\dot z(t)\|}=(0,-1),\qquad \lim_{t\to 0+}{\dot z(t)\over\|\dot z(t)\|}=(0,1)\ .$$

Answer 3

The assertion is logically equivalent to:

If a curve starts to double back on itself at some point and it's differentiable there, then both derivatives must simultaneously equal zero.

Proof: Let $C$ be a curve specified parametrically by the equation $r(t)=(f(t),g(t))$ for $a≤t≤b$, which starts to double back on itself at $t=t_0$. That is, one can find some $\delta>0$ and a decreasing function $p:[t_0, t_0+\delta]\to [t_0-\delta', t_0]$ (i.e $p$ is bijective) such that $r(t)=r(p(t))$. (Imagine as if the value of $t$ goes backward)

Suppose the curve is differentiable at $t=t_0$: $f’(t_0)=\lim_{t\to t_0} \frac{f(t)-f(t_0)}{t-t_0}$.

If $f(t)=f(t_0)$ infinitely many times in an ϵ-right neighborhood of $t_0$, there is a sequence $t_n\to t_0^+$ which $\lim_{n\to \infty}\frac{f(t_n)-f(t_0)}{t-t_0}=\lim_{n\to \infty} 0=0$. Thus $f'(t_0)=0$.

Suppose otherwise, that there is an ϵ-right neighbourhood of $t_0$ which $f(t)\ne f(t_0)$. Since $f$ is continuous, either $f(t)>f(t_0)$ or $f(t)<f(t_0)$ for all $t$. For $t'\in [t_0-\delta', t_0]$, there always exists $t\in [t_0, t_0+\delta]$ such that $f(t')=f(p(t))=f(t)$.

The two one-sided derivatives are equal: $\lim_{t\to t_0^+} \frac{f(t)-f(t_0)}{t-t_0}=\lim_{t'\to t_0^-} \frac{f(t')-f(t_0)}{t'-t_0}$. The numerators are of same sign while the denominators are of opposite signs. So, $0\le f'(t_0)\le 0$, i.e $f'(t_0)=0$. Alternatively, one can note that $f(t_0)$ is local extremum and apply Fermat's theorem to get the same result.

By the same argument, $g'(t_0)=0$. This completes the proof.