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.

  • $\begingroup$ What's $f$? What's $g$? Must be understood by context? $\endgroup$ Apr 7, 2017 at 10:26
  • $\begingroup$ @RafaBudría I apologise for not specifying this: $C$ is a curve specified parametrically by the equation $x = f(t)$ and $y = g(t)$ for $a \le t \le b$. $\endgroup$ Apr 7, 2017 at 10:30
  • $\begingroup$ The book seems vague on the meaning of "double-back". $\endgroup$ May 16, 2017 at 20:50

3 Answers 3


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 $$\lim_{t\to 0-}{\dot z(t)\over\|\dot z(t)\|}=(0,-1),\qquad \lim_{t\to 0+}{\dot z(t)\over\|\dot z(t)\|}=(0,1)\ .$$

  • $\begingroup$ Thanks for the enlightening answer. Your post certainly clarifies a critical misunderstanding I had about smooth functions and smooth curves; thank you for that. However, it doesn't address my main point of confusion, which is specifically regarding why the curve fails to be differentiable or both derivatives must simultaneously equal zero. Nonetheless, as I said, your post was certainly enlightening and taught me something new! $\endgroup$ Apr 7, 2017 at 20:12

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.

  • $\begingroup$ But the fact that they can reach a local extremum doesn't explain the non differentiability aspect, right? Going from increasing to decreasing (or vice-versa) would require the derivatives of the vectors to be equal to $0$ at some point, but it does not explain the non differentiability aspect of the explanation. $\endgroup$ Apr 7, 2017 at 17:47

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.


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