I just solved question #2 on p. 248 from Spivak's Calculus Fourth Edition (2008). Solving it wasn't the issue. I'm trying to understand the idea behind it. This is a screenshot of the question:

enter image description here

I'm trying to understand what is meant by "reparameterizing hides a corner". What does the author mean by "hide" in what sense is it being hidden? For reference, the function that this is being applied to is the following:

$$ f(x) = \left\{ \begin{array}\ x^{2},\ x \geq 0 \\ -x^{2},\ x \leq 0 \\ \end{array} \right.$$

EDIT: Image of fig 21 as requested

enter image description here

  • $\begingroup$ Please add screenshot of Fig.21 ch.9. Thanks... $\endgroup$
    – coffeemath
    Aug 15, 2020 at 22:51
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    $\begingroup$ One way to detect a corner is to see if $c'(t)$ is discontinuous or undefined. This example shows that this test will only work if you add $c'(t) \ne 0$. $\endgroup$ Aug 15, 2020 at 22:52
  • $\begingroup$ What does Spivak mean by $c'(0) $? Perhaps he should have added some clarity instead of trying to surprise the reader unnecessarily. $\endgroup$
    – Paramanand Singh
    Aug 16, 2020 at 13:59
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    $\begingroup$ why didn't you cite edition, question number, page number? math.meta.stackexchange.com/q/656 $\endgroup$
    – user53259
    Aug 16, 2020 at 15:27
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    $\begingroup$ @ParamanandSingh It is widely understood that when speaking of a function $(f_1(t), f_2(t) ... f_n(t))$, the term "derivative" refers to $(f_1'(t), f_2'(t) ... f_n'(t))$, and that "0" can be used to refer $(0, 0 ... 0)$, although it is usually bolded to distinguish it from the real number 0, and if one is speaking precisely, it is said to be the "zero vector" or otherwise qualified. It's hardly a "surprise" for the author to rely on these shared understandings. $\endgroup$ Aug 17, 2020 at 16:11

3 Answers 3


@IsaacBrown 's answer is concise and correct. Here's another way to understand what is going on.

If you think about physically tracing the graph of $|x|$ between $(-1,1)$ and $(1,1)$ as you move at unit speed along the $x$-axis then the corner at the origin requires an instantaneous change of direction.

If you slow down so that your speed approaches $0$ as you approach the origin then there's no sudden change of direction there, and you speed up afterwards. That's how the given reparameterization "hides the corner".

In this animation contributed by @leftaroundabout the red vertical line moves at uniform speed from left to right; the blue one slows down to cross the origin at $0$ speed.

Animation showing the slow-down at the kink

What's going on here: time runs from $t=-5$ to $t=5$. The red vertical shows that time on the x-axis ($\color{red}{x=t}$), the green horizontal $\color{green}{y=t}$. The green curve gives the parametrisation of $\color{green}{x=f(t)}$, and the red parabola represents $\color{red}{(t,t^2)}$. Taking at each $t$ the $x$-values from $\color{blue}{f(t)}$ and the $y$-values from $\color{blue}{t^2}$ gives the yellow graph, which is identical to absolute-value function $\color{yellow}{|x|}$.

Animation source code: https://gist.github.com/leftaroundabout/2a19aea0e8dcb7b63d919406ecdb8c4a

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    $\begingroup$ @leftaroundabout Thanks for the animation. It took me a while to parse and I'm still not sure how best to describe the various parts - I've added some text, and should add more. I think that moving more slowly over a shorter interval (length $2$ or $4$ rather than $8$) might be clearer. $\endgroup$ Aug 16, 2020 at 13:44
  • $\begingroup$ +1 Wow! ........ $\endgroup$ Aug 16, 2020 at 14:52
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    $\begingroup$ This is impressive @EthanBolker so if I understand your explanation correctly, the red line is tracing out the reparameterization of the curve while the blue is showing the instantaneous change? $\endgroup$ Aug 16, 2020 at 16:37
  • $\begingroup$ The vertical red line is moving at a constant rate. That's the parameterization of the orange vee that has a sudden change of direction. The blue line is moving more and more slowly as it reaches the origin, so following the orange vee you slowsdown to $0$ just at the origin. You can see how the blue line catches up to the red line at $-1$. I haven't thought through yet what the other lines/curves do. $\endgroup$ Aug 16, 2020 at 18:25
  • $\begingroup$ @EthanBolker yeah, it's not as self-descriptive as I hoped it would be. I tried adding a legend and axis labels, but that just makes it more confusing since I'm plotting different things on each axes. I added also an explanation to the answer, does that clear it up? $\endgroup$ Aug 16, 2020 at 19:51

The corner here is the point $(0,0)$, as it is a non-differentiable corner of the curve represented by the graph of $h(x)=|x|$ (parametrized by $(t,h(t))$).

This non-differentiability is what is being hidden by this reparametrization of the curve as both $f(t)$ and $t^2$ are differentiable functions. The way it does this is by having derivative $0$ for both the $f(t)$ and $t^2$ at $(0,0)$.

  • $\begingroup$ But the graph of h(x) only appears after we take the derivative of the original function. So what you're saying is that if I "wanted" to take the derivative of my derivative (i.e $c''(t)$) it is here where trouble will come about ? $\endgroup$ Aug 15, 2020 at 22:58
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    $\begingroup$ The point is not that $h(x)$ is the derivative of the parametrization. That is merely a coincidence. The point is that $(t,h(t))$ and $(f(t),t^2)$ represent the same line. I'll clarify this $\endgroup$ Aug 15, 2020 at 23:06
  • $\begingroup$ Ah!!!....OK I see what you mean now. Thanks for your help. $\endgroup$ Aug 15, 2020 at 23:08

The other answers cover this nicely. I offer a somewhat different perspective. Notice that the function $h$ is a map from $\mathbb R$ to $\mathbb R$, whereas the parameterization $c$ is a map from $\mathbb R$ to $\mathbb R^2$. This means a priori that the $meaning$ of $h'$ and $c'$ are different. Thus, there is no reason to expect that differentiability of one of them implies differentiablity of the other.

Here is an even worse situation that can occur: take $f(t)=(\cos t,\sin 2t):\ \frac{-\pi}{2}< t< \frac{3 \pi}{2}$. Now, $f$ as defined is injective and differentiable on its domain. The curve (image of $f$) is

enter image description here

If we restrict the curve to $AB$, then the inverse image contains the isolated point $\frac{\pi}{2},$ and so is not open, but $AB$ may be considered as the $graph$ of some $real-valued$ function which is evidently continuous (even differentiable).

The underlying problem is that when we consider continuity and differentiability of $f$ the codomain $\mathbb R^2$ is the space we work in. But when we consider the curve to be the graph of a relation, and restrict it to $AB$ so that is becomes a function, we are changing the codomain to a subset of $f((\frac{-\pi}{2} , \frac{3 \pi}{2})),$ which is quite a different thing. Although $f$ is smooth as a function $(\frac{-\pi}{2}, \frac{3 \pi}{2})\to \mathbb R^2$, $AB$ is not even an open set in the topology induced by $f$, because $f^{-1}(AB)$ is not open!

  • $\begingroup$ Hmmm.....I like this explanation. What I get from it is that we need to take care of the space we're working on. $\endgroup$ Aug 16, 2020 at 16:45
  • $\begingroup$ @dc3rd Right, And the topologies that we use. Considering $f$ as a map into $\mathbb R^2$, it is nice. But if we consider it to be a map onto its image call it $N$, then if we give $N$ the topology induced by $f$, then the figure eight is an immersed manifold and yet in this topology $AB$ is not open. Now, if we give $N$ the subspace topology, then the figure eight is not even a manifold at all (geometrically, the cross at the origin makes it not locally Euclidean there). $\endgroup$ Aug 16, 2020 at 22:50

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