Idea behind "reparameterization hiding a corner" in single variable calculus

Question

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:

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

Answer 1

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 $(0,0)$.

Answer 2

@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.

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}}

Answer 3

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

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!