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I have a puzzling situation involving derivatives. I want to derivate: $$ \frac{d}{dx}| \mathbf F(x)| $$

This was actually something involving physics. Lets be 2-dimensional for simplicity. Let a particle be at position $\mathbf r = (x, y)$. The distance $s$ of the particle from point $(0, 0)$ is simply $s = |\mathbf r|$. I want to calculate how that distance changes over time. $$ \frac{ds}{dt} = \frac{d}{dt}|\mathbf r| = \frac{d}{dt}\sqrt{x(t)^2 + y(t)^2} = \frac{1}{\sqrt{x(t)^2 + y(t)^2}}\left(x\frac{dx}{dt} + y\frac{dy}{dt}\right) = \frac{1}{|\mathbf r|}\left(x\frac{dx}{dt} + y\frac{dy}{dt}\right) $$

As you can see, $ds/dt$ is not defined when $|\mathbf r| = 0$. I can't see why. On physics point of view, the particle should always travel continuously in the plane (assuming the path it makes is continuous and fully differentiable). Why is the distance variation undefined? Assume for instance, I have a table, and $(x, y)$ is the position of my fingers. I can't see why it wouldn't exist.

Hypothesis: Notice that, by description I told, the curve $(x, y)$ is continuous on all points, and smooth/differentiable on all points. Thus, $x(t), y(t), x'(t), y'(t)$ is well defined, for all points. If you want, consider them to be class $C^\infty$.

My question: Does this derivative exist or not when $|\mathbf r| = 0$? What is the value/evaluation of such derivative in an arbitrary given period $t_0$ when $|\mathbf r| = 0$?

Considering $x(t) = t^2$ and $y(t) = t^2$, we get $s$ proportional to $t^2$, and thus its derivative exists at $t=0$ with the derivative having a well defined value of zero.

Michael Hardy
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Physicist137
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  • If $\lvert \textbf{r} \rvert = 0$, does $d \lvert \textbf{r} \rvert /dt$ not equal $0$ to begin with? – Matthew Cassell Jun 25 '16 at 04:12
  • @Mattos No. You're forgetting that $|\mathbf r|$ is time-dependent. It might just pass through the origin. –  Jun 25 '16 at 04:13
  • @Bye_World So I was, I knew there was something wrong with my argument. Thanks. – Matthew Cassell Jun 25 '16 at 04:14
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    "Why is this distance undefined?" This is not a distance. At a moment before it, $|\vec{r}|$ was decreasing. Now, it is immediately increasing. It cannot switch from negative to positive without breaking off, making it undefined. – Kenny Lau Jun 25 '16 at 04:17
  • @KennyLau You are right. I just edited. (But, $|\vec r|$ cannot go negative...). – Physicist137 Jun 25 '16 at 04:28
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    It's for the same reason that the real function $f(x) = |x|$ is not differentiable at $x = 0$. However, $|\mathbf{F}(t)|$ is absolutely continuous, hence differentiable almost everywhere, and the fundamental theorem of calculus is applicable. Hence it poses no difficulty in practice. – Ningxin Jun 25 '16 at 04:29
  • @Physicist137 But its slope can. – Kenny Lau Jun 25 '16 at 04:29
  • @QiyuWen But I just gave an example in the end of the question of functions $x, y$ st the derivative exists. Thus, its not because of the same reason of $|x|$. In case of $(t^2, t^2)$, everything is $C^\infty$, even the distance to the center. Thus there are derivatives everywhere (instead of almost everywhere). – Physicist137 Jun 25 '16 at 04:44
  • The chain rule states: If $f'(x)$ and $g'(f(x))$ exists, then $(g\circ f)'(x)$ exists, and $(g\circ f)'(x) = g'(f(x))f'(x)$. Note that it's doesn't say anything about the case when $g'(f(x))$ doesn't exist. In this case, it may still be that $(g\circ f)'(x)$ exists, but you can't get it from the formula. – Ningxin Jun 25 '16 at 04:48
  • @QiyuWen Indeed. Alright, the formula I derived is only valid for $|\vec r|\ne 0$. Is there a way to get a formula for some $t$ such that $|\vec r| = 0$? How? – Physicist137 Jun 25 '16 at 05:21
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    If $|\vec{r}(t)| = 0$ and $|\vec{r}|'(t)$ exists, $|\vec{r}|$ is at its minimum. Hence $|\vec{r}|'(t)$ is necessarily $0$. – Ningxin Jun 25 '16 at 05:23
  • @QiyuWen You are a genius. This is brilliant.. and so ridiculously simple. Amazing. If you write down that answer, I'll give you the best answer. – Physicist137 Jun 25 '16 at 05:34
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    Aside: a shorter calculation is $$2s \mathrm{d}s = \mathrm{d}(s^2) = \mathrm{d}(x^2 + y^2) = 2x \mathrm{d} x + 2 y \mathrm{d} y $$ –  Jun 25 '16 at 11:50

4 Answers4

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CASE $1$: $\vec r(t)\ne 0$

Note that for $\vec r \ne 0$, we can write

$$\begin{align} \frac{ds(t)}{dt}&=\frac{\vec r(t)\cdot \frac{d\vec r(t)}{dt}}{|\vec r(t)|}\\\\ &=\bbox[5px,border:2px solid #C0A000]{\hat r(t)\cdot \frac{d\vec r(t)}{dt}} \tag 1 \end{align}$$

where in $(1)$, $\hat r(t)=\frac{\vec r(t)}{|\vec r(t)|}$ is the position unit vector. However, the unit vector $\hat r$ is undefined at the origin.

This fact does not automatically imply that the derivative $s'(t)$ fails to exist at $\vec r=0$. In the ensuing analysis, we will explore whether $s'(t)$ exists at $\vec r=0$.

CASE $2$: $\vec r(t)=0$

Assume that at $t_0$, $\vec r(t_0)=0$. We assume that $\vec r''(t)$ exists. Then, the derivative of $s(t)$ at $t_0$, if it exists, is given by

$$\begin{align} s'(t_0)&=\lim_{h\to 0}\left(\frac{\left|\vec r(t_0+h)\right|-\left|\vec r(t_0)\right|}{h}\right)\\\\ &=\lim_{h\to 0}\frac{\left|\vec r(t_0+h)\right|}{h} \\\\ &=\lim_{h\to 0}\frac{\left|\vec r'(t_0)h+O(h^2)\right|}{h} \\\\ &=\bbox[5px,border:2px solid #C0A000]{\lim_{h\to 0}\left(\frac{|h|}{h}\,\left|\vec r'(t_0)+O(h)\right|\right)} \tag2 \end{align}$$

If $\vec r'(t_0)=0$, then from $(2)$ we see that $s'(t_0)=0$ also. However, if $\vec r'(t_0) \ne 0$, then the limit fails to exist since the limits from the right-hand side and left-hand side are unequal.

Putting everything together, we find that

$$s'(t)=\begin{cases}\hat r(t)\cdot \frac{d\vec r(t)}{dt} &,\vec r(t) \ne 0\\\\0&,\vec r(t)=\vec r'(t)=0\\\\\text{fails to exist}&,\vec r(t) =0,\vec r'(t)\ne 0\end{cases}$$

Mark Viola
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  • Yes. But I asked the value of the derivative when $|\vec r| = 0$, not otherwise. – Physicist137 Jun 25 '16 at 05:10
  • @Physicist137 : In your long display equation, when $\mathbf{r} = 0$, the last expression is $\dfrac{1}{0} \left(0+0\right)$, or $\dfrac{0}{0}$. Thus any answer other than "undefined" will use limits. Even then, the derivative may be undefined, for example one dimensional unit velocity motion through the origin gives $|t|$, which has no derivative at $t=0$. – Eric Towers Jun 25 '16 at 09:17
  • @Physicist137 I've edited significantly to address the question as to the existence of $s'(t)$. Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Jun 25 '16 at 15:12
  • You only need once differentiable, as that's enough to imply $f(x+h) = f(x) + f'(x) h + o(h)$. –  Jun 26 '16 at 04:00
  • Thanks! Its indeed a nice answer. =). – Physicist137 Jul 22 '16 at 22:59
  • @Hurkyl Yes, we can use the Peano form of the remainder and proceed as you have suggested. – Mark Viola Jul 22 '16 at 23:14
  • @Physicist137 You're welcome! My pleasure. And thank you. – Mark Viola Jul 22 '16 at 23:17
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Your confusion stems from a subtlety regarding differentiation of composite functions. In particular, your calculation of $s'(t)$ uses the chain rule, but the chain rule has hypotheses.

Define $s$ to be $f\circ g$ where $g$ sends $t\mapsto\mathbf{x}(t)$ and $f$ sends $\mathbf{x}\mapsto\|\mathbf{x}\|$. The chain rule allows us to conclude that $s'(t)$ exists and is what you say it is, but only under the hypotheses that $g$ is differentiable at $t$ and $f$ is differentiable at $\mathbf{x}(t)$. Unfortunately, the function $f$ is not differentiable when $\mathbf{x}=\mathbf{0}$, so the chain rule does not apply in the case where the particle is at the origin. Are we to conclude that $s'(t)$ must not exist when $\mathbf{x}=\mathbf{0}$?

No! There is no such converse to the chain rule; the derivative of the composite may still exist. In other words, the chain rule supplies sufficient but not necessary conditions for the derivative of a composite to exist. See this question, which involves an example similar to your example with $x=y=t^2$.

Community
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symplectomorphic
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As you can see, $ds/dt$ is not defined when $|\mathbf r| = 0$. I can't see why. On physics point of view, the particle should always travel continuously in the plane (assuming the path it makes is continuous and fully differentiable). Why is the distance variation undefined? Assume for instance, I have a table, and $(x, y)$ is the position of my fingers. I can't see why it wouldn't exist.

The same phenomenon happens in one dimension, and exemplifies what happens in larger dimension: The distance from $0$ to $x$ is $|x|$. If a particle travels at unit speed along a smooth path through the origin, the rate of change of distance from the origin jumps from $-1$ to $1$ at the instant the particle reaches the origin. (If you want to split hairs, this qualitative language has to be tightened up: If $f(t) = |t|$, then $f'(t) = -1$ for all $t < 0$ and $f'(t) = 1$ for all $t > 0$, but $f'(0)$ does not exist.)

The distance to the origin for a specific particle may still be differentiable as a function of time (for instance, if the particle sits at the origin for all time), but that's not the same thing as saying the "scalar field" $\mathbf{r}$ is differentiable at the origin.

This is why differential geometers almost always work with regular curves, those whose velocity is non-vanishing: Even a real-analytic mapping, such as a cycloid, or hypocycloid such as the astroid, can trace a curve with corners or cusps.

Andrew D. Hwang
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Well.. I decided to answer my own question, because user @QiyuWen was apparently too shy to write down an answer about his brilliant insights. What he said can be found in the comments write below the question.

The chain rule states: If $f′(x)$ and $g′(f(x))$ exists, then $(g∘f)′(x)$ exists, and $(g∘f)′(x)=g′(f(x))f′(x)$. Note that it's doesn't say anything about the case when $g′(f(x))$ doesn't exist. In this case, it may still be that $(g∘f)′(x)$ exists, but you can't get it from the formula.

Thus, $\displaystyle\frac{ds}{dt} = \mathbf{\hat r}\cdot\frac{d\mathbf r}{dt}$ is only valid if $|\mathbf r|\neq 0$.

To find evaluation of such derivative at $|\mathbf r| = 0$, Qiyu Wen (brilliantly) reasoned the following:

If $|\vec r(t)| = 0$ and $|\vec r|'(t)$ exists, then $|\vec r|$ is at its minimum. Hence $|\vec r|'(t)$ is necessarily $0$.

Thus, at $|\mathbf r| = 0$, derivative either do not exist, or exists with value zero.

Physicist137
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