The norm of the derivative of a vector

[redtree]

TL;DR

Is the norm of the derivative of a vector the same as taking the derivative of the norm of the vector with respect to the norm of the parameterization variable?

Is the following true?
$\left| \frac{d\vec{u}}{d t} \right| \overset{?}{=} \frac{d |\vec{u}|}{d |t|}$

 

[fresh_42]

Summary:: Is the norm of the derivative of a vector the same as taking the derivative of the norm of the vector with respect to the norm of the parameterization variable?

Is the following true?
$\left| \frac{d\vec{u}}{d t} \right| \overset{?}{=} \frac{d |\vec{u}|}{d |t|}$

This doesn't make much sense. The LHS looks as if you meant the norm of the function $p\longmapsto \left. \dfrac{d}{d t}\right|_{t=p} u(t)$ but it could as well mean the norm of the (unbounded) differential operator $u \longmapsto u'$. The RHS makes no sense at all. The nominator relates to the function $t\longmapsto |u(t)|,$ which isn't differentiable, but what is $d|t|?$ A one-sided limit?

The Leibniz notation is an abbreviation for a limit, it cannot be handled like an ordinary quotient.

 

[redtree]

I do mean the function $\vec{u}(t)$, not the differential operator. My understanding of norms is that they are differentiable; see https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm

 

[redtree]

Instead of $\vec{u}(t)$, what if one considers a function $\vec{u}(|t|) = \big( (|t|)^2 + C \big)^{1/2}$, where $C$ is a constant, why would that function not be differentiable by $|t|$?

 

[fresh_42]

I do mean the function $\vec{u}(t)$, not the differentiable operator. My understanding of norms is that they are differentiable; see https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm

$|\cdot| \, : \,\mathbb{R}\longrightarrow \mathbb{R}\, , \,x\longmapsto |x|$ is a norm, but is not differentiable at $x=0.$

Differentiability is a local property that qualifies and quantifies a function by a linear approximation.

Local means we need a topology. Quantification and approximation require a concept of distance, usually a metric induced by a norm.

 

[redtree]

$|\cdot| \, : \,\mathbb{R}\longrightarrow \mathbb{R}\, , \,x\longmapsto |x|$ is a norm, but is not differentiable at $x=0.$

Why is it not differentiable at $x=0$?

 

[fresh_42]

Why is it not differentiable at $x=0$?

Because the linear approximation from the left is $-1$ and from the right, it is $+1$, so they cannot be matched locally at $x=0.$

 

[redtree]

Even for an even function?

 

[redtree]

Also why can't one substitute $\big( (\vec{u}(t))^2 \big)^{1/2}$ for $|\vec{u}(t)|$ in the derivative such that $\frac{d |\vec{u}(t)|}{dt} = \frac{d }{dt}\left[ \big( (\vec{u}(t))^2 \big)^{1/2}\right]$?

 

[fresh_42]

Do you mean $\left(\left(\vec{u}(t)\right)^2\right)^{1/2}=+\sqrt{\vec{u}^2(t)}$ or $\left(\left(\vec{u}(t)\right)^2\right)^{1/2}=-\sqrt{\vec{u}^2(t)}$?

Edit: $\dfrac{d}{dt}f(t)^{1/2}=\dfrac{1}{2}\dfrac{f'(t)}{f(t)^{1/2}}$ is not defined at $f(t)=0$. Same result.

 

[redtree]

If $f(t) = (t^2 + C)$, $f(0) = C$.

 

[Delta2]

I basically have problem interpreting the RHS of your equation. What that $d|t|$ means?

Well anyway I think we can use chain rule to write it is $$\frac{d|\vec{u}|}{d|t|}=\frac{d|\vec{u}|}{dt}\frac{dt}{d|t|}$$ where $$\frac{dt}{d|t|}=1, t>0$$ $$\frac{dt}{d|t|}=-1, t<0$$ and it is undefined at t=0.

So to answer your question :
I think it holds iff $\vec{u}=u(t)\hat u$ where $\hat u$ must be a constant vector that does not depend on t. Additionally
if t<0 it must be $\frac{d|u(t)|}{dt}<0$ that is $|u(t)|$ is a decreasing function of t
if t>0 it must be $\frac{d|u(t)|}{dt}>0$ that is $|u(t)|$ is an increasing function of t.

So , in the general case, the equation doesn't hold.

 

[wrobel]

Summary:: Is the norm of the derivative of a vector the same as taking the derivative of the norm of the vector with respect to the norm of the parameterization variable?

Is the following true?
$\left| \frac{d\vec{u}}{d t} \right| \overset{?}{=} \frac{d |\vec{u}|}{d |t|}$

just consider a vector $u=(\cos t,\sin t)$

 

[redtree]

Thank you for your help. I have a much better understanding now.