B The norm of the derivative of a vector

redtree
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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|}##
 
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redtree said:
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.
 
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|##?
 
redtree said:
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.
 
fresh_42 said:
##|\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##?
 
redtree said:
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.##
 
Even for an even function?
 
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]##?
 
  • #10
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.
 
  • #11
If ##f(t) = (t^2 + C)##, ##f(0) = C##.
 
  • #12
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.
 
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  • #13
redtree said:
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)##
 
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  • #14
Thank you for your help. I have a much better understanding now.
 
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