The norm of the derivative of a vector

In summary, the conversation discusses the relationship between the norm of a vector and its derivative with respect to a parameterization variable. It is determined that the equation ##\left| \frac{d\vec{u}}{d t} \right| \overset{?}{=} \frac{d |\vec{u}|}{d |t|}## is not always true, and only holds in special cases such as when the vector is of the form ##\vec{u}=u(t)\hat u## where ##\hat u## is a constant vector and ##|u(t)|## is an increasing or decreasing function of t.
  • #1
redtree
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TL;DR 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|}##
 
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  • #2
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.
 
  • #3
  • #4
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|##?
 
  • #5
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.
 
  • #6
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##?
 
  • #7
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.##
 
  • #8
Even for an even function?
 
  • #9
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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FAQ: The norm of the derivative of a vector

What is the norm of the derivative of a vector?

The norm of the derivative of a vector is a measure of the rate of change of the vector. It represents the magnitude of the vector's change over a small interval of time or space.

How is the norm of the derivative of a vector calculated?

The norm of the derivative of a vector is calculated by taking the derivative of each component of the vector and then finding the square root of the sum of the squares of these derivatives.

What does a high norm of the derivative of a vector indicate?

A high norm of the derivative of a vector indicates a large rate of change, meaning that the vector is changing quickly in a short period of time or space.

How is the norm of the derivative of a vector used in physics?

In physics, the norm of the derivative of a vector is used to calculate the velocity and acceleration of an object. It is also used in differential equations to model the behavior of physical systems.

Can the norm of the derivative of a vector be negative?

No, the norm of the derivative of a vector is always a positive value. This is because it represents the magnitude of change and cannot have a negative value.

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