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Trying to calculate the time derivative of a position differential

  • Thread starter Apashanka
  • Start date
350
13
Problem Statement
Trying to calculate ##\frac{d}{dt}dx##
Relevant Equations
Trying to calculate ##\frac{d}{dt}dx##
here I am trying to find ##\frac{d}{dt}dx## where ##x(t)## is the position vector
Now ##\frac{d}{dt}(v_x(x,y,z,t)dt)=\frac{dv_x}{dt}dt=\frac{\partial v_x}{\partial t}dt+\frac{\partial v_x}{\partial x}dx+\frac{\partial v_x}{\partial y}dy+\frac{\partial v_x}{\partial z}dz##
Now dividing by ##dx##
##\frac{\partial v_x}{\partial t}\frac{dt}{dx}+\frac{\partial v_x}{\partial x}##
Other terms goes to zero.
It therefore becomes ##\frac{\partial v_x}{\partial t}\frac{\partial t}{\partial x}+\frac{\partial v_x}{\partial x}=\frac{\partial v_x}{\partial x}+\frac{\partial v_x}{\partial x}=2\frac{\partial v_x}{\partial x}##
Am I right in doing so??
 
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Problem Statement: Trying to calculate ##\frac{d}{dt}dx##
Relevant Equations: Trying to calculate ##\frac{d}{dt}dx##

here I am trying to find ##\frac{d}{dt}dx## where ##x(t)## is the position vector
As stated, your problem doesn't make sense to me.
One thing that is confusing is that if ##x(t)## is a position vector, are its components x, y, and z values?
If so, the usual way to describe it is as ##\vec r(t)## where ##\vec r(t) = <x(t), y(t), z(t)>## if we're talking about a vector in ##\mathbb R^3##.
In this case, ##\frac{d}{dt}\left(\vec r(t)\right) = <x'(t), y'(t), z'(t)>##, with the primes indicating the derivative with respect to t.
Apashanka said:
Now ##\frac{d}{dt}(v_x(x,y,z,t)dt)=\frac{dv_x}{dt}dt=\frac{\partial v_x}{\partial t}dt+\frac{\partial v_x}{\partial x}dx+\frac{\partial v_x}{\partial y}dy+\frac{\partial v_x}{\partial z}dz##
Now dividing by ##dx##
##\frac{\partial v_x}{\partial t}\frac{dt}{dx}+\frac{\partial v_x}{\partial x}##
Other terms goes to zero.
It therefore becomes ##\frac{\partial v_x}{\partial t}\frac{\partial t}{\partial x}+\frac{\partial v_x}{\partial x}=\frac{\partial v_x}{\partial x}+\frac{\partial v_x}{\partial x}=2\frac{\partial v_x}{\partial x}##
Am I right in doing so??
 

BvU

Science Advisor
Homework Helper
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In physics (maybe also in math 😉) ##\mathrm d## has a special meaning: take a small chunk of what follows and let that chunk shrink until it is infinitesimally small. When we speak of ##{\mathrm d} x## we use the term 'infinitesimal'.

If ##x## is a function of ##t##, then ##{\mathrm d} x## is also a function of ##t## and you can form a mental idea by thinking of ##{\mathrm d} x(t)## as ##\ \displaystyle \lim_{h_\downarrow 0} \Bigl ( x(t+h) - x(t) \Bigr )##. But only as a reminder (basically it is zero).

Because that way, for ##{\mathrm d} t ## you would get ##\ \displaystyle\lim_{h_\downarrow 0} t+h \ - t= 0\ ##. So we write ##{\mathrm d} t ## as a reminder, like: something we will divide by later on and then take a limit letting it go to zero.

Things only becomes non-infinitesimal in the division -- where we write a fraction and the (single) limit of the ratio is taken and not the ratio of two limits:$${{\mathrm d} x\over dt }\equiv \lim_{h_\downarrow 0} {x(t+h) - x(t) \over h} $$


Continuing in that jargon, ##\ {{\mathrm d} x\over {\mathrm d} t}\ = {{\mathrm d} \over {\mathrm d} t }(x) \ = v(t) ## is a derivative.



One can ask for a second derivative, but there is no point in asking for a derivative of an infinitesimal. Can you provide some more context of what you try to do ?
 
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As stated, your problem doesn't make sense to me.
One thing that is confusing is that if ##x(t)## is a position vector, are its components x, y, and z values?
If so, the usual way to describe it is as ##\vec r(t)## where ##\vec r(t) = <x(t), y(t), z(t)>## if we're talking about a vector in ##\mathbb R^3##.
In this case, ##\frac{d}{dt}\left(\vec r(t)\right) = <x'(t), y'(t), z'(t)>##, with the primes indicating the derivative with respect to t.
Ok the thing is it ,I came across an equation which is ##dx(t)dy(t)dz(t)=JdXdYdZ....(1)## where ##x,y,z## are cartesian components of the position vector and ##X,Y,Z## are constt of time.
and it is written that ##\dot J=J\theta## where ##\theta=\vec \nabla•\vec v## and ##\vec v## is the velocity vector.
In order to prove it I took time derivative of both sides of (1) for which Rhs become ##\dot J dXdYdZ## and for one part of LHS out of three parts it came as ##\frac{\partial v_x}{\partial x}+\frac{\partial v_x}{\partial t}\frac{\partial t}{\partial x}## and combining three terms it came as ##\theta##+extra terms
Then how to prove it??
 
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In physics (maybe also in math 😉) ##\mathrm d## has a special meaning
Yeah, it has the same meaning in mathland -- the differential of something.

BvU said:
One can ask for a second derivative, but there is no point in asking for a derivative of an infinitesimal.
I agree.
 
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Ok the thing is it ,I came across an equation which is ##dx(t)dy(t)dz(t)=JdXdYdZ....(1)## where ##x,y,z## are cartesian components of the position vector and ##X,Y,Z## are constt of time.
constt (sic) of time?
What this looks like to me is a change of coordinates using the Jacobian. On the left side is a volume element ##dx(t)dy(t)dz(t)##. The corresponding volume element in a different coordinate system would be ##JdX(t)dY(t)dZ(t)##.
If X, Y, and Z are constants, as you seem to be saying, then dX = dY = dZ = 0.
Apashanka said:
and it is written that ##\dot J=J\theta## where ##\theta=\vec \nabla•\vec v## and ##\vec v## is the velocity vector.
 
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constt (sic) of time?
What this looks like to me is a change of coordinates using the Jacobian. On the left side is a volume element ##dx(t)dy(t)dz(t)##. The corresponding volume element in a different coordinate system would be ##JdX(t)dY(t)dZ(t)##.
Here is the snap where I tried to prove equation 3
IMG_20190504_221933.jpg
 

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Last edited by a moderator:
350
13
In physics (maybe also in math 😉) ##\mathrm d## has a special meaning: take a small chunk of what follows and let that chunk shrink until it is infinitesimally small. When we speak of ##{\mathrm d} x## we use the term 'infinitesimal'.

If ##x## is a function of ##t##, then ##{\mathrm d} x## is also a function of ##t## and you can form a mental idea by thinking of ##{\mathrm d} x(t)## as ##\ \displaystyle \lim_{h_\downarrow 0} \Bigl ( x(t+h) - x(t) \Bigr )##. But only as a reminder (basically it is zero).

Because that way, for ##{\mathrm d} t ## you would get ##\ \displaystyle\lim_{h_\downarrow 0} t+h \ - t= 0\ ##. So we write ##{\mathrm d} t ## as a reminder, like: something we will divide by later on and then take a limit letting it go to zero.

Things only becomes non-infinitesimal in the division -- where we write a fraction and the (single) limit of the ratio is taken and not the ratio of two limits:$${{\mathrm d} x\over dt }\equiv \lim_{h_\downarrow 0} {x(t+h) - x(t) \over h} $$


Continuing in that jargon, ##\ {{\mathrm d} x\over {\mathrm d} t}\ = {{\mathrm d} \over {\mathrm d} t }(x) \ = v(t) ## is a derivative.



One can ask for a second derivative, but there is no point in asking for a derivative of an infinitesimal. Can you provide some more context of what you try to do ?
IMG_20190504_221933.jpg

Trying to prove eq.3 ....
Any hints to arrive at the result??
 
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What I have tried is taking time derivative of ##d^3x=Jd^3X## which gives RHS ##\dot J d^3X##,for LHS simplying one term out of three e.g ##\frac{d(dx(t))}{dt}dy(t)dz(t)=dv_xdydz=(\frac{\partial v_x}{\partial x}dx+\frac{\partial v_y}{\partial y}dy+\frac{\partial v_z}{\partial z}dz+\frac{\partial v_x}{\partial t}dt)dydz##
Multiplying throughout by ##d^3x=Jd^3X##
Rhs becomes ##\frac{\dot J}{J}## and LHS becomes ##\frac{(\frac{\partial v_x}{\partial x}dx+\frac{\partial v_y}{\partial y}dy+\frac{\partial v_z}{\partial z}dz+\frac{\partial v_x}{\partial t}dt)}{dx}=\frac{\partial v_x}{\partial x}+\frac{\partial v_x}{\partial t}\frac{\partial t}{\partial x}##
Combining all three terms we get
##\frac{\partial v_x}{\partial x}+\frac{\partial v_x}{\partial t}\frac{\partial t}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_y}{\partial t}\frac{\partial t}{\partial y}+\frac{\partial v_z}{\partial z}+\frac{\partial v_z}{\partial t}\frac{\partial t}{\partial z}##
Hence ##\dot J=J(\nabla•\vec v+##extra terms).
Can anyone please help me out to get rid of these extra terms??
 
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From post #1:
here I am trying to find ##\frac{d}{dt}dx## where ##x(t)## is the position vector
Now ##\frac{d}{dt}(v_x(x,y,z,t)dt)=\frac{dv_x}{dt}dt=\frac{\partial v_x}{\partial t}dt+\frac{\partial v_x}{\partial x}dx+\frac{\partial v_x}{\partial y}dy+\frac{\partial v_x}{\partial z}dz##
Now dividing by ##dx##
##\frac{\partial v_x}{\partial t}\frac{dt}{dx}+\frac{\partial v_x}{\partial x}##
Other terms goes to zero.
It therefore becomes ##\frac{\partial v_x}{\partial t}\frac{\partial t}{\partial x}+\frac{\partial v_x}{\partial x}=\frac{\partial v_x}{\partial x}+\frac{\partial v_x}{\partial x}=2\frac{\partial v_x}{\partial x}##
Am I right in doing so??
I don't think you're on the right track here.
You are trying to prove that ##\dot J = J \nabla \cdot \vec v##, right?
What are ##J##, ##\dot J##, and ##\nabla \cdot \vec v## in this case?

The reason I believe you're on the wrong track is that you are workiing with the volume elements ##dx~dy~dz## and ##JdX~dY~dZ##, and trying to find the derivatives or differentials of them, when you should be working directly with the Jacobian J and showing that its derivative ##\dot J## is equal to J times the divergence of ##\vec v##; i.e. ##J \nabla \cdot \vec v##. See https://en.wikipedia.org/wiki/Del.
 
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From post #1:I don't think you're on the right track here.
You are trying to prove that ##\dot J = J \nabla \cdot \vec v##, right?
What are ##J##, ##\dot J##, and ##\nabla \cdot \vec v## in this case?

The reason I believe you're on the wrong track is that you are workiing with the volume elements ##dx~dy~dz## and ##JdX~dY~dZ##, and trying to find the derivatives or differentials of them, when you should be working directly with the Jacobian J and showing that its derivative ##\dot J## is equal to J times the divergence of ##\vec v##; i.e. ##J \nabla \cdot \vec v##. See https://en.wikipedia.org/wiki/Del.
##J=\frac{\partial f_1(X_1)}{\partial X_1}\frac{\partial f_2(X_2)}{\partial X_2}\frac{\partial f_3(X_3)}{\partial X_3}## where ##f_1,f_2## and ##f_3## changes with time, but they doesn't have any explicit time dependence.
And ##X_1,X_2## and ##X_3## doesn't have time dependence,they are initial coordinates at time ##t_0## fixed.
Then how to prove it??
 
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Thread closed as being a duplicate of this thread:
@Apashanka , please don't start new threads with the same question.
 

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