Trying to calculate the time derivative of a position differential

In summary, according to the equation, the velocity is a function of the position and time. The problem is that the equation is not clear and it is difficult to understand.
  • #1
Apashanka
429
15
Homework 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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  • #2
Apashanka said:
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??
 
  • #3
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 ?
 
  • #4
Mark44 said:
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??
 
  • #5
BvU said:
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.
 
  • #6
Apashanka said:
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.
 
  • #8
Mark44 said:
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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  • #9
BvU said:
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??
 
  • #10
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??
 
  • #11
From post #1:
Apashanka said:
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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  • #12
Mark44 said:
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??
 

1. What is a position differential?

A position differential is a measurement of the change in position of an object over a certain period of time. It is often represented as ∆x, where ∆ represents the change in position and x represents the position itself.

2. Why is it important to calculate the time derivative of a position differential?

Calculating the time derivative allows us to determine the rate at which an object is changing position. This information is crucial in understanding the motion of objects and can help us make predictions about future positions.

3. How do you calculate the time derivative of a position differential?

The time derivative of a position differential is calculated by dividing the change in position (∆x) by the change in time (∆t). This is represented by the formula ∆x/∆t, which is also known as the average velocity.

4. What is the difference between a position differential and a displacement differential?

A position differential measures the change in position of an object over a period of time, while a displacement differential measures the total distance an object has moved from its starting position to its ending position.

5. Can the time derivative of a position differential be negative?

Yes, the time derivative of a position differential can be negative. This indicates that the object is moving in the opposite direction of its initial motion. For example, if an object initially moves in a positive direction and then changes direction to move in the negative direction, the time derivative would be negative.

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