Is dx/dv Equal to Time When Considering Small Changes in Velocity and Position?

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Discussion Overview

The discussion revolves around the relationship between small changes in position (dx) and velocity (dv), and whether dx/dv can be equated to time (t). Participants explore the manipulation of equations involving derivatives and the validity of certain mathematical operations in this context.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether dx/dv can equal time, suggesting that manipulating equations in this way may not be valid.
  • Another participant proposes setting v equal to x and vice versa to explore the implications of this substitution.
  • A participant expresses concern about the validity of operations involving derivatives, specifically cautioning against multiplying and dividing by time inappropriately.
  • One participant argues that integrating both sides leads to a conclusion that dx/dv equals time, presenting it as a correct outcome.
  • Another participant comments on the simplicity of the derivatives involved and suggests that learning the process independently would be beneficial.
  • A participant clarifies their intent was to illustrate a point about derivatives rather than seeking direct assistance with calculations.
  • One participant states that they were measuring tiny changes in velocity and position rather than time, indicating a different approach to the problem.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the validity of equating dx/dv to time, with multiple competing views and interpretations of mathematical operations present in the discussion.

Contextual Notes

There are unresolved assumptions regarding the manipulation of derivatives and the definitions of the variables involved, particularly concerning the treatment of time in relation to changes in position and velocity.

splatcat
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Does dx/dv = t ? Can you just manipulate equations like this?
[ x is position and v is velocity, t time :P ]

Stu
 
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set
v=x
x=v
see what you get
 
How did you get that?

dx/dt = v. I hope you didn't multiply both sides by t and divide both sides by v to get the equation above. If that's what you did (just a guess), that's not a valid operation. dt is not the product of d and t.
 
surely you just get = 1/t then intergrate both sides you get v = x/t which is correct, so dx/dv must be = to t ??

(thank you for your reply, :) )
 
Last edited:
these derivatives are trivial. you can easily learn to do them in a single day.
you would be better off learning to do this yourself than asking us to do it for you
 
They are, I was trying to show someone that they could do what I asked in the initial question with a different derivative, they were confused and I was struggling to justify that it was true, getting myself in a loop of confusion. I was not really asking people to do it for me.
 
No that is not what I did, I assumed you were measuring a tiny change in v and x instead of x and t.
 

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