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

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The discussion centers on whether the equation dx/dv equals time (t) when considering small changes in position (x) and velocity (v). Participants clarify that manipulating derivatives requires careful consideration, as operations like multiplying and dividing by variables can lead to invalid conclusions. The correct relationship derived is v = x/t, leading to the conclusion that dx/dv can indeed be interpreted as time. There is an emphasis on the importance of understanding derivatives rather than relying solely on others for explanations. The conversation highlights the complexities involved in calculus and the need for clarity in mathematical manipulations.
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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, :) )
 
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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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