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

[splatcat]

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

Stu

 

[granpa]

set
v=x
x=v
see what you get

 

[Mark44]

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.

 

[splatcat]

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, :) )

 

[granpa]

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

 

[splatcat]

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.

 

[splatcat]

No that is not what I did, I assumed you were measuring a tiny change in v and x instead of x and t.