I am not understanding this differential relationship

Click For Summary

Discussion Overview

The discussion revolves around understanding the relationship between velocity, acceleration, and displacement in the context of engineering dynamics. Participants are exploring the derivation of the equation "a ds = v dv" from the fundamental definitions of velocity and acceleration, specifically questioning the process of eliminating the differential time element (dt).

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Homework-related

Main Points Raised

  • One participant expresses confusion about the derivation of the equation "a ds = v dv" and requests clarification on the intermediate steps involved.
  • Another participant suggests that eliminating dt can be done by manipulating the equations to express them in terms of dt, leading to the equation vdv = ads.
  • A third participant introduces the chain rule as a method to eliminate dt, explaining that acceleration can be expressed in terms of velocity and displacement, leading to the same equation.

Areas of Agreement / Disagreement

Participants are engaged in a discussion with varying approaches to the derivation, and while some methods are presented, there is no consensus on a single clear explanation or method that resolves the initial confusion.

Contextual Notes

The discussion includes assumptions about the familiarity with calculus concepts such as the chain rule and differential equations, which may not be explicitly stated by all participants.

Who May Find This Useful

Students studying engineering dynamics, particularly those struggling with the concepts of velocity, acceleration, and their mathematical relationships.

mindheavy
Messages
62
Reaction score
0
I'm studying engineering dynamics. The first chapter is discussing the velocity and acceleration equations; v = ds/dt and a = dv/dt. It then goes on to show a third equation that is stated as "a ds = v dv". They say they derived this equation by combining the two previous and 'eliminating dt'. I am just not seeing how they arrived at this, what are the intermediate steps? I also am not understanding the reason for just 'eliminating' dt. Could anyone develop this or help me along my way of understanding how this third equation was reached, I feel very uncomfortable just memorizing it without understanding where it came from...
 
Physics news on Phys.org
Pretty much solve as you would any other equation, the like term is dt, so it can be eliminated.

get both in terms of dt = something, then set equal to eliminate the term.

\frac{dv}{a} = \frac{ds}{v}
vdv = a ds
 
You can "eliminate the dt" by using the chain rule, a= dv/dt= (dv/ds)(ds/dt)= (dv/ds)v

From a/v= dv/ds, we get ads= vdv or, equivalently, ds/v= dv/a
 
Thanks for the reply saminator910 & HallsofIvy, very helpful!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K