Solving Derivatives Problems: Understanding dv/dt and dv/ds Differences

  • Thread starter Thread starter Mathnewbie
  • Start date Start date
  • Tags Tags
    Derivatives
Click For Summary
SUMMARY

The discussion focuses on the relationship between derivatives in the context of particle motion, specifically addressing the equation a(t) = v(t) dv/ds. It clarifies the meanings of dv/dt and dv/ds, where dv/dt represents the change in velocity over time, while dv/ds indicates the change in velocity over displacement. The chain rule is highlighted as a method to connect these derivatives, emphasizing the distinction between rates of change with respect to time and displacement.

PREREQUISITES
  • Understanding of basic calculus concepts, particularly derivatives
  • Familiarity with particle motion equations: displacement s(t), velocity v(t), and acceleration a(t)
  • Knowledge of the chain rule in calculus
  • Basic physics principles related to motion
NEXT STEPS
  • Study the application of the chain rule in calculus
  • Explore the relationship between velocity and acceleration in physics
  • Learn about higher-order derivatives and their physical interpretations
  • Investigate real-world applications of derivatives in motion analysis
USEFUL FOR

Students and educators in physics and mathematics, particularly those studying calculus and mechanics, as well as anyone interested in understanding the dynamics of particle motion.

Mathnewbie
Messages
14
Reaction score
0
Hello can someone point in the right direction on this one.

A particle moves along a strainght line with displacement s(t), velovity v(t), and acceleration a(t). Show that

a(t) = v(t) dv/ds

Explain the difference between the meanings of the derivatives dv/dt and dv/ds.

Does dv/dt mean difference of velocity over the difference time ?

Does dv/ds mean difference of velocity over the difference displacement ?

Any help would be great? Thanks
 
Physics news on Phys.org
Basically yes: how fast the speed changes "per foot" rather than "per second", for example.
To do the first part, use the chain rule:
[tex]\frac{dv}{dt}= \frac{dv}{ds}\frac{ds}{dt}[/tex]
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
5
Views
5K
  • · Replies 42 ·
2
Replies
42
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K