V dot (dv/dt) = (0.5)*(d/dt)*(v^2) ?

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SUMMARY

The discussion centers on the derivation of the equation \( \frac{d}{dt} \frac{1}{2} mv^2 \) in the context of classical mechanics. The key steps involve starting from Newton's second law \( F = ma \) and applying the relationship between force and velocity. The factor of \( \frac{1}{2} \) arises from the kinetic energy formula \( K = \frac{1}{2} mv^2 \), which is differentiated to yield power. The discussion clarifies the application of the chain rule in differentiation, specifically how \( \frac{d}{dt} v^2 \) is computed as \( 2v \frac{dv}{dt} \).

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Familiarity with differentiation and the chain rule in calculus
  • Knowledge of kinetic energy formula \( K = \frac{1}{2} mv^2 \)
  • Basic concepts of power in physics
NEXT STEPS
  • Study the derivation of kinetic energy from work-energy principles
  • Learn advanced differentiation techniques, including the chain rule
  • Explore applications of power in mechanical systems
  • Investigate the relationship between force, velocity, and acceleration in various contexts
USEFUL FOR

This discussion is beneficial for physics students, educators teaching classical mechanics, and anyone interested in the mathematical foundations of motion and energy concepts.

wavingerwin
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This is not a homework question, but a derivation in my class which I can't get around.

Homework Statement


Step1 F=ma
Step2 \vec{F} = m\frac{d\vec{v}}{dt}
Step3: Multiply both side by v \vec{F}.\vec{v} = m\vec{v}.\frac{d\vec{v}}{dt}
Step4 Power = \frac{d}{dt}\frac{1}{2}m\vec{v}.\vec{v}
Step5 Power = \frac{d}{dt}\frac{1}{2}mv^{2} = \frac{dK}{dt}<br />


Homework Equations



The Attempt at a Solution


I just can't figure out where did the \frac{1}{2} come from in step 4.

Please help & thanks in advance!
 
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v2=v.v

So that d/dt(v2)=2v*dv/dt

They just wrote v2 as v.v
 
rock.freak667 said:
So that d/dt(v2)=2v*dv/dt

Aha, I see.
Or more elaborately:

\frac{d}{dt}v^{2} = \frac{dv}{dt}\frac{d}{dv}v^{2} = \frac{dv}{dt}2v

Thanks rock!
 

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