Author 
Message 
G.E. Ivey science forum Guru
Joined: 29 Apr 2005
Posts: 308

Posted: Sun Jul 16, 2006 11:58 am Post subject:
Re: derivation of relativistic kinetic energy



Do you understand that "D" here means "the derivative with respect to t"? If you do, then the quotient rule (together with the chain rule), as Narcoleptic Insomniac said, gives the result. 

Back to top 


Narcoleptic Insomniac science forum Guru
Joined: 02 May 2005
Posts: 323

Posted: Sun Jul 16, 2006 11:06 am Post subject:
Re: derivation of relativistic kinetic energy



On Jul 15, 2006 7:22 PM CT, castertroy14@hotmail.com wrote:
Quote:  K = integral(vDp) = integral(vD(mv / (1v^2/c^2)^0.5))
D(mv / (1v^2/c^2)^0.5) = m(1  v^2/c^2)^1.5 Dv
therefore
K = integral(m(1v^2/c^2)^1.5 vDv) =
mc^2(1/(1(v^2/c^2)^0.5  1)
so my question is how do you get from
D(mv / (1v^2/c^2)^0.5) to m(1  v^2/c^2)^1.5 Dv

We can use the the quotient rule to differentiate from
elementary calculus...
d[mv / (1  v^2 / c^2)^(1/2)] / dv =
[m * (1  v^2 / c^2)^(1/2) + (mv^2 / c^2) *
(1  v^2 / c^2)^(1/2)] / (1  v^2 / c^2) =
m * [1  v^2 / c^2 + v^2 / c^2] / (1  v^2 / c^2)^(3/2) =
m * (1  v^2 / c^2)^(3/2).
Regards,
Kyle Czarnecki 

Back to top 


castertroy14@hotmail.com1 science forum beginner
Joined: 15 Jul 2006
Posts: 5

Posted: Sun Jul 16, 2006 12:22 am Post subject:
derivation of relativistic kinetic energy



K = integral(vDp) = integral(vD(mv / (1v^2/c^2)^0.5))
D(mv / (1v^2/c^2)^0.5) = m(1  v^2/c^2)^1.5 Dv
therefore
K = integral(m(1v^2/c^2)^1.5 vDv) = mc^2(1/(1(v^2/c^2)^0.5  1)
so my question is how do you get from
D(mv / (1v^2/c^2)^0.5) to m(1  v^2/c^2)^1.5 Dv 

Back to top 


Google


Back to top 


