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Hey,

In a derivation of relativistic energy (in Physics for Scientists and Engineers, 5th edition, Serway and Beichner) they use a method of integration by substitution:

Given that

[itex]F=\frac{dp}{dt}[/itex]

and relativistic momentum is given by

[itex]p=\frac{mv}{\sqrt(1-(v^2/c^2))}[/itex]

[itex]W=∫F dx=∫\frac{dp}{dt} dx[/itex]

which is fine then they say

[itex]\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{\sqrt(1−(\frac{(v^2)}{c^2}))}=\frac{m\frac{dv}{dt}}{(1−(\frac{v^2}{c^2}))^(\frac{3}{2})}[/itex]

(This is about halfway through the derivation.)

its that last step that i dont understand. why does the denominator term now have a 3/2 index? and why is du/dt given like that? Any help would be appreciated.

In a derivation of relativistic energy (in Physics for Scientists and Engineers, 5th edition, Serway and Beichner) they use a method of integration by substitution:

Given that

[itex]F=\frac{dp}{dt}[/itex]

and relativistic momentum is given by

[itex]p=\frac{mv}{\sqrt(1-(v^2/c^2))}[/itex]

[itex]W=∫F dx=∫\frac{dp}{dt} dx[/itex]

which is fine then they say

[itex]\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{\sqrt(1−(\frac{(v^2)}{c^2}))}=\frac{m\frac{dv}{dt}}{(1−(\frac{v^2}{c^2}))^(\frac{3}{2})}[/itex]

(This is about halfway through the derivation.)

its that last step that i dont understand. why does the denominator term now have a 3/2 index? and why is du/dt given like that? Any help would be appreciated.

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