In my textbook there's the definition of the kinetic energy in analogy with classical mechanics

$$dT=dW= \bf{F} \cdot d \bf{x}$$ where dW is the infinitesimal work done by the force $\bf{F}= \frac{d}{dt}(m \gamma v)$

After that it states

$$ \frac{dT}{dt}= \bf{F} \cdot \frac{d\bf{x}}{dt}$$ while I think it should be $$ \frac{dT}{dt}= \bf{F} \cdot \frac{d\bf{x}}{dt} + \frac{dF}{dt} \cdot d\bf{x} $$

Am I missing something here?

  • 2
    $\begingroup$ Hint: Consider a relation of the form $dy = f dx$. What should $dy/dx$ be? $\endgroup$ – Raziman T V Jan 13 '17 at 11:57
  • $\begingroup$ no, Run Like Hell is right in the general mathematics of calculus, it must just be that the force is not time varying $dF/dt=0$, or is small and can be neglected. It's a bit difficult without the context to validate that assumption. What is your $dT/dt$, (rate of change of k.e.) used for? $\endgroup$ – JMLCarter Jan 13 '17 at 12:18
  • $\begingroup$ We use the $dT$ to find $T$ after an integration. It writes down that derivative in an explicit way and then integrates to find $T=mc^2(\gamma -1)$ $\endgroup$ – Run like hell Jan 13 '17 at 13:53

You just devide $dT=Fdx $ by $dt$ to get the result.


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