
#1
Aug409, 12:02 AM

P: 28

I see in many text that a = dv/dt implies that
dv = a dt How is that possible, can anybody please explain me. As far as i know dv/dt is a symbol for derivative of v w.r.t t and not ratio between dv and dt. 



#3
Aug409, 12:19 AM

P: 1,623

Rather than offer a subpar explanation, this thread  with input from several knowledgeable members  should answer your questions: http://www.physicsforums.com/showthread.php?t=328193




#4
Aug409, 12:21 AM

P: 77

a= dv/dt
dv/dt is not a symbol. It is the mathematical formula for the derivative of v with respect to t. This is how one would show any derivative of a dependent variable with respect to an independent variable.




#5
Aug409, 09:30 AM

P: 28





#6
Aug409, 09:31 AM

P: 28





#7
Aug409, 09:56 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

Now, WHY can we utilize at times the dv=adt formula, in particular, WHERE is it usable?
Answer: When doing integration with the technique called substitution of variables (i.e the "inverse" of the chain rule): Given a=dv/dt, we have, trivially: [tex]\int_{t_{1}}^{t_{2}}adt=\int_{t_{1}}^{t_{2}}\frac{dv}{dt}dt[/tex] But the righthand side can, by the theorem of substitution of variables, be reformulated, giving the identity: [tex]\int_{t_{1}}^{t_{2}}adt=\int_{v(t_{1})}^{v(t_{2})}dv=\int_{v_{1}}^{v_{2 }}dv[/tex] Now, by IGNORING that the limits of integration actually refers to the limits of DIFFERENT variables, we "may say" that the "integrands" are equal, i.e, adt=dv! Thus, adt=dv should, at this stage of your education, be regarded as notational garnish (or garbage, if you like!) 


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