a= dv/dt

vijay_singh

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.

Cyrus

What you wrote is in differential form.

jgens

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

bucher

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.

vijay_singh

And what did I say :-)

vijay_singh

Thanks Jgens, for pointing me to right direction.

arildno

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 right-hand 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!)