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    • Thread Starter

    During my mechanics course the professor had an equation like:

    \displaystyle \frac{F}{m}=\frac{dv}{dt} he then split the fraction up like this and integrated \displaystyle \int \frac{F}{m} ~dt=\int dv like so to find a formula for \displaystyle v. (Can't remember if this was the expression but it is irrelevant.)

    I was wondering how you can do such a thing (what is really going on and what justifications have to be made in order to do such a thing)?

    I remember treating the derivative as a fraction before at A-Level and not knowing what I was actually doing so I have decided to try and fix that.

    I have know the formal definitions of derivatives, riemann integrals and limits (and just now seeing the fundamental theorem of calculus.)

    I haven't taken any modules in differential equations and I'm not really sure if it is beyond my scope at the moment.

    Could anyone explain or give me a rough idea of what is actually going on or point me in the direction of some resources I can look at?


    I think this is called the variable separable method of integration. (I learnt about it a while ago). Pretty useful in C4 and FP2-3. Sorry I can't help more!

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    (Original post by alex2100x)
    Could anyone explain or give me a rough idea of what is actually going on or point me in the direction of some resources I can look at?

    From what I can tell, he's simply integrating both sides of the equation with respect to t. Think back to implicit differentiation: because both sides are equal, their differentials must be equal, too. This is the same concept, but with integrals instead.
    The integral of \displaystyle \frac{F}{m} with respect to t is just \displaystyle \int \frac{F}{m} ~dt.
    As for the right hand side of the equation, it's pretty much the same story, except the dt essentially cancels out.

    Look at that bit as a fraction, and consider what's actually going on in terms of gradients/areas. Put simply, the fraction's got dt on the bottom and by integrating, you're multiplying by dt, so those terms cancel out because, obviously, dt/dt is just 1.

    That's probably the simplest way of looking at it.

    It can be justified by considering that:

    \displaystyle \int \frac{dv}{dt} dt = v + c

    since we have just differentiated then integrated the result of that differentiation - these are inverse operations (up to the constant c) so we end up where we started. Also:

    \displaystyle \int dv = \int 1 dv = v+c =  \int \frac{dv}{dt} dt

    so these expressions are equivalent
    • Thread Starter

    Ok thanks I guess it relies on FTOC to justify that differentiating and integrating are inverses of each other.

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