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Mv dv = 1/2m d(v^2)?

  1. Oct 30, 2012 #1
    I am a physics novice (no formal science education beyond age 16) reading "Space Time and Quanta" by Robert Mills. At one point in the book whislt demonstrating the derivation of 1/2 mv^2 he says "mv dv = 1/2m d(v^2) = d(1/2mv^2)". I have been looking at this for quite a while and consulted other books on calculus but am unable to follow these steps. He seems to be using the reverse power rule to do something similar to integrating it. Could someone please explain these steps to me?
     
  2. jcsd
  3. Oct 30, 2012 #2
    It's in fact simple. Here m is taken constant, and v is a variable.

    d(v^2) = 2v dv
    Thus 1/2m d(v^2) = 1/2m * 2v dv = mv dv.

    Next, the constants can be put inside the brackets.
     
  4. Oct 30, 2012 #3
    Thanks for your reply. If I understand you correctly you are saying:
    mv dv
    = 1/2m 2v dv (the 1/2 and the 2 are simply inserted to yield the d(v^2) in the following, so no calculus involved in this step)
    = 1/2m d(v^2)
    = d(1/2mv^2)

    So why does d(v^2) = 2v dv? As I understand it dv is an infintesimal amount of v, so d(v^2) is presumably an infintesimal amount of v^2. 2v dv looks like an integral but without the summation symbol. Again I must reiterate, I'm a total novice, so I'm sure this is completeley rudimentary stuff.
     
  5. Oct 30, 2012 #4

    boneh3ad

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    It is an application of the product rule.
    [tex]d(v^2) = d(vv) = v\;dv + v\;dv = 2v\;dv[/tex]
    Therefore,
    [tex]mv\;dv = \dfrac{1}{2}m\;d(v^2)[/tex]
    Assuming the mass is constant, that gives
    [tex]mv\;dv = \dfrac{1}{2}m\;d(v^2) = d\left(\dfrac{1}{2}mv^2\right)[/tex]
     
  6. Oct 30, 2012 #5

    sophiecentaur

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    You can also work this out by the 'function of a function' rule, iirc.
     
  7. Oct 30, 2012 #6

    boneh3ad

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    True, you could chain rule it, giving
    [tex]d(v^2) = \dfrac{d}{dv}(v^2) \;dv = 2v\;dv[/tex]
     
  8. Oct 30, 2012 #7

    sophiecentaur

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    @Appleton
    You need to get up to speed on Calculus at this level. There is little point in just finding out about one particular instance like this one. The overall principles of differential calculus are fairly straightforward and there are a number of smart little tricks that allow us to simplify problems that can look daunting at first.
     
  9. Oct 30, 2012 #8
    In fact that is what I did; but I found your method much more elegant. :smile:
     
  10. Nov 1, 2012 #9
    I think I follow your arguments. I still feel quite uncomfortable with a couple of steps though:

    1/ Multiplying m by 1/2 and v by 2 to get 1/2m 2v dv.
    I suppose you are taking advantage of the commutability of multiplication in order to organise the expression in such a form that you can product rule it to yield 1/2m d(v^2). Correct? If so, it's an operation I haven't seen much in my admittedly rather meagre studies. Has this sort of operation (ab = 1/2a 2b) got a name?

    2/ 2v dv is yielded by product ruling or chain ruling d(v^2)
    Wouldn't it be simpler to say that d(v^2)/dv = 2v therefore d(v^2) = 2v dv and forget about chain and product rules?
     
  11. Nov 1, 2012 #10
    I would not know how to call such things, that goes too far back in time for me!
    ab=1/2*2a*b= 1/2*a*2*b= 1/2*b*2*a etc.
    That is exactly what I did, and also what was said in post #8 (just written in a more complex way).
     
  12. Nov 1, 2012 #11

    jtbell

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    Not that I know of. I don't use it very often, but it's not unusual either.
     
  13. Nov 1, 2012 #12

    boneh3ad

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    If that had a name it would probably just be called multiplying by one. Be that as it may, no one is explicitly doing that step. Your original goal was to prove that [itex]mv\;dv = d(1/2\;mv^2)[/itex]. The [itex]m[/itex] is assumed constant, so what you are really trying to relate is the [itex]1/2\;d(v^2)[/itex] to the [itex]v\;dv[/itex] on the other side, and that is simply the chain rule or the product rule depending on how you feel most comfortable with it. There need not be any special tricks. The only real trick is knowing that this problem is easier to work backwards.

    What you just did was the chain rule.
     
  14. Nov 1, 2012 #13
    Ok, I think I understand better now. I think the root of my confusion was my disorientation with regard to manipulating the infinitesimals when they are detached from the derivative (oh and my infinitesimal knowledge of maths). Thanks for setting me on the correct path.
     
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