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Quick derivative question

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data
    I just have a quick question concerning the derivative of both sides of the equation shown below.

    2. Relevant equations
    Starting with [itex]e^\frac{t}{m}=(\frac{v}{u})^2[/itex]


    3. The attempt at a solution
    Would it be correct that: [itex]dt=(2m)(\frac{dv}{v}-\frac{du}{u})[/itex]
     
  2. jcsd
  3. Oct 18, 2012 #2

    Zondrina

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    Wait... so you have a multivariate, implicit function right? Where you want to treat t, m, v and u all as variables.

    Are you trying to find [itex]\frac{∂}{∂t}[/itex] of both sides?
     
  4. Oct 18, 2012 #3
    I am trying to isolate for dt so that I can use it in another equation
     
  5. Oct 18, 2012 #4

    Zondrina

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    So you want to get t to be an explicit function in terms of m, v and u and then find dt with respect to what?
     
  6. Oct 18, 2012 #5
    v and u
     
  7. Oct 18, 2012 #6

    Zondrina

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    Okay, so first off, what is t in terms of m, v and u ( Hint : Use ln to simplify this nicely ).

    Then find [itex]\frac{∂t}{∂v}[/itex] and [itex]\frac{∂t}{∂u}[/itex]
     
  8. Oct 18, 2012 #7
    well if t=2m(ln(v)-ln(u)), then I would conclude that dt=2m(dv/v) or dt=-2m(du/u). However, does this mean that dt cannot be 2m(dv/v-du/u) ?
     
  9. Oct 18, 2012 #8

    SammyS

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    If u and v are functions of t, then yes, you are correct.
     
  10. Oct 18, 2012 #9
    This is something that was shown in one of my classes but I am unsure of the steps to show it? Would you mind showing me just from the line t=2m(ln(v)-ln(u))?
     
  11. Oct 18, 2012 #10

    Zondrina

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    I was under the impression that you meant u, v and m were variables, but if u and v are functions of t, then you have to remember to use the chain rule here.
     
  12. Oct 18, 2012 #11
    my apologies for not being clear, I think I have things figured out though, thanks!
     
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