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Derivative for energy

  1. Feb 10, 2016 #1
    1. The problem statement, all variables and given/known data
    1/2mz^2 +mgh=mgh-zero , get g
    3. The attempt at a solution
    z= velocity

    z^2=g(2h0-2h)

    if i set z^2=a
    2h0=b (nonvariable)
    2h=c

    a=g(b-c)
    y'=-g
    Can i then say that

    dz^2/d2h = -g




    I wonder if every step is correct, The writing inbetween is not very important! I mostly wonder about the math. That if i do the substitution it actually work that way! x0 is not a variable!

    dv^2/d2h feels like second derivate, but i use it as a first derivate or something, that was makes me clueless
     
  2. jcsd
  3. Feb 10, 2016 #2

    haruspex

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    No, it's a first derivative. A second derivative would have a power of 2 on the d itself (in the numerator).
    It is correct, though you can take the 2 outside the 'd', in the denominator: ##\frac{d(v^2)}{2dh}##.
    The d(v2) can also be simplified.
     
  4. Feb 10, 2016 #3
    2v/2dh?

    But y-axis has to be z^2 for it to show -g as a graph and not 2z hmmm

    I was sortof proving that if y-axis is z^2 and x-axis 2h it will represent

    Checked with theory numbers and it shows 9,81 excaltly so it works
     
  5. Feb 10, 2016 #4

    haruspex

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    If the d(v2) form gives you all you need, that's fine. But it can also be written as 2vdv, giving ##v\frac{dv}{dh}##.
     
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