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Splitting the derivative?

  1. May 26, 2012 #1
    1. The problem statement, all variables and given/known data

    From
    [itex]\frac{de_{s}}{dT} = \frac{L_{v}e_{s}}{R_{v}T^{2}} [/itex]

    derive

    [itex] e_{s}(T) = 6.11 e^{\frac{L}{RV}(\frac{1}{T}-\frac{1}{273})} [/itex]


    2. Relevant equations



    3. The attempt at a solution

    The way my lecturer derived it was he 'split' the derivative and took them to their respective sides and integrated.

    So he got

    [itex] \frac{de_{s}}{e_{s}} = \frac{LdT}{R_{v}T^{2}} [/itex]

    However I was under the impression that you can't 'split' a derivative like that. Is this just a shortcut some physicists take to make the maths more simple? If it is what is the correct way of deriving this?
     
  2. jcsd
  3. May 26, 2012 #2

    Curious3141

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  4. May 26, 2012 #3
    How accepted is it? I am doing a double degree in pure mathematics and physics. If I was to do this in my pure maths classes would I be stoned?
     
  5. May 26, 2012 #4
    I've taken calculus 1 and I've seen it before? :s
     
  6. May 26, 2012 #5
    Sorry I wasn't very clear. I have seen it before and done it countless times I just always have a memory of a teacher saying don't tell a pure mathematician about it.
     
  7. May 26, 2012 #6

    SammyS

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    In my experience, physicists & engineers are notorious for treating Leibniz's notation for the derivative as if it were a fraction.

    A more rigorous handling of this derivative equation might be as follows:
    Treating es as a function of T we have

    [itex]\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}[/itex]

    Rewriting this equation gives us

    [itex]\displaystyle \frac{1}{e_{s}}\frac{d}{dT}(e_{s}) = \frac{L}{R_{v}T^{2}} [/itex]

    Integrating w.r.t. T gives

    [itex]\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int{\frac{L}{R_{v}T^{2}}}dT [/itex]

    We can rewrite the integral on the left hand side.

    [itex]\displaystyle \int {\frac{1}{e_{s}}\frac{d}{dT}(e_{s})}dT = \int {\frac{1}{e_{s}}}\,de_{s}[/itex]​

    Alternatively, if [itex]\displaystyle \frac{d}{dT}(e_{s}) = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,,[/itex]
    then the differential of es is given by [itex]\displaystyle d\,e_{s} = \frac{L_{v}e_{s}}{R_{v}T^{2}}\,dT[/itex]
     
  8. May 27, 2012 #7
    exactly what I wanted. thank you
     
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