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Error analysis of a function

  1. Dec 2, 2014 #1
    Hi guys,

    I have a silly question, but I seem to be confused about it. Let's say I have a function of Newton's cooling law. I measured the exponential drop of the temperature of some system and now I want to make an error analysis. Do I treat this function as a two variable function of tau (time constant in exponent) and time ? If yes, according to the error analysis I need to take partial derivatives over time and tau. This results in having time as a product in df/dtau result. Shall this be a measured time of the temperature falling to the constant level ?

    In equations :

    ## \frac{df(\tau,t)}{d\tau} = \tau e^{-\tau t} ##
    ## \frac{df(\tau,t)}{dt} = t e^{-\tau t }##
     
  2. jcsd
  3. Dec 2, 2014 #2

    Mark44

    Staff: Mentor

    Since ##\tau## is a time constant (your words), it's not a variable, so don't differentiate with respect to it. It could be that here ##\tau## is a parameter, a value that can have different values for different scenarios, but isn't considered to be a variable.
     
  4. Dec 3, 2014 #3
    I asked my question in a wrong way. The uncertainty i want to measure is the one if the time constant. Then it's a function of tine and temperature. The question remains, is the t value that stays in derivative a value over i made my calculations? It makes sense if i take the uncertainty of the arguments and multiply them by time so kind of number of samples.
     
  5. Dec 3, 2014 #4

    Mark44

    Staff: Mentor

    If the time constant is what? This isn't a complete thought.
    I don't understand what you're asking.
    I don't understand this either.
     
  6. Dec 3, 2014 #5
    Ok, maybe more context will help : )

    I have samples of a solid cooling down. I do data fitting with Levenberg Marquardt algorithm to find ##\tau## - time constant of the exponential function. Now I want to calculate the uncertainty of the time constant ##\tau##. Let's put it simple - what are the possibilities to do that ?
     
  7. Dec 3, 2014 #6

    Mark44

    Staff: Mentor

    OK, that's clearer. The bit about a time constant threw me off, since it's not actually known.

    I guess I would approach this as f being a function of both t and ##\tau##.

    The total differential, df, would be
    $$df \approx \frac{\partial f}{\partial \tau}\cdot \Delta \tau + \frac{\partial f}{\partial t} \cdot \Delta t$$
     
  8. Dec 3, 2014 #7
    Yes, sorry, sometimes I forget to keep things straight forward : )

    Ok, great - this is exactly where I got stuck: the derivative with respect to ##\tau## will in the end include time wouldn't it ? So :
    ## f = T_{ambient} + (T(0)-T_{ambient})e^{-kt}##
    Now I keep the t as a constant (if I understand correctly) and :
    ##\frac{df}{d\tau} \Delta\tau= -t(T(0)-T_{ambient})\Delta\tau##

    I am confused what t is in my example ?

    But the question still remains how to calculate the ##\Delta\tau##. I got this parameter from data fitting. I just wonder if I can turn the function around so that the ##\tau(T,t)## and then calculate ##\Delta\tau##
     
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