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What is the uncertainty in the derivative of a function?

  1. Dec 7, 2017 #1


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    1. The problem statement, all variables and given/known data

    I was working on a kinematics experiment using Tracker to do a video analysis. I obtained a graph of displacement against time for the body under constant acceleration and the software also gives me the rms error between the parabolic trend line and the data points representing the position of the object over time.

    However, I actually want the velocity-time function of the object but I do not want to rely on the software's algorithms to determine it. As such, I plan to find the derivative of the displacement-time trend line by hand however, is there a way to calculate the uncertainty in the derivative using the rms value of the original function?

    2. Relevant equations

    s(t) = At^2 + Bt + c
    v = s'(t) = 2At + B
  2. jcsd
  3. Dec 7, 2017 #2

    Dr. Courtney

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    There are several approaches which I view as complementary. Which error estimate I favor depends on how well the frame rate and pixel resolution match the original experiment and whether more uncertainty is likely to be introduced by 1) a relatively slow frame rate (velocity changing significantly between frame intervals) 2) a relatively coarse pixels (movement of 100 pixels per frame is better than 8 pixels per frame) 3) motion not being all in the same plane perpendicular to the line of sight 4) optical effects 5) the reasonableness of the models used in determining the displacement function.

    Assigning the displacement to be a quadratic assumes a constant force, which is usually not true in most real experiments. It can only lead to a linear change in velocity (constant acceleration). But it does allow a rough estimate of uncertainties in a couple of ways: 1) propagate the error from the uncertainties in the fitting parameters A, B, and C. 2) repeat the analysis with a higher order polynomial and compute the resulting differences in velocity estimates.
  4. Dec 7, 2017 #3


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    Alright, thanks so much for the help!
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