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Homework Help: Estimating the maximum possible percentage error

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi all! Im new to the forum, came across it from googling the problem which I had and someone was having difficuilty with a question like this on here! Here's the question I have...

    The coefficient of rigidity, n, of a wire of length L and uniform diameter d is given by


    If A is a constant and the parameters L and d are known to within ± 2% of their correct values, estimate the maximum possible percentage error in the calculated value of n.

    2. Relevant equations


    3. The attempt at a solution

    dn = dn/dL x dL + dn/dA x dA

    dn = A/d^4 dL + L/d^4 dA

    n = A/d^4 x ΔL + L/d^4 x ΔA

    n = ΔL/L + ΔA/A

    Where do I go from here? Any help will be greatly appreciated :)
  2. jcsd
  3. Aug 10, 2010 #2


    Staff: Mentor

    Welcome, benji. You have come to a good place for help.
    A is a constant, so dn/dA doesn't make any sense. Instead, you want dn/dd. For the record, dn/dL and dn/dd are really the partial derivatives of n. I.e.,
    [tex]\frac{\partial n}{\partial L} \text{and} \frac{\partial n}{\partial d}[/tex]
    For the percent change in n, you want dn/n.
  4. Aug 10, 2010 #3
    Thanks for the reply Mark :) and correcting me with the dn/dA and dn/dd situation!

    I was wondering, how do I get the estimation when I have no values to work from for L and d?
  5. Aug 10, 2010 #4


    Staff: Mentor

    You don't have L and d, but you have their relative errors: ΔL/L and Δd/d.
  6. Aug 10, 2010 #5
    What do I do with the relative error? As "Δd" will be 2% what will I use for "d" ?
  7. Aug 10, 2010 #6


    Staff: Mentor

    No, Δd is not a percent value; it is the error in determining d. The relative error, which is what you are given, is Δd/d, and that is .02. Actually, what is given is that |Δd|/d <= .02. Same with L and ΔL.
  8. Aug 11, 2010 #7
    So would it be...

    % change in n = % change in L - (4 times % change in d)
    % change in n = -2 - (4 times -2)
    % change in n = -2 +8
    % change in n = + 6%

    Therefore, ±2% change in L and d = ±6% change in n?
  9. Aug 11, 2010 #8


    Staff: Mentor

    No. Where are you getting the -2 numbers?

    You started off in the right direction, but didn't fix the errors I pointed out earlier, so let's start from the beginning.

    n = AL/d4
    [tex]dn = A(\frac{\partial n}{\partial L}~dL + \frac{\partial n}{\partial d}~dd)[/tex]

    The percent change in n is Δn/n. Fill in the partial derivatives in the equation above and divide both sides by n.
  10. Aug 11, 2010 #9
    There is a more elementary way.

    n \propto L

    means that n increases with increasing L, while

    n \propto d^{-4}

    means that n decreases with increasing d.

    Therefore, the smallest value for n is obtained by taking the smallest value for L and the largest value for d:

    n_{\mathrm{min}} = A \, \frac{L_{\mathrm{min}}}{d^{4}_{\mathrm{max}}}

    The largest value for n, on the other hand, is obtained by taking the largest value for L and the smallest value for d:

    n_{\mathrm{max}} = A \, \frac{L_{\mathrm{max}}}{d^{4}_{\mathrm{min}}}

    In this way, you obtain an interval for the possible values of n:

    n \in [n_{\mathrm{min}}, n_{\mathrm{max}}]

    Instead of the interval notation, one usually uses the "techincal notation":

    n = \bar{n} \pm \Delta n

    which actually means:

    n_{\mathrm{min}} = \bar{n} - \Delta n \\

    n_{\mathrm{max}} = \bar{n} + \Delta n
    \end{array}\right. \Leftrightarrow \left\{\begin{array}{l}
    \bar{n} = \frac{1}{2} \, (n_{\mathrm{min}} + n_{\mathrm{max}}) \\

    \bar{n} = \frac{1}{2} \, (n_{\mathrm{max}} - n_{\mathrm{min}}) \\

    Then, of course, the relative uncertainty, expressed in percent, is defined as:

    \delta_{n} \equiv \frac{\Delta n}{\bar{n}} \cdot 100\%

    It is up to you to:
    1. Find Lmin and Lmax by knowing [itex]\bar{L}[/itex] (the nominal value) and [itex]\Delta L = \delta_{L}/{100 \%} \, \bar{L}[/itex] (the absolute uncertainty);

    2. Do the same for dmin and dmax;

    3. Find nmin and nmax according to the above formulas;

    4. Find [itex]\bar{n}[/itex] (nominal value) and [itex]\Delta n[/itex] (absolute uncertainty) according to the above formulas;

    5. Find the relative uncertainty [itex]\delta_{n}[/itex].
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