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Relative error problem in vector calculus gradient intro

  1. Feb 18, 2008 #1
    (a) Write a formula for the number in terms of the perimeter L and the area A of a circle.

    (b) Write the differential for your answer in part (a).

    (c) Suppose that L and A are determined experimentally. Write the resulting relative error in using your answer in part (b).


    I've tried several different values for part c, but can't seem to find a convincing answer. I know that the equation begins with dpi/pi and each part has some value with a coefficient of dL/L + dA/A
  2. jcsd
  3. Feb 18, 2008 #2


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    I can think of two approaches:

    1. Suppose experimenting introduces measurement error into each of L and A. Let L* = L + dL and A* = A + dA be the measured quantities of each of L and A respectively (dL and dA are the absolute errors; dL/L and dA/A are the relative errors). Substitute into dpi(A*,L*)/pi(A*,L*).

    2. Write dpi = (L^2/2A)dL/L + (L^2/4A)dA/A; divide through by pi = L^2/4A to obtain dpi/pi.
    Last edited: Feb 18, 2008
  4. Feb 19, 2008 #3
    I made a boo boo and seem to have done something wrong on the webassign program. But thank you for the help. I understand the concept a little better now.
  5. Feb 20, 2008 #4
    Haha, I have the same homework problem on webassign right now. I was a little lost as well, crazy that I found this here.
  6. Feb 20, 2008 #5
    Turns out that the problem is bugged. Even the teacher couldn't get the right answer.
  7. Feb 20, 2008 #6
    Do you also go to UA or is this a common vector calc problem that happened to be assigned at nearly the same exact time lol?
  8. Feb 12, 2009 #7
    So what is the correct answer to this question ? I am also doing the same problem and didn't get the answer
  9. Feb 9, 2012 #8
    I go to UA and this problem gave me so much grief. However, I found the right answer!
    so d(pi)/pi is the same as taking the differential function and dividing by the original:
    Simplify, and this gives you:

    Hope no one else has to struggle like I did!
  10. Feb 9, 2012 #9
    After coming down to my last submission you're post was extremely helpful! Also a UA student here.
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