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Percent Uncertainty Problem -NEED HELP!

  1. May 24, 2009 #1
    Percent Uncertainty Problem -NEED URGENT HELP!

    1. The problem statement, all variables and given/known data
    The question is: What, approximately, is the percent uncertainty for the measurement given as 1.57 m^2.


    2. Relevant equations
    I know the formula that this textbook: 6th edition: Physics - Giancoli gave. It is simply, the ratio of the uncertainty to the measured value, multiplied by 100.


    3. The attempt at a solution
    I have tried numerous attempts and for some reason, I couldn't seem to come up with the right answer the book gives in the back. I'm really not sure what other way to do it.

    I tried, simply to put:
    (.01)/1.57 * 100 = .64%

    (.01)/(1.57^2) * 100 = .41%

    (1.58-1.57)/1.57 *100 = .64%

    ...the correct answer in the back of the book is 1%. I kept using .01 because the measured value they gave me 1.57m^2. I am not sure whether I messed up there, I was hoping someone could help me. Thanks
     
    Last edited: May 24, 2009
  2. jcsd
  3. May 24, 2009 #2
    Re: Percent Uncertainty Problem -NEED URGENT HELP!

    I'm very new to both calculus and physics but I am having trouble understanding how any value is to be found with such little information.

    The final measurement 1.57 is in m^2 and maybe is thought of as an area?

    I think the percent uncertantity is related to the differential or dy.

    dy is equal to f'(x)dx.

    So I guess the equation f(x) could be A=X^2 and A' would be 2X. A' could be thought of as delta A because if there is error in the measurements (delta X or Y) it would be propagated into the calculation of the area. For small values of error, delta X=dX and delta Y=dy and dx*dy=dA ( I wonder if this thinking is correct).

    Anyway then you would need to be given the estimated deviation of any measurement taken. For instance X is found to be correct to within dX m.

    So with this I did

    A=x^2
    1.57=X^2
    X=1.25

    dA=2Xdx
    =2(1.25)(dX)
    =2.5dX m^2 propagated error

    to find the percent error the ratio of the change to the actual areas multiplied by 100 should be preformed...

    dA/A
    = (2Xdx/X^2) 100
    = (2dx/x) 100
    =[2(dX)/1.25] 100
    =160 dX% error.

    ....thats all I think after having taken calculus one....
     
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