using differentials to calculate area percentage error

Posted on 2011-10-11
Last Modified: 2012-05-12
The length and the width of a rectangle are measured with error at most 3% and 5% respectively. Use differentials to approximate percentage error in calculating area.

I got stuck after I got to:

A = L*W
dA = L*dw + w*dL, where dL = 0.03, and dw = 0.05, I'm not sure how much further this problem can go without some more numbers to plug into the variables. Unless I'm missing something, wouldn't this be the "final answer"?

I asked my professor and she was saying to divide through by A, but that does not change anything as you will still end up with dA/(L*w) = 0.05/w + 0.03/L... Any guidance is much appreciated it. Kind of need this answered within the next couple of hours, so sooner the btter. Thanks!
Question by:Zenoture
    LVL 26

    Accepted Solution

    but dL does not equal .03     the .03 is the percentage      dL = .03 * L
    same for dw                                                                       dw = .05 * w
    dA/A = 0.03 + 0.05

    Author Comment

    if that is the case, then maybe my understanding of the problem is flawed. After reading the problem, I came up with the equation by differentiating sides and coming up with what I posted above. Why are we assuming dL = 0.03*L and dw = 0.05*w?
    LVL 26

    Expert Comment

    Why are we assuming dL = 0.03*L and dw = 0.05*w?

    The problem did not say that dL = 0.03.  The problem said that the error percentage was 3 percent
    Therefore the value of the error itself is 0.03 * L         ie dL = 0.03 * L as I said
    LVL 26

    Expert Comment

    your differentiation is all right.  It is just that there is a difference between a percent and the value itself

    Author Comment

    Yes, I understand my differential is right, but I'm trying to see the physical meaning behind it if that makes any sense. I know how to do it mathematically/mechanically, but I wanted to know what the thinking process behind it was to get to the answer you did.
    LVL 26

    Assisted Solution

    the dL is a small change in the L.  If the real value of L was a bit off from the measured value, it would change the area by a little bit (a dL) Your possible L might be off by 3%. That means that your dA might be a bit off too. You calculated the possible bit off.

    Author Closing Comment

    ah, ok, I get it now. Thanks!
    LVL 37

    Expert Comment

    So did you end up with the area having a maximum of 8.15% error?

    1.03*1.05 = 1.0815

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