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Help! Partial Derivation to calculate percentage relative uncertainty

  1. Oct 22, 2006 #1
    Hello.

    I've encountered a problem when trying to attack this problem:

    A right triangle has sides of length x, y and z. Measurements show that legs x and y have lengths of 119ft and 120ft, respectively. The percentage relative uncertainty in each measurement is 1%. For x, this means that 100 * dx / x = 1. (That 1 would be reported as 1%)

    A)Use the pythagorean theorem to calculate the percentage relative uncertainty in z

    B)Use the law of cosines and your result from part a to calculate the percentage relative uncertainty in the right angle.

    So for part A, z = (x^2 + y^2)^.5

    therefore,

    dz = (.5(2x)^-.5)dx + (.5(2y)^-.5)dy

    if I simply substitute the values in for x, dx, y and dy I get a bizzare answer for the percentage relative unceratainty, and 100*dz/z should be 1. Where am I going wrong with this problem?

    and for part B)

    z^2 = x^2 + y^2 - 2xycos(theta)

    so the partial derivation would be

    2z = -2xy(-sin(theta) * curlyd(theta)/curlydz)

    but I am completely stuck where to go from here.

    A help will be greatly appriciated!
     
  2. jcsd
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