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Partial Derivation to calculate percentage relative uncertainty. Plz help

  1. Oct 22, 2006 #1

    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


    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
  3. Oct 23, 2006 #2


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    Science Advisor

    Why should 100*dz/z be 1? That's what you are asked to find. Also notice that dx and dy are the actual "uncertainties", not the "percentage relative" but you are given "percentage relative" uncertainties when you are told 1%. That is what is meant by 100 dx/x= 1. dx= 0.1(x) and dy= 0.1(y). Finally, you are differentiating wrong there should be a factor of [itex](x^2+ y^2)^{-0.5}[/itex]. It's simpler to use implicit differentiation: if [itex]x^2+ y^2= z^2[/itex], then [itex]2x= 2z\frac{\partial z}{\partial x}[/itex] so that [itex]\frac{\partial z}{\partial x}= \frac{x}{z}[/itex].
    Of course, [itex]\frac{\partial z}{\partial y}= \frac{y}{z}[/itex] so that
    [tex]dz= \frac{x}{z}dx+ \frac{y}{z}dy[/tex].

    Why are you using only the derivative with respect to z?
    [itex]0= 2x- 2ycos(\theta)+ 2xy sin(\theta)\frac{\partial \theta}{\partial x}[/itex] so
    [tex]\frac{\partial \theta}{\partial x}= \frac{2x- 2y cos(\theta)}{2xy sin(\theta)}[/tex]
    [tex]\frac{\partial \theta}{\partial y}= \frac{2y- 2x cos(\theta)}{2xy sin(\theta)}[/tex]
    [tex]\frac{\partial \theta}{\partial z}= \frac{2z}{-2xy sin(\theta)}[/tex]

    [tex]d\theta= \frac{\partial \theta}{\partial x}dx+ \frac{\partial \theta}{\partial y}dy+ \frac{\partial \theta}{\partial z}dz[/tex]
  4. Jan 21, 2012 #3
    Vo =x√(g/2h)
    can you please help me ...i have to find the absolute error using the derivate .Does anyone knows the formula???
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