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Desperate Help Needed

  1. Mar 9, 2009 #1
    Can someone tell me how to claculate the uncretinty on

    (R^2 + z^2) ^3/2
     
  2. jcsd
  3. Mar 9, 2009 #2

    gabbagabbahey

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    Assuming R and z are independent variables, you do it the same way you calculate the uncertainty in any multi variable function:

    [tex]\Delta f(R,z) \approx \left| \frac{\partial f}{\partial R} \right| \Delta R+\left| \frac{\partial f}{\partial z} \right| \Delta z[/tex]
     
  4. Mar 9, 2009 #3
    Isn't that just for (z^2 + R^2)?
     
  5. Mar 9, 2009 #4
    Also if you have a cos/sin function in an equation(multiplying) how would you do it for cos/sin?

    is it just for cos x = sinxdx/cosx and cosxdx/sinx for sin x?
     
  6. Mar 9, 2009 #5

    gabbagabbahey

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    Why would it apply to one multi variable function, but not the others?
     
  7. Mar 9, 2009 #6
    So would I just take the uncertainty on z^2 + R^2, then multiply in by 3/2 and squareroot z^2 + R^2?
     
  8. Mar 9, 2009 #7

    gabbagabbahey

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    Do you know how to take derivatives? And what happened to the absolute value brackets?....Errors are never negative.

    [tex]\left| \frac{\partial}{\partial x} (\cos x) \right|\Delta x= |-\sin(x)|\Delta x=|\sin(x)|\Delta x \neq \frac{\sin x}{\cos x}dx[/tex]

    Also, [itex]\Delta x[/itex] is the uncertainty in x, not the differential 'dx'.
     
  9. Mar 9, 2009 #8

    gabbagabbahey

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    No, you would compute the absolute value of the partial derivative of [itex](R^2+z^2)^{3/2}[/itex] with respect to R and multiply it by the uncertainty in R; then do the same with respect to z, and then add them together...
     
  10. Mar 9, 2009 #9
    Ok thats for cos x, but I need (unceratinty in cosx)/cosx. The relative over the value.
     
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