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Random error calculation

  1. May 16, 2008 #1
    1. Data
    Outer Diameter of tube=(64 +/- 2) and internal diameter= (47 +/- 1). Find cross-sectional area of tube and find percentage uncertainty.

    2. Equations

    3. The attempt at a solution

    I got (Delta A)/A = (Delta D2-d2)/(D2-d2) where D2-d2=1887 +/- 198. Hence my A= 1500 and my delta A is 157.
    But the correct answer is 1500 +/- 300. And percentage uncertainty is 20%.

    Can anyone give me the equations of the uncertainties??? For example there is another question about z=2x2 + y. Can you give me the equation? Im still very blur about this topic.
  2. jcsd
  3. May 16, 2008 #2
    Perhaps you mean:

    [tex]\Delta_{f(x,y,z)} = \left| \frac{ \partial f}{\partial x} \right| \Delta_x + \left| \frac{ \partial f}{\partial y} \right| \Delta_y + \left| \frac{ \partial f}{\partial z} \right| \Delta_z[/tex]
    Here, [tex]\left| \frac{ \partial f}{\partial x} \right|[/tex] denotes the absolute value of the partial derivative of f to x.
    [tex]\Delta_x[/tex] denotes the error in x.

    For your question:
    [tex]\text{Outer diameter} = d_o = 64 \pm 2[/tex]
    [tex]\text{Inner diameter} = d_i = 47 \pm 1[/tex]

    [tex]\text{Area} = A = \frac{1}{4} \pi ( d_o^2 - d_i^2 )[/tex]

    This only depends on two variables so just 'ignore' the z part in above general equation..

    [tex]\Delta_A = \left| \frac{ \partial A}{\partial d_o} \right| \Delta_{d_o} + \left| \frac{ \partial A}{\partial d_i} \right| \Delta_{d_i}[/tex]

    Computing this I got a value for delta_A of 274, which you have to round to one significant number, so the answer is:
    [tex]A = (1.5 \pm 3) \times 10^2 \text{(whatever units your using)}[/tex]

    I can't make sense out of z = 2x2 + y. Which do you mean?:
    [tex]\text{a) } z = 2x (2+y)[/tex]
    [tex]\text{b) }z = 2 \times 2 + y = 4 + y[/tex]
    [tex]\text{c) }z = 2x2 + y = 4x + y[/tex]
    Last edited: May 16, 2008
  4. May 16, 2008 #3
    Hmmm. Im still a beginner so I may not really understand that. (Im in junior high sch by the way) My teacher taught me 3 basic ones such as the addition one such as y=x+z and delta y= delta x + delta z . Multiplication and divisions are done using fractional uncertainties. Exponential just shift the power times the fractional uncertainty (e.g. y=x3 so delta y=3 delta x).

    Because most of my questions (which are multiplications) , I tend to ignore constants and just use fractional uncertainty method. But does it mean that for additions, coefficients of the variables need to be considered? I know that for z=2x2 +y (which is the option c) or z=2 'x' squared, is equal to z=x2+x2+y so the uncertainty of the x2 must be considered twice. So like if z=π x , does it mean that uncertainty of z = π times uncertainty of x? And am I right in saying that when using fractional uncertainty method, the coefficients need not be considered?
  5. May 18, 2008 #4
    Anyone? Is my question not clear or anything?
  6. May 18, 2008 #5
    No, I don't really understand your question...

    I understand you don't need to use the formula I listed, but you merely use a few rules?
    I don't understand the questions underlined at all...
    Can you give a clear example of what you mean, how you tried to solve it and why it is wrong?
    Can you try using latex or at least make it clear which formulas you are using, because I'm still not sure...

    You say the formula z = 2x2 + y means option c, or [itex]z = 2x2 + y[/itex] which is the same as [itex]z = 4x + y[/itex] and which doesn't make any sense (why do you write 2x2 instead of 4x?
    Then you say you mean either [itex]z = 2x^2 + y[/itex] or [itex] z = (2x)^2 + y = 4x^2 + y[/itex], no?
  7. May 19, 2008 #6
    I cant do the square option..... I dont use 'x' as multiplication, i dont use the multiplication sign, sry if you misunderstood. All numbers at the back are powers.
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