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Homework Help: Finding the error in z

  1. Oct 12, 2014 #1
    1. The problem statement, all variables and given/known data

    For the equation [tex] q = \frac{z(z+2) - 2DH}{z^2} [/tex] [itex] q = -0.6 \pm 10\% [/itex] , and z = 0.2.
    D and H are known exactly.
    I have to find the error in z that will give an answer of [itex] q = -0.6 \pm 10\% [/itex]

    2. Relevant equations

    3. The attempt at a solution

    I have considered rewriting the equation in terms of z, which gives
    [tex] z = \frac{1 \pm \sqrt{1-2DH(q-1)}}{q-1} [/tex]
    but I'm not sure where to go with that, how the plus/minus affects it, and what to do with D and H (if anything).

    I've tried to do (error in q * 0.5) / (error in q) which gives an answer for z of [itex] 0.2 \pm 10\% [/itex] but that seems a bit too simple
  2. jcsd
  3. Oct 12, 2014 #2


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    Homework Helper

    It is not needed. You know that z=0.2. From that, you can determine DH. You know also q and its error.
    The error of z can be obtained by differentiating both sides of the equation
    [tex]q = \frac{z(z+2) - 2DH}{z^2}[/tex]
    [tex]\Delta q =\frac {dq}{dz}\Delta z[/tex] Substitute z, DH, and ##\Delta q##. Solve for ##\Delta z##.

  4. Oct 12, 2014 #3
    so I get this:

    [tex] DH = \frac{z(z+2) - z^2 q}{2} = 0.232 [/tex]
    [tex] \frac{dq}{dz} = \frac{4DH - 2z}{z^3} = 66 [/tex]
    [tex] \Delta q = \frac{dq}{dz} \Delta z [/tex]
    [tex] \Delta z = \frac{0.06}{66} = 9 \times 10^{-4} [/tex]

    The error seems really small..?
  5. Oct 13, 2014 #4


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    Homework Helper

    It is almost right. You made some small mistake when calculating DH. Check.

    The error is small, but z=0.2, so its relative error is about 0.5 %.

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