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Parabolic Curve

  1. Dec 19, 2008 #1
    Hi,

    Just wondering if anyone could tell me how I would go about plotting the parabolic curve for the equation:

    sigmaY*(1-(L/k)^2/2*(L/k)c^2)

    Any help most appreciated

    Will
     
  2. jcsd
  3. Dec 19, 2008 #2

    HallsofIvy

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    First of all, that's not an equation. Second, your notation is ambiguous.
    Do you mean
    [tex]f(c)= sigmaY*(1- \frac{(L/k)^2}{2}(L/k) c^2)= sigmaY*(1- (L/k)^3c^2/2)[/tex]
    or
    [tex]f(c)= sigmaY(1- \frac{(L/k)^2}{2(L/k)} c^2= sigmaY*(1- (L/k)c^2/2)[/tex]
    or
    [tex]f(x)= sigmaY(1- \frac{L/k)^2}{2(L/k)c^2}= sigmaY*(1- L/(2kc^2))[/tex]
    Since the last is not a quadratic and its graph is not a parabola, you must mean one of the first two. In either case, that is f(c)= A- Bc2 (what A and B are depends on which of the two you meant) which has vertex at (0, A) and opens downward.
     
  4. Dec 19, 2008 #3
    I will try and write the equation properly next time but I dont know how to use latex.

    Ok the equation I have is the third one? but with sigma instead of f(x).

    It is from a lab report and have been asked to draw the parabola for it. The original thread is here,

    https://www.physicsforums.com/showthread.php?t=280106

    Thanks for your help.
     
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