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Math Help - Curvature of a rectangular hyperbola

  1. #1
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    Curvature of a rectangular hyperbola

    I'm trying to figure out what the curvature of the hyperbola given by y = 1/x is, and I already know the answer. Unfortunately, I'm getting a strange factor of x^3 in my answer.

    Essentially:

    y = 1/x

    Curvature formula:

    |f''(x)|/((1+(f'(x))^2)^(3/2))

    So

    f'(x) = -1/(x^2)
    f''(x) = 2/(x^3)

    And therefore

    k = |2/(x^3)|/((1/(x^4) + 1)^(3/2))
    = 2/(|t^3| * (1 + 1/(t^4))^(3/2)) (as x is the same as the parameter t, (x, y) = (t, 1/t))

    and p = 1/t = 1/2 * (1 + 1/(t^4))^(3/2) * |t^3|

    But, the curvature which leads to the correct evolute for the curve (which is what I'm really looking for) is simply 1/2(1 + 1/(t^4))^(3/2), without the |t^3|. I've been working on this for an hour, and I still can't figure out why it disappears. Thanks for any help.
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  2. #2
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    Quote Originally Posted by StarWrecker View Post
    I'm trying to figure out what the curvature of the hyperbola given by y = 1/x is, and I already know the answer. Unfortunately, I'm getting a strange factor of x^3 in my answer.

    ... [SNIP] ...

    And therefore

    k = |2/(x^3)|/((1/(x^4) + 1)^(3/2))
    = 2/(|t^3| * (1 + 1/(t^4))^(3/2)) (as x is the same as the parameter t, (x, y) = (t, 1/t))

    and p = 1/t = 1/2 * (1 + 1/(t^4))^(3/2) * |t^3|

    But, the curvature which leads to the correct evolute for the curve (which is what I'm really looking for) is simply 1/2(1 + 1/(t^4))^(3/2), without the |t^3|. I've been working on this for an hour, and I still can't figure out why it disappears.
    Your formula k = \frac2{|t^3|(1 + 1/(t^4))^{3/2}} can be written k = \frac2{(t^2 + t^{-2})^{3/2}} (by writing |t|^3=(t^2)^{3/2}). That is the correct answer. It is symmetric in t and t^{-1}, as it has to be on geometric grounds (just think of the graph of the hyperbola: it is symmetric about the line y=x, so it must have the same curvature at x=t^{-1} as it does at x=t).

    So your calculation is absolutely correct, but unfortunately for you "the curvature which leads to the correct evolute" is not the correct curvature.
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  3. #3
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    Oh, I see what I did there. I must have made a mistake elsewhere, because 2/(t^2 + t^-2)^(3/2) works for the evolute now. Thanks a lot. I didn't think of dealing with the |t^3| that way.
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