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Curvature of a rectangular hyperbola

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data
    The hyperbola y = 1/x in the first quadrant can be given the parametric definition (x, y) = (t, 1/t), t>0.

    Find the corresponding parametric form of its evolute, and sketch both curves in the region 0<x<10, 0<y<10

    2. Relevant equations

    Curvature formula:

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

    3. The attempt at a solution

    I've worked through this in full, but there is a single term in my curvature that is throwing off my end result. My calculation of this curvature is as follows:

    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.
     
  2. jcsd
  3. Sep 12, 2009 #2
    Hi,StarWrecker,I am 100 percent sure "p = 1/t = 1/2 * (1 + 1/(t^4))^(3/2) * |t^3|"
    is correct.
    And I am also sure that p(t) leads to the correct evolute for the curve.
    Check your answer again and again,you will find out your errors on computation.
     
  4. Sep 14, 2009 #3
    Yeah, I made a mistake in my later calculations for the evolute that I ironed out after trying my solution for the curvature again. Thanks.
     
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