Results 1 to 3 of 3

Thread: Curvature of a rectangular hyperbola

  1. #1
    Newbie
    Joined
    Sep 2009
    From
    Dunedin
    Posts
    3

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    From
    Dunedin
    Posts
    3
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rectangular Hyperbola Problem
    Posted in the Geometry Forum
    Replies: 0
    Last Post: Apr 17th 2010, 06:04 AM
  2. Rectangular Hyperbola - Equ. of Tangents
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Apr 9th 2010, 09:27 PM
  3. rectangular hyperbola
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Aug 7th 2009, 05:34 AM
  4. Rectangular Hyperbola
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Apr 20th 2008, 03:54 AM
  5. Rectangular Hyperbola
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Feb 12th 2008, 12:54 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum