Results 1 to 5 of 5

Math Help - Osculating Circle for a Curve

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    42

    Osculating Circle for a Curve

    I lost trying to find the osculating circle for the curve defined by r(t) = <sint, cost, -pi + t> at the point t=pi.

    Ive calculated the radius of the circle to be 1 and I know I need to somehow use the unit normal vector at point t=pi, which is <0, 1/sqrt(2), 0> to solve for the osculating circle's centre, but I do not know how to do that.

    Thans for your help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by JoAdams5000 View Post
    I lost trying to find the osculating circle for the curve defined by r(t) = <sint, cost, -pi + t> at the point t=pi.

    Ive calculated the radius of the circle to be 1 and I know I need to somehow use the unit normal vector at point t=pi, which is <0, 1/sqrt(2), 0> to solve for the osculating circle's centre, but I do not know how to do that.

    Thans for your help!
    That is not a unit vector.

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2009
    Posts
    42
    Right...i calculated that wrong...

    The unit normal vector is actually [0, -1, 0] from the point where t=pi.

    How can I use this to calculate the center of the osculating circle if possible?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,812
    Thanks
    117
    Quote Originally Posted by JoAdams5000 View Post
    I lost trying to find the osculating circle for the curve defined by r(t) = <sint, cost, -pi + t> at the point t=pi.

    Ive calculated the radius of the circle to be 1 and I know I need to somehow use the unit normal vector at point t=pi, which is <0, 1/sqrt(2), 0> to solve for the osculating circle's centre, but I do not know how to do that.

    Thans for your help!
    1. The stationary vector of point P when t = \pi is

    \vec p=(0,-1,0).

    2. I've got the normal unit vector as \vec n = (0,1,0)

    3. The sum of these two vectors will yield the stationary vector of the center of the circle:

    \vec p + \vec n = (0,0,0)

    4. The tangent at the curve in P and the normal vector produce a plain in which the osculating circle must be placed. The equation of the tangent is:

    t: \overrightarrow{r(t)}=(0,-1,0)+t \cdot (-1,0,1)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,812
    Thanks
    117
    In addition to my previous post I've attached a drawing containing

    the curve: \overrightarrow{r(t)}=(\sin(t), \cos(t), -\pi+t)

    the tangent: <br />
t: \overrightarrow{r(t)}=(0,-1,0)+t \cdot (-1,0,1)<br />

    the circle: \overrightarrow{r(t)}=(\sin(t), \cos(t), -\sin(t))
    Attached Thumbnails Attached Thumbnails Osculating Circle for a Curve-raumkurve_tang_krummkreis.png  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Osculating circle at given points
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 30th 2010, 03:15 AM
  2. Find the Osculating Circle
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 24th 2010, 12:23 AM
  3. determining radius of osculating circle
    Posted in the Geometry Forum
    Replies: 2
    Last Post: October 22nd 2009, 11:59 AM
  4. Vector Calculus (osculating circle)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 21st 2009, 04:29 AM
  5. osculating circle problem please!!!
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 26th 2009, 12:26 PM

Search Tags


/mathhelpforum @mathhelpforum