Parametrization of the osculating circle to a space curve?











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Find a parametrization of the osculating circle to r(t)= at t=0



So I found the center of the osculating circle by calculating the radius of curvature and the normal vector. I've also found the equation of the osculating plane in which the circle will lie, but I'm stuck on how to parametrize the circle on this plane.



For time's sake, here are the unit tangent, unit normal, and binormal vectors, the center of the osculating circle, and the equation for the plane:



T(t)= (1/sqrt(2))<-sin(7t),cos(7t),1>
N(t)=<-cos(7t),-sin(7t),0>
B(0)=<0,1/sqrt(2),1/sqrt(2)>



center of osculating circle at t=0, = <-1,0,0>
equation for osculating plane at t=0, = (1/sqrt(2))y+(1/sqrt(2))z=0



at this step, I tried to find the equation of a sphere with the same center and radius as the osculating circle, which gave me (x+1)^2 + y^2 + z^2 =4 and then find the intersection of this sphere with the osculating plane. Solving for y in terms of z in the osculating plane equations yields y=-z, and when plugged into the equation for the sphere, yields (x+1)^2 + 2y^2 = 4. I don't know if i'm on the right track here, but this is as far as i've been able to get.



If anyone could help me out, or show me an entirely different way of approaching the problem, it would be greatly appreciated.



Thanks!










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    up vote
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    down vote

    favorite












    Find a parametrization of the osculating circle to r(t)= at t=0



    So I found the center of the osculating circle by calculating the radius of curvature and the normal vector. I've also found the equation of the osculating plane in which the circle will lie, but I'm stuck on how to parametrize the circle on this plane.



    For time's sake, here are the unit tangent, unit normal, and binormal vectors, the center of the osculating circle, and the equation for the plane:



    T(t)= (1/sqrt(2))<-sin(7t),cos(7t),1>
    N(t)=<-cos(7t),-sin(7t),0>
    B(0)=<0,1/sqrt(2),1/sqrt(2)>



    center of osculating circle at t=0, = <-1,0,0>
    equation for osculating plane at t=0, = (1/sqrt(2))y+(1/sqrt(2))z=0



    at this step, I tried to find the equation of a sphere with the same center and radius as the osculating circle, which gave me (x+1)^2 + y^2 + z^2 =4 and then find the intersection of this sphere with the osculating plane. Solving for y in terms of z in the osculating plane equations yields y=-z, and when plugged into the equation for the sphere, yields (x+1)^2 + 2y^2 = 4. I don't know if i'm on the right track here, but this is as far as i've been able to get.



    If anyone could help me out, or show me an entirely different way of approaching the problem, it would be greatly appreciated.



    Thanks!










    share|cite|improve this question
























      up vote
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      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Find a parametrization of the osculating circle to r(t)= at t=0



      So I found the center of the osculating circle by calculating the radius of curvature and the normal vector. I've also found the equation of the osculating plane in which the circle will lie, but I'm stuck on how to parametrize the circle on this plane.



      For time's sake, here are the unit tangent, unit normal, and binormal vectors, the center of the osculating circle, and the equation for the plane:



      T(t)= (1/sqrt(2))<-sin(7t),cos(7t),1>
      N(t)=<-cos(7t),-sin(7t),0>
      B(0)=<0,1/sqrt(2),1/sqrt(2)>



      center of osculating circle at t=0, = <-1,0,0>
      equation for osculating plane at t=0, = (1/sqrt(2))y+(1/sqrt(2))z=0



      at this step, I tried to find the equation of a sphere with the same center and radius as the osculating circle, which gave me (x+1)^2 + y^2 + z^2 =4 and then find the intersection of this sphere with the osculating plane. Solving for y in terms of z in the osculating plane equations yields y=-z, and when plugged into the equation for the sphere, yields (x+1)^2 + 2y^2 = 4. I don't know if i'm on the right track here, but this is as far as i've been able to get.



      If anyone could help me out, or show me an entirely different way of approaching the problem, it would be greatly appreciated.



      Thanks!










      share|cite|improve this question













      Find a parametrization of the osculating circle to r(t)= at t=0



      So I found the center of the osculating circle by calculating the radius of curvature and the normal vector. I've also found the equation of the osculating plane in which the circle will lie, but I'm stuck on how to parametrize the circle on this plane.



      For time's sake, here are the unit tangent, unit normal, and binormal vectors, the center of the osculating circle, and the equation for the plane:



      T(t)= (1/sqrt(2))<-sin(7t),cos(7t),1>
      N(t)=<-cos(7t),-sin(7t),0>
      B(0)=<0,1/sqrt(2),1/sqrt(2)>



      center of osculating circle at t=0, = <-1,0,0>
      equation for osculating plane at t=0, = (1/sqrt(2))y+(1/sqrt(2))z=0



      at this step, I tried to find the equation of a sphere with the same center and radius as the osculating circle, which gave me (x+1)^2 + y^2 + z^2 =4 and then find the intersection of this sphere with the osculating plane. Solving for y in terms of z in the osculating plane equations yields y=-z, and when plugged into the equation for the sphere, yields (x+1)^2 + 2y^2 = 4. I don't know if i'm on the right track here, but this is as far as i've been able to get.



      If anyone could help me out, or show me an entirely different way of approaching the problem, it would be greatly appreciated.



      Thanks!







      multivariable-calculus parametrization osculating-circle






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      asked Oct 2 '16 at 18:32









      Cole Bisaccia

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          $newcommand{Vec}[1]{mathbf{#1}}$Generally, if $Vec{v}_{1}$ and $Vec{v}_{2}$ are orthogonal unit vectors in $Vec{R}^{n}$, then
          $$
          Vec{x}(t) = Vec{p}_{0} + rbigl((cos t)Vec{v}_{1} + (sin t)Vec{v}_{2}bigr)
          $$
          parametrizes the circle of center $Vec{p}_{0}$ and radius $r$ in the plane spanned by $Vec{v}_{1}$ and $Vec{v}_{2}$.






          share|cite|improve this answer





















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            $newcommand{Vec}[1]{mathbf{#1}}$Generally, if $Vec{v}_{1}$ and $Vec{v}_{2}$ are orthogonal unit vectors in $Vec{R}^{n}$, then
            $$
            Vec{x}(t) = Vec{p}_{0} + rbigl((cos t)Vec{v}_{1} + (sin t)Vec{v}_{2}bigr)
            $$
            parametrizes the circle of center $Vec{p}_{0}$ and radius $r$ in the plane spanned by $Vec{v}_{1}$ and $Vec{v}_{2}$.






            share|cite|improve this answer

























              up vote
              0
              down vote













              $newcommand{Vec}[1]{mathbf{#1}}$Generally, if $Vec{v}_{1}$ and $Vec{v}_{2}$ are orthogonal unit vectors in $Vec{R}^{n}$, then
              $$
              Vec{x}(t) = Vec{p}_{0} + rbigl((cos t)Vec{v}_{1} + (sin t)Vec{v}_{2}bigr)
              $$
              parametrizes the circle of center $Vec{p}_{0}$ and radius $r$ in the plane spanned by $Vec{v}_{1}$ and $Vec{v}_{2}$.






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                $newcommand{Vec}[1]{mathbf{#1}}$Generally, if $Vec{v}_{1}$ and $Vec{v}_{2}$ are orthogonal unit vectors in $Vec{R}^{n}$, then
                $$
                Vec{x}(t) = Vec{p}_{0} + rbigl((cos t)Vec{v}_{1} + (sin t)Vec{v}_{2}bigr)
                $$
                parametrizes the circle of center $Vec{p}_{0}$ and radius $r$ in the plane spanned by $Vec{v}_{1}$ and $Vec{v}_{2}$.






                share|cite|improve this answer












                $newcommand{Vec}[1]{mathbf{#1}}$Generally, if $Vec{v}_{1}$ and $Vec{v}_{2}$ are orthogonal unit vectors in $Vec{R}^{n}$, then
                $$
                Vec{x}(t) = Vec{p}_{0} + rbigl((cos t)Vec{v}_{1} + (sin t)Vec{v}_{2}bigr)
                $$
                parametrizes the circle of center $Vec{p}_{0}$ and radius $r$ in the plane spanned by $Vec{v}_{1}$ and $Vec{v}_{2}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 2 '16 at 18:44









                Andrew D. Hwang

                52.2k447111




                52.2k447111






























                     

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