Osculating Circle for vector-valued function?

  1. Hi! This is my first time on Physics Forum. (which shows how desperate I am on figuring out this question).

    1. The problem statement, all variables and given/known data

    r(t) = <3sin(t),4cos(t)>

    There is a unique circle with the following properties:

    1. It passes through the point r(∏/2)
    2. At the point r(∏/2), the tangent line to r(t) and the tangent line to the circle are the same.
    3. The radius of the circle is 1/κ, where κ is the curvature of r(t) at t = ∏/2
    4. The center of the circle lies on the concave side of the surve

    2. Relevant equations

    κ = |dT/ds|
    T = r'(t)/||r'(t)||
    N = T'(t)/||T'(t)||
    Not sure if I need more equations for this??

    3. The attempt at a solution

    So I know that the graph they gave is an elipse, so I can easily visualize what the hint #2 says. I know how to find curvature, so that gives me the radius of the circle. I just need the center. Based on the info above, the center lines on the concave side of the curve. So that means it's in the same direction as N(t). But not sure where to go from there. Any help would be greatly appreciated.
     
  2. jcsd
  3. LCKurtz

    LCKurtz 8,269
    Homework Helper
    Gold Member

    Did you graph your ellipse? You know r(1) and r'(1) and you know N(1) is perpendicular to r'(1) and points to the concave side. Draw a picture and if you know the radius, it will be obvious where the center is.
     
  4. Yeah I graphed the ellipse. By 1, do you mean pi/2?

    I guess the main doubt I have is that I'm not sure what the concave side means. The whole ellipse is looks convex.
     
  5. So I have the radius. I tried to solve N(t), but that yielded something hopelessly convoluted, and I think I only need to know the direction of N(t) (which is perpendicular to the tangent vector). So does the center lie on the bottom of the ellipse? This is confusing me.
     
  6. LCKurtz

    LCKurtz 8,269
    Homework Helper
    Gold Member

    Yes I meant ##\pi/2##, sorry. An ellipse looks convex if you view it from its exterior, but concave if you view it from its interior. So the concave side of the curve is the inside of it. Your normal points inward.


    Where is the point when ##t=0## and ##t=\pi/2##? I don't think it is at the "bottom"" of the ellipse. At ##t=\pi/2## what direction is the tangent vector? At ##t=\pi/2##, the normal is perpendicular to the tangent and points inward. What direction is that on this graph? You don't need to do those complicated calculations to figure that out.
     
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