# Center of osculating circle

Gold Member
Ok so I have a T vector and N vector...

$$\begin{array}{l} T(1) = \langle \frac{2}{3},\frac{{ - 1}}{3},\frac{2}{3}\rangle \\ N(1) = \langle \frac{2}{3},\frac{2}{3},0\rangle \\ B(1) = T(1) \times N(1) \\ B(1) = \langle \frac{{ - 1}}{3},\frac{2}{3},\frac{2}{3}\rangle \\ \end{array}$$

I also have the coordinate of the original equation at 1...$$r(1) = \langle 2,1,0\rangle$$
This left me with the standard equation of the osculating plane...

$$- x + 2y + 2z = 0$$

Now I need to find the coordinates of hte center of this circle where t=1. How am i suppose to do this?

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HallsofIvy
Homework Helper
$$T(1) = \langle \frac{2}{3},\frac{{ - 1}}{3},\frac{2}{3}\rangle$$
Okay, T, I assume, is the unit normal vector at t= 1.
$$N(1) = \langle \frac{2}{3},\frac{2}{3},0\rangle$$
This, I have a problem with! I would guess that N is supposed to be a normal vector but it neither has unit length nor is normal to T(1).
Did you intend $$N(1) = \langle \frac{2}{3},\frac{2}{3},-frac{1}{3}\rangle$$?

If that was what you intended, then you cannot find the center of the osculating circle without know its radius: the radius of curvature which is 1 over the curvature. The curvature is the length of the derivative of the unit tangent vector with respect to arclength. If you are only given the unit tangent and normal vectors at t=1, you cannot find that directly. There exist many curves having the given tangent and normal vectors at a point with different curvatures and different osculating circles.

Gold Member
Uhh... it should be normal to T, i ran this all through mathematica. I also have the radius at 9/2, didnt' know that was important.

haha after finding an equation online for the center.. im amazed how I couldn't figure it out on my own. If you have a point on that circle and you have teh radius and the normal vector at that point... it should be a tad bit obvious as to how you find that center point.

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