Find parametric equations of the osculating circle to the helix r(t)=cos(t)i+sint(t)j+tk corresponding to t=pi. Recall that the osculating circle is the best possible circle approximating a curve C at a given point P. It lies in the osculating plane (i.e., the plane containing T and N) of the curve at the point P, touches the curve at P, and its radius is equal to the reciprocal of the curvature of C at point P.
T(t) = r'(t) / [length of r'(t)]
N(t) = T'(t) / [length of T'(t)]
curvature k(t) = [r'(t)Xr''(t)] / [length of r'(t)]^3
The Attempt at a Solution
(I think these are correctly calculated! Please check!)
So r(t) = <cos t, sin t, t>. At t=pi, r(t)= <-1, 0, pi>
r'(t) = <-sin t, cos t, 1>. At t=pi, r'(t)= <0, -1, 1>
length of r'(t) = sqrt(2)
r''(t) = <-cos t, -sin t, 0>. At t=pi, r''(t)= <1, 0, 0>
So since the radius is 1/curvature, then the radius is 2.
T(t)= 1/sqrt(2) <-sin t, cos t, 1>. At t=pi, T(t)= <0,-1/sqrt(2), 1/sqrt(2)>
N(t)=<-cos t, -sin t, 0>. At t=pi, N(t)= <-1, 0, 0>
Now... how do I find the center? I believe I must use the normal vector to the curve somehow, but can someone explain how exactly?
After finding the center and radius, how do I get the parametric equations of the circle?
Please help ASAP! Thank you!!!