   Frenet Frame
  It is said that for a curve c(t), the tangent vector T = c'(t) and normalis N = T' = c''(t).Say we have c(t) = ( t, sin(t), 0 ).  T = ( 1, cos(t), 0 ).  N = ( 0,-sin(t) 0 ).  Based on this, it looks like the normal is always in the ydirection, which seems quite untrue for a sine function. Am I missingsomething?Thanks.   
  I think, the normal is the first derivative,  linear: 2 = 2x - ythe normal is (2,-1), un-nomalizedsphere: r^2 = x^2 + y^2 + z^2normal: (2x,2y,2z)=(x,y,z);  If you look at any reference, it say the tangent is the first derivative, which make sense since it's the velocity vector.   ;  Yeah that's true, i jus never heard of the normal being a derivative of the tangent.The tangent is perpendicular to the normal, which the tangent is a plane(or 2D Line).The normal, i believe, is aquired when you plug the numbers into the first derivative. Then you atttain a vector, which is the normal to the tanget plane. ;  The normal N to the curve is given by the derivative of the *unit-length* tangent. So in your case, you need to take the derivative of <1, cos(t), 0> / sqrt(1 + cos^2(t)), which isN(t) = sin(t) / (1 + cos^2(t))^(3/2) * <cos(t), -1, 0>