Finding the unit normal vector for a parametric curve at t=π

[kasse]

I don't know if I've used the correct english terms in this text. Anyway, I'm working with the curve

x(t)=cos(t)
y(t)=sin(t)
z(t)=t

I have found the position and velocity vectors and scalars at t=pi. I've also calculated the unit tangent vector by dividing the velocity vector by the instant speed.

Next, I found the acceleration vector at t=pi and the curvature, and I'm now about to find the unit normal vector at t=pi. How can I do that without calculating the tangent and normal component of the acceleration first? (I'm not allowed to to it that way, and I'm going to use the information to find the osculating circle).

 

[HallsofIvy]

The normal to the curve is the derivative of the unit tangent vector. To get the unit normal curve, of course, divide by its length. It happens to be particularly simple in this case!

 

[Dick]

Who says you 'aren't allowed'? The only other way I can think of to calculate an osculating circle is to take three points on the curve, say at t=pi, pi+delta, pi-delta, calculate the circle thru them and then let delta->0.