[VectorCalc] Tangent and normal vectors of r=<x(t),y(t),z(t)> appear to have the same formula?!

>If we think of T as a curve of its own, then N=T' / ||T'|| should literally calculate the tangent vector to T, no?

Yes, precisely.

>And even if we think of T as being the tangent vector to some curve r, this would mean that N is comprised of the second derivatives of r, which still doesn't make intuitive sense because in 2D we don't get perpendicular lines by differentiating twice.

Yes, the normal vector is related to the second derivative of the original curve.

You can't quite say that N=r''. You're right that r'' could point in any direction, but you can always decompose it into a tangential component plus another component orthogonal to that. The tangential component of the acceleration tells you how much you are speeding up or slowing down: it is a force acting along the direction of movement, so it is adding or removing energy in the usual work=force\*distance way.

The other component is the normal component of the acceleration, which causes your direction to change but does zero work. The unit vector in this direction is the standard normal vector N.

For example, the normal vector to a circle points toward the center of a circle. If you think of that circle as an orbit, it should make sense: gravity (a force, which causes acceleration, ie the second derivative) is pulling you towards the thing you are orbiting, which is at the center of the circle.

Also, to revisit this line again:

>in 2D we don't get perpendicular lines by differentiating twice.

You're right: you get perpendicular lines by differentiating *once*! I mean, not always of course, but certainly for something like T that satisfies |T|=1. Think about why this is. Hint: if T' was *not* perpendicular to T, then T'  would have a nonzero tangential component...
>in 2D we don't get perpendicular lines by differentiating twice.

You can go through this process for a plane curve and you will get perpendicular lines.

For example, if we take the curve r(t) = < cos(t) , sin(t) > then

r'(t) = < -sin(t), cos(t) >
||r'(t)|| = sqrt(sin^2 t + cos^2 t) = 1
T(t) = < -sin(t), cos(t) >
T'(t) = < -cos(t), -sin(t) >
||T'(t)|| = sqrt(cos^2 t + sin^2 t) = 1
N(t) = < -cos(t), -sin(t) >

Then T *is* perpendicular to N because we can calculate

T * N = sin(t) cos(t) - cos(t) sin(t) = 0

I chose a curve where things turned out nicely--everything was magnitude one which simplified the calculations a lot. In fact, the curve I chose traces out the unit circle, parametrized counter-clockwise, starting from the point (1, 0), at speed 1. It's a good example to keep in mind.

But you can try going through the same process for any example you want and you'll get the same result. You can try just using the general curve r(t) = < x(t), y(t) > and you'll see that everything cancels and you get zero when you compute the dot product. It always checks out.

As the other commenter said, this is a general thing that always happens for any **unit length** vector function of t. r'(t) has a component parallel to r(t) and a component perpendicular to it. But the component parallel to it represents the rate of change of the *length* of r(t). So if r(t) is a *unit* vector, so its length is always one, then the parallel component is always zero, and r'(t) is perpendicular to r(t). In this case, we're doing this with T(t). T is the *unit* tangent vector. So T'(t) is always perpendicular to T(t). And N(t) is the normal vector of T'(t) so it is too.

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