# Normal/Tangent to space curve

1. Jun 18, 2010

Hello all,

Given a 3d vector function f(t) that traces out a path in space, how can I find the normal and tangent vectors at any location along the curve?

Cheers,

2. Jun 18, 2010

### Gib Z

Well, for us to have a tangent at $P(x(t_1), y(t_1), z(t_1) )$ , the direction the tangent should be pointing in the same direction as the point P is "going" right at that instant when t= t_1. We could break it into dimensions and think about how fast P is going in just the x-direction, how fast it's going in the y-direction and how fast in the z-direction. How could we figure that out?

3. Jun 18, 2010

Right, so it would be the first derivative, or f'(t). But the normal?

4. Jun 18, 2010

### Gib Z

Well theres no first derivative, you have a parametrically defined function of 3 variables. So in the x direction, its x'(t), then in y, y'(t) and in z axis, z'(t).

For the normal, I'm not sure. How would you define the normal vector in this case? Normally its the vector perpendicular to the Tangent, but we get a whole Plane that has that for this case.

5. Jun 18, 2010

### HallsofIvy

If $\vec{f}(t)= u(t)\vec{i}+ v(t)\vec{j}+ w(t)\vec{k}$, then a tangent vector is given by $\vec{f}'(t)= u'(t)\vec{i}= v'(t)\vec{j}+ w'(t)\vec{k}$. The unit tangent vector is that vector divided by its length and the normal vector is the derivative of the unit tangent vector with respect to t.