# Unit tangent vector in 3D

1. Feb 18, 2006

### Stevecgz

Unit tangent vector in 3D - and what am I doing wrong with latex?

Question: At a given point on a curve, does the unit tangent vector given by
$$\frac{\vec{r}'(t)}{|\vec{r}'(t)|}$$
depend on the direction in which the curve is being swept out?

My initial thought on this was that the unit tangent vector didn't depend on direction because it only depends on $$t$$. But now I am thinking that it would matter, but only in the case that $$\vec{r}(t)$$ was not continuous at $$t$$. Any thoughts or advice?

Thanks,
Steve

And what am I doing wrong that latex isn't working?

Last edited by a moderator: Feb 18, 2006
2. Feb 18, 2006

### dicerandom

Think about what you mean by "direction in which the curve is swept out," if you mean what I think you mean (or at least what *I* think it means ) then it's basically the definition of the derivative, i.e. how r(t) is changing with time.

3. Feb 18, 2006

### Stevecgz

The derivative is independent of the direction which the curve is traced out, right?

4. Feb 18, 2006

### dicerandom

No, that's equivalent to saying that velocity is independent of the direction of motion. The magnitude of the derivative, i.e. the speed, doesn't depend on direction, but the actual vector quantity of velocity does.

Consider the simple example of a ball which is on the end of a string fixed to the origin. If the ball is going around the origin counter-clockwise with a frequency $\omega$ then its position is:

$$\vec{r}(t) = \cos(\omega t) \hat{x} + \sin(\omega t) \hat{y}$$

And its velocity is:

$$\vec{r^\prime}(t) = - \omega \sin(\omega t) \hat{x} + \omega \cos(\omega t) \hat{y}$$

If you want the ball to go around the other direction just replace t with -t and I think you will be able to see that the velocity vectors point in opposite directions.

Last edited: Feb 18, 2006
5. Feb 18, 2006

### HallsofIvy

Stevecgz, I corrected your Latex- you were using [\tex] to end when it should be [/tex].

6. Feb 18, 2006

### Stevecgz

Thanks for the reply dicerandom. I think I understand your analogy. So would I be correct in saying that for any given point on a space curve if the direction that the curve is being swept out changes than the new unit tangent vector for that point will be parallel to and of the same magnitude as the previous unit vector, but in the opposite direction?

Thanks HallsofIvy for fixing my tex tags.

7. Feb 18, 2006

### 0rthodontist

Yes, although you already know it has the same magnitude because it is a unit vector. If it weren't a unit vector then it might not have the same magnitude if the curve is not parametrized with the same speed, but it would still be parallel and in the opposite direction.

8. Feb 18, 2006

### Stevecgz

Right... unit vectors always have a magnitude of one. Thanks.