What Factors Affect the Direction of a Unit Tangent Vector in 3D?

[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?

 

[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.

 

[Stevecgz]

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.

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

 

[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.

 

[HallsofIvy]

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

 

[Stevecgz]

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.

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.

 

[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.

 

[Stevecgz]

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.

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