# Unit tangent vector

As I want you to learn a bit of differential geometry, which is not as complicated as they are painted or how it sounds and since I imagine that you have seen my previous posts about the nabla operator and the fields, I am going to start a series of post that ended with The Frenet reference system, a very cool thing.

Let’s start by defining what is a smooth curve and a parametric motion.

When something is moving in space, what it is doing is to vary its position in relation to time, that is, a point, an object, varies where it is as time goes by.

The points where the object (or point from here on) changes its position is called trajectory. In a two-axis world (X, Y or i and j) the position of the particle, the path, can be of any type. We will not go into whether there are forces that force it or not, only if a particle is observed in a defined time interval t , for example between two values ​​each of the coordinates we can correspond with a function. So:

$x=x(t)$
$y=y(t)$

From what we could say that the trajectory we could define it so that:

$\overrightarrow{r}=\overrightarrow{r}(t)$

For that reason we would also have, in the trajectory, a thing called direction. And so r would be a function according to a parameter that would be t. Let’s take an example in three dimensions.

Imagine a particle that follows the next path that we can resemble this function.

In short, it is a particle that follows in two of its axes (i and j) draws a circle of amplitude to and on the axis k (the Z for friends) At constant speed according to b. The good thing is that we know the direction by how time increases.

The case is that to that path we call curve and say (by definition) that it is a smooth curve since its partial derivatives are not all zero at the same time. This, which seems silly is fundamental to know that it is continuous, simply and does not have rare “bits”.

From the trajectory of the example, for example (worth the redundancy) we can calculate the speed (either vector or quantity) knowing that:

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

And that the speed (in quantity) is not more than:

$v=| \frac{d\overrightarrow{r}}{dt} |$

If you look at how we have calculated the speed (vector) we will realize one thing. The speed is always tangential to the curve. That is, by the mere fact of being the (partial) differential we have that it is always a vector tangential to the curve, which gives us the definition of today’s post the tangent unit vector.

$\widehat{T}(t)=\frac{\overrightarrow{v}(t)}{v(t)}=\frac{\frac{d\overrightarrow{r}}{dt}}{| \frac{d\overrightarrow{r}}{dt} |}$

Of course, if the curve has a continuous velocity, the tangent vector is a continuous function of t as can be seen.

If we now consider the angle of the unit tangent vector and any unit vector of the chosen base, we will now have:

$\phi (t)=cos^{-1}\widehat{T}(t)\cdot \overrightarrow{u}$

Where $\overrightarrow{u}$ is one of the unit vectors of the chosen base (which we like) and therefore the curve will be smooth if the tangent varies continuously. How nice!

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