Answer to Question #122825 in Vector Calculus for Ojugbele Daniel

Question

What do you understand by unit tangent vector

Answer 1

Given a smooth vector-valued function  $\overrightarrow{r}(t)$. Any vector parallel to $\overrightarrow{r}'(t_0)$ is tangent to the graph of $\overrightarrow{r}(t)$ at $t=t_0.$ It is often useful to consider just the direction of $\overrightarrow{r}'(t)$ and not its magnitude. 

Therefore we are interested in the unit vector in the direction of $\overrightarrow{r}'(t)$

This leads to a definition.

Let $\overrightarrow{r}(t)$ be a smooth function on an open interval $I.$ The unit tangent vector $\overrightarrow{T}(t)$ is

$$\overrightarrow{T}(t)={1\over ||\overrightarrow{r}'(t)||}\overrightarrow{r}'(t), ||\overrightarrow{r}'(t)||\not=0$$

Let $\overrightarrow{v}(t)=\overrightarrow{r}'(t)$ denote the velocity vector. Then we define the unit tangent vector by as the unit vector in the direction of the velocity vector.

$$\overrightarrow{T}(t)={1\over ||\overrightarrow{v}(t)||}\overrightarrow{v}(t), ||\overrightarrow{v}(t)||\not=0$$

The tangential component of acceleration is in the direction of the unit tangent vector

$$\overrightarrow{a}_\tau(t)=a\cdot\overrightarrow{T}(t)=a\cdot{1\over ||\overrightarrow{v}(t)||}\overrightarrow{v}(t), ||\overrightarrow{v}(t)||\not=0$$