Question

Given a differentiable vector-valued function $r\left(t\right)$, what is the relationship between $r\text{'}\left(t_0\right)$and $T\left(t_0\right)$at a point$t_0$in the domain of $r\left(t\right)$?

Short answer

The relationship between$r\text{'}\left(t_0\right),T\left(t_0\right)$ is$T\left(t\right)=\frac{r\text{'}\left(t\right)}{||r\text{'}\left(t\right)||}$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

A differentiable vector-valued function is$r\left(t\right)$.

Step 2. Write the relationship between r't0,Tt0.

Assume $t_0$be any point in the domain of $r\left(t\right)$. The unit tangent vector of $r\left(t\right)$is denoted by $T\left(t\right)$.

According to the definition of unit tangent vector function on some interval $I\subseteq R$such that $r\text{'}\left(t\right)\ne 0$ on I.

Then the unit tangent function can be defined as,

$T\left(t\right)=\frac{r\text{'}\left(t\right)}{||r\text{'}\left(t\right)||}$

Where,$r\text{'}\left(t_0\right)\ne 0$

Step 3. Consider the definition.

Consider the definition it is clear that,

Both $T\left(t_0\right),r\text{'}\left(t_0\right)$have same direction.

Also, role="math" localid="1649564983175" $r\text{'}\left(t_0\right),T\left(t_0\right)$may differ in lengths unless $||r\text{'}\left(t_0\right)||=1$, as $T\left(t_0\right)$always have the magnitude of $1$unit because it is a unit vector of $r\left(t\right)$.