Unit tangent and unit normal vectors

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We introduce two important unit vectors.

Given a smooth vector-valued function $\vec {p}(t)$ , any vector parallel to $\vec {p}'(t_0)$ is tangent to the graph of $\vec {p}(t)$ at $t=t_0$ . It is often useful to consider just the direction of $\vec {p}'(t)$ and not its magnitude. Therefore we are interested in the unit vector in the direction of $\vec {p}'(t)$ . This leads to a definition.

Let $\vec {p}(t)$ be a smooth function on an open interval $I$ . The unit tangent vector $\uvec {t}(t)$ is $\utan (t) = \frac {\vec {p}'(t)}{|\vec {p}'(t)|}.$

Let $\vec {p}(t) = \vector {3\cos (t), 3\sin (t), 4t}$ . Find $\utan (t)$ . $\utan (t) = \vector {\answer {\frac {-3}{5}\sin (t)},\answer {\frac {3}{5}\cos (t)},\answer {\frac {4}{5}}}$

The unit tangent vector $\utan (t)$ always has a constant magnitude of $1$ .

Just as knowing the direction tangent to a path is important, knowing a direction orthogonal to a path is important. When dealing with real-valued functions, one defined the normal line at a point to the be the line through the point that was perpendicular to the tangent line at that point. We can do a similar thing with vector-valued functions. Given $\vec {p}(t)$ in $\R ^2$ , we have $2$ directions perpendicular to the tangent vector

The young mathematician wonders “Is one of these two directions preferable over the other?” This question only gets harder in higher dimensions. Given $\vec {p}(t)$ in $\R ^3$ , there are infinite vectors orthogonal to the tangent vector at a given point. Again, we might wonder “Is one of these infinite choices preferable over the others? Is one of these the ‘right’ choice?”

The answer in both $\R ^2$ and $\R ^3$ is “Yes, there is one vector that is preferable, and it is the ‘right’ one to choose!” Recall:

If $\vec {p}(t)$ has constant length, then $\vec {p}(t)$ is orthogonal to $\vec {p}'(t)$ for all $t$ .

Since $\utan (t)$ , the unit tangent vector, it necessarily has constant length. Therefore $\utan (t)\text { is orthogonal to }\utan '(t).$

The vector-valued function $\utan '(t)$ is more than just a convenient choice of vector that is orthogonal to $\vec {p}'(t)$ ; rather, it is the “right” choice. We will use this to construct our unit normal vector:

Let $\vec {p}(t)$ be a vector-valued function where the unit tangent vector, $\utan (t)$ , is smooth on an open interval $I$ . The unit normal vector $\unormal (t)$ is $\unormal (t) = \frac {\utan '(t)}{|\utan '(t)|}.$ Some folks call this the principal unit normal vector.

Even though $\utan (t)$ is a unit vector, this does not imply that $\utan '(t)$ is also a unit vector.

Let $\vec {p}(t) = \vector {3\cos t, 3\sin t, 4t}$ as before. Find $\unormal (t)$ . $\unormal (t) =\vector {\answer {-\cos (t)},\answer {-\sin (t)},\answer {0}}.$

As a gesture of friendship, we present you with the following graph of the situation.