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 $1$ .

In previous courses, we found tangent lines to curves at given points. 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 defines the normal line at a point to the be the line through the point 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?” Well, we have several options for finding vectors normal to curves. In $\R ^2$ if $\vec {p}(t) = \vector {x(t),y(t)}$ we could write the tangent vector as: $\vec {p}'(t) =\vector {x'(t),y'(t)}$ and then a normal vector as $\vec {n}(t) = \vector {y'(t),-x'(t)}$ for a vector normal to $\vec {p}'(t)$ . You can check for yourself that this vector is normal to $\vec {p}'(t)$ using the dot product. In two-dimensions, the vector $\vec {n}$ defined above will always point “outward” for a closed curve drawn in a counterclockwise fashion. Below we see a closed curve drawn in a counterclockwise fashion with some normal vectors:

On the other hand, there is no analogous trick for vector-valued functions in higher dimensions. In this case, recall:

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

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

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

Since constructing normal vectors using the derivative works in all dimensions, we will often 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.

The principal unit normal vector will always point toward the “inside” of how a curve is curving.

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

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