# Subsection11.4.1Unit Tangent Vector

Given a smooth vector-valued function $\vrt\text{,}$ we defined in Definition 11.2.16 that any vector parallel to $\vrp(t_0)$ is tangent to the graph of $\vrt$ at $t=t_0\text{.}$ It is often useful to consider just the direction of $\vrp(t)$ and not its magnitude. Therefore we are interested in the unit vector in the direction of $\vrp(t)\text{.}$ This leads to a definition.

##### Definition11.4.1Unit Tangent Vector

Let $\vrt$ be a smooth function on an open interval $I\text{.}$ The unit tangent vector $\vec T(t)$ is \begin{equation*} \vec T(t) = \frac{1}{\norm{\vrp(t)}}\vrp(t). \end{equation*}

##### Example11.4.2Computing the unit tangent vector

Let $\vrt = \la 3\cos(t) , 3\sin(t) , 4t\ra\text{.}$ Find $\vec T(t)$ and compute $\vec T(0)$ and $\vec T(1)\text{.}$

Solution

In many ways, the previous example was “too nice.” It turned out that $\vrp(t)$ was always of length 5. In the next example the length of $\vrp(t)$ is variable, leaving us with a formula that is not as clean.

##### Example11.4.4Computing the unit tangent vector

Let $\vrt=\la t^2-t,t^2+t\ra\text{.}$ Find $\vec T(t)$ and compute $\vec T(0)$ and $\vec T(1)\text{.}$

Solution

# Subsection11.4.2Unit Normal Vector

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, we 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 $\vrt$ in $\mathbb{R}^2\text{,}$ we have 2 directions perpendicular to the tangent vector, as shown in Figure 11.4.6. It is good to wonder “Is one of these two directions preferable over the other?”

Given $\vrt$ in $\mathbb{R}^3\text{,}$ 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 $\mathbb{R}^2$ and $\mathbb{R}^3$ is “Yes, there is one vector that is not only preferable, it is the right' one to choose.” Recall Theorem 11.2.26, which states that if $\vrt$ has constant length, then $\vrt$ is orthogonal to $\vrp(t)$ for all $t\text{.}$ We know $\vec T(t)\text{,}$ the unit tangent vector, has constant length. Therefore $\vec T(t)$ is orthogonal to $\vec T\,'(t)\text{.}$

We'll see that $\vec T\,'(t)$ is more than just a convenient choice of vector that is orthogonal to $\vrp(t)\text{;}$ rather, it is the “right” choice. Since all we care about is the direction, we define this newly found vector to be a unit vector.

$\vec T(t)$ is a unit vector, by definition. This does not imply that $\vec T\,'(t)$ is also a unit vector.

##### Definition11.4.7Unit Normal Vector

Let $\vrt$ be a vector-valued function where the unit tangent vector, $\vec T(t)\text{,}$ is smooth on an open interval $I\text{.}$ The unit normal vector $\vec N(t)$ is \begin{equation*} \vec N(t) = \frac1{\norm{\vec T\,'(t)}}\vec T\,'(t). \end{equation*}

##### Example11.4.8Computing the unit normal vector

Let $\vrt = \la 3\cos(t) , 3\sin(t) , 4t\ra$ as in Example 11.4.2. Sketch both $\vec T(\pi/2)$ and $\vec N(\pi/2)$ with initial points at $\vec r(\pi/2)\text{.}$

Solution

The previous example was once again “too nice.” In general, the expression for $\vec T(t)$ contains fractions of square roots, hence the expression of $\vec T\,'(t)$ is very messy. We demonstrate this in the next example.

##### Example11.4.10Computing the unit normal vector

Let $\vrt=\la t^2-t,t^2+t\ra$ as in Example 11.4.4. Find $\vec N(t)$ and sketch $\vrt$ with the unit tangent and normal vectors at $t=-1,0$ and 1.

Solution

The final result for $\vec N(t)$ in Example 11.4.10 is suspiciously similar to $\vec T(t)\text{.}$ There is a clear reason for this. If $\vec u = \la u_1,u_2\ra$ is a unit vector in $\mathbb{R}^2\text{,}$ then the only unit vectors orthogonal to $\vec u$ are $\la -u_2,u_1\ra$ and $\la u_2,-u_1\ra\text{.}$ Given $\vec T(t)\text{,}$ we can quickly determine $\vec N(t)$ if we know which term to multiply by $(-1)\text{.}$

Consider again Figure 11.4.11, where we have plotted some unit tangent and normal vectors. Note how $\vec N(t)$ always points “inside” the curve, or to the concave side of the curve. This is not a coincidence; this is true in general. Knowing the direction that $\vec r(t)$ “turns” allows us to quickly find $\vec N(t)\text{.}$

# Subsection11.4.3Application to Acceleration

Let $\vrt$ be a position function. It is a fact (stated later in Theorem 11.4.13) that acceleration, \vat, lies in the plane defined by $\vec T$ and $\vec N\text{.}$ That is, there are scalars $a_{\text{T} }$ and $a_{\text{N} }$ such that \begin{equation*} \vat = a_{\text{T} }\vec T(t) + a_{\text{N} }\vec N(t). \end{equation*}

The scalar $a_{\text{T} }$ measures “how much” acceleration is in the direction of travel, that is, it measures the component of acceleration that affects the speed. The scalar $a_{\text{N} }$ measures “how much” acceleration is perpendicular to the direction of travel, that is, it measures the component of acceleration that affects the direction of travel.

We can find $a_{\text{T} }$ using the orthogonal projection of $\vec a(t)$ onto $\vec T(t)$ (review Definition 10.3.17 in Section 10.3 if needed). Recalling that since $\vec T(t)$ is a unit vector, $\vec T(t)\cdot\vec T(t)=1\text{,}$ so we have \begin{equation*} \proj{a(t)}{T(t)} = \frac{\vec a(t)\cdot\vec T(t)}{\vec T(t)\cdot\vec T(t)}\vec T(t) = \underbrace{\big(\vec a(t)\cdot\vec T(t)\big)}_{a_{\text{T} }}\vec T(t). \end{equation*}

Thus the amount of \vat in the direction of $\vec T(t)$ is $a_{\text{T} }=\vat\cdot\vec T(t)\text{.}$ The same logic gives $a_{\text{N} } = \vat\cdot\vec N(t)\text{.}$

While this is a fine way of computing $a_{\text{T} }\text{,}$ there are simpler ways of finding $a_{\text{N} }$ (as finding $\vec N$ itself can be complicated). The following theorem gives alternate formulas for $a_{\text{T} }$ and $a_{\text{N} }\text{.}$

Keep in mind that both $a_\text{T}$ and $a_\text{N}$ are functions of $t\text{;}$ that is, the scalar changes depending on $t\text{.}$ It is convention to drop the “$(t)$” notation from $a_\text{T} (t)$ and simply write $a_\text{T}\text{.}$

Note the second formula for $a_\text{T}\text{:}$ $\ds \frac{d}{dt}\Big(\norm{\vvt}\Big)\text{.}$ This measures the rate of change of speed, which again is the amount of acceleration in the direction of travel.

##### Example11.4.14Computing $a_T$ and $a_N$

Let $\vrt = \la 3\cos(t) , 3\sin(t) , 4t\ra$ as in Examples 11.4.2 and 11.4.8. Find $a_\text{T}$ and $a_\text{N}\text{.}$

Solution
##### Example11.4.15Computing $a_T$ and $a_N$

Let $\vrt=\la t^2-t,t^2+t\ra$ as in Examples 11.4.4 and 11.4.10. Find $a_\text{T}$ and $a_\text{N}\text{.}$

Solution
##### Example11.4.17Analyzing projectile motion

A ball is thrown from a height of 240ft with an initial speed of 64ft/s and an angle of elevation of $30^\circ\text{.}$ Find the position function $\vrt$ of the ball and analyze $a_\text{T}$ and $a_\text{N}\text{.}$

Solution

Our understanding of the unit tangent and normal vectors is aiding our understanding of motion. The work in Example 11.4.17 gave quantitative analysis of what we intuitively knew.

The next section provides two more important steps towards this analysis. We currently describe position only in terms of time. In everyday life, though, we often describe position in terms of distance (“The gas station is about 2 miles ahead, on the left.”). The arc length parameter allows us to reference position in terms of distance traveled.

We also intuitively know that some paths are straighter than others — and some are curvier than others, but we lack a measurement of “curviness.” The arc length parameter provides a way for us to compute curvature, a quantitative measurement of how curvy a curve is.

# Subsection11.4.4Exercises

Terms and Concepts

In the following exercises, given $\vrt\text{,}$ find $\vec T(t)$ and evaluate it at the indicated value of $t\text{.}$

In the following exercises, find the equation of the line tangent to the curve at the indicated $t$-value using the unit tangent vector. Note: these are the same problems as in Exercises 11.4.4.511.4.4.8.

In the following exercises, a position function $\vrt$ is given along with its unit tangent vector $\vec T(t)$ evaluated at $t=a\text{,}$ for some value of $a\text{.}$

1. Confirm that $\vec T(a)$ is as stated.

2. Using a graph of $\vrt$ and Theorem 11.4.12, find $\vec N(a)\text{.}$

In the following exercises, find $\vec N(t)\text{.}$

In the following exercises, find $a_{\text{T} }$ and $a_{\text{N} }$ given $\vrt\text{.}$ Sketch $\vrt$ on the indicated interval, and comment on the relative sizes of $a_{\text{T} }$ and $a_{\text{N} }$ at the indicated $t$ values.