12.4 Unit Tangent and Normal Vectors

Given a smooth vector-valued function $\overrightarrow{r}(t)$, we defined in Definition 12.2.4 that any vector parallel to ${\overrightarrow{r}}^{\prime }(t_0)$ is tangent to the graph of $\overrightarrow{r}(t)$ at $t=t_0$. It is often useful to consider just the direction of ${\overrightarrow{r}}^{\prime }(t)$ and not its magnitude. Therefore we are interested in the unit vector in the direction of ${\overrightarrow{r}}^{\prime }(t)$. This leads to a definition.

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

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 $\overrightarrow{r}(t)$ in ${ℝ}^2$, we have 2 directions perpendicular to the tangent vector, as shown in Figure 12.4.3. It is good to wonder “Is one of these two directions preferable over the other?”

Given $\overrightarrow{r}(t)$ 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.” Recall Theorem 12.2.5, which states that if $\overrightarrow{r}(t)$ has constant length, then $\overrightarrow{r}(t)$ is orthogonal to ${\overrightarrow{r}}^{\prime }(t)$ for all $t$. We know $\overrightarrow{T}(t)$, the unit tangent vector, has constant length. Therefore $\overrightarrow{T}(t)$ is orthogonal to ${\overrightarrow{T}}^{\prime }(t)$.

We’ll see that ${\overrightarrow{T}}^{\prime }(t)$ is more than just a convenient choice of vector that is orthogonal to ${\overrightarrow{r}}^{\prime }(t)$; 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.

The previous example was once again “too nice.” In general, the expression for $\overrightarrow{T}(t)$ contains fractions of square-roots, hence the expression of ${\overrightarrow{T}}^{\prime }(t)$ is very messy. We demonstrate this in the next example.

The final result for $\overrightarrow{N}(t)$ in Example 12.4.4 is suspiciously similar to $\overrightarrow{T}(t)$. There is a clear reason for this. If $\overrightarrow{u}=⟨u_1,u_2⟩$ is a unit vector in ${ℝ}^2$, then the only unit vectors orthogonal to $\overrightarrow{u}$ are $⟨-u_2,u_1⟩$ and $⟨u_2,-u_1⟩$. Given $\overrightarrow{T}(t)$, we can quickly determine $\overrightarrow{N}(t)$ if we know which term to multiply by $(-1)$.

Consider again Figure 12.4.5, where we have plotted some unit tangent and normal vectors. Note how $\overrightarrow{N}(t)$ 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 $\overrightarrow{r}(t)$ “turns” allows us to quickly find $\overrightarrow{N}(t)$.

Let $\overrightarrow{r}(t)$ be a position function. It is a fact (stated later in Theorem 12.4.2) that acceleration, $\overrightarrow{a}(t)$, lies in the plane defined by $\overrightarrow{T}$ and $\overrightarrow{N}$. That is, there are scalars $a_{\mathrm{T}}$ and $a_{\mathrm{N}}$ such that

The scalar $a_{\mathrm{T}}$ measures “how much” acceleration is in the direction of travel, that is, it measures the component of acceleration that affects the speed. The scalar $a_{\mathrm{N}}$ 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 $a_{\mathrm{T}}$ using the orthogonal projection of $\overrightarrow{a}(t)$ onto $\overrightarrow{T}(t)$ (review Definition 11.3.3 in Section 11.3 if needed). Recalling that since $\overrightarrow{T}(t)$ is a unit vector, $\overrightarrow{T}(t)\cdot \overrightarrow{T}(t)=1$, so we have

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

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

Note the second formula for $a_{\mathrm{T}}$: ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}\left(\parallel \overrightarrow{v}(t)\parallel \right)$. This measures the rate of change of speed, which again is the amount of acceleration in the direction of travel.

• Proof

  
  We see that

  	$\overrightarrow{a}(t)$	$={\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}\overrightarrow{v}(t)={\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}\left(\parallel \overrightarrow{v}(t)\parallel \overrightarrow{T}(t)\right)=\left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}\parallel \overrightarrow{v}(t)\parallel \right)\overrightarrow{T}(t)+\parallel \overrightarrow{v}(t)\parallel {\overrightarrow{T}}^{\prime }(t)$
  		$=\left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}\parallel \overrightarrow{v}(t)\parallel \right)\overrightarrow{T}(t)+\parallel \overrightarrow{v}(t)\parallel \parallel {\overrightarrow{T}}^{\prime }(t)\parallel \overrightarrow{N}(t).$

  Since $\overrightarrow{T}(t)$ and $\overrightarrow{N}(t)$ are not parallel, this decomposition is unique and the coefficients tell us $a_{\mathrm{T}}$ and $a_{\mathrm{N}}$.

  Because $\parallel \overrightarrow{T}\parallel =1$, Theorem 12.2.5 tells us that $\overrightarrow{T}$ and ${\overrightarrow{T}}^{\prime }=\parallel {\overrightarrow{T}}^{\prime }\parallel \overrightarrow{N}$ are orthogonal. This means that

  $$\parallel \overrightarrow{a}(t)\times \overrightarrow{v}(t)\parallel =\parallel a_{\mathrm{N}}\overrightarrow{N}(t)\times \parallel \overrightarrow{v}(t)\parallel \overrightarrow{T}(t)\parallel =a_{\mathrm{N}}\parallel \overrightarrow{v}(t)\parallel .$$

  Also, the Pythagorean theorem tells us that

  $${\parallel \overrightarrow{a}(t)\parallel }^2={\parallel a_{\mathrm{T}}\overrightarrow{T}(t)\parallel }^2+{\parallel a_{\mathrm{N}}\overrightarrow{N}(t)\parallel }^2=a_{\mathrm{T}}^2+a_{\mathrm{N}}^2.\mathit{∎}$$

Our understanding of the unit tangent and normal vectors is aiding our understanding of motion. The work in Example 12.4.7 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.