Tangent Vectors and Unit Tangent Vectors

Learning Outcomes

Find the tangent vector at a point for a given position vector
Find the unit tangent vector at a point for a given position vector and explain its significance

Recall from the Introduction to Derivatives that the derivative at a point can be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function, the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function ${\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}$ . the derivative of this function is $-\sin{t\,{\bf{i}}}+\cos{t\,{\bf{j}}}$ . If we substitute the value $t=\frac{\pi}{6}$ into both functions we get

${\bf{r}}\,\big(\frac{\pi}{6}\big)=\frac{\sqrt{3}}{2}\,{\bf{i}}+\frac{1}{2}\,{\bf{j}}$ and ${\bf{r}}'\,\big(\frac{\pi}{6}\big)=-\frac{1}{2}\,{\bf{i}}+\frac{\sqrt{3}}{2}\,{\bf{j}}$ .

The graph of this function appears in Figure 1, along with the vectors ${\bf{r}}\,\big(\frac{\pi}{6}\big)$ and ${\bf{r}}'\,\big(\frac{\pi}{6}\big)$ .

This figure is the graph of a circle represented by the vector-valued function r(t) = cost i + sint j. It is a circle centered at the origin with radius of 1, and counter-clockwise orientation. It has a vector from the origin pointing to the curve and labeled r(pi/6). At the same point on the circle there is a tangent vector labeled “r’(pi/6)”.

Figure 1. The tangent line at a point is calculated from the derivative of the vector-valued function ${\bf{r}}(t)$ .

Notice that the vector ${\bf{r}}'\,\big(\frac{\pi}{6}\big)$ is tangent to the circle at the point corresponding to $t=\frac{\pi}{6}$ . This is an example of a tangent vector to the plane curve defined by ${\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}$ .

Definition

Let $C$ be a curve defined by a vector-valued function $\bf{r}$ , and assume that ${\bf{r}}'\,(t)$ exists when $t=t_{0}$ . A tangent vector $\bf{v}$ at $t=t_{0}$ is any vector such that, when the tail of the vector is placed at point ${\bf{r}}\,(t_{0})$ on the graph, vector $\bf{v}$ is tangent to curve $C$ . Vector ${\bf{r}}'\,(t_{0})$ is an example of a tangent vector at point $t=t_{0}$ . Furthermore, assume that ${\bf{r}}'\,(t)\neq{\bf{0}}$ The principal unit tangent vector at $t$ is defined to be

T (t) = \frac{r^{'} (t)}{∥ r^{'} (t) ∥},

${\bf{T}}\,(t)=\frac{{\bf{r}}'\,(t)}{\left\Vert{\bf{r}}'\,(t)\right\Vert},$

provided $\left\Vert{\bf{r}}'\,(t)\right\Vert\neq{0}.$

The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangent vector, first find the derivative ${\bf{r}}'\,(t)$ . Second, calculate the magnitude of the derivative. The third step is to divide the derivative by its magnitude.

Example: Finding a Unit Tangent Vector

Find the unit tangent vector for each of the following vector-valued functions:

${\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}$
${\bf{u}}\,(t)=(3t^{2}+2t)\,{\bf{i}}+(2-4t^{3})\,{\bf{j}}+(6t+5)\,{\bf{k}}$

Show Solution

$\begin{alignat}{2} \hspace{2cm}&\text{First step:} &\quad {\bf{r}}'(t)&=-\sin{t{\bf{i}}}+\cos{t{\bf{j}}}\\ &\text{Second step:} &\quad ||{\bf{r}}'(t)||&=\sqrt{(-\sin{t})^2+(\cos{t})^2}=1\\ &\text{Third step:} &\quad {\bf{T}}(t)&=\frac{{\bf{r}'(t)}}{||{\bf{r}'(t)||}}=\frac{-\sin{t{\bf{i}}+\cos{t{\bf{j}}}}}1=-\sin{t{\bf{i}}}+\cos{t{\bf{j}}}&\quad\\ \end{alignat}$

$\begin{alignat}{2} &\text{ First step:} &\quad {\bf{u}}'(t)&=(6t+2){\bf{i}}-12t^2{\bf{j}}+6{\bf{k}}\\ \hspace{2cm}&\text{Second step:} &\quad ||{\bf{u}}'(t)||&=\sqrt{(6t+2)^2+(-12t^2)^2+6^2}\\ \text{ }& &\quad &=\sqrt{144t^4+36t^2+24t+40}\\ \text{ }& &\quad &=2\sqrt{36t^4+9t^2+6t+10}\\ &\text{Third step:} &\quad {\bf{T}}(t)&=\frac{{\bf{u}}'(t)}{||{\bf{u}}'(t)||}=\frac{(6t+2){\bf{i}}-12t^2{\bf{j}}+6{\bf{k}}}{2\sqrt{36t^4+9t^2+6t+10}}\\ \text{ }& &\quad &=\frac{3t+1}{\sqrt{36t^4+9t^2+6t+10}}{\bf{i}}-\frac{6t^2}{\sqrt{36t^4+9t^2+6t+10}}{\bf{j}}+\frac{3}{\sqrt{36t^4+9t^2+6t+10}}{\bf{k}}&\quad\\ \end{alignat}$

TRY IT

Find the unit tangent vector for the vector-valued function

${\bf{r}}\,(t)=(t^{2}-3)\,{\bf{i}}+(2t+1)\,{\bf{j}}+(t-2)\,{\bf{k}}$

Show Solution

${\bf{T}}\,(t)=\frac{2t}{\sqrt{4t^{2}+5}}\,{\bf{i}}+\frac{2}{\sqrt{4t^{2}+5}}{\bf{j}}+\frac{1}{\sqrt{4t^{2}+5}}\,{\bf{k}}$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 3.7” here (opens in new window).

Module 3: Vector-Valued Functions