Tangent Vectors and Unit Tangent Vectors

Recall in single-variable calculus, the derivative at a point gives the slope of the tangent line to the graph at that point. For vector-valued functions, the derivative provides a tangent vector to the curve.

Consider the vector-valued function [latex]{\bf{r}}(t)=\cos t{\bf{i}}+\sin t{\bf{j}}[/latex]. The derivative is [latex]{\bf{r}}'(t)=-\sin t{\bf{i}}+\cos t{\bf{j}}[/latex]. If we substitute [latex]t=\frac{\pi}{6}[/latex] into both functions, we get:

[latex]{\bf{r}}\big(\frac{\pi}{6}\big)=\frac{\sqrt{3}}{2}{\bf{i}}+\frac{1}{2}{\bf{j}}[/latex] and [latex]{\bf{r}}'\big(\frac{\pi}{6}\big)=-\frac{1}{2}{\bf{i}}+\frac{\sqrt{3}}{2}{\bf{j}}[/latex]

Figure 1 shows the graph of this function along with the vectors [latex]{\bf{r}}\big(\frac{\pi}{6}\big)[/latex] and [latex]{\bf{r}}'\big(\frac{\pi}{6}\big)[/latex].

Notice that [latex]{\bf{r}}'\big(\frac{\pi}{6}\big)[/latex] is tangent to the circle at the point corresponding to [latex]t=\frac{\pi}{6}[/latex]. This is a tangent vector to the curve defined by [latex]{\bf{r}}(t)[/latex].

tangent vector and principal unit tangent vector

Let [latex]C[/latex] be a curve defined by a vector-valued function [latex]{\bf{r}}[/latex], and assume that [latex]{\bf{r}}'(t)[/latex] exists when [latex]t=t_{0}[/latex].

[latex]\\[/latex]

A tangent vector [latex]{\bf{v}}[/latex] at [latex]t=t_{0}[/latex] is any vector such that, when the tail of the vector is placed at point [latex]{\bf{r}}(t_{0})[/latex] on the graph, vector [latex]{\bf{v}}[/latex] is tangent to curve [latex]C[/latex]. [latex]\\[/latex]The derivative [latex]{\bf{r}}'(t_{0})[/latex] is an example of a tangent vector at point [latex]t=t_{0}[/latex].

[latex]\\[/latex]

The principal unit tangent vector at [latex]t[/latex] is defined as:

[latex]{\bf{T}}(t)=\frac{{\bf{r}}'(t)}{\left\Vert{\bf{r}}'(t)\right\Vert}[/latex]

provided [latex]\left\Vert{\bf{r}}'(t)\right\Vert\neq 0[/latex].

The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. It points in the direction of motion along the curve and always has magnitude [latex]1[/latex].

How to: Find the Unit Tangent Vector

1. Find the derivative: Calculate [latex]{\bf{r}}'(t)[/latex]
2. Find the magnitude: Calculate [latex]\left\Vert{\bf{r}}'(t)\right\Vert[/latex]
3. Divide: Compute [latex]{\bf{T}}(t)=\frac{{\bf{r}}'(t)}{\left\Vert{\bf{r}}'(t)\right\Vert}[/latex]

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

1. [latex]{\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}[/latex]
2. [latex]{\bf{u}}\,(t)=(3t^{2}+2t)\,{\bf{i}}+(2-4t^{3})\,{\bf{j}}+(6t+5)\,{\bf{k}}[/latex]

Show Solution

1.

[latex]\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}[/latex]

2.

[latex]\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}[/latex]