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 [latex]{\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}[/latex]. the derivative of this function is [latex]-\sin{t\,{\bf{i}}}+\cos{t\,{\bf{j}}}[/latex]. If we substitute the value [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].

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

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 [latex]{\bf{r}}(t)[/latex].

Notice that the vector [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 an example of a tangent vector to the plane curve defined by [latex]{\bf{r}}\,(t)=\cos{t\,{\bf{i}}}+\sin{t\,{\bf{j}}}[/latex].

Definition

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]. 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]. Vector [latex]{\bf{r}}'\,(t_{0})[/latex] is an example of a tangent vector at point [latex]t=t_{0}[/latex]. Furthermore, assume that [latex]{\bf{r}}'\,(t)\neq{\bf{0}}[/latex] The principal unit tangent vector at [latex]t[/latex] is defined to be

[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. To calculate a unit tangent vector, first find the derivative [latex]{\bf{r}}'\,(t)[/latex]. 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:

  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]

TRY IT

Find the unit tangent vector for the vector-valued function

[latex]{\bf{r}}\,(t)=(t^{2}-3)\,{\bf{i}}+(2t+1)\,{\bf{j}}+(t-2)\,{\bf{k}}[/latex]

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).