Unit Tangent Vectors to a Space Curve

Recall from the Derivatives of Vector-Valued Functions page that if $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function defined for $t$ an the interval $I = [a, b]$ and is differentiable, then the derivative $\vec{r'}(t)$ gives us the tangent vector corresponding to each $t \in [a, b]$. We also noted that $\vec{r'}(t)$ could represent a velocity vector-valued function for a point travelling along the curve traced by $\vec{r}(t)$ at time $t$. The length/magnitude of these vectors represent the speed of such a particle at time $t$. Of course, the speed need not be constant, and so if we were to make these velocity vectors have unit length/magnitude $1$, then the result is what is known as a unit tangent vector.

Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function for $t \in [a, b]$ that is differentiable. Then the Unit Tangent Vector at $t$ denoted $\hat{T}(t)$ is the tangent vector at the point $\vec{r}(t)$ that has magnitude/length $1$, that is $\hat{T} = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} = \frac{\vec{v}(t)}{\| \vec{v}(t) \|}$.

The following graph represents some unit vectors for an arbitrary curve $C$. Notice that the length of each vector is equal to the unit length, $1$.

Let's now look at an example of computing a unit tangent vector.

Example 1

Find the unit tangent vector to the curve defined by the vector-valued function $\vec{r}(t) = (t^2, 2t, 3)$ at $t = 2$.

We must first differentiate $\vec{r}(t)$ to get $\vec{r'}(t)$. Differentiating component by component we have that $\vec{r'}(t) = (2t, 2, 0)$. Now to compute $\hat{T}$, we will divide this vector by its magnitude. We must first compute the magnitude of $\vec{r'}(t)$ though, which isn't too hard as $\| \vec{r'}(t) \| = \sqrt{(2t)^2 + (2)^2 + (0)^2} = \sqrt{4t^2 + 4}$, and so:

(1)

\begin{align} \hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r} \|} = \frac{1}{\sqrt{4t^2 + 4}} (2t, 2, 0) \end{align}

Now to figure out the unit tangent vector for $t = 2$, we just need to plug in $2$ into the above equation and so $\hat{T}(2) = \frac{1}{\sqrt{20}} (4, 2, 0) = \frac{1}{2 \sqrt{5}} (4, 2, 0) = \left ( \frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}, 0 \right )$ is the unit tangent vector at $t = 2$.

Example 2

Find the unit tangent vector to the curve defined by the vector-valued function $\vec{r}(t) = (3\sin t, 2 \cos t, 3t^2 - 1)$ at $t = \pi$.

Once again, we must first differentiate $\vec{r}(t)$ to get $\vec{r'}(t)$. Differentiating component by component we have that $\vec{r'}(t) = (3 \cos t, -2 \sin t, 6t)$. Now to compute $\hat{T}$ we must divide this vector by its magnitude. We note that $\| \vec{r'}(t) \| = \sqrt{(3 \cos t)^2 + (-2 \sin t)^2 + (6t)^2} = \sqrt{9 \cos ^2 t + 4 \sin ^2 t + 36t^2}$, and so:

(2)

\begin{align} \quad \quad \hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} = \frac{1}{\sqrt{9 \cos ^2 t + 4 \sin ^2 t + 36t^2}} (3 \cos t, -2 \sin t, 6t) = \frac{1}{\sqrt{5 \cos ^2 t + 4 + 36t^2}} (3 \cos t, -2 \sin t, 6t ) \end{align}

Plugging in $t = \pi$ we have that $\hat{T}(\pi) = \frac{1}{\sqrt{9 + 36\pi^2}} (-3, 0, 6\pi) = \left ( \frac{-3}{\sqrt{9 + 36\pi^2}}, 0, \frac{6 \pi}{\sqrt{9 + 36\pi^2}} \right )$.