Unit Tangent Vectors to a Space Curve Examples 1

Recall that we can calculate the unit tangent vector for a point on a space curve $\vec{r}(t) = (x(t), y(t), z(t))$ with the formula $\hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|}$. We are now going to look at some examples of finding unit tangent vectors.

Example 1

Find the unit tangent vector to the space curve $\vec{r}(t) = (a\cos t \sin t, b^2(t - t^2), c^3 e^t \ln t )$ where $a, b, c \in \mathbb{R}$, not all equal to zero, and $t > 0$.

We have that the derivative of $\vec{r}(t)$ is given by:

(1)

\begin{align} \quad \vec{r'}(t) =\left (a(\cos^2 t - \sin^2 t), b^2(1 - 2t), c^2 \left (\frac{e^t}{t} + e^t \ln t \right ) \right ) \end{align}

The norm of $\vec{r'}(t)$ is given by:

(2)

\begin{align} \quad \| \vec{r'}(t) \| = \sqrt{\left (a(\cos^2 t - \sin^2 t) \right )^2 + \left (b^2(1 - 2t) \right )^2 + \left ( c^2 \left (\frac{e^t}{t} + e^t \ln t \right ) \right )^2} \end{align}

Therefore the unit tangent vector to the space curve for $t > 0$ is given by:

(3)

\begin{align} \quad \hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} = \frac{\left (a(\cos^2 t - \sin^2 t), b^2(1 - 2t), c^2 \left (\frac{e^t}{t} + e^t \ln t \right ) \right )}{\sqrt{\left (a(\cos^2 t - \sin^2 t) \right )^2 + \left (b^2(1 - 2t) \right )^2 + \left ( c^2 \left (\frac{e^t}{t} + e^t \ln t \right ) \right )^2}} \end{align}

Example 2

Find the unit tangent vector to the elliptical-helix $\vec{r}(t) = (a \cos t, b \sin t, t)$ where $a, b \in \mathbb{R}$, $a, b \neq 0$.

We have that the derivative of $\vec{r}(t)$ is given by:

(4)

\begin{align} \quad \vec{r'}(t) = (-a\sin t, b \cos t, 1) \end{align}

The norm of $\vec{r'}(t)$ is given by:

(5)

\begin{align} \quad \| \vec{r'}(t) \| = \sqrt{(-a \sin t)^2 + (b \cos t)^2 + 1} = \sqrt{a^2 \sin^2 t + b^2 \cos^2 t + 1} \end{align}

Therefore the unit tangent is given by:

(6)

\begin{align} \quad \hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} = \frac{(-a\sin t, b \cos t, 1) }{\sqrt{a^2 \sin^2 t + b^2 \cos^2 t + 1}} \end{align}