Unit Tangent Vectors to a Space Curve Examples 1

# 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

Let $\vec{r}(t) = (2 \cos t, e^t)$. Find $\hat{T}(t)$ and use this to determine the unit tangent vector at $t = 0$.

First let's find the derivative of $\vec{r}(t)$ by differentiating each component of this vector-valued function. Upon doing so, we get $\vec{r'}(t) = (-2 \sin t, e^t)$.

We now need to calculate the magnitude of the derivative $\vec{r'}(t)$ as $\| \vec{r'}(t) \| = \sqrt{(-2 \sin t)^2 + (e^t)^2} = \sqrt{4 \sin ^2 t + e^{2t}}$. Putting this together and we have that:

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

So for $t = 0$ we have $\hat{T}(0) = \frac{1}{\sqrt{1}} (0, 1) = (0, 1)$.

## Example 2

Let $\vec{r}(t) = (t^3, 2t^2 - 4t, 1 + t + t^2)$. Find $\hat{T}(t)$ and use this to determine the unit tangent vector at $t = 1$.

First let's find the derivative of $\vec{r}(t)$ by differentiating each component of this vector-valued function. Upon doing so, we get $\vec{r'}(t) = (3t^2, 4t - 4, 1 + 2t)$.

We now need to calculate the magnitude of the derivative $\vec{r'}(t)$ as $\| \vec{r'}(t) \| = \sqrt{(3t^2)^2 + (4t - 4)^2 + (1 + 2t)^2} = \sqrt{9t^4 + 16t^2 -32t + 16 + 1 + 4t + 4t^2} = \sqrt{9t^4 + 20t^2 -28t + 17}$. Putting this together and we have that:

(2)
\begin{align} \quad \hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} = \frac{1}{\sqrt{9t^4 + 20t^2 -28t + 17}} (3t^2, 4t - 4, 1 + 2t) \end{align}

So for $t = 1$ we have $\hat{T}(1) = \frac{1}{\sqrt{9 + 20 - 28 + 17}} (3, 0, 3) = \frac{1}{\sqrt{18}} (3, 0, 3) = \left ( \frac{3}{\sqrt{18}}, 0, \frac{3}{\sqrt{18}} \right )$.

## Example 3

Let $\vec{r}(t) = (1, 2t, 3t^2, 4t^3, 5t^4)$. Find $\hat{T}(t)$ and use this to determine the unit tangent vector at $t = 1$.

We first compute the derivative of $\vec{r}(t)$ as $\vec{r'}(t) = (0, 2, 6t, 12t^2, 20t^3)$.

Now we compute the magnitude of $\vec{r}(t)$ as $\| \vec{r'}(t) \| = \sqrt{0^2 + 2^2 + (6t)^2 + (12t^2)^2 + (20t^3)^2} = \sqrt{4 + 36t^2 + 144t^4 + 400t^6}$.

Therefore we have that:

(3)
\begin{align} \hat{T}(t) = \frac{\vec{r'}(t)}{\| \vec{r'}(t) \|} = \frac{1}{ \sqrt{4 + 36t^2 + 144t^4 + 400t^6}} (0, 2, 6t, 12t^2, 20t^3) \end{align}

So for $t = 1$ we have that $\hat{T}(1) = \frac{1}{\sqrt{4 + 36 +144 + 400}} (0, 2, 6, 12, 20) = \frac{1}{\sqrt{584}} (0, 2, 6, 12, 20) = \left (0, \frac{2}{\sqrt{584}}, \frac{6}{\sqrt{584}}, \frac{12}{\sqrt{584}}, \frac{20}{\sqrt{584}} \right )$.