Problem 99 - Calculus 3

Question

Find the unit tangent vector \(\mathbf{T}(t)\) for the following vector-valued functions. $$\mathbf{r}(t)=\langle t+1,2 t+1,2 t+2\rangle$$

Step-by-Step Solution

Answer

The unit tangent vector is \( \mathbf{T}(t) = \langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \rangle \).

Step 1: Find the derivative of the vector function

The first step is to find the derivative of the given vector function \( \mathbf{r}(t) = \langle t+1, 2t+1, 2t+2 \rangle \). Differentiate each component with respect to \( t \):\[ \mathbf{r}'(t) = \langle \frac{d}{dt}(t+1), \frac{d}{dt}(2t+1), \frac{d}{dt}(2t+2) \rangle = \langle 1, 2, 2 \rangle \]

Step 2: Find the magnitude of the derivative

The next step is to find the magnitude of the derivative \( \mathbf{r}'(t) = \langle 1, 2, 2 \rangle \).Calculate the magnitude:\[ \|\mathbf{r}'(t)\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

Step 3: Calculate the unit tangent vector

To find the unit tangent vector \( \mathbf{T}(t) \), divide the derivative \( \mathbf{r}'(t) \) by its magnitude:\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{\langle 1, 2, 2 \rangle}{3} = \langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \rangle \]

Key Concepts

Vector-Valued Functions

Vector-valued functions assign a vector to each value in their domain, typically denoted as time \( t \). Whereas regular functions map a single number to another number, vector-valued functions map each input to a vector in space.
For example, the given function \( \mathbf{r}(t) = \langle t+1, 2t+1, 2t+2 \rangle \) is a three-dimensional vector-valued function. Here, each component of the vector depends on \( t \).

The function \( \mathbf{r}(t) \) essentially traces a path in three-dimensional space as \( t \) changes. This concept is important in physics and engineering for describing motion, among other things.

• First component \( t+1 \)
• Second component \( 2t+1 \)
• Third component \( 2t+2 \)

A good way to visualize vector-valued functions is to think about them as arrows moving through space as \( t \) varies. This allows us to better understand paths and trajectories.

Derivative of Vector Function

The derivative of a vector function helps to find the rate of change of the vector with respect to time \( t \). It's similar to taking the derivative of a normal function, but it applies to each component of the vector separately.
For our function \( \mathbf{r}(t) = \langle t+1, 2t+1, 2t+2 \rangle \), the derivative \( \mathbf{r}'(t) \) is found by differentiating each component:

• \( \frac{d}{dt}(t+1) = 1 \)
• \( \frac{d}{dt}(2t+1) = 2 \)
• \( \frac{d}{dt}(2t+2) = 2 \)

This results in \( \langle 1, 2, 2 \rangle \), which represents how fast and in which direction the vector \( \mathbf{r}(t) \) is changing at any moment. The derivative vector is vital in understanding the velocity or instantaneous rate of change for the given path.

Magnitude of a Vector

The magnitude of a vector measures its length, providing insight into its size regardless of direction.
For a vector \( \mathbf{a} = \langle x, y, z \rangle \), the magnitude is calculated using the formula \( \|\mathbf{a}\| = \sqrt{x^2 + y^2 + z^2} \).

When determining the unit tangent vector, finding the magnitude of the derivative is essential. For \( \mathbf{r}'(t) = \langle 1, 2, 2 \rangle \), the magnitude is:
\[ \|\mathbf{r}'(t)\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \]
This value helps normalize the derivative vector, scaling it to unit length. Thus, the unit tangent vector \( \mathbf{T}(t) \) represents only the direction of change, not its speed. This simplification is crucial in many applications to focus on direction rather than magnitude.