Problem 60 - Calculus III

Question

Find the unit tangent vector \(\mathrm{T}(\mathrm{t})\) 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 \( \mathbf{T}(t) \) is \( \left\langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right\rangle \).

Step 1: Find the Derivative of the Function

First, we need to find the derivative of the vector-valued function \( \mathbf{r}(t) = \langle t+1, 2t+1, 2t+2 \rangle \). The derivative \( \mathbf{r}'(t) \) is:\[ \mathbf{r}'(t) = \left\langle \frac{d}{dt}(t+1), \frac{d}{dt}(2t+1), \frac{d}{dt}(2t+2) \right\rangle = \langle 1, 2, 2 \rangle \]

Step 2: Calculate the Magnitude of the Derivative

Next, we calculate the magnitude of the derivative \( \mathbf{r}'(t) = \langle 1, 2, 2 \rangle \):\[\| \mathbf{r}'(t) \| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3\]

Step 3: Form the Unit Tangent Vector

Finally, we form the unit tangent vector \( \mathbf{T}(t) \) by dividing the derivative \( \mathbf{r}'(t) \) by its magnitude:\[\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{\langle 1, 2, 2 \rangle}{3} = \left\langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right\rangle\]

Key Concepts

Vector-Valued Functions

A vector-valued function is a function that takes a real number as input and produces a vector as output. This is different from a regular function, which typically returns a single numerical value. With a vector-valued function, you receive a combination of numerical components that define a vector. Each element of the vector is itself dependent on the real number input, usually denoted by \( t \).
For the problem at hand, the vector-valued function given is \( \mathbf{r}(t) = \langle t+1, 2t+1, 2t+2 \rangle \). This means:

• The first component is \( t+1 \),
• The second component is \( 2t+1 \),
• The third component is \( 2t+2 \).

Understanding vector-valued functions is crucial as it lays the groundwork for differentiating and analyzing these vector outputs, often representing paths or trajectories in space. They can commonly be found in physics and engineering disciplines where motion and force are studied over time.

Derivative Calculation

Taking the derivative of a vector-valued function involves calculating the derivative of each of its components separately. This process applies the standard rules of derivatives to each independent component of the vector.
For \( \mathbf{r}(t) = \langle t+1, 2t+1, 2t+2 \rangle \), the derivative is determined as follows:
\( \mathbf{r}'(t) = \left\langle \frac{d}{dt}(t+1), \frac{d}{dt}(2t+1), \frac{d}{dt}(2t+2) \right\rangle \).
This simplifies to:

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

Thus, the derivative vector \( \mathbf{r}'(t) \) is calculated as \( \langle 1, 2, 2 \rangle \). This process not only helps in understanding the rate of change in each component individually, but also is the first step toward determining characteristics such as speed and direction of a path traced by the vector.

Magnitude Calculation

Once the derivative of a vector function is found, the next step is calculating its magnitude. The magnitude gives the length or size of the vector, without regard to its direction, essentially depicting its speed if the vector represents a movement.
For a vector \( \langle x, y, z \rangle \), its magnitude \( \| \mathbf{v} \| \) is given by the formula:\[ \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \]
Applying this to \( \mathbf{r}'(t) = \langle 1, 2, 2 \rangle \), we compute:
\[ \| \mathbf{r}'(t) \| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
This magnitude is crucial as it is used to scale the derivative vector to form the unit tangent vector.
Understanding how to calculate the magnitude is essential in vector analysis as it aids in transforming any vector into its unit form, which provides direction without affecting its sense of speed.