Problem 46 - Essential Calculus Early Transcendentals

Question

\(45-46\) . Find the unit tangent vector \(\mathbf{T}(t)\) at the point with the given value of the parameter \(t\) . $$\mathbf{r}(t)=\left\langle t^{3}+3 t, t^{2}+1,3 t+4\right\rangle, \quad t=1$$

Step-by-Step Solution

Answer

The unit tangent vector at \( t = 1 \) is \( \left\langle \frac{6}{7}, \frac{2}{7}, \frac{3}{7} \right\rangle \).

Step 1: Find the Derivative

First, calculate the derivative of the vector function \( \mathbf{r}(t) \). For \( \mathbf{r}(t) = \langle t^3 + 3t, t^2 + 1, 3t + 4 \rangle \), the derivative is found by differentiating each component with respect to \( t \). Thus, \( \mathbf{r}'(t) = \langle 3t^2 + 3, 2t, 3 \rangle \).

Step 2: Evaluate the Derivative at t=1

Substitute \( t = 1 \) into the derivative \( \mathbf{r}'(t) \) to find \( \mathbf{r}'(1) \). This gives: \( \mathbf{r}'(1) = \langle 3(1)^2 + 3, 2(1), 3 \rangle = \langle 6, 2, 3 \rangle \).

Step 3: Calculate the Magnitude

Find the magnitude of \( \mathbf{r}'(1) \), denoted as \(||\mathbf{r}'(1)||\). The magnitude is \(||\mathbf{r}'(1)|| = \sqrt{(6)^2 + (2)^2 + (3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7\).

Step 4: Compute the Unit Tangent Vector

Divide the derivative \( \mathbf{r}'(1) \) by its magnitude to find the unit tangent vector \( \mathbf{T}(1) \). Thus, \( \mathbf{T}(1) = \frac{1}{7} \langle 6, 2, 3 \rangle = \langle \frac{6}{7}, \frac{2}{7}, \frac{3}{7} \rangle \).

Key Concepts

Derivative of Vector Function

The derivative of a vector function is a fundamental concept in calculus, especially when dealing with curves in a multidimensional space. For any vector function \( \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \), differentiating each component separately with respect to the parameter \( t \) gives the derivative vector \( \mathbf{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle \). This process is akin to taking the derivative of each standard function, just applied to each vector component.

In our specific example, \( \mathbf{r}(t) = \langle t^3 + 3t, t^2 + 1, 3t + 4 \rangle \), we differentiate each element as follows:

• \( f(t) = t^3 + 3t \) becomes \( f'(t) = 3t^2 + 3 \)
• \( g(t) = t^2 + 1 \) becomes \( g'(t) = 2t \)
• \( h(t) = 3t + 4 \) becomes \( h'(t) = 3 \)

Therefore, the derivative vector is \( \mathbf{r}'(t) = \langle 3t^2 + 3, 2t, 3 \rangle \). This derivative vector provides the tangent direction to the curve at any point \( t \).

For \( t = 1 \), the derivative at that point is \( \mathbf{r}'(1) = \langle 6, 2, 3 \rangle \). This specific vector tells us precisely how the curve is heading at \( t = 1 \).

Magnitude of a Vector

The magnitude, often referred to as the length or norm, is a measure of how long a vector is, and it's a crucial concept in understanding the behavior of vectors. For a vector \( \mathbf{v} = \langle a, b, c \rangle \), the magnitude \( ||\mathbf{v}|| \) is calculated using the formula:

• \( ||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2} \)

This formula is essentially an extension of the Pythagorean theorem to three dimensions.

To find the magnitude of \( \mathbf{r}'(1) = \langle 6, 2, 3 \rangle \), we substitute the components into our formula:

• \( ||\mathbf{r}'(1)|| = \sqrt{6^2 + 2^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \)

The result, 7, represents the length of the derivative vector at \( t = 1 \). This length is paramount when scaling the vector to make it a unit vector, as you will see next.

Vector Calculus

Vector calculus extends the principles of calculus to vector fields, dealing with physical and geometric interpretations in space. A unit tangent vector is one aspect of vector calculus which describes the direction of the curve without considering its length. It's crucial for understanding paths and motion in physics and engineering.

To form a unit vector from any vector \( \mathbf{v} \), divide each component of \( \mathbf{v} \) by its magnitude. This operation adjusts the vector's length to one unit while maintaining its direction. It's a common practice to represent direction in space.

For the unit tangent vector \( \mathbf{T}(t) \), we take \( \mathbf{r}'(t) \) and scale it by its magnitude:\[ \mathbf{T}(1) = \frac{1}{7} \langle 6, 2, 3 \rangle = \langle \frac{6}{7}, \frac{2}{7}, \frac{3}{7} \rangle \]
This new vector, \( \mathbf{T}(1) \), points in the same direction as \( \mathbf{r}'(1) \) but has a magnitude of 1, known as a unit vector. It allows us to understand the direction of motion at \( t = 1 \) without involving the speed, providing a pure directional insight.