Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_{1}\). \(\mathbf{r}(t)=e^{-2 t} \mathbf{i}+e^{2 t} \mathbf{j}+2 \sqrt{2} t \mathbf{k} ; t_{1}=0\)

Step-by-Step Solution

Answer

Curvature: \(\kappa = \sqrt{2}\), \(\mathbf{T} = -\frac{1}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \frac{\sqrt{2}}{2}\mathbf{k}\), \(\mathbf{N} = \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j}\), \(\mathbf{B} = \frac{\sqrt{2}}{2}\mathbf{i} + \frac{\sqrt{2}}{2}\mathbf{j} - \frac{1}{\sqrt{2}}\mathbf{k}\).

Step 1: Find the Derivative of the Position Vector

First, we need to compute the derivative of the position vector \( \mathbf{r}(t) = e^{-2t} \mathbf{i} + e^{2t} \mathbf{j} + 2\sqrt{2}t \mathbf{k} \). The derivative \( \mathbf{r}'(t) \) is found by differentiating each component with respect to \( t \).\[ \mathbf{r}'(t) = \left( \frac{d}{dt} e^{-2t} \right) \mathbf{i} + \left( \frac{d}{dt} e^{2t} \right) \mathbf{j} + \left( \frac{d}{dt} 2\sqrt{2}t \right) \mathbf{k} \]Simplifying, we get:\[ \mathbf{r}'(t) = -2e^{-2t} \mathbf{i} + 2e^{2t} \mathbf{j} + 2\sqrt{2} \mathbf{k} \].

Step 2: Evaluate the Derivative at \(t=t_1\)

Substitute \( t_1 = 0 \) into \( \mathbf{r}'(t) \) to find \( \mathbf{r}'(t_1) \):\[ \mathbf{r}'(0) = -2e^{0} \mathbf{i} + 2e^{0} \mathbf{j} + 2\sqrt{2} \mathbf{k} \]This simplifies to:\[ \mathbf{r}'(0) = -2 \mathbf{i} + 2 \mathbf{j} + 2\sqrt{2} \mathbf{k} \].

Step 3: Calculate the Magnitude of the Derivative

Find the magnitude \( \| \mathbf{r}'(0) \| \) needed for the unit tangent vector \( \mathbf{T} \):\[ \| \mathbf{r}'(0) \| = \sqrt{(-2)^2 + 2^2 + (2\sqrt{2})^2} \]Calculating further:\[ \| \mathbf{r}'(0) \| = \sqrt{4 + 4 + 8} = \sqrt{16} = 4 \].

Step 4: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T} \) is given by:\[ \mathbf{T} = \frac{\mathbf{r}'(t_1)}{\| \mathbf{r}'(t_1) \|} \]By substituting the values we found:\[ \mathbf{T} = \frac{-2 \mathbf{i} + 2 \mathbf{j} + 2\sqrt{2} \mathbf{k}}{4} = -\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} + \frac{\sqrt{2}}{2} \mathbf{k} \].

Step 5: Find the Derivative of the Tangent Vector

Compute \( \mathbf{T}'(t) \). First, differentiate each component of \( \mathbf{r}'(t) \).\[ \mathbf{T}'(t) = \frac{d}{dt}\left( -2e^{-2t} \right) \mathbf{i} + \frac{d}{dt}\left( 2e^{2t} \right) \mathbf{j} + \frac{d}{dt}\left( 2\sqrt{2} \right) \mathbf{k} \]This gives:\[ \mathbf{T}'(t) = 4e^{-2t} \mathbf{i} + 4e^{2t} \mathbf{j} \].

Step 6: Evaluate \(\mathbf{T}'(t)\) at \(t=t_1\) and Find the Unit Normal Vector

Evaluate \( \mathbf{T}'(t_1) \) at \(t_1 = 0\):\[ \mathbf{T}'(0) = 4e^{0} \mathbf{i} + 4e^{0} \mathbf{j} = 4 \mathbf{i} + 4 \mathbf{j} \]Find the magnitude \( \| \mathbf{T}'(0) \| \):\[ \| \mathbf{T}'(0) \| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \].Unit normal vector \( \mathbf{N} \):\[ \mathbf{N} = \frac{\mathbf{T}'(t_1)}{\| \mathbf{T}'(t_1) \|} = \frac{4 \mathbf{i} + 4 \mathbf{j}}{4\sqrt{2}} = \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}} \mathbf{j} \].

Step 7: Calculate the Binormal Vector

The binormal vector \( \mathbf{B} \) is the cross product of \( \mathbf{T} \) and \( \mathbf{N} \):\[ \mathbf{B} = \mathbf{T} \times \mathbf{N} \]\[ \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -\frac{1}{2} & \frac{1}{2} & \frac{\sqrt{2}}{2} \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{vmatrix} \]Calculate the determinants:\[ \mathbf{B} = \left( \frac{\sqrt{2}}{2} \right) \mathbf{i} + \left( \frac{\sqrt{2}}{2} \right) \mathbf{j} + \left( -\frac{1}{\sqrt{2}} \right) \mathbf{k} \].

Step 8: Calculate Curvature

Use the formula \( \kappa = \frac{\| \mathbf{T}'(t_1) \|}{\| \mathbf{r}'(t_1) \|} \):\[ \kappa = \frac{4\sqrt{2}}{4} = \sqrt{2} \].

Key Concepts

Unit Tangent Vector

The unit tangent vector, often denoted as \( \mathbf{T} \), is an essential concept in vector calculus. It provides a direction and orientation for a curve at a given point. To find the unit tangent vector, we first compute the derivative of the position vector \( \mathbf{r}(t) \). This derivative, also known as the velocity vector, is then normalized to give the unit tangent vector.

In this exercise, the position vector is \( \mathbf{r}(t) = e^{-2t} \mathbf{i} + e^{2t} \mathbf{j} + 2\sqrt{2}t \mathbf{k} \). By differentiating, we obtain \( \mathbf{r}'(t) = -2e^{-2t} \mathbf{i} + 2e^{2t} \mathbf{j} + 2\sqrt{2} \mathbf{k} \), and evaluating at \( t = 0 \) gives \( \mathbf{r}'(0) = -2 \mathbf{i} + 2 \mathbf{j} + 2\sqrt{2} \mathbf{k} \).

To convert this into a unit vector, we calculate its magnitude: \( \| \mathbf{r}'(0) \| = 4 \). Therefore, the unit tangent vector \( \mathbf{T} \) is \( -\frac{1}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} + \frac{\sqrt{2}}{2} \mathbf{k} \). This vector essentially lays out the path of the curve at \( t = 0 \), providing insight into the curve's trajectory.

Unit Normal Vector

The unit normal vector, represented as \( \mathbf{N} \), is perpendicular to the unit tangent vector \( \mathbf{T} \). It indicates how a curve is bending at a point on the curve. To find \( \mathbf{N} \), we need to differentiate the tangent vector and normalize the result.

To compute \( \mathbf{T}'(t) \), we differentiate each component of \( \mathbf{r}'(t) \) with respect to \( t \), giving us \( \mathbf{T}'(t) = 4e^{-2t} \mathbf{i} + 4e^{2t} \mathbf{j} \). Evaluating at \( t = 0 \), we get \( \mathbf{T}'(0) = 4 \mathbf{i} + 4 \mathbf{j} \).

Finding its magnitude \( \| \mathbf{T}'(0) \| = 4\sqrt{2} \), the unit normal vector \( \mathbf{N} \) is then \( \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}} \mathbf{j} \). This vector shows the direction in which the curve is turning, further explaining the curvature of the path.

Binormal Vector

The binormal vector \( \mathbf{B} \) completes the Frenet-Serret frame, an orthonormal basis consisting of \( \mathbf{T} \), \( \mathbf{N} \), and \( \mathbf{B} \). It is computed as the cross product of the unit tangent vector \( \mathbf{T} \) and the unit normal vector \( \mathbf{N} \).

The cross product calculation is executed using a determinant format:
\[ \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -\frac{1}{2} & \frac{1}{2} & \frac{\sqrt{2}}{2} \ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{vmatrix} \]
This results in the binormal vector \( \mathbf{B} = \left( \frac{\sqrt{2}}{2} \right) \mathbf{i} + \left( \frac{\sqrt{2}}{2} \right) \mathbf{j} + \left( -\frac{1}{\sqrt{2}} \right) \mathbf{k} \).

The binormal vector is crucial as it defines a plane perpendicular to the tangent plane of the curve, adding another dimension to understanding how the curve is situated in space.

Vector Calculus

Vector calculus is an area of mathematics that focuses on vector fields and differentiable functions. This exercise is a practical application of vector calculus concepts involving differentiation and operations on vectors like the cross product.

By differentiating the position vector, we determine how the position changes with respect to time, giving us the velocity vector or the path's direction at any point. The cross and dot products are fundamental operations that help in calculating perpendicularity and parallelism between vectors.

In computing the Frenet-Serret frame components, such as \( \mathbf{T} \), \( \mathbf{N} \), and \( \mathbf{B} \), vector calculus tools allow us to fully understand the geometric properties of the curve. This lays the groundwork for solving more complex three-dimensional calculus problems and is essential for fields like physics and engineering where spatial reasoning is crucial.

Differentiation

Differentiation is a fundamental concept in calculus that helps compute a function's derivative, showing the rate of change of a variable with respect to another. For vector functions, differentiation helps us explore the dynamics of vectors with respect to time or another parameter.

In our exercise, we started with differentiating the position vector \( \mathbf{r}(t) \), leading us to understand the velocity vector \( \mathbf{r}'(t) \) and further, its magnitude. Differentiation doesn't just stop there; it lets us retrieve the unit tangent vector by normalizing the differentiated result, providing a sense of directionality.

It also enables the determination of the unit normal vector \( \mathbf{N} \) through the derivative of the unit tangent vector, reflecting how sharply the path curves. Differentiation, thereby, serves as a foundational tool for examining and interpreting the behavior of vectors within vector calculus, ultimately unveiling the iterative process of change.