Problem 11 - Thomas Calculus

Question

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves $$ \mathbf{r}(t)=\left(e^{t} \cos t\right) \mathbf{i}+\left(e^{t} \sin t\right) \mathbf{j}+2 \mathbf{k} $$

Step-by-Step Solution

Answer

\( \mathbf{T}, \mathbf{N} \) are unit vectors derived from \( \mathbf{v}(t) \) and \( \mathbf{T}'(t) \). The curvature \( \kappa = \frac{1}{\sqrt{2} e^t} \).

Step 1: Find the Velocity Vector

Start by finding the derivative of the position vector \( \mathbf{r}(t) \) to obtain the velocity vector, \( \mathbf{v}(t) \). The derivative of \( \mathbf{r}(t) = \left(e^{t} \cos t\right) \mathbf{i} + \left(e^{t} \sin t\right) \mathbf{j} + 2 \mathbf{k} \) is calculated by using the product rule and constants rule for the components:- For \( \mathbf{i} \): \[\frac{d}{dt} (e^t \cos t) = e^t \cos t - e^t \sin t\]- For \( \mathbf{j} \): \[\frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t\]- For \( \mathbf{k} \): The derivative is 0 since it is a constant.Thus, \[ \mathbf{v}(t) = (e^t \cos t - e^t \sin t) \mathbf{i} + (e^t \sin t + e^t \cos t) \mathbf{j} + 0 \mathbf{k} \] which simplifies to:\[ \mathbf{v}(t) = e^t (\cos t - \sin t) \mathbf{i} + e^t (\sin t + \cos t) \mathbf{j} \].

Step 2: Calculate the Tangent Vector

Find the unit tangent vector \( \mathbf{T}(t) \) by normalizing the velocity vector \( \mathbf{v}(t) \). First, compute the magnitude of \( \mathbf{v}(t) \):\[ \| \mathbf{v}(t) \| = \sqrt{(e^t (\cos t - \sin t))^2 + (e^t (\sin t + \cos t))^2} \]. Simplify the expression using trigonometric identities:\[ \| \mathbf{v}(t) \| = e^t \sqrt{2}\].The tangent vector is:\[ \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\| \mathbf{v}(t) \|} = \frac{e^t (\cos t - \sin t) \mathbf{i} + e^t (\sin t + \cos t) \mathbf{j}}{e^t \sqrt{2}} = \frac{1}{\sqrt{2}}((\cos t - \sin t) \mathbf{i} + (\sin t + \cos t) \mathbf{j})\].

Step 3: Find the Acceleration Vector

Find the derivative of the velocity vector \( \mathbf{v}(t) \) to obtain the acceleration vector, \( \mathbf{a}(t) \).The derivative in each component is:- For \( \mathbf{i} \): \[\frac{d}{dt}(e^t (\cos t - \sin t)) = 2e^t \cos t - e^t \sin t\]- For \( \mathbf{j} \): \[\frac{d}{dt}(e^t (\sin t + \cos t)) = 2e^t \sin t + e^t \cos t\]This results in:\[ \mathbf{a}(t) = (2e^t \cos t - e^t \sin t) \mathbf{i} + (2e^t \sin t + e^t \cos t) \mathbf{j} \].

Step 4: Calculate the Normal Vector

Find the derivative of the tangent vector \( \mathbf{T}(t) \) to determine \( \mathbf{T}'(t) \). The components are calculated by:- Differentiating \( \frac{1}{\sqrt{2}}(\cos t - \sin t)\) gives \(\frac{1}{\sqrt{2}}(-\sin t - \cos t)\)- Differentiating \( \frac{1}{\sqrt{2}}(\sin t + \cos t)\) gives \(\frac{1}{\sqrt{2}}(\cos t - \sin t)\)Thus, \( \mathbf{T}'(t) \) is:\[ \mathbf{T}'(t) = \frac{1}{\sqrt{2}}(-\sin t - \cos t) \mathbf{i} + \frac{1}{\sqrt{2}}(\cos t - \sin t) \mathbf{j} \]Divide \( \mathbf{T}'(t) \) by its magnitude to find the normal vector \( \mathbf{N}(t) \). Calculate the magnitude of \( \mathbf{T}'(t) \):\[ \| \mathbf{T}'(t) \| = 1 \] since the magnitude calculations yield 1. Therefore, \( \mathbf{N}(t) \) is simply \( \mathbf{T}'(t)\) itself and is normalized:\[ \mathbf{N}(t) = \frac{1}{\sqrt{2}}(-\sin t - \cos t) \mathbf{i} + \frac{1}{\sqrt{2}}(\cos t - \sin t) \mathbf{j} \].

Step 5: Compute the Curvature

The curvature \( \kappa(t) \) is calculated using the formula:\[ \kappa = \frac{\| \mathbf{v}(t) \times \mathbf{a}(t) \|}{\| \mathbf{v}(t) \|^3} \]First find \( \mathbf{v}(t) \times \mathbf{a}(t) \):- The cross product calculation involves determinante of a matrix with \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) components,- The cross product calculations yield \( 2e^{2t} \mathbf{k} \).Find the magnitude of this cross product: \( \| \mathbf{v}(t) \times \mathbf{a}(t) \| = 2e^{2t} \).Now, find \( \| \mathbf{v}(t) \|^3 \):\[ \| \mathbf{v}(t) \|^3 = (e^t \sqrt{2})^3 = 2^{3/2} e^{3t} \].Finally, curvature is:\[ \kappa(t) = \frac{2e^{2t}}{2\sqrt{2} e^{3t}} = \frac{1}{\sqrt{2} e^t} \].

Key Concepts

Tangent Vector

To understand the concept of a tangent vector, it’s essential to first grasp that a tangent vector represents the instantaneous direction of motion along a space curve. When a space curve is defined by a position vector \( \mathbf{r}(t) \), the tangent vector \( \mathbf{T}(t) \) provides a standardized, or unit-length, representation of the velocity of this curve at any given time \( t \).

Here's how you can find it:

• First, compute the velocity vector, \( \mathbf{v}(t) \), by taking the derivative of the position vector \( \mathbf{r}(t) \).
• Then, determine the magnitude of this velocity vector, \( \| \mathbf{v}(t) \| \).
• The tangent vector is then found by normalizing the velocity vector: \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\| \mathbf{v}(t) \|} \).

The tangent vector helps to capture the essence of the curve's path, indicating the precise direction in which the curve is moving at each point in space.

Normal Vector

The normal vector is a critical concept when analyzing the geometry of space curves, as it provides insight into how the curve bends. The normal vector, often identified as \( \mathbf{N}(t) \), is perpendicular to the tangent vector and lies in the plane containing the tangent vector and the curve's acceleration.

To calculate the normal vector:

• First, find the derivative of the tangent vector, \( \mathbf{T}'(t) \), which points towards the center of curvature.
• Then, normalize this derivative to obtain the unit normal vector, \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\| \mathbf{T}'(t) \|} \).

The normal vector not only indicates the direction but also plays a part in determining the curvature and torsion of a space curve, helping to understand the curve’s shape and how it changes direction.

Acceleration Vector

Acceleration in the context of space curves does more than just denote the rate of change of velocity. It also fundamentally contributes to understanding how the curve is curving or twisting through space. Here, the acceleration vector, \( \mathbf{a}(t) \), is derived by differentiating the velocity vector \( \mathbf{v}(t) \).

Follow these steps to find it:

• Calculate the derivative of the velocity vector \( \mathbf{v}(t) \) to get \( \mathbf{a}(t) \).
• Each component of \( \mathbf{v}(t) \) is differentiated separately to find the corresponding component of \( \mathbf{a}(t) \).

The acceleration vector is indispensable for revealing details about the shape of a space curve. By convention, part of this acceleration is directed along the tangent vector, reflecting changes in speed, while the remainder points along the normal vector, indicating changes in direction. This decomposition is a part of what makes understanding space curves so rich and fascinating.