Problem 11 - Thomas Calculus

Question

Section 13.4, you found \(\mathbf{T}, \mathbf{N},\) and \(\kappa .\) Now, in the following Exercises 9-16, find \(\mathbf{B}\) and \(\tau\) for these 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

\( \mathbf{B} = \mathbf{k} \) and \( \tau = 0 \).

Step 1: Compute the Tangent Vector T

The tangent vector \( \mathbf{T} \) is given by \( \mathbf{T} = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \). First, find \( \mathbf{r}'(t) \) by differentiating \( \mathbf{r}(t) \):\[\mathbf{r}'(t) = \left( e^t \cos t - e^t \sin t \right) \mathbf{i} + \left( e^t \sin t + e^t \cos t \right) \mathbf{j} \]Thus, \[\|\mathbf{r}'(t)\| = \sqrt{(e^t \cos t - e^t \sin t)^2 + (e^t \sin t + e^t \cos t)^2} = e^t \sqrt{2}\]Now, compute the unit tangent vector \( \mathbf{T} \):\[\mathbf{T} = \frac{1}{\sqrt{2}} \left( \cos t \mathbf{i} + \sin t \mathbf{j} + \sqrt{2} \mathbf{k} \right)\]

Step 2: Compute the Normal Vector N

The normal vector \( \mathbf{N} \) is obtained by differentiating \( \mathbf{T}(t) \) and normalizing it. First, find \( \mathbf{T}'(t) \):\[\mathbf{T}'(t) = \frac{1}{\sqrt{2}} \left( -\sin t \mathbf{i} + \cos t \mathbf{j} \right)\]Now, compute the magnitude of \( \mathbf{T}'(t) \) as \( \|\mathbf{T}'(t)\| = \frac{1}{\sqrt{2}} \) (since only one component is considered for unit vector component magnitude one).Normalize \( \mathbf{T}'(t) \) to get \( \mathbf{N}(t) \):\[\mathbf{N} = -\sin t \mathbf{i} + \cos t \mathbf{j}\]

Step 3: Calculate the Binomial Vector B

The binomial vector \( \mathbf{B} \) is the cross product of \( \mathbf{T} \) and \( \mathbf{N} \):\[\mathbf{B} = \mathbf{T} \times \mathbf{N} = \left( \frac{1}{\sqrt{2}} \right) (\cos t \mathbf{i} + \sin t \mathbf{j} + \sqrt{2} \mathbf{k}) \times (-\sin t \mathbf{i} + \cos t \mathbf{j})\]This gives: \[\mathbf{B} = \mathbf{k}\]

Step 4: Find the Torsion τ

Torsion \( \tau \) requires the derivative of the binormal vector \( \mathbf{B} \) and considering curvature \( \kappa \). If \( \mathbf{B} \) is constant (here \( \mathbf{k} \) is constant), then \( \mathbf{B}'(t) = 0 \), resulting in:\[\tau = 0 \]

Key Concepts

Tangent Vector

The tangent vector is essential when analyzing a curve because it tells us the direction of the curve at any particular point. Imagine you're tracing a curve with a pencil—wherever the point of the pencil is heading, that's what the tangent vector shows at that instant.

Mathematically, the tangent vector \( \mathbf{T} \) is calculated by taking the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \) and normalizing it. This can be written as:

• \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \)
• \( \mathbf{r}'(t) \) is the derivative of the position vector \( \mathbf{r}(t) \).
• \( \|\mathbf{r}'(t)\| \) is the magnitude of \( \mathbf{r}'(t) \).

For the given position vector \( \mathbf{r}(t) = (e^{t} \cos t) \mathbf{i} + (e^{t} \sin t) \mathbf{j} + 2 \mathbf{k} \), the derivative \( \mathbf{r}'(t) \) becomes \(( e^t \cos t - e^t \sin t ) \mathbf{i} + ( e^t \sin t + e^t \cos t ) \mathbf{j} \). By finding its magnitude and normalizing, you get the tangent vector \( \mathbf{T}(t) = \frac{1}{\sqrt{2}} ( \cos t \mathbf{i} + \sin t \mathbf{j} + \sqrt{2} \mathbf{k} ) \).

In summary, the tangent vector is perpendicular to the surface of the curve and plays a critical role in understanding the curve's geometrical behavior.

Normal Vector

The normal vector, often represented as \( \mathbf{N} \), is important because it provides insight into the curve's directional change, acting as a perpendicular component to the tangent vector. While the tangent vector shows where the curve is heading, the normal vector helps us understand the curve's tendency to "turn."

To determine the normal vector, differentiate the tangent vector \( \mathbf{T}(t) \) with respect to \( t \), and then normalize it. Here's how it's done:

• Compute \( \mathbf{T}'(t) \), which involves differentiating each component of \( \mathbf{T}(t) \).
• Calculate the magnitude \( \|\mathbf{T}'(t)\| \) and use it to normalize \( \mathbf{T}'(t) \) to obtain \( \mathbf{N}(t) \).

For the curve in question, the differentiation of the tangent vector results in \( \mathbf{T}'(t) = \frac{1}{\sqrt{2}} ( -\sin t \mathbf{i} + \cos t \mathbf{j} ) \). Normalizing it gives \( \mathbf{N}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} \).

The normal vector is crucial in further calculations, like finding the curvature, and is essential in describing curve properties such as torsion.

Binomial Vector

The binomial vector, denoted as \( \mathbf{B} \), completes the trio of vectors that describe a curve's behavior in space. It is crucial in providing a perpendicular component to both the tangent vector \( \mathbf{T} \) and the normal vector \( \mathbf{N} \). This is achieved through their cross product, encapsulating the three-dimensional orientation of the curve.

To find the binomial vector:

• Calculate the cross product \( \mathbf{B} = \mathbf{T} \times \mathbf{N} \).
• This vector ensures the establishment of an orthonormal basis, meaning all vectors are mutually perpendicular.

In the provided problem, by crossing \( \mathbf{T}(t) = \frac{1}{\sqrt{2}} ( \cos t \mathbf{i} + \sin t \mathbf{j} + \sqrt{2} \mathbf{k} ) \) with \( \mathbf{N}(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} \), you determine that \( \mathbf{B} = \mathbf{k} \).

The binomial vector is vital for defining the curve's orientation in three-dimensional space and plays a role in calculating torsion, a measure of how sharply a curve twists out of its osculating plane.