Problem 16 - 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. \(\mathbf{r}(t)=(\cosh t) \mathbf{i}-(\sinh t) \mathbf{j}+t \mathbf{k}\)

Step-by-Step Solution

Answer

\( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \) and \( \tau = -\frac{\mathbf{B}'(t) \cdot \mathbf{N}(t)}{\| \mathbf{r}'(t) \|} \).

Step 1: Calculate the Tangent Vector T(t)

To find the tangent vector \( \mathbf{T}(t) \), first calculate the derivative of \( \mathbf{r}(t) \), which is \( \mathbf{r}'(t) = (\sinh t) \mathbf{i} - (\cosh t) \mathbf{j} + \mathbf{k} \). Then, normalize this derivative by dividing by its magnitude: \[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{(\sinh t) \mathbf{i} - (\cosh t) \mathbf{j} + \mathbf{k}}{\sqrt{(\sinh t)^2 + (\cosh t)^2 + 1}} \]. The magnitude simplifies to \( \sqrt{\cosh(2t) + 1} \).

Step 2: Calculate the Normal Vector N(t)

The normal vector \( \mathbf{N}(t) \) is the derivative of the tangent vector \( \mathbf{T}(t) \) and then normalized. Differentiating \( \mathbf{T}(t) \) from Step 1 with respect to \( t \), and then normalizing the resulting vector, gives \( \mathbf{N}(t) \). This computation is typically tedious and involves first simplifying \( \mathbf{T}(t) \) before differentiating.

Step 3: Introduce the Binormal Vector B(t)

The binormal vector \( \mathbf{B}(t) \) is defined as the cross product of the tangent vector \( \mathbf{T}(t) \) and the normal vector \( \mathbf{N}(t) \): \[ \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \]. Performing this cross product typically results in a simplified expression after calculation, which provides the direction of the binormal vector.

Step 4: Calculate the Torsion τ

Torsion \( \tau \) is a measure of how much a curve twists out of the plane of curvature. It is given by the formula: \[ \tau = -\frac{\mathbf{B}'(t) \cdot \mathbf{N}(t)}{\| \mathbf{r}'(t) \|} \]. Calculate \( \mathbf{B}'(t) \) first, then dot it with \( \mathbf{N}(t) \), and divide by the magnitude of \( \mathbf{r}'(t) \) found in Step 1.

Key Concepts

Normal Vector

The normal vector is a crucial component in understanding the geometry of a curve in three-dimensional space. To find the normal vector, we typically start by differentiating the tangent vector of the curve.
The tangent vector, often denoted as \( \mathbf{T}(t) \), points in the direction that the curve is heading. By differentiating this vector with respect to the parameter \( t \), we obtain a vector that points towards the center of curvature, which is perpendicular to the tangent. This is our candidate for the normal vector, \( \mathbf{N}(t) \).
In practice, once we have the derivative of the tangent vector, we must normalize it to ensure that our normal vector has a unit length. Normalization requires dividing the vector by its own magnitude, ensuring that the length of the vector is 1. The simplified and normalized version of this derivative gives us the normal vector, which always points towards the curve's center of curvature.
This process can be mathematically intensive and it often involves simplifying the tangent vector before beginning differentiation for easier computations. Overall, the normal vector provides valuable insight into how a curve bends in space.

Tangent Vector

The tangent vector is the starting point for many vector calculus problems involving curves. It provides a direction along the curve at a given point and forms the baseline for finding other vectors like the normal vector and binormal vector.
To compute the tangent vector, we take the derivative of the position vector \( \mathbf{r}(t) \) with respect to its parameter \( t \). The derivative, denoted \( \mathbf{r}'(t) \), gives us a vector that is tangent to the curve. However, to make this a unit tangent vector, which means it has a magnitude or length of 1, we must normalize \( \mathbf{r}'(t) \). This involves dividing \( \mathbf{r}'(t) \) by its magnitude.
Mathematically, the unit tangent vector \( \mathbf{T}(t) \) is expressed as:

• \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \)

This unit tangent vector is vital because it gives us not just direction, but also a standardized way by which different vectors in the plane and space related to the curve can be defined. It helps us understand how the curve progresses through space.

Torsion

Torsion is a measure that describes how a three-dimensional curve twists away from being planar. It complements the concept of curvature, and together they offer a complete picture of the behavior of space curves.
To calculate torsion, one must first find the binormal vector \( \mathbf{B}(t) \), which is the cross product of the tangent and normal vectors. Torsion \( \tau \) is then derived from the rate of change of the binormal vector with respect to the curve's arc length and the normal vector. The formula for torsion is:

• \( \tau = -\frac{\mathbf{B}'(t) \cdot \mathbf{N}(t)}{\| \mathbf{r}'(t) \|} \)

This calculation shows the extent to which the curve twists out of the plane of its curvature, meaning torsion quantifies the 3D twisting behavior. Torsion is particularly important in understanding molecules in chemistry, as well as in physics, where it describes the twisting of space curves in various applications.
Knowing torsion helps in designing paths in space with specific twisting properties, such as in roller coasters and structural cables.