Problem 16 - Thomas Calculus

Question

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves. \(\mathbf{r}(t)=(\cosh t) \mathbf{i}-(\sinh t) \mathbf{j}+t \mathbf{k}\)

Step-by-Step Solution

Answer

\(\mathbf{T}(t) = \frac{(\sinh t) \mathbf{i} - (\cosh t) \mathbf{j} + \mathbf{k}}{\sqrt{2}}\); \(\mathbf{N}(t) = \frac{(\cosh t) \mathbf{i} + (\sinh t) \mathbf{j}}{\sqrt{2\cosh(2t)}}\); \(\kappa = \frac{\sqrt{\cosh(2t)}}{\sqrt{2}}\).

Step 1: Find the Derivative of the Position Vector

First, we need to find the derivative of the position vector \( \mathbf{r}(t) \). The position vector is given by \( \mathbf{r}(t) = (\cosh t) \mathbf{i} - (\sinh t) \mathbf{j} + t \mathbf{k} \). Calculate the derivative to get the velocity vector \( \mathbf{v}(t) \). \[ \mathbf{v}(t) = \frac{d}{dt}((\cosh t) \mathbf{i} - (\sinh t) \mathbf{j} + t \mathbf{k}) = (\sinh t) \mathbf{i} - (\cosh t) \mathbf{j} + \mathbf{k} \]

Step 2: Calculate the Unit Tangent Vector \( \mathbf{T} \)

The unit tangent vector \( \mathbf{T} \) is found by normalizing the velocity vector \( \mathbf{v}(t) \). This means we divide \( \mathbf{v}(t) \) by its magnitude.The magnitude of \( \mathbf{v}(t) \) is\[ \|\mathbf{v}(t)\| = \sqrt{(\sinh t)^2 + (-\cosh t)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]Hence, \[ \mathbf{T}(t) = \frac{(\sinh t) \mathbf{i} - (\cosh t) \mathbf{j} + \mathbf{k}}{\sqrt{2}} \]

Step 3: Find the Derivative of the Unit Tangent Vector \( \mathbf{T} \)

Differentiate \( \mathbf{T}(t) \) to find its derivative \( \mathbf{T}'(t) \).\[ \mathbf{T}'(t) = \frac{d}{dt} \left( \frac{(\sinh t) \mathbf{i} - (\cosh t) \mathbf{j} + \mathbf{k}}{\sqrt{2}} \right) = \frac{(\cosh t) \mathbf{i} + (\sinh t) \mathbf{j}}{\sqrt{2}} \]

Step 4: Calculate the Principal Normal Vector \( \mathbf{N} \)

The principal normal vector \( \mathbf{N} \) is the normalized version of \( \mathbf{T}'(t) \).The magnitude of \( \mathbf{T}'(t) \) is\[ \| \mathbf{T}'(t) \| = \sqrt{(\cosh t)^2 + (\sinh t)^2 } = \sqrt{\cosh^2 t + \sinh^2 t} = \sqrt{\cosh(2t)} \]Thus, \[ \mathbf{N}(t) = \frac{\frac{(\cosh t) \mathbf{i} + (\sinh t) \mathbf{j}}{\sqrt{2}}}{\sqrt{\cosh(2t)}} = \frac{(\cosh t) \mathbf{i} + (\sinh t) \mathbf{j}}{\sqrt{2\cosh(2t)}} \]

Step 5: Compute the Curvature \( \kappa \)

Curvature \( \kappa \) is given by the formula \( \kappa = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{v}(t) \|} \).Substitute the magnitudes previously calculated,\[ \kappa = \frac{\sqrt{\cosh(2t)}}{\sqrt{2}} \]

Key Concepts

Space Curves

Space curves are a fascinating concept in mathematics. They refer to curves that extend into three-dimensional space. Unlike flat curves on a plane, space curves have components that can move freely in multiple directions. In mathematical terms, space curves are represented by vector functions that depend on a parameter—usually denoted as \( t \). For instance, the curve given by \( \mathbf{r}(t) = (\cosh t) \mathbf{i} - (\sinh t) \mathbf{j} + t \mathbf{k} \) is a space curve. It's represented in terms of the basis vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in a 3D space, where each component is a function of \( t \), indicating how the curve evolves with changes in \( t \).
Space curves are essential for visualizing and understanding motions, pathways, or shapes in three dimensions, making them crucial in fields like physics, engineering, and computer graphics.

Unit Tangent Vector

The unit tangent vector, symbolized as \( \mathbf{T}(t) \), plays an important role in understanding the direction of a curve at any given point. When you have a vector function defining a space curve, the derivative of that function, called the velocity vector \( \mathbf{v}(t) \), tells you the immediate direction in which the curve is moving.
However, this velocity vector may not always be a unit vector. To find the unit tangent vector, you normalize this velocity vector by dividing it by its magnitude. This ensures the length of the tangent vector is always 1, making it easy to understand just the direction, independent of the speed or length of the curve segment. For example:

• Initially, calculate \( \mathbf{v}(t) = (\sinh t) \mathbf{i} - (\cosh t) \mathbf{j} + \mathbf{k} \).
• Normalize with its magnitude: \( \| \mathbf{v}(t) \| = \sqrt{2} \).
• So, \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\sqrt{2}} \).

This gives a consistent and comparable way to talk about the "direction" of the curve at any point.

Normal Vector

The normal vector, specifically the principal normal vector \( \mathbf{N} \), is a vector perpendicular to the tangent of the curve at a specific point. To find this normal vector, you first differentiate the unit tangent vector \( \mathbf{T}(t) \) with respect to the parameter \( t \).
The derivation results in a new vector \( \mathbf{T}'(t) \), which still needs to be normalized to become the principal normal vector. This process is similar to finding the unit tangent vector but applied to \( \mathbf{T}'(t) \).
This normal vector is instrumental in understanding how the curve is curving or bending at a particular point. Here's how it is calculated:

• Find \( \mathbf{T}'(t) = \frac{(\cosh t) \mathbf{i} + (\sinh t) \mathbf{j}}{\sqrt{2}} \).
• Calculate its magnitude: \( \| \mathbf{T}'(t) \| = \sqrt{\cosh(2t)} \).
• Normalize to obtain \( \mathbf{N}(t) = \frac{(\cosh t) \mathbf{i} + (\sinh t) \mathbf{j}}{\sqrt{2\cosh(2t)}} \).

This vector lies in the plane of curvature and indicates which way the curve is veering, helping us better understand the three-dimensional behavior of curves.

Differential Calculus

Differential calculus is a branch of mathematics that deals with how things change. It includes the study of rates of change in variables and how these changes are connected. When studying space curves, differential calculus is crucial because it allows us to determine vectors that describe the curve's direction and how it bends.
The key tools in differential calculus are derivatives, which tell us the slope of a function at any given point and thus provide information on the rate and direction of change. In the context of space curves:

• First, the derivative of the position vector \( \mathbf{r}(t) \) gives us the velocity vector \( \mathbf{v}(t) \), which is the rate of change of position with time.
• Further differentiation of \( \mathbf{T}(t) \) gives insight into the change in direction or curvature of the space curve.

Through these processes, differential calculus provides the tools to deeply analyze and comprehend the intricacies of space curves, their tangents, normals, and curvature.